Show that the equation $z^{n}+z+1=0$ has a solution with $|z|=1$ if and only if $n-2$ is divisble by $3$.

If $|z| = 1$ then $|z^{n}| = |z+1| = 1$, hence $z$ is a primitive third root of unity, and of course $\omega^{n}+\omega+1 = 0 \Leftrightarrow 3 | n-2$.

This must have been the most trivial problem ever to be proposed in the NMO


algebra
function
domain
complex analysis
complex analysis unsolved

Let $ D$ be a bounded domain, and let $ f(z)$ and $ h(z)$ be meromorphic functions on $ D$ that extend to be analytic on $ \partial D$. Suppose that $ |h(z)|< |f(z)|$ on $ \partial D$. Show by example that $ f(z)$ and $ f(z)\plus{}h(z)$ can have different numbers of zeros on $ D$. What can be said about $ f(z)$ and $ f(z)\plus{}h(z)$? Prove your assertion.

I do not see what example I can use here. I think what we can say is that they have the same number of zeros minus poles. I just don't see an example here. Thanks.

You are almost done. Since the number (zeros-poles) is the same for both functions, take two functions that have different number of poles, and they'll have different number of zeroes.


Show us what you've done so far.

In particular, can you give an equation relating the side and area of the triangle?

[img]http://www.mathwords.com/a/a_assets/area%20equilateral%20triangle%20formula.gif[/img]

I used that formula and found out that each side is ~15.1967...m
Now I am stuck on how to find the rate

That specific value of $s$ is "snapshot" information: it's only true at one instant, and does not remain true as things change. So it's not something you can differentiate. However, the equation $A=\frac{s^{2}\sqrt{3}}4$ does remain true even as things change. Both $A$ and $s$ depend on $t,$ time. Take $\frac{d}{dt}$ of both sides of that equation. Once you've done that, what pieces of information do you already know and what are you trying to find?

[quote="V4Vendetta"]The side of an Equilateral Triangle decreases at the rate of 2m/s
At what rate is the Area Decreasing if the area is 100 m^2?[/quote]
We know: $\frac{ds}{dt}=-2 \frac{m}{sec}$ and that $A=\frac{s^{2}\sqrt{3}}{4}$.

[hide]$A=\frac{s^{2}\sqrt{3}}{4}$
$\iff 100=\frac{s^{2}\sqrt{3}}{4}$
$\iff \frac{400}{ \sqrt{3}}=s^{2}\implies s=\frac{20}{3^{\frac{1}{4}}}$

Now:

$A'=\frac{s \sqrt{3}}{2}\cdot \frac{ds}{dt}$
$\iff A'=\frac{\left( \frac{20}{3^{\frac{1}{4}}}\right) \sqrt{3}}{2}\cdot-2$
${\iff \frac{dA}{dt}=-20 \cdot 3^{\frac{1}{4}}}$
$\implies$ that $A$ is decreasing at a rate of $\boxed{20 \cdot 3^{\frac{1}{4}}\frac{m^{2}}{sec}}$[/hide]

I think Skimnc hit the spot. Though i think he made a mistake on the last step.
Shouldn't it be
20/ 3^(1/2)

skinmc's calculation is correct (although I wish he'd held off and made V4Vendetta work a little more). Note that $\frac{20}{\sqrt[4]3}$ is the approximately $15.197$ that V reported earlier.

[quote="V4Vendetta"]The side of an Equilateral Triangle decreases at the rate of 2m/s
At what rate is the Area Decreasing if the area is 100 m^2?[/quote]

[hide="slightly different"][size=200]$\frac{dA}{dt}=\frac{\frac{ds}{dt}}{\frac{ds}{dA}}$[/size]

[color=blue]$\frac{ds}{dt}=-2$[/color]

$A=\frac{s^{2}\sqrt 3}{4}$ it follows that $s=\frac 2{\sqrt[4]3}\sqrt{A}$

$\frac{ds}{dA}=\frac{1}{\sqrt[4]3\sqrt A}$  $A=100$ so, $\frac{ds}{dA}=\frac 1{10\sqrt[4]3}$

so the answer is $\frac{dA}{dt}=\frac{2}{\frac 1{10\sqrt[4]3}}=20\sqrt[4]3$ $m^{2}/s$[/hide]


An ice cream shop offers three toppings. If a customer's order may include none, one, two or all of the toppings, how many combinations of the toppings can a customer choose?

0 toppings- 1combo
1 topping-   3combo
2 toppings- 3combo
3 toppings- 1combo

1+3+3+1= 8

 :surf:

or you could notice that it fits the description of a set and just apply $ 2^3\equal{}\boxed{8}$


These types questions have been bothering me:
What is the remainder when 13^(13)+5 is divided by 6.

Thanks

[hide="Solution"]
$ 13^{13} \plus{} 5\equiv 1^{13} \plus{} 5\equiv 0\mod 6$

Answer: 0[/hide]

[url=http://en.wikipedia.org/wiki/Modular_arithmetic]Modular[/url] [url=http://www.artofproblemsolving.com/Wiki/index.php/Modular_arithmetic/Introduction]arithmetic[/url] is what is used to solve these problems.

In this particular one, $ 13^{13}\plus{}5\equiv1^{13}\plus{}(\minus{}1)\equiv1\minus{}1\equiv\boxed0\pmod6$.


Here's an LCM (Least common multiple) question.

What is lcm(1,2,3,4,5,6,7,8,9,10)?

sort by prime factors; the biggest exponent on the factors of 2 is 3, on 3 it's 2, on 5, 1, on 7, 1.  This gives 8*9*5*7


What is your favorite team and who do you think will win the BCS Championship?

I like Tennessee the most (I know they're not that great right now).
Other teams that I like include UTEP, Colorado State, and Ohio State.

I think that Ohio State has a chance at winning the BCS Championship? What do you think?

I don't like football, but I'll say this
OSU WILL PWN EVERYONE

In my opinion football is great.
Why do you like OSU?

My daddy and mommy went there :P
AND they had the BEST chinese food at the cafeteria

Go OSU!!!

They re beating Texas!!!

YAY!

I used to live in Ohio.


People need to stop using periods as multiplication signs; the code is either \cdot or \times.

This doesn't even look like it's true.  You can't multiply together odd numbers and expect to get back an even one, as you have.  If this is not what you meant, it's even more reason not to use periods since no one can understand what you're trying to say.

proof:
$ 1\times 3^2\times 5^2\times 7^2\times . . . \times (2n\plus{}1)^2\times . . . \equal{} (2\times 4)(4\times 2^2)(4\times 3^2)\times . . . \times (4\times n^2)\times . ..$.
help me. Thanks

As JRav said, this is clearly impossible. Let $ n\equal{}1$, for example. You get that $ 3 \equal{} 4$, which is false. In general, the LHS is odd, but the RHS is even.

I suspect the LHS numbers are all supposed to be even, in which case the proof is trivial.


By the new rules, if we only send the PDF and not the TEX, how will solutions get published then? (i.e. on the solutions page [url=http://www.usamts.org/Problems/U_ProblemsPast.php]here[/url])

Somewhere on the changes page, it says that they will no longer be publishing students' solutions, and will just publish their own official solutions. This, along with no longer commending solutions, will supposedly speed up the process of grading. (Yes!!!)

edit: Here is the link. [url]http://usamts.org/year20changes.php[/url]

Indeed, we will not be publishing students' solutions anymore, for 2.5 reasons:

1) It was rather time-consuming for the USAMTS staff to select solutions and edit them for publication.  Plus, we had to keep track of all the TeX source files.

2) Instead, we're going to post "official" solutions a few days after each round is due.  That way, you'll get to see more quickly what the solutions to the problems are, instead of having to wait a month or more.

2.5) If you think you have a nice solution, you can always post it on this message board.


Solve the equation for any integer x
x³ + x² + 5 is congruent to 0 (mod 7³)

Interesting problem!  All I can think of is using the cubic formula but it requires TONS of time as is extremely UGLY.

Short solution without ugly computation:
Find all of them $\mod 7$ and use Hensel.

Well, you could also use Tschirnhaus and Vieta substitutions; it doesn't require much ugly computations. :)


I imported the last six years (2003-2008) list into a database and ran some queries.

 count |                      school
-------+-------------------------------------------------
    76 | Phillips Exeter Academy
    75 | Thomas Jefferson HS Sci/Tech
    47 | Montgomery Blair HS
    43 | Academy Advancement Sci/Tech
    40 | Stuyvesant HS
    31 | Lynbrook HS
    26 | Lexington HS
    26 | Phillips Academy
    24 | Henry M. Gunn HS
    24 | Saratoga HS
    22 | Illinois Math & Sci Academy
    17 | Clements HS
    17 | Mission San Jose HS
    17 | The Harker School
    16 | Palo Alto HS
    16 | Detroit Country Day School
    14 | Naperville North HS
    14 | Vestavia Hills HS
    12 | East Chapel Hill HS
    11 | James Rickards HS
    11 | Lakeside School
    10 | Vincent Massey Secondary School
    10 | Choate Rosemary Hall

Note, the same person qualifying for 3 years in a row count as 3.

This has basically already been posted, in the form of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=199100]this thread[/url]. Those tables also incorporate some details that the automatic system would have missed.


i think this is a Lemoine P point with Soddy circle but i don't know how  :(


geometry
angle bisector
similar triangles
power of a point
radical axis
geometry proposed

Let $\triangle ABC$ and two mobile points $M\in (BA,\ N\in (CA,\ BM=CN.$
Ascertain the geometrical locus of the intersection $L\in CM\cap BN$.

[u]Remark.[/u] $X\in (UV\Longleftrightarrow X\in (UV]\ \vee \ V\in (UX)$, i.e. $(UV$ is a [b]ray[/b].

Let $AD$ be the interior angle bisector of $\triangle ABC$.

The perallel lines from $M,N,L$ to $AD$ intersect the line $BC$ at the points $M',N',L'$ respectively.
We will show that point $L'$ is fixed point

$\triangle ABD \sim \triangle MBM' \Rightarrow \frac{AB}{BM} = \frac{BD}{BM'} = \frac{AD}{MM'}$

$\triangle ACD \sim \triangle NCN' \Rightarrow \frac{AC}{CN} = \frac{CD}{CN'} = \frac{AD}{NN'}$

From these equations we get

$BM' = CN'$ and

$\frac{MM'}{NN'} = \frac{AC}{BC}$

$BM' = CN'$ means that the points $M',N'$ are symmetric with respect to the midpoint of $BC$
[we are sure for this because $M' \in (BC$ and $N' \in (CB$]

So we have also $BN' = CM'$

From the other similar triangles :

$\triangle BNN' \sim \triangle BLL' \Rightarrow \frac{LL'}{NN'} = \frac{BL'}{BN'}$

$\triangle CMM' \sim \triangle CLL' \Rightarrow \frac{LL'}{MM'} = \frac{CL'}{CM'}$

dividing these two equations we get

$\frac{MM'}{NN'} = \frac{BL'}{CL'} \cdot \frac{CM'}{BN'} = \frac{BL'}{CL'}$

finally $\frac{BL'}{CL'} = \frac{AC}{BC} = \frac{CD}{BD}$

Hence $L'$ is symmetric of $D$ through the midpoint of $BC$

Pontios, I erred the address. Here is the shortest solution of the proposed problem from the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=72366
$ML\parallel PN\Longrightarrow \widehat {LMP}\equiv \widehat {MPN}\equiv \widehat {RPN}\equiv \widehat {RAN}\equiv \widehat {LAN}\Longrightarrow$
$\widehat {LMP}\equiv \widehat {LAN}\Longrightarrow \triangle LMR\sim \triangle LAM\Longrightarrow \frac{LM}{LA}=\frac{MR}{AM}=\frac{RL}{ML}\Longrightarrow$
$LM^2=LR\cdot LA\ .$ Therefore, the point $L$ has same power w.r.t. the crcles
$w_1$ and $w_2$, i.e. the point $L$ belongs to the radical axis between these circles.


There are positive integers that have these properties:\\
I. The sum of the squares of their digits is 50, and\\
II. Each digit is larger than the one on its left.\\
What is the product of the digits of the largest integer with both
properties?

[quote="AIME15"]The official solution found a larger one...[/quote]
I have found the smallest number in a big digit...
Do we found the largest number in the largest digit?

If so, The answer is $ \boxed{543}$.

What about 6321?  6^2+3^2+2^2+1^2=36+9+4+1=45+5=50

Also, we're looking for the product of the digits so it's 36.

How can we prove it's the largest, though?

We know it can't be 10 digits (only 9 numbers! we cant use 0 here)

9-Sum of squares is too large. (1^2+2^2....+9^2>50)

Let's keep with this process.

1^2+2^2+3^2+4^2+5^2>50. Thus it must be 4 digits.

The first digit MUST be 1. Can you see why?

Thus we have 3 numbers whose squares add up to 49.

In order to maximize we want the smallest of the three to be as large as possible.

The largest can at most be 7, meaning the smallest can at most be 5.

5-MInimum possible is 5,6,7 too big

4-4^2+5^2+6^2 is too big

3-3^2+4^2+5^2 is 50! So close!

2-2^2+3^2+4^2 fits in our range!

Thus if its 4 digits our number is 12xx

Now the sum must be 45. Let's use a quick trick! 45 is 0mod3. All squares are 0 or 1 mod 3. The only way a sum of two squares is 0 mod 3 is if both squares are 0mod3. Thus our only possiblities contain 3,6,9. But 9^2 is already greater than 45. Trying 3^2+6^2 we see it work so our number is 1236, whose digits product is 36.

I'm sorry.
I jumped to a misleading conclusion and seem to have done big misunderstanding.

I correct the answer as follows.

$ x^2 \le 50$  i.e.  $ x\equal{}1, 2, 3, 4, 5, 6, 7$.
$ 50\minus{}7^2\equal{}1$.
$ 50\minus{}6^2\equal{}14\equal{}3^2\plus{}2^2\plus{}1$.  (Because, $ x^2<14$ and $ x \neq 6$  i.e.  $ x\equal{}1, 2, 3$. similar after this...)
$ 50\minus{}5^2\equal{}25\equal{}4^2\plus{}3^2$.  (Because, $ x^2<25$ and $ x \neq 5$  i.e.  $ x\equal{}1, 2, 3, 4$. similar after this...)

Therefore, $ 17$, $ 1236$, $ 345$.  (Because property II.)
Thus, $ \boxed{1236}$.

Then we read the problem, and see that it asks for the product of the digits, so we do [hide]1*2*3*6=36[/hide].

[hide=solution]Since we want the largest number possible, we also want the number with the most digits. Thus, we want to use the smallest digits possible.
Let's start with $1$, leaving $49$ "points" left over. We could place a $7$, but it is possible that we could get more digits.
Then, place $2$, leaving $45$ points.
Then, place $3$, leaving $36$ points.
We know we can't place a $4$ and $5$, because this would cost more than $36$ points. Thus, we place a $6$.
Our number is $1236$, but let's make sure that there aren't any larger numbers. As we proved, there aren't enough points to have a 5 digit number. Additionally, we can't do $1245$, or any $4$ digit number containing $4,5$, as there aren't any other numbers to switch such that it will sum to $50$ points. Finally, we know we can't do $13xy$, since we wouldn't reach exactly $50$ points.[/hide]


modular arithmetic
ceiling function
number theory unsolved
number theory

Find all pairs if positive integers $ (m,n)$ so that $ m|a^n\minus{}1$ $ \forall$ $ a \in \{1,2,...,n\}$.

Solutions are $ (m,1)$ for any $ m$ and $ (p,p \minus{} 1)$ for any prime $ p$.

Suppose that $ (m,n)$ is such that $ m|a^n \minus{} 1$ $ \forall a \in \{1,2,\dots,n\}$ and $ n > 1$. Then:
i) $ n < p$ where $ p$ is the smallest prime divisor of $ m$ (since $ m$ does not divide $ p^n \minus{} 1$);
ii) for every prime divisor $ q|m$, $ n \equal{} q \minus{} 1$.

To prove ii) one needs to notice that according to i), $ n < q$, and the number of solutions to the congruence $ x^n\equiv 1\pmod{q}$ equals $ \gcd(n,q \minus{} 1)$. Since $ 1,2,\dots,n$ represent distinct solutions to this congruence, $ n\leq \gcd(n,q \minus{} 1)$, implying that $ n$ divides $ q \minus{} 1$. But if $ n < q \minus{} 1$ (i.e., $ n\leq \frac {q \minus{} 1}{2}$) then $ 2\cdot \left\lceil \frac {n \plus{} 1}{2} \right\rceil > n$ represents yet another solution to the aforementioned congruence. This contradiction proves that $ n \equal{} q \minus{} 1$.

The statement ii) implies that $ m$ cannot have more then one distinct prime divisor. Therefore, $ m \equal{} p^k$ for some prime $ p$ and integer $ k > 0$. According to ii), we have $ n \equal{} p \minus{} 1$. Since the number of solutions to $ x^{p \minus{} 1}\equiv 1\pmod{p^k}$ equals $ p \minus{} 1$, the numbers $ 1,2,\dots,p \minus{} 1$ represent all such solutions. But for $ k > 1$, there is yet another solution $ 2\cdot \left\lceil \frac {p}{2} \right\rceil > p \minus{} 1$. The contradiction proves that $ k \equal{} 1$.


factorial
probability
number theory open
number theory

Hi everybody,

Let a integer >2

s=int(sqrt(a))
s! is factorial s

P= a + s!

If P is prime then a is prime.

How to prove it?
If a is composite we know for sure that P is composite but is there another way to prove it?

That's by far the most straightforward way.  Are you asking for a way to "reverse" the proof so that it doesn't rely on proving the contrapositive or are you asking for a genuinely different proof?

I'm just asking if there is another way to prove it.

I'm thinking to Wilson theorem.
Maybe we could use it...

[quote="Matharith"]
Let a integer >2

s=int(sqrt(a))
s! is factorial s

P= a + s!

If P is prime then a is prime.

How to prove it?
[/quote]
OK, I'm rewriting:
If $ a \plus{} ([\sqrt {a}])!$ is prime, then $ a$ is prime.
Solution...
[hide]
Suppose that it isn't true, so $ a \equal{} d_{1}d_{2}$, where $ 1 < d_1 \leq \sqrt {a}$.
It's easy to see that $ d_1 \leq [\sqrt {a}]$.
We have that $ ([\sqrt {a}])!$ is divisible by $ d_1$.
So, $ a \plus{} ([\sqrt {a}])! \equal{} d_{1} d_{2} \plus{} 1*2*...*d_1*...*[\sqrt {a}]$ is divisible by $ d_1 \neq 1$, but not equal to $ d_1$.
Contradiction.[/hide]

P=a+s!

where s=int(sqrt(a))

If we express
a= s^2 + r

P= s(s+(s-1)!) + r

Can we find forms of s and r such as P will have a high probability to be prime?


Let $ S^{n}=\{(x_{0},\ \cdots ,\ x_{n})\in{{\mathbb{R}}^{n+1}| x_{0}^{2}+\cdots x_{n}^{2}=1\}}$ be a n-dimensional sphere.

(1) Let $ f: S^{n}\rightarrow\mathbb{R}$ be a continuous mapping for $ n\geq 1$. Prove that there exists $ x\in{S^{n}}$ satisfying $ f(x)=f(-x)$.

(2) Let $ f: S^{n}\rightarrow S^{1}$ be a continuous mapping for $ n\geq 2$. Prove that there exists $ x\in{S^{n}}$ satisfying $ f(x)=f(-x)$.

For (1), let $ g(x) \equal{} f(x)\minus{}f(\minus{}x)$. Then $ g$ is a continuous mapping $ S^{n}$ into $ \mathbb{R}$.

Choose $ a\in S^{n}$. If $ g(a) \equal{} 0$, then it is done. If not, then $ g(\minus{}a) \equal{}\minus{}g(a)$ and one of $ g(a)$ and $ \minus{}g(a)$ is positive, and the other is negative.

Since continuous mapping preserves connectedness and $ S^{n}$ is connected, $ g(S^{n})$ is connected and hence $ g(S^{n})$ is an interval containing both $ g(a)$ and $ \minus{}g(a)$. So $ [\minus{}g(a), g(a)]\subseteq g(S^{n})$ and $ 0\in g(S^{n})$, which proves (1).


I'm sorry this was asked before but can anyone tell me where I can get 2005-2006 COMC? I don't participate in COMC but I'll be adding the problems to the resources page so if anyone can give them to me, please PM or post here.

If you want some translation of Korean Mathematics Olympiad, I could help you out. I already translated some of them. :)

[quote="Singular"]http://cemc.uwaterloo.ca/english/contests/past_contest/2006/comc.pdf[/quote]

So, 2006 COMC hasn't taken place yet (thanks for 2005 test)? I know the great deals about the contests around the world but time is something I always don't remember.

Also, I'm pretty sure that COMC goes before 1996. If so, can you see if you can find any extra tests from your school or teachers? I remember you knowing a lot of CMO and COMC stuffs so that's why I'm asking. Thanks.

To lightrhee, yes, I'll very appreciate them. Shoot me a private message and I'll give you my "math" email address. I can't guarantee on when the problems will be added but I'll do them ASAP so send them if you translated them.

COMC takes place in November. And the 1996 COMC is the first COMC there ever was.

I thought there was a contest before 1996 that determined qualification to the CMO.  Wasn't it like the Canadian 11th Grade...I ono...


Let n blue lines, no two of which are parallel and no three concurrent, be drawn on a plane. An intersection of two blue lines is colored blue. Through any two blue points that have not alreadz been joined by a blue line, a red line is drawn. An intersection of two red points is colored red, and an intersection of a red line and a blue line is colored purple. What is the maximum possible number of purple points?

This could be wrong as I am in a hurry, but I have calculated it to be $ \frac {n{(n \minus{} 2)^{2}}(n \minus{} 1)(n \plus{} 1)}{8}\minus{}n(n\minus{}2)$

Now that I have some more time I looked at the problem again and I'll give the sketch of a solution:

Step 1: The blue points are $ \frac {n(n \minus{} 1)}2$

Step 2: Every red point "creates" $ \frac {n(n \minus{} 1)}2 \minus{} (n \minus{} 1)$

Step 3: The number of red lines is $ \frac {n{(n \minus{} 1)^{2}}(n \minus{} 2)}{8}$

Step 4: Red-blue lines multiplication principle

Step 5: Final answer $ \frac {{n^2}{(n \minus{} 1)^{2}}(n \minus{} 2)}{8}$ and not $ \frac {n{(n \minus{} 2)^{2}}(n \minus{} 1)(n \plus{} 1)}{8} \minus{} n(n \minus{} 2)$


Thursday night, same time? Tinytim please comment!

Where are u guys going lol? Top-secret Mathcounts meeting???

*007math gets mew's address and plans an mission to make them fail along with his friends who failed at states*


category theory
advanced fields
advanced fields theorems

let $ S$ be a scheme; a $ \mathcal{O}_S$-algebra $ A$ is called quasi-coherent if it is a sheaf and for every open affine $ U \subseteq S$ there is a $ \mathcal{O}_S(U)$-algebra $ B(U)$, whose associated $ \mathcal{O}_U$-algebra is isomorphic to $ A|_U$, that is $ A|_U \cong \widetilde{B(U)}$. obviously, we may assume $ B(U) = A(U)$.

I try to understand a construction of $ Spec(A)$ in a book, which glues the ordinary spectra $ Spec(A(U))$ together. when I try to fill in the details, many problems arise, which the author does not mention at all:

let $ S = \cup_{i \in I} S_i$ be an open affine covering of $ S$. put $ X_i = \text{Spec~}A(S_i)$. now $ \mathcal{O}_S(S_i) \to A(S_i)$ induces $ p_i : X_i \to \text{Spec~}\mathcal{O}_S(S_i)  \cong S_i$. put $ X_{ij} = p_i^{-1}(S_i \cap S_j)$. the first claim is that there is a canonical ([i]what does canonical mean _here_?[/i]) isomorphism $ X_{ij} \to X_{ji}$. first choose an open affine covering $ S_i \cap S_j = \cup_{\lambda \in \Lambda_{ij}} S_{ij \lambda}$ such that $ S_{ij \lambda} \cap S_{ij \mu}$ is affine for all $ \lambda, \mu \in \Lambda_{ij}$ (for example, we may take basic-open subsets $ D_{S_i}(f)$). for $ f \in \mathcal{O}_S(S_i)$, we have

(1) $ ~~~p_i^{-1}(D_{S_i}(f)) = D_{\text{Spec~}A(S_i)}(f) \cong \text{Spec~}A(S_i)_f$
$ ~~~ \cong \text{Spec~}A(S_i) \otimes_{\mathcal{O}_S(S_i)} \mathcal{O}_S(S_i)_f = \text{Spec~}\widetilde{A(S_i)}(D_{S_i}(f)) \cong \text{Spec~}A(D_{S_i}(f))$

[i]honestly, what justifies the author to write always equalities instead of isomorphisms here?[/i]

in particular, $ p_i^{-1}(S_{ij \lambda}) \cong \text{Spec~}A(S_{ij \lambda})$. [i]why do we have the same result with $ p_j$[/i]? I think we may assume $ \Lambda_{ij} = \Lambda_{ji}$, $ S_{ij \lambda} = S_{ji \lambda}$, and that the $ S_{ij \lambda}$ are basic-open simultanuously for $ S_i$ and $ S_j$ (I omit the proof for that). but that's not what the author had in mind.

there is an isomorphism $ \phi_{ij \lambda} : p_i^{-1}(S_{ij \lambda}) \to p_j^{-1}(S_{ij \lambda})$. now the author claims that the same holds for the intersections $ S_{ij \lambda} \cap S_{ij \lambda'}$. [i]how do you see this directly? probably (1) can be extended to arbitrary open affines in $ S_i$?[/i] I wrote $ S_{ij \lambda} = D_{S_i}(f_\lambda)$, and checked by the explicit definitions that the isomorphism $ p_i^{-1}(D_{S_i}(f_\lambda) \to \text{Spec~}A(D_{S_i}(f_{\lambda'}))$ restricts to the isomorphism for $ f_\lambda * f_\lambda'$, which comes down to the fact that the $ \phi_{ij \lambda}$ restrict to isomorphisms $ p_i^{-1}(S_{ij \lambda} \cap S_{ij \lambda'}) \to p_j^{-1}(S_{ij \lambda} \cap S_{ij \lambda'})$. thus the $ \phi_{ij \lambda}$ glue together to an isomorphism $ \phi_{ij} : X_{ij} \to X_{ji}$. clearly, we have $ \phi_{ii} = id$ and $ \phi_{ij} = \phi_{ji}^{-1}$. in order to apply the gluing lemma and to optain $ Spec(A)$, [i]why does the cocycle-condition hold?[/i] that is, why does $ \phi_{ij}$ restrict to an isomorphism $ \phi_{ij}^{(k)} : X_{ij} \cap X_{ik} \to X_{ji} \cap X_{jk}$ such that $ \phi_{jk}^{(i)} \circ \phi_{ij}^{(k)} = \phi_{ik}^{(j)}$? the author gives the following reason: over open affine parts, the isomorphisms are induced by identity maps on sections of $ A$. well, I don't see any identity maps, have no idea why this justifies the cocycle condition and my calculation will be nasty.

finally, the author writes that is easy to see that $ Spec(A)$ - up to [i]canonical[/i] (?) isomorphy - does not depend on the covering of $ S$. I have not really thought over this, but probably we become the same $ Spec(A)$ if we refine $ (S_i)$. then choose a common refinement of two open coverings of $ S$. but at the moment, I'm very unsure if this works and I would have to do redundant,   tedious calculations. instead, I ask if someone knows some sort of [i]universal property[/i] of $ Spec(A)$ and if it is possible to avoid these calculations. there are many other gluing constructions which work similar, which start with a functor {affine schemes} ->  {schemes}. can we trace a general condition which yields by gluing a functor {schemes} -> {schemes}?

the experienced reader will probably lough that I have difficulties with such natural constructions, but I have not seen this naturality yet ;)


function
conics
ellipse
rotation
ratio
geometry
geometric transformation

Hello all,

I'm baffled over this problem...

For a webproject I'm trying to move objects in a trajectory like a planet: an elliptic track that slowly rotates. The ellipse is defined by the following functions:\[x(t) = sin(t) \]\[y(t) = .2cos(t) \]These equations deliver a flat ellipse, now I would like this ellipse to rotate while keeping it's height-width ratio the same. A rotation should have a period of 20pi.
I've tried all I can, I know it should be possible, but I can't figure it out.
Does anyone here know the trick?

Thanks for any help,
Smakkie.

Rotating how? Around one of its axes, making a 3D solid? Or within its plane - if that, whether around the center or around a focus or around some other point?

I would like a 'flat' rotation around the center. Somewhat like a flower's leaves.

Then just put $x=\sin(t-\phi), y=0.2\cos(t-\phi)$ where $\phi$ is the angle of rotation. Creating a loop with increasing values of $\phi$ would make a "flower".

That's not quite what I mean or I misunderstand the use of $\phi$.
With those equations, all I get is points ranging between -.2 and .2 on the $y$-axis, creating a 'flattened' flower.
I would like it to look like a flower from above, so the tips of the 'leaves' are in a circle.

Oh, sorry, my bad.

The correct equations are

$x=\sin t\cos \phi-0.2\cos t\sin\phi$
$y=\sin t\sin\phi+0.2\cos t\cos\phi$

That does the trick perfectly!
Thank you very much.


[quote]The New 'Da Vinci Code' Code

Updated 1:30 AM ET April 28, 2006

A British High Court judge who rejected claims that "The Da Vinci Code" author Dan Brown had plagiarized his novel managed to inject a bit more mystery into the international bestseller. It emerged on Wednesday that the judge at the center of the copyright trial has hidden his own code in his judgment -- and he seems to be daring anyone to try to crack it. Dan Tench, an observant lawyer at the London-based law firm Olswang, noticed 38 random italicized letters in the 71-page court document. He said at first he didn't believe it was anything more than a problem with the judge's computer. "One doesn't really think that a High Court judge is going to encipher a message in the text of a judgement, so I thought it was probably just a word processing error," Tench told ABC News. Justice Peter Smith sent Tench an e-mail confirming his suspicions and told him to go back and look at the first paragraph of the judgement. When Tench did, he realized that the first italicized letters spell out "Smith Code." For three weeks in March, Smith had presided over the case brought by authors Michael Baigent and Richard Leigh, who claimed that Brown plagiarized their book, "The Holy Blood and Holy Grail." "It's clear from the judgement itself, the text of the judgement, that the judge got very much into the substance of the two books in question and it was very much in the spirit of those two books that he's obviously encoded this message," Tench said. If Tench's reading of the judge's ruling is correct, it would suggest that the code the jusge worked into his decision is related to Brown's theory that the descendents of Jesus Christ and Mary Magdalene are still alive. Although the judge would not reveal his code and its meaning, he did say he would probably confirm it if someone cracked it, which was "not a difficult thing to do." The first nine letters have been deciphered, but beyond that is anyone's guess. These are the letters: s, m, i, t, h, c, o, d, e, J, a, e, i, e, x, t, o, s, t, p, s, a, c, g, r, e, a, m, q, w, f, k, a, d, p, m, q, z. It can only come as welcome public relations not only for the book, which has now sold more than 40 million copies worldwide, but also for the much-anticipated movie version of the novel, starring Tom Hanks as historian Robert Langdon, which is due to be released next month. [/quote]


In triangle ABC, AC=12, AB=8, and BC=10. Point T is on BC, AT bisects angle BAC, and AM is the median from A to side BC. What is the area of triangle AMT?

since the point T bisects the angle BAC, BT:TC = 8:12= 4:6
and M is the median of BC so BM:MC=5:5

consequently, point T and M is on the line BC in order
and BT=4, TM=1, MC=5
→area of the triangle AMT is one tenth of triangle ABC
and the area of triangle ABC can be found by the Heron's formula.


Hi guys,I remember an interesting and wery easy exersise :D
Her it.
[color=blue]Prove that $f(x)=\sqrt[3]{x}$ is elementary function[/color]  :rotfl:.
I am don't joking :blush:

[quote="ZetaX"]Do you mean $\sqrt [3] x = e^{\frac{1}{3}ln(x)}$ ¿[/quote]
This equaliti is true only if $x>0$,isn't it?

It's also true for $x<0$ (or can easily be adapted to include this).
But by complex continuation thats all neglegtable :P

But why even prove it, the definition from MathWorld says:
[quote="MathWorld"]A function built up of a finite combination of constant functions, field operations (addition, multiplication, division, and root extractions--the elementary operations)--and algebraic, exponential, and logarithmic functions and their inverses under repeated compositions (Shanks 1993, p. 145; Chow 1999). Among the simplest elementary functions are the logarithm, exponential function (including the hyperbolic functions), power  function, and trigonometric functions.[/quote]
So the inverse of $x^3$ is allowed.

By the way, is $f(x)=\sqrt{x^2}, f: \mathbb{R}\to\mathbb{R}$ elementary ?:)

How I know $\sqrt[n]{x}={x}^{\frac{1}{n}}$ in real plane dose holde if and only if$x\geq 0$.
So we can say that $\sqrt{x^2}=(x^2)^{\frac{1}{2}}$
But $x^2=x*x$ is elementary =>$\sqrt{x^2}={x^2}^{\frac{1}{2}}$ is elementary too.=>$\sqrt{x^2}=|x|$is elementary=>$\frac{x+|x|}{2}$ and $\frac{x-|x|}{2}$
is elementary functions,also we can easy see that$\frac{x+|x|}{2}\geq 0$ and $\frac{x-|x|}{2}\geq 0$for all $x\in R$=>$\sqrt[3]{x}=\sqrt[3]{\frac{x+|x|}{2}}+\sqrt[3]{\frac{x-|x|}{2}}=(\frac{x+|x|}{2})^{\frac{1}{3}}+(\frac{x-|x|}{2})^{\frac{1}{3}}$
is elementary function :)

[quote="ZetaX"]You also use $x^{\frac{1}{3}}$ here...
How do you define $\sqrt[n] x$ and $x^\frac{1}{n}$¿[/quote]
What do you mean? :?


I'm going on holiday to italy in a few weeks (going to rome then some other places then staying in Chianciano in Tuscany for 9 days)... is maths quite a big thing in chianciano... not that im gonna do any maths on holiday :P

NO you won't find anything about maths in chianciano. Enjoy Italy :)


A quite beautiful and challenging problem:
Three magicians show a trick. They give you a deck of cards labelled from $ 1$ to $ 2n \plus{} 1$ ($ n > 6$). You pick one of them and give the remaining ones [i]equally[/i] to the first $ 2$ magicians. Each of these $ 2$ magicians independently of each other and independently of the third one picks some $ 2$ of his $ n$ cards, and gives them as an ordered pair to the third one. This magician, after looking at the received $ 4$ cards, correctly guesses the number of the card you have taken. Explain how this trick may be shown.

[hide] According to the statistics from the book, nobody solved the problem at the contest.
Also, you might want to try the following good problem [unsolved at the moment this problem is posted]:
http://www.mathlinks.ro/Forum/viewtopic.php?t=162659[/hide]

I know two similar problems,both of them can be solved by [b]hall theory[/b] in graphs...
so this one might also have the same idea...


ratio
geometry
perimeter
geometry unsolved

[quote="mustafa-yafes"]In a triangle $ ABC$ the excircle of $ [BC]$ is tangent to the lines $ BC$ , $ CA$ , $ AB$ at $ A_1$ , $ B_1$ , $ C_1$. The excircle of $ [CA]$ is tangent to the lines $ BC$ , $ CA$ , $ AB$ at $ A_2$ , $ B_2$ , $ C_2$. the excircle of $ [AB]$ is tangent to the lines $ BC$ , $ CA$ , $ AB$ at $ A_3$ , $ B_3$ , $ C_3$. Find the greatest ratio between the sum of perimeters of triangles $ A_1B_1C_1 , A_2B_2C_2 , A_3B_3C_3$ with the circumradii of triangle $ ABC$.

Sorry for my bad english.[/quote]
It's easy
$ p_A\equal{}2R_A(sin(\frac{C}{2})\plus{}sin(\frac{b}{2})\plus{}cos(\frac{A}{2})$
The expression that we need to find the greatest value is
$ 2(sinA\plus{}sinB\plus{}sinC)(sin(\frac{C}{2})\plus{}sin(\frac{b}{2})\plus{}sin(\frac{A}{2}))\plus{}2(sinA\plus{}sinB\plus{}sinC)\{tan(\frac{A}{2})[sin(\frac{C}{2})\plus{}sin(\frac{B}{2})]\plus{}tan(\frac{B}{2})[sin(\frac{A}{2})\plus{}sin(\frac{C}{2})]\plus{}tan(\frac{C}{2})[sin(\frac{A}{2})\plus{}sin(\frac{B}{2})]\}$ $ \leq 2.\frac{{3\sqrt 3 }}{2}.\frac{3}{2} \plus{} 2.\frac{{3\sqrt 3 }}{2}.\sqrt 3  \equal{} 9 \plus{} \frac{{9\sqrt 3 }}{2}$


Solve the equation for the reals (x,y):
  $ {x}^{2}\plus{}20x\plus{}351\equal{}\frac{2008}{{y}^{2}\plus{}30y\plus{}233}$

L.H.S=>251 for all x, R.H.S<= 251 for all y. So the only possible solution is L.H.S=R.H.S=251, which is the case for x=-10, y=-15.
(x,y)=(-10,-15) is the solution.
Good and easy question


For what value of $ b$ is $ \frac{b}{13} \equal{} \frac{16}{52}$?

$ \frac{b}{13} = \frac{16}{52}$
\begin{eqnarray*}
b
& = & \frac{16}{52} \times 13 \\
& = & \frac{4 \times 4}{4 \times 13} \times 13 \\
& = & \frac{\color{red}\not\color{black}4 \times 4}{\color{red}\not\color{black}4 \times \color{blue}\not\color{black}{13}} \times \color{blue}\not\color{black}{13} \\
& = & \boxed{4} \\
\end{eqnarray*}

Or, we can simply notice that $ 13 \times 4\equal{}52$.  Thus, we divide $ 16$ by $ 4$ to get $ 16 \div 4\equal{}\boxed{4}$.


A trapezoid has base angles of $ 45^\circ$ and bases that measure 8 inches and 18 inches. The trapezoid is folded as shown. What is the ratio of the area of the black section to the area of the white section? Express your answer as a common fraction.

We can't solve this, due to the lack of a picture.

On FTW! there is a picture.
[asy]
draw((1,0)--(4,0)--(3,1)--(1,1)--cycle);
draw((2,0)--(1,1));
fill((1,0)--(2,0)--(1,1)--cycle);
draw((0,0)--(1,0)--(1,1)--cycle,dashed);
[/asy]
This is the picture. Since we have the top base $8$ and bottom base $18$, we can see that the ratio of the shaded area to the white area is \[\boxed{\frac{5}{16}}\]


α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ σ τ υ φ χ ψ ω
Α Β Γ Δ Ε Ζ Η Θ Ι Κ ∧ Μ Ν Ξ Ο ∏ Ρ ∑ Τ Υ Φ Χ Ψ Ω
•  ¨  …  ∶   ∷   ∴   ∵
∫ ∮ ÷ ± ∈ ∽ ≈ ≌ ∝ ≠ ∞ ∑ ∏ ∪ ∩
〓！≈ ≠ ≡ ≤ ≥ ≤ ≥ ≮ ≯『』〖〗【】〔〕〔〕｛｝
‖ ⊥ ⊿ ∟ ∠ ‖ ∧ ∨ ∩ ∪ ∣
√ ＊ ＠ ＠ ∞ ⊕ ⌒ ⊙ ‰ ＆ ？！ ％ ℅ ℉
℃ ＄ ￠ ￡ ‰ § № ☆ ★ ※ ￥ ％ ＆

[quote="dingdongdog"]α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ σ τ υ φ χ ψ ω
Α Β Γ Δ Ε Ζ Η Θ Ι Κ ∧ Μ Ν Ξ Ο ∏ Ρ ∑ Τ Υ Φ Χ Ψ Ω
How to print them in Latex?[/quote]
Ξ: $\equiv$
∏: $\prod$
∑: $\sum$
ξ: $\zeta$
Γ: $\Gamma$
λ: $\lambda$
$\cdots \cdots$

[quote="shobber"][quote="dingdongdog"]α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ σ τ υ φ χ ψ ω
Α Β Γ Δ Ε Ζ Η Θ Ι Κ ∧ Μ Ν Ξ Ο ∏ Ρ ∑ Τ Υ Φ Χ Ψ Ω
How to print them in Latex?[/quote]
Ξ: $\equiv$
[/quote]
I think you are mistaken...
Ξ is the capital of ξ...


modular arithmetic
number theory unsolved
number theory

$ a$ and $ n$ are natural numbers such that $ a>1$ and $ n|a^n\minus{}1$. Prove that $ (a\minus{}1)n|a^n\minus{}1$.

this follows very easily from the lemma

$ a\equiv b\pmod{p^k}\Rightarrow a^p\equiv b^p\pmod{p^{k\plus{}1}}$.

well, pick a prime and let $ p^k$ be the highest power of $ p$ dividing $ a\minus{}1$ and $ p^l$ the highest power of $ p$ dividing $ n$. then $ a\equiv 1\pmod{p^k}$. now applying the lemma $ l$ times, we get $ a^{p^l}\equiv 1\pmod{p^{k\plus{}l}}$. since $ p^l|n$, it then follows that $ a^n\equiv 1\pmod{p^{k\plus{}l}}$. since $ p^{k\plus{}l}$ is the highest power of $ p$ dividing $ n(a\minus{}1)$, the desired result follows.


Let $f(x)=ax^{2}+bx+c$ be a non constant polynomial, and assume that $a$, $b$, and $c$ are integers.

Prove that there are infinitely many integers $n$ such that $f(n)$ is composite.

(I'm shameless -- this is a homework problem I'm stuck on >_<. Any and all help is appreciated!)

[quote="G-UNIT"][hide="hint 1"] Quadratic Equation [/hide]

[hide="hint 2"] Factoring [/hide][/quote]
I don't understand how your hints help: consider $x^2+x+1$. This cannot be factored and has integer coefficients and it is composite for infinitely many values of $n\in\mathbb{Z}$

I may have found a way to prove it, but I don't really know much about setting up proofs, so it'll be a bit sloppy.

Let $n=kc$, where $c$ is the constant term in the polynomial, and k is any integer $\ge 1$.

So, $f(kc) = ak^{2}c^{2} + bkc + c\\f(kc) = c(ak^{2}c+bk+1)$

It follows that $c|c(ak^{2}c+bk+1)$, so $f(kc)$ must be composite.

Because $k$ can be any positive integer, $f(kc)$ or $f(n)$ has an infinite amount of $n$ values that make it composite.

It's really sloppy, but I think it works. Did I make any wrong/illegal moves? Is this sufficient form for a proof?

[quote] Is this sufficient form for a proof?  [/quote]
Good work but you may have some more work to do.. what if c=1? Hope that helps.
[hide](Goes without saying that if c divdes a number , the number need not be composit if c=1)[/hide]

well since it goes through all numbers, consider the numbers that are a multiple of c, so $x=nc$, that gives us
a(nc)^2 +b(nc)+c
c(acn+bn+1), and clearly that has 2 factors unless the second part is 1 or 0, but that cannot happen for every value of n, therefore we have infinitely many numbers that are composite because there are infinite values of n that can be input, and there may be 2 y values for a given composite, but there cannot be more repeates than that due to the nature of the function

But you still have the problem for c=1 and $an^2+bn+1$ a prime.

[quote]c(acn+bn+1), and clearly that has 2 factors unless the second part is 1 or 0, but that cannot happen for every value of n,[/quote]
Again question is why? To start with let $c=1$ , now you have only one factor :)  You have to show/prove  " but that cannot happen" (remember c is given and it could be 1 for all values of n) ..... Hope this points  in the right direction.. Yes, still there is some work needed before the proof can be counted  as a good proof.


vector
linear algebra
linear algebra unsolved

I need some help with this one.

Let V and W be isomorphic vector spaces. Prove that if V' is a subspace of V, then V' is isomorphic to a subspace W' of W.

since V and W are isomorphic we can find an isomorphism T from V into V. now choose basis vectors for V'. look at the images of these basis vectors. the space spanned by these vectors is the W' u r lookin for

What if V hasn't a finite dimension ?
Does this result remain true ?

Yes. The restriction of an isomorphism is an isomorphism, so $T(V')$ is isomorphic to $V'$. $T(V')$ is a subspace of $W'$, and we are done.


Solve system in positive reals:

$2\sqrt{a}+3b=4$;
$4\sqrt{2(2-b)}=5+3a$.

From the first equation, $a=\frac{9b^{2}-24b+16}{4}$. Inserting into the second gives $2(2-b)=\frac{1}{256}\left(4624-9792b+8856b^{2}-3888b^{3}+729 b^{4}\right)$, and tidying this up we get
$\frac{1}{256}(9b-10)^{2}\left(9b^{2}-28b+36\right)=0$. The discriminant of the quadratic term is negative, so the only solution in positive reals is $b=\frac{10}{9}$, which gives $a=\frac{1}{9}$.

[quote="Diarmuid"]..., and tidying this up we get ...[/quote]
Which trick did you use to factor it?

I think he found it searching for rational roots  :wink:


prove that:
$ \frac{1\minus{}sin(\frac{\pi}{14})}{2*sin(\frac{\pi}{14})}>\sqrt{3*cos(\frac{\pi}{7})}$

after 2 hours.I can solve this ineq. :D
use this lemma;
sin(pi/14) is the root of this equation:
$ 8x^3\minus{}4x^2\minus{}4x\plus{}1\equal{}0$


induction
number theory
number theory proposed

This recurrence reminds me Aperi's proof of the irrationality of $\zeta(3)$  :)

[quote="Arne"]Oh well.

It's easy to prove by induction that $ny_n = (n - 1)! + 1$.

By Wilson's theorem, this implies that $n$ must be prime.[/quote]
  Great !  ;)  My solution is the same with yours  :)


group theory
abstract algebra
superior algebra
superior algebra unsolved

This picture is a part of Hungerford Book . In above lemma needs the $ |S|\equal{}p^n,p \ is \ prime$, but in the proof of Cauchy , the order of $ |S|\equal{}n^{p\minus{}1}$, and $ n$ needs not be a prime number . Could u clarify this point for me please?

Thank you in advanced  :)

We have $ |S| \equal{} n^{p \minus{} 1}$ as we have no matter how do we choose the first $ p \minus{} 1$ elements. The last is fixed by the uniqueness of the inverse. I don't see how Lemma 5.1 is used in Hungerford's book. I have learned the proof which uses orbit-stabilizer theorem.

Okay

In Hungerford's book , The lemma in the picture depends on finite $ p\minus{}group$ , for some prime $ p$ , in the proof of Cauchy , Hungerford depended on that lemma , and he considered $ S\equal{}\Omega$ as u wrote it . Now, $ |\Omega|$ which is $ |S|$ in hungerford is $ n^{p\minus{}1}$ elements ( No problem here for me ), but he used the lemma during the proof on the set $ |\Omega|$.

How he used that lemma since $ |\Omega|$ is not necessary to be of order prime to some power , i.e. $ n$ is not necessary to be prime ?

 :)

Does he uses the theorem at all? I don't see where it is required.

Oh, sorry  :blush: , I forgot to put the rest of a proof  :D

Here is the full proof :

I can't see where he defines the set $ S_0$. However, $ |S|$ is divisible by $ p$ as the order of $ G$ is divisible by $ p$. The sentence
[quote]$ (a_1,\ldots,a_p)\in S_0$ if and only if $ a_1\equal{}a_2\equal{}\ldots \equal{}a_p$[/quote] implies that the number of elements of $ S_0$ is a multiple of $ p$, again as the order of $ G$ is divisible by $ p$.

Guys , I understood what are u said    :whistling:

To see the definition of $ S_0$ , take a look at first picture .

[b]MY QUESTION :[/b] How he used the lemma , when $ |S|$ [b]is not necessary[/b] to be some power of prime , i.e. $ n$ is not necessary to be a prime , so we couldn't say $ |S|$ is $ p$-subgroup .

The first necessary condition to use lemma is the group or subgroup is a $ p \minus{} group$.

 :)

$ S$ is not even a group, much less a p-group!  The $ H$ of the lemma is $ Z_p$, $ S$ is merely the set on which it acts.

Oh, I should pay attention more in next time  :cleaning:

Thank you LydianRain  :)


topology
advanced fields
advanced fields unsolved

Given a connected, weighted graph G with n nodes and m edges, and the cost value for each edge is also given.  Then we can construct the Minimum Spanning Tree T for this graph G.

Now the question is:

1) If a new node added into graph G, with some new edges connected with some nodes in the old graph. Then how to simply modify the constructed MST T to get a MST for the new graph.

2) If one node of graph G is deleted, the remaining n-1 nodes still connected. Then how to simply construct the MST for the new graph.

The similar problem for the directed graph.

Please give me any hints, ideas or reference papers if you have.
Thanks.

1) I would add the edges to the new vertex one by one in order of weight, and always remove the edge of largest weight in the unique emerging cycle.

2) The original ST is still good, but maybe not connected, so continue adding the cheapest edges until it does.

Exactly, the solution you give is practical and will produce a good ST.
But the solution is not necessarily optimal, right? i.e. it is not Minimum Spanning Tree any more.

So I wonder what is the simple way to generate the Minimum Spanning Tree for the new graph by modifying the old MST to adapt the topology change due to vertex inserting or vertex removing.

The method should be much simple (from the point of computation complexity) than generating the MST for the new graph without using the old MST information.

My spanning trees are optimal, at least if the original ones have been derived with the greedy algorithm.

Yes, I aggre with you that your method is optimal. Thanks for your answering my questions.

Would you please consider the following question further:

What I exactly want to do is move one of the vertex in the graph, thus weights of the edges corresponding to this vertex will be changed, especially we consider the Euclidean graph (the weight of each edge is the squared distance between the two end vertex).

One of the way to solve the problem is to remove the vertex which will be moved, and construct the MST, and then add a vertex at the new position where the vertex will move to, and construct the MST. This is why I asked the two questions before.

So now my question is whether there is a simple method to directly modify the original MST to take into account position and weights changeing of one vertex.

Furthermore, if we can move the given vertex to anywhere, where is the optimal position to minimize the total weights of the MST tree?


trigonometry
Taylor series
calculus
Inequality
Trigonometric inequality
IMO Shortlist
IMO Longlist

Prove the trigonometric inequality $\cos x < 1 - \frac{x^2}{2} + \frac{x^4}{16},$ when $x \in \left(0, \frac{\pi}{2} \right).$

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Since $\sin x<x<\frac{3}{2}x\ \left(0<x<\frac{\pi}{2}\right),$ we have

$\int_0^x \sin x\ dx<\int_0^x \frac{3}{2}x\ dx \Longleftrightarrow \left[-\cos x\right]_0^x <\left[\frac{3}{4}x^2\right]_0^x \Longleftrightarrow 1-\cos x<\frac{3}{4}x^2.$

$\therefore \int_0^x (1-\cos x)\ dx<\int_0^x\frac{3}{4}x^2\ dx \Longleftrightarrow \left[x-\sin x\right]_0^x <\left[\frac{1}{4}x^3\right]_0^x$

$\Longleftrightarrow x-\sin x<\frac{1}{4}x^3.$

$\therefore \int_0^x (x-\sin x)dx<\int_0^x \frac{1}{4}x^3\ dx \Longleftrightarrow \left[\frac{1}{2}x^2+\cos x\right]_0^x <\left[\frac{1}{16}x^4\right]_0^x$

$\Longleftrightarrow \frac{1}{2}x^2+\cos x-1<\frac{1}{16}x^4.$

$\therefore \cos x<1-\frac{1}{2}x^2+\frac{1}{16}x^4\ \left(0<x<\frac{\pi}{2}\right)\ Q.E.D.$

[quote="lastquincy"]yo yo yo man i apreceate yours solution it a very good one but we can apply the development limited[/quote]

Thanks. :) but, what is the development limited?

kunny

[quote="orl"]Prove the trigonometric inequality $\cos x < 1 - \frac{x^2}{2} + \frac{x^4}{16},$ when $\dsp x \in \left(0, \frac{\pi}{2} \right).$[/quote]

Using 4-derrivative is also [hide="avallable"] but ugly [/hide]

[quote="me@home"]is that a common problem solving technique (in upper math...kindof...) to take an expression, then make an integral from 0 to x of both sides, then equate , then repeat? or is it a really original way to attack a problem?? it seems like it would only work (practicably) on sin/cosine/(?tangent?etc.) or natural logs (maybe?). Anyway, is it useful or does it only work on this problem.[/quote]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=130777#p130777]How about this one?[/url]

ok, so I guess it is fairly common when you're stuck.
also what do people have to say about using calculus to solve some AIME problems without really knowing the "proper" solution?


quadratics
modular arithmetic
number theory unsolved
number theory

Find all integers x,y
$ x^2\plus{}6\equal{}y^5$

[quote="Athinaios"]hint
[hide]$ Z[i]$ !![/hide][/quote]
i think it shouldn't be an elementary method...

i think we can use quadratic resiprosity and legendre symbol

I'm not sure but I'am posting my ideas
$ 2|x$ $ \implies$  $ 4|6$ contradiction!
$ 2\nmid x$  $ \implies$  $ y^{5}\equiv \minus{}1\pmod{4}$ $ \implies$ $ y\equiv \minus{}1\pmod{4}$
$ x^{2}\plus{}2\equal{}y^{5}\minus{}4$,$ y^{5}\minus{}4\equiv \minus{}1\pmod{4}$ and $ y^{5}\minus{}4$ has prime divisor $ 4k\minus{}1$ form
$ p|x^{2}\plus{}2$ and $ p\equiv \minus{}1\pmod{4}$
$ x^{2}\equiv \minus{}2\pmod{p}$ $ \implies$  $ (\frac{\minus{}2}{p})\equal{}(\frac{\minus{}1}{p})(\frac{2}{p})$ $ \implies$
we know $ (\frac{\minus{}1}{p})\equal{}\minus{}1$ for $ p\equiv \minus{}1\pmod{4}$  $ \implies$
$ (\frac{2}{p})\equal{}\minus{}1$ $ \implies$  $ x^{2}\equiv 2\pmod{p}$ has not any solution.
but $ p\equal{}7$ is a solution, contradiction!

[quote="Athinaios"]hint
[hide]$ Z[i]$ !![/hide][/quote]
could one show a solution using this ring?


[quote="chien than"]Prove that:$x^{4}y^{4}+y^{4}z^{4}+z^{4}x^{4}\leq 3$
where $x;y;z$ are non-negative and satisfying $x^{3}+y^{3}+z^{3}=1$[/quote]
Maybe $x^{3}+y^{3}+z^{3}=3$ $?$ If so, see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34428
the Namdung's post. :wink:


Let's keep things simple and start with a number, $ 100$. \[ 100\minus{}\frac{10}{100}\cdot100\equal{}90\]Next\[ 90\minus{}\frac{20}{100}\cdot90\equal{}78\]Finally, $ 78$ is $ \boxed{22}$ percent of $ 100$.

It's just $ 1\minus{}0.9\cdot0.8\equal{}0.28\implies\boxed{28\%}$...

[quote="GameBot"]What is the single discount that is equivalent to the two successive discounts of $ 10\%$ off followed by $ 20\%$ off the discounted price?[/quote]

Unless I'm missing something, this is $ .9 \cdot .9 \cdot .8\equal{}.648$, so $ 64.8 \%$

The official answer is $ \boxed{28\%}$ (I have the handbook). That was my first interpretation too, BOGTRO, but I think they mean "What is the single discount that is equivalent to the two successive discounts (of $ 10\%$ off followed by $ 20\%$ off) the discounted price?" and not "What is the single discount that is equivalent to the (two successive discounts of $ 10\%$ off) followed by [a] $ 20\%$ off the discounted price?" because the "[a]" is not there. Wording could be better, but whatever.


If I have a matrix $ A$ which satisfies \[ A^2\equal{}\minus{}A\] what can be the values of $ trace(A)$?
regards
bhupala

What are the eigenvalues? How many of each are there?

Either $ A$ is the zero matrix or $ \text{trace}A$ can be any negative integer. However, if $ A$ is $ n\times n,$ then $ 0>\text{trace}A\ge\minus{}n.$

The specific thing I want to comment on is that if $ A\ne0,$ then $ A$ cannot be nilpotent. We get this from the iterating the equation above: $ A^3\equal{}A,A^4\equal{}\minus{}A,$ and in general, $ A^k\equal{}(\minus{}1)^kA,$ which is assumed not to be zero.


If the equation $ ax^2+(c-b)x+(e-d)=0$ has real roots greater than $1$, prove that the equation $ax^4+bx^3+cx^2+dx+e=0$ has at least one real root.

Let $ r$ be the root of the first equation and $ f(x)$ be the second polynomial.
It is easy to verify $ f(\sqrt r)\equal{}(br\plus{}d)(\sqrt r\plus{}1)$ and $ f(\minus{}\sqrt r)\equal{}(br\plus{}d)(\minus{}\sqrt r\plus{}1)$.
Hence, $ f(\sqrt r)f(\minus{}\sqrt r)\le0$.
Then there is a root on $ [\minus{}\sqrt r,\sqrt r]$ to $ f(x)\equal{}0$.


Given a circumference D, a circular arc G joins two distincts points on D and is inside the circle C contained in D. Prove that, if the two regions in which G divides C have equal area, then G's lenght is greater then D's diameter.

Disclaimer: the usual user of this account is not the author of the following (which basically means it will be not very well done).

We fix two points on the circle, seperated by angle $ \theta$. We must determine the curve for which area is split in half. WLOG, raduis is 1 hence we want a  bounded area of $ \frac{\pi}{2}$. We divide the cut off (inner) region into two sectorish parts of a circle - each, when viewed with correct circle, is a sector minus a triangle. Plugging through the algebra (and considering a new circle with arc on the circle acting as critical boundary) we find the constraint to be:
\[ \frac{\pi}{2}=\left(\frac{\sin \frac{\theta}{2}}{\sin \frac{\phi}{2}}\right)^{2}\cdot\left(\frac{\phi-\sin \phi}{2}\right)+\frac{\theta-\sin \theta}{2}\]
It remains to prove, under this constraint, that $ \phi\sin \frac{\theta}{2}\geq 2\sin \frac{\phi}{2}$
The last two statements are equivalent to the problem.


abstract algebra
superior algebra
superior algebra solved

let $G$ be a finite semigroup such that:
if $xy=yz$ then $x=z$
prove that $G$ is a abelian group.

We have to assume $G$ to be non-empty. Pick an element $a$. There are $i>j>0$ so that $a^{i}= a^{j}$, since $G$ is finite. Consider $e : = a^{i-j}$. We have $ae a^{j}= a a^{i}= a a^{j}= a^{j}a$, hence $ae = a$ by assumption. Now let $x \in G$ be arbitrary. Then $ae = a \Rightarrow axa = aexa \Rightarrow xa = aex \Rightarrow x = ex$ by applying twice the assumption. Similary, we optain $ea = a$ and then $x = xe$. Thus, $e$ is an identity element.

To find an inverse element for $x$, choose again $i>j>0$ so that $x^{i}= x^{j}$. Then $x^{i-j}x^{j}= x^{i}= x^{j}= x^{j}e$, hence $x^{i-j}= e$ by assumption. Since $i-j-1 \geq 0$, we can consider $y : = x^{i-j-1}$. It follows $x y x^{j+1}= x x^{i}= x x^{j}= x^{j+1}= x^{j+1}e$, thus $x y = e$ by assumption.

We conclude, that $G$ is a group. Finally, for all $x,y \in G$, we have $xy = y(y^{-1}xy) \Rightarrow x = y^{-1}xy \Rightarrow yx = xy$, so that $G$ is abelian.


What is the function $ v(t)$? What does it describe?

Assuming $ t$ is time and $ x$ is money, $ x>100\implies 1.04^x>1.04^100\approx 50.5\implies v(t)>5050.5$.
And $ v(t)$ diverges as $ x\to\infty$ (albeit slowly) so I guess your answer is it starts at about 5050 and then increases.

When you were born, your Grandmother gives you 100 dollars to put in the bank. Fortunately, at that time, Nick Bank is offering a 4% annual interest rate.
1.	Identify the multiplier.  _____________   Why?
2.	Write an Exponential Model, V(t), for the given information.
Model: ________________________  Explain:
3.	Identify the first 6 terms of the function to the nearest cent if t=0 in 1992.
t		V(t)
               t=0
               t=1

4.	Prove these values belong to the function V(t).

5.	Algebraically determine the value of your $ 100 today.
6.	Determine with the aid of your calculator when your money will be worth hundred times its initial value.
7.	How would the model change if your Grandmother told you that your money will never be worth$100 or less?

......................Does number seven make sense?

These are clearly homework problems.
Stop trying to cheat using AoPS and try asking about the concepts rather than just getting answers to problems.

I do not want the answers to the homework problems, i am not trying to cheat, i am giving you the other questions in order to help me on #7. Those other questions serve as a reference in order to answer number #7. All i want is information on #7, the others were just clarify the steps in which i had to solve to get to the last one.

Number seven is the last question. Please note the homework questions i put are there just there to show you the work i had to do in order to get to #7. Essentially, showing where the function was dervived.

[quote="ydog"]How would the model: v(t) = 100(1.04)^x change if your money will never be worth 100 or less.?[/quote]

It's unclear what your variables, or your constants for that matter, represent in this equation. For instance, $ v$ is a function of $ t$, but there is no $ t$ on the right hand side of the equals symbol.

Despite its defects, the equation bears a vague resemblance to the formula for the future value of a single deposit, held at 4% interest compounded annually. If that is so then the equation should be $ v(x) \equal{} 100(1.04)^x$, where $ x$ is the number of years the deposit is held.

Also this means the answer to your question is that the model would not have to change in any way for your money to never be worth less than $ 100.$ It will be worth exactly $ 100$ at $ x \equal{} 0$, i.e. when you first put it in the bank. Further, as far as this model is concerned $ x$ can only take on whole number values.


Let $p_i$ denote the $i$-th prime. Show that $p_1p_2...p_n + 1$ is never a square!

[quote="TomciO"]Hmm... it's just $\mod{4}$?[/quote]
Yes, and that shows that it's even not the sum of two squares ;)


quadratics
algebra
polynomial
sum of roots
product of roots

[color=red][size=200]In the name of GOD[/size][/color]

[color=blue][size=150]Dear Brothers and sisters :

If A is the root of the equation X^2 + 5X + 7 = 0. Find an equation whose new root is A+1 .

(please attach the explanation )[/size][/color]

Have $ P(x) \equal{} x^2 \plus{} 5x \plus{} 7 \equal{} (x \minus{} A)(x \minus{} B) \equal{} ((x\plus{}1) \minus{} (A\plus{}1))((x\plus{}1) \minus{} (B\plus{}1))$. Then $ A\plus{}1$ is a root of $ Q(x) \equal{} P(x\plus{}1) \equal{} (x\plus{}1)^2 \plus{} 5(x\plus{}1) \plus{} 7 \equal{} x^2 \plus{} 7x \plus{} 13 \equal{} 0$.

yes
thank you mavropnevma
this is true answer

You are a respectable

[quote="Layla Fu"]But...the equation X^2+5X+7=0 doesn't have any roots....[/quote]

$ x^2 \plus{} 5x \plus{} 7 \equal{} 0$
$ x \equal{} \frac {\minus{}5\plus{}\sqrt {3}i}{2}$ or$ \frac{\minus{}5\minus{}\sqrt{3}i}{2}$

[quote="mavropnevma"]Have $ P(x) \equal{} x^2 \plus{} 5x \plus{} 7 \equal{} (x \minus{} A)(x \minus{} B) \equal{} ((x \plus{} 1) \minus{} (A \plus{} 1))((x \plus{} 1) \minus{} (B \plus{} 1))$. Then $ A \plus{} 1$ is a root of $ Q(x) \equal{} P(x \plus{} 1) \equal{} (x \plus{} 1)^2 \plus{} 5(x \plus{} 1) \plus{} 7 \equal{} x^2 \plus{} 7x \plus{} 13 \equal{} 0$.[/quote]

$ P(A) \equal{} 0$
$ A \plus{} 1$ is a root of $ Q(x) \equal{} P(x \minus{} 1)$ instead of $ P(x \plus{} 1)$
$ P(x \minus{} 1) \equal{} (x \minus{} 1)^2 \plus{} 5(x \minus{} 1) \plus{} 7 \equal{} x^2 \plus{} 3x \plus{} 3$

Thought X belonged to R...
Thx though :oops:

let's think about this situation with an equation in the general form: ax^2 +bx+c=0 ???
Maybe or maybe not?
please instruct me :D

If the general quadratic equation $ ax^{2} \plus{} bx\plus{}c\equal{}0$ has roots $ x\equal{}\alpha$ and $ x\equal{}\beta$, then the equation with roots $ y\equal{}\alpha \plus{} 1$ and $ y\equal{}\beta \plus{}1$ is obtained by writing $ y\equal{}x\plus{}1$, that is, $ x\equal{}y\minus{}1$, giving $ a(y\minus{}1)^{2} \plus{} b(y\minus{}1) \plus{} c \equal{} 0$, that is $ ay^{2} \plus{}(b\minus{}2a)y \plus{} a \minus{} b\plus{}c\plus{}0$ or simply $ ax^{2} \plus{} (b\minus{}2a)x \plus{} a \minus{}b \plus{}c\equal{}0$.

In terms of the graph of $ ax^{2} \plus{} bx \plus{} c$, we want to move it $ 1$ unit to the right; this is equivalent ot replacing $ x$ by $ x\minus{}1$.

Alternatively, if the roots of the original equation are $ \alpha$ and $ \beta$, the the roots of the new equation are $ \alpha\plus{}1$ and $ \beta \plus{}1$. The sum of the new roots is $ \alpha \plus{} \beta \plus{}2$ and the product of tghe new roots is $ (\alpha\plus{}1)(\beta\plus{}1)$ and we can substitute these into $ x^{2} \minus{} Sx \plus{} P \equal{} 0$, where $ S$ is the sum of the roots and $ P$ is the product of the roots.

x^2 + 5x + 7 = 0

Let the other root be B

Sum of roots= A+B= -5
Product of roots= AB= 7

Let the new equation have roots A+1 and B+1

New sum of roots= A+1+B+1= -3
New product of roots= (A+1)(B+1)= AB+A+B+1=3

Then the new equation is x^2+3x+3=o

Is my answer acceptable?


geometry
perimeter
pigeonhole principle
geometry unsolved

I am having trouble with this problem.

n points are chosen from the perimeter or the interior of a regular hexagoin with sides of unit length, so that the distance between any two of them is not less than $\sqrt{2}$.  What is the largest natural number n for which this is possible?

Could you please help?

Thanks.

PS, no need for pigeonhole right?

[quote="yagaron"]what do you mean by not less?

it's greater than it? i suposse so, if it the opposite we will have a strange problem

what's the perimeter of this polygon?

split the poligon to $\sqrt{2}$ and take the integer part. you will get n[/quote]

That means that the distance between any two points has to be greater than or equal to $\sqrt{2}$.

Yeah, so you were right.

The perimeter of this polygon is just the circumference of the polygon, as in the points lie on the sides of the polygon.

That should be good enough.


a)     Howard and Samantha are sitting in a boat in a small lake.  There is a large rock in the boat.  Howard throws the rock overboard and it sinks to the bottom of the lake.  Does the water level in the lake rise, fall or remain the same?
b)     Samantha throws Howard overboard, but he’s wearing a lifejacket and thus he floats.  Does the water level in the lake rise, fall or remain the same?

a) seems straightforward but theres probably a trick to this... is therE?

It is worth mentioning that question a) was once proposed to proeminent physicists, including Robert Oppenheimer, George Gamow, and Felix Bloch, and all them gave wrong answers. However, the answer can be easily reached by comparing the volume of water displaced by the rock when it is 1) in the boat, and 2) in the water.

For question b), a similar approach can be used.


Can any one help me about this topic? My project is about this topic. I got stuck about quantization. What is it? How it work?. And about arithmetic coding, can any one give some useful link about this?
THANK YOU :)


geometry
parallelogram
geometry proposed

Let $ ABC$ be a triangle.$ M$ is a point in this triangle.Let $ A_o;B_o;C_o$ be midpoints of $ BC;CA: AB$,respectively.$ AA_o;BB_o;CC_o$ concurrency at $ M$. Take the points $ A';B';C'$ such that $ A_o;B_o;C_o$ be midpoints of $ MA';MB';MC'$,res.Prove that $ AA';BB';CC'$  concurrency  at $ M'$ and $ GM'\equal{}\frac{GM}{2}$

[quote="chien than"]Let $ ABC$ be a triangle.$ M$ is a point in this triangle.Let $ A_o;B_o;C_o$ be midpoints of $ BC;CA: AB$,respectively.[u]$ AA_o;BB_o;CC_o$ concurrency at $ M$[/u]. Take the points $ A';B';C'$ such that $ A_o;B_o;C_o$ be midpoints of $ MA';MB';MC'$,res.Prove that $ AA';BB';CC'$  concurrency  at $ M'$ and $ GM' \equal{} \frac {GM}{2}$[/quote]

Could you explain if $ M$ is a random point or is it the centroid of the triangle?(then what do you mean by the phrase I underlined?). Also is G the centroid or another point?
Please make more clear these points...

It 's not so difficult.

A', B', C' are images of M after the symetry transformation about $ A_0, B_0, C_0$

We have:

$ AC' \equal{} BM$ ( $ AMBC'$ is a parallelogram)
$ BM \equal{} CA'$ ( $ BMCA'$ is a parallelogram)

Then $ AC' \equal{} CA'$ or $ AC'A'C$ is a parallelogram. Therefore, $ CC'$ passes through the midpoint of $ AA'$ (1)

Similarly, $ BB'$ passes through the midpoint of $ AA'$. (2)

(1) and (2) means the three segments $ AA', BB', CC'$ intersects each other at the midpoint or they are concurent at the common midpoint.

$ M'$ is the midpoint of AA' then $ MM'$ and $ AA_0$ are 2 medians of $ \Delta MAA'$. So they intersects at the centroid $ G$ of $ \Delta MAA'$ which means $ GM' \equal{} \frac{GM}{2}$

[quote="chien than"]Let $ ABC$ be a triangle.$ M$ is a point in this triangle.Let $ A_o;B_o;C_o$ be midpoints of $ BC;CA: AB$,respectively.[u]$ AA_o;BB_o;CC_o$ concurrency at $ G$[/u]. Take the points $ A';B';C'$ such that $ A_o;B_o;C_o$ be midpoints of $ MA';MB';MC'$,res.Prove that $ AA';BB';CC'$  concurrency  at $ M'$ and $ GM' \equal{} \frac {GM}{2}$[/quote]
1.We have $ \vec{MA} \plus{} \vec{MA'} \equal{} \vec{MB} \plus{} \vec{MB'} \equal{} \vec{MC} \plus{} \vec{MC'}( \equal{} \vec{MA} \plus{} \vec{MB} \plus{} \vec{MC})$
So $ AA';BB';CC'$  concurrency at $ M'$ in there $ M'$ be midpoints of $ AA';BB';CC'$
2.We have $ 3\vec{MG} \equal{} \vec{MA} \plus{} \vec{MB} \plus{} \vec{MC} \equal{} 2\vec{MM'}$
So $ GM' \equal{} \frac {GM}{2}$
 :blush:  :D


search Harvard college percent geometry quadratics algebra

I think that homeworks are often useless(not just to top but also to other students).

They don't teach you anything new and they only 'review' of school stuff.

So do you think that homeworks are good?Do you think that some things about homeworks are good and some are not?

I voted for the third option, because in my experience, I've had homework that was extremely helpful in continuing my understanding of a particular subject, or to introduce a new concept for another day.

Of course, busiwork is never a good thing, because the sole purpose is to keep someone busy, and therefore it's unlikely you'll learn something.

I agree if homework means re-thinking and doing similar problems to make sure you really understand something ... but sometimes you need to have that sort of review and sometimes not. And in the most textbooks tasks are boring . Often the tasks are bacisly the same exept they put in different numbers . I think that it's easy to make  someone hate maths if you make him doing homework.

I also voted for the third option, but I have never heard of this "busiwork" you speak of...Heres the rest of my post from the other topic.  :wink:

[quote]Homework can help boost your status of the current unit, as example, fractions. At a very young age fractions can be difficult to master and certainly make you confused. My math teacher thinks it as a review, to make sure a few hours later that you still obtain the knowledge. I think that they shouldn't have as much homework unless it's reading a chapter, taking notes, finishing up classwork because it would only waste time. You could cover the worksheet or the book work in class and ask your teachers or classmates about any questions or concerns you might have. Also, my math/social studies teacher believes that homework would only clutter on the stress of after school activities, going to the park or playing a sport with friends. Maybe that's why there is obesity in the U.S, too much homework and not enough time to run around.
I don't care much about homework unless it's a whole lot to be done in one day. I'd be happy to do it as long as I'm sure I still got the information in my head.[/quote]

WELL.. homework is good if it's useful homework..
USEFUL is like: helps you understand a concept.

However, whenever homework is gone over during class, it usually takes up half the class to explain the homework and basically go over what was supposed to be understood the day before.
That's one of the reasons I don't want "stupid" people in my class.
i don't mind if homework is gone over in 5 minutes, and the rest is used for learning.

and homework can be extremely stupid such as word scrambles and word searches.
How are they supposed to help?

With word searches I understand if you give the definition and you search for the word, but for word scrambles, you are given a bunch of letters and you know the list of words it comes from.

and for reading passages, most people, like me, look only for the answers. never for the story itself.

All the work I do at school can be done at home in 10 minutes with a textbook.
:wink:

Meh. What everyone else said.

...Sometimes, it makes me wish I could just be homeschooled. It would go a lot faster, even if I did every problem.

I believe that homework is a good thing. It has been assigned to students for quite a while, and it seems to be one of those things that children with their ARGH RAGE AGAINST THE MACHINE IF IT IS CONVENIENT TO DO SO mindset enjoy viewing as worthless. It rarely takes as long as one claims it does, and it has absolutely no negative effects. (Procrastinate less if it cuts into your sleep?) It additionally has the benefit of getting kids who care somewhat about grades, but are not motivated enough to self-study - like myself - to review materials and learn something outside of the limited time that is given in class.

Also, the "That's one of the reasons I don't want 'stupid' people in my class" argument kinda bothers me. You are the unusual one for learning at an accelerated pace with respect to your classmates, yet you are placing blame upon the average students while expecting special treatment? Other kids do not learn at the same pace as the rare talented and hardworking child. Sometimes the learning basic action is difficult enough without simultaneously being force-fed the abstract concepts behind this action - just so the "stupid" people can learn at the same pace as the student multiple standard deviations above the norm.

Just my two cents. Sorry kimmystar94 if it sounds a little ad hominem; don't take it that way.

[quote="Lazarus"]Also, the "That's one of the reasons I don't want 'stupid' people in my class" argument kinda bothers me. You are the unusual one for learning at an accelerated pace with respect to your classmates, yet you are placing blame upon the average students while expecting special treatment? Other kids do not learn at the same pace as the rare talented and hardworking child. Sometimes the learning basic action is difficult enough without simultaneously being force-fed the abstract concepts behind this action - just so the "stupid" people can learn at the same pace as the student multiple standard deviations above the norm.[/quote]

I think you're missing the point. There's a reason "stupid" is in quotes. It's not because the people are actually stupid. Some people who have little academic motivation think that it's the teacher's job to lay a mound of knowledge in front of them on a silver platter, with fine champagne and a side of caviar. They are so apathetic and unmotivated that they do not want to invest the thought to determine the value or the point of the homework, and they make the teacher tell them.

Secondly, I don't think an education should focus on the average student at the expense of the advanced students. Perhaps I'm biased, but the notion that advanced students are there to help the teachers teach the average students is somewhat bogus. This is because, often, the average students are in that position because they really don't care to do better. They have other priorities.

[quote]I think you're missing the point. There's a reason "stupid" is in quotes. It's not because the people are actually stupid. Some people who have little academic motivation think that it's the teacher's job to lay a mound of knowledge in front of them on a silver platter, with fine champagne and a side of caviar. They are so apathetic and unmotivated that they do not want to invest the thought to determine the value or the point of the homework, and they make the teacher tell them. [/quote]

This remainded me abot another disatventage of homeworks.Many people say that they teach you the working habits. I don't agree. Because homework doesn't teach you that you need to do something to get what you want, but it teaches you that you'll have to do things if an only if someone tells you to do them.

[quote]
Secondly, I don't think an education should focus on the average student at the expense of the advanced students. Perhaps I'm biased, but the notion that advanced students are there to help the teachers teach the average students is somewhat bogus. This is because, often, the average students are in that position because they really don't care to do better. They have other priorities.[/quote]

$ 100$% true.

[quote="The Zuton Force"]I don't think an education should focus on the average student at the expense of the advanced students.[/quote]

It's not clear why an educational system should focus on the needs of any particular group at the expense of any other group, but if it must do so, I'd prefer it focus on "average students" (the largest group, by far) at the expense of people who are likely to succeed even with a less-than-ideal education.  I also wonder whether you think you will feel this way when you are no longer one of the smartest people around.

This latter musing is phrased in the future tense because I think it extremely likely that you will find yourself in this situation at some point (regardless of how smart you actually are -- in particular, please don't think this a slight on your intelligence).  I think this because, for example, a very large number of undergraduates at Harvard were essentially the smartest person they'd ever met until the first day of freshman year, and most of them found themselves in the middle of the pack once they arrived.

And, to give this post a little on-topic veneer: I'm hard-pressed to imagine a successful educational system that does not include required work of some sort, and I imagine this is true of the author as well.  Then the question is whether "homework in its current form" is good or bad, and the major objection here seems to be that it's not very interesting.  But since homework doesn't prevent you from going to study interesting questions in your own free time, this doesn't really strike me as compelling.  (That isn't to say, of course, that it wouldn't be nice if homework were more interesting, just that this does not seem to be a fault of homework in general but rather of the particular homework that you get assigned.)

[quote]Then the question is whether "homework in its current form" is good or bad,[/quote]

that was also meant (however I think that you can learn people some stuff without making them do it)

[quote]But since homework doesn't prevent you from going to study interesting questions in your own free time,[/quote]

here i don't agree. The main problem is that it waists your time.That time could be used to do that interesting stuff....

[color=green]Moderator says: there is [i]never[/i] a good reason to block-quote the post immediately preceding yours.[/color]

In all honesty, how many people are going to spend the time that would have been spent doing homework on something both interesting and educational?

[color=green]Moderator says: there is [i]never[/i] a good reason to block-quote the post immediately preceding yours.  (Although you get half-credit for cutting out the older quote in the post you were quoting.  :P )[/color]

Agreed, I mean if it was me, I'd spend the time on my Xbox on most days, and possibily doing AoPS on others.

Depends how new it is.

For example, we covered partial fractions a few weeks back at school, and since I'd already done some work on them I didn't do any re-enforcing exercises. (And still fare fine with them)

[quote="The Zuton Force"]I think you're missing the point. There's a reason "stupid" is in quotes. It's not because the people are actually stupid. Some people who have little academic motivation think that it's the teacher's job to lay a mound of knowledge in front of them on a silver platter, with fine champagne and a side of caviar. They are so apathetic and unmotivated that they do not want to invest the thought to determine the value or the point of the homework, and they make the teacher tell them.[/quote]

I thought of this immediately today when talking to such a student. He sleeps through class but complains about teachers not teaching. He was talking about how one of his teachers is "the worst teacher ever" (probably with a swear word in there somewhere) for having the students create a presentation on the subjects instead of him just teaching the information.

In fact, he was sleeping through our English class's discussion on the previous night's homework, which was actually had some pretty difficult analytical questions about the book we were reading. I would rate the worksheet on the "better" side of homework because it forced you to think about the book, instead of being just busywork, despite that it was annoying to do (but the question of how I feel about the usefulness of studying these books is another topic...)

I believe that if homework were to be removed completely, school performance would go up in general. Why? Because now, homework is a quite big part of one's grade (and projects). If homework were to be abolished, then the only thing left for the teacher to fill in grades with is tests. If one does need to reinforce what they have learned, then they will be forced to seek practice eventually, or fail due to believing more in "other priorities", like The Zuton Force said. I don't think anyone will be not motivated, because they will be forced to realize that they do need to know the material to do well. Anyone not motivated to earn a good test grade is not motivated to do homework either.
The only problem I see with that would be the "unfairness" of certain testing methods, or the likes, but that is not too hard to do in my opinion.

Your reasoning is solid, serialk11r, but in fact, I believe school performance would go [i]down[/i]. Not everyone would spend their time in education, in fact many are only at school because they are "forced" too. Taking tests would only result in their failure. Only the people that will take time in their education will be doing well.

If you took away homework, school performance would go [i]up[/i]? That would be true in some cases except that my teachers check if you actually [i]tried[/i], and not wrote down random numbers. That should be the case for every school.

@serialk11r's second paragraph: I agree.

But I think they should spend more money on those less gifted than the more advanced students, because advanced students can progress on their own and independence is a factor of life, is it not?

[quote="serialk11r"]
If you believe that the purpose of education is to educate society as a whole, then no homework would be bad. However I don't think that the people who aren't motivated are doing well in school anyways...
If you believe that the purpose of education is to provide for those who are motivated, then I see no reason why abolishing homework would be bad.

Personally, I would really love to have no homework.[/quote]

And that's not a selfish attitude?

The reason education is state funded is mainly due to the economic positive externalities of educating the [i]whole[/i] of society. Therefore homework has some value

WRT to the performance argument.

Please can we have some studies or leave it, hypothesis mean nothing.

[quote]In all honesty, how many people are going to spend the time that would have been spent doing homework on something both interesting and educational?[/quote]
 not many. :lol:

 [quote]How can you be so sure of new material that you don't need any reinforcing exercises?[/quote]

beacuse I often already know this stuff + whenever I learn something new i often do it by solving some problem

[quote]Well that may also happen, but I think it depends on each person's view on education and the purposes of it.
At my school I see many many people who may not be very smart, and may not get good grades, but they still try to some extent. They DO study for tests, and they study very much for them. The people failing school are pretty much all the people who sit by the back of the school and smoke weed, and I don't see any hope in them already.
If you believe that the purpose of education is to educate society as a whole, then no homework would be bad. However I don't think that the people who aren't motivated are doing well in school anyways...
If you believe that the purpose of education is to provide for those who are motivated, then I see no reason why abolishing homework would be bad.

Personally, I would really love to have no homework. I pick up on info really fast, and homework is pretty useless for me. In math I already know everything, in history there really isn't that much info in the book to memorize, bio I already know everything, etc. so it would be really great if I had no homework, then I could focus on math more and piano more. I really hate this effort thing on homework, I really don't put a whole lot of effort into homework/projects because I think it's redundant, and it hurts my grade. If grades were completely based off test grades (with the exception of english) then my grades would be a lot higher. I think a lot of people here possibly feel the same way.[/quote]

[quote]
I'm not talking about advanced students. Students who are advanced enough at math for example, will not benefit at all from going to math class. What I see is that anyone who is motivated, and can learn from class, will benefit. I really don't see any way to convince those people who look like they'll be dropping out of school to try harder and learn, so I think it is pointless. The average student seems to care to a good extent about doing well in school. Optional, light homework may be a solution to this, so those who feel like they do not have a good grasp on the material may do some problems to study. In bio for example, almost everyone in my class thinks it's a hard class, so when the teacher says "do you guys want the review sheet to be graded?" they all say "YES!!!!" and I'm the only one saying "NOOOOOO!!!". Well another reason is because they want the extra points to boost their grade, but its only a measly 2 points. Maybe they just don't realize that it won't increase their grade much, but still, they WANTED more homework.[/quote]

this is is what i'm talking about!!!

school is meant to educate avrege student. Ok I can live with that. However I really don't like when school forces me to do stuff i don't need to do.

I'm not form USA and I was completly shocked when I heard that a part of grade in USA come from these homeworks and projects.

I really don't know why school is more happy if you're doing something , and it doesn't matter wether it is going to bring any result, instead if you do a little less but there are some results at the end.

for example my teacher is always insainly happy when she sees that one of my school frineds copys every note from lesson in another notebook. this student still gets C in that subject...

i don't even write things said at lessons and I still get an A ...{but the teacher doesn't like me not doing anything}

yes I know making a good example however I don't know form where came this assumption that people all learn in the same way ?

I don't know about elsewhere, but homework is only 10-20% of our grade. And this is taking in the standard "gifted" course.

Now if you're in an AP class, there is absolutely no homework at all, and your grade is based on tests.

In my experience, having homework being a marginally small percent of your overall grade lets you "choose" which homework is needed. For example, in my geometry class, every homework is assigned a 5 point value, with a test being a 100 point value. The final grade is dividing the points you recieved over the total points possible. I have found that rarely doing your homework, but acing the tests, can still give you a good grade. We recieved a small report of our geometry grades today. I still have an A in that class, but my homework is ranked as a C.

so it isn't standard ... it can be between 10-20 %

What would be your grade if you wouldn't do any homework?

Well assuming that homework is 15% of your grade, and assuming you didn't do any of it, your maximum grade would be an 85%. Now, if your teacher uses a point system like my geometry teacher, then it is different.

[quote="Lazarus"]
Also, the "That's one of the reasons I don't want 'stupid' people in my class" argument kinda bothers me. You are the unusual one for learning at an accelerated pace with respect to your classmates, yet you are placing blame upon the average students while expecting special treatment? Other kids do not learn at the same pace as the rare talented and hardworking child. Sometimes the learning basic action is difficult enough without simultaneously being force-fed the abstract concepts behind this action - just so the "stupid" people can learn at the same pace as the student multiple standard deviations above the norm.[/quote]

I understand about the hard-working people. I can't stand the people who go to school because their parents force them to.
and I understand that the stupid thing was a bit mean of me. I really do. Sometimes I post things without even rechecking.

and the thing is, I have to say, I am very impatient. I try to do things as fast as possible, and that, I think, is what drives me. It drives me to accelerate.

BACK TO THE TOPIC!

[quote]All the work I do at school can be done at home in 10 minutes with a textbook. [/quote]

exactly my point. We need to have homework that's useful. Ten minutes with a textbook isn't useful. You search for the answer and that's about it.

[quote]Personally, I would really love to have no homework. I pick up on info really fast, and homework is pretty useless for me.[/quote]

Yeah. My case.

:\

I am not quite sure if you guys are getting the fact that homework - as trivial as it may seem to you - is [i]actually difficult for average students[/i].

Let this sink in, and reevaluate your arguments.

You all are generally pushing for homework that is challenging. However, for most students, not all, not you, homework is actually challenging.

Making homework even more difficult for the other students is unrealistic.

Similarly, although it is sad that this is the case, giving gifted students specially-tailored homework assignments to fit their needs is often unrealistic as well. Homeschooled students and privately schooled students are more lucky in this respect. It is a shame that public schools cannot do this, for funding is more necessary in other areas, but I mean this is something you will have to live with.

Take up a school sport if you want to see taxpayer money put to good use. Academics? Funding?

and on a totally unrelated note, GOOD LUCK with exeter, kimmystar94. from what i hear it's a cluster**** of homework assignments, impossibly difficult classes, and actual mental stimulation. who would have known.

[quote]I am not quite sure if you guys are getting the fact that homework - as trivial as it may seem to you - is actually difficult for average students. [/quote]

School is meant for avrege student. I can live with that.
School can't provide more interesting taks for me. I can live with that either.

But i really don't like that school is making me do this trivial tasks. Ok they can't find anything interesting so why not just leave me alone?

Homework is 70% of our grade in history :O Which is why my grade sucks ***. Homework is >80% of our grade in english, but that's because everything pretty much has to be done at home.

Yes I realize that I'm one of the few people in the whole grade at my school who can easily pull As on bio tests, and that many people are struggling to get a B. Still, there should be some kind of option to not do homework, like if your grade is below a B or something, homework is mandatory, then if your grade is higher, you can skip it. A lot of people in my class would end up doing all the homework anyways, even people with As (hard earned ones...).
I've heard from some seniors that they have teachers who don't believe in homework, or they have optional homework, and I really think this is good, because it allows each student to be able to choose the way they want to study.

[quote]Yes I realize that I'm one of the few people in the whole grade at my school who can easily pull As on bio tests, and that many people are struggling to get a B. Still, there should be some kind of option to not do homework, like if your grade is below a B or something, homework is mandatory, then if your grade is higher, you can skip it. A lot of people in my class would end up doing all the homework anyways, even people with As (hard earned ones...).
I've heard from some seniors that they have teachers who don't believe in homework, or they have optional homework, and I really think this is good, because it allows each student to be able to choose the way they want to study.[/quote]

optional homework is a very good idea(and I have it at some subjects!)  :)

This is why external examination is a much better idea. In the UK it doesn't really matter what grades you get at school (to a certain extent) since the only results which universities will care about are examined externally so all that homework I missed 10 years ago wont have too much of an effect when I apply to universities.

well the main problem with the American education system is that good students are paired with bad students- and it makes the bad students lag behind while the good ones are bored.

Giving more/less homework isn't the answer. Many times, it is helpful to lots of people, and helps the re-enforce topics. The problem is that students just skim through homework for the answers, while not really finding the meaning in it. Another problem is that while advanced students think it is easy, bad students find they can't even do it and many times don't even try.

The solution to this is to have more levels. If you have 8 periods including lunch, there would be about 5-6 academic classes. They should have a honors class, a level 2 class, a level 3 class,etc. in each subject so the best students get what they need to expand their horizons and the worst students can get extra help so they will actually understand the material.

Several thoughts:

Firstly, I've had an English/History teacher that gave us the following option for making study materials for tests that is somewhat close to optional homework. Essentially, if you get at least 85% on the test, she'll assume that you studied for it and give you an A on the study materials. If you didn't get at least 85%, she looks at your study materials and talks with you about them.

Secondly, what would really be nice is if teachers could give an independant study option for the class, and have the people using it only take the tests. Then, they couldy study/work on problems without distracting everyone around them while being bored, but the teacher could still confirm that they knew the basic curriculum.

My favorite math teacher of all time had a nice method: she would tell you the homework at the beginning of class, and then ask people what problems from the last homework they had trouble on, and go over them on the projector. Meanwhile, she didn't really care if we just did the next day's homework instead of actually listening to her, especially if we didn't have any issues with the homework she was talking about. (We didn't have any difference between homework and classwork; whatever you didn't get done in class became homework.) It was really great because she didn't care too much about what she did as long as we got the work done. We probably spent only about 7 or 8 days out of 180 taking notes, and most of the time actually working on math.  :o

One of my pet peeves is teachers (especially math) that feel like they have to force-feed you every part of the curriculum, and will waste their class time going over the previous homework even if everybody understands it anyways.

Just because homework is boring doesn't mean that it's necessarily useless (although it should be part of a teacher's job to make their subject interesting). Especially with basics in math (i.e. factoring quadratic equations, arithmetic), it is important to be able to run through it quickly so that you can focus on the more interesting parts of problems, instead of getting bogged down. It's even worse when they actually expect everyone to be paying attention.  :P

[quote]well the main problem with the American education system is that good students are paired with bad students- and it makes the bad students lag behind while the good ones are bored. [/quote]

I wouldn't quite agree with that. Especially in my experience, teachers tend to teach down to the slowest student in the class. Even as one of the "smarter" students in my class, I'm not particularly bothered if other people ask questions about things that seem easy to me. What bothers me is when the teacher has to go over everything multiple times because people aren't paying attention.

I have willingly spent a great deal of time working on homework and preparing for tests in some of my history classes in middle school, because I felt like the topics were interesting and I was actually getting out of it. However, I think one of the issues for us nowadays is that a majority of the homework we get [i]is[/i] busywork, so that even when we get non-busywork, nobody's really willing to do it.

That comes back to how to define busywork. What may be busywork for one person might not be busywork for somebody else. One possibility: as long as you get above a certain percentage on tests, you don't have to do the homework. Once you drop below that percentage, you have to do it.

BTW, I've survived class by making origami, playing with embroidery floss, and doodling wallpaper designs (http://simathcircle.org/extras_wallpaper.aspx). I know friends that have actually worked on cross-stitching in class and the teachers don't care.

This post is getting ridiculously long now, so I'll stop.

personally, I think SCHOOL is pretty pointless. everything that we learn in a year at school, I could probably master in a month or so just from a textbook.

I wish instead of giving homework, they extended the school day.


Jemima was given a spherical popsicle with radius 3. She decided to evenly remove popsicle from all sides of the popsicle until only 1/2 the popsicle was left (by volume) and give brother Jack the rest. What is the radius of the popsicle given to Jack? Express your answer in simplest radical form.

[hide]The volume of the original popsicle sphere is $3^3(\frac{4}{3})\pi=36\pi$. THus half of the sphere's volume is $18\pi$. The radius of this is $\sqrt[3]{18(\frac{3}{4})}=\frac{3}{\sqrt[3]{2}}=\frac{3\sqrt[3]{4}}{2}$. [/hide]

[hide]The area of the total popsicle is $3^3\cdot\frac43\pi=36\pi$.  Half of that area is $\frac{36\pi}{2}=18\pi$.  The radius of the small popsicle is $x$.  That means that $18\pi=\frac{4\pi r^3}{3}$
$54=4r^3$
$4r^3=54$
$r^3=\frac{27}{2}$
$r=\sqrt[3]{\frac{27}{2}}$
$r=\frac{3\sqrt[3]{4}}{2}$[/hide]

236factorial, I think you have a typo.

Pay no attention to this post. :whistling:

[quote="JesusFreak197"]Yes, he has the correct answer, but one of his fractions is flipped partway through.[/quote]

Are you talking about the $\frac{3}{4}$?

Err... I don't know what you're talking about. :whistling:

[hide]
The area of the lollipop is $\frac{4}{3}\pi27=36\pi$
Half of $36\pi=18\pi$
$\sqrt[3]{18\cdot\frac{3}{4}}=\sqrt[3]{\frac{27}{2}}=\boxed{\boxed{\frac{3\sqrt[3]{4}}{2}}}$[/hide]


let  explain contradiction:
  solve  equation :
solve :
${ x^{x^{{x}^...^{x}}}} = a$   or $ x^a = a$  hence :  $ x = \sqrt [a]{a}$
   if where $ a = 2$ or$ a = 4$then  we  accepted $ x = \sqrt {2}.$
  but retry roots then  appear  1 contracdiction:
$ \sqrt {2}^{{\sqrt {2}}^{...^{\sqrt {2}}}} = 2 = 4$
 let explain why ?

Well $ \sqrt{2}^{\sqrt{2}^{\sqrt{2}^...}}\neq 2\neq 4$.  If you have an infinite number of powers, then the expression will be infinitely large.

[quote="JRav"]Well $ \sqrt {2}^{\sqrt {2}^{\sqrt {2}^...}}\neq 2\neq 4$.  If you have an infinite number of powers, then the expression will be infinitely large.[/quote]
  maybe  explain for contradiction  problem's  not right

I'm pretty sure your solution is correct.  However, for $ x\geq1$, the exponent will blow up.  If $ x\neq1$, then the exponent will converge to 1.  So I believe $ x\equal{}a\equal{}1$ is the only value that works.

Actually since $ (\sqrt {2}) ^{\sqrt {2}} < (\sqrt {2})^2 \equal{} 2$ , you can prove that the expression with powers of $ \sqrt2$ taken finite times is less than 2. In the limit it becomes 2

[quote="JRav"]Well $ \sqrt {2}^{\sqrt {2}^{\sqrt {2}^...}}\neq 2\neq 4$.  If you have an infinite number of powers, then the expression will be infinitely large.[/quote]

This is incorrect JRav. This expression will converge for $ 1/e \leq x \leq e$ and diverge otherwise. The proof of this can be found (somewhere) on this forum.

The mistake in your original solution Hong Quy is that $ x \equal{} \sqrt[a]{a}$ is a solution only if $ a$ exists in the first place (i.e. the series converges). It would be like saying $ 1\minus{}1\plus{}1\minus{}1\plus{}... \equal{} a$ so $ a\minus{}1 \equal{} \minus{}a$, and therefore $ a \equal{} 1/2$ (which is not true).

Consider the sequence $ a_1 \equal{} x$ and $ a_{n \plus{} 1} \equal{} x^{a_n}$

For $ 0 < x < 1$, the sequence is bounded above by $ 1$. Also, we can see that the sequences $ \left\langle a_{2n}\right\rangle$ and $ \left\langle a_{2n \plus{} 1}\right\rangle$ are seperately monotonic, and I believe that they have the same limit.

For $ 2 < x$, we can write $ a_{n \plus{} 1} \equal{} (1 \plus{} h)^{a_n}\geq 1 \plus{} a_nh$
i.e. $ a_{n \plus{} 1} \minus{} a_n\geq 1 \plus{} (h \minus{} 1)a_n\geq 1$, ad hence, sequence is not convergent.

It seems that $ \exists \alpha$ such that the sequence is convergent if and only if $ 0<x < \alpha$
By actual calculation $ \alpha\approx 1.444$

Sorry, I made a mistake in my last post. The series $ x^{x^{{x}...}}$ actually converges for $ 1/e^e < x < \sqrt[e]{e}$. I know a discussion of this occured on this forum somewhere, and a proof can be found on the internet, probably by searching for "tetration".

[quote="shreyas_patankar"]Also, we can see that the sequences $ \left\langle a_{2n}\right\rangle$ and $ \left\langle a_{2n \plus{} 1}\right\rangle$ are seperately monotonic, and I believe that they have the same limit.[/quote]
This is the part that fails for $ x < e^{\minus{}e}$: for $ x \equal{} \frac{1}{20}$, for example, the odd and even sequences approach different values.


A person was looking at the statistics on the number of pedestrian fatalities in the past year. Of a total of 12 deaths, 6 were killed while crossing with the proper light and 6 were killed while crossing with the wrong light. Could this person conclude that it was just as dangerous to obey traffis signals in crossing the street as it is to disregard them? Explain

Such a conclusion would be wrong. The risk of crossing a street at a particular time and place is measured not by the number of fatalities, but by the ratio $ \frac{\text{number of fatalities}}{\text{number of pedestrians}}$. In this case the numerators are the same, but denominators are vastly different.

Aside: I find few things as frustrating as seeing (supposedly) educated people misusing statistics, in particular when it concerns education.

[quote="mlok"]Aside: I find few things as frustrating as seeing (supposedly) educated people misusing statistics, in particular when it concerns education.[/quote]
Well, has it ever crossed your mind that it may be done on purpose rather than as a result of ignorance ;)

As to the problem itself, the more familiar version of it is "about 99% of traffic accidents involve vehicles going at speeds less than 100 miles per hour and only 1% involves cars travelling above that speed, so is it 99 times safer to drive fast?"