limit
integration
trigonometry
absolute value
calculus
calculus computations

$ \lim_{ n\to \infty  }  \int_{\mathbb{R}} e ^{\minus{}|x|} (\sin x) ^{n}dx$

I'm not sure how to take arbitrary powers of negative numbers, but pretending there's an absolute value sign on the sine, the integrand pointwise converges to zero almost everywhere and we may apply Lebesgue's dominated convergence theorem.

sure :
we consider $ \int_{J}e^{-|x|}(\sin x)^{n}$
we know that $ |\sin x|\leq 1$ hence we need point' such that $ \sin x \approx 1$ .Note that :
when $ x \to \frac{\pi}{2}+2\pi k$ then $ 1- \sin x \approx \frac{1}{2} (x-\frac{\pi}{2}+2\pi k)^{2}$
hence : say $ J=[a,b]$
$ \int_{J}(\sin x)^{n}dx\approx \int_{a}^{a+\beta } (\sin x)^{n}+  \int_{b-\alpha}^{b } (\sin x)^{n}+ \sum_{j=p}^{q} \int_{\frac{\pi}{2}+2\pi k }^{\frac{\pi}{2}+2\pi (k+1)}(\sin x)^{n}$
where $ (a+\beta =0(mod \frac{\pi}{2}))$ and $ b-\alpha =0 (mod \frac{\pi}{2}))$ and $ \alpha, \beta$ is smallest .
hence (by laplace asymptotical formula ) it's $ \approx 2(q-p+2)B\frac{1}{\sqrt{n}}$ as $ \n \to \infty$
hence $ |\int_{J}e^{-|x|}(\sin x)^{n}|=O(\frac{1}{\sqrt{n}})$

[quote="Extremal"]if $ R$ don't include $ \infty$ then I think limit is equal to $ 0$[/quote]

[b]I think the limit is equal to 0 even if R include $ \infty$[/b] :D

By using [b]Lebesgue's Dominate Convergence Theorem[/b], we'll prove that easily

Put :
$ B = \{ \frac{\pi}{2} + k \pi : k \in Z \}$
$ f_{_n} \left( x \right) =  e^{- \left| x \right|}.sin^{n}x$

we see : $ \forall x \in \Re 	\backslash B ,  f_{_n} \left( x \right) \longrightarrow 0$

Because the measure of B is equal to 0, we can say that
$ f_{_n} \left( x \right) \longrightarrow 0$ almost everywhere on R (a.e. on R).

Use [b]Lebesgue's Dominate Convergence Theorem[/b].

We have :
*$ f_{_n} \left( x \right) \longrightarrow 0$ a.e on R.
*${ \left| f_{_n} \left( x \right) \right| \le  f \left( x \right) = e^{- \left| x \right|}}$ a.e. on R.
*$ f \in L^{1}\left( \Re \right)$.

$ \Rightarrow \lim_{n\rightarrow \propto}\int_{R}^{}f_{_n}\left(x \right)dx = \int_{R}^{}\left( \lim_{n\rightarrow \propto}f_{_n}\left(x \right)\right)dx = 0$


Some one pls tell how to delete posts????????????

I think it can be deleted only within a few minutes of posting it.


discussion has begun  :lol:

I'll edit this post, since I haven't thought of anything yet.

Hopefully, I will be able to retire.
 :lol:

I'll be lucky to still be alive with my legs, my heart, my head, my brain, my hands... yeah about that

Since i know i'll be rich with 2.39820997095809384092830948 x 10^83270948903850948509348 dollars, and Jesus is coming back to take with him his church, i'll donate five sixth of my money to charity and spend one twelfth on making myself comfortable, one twenty-fourth on making my family comfortable, and i'll still have around 9 x 10^83270948903850948509346 dollars left. then i'll just wait for Lord Jesus Christ to come, even if it takes one thousand years, since i have enough money to make people keep me alive to see Jesus come back the second time.

i'll also be praying that everyone on earth will trust in God.

when I'm 55 I will do 56.

if u didnt understrand, read a calc book or visit http://www.senhorcarteiro.blogspot.com

by symmetry

because 5=5

i think

i will have 666.666.666 posts on AoPS

hmm....

but if x<0

i died?

oh god! santa class


26)  Acetic acid, C2H4O2, is the substance that makes vinegar taste sour. What is the % mass of carbon in acetic acid?

1)20%

2)25%

3)35%

4)40%

5)none of these


[quote="sinajackson"]For real numbers $ a,b,c$ prove that:

$ \sum_{(a,b,c)}a\left(b^{2}+c^{2}-bc\right)\geq 3abc\geq\sum_{(a,b,c)}a\left(b^{2}+c^{2}-a^{2}\right)$[/quote]
Maybe $ a,$ $ b$ and $ c$ are non-negative numbers? :wink:

[quote="sinajackson"]For [b]nonnegative[/b] real numbers $ a,b,c$ prove that:

$ \sum_{(a,b,c)}a\left(b^{2}+c^{2}-bc\right)\geq 3abc\geq\sum_{(a,b,c)}a\left(b^{2}+c^{2}-a^{2}\right)$[/quote]

first part: $ \sum a\left(b^{2}+c^{2}-bc\right)\ge \sum a(bc)=3abc$, AM-GM

second part: $ 3abc\ge a\left(b^{2}+c^{2}-a^{2}\right)\iff a^{3}+b^{3}+c^{3}+3abc\ge \sum a(b^{2}+c^{2})$, schur.


Consider a flat washer with inner radius a, outer radius b, and thickness t.

You cut the washer in half, throw away one halve and so now you have a hemispherical washer-half with inner radius a, outer radius b, and thickness t

The resistivity is a given, p

What is the resistance from one end of the washer to the other end of the washer?

R = pL/A. I was thinking of using calculus but I end up with R = (p*2*pi*r)/(t*dr). Having dr in the denominator makes it impossible to integrate. Then I was thinking maybe I should just set the Area = (b-a)*t and setting the length to (b-a)*pi, the average of the two half-circumferences, but I don't think I can just set the length to the average...there's no logic to it at least, just intuition.

Help me!

I think I figured it out now, but there's something strange about it. If you look at it as infinitely many resistors in parallel (slice it into infinitely many semicircles) then you take the reciprocal of every individual resistor which brings the dr to the top to be integrated. The problem with this is that you assume that the current will go through each one of these "infinitely thin" resistors. But the path of charge flow, I don't think, will be around in a perfect semicircle. Wouldn't the charges on the outside radius try to move in and hug the inside radius because the inside radius is the shortest path to potential decrease? Which means the current flow will not always be through a cross-sectional area of (b-a) by t, right?


Identify the FINAL( :maybe: ) product  :wink:

If you post the subject as hoffman, doesnt that kinda ruin the fun?    :P

[hide="Answer"]The final product is phenylacetaldehyde ($ \ce{C6H5\minus{}CH2\minus{}CHO}$).[/hide]

How about if, instead of the double bond we have a triple bond?

Should I post an hint regarding my last question?

I have not been able to figure out the mechanism for the double bond case. Can someone help me?

That's right, phenylacetonitrile is the correct answer.

doesnt it get hydrolysed to the corresponding acid?  :maybe:

No, if the temperature is not high enough and if you use just one equivalent of NaHO.


[hide="2 line solution"]

The polynomial must be of degree at least 4 (to get rid of the fourth root, easy manipulation shows this)

Minimal spanning polynomial is the unique irreducible polynomial over the integers containing (x-this number) as a factor, which just so happens to be (x-1)^4-2.

Since every integer polynomial having this number as a factor has this minimal spanning polynomial as a factor, f(x) has the form ((x-1)^4-2)g(x) for some g(x). Therefore, $ f(1\minus{}2^{0.25})\equal{}((1\minus{}2^{0.25}\minus{}1)^4\minus{}2)g(1\minus{}2^{0.25})\equal{}((\minus{}2^{0.25})^4\minus{}2)g(1\minus{}2^{0.25})\equal{}(2\minus{}2)g(1\minus{}2^{0.25})\equal{}0.$[/hide]

[quote="4everwise"]Anybody have a clean proof for this?

A polynomial $ f(x)$ has integer coefficients. Prove that if $ f(1 \plus{} \sqrt [4]{2}) \equal{} 0,$ then $ f(1 \minus{} \sqrt [4]{2}) \equal{} 0.$

I just did this:
[hide]For an nth degree polynomial, use viete's formulas and some algebra on the n-1 degree coefficient and on the constant term.[/hide][/quote]
I'm not sure if this works:
[hide="Solution"]
Call all numbers in the form of $ a \plus{} b\sqrt {2}$ good where $ a$ and $ b$ are rational.  All numbers that cannot be represented in this form are bad.  Notice that the good part of $ (1 \plus{} \sqrt [4]{2})^n$ is equal to the good part of $ (1 \minus{} \sqrt [4]{2})^n$ and the bad part of $ (1 \plus{} \sqrt [4]{2})^n$ is equal to $ \minus{} 1$ times the bad part of $ (1 \minus{} \sqrt [4]{2})^n$ by a binomial expansion argument.  Thus, the good part of $ f(1 \plus{} \sqrt [4]{2})$ is equal to the good part of $ f(1 \minus{} \sqrt [4]{2})$ and the bad part of $ f(1 \minus{} \sqrt [4]{2})$ is equal to $ \minus{} 1$ times the bad part of $ f(1 \minus{} \sqrt [4]{2})$ since $ f(x)$ is a polynomial and thus can be written as $ \sum_{i \equal{} 0}^{m} {a_ix^i}$.  Yet, since $ f(1 \plus{} \sqrt [4]{2}) \equal{} 0$, we see that the good and bad part of $ f(1 \plus{} \sqrt [4]{2})$ are $ 0$, so the good and bad part of $ f(1 \minus{} \sqrt [4]{2})$ are $ 0$.  We conclude that $ f(1 \minus{} \sqrt [4]{2}) \equal{} 0$.   [/hide]

[quote=OlympusHero][quote=4everwise]Anybody have a clean proof for this?

A polynomial $ f(x)$ has integer coefficients. Prove that if $ f(1\plus{}\sqrt[4]{2})\equal{}0,$ then $ f(1\minus{}\sqrt[4]{2})\equal{}0.$

I just did this:
[hide]For an nth degree polynomial, use viete's formulas and some algebra on the n-1 degree coefficient and on the constant term.[/hide][/quote]

This is a special case of the rule that if a polynomial $f(x)$ has rational coefficients, then if any number is a root, then its conjugate is also a root. I would say that your method (@4everwise's) would work to prove this general case, but I'm not sure.[/quote]

not exactly, like if we had $1 + \sqrt[3]{2}$ as a root of polynomial $P$ then $1-\sqrt[3]{2}$ is not nessecarily a root.

in general if $a + \sqrt[n]{b}$ is a root then $a + \sqrt[n]{b}\cdot \omega$ MUST be a root where $\omega$ is a $n$'th root of unity
as usual $\sqrt[n]{b}$ cannot be written as like $\sqrt[m]k$ where $m > b$ and $a, b, n$ reals


limit
integration
logarithms
calculus
calculus computations

Prove that
\[ \lim_{n\rightarrow \infty} n^2 \int_{0}^{\frac {1}{n}} x^{x \plus{} 1}dx \equal{} \frac {1}{2}
\]
.

[hide]
Use L'Hopital's rule.  First let $ u \equal{} 1/n$, so that

$ L \equal{} \lim_{n\rightarrow\infty} n^2 \int_0^{\frac{1}{n}} x^{x\plus{}1} \, dx \equal{} \lim_{u \rightarrow 0^\plus{}} \frac{1}{u^2} \int_0^u x^{x\plus{}1} \, dx$.

Then since this is clearly an indeterminate form of 0/0 type, we let

$ F'(x) \equal{} f(x) \equal{} x^{x\plus{}1}$

and write

$ L \equal{} \lim_{u \rightarrow 0^\plus{}} \frac{\frac{d}{du}\left[F(u) \minus{} F(0)\right]}{\frac{d}{du}\left[u^2\right]} \equal{} \lim_{u \rightarrow 0^\plus{}} \frac{u^{u\plus{}1}}{2u} \equal{} \frac{1}{2} \lim_{u \rightarrow 0^\plus{}} u^u$.

Now, it is relatively straightforward to compute the last limit:

$ \lim_{u \rightarrow 0^\plus{}} u^u \equal{} \exp \lim_{u \rightarrow 0^\plus{}} u \log u \equal{} \exp \lim_{u \rightarrow 0^\plus{}} \frac{1/u}{\minus{}1/u^2} \equal{} \exp 0 \equal{} 1$.

Therefore, L = 1/2, as claimed.
[/hide]

very interesting. that was a somewhat different method than what i used.

Hmmm.  So then, what method did you use?

First i proved  $ \lim_{x\rightarrow0^ \plus{} } x^x \equal{} 1$
Now let $ \epsilon > 0$ and choose a particular $ \delta > 0$ such that for $ 0 < x < \delta$ we have $ |x^x \minus{} 1| < \epsilon$. So for $ n \geq \frac {1}{\delta}$ we have the following:
$ \left| n^2 \int_{0}^{1/n} (x^{x \plus{} 1} \minus{} x) dx \right| \leq n^2 \int_{0}^{1/n} |x^{x \plus{} 1} \minus{} x| dx$
$ \equal{} n^2 \int_{0}^{1/n} x |x^x \minus{} 1| dx < \epsilon n^2 \int_{0}^{1/n} x dx \equal{} \frac {\epsilon}{2}$
And this implies that
$ \lim_{n\rightarrow \infty} n^2 \int_{0}^{1/n} (x^{x \plus{} 1} \minus{} x) dx \equal{} 0$
so we get
$ \lim_{n\rightarrow \infty} n^2 \int_{0}^{1/n} x^{x \plus{} 1} dx \equal{} \lim_{n\rightarrow \infty} n^2 \int_{0}^{1/n} x dx \equal{} \frac {1}{2}$.


limit
induction
algebra unsolved
algebra

Replace $2004$ by $m$.

We will construct the desired sequences by induction on $m \geq 1$.

- For $m=1$, just choose $x_{1,n} = n$

- For $m=2$, let $\alpha + \beta = 1$, where $\alpha, \beta > 0$.
If $\displaystyle \alpha = \frac p q $ is rational, where $1 \leq p < q$ are positive integers.
For $n = pi+r$, with $ 0 \leq r \leq p-1$, let $x_{1,n} = (i-1)q+r$, that is the sequence $\{x_{1,n} \}$ goes like this :
$0,1, \cdots, p-1,q,q+1, \cdots, q+p-1, 2q,2q+1, \cdots $.
Clearly $\displaystyle \frac n {x_{1,n}} \sim \frac {pi} {q(i-1)} \sim \frac p q = \alpha$, as desired.
Similarly, $\displaystyle \frac n {x_{2,n}}  \sim \frac {p-q} q = \beta$.
Thus, we are done in that case.

If $\alpha$ is not rational, then $\beta$ is not rational too. Since $\alpha + \beta = 1$, the well-known theorem of Beatty states that the sets ${\displaystyle S(\alpha) = \{ [ \frac n {\alpha} ]  / n=1,2,3, \cdots }$ and ${\displaystyle S(\beta) = \{ [ \frac n {\beta} ] / n=1,2,3, \cdots }$ form a partition of $N^{*}$.
Now, add $0$ to $S(\alpha)$ and define $\displaystyle x_{1,n} =  [ \frac n {\alpha} ]$ and $\displaystyle x_{2,n} =  [ \frac n {\beta} ]$.
Clearly, $\displaystyle \frac n {x_{1,n}} \sim \alpha$ and $\displaystyle \frac n {x_{2,n}} \sim \beta$, as desired.
Thus, the result is proved for $m=2$.

Now assume that it holds for some $m- 1 \geq 2$.
Let $\alpha_1, \cdots, \alpha_m > 0$ such that $\displaystyle \sum_{i=1}^m \alpha_i = 1$.
Let's use the result for $m=2$ to $\alpha = \alpha_m$ and $\displaystyle \beta = 1 - \alpha_m = \sum_{i=1}^{m-1} \alpha_i$.
Then, we may consider a sequence $\{x_{m,n} \}$ such that $\displaystyle \frac n {x_{m,n}} = \alpha_m$ and a sequence $\{x_n \}$ such that $\displaystyle \frac n {x_n} = \beta$.

Now let's consider the canonical bijection $f$ such that $f(x_n)=n$ for all $n$.
Not that $\displaystyle \frac {f(x_n)} {x_n} = \frac n {x_n} \sim \beta$. (1)

Then use the induction hypothesis  for the numbers $\displaystyle \frac {\alpha_1} {\beta}, \cdots, \frac {\alpha_{m-1}} {\beta}$ which are positive and add up to $1$.
Then, we may consider sequences $\{X_{k,n} \}$, where $k=1, \cdots, m-1$, which form a partition of $N$ and such that $\displaystyle \frac n {X_{k,n}} = \frac {\alpha_k} {\beta}$. (2)
Now, for $k=1, \cdots, m-1$ and all $n$, let's define $x_{k,n}$ to be the integer (recall that $f$ is bijective) such that $f(x_{k,n}) = X_{k,n}$.
Thus, using (1) and (2), we deduce that :
$\displaystyle \frac n {x_{k,n}} = \frac n {X_{k,n}} \cdot \frac {X_{k,n}} {x_{k,n}} = \frac n {X_{k,n}} \cdot \frac {f(x_{k,n})} {x_{k,n}} \sim \frac  {\alpha_k} {\beta} \cdot  \beta = \alpha_k$ and we are done.

Pierre.

Thanks Pierre,very nice solution.Here is my approach:
*) The set density:
Suppose s is the subset of N, we call $\alpha$ is the density of S if:
  $\lim_{n\to\infty}\frac{\#\{x\in S;x\geq n\}}{n} = \alpha (0\leq \alpha \leq 1)$
Note: Dirichlet' theorem
c(n)=The number of prime number $p=a+bn$;
d(n)=The number of prime number $p\leq n$
    $\lim_{n\to\infty}\frac{c(n)}{d(n)}=\frac{1}{\varphi(b)}$
*)Theorem:
Assume S is infinite set,$\alpha$ is the density of S,$x_1,...,x_n $ are elemets of S s.t.:$x_1<x_2<...<x_n$
Prove that: $ \lim_{n\to\infty}\frac{n}{x_n}= \alpha$
Of course, here is a question! Try to find a proof!
*)I think if we use this theorem we'll find the problem13' solution


I need this to be factored, I CAN get the answer, but the problem is it takes too long, and it revolves around guessing around, put things here and their, so here the problem..

12y^6 +23y^3 + 10

thanks!


inequalities
real analysis
real analysis solved

Let $ a_n>0$ such that $\sum_{n>0} {\frac{1}{na_n}}$ converges. Then $\sum_{n>0}{\frac{1}{a_1+a_2+...+a_n}}$ also converges.

First, by the arithmetic mean-harmonic mean inequality, $\frac1{a_1+a_2+\dots+a_n}\le \frac1{n^2}(\frac1{a_1}+\frac1{a_2}+\dots+\frac1{a_n})$
Now, $\sum_n \frac1{n^2}(\frac1{a_1}+\frac1{a_2}+\dots+\frac1{a_n}) =\sum_j \sum_{n=j}^{\infty}\frac1{n^2}\cdot\frac1{a_j}<\sum_j \frac2j\cdot\frac1{a_j}$
(using the estimate $\frac1{n^2}\le\frac2{n(n+1)}$) and the final series converges by hypothesis.

This also gives the estimate $\sum_{n>0}\frac{1}{a_1+a_2+...+a_n}<2\sum_{n>0} \frac{1}{na_n} $.


probability
geometry
geometric transformation
reflection

I have a class (Java language) that can have three types of errors, Severe, Medium and Informational error. There can be any combination of these 3 types of errors in all classes. There is also the possibility that none of the errors might occur. I want a formula for calculating the “probability of error” for each class based on how many errors of different types each class has.

Class a has 2 severe error, 3 medium and 5 informational
Class B has 4 severe error, 2 medium and 3 informational
……..
…….
Obviously class B will be more error prone than class A as it has more severe error. I need a formula that will consistently reflect this idea. Greatly appreciate your time.

Please read the forum descriptions before you post -- the Round Table is [i]not[/i] a forum for solving math problems.


Let $ABC$ be a triangle and $D,E$ are points on the segment $BC,CA$ respectively such that $AE=xAC$ and $BD=yBC$ Find in terms of $x,y$ the ratio $AF:FD$ ($AD$ intersects $BE$ in $F$)

if this can help for solving the problem it was in the exercises about the theorem:
If the lines $AB$ and $PQ$ intersect at $M$ then $\frac{(ABP)}{(ABQ)}= \frac{PM}{QM}$

Since nobody else seems to reply, let me do:

[quote="Labiliau"]Perhaps you can use the Van Aubel's theorem. Let $J = AB \cap FC$

You obtain that :

$\frac{AF}{FD}=\frac{AE}{EC} + \frac{AJ}{BJ} = \frac{x}{1-x} +\frac{AJ}{BJ}$

But , after that???[/quote]

After that, you can apply Ceva and you're done. But there is still a simpler solution:

Apply the Menelaos theorem to the triangle ADC and the collinear points F, B, E on its sides AD, DC, CA. You get

$\frac{AF}{FD}\cdot\frac{DB}{BC}\cdot\frac{CE}{EA}=-1$

(with directed segments). Now, DB = - BD = - y BC, CE = AE - AC = x AC - AC = (x - 1) AC and EA = - AE = - x AC. Thus, this becomes

$\frac{AF}{FD}\cdot\frac{-yBC}{BC}\cdot\frac{\left(x-1\right)AC}{-xAC}=-1$,

what simplifies to $\frac{AF}{FD}=\frac{-x}{\left(x-1\right)y}=\frac{x}{\left(1-x\right)y}$.

  Darij

Use Menelaus' theorem $\frac{AE}{EC}\cdot \frac{CB}{BD}\cdot \frac{DF}{FA}=1\Longleftrightarrow \frac{x}{1-x}\cdot \frac{1}{y}\cdot \frac{DF}{FA}=1$,yielding $AF:FD=x:y(1-x)$


Prove that there are no 4 coplanar points, such that the distance between any two of them is an odd integer.


geometry
3D geometry
sphere
topology
Euler
advanced fields
advanced fields unsolved

I'm just curious to know how we could prove that a cup of the tea is same as Donut?

I know that how to convert cylinder to tours starting with rectangular  . Is this has a different transformation in Quotient space.?

and if so , give me such an example please?

 :)

There is no good way to give a topological model of a tea-cup, so you have to somehow visualise the deformation.
The point is that the cup can be continuously deformed into a sphere with a handle, which is a homeomorphic to the torus.
But of course you don't prove anything like without using e.g the classification of topological $ 2\minus{}$manifolds.

Thus, you meant we need something different to visulaize the deformation of the cup .

For me, need more detailes ..... :maybe:

Could you give such an example in 2-mainfield please..

Thank you :)

I meant that we need to formalize the concept of a tea-cup in order to prove something mathematically.  If we don't have a model for something, all we have is the intuitive feeling of the proof.
A manifold is a topological space which looks like $ \mathbb{R}^n$ locally around each point. The $ 2\minus{}$ indicates the dimension of $ M$, i.e $ n\equal{}2$.
Thus a sphere is a 2-manifold, since around each point we have an open set which can be identified by an open set in $ \mathbb{R}^2$ (for example by projection).
Also, its not hard to convince oneself that the torus is also a 2-manifold. The point is that we have classified all compact 2-manifolds, and it turns out that they are one of the following types:

1) T # T # … # T, a connected sum of tori;
2) P # P # … # P, a connected sum of projective planes.

The point is that if you accept that a tea-cup has certian topological properties, like being a orientable compact 2-manifold, the classification theorem gives you the equivalence..
So this is a way of going around the problem of a concrete model for the tea-cup.

That's more clear dude.

Thank you ... :D

http://en.wikipedia.org/wiki/Topology
gives the actual deformation
(also a picture of this deformation is given in Kinsey's "Introduction to Topology")

To rigorously prove, though, after establishing the structure of a tea-cup (i.e. give a simplical complex for it), we would use classification theorem (in this case, showing that tea-cup is equivalent to 1 Torus)

to Ottem: also don't forget compact surface can be classified as simply a sphere (we can't forget about all of those three-dimensional solids with euler characteristic 2 !)

I saw that deformation , do you know these subjects are very interested in Topology such as to understand why the Topologist couldn't distinguish between a cup of tea and a donut  :D  :D  :D

Classification theorem  one of my favorite theorem .

Thank you


Find all integers x such that

(3 - (2 - (1 - x2)2)2)2 = 1

i don't think there are more but -1 and 1 work. i plugged it in. hehe.

Alternative solution :

Equation is f(x) = x^18 + ...(doesn't matter) ... + 3 = 0 so if x is an integer solution then x | 3 i.e. x = +-1 or +-3.

+-1 is easily checked to be good, f(+-3) is obviously too big to be 0.

And DUDE [hide]Spoiler[/hide] man. Now there's no fun for other people.

[quote="JBL"]churchill -- don't use Mathematica!  This is a non-computer (or calculator) -aided challenge :)

Scrambled:  You're on the right track, except for one mistake.  At EACH square root, you have to make the positive or negative choice.  You can't just take all positives or all negatives -- you also have to consider the case where, say, the first is negative, the second is positive, and the third is negative.[/quote]

i don't have mathematica! why would i use mathematica?

[quote="TripleM"]OK, mathematica, maple, blah blah, you used a computer/calculators solving power. Please don't do that.[/quote]
not exactly maple's motto is COMMAND THE BRILLIANCE OF A THOUSAND MATHEMATICIANS. according to Waterloo i asked alot of smart of people to help me solve this LOL.

I have no idea what you're talking about, but you obviously used a calculator of some form.  Please don't do that (especially on problems where it triviallizes it entirely).


Given an integer $ a_0 > 0$, let $ a_{n \plus{} 1}$ be either $ 2a_n \plus{} 1$ or $ 3a_n \plus{} 1$. Can we always make the choices between $ 2a_n \plus{} 1$ and $ 3a_n \plus{} 1$ such that $ a_i$ is a square for some $ i$.

I don't understand what you mean by "Can we always make the choices between...". Would you mind clarifying?

I think he means "for any $ a_0$, is there always a choice of transitions $ a_i \to 2a_i \plus{} 1$ or $ a_i \to 3a_i \plus{} 1$ for each $ i$ such that some term in the resulting sequence is a square?"

on second thought this definitely belongs in olympiad


Each of the 15 student council members of Alamo Mid-school is responsible for telling exactly two other students about the upcoming school dance. Each of those two, in turn, is responsible for telling two other previously uniformed students. It takes 15 minutes for each round of students to be told. How many minutes will it take for 1,000 students to receive the information about the school dance?

I believe the answer is the smallest possible value for $ n \times 15$ where $ n$ is $ 15 \times 2^n > 1000$.

Thus, the answer is $ 7 \times 15$, or $ \boxed{105}$.


function
trigonometry
algebra
domain
calculus
derivative
calculus computations

Find the maximal and minimal values of the function $u=2sinx+2siny+sin(x+y)$ on the ste of points $0\leq x,y\leq \frac{\pi}{2}$.
 :?

I take it we are to use calculus.

[hide="Well, then... "]
$u_x = 2 \cos x + \cos (x + y)$
$u_y = 2 \cos y + \cos (y + x)$

We seek $u_x = u_y = 0$.  Then $\cos x = \cos y$.  On the given domain $\cos$ is a one-to-one function, so $x = y$.  Then we wish to find $x$ such that

$2 \cos x + \cos 2x = 0$
$2 \cos^2 x + 2 \cos x - 1 = 0$
$\cos x = \frac{-1 \pm \sqrt{3}}{2}$

Obviously only the positive solution is possible.  Thus

$(x, y) = \left( \arccos \frac{-1 + \sqrt{3}}{2}, \arccos \frac{-1 + \sqrt{3}}{2} \right)$

Is the only critical point.  The rest is calculation.[/hide]

[quote="t0rajir0u"]I take it we are to use calculus.  [/quote]
Yes.
Thanx.

One more question:
what value given function would have in this point $(x, y) = \left( \arccos \frac{-1 + \sqrt{3}}{2}, \arccos \frac{-1 + \sqrt{3}}{2} \right)$
maximal or minimal?

Ah, sorry, my solution actually far from complete.

[hide="Rest of solution"]
We need to apply the Second Derivative Test.

$u_{xx} = -2 \sin x - \sin (x + y)$
$u_{yy} = -2 \sin y - \sin (y + x)$
$u_{xy} = - \sin (x + y)$

$D = u_{xx} u_{yy} - ( u_{xy} )^2 = 2 \sin x \sin y + 2 \sin y \sin (x + y) + 2 \sin x \sin (x + y)$

$D$ must be positive because each individual term is positive given our domain.  That means we have an extrema.

Now, $u_{xx}$ is negative.  That means we have a maxima.

Now, the other thing I neglected to mention is that the entire border region (a square) needs to be tested for maxima/minima.  Because the sides of the square are parallel to the axes, this only requires that you maximize/minimize on each side individually, which is a 2-dimensional problem and much easier.

The calculations for this step are extremely tedious.  :blush:
[/hide]

write it as

$T=4\sin(\frac{x+y}{2})\cos(\frac{x-y}{2})+2\sin(\frac{x+y}{2})\cos(\frac{x+y}{2})\\=2\sin(\frac{x+y}{2})(2\cos(\frac{x-y}{2})+\cos(\frac{x+y}{2}))$

Since
$0\leq \frac{x+y}{2}\leq \frac{\pi}{2}$ and $-\frac{\pi}{4}\leq \frac{x-y}{2}\leq\frac{\pi}{4}$

so $T\geq 0$ . So the minimum is when $x=y=0$ .

Now for the maximum , we see that if $x+y$ is fixed , then $\cos\frac{x-y}{2}$ is maximize when $x=y$ . So maximum occurs when $x=y$ .

So we rewrite $T=4\sin x+\sin 2x$

$\frac{dT}{dx}=4\cos x+2\cos 2x$

$\frac{dT}{dx}=0$ give us $2\cos^2x+2\cos x-1=0$ which is

$\cos x=\frac{-1+\sqrt{3}}{2}$

Now $\frac{d^2T}{dx^2}=-4\sin x-4\sin 2x\leq 0$ for $x\in[0,\pi/2]$

So $x=y=\arccos\frac{-1+\sqrt{3}}{2}$ is indeed the maximum .


Let $ A,B \in \mathcal{M}_n(\mathbb{R})$, s.t. $ A^2\plus{}B^2\equal{}AB$.
If $ BA\minus{}AB \in \mbox{GL}_n(\mathbb{R})$, what could we say about $ n$?

See maky's third post [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=816016#816016]here[/url].


ceiling function
floor function
number theory unsolved
number theory

With$ a,b\in Z_{\plus{}}$ Denote $ d(a,b)$ is the number of divisors of $ a$ which not less than $ b$. Find  all positive integer
$ n$ such that: $ d(3n\plus{}1,1)\plus{}d(3n\plus{}2,2)\plus{}...\plus{}d(4n,n) \equal{} 2006$

The answer is $ n\equal{}708$.

Call the expression $ D(n)$. Count the number of times a certain divisor $ k$ is counted in the sum. It may only be counted in the numbers $ 3n\plus{}1$ through $ 3n\plus{}k$, where it occurs as a divisor exactly once. Hence each possible divisor $ 1$ through $ n$ is counted exactly once, so $ D(n)\equal{}n\plus{}d(3n\plus{}1,n\plus{}1)\plus{}d(3n\plus{}2,n\plus{}1)\plus{}...\plus{}d(4n,n\plus{}1)$. The remaining count includes the numbers $ 3n\plus{}1$ through $ 4n$ (there are another $ n$), as well as a half and a third of each, if possible. The number which are even is $ \lceil\frac{n}{2}\rceil$, and the number which are odd is $ \lfloor\frac{n}{3}\rfloor$. Therefore $ D(n)\equal{}2n\plus{}\lceil\frac{n}{2}\rceil\plus{}\lfloor\frac{n}{3}\rfloor\equal{}2006$. Clearly, D(n) is increasing, since $ 2n$ is increasing and $ \lceil\frac{n}{2}\rceil\plus{}\lfloor\frac{n}{3}\rfloor$ is nondecreasing. Hence we may check that $ n\equal{}708$ is an answer, and we are done.

Where does this problem come from?

It comes from http://www.mathlinks.ro/Forum/viewtopic.php?p=524249#524249


1.  Consider two concentric circle of radii 2 and 5.  Let S be the set of all points which are the midpoint of a segment with one endpoint on each of the two circles.  Compute the area of S.

2.  Do the same thing for circles of radii R1 and R2.

3. Two concentric circles are given.  A fixed point P is chosen on the larger of the two circles.  A non-fixed point S is chosen on the smaller of the two circles, and points Q and R are chosen on segment PS so that PQ = QR = RS (they trisect PS).  It is found that as S is moved about its circle, Q and R both trace out circles.  It is found that these two circles are tangent to each other.  Compute the ratio of the radii of the circles.

I'm not totally sure because I am unable to get some paper right now, but i think i have #1 and #2:

[hide]

1. The boundaries of S are 1) the midpoints of the lines from the outer circle to the inner circle that are of the shortest distance, 3, and 2) the midpoints of the lines from the outer circle to the inner circle that are of the longest distance, 7. Any point inside these boundaries can be reached by adjusting the line.

So this makes two circles, one of radius 3.5 and one of radius 1.5. The area is  :pi: 3.5 :^2:  -  :pi: 1.5 :^2:  = 10 :pi: .

2. By the same method, we get one circle of radius (R_1 + R_2)/2 and another of radius (R_2 - R_1)/2. The area between is thus  :pi: [(R_1 + R_2) :^2: /4 - (R_2 - R_1) :^2: /4] = R_2 * R_1 :pi: .

[/hide]

Is it harder than this maybe?


I have a question:
How come i cant activate my avatar ... it has 2.37 kb and is regular size ... its all ok
plz sum help ^^

e-mail me with your avatars. I will upload them.

i sent you my avatar some time ago(kuba, Fratila Dragos)
did u receive it??
if not i'll send it to you again


vector
calculus
derivative
function
Pythagorean Theorem
geometry
absolute value

Let a particle on a plane have its path described by the vector $f(t) = x(t)i + y(t)j$. Then its velocity vector is $f^{'}(t) = x^{'}(t)i + y^{'}(t)j$ and its acceleration vector is $f^{''}(t) = x^{''}(t)i + y^{''}(t)j$.

As I understand it, the speed of the particle can be calculated as $\sqrt{x^{'}(t)^2 + y^{'}(t)^2}$, and so the rate of acceleration is the derivative of the speed: $\frac{d}{dt} \sqrt{x^{'}(t)^2 + y^{'}(t)^2}$.

However, can the rate of acceleration not also be calculated as $\sqrt{x^{''}(t)^2 + y^{''}(t)^2}$ using the Pythagorean theorem on $f^{''}(t)$ (in the same way speed was calculated using the Pythagorean theorem on $f^{'}(t)$)? But these two methods of calculating the rate of acceleration yield different values. Why or why not?

Any help would be appreciated. Thanks.

Think of a constant speed circular motion: you have some non-zero acceleration directed towards the center of the circle because the velocity vector changes its direction all the time. On the other hand, the speed (the absolute value of the velocity) does not change and an attempt to differentiate it gives zero. So, it is not true that the absolute value of the acceleration is the derivative of the absolute value of the velocity and, in general, the absolute value of the derivative of a vector-valued function is not the same as the derivative of its absolute value.


${\left(\ldots \left(\left(\left(\left(A_{1}\times A_{2}\right)\times A_{3}\right)\times A_{4}\right)\times A_{5}\right)\times \ldots \right)}$

someone have a good formula for this using Levi-Civita tensor?!

i know that this is a stupid problem.

but i cannot make a good equation using this...

thanks for every body!!!!!

I don't know... iterated cross products are mysterious. The statement
[quote]Given two parenthesizations of the iterated cross product $v_{1}\times v_{2}\times \cdots\times v_{n}$, there is a way of choosing each vector $v_{1},\dots,v_{n}$ from the set of basis vectors $\imath,\jmath,k$ so that both parenthesizations give the same nonzero result[/quote]
is [url=http://math.ucr.edu/home/baez/qg-fall2000/qg8.2.html]equivalent to the four-color theorem[/url]. :what?:
If there was a simple way to represent such products, perhaps the four-color theorem would not be that hard.

wooow..

sorry but i'm too stupid to understant this theorem.. but thanks

in my problem we can use the Levi-Civita tensor a lot of time.
we will have a Product. but. i think this product is a permutation. and i dont know how to interpret this.

Example: with have a Symmetrical Sum, a good notation of this. and anothers sums.

but i never see a Symmetrical Product or similar, this exist?!

if yes the problem can be solved easily.

do someone knows a symmetrical product? this exist?!

thanks everybody again


Let A be an nxn mx with n distinct eigenvalues and let B be an nxn mx with AB=BA. if X is an eigenvector of A, show that BX is zero or is an eigenvector of A with the same eigenvalue. Conclude that X is also an eigenvector of B.

Thanks!

Does anyone know how how to do it? Thanks!

Please help me out for this question. It is very important for me. Thanks!

Thanks, that is great.
[quote="fredbel6"]Okay I wil help you out

first note that if A has n different eigenvalues, for every eigenvalue, the eigenspace is just onedimensional
let $x$ be an eigenvector of $A$ with eigenvalue $\lambda$

Now we have $ABx=BAx$
or $ABx=B(Ax)$
or $ABx=B(\lambda x)$
or $A(Bx)=\lambda(Bx)$
now either $Bx$=0, or it is not, but in that case it is obviously another eigenvector of $A$ with eigenvalue $\lambda$, thus it is a multiple of $x$, thus $Bx=kx$
We conclude that in the last case $x$ is also an eigenvector of $B$[/quote]


abstract algebra
group theory
superior algebra
superior algebra unsolved

I came across this (probably very simple) exercise:

Let $G$ be a group, let $A$ be an abelian subgroup of $G$. Show that $A \lhd C(A)$.

(Here $C(A)$ is the centralizer of $A$, the set of all elements of $G$ which commute with all elements of $A$.)

Now, in my "proof" (?) I do not use the fact that $A$ is ABELIAN. Am I doing something wrong?
Or doesn't $A$ need to be Abelian?

[quote="Arne"]I came across this (probably very simple) exercise:

Let $G$ be a group, let $A$ be an abelian subgroup of $G$. Show that $A \lhd C(A)$.

(Here $C(A)$ is the centralizer of $A$, the set of all elements of $G$ which commute with all elements of $A$.)[/quote]

Looks rather trivial to me, but please correct me if I am mistaken.

First, since A is abelian, every element of A commutes with all elements of A, so every element of A is an element of C(A), and thus A is a subgroup of C(A).

Now, if a is an element of A, and b is an element of C(A), then a and b commute (by the definition of C(A)), so that ba = ab and thus $bab^{-1}=abb^{-1}=a$. As this is true for every $a\in A$, we thus get $bAb^{-1}=A$, and since this holds for every $b\in C\left(A\right)$, it follows that A is a normal subgroup of C(A).

[quote="Arne"]Now, in my "proof" (?) I do not use the fact that $A$ is ABELIAN. Am I doing something wrong?
Or doesn't $A$ need to be Abelian?[/quote]

Hmm, this sounds very strange to me, as already the fact that A is a subset of C(A) implies that every element of A commutes with every element of A, and thus A must be Abelian. So, for non-Abelian A, simply A is not a subgroup of C(A) (the "normal" condition is satisfied).

  Darij

I agree the exercise is completely trivial, but I didn't "notice" that the fact that A is abelian implies that C(A) is a subgroup of A.

[quote="darij grinberg"]
Hmm, this sounds very strange to me, as already the fact that A is a subset of C(A) implies that every element of A commutes with every element of A, and thus A must be Abelian. So, for non-Abelian A, simply A is not a subgroup of C(A) (the "normal" condition is satisfied).
[/quote]

So that's the clue. Thanks, I should have seen that.


Show that there cannot be four squares in arithmetical progression.

let $ a^{2}$ $ b^{2}$ $ c^{2}$ $ d^{2}$  be $ 4$ squares in an ap
therefore $ c^{2}$ $ \equal{}$ $ 2$ $ \cdot b^{2}$ $ \minus{}$ $ a^{2}$
and $ d^{2} \equal{}3 \cdot b^{2} \minus{} 2 \cdot a^{2}$
thus $ b^{2} \plus{} d^{2} \equal{} 2 \cdot c^{2}$
this equation has no solution {trival by infinite desent}

This equation has infinitely many solutions, for example $ (b, d, c)\equal{}(1, 1, 1)$.

A classic by Fermat ! Another solution [url=https://arxiv.org/pdf/0712.3850]here[/url].


function
calculus
derivative
geometry
3D geometry
limit
calculus computations

Find all functions f such
      $ \displaystyle 2{f^3}\left( x \right) \plus{} 1 \equal{} f\left( {2{x^3} \plus{} 1} \right),\forall x \in R$

Let $ g(x) \equal{} 2x^3 \plus{} 1$.  This is a bijection from $ \bf R$ to itself.  Define an equivalence relation $ \sim$ on $ \bf R$ by $ x \sim y$ if and only if there is some $ n \in \bf Z$ such that $ x \equal{} g^{n}(y)$, where this means $ g$ composed with itself $ n$ times (or $ g^{ \minus{} 1}$ composed with itself $ |n|$ times, if $ n$ is negative).  From each equivalence class, choose a representative and define $ f$ arbitrarily on this representative.  This extends uniquely to a function $ f$ on $ \bf R$ with the desired property.

Now, if you wanted $ f$ to be continuous at some point or something ....

thx JBL.

I am interested in analytic solutions

an obvious solution is $ f(x) \equal{} x$.

Yup, obvious is x and inverse too. Are there others analyticals?

From the notation of my previous post, any function of the form $ g^n(x)$ for $ n \in \bf Z$ is a solution.  If $ n \geq 0$, these are all polynomials.  If $ n < 0$, they are continuous and differentiable almost everywhere but have infinite derivative in some points (they involve cube roots and whatnot).

To get a continuous function, we again get lots of arbitrary free choices: define $ f(\minus{}1)$ arbitrarily, then define $ f$ on $ [1, 3]$ and $ [\minus{}15, \minus{}2]$ as an arbitrary continuous function so that $ f(3) \equal{} g(f(1))$ and $ f(\minus{}15) \equal{} g(f(\minus{}2))$ and $ \lim_{n \to \infty} g^{\minus{}n}(f(1)) \equal{} \lim_{n \to \infty} g^{\minus{}n}(f(\minus{}2)) \equal{} f(\minus{}1)$.  Then $ f$ extends uniquely to a continuous function on all of $ \bf R$.

To get derivatives to work out is obviously going to require a bit more care: we need that when we extend $ f$ by using the defining relation, the derivatives patch together at the appropriate points and also that the derivative exists at $ \minus{}1$.


$\bigstar$ Let $a,b,c,d$ are real number such that $ad-bc=\sqrt{3}$. Prove that:
$a^{2}+b^{2}+c^{2}+d^{2}+ac+bd\ge 3$


Suppose that among any four participants of a competition there is at
least one who knows the other three. Prove that there is a participant
who knows all the others.

Let's take four people : M, A, T and H.
Hypothesis : M doesn't know A.
Then, if we take T and H, T or H knows everybody else.
So T knows H and T or H knows both M and A.
Hence all the $n-2$ people apart from M and A know each other.
And since one of those knows M and A, conclusion.

One can also easily prove this result via induction.


I hate having to get back into all this again grrrr.
I'll try to shorten it:
Q: A company has inspectors to find defective products. Inspectors A, B, and C have the probabilities of detecting an existing defect at 0.9, 0.8, and 0.7 respectively. They operate independently.

1. "A" inspects item and either rejects it or passes it on to "B" who either rejects or passes it on to "C" and either rejects it or assumes its good. What is the probability that a defective item is detected?

AND

2. All three inspect the item at the same time for defects, and the product is either accepted or rejected. What is the probability that a defective item is detected?

My take is for #2 you would just multiply .9*.8*.7 since they are all working on the same number of products?

For #1 though I'm not sure how to take into account the ones that A rejects before B and the ones that A&B reject before C has a chance to reject. (I assume that #1 would be more accurate but am confused a little)

it should be 1 - 0.1*0.2*0.3 or 0.994 because it's the probability that they all screw up subtracted from 1


Buenos dias,

Durante esta semana se esta realizando el Examen de admision a
la Universidad Nacional de Ingenieria
aqui en Lima-PERU,
es considerada como una de las más exigentes en aspecto academico...
esta organizada en tres examenes:
Lunes 13 Ago: Aptitud Academica y Cultura General.
Miercoles 15 Ago: Matematicas.
Viernes 17 Ago: Fisica Quimica.

Aqui les pongo unos links donde podran encontrar los examenes
y las soluciones de los exams..

http://www.aduni.com.pe/eventos/uni_2007_ii/portada.html
Aqui pueden encontrar los exams, solucionarios

http://www.aduni.com.pe/eventos/seminariosuni/portada.html
Aqui inclusive pueden encontrara los seminarios
que organiza la ICH especialmente para este proceso de admision.. revisar.

Saludos,

¿como es el sistema de ingreso a la educacion superior en Peru?

interesante la prueba de matemáticas, me doy cuenta de la diferencia en el nivel de educación y aprendizaje comparado con otros paises.  :o


Find all $ f: \mathbb{R}\to\mathbb{R}$ such that for all $x,y \in \mathbb R$,
\[ f(x)f(y) - f(x + y) = \sin(x) \sin(y).\]

[quote="TRAN THAI HUNG"]Find $ f(x)$ continuous in $ R$ and
$ f(x)f(y) \minus{} f(x \plus{} y) \equal{} sin(x)sin(y)$ for all $ x,y \in R$:?:[/quote]

Let $ P(x,y)$ be the assertion $ f(x)f(y)\minus{}f(x\plus{}y)\equal{}\sin(x)\sin(y)$

$ P(x,0)$ $ \implies$ $ f(x)(f(0)\minus{}1)\equal{}0$ and, since $ f(x)$ cant be always zero, $ f(0)\equal{}1$

$ P(\frac{\pi}2,\minus{}\frac{\pi}2)$ $ \implies$ $ f(\frac{\pi}2)f(\minus{}\frac{\pi}2))\equal{}0$ and so

Either $ f(\frac{\pi}2)\equal{}0$ and then $ P(x\minus{}\frac{\pi}2,\frac{\pi}2)$ $ \implies$ $ f(x)\equal{}\cos(x)$
Either $ f(\minus{}\frac{\pi}2)\equal{}0$ and then $ P(x\plus{}\frac{\pi}2,\minus{}\frac{\pi}2)$ $ \implies$ $ f(x)\equal{}\cos(x)$

And it's easy to check back that this function indeed is a solution.

Hence the unique solution : $ \boxed{f(x)\equal{}\cos(x)}$

And, btw, the "continuity" constraint is useless.


is there any principle on the left site of this inequality?  :huh:

well, this can be done by induction:

This was done in school and I wasn't feeling like writing it all again.  :oops:


Anyone else depressed about this?

Get moral support here. :D

 :( Guess I just have to wait until USAMTS comes out in August.

I'm depressed as well, as I didn't do too well this year  :(

I feel a mixture of 2 feelings :

for 3 months,

1) No more fun math competitions... :(
2) No more failure (except swimming)... :)

I will just be studying for the whole summer for the contests next year(yeah I need a life). This year was awesome...too bad its gone.

@dynamo729, you swim? Awesome, you're the only other Aopser that I know that swims.

yeah... what are your times at each stroke???? pm me.

ps I bet swimfreak22 swims. :P

Ehmm... In Colombia (and in a great part of South and Center America, I think :|) the most active months are May, June (OMCC, IMO) and September (IbMO), so I don't feel very depressed :| .


function
algebra
polynomial
number theory
relatively prime
IMC
college contests

Let $f: \mathbb{R}\to\mathbb{R}$ be a function such that  $(f(x))^{n}$ is a polynomial for every integer $n\geq 2$. Is $f$ also a polynomial?

What does $f^n$ here mean? Does it mean the $n$-th power of $f$?

Furthermore for what values of $n$ that takes place? For a certain fixed $n\geq 2$, or for an infinity of values of $n$?

The problem says that $(f(x))^{n}$ is a polynomial for any $n{\geq}2$.

In this case, the answer is positive, right?

First of all, the irreducible factors of $f^2$ have even exponents, since this must hold for $f^6=(f^3)^2=(f^2)^3$. This means that there is a real (I suppose we're working with real functions? always specify these "details", guys :)) polynomial $P$ s.t. $f^2=P^2$.

Now, since $f=\sqrt[3]{f^3}$, and both $f^3$ and $\sqrt[3]{\cdot}$ are continuous, $f$ is continuous, so $f$ is either $P(x)$ or $-P(x)$ on every interval on which $P$ does not vanish. The only problem (making $f$ different from both $\pm P$ on the real line) might arise in this manner: for some zero $t$ of $P$, $f(x)=P(x)$ on some interval of the form $[t,t+\varepsilon)$ and $f(x)=-P(x)$ on the interval $(t-\varepsilon,t]$ (or vice-versa, it doesn't really matter; take $\varepsilon$ so small that there are no other zeros of $P$ in $I=(t-\varepsilon,t+\varepsilon)$), but in this case it's easy to see that $f^3$ is not polynomial on $I$.

At grobber: yes it is real, but it does not matter does it? You can just prove it generally?

My sol:

$f^2=\prod^m_{i=1}(x-\lambda_i)^{a_i}$, $a_i$ positive integers
$f^4=\prod^m_{i=1}(x-\lambda_i)^{2a_i}$
therefor
$f^3=\prod^m_{i=1}(x-\lambda_i)^{\frac32a_i}$ polynomial, thus $a_i$ even.

So $f:=\prod^m_{i=1}(x-\lambda_i)^{\frac{a_i}2}$ is a valid polynomial f, now prove its uniqueness:

Where $f^2(x)\not=0$, we have $f=f^3/f^2$, so unique. Where $f^2(x)=0$, we have that the multiplicity of the root is higher in $f^3$, while $f^3=f^2.f$, so $f(x)=0$ there, thus f is unique everywhere.

f = f^3/f^2 is a rational fraction.
So f = A/B, where A, B are relatively prime polynomials; as A^2 = B^2.f^2, B divides A and B is relatively prime with A; hence B is constant and f is a polynomial

for any $x$, there exists $n_x \in \mathbb{N}$ such that: $f^{(n_x)}(x)=0$. Hence, $f$ is a polynoimial  :D


calculus
integration
calculus computations

Show that for $ a,b,c>0$:
\[ \int_{0}^{1}x^a(1\minus{}x^b)^cdx\equal{}\Gamma{(\frac{a\plus{}1}{b})}\frac{\Gamma{(c\plus{}1)}}{b}\Gamma{(c\plus{}1\plus{}\frac{(a\plus{}1)}{b})}\]

Substitute $ u\equal{}x^b,$ then you'll get a beta integral.


Prove that $(x-a)(x-b)=c^{2}$ has real roots

Edit:  Latex stuff fixed.
I would assume that c is a real number because if it were a complex number the equation is not satisfied for all $a$ and $b$.

[hide]
Expand LHS, bring over $c^{2}$

$x^{2}-(a+b)x+ab-c^{2}= 0$

Use the quadratic formula.  The discriminant must be positive because it is under a radical, so proving that it is would be enough for the proof.  The discriminant is
$b^{2}-4ac$ where $b$ is the x coefficient, $a$ is the $x^{2}$ coefficient and $c$ is the constant.

$(a+b)^{2}-4(ab-c^{2}) = a^{2}+2ab+b^{2}-4ab+4c^{2}= a^{2}-2ab+b^{2}+4c^{2}= (a-b)^{2}+4c^{2}$

all terms are positive since they are squared. QED[/hide]

[quote="Aegor"]My Latex skills do not exist, so please forgive me. :P

[hide]
Expand LHS, bring over c^2

x^2 - (a+b)x + ab - c^2 = 0

Use the quadratic formula.  The discriminant must be positive, so proving that is enough

(a+b)^2 - 4(ab-c^2) = a^2 + 2ab + b^2 - 4ab + c^2 =
a^2 - 2ab + b^2 + c^2 =
(a-b)^2 + c^2

all terms are positive since they are squared. QED[/hide][/quote]

To use Latex, just put the text you want to write in Latex between dollar signs ($\$$).  It's that simple!

For example:

My Latex skills do not exist, so please forgive me. :P

[hide]
Expand LHS, bring over $c^{2}$

$x^{2}-(a+b)x+ab-c^{2}= 0$

Use the quadratic formula.  The discriminant must be positive, so proving that is enough

$(a+b)^{2}-4(ab-c^{2}) = a^{2}+2ab+b^{2}-4ab+c^{2}= a^{2}-2ab+b^{2}+c^{2}= (a-b)^{2}+c^{2}$

all terms are positive since they are squared. QED[/hide]


conics
parabola
rotation
geometry
geometric transformation
search

Let V be the vertex of $ y\equal{}k\minus{}x^2$ and let A lie on the parabola so that $ AV\equal{}sqrt30$. The parabola is rotated around the y-axis. Determine the area of the circle that A sweeps out.

Can someone help?

[hide]So you let A be $ (m, k\minus{}m^2)$. Using the distance formula, you find that $ m\equal{}\sqrt{5}$. Then, I [i]think[/i] that rotating around the y-axis means you basically find draw the perpendicular from A to the y-axis, and rotate A about that point. The answer would then be $ m^2\pi\equal{}5\pi$. Remember, I'm not sure about what rotating about the y-axis means.[/hide]

yeah, i got to that, but then, i didnt know what rotation around the y axis meant.

thanks though =].

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=243470]This was just posted.[/url]

In the future, search problems before posting.


inequalities
AMC
USA(J)MO
USAMO
symmetry
three variable inequality
Hi

We have x³ + y³ \geq x²y + xy² for all x, y > 0 by rearrangement. So
\sum 1/(a³ + b³ + abc) \leq \sum 1/(a²b + ab² + abc) = 1/abc by simple computation.

Out of curiosity, why would you even want to let $abc=1$?  You can apply Jensen's in the exact same fashion without de-homogenizing.

That's a good point. I don't know why; I was just working with it and decided to set $abc=1$. Then I saw the solution with Jensen's after. So it was merely a detour in my thought process. Thanks for pointing that out. It's probably a clearer solution that way.

this has been discussed a couple times  ;)  I want to say it's  USAMO 97, not sure. the idea is to say a^3 + b^3 >= ab(a+b) to make each denominator smaller.

[Moderator edit: For this problem, see also
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=58140
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19469
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=69177
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=55677
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=49889
]

I came across this thread looking for practice USAMO inequalities so I apologize for reviving such an old thread, but here's a slightly different solution using symmetry and bunching:

[hide]$\frac{1}{a^3 + b^3 + abc} + \frac{1}{b^3 + c^3 + abc} + \frac{1}{c^3 + a^3 + abc} \leq \frac{1}{abc}$

Clearing out the denominators and multiplying by $2$, we obtain:

$\displaystyle{\sum_{\text{sym}} (a^3+b^3+abc)(b^3+c^3+abc)abc \le 2(a^3+b^3+abc)(b^3+c^3+abc)(c^3+a^3+abc)}$

This simplifies to:

$\displaystyle{\sum_{\text{sym}} a^7bc+3a^4b^4c+4a^5b^2c^2+a^3b^3c^3 \le \displaystyle{\sum_{\text{sym}} a^3b^3c^3+2a^6b^3+3a^4b^4c+2a^5b^2c^2+a^7bc}}$

We can simplify this even further to:

$\displaystyle{\sum_{\text{sym}} 2a^6b^3 - 2a^5b^2c^2 \ge 0}$, which holds by bunching.[/hide]

[i]Some motivation.[/i] The summands are just so 'nice'. This makes me feel the urge to separate right hand side into $\frac{a}{abc(a+b+c)},$ etc.
[i]Solution.[/i] And miraculously it works. Of course by muirhead and $[3,0] \succ [2,1]$ one has $a^3 + b^3 \geq a^2b + ab^2.$ Then, the left hand side is at most $\sum_{cyc} \frac{a}{abc(a+b+c)} = \frac{1}{abc}.$

Does this count as a one liner  :D ?
[hide=solution]
By Muirhead, \[ \frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \leq \frac{1}{a^2b+ab^2+abc}+\frac{1}{b^2c+bc^2+abc}+\frac{1}{c^2a+a^2c+abc} = \frac{1}{a+b+c}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right) = \frac{1}{(abc)} \]
[/hide]

Well I will add another one using Cauchy- Schwarz Inequality

We have using that $\dfrac{a^2}{bc}+\dfrac{b^2}{ca}\ge \dfrac{(a+b)^2}{c(a+b)}\Rightarrow  \dfrac{a^2}{bc}+\dfrac{b^2}{ca}+1\ge \dfrac{a+b+c}{c}$ $\Rightarrow \dfrac{1}{\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+1}\le \dfrac{c}{a+b+c}\Rightarrow \sum_{cyc}\dfrac{1}{\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+1}\le 1$

Equality holds iff $a=b=c$.

[quote=Maverick]Prove that, for all positive real numbers $ a$, $ b$, $ c$, the inequality
\[ \frac {1}{a^3 \plus{} b^3 \plus{} abc} \plus{} \frac {1}{b^3 \plus{} c^3 \plus{} abc} \plus{} \frac {1}{c^3 \plus{} a^3 \plus{} abc} \leq \frac {1}{abc}
\]
holds.[/quote]

Prove that, for all positive real numbers $ a$, $ b$, $ c$, the inequality
\[ \frac {1}{a^3 \plus{} b^3 \plus{} abc} \plus{} \frac {1}{b^3 \plus{} c^3 \plus{} abc} \plus{} \frac {1}{c^3 \plus{} a^3 \plus{} abc} \geq  \frac {5}{2abc+a^3+b^3+c^3}\]
holds.

[quote=Maverick]Prove that, for all positive real numbers $ a$, $ b$, $ c$, the inequality
\[ \frac {1}{a^3 \plus{} b^3 \plus{} abc} \plus{} \frac {1}{b^3 \plus{} c^3 \plus{} abc} \plus{} \frac {1}{c^3 \plus{} a^3 \plus{} abc} \leq \frac {1}{abc}
\]
holds.[/quote]

Note

\[{a}^{3}+{b}^{3}+abc=ab \left( a+b+c \right) + \left( a+b \right)
 \left( a-b \right) ^{2}\]

\[{\frac {1}{ab \left( a+b+c \right) }}+{\frac {1}{bc \left( a+b+c
 \right) }}+{\frac {1}{ca \left( a+b+c \right) }}={\frac {1}{abc}}
\]

Done.

Not sure if this is what splintercell990 meant by bunching:

Clearing out the denominators and rearranging yields:

$\sum_{sym} a^6b^3 \geq 2\sum_{sym} a^5b^2c^2 $

We can split this into two inequalities:

$\sum_{cyc} a^6b^3 \geq \sum_{sym} a^5b^2c^2 $
and
$\sum_{cyc} a^6c^3 \geq \sum_{sym} a^5b^2c^2
$

Both follow from weighted am-gm.

Clearing the denominator, we get
$S(5,2,2) \leq S(6,3,0)$
This is just Muirhead inequality :)

By CS we have,
\[(a^3+b^3+abc)\left(\frac1a+\frac1b+\frac{c}{ab}\right)\geq (a+b+c)^2\]
\[\implies \frac{1}{a^3+b^3+abc}\leq \frac{c}{abc(a+b+c)}\]
Summing the three inequalities, we get
\[\sum_{cyc} \frac{1}{a^3+b^3+abc} \leq \sum_{cyc}  \frac{c}{abc(a+b+c)} = \frac{1}{abc}\]

lets use the following inequality

$a^3+b^3 \geq ab(a+b)$ then

$\sum_{cyc} \frac{1}{a^3+b^3+abc} \leq \sum_{cyc} \frac{1}{ab(a+b)+abc} \leq \frac {1}{abc}$

[quote=Arne]We have$ x^3 + y^3 \geq x^2y + xy^2$ for all $ x, y > 0$ by  $ rearrangement.$ So
$\sum \dfrac {1}{(a^3 + b^3 + abc)} \le \sum \dfrac {1}{(a^2b + ab^2 + abc)} = \dfrac {1}{abc} $ by simple computation.[/quote]
Fixed Arne's solution  :)

$a^2–2ab+b^2\ge 0$
$\implies a^2–ab+b^2\ge ab$
$\implies (a+b)(a^2–ab+b^2)\ge (a+b)ab$
$a^3+b^3\ge ab(a+b)$
$\implies ab(a+b)+abc=ab(a+b+c)$
 [color=pink]And thus follows I think everyone has the same conclusion............. [/color]

We know that $a, b, c > 0$.

By Chebyshev's inequality we have $(\sum_{cyc}\frac{1}{a^3+b^3+abc})(2(a^3+b^3+c^3) + 3abc) \leq 9$

Notice that $(2(a^3+b^3+c^3) + 3abc) \leq 3(a^3+b^3+c^3)$ by AM GM

$(\sum_{cyc}\frac{1}{a^3+b^3+abc}) \leq \frac{3}{a^3+b^3+c^3}$

By AM GM $a^3+b^3+c^3 \geq 3abc$ so it follows that $(\sum_{cyc}\frac{1}{a^3+b^3+abc}) \leq \frac{3}{a^3+b^3+c^3} \leq \frac{1}{abc}$. $\blacksquare$

Wait in the Chebyshev's Inequality part, shouldn't it be the other way around, because Cauchy Scwarz says that is greater than or equal to 9???

[quote=fukano_2]Solved with [b]nprime06[/b]

Since the inequality is homogenous, WLOG let $a+b+c = 1$ and since $a^3+b^3 \ge a^2b+b^2a$ we have
\[\sum_{\text{cyc}} \frac{1}{a^3+b^3+abc} \le \sum_{\text{cyc}} \frac{1}{a^2b+b^2a+abc} = \frac{1}{a+b+c}\sum_{\text{cyc}} \frac{1}{ab} = \sum_{\text{cyc}} \frac{1}{ab} \le \frac{1}{abc} \]
which is trivial by expanding. $\square$[/quote]
----------------------
Why do you even need to hold that ? pls correct me if I am wrong .--------------------------
[quote]let $a+b+c=1$[/quote]
----------------------------------
Since by Muirhead Inequality,
\[\sum_{\text{cyc}} \frac{1}{a^3+b^3+abc} \le \sum_{\text{cyc}} \frac{1}{a^2b+b^2a+abc} \]
$$\sum_{\text{cyc}} \frac{1}{a^2b+b^2a+abc}=\frac{1}{a^2b+b^2a+abc}+\frac{1}{b^2c+c^2b+abc}+\frac{1}{c^2a+a^2c+abc}$$
$$=\frac{1}{a+b+c}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)$$
$$\left(\frac{1}{a+b+c}\right)\left(\frac{a+b+c}{abc}\right)=\frac{1}{abc}$$
So it true that,
$$ \frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \leq \frac {1}{abc}
$$
Proved $\square$

Here is mine solution just for storage and make things simple if something wrong in it please tell me.  :roll:
[b][color=#00f]My solution:[/color][/b]
Here we can use that \[a^3+b^3\geq ab(a+b)\]this can be seen by the $(a+b)^3$ formula and proved by muirhead because we know that $(3,0)\succ(2,1)$\\
Now, We can see that \[\sum_{cyc}\frac{1}{a^3+b^3+abc}\leq \sum_{cyc}\frac{1}{ab(a+b)+abc}=\sum_{cyc}\frac{1}{ab(a+b+c)}\]After multiplying but $c$ in numerator and denominator we can get\[\sum_{cyc}\frac{1}{a^3+b^3+abc}\leq \sum_{cyc}\frac{c}{abc(a+b+c}=\frac{1}{abc}\sum_{cyc}\frac{c}{a+b+c}=\frac{1}{abc}\] And we are done ! :D

By Rearrangement, $a^3+b^3\ge a^2b+ab^2$. Then we homogenize by assuming $a+b+c=1$. Finally:
$$\sum_{\text{cyc}}\frac1{a^3+b^3+abc}\le\sum_{\text{cyc}}\frac1{a^2b+ab^2+abc}=\sum_{\text{cyc}}\frac1{ab}=\frac1{abc}.~\square$$

Let $ a,  b,  c$ be  positive real numbers .Prove that
$$ \frac{1}{a^3+b^3+kabc} + \frac{1}{b^3+c^3+kabc} + \frac{1}{a^3+c^3+kabc} \leq \frac{3}{(k+2)abc}$$
Where $k\geq 1.$

[quote=blahblahblah]$a^3+b^3\geq ab^2+a^2b$, so we get that:

$\sum \frac{abc}{a^3+b^3+abc}\leq \sum\frac{abc}{ab(a+b+c)} = \sum\frac{c}{a+b+c}=1$[/quote]

very effective

Let $ a,  b,  c$ be  positive real numbers .Prove that
$$ \frac {a}{b + c +ka} + \frac {b}{c + a +kb}+\frac {c}{a + b +kc} \leq \frac{3}{k+2}$$
Where $k\geq 1.$

The key is to notice that $$\frac{abc}{a^3+b^3+abc} \leq \frac c{a+b+c} \iff a^3+b^3 \geq a^2b + ab^2$$ is true. Now just sum cyclically and divide by $abc$.

Let $a^3=m$, $b^3=n$, $c^3=p$, WLOG(homogeneous)let $mnp=1$, we need to prove that
\[\frac1{m+n+1}+\frac1{n+p+1}+\frac1{p+m+1}\le1,\]
(expand) or $\sum(m^2n+mn^2)\ge2\sum m=2\sum m^{4/3}n^{1/3}p^{1/3}$ which is true by Murihead.

We see by Muirhead that $x^3 + y^3 \ge x^2y + xy^2$ for any positive reals $x,y$. Hence \begin{align*}
\sum_{cyc} \frac{1}{a^3 + b^3 + abc} \\
\leq  \sum_{cyc} \frac{1}{a^2b + ab^2 + abc}  = \sum_{cyc} \frac{1}{ab(a + b + c)} \\
= \frac{1}{abc} ,\\
\end{align*} as desired.

WLOG, to homogenize, let $abc=1$. Note that since by Muirhead's, we have that $a^2b+ab^2\geq a^3+b^3$. Using this, we get that
\[\sum_{\text{cyc}} \frac{1}{a^3+b^3+abc}=\sum_{\text{cyc}}\frac{abc}{a^3+b^3+abc} \leq \sum_{\text{cyc}} \frac{abc}{a^2b+ab^2+abc} = \sum_{\text{cyc}} \frac{c}{a+b+c}=1=\frac{1}{abc},\]
finishing the problem.

By Murihead's inequality:
\[a^3+b^3\geq a^2b+ab^2.\]
Therefore, the LHS is $\leq$:
\begin{align*}
\sum_{\text{cyc}}\frac{1}{ab(a+b+c)}&=\frac{1}{a+b+c}\sum_{\text{cyc}}\frac{1}{ab}\\
&\leq\frac{a+b+c}{abc(a+b+c)}\\
&=\frac{1}{abc}.\\
\end{align*}

The equation is homogeneous so assume $abc=1$. Then it suffices to prove

\[\sum_{\text{cyc}} \frac{1}{a^3+b^3+abc} \le 1.\]

Notice that $(3,0)$ majorizes $(2,1)$ so

\[x^3+y^3 \ge x^2y+xy^2.\]

by Muirhead. Thus, we have

\begin{align*}
\sum_{\text{cyc}} \frac{1}{a^3+b^3+abc} &= \sum_{\text{cyc}} \frac{abc}{a^3+b^3+abc} \\
&\le \sum_{\text{cyc}} \frac{abc}{a^2b+ab^2+abc} \\
&= \sum_{\text{cyc}} \frac{c}{a+b+c} = 1. \ \square
\end{align*}

$a^3+b^3=(a+b)(a^2-ab+b^2)\geq(a+b)(ab)$     as $a^2-2ab+b^2\geq 0$.

Thus $LHS\leq \sum_{\mathrm{cyc}}\frac{1}{ab(a+b+c)}= \frac{a+b+c}{abc(a+b+c)}= RHS$

We multiply both sides by $(a^3+b^3+abc)(a^3+c^3+abc)(b^3+c^3+abc)abc$. We cancel terms and get \[\sum_{\text{sym}} a^6b^3 \geq \sum_{\text{sym}} a^5b^2c^2\] This is obvious from Muirhead. $\blacksquare$


You could not be more wrong.  Just go to a university website, look at the faculty of the department that interests you and look at what the professors are doing research in, the papers they publish.

[quote]I am interested in math, computer programming, and physics. What career can I have with those interests ? I have seen a lot of people interested in biology and chemistry doing amazing things. Just wonder what I can do with my interests. Is it true that math, computer science, and physics became old ? And this is the age of biology and chemistry.[/quote]

Why not look at Computational Biology?  One of the [b]big[/b] reasons biology is doing amazing things is that mathematicians and computer scientists have been developing new statistical and software tools that biologists can use.  In the area I work in (phylogeny reconstruction), there is just as much fun and important work to be done on the math and CS as there is on the biology-- my supervisor, in fact, is a computer scientist!  (I'm not!)  You don't have to be a biology whiz to do really, really good work.

Another career choice.  If you like economics, become an investment banker.  If you get hired by a good company, you'll get more money in a year than most people see in their lives.

[quote="Dominic Connor"]I was just about to agree 100% with JRav, when I did some arithmetic (this is a maths forum after all).

langtupanda is presumably pre-university and therefore late teens (say 18), maybe going to college this Autumn.
That means hitting the job market in 2012 to 2016 then working until the 2050s / 2070s

I'm not smart enough or arrogant enough to call the demand skills in 2030.

The obvious career [i]today[/i] for Maths/Physics/Programming is Quantitative Analyst. Good money, sometimes [i]very[/i] good money. The right sort of PhD gets 150-200K in first year, but I emphasise [u]right[/u], and certainly there is no guarantee. It's high end problem solving, with a competitve element.

I'm in two minds about CompSci as a career degree, and some major wells in this topic have run dry. But today there is enormous amounts of stuff going on. Physics and IT intersect deeply in chip design which is continuously evolving playgroud. Sorting out the energy mess will require physics.

As  for chemistry, again you've seen too much TV. Nanotechnology is deeply cool and doing big stuff, and is basically chemistry with a bit of physics spun in.

But I fear that ultimately the problem is that Creationists run your media. When dumb, dishonest people portray science they hook onto celebrities and stuff they can understand, which is a small set. Hence you get the same things repeated over and over. Hence lots of dull trash about cute animals, but whilst trying to dodge the "e*******n" word.

You need to work out what you are better than other people at.
Once you've done that, research into areas that use those talents.[/quote]

What does right sort of PhD mean?

I don't think it matters this much. I think it's not the words on your diploma but the fact that you have one and what sort of stuff you actually did while getting it. If you are interested in a particular thing then you will probably do research projects on that or build something related to it or whatever, and then you can talk about that when you are applying for a job in a related field. They won't care about what exactly are the courses you took if you are competent and motivated.

So basically with all of those it sounds like you could get some kind of EE type job in industry (like for a computer hardware firm), a programming job, a tech job with any kind of company, a research position (at a lab or university), or you could teach high school CS or computer maintenance or something. and that may seem like a lot of different stuff but it's probably like 1/10 of the possibilities.

[quote="§outh§tar"][quote="langtupanda"]To §outh§tar: "PhD" stands for "Doctor of Philosophy". It's considered as an advanced academic degree in university.

Can anyone tell me what I can do with:

1. A computer science degree and a computer engineering degree.

2. A computer science degree and a electrical engineering degree.

3. A computer science degree and a electronic engineering degree.

I just don't know which engineering degree would make up a good combination with computer science degree.

Thanks in advance.[/quote]

I know what  PhD is. I was asking what "right sort" means.[/quote]

I guess "Right sort" = "Right kind"

Thank Xevarion for the advice.

In my quant careers guide, I go on at some length about choosing PhDs.

Currently signal processing is deeply valued by banks, as is applied maths in general, time series analysis is good.
But you have to be good at them, and doing real programming (C++) is necessary to get the most value.
But that's now...
In 5 years ?

[quote="langtupanda"]
Can anyone tell me what I can do with: ...

[/quote]
Roughly speaking, if you want to be a software developer, develop software systems or algorithms then Computer Science and Computer Engineering degrees are better choices. If you want to design hardware (e.g. processors), an Electrical Engineering degree might be a better choice. Computer Engineering is more focused on practical aspects of software development comparing to Computer Science. So if you want to do research or work in academia, you should choose a Computer Science degree.

im an economics/mathematics graduate from UT Austin, and for the past several years I've been teaching high school-level mathematics. now im looking to change careers, and I have no idea what I want to do.
I know I don't want to be a banker, an actuary, or a sales guy in anyway. i like problem solving, statistics, numbers in general. i don't have a graduate degree yet either. maybe in the future but first i want some relevant work experience.the problem is i have no idea though where any of this stuff fits into any particular career. anybody got any suggestions? please reply when you've got a chance.

loco


Use the figure below to obtain exact expressions for sin 18 and cos 18:

any help would be greatly appreciated!

thank you.

Well, if you want to find an expression for sine, I'd suggest drawing a perpendicular from a vertex.

[hide="Stuff"]
We assume that $ AB\equal{}1$, since it doesn't matter.

Since $ m\angle ABC\equal{}m\angle ACB\equal{}72^\circ$, we have $ AB\equal{}AC\equal{}1$ as well, and drawing the altitude/angle bisector from $ A$ to $ \overline{BC}$, we see that $ BC\equal{}2\sin18^\circ$.

We then draw the angle bisector of $ \angle ABC$ to point $ D$ on $ \overline{AC}$. We find that $ \angle BDC\equal{}180^\circ\minus{}36^\circ\minus{}72^\circ\equal{}72^\circ$, so $ \triangle BCD$ is isosceles and $ BC\equal{}BD\equal{}2\sin18^\circ$.

Also, we have $ \angle BAD\equal{}\angle ABD\equal{}36^\circ$, so $ AD\equal{}BD\equal{}2\sin18^\circ$, which means that $ DC\equal{}AC\minus{}AD\equal{}1\minus{}2\sin18^\circ$.

Finally, by the Angle Bisector Theorem on $ \angle ABC$, we have

$ \frac{BC}{CD}\equal{}\frac{AB}{AD}\implies\frac{2\sin18^\circ}{1\minus{}2\sin18^\circ}\equal{}\frac1{2\sin18^\circ}$

$ \implies4\sin^2 18^\circ\equal{}1\minus{}2\sin18^\circ\implies4\sin^2 18^\circ\plus{}2\sin18^\circ\minus{}1\equal{}0$

$ \implies\sin18^\circ\equal{}\frac{\minus{}2\pm\sqrt{20}}8\equal{}\frac{\minus{}1\pm\sqrt5}4$.

But $ \sin18^\circ>0$, so we conclude that $ \sin18^\circ\equal{}\boxed{\frac{\sqrt5\minus{}1}4}$ and

$ \cos18^\circ\equal{}\sqrt{1\minus{}\left(\frac{\sqrt5\minus{}1}4\right)^2}\equal{}\boxed{\frac{\sqrt{5\plus{}\sqrt5}}{2\sqrt2}}$.
[/hide]

but how would i use the figure that i drew?

[quote="mewto55555"]Look at triangle ABC, now look at your original triangle.[/quote]

in my figure, angle A is 36, B is 36, and C is 108.

But, in math154's figure, A is 36, B is 72, and C is 72.

Do as mewto55555 said...you can simply ignore $ \overline{CD}$...maybe you don't want to, though. :wink:

OK...do you see the resemblance now? :wink:

no...im really slow today and everyday...

Blargh...if you don't want to use my diagram, which contains your diagram and more, then...

[hide="Trig Bash"]
Most of the diagram is self-explanatory, with the right angles, isosceles triangles, etc.

The clever part is finding the $ \frac {2\sin18^\circ}{1 \plus{} 2\sin18^\circ}$ length.

Draw the angle bisector/altitude/median from angle $ C$ to side $ \overline{AB}$. The $ 108^\circ$ angle is divided into two $ 54^\circ$ angles. From this, we can see that $ \sin54^\circ \equal{} \frac {\frac {AB}2}1 \equal{} \frac {1 \plus{} 2\sin18^\circ}2$.

Let the point at which the angle bisector of $ \angle ABC$ hits $ \overline{AC}$ be $ B'$. Then $ CB'B \equal{} 90^\circ \minus{} 36^\circ \equal{} 54^\circ$, and $ \sin54^\circ \equal{} \frac {\sin18^\circ}{CB'}$.

But $ \sin54^\circ \equal{} \frac {1 \plus{} 2\sin18^\circ}2$!

Solving $ \frac {\sin18^\circ}{CB'} \equal{} \frac {1 \plus{} 2\sin18^\circ}2$ gives $ CB' \equal{} \frac {2\sin18^\circ}{1 \plus{} 2\sin18^\circ}$, and since $ AC \equal{} 1$, $ AB' \equal{} AC \minus{} B'C \equal{} \frac1{1 \plus{} 2\sin18^\circ}$.

Finally, by the Angle Bisector Theorem on $ \angle ABC$, we have $ \frac {BC}{CB'} \equal{} \frac {BA}{AB'}$, so

$ \frac {1 \plus{} 2\sin18^\circ}{2\sin18^\circ} \equal{} (1 \plus{} 2\sin18^\circ)^2$.

Expanding gives us $ 4\sin^218^\circ \plus{} 2\sin18^\circ \minus{} 1 \equal{} 0$, which gives the same result as my post above.
[/hide]

*Note that you could have cheated and drew a bisector in the Golden Triangle in your diagram, and then used my other solution, ignoring the irrelevant parts in your diagram :P I think that solution is more elegant. :wink:


calculus
integration
trigonometry
real analysis
real analysis unsolved

Find: $ \int e^{arcsinx} \,dx$

P.S. Do not use integration software :-)

[color=darkblue]Define $ \{\begin{array}{c} \mathrm M\equiv\int e^t\cos t\ \mathrm {dt}\ . \\
 \\
\mathrm N\equiv\int e^t\sin t\ \mathrm {dt}\ .\end{array}$ Prove easily that $ \{\begin{array}{c} M + N = e^t\sin t + \mathcal C\ . \\
 \\
M - N = e^t\cos t + \mathcal C\ .\end{array}$

Indeed, integrate the relations $ \|\begin{array}{ccc}
(e^t\sin t)' & = & e^t(\sin t+\cos t)\ .\\\
(e^t\cos t)' & = & e^t(\cos t-\sin t)\ .\end{array}$

$ \mathrm I\equiv\int e^{\arcsin x}\ \mathrm {dx} =$ $ \mathrm M(\arcsin x) + \mathcal C$, where $ \mathrm M = \frac 12\cdot e^t(\sin t + \cos t) + \mathcal C\ .$

In conclusion, $ \boxed {I = \frac 12\cdot e^{\arcsin x}\cdot(x + \sqrt {1 - x^2}) + \mathcal C}$.

[b][u]Analogous exercise.[/u][/b] $ \int e^{\arcsin \sqrt x}\ \mathrm {dx} = K(\arcsin\sqrt x) + \mathcal C$, where $ K = \int e^t\sin 2t\ \mathrm {dt}\ .$[/color]

You may integrate twice by parts and to get equation for the starting integral. I like such kind of integrals.


I took the 2001 Chapter Sprint Round and missed 4, couldn't figure out this one.

A combination lock uses three of the digits 0,1,2,3. If consecutive numbers in the combination cannot be the same, how many possible combinations are there?

I keep getting 24.

The answer key says the answer is...
[hide]36[/hide]

[hide]
For the first number, we obviously have $ 4$ choices.
For the second number, we have $ 3$ choices since it can't match the first one.
For the third number, we still have $ 3$ choices since it just doesn't have to match the second one, but still can match the first one. Note that it only says [b]consecutive[/b] numbers can't be the same ;)
So, our answer is $ 4*3*3\equal{}\boxed{36}$[/hide]


some pictures about the great friendship between the liberators and the liberat-ees in Iraq, for example, the Iraqi greet US troop's entering Baghdad.
By the name of Yahveh, I swear that there won't be any faked pictures by those infamous reporters in the post, like the SM or the torture.
Can I do it?
Waitting for your permittion.

Post whatever pictures you want -- if they're inappropriate, I'll delete them and explain why, just as I have done recently in another thread.

I'm not sure what the point of this is, though.  Whatever stand you're taking, either against American imperialism or board-moderator imperialism, this seems like a silly way to begin it.

I am a lazy man, so it will displease me much if someone delete my post just because he think it's [b]inappropriate[/b]. To make sure, I need the written-down definition of "inappropriate", where can I find it? Or it's just something in the moderators' mind?
[quote="JBL"]this seems like a [b]silly[/b] way to begin it.[/quote]
My ignorance on English make me confused, I'm not quite sure whether I could take this as "personally insulting".

The use of the word silly indicates that I am not taking this very seriously.  If I were you, I might be annoyed about it but probably would not consider it a personal insult.

Well, it's good to know that the topic is switched to board-moderator imperialism.  This topic has most probably been discussed in various places in the past, but I'll give it a go anyhow:

The definition of inappropriate is naturally subjective.  However, that does not mean that there are no guidelines whatsoever.  There are large numbers of members of this forum in the American middle-school age range, 14 years old or younger.  As such, posts concerning violence, sex, and illegal acts are likely to make me consider them closely.  A good guide here might be the [url=http://www.mpaa.org/FlmRat_Ratings.asp]guidelines for what qualifies a movie to be rated PG-13[/url]:
[quote="MPAA"]A PG-13 film is one which, in the view of the Rating Board, leaps beyond the boundaries of the PG rating in theme, violence, nudity, sensuality, language, or other contents, but does not quite fit within the restricted R category. Any drug use content will initially require at least a PG-13 rating. In effect, the PG-13 cautions parents with more stringency than usual to give special attention to this film before they allow their 12-year-olds and younger to attend. If nudity is sexually oriented, the film will generally not be found in the PG-13 category. If violence is too rough or persistent, the film goes into the R (restricted) rating. A film’s single use of one of the harsher sexually derived words, though only as an expletive, shall initially require the Rating Board to issue that film at least a PG-13 rating. More than one such expletive must lead the Rating Board to issue a film an R rating, as must even one of these words used in a sexual context. These films can be rated less severely, however, if by a special vote, the Rating Board feels that a lesser rating would more responsibly reflect the opinion of American parents. [/quote]
There is behavior (rather than content) which is also unacceptable; direct insults of other board members are an example.  Other examples can be found in the "Usage of this Forum" thread.  However, this does not concern the question that you asked.

No violence, sex, and illegal acts, no direct(I don't know why you specify direct here, does this imply that indirect insults are allowed? I don't think so) insults of other board members, that's all?

Does your "subjective" include something like the switching of individual emotions, for example, sometimes a seller works in bad mood simply because he quarrelled with his wife in the morning?

May I know that under what conditions will you delet posts with no violence, sex, and illegal acts, no direct insults? In other words, how subjective you could be.

Thank you, rrusczyk.

I will not give you a complete list of every single word, phrase, idea or image that I am inclined to delete -- I have no interest in setting up rules for the sole purpose of allowing you to push their boundaries and/or annoy me.  I will not pre-emptively defend my decisions to delete, edit or lock threads with notice or explanation.

I think you should reconsider what you are doing here.  You don't have any leverage in this situation, so I see no way that being purposefully troublesome could possibly serve your ends, whatever they may be.  If you think I'm acting inappropriately, the proper response would be to PM or e-mail one of the administrators.  If, on the other hand, you have something worth saying then by all means, say it.  If you think you can convince me that I am an unreasonable tyrant violating your fundamental rights as a member of the human race, go ahead.  Your current tactic, however, is extremely annoying, especially because I have had a similar discussion with you in the past via PM, and provides no content for discussion by forum-members in general.

[quote="bleumoose"][quote="rrusczyk"]Justice Potter Stewart, in a ruling on pornography: 'I shall not today attempt further to define the kinds of material … but I know it when I see it.'

I think that's a fair enough way to define offensive for the board.[/quote]

Or you could do that three-part test that he also defined, in the same case. ^_^ Sorry, just had my government exam![/quote]  Government exam?

Well Buffalo, are you going to post it/them or not?

[quote="aznphatso"]yeah just post the pics, or don't post the pics, but don't keep leading us on.[/quote]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=88258[/url]
he has?


blogs
search
Support
geometry
LaTeX
linear algebra
\/closed

Just a notice for all to know that the new album is operational. You can check it out by clicking on my personal album link (near to the Blog link).

For now, there is a limit of 512 KB / image to upload. If your image is larger than this, please shrink it using MS Paint ( Menu >> Image >> Strech/Skew >> 50%/50% ) or any other imaging editing program and save it as a .jpg (for the best compression / quality ratio).

We also will be needing moderators for the album, so please sign up in this thread.

The address is: http://www.mathlinks.ro/Forum/galleries.php

[b]Gallery Moderators Signup list[/b]
[list]anirudh
besttate
chess64[/list]

To add an image from the Gallery in your post, just use the code
[code][img]5761[/img][/code] for example to display the image with id 5761:

[img]5761[/img].

Search also supports searching withing images titles, descriptions, and comments. For example here's a result when searching [url=http://www.mathlinks.ro/Forum/search.php?ppage=50&search_id=971567896&highlight=zetax&sort_dir=desc]for Zetax[/url] ;)

(PS if you sign up, that doesn't mean that you will be chosen as a Gallery Moderator).

[quote="Valentin Vornicu"]Just a notice for all to know that the new album is operational. You can check it out by clicking on my personal album link (near to the Blog link).

We will put up a header link also soon. We also will be needing moderators for the album, so please sign up in this thread.

The address is: http://www.mathlinks.ro/Forum/galleries.php[/quote]
where is it in ML board style?

I think he's asking this: When I enter ML, how can I go to http://www.mathlinks.ro/Forum/galleries.php without typing it? Is there a shortcut somewhere that I've missed?

Don't you guys read all the lines in a post? :) [quote]We will put up a header link also soon.[/quote]

[quote="solafidefarms"]What happened to the old pictures in the old album??[/quote]We have the old pictures stored on our server, however it's not easy to re-add them since the structure of the album is totally changed (the old album was something I added, this one I built myself from scratch).

[quote="anirudh"]I'll mod it. :D[/quote]

WTF? OBJECTION

By signing up you don't necessarily become a Galleries Moderator. You [i]might[/i] become one :)

Also, if any of you guys (and by this I mean you ccy :D) have some CSS improvements (especially on the AoPS skin) please let me know.

Improvements to the AoPS skin? YEAH! I'll make a new stylesheet for AoPS skin and send it to you soon...

[quote="Valentin Vornicu"]Yeah the long term plan is to make the site validate in XHTML 1.1 :)[/quote]

I could help with that too :D

(btw: wow, this is css is really hard to read... i'm still reformatting it :D)

[quote="chess64"](btw: wow, this is css is really hard to read... i'm still reformatting it :D)[/quote]I know ...

you're already a mod :wink: for the galleries
-jorian

Uhh... I am?

I don't think so...

I never signed up either.

I'll be a mod if Mr. Vornicu wants me to... but I never signed up...

[quote="Valentin Vornicu"]
[b]Gallery Moderators Signup list[/b]
[list]anirudh
besttate
chess64[/list]

(PS if you sign up, that doesn't mean that you will be chosen as a Gallery Moderator).[/quote]

hmm...i think he thought you wanted to be a mod hence your posts
-jorian

[quote="bubka"]Ok if he doesn't want to be a mod, I'd love to replace him.
I'm quite old enough - why don't you make me one? :([/quote]

So am I :wink:

[quote="Valentin Vornicu"]Patience is a virtute[/quote]

[quote="anirudh"][quote="bubka"]Ok if he doesn't want to be a mod, I'd love to replace him.
I'm quite old enough - why don't you make me one? :([/quote]

So am I :wink:
[/quote]
huuuuuuuhh how does that work?
you=8th grader=12,13,14years old?
bubka=21years old.

uhm.... 12-14 year-old people usually don't count as old enough for...um bad pictures

Oh my my junggi is in a particularly foul mood nowadays.

Anyway, I'm 18 not 21 but that should be good enough. Peter is only like 19 years old.

[quote="bubka"]Oh my my junggi is in a particularly foul mood nowadays.
[/quote]
well yea all this emo music is making me go emo
[quote="bubka"]
Anyway, I'm 18 not 21 but that should be good enough. Peter is only like 19 years old.[/quote]
sorry :blush:

If my AoPS karma isn't too low, I'll be a moderator.

hmmm... I tried to put a .bmp on it, but it said that .bmp's weren't acceptable...

P.S. All my .jpg's are personal...

[quote="nutz_for2.718281828"]

P.S. All my .jpg's are personal...[/quote]

well save your .bmp as a .jpg

Check the address bar! You should have something like [code]http://www.mathlinks.ro/...&id=666[/code]

That one I've already gotten yesterday, but sent cyprusad as a link.

It want work as Valentin said in his first post, i.e.with
[img}9490[/img].

Now I'm trying putting it "into" LaTex, but it comes out weird


geometry
3D geometry
sphere
trigonometry
inradius
IMO
IMO 1960

Consider a cone of revolution with an inscribed sphere tangent to the base of the cone. A cylinder is circumscribed about this sphere so that one of its bases lies in the base of the cone. let $V_1$ be the volume of the cone and $V_2$ be the volume of the cylinder.

a) Prove that $V_1 \neq V_2$;

b) Find the smallest number $k$ for which $V_1=kV_2$; for this case, construct the angle subtended by a diamter of the base of the cone at the vertex of the cone.

[color=blue][b](1)[/b][/color] Let $\triangle ABC$ be an isosceles triangle with the base a = BC = 2R and the A-altitude to the base equal to h. The shoulder sides AB = AC are equal to $b = \sqrt{R^2 + h^2}$ and the angle $\alpha$ opposite to the base is given by $\tan \frac \alpha 2 = \frac R h$. The triangle semiperimeter is $s = b + R$ and its inradius r is equal to

$r = (s - a) \tan \frac \alpha 2 = (b + R - 2R) \tan \frac \alpha 2 = \left(\sqrt{R^2 + h^2} - R\right) \frac R h$

Let $\square PQRS$ be a square circumscribed around the incircle (I), such that $P, Q \in BC$ and let the square sides SP, QR intersect the triangle sides AB, AC at points K, L and let the square side RS intersect the triangle sides AB, AC at points M, N. The right angle triangles $\triangle MSK \sim \triangle BPK$ are similar, having equal vertical angles at the common vertex K. Since the lines AB, SP intersecting at the point K are tangent to the incircle (I), the line KI bisects the angle $\angle AKP = 180^\circ - \frac \alpha 2$. Therefore, $\angle PKI = \frac{\angle AKP}{2} = 90^\circ - \frac \alpha 4 < 90^\circ$, $KP > r = \frac{SP}{2}$ and since SK + KP = SP, $SK < r = \frac{SP}{2}$. Thus the areas $S(\triangle MSK) < S(\triangle BPK)$ and similarly the areas $S(\triangle NRL) < S(\triangle CQL)$. As a result, the area of the square $\square PQRS$ is less than the area of the triangle $\triangle ABC$:

$S(\square PQRS) = S(\triangle MSK) + S(\triangle NRL) + S(PQLNMK) <$

$< S(\triangle BPK) + S(\triangle CQL) + S(PQLNMK) <$

$< S(\triangle BPK) + S(\triangle CQL) + S(PQLNMK) + S(\triangle AMN) = S(\triangle ABC)$

If the isosceles triangle $\triangle ABC$ and the square $\square PQRS$ are rotated around the A-altitude, they result in the defined cone and cylinder. Since all interior points of the triangles $\triangle BPK, \triangle CQL$ are farther away from the axis of rotation than the interior points of the triangles $\triangle MSK, \triangle NRL$ and the 1st pair has larger areas than the 2nd pair, the 1st pair makes a larger contribution to the cone volume $V_1$ than the 2nd pair to the cylinder volume $V_2$. Asa result, $V_1 > V_2$.

[color=blue][b](2)[/b][/color] Denote $x = \tan \frac \alpha 2 = \frac R h$. The cone and cylinder volumes are

$V_1 = \frac{\pi hR^2}{3} = \frac{\pi h^3x^2}{3}$

$V_2 = 2 \pi r^3 = 2\pi h^3x^3\left(\sqrt{x^2 + 1} - x\right)^3$

and their ratio is

$q = \frac{V_2}{V_1} = 6x\left(\sqrt{x^2 + 1} - x\right)^3$

It is clear that $q \longrightarrow 0$ for both $x \longrightarrow 0$ and for $x \longrightarrow \infty$. The maximum q is given by

$\frac{\text dq}{\text dx} = 0$

$6\left(\sqrt{x^2 + 1} - x\right)^3 + 18x\left(\sqrt{x^2 + 1} - x\right)^2 \left(\frac{x}{\sqrt{x^2 + 1}} - 1\right) = 0$

$\sqrt{x^2 + 1} - x + \frac{3x^2}{\sqrt{x^2 + 1}} - 3x = 0$

$4x^2 + 1 = 4x \sqrt{x^2 + 1}$

$16x^4 + 8x^2 + 1 = 16x^4 + 16x^2$

$8x^2 = 1,\ \ \ x = \tan \frac \alpha 2 = \frac{\sqrt 2}{4},\ \ \ \alpha \doteq 38.94^\circ$

The angle $\frac \alpha 2$ can be constructed by constructing a right angle triangle with the legs $\sqrt 2$ and 4. The maximum volume ratio is then:

$q_{\text{max}} = \frac{V_2}{V_1} = \frac{6\sqrt 2}{4}\left(\sqrt{\frac 1 8 + 1} - \frac{\sqrt 2}{4}\right)^3 = \frac{3 \sqrt 2}{2 (2 \sqrt 2)^3} (3 - 1)^3 = \frac 3 4$

and the minimum reciprocal volume ratio $k_{\text{min}} = \frac{V_1}{V_2} = \frac{1}{q_{\text{max}}} = \frac 4 3$

[u]Solution.[/u] Let $ O$, $ A$, $ B$, $ \theta$, and $ r$ be the cone's vertex, cone's base's center, a point on the cone's base's circle, $ \angle AOB$, and the radius of the sphere, respectively.  Then it follows, from basic trigonometry, that $ OA \equal{} \frac {r(1 \plus{} \sin\theta)}{\sin\theta}$ and $ AB \equal{} \frac {r(1 \plus{} \sin\theta)}{\cos\theta}$.  Hence, $ V_1 \equal{} \frac {\pi AB^2OA}{3} \equal{} \frac {\pi\left(\frac {r(1 \plus{} \sin\theta)}{\cos\theta}\right)^2\left(\frac {r(1 \plus{} \sin\theta)}{\sin\theta}\right)}{3}$ and $ V_2 \equal{} 2\pi r^3$.  Ergo,
\[ k \equal{} \frac {\pi\left(\frac {r(1 \plus{} \sin\theta)}{\cos\theta}\right)^2\left(\frac {r(1 \plus{} \sin\theta)}{\sin\theta}\right)}{6\pi r^3}\implies (1 \plus{} 6k)\sin^2\theta \plus{} (2 \minus{} 6k)\sin\theta \plus{} 1 \equal{} 0.
\]
However, the discriminant of the above quadratic equation must be nonnegative, that is, $ (1 \minus{} 3k)^2\geq 1 \plus{} 6k\implies k\geq\frac {4}{3}$.

And the conclusion follows. $ \Box$


Prove that for all natural numbers $ n$, the number

$ (n^3\minus{}n)(5^{8n\plus{}4}\plus{}3^{4n\plus{}2})$

is a multiple of $ 3804$.

$ E\equal{}(n\minus{}1)n(n\plus{}1)(625^{2n\plus{}1}\plus{}9^{2n\plus{}1})$

Amongst three consecutive numbers, at least one is even and exactly one is divisible by $ 3$, and the last term is divisible by $ 625\plus{}9\equal{}634$, so $ E$ is a multiple of $ 2\cdot 3\cdot 634\equal{}3804$.


Consider the infinite decimal 0.123456789011223344..9900111222..9990001111 ... [pattern is very clear]
Between every two consecutive digits in the decimal insert some number of digits (this number is not constant) which is at least 1 and at most 9. Prove the resulting number obtained is always irrational.

Same method as here: http://www.mathlinks.ro/Forum/viewtopic.php?t=89010 (just a bit more work to do).

hi,
sorry could you elaborate on that solution? i looked at the other thread with a very easy problem where you had arbitrarily long strings of 0's which proves irrationality. the same method doesn't apply here because the digits you insert can be pretty much anything so you won't necessarily get arbitrarily strings of any 1-digit number .. could you elaborate on  'a little more work'?


vector
calculus
derivative
calculus computations

A child twirls a wieght at the end of a 2 meter string at a rate of 1 rps. Find $a_{T}$ and $a_{N}$.

Okay, so the speed= 1 revolution per second which equals $4\pi$ per second.
Speed $=\left\|\vec{v}(t)\right\|=4\pi$
Centripetal force position vector=
$\vec{r}(t)\left\langle -\rho\omega sin(\omega t), \rho\omega cos(\omega t) \right\rangle$
where $\rho$ = radius and $\omega$ = speed, so:
$\vec{r}(t)\left\langle -8\pi sin(4\pi t), 8\pi cos(4\pi t) \right\rangle$
$a_{T}=\left\|\vec{v} '(t)\right\|$ and $a_{N}=\kappa\left\|\vec{v}(t)\right\|^{2}$
where $\kappa$=curvature=$\frac{1}{\rho}=\frac{1}{4\pi}$ therefore $a_{N}=\frac{(4\pi)^{2}}{4\pi}=4\pi$
here is where it breaks down for me. The derivative of $4\pi$ is 0. Does this mean that there is no acceleration on the tangent?
(And let me know if I made any other mistakes, please)

The tangential speed is constant $4\pi \ m/s$, so there is no tangential acceleration.

Also, the curvature is $\frac{1}{2}$, and thus $a_N = 8\pi^2$.

Any time the tangential speed is constant (in a perfect circle), all of the acceleration is centripetal.

Can someone move this to the phyiscs forum? :)

The fact that the acceleration is "centripetal" already indicates that the acceleration doesn't have a tangential component, only a normal component, isn't it? ;)

[quote="apmcavoy"]The total acceleration has both centripetal and tangential components:

$\vec{a}=a_t\hat{\theta}+a_r\hat{r}=\frac{d\vec{v}}{dt}\hat{\theta}-\frac{v^2}{r}\hat{r}$[/quote]

Right, but in this case (and whenever you have constant tangetial velocity around a perfect circle) the tangetial acceleration is equal to zero and all of the acceleration is centripetal.


The product of two inches and the circumference of a circle, in inches, is equal to the circle's area. What is the length of the radius of the circle, in inches?

$ 2*\pi*r*2\equal{}\pi*r^2$. So r=4 or 0. 0 is not a valid answer, so the answer is [b]4[/b].

[hide="answer"]1.turning the words to math we get    2*2pi(r)=4pi(r)

2.the area for a circle is pi(r)^2 and that equals to what i wrote on the top

3.so we get an equation of 4pi(r)=pi(r)^2

4.canceling and calculating we get an answer of

5.$\boxed{4}$[/hide]

[b][i][u]please rate my post thank you so much! [/u][/i][/b]

please be fair though [=

 :cleaning:  :cleaning:   :cleaning:  :cleaning:


induction
inequalities unsolved
inequalities

Let $\{x_n\}$ be a sequence such that $x_1=\frac 12, \ x_{n+1}=x_n+\left(\frac{x_n}n\right)^2$. Prove that:
(1)  $x_n>\frac{6n}{5n+6}, \forall n\geq 3$.
(2)  $x_n<\frac 32, \forall n\geq1$.

1- is just proved by induction.

2- if my computations are correct, the trick may be to prove by induction that :

$x_n \leq \frac 3 2 - \frac{1} {\sqrt n}$. Induction works for $n$ big enough ($n > 10$ works).

Computations are terrific ...

tµtµ could you post the full solution ?
thank you


let $ \psi: V\rightarrow V$ on an inner product space $ V$. Using the fact that $ ker(\psi^*)\equal{}(im(\psi))^\bot$,
Show that a vector $ x$ is a least squares approximation to the matrix equation $ A\underline{x}\equal{}\underline{\beta}$ if and only if $ x$ is a solution to the matrix equation $ A^*A\underline{x}\equal{}A^*\underline{\beta}$


A student is writing the numbers 1, 2, 3,...,102000 in standard decimal form.
Exactly how many times will she write the digit 1?

[hide="Solution"]
In general the number of times we count 1 in $ 1,2,...,10^{n}$ is equal to:
$ n10^{n-1}$. I can prove this to you, and anyway its pretty straightforward.
So you need to count the $ 1$ found $ 100001,...,102000$.

The problem is I got to run. But those are the details. Sorry for not being specific enough.
[/hide]

the number of times we write the digit 1 in $ 1,2,3...,a10^{n}$ is $ an10^{n-1}+10^{n}(nembre\ of\ times\ we\ write\ the\ 1 in\ a)$ then then answer is $ 102.3.10^{2}+10^{2}(numbre\ of\ times\ we\ write\ the\ 1 in\ 102)$
and $ (numbre\ of\ times\ we\ write\ the\ 1 in\ 102)=2.10^{1}+1+2.1$