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The Proof is Trivial!

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Reply 500
Avatar for Mladenov
Mladenov
1
Original post by bananarama2
...


Yup, it is about the square brackets.

Spoiler



As we speak, let me suggest something in this direction.

Problem 68**

Evaluate ∫01∫01...∫01[x1+x2+...+xn]dx1dx2...dxn\displaystyle \int_{0}^{1}\int_{0}^{1}...\int_{0}^{1} [x_{1}+x_{2}+...+x_{n}]dx_{1}dx_{2}...dx_{n}.
Of course, [x][x] is the largest integer not greater than xx.
Original post by dknt
Another physicsy maths one :awesome:

Problem 65 **/***

A cone of base radius R, height H, mass M and homogeneous mass density ϱ is orientated as shown below.

Untitled.jpg

The angle the cone makes to the xy plane is α. The cone is then attached to an axis along its outer mantle in the direction of u^ \hat{\mathbf{u}}.

Calculate the moment of inertia of the cone, with respect to this axis.


I don't *know* how to do this. Is there anywhere you can point me to to read about it. (Inertia tensor? )
Reply 502
Avatar for dknt
dknt
12
Original post by bananarama2
I don't *know* how to do this. Is there anywhere you can point me to to read about it. (Inertia tensor? )


Spoiler

Problem 69***

For x≠0x \neq 0

evaluate

∑r=1∞1r2+x2\displaystyle \sum_{r=1}^{\infty} \dfrac{1}{r^2 + x^2}
(edited 11 years ago)
Reply 504
Avatar for Mladenov
Mladenov
1
Solution 69

Consider the function f(z)=(−1)zz2+x2\displaystyle f(z)=\frac{(-1)^{z}}{z^{2}+x^{2}}, which has poles at ixix and −ix-ix

We have Res(πcsc⁡(πz)f(z),ix)=(−1)ixπcsc⁡(πxi)2ix\displaystyle Res(\pi \csc (\pi z)f(z),ix)= \frac{(-1)^{ix}\pi \csc (\pi x i)}{2ix} and Res(πcsc⁡(πz)f(z),−ix)=(−1)−ixπcsc⁡(πxi)2ix\displaystyle Res(\pi \csc( \pi z)f(z),-ix)= \frac{(-1)^{-ix} \pi \csc (\pi x i)}{2ix}

Notice csc⁡(πxi)=−icsch(πx)\displaystyle \csc (\pi x i)= -i csch (\pi x) and csch(πx)=2eπxe2πx−1\displaystyle csch (\pi x)= \frac{2e^{\pi x}}{e^{2\pi x}-1}

So,

Res(πcsc⁡(πz)f(z),ix)=(−1)ixπcsc⁡(πxi)2ix=−e−πxπeπxx(e2πx−1)=−πx1e2πx−1\begin{aligned}\displaystyle Res(\pi \csc( \pi z)f(z),ix)&= \frac{(-1)^{ix}\pi \csc (\pi x i)}{2ix} \\ &= -\frac{e^{-\pi x}\pi e^{\pi x}}{x(e^{2\pi x}-1)}\\ &= -\frac{\pi}{x}\frac{1}{e^{2\pi x}-1} \end{aligned}

Res(πcsc⁡(πz)f(z),−ix)=−πx(1+1e2πx−1)\begin{aligned}\displaystyle Res(\pi\csc( \pi z)f(z),-ix) = -\frac{\pi}{x}\left(1+\frac{1}{e^{2\pi x}-1}\right) \end{aligned}


Therefore −∑akRes(πcsc⁡(πz)f(z),ak)=πx(1+2e2πx−1)=πxcoth⁡(πx)\displaystyle -\sum_{a_{k}} Res(\pi \csc( \pi z)f(z),a_{k}) = \frac{\pi}{x}\left(1+\frac{2}{e^{2\pi x}-1}\right) = \frac{\pi}{x} \coth (\pi x)

Now, since ∑r=−∞∞(−1)rf(r)=−∑akRes(πcsc⁡(πz)f(z))\displaystyle \sum_{r= -\infty}^{\infty} (-1)^{r}f(r) = -\sum_{a_{k}} Res(\pi\csc (\pi z) f(z)), where aka_{k} are the poles of ff, we obtain ∑r=−∞∞1r2+x2=πxcoth⁡(πx)\displaystyle \sum_{r=-\infty}^{\infty} \frac{1}{r^{2}+x^{2}}= \frac{\pi}{x}\coth( \pi x)
∑r=−∞−11r2+x2+∑r=1∞1r2+x2=1x2(πxcoth⁡(πx)−1)\displaystyle \sum_{r= -\infty}^{-1} \frac{1}{r^{2}+x^{2}} + \sum_{r=1}^{\infty} \frac{1}{r^{2}+x^{2}} = \frac{1}{x^{2}}(\pi x\coth(\pi x)-1)

Hence ∑r=1∞1r2+x2=12x2(πxcoth⁡(πx)−1)\displaystyle \sum_{r=1}^{\infty} \frac{1}{r^{2}+x^{2}} = \frac{1}{2x^{2}}(\pi x\coth(\pi x)-1)
Original post by Mladenov
Solution 69

Consider the function f(z)=(−1)zz2+x2\displaystyle f(z)=\frac{(-1)^{z}}{z^{2}+x^{2}}, which has poles at ixix and −ix-ix

We have Res(πcsc⁡(πz)f(z),ix)=(−1)ixπcsc⁡(πxi)2ix\displaystyle Res(\pi \csc (\pi z)f(z),ix)= \frac{(-1)^{ix}\pi \csc (\pi x i)}{2ix} and Res(πcsc⁡(πz)f(z),−ix)=(−1)−ixπcsc⁡(πxi)2ix\displaystyle Res(\pi \csc( \pi z)f(z),-ix)= \frac{(-1)^{-ix} \pi \csc (\pi x i)}{2ix}

Notice csc⁡(πxi)=−icsch(πx)\displaystyle \csc (\pi x i)= -i csch (\pi x) and csch(πx)=2eπxe2πx−1\displaystyle csch (\pi x)= \frac{2e^{\pi x}}{e^{2\pi x}-1}

So,

Res(πcsc⁡(πz)f(z),ix)=(−1)ixπcsc⁡(πxi)2ix=−e−πxπeπxx(e2πx−1)=−πx1e2πx−1\begin{aligned}\displaystyle Res(\pi \csc( \pi z)f(z),ix)&= \frac{(-1)^{ix}\pi \csc (\pi x i)}{2ix} \\ &= -\frac{e^{-\pi x}\pi e^{\pi x}}{x(e^{2\pi x}-1)}\\ &= -\frac{\pi}{x}\frac{1}{e^{2\pi x}-1} \end{aligned}

Res(πcsc⁡(πz)f(z),−ix)=−πx(1+1e2πx−1)\begin{aligned}\displaystyle Res(\pi\csc( \pi z)f(z),-ix) = -\frac{\pi}{x}\left(1+\frac{1}{e^{2\pi x}-1}\right) \end{aligned}


Therefore −∑akRes(πcsc⁡(πz)f(z),ak)=πx(1+2e2πx−1)=πxcoth⁡(πx)\displaystyle -\sum_{a_{k}} Res(\pi \csc( \pi z)f(z),a_{k}) = \frac{\pi}{x}\left(1+\frac{2}{e^{2\pi x}-1}\right) = \frac{\pi}{x} \coth (\pi x)

Now, since ∑r=−∞∞(−1)rf(r)=−∑akRes(πcsc⁡(πz)f(z))\displaystyle \sum_{r= -\infty}^{\infty} (-1)^{r}f(r) = -\sum_{a_{k}} Res(\pi\csc (\pi z) f(z)), where aka_{k} are the poles of ff, we obtain ∑r=−∞∞1r2+x2=πxcoth⁡(πx)\displaystyle \sum_{r=-\infty}^{\infty} \frac{1}{r^{2}+x^{2}}= \frac{\pi}{x}\coth( \pi x)
∑r=−∞−11r2+x2+∑r=1∞1r2+x2=1x2(πxcoth⁡(πx)−1)\displaystyle \sum_{r= -\infty}^{-1} \frac{1}{r^{2}+x^{2}} + \sum_{r=1}^{\infty} \frac{1}{r^{2}+x^{2}} = \frac{1}{x^{2}}(\pi x\coth(\pi x)-1)

Hence ∑r=1∞1r2+x2=12x2(πxcoth⁡(πx)−1)\displaystyle \sum_{r=1}^{\infty} \frac{1}{r^{2}+x^{2}} = \frac{1}{2x^{2}}(\pi x\coth(\pi x)-1)


Don't you just love elementary solutions?
Reply 506
Avatar for Mladenov
Mladenov
1
Original post by ben-smith
Don't you just love elementary solutions?


As Nikolai Nikolov, the leader of our mathematics team, says: Why simple, when it could be complicated?!

Spoiler

Reply 507
Avatar for shamika
shamika
16
Original post by Mladenov
As Nikolai Nikolov, the leader of our mathematics team, says: Why simple, when it could be complicated?!

Spoiler



Out of curiosity, how old are you and are you planning on studying at university here? (Take to PM if you'd like)
This one is a bit easy compared to some of what's been posted but it's quite a nice question;

Problem 70 */**

Consider the set {1,2,3,...,N}\{1, 2, 3,..., N\}. Two sets A and B are chosen independently and equally likely among all the subsets of {1,2,3,...,N}\{1, 2, 3,..., N\}.

Find P(A⊆B)P(A \subseteq B).

I hope it's not already been done.
Original post by Mladenov
...


Which IMO team are you on then?
Reply 510
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Mladenov
1
Original post by shamika
Out of curiosity, how old are you and are you planning on studying at university here? (Take to PM if you'd like)


Spoiler



Original post by bananarama2
Which IMO team are you on then?


Spoiler

Original post by Mladenov

Spoiler



That's literally amazing. Good luck!
Reply 512
Avatar for Mladenov
Mladenov
1
Original post by bananarama2
...

Thanks.


Solution 70

Note that there are 2N2^{N} subsets of {1,2,..,N}\{1,2,..,N\}.
Let card(B)=icard(B)=i, i∈{0,1,..,N} i \in \{0,1,..,N\}. Then P(card(B)=i)=(Ni)2N\displaystyle P(card(B)=i)= \frac{\dbinom{N}{i}}{2^{N}}. So, since there are 2i2^{i} subsets of BB, we have P(A⊆B)=∑i=0N(Ni)2N12N−i=122N∑i=0N(Ni)2i=3N22N\displaystyle P(A \subseteq B)= \sum_{i=0}^{N} \frac{\dbinom{N}{i}}{2^{N}}\frac{1}{2^{N-i}}= \frac{1}{2^{2N}}\sum_{i=0}^{N} \dbinom{N}{i} 2^{i}=\frac{3^{N}}{2^{2N}} .
Original post by Mladenov
**Complex analysis awesomeness**

My way:
sinxx=Π(1−x2(πr)2)[br]⇒sin(iπx)iπx=Π(1+x2r2)[br]⇒ln(sin(iπx)iπx)=Σln(1+x2r2)\dfrac{sinx}{x}= \Pi (1-\dfrac{x^2}{(\pi r)^2})[br]\Rightarrow \dfrac{sin(i \pi x)}{i \pi x}= \Pi (1+\dfrac{x^2}{ r^2})[br]\Rightarrow ln(\dfrac{sin(i \pi x)}{i \pi x})= \displaystyle \Sigma ln(1+\dfrac{x^2}{r^2})
Differentiating both sides wrt x:
Σ2xr2+x2=ddx[ln(sin(iπx)iπx)]=ddx[ln(sinh(πx)πx)]=1/x(πxcoth(πx)−1)[br]⇒∑r=1∞1r2+x2=12x2(πxcoth⁡(πx)−1)\displaystyle \Sigma \dfrac{2x}{r^2+x^2}= \dfrac{d}{dx}[ln(\dfrac{sin(i \pi x)}{i \pi x})]=\dfrac{d}{dx}[ln(\dfrac{sinh(\pi x)}{\pi x})]=1/x(\pi x coth(\pi x)-1)[br]\Rightarrow \displaystyle \sum_{r=1}^{\infty} \frac{1}{r^{2}+x^{2}} = \frac{1}{2x^{2}}(\pi x\coth(\pi x)-1)
Reply 514
Avatar for shamika
shamika
16
Original post by Mladenov

Spoiler





Spoiler



:smile:

That really is incredible. Make sure you have a look at STEP and MAT, assuming you apply to Oxford / Cambridge / Imperial / Warwick.

Also, you'll be pleased to know that there is no Euclidean plane geometry of the kind that's useful for IMO at university!
Reply 515
Avatar for joostan
joostan
13
Original post by Mladenov

Spoiler



Original post by shamika
:smile:
Also, you'll be pleased to know that there is no Euclidean plane geometry of the kind that's useful for IMO at university!


This may be a silly question but does anybody like geometry?? Even Mladenov who appears to be a maths genius dislikes it :s-smilie:
Problem 71**

Find all x,y∈Cx,y \in \mathbb{C} such that xy=yxx^y = y^x.

EDIT: The question is asking for a way to generate solutions of that kind. To be precise, define the pair of parametric functions x,y:S→Cx,y : S \to \mathbb{C} such that x(a)y(a)=y(a)x(a)x(a)^{y(a)} = y(a)^{x(a)} for all aa of some set SS (for example, R\mathbb{R} or C\mathbb{C}), and the functions must be able to generate all the possible solutions in C\mathbb{C}. You shouldn't need to use any function definitions beyond A-level.
(edited 11 years ago)
Reply 517
Avatar for shamika
shamika
16
Original post by joostan
This may be a silly question but does anybody like geometry?? Even Mladenov who appears to be a maths genius dislikes it :s-smilie:


At least in the UK, the problem is that it's not taught properly. I remember at GCSE I had to randomly learn the circle theorems a day before the exam because no one had bother teaching them at school. Similarly with vectors; in retrospect my teacher had really confused the hell out of us, despite spending weeks trying to explain. The fact that a vector could be 'free' just made no sense to me until several months into my A-Levels. Needless to say, my mind was blown :smile:
Original post by joostan
This may be a silly question but does anybody like geometry?? Even Mladenov who appears to be a maths genius dislikes it :s-smilie:

I don't :biggrin: Dunno if they like it or not but the Chinese have some really freaky Euclidean geometry in their syllabus - someone mus like it :tongue:
Original post by joostan
This may be a silly question but does anybody like geometry?? Even Mladenov who appears to be a maths genius dislikes it :s-smilie:


I quite like it :smile: The only real disadvantage is trying to get a compass that doesn't screw up your circle.

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