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The Proof is Trivial!

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Original post by Renzhi10122
Problem 583

xi=±1 x_i=\pm 1 and x1x2x3x4+x2x3x4x5+...+xnx1x2x3=0 x_1x_2x_3x_4+x_2x_3x_4x_5+...+x_nx_1x_2x_3=0
Prove that n n is divisible by 4.


A beauty, only invariant question I got off that problem sheet. I was so pleased.
Original post by physicsmaths
A beauty, only invariant question I got off that problem sheet. I was so pleased.


Lol, I quite like it.
Original post by Renzhi10122
Problem 583

xi=±1 x_i=\pm 1 and x1x2x3x4+x2x3x4x5+...+xnx1x2x3=0 x_1x_2x_3x_4+x_2x_3x_4x_5+...+x_nx_1x_2x_3=0
Prove that n n is divisible by 4.


Solution 583:
Method 1:

Spoiler

Method 2:

Spoiler

(edited 7 years ago)
Original post by joostan
Solution 583:
Method 1:

Spoiler

Method 2:

Spoiler



Yup, method 2 was the one I knew, but I quite like method 1.
Original post by joostan
Solution 583:
Method 1:

Spoiler

Method 2:

Spoiler



Yep, this is from a problem sheet on invariants so yeh changibg the 1s to -1s etc is a nice way but Method 1 is very nice aswell.
This problem is for any younger lurkers, as it only requires basic trig, calculus and geometry. I'd post it on the STEP thread, but they have better things to do than do what i spent the better part of a few hour messing around with :wink:

Problem 584:

A clean rainbow appears due to an object of sufficient symmetry scattering light from a source. Such light will appear at a certain angle along the line of the observer and the light source due to a focusing effect, such that a complete circle would form (if the ground isn't in the way). Let the order of a rainbow be related to the number of reflections inside such an object.

Let the light source be the sun and the scattering object be perfectly spherical raindrops (refractive index = 4/3). Explain mathematically why the focusing of light occurs at a particular angle for primary rainbows. Why would secondary rainbows look so different to primary rainbows? Find a general solution to angles which rainbows can exist along the line connecting the sun to the observer due to such focusing. Why do primary rainbows not have the expected width of 1.9 degrees, but rather one closer to 2.5 degrees?

Violet light has a refractive index of ~1.344
The sun is sufficiently far away that incident rays are parallel

Rough diagram of what is going on in a primary rainbow for anyone completely lost:

Spoiler

Problem 585**

i=1nxi=n,i=1nxi32=n,xiR \sum^n_{i=1}x_i=-n, \sum^n_{i=1}x_i^{32}=n, x_i \in \mathbb{R} . Prove that xi=1 x_i=-1 .
Original post by Renzhi10122
Problem 585**

i=1nxi=n,i=1nxi32=n,xiR \sum^n_{i=1}x_i=-n, \sum^n_{i=1}x_i^{32}=n, x_i \in \mathbb{R} . Prove that xi=1 x_i=-1 .


failed attempt.
Ew :lol: think this is okay? Not ever good at proofs :P

Spoiler

(edited 7 years ago)
Original post by The-Spartan
Solution 585
Ew :lol: think this is okay? Not ever good at proofs :P

Spoiler



You seem to have assumed each xix_i is the same. . .
Original post by The-Spartan
Solution 585
Ew :lol: think this is okay? Not ever good at proofs :P

Spoiler



Original post by joostan
You seem to have assumed each xix_i is the same. . .


Ah, I was wondering what was happening... Yeah, it's a bit more complicated than that.
Original post by Renzhi10122
Ah, I was wondering what was happening... Yeah, it's a bit more complicated than that.


Damn :lol:
Didnt notice that assumption. Was wondering why it was a ** label :biggrin:
Thanks @joostan
Problem 583

For anyone as fed up of studying as I am. Evaluate

(n+1)+2n+22(n1)+...+2k(n+1k)+....+2n(n+1) + 2n + 2^2 (n - 1) + ... + 2^k (n + 1 - k) + .... + 2^n
Original post by 16Characters....
Problem 583

For anyone as fed up of studying as I am. Evaluate

(n+1)+2n+22(n1)+...+2k(n+1k)+....+2n(n+1) + 2n + 2^2 (n - 1) + ... + 2^k (n + 1 - k) + .... + 2^n

Why not, note:

k=0nxk=1xn+11x\displaystyle\sum_{k=0}^{n} x^k = \dfrac{1-x^{n+1}}{1-x}.

Differentiating this w.r.t. xx, also have:

k=0nkxk1=1(n+1)xn+nxn+1(1x)2\displaystyle\sum_{k=0}^{n} kx^{k-1} = \dfrac{1-(n+1)x^n +nx^{n+1}}{(1-x)^2}

So,

Unparseable latex formula:

[br]\begin{array}{rcl}[br]\displaystyle\sum_{k=0}^{n} 2^k(n+1-k) &=& (n+1)\displaystyle\sum_{k=0}^{n} 2^k - 2\displaystyle\sum_{k=0}^{n} k2^{k-1} \\[br]&=& (n+1)(2^{n+1}-1) - 2(1-(n+1)2^n+n2^{n+1})\\[br]\\[br]&=& 2^{n+2}-n-3[br]\end{array}

(edited 7 years ago)
Reply 3693
Solution 585

Spoiler



So the 32-exponent was entirely arbitrary, a bit unsurprisingly. Any even non-zero exponent would've done the job.
(edited 7 years ago)
I've just figured these results out. Not sure how well known they are, so they may be easier than I think:

Show that:

n=1sinnπbsinnπan=12(πaπb)\displaystyle \sum_{n=1}^\infty \frac{\sin \frac{n \pi}{b}-\sin \frac{n \pi}{a}}{n} = \frac{1}{2}(\frac{\pi}{a}-\frac{\pi}{b})

n=1cosnπbcosnπan=ln(sinπ2asinπ2b)\displaystyle \sum_{n=1}^\infty \frac{\cos \frac{n \pi}{b}-\cos \frac{n \pi}{a}}{n} = \ln(\frac{\sin \frac{\pi}{2a}}{\sin \frac{\pi}{2b}})
Original post by atsruser
n=1sinnπbsinnπan=12(πaπb)\displaystyle \sum_{n=1}^\infty \frac{\sin \frac{n \pi}{b}-\sin \frac{n \pi}{a}}{n} = \frac{1}{2}(\frac{\pi}{a}-\frac{\pi}{b})

n=1cosnπbcosnπan=ln(sinπ2asinπ2b)\displaystyle \sum_{n=1}^\infty \frac{\cos \frac{n \pi}{b}-\cos \frac{n \pi}{a}}{n} = \ln(\frac{\sin \frac{\pi}{2a}}{\sin \frac{\pi}{2b}})


The first follows from this. A parallel manipulation gives v1cosvx=log[2sinx2]\sum v^{-1}\cos vx=-\log \left[2\sin \frac{x}{2}\right].
Original post by Lord of the Flies
The first follows from this. A parallel manipulation gives v1cosvx=log[2sinx2]\sum v^{-1}\cos vx=-\log \left[2\sin \frac{x}{2}\right].


Damn, I'm behind the times, aren't I. There I was, thinking I'd found a nice, fancy result, designed to challenge the finest minds of TSR.

It would take me quite a while to follow your derivation - looks very sophisticated. I got to it via a rather simpler route (and easier than those others posted, I think). I'll put it up later.

Oh, and having followed that thread a bit from there - cimer? Verlan?
Original post by atsruser
It would take me quite a while to follow your derivation - looks very sophisticated. I got to it via a rather simpler route (and easier than those others posted, I think). I'll put it up later.


Cool - and nah it isn't too sophisticated don't worry. It looks that way because I generalised and went into some nice consequences, but the evaluation of the initial series isn't too long [starts at "solution 142" obvz].

Oh, and having followed that thread a bit from there - cimer? Verlan?


Yup! Ah I miss the good old days when this thread was on fire.
(edited 7 years ago)
Original post by Lord of the Flies
The first follows from this. A parallel manipulation gives v1cosvx=log[2sinx2]\sum v^{-1}\cos vx=-\log \left[2\sin \frac{x}{2}\right].


This is great! :biggrin:

Though can you explain this step; not sure I quite follow (in particular the second term on the last line):


Furthermore, observe that:

Unparseable latex formula:

\displaystyle\begin{aligned} \int_0^x\int_0^{t_1}\cdots \int_0^{t_{2k-1}} \sum_{v\geq 1}\frac{\sin vt_{2k}}{v}\,dt_{2k}\cdots dt_1=\frac{1}{2}\int_0^x\int_0^{ t_1}\cdots \int_0^{t_{2k-1}} \pi -t_{2k}\, dt_{2k}\cdots\, dt_1



gives

Unparseable latex formula:

\displaystyle\begin{aligned} (-1)^k\sum_{v\geq 1} \frac{\sin vx}{v^{2k+1}}+\sum_{i=1}^{k} \left( \frac{(-1)^{k+i+1}x^{2(k-i)+1}}{(2(k-i)+1)!}\sum_{v\geq 1}\frac{1}{v^{2i}} \right)= \frac{x^{2k}}{2}\left(\frac{\pi} {(2k)!}-\frac{x}{(2k+1)!}\right)

Original post by Lord of the Flies
Cool - and nah it isn't too sophisticated don't worry. It looks that way because I generalised and went into some nice consequences, but the evaluation of the initial series isn't too long [starts at "solution 142" obvz].


I did this: we have:

I=πaπbdxeix1=πaπb(12+isinx2(cosx1)) dx=12(πaπb)+iln(sinπ2asinπ2a)I = \int_{\frac{\pi}{a}}^{\frac{\pi}{b}} \frac{dx}{e^{ix}-1} = \int_{\frac{\pi}{a} }^{ \frac{\pi}{b} } \big( -\frac{1}{2} + i \frac{\sin x}{2(\cos x-1)} \big) \ dx = \frac{1}{2}(\frac{\pi}{a} - \frac{\pi}{b}) + i \ln(\frac{\sin \frac{\pi}{2a}}{\sin \frac{\pi}{2a}})

and:

I=πaπbn=1einx dx=n=1πaπb(cosnxisinnx) dx=\displaystyle I = \int_{\frac{\pi}{a}}^{\frac{\pi}{b}} \sum_{n=1}^{\infty} e^{-inx} \ dx = \sum_{n=1}^{\infty} \int_{\frac{\pi}{a}}^{\frac{\pi}{b}} \big( \cos nx - i \sin nx \big) \ dx =

n=1sinnπbsinnπbn+in=1cosnπbcosnπbn\displaystyle \sum_{n=1}^{\infty} \frac{\sin \frac{n\pi}{b} - \sin \frac{n\pi}{b}}{n} + i \sum_{n=1}^{\infty} \frac{\cos \frac{n\pi}{b} - \cos \frac{n\pi}{b}}{n}

(which I think is OK...)

Yup! Ah I miss the good old days when this thread was on fire.


Right. I was wondering if "cimer" was what people did when they got to the top of a mountain or something. Are you bilingual?

As for the good old days, yeah, those days are gone - now you've just got to slum it with second raters like me (and Zacken, of course).

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