The Proof is Trivial!

Original post by ben-smith

I've had a shocking day mathswise, am I close?

Can't be worse than me. I've been wrestling with "games" that are ironically not fun at all for the whole day :coffee:

Reply 722

SParm

8

Original post by Lord of the Flies

I thought it had been solved! Here we go:

Solution 98

\dfrac{a_{n+1}-a_n}{\sqrt{n+1}-\sqrt{n}}=\dfrac{1}{a_{n}\big( \sqrt{n+1}-\sqrt{n}\big)}\to 0\;\;\Rightarrow \;\;\dfrac{a_n}{\sqrt{n}}\to 0

by Stolz-Cesaro.

Hmmm...never heard of that test. It looks pretty cool. However; could you walk me through your justification as to why the original limit exists?

Edit: The one with the a_n and roots on the denominator.

Reply 723

Dirac Spinor

10

Original post by ukdragon37

Can't be worse than me. I've been wrestling with "games" that are ironically not fun at all for the whole day :coffee:

Part III sounds tough :tongue:

Original post by Lord of the Flies

Yep - there is an arctan missing but that is just a typo.

You have me intrigued as to what the slick way is. Complex numbers?

Reply 724

ukdragon37

Original post by ben-smith

Part III sounds tough :tongue:

Thankfully I only need a pass to move on to PhD. :tongue:

And if my exams go as I planned I should only need 65% in my dissertation to get a distinction (75%) overall (60% is the passmark).

Reply 725

Dirac Spinor

10

Original post by ukdragon37

Thankfully I only need a pass to move on to PhD. :tongue:

And if my exams go as I planned I should only need 65% in my dissertation to get a distinction (75%) overall (60% is the passmark).

Are you going to minor in anything cool like, I don't know... maths?

Reply 726

ukdragon37

Original post by ben-smith

Are you going to minor in anything cool like, I don't know... maths?

I was thinking of minor-ing in Law or Music actually :tongue:

Law was my first choice for applying to Cambridge before I changed my mind and went with my second choice, CompSci.

Reply 727

Original post by SParm

Hmmm...never heard of that test. It looks pretty cool. However; could you walk me through your justification as to why the original limit exists?

Edit: The one with the a_n and roots on the denominator.

a_{n+1}-a_n>0

hence either

a_n\to \ell

or

a_n\to \infty

as

n\to\infty

. A quick contradiction shows that there is no such

\ell

Original post by ben-smith

You have me intrigued as to what the slick way is. Complex numbers?

You're going to be disappointed - essentially the same as what you did, but set up differently:

\displaystyle\begin{aligned} \int_0^{\infty} \frac{\arctan ex-\arctan x}{x}\,dx &= \int_0^{\infty}\int_x^{ex}\frac{1}{x(1+t^2)}\,dt \,dx\\&=\int_0^{\infty}\int_{t/e}^{t}\frac{1}{x(1+t^2)}\,dx \, dt=\int_0^{\infty}\frac{dt}{1+t^2}=\frac{\pi}{2}\end{aligned}

Another? You TSR folk are too good!

Problem 105* (technically, if we are rigorous, this is a ***)

\displaystyle \int_0^{\infty} \left\{x-\frac{x^3}{2}+\frac{x^5}{2\cdot 4}-\frac{x^7}{2\cdot 4\cdot 6}+\cdots\right\}\left\{1+\left( \frac{x}{2}\right)^2+\left(\frac{x^2}{2 \cdot 4}\right)^2+\left(\frac{x^3}{2 \cdot 4 \cdot 6}\right)^2+ \cdots\right\}\,dx

(the curly braces don't mean anything in particular - I just thought they looked nice)

(edited 10 years ago)

Reply 728

SParm

8

Original post by Lord of the Flies

a_{n+1}-a_n>0

hence either

a_n\to \ell

or

a_n\to \infty

as

n\to\infty

. A quick contradiction shows that there is no such

\ell

That bit is fine. I'm just wondering about what happens to the roots at n goes to infinity. Mind it's late and I'm tired but I certainly didn't get 0 for this (Well I didn't then I did by second guessing myself and then I didn't again once I realised I was being a tit.)

Reply 729

FireGarden

Problem 106

Prove the formula

\displaystyle\sum_{k=0}^n \prod_{\mu=1}^a(k+\mu) = \dfrac{1}{a+1} \displaystyle\prod_{\mu=1}^{a+1}(n+\mu)

Problem 107

Evaluate the series

\displaystyle\sum_{n=0}^{\infty}\dfrac{(n+\mu)!}{2^n n!\mu!}

(edited 10 years ago)

Reply 730

Original post by SParm

...

Oh no, you're absolutely correct. I am the one being an idiot. I somehow omitted the roots in my working leaving (n+1) - n = 1. My sincere apologies.

Edit: amended below.

(edited 10 years ago)

Reply 731

SParm

8

Original post by Lord of the Flies

Oh no, you're absolutely correct. I am the one being an idiot. I somehow omitted the roots in my working leaving (n+1) - n = 1. My sincere apologies.

It's fine, I'm sure we've all had that one (or few dozen for me) shockers when writing maths down on paper. I'm reluctant to post my solution to this, even if I'm confident it works, simply because this thread is brilliant for pre-undergrad level peeps to have a go at some cool maths. My solving what I think is a lovely Analysis problem with a first year Analysis course under my belt kind of ruins that a bit.

I'm sure Mladenov would be happy to give anyone who wants to do it some hints, as would I.

Edit: Also my solution's laborious to the max. I'm sure one of you guys can come up with a lovely 4 or 5 liner.

(edited 10 years ago)

Reply 732

A lot of big terrifying things flying around this thread

Spoiler

so how about something simple:

Problem 108*

Find all pairs of integers (p,q) such that the the roots of the equation,

\displaystyle(px-q)^2+(qx-p)^2=x

, are integers.

Spoiler

-----------------

Btw, can people make sure they put 'assumed knowledge' ratings on problems please.

(edited 10 years ago)

Reply 733

Solution 98

a_{n+1}^2=a_n^2+\dfrac{1}{a_n^2}+2\Rightarrow a_{n+1}^2 =a_0^2+2(n+1)+\displaystyle\sum_{k=0}^n \dfrac{1}{a_k^2}

Square the limit:

\displaystyle\left(\lim_{n\to \infty} \frac{a_n}{\sqrt{n}}\right)^2 =\lim_{n\to\infty} \frac{a_n^2}{n}=2+\frac{a_0^2}{n}+\sum_{k=0}^{n-1}\frac{1}{a_k^2n}

w_n=\displaystyle\sum_{k=0}^{n-1}\dfrac{1}{a_k^2}

then

\dfrac{w_{n+1}-w_n}{(n+1)-n}=\dfrac{1}{a_n^2}\to 0

hence

\dfrac{w_n}{n}\to 0

by Stolz Cesaro.

Hence

\displaystyle\lim_{n\to\infty} \frac{a_n}{\sqrt{n}}=\sqrt{2}

Solution 106

\displaystyle\binom{\alpha}{ \beta}+\binom{\alpha}{\beta+1}= \binom{\alpha+1}{\beta+1} \Rightarrow\displaystyle\sum_{k=0}^n \binom{k+a}{k}=\binom{n+a+1}{n}

inductively.

Multiply by

a!:

\displaystyle\sum_{k=0}^n \frac{(k+a)!}{k!}=\frac{1}{a+1} \frac{(n+a+1)!}{n!} \iff \sum_{k=0}^n \prod_{\mu=1}^a (k+\mu)=\frac{1}{a+1}\prod_{\mu=1}^{a+1}(n+\mu)

Solution 107

\displaystyle \frac{1}{(1-x)^{\mu+1}}=\sum_{k=0}^{\infty} \binom{n+\mu}{n} x^n\Rightarrow 2^{\mu +1}=\sum_{k=0}^{\infty}\frac{1}{2^n}\binom{n+\mu}{n}

(edited 10 years ago)

Reply 734

Original post by FireGarden

Problem 106
Prove the formula

\displaystyle\sum_{k=0}^n \prod_{\mu=1}^a(k+\mu) = \dfrac{1}{a+1} \displaystyle\prod_{\mu=1}^{a+1}(n+\mu)

...for all non-negative integers n.

Solution 106

When n=0, LHS

\displaystyle =\prod_{\mu=1}^a(\mu)

. RHS

\displaystyle =\frac{1}{a+1} \prod_{\mu=1}^{a+1}(\mu) = \prod_{\mu=1}^a(\mu)=LHS

. So true for n=0.

Assume true for n=m. This gives...

\displaystyle\sum_{k=0}^{m+1} \prod_{\mu=1}^a(k+\mu) =\frac{1}{a+1}\prod_{\mu=1}^{a+1}(m+\mu) + \prod_{\mu=1}^a(m+1+\mu)

\displaystyle =\frac{1}{a+1} \prod_{\mu=0}^a(m+1+\mu) + \prod_{\mu=1}^a(m+1+\mu)

\displaystyle =(\frac{m+1}{a+1}+1)\prod_{\mu=1}^{a}(m+1+\mu)

\displaystyle =\frac{1}{m+a+2}(\frac{m+1}{a+1}+1)\prod_{\mu=1}^{a+1}(m+1+\mu)

\displaystyle =\frac{1}{a+1}\prod_{\mu=1}^{a+1}((m+1)+\mu)

.

Therefore, if true for n=m, true too for n=m+1. As true for n=0, true for all non-negative integers n by mathematical induction. QED.

Spoiler

(edited 10 years ago)

Reply 735

Original post by Lord of the Flies

Solution 98
\lim_{n\to\infty} \frac{a_n^2}{n}=2+\frac{a_0}{n}+\sum_{k=0}^{n-1}\frac{1}{a_kn}

Awesome solution, Stolz Cesaro seems pretty useful. Where did the squares go though? (even if they were squares the proof would still work so I'm just wondering)

Reply 736

Original post by Jkn

Awesome solution, Stolz Cesaro seems pretty useful. Where did the squares go though? (even if they were squares the proof would still work so I'm just wondering)

I forgot to type them in! :lol:

All fixed now, thanks for noticing!

(edited 10 years ago)

Reply 737