The Proof is Trivial!

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Reply 1080

Jkn

Original post by bananarama2

By parts...?

Original post by joostan

As above, by parts and evaluating the limits around uv. :smile:

Way too tired to be on TSR

Original post by bogstandardname

Is this differentiation under the integral sign? Or if not what is this technique, is it useful?

Certainly is! You are essentially generalising the expression in order to allow taking the partial derivative of an integral with respect to a variable (that is independent of that which you are integrating with respect to) and, hence, altering the problem to one that is more manageable. Re-intergrating in order to find the original function (before it was partially differentiated) can be seen as being analogous to the inverse Laplace Transformation and other such transformations of this nature :tongue:

It was set in the "Putnam" competition as one of the harder problems (University level competition for top few thousand university students in the US) and almost no-one managed to do it! I thought it was a really nice example of how powerful the technique is! I looked up the solutions after I tried it and this particular method was listed as "Solution 3" (with the other ones looking insane!)

Btw, If you've just learnt the technique, try some of the problems I posted the other day, they are a nice introduction :smile:

Alternate solutions can be found under "Problem A5" for those who are interested.

(edited 10 years ago)

Reply 1081

Lord of the Flies

Solution 166

Letting

x\to x^{-1}

and adding the integrals reveals that:

\displaystyle\int_0^{\infty} \frac{x\ln x\,dx}{(x^2+1)(x^3+1)^2}=\frac{1}{2}\left(\int_0^{\infty} \frac{x\ln x\,dx}{(x^3+1)^2}-\int_0^{\infty} \frac{x^3\ln x\,dx}{(x^3+1)^2}\right)

Observe also both these integrals are the same

(

let

x\to x^{-1})

hence we have:

\displaystyle\int_0^{\infty} \frac{x\ln x\,dx}{(x^2+1)(x^3+1)^2}=\int_0^{\infty} \frac{x\ln x\,dx}{(x^3+1)^2}

Set

\displaystyle f(\lambda)=\int_0^{\infty} \frac{x^{\lambda}\,dx}{(x^3+1)^2}

and let

t=(x^3+1)^{-1}:

\begin{aligned} f(\lambda)=\displaystyle\frac{1}{3}\int_0^{1}\left(\frac{1}{t}-1\right)^{\frac{\lambda-2}{3}}\,dt &= \frac{1}{3}\int_0^{1}t^{\frac{2-\lambda}{3}}\left(1-t\right)^{\frac{\lambda-2}{3}}\,dt \\&=\frac{1}{3}\text{B}\left( \frac{5-\lambda}{3},\,\frac{1+\lambda}{3}\right)\\ &=\frac{1}{3}\Gamma \left(\frac{5-\lambda}{3} \right)\Gamma\left(\frac{1+ \lambda}{3}\right)\\&=\frac{\pi(2-\lambda)}{9}\csc\left(\frac{1+ \lambda}{3}\pi\right)\end{aligned}

Differentiate and set

\lambda =1:

\displaystyle\int_0^{\infty} \frac{x\ln x\,dx}{(x^2+1)(x^3+1)^2}=\frac{2\pi}{27} \left(\frac{\pi}{3}-\sqrt{3}\right)

Original post by joostan

Or rather, * isn't really A-Level knowledge :cool:

The solutions people post to (*) problems aren't necessarily (*) - a single star only indicates that the question can be done with A-level knowledge.

(edited 10 years ago)

Hint for my problem

Original post by Jkn

Alternate solutions can be found under "Problem A5" for those who are interested.

"The Mathematica computer algebra system (version 5.2) successfully computes this integral, but we do not know how."

That's right :wink:

Reply 1084

joostan

Original post by Lord of the Flies

The solutions people post to (*) problems aren't necessarily (*) - a single star only indicates that the question can be done with A-level knowledge.

Is this some kind of sneaky quote where you don't get notified o.O
It's not just that, I'm not convinced that some of these problems are really * but hey ho :rolleyes:

Reply 1085

Jkn

Original post by ukdragon37

"The Mathematica computer algebra system (version 5.2) successfully computes this integral, but we do not know how."

That's right :wink:

Hahahaha

**** Compsci! Pure maths strikes back with solutions 4, and especially, 5 :biggrin:

It's so nice! It reminds me of the irrationality proofs for

\zeta(2)

(which then leads to the irrationality of

\pi^2

and

\pi

) and

\zeta(3)

(which is a problem I set on here that no-ones been able to solve yet!) I think it needed a hint because no-one pre-PhD-level would likely be able to solve it unless they had seen the approach used on other things like this and

\zeta(2)

Reply 1086

Jkn

Original post by Lord of the Flies

Differentiate and set

\lambda =1:

How did you get from

f(\lambda)

to having evaluated the thing with the log in it? :confused:

Nice solution btw! Do you know of any similar problems that would be nice to practice the use of all of the little Beta/Gamma/Zeta function theorems? :smile:

Reply 1087

bogstandardname

Original post by Jkn

Way too tired to be on TSR

Certainly is! You are essentially generalising the expression in order to allow taking the partial derivative of an integral with respect to a variable (that is independent of that which you are integrating with respect to) and, hence, altering the problem to one that is more manageable. Re-intergrating in order to find the original function (before it was partially differentiated) can be seen as being analogous to the inverse Laplace Transformation and other such transformations of this nature :tongue:

Alternate solutions can be found under "Problem A5" for those who are interested.

I haven't even learnt partial derivatives yet so I'll give this a miss until summer I think.

Reply 1088

Jkn

Original post by bogstandardname

I haven't even learnt partial derivatives yet so I'll give this a miss until summer I think.

They're nothing difficult! The justification/adjusting is often excruciatingly tedious and difficult but with theorem like the Leibniz Integral Rule (aka. Feynman Integration aka. "Differentiating under the integral sign") you need only apply it without much thought (though you must assure the function is continuous in the relevant range - as usual) so it's definitely worth learning! Wikipedia is a good place to start, that's where I found a lot of these problems in fact :lol:

In a nut shell:

\frac{\partial}{\partial \alpha} f(\alpha, x)

is done by assuming all other variables except the subject of your partial differentiation are constant.

Spoiler

Reply 1089

Lord of the Flies

Original post by joostan

It's not just that, I'm not convinced that some of these problems are really * but hey ho :rolleyes:

Disprove my claim by counterexample then :wink:

Original post by Jkn

How did you get from

f(\lambda)

to having evaluated the thing with the log in it? :confused:

I don't understand your question.

Reply 1090

username937760

Original post by Lord of the Flies

Disprove my claim by counterexample then :wink:

Well, if Mladenov's solution to 168 counts as A Level knowledge...

Reply 1091

Felix Felicis

Original post by DJMayes

Well, if Mladenov's solution to 168 counts as A Level knowledge...

But that's only his solution - LotF must have labelled it as * if a solution with A-level knowledge exists...give me a few hours :L

Reply 1092

MathsNerd1

I wish I could actually solve some of the questions that are posted here, however the sight of some are just too much for me to even give it a go! I'm curious to learn all the new techniques but that'll all have to wait until the holidays after exams then hopefully I'll be able to attempt some of these questions :biggrin:

Reply 1093

Jkn

Original post by Lord of the Flies

I don't understand your question.

Well how did you get from the integral to

f(\lambda)

i.e. where did the log go? :tongue:

Reply 1094

shamika

Original post by ukdragon37

"The Mathematica computer algebra system (version 5.2) successfully computes this integral, but we do not know how."

That's right :wink:

LOLOLOLOLOLOL!

I actually don't understand the sentence though. If it's referring to the Gaussian error integral, surely the authors do know how Mathematica (successfully) computes the integral? :confused:

Reply 1095

Lord of the Flies

Original post by Jkn

Well how did you get from the integral to

f(\lambda)

I didn't, I defined

f

and evaluated it. Then, as stated,

f'(1)

is the integral.

Reply 1096

bananarama2

Original post by MathsNerd1

I want to learn the techniques too. But with my friends telling me to get out more I don't think it's going to happen :rolleyes:

Reply 1097

Jkn

Original post by Lord of the Flies

I didn't, I defined

f

and evaluated it. Then, as stated,

f'(1)

is the integral.

But

\displaystyle f'(\lambda)=\int_0^{\infty} \lambda \frac{x^{\lambda-1}\,dx}{(x^3+1)^2 }

But we havent proved that this would be equal to the thing with the log in? Have I missed something? :confused:

On your marks...
...
Get set...
...

Problem 169*/**/*** (Looks friendly... at first)

Evaluate

\displaystyle\int_0^{\frac{\pi}{2}}x\cot(x) dx

Problem 170***

Prove that

\displaystyle \int_0^{\frac{\pi}{2}} (\sin \theta)^{2x-1} (\cos \theta)^{2y-1} d\theta = \frac{x+y}{2xy} \prod_{n=1}^{\infty} \left(1+\frac{xy}{n(x+y+n)} \right)^{-1}

(edited 10 years ago)

Reply 1098

Zakee

Problem 169*/** (Finally a question that normal people can do without Stokes' theorem or Lagrange multipliers).

This is a short but simple problem for you guys (your mathematical ability is far more superior to mine). Did take me a fair time though to realize what to do, but it was nice.

Prove the following result:

[br][br][br]\cos \frac{\pi}{7} - \cos \frac{2\pi}{7}+\cos \frac{3\pi}{7} = \frac{1}{2}.

(edited 10 years ago)

Reply 1099

shamika

Suggestion for people - if a problem can theoretically be done with only * (A-Level) knowledge but requires a lot of cleverness compared to the ** version, make that explicit, because it saves people from trying questions and then getting horribly frustrated. Having said that, most of you left on the thread are dedicated mathmos so go ahead and keep posting I guess...