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The Proof is Trivial!

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Original post by Jkn

It seemed odd that it could be stated in such a simple form, especially when he asked for a series solution ;0

Yeah, but then again there were a couple of little tricks to perform.
I realise that in the solution I was a bit lazy with the tex, so some people may have trouble following, but hey ho :tongue:

Spoiler

Reply 1981
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Jkn
11
Original post by joostan
Yeah, but then again there were a couple of little tricks to perform.
I realise that in the solution I was a bit lazy with the tex, so some people may have trouble following, but hey ho :tongue:

Spoiler


I thought your technical details seemed sufficiently thorough (I may have used a similar amount if not slightly less?) though I am trying to get into the habit of naming all the theorems and representations that I am using (i.e. your series representation etc..) :tongue:
Original post by Jkn
I thought your technical details seemed sufficiently thorough (I may have used a similar amount if not slightly less?) though I am trying to get into the habit of naming all the theorems and representations that I am using (i.e. your series representation etc..) :tongue:


I tend to like to spell things out, for readers, though I often do less working on paper :tongue:
Reply 1983
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Jkn
11
Would anyone benefit from small hints to some of the problems I've set recently? A few of them follow trivially from theorems that can be quoted (once found). Also, on the problems where I have said that you must prove all theorems used, do feel that you can 'look them up' so long as they seem like the kind of thing that you would be expected to memorize in a course (the so-called 'book work').

Yet another that requires a simple yet rather advanced 'trick':

Problem 288***

Evaluate ∫0∞γx+log⁡Γ(1+x)x5/2 dx\displaystyle \int_0^{\infty} \frac{\gamma x + \log \Gamma(1+x)}{x^{5/2}} \ dx

Problem 289*/**

Let x, y and z denote positive integers (which can be extended without any loss of generality from the case whereby they denote non-zero integers).

Find the general solution to the equation x2+y2=z2\displaystyle x^2+y^2=z^2 (the 'Pythagorean Triples').

Hence show that x4+y4≠z4\displaystyle x^4+y^4 \not= z^4 ('Fermat's Last Theorem' in the case n=4).

Can you generalize the result to other classes of exponents in the set of natural numbers? How many other ways can you find to prove the above result? (bonus).
(edited 10 years ago)
Problem 290*/**

Suppose a,ba,b are fixed points of f(x)f(x) (that is, f(a)=a, f(b)=b)). Prove

∫abf(x)+f−1(x) dx=b2−a2 \displaystyle\int_{a}^{b}f(x) + f^{-1}(x)\ dx = b^2 - a^2
Reply 1985
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Jkn
11
Original post by FireGarden
Problem 290*/**

Suppose a,ba,b are fixed points of f(x)f(x) (that is, f(a)=a, f(b)=b)). Prove

∫abf(x)+f−1(x) dx=b2−a2 \displaystyle\int_{a}^{b}f(x) + f^{-1}(x)\ dx = b^2 - a^2

Have you had a go at deriving the fractional derivative of xax^a (formally) yet? :colone:

Solution 290

Using the property that a function's inverse can be found by reflecting the function in the line y=x in the Cartesian plane, we deduce that the area between each function and the line x=y are equal between the values for which f(x)=f−1(x)f(x)=f^{-1}(x):

∫ab(x−f(x)) dx=∫ab(f−1(x)−x) dx  ⟺  ∫ab(f(x)+f−1(x)) dx=∫ab2x dx=b2−a2 □\displaystyle \begin{aligned} \int_a^b (x-f(x)) \ dx = \int_a^b (f^{-1}(x)-x) \ dx \iff \int_a^b ( f(x)+f^{-1}(x) ) \ dx = \int_a^b 2x \ dx = b^2-a^2 \ \square \end{aligned}

Also, there are some conditions that need to be placed on f for this to be true. The inverse must exist, for example.
(edited 10 years ago)
Original post by Jkn


Solution 290

Using the property that a function's inverse can be found by reflecting the function in the line y=x in the Cartesian plane, we deduce that the area between each function and the line x=y are equal between the values for which f(x)=f−1(x)f(x)=f^{-1}(x):

∫ab(x−f(x)) dx=∫ab(f−1(x)−x) dx  ⟺  ∫ab(f(x)+f−1(x)) dx=∫ab2x dx=b2−a2 □\displaystyle \begin{aligned} \int_a^b (x-f(x)) \ dx = \int_a^b (f^{-1}(x)-x) \ dx \iff \int_a^b ( f(x)+f^{-1}(x) ) \ dx = \int_a^b 2x \ dx = b^2-a^2 \ \square \end{aligned}

Also, there are some conditions that need to be placed on f for this to be true. The inverse must exist, for example.


Nice. Now generalise by finding the result if there are no fixed points (or at least a and b are not such points).
Reply 1987
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henpen
Original post by FireGarden
Problem 290*/**

Suppose a,ba,b are fixed points of f(x)f(x) (that is, f(a)=a, f(b)=b)). Prove

∫abf(x)+f−1(x) dx=b2−a2 \displaystyle\int_{a}^{b}f(x) + f^{-1}(x)\ dx = b^2 - a^2


Geometrical proof:
89490965d7a87426cf38a5002807f20f.png

∫abf(x)dx=A5+A4\displaystyle \int_a^b f(x) dx=A_5+A_4

∫abf−1(x)dx=A5+A4+A3+A2\displaystyle \int_a^b f^{-1}(x) dx=A_5+A_4+A_3+A_2

Obviously, as the inverse function is a reflection of the original function in the line y=xy=x, A2=A3A_2=A_3

Thus ∫abf(x)+f−1(x)dx=2A5+2(A4+A3)=2(b−a)a+(b−a)2=2ba−2ba+a2−2a2+b2=b2−a2\displaystyle \int_a^b f(x)+f^{-1}(x) dx= 2A_5+2(A_4+A_3)=2(b-a)a+(b-a)^2=2ba-2ba+a^2-2a^2+b^2=b^2-a^2


Where have I not been rigorous here (I assume I've been non rigorous somewhere).

On reflection, this is pretty much the same as Jkn's.
(edited 10 years ago)
Reply 1988
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Jkn
11
Original post by FireGarden
Nice. Now generalise by finding the result if there are no fixed points (or at least a and b are not such points).

Hmm, well you lose the simplicity. We could do some clever things involving differentiating with respect to y in the Cartesian place (where we set y=f(x) etc..) but it didn't look as though it was likely to simplify to the given result

Problem 290 (extension):

We can add and subtract the additional areas in order to 'adjust' and use the fixed point property:

∫ab(f(x)+f−1(x)) dx=(b2+∫bf(b)(x−f(x)) dx)−(a2+∫af(a)(x−f(x)) dx)\displaystyle \begin{aligned}\int_a^b (f(x)+f^{-1}(x)) \ dx = \left(b^2+\int_b^{f(b)} (x-f(x)) \ dx \right) - \left(a^2+\int_a^{f(a)} (x-f(x)) \ dx \right) \end{aligned}

which simplifies to any of the special cases when we have that a and b are fixed points.
Reply 1989
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henpen
Original post by Jkn


Problem 286***

Evaluate ∏n=19Γ(n10)\displaystyle \prod_{n=1}^{9} \Gamma \left( \frac{n}{10} \right)

Can the result be easily generalised in any interesting way? If so, provide proof of your assertions.


Solution 286

Γ(z)Γ(1−z)=πsin⁡(πz)\displaystyle \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}
So almost all of the terms disappear, giving

Γ(510)∏m=14πsin⁡(πm10)\Gamma(\frac{5}{10}) \prod_{m=1}^4\frac{\pi}{\sin(\pi \frac{m}{10})}

Using the result again, we have

Γ(12)2=π\displaystyle \Gamma(\frac{1}{2})^2=\pi, so

π∏m=14πsin⁡(πm10)\sqrt{\pi} \prod_{m=1}^4\frac{\pi}{\sin(\pi \frac{m}{10})}

In general,

∏m=12n−1Γ(m2n)=π∏m=1n−1πsin⁡(πm2n)\prod_{m=1}^{2n-1}\Gamma(\frac{m}{2n})= \sqrt{\pi} \prod_{m=1}^{n-1}\frac{\pi}{\sin(\pi \frac{m}{2n})}

Proof:
∏m=12n−1Γ(m2n)=π∏m=1n−1Γ(m2n)∏m=n+12n−1Γ(m2n)=π∏m=1n−1Γ(m2n)∏m=2n−(n−1)2n−(1)Γ(m2n)\prod_{m=1}^{2n-1}\Gamma(\frac{m}{2n})= \sqrt{\pi}\prod_{m=1}^{n-1}\Gamma(\frac{m}{2n})\prod_{m=n+1}^{2n-1}\Gamma(\frac{m}{2n})= \sqrt{\pi}\prod_{m=1}^{n-1}\Gamma(\frac{m}{2n})\prod_{m=2n-(n-1)}^{2n-(1)}\Gamma(\frac{m}{2n})
=π∏m=1n−1Γ(m2n)Γ(2n−m2n)= \sqrt{\pi}\prod_{m=1}^{n-1}\Gamma(\frac{m}{2n})\Gamma( \frac{2n-m}{2n})
The result follows
(edited 10 years ago)
Reply 1990
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henpen
By the way, here http://dlmf.nist.gov/5 is an excellent selection of properties of the Gamma and Beta function (not proved).
Original post by henpen
On reflection…

:colondollar:
Reply 1992
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henpen
Original post by Smaug123
:colondollar:


I have a knack for unintentional puns.
Reply 1993
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Jkn
11
Original post by henpen
Problem 228**

Calculate ∑n=4∞1nHnHn−1\large \sum_{n=4}^{\infty}\frac{1}{n H_nH_{n-1}}, where Hn=∑k=1n1kH_n=\sum_{k=1}^{n}\frac{1}{k}.

Hint:

Spoiler


Not sure how I didn't spot that this one hadn't been solved (has it?)

Solution 228

Using the method of differences and the given definition of the nth harmonic number:

∑n=4∞1nHnHn−1=∑n=4∞Hn−Hn−1HnHn−1=∑n=4∞1Hn−1−1Hn=1H3−lim⁡n→∞1Hn⏟α=611 \displaystyle \sum_{n=4}^{\infty} \frac{1}{n H_n H_{n-1}}= \sum_{n=4}^{\infty} \frac{H_n - H_{n-1}}{H_n H_{n-1}} = \sum_{n=4}^{\infty} \frac{1}{H_{n-1}}-\frac{1}{H_{n}}=\frac{1}{H_3}- \underbrace{ \lim_{n \to \infty} \frac{1}{H_n}}_{\alpha} = \frac{6}{11}

For the sake of rigour:

∣H2n−Hn∣=∣12n+12n−1+⋯+1n+1∣>∣n×12n∣=12\displaystyle \left|H_{2n}-H_n \right| = \left|\frac{1}{2n}+\frac{1}{2n-1}+ \cdots + \frac{1}{n+1} \right|> \left|n \times \frac{1}{2n} \right|=\frac{1}{2} and so HnH_n is not a Cauchy sequence and hence diverges.
We now deduce that α=lim⁡n→∞1Hn=0\displaystyle \alpha = \lim_{n \to \infty} \frac{1}{H_n} = 0. The result follows □\square
Reply 1994
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Jkn
11
Original post by henpen
Solution 286

This is not what I was expecting -the result can be simplified nicely.

Also, an extremely simple closed form for ∏k=1n−1Γ(kn)\displaystyle \prod_{k=1}^{n-1} \Gamma \left( \frac{k}{n} \right) exists! :tongue:
Original post by Jkn
Hmm, well you lose the simplicity. We could do some clever things involving differentiating with respect to y in the Cartesian place (where we set y=f(x) etc..) but it didn't look as though it was likely to simplify to the given result

Problem 290 (extension):

We can add and subtract the additional areas in order to 'adjust' and use the fixed point property:

∫ab(f(x)+f−1(x)) dx=(b2+∫bf(b)(x−f(x)) dx)−(a2+∫af(a)(x−f(x)) dx)\displaystyle \begin{aligned}\int_a^b (f(x)+f^{-1}(x)) \ dx = \left(b^2+\int_b^{f(b)} (x-f(x)) \ dx \right) - \left(a^2+\int_a^{f(a)} (x-f(x)) \ dx \right) \end{aligned}

which simplifies to any of the special cases when we have that a and b are fixed points.


This is not what should be found. The answer, if it helps, is of the form xb+ya; the fixed points were the special case where x=b and y=a.
Reply 1996
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Jkn
11
Original post by FireGarden
This is not what should be found. The answer, if it helps, is of the form xb+ya; the fixed points were the special case where x=b and y=a.

I don't think that's right. If what I have is right then what you're saying is that we have an expression for the definite integral of any function between the limits a and f(a).

Also, if it can be properly evaluated for any limits, then we could immediately find things like ∫sin⁡x+arcsin⁡x dx\int \sin x + \arcsin x \ dx which seems unlikely (not to mention so powerful and fundamental that I would be surprised that I don't know it already). :P
Original post by Jkn
I don't think that's right. If what I have is right then what you're saying is that we have an expression for the definite integral of any function between the limits a and f(a).

Also, if it can be properly evaluated for any limits, then we could immediately find things like ∫sin⁡x+arcsin⁡x dx\int \sin x + \arcsin x \ dx which seems unlikely (not to mention so powerful and fundamental that I would be surprised that I don't know it already). :P

Well, the arcsine integral's not too bad :P but I see what you mean - it's pretty likely not linear.
Reply 1998
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Jkn
11
Original post by Smaug123
Well, the arcsine integral's not too bad :P but I see what you mean - it's pretty likely not linear.

Of course it's easy to do (IBP) but it's an example of an integral of that form with a 'messy' answer. :tongue:
Reply 1999
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henpen
Original post by Jkn
This is not what I was expecting -the result can be simplified nicely.

Also, an extremely simple closed form for ∏k=1n−1Γ(kn)\displaystyle \prod_{k=1}^{n-1} \Gamma \left( \frac{k}{n} \right) exists! :tongue:




Solution 286

Standing on the shoulders of giants (here, Mladenov, solution 24), we have

∏m=1rsin⁡(πm2r+1)=2r+12r\displaystyle \prod_{m=1}^{r}\sin\left(\frac{ \pi m}{2r+1} \right)=\frac{\sqrt{2r+1}}{2^{r}}



For even n,

Unparseable latex formula:

\displaystyle \prod_{m=1}^{2r}\Gamma \left(\frac{m}{2r+1} \right)=\displaystyle \prod_{m=1}^{r}\frac{\pi}{\sin \left(\frac{\pi m}{2r+1} \right)} \right)= \frac{\pi^r 2^{r}}{\sqrt{2r+1}}



For odd n, I'll try to generalise 24 later, but I need to sleep now.

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