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The Proof is Trivial! Watch

1. (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
Spoiler:
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I know why LOTF does it
2. (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
Spoiler:
Show
I know why LOTF does it
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..)
3. (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..)
I tend to like to spell things out, for readers, though I often do less working on paper
4. 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

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 (the 'Pythagorean Triples').

Hence show that ('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).
5. Problem 290*/**

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

6. (Original post by FireGarden)
Problem 290*/**

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

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

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 :

Also, there are some conditions that need to be placed on f for this to be true. The inverse must exist, for example.
7. (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 :

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).
8. (Original post by FireGarden)
Problem 290*/**

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

Geometrical proof:

Obviously, as the inverse function is a reflection of the original function in the line ,

Thus

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.
9. (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:

which simplifies to any of the special cases when we have that a and b are fixed points.
10. (Original post by Jkn)

Problem 286***

Evaluate

Can the result be easily generalised in any interesting way? If so, provide proof of your assertions.
Solution 286

So almost all of the terms disappear, giving

Using the result again, we have

, so

In general,

Proof:

The result follows
11. By the way, here http://dlmf.nist.gov/5 is an excellent selection of properties of the Gamma and Beta function (not proved).
12. (Original post by henpen)
On reflection…
13. (Original post by Smaug123)
I have a knack for unintentional puns.
14. (Original post by henpen)
Problem 228**

Calculate , where .

Hint:
Spoiler:
Show
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:

For the sake of rigour:

and so is not a Cauchy sequence and hence diverges.
We now deduce that . The result follows
15. (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 exists!
16. (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:

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.
17. (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 which seems unlikely (not to mention so powerful and fundamental that I would be surprised that I don't know it already). :P
18. (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 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.
19. (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.
20. (Original post by Jkn)
This is not what I was expecting -the result can be simplified nicely.

Also, an extremely simple closed form for exists!

Solution 286

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

For even n,

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

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