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# The Proof is Trivial! watch

1. Unless ive missed something terribly obvious 😁 i havent done integration in a while.. Since step III actually.

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2. (Original post by atsruser)
Problem 546 (*/**)

Evaluate
Solution 546:

Sending yields:
.

Oh snap, 2nd place .
3. I missed the obvious steps as its not like we all don't know what is going on.

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4. (Original post by joostan)
Solution 546:

Sending yields:
.
Thanks, someone did it in latex. I really need to learn how to use this..

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5. (Original post by physicsmaths)
Thanks, someone did it in latex. I really need to learn how to use this..

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It ain't all that hard although on TSR it can be quite buggy at times.
6. (Original post by joostan)
It ain't all that hard although on TSR it can be quite buggy at times.
Very true.

Also, love the solution.
7. (Original post by physicsmaths)

Unless ive missed something terribly obvious 😁 i havent done integration in a while.. Since step III actually.

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Yes, that's fine. It's very straightforward if you know what to do, but it's a pretty nice result anyway.
8. (Original post by atsruser)
Yes, that's fine. It's very straightforward if you know what to do, but it's a pretty nice result anyway.
Very pretty.
9. (Original post by Indeterminate)
Nicely done, and I don't blame you for that second residue
As I guess you know, that substitution is standard for integrals involving sin and cos between 0 and 2pi.

And as for the second residue, yes, I simply brute forced it rather than thinking about any cleaner approach. And speaking of cleaner approaches ..

Besides finding the second derivative, another way of calculating it would involve considering a small circle around the origin. The residue would then be the coefficient of in the expansion of

but that's hardly mess-free
.. I'm not sure I'm following you. Where does the coefficient of z^2 come into the picture? I'm trying to imagine a Laurent series method here, but that doesn't seem to be what you're doing.
10. (Original post by Zacken)
Very pretty.
I guess an added bonus question would be: come up with an intuitive explanation for why all of those integrals yield the same result.

I don't have one, but it seems to me that there must be a more revealing way of getting to that answer, apart from just doing the integral.
11. Problem 547

Evaluate for

Spoiler:
Show
Can you spot a general pattern?
Spoiler:
Show
DUTIS.
12. (Original post by atsruser)
As I guess you know, that substitution is standard for integrals involving sin and cos between 0 and 2pi.

And as for the second residue, yes, I simply brute forced it rather than thinking about any cleaner approach. And speaking of cleaner approaches ..

.. I'm not sure I'm following you. Where does the coefficient of z^2 come into the picture? I'm trying to imagine a Laurent series method here, but that doesn't seem to be what you're doing.
Well here's my method:
Spoiler:
Show
As long as we have

and by expanding we get

13. (Original post by Indeterminate)
Well here's my method:
Spoiler:
Show
As long as we have

and by expanding we get

Right. Yes, that works due to the that I'd forgotten about. That's a lot easier.

I have it in my mind that the "compute a series" approach is usually a lot more hassle than just using the formulae, but this is a nice counterexample.
14. Problem 548**

By considering a suitable integral, or otherwise, prove that

.
15. (Original post by 16Characters....)
Problem 548**

By considering a suitable integral, or otherwise, prove that

.
A proof using series, because integrals are boring.
Spoiler:
Show
I'm kidding.
First note that:
.
A reasonably well known fact, which can be checked using Parseval on .
(OK, that uses an integral, but there's a series too so shush yo. )
Thus:

Also:

converges for and for . So:
.

But with equality if, and only if, .
Hence:

Now observe that: .
Note further that: with equality if, and only if, .
To prove this: Note the result is true for , and that if true for , then:
.
The last inequality holds as:
.

Thus we have that:

So: .
16. Problem 549***

Suppose that is a non-negative matrix with dominant eigenvalue . Let . Using the matrix expansion

Show that if (the 1-norm of is 1) then

where and are the minimum and maximum column sums of respectively.
17. (Original post by joostan)
A proof using series, because integrals are boring. .
Spoiler:
Show

Significantly more rigorous than the integral method but I included that comment because of a recent thread hanging around here concerning the integral

Which does of course simplify into your series. Inequalities follow by considering the areas of suitable rectangles.
18. Problem 550

Show that

19. (Original post by Indeterminate)
Problem 550

Show that

Solution:

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20. (Original post by physicsmaths)
Solution:

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That's the solution I was looking for

The complex analysis route is really quite messy in this case.

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Updated: February 22, 2018
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