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The Proof is Trivial!

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Original post by james22
How can you evaluate that integral though?

Doesn't seem impossible, right? It's e^z around the unit semicircle. Currently practically asleep, though, so I might be talking nonsense.
Original post by james22
How can you evaluate that integral though?


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Original post by Smaug123
Doesn't seem impossible, right? It's e^z around the unit semicircle. Currently practically asleep, though, so I might be talking nonsense.


Surely then the integral would be 0 (no poles), which is not correct?

(I haven't done contour integrals in over a year btw.)
Original post by james22
Surely then the integral would be 0 (no poles), which is not correct?

(I haven't done contour integrals in over a year btw.)


It's e^z around the unit semicircle, which isn't a closed loop. We just need to evaluate the integral along the x-axis to complete the semicircular closed contour: 11ezdz-\int_{-1}^1 e^z dz. OK, I've definitely got something horribly wrong :P
Original post by Noble.
I can't be bothered to dig out some of my notes, so this may not be rigorous:


Nice, I also got it into exponential form but didn't think writing it as a sum would lead anywhere
Original post by Smaug123
It's e^z around the unit semicircle, which isn't a closed loop. We just need to evaluate the integral along the x-axis to complete the semicircular closed contour: 11ezdz-\int_{-1}^1 e^z dz. OK, I've definitely got something horribly wrong :P


The answer is not real, and that integral is real, so something has gone very wrong. I cannot see waht though which is really worrying because I should be pretty good at complex analysis.
Solution 483

This is an elementary solution to the problem (here meaning not using complex analysis). For full disclosure, I used ThatPerson's hint.

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(edited 9 years ago)
Original post by DJMayes
Solution 483

This is an elementary solution to the problem (here meaning not using complex analysis). For full disclosure, I used ThatPerson's hint.

Spoiler



I'm not sure that

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. I got a different answer for that part, though it doesn't affect the end since we're only interested in the real part.

Also, apparently this was set in a Cambridge exam over a hundred years ago:

Problem 484**

04ln(x)4xx2 dx \displaystyle \int^{4}_{0} \dfrac{\ln (x)}{\sqrt{4x-x^2}} \ dx
(edited 9 years ago)
Original post by ThatPerson
I'm not sure that

Spoiler

. I got a different answer for that part, though it doesn't affect the end since we're only interested in the real part.

Also, apparently this was set in a Cambridge exam over a hundred years ago:

Problem 484**

04ln(x)4xx2 dx \displaystyle \int^{4}_{0} \dfrac{\ln (x)}{\sqrt{4x-x^2}} \ dx


Good call; I was being very careless with my constants. Will fix that.
Original post by Flauta
Is it

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?

I'll type up my "solution" tomorrow if I'm right, I've only done a little probability though so the chances are I'm far off.


Indeed it is
Original post by ThatPerson
I'm not sure that

Spoiler

. I got a different answer for that part, though it doesn't affect the end since we're only interested in the real part.

Also, apparently this was set in a Cambridge exam over a hundred years ago:

Problem 484**

04ln(x)4xx2 dx \displaystyle \int^{4}_{0} \dfrac{\ln (x)}{\sqrt{4x-x^2}} \ dx


I get my answer as:

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Original post by newblood
I get my answer as:

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Spoiler

Original post by ThatPerson

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In which case, that's a neat little integral :tongue:
[QUOTE="newblood;52061983"]Problem 480 ** (not much probability theory in A-Level)

Let X1,X2,...,Xn X_1, X_2,...,X_n be i.i.d r.v's with uniform distribution on [1,2] [1,2] .

Determine a unique value, Δ \Delta , such that limnPr(a<(X1X2...Xn)1/n<b)=1\lim_{n\to \infty} Pr(a <(X_1X_2...X_n)^{1/n} < b) = 1 iff a<Δ a< \Delta

Solution 480

14202228825781009771754.jpg

There's another method involving how the arithmetic and geometric means are related
(edited 9 years ago)
Original post by jjpneed1
Pretty average tbh :wink:

Problem 471
Determine all sets of non-negative integers x,y x, y and zz that satisfy 2x+3y=z2 2^x + 3^y = z^2 .

Problem 472
Evaluate 0ln(1+x)ln(1+x2)x3dx \displaystyle \int_0^{\infty} \frac{\ln(1 + x)\ln(1 + x^2)}{x^3} dx

Last ones for a while now while I sit some tasty exams, gl on that integral, good job if someone gets it out before exams finish :smile:


Just was looking back a few pages and saw 472. Does that integral have a nice answer? I plugged it into Mathematica and the indefinite integral is horrific.

Edit: Surprisingly, I'm making some progress :tongue:

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(edited 9 years ago)
Reply 2975
Problem 485**

Prove that every open set GG on the real line is the union of a finite or countable system of pairwise disjoint open intervals, where we regard (,x),(,),(y,)(-\infty,x),(-\infty,\infty),(y,\infty) as open intervals.
Original post by ThatPerson
Just was looking back a few pages and saw 472. Does that integral have a nice answer? I plugged it into Mathematica and the indefinite integral is horrific.

Edit: Surprisingly, I'm making some progress :tongue:

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Honesty I can't remember the answer, I think it's something like that though.
Original post by 0x2a
Problem 485**

Prove that every open set GG on the real line is the union of a finite or countable system of pairwise disjoint open intervals, where we regard (,x),(,),(y,)(-\infty,x),(-\infty,\infty),(y,\infty) as open intervals.


Solution 485

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(edited 9 years ago)
Original post by 0x2a
Problem 485**

Prove that every open set GG on the real line is the union of a finite or countable system of pairwise disjoint open intervals, where we regard (,x),(,),(y,)(-\infty,x),(-\infty,\infty),(y,\infty) as open intervals.


For each xGx \in G, but a rational pxp_x in the same path connected component as xx (existance is obvious from the definition of openness). For each pxp_x let ax,bxa_x,b_x be such that (ax,bx)G(a_x,b_x) \subseteq G but ax,bx∉Ga_x,b_x \not\in G (existance comes from GG being open). We then have countable many open intervals, and they clearly cover all of GG
Original post by DJMayes
Solution 485

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I'm not sure about this, since you seem to be trying to perform uncountably many operations. I think it needs more justification to be correct.

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