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The Proof is Trivial!

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Original post by atsruser
Problem 540 (**/***)

a) Show that:

02πr2sin22θ12r2cos2θ+r4 dθ=πr2\displaystyle \int_0^{2\pi} \frac{r^2\sin^2 2\theta}{1-2r^2\cos 2\theta+r^4} \ d\theta = \pi r^2

stating any conditions that must hold for this to be valid.

b) Find the value of 02π4sin22θ178cos2θ dθ\displaystyle \int_0^{2\pi} \frac{4\sin^2 2\theta}{17-8\cos 2\theta} \ d\theta

(Hints available if necessary)


Original post by atsruser
Bumping with hints:

Spoiler



Nice ones ... I spent 5 minutes on each and beat me.
I will be back ...
Original post by Zacken
Sorry to be a buzzkill, but it's specified in the OP to Latex all solutions/problems (I understand that you're on an iPad, but it's not hard to use Latex) - just to maintain austerity. :smile:

Also - it's fine to post a solution to your own problem once somebody else has already answered it, but not before.


I need to learn how to use it then. What if i just post a picture of the solution. I have some great problems with me that i would like to share. Also how do i spoiler things again?


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Original post by physicsmaths
I need to learn how to use it then. What if i just post a picture of the solution. I have some great problems with me that i would like to share. Also how do i spoiler things again?


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Latex guide.

If possible, problems in Latex and solutions in pictures works out fine! :smile:

Eh, you don't really need to spoiler your answers, but you can spoiler things by using spoiler thing[ /spoiler] without a space in the last tag.
Original post by DFranklin
x


Ah, thanks very much for that! I was looking for the 'preformatted text' option in the toolbar but didn't see it.
Original post by atsruser
Problem 540 (**/***)

a) Show that:

02πr2sin22θ12r2cos2θ+r4 dθ=πr2\displaystyle \int_0^{2\pi} \frac{r^2\sin^2 2\theta}{1-2r^2\cos 2\theta+r^4} \ d\theta = \pi r^2

stating any conditions that must hold for this to be valid.

b) Find the value of 02π4sin22θ178cos2θ dθ\displaystyle \int_0^{2\pi} \frac{4\sin^2 2\theta}{17-8\cos 2\theta} \ d\theta

(Hints available if necessary)


Hints:

Spoiler

Problem 542 (*/**)

1. A line is of length LL. Point P lies ll distant from one end of the line. Find the average distance from P to all other points on the line.

2. A disc is of radius aa. Find the average distance from the centre of the disc to all other points in the disc.

3. For the same disc, find the average distance from a fixed point P on the circumference to all other points in the disc.
Original post by atsruser
Problem 542 (*/**)
2. A disc is of radius aa. Find the average distance from the centre of the disc to all other points in the disc.


Roughly speaking, is this equivalent to finding the distance of one independently chose point on the unit disk. I've heard something similar about the mean distance between points on a disk, does it apply to the distance between points on a disk and it's centre?
Original post by atsruser
Problem 542 (*/**)

1. A line is of length LL. Point P lies ll distant from one end of the line. Find the average distance from P to all other points on the line.

2. A disc is of radius aa. Find the average distance from the centre of the disc to all other points in the disc.

3. For the same disc, find the average distance from a fixed point P on the circumference to all other points in the disc.


For part 3.
4aπ0π(πθ)sin2(θ2)cos(θ2)dθ=32a9π\displaystyle \frac{4a}{\pi} \int_0^{\pi} {(\pi - \theta) sin^2 \left( \frac{\theta}{2} \right) cos \left( \frac{\theta}{2} \right) } d\theta = \frac{32a}{9 \pi}
(edited 8 years ago)
Original post by EricPiphany
For part 3.
4aπ0π(πθ)sin2(θ2)cos(θ2)dθ=32a9π\displaystyle \frac{4a}{\pi} \int_0^{\pi} {(\pi - \theta) sin^2 \left( \frac{\theta}{2} \right) cos \left( \frac{\theta}{2} \right) } d\theta = \frac{32a}{9 \pi}


That looks to be correct, but bear in mind that the value of these questions isn't in knowing the answer, it's in seeing the argument, and you haven't given one. For example, I approached this by doing a double integral, which is easy to justify; how did you write down that single integral?
Original post by Zacken
Roughly speaking, is this equivalent to finding the distance of one independently chose point on the unit disk.


I'm not sure what you're asking here: the distance of what from what? It looks like you are asking about some kind of probabilistic question, but this isn't a problem in probability (which is good, for probability is the work of Beelzebub)

I've heard something similar about the mean distance between points on a disk, does it apply to the distance between points on a disk and it's centre?


Again you have me baffled here. You haven't told me what the "something similar" is, so I can't tell if it applies to anything, I'm afraid.
Original post by atsruser
That looks to be correct, but bear in mind that the value of these questions isn't in knowing the answer, it's in seeing the argument, and you haven't given one. For example, I approached this by doing a double integral, which is easy to justify; how did you write down that single integral?


Let PP be a point on the circumference of the disk, and OO be the center of the disk.
Draw a circle around PP intersecting the edge of the disk atS S and T T, let θ=POS\theta = \angle POS, r=PSr = |PS| and ϕ=SPT=πθ\phi = \angle SPT = \pi - \theta.
Now let another slightly larger concentric circle be drawn around PP intersecting the edge of the disk at UU such that POU=θ+δθ\angle POU = \theta + \delta \theta, the thickness of the resulting shell being δt\delta t.
Now we have r=2asin(θ2)r = 2a sin \left( \frac{\theta}{2} \right) and δtacos(θ2)δθ\delta t \approx a cos \left( \frac{\theta}{2} \right) \delta \theta and arc length
Unparseable latex formula:

\arc ST = r \phi

.

Someone finish this proof for me, I can't type any more Latex on my phone.
(edited 8 years ago)
Problem 543 *

Since this thread has been a bit quiet I'll post a relatively easy question which would have made OCR MEI'S S2 a lot more interesting!

Let the DRV X follow a Poisson distribution with parameter λ\lambda Prove that:

E(X)=Var(X)=λ E(X) = Var(X) = \lambda
Original post by 16Characters....
Problem 543 *

Since this thread has been a bit quiet I'll post a relatively easy question which would have made OCR MEI'S S2 a lot more interesting!

Let the DRV X follow a Poisson distribution with parameter λ\lambda Prove that:

E(X)=Var(X)=λ E(X) = Var(X) = \lambda


Solution 543:
Since generating functions are a bit of a faff I'll do it directly.

E(X):=xIm(X)xP(X=x)E(X):=\displaystyle\sum_{x \in Im(X)} xP(X=x)
For a Poisson variable, this means that:
E(X)=k=0keλλkk!=λeλk=1λk1(k1)!=λeλk=0λkk![br]E(X)=λeλeλ=λE(X)= \displaystyle\sum_{k=0}^{\infty} k\dfrac{e^{-\lambda} \lambda ^k}{k!}=\lambda e^{-\lambda}\displaystyle\sum_{k=1}^{\infty} \dfrac{\lambda ^{k-1}}{(k-1)!}= \lambda e^{-\lambda} \displaystyle\sum_{k=0}^{\infty}\dfrac{\lambda ^k}{k!}[br]\Rightarrow E(X)=\lambda e^{-\lambda} \cdot e^{\lambda}=\lambda

Var(X):=E(X2)(E(X))2Var(X):=E(X^2)-(E(X))^2
Observe that E(X2X)=E(X2)E(X)E(X^2-X)=E(X^2)-E(X) as expectation is a linear operator.
Thus:
Unparseable latex formula:

\Var(X)=E(X^2-X)+E(X)-(E(X))^2

.
But:
E(X2X)=k=0(k2k)k!eλλk=λ2eλk=2λk2(k2)!=λ2eλk=0λkk!E(X^2-X)=\displaystyle\sum_{k=0}^{ \infty} \dfrac{(k^2-k)}{k!}e^{-\lambda} \lambda^k=\lambda ^2 e^{-\lambda} \displaystyle\sum_{k=2}^{\infty} \dfrac{\lambda ^{k-2}}{(k-2)!}=\lambda^2 e^{-\lambda} \displaystyle\sum_{k=0}^{\infty} \dfrac{\lambda^{k}}{k!}.

E(X2X)=λ2[br]eλeλ=λ2[br]\Rightarrow E(X^2-X) = \lambda^2[br] e^{-\lambda} \cdot e^{\lambda}=\lambda ^2[br]
Thus: Var(X)=(λ2+λ)λ2=λVar(X)=(\lambda^2+\lambda)-\lambda^2=\lambda.

Hence E(X)=Var(X)=λE(X) = Var(X)= \lambda.      \ \ \ \ \Box
(edited 8 years ago)
Problem 544***

Show that

02πcos3x54cosx dx=π12\displaystyle \int_{0}^{2\pi} \dfrac{\cos 3x}{5 - 4\cos x} \ dx = \dfrac{\pi}{12}
Original post by Indeterminate
Problem 544***

Show that

02πcos3x54cosx dx=π12\displaystyle \int_{0}^{2\pi} \dfrac{\cos 3x}{5 - 4\cos x} \ dx = \dfrac{\pi}{12}


I've done this, via a contour integral. However, calculating one of the residues was messy (at least the way I did it) so I'm wondering if there's some slicker approach that you had in mind. Any hints?
Original post by atsruser
I've done this, via a contour integral. However, calculating one of the residues was messy (at least the way I did it) so I'm wondering if there's some slicker approach that you had in mind. Any hints?


I haven't tried it myself, but perhaps working with 154cosx\dfrac{1}{5 - 4\cos x} and cosx\cos x in its exponential form and then partial fractions and geometric series may prove useful.
Original post by Zacken
I haven't tried it myself, but perhaps working with 154cosx\dfrac{1}{5 - 4\cos x} and cosx\cos x in its exponential form and then partial fractions and geometric series may prove useful.


Would you like to expand on that a bit? Not sure I see where the partial fractions and geometric series will arise. (Though if you can end up with a series in cosnx\cos nx that will be very useful)
Original post by atsruser
Would you like to expand on that a bit? Not sure I see where the partial fractions and geometric series will arise. (Though if you can end up with a series in cosnx\cos nx that will be very useful)


Maybe he is talking about doing a GP on the reals of tbe imaginary numbers. I have had a try and came up as problematic as i found cos3x(pretty easy) 4cos^3(x)-3cosx now divide this to get a quadratic in cosx and the fraction A/(5-4cosx) problem is i get the integral of that through u=tan(x/2) sub but the limits are problematic as they are both zero it seems.


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