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The Proof is Trivial!

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Original post by DJMayes
I find it telling how the OP hasn't submitted a solution since about question 50, and we're now past question 120. This thread stopped being interesting (Or even accessible) for the vast majority of people quite a long time ago and now it seems very much to just be a battle of wits between Mladenov and Lord of the Flies.

(If this comes across as jealous, then that's probably because it is - I'd love to be able to even attempt the questions that are being thrown around lately! :lol: Unfortunately, the questions that are being thrown out are aimed at a very small audience indeed and just looking at the list in the OP it's very easy to see. It reminds me very strongly of what the STEP thread was like before Christmas.)


Well, the last question posted by Mladenov could be attemped by an A-level student.. (Though establishing the inequality between the 2nd and 3rd terms has got me in a bind at the moment! 1 < 2 and 1 < 3 were actually very easy (and im hoping the one im stuck on doesnt have an obvious solution im not missing :colondollar: though I don't think so, due to the three stars). Everyone could give the two sub-question inequalities i just outlined a go!
Original post by shamika
LOL! At least you could stick my curve into Wolfram Alpha easily (which is what I was intending :tongue:)

What is it?


:flute: Na na na na na na na na na na na na na na na na....
Original post by FireGarden
Well, the last question posted by Mladenov could be attemped by an A-level student.. (Though establishing the inequality between the 2nd and 3rd terms has got me in a bind at the moment! 1 < 2 and 1 < 3 were actually very easy (and im hoping the one im stuck on doesnt have an obvious solution im not missing :colondollar: though I don't think so, due to the three stars). Everyone could give the two sub-question inequalities i just outlined a go!


The triple stars means that I somewhat doubt it'll be simple either, although I do agree that 1 < 2 and 1 < 3 seem fair enough. However, the questions on here for the most part stopped being accessible a long time ago.
Reply 843
Original post by ukdragon37
:flute: Na na na na na na na na na na na na na na na na....


I'm really annoyed with myself I should've got that :smile:
Problem 126:
I=01f(x)(x1/2)dx=01/2f(x)(x1/2)dx+1/21f(x)(x1/2)dx[br]1/21f(x)(x1/2)dx=01/2f(x+1/2)xdx[br]01/2f(x)(x1/2)dx=01/2f(1/2x)xdx[br]I=01/2(f(x+1/2)f(1/2x))xdx0[br][br]01xf(x)dx=01/2xf(x)dx+1/21xf(x)dx[br]=1/21(x1/2)f(x1/2)+1/21xf(x)dxI=\displaystyle \int^1_0 f(x)(x-1/2)dx=\displaystyle \int^{1/2}_0 f(x)(x-1/2)dx+\displaystyle \int^{1}_{1/2}f(x)(x-1/2)dx[br]\displaystyle \int^{1}_{1/2}f(x)(x-1/2)dx=\displaystyle \int^{1/2}_0f(x+1/2)xdx[br]\displaystyle \int^{1/2}_{0}f(x)(x-1/2)dx=-\displaystyle \int^{1/2}_{0}f(1/2-x)xdx[br]\therefore I=\displaystyle \int^{1/2}_{0}(f(x+1/2)-f(1/2-x))xdx \geq 0[br][br]\displaystyle \int^1_0 xf(x)dx=\displaystyle \int^{1/2}_0 xf(x)dx+\displaystyle \int^1_{1/2} xf(x)dx[br]=\displaystyle \int^1_{1/2}(x-1/2)f(x-1/2)+\displaystyle \int^1_{1/2} xf(x)dx
So:
1/21f(x)dx01xf(x)dx[br]=1/21f(x)dx1/21(x1/2)f(x1/2)dx1/21xf(x)dx[br]=1/21f(x)(1x)dx1/21(x1/2)f(x1/2)dx[br]1/21f(x)(1x)dx=1/21f(3/2x)(x1/2)dx\displaystyle \int^1_{1/2}f(x)dx-\displaystyle \int^1_0 xf(x)dx[br]=\int^1_{1/2} f(x)dx-\displaystyle \int^1_{1/2}(x-1/2)f(x-1/2)dx-\displaystyle \int^1_{1/2} xf(x)dx[br]=\displaystyle \int^1_{1/2} f(x)(1-x)dx-\displaystyle \int^1_{1/2}(x-1/2)f(x-1/2)dx[br]\displaystyle \int^1_{1/2} f(x)(1-x)dx=\displaystyle \int^1_{1/2}f(3/2-x)(x-1/2)dx
So:
1/21f(x)dx01xf(x)dx[br]=1/21(x1/2)(f(3/2x)f(x1/2))dx0\displaystyle \int^1_{1/2}f(x)dx-\displaystyle \int^1_0 xf(x)dx[br]=\displaystyle \int^1_{1/2}(x-1/2)(f(3/2-x)-f(x-1/2))dx \geq 0
Problem 127 *

Two cards are chosen (randomly, without replacement) from a standard deck (52). Find the probability that

i) both cards are aces given that at least one ace is chosen

ii) both cards are aces given that the ace of spades is chosen

and comment on your answers.
Original post by Mladenov
Problem 44***

Let P(n)P(n) be a polynomial with coefficients in Z\mathbb{Z}. Suppose that deg(P)=p\deg(P)=p, where pp is a prime number. Suppose also that P(n)P(n) is irreducible over Z\mathbb{Z}. Then there exists a prime number qq such that qq does not divide P(n)P(n) for any integer nn.



Assume for all primes qq there exists nZn \in \mathbb{Z} such that P(n)0 (q)P(n) \equiv 0\ (q). Hence P(x)P(x) is reducible in Z/qZ[X]\mathbb{Z}/q\mathbb{Z}[X] for all primes qq. By Chebotarev's density theorem (and a fair bit of kung fu), this implies that P(x)P(x) is reducible in Z[X]\mathbb{Z}[X]; thus a contradiction.
Reply 847
Original post by DJMayes
However, the questions on here for the most part stopped being accessible a long time ago.

I'm thinking people miss the problems I was setting :wink:
Problem 127**/***

Undergrads, you should be able to do this easily so give A-level people a go :P

Prove analytically that the area under the curve E^-(x^2)=sqrt(pi)
Reply 849
Solution 127

The probability that both cards are aces is (42)(522)\displaystyle \frac{\dbinom{4}{2}}{\dbinom{52}{2}}. The probability that at least one card is an ace is 1(482)(522)\displaystyle 1-\frac{\dbinom{48}{2}}{\dbinom{52}{2}}.
Bayes gives (42)(522)1(482)(522)\displaystyle \frac{ \frac{\dbinom{4}{2}}{\dbinom{52}{2}}}{1-\frac{\dbinom{48}{2}}{\dbinom{52}{2}}}.

For ii) we have three possibilities such that one of the aces is spade. Hence the probability that two cards are aces and one is the ace of spades is 3(522)\displaystyle \frac{3}{\dbinom{52}{2}}. The probability that one of the cards is the ace of spades and the oder is random is 51(522)\displaystyle \frac{51}{\dbinom{52}{2}}.
Again we use Bayes formula to obtain 3(522)51(522)\displaystyle \frac{\displaystyle \frac{3}{\dbinom{52}{2}}}{ \displaystyle \frac{51}{\dbinom{52}{2}}}.


Original post by jack.hadamard
Assume for all primes qq there exists nZn \in \mathbb{Z} such that P(n)0 (q)P(n) \equiv 0\ (q). Hence P(x)P(x) is reducible in Z/qZ[X]\mathbb{Z}/q\mathbb{Z}[X] for all primes qq. By Chebotarev's density theorem (and a fair bit of kung fu), this implies that P(x)P(x) is reducible in Z[X]\mathbb{Z}[X]; thus a contradiction.


This is exactly what I had in mind. :tongue:
By the way, I found this result, when I was trying to prove a special case of it, and after annoying attempts to solve my problem using cyclotomic polynomials, I decided to employ Chebotarev's theorem.
Edit: For the sake of completeness, I would like to add that we can use Chebotarev's density theorem to show that there are many primes qq which satisfy the condition. In other words, we look for those primes qq for which the degree of the splitting field of PP over FqF_{q} is pp.
What do you think about the case when deg(P)=pα\deg (P) = p^{\alpha}, where α>1\alpha > 1.

Original post by ben-smith
Problem 126


Yup, it is correct.

Spoiler

(edited 10 years ago)
Original post by natninja
Problem 127**/***

Undergrads, you should be able to do this easily so give A-level people a go :P

Prove analytically that the area under the curve E^-(x^2)=sqrt(pi)


Solution 127

(ex2dx)2 \left(\int_{-\infty}^{\infty} e^{-x^2}dx \right)^2

=(ex2dx)(ey2dy) =\left(\int_{-\infty}^{\infty} e^{-x^2}dx \right)\left(\int_{-\infty}^{\infty} e^{-y^2}dy \right)

=e(x2+y2)dxdy =\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}dx dy

Polar.

=002πrer2drdθ =\int_{0}^{\infty}\int_{0}^{2\pi} re^{-r^2}dr d\theta

=02πrer2dr =\int_{0}^{\infty}2\pi re^{-r^2}dr

=π = \pi and the result follows.
(edited 10 years ago)
Original post by bananarama2
Solution 127


Very good, it's a rather neat trick imo :P

Problem 129 **

Show that:

y1=(1/k)*integral [0,x] f(x')sinh{k(x-x')} dx'

Is a particular solution to the second order differential equation:

y''-(k^2)y=f(x)
Reply 852
Original post by bananarama2

=002πrer2drdθ =\int_{0}^{\infty}\int_{0}^{2\pi} re^{-r^2}dr d\theta

=02πrer2dr =\int_{0}^{\infty}2\pi re^{-r^2}dr

How did you do this step? Where did the theta go? :tongue: And where do the new limits keep coming from?

Nice proof btw, did you come up with it yourself? (As opposed to having seen it derived before)
Original post by Jkn
How did you do this step? Where did the theta go? :tongue: And where do the new limits keep coming from?

Nice proof btw, did you come up with it yourself? (As opposed to having seen it derived before)


I just integrated respects to theta. The new limits are because the range of r in polar coords is from 0 to infinity and the range of theta is 0 to 2pi. I just integrating over R2 \mathbb{R}^2 .

Unfortunately not, well sort of. It was hinted at in a question I did once. (Kinda like a STEP questions I suppose)
Reply 854
Original post by bananarama2
I just integrated respects to theta. The new limits are because the range of r in polar coords is from 0 to infinity and the range of theta is 0 to 2pi. I just integrating over R2 \mathbb{R}^2 .

Unfortunately not, well sort of. It was hinted at in a question I did once. (Kinda like a STEP questions I suppose)

Ohhhh I see I thought you had integrated with respect to the one on the left :tongue:

But why is it valid to say the total range of values in polar co-ordinates is exactly equivalent to the range in cartesian co-ordinates? You need some sort of formal derivation of these limits or a rigorous reason as to why this is a valid deduction.

Haha fair enough :tongue: Were you originally going to apply to do maths then?

I've been trying to come up with a solution of my own. I got as far as using repeated applications of integration by parts to equate it to ex2e^{-x^2} multiplied by an infinite polynomial with coefficients in a well-defined (but complicated!) pattern. Now it gets a bit trickier. The Maclaurin's expansion doesn't seem to help. The only thing that seems promising is to consider the possible convergence of terms in the form ex2xn\frac{e^{x^2}}{x^n} which is pretty tricky!

Any ideas? :tongue:
(edited 10 years ago)
Reply 855
[QUOTE="Flies;42427722" Lord="Lord" of="of" the="the"][br][br]Youregoingtobedisappointedessentiallythesameaswhatyoudid,butsetupdifferently:[br][br][tex]0arctanexarctanxxdx=0xex1x(1+t2)dtdx=0t/et1x(1+t2)dxdt=0dt1+t2=π2[/tex][br][br][br][br]You're going to be disappointed - essentially the same as what you did, but set up differently:[br][br][tex]\displaystyle\begin{aligned} \int_0^{\infty} \frac{\arctan ex-\arctan x}{x}\,dx &= \int_0^{\infty}\int_x^{ex}\frac{1}{x(1+t^2)}\,dt \,dx\\&=\int_0^{\infty}\int_{t/e}^{t}\frac{1}{x(1+t^2)}\,dx \, dt=\int_0^{\infty}\frac{dt}{1+t^2}=\frac{\pi}{2}\end{aligned}[/tex][br][br]

Is this the (***) solution? If not, I don't have any knowledge of double integrals at A Level! :lol:
Original post by natninja
Very good, it's a rather neat trick imo :P

Problem 129 **

Show that:

y1=(1/k)*integral [0,x] f(x')sinh{k(x-x')} dx'

Is a particular solution to the second order differential equation:

y''-(k^2)y=f(x)


Solution 129

ddx1k0xf(x)sinh(k(xx))dx \frac{d}{dx} \frac{1}{k} \int_0^{x} f(x')\sinh (k(x-x')) dx'

=1kf(x)sinh(k(xx))f(0)1ksinh(kx)×0+1k0xkf(x)cosh(k(xx))dx= \frac{1}{k} f(x) \sinh (k(x-x)) - f(0) \frac{1}{k}\sinh(kx) \times 0 + \frac{1}{k} \int_0^x k f(x') \cosh (k(x-x')) dx'

=0xf(x)cosh(k(xx))dx= \int_0^x f(x') \cosh (k(x-x')) dx'

ddx0xf(x)cosh(k(xx))dx \frac{d}{dx} \int_0^x f(x') \cosh (k(x-x')) dx'

Unparseable latex formula:

= f(x)\cosh (k(x-x)) - 0 + \int_0^x kf(x')\shin (k(x-x')) dx'



=y1 = y_1''

SO

y1k2y y_1'' - k^2y

Unparseable latex formula:

= f(x)\cosh (k(x-x)) + \int_0^x kf(x')\shin (k(x-x')) dx' - \int_0^x kf(x')\shin (k(x-x')) dx'



=f(x) = f(x)
Original post by Jkn
Ohhhh I see I thought you had integrated with respect to the one on the left :tongue:

But why is it valid to say the total range of values in polar co-ordinates is exactly equivalent to the range in cartesian co-ordinates? You need some sort of formal derivation of these limits or a rigorous reason as to why this is a valid deduction.


Rigorous? I'm a Natsci :rofl: Well I don't see the need for it here. I'm just evaluating the integral over all possible coords in the x-y plane. I don't think I understand you problem?


Haha fair enough :tongue: Were you originally going to apply to do maths then?

I've been trying to come up with a solution of my own. I got as far as using repeated applications of integration by parts to equate it to ex2e^{-x^2} multiplied by an infinite polynomial with coefficients in a well-defined (but complicated!) pattern. Now it gets a bit trickier. The Maclaurin's expansion doesn't seem to help. The only thing that seems promising is to consider the possible convergence of terms in the form ex2xn\frac{e^{x^2}}{x^n} which is pretty tricky!

Any ideas? :tongue:


No, I've never wanted to apply to do maths :smile:

I don't at the minute, give me a few minutes.
(edited 10 years ago)
Reply 858
Original post by bananarama2
Rigorous? I'm a Natsci :rofl: Well I don't see the need for it here. I'm just evaluating the integral over all possible coords in the x-y plane. I don't think I understand you problem?

For ****s sake :lol: classic physicist :lol:

Well why is it that the cartesian domain (,)(-\infty,\infty) translates exactly to [0,2π)[0,2\pi) and [0,)[0,\infty) in each parameter of the polar plane respectively? You must derive them using the standard techniques for variable-changing substitutions. :tongue: You've plucked the limits out of nowhere without thought as to the size of each numerical set. For example: when substituting r2=x2+y2r^2=x^2+y^2, in order to ensure you don't have r(,)r \in (\infty,\infty), which is nonsense, you would have to split the limit exactly into two halves. This means you now have a "2" and the fact this cancels out is merely fortunate as per the elegance of the final result, rather than being an immediate deduction (though there might be an obvious way to deduce it that I'm missing :tongue: ...)

I don't at the minute, give me a few minutes.

Hahaha :lol:

Edit: wait.... "integrating over R squared"?! How/why have you learnt this already? ;o Why have you been doing STEP then? And how good are you at the questions? :biggrin:
(edited 10 years ago)
Reply 859
As you want another proof for problem 128, try to prove the following result:

Problem 130*

n2n+1((2n)!!)2((2n1)!!)2(2n+1)<(0ex2dx)2<n2n1(2n3)!!)2(2n1)((2n2)!!)2(π2)2\begin{aligned} \displaystyle \frac{n}{2n+1}\frac{((2n)!!)^{2}}{((2n-1)!!)^{2}(2n+1)} < \left(\int_{0}^{\infty} e^{-x^{2}}dx \right)^{2} < \frac{n}{2n-1}\frac{(2n-3)!!)^{2}(2n-1)}{((2n-2)!!)^{2}}\left(\frac{\pi}{2} \right)^{2} \end{aligned}.
Then note that ex2e^{-x^{2}} is even.
(edited 10 years ago)

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