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! 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 though I don't think so, due to the three stars). Everyone could give the two sub-question inequalities i just outlined a go!
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 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.
Problem 126: I=∫01f(x)(x−1/2)dx=∫01/2f(x)(x−1/2)dx+∫1/21f(x)(x−1/2)dx[br]∫1/21f(x)(x−1/2)dx=∫01/2f(x+1/2)xdx[br]∫01/2f(x)(x−1/2)dx=−∫01/2f(1/2−x)xdx[br]∴I=∫01/2(f(x+1/2)−f(1/2−x))xdx≥0[br][br]∫01xf(x)dx=∫01/2xf(x)dx+∫1/21xf(x)dx[br]=∫1/21(x−1/2)f(x−1/2)+∫1/21xf(x)dx So: ∫1/21f(x)dx−∫01xf(x)dx[br]=∫1/21f(x)dx−∫1/21(x−1/2)f(x−1/2)dx−∫1/21xf(x)dx[br]=∫1/21f(x)(1−x)dx−∫1/21(x−1/2)f(x−1/2)dx[br]∫1/21f(x)(1−x)dx=∫1/21f(3/2−x)(x−1/2)dx So: ∫1/21f(x)dx−∫01xf(x)dx[br]=∫1/21(x−1/2)(f(3/2−x)−f(x−1/2))dx≥0
Let P(n) be a polynomial with coefficients in Z. Suppose that deg(P)=p, where p is a prime number. Suppose also that P(n) is irreducible over Z. Then there exists a prime number q such that q does not divide P(n) for any integer n.
Assume for all primes q there exists n∈Z such that P(n)≡0(q). Hence P(x) is reducible in Z/qZ[X] for all primes q. By Chebotarev's density theorem (and a fair bit of kung fu), this implies that P(x) is reducible in Z[X]; thus a contradiction.
The probability that both cards are aces is (252)(24). The probability that at least one card is an ace is 1−(252)(248). Bayes gives 1−(252)(248)(252)(24).
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 (252)3. The probability that one of the cards is the ace of spades and the oder is random is (252)51. Again we use Bayes formula to obtain (252)51(252)3.
Assume for all primes q there exists n∈Z such that P(n)≡0(q). Hence P(x) is reducible in Z/qZ[X] for all primes q. By Chebotarev's density theorem (and a fair bit of kung fu), this implies that P(x) is reducible in Z[X]; thus a contradiction.
This is exactly what I had in mind. 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 q which satisfy the condition. In other words, we look for those primes q for which the degree of the splitting field of P over Fq is p. What do you think about the case when deg(P)=pα, where α>1.
How did you do this step? Where did the theta go? 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.
Unfortunately not, well sort of. It was hinted at in a question I did once. (Kinda like a STEP questions I suppose)
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.
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
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 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 e−x2 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 xnex2 which is pretty tricky!
Ohhhh I see I thought you had integrated with respect to the one on the left
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 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 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 e−x2 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 xnex2 which is pretty tricky!
Rigorous? I'm a Natsci 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 classic physicist
Well why is it that the cartesian domain (−∞,∞) translates exactly to [0,2π) and [0,∞) in each parameter of the polar plane respectively? You must derive them using the standard techniques for variable-changing substitutions. You've plucked the limits out of nowhere without thought as to the size of each numerical set. For example: when substituting r2=x2+y2, in order to ensure you don't have r∈(∞,∞), 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 ...)
I don't at the minute, give me a few minutes.
Hahaha
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?