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    Problem 195*/**

    Prove that there exists infinitely many arithmetic progressions of 3 distinct perfect squares.

    Problem 196**/***

    Prove that there exists no arithmetic progressions of 4 distinct perfect squares.

    Problem 197*/**

    For a non-negative integer n, is 5^{5^{n+1}}+5^{5^{n}}+1 ever prime? Prove your assertions.

    Problem 198**

    Let the real numbers a,b,c,d satisfy the relations a+b+c+d=6 and a^2+b^2+c^2+d^2=12.

    Prove that 36 \le 4(a^3+b^3+c^3+d^3)-(a^4+b^4+c^4+d^4) \le 48

    Problem 199***

    Prove that \displaystyle \int_0^1 t^x (1-t)^y dt = \frac{\Gamma(x+1) \Gamma(y+1)}{\Gamma(x+y+2)}

    Hint:
    Spoiler:
    Show
    Convolution


    Problem 200*/***

    Evaluate \displaystyle\int_0^1 \ln(t) t^2 dt
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    (Original post by Jkn)
    Problem 200***

    Evaluate \displaystyle\int_0^1 \ln(t) t^2 dt
    Solution 200

    \displaystyle\int_{0}^{1} \ln t \cdot t^{2} \ dt \overset{ \text{IBP}}= \displaystyle\underbrace{ \dfrac{1}{3} t^{3} \ln t \bigg|_{0}^{1}}_{*} \ - \ \dfrac{1}{3} \displaystyle\int_{0}^{1} t^{2} \ dt = - \dfrac{1}{9}

    Why is this ***? :confused: I presume you want justification of the * limit...

    \displaystyle \begin{aligned} \dfrac{1}{3} t^{3} \ln t \bigg|_{0}^{1} & = - \dfrac{1}{3} \left( \displaystyle\lim_{t \to 0} t^{3} \ln t \right) \\& = \dfrac{1}{3} \cdot \left( \displaystyle\lim_{t \to 0} \dfrac{t^{3}}{3} \right) & (\text{L'H\^{o}pital's rule}) \\& = 0 \end{aligned}

    \therefore \displaystyle\int_{0}^{1} \ln t \cdot t^{2} \ dt = - \dfrac{1}{9}
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    (Original post by Implication)
    Isn't that because mathematical induction is actually deduction haha?
    Yes, that's what I was saying.

    (Original post by Implication)
    Interestingly, I'm not sure if that is a solution the Problem of Induction: you still need to find a way to justify your confidence level, do you not? And how are you going to do that, if not by reasoning inductively in the first place? And for this you will need another confidence interval. And so on, ad infinitum.
    No a confidence level constructed correctly is deductively justified, for example through interpretations of probability or statistical inference, which are sound albeit open to disputes of opinion.

    (Original post by Implication)
    There's a problem very closely linked to the problem of induction that I've always found fascinating and extremely frustrating. Again, I think it is a pretty old di/tri/multilemma, but it's always fun to think about!

    Any valid argument must rely on either a circular chain of reasons, an infinite chain of reasons or on reasons that are not themselves justified.
    An argument cannot be justified by a circular chain of reasons.
    That depends on the nature of your argument. For example Axiom of Choice => Zorn's Lemma => Well-Ordering Theorem => Axiom of Choice is a well-known circular set of results that stand independently from traditional foundations of set theory, and is unprovable from them unless you assume one of the equivalent (or stronger) forms. But that just reduces to the matter of what you pick as your axioms.

    (Original post by Implication)
    An argument cannot be justified by an infinite chain of reasons.
    Yes they can.

    (Original post by Implication)
    An argument cannot be justified if it relies upon unjustified assumptions.
    Again, depends on the argument. A vacuous or tautological argument can have false assumptions.

    (Original post by Implication)
    Thus, no valid argument can be justified.
    Essentially the whole thing is ignoring the key foundation of deduction that justification is only relative to some set of axioms - within that axiomatic system the deductions could be perfectly sound, albeit perhaps open to dispute of matters of opinion (choice of axioms). In a carefully constructed logical system even infinite proofs can be shown to sound, and there are plenty of philosophical justifications for why one may like to choose to adopt that instead of the more classical systems we are used to that does not allow infinite proof.

    Tl;dr: What you said appears to be a trilemma only because you are confining yourself to classic logic, and even then it's a false one since it ignores the fact that arguments are justified only relative to some axioms.
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    (Original post by Felix Felicis)
    Solution 200

    \displaystyle\int_{0}^{1} \ln t \cdot t^{2} \ dt \overset{ \text{IBP}}= \displaystyle\underbrace{ \dfrac{1}{3} t^{3} \ln t \bigg|_{0}^{1}}_{*} - \dfrac{1}{3} \displaystyle\int_{0}^{1} t^{2} \cdot \dfrac{1}{t} \ dt = - \dfrac{1}{9}

    Why is this ***? :confused: I presume you want justification of the * limit...

     \dfrac{1}{3} t^{3} \ln t \bigg|_{0}^{1} = - \dfrac{1}{3} \left( \displaystyle\lim_{t \to 0} t^{3} \ln t \right)

    \overset{ \text{L'Hopital's rule}}= \dfrac{1}{3} \cdot \left( \displaystyle\lim_{t \to 0}  \dfrac{t^{3}}{3}  \right) = 0

    \therefore \displaystyle\int_{0}^{1} \ln t \cdot t^{2} \ dt = - \dfrac{1}{9}
    Actually I found it in a list of problems to practice using the Gamma Function and results such as that obtained in 199 and so hadn't really stopped to consider how it might be approached outside of this context (the advanced method is still worth doing though if you're up for it!) Your solution is far better:lol:

    Spoiler:
    Show
    Btw, typo in your first line
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    (Original post by Jkn)
    Problem 195*/**

    Prove that there exists infinitely many arithmetic progressions of 3 distinct perfect squares.

    Problem 196**/***

    Prove that there exists no arithmetic progressions of 4 distinct perfect squares.

    Problem 197*/**

    For a non-negative integer n, is 5^{5^{n+1}}+5^{5^{n}}+1 ever prime? Prove your assertions.

    Problem 198**

    Let the real numbers a,b,c,d satisfy the relations a+b+c+d=6 and a^2+b^2+c^2+d^2=12.

    Prove that 36 \le 4(a^3+b^3+c^3+d^3)-(a^4+b^4+c^4+d^4) \le 48

    Problem 199***

    Prove that \displaystyle \int_0^1 t^x (1-t)^y dt = \frac{\Gamma(x+1) \Gamma(y+1)}{\Gamma(x+y+2)}

    Hint:
    Spoiler:
    Show
    Convolution


    Problem 200***

    Evaluate \displaystyle\int_0^1 \ln(t) t^2 dt
    Solution 197

    Let  x = 5^{5^n}

    then  x^5 + x + 1 = (x^2+x+1)(x^3-x^2+1) is a prime. Since n is a non-negative integer 5^{5^n} is also an integer. Hence, both brackets will be integers, so one of the brackets must be equal to 1 for the expression to be a prime. Since n is non-negative, 5^n will be a positive integer, and it follows that 5^5^n is also a positive integer. Hence, all terms in the first bracket are positive and the sum must be > 1, so the first bracket cannot = 1. Leaving:

     x^3 - x^2 + 1 = 1
     x^3 = x^2

    As above x will be a positive integer so not 0.

     x = 5^{5^n} = 1 which is impossible since n would have to be negative infinity for the power to be 0, which is impossible firstly because n is a non-negative integer. Hence, no such prime exists.
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    (Original post by Jkn)
    Actually I found it in a list of problems to practice using the Gamma Function and results such as that obtained in 199 and so hadn't really stopped to consider how it might be approached outside of this context (the advanced method is still worth doing though if you're up for it!) Your solution is far better:lol:

    Spoiler:
    Show
    Btw, typo in your first line
    Damn! Amended.

    Ehh, I don't know nearly enough about the Beta/ Gamma functions past their definitions to attempt questions with them :lol: I'll save further reading 'til my exams are over.
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    (Original post by Zakee)
    You say 'no valid argument' can be justified. Does that mean your argument is invalid as you have justified it?
    No; not quite, I don't think. The argument I presented is clearly valid in that its premises entail its conclusion. Whether or not it is justified is another matter! If the conclusion is justified, then it is correct and the conclusion cannot be justified or we have a contradiction. So the conclusion cannot be justified. If the conclusion is not justified, then everything is fine i.e. there is no contradiction. So we must conclude that the conclusion is unjustified (though not necessarily false) as it must either be justified or unjustified and we have shown that it is not justified.

    So we need to look at where the problem might lie, and I think the answer is clear: the argument rests upon unjustified assumptions (axioms) as, I think, do most/all arguments. And so it boils down to the idea behind the problem in the first place: if an argument rests on assumptions that are not inferentially justified, then how can that argument ultimately be justified?

    Now I think an axiom (one of the initial assumptions we don't justify) can be of two types; it can be a defining property of something (such as with most axioms in mathematics, I think), or it can be "self-evidently true". And I think, aside from tautologies that give no new information, the idea of anything about reality being "self-evidently" true is unrealistic. We could always be wrong about what is "self-evidently" true.


    But hey, I'm just a young man trying to sound cleverer than he is I'm not a philosopher; nor do I have a mathematics degree. I could be entirely wrong
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    (Original post by Implication)
    No; not quite, I don't think. The argument I presented is clearly valid in that its premises entail its conclusion. Whether or not it is justified is another matter! If the conclusion is justified, then it is correct and the conclusion cannot be justified or we have a contradiction. So the conclusion cannot be justified. If the conclusion is not justified, then everything is fine i.e. there is no contradiction. So we must conclude that the conclusion is unjustified (though not necessarily false) as it must either be justified or unjustified and we have shown that it is not justified.

    So we need to look at where the problem might lie, and I think the answer is clear: the argument rests upon unjustified assumptions (axioms) as, I think, do most/all arguments. And so it boils down to the idea behind the problem in the first place: if an argument rests on assumptions that are not inferentially justified, then how can that argument ultimately be justified?

    Now I think an axiom (one of the initial assumptions we don't justify) can be of two types; it can be a defining property of something (such as with most axioms in mathematics, I think), or it can be "self-evidently true". And I think, aside from tautologies that give no new information, the idea of anything about reality being "self-evidently" true is unrealistic. We could always be wrong about what is "self-evidently" true.


    But hey, I'm just a young man trying to sound cleverer than he is I'm not a philosopher; nor do I have a mathematics degree. I could be entirely wrong

    There are probably inferences and statements which are so profoundly conspicuous, to justify them would be mere tautology.

    For example, "This is", is a statement which is obvious and the only flaws viewed would have to be in the structure of language and how language is used. If we try and investigate the plights met in language, then well, I'd prefer to refrain from such as it'd probably drive me to insanity. :rolleyes:


    Also with respect to your last line: doesn't everyone try to do that? We all delude ourselves some way or another into thinking something that isn't true: that we're loved, or that we understand something or that we have some ineffable, majestic purpose in the Universe. :moon:
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    (Original post by ukdragon37)
    No a confidence level constructed correctly is deductively justified, for example through interpretations of probability or statistical inference, which are sound albeit open to disputes of opinion.
    I think I got a little confused in my last post, but I'll try to explain my thoughts again now. It may well be that I just don't understand, though.

    Say we've measured the stress levels of managers and floor staff in a sample of shops and found that, in general, managers are more stressed than floor staff. We can do our stats to compute the probability that the difference in stress measurements between managers and floor staff is merely due to random chance. But, without any inductive inferences, we can't make any judgement about what did cause this difference: we can only decide whether it was likely to be due to random chance or a real variable difference. If it is likely to be due to a real variable, the stats still hasn't told anything about that variable beyond the fact that it isn't random chance.

    With regard to the icebergs, wouldn't our confidence interval necessarily rely on assumptions and inductive inferences? We may well have measured 1 000 000 icebergs and done our stats to find that we can be 99.7% certain that all icebergs are cold, for example... But what if there is a different type of underground iceberg that forms under pressure - and at higher temperatures - that we didn't know about then and is subsequently discovered? The stats may well have told us that the probability was 99.7%, but, without sounding like an excitable philosophy freshman with an obsession with Descartes and his method of doubt, were we really in a position to make that call when we had no way of knowing what information and data we were missing regarding icebergs in the first place? Our statistical calculation didn't (and couldn't have) taken this into account, so how can we ever have really been so sure in the first place?

    I really don't know that much about stats, as you can probably tell, so I'm struggling to see how we can be so confident in a calculation that doesn't use so much data


    (Original post by ukdragon37)
    That depends on the nature of your argument. For example Axiom of Choice => Zorn's Lemma => Well-Ordering Theorem => Axiom of Choice is a well-known circular set of results that stand independently from traditional foundations of set theory, and is unprovable from them unless you assume one of the equivalent (or stronger) forms. But that just reduces to the matter of what you pick as your axioms.



    Yes they can.



    Again, depends on the argument. A vacuous or tautological argument can have false assumptions.



    Essentially the whole thing is ignoring the key foundation of deduction that justification is only relative to some set of axioms - within that axiomatic system the deductions could be perfectly sound, albeit perhaps open to dispute of matters of opinion (choice of axioms). In a carefully constructed logical system even infinite proofs can be shown to sound, and there are plenty of philosophical justifications for why one may like to choose to adopt that instead of the more classical systems we are used to that does not allow infinite proof.

    Tl;dr: What you said appears to be a trilemma only because you are confining yourself to classic logic, and even then it's a false one since it ignores the fact that arguments are justified only relative to some axioms.
    I think that is exactly the point of the problem, aside from a difference in usage of the term "justification". Any argument we make necessarily relies upon some assumptions; axioms, doesn't it? So whether or not an argument is valid is entirely relative to these axioms.

    All cheeses are purple.
    All iPods are cheese.
    Therefore all iPods are purple.

    This is a valid argument and is "true" relative to the assumptions that all cheeses are purple and all ipods are cheese... But wouldn't it be absurd to claim that the conclusion was ultimately justified (that is, it is appropriate to believe it)? If any argument we make relies upon something that we don't check to be true - we just assume it - then how can we claim to be justified when saying that any proposition is true or false? Our argument for its truth or falsity is necessarily based upon premises that we haven't verified.

    If we have a reason to believe that one axiom is "better" or "truer" than another, then aren't we really justifying it? If so, then it isn't really an axiom, is it?
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    Problem 200 * / **

    Three positive integers are chosen from the first 2k positive integers. What is the probability that the numbers chosen can form the sides of a triangle? What is the limiting probability as k tends to infinity?
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    (Original post by jack.hadamard)
    The second part can be done in three lines by using only value of Hurwitz's zeta function.
    I missed such a nice approach, and used character functions and L-functions...


    Solution 193

    Rewrite \displaystyle x^{2}+y^{2}= p^{2}-z^{4} = (p-z^{2})(p+z^{2}).
    Suppose v_{2}(z)=1. Then, p+z^{2} \equiv 3 \pmod 8, and p-z^{2} \equiv 3 \pmod 8. This implies that there exists prime q \equiv 3 \pmod 4 such that p-z^{2} \equiv p+z^{2} \equiv 0 \pmod q, which is a contradiction.
    Suppose v_{2}(z) > 1. Clearly, p+z^{2} \equiv p-z^{2} \equiv 7 \pmod 8, which is a contradiction again.
    Now, let v_{2}(z) =0. Hence, p-z^{2} \equiv 6 \pmod 8; thus there exists prime q \equiv 3 \pmod 4 such that p-z^{2} \equiv p+z^{2} \equiv 0 \pmod q - contradiction.
    Therefore, there are no non-trivial solutions to this diophantine equation.
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    Oh btw if anyone is thinking about life afar STEP, summer maths or something to do right now, then I've just found this really great website talking through all of the awesome integration techniques!

    It's quite rare to get something that explains these techniques in a functional rare rather than via rigorous logic (which isn't particularly necessary for practical purposes) so enjoy!
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    (Original post by Implication)
    I think I got a little confused in my last post, but I'll try to explain my thoughts again now. It may well be that I just don't understand, though.

    Say we've measured the stress levels of managers and floor staff in a sample of shops and found that, in general, managers are more stressed than floor staff. We can do our stats to compute the probability that the difference in stress measurements between managers and floor staff is merely due to random chance. But, without any inductive inferences, we can't make any judgement about what did cause this difference: we can only decide whether it was likely to be due to random chance or a real variable difference. If it is likely to be due to a real variable, the stats still hasn't told anything about that variable beyond the fact that it isn't random chance.
    But you do have a real variable there - the variable of whether the workers are managers or not. The hypothesis test has told you only that managers are more stressed in general, but not exactly why it is that way, because that's the exact variable you tried to test. In other words, your two hypotheses ended up being:

    H0: There is no evidence to suggest being a manager affects stress levels any more than random chance.
    H1: Being a manger does significantly affect stress levels.

    And you can then carry out a statistical test which determines which hypothesis is correct, as it was set out to do, but that as it should does not guarantee at all why you would find out anything beyond the hypotheses that are stated. If you then conjecture that stress levels are to do with shouting at subordinates, you could then frame that as appropriate hypotheses and you test again. What the statistical test guarantees you is that once you have found the a variable which you can show it is independent of random chance at , say, the 95% confidence level, then you know that there is a 95% chance that particular variable does indeed cause the phenomenon. If that wasn't the case then you just have to keep trying with another variable.

    (Note that if you find H0 is the acceptable hypothesis then that does not mean being a manager does not affect stress levels, only there wasn't evidence of such from the data.)

    XKCD demonstrates it quite nicely:

    Spoiler:
    Show



    What's the flaw here?


    Essentially, yes you are right that you have to use some inductive experience to guess the correct variable to test, but there is nothing wrong with that - once you performed the test that gives you a purely deductive guarantee it is correct up to some confidence level.

    EDIT: To those that are more pedantic than me, yes I know I am being sloppy with regards to the null hypothesis and interpretation of probability.

    (Original post by Implication)
    With regard to the icebergs, wouldn't our confidence interval necessarily rely on assumptions and inductive inferences? We may well have measured 1 000 000 icebergs and done our stats to find that we can be 99.7% certain that all icebergs are cold, for example... But what if there is a different type of underground iceberg that forms under pressure - and at higher temperatures - that we didn't know about then and is subsequently discovered? The stats may well have told us that the probability was 99.7%, but, without sounding like an excitable philosophy freshman with an obsession with Descartes and his method of doubt, were we really in a position to make that call when we had no way of knowing what information and data we were missing regarding icebergs in the first place? Our statistical calculation didn't (and couldn't have) taken this into account, so how can we ever have really been so sure in the first place?
    I think you are confusing the difference between the confidence of whether all icebergs are cold or whether the next one discovered would be cold.

    In the former case, there is no statistical method which would have allowed you to obtain the figure anyway from measuring a subset of the population of icebergs, unless that population is finite and we know it (which would take into account the ones underground) or estimate it (which would exclude the possibility of the ones underground). The accuracy of the 99.7% figure is dependent on the accuracy of that estimate. If the population is infinite then naturally it is impossible to deduce (confidently or otherwise using statistics) that all icebergs are cold, unless there is some additional deductive evidence (such that the underground of Earth is proven to be inhospitable to icebergs).

    If instead you meant that we are 99.7% sure that the next one we discover will be cold (i.e. a probability) then you should be aware that probability based on prior trials is only a relative probability based on exactly that, pre-existing results. It could well go down if new information comes to light (a hot iceberg gets discovered, a new way of discovering icebergs is discovered, etc. etc.), and further it presumes that what we have observed is a fair sample of the population (here it may be not as underground icebergs are more difficult to observe). This is the case that usually gets twisted in the media as "we are X% sure every Y is Z" because they ignore the crucial assumptions.

    In other words you are accusing statistics of something it's not guilty of.

    (Original post by Implication)
    I think that is exactly the point of the problem, aside from a difference in usage of the term "justification". Any argument we make necessarily relies upon some assumptions; axioms, doesn't it? So whether or not an argument is valid is entirely relative to these axioms.

    All cheeses are purple.
    All iPods are cheese.
    Therefore all iPods are purple.

    This is a valid argument and is "true" relative to the assumptions that all cheeses are purple and all ipods are cheese... But wouldn't it be absurd to claim that the conclusion was ultimately justified (that is, it is appropriate to believe it)? If any argument we make relies upon something that we don't check to be true - we just assume it - then how can we claim to be justified when saying that any proposition is true or false? Our argument for its truth or falsity is necessarily based upon premises that we haven't verified.
    You are right, we cannot justify something absolutely, except the proposition "True" itself and its equivalents (things that can be demonstrated as tautologies). However what we can do is fix the assumptions that we prefer (either ones that worked well inductively from experience, or we "just believe" through faith) and once we have those, everything that follows is deductive.

    However it sounds like again you are accusing deduction for something which it is not. Justifiable (i.e. absolutely true) results from deductive reasoning is always in the form of "If X then Y", with X possibly being some axioms, or just "True", and this is justified absolutely because the task of the reasoning is to actually demonstrate it is a tautology. Note that all deductive reasoning is claiming to do is assert "if you believe in this, perhaps inductively, then you get that". It does not say "you should believe in this", if "this" is not a tautology.

    In fact your example is actually defective because you have not "discharged all the assumptions". You have actually deduced "If all cheeses are purple, then if all iPods are cheese, then all iPods are purple". This as a whole is justified perfectly well.

    (Original post by Implication)
    If we have a reason to believe that one axiom is "better" or "truer" than another, then aren't we really justifying it? If so, then it isn't really an axiom, is it?
    You are assuming a contradiction. There is never a reason to believe one axiom is "truer" than another, unless as you said you are not treating them properly as axioms. In mathematics, one favours a certain set of axioms because they are "convenient" for the task at hand, or demonstrate properties that can be considered to be elegant or beautiful, or seem to model the real world quite well. At no time does one assert some axioms are worth more than others in truth.
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    (Original post by Lord of the Flies)
    Spoiler:
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    Woah! Just realised that problem 142 can be tweaked to form an algorithm which yields the successive values of \zeta (2n). Firstly, we can solve the Basel problem using the result:

    \displaystyle \begin{aligned}\sum_{v\geq 1}\frac{\sin vx}{v}=\frac{\pi-x}{2} &\Rightarrow \int_0^{\pi} \sum_{v\geq 1}\frac{\sin vx}{v}\,dx=\int_0^{\pi}\frac{\pi-x}{2}\,dx\\&\Rightarrow \sum_{v\geq 1}\frac{1}{v^2}-\sum_{v\geq 1}\frac{(-1)^v}{v^2}=\frac{\pi^2}{4}\iff 2\sum_{k\geq 1}\frac{1}{(2k-1)^2}=\frac{\pi^2}{4}\end{aligne  d}

    Noting that \displaystyle \sum_{k\geq 1}\frac{1}{(2k-1)^2}=\sum_{v\geq 1}\frac{1}{v^2}-\sum_{v\geq 1}\frac{1}{(2v)^2}=\frac{3}{4} \sum_{v\geq 1}\frac{1}{v^2} we get \displaystyle \frac{3}{2}\sum_{v\geq 1}\frac{1}{v^2}=\frac{\pi^2}{4} \Rightarrow\sum_{v\geq 1} \frac{1}{v^2}=\frac{\pi^2}{6}

    Now for the general step define for x\in (0,2\pi): \displaystyle\;p_k(x)=\sum_{v \geq 1}\frac{\sin vx}{v^{2k-1}}\,dx and \displaystyle \pi^{2k}\eta_k=\int_0^{\pi} p_k(x)\,dx where \eta_k\in\mathbb{Q}
    (the fact that \displaystyle\int_0^{\pi}p_k \,dx can be written as \eta_k \pi^{2k} follows from the formula in solution 142 + induction)

    \displaystyle\begin{aligned}\int  _0^{\pi}\sum_{v\geq 1}\frac{\sin vx}{v^{2k-1}}\,dx=\int_0^{\pi} p_k(x)\,dx&\iff \sum_{v\geq 1}\frac{1}{v^{2k}}-\sum_{v\geq 1}\frac{(-1)^v}{v^{2k}}=\pi^{2k}\eta_k \\&\iff\left(\frac{4^k-1}{4^k}\right)\sum_{v\geq 1}\frac{2}{v^{2k}}=\pi^{2k}\eta_  k \\&\iff \zeta (2k)=\frac{ \pi^{2k}\eta_k}{2}\left(\frac{4^  k}{4^k-1}\right) \end{aligned}

    (in other words, the only thing we need to do to evaluate the next value of \zeta (2n) is integrate a polynomial to get \eta_n)

    For instance, integrating p_2 gives \eta_2 =\dfrac{1}{48} hence \zeta (4)=\displaystyle \frac{ \pi^{4}\eta_2}{2}\left(\frac{4^2 }{4^2-1}\right)=\frac{\pi^4}{90}

    Incidentally this also shows that for any n,\; \zeta (2n)=q\pi^{2n} for some q\in\mathbb{Q}.
    Brilliant result! I've lately realised that, if you can find two ways to do a problem and one of those ways avoids using a standard result, you tend to be able to compare the two and deduce that result! Whist you have an alternative to Euler's work, I have come up with a different way to analyse the Gaussian Integral (needless to say, I'm sure these derivation are trivial common knowledge in some circles but it's still, of course, very enjoyable to realise such things for yourself!)

    Let u=x^2,

    \displaystyle \Rightarrow \int_{-\infty}^{\infty} e^{-x^2} dx=2\int_0^{\infty} e^{-x^2} dx = \int_0^{\infty} u^{-\frac{1}{2}} e^{-u} du = \Gamma(\frac{1}{2}) (1)

    First we observe that \displaystyle \int_0^{\infty} \frac{\ln(x)}{(1+x^2)} dx = 0 (2) by letting x \to x^{-1}.

    \displaystyle \int_0^{\infty} \frac{\ln(x)}{(1+x^2)} dx = \left[\frac{x(\ln(x)-1)}{(1+x^2)} \right]_0^{\infty} -2\int_0^{\infty} \frac{x^2(\ln(x)-1)}{(1+x^2)^2} dx

    By using (2) as well as making trigonometric substitutions,

    \displaystyle \begin{aligned} \Rightarrow \int_0^{\infty} \frac{\ln(x)}{(1+x^2)^2} dx =\frac{3}{2} \int_0^{\infty} \frac{\ln(x)}{(1+x^2)} dx - 2\int_0^{\infty} \frac{1}{1+x^2} dx + 2\int_0^{\infty} \frac{1}{(1+x^2)^2} dx = -\frac{\pi}{4} \end{aligned} (3)

    Now, notice that in the proceeding working, we do not use any sophisticated techniques but only notation for special functions. Note that \displaystyle\psi(z)=\frac{d}{dz  }{\ln( \Gamma(z) )} where \psi(z) is the digamma function.

    Let \displaystyle f(\alpha)=\int_0^{\infty} \frac{x^{\alpha}}{(1+x^2)^2} dx,

    Let x=\sqrt{u},

    \displaystyle \begin{aligned} \Rightarrow f(\alpha) = \frac{1}{2} \int_0^{\infty} \frac{u^{\frac{\alpha -1}{2}}}{(1+u)^2} du = \frac{1}{2} B(\frac{\alpha +1}{2}, 2-\frac{\alpha +1}{2})=\frac{1}{2}\Gamma(\frac{  \alpha +1}{2}) \Gamma(2-\frac{\alpha +1}{2}) \end{aligned}

    \displaystyle \Rightarrow \frac{\partial}{\partial \alpha} f(\alpha) = \frac{1}{4} \Gamma(\frac{\alpha +1}{2}) \Gamma(2-\frac{\alpha+1}{2}) \left(\psi(\frac{\alpha +1}{2})-\psi(2-\frac{\alpha +1}{2}) \right) (4)

    Now we note that \displaystyle \frac{\partial}{\partial \alpha} f(\alpha)=\int_0^{\infty}\frac{\  ln(x) x^{\alpha}}{(1+x^2)^2} dx (5)

    Equating (4) and (5) and letting \displaystyle \alpha=0 \Rightarrow \int_0^{\infty} \frac{\ln(x)}{(1+x^2)^2} dx = \frac{1}{4} \Gamma(\frac{1}{2}) \Gamma(\frac{3}{2}) \left(\psi(\frac{1}{2})-\psi (\frac{3}{2}) \right)

    By differentiating \displaystyle \frac{\Gamma(z+1)}{\Gamma(z)}=z we get that \psi(z+1)-\psi(z)=\frac{\Gamma(z)}{\Gamma(  z+1)} (6)

    Using (6) and equating with (3),

    \displaystyle \Rightarrow -\frac{\pi}{4} =\frac{1}{4} \Gamma(\frac{1}{2}) \Gamma(\frac{3}{2}) \left( -\frac{\Gamma(\frac{1}{2})}{\Gamm  a(\frac{3}{2})} \right) = -\frac{1}{4} \Gamma^2 (\frac{1}{2})



\displaystyle \Rightarrow \Gamma^2 (\frac{1}{2}) =\pi

    Using (1),

    \displaystyle \Rightarrow \left( \int_{-\infty}^{\infty} e^{-x^2} dx \right)^2 = \pi



\displaystyle \Rightarrow \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi} \ \square

    Note also that, from this result, combined with the fact that \displaystyle \Gamma(z+1)=z\Gamma(z), we get the delightful result that \Gamma(\frac{1}{2} +n)=\frac{(2n-1)!!}{2^n} \sqrt{\pi}\ \forall n \in \mathbb{N}.
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    (Original post by Jkn)
    Problem 195*/**

    Prove that there exists infinitely many arithmetic progressions of 3 distinct perfect squares.
    Solution 195:

    We must have:

     (n+b)^2-(n+a)^2 = (n+a)^2 - n^2

     n^2+2bn+b^2-n^2-2an-a^2 = n^2+2an+a^2-n^2

     2a^2+4an = b^2+2bn

     2a(a+2n) = b(b+2n)

    Now, let:

     2a = b+2n  b = a+2n

     2a-b = b-a

     3a = 2b

    Now, we can clearly choose an infinite number of even integers a and b such that this equation is satisfied, and if both a and b are even integers then n will also be an even integer. This satisfies all the conditions and so there are infinitely many arithmetic progressions of three perfect squares.

    Problem 201: *

    A stick of unit length is bent at a point along its length, with each such point being equally likely. The stick is then placed on a horizontal surface so that it forms a right-angled triangle, with the shorter edge of the stick being perpendicular to the ground. The angle of elevation of the longer edge of the stick with the ground is  \alpha . Prove that the expected value of  \alpha is given by:

     \dfrac{4- \pi }{2}
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    (Original post by DJMayes)

    Problem 201:

    A stick of unit length is bent at a point along its length, with each such point being equally likely. The stick is then placed on a horizontal surface so that it forms a right-angled triangle, with the shorter edge of the stick being perpendicular to the ground. The angle of elevation of the longer edge of the stick with the ground is  \alpha . Prove that the expected value of  \alpha is given by:

     \dfrac{4- \pi }{4}
    I didn't expect this to be so pure
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    (Original post by DJMayes)

    Problem 201:

    A stick of unit length is bent at a point along its length, with each such point being equally likely. The stick is then placed on a horizontal surface so that it forms a right-angled triangle, with the shorter edge of the stick being perpendicular to the ground. The angle of elevation of the longer edge of the stick with the ground is  \alpha . Prove that the expected value of  \alpha is given by:

     \dfrac{4- \pi }{4}

    First question - Is this a */**/*** problem?

    Second question - Is this stick being dropped from the Eiffel Tower? :ahee:
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    (Original post by Zakee)
    First question - Is this a */**/*** problem?

    Second question - Is this stick being dropped from the Eiffel Tower? :ahee:
    It's a * I've just finished it I think

    This is going to become a TSR Meme
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    (Original post by bananarama2)
    I didn't expect this to be so pure
    I think it's a pretty nice problem. Once you take away the stats dressing it's nothing but Pure.

    (I like trying to make the few questions I post here feel fairly similar to the sort of thing you'd get on STEP. Have I managed that with this one?)

    (Original post by Zakee)
    First question - Is this a */**/*** problem?

    Second question - Is this stick being dropped from the Eiffel Tower? :ahee:
    It's a * problem, all of the ones I post pretty much are. Nothing beyond A Level knowledge is assumed for this.
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    Solution 201

    Let x be the length of the short piece.

    We have  (1-x)\sin \alpha = x

     \displaystyle \langle \alpha \rangle =  \left\langle \arcsin \left( \frac{x}{1-x} \right) \right\rangle

     = \int_0^{1/2} \arcsin \left( \frac{x}{1-x} \right) dx IBP

    = \left[x\arcsin \left( \frac{x}{1-x} \right) \right]_0^{1/2} - \int_0^{1/2} \frac{x}{\sqrt{1-2x}(x-1)}dx

    I left out the sub and rearrangement. I completely understand why LOTF does it now.  u^2=1-2x

    = \left[x\arcsin \left( \frac{x}{1-x} \right) \right]_0^{1/2} + \int_1^0 -1+ \frac{2}{1+u^2} du

     = \frac{\pi}{4} +\left[2 \arctan (u) - u \right]_1^0

     = \frac{\pi}{4} + 0 -0 -\frac{pi}{2} +1

     = \frac{4- \pi}{4}

    And then you multiply by two.

    Footnote: The only stats I know is S1 and that from reading about QM, so I just basically made it up from what I know :teehee:
 
 
 
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