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    (Original post by metaltron)
    Here's a question everybody should be able to have a go at:

    Problem 131 (*)

     Find \ the \ sum \ of \ squares \ of \ all \ real \ roots \ of \ the \ polynomial:

     f(x) = x^5 - 7x^3 + 2x^2 -30x + 6

    If you find it easy, please wait for somebody else to post an answer!
    x^5 - 7x^3 + 2x^2 -30x + 6=(x^2+3) (x^3-10 x+2)
    a,b,c be the real roots:
    (a^2+b^2+c^2)=(a+b+c)^2-2(ab+ac+bc)
    Vieta tells us:
    =0^2-2(-10)=20
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    (Original post by ben-smith)
    x^5 - 7x^3 + 2x^2 -30x + 6=(x^2+3) (x^3-10 x+2)
    a,b,c be the real roots:
    (a^2+b^2+c^2)=(a+b+c)^2-2(ab+ac+bc)
    Vieta tells us:
    =0^2-2(-10)=20
    Well done, the trick was in factorising the polynomial, though you didn't show how you did it in the solution. Also, how can you guarantee that all the roots of the cubic are real?
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    (Original post by shamika)
    When you say algebraic number theory, do you mean things like number fields, calculating class numbers, cyclotomic fields and that kind of thing? My experience is that the hard part is learning the prerequisite algebra, the algebraic number theory part is very straightforward.

    To that end, you need a good introduction to rings. I reckon the best way to get that (and algebraic number theory) is via online lecture notes. Search for some and see what comes up
    Number fields - that is it. Broadly speaking, my aims are understanding of the techniques of Galois cohomology, and their applications in global class field theory; understanding the axioms of class formation; and, to some extent, the Brauer group.
    In addition, Riemann - Roch theory, more specifically Grothendieck groups and his theorem.

    I have been told that Commutative Ring Theory by Matsumura is more than sufficient. What is your view?



    (Original post by metaltron)
    Well done, the trick was in factorising the polynomial, though you didn't show how you did it in the solution. Also, how can you guarantee that all the roots of the cubic are real?
    Let f(x) = x^{3}-10x+2. Note that f is continuous in \mathbb{R}, and f(-4)<0, f(0)>0, f(2)<0, and f(4)>0.



    Problem 136*

    Let f: \mathbb{R} \to \mathbb{R} be a strictly increasing invertible function such that for all x \in \mathbb{R} we have f(x)+f^{-1}(x)=e^{x}-1. Prove that f has at most one fixed point.

    Problem 137**

    Find all continuous functions f: \mathbb{R} \to \mathbb{R} which satisfy f(f(x))-2f(x)+x=0 for all x \in \mathbb{R}.

    Problem 138***

    Let P(X) be an irreducible polynomial over \mathbb{Z}[X]. Show that P(XY) is irreducible over \mathbb{Z}[X,Y].
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    (Original post by Mladenov)
    For number theory, I use Niven's An Introduction to the Theory of Numbers, Vinogradov's Elements of Number Theory, Borevich and Shafarevic's Number Theory, and I am to buy Lang's Algebraic Number Theory. Would you suggest anything else for algebraic number theory?

    Algebra - Lang's Undergraduate Algebra and Lang's Algebra, Waerden's Algebra, and Kostrikin's An Introduction to Algebra. I am planning to buy Weibel's An Introduction to Homological Algebra, since it is essential when it comes to algebraic number theory, and more specifically - class field theory.

    Would you advise some books, I should be grateful.
    You have some serious literature (which is good). I have used Niven's book too; it is very good. I cannot say much about Vinogradov's and Borevich and Shafarevich's books. My choices were Ireland and Rosen's book, for a more algebraic approach, and Neukirch's book (which you can possibly combine with Murty's book). I have copies of Lang's Algebra and his Undergraduate Algebra. I also try to use books that do not overlap (much) in content. I am not sure you will be needing Homological Algebra in the near future.

    Take a look here (taken from) and here (taken from) for structural information. This can be useful too. As Shamika says, you will need good foundation in rings, modules and Galois theory. Lecture notes are available here and here (EDIT: and here).

    I can't think of anything more right now.

    EDIT: Most of these books cost a fortune. It may be best if you borrow them from a (university) library. For example, all of these are available in Cam's library.
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    (Original post by jack.hadamard)
    You have some serious literature (which is good). I have used Niven's book too; it is very good. I cannot say much about Vinogradov's and Borevich and Shafarevich's books. My choices were Ireland and Rosen's book, for a more algebraic approach, and Neukirch's book (which you can possibly combine with Murty's book). I have copies of Lang's Algebra and his Undergraduate Algebra. I also try to use books that do not overlap (much) in content. I am not sure you will be needing Homological Algebra in the near future.

    Take a look here (taken from) and here (taken from) for structural information. This can be useful too. As Shamika says, you will need good foundation in rings, modules and Galois theory. Lecture notes are available here and here (EDIT: and here).

    I can't think of anything more right now.

    EDIT: Most of these books cost a fortune. It may be best if you borrow them from a (university) library. For example, all of these are available in Cam's library.
    I am grateful to you; this is a considerable amount of resources.

    Vinogradov's book treats analytic number theory (hence not quite relevant). It is not as deep as Apostol's book, yet there are lots of good exercises.
    Borevich and Shafarevich's book is one of the best. It is a shame that this book is currently out of print. I have the Russian edition.

    I consider Lang's Algebra a bit demanding; also he never explains his motivation.

    By the way, Murty's book will be very beneficial.

    Unfortunately, I will have to buy these books, as I do not have permission to borrow books from university libraries.
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    This has been a terrible day for me, maths-wise. I shall leave 136 for someone else, since people have been asking for (*) q's.

    Solution 137

    f(x)=x+h(x)\Rightarrow h(x+h(x))=h(x)\Rightarrow h(x+nh(x))=h(x)\Rightarrow h=\text{cnst.}^{(*)}

     ^{(*)} if h were non-constant there is an interval [x_0,x_1] for which h(x_0)<h(x_1) and h\neq h(x_0),h(x_1) for x\in (x_0,x_1). Define y_n=x_0+nh(x_0) and z_n=x_1+nh(x_1). Then we have h\neq h(x_0),h(x_1) for x\in (y_n,z_n) . But since z_n-y_n\to\infty there will exist i such that y_{i+1}\in (y_i,z_i)\Rightarrow h(y_{i+1})\in (h(x_0),h(x_1))\Rightarrow \Leftarrow
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    (Original post by Mladenov)
    I consider Lang's Algebra a bit demanding; also he never explains his motivation.
    You are welcome. This blog also provides a useful guide: the missing article by Sam Ruth can be found here (he has removed it for some reason). Lang's Algebra becomes useful only after you have learnt most of the basic and intermediate stuff and have gained enough intuition/maturity; it is often used as a reference and not a book to learn from. Are you going to university this year? If this is the case, then you could wait a couple of months (for some of the books).


    Problem 139 / **

    Given an integer-valued polynomial, can you always find some values such that their sum is divisible by n \in \mathbb{N}?
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    (Original post by ben-smith)
    Problem 132
    Is this Feynman integration?
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    Solution 139

    Fix n. We work in \mathbb{Z}/n\mathbb{Z}.

    Let P(m) be an integer-valued polynomial. Clearly, there are infinitely many m_{i} \in \mathbb{Z}, i\in \mathbb{Z^{+}} such that P(m) is constant over the set \{ m_{i} | i \in \mathbb{Z^{+}} \}. Hence, we can always choose n elements from the set \{ m_{i} | i \in \mathbb{Z^{+}} \}, and \displaystyle \sum_{i=1}^{n} P(m_{i}) = 0.

    (Original post by jack.hadamard)
    You are welcome. This blog also provides a useful guide: the missing article by Sam Ruth can be found here (he has removed it for some reason). Lang's Algebra becomes useful only after you have learnt most of the basic and intermediate stuff and have gained enough intuition/maturity; it is often used as a reference and not a book to learn from. Are you going to university this year? If this is the case, then you could wait a couple of months (for some of the books).
    I had found Lang's Undergraduate Algebra quite easy to comprehend, and thus I decided to buy his Algebra.

    Yep, I am finishing high school this year.

    Problem 140*

    Evaluate \displaystyle \sum_{v=1}^{\infty} \frac{1}{2^{v}v^{2}}.

    Problem 141**

    Let f \in C^{\infty} for x>0, and 0 \le (-1)^{n}f^{(n)}(x) \le e^{-x} for all x >0 and n \in \mathbb{Z^{+}} \cup \{0\}. Find f.

    Remark: f^{(n)}(x) is the nth derivative of f.
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    Solution 140

    Right, after the recent discussion on the STEP solutions thread about integral notation, I need to be careful here not to mix up bound and free variables.

    Lemma (is this the right time to use a lemma? )

    \displaystyle\sum_{r=1}^{\infty} \dfrac{1}{r^{2}} = \dfrac{\pi^{2}}{6} (the Basel problem for those who haven't seen it)

    \displaystyle\sum_{v=1}^{\infty} \dfrac{1}{2^{v} v^{2}}

    Consider \displaystyle\sum_{v=1}^{\infty} x^{v-1} for |x| < 1 = \dfrac{1}{1-x}

    \Rightarrow \displaystyle\sum_{v=1}^{\infty} \dfrac{x^{v}}{v} = \displaystyle\int_{0}^{x} \dfrac{1}{1-t} dt = - \ln |1-x| (*) \Rightarrow \displaystyle\sum_{v=1}^{\infty} \dfrac{x^{v-1}}{v} = -\dfrac{\ln |1-x|}{x}

    \Rightarrow \displaystyle\sum_{v=1}^{\infty} \dfrac{x^{v}}{v^{2}} = - \displaystyle\int_{0}^{x} \dfrac{\ln |1-t|}{t} dt

    The sum in question, \displaystyle\sum_{v=1}^{\infty} \dfrac{1}{2^{v} v^{2}} (call it \mathcal{S}) is for x=\frac{1}{2}, thus:

    \mathcal{S} = - \displaystyle\int_{0}^{1/2} \dfrac{\ln |1-x|}{x} dx. Applying 1-x \to x we get:

    - \displaystyle\int_{0}^{1/2} \dfrac{\ln |1-x|}{x} dx = - \displaystyle\int_{1/2}^{1} \dfrac{ \ln x}{1-x} dx

    - \displaystyle\int_{1/2}^{1} \dfrac{ \ln x}{1-x} dx = - \displaystyle\sum_{r=0}^{\infty} \underbrace{\displaystyle\int_{1/2}^{1} x^{r} \ln x dx}_{\mathcal{I}}

    By parts, \mathcal{I} = \left[ \dfrac{x^{r+1} \ln x}{r+1} - \dfrac{x^{r+1}}{(r+1)^{2}} \right]_{1/2}^{1} = \left[ \dfrac{x^{r+1} \ln x (r+1) - x^{r+1}}{(r+1)^{2}} \right]_{1/2}^{1}

    = \dfrac{-1}{(r+1)^{2}} + \dfrac{\ln 2}{2^{r+1} (r+1)} + \dfrac{1}{2^{r+1} (r+1)^{2}}

    \Rightarrow \mathcal{S} = -\displaystyle\sum_{r=0}^{\infty} \left( \underbrace{\dfrac{-1}{(r+1)^{2}}}_{-\frac{\pi^{2}}{6} \text{by Lemma}} + \underbrace{\dfrac{\ln 2}{2^{r+1} (r+1)}}_{\mathcal{S}_{2}} + \underbrace{\dfrac{1}{2^{r+1} (r+1)^{2}}}_{\mathcal{S}} \right)

    \mathcal{S}_{2} = \ln2 \cdot \displaystyle\sum_{r=0}^{\infty} \dfrac{1}{2^{r+1} (r+1)} = \ln 2 \cdot -\ln \left(1 - \frac{1}{2} \right) (by (*) applying x = \frac{1}{2})  = \ln^{2} 2

    \mathcal{S} = -\left( -\dfrac{\pi^{2}}{6} + \ln^{2} 2 + \mathcal{S} \right) \Rightarrow \mathcal{S} = \dfrac{\pi^{2}}{12} - \dfrac{\ln^{2} 2}{2}
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    (Original post by Mladenov)
    I had found Lang's Undergraduate Algebra quite easy to comprehend, and thus I decided to buy his Algebra.
    In the foreword, he says that his Undergraduate Algebra and Linear Algebra provide more than enough background for a graduate course, but does not say that they are sufficient for reading his Algebra book. Nevertheless, many of the examples come from different areas of the undergraduate curriculum; to make full use of this book you need an undergraduate degree.


    Let's now make the problem more concrete.

    Extension to Problem 139 / **

    i) Let a_1, a_2, ..., a_n \in \mathbb{Z}. Prove that there exists I \subset \{1, 2, ..., n\} such that

    \displaystyle \sum_{i \in I} a_i \equiv 0\ (n).

    ii) How much more difficult does it become (and how many integers do we need to start with) in order to guarantee that there will be a subset, I, of cardinality exactly n having the same property?
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    Solution 139 (2)

    Part i)
    We consider the numbers 0,a_{1}, a_{1}+a_{2}, \cdots, a_{1}+a_{2}+\cdots+a_{n}. These are n+1. Hence at least two of them are congruent modulo n. Therefore, there exists I such that \displaystyle \sum_{i \in I} a_{i} \equiv 0 \pmod n.
    Note that I=\{1,2,\cdots,n\} is also possible. For example, when all a_{i} are \equiv 1 \pmod n.
    The following is also true:
    Let n be an arbitrary positive integer. Suppose a_{1},\cdots, a_{k} are given integers, which give at least n-k+1 distinct remainders \pmod n. Then there is I \subseteq \{1,2,\cdots, k\} such that \displaystyle \sum_{i\in I} a_{i} \equiv 0 \pmod n.

    Part ii)

    It is quite well-known that if G is an additively written abelian group, card(G)=n, for some positive integer n, then, if a_{i} \in G, for i \in \{1,2, \cdots, 2n-1\}, we can represent e - the unit element of G, as a sum of the elements of S, where S \subset \{a_{1},a_{2}, \cdots, a_{2n-1}\}, and card(S) = n.
    Hence, if we commence with 2n-1 integers, there always will be I \subset \{1,2, \cdots , 2n-1\}, card(I)=n, such that \displaystyle \sum_{i \in I} a_{i} \equiv 0 \pmod n.
    Another way to prove this is the following:
    First: it is true when n is a prime number.
    Second: If the result is true for n=a,b, where a,b \in \mathbb{Z^{+}}, then it is true for n=ab.
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    (Original post by Mladenov)
    Part ii)

    It is quite well-known that...
    You have got quite a few rabbits in your hat. I intended to ask for a general comment. For the people who don't know the result, this is known as Erdős–Ginzburg–Ziv theorem (EGZ). The proof you referred to is a beautiful application of the Chevalley-Warning theorem.

    In fact, John Olson generalised EGZ to any finite group (abelian or not).
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    (Original post by jack.hadamard)
    The proof you referred to is a beautiful application of the Chevalley-Warning theorem.
    Precisely.

    (Original post by jack.hadamard)
    In fact, John Olson generalised EGZ to any finite group (abelian or not).
    Apropos, I knew not that the result is still true in the non-abelian case.

    Here are some * questions.

    Problem 142*

    Evaluate \displaystyle \sum_{v=1}^{\infty} \frac{x^{v} \sin v\alpha}{v}, (|x| <1).
    Hence, find \displaystyle \sum_{v=1}^{\infty} \frac{\sin vx}{v}, (0<x< 2\pi). Can you evaluate \displaystyle \sum_{v=1}^{\infty} \frac{\sin vx}{v^{3}}, (0 \le x \le 2\pi)?

    Problem 143*

    Prove that \displaystyle \frac{1}{n}\sum_{v=1}^{n} \left(1+ \frac{1}{2v} \right)^{2v} \le \left(1+ \frac{1}{n+1} \right)^{n+1}.

    Problem 144*

    There are several castles in one country and three roads lead from every castle. A knight leaves his castle. Traveling around the country he leaves every new castle via the road that is either to the right or to the left of the one by which he arrived. According to The Rule the knight never takes the same direction (right or left) twice in a row. Prove that some day he will return to his own castle.
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    (Original post by Mladenov)

    Problem 142*

    Evaluate \displaystyle \sum_{v=1}^{\infty} \frac{x^{v} \sin v\alpha}{v}, (|x| <1).
    Hence, find \displaystyle \sum_{v=1}^{\infty} \frac{\sin vx}{v}, (0<x< 2\pi). Can you evaluate \displaystyle \sum_{v=1}^{\infty} \frac{\sin vx}{v^{3}}, (0 \le x \le 2\pi)?
    Solution 142 (so far)

    Consider that, summing the geometric sequence, we have:

     \displaystyle \sum_{v=1}^\infty x^ {v-1} e^{iva} = \frac {e^{ia}}{1- x e^{ia}}



\Rightarrow \displaystyle\sum_{v=1}^{\infty} \dfrac{x^{v}e^{iva}}{v} = \displaystyle\int_{0}^{x} \frac{e^{ia}}{1-ze^{ia}} dz



\Rightarrow \displaystyle\sum_{v=1}^{\infty} \dfrac{x^{v}e^{iva}}{v} =  e^{ia} \left[ \frac {-1}{e^{ia}} ln (1- z e^{ia}) \right]_0^x



\Rightarrow \displaystyle\sum_{v=1}^{\infty} \dfrac{x^{v}e^{iva}}{v} = - ln (1- x e^{ia})

    Now, suppose we have an imaginary number z, such that,

     z= x+ iy= r e^{it} (where t is the argument of z.}
    



\Rightarrow ln (x+iy) = ln (r) + it



\Rightarrow Im \left[ ln (x+iy) \right] = t= arctan (\frac{y}{x})

    Now, consider that,

    



1- xe^{ia} = (1- x \cos a) - i \sin a



\Rightarrow Im \left[\displaystyle\sum_{v=1}^{\infty} \dfrac{x^{v}e^{iva}}{v} \right] = - Im \left[ln (1- x e^{ia}) \right]



\Rightarrow \displaystyle\sum_{v=1}^{\infty} \dfrac{x^{v} \sin {av} }{v} = - arctan (\frac{-x \sin a}{1- x \cos a}) = arctan (\frac{x \sin a}{1- x \cos a})

    Now, substituting 1 in place of x in this formula and x instead of a, we get,

    



\displaystyle\sum_{v=1}^{\infty} \dfrac{ \sin {vx} }{v} = arctan ( \frac { \sin x}{1- \cos x })= arctan( \cot { \frac {x}{2}})

    Let, t= pi/2 - x/2

    



\Rightarrow arctan ( cot \frac {x}{2}) = arctan ( \tan t) = t = \frac { \pi - x}{2}
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    (Original post by MW24595)
    Solution 142 (so far)
    Well. There is only one point - you can't simply substitute x=1, as the radius of convergence is |x| <1. I mean, you need to justify it.
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    Ach, doing my best to ignore this thread but I love series and integrals too much. Here are alternative solutions to 140 and 142:

    Solution 140

    Differentiation under the integral sign!

    \displaystyle \sum_{v\geq 1}x^{v-1}=\frac{1}{1-x}\Rightarrow \sum_{v\geq 1}\frac{1}{2^vv^2}=-\int_0^{\frac{1}{2}} \frac{\ln (1-x)}{x}\,dx

    \displaystyle \begin{aligned}\int_0^{\frac{1}{  2}} \frac{\ln (1-x)}{x}\,dx\overset{x\to \frac{x}{2}}=\int_0^1 \frac{\ln (1-\frac{x}{2})}{x}\,dx \overset{ \text{IBP}}=\int_0^1 \frac{\ln x}{2-x}\,dt \overset{x\to 1-x}=\int_0^1 \frac{\ln (1-x)}{1+x}\,dx

    \displaystyle \begin{aligned}f(t)=\int_0^1 \frac{\ln (1-tx)}{1+x}\,dx\Rightarrow f'(t) &= \int_0^1 \frac{x\,dx}{(tx-1)(1+x)} \\&= \int_0^1 \frac{dx}{tx-1}+\int_0^1 \frac{dx}{(t+1)(1+x)}-\int_0^1 \frac{t\,dx}{(tx-1)(t+1)}\\& =\frac{\ln (1-t)}{t}+\frac{\ln 2}{t+1} -\frac{\ln (1-t)}{1+t}\end{aligned}

    Everything falls into place now:

    \displaystyle f(1) =\int_0^1 \frac{\ln (1-t)}{1+t}dt=\int_0^1 \frac{\ln (1-t)}{t}\,dt+\int_0^1\frac{\ln 2}{t+1}\,dt-\int_0^1 \frac{\ln (1-t)}{1+t}\,dt

    \displaystyle -f(1) =\frac{1}{2}\left(-\int_0^1 \frac{\ln (1-t)}{t}\,dt-\int_0^1\frac{\ln 2}{t+1}\right)\,dt=\frac{\pi^2}{  12}-\frac{\ln^2 2}{2}

    Solution 142

    We can evaluate the sum directly:

    \displaystyle\begin{aligned} g_n = \sum_{v=1}^n \frac{\sin vx}{v}=\sum_{v=1}^n \int_0^x \cos vt \,dt =\frac{1}{2}\int_0^x \csc \frac{t}{2}\sin \left(\frac{t}{2}(2n+1) \right)-1\,dt \end{aligned}

    The trick is now to use that \csc \frac{t}{2}\sim \frac{2}{t} near  0 to get rid of a bunch of stuff:

    \displaystyle\begin{aligned} g_n = \underbrace{\int_0^x \frac{1}{t}\sin \left(\frac{t}{2}(2n+1) \right) dt}_{\alpha_n}+\frac{1}{2} \underbrace{\int_0^x \left(\csc \frac{t}{2}-\frac{2}{t}\right)\sin \left(\frac{t}{2}(2n+1) \right)dt}_{\beta_n}-\frac{x}{2}\end{aligned}

    As n\to\infty:

    \alpha_n= \displaystyle\int_0^x \frac{1}{t}\sin \left(\frac{t}{2}(2n+1) \right) dt\overset{\frac{t}{2}(2n+1)\to t}=\int_0^{\frac{x}{2}(2n+1)} \frac{\sin t}{t}\,dt\to \frac{\pi}{2}

    \begin{aligned} \beta_n = \displaystyle \left[\frac{\cos \left(\frac{t}{2}(2n+1)\right) \left(\csc\frac{t}{2}-\frac{2}{t}\right)}{2n+1}\right]_0^{x} +\frac{1}{2n+1}\int_0^x\cos \left( \frac{t}{2}(2n+1)\right)\left( \cot \frac{t}{2} \csc\frac{t}{2}-\frac{4}{t^2}\right)\,dt\to 0

    Hence \displaystyle g_n\to \sum_{v\geq 1}\frac{\sin vx}{v}=\frac{\pi-x}{2}

    Furthermore, observe that:

    \displaystyle\begin{aligned} \int_0^x\int_0^{t_1}\cdots \int_0^{t_{2k-1}} \sum_{v\geq 1}\frac{\sin vt_{2k}}{v}\,dt_{2k}\cdots dt_1=\frac{1}{2}\int_0^x\int_0^{  t_1}\cdots \int_0^{t_{2k-1}} \pi -t_{2k}\, dt_{2k}\cdots\, dt_1

    gives:

    \displaystyle\begin{aligned} (-1)^k\sum_{v\geq 1} \frac{\sin vx}{v^{2k+1}}+\sum_{i=1}^{k} \left( \frac{(-1)^{k+i+1}x^{2(k-i)+1}}{(2(k-i)+1)!}\sum_{v\geq 1}\frac{1}{v^{2i}} \right)= \frac{x^{2k}}{2}\left(\frac{\pi}  {(2k)!}-\frac{x}{(2k+1)!}\right)

    \displaystyle \sum_{v\geq 1} \frac{\sin vx}{v^{2k+1}}=(-1)^k\left[ \frac{x^{2k}}{2}\left(\frac{\pi}  {(2k)!}-\frac{x}{(2k+1)!}\right)+\sum_{i  =1}^{k} \frac{(-1)^ix^{2(k-i)+1}\zeta (2i)}{(2(k-i)+1)!}\right]\;\;(\clubsuit)

    So, for instance:

    \displaystyle\sum_{v\geq 1}\frac{\sin vx}{v^{3}}=-\left[\frac{x^2}{2}\left(\frac{\pi}{2!  }-\frac{x}{3!}\right)-x\zeta(2) \right]=\frac{\pi^2}{6}x-\frac{\pi}{4}x^2+\frac{1}{12}x^3

    \displaystyle\sum_{v\geq 1}\frac{\sin vx}{v^{5}}=\left[\frac{x^4}{2}\left(\frac{\pi}{4!  }-\frac{x}{5!}\right)+ x\zeta(4) -\frac{x^3\zeta(2)}{3!}\right]=\frac{\pi^4}{90}x-\frac{\pi^2}{36}x^3+\frac{\pi}{4  8}x^4-\frac{1}{240}x^5

    If we ignore knowledge of the \zeta values, we can in fact construct an algorithm to derive them using the results above. For \zeta (2) (thus solving the Basel problem along the way):

    \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 (\clubsuit) + 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}.
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    Yet another way to do problem 140 without integration :

    Spoiler:
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    We denote \displaystyle f(x) = \sum_{v=1}^{\infty} \frac{x^{v}}{v^{2}}. Set \varphi (x) =f(x)+f(1-x)+ \ln x \times \ln(1-x). This function is constant, call it C, over \left(0,1 \right). We can find this constant by letting x \to 0. Thus, C= f(0)+f(1) - this is so, since f is continuous over \left[-1,1 \right], and differentiable over \left(-1,1 \right). Hence, \displaystyle C=\frac{\pi^{2}}{6}. Therefore \displaystyle 2f(\frac{1}{2}) = \frac{\pi^{2}}{6}- \ln^{2}(2).
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    (Original post by Lord of the Flies)
    ...
    Mother of god... :adore:
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    (Original post by Felix Felicis)
    Mother of god... :adore:

    This is the post I congratulated him for. My pantaloons are drenched from carnal delight.
 
 
 
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