Hey there! Sign in to join this conversationNew here? Join for free
    Offline

    18
    ReputationRep:
    (Original post by Felix Felicis)
    ...
    (Original post by Zakee)
    ...
    Thanks guys. :lol:

    Now back to STEP.
    Offline

    2
    ReputationRep:
    (Original post by Lord of the Flies)
    Spoiler:
    Show
    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]

    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
    Lol, my solution involves inverse fourier transforming everything in sight. Needless to say, I am not an elegant mathematician.
    Offline

    16
    ReputationRep:
    (Original post by Lord of the Flies)
    ...


    To top it off, you presented the answers really well. Très bien!
    Offline

    18
    ReputationRep:
    (Original post by ben-smith)
    Needless to say, I am not an elegant mathematician.
    Pf, nonsense Ben! Your solution to problem 69 was very neat for instance.

    (Original post by shamika)
    To top it off, you presented the answers really well. Très bien!
    If only I could present things as neatly on paper... Mais cimer!
    Offline

    14
    ReputationRep:
    Problem 145**

    Let \mathcal{P}_2(x) be the set of all polynomials of degree at most 2.

    Find q(x) \in \mathcal{P}_2(x) such that  \displaystyle\int_0^1 p(x)q(x)\ dx = p(\frac{1}{2}) \ for all p(x) \in \mathcal{P}_2(x)
    Offline

    2
    ReputationRep:
    I'm back with an awesome problem... ENJOY!

    Problem 146**


    Evaluate \displaystyle\sum_{p\in\mathmm{P  }}^{\infty}\frac{1}{p}
    Offline

    14
    ReputationRep:
    (Original post by Jkn)
    I'm back with an awesome problem... ENJOY!

    Problem 146**


    Evaluate \displaystyle\sum_{p\in\mathmm{P  }}^{\infty}\frac{1}{p}
    So.. you mean the sum of the reciprocals of all the primes? 'cause if so..

    Spoiler:
    Show
    that sum diverges..
    Offline

    1
    ReputationRep:
    (Original post by FireGarden)
    Problem 145**
    I like the problem, so...

    Let P_{\leq n}(x) denote the set of all univariate polynomials, in x, of degree atmost n.

    Define \displaystyle \Phi_q(p) = \int_{0}^{1} p(x) q(x)\ dx for p,q \in P_{\leq n}(x).

    The goal is to find q such that \Phi_q(p) = p(a) for all p \in P_{\leq n}(x), where a \in \mathbb{R}.

    Notice that \Phi_q(x^n) = a^n by assumption. In addition, we have:

    i) \Phi_{a q} = a\Phi_{q} for a constant a and polynomial q.
    ii) \Phi_{q + p} = \Phi_{q} + \Phi_{p} for polynomials q and p.

    and \displaystyle \Phi_{x^a}(x^b) = \frac{1}{a + b + 1}.

    Spoiler:
    Show

    Let q(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0.

    Hence, \Phi_{q}(p) = a_n \Phi_{x^n}(p) + a_{n-1} \Phi_{x^{n-1}}(p) + \cdots a_1 \Phi_{x}(p) + a_0 \Phi_{1}(p).

    Using, for p, the polynomials 1, x, x^2, ..., x^n, we get the following system.

    \begin{bmatrix} \Phi_{x^n}(1) & \Phi_{x^{n-1}}(1) & \cdots & \Phi_{1}(1) \\ \Phi_{x^n}(x) & \cdots &  & \\ \vdots &  &  & \\ \Phi_{x^n}(x^n) & \cdots & &  \end{bmatrix} \begin{bmatrix} a_n \\ a_{n-1} \\ \vdots \\ a_0 \end{bmatrix} = \begin{bmatrix} 1 \\ a \\ \vdots \\ a^n \end{bmatrix}

    The symmetry in the values of \Phi_{x^a}(x^b) implies that this is a Toeplitz matrix.

    The special case, n = 2 and a = \frac{1}{2}, produces the following system

    \displaystyle \begin{bmatrix} \frac{1}{n+1} & \frac{1}{n} & \frac{1}{n - 1} \\ \frac{1}{n+2} & \frac{1}{n+1} & \frac{1}{n} \\ \frac{1}{n+3} & \frac{1}{n+2} & \frac{1}{n + 1} \end{bmatrix} \begin{bmatrix} a_n \\ a_{n-1} \\ a_0 \end{bmatrix} = \begin{bmatrix} 1 \\ a^{n-1} \\ a^n \end{bmatrix}

    which has the solution \displaystyle \begin{bmatrix} -15 \\ 15 \\ -\frac{3}{2} \end{bmatrix}.
    Offline

    1
    ReputationRep:
    Problem 147 / ***

    Assume s is a set containing numbers. Let \sigma(s) and \pi(s) denote the sum and product, respectively, of all the elements in the set s.

    Let S be the set \{1, 2, ..., n\}.

    i) Find an expression for \displaystyle \sum_{s \in \mathcal{P}(S)} \frac{1}{\pi(s)}.

    ii) Prove that \displaystyle \sum_{s \in \mathcal{P}(S), s \not= \varnothing } \frac{\sigma(s)}{\pi(s)} = n(n+2) - (n+1)\sum_{k=1}^{n} \frac{1}{k}.

    Note that \pi(\varnothing) = 1.
    Offline

    2
    ReputationRep:
    (Original post by FireGarden)
    So.. you mean the sum of the reciprocals of all the primes? 'cause if so..

    Spoiler:
    Show
    that sum diverges..
    Spoiler:
    Show
    Yes. No prove it ...don't look it up though!
    Offline

    0
    ReputationRep:
    (Original post by Jkn)
    Spoiler:
    Show
    Yes. No prove it ...don't look it up though!
    Spoiler:
    Show
    I'm thinking proof by contradiction? I feel like the you write the LHS as a fraction and the right is a real number...erm :lol: I'm so bad at maths
    Offline

    2
    ReputationRep:
    (Original post by bananarama2)
    Spoiler:
    Show
    I'm thinking proof by contradiction? I feel like the you write the LHS as a fraction and the right is a real number...erm :lol: I'm so bad at maths
    Spoiler:
    Show
    You need to find a property of convergent series that can be applied in order to create the contradiction! Writing the left hand side as a fraction would require 2 sigmas and a capital pi so I don't know whether or not this would simplify the problem. This problem is extremely difficult though so don't feel too downtrodden (I found the proof online but didn't actually do it myself :lol:) There may exist an easy proof using sophisticated methods but the proof I read is elementary and incredibly beautiful.
    Offline

    1
    ReputationRep:
    Solution 147

    Part i). Consider \displaystyle \left(1+1 \right)\times \cdots \times \left(1+\frac{1}{n} \right) = n+1. Hence, \displaystyle \sum_{s \in \mathcal{P(S)}} \frac{1}{\pi(s)} = n+1. I include s = \O.

    Part ii). We use induction. The base case is obvious. Suppose that our result is true for all i \in \{1,2, \cdots, n\}. We prove it for n+1.
    Split our sum into two parts:
    \begin{aligned} \displaystyle \sum_{s \in \mathcal{P}(S_{n+1}), s \not= \O} \frac{\sigma(s)}{\pi(s)} & = \sum_{s \in \mathcal{P}(S_{n}), s \not= \O} \frac{\sigma(s)}{\pi(s)} + \sum_{s \in \mathcal{P}(S_{n} )} \frac{n+1 + \sigma(s)}{(n+1)\pi(s)} \\&= n(n+2) - (n+1)H_{n} + 1+n + \frac{1}{n+1}(n(n+2) - (n+1)H_{n}) \\& = n(n+2) - (n+1)H_{n} + 2n+2 - \frac{1}{n+1} -H_{n} \\&= (n+1)(n+3) - (n+2)H_{n+1} \end{aligned} .

    Remark: I denoted S_{n} = \{1,2,\cdots, n\} in part ii).

    Solution 146

    The kth prime is less than 2k\ln k +2 for k \ge 2. Hence the series diverges.
    Offline

    2
    ReputationRep:
    (Original post by Mladenov)
    Solution 146

    The kth prime is greater than k(\ln k + \ln(\ln k-1)) for k \ge 2. Hence the series diverges.
    Reminds me of this:
    Name:  And_then_a_miracle_happens_cartoon.jpg
Views: 164
Size:  41.0 KB
    Spoiler:
    Show
    You're going to have explain how you got that and prove it :lol: And also why it implies that a series diverges :lol:



    Here's an easier one. Undergrads will find it child's play but it's a really nice result so I'll post it anyway:

    Problem 148**

    Evaluate \displaystyle\int_{-\infty}^{\infty}\frac{sin^2x}{x^  2}dx
    Offline

    18
    ReputationRep:
    I shall use the same method as when I did \displaystyle \int_{-\infty}^{\infty} \frac{\sin x}{x}\,dx in a previous solution.

    Solution 148

    \begin{aligned}  f(t)=\displaystyle \int_{0}^{\infty} \left(\frac{\sin tx}{x}\right)^2\,dx\Rightarrow \mathcal{L}\{ f(t)\} &=\int_0^{\infty}\frac{1}{x^2  } \int_0^{\infty}e^{-ts} \sin^2 tx\,dt\,dx \\ & =\int_0^{\infty}\frac{1}{x^2} \left( \frac{2x^2}{s(s^2+4x^2)}\right) \,dx\\&=\frac{\pi}{2s^2}\;\;\; \Rightarrow\; \mathcal{L}^{-1}\left\{\dfrac{\pi}{2s^2} \right\}=\dfrac{\pi t}{2}\;\Rightarrow\;\int_{-\infty}^{\infty} \left( \frac{\sin x}{x}\right)^2\,dx=\pi\end{align  ed}

    Spoiler:
    Show
    In fact we can carry on to higher powers, however getting to the 2nd line demands considerably more algebra.

    \begin{aligned}  f(t)=\displaystyle \int_{0}^{\infty} \left(\frac{\sin tx}{x}\right)^3\,dx\Rightarrow \mathcal{L}\{ f(t)\} &=\int_0^{\infty}\frac{1}{x^3  } \int_0^{\infty}e^{-ts} \sin^3 tx\,dt\,dx \\ & =\int_0^{\infty}\frac{1}{x^3} \left( \frac{6x^3}{9x^4+10s^2x^2+s^4} \right) \,dx\\&=\frac{3\pi}{4s^3}\;\;\; \Rightarrow\; \mathcal{L}^{-1}\left\{\dfrac{3\pi}{4s^3} \right\}=\dfrac{3\pi t^2}{8}\;\Rightarrow\;\int_{-\infty}^{\infty} \left( \frac{\sin x}{x}\right)^3\,dx=\frac{3\pi}{4  }\end{aligned}

    It would be interesting if we could figure out a closed form for \displaystyle \int_{-\infty}^{\infty} \left(\frac{\sin x}{x}\right)^n \,dx
    Offline

    0
    ReputationRep:
    I really must learn about Laplace transforms.
    Offline

    18
    ReputationRep:
    (Original post by bananarama2)
    I really must learn about Laplace transforms.
    Have a look at this. It's a bit slow at times but it is clear and he's quite a fun lecturer!
    Offline

    1
    ReputationRep:
    (Original post by Jkn)
    Spoiler:
    Show
    Reminds me of this:
    Name:  And_then_a_miracle_happens_cartoon.jpg
Views: 164
Size:  41.0 KB
    Spoiler:
    Show
    You're going to have explain how you got that and prove it :lol: And also why it implies that a series diverges :lol:
    I am too busy these days, and I considered the problem superficially.
    I need an upper bound.
    Here is one sufficient upper bound - for k>1, we have p_{k} < 2k \ln k +2. Now, it is obvious that \displaystyle \sum_{p \in \mathbb{P}} \frac{1}{p} > \sum_{k \ge 2} \frac{1}{2k \ln k +2}; the last series diverges.
    Offline

    0
    ReputationRep:
    (Original post by Lord of the Flies)
    Have a look at this. It's a bit slow at times but it is clear and he's quite a fun lecturer!
    Indeed, that was quite good! I skipped bits of working and examples though
    Offline

    18
    ReputationRep:
    (Original post by bananarama2)
    Indeed, that was quite good! I skipped bits of working and examples though
    Ha, yes that's what I meant by "slow"
 
 
 
Poll
Who is your favourite TV detective?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.