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

    2
    ReputationRep:
    Just came up with a really nice result (can I blag it as a theorem? :lol:)

    Problem 152**

    Given that \displaystyle\sum_{i=1}^n x_i=S and \displaystyle max(\prod_{i=1}^n x_i) =P for real numbers x_i such that x_i >0 \ \forall i, find a condition on S for which P is irrational.

    Spoiler:
    Show
    Note: To find a complete condition on P would require solving one of "the great unsolved problems" in mathematics. For this problem, you must find all conditions on P that are you can and then prove this assertion from first principles
    Offline

    2
    ReputationRep:
    For anyone taking Edexcel Statistics 4...

    Problem 153
    */**

    Prove that a t-distribution with degree of freedom \nu is asymptotically equivalent a the Normal distribution for sample size n such that n \to \infty, stating the parameters of the Normal distribution to which it is equivalent.

    Hint:
    Spoiler:
    Show

    Would be awesome if someone derived this from definitions, but if not:

    It can be shown that \displaystyle f(x)=\frac{\Gamma{(  \frac{\nu+1}{2})}}{\sqrt{\nu \pi} \Gamma{( \frac{\nu}{2}} )}} \left(\frac{1+\frac{x^2}{\nu}} \right)^{-\frac{\nu +1}{2}}, where f(x) denotes the probability density function of a t-distribution with degree of freedom \nu

    Note: You may not use standard values/results for the gamma function. First principles por favor

    Spoiler:
    Show
    Also... Is this so-called hint just an over complication? Perhaps....
    Offline

    1
    ReputationRep:
    Problem 154 / *

    Let a_1 < a_2 < a_3 < ... be positive, non-consecutive integers. Let

    s_m = a_1 + a_2 + a_3 + ... + a_m for m \in \mathbb{N}.

    Prove that, for all n \in \mathbb{N}, the interval [s_n, s_{n+1}) contains a perfect square.
    Offline

    2
    ReputationRep:
    I did some really nice problems today and thought I would share them in an attempt to get people interested in this thread again:

    Problem 155*

    Find (without calculus) a fifth degree polynomial p(x) such that p(x)+1 is divisible by (x-1)^3 and p(x)-1 is divisible by (x+1)^3.

    Spoiler:
    Show
    Not too bad but is very reminiscent of a creation STEP question, but without being permitted (let alone led into) the simple differentiation approach. So is going to be good STEP preparation!


    Problem 156*/**/***

    Evaluate \displaystyle\int_0^{\pi} \ln(1-2a\cos(x)+a^2) dx, where a is a real number, in the case such that |a|>1.

    Problem 157*/**/***

    Evaluate \displaystyle\int_0^1 \frac{x-1}{\ln(x)} dx

    Problem 158*/**/***

    Evaluate \displaystyle\int_0^{\infty} \frac{\ln(1+x^2)}{1+x^2} dx

    and finally, because there's no point in me simply typing my solution up (especially since no-one seems too interested anyway)...

    Problem 159**

    Evaluate \displaystyle\int_{-\infty}^{\infty} \left(\frac{\sin(x)}{x} \right)^n dx, where n is a positive integer, in terms of a combination of a finite number of elementary functions.
    Offline

    2
    ReputationRep:
    (Original post by jack.hadamard)
    Problem 154 / *

    Let a_1 < a_2 < a_3 < ... be positive, non-consecutive integers. Let

    s_m = a_1 + a_2 + a_3 + ... + a_m for m \in \mathbb{N}.

    Prove that, for all n \in \mathbb{N}, the interval [s_n, s_{n+1}) contains a perfect square.
    Solution 154

    As they are non-consecutive integers, the difference between to consecutive terms in at least 2.

    For the size of the interval, s_{n+1}-s_{n}=a_{n+1} \ge 2(n+1)-1=2n+1

    Similarly, s_{n} \ge (2(n)-1)+(2(n-1)-1)+(2(n-2)-1)+...+(2(1)-1)=2(1+2+3+...+n)-n=n(n+1)-n=n^2.

    From this we can deduce that s_{n+1}=a_{n+1}+s_{n} \ge n^2+2n+1=(n+1)^2.

    Therefore, as the interval is always greater than or equal to (n+1)^2-n^2-1 and the smallest term in the interval is always greater than or equal to n^2, it follows trivially that the interval is always larger than or equal to the difference between any two square numbers.

    Therefore, the interval [s_n,s_{n+1}) must contain a perfect square. \square

    Edit: Where's this problem from btw? It was rather neat
    Offline

    1
    ReputationRep:
    Solution 156

    We denote \displaystyle I(a) = \int_{0}^{\pi} \ln(1-2a \cos x+a^{2})dx. Clearly, I(a)=I(-a). Furthermore, for  a \to 0, I(a) \to 0. Now, \displaystyle 2I(a)= I(a)+I(-a) = \frac{1}{2} \int_{0}^{2\pi} \ln(1-2a^{2} \cos 2x+a^{4})dx = I(a^2). Hence, we obtain \displaystyle I(a)= \frac{I(a^{2^{n}})}{2^{n}}, implying I(a)=0 for |a| <1.
    Let |a|>1. So, \displaystyle \ln(1-2a \cos x + a^{2}) = \ln(1 - 2\frac{1}{a} \cos x + \frac{1}{a^{2}}) + 2\ln|a|. Therefore, for |a|>1, we have I(a) = 2\pi\ln|a|.

    Solution 157

    We let \ln t \mapsto t, and then t \mapsto -t. Thus obtaining \displaystyle \int_{0}^{1} \frac{x-1}{\ln x}dx= \int_{0}^{\infty} \frac{e^{-x}-e^{-2x}}{x}dx = \ln2. More generally, \displaystyle \int_{0}^{1} \frac{x^{a-1}-x^{b-1}}{\ln x}dx = \ln\frac{b}{a}.

    Solution 158

    For a,b >0, we define \displaystyle I(a,b) = \int_{0}^{\infty} \frac{\ln(1+a^{2}x^{2})}{b^{2}+x  ^{2}}dx. I(a,b) is continuous, with respect to a, for a >0. Thus \displaystyle \frac{\partial I(a,b)}{\partial a} = \int_{0}^{\infty} \frac{2ax^{2}}{(b^{2}+x^{2})(1+a  ^{2}x^{2})}dx = \frac{\pi}{ab+1}. Thence, we have \displaystyle I(a,b) = \frac{\pi}{b}\ln(ab+1).

    Solution 159

    Notice \displaystyle \int_{0}^{\infty} \frac{\sin^{n}x}{x^{m}}dx= \frac{1}{(m-1)!}\int_{0}^{\infty} \frac{d^{m-1}}{dx^{m-1}}\sin^{n}x\frac{dx}{x}.
    We need to treat the cases m \equiv n \pmod 2 only.
    If m \equiv n \equiv 1 \pmod 2, differentiating \sin^{n}x (m-1 times), we get \displaystyle \int_{0}^{\infty} \frac{\sin^{n}x}{x^{m}}dx = \frac{(-1)^{\frac{(n-1)(m-1)}{4}}\pi}{2^{n}(m-1)!} \times \left(n^{m-1}-n(n-2)^{m-1}+ \cdots \right). The case m \equiv n \equiv 0 \pmod 2 is done analogically.
    Hence, \displaystyle \int_{0}^{\infty} \frac{\sin^{n}x}{x^{n}}dx = \frac{\pi}{2^{n}(n-1)!} \times \left(n^{n-1}-n(n-2)^{n-1}+\cdots \right).


    By the way, Borwein have proved far more general result.
    Fix n+1 complex numbers z_{0},\cdots,z_{n}.
    Define a= (a_{1},\cdots,a_{n}), a_{i} \in \{-1,1\}; \displaystyle b_{a} = z_{0}+ \sum_{i} z_{i}a_{i}, and \displaystyle c_{a} = \prod_{i} a_{i}.
    It follows that \displaystyle \int_{0}^{\infty} \prod_{n=0}^{n} \frac{\sin z_{i}x}{x}dx = \frac{\pi}{2^{n+1}n!}\sum_{a} c_{a}b_{a}^{n}{\rm sign b_{a}}.
    Hence in our case, when z_{0}=z_{1}=\cdots=z_{n-1}=1, we have \displaystyle \int_{0}^{\infty} \left(\frac{\sin x}{x} \right)^{n}dx= \frac{\pi}{2^{n}(n-1)!}\sum_{0 \le i \le \frac{n}{2}} (-1)^{i} \dbinom{n}{i} (n-2i)^{n-1}.

    May I ask, why are we not allowed to use calculus in problem 155 in order to solve it in a minute?
    Spoiler:
    Show
    It is enough to note that p'(x) is divisible by (x-1)^{2} and (x+1)^{2}.
    Offline

    12
    ReputationRep:
    (Original post by Mladenov)
    May I ask, why are we not allowed to use calculus in problem 155 in order to solve it in a minute?
    Spoiler:
    Show
    It is enough to note that p'(x) is divisible by (x-1)^{2} and (x+1)^{2}.
    Because the way which uses calculus was part of/similar to a previous STEP question that has obviously been solved many times on this forum. :laugh:
    Offline

    18
    ReputationRep:
    Accccch why did I have to miss the integrals? Well done on 159 Mlad, that's a nice solution.
    Offline

    2
    ReputationRep:
    (Original post by Mladenov)
    Spoiler:
    Show
    Solution 156

    We denote \displaystyle I(a) = \int_{0}^{\pi} \ln(1-2a \cos x+a^{2})dx. Clearly, I(a)=I(-a). Furthermore, for  a \to 0, I(a) \to 0. Now, \displaystyle 2I(a)= I(a)+I(-a) = \frac{1}{2} \int_{0}^{2\pi} \ln(1-2a^{2} \cos 2x+a^{4})dx = I(a^2). Hence, we obtain \displaystyle I(a)= \frac{I(a^{2^{n}})}{2^{n}}, implying I(a)=0 for |a| <1.
    Let |a|>1. So, \displaystyle \ln(1-2a \cos x + a^{2}) = \ln(1 - 2\frac{1}{a} \cos x + \frac{1}{a^{2}}) + 2\ln|a|. Therefore, for |a|>1, we have I(a) = 2\pi\ln|a|.

    Solution 157

    We let \lnx \mapsto t, and then t \mapsto -t. Thus obtaining \displaystyle \int_{0}^{1} \frac{x-1}{\ln x}dx= \int_{0}^{\infty} \frac{e^{-x}-e^{-2x}}{x}dx = \ln2. More generally, \displaystyle \int_{0}^{1} \frac{x^{a-1}-x^{b-1}}{\ln x}dx = \ln\frac{b}{a}.

    Solution 158

    For a,b >0, we define \displaystyle I(a,b) = \int_{0}^{\infty} \frac{\ln(1+a^{2}x^{2})}{b^{2}+x  ^{2}}dx. I(a,b) is continuous, with respect to a, for a >0. Thus \displaystyle \frac{\partial I(a,b)}{\partial a} = \int_{0}^{\infty} \frac{2ax^{2}}{(b^{2}+x^{2})(1+a  ^{2}x^{2})}dx = \frac{\pi}{ab+1}. Thence, we have \displaystyle I(a,b) = \frac{\pi}{b}\ln(ab+1).

    Solution 159

    Notice \displaystyle \int_{0}^{\infty} \frac{\sin^{n}x}{x^{m}}dx= \frac{1}{(m-1)!}\int_{0}^{\infty} \frac{d^{m-1}}{dx^{m-1}}\sin^{n}x\frac{dx}{x}.
    We need to treat the cases m \equiv n \pmod 2 only.
    If m \equiv n \equiv 1 \pmod 2, differentiating \sin^{n}x (m-1 times), we get \displaystyle \int_{0}^{\infty} \frac{\sin^{n}x}{x^{m}}dx = \frac{(-1)^{\frac{(n-1)(m-1)}{4}}\pi}{2^{n}(m-1)!} \times \left(n^{m-1}-n(n-2)^{m-1}+ \cdots \right). The case m \equiv n \equiv 0 \pmod 2 is done analogically.
    Hence, \displaystyle \int_{0}^{\infty} \frac{\sin^{n}x}{x^{n}}dx = \frac{\pi}{2^{n}(n-1)!} \times \left(n^{n-1}-n(n-2)^{n-1}+\cdots \right).


    By the way, Borwein have proved far more general result.
    Fix n+1 complex numbers z_{0},\cdots,z_{n}.
    Define a= (a_{1},\cdots,a_{n}), a_{i} \in \{-1,1\}; \displaystyle b_{a} = z_{0}+ \sum_{i} z_{i}a_{i}, and \displaystyle c_{a} = \prod_{i} a_{i}.
    It follows that \displaystyle \int_{0}^{\infty} \prod_{n=0}^{n} \frac{\sin z_{i}x}{x} = \frac{\pi}{2^{n+1}n!}\sum_{a} c_{a}b_{a}^{n}{\rm sign b_{a}}.
    Hence in our case, when z_{0}=z_{1}=\cdots=z_{n-1}=1, we have \displaystyle \int_{0}^{\infty} \left(\frac{\sin x}{x} \right)^{n}dx= \frac{\pi}{2^{n}(n-1)!}\sum_{0 \le i \le \frac{n}{2}} (-1)^{i} \dbinom{n}{i} (n-2i)^{n-1}.

    May I ask, why are we not allowed to use calculus in problem 155 in order to solve it in a minute?
    Spoiler:
    Show
    It is enough to note that p'(x) is divisible by (x-1)^{2} and (x+1)^{2}.
    He returns

    You used the same method I did for Problem 158 but the rest I did differently. I used differentiation under the integral sign for 156 which pops the solution out almost instantly (yours looks a little complicated!) Really like the elegance in 157 (I simply used the same approach for that one - was fairly simple but surely took a little longer than yours!)

    And I don't see what's going on in 159. Where does the first line come from? And how does it link to the second and third? The differentiation thing doesn't seem to make sense (though I'm sure it does), could you clarify it for me My solution looks so different to yours, it's in you form of two symmetrical series expression involving all the square numbers (and no factorials!) :lol: (I checked it though and it does work!)

    And yeah, as said above, it's far too easy and boring to do it that way

    Please post a load of really nice integrals! (None of that crap involving 2013s and stuff where you're never going to come across a similar thing :lol:)

    This one is really nice:

    Problem 160*/**/***

    Evaluate \displaystyle\int_0^1 \frac{\ln(x+1)}{x^2+1} dx
    Offline

    2
    ReputationRep:
    (Original post by Lord of the Flies)
    Accccch why did I have to miss the integrals? Well done on 159 Mlad, that's a nice solution.
    Post some! Post loads! I hereby declare the 29th May TSR Integration Day :pierre:
    Offline

    2
    ReputationRep:
    Very nice Mlad! I'm too slow again
    solution 156
    \displaystyle \int_{0}^{\pi} \ln(1-2a \cos x+a^{2})dx=\displaystyle \int_{0}^{\pi}ln((a-e^{ix})(a-e^{-ix}))dx=\displaystyle \int_{-\pi}^{\pi}ln(a-e^{ix})dx=\displaystyle \int_{-\pi}^{\pi}lnadx+\displaystyle \int_{-\pi}^{\pi}ln(1-e^{ix}/a)dx=2\pi lna
    the second integral can be seen to be zero by expanding the logarithm (|a|>1 guarantees convergence) and each term integrated is still just a function of e^ix which is 2pi periodic.
    <insert rigour here>
    Not sure if this works but I thought I'd give it a go
    Offline

    18
    ReputationRep:
    Solution 160

    \displaystyle f(t)=\int_0^1 \frac{\ln (1+tx)}{1+x^2}\,dx

    \displaystyle \begin{aligned}f'(t)=\int_0^1 \frac{x}{(1+tx)(1+x^2)}\,dx &= \frac{1}{t^2+1}\left[\int_0^1 \frac{t}{x^2+1}\,dx+\int_0^1 \frac{x}{x^2+1}\,dx-\int_0^1 \frac{t\,dx}{1+tx}\right]\\&=\frac{1}{t^2+1}\left[\frac{\pi t}{4}+\frac{1}{2}\ln 2 -\ln (1+t)\right]\end{aligned}

    \displaystyle f(1)= \int_0^1 \frac{\ln (1+t)}{1+t^2}\,dt=\frac{1}{2} \int_0^1 \frac{\pi t}{4(t^2+1)}+\frac{\ln 2}{2(t^2+1)}\,dt=\frac{\pi \ln 2}{8}

    Alternate solution to previous integrals:

    Solution 158

    x=\tan t:\;\;\displaystyle \int_0^{\infty} \frac{\ln (1+x^2)}{1+x^2}\,dx= -2\int_0^{\frac{\pi}{2}}\ln \cos t\,dt =\pi \ln 2

    Solution 157

    f(s)=\displaystyle \int_0^1 \frac{x^s-1}{\ln x}\,dx\Rightarrow f'(s)= \int_0^1 x^s\, dx=\frac{1}{s+1}\Rightarrow f(1)=\int_0^1 \frac{ds}{s+1}=\ln 2
    Offline

    2
    ReputationRep:
    (Original post by ben-smith)
    Very nice Mlad! I'm too slow again
    solution 156
    \displaystyle \int_{0}^{\pi} \ln(1-2a \cos x+a^{2})dx=\displaystyle \int_{0}^{\pi}ln((a-e^{ix})(a-e^{-ix}))dx=\displaystyle \int_{-\pi}^{\pi}ln(a-e^{ix})dx=\displaystyle \int_{-\pi}^{\pi}lnadx+\displaystyle \int_{-\pi}^{\pi}ln(1-e^{ix}/a)dx=2\pi lna
    the second integral can be seen to be zero by expanding the logarithm (|a|>1 guarantees convergence) and each term integrated is still just a function of e^ix which is 2pi periodic.
    <insert rigour here>
    Not sure if this works but I thought I'd give it a go
    Looks decent to me!

    I might as well add my solution into the soup :lol: ...

    Solution 156 (3)

    Let \displaystylef(a)=\int_{0}^{\pi} \ln(1-2a \cos x+a^{2})dx

    \displaystyle\Rightarrow \frac{\partial}{\partial a} f(a) = \int_{0}^{\pi} \frac{2a-2\cos(x)}{1-2a \cos x+a^{2}}dx

    Using the substitution u=\tan(\frac{1}{2}x):
    \displaystyle\Rightarrow \frac{\partial}{\partial a} f(a) =\frac{1}{a}\int_0^{\pi} dx + \frac{2(a-1)}{a(a+1)} \int_{0}^{\infty}\frac{1}{u^2+ \left(\frac{a-1}{a+1} \right)^2} dx=\frac{2\pi}{a},
    for |a|>1.

    ...also, \frac{\partial}{\partial a} f(a) = 0, for |a|<1 (as the arctangent would take a different value).
    Spoiler:
    Show

    Or I suppose you could just say f(1)=\int_0^{\pi} \ln(4\sin^2(\frac{x}{2})) dx = 0


    So as f(1)=0, \displaystyle f(a)=\int_1^a \frac{2\pi}{t} dt =2\pi \ln(a) for |a|>1. \square
    Offline

    2
    ReputationRep:
    (Original post by Lord of the Flies)
    Solution 160

    Solution 157
    Exactly what I did

    160 is from the Putnam exam (undergrad comp for the USA) and only 20 of the top 200 scorers got more than 2/10 marks on that question! :eek: No idea why this method isn't widely taught/used!

    POST SOME INTEGRALS!!!!!!!!!! :eek:
    Offline

    18
    ReputationRep:
    (Original post by Jkn)
    As f(0)=0, \displaystyle f(a)=\int_0^a \frac{2\pi}{t} dt =2\pi \ln(a) for |a|>1. \square
    This doesn't work, the integral clearly does not converge. You need to show that f(1) = 0 and integrate from 1 to a.
    Offline

    2
    ReputationRep:
    (Original post by Lord of the Flies)
    This doesn't work, the integral clearly does not converge. You need to show that f(1) = 0 and integrate from 1 to a.
    Typo fixed!
    Offline

    2
    ReputationRep:
    Problem 161**/***

    Let k be an integer greater than 1. Suppose a_0&gt;0 and \displaystyle a_{n+1}=a_{n}+\frac{1}{\sqrt[k]{a_{n}}} for n&gt;0

    Evaluate \displaystyle\lim_{n \to {\infty}} \frac{a^{k+1}_n}{n^k}
    Offline

    0
    ReputationRep:
    Problem 162 *

    If I drop a ball off the Eiffel tower facing east. Where does it land? (Neglect air resistance)
    Offline

    2
    ReputationRep:
    (Original post by bananarama2)
    Problem 162 *

    If I drop a ball off the Eiffel tower facing east. Where does it land? (Neglect air resistance)
    On the ground below. (?!?!?)
    Offline

    0
    ReputationRep:
    (Original post by Jkn)
    Solution 162

    On the ground below. (?!?!?)
    Directly below?
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you like to hibernate through the winter months?
    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
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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

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