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

    11
    ReputationRep:
    (Original post by Jkn)
    It seemed odd that it could be stated in such a simple form, especially when he asked for a series solution ;0
    Yeah, but then again there were a couple of little tricks to perform.
    I realise that in the solution I was a bit lazy with the tex, so some people may have trouble following, but hey ho
    Spoiler:
    Show
    I know why LOTF does it :ahee:
    Offline

    2
    ReputationRep:
    (Original post by joostan)
    Yeah, but then again there were a couple of little tricks to perform.
    I realise that in the solution I was a bit lazy with the tex, so some people may have trouble following, but hey ho
    Spoiler:
    Show
    I know why LOTF does it :ahee:
    I thought your technical details seemed sufficiently thorough (I may have used a similar amount if not slightly less?) though I am trying to get into the habit of naming all the theorems and representations that I am using (i.e. your series representation etc..)
    Offline

    11
    ReputationRep:
    (Original post by Jkn)
    I thought your technical details seemed sufficiently thorough (I may have used a similar amount if not slightly less?) though I am trying to get into the habit of naming all the theorems and representations that I am using (i.e. your series representation etc..)
    I tend to like to spell things out, for readers, though I often do less working on paper
    Offline

    2
    ReputationRep:
    Would anyone benefit from small hints to some of the problems I've set recently? A few of them follow trivially from theorems that can be quoted (once found). Also, on the problems where I have said that you must prove all theorems used, do feel that you can 'look them up' so long as they seem like the kind of thing that you would be expected to memorize in a course (the so-called 'book work').

    Yet another that requires a simple yet rather advanced 'trick':

    Problem 288
    ***

    Evaluate \displaystyle \int_0^{\infty} \frac{\gamma x + \log \Gamma(1+x)}{x^{5/2}} \ dx

    Problem 289*/**

    Let x, y and z denote positive integers (which can be extended without any loss of generality from the case whereby they denote non-zero integers).

    Find the general solution to the equation \displaystyle x^2+y^2=z^2 (the 'Pythagorean Triples').

    Hence show that \displaystyle x^4+y^4 \not= z^4 ('Fermat's Last Theorem' in the case n=4).

    Can you generalize the result to other classes of exponents in the set of natural numbers? How many other ways can you find to prove the above result? (bonus).
    Offline

    3
    ReputationRep:
    Problem 290*/**

    Suppose a,b are fixed points of f(x) (that is, f(a)=a, f(b)=b)). Prove

     \displaystyle\int_{a}^{b}f(x) + f^{-1}(x)\ dx = b^2 - a^2
    Offline

    2
    ReputationRep:
    (Original post by FireGarden)
    Problem 290*/**

    Suppose a,b are fixed points of f(x) (that is, f(a)=a, f(b)=b)). Prove

     \displaystyle\int_{a}^{b}f(x) + f^{-1}(x)\ dx = b^2 - a^2
    Have you had a go at deriving the fractional derivative of x^a (formally) yet?

    Solution 290

    Using the property that a function's inverse can be found by reflecting the function in the line y=x in the Cartesian plane, we deduce that the area between each function and the line x=y are equal between the values for which f(x)=f^{-1}(x):

    \displaystyle \begin{aligned} \int_a^b (x-f(x)) \ dx = \int_a^b (f^{-1}(x)-x) \ dx \iff \int_a^b ( f(x)+f^{-1}(x) ) \ dx = \int_a^b 2x \ dx = b^2-a^2 \ \square \end{aligned}

    Also, there are some conditions that need to be placed on f for this to be true. The inverse must exist, for example.
    Offline

    3
    ReputationRep:
    (Original post by Jkn)

    Solution 290

    Using the property that a function's inverse can be found by reflecting the function in the line y=x in the Cartesian plane, we deduce that the area between each function and the line x=y are equal between the values for which f(x)=f^{-1}(x):

    \displaystyle \begin{aligned} \int_a^b (x-f(x)) \ dx = \int_a^b (f^{-1}(x)-x) \ dx \iff \int_a^b ( f(x)+f^{-1}(x) ) \ dx = \int_a^b 2x \ dx = b^2-a^2 \ \square \end{aligned}

    Also, there are some conditions that need to be placed on f for this to be true. The inverse must exist, for example.
    Nice. Now generalise by finding the result if there are no fixed points (or at least a and b are not such points).
    Offline

    0
    ReputationRep:
    (Original post by FireGarden)
    Problem 290*/**

    Suppose a,b are fixed points of f(x) (that is, f(a)=a, f(b)=b)). Prove

     \displaystyle\int_{a}^{b}f(x) + f^{-1}(x)\ dx = b^2 - a^2
    Geometrical proof:
    Click image for larger version. 

Name:	89490965d7a87426cf38a5002807f20f.png 
Views:	204 
Size:	7.4 KB 
ID:	234458

    \displaystyle \int_a^b f(x) dx=A_5+A_4

    \displaystyle \int_a^b f^{-1}(x) dx=A_5+A_4+A_3+A_2

    Obviously, as the inverse function is a reflection of the original function in the line y=x, A_2=A_3

    Thus \displaystyle \int_a^b f(x)+f^{-1}(x) dx= 2A_5+2(A_4+A_3)=2(b-a)a+(b-a)^2=2ba-2ba+a^2-2a^2+b^2=b^2-a^2


    Where have I not been rigorous here (I assume I've been non rigorous somewhere).

    On reflection, this is pretty much the same as Jkn's.
    Offline

    2
    ReputationRep:
    (Original post by FireGarden)
    Nice. Now generalise by finding the result if there are no fixed points (or at least a and b are not such points).
    Hmm, well you lose the simplicity. We could do some clever things involving differentiating with respect to y in the Cartesian place (where we set y=f(x) etc..) but it didn't look as though it was likely to simplify to the given result

    Problem 290 (extension):


    We can add and subtract the additional areas in order to 'adjust' and use the fixed point property:

    \displaystyle \begin{aligned}\int_a^b (f(x)+f^{-1}(x)) \ dx = \left(b^2+\int_b^{f(b)} (x-f(x)) \ dx \right) - \left(a^2+\int_a^{f(a)} (x-f(x)) \ dx \right) \end{aligned}

    which simplifies to any of the special cases when we have that a and b are fixed points.
    Offline

    0
    ReputationRep:
    (Original post by Jkn)

    Problem 286***

    Evaluate \displaystyle \prod_{n=1}^{9} \Gamma \left( \frac{n}{10} \right)

    Can the result be easily generalised in any interesting way? If so, provide proof of your assertions.
    Solution 286

    \displaystyle \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}
    So almost all of the terms disappear, giving

    \Gamma(\frac{5}{10}) \prod_{m=1}^4\frac{\pi}{\sin(\pi \frac{m}{10})}

    Using the result again, we have

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

    \sqrt{\pi} \prod_{m=1}^4\frac{\pi}{\sin(\pi \frac{m}{10})}

    In general,

    \prod_{m=1}^{2n-1}\Gamma(\frac{m}{2n})= \sqrt{\pi} \prod_{m=1}^{n-1}\frac{\pi}{\sin(\pi \frac{m}{2n})}

    Proof:
    \prod_{m=1}^{2n-1}\Gamma(\frac{m}{2n})= \sqrt{\pi}\prod_{m=1}^{n-1}\Gamma(\frac{m}{2n})\prod_{m=n  +1}^{2n-1}\Gamma(\frac{m}{2n})= \sqrt{\pi}\prod_{m=1}^{n-1}\Gamma(\frac{m}{2n})\prod_{m=2  n-(n-1)}^{2n-(1)}\Gamma(\frac{m}{2n})
    = \sqrt{\pi}\prod_{m=1}^{n-1}\Gamma(\frac{m}{2n})\Gamma( \frac{2n-m}{2n})
    The result follows
    Offline

    0
    ReputationRep:
    By the way, here http://dlmf.nist.gov/5 is an excellent selection of properties of the Gamma and Beta function (not proved).
    • PS Helper
    • Study Helper
    Offline

    13
    ReputationRep:
    PS Helper
    Study Helper
    (Original post by henpen)
    On reflection…
    Offline

    0
    ReputationRep:
    (Original post by Smaug123)
    I have a knack for unintentional puns.
    Offline

    2
    ReputationRep:
    (Original post by henpen)
    Problem 228**

    Calculate \large \sum_{n=4}^{\infty}\frac{1}{n H_nH_{n-1}}, where H_n=\sum_{k=1}^{n}\frac{1}{k}.

    Hint:
    Spoiler:
    Show
    The answer's a rational number.
    Not sure how I didn't spot that this one hadn't been solved (has it?)

    Solution 228

    Using the method of differences and the given definition of the nth harmonic number:

     \displaystyle \sum_{n=4}^{\infty} \frac{1}{n H_n H_{n-1}}= \sum_{n=4}^{\infty} \frac{H_n - H_{n-1}}{H_n H_{n-1}} = \sum_{n=4}^{\infty} \frac{1}{H_{n-1}}-\frac{1}{H_{n}}=\frac{1}{H_3}- \underbrace{ \lim_{n \to \infty} \frac{1}{H_n}}_{\alpha} = \frac{6}{11}

    For the sake of rigour:

    \displaystyle \left|H_{2n}-H_n \right| = \left|\frac{1}{2n}+\frac{1}{2n-1}+ \cdots + \frac{1}{n+1} \right|> \left|n \times \frac{1}{2n} \right|=\frac{1}{2} and so H_n is not a Cauchy sequence and hence diverges.
    We now deduce that \displaystyle \alpha = \lim_{n \to \infty} \frac{1}{H_n} = 0. The result follows \square
    Offline

    2
    ReputationRep:
    (Original post by henpen)
    Solution 286
    This is not what I was expecting -the result can be simplified nicely.

    Also, an extremely simple closed form for \displaystyle \prod_{k=1}^{n-1} \Gamma \left( \frac{k}{n} \right) exists!
    Offline

    3
    ReputationRep:
    (Original post by Jkn)
    Hmm, well you lose the simplicity. We could do some clever things involving differentiating with respect to y in the Cartesian place (where we set y=f(x) etc..) but it didn't look as though it was likely to simplify to the given result

    Problem 290 (extension):


    We can add and subtract the additional areas in order to 'adjust' and use the fixed point property:

    \displaystyle \begin{aligned}\int_a^b (f(x)+f^{-1}(x)) \ dx = \left(b^2+\int_b^{f(b)} (x-f(x)) \ dx \right) - \left(a^2+\int_a^{f(a)} (x-f(x)) \ dx \right) \end{aligned}

    which simplifies to any of the special cases when we have that a and b are fixed points.
    This is not what should be found. The answer, if it helps, is of the form xb+ya; the fixed points were the special case where x=b and y=a.
    Offline

    2
    ReputationRep:
    (Original post by FireGarden)
    This is not what should be found. The answer, if it helps, is of the form xb+ya; the fixed points were the special case where x=b and y=a.
    I don't think that's right. If what I have is right then what you're saying is that we have an expression for the definite integral of any function between the limits a and f(a).

    Also, if it can be properly evaluated for any limits, then we could immediately find things like \int \sin x + \arcsin x \ dx which seems unlikely (not to mention so powerful and fundamental that I would be surprised that I don't know it already). :P
    • PS Helper
    • Study Helper
    Offline

    13
    ReputationRep:
    PS Helper
    Study Helper
    (Original post by Jkn)
    I don't think that's right. If what I have is right then what you're saying is that we have an expression for the definite integral of any function between the limits a and f(a).

    Also, if it can be properly evaluated for any limits, then we could immediately find things like \int \sin x + \arcsin x \ dx which seems unlikely (not to mention so powerful and fundamental that I would be surprised that I don't know it already). :P
    Well, the arcsine integral's not too bad :P but I see what you mean - it's pretty likely not linear.
    Offline

    2
    ReputationRep:
    (Original post by Smaug123)
    Well, the arcsine integral's not too bad :P but I see what you mean - it's pretty likely not linear.
    Of course it's easy to do (IBP) but it's an example of an integral of that form with a 'messy' answer.
    Offline

    0
    ReputationRep:
    (Original post by Jkn)
    This is not what I was expecting -the result can be simplified nicely.

    Also, an extremely simple closed form for \displaystyle \prod_{k=1}^{n-1} \Gamma \left( \frac{k}{n} \right) exists!


    Solution 286

    Standing on the shoulders of giants (here, Mladenov, solution 24), we have

    \displaystyle \prod_{m=1}^{r}\sin\left(\frac{ \pi m}{2r+1} \right)=\frac{\sqrt{2r+1}}{2^{r}  }



    For even n,

    \displaystyle \prod_{m=1}^{2r}\Gamma \left(\frac{m}{2r+1} \right)=\displaystyle \prod_{m=1}^{r}\frac{\pi}{\sin \left(\frac{\pi m}{2r+1} \right)} \right)= \frac{\pi^r 2^{r}}{\sqrt{2r+1}}

    For odd n, I'll try to generalise 24 later, but I need to sleep now.
 
 
 
  • 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.