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    (Original post by Jkn)
    Problem 278***

    Evaluate \displaystyle \lim_{x \to \infty} \frac{\Gamma(x+1)}{x^x e^{-x} \sqrt{2 \pi x}} for x \in \mathbb{R}.

    Comment also on what happens when x \in \mathbb{C} and |x| \to \infty.
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    The denominator is an asymptotic expression for x!, while the numerator is exactly x!, so the limit is 1.

    (Sorry, doing this in a rush, I saw a question I could do with no thought at all and did that bit :P)
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    (Original post by bogstandardname)
    Is the supposed to be a natural log?
    I assume so.
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    I derived an ugly closed-form series representation for it as well as a link with the incomplete Gamma function but will leave it to someone else as I am not quite sure I have the form that he intended to be found.
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    (Original post by Smaug123)
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    The denominator is an asymptotic expression for x!, while the numerator is exactly x!, so the limit is 1.

    (Sorry, doing this in a rush, I saw a question I could do with no thought at all and did that bit :P)
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    Of course the question is to prove that very stirling formula

    Oh and the denominator is not exactly x! The relation is merely asymptotic (though it can, as you know, be modified to give upper and lower bounds).

    Perhaps it would have been clearer if I gave the limit and said to prove it... or if I gave the factorial function rather than the Gamma function.. but I thought it was better to leave it in a more 'raw' state.
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    (Original post by Jkn)
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    Of course the question is to prove that very stirling formula

    Oh and the denominator is not exactly x! The relation is merely asymptotic (though it can, as you know, be modified to give upper and lower bounds).

    Perhaps it would have been clearer if I gave the limit and said to prove it... or if I gave the factorial function rather than the Gamma function.. but I thought it was better to leave it in a more 'raw' state.
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    The proof of Stirling was marked as non-examinable in Probability lectures, so I didn't learn it, being swamped with analysis and groups theorems at the time :P I'd give it a go at re-deriving it, but I've got too much calculus of variations homework to do at the moment
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    (Original post by Jkn)


    Problem 281
    *

    Find an infinite series representation for \displaystyle \cos \frac{\pi}{x} and \displaystyle \sin \frac{\pi}{x} involving non-trancendental numbers (ignoring, of course, the possible transcendence of x). A transcendental number is a number that cannot be expressed as the root of a polynomial equation with integer coefficients.
    I think you mean rational coefficients
    Took a crack at the limits one but copied out the question wrong. :facepalm: Back to square one.
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    (Original post by Smaug123)
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    The proof of Stirling was marked as non-examinable in Probability lectures, so I didn't learn it, being swamped with analysis and groups theorems at the time :P I'd give it a go at re-deriving it, but I've got too much calculus of variations homework to do at the moment
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    Good luck bro. I'd say it is probably one of the harder problems in my fiendish list

    Oo is that that 'Variational Principles' stuff?
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    (Original post by joostan)
    I think you mean rational coefficients
    Took a crack at the limits one but copied out the question wrong. :facepalm: Back to square one.
    I conjecture that any number that is a root of a polynomial with rational coefficients is also the root of a polynomial with integer coefficients. Prove me wrong?
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    (Original post by Jkn)
    <this was spoilered>Oo is that that 'Variational Principles' stuff? </spoilered>
    Yep, it is - the stuff done in Part IB is pretty basic, but I just can't get the Euler-Lagrange equations to stick in my head, and they take me about two minutes to derive (during which time I could have solved a question if I knew the E-L equations). I'm resorting to Anki to learn them by heart :P
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    (Original post by Jkn)
    I conjecture that any number that is a root of a polynomial with rational coefficients is also the root of a polynomial with integer coefficients. Prove me wrong?
    Maybe joostan thought you meant "monic polynomial" or something
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    (Original post by Jkn)
    I conjecture that any number that is a root of a polynomial with rational coefficients is also the root of a polynomial with integer coefficients. Prove me wrong?
    Good point . . .

    (Original post by Smaug123)
    Maybe joostan thought you meant "monic polynomial" or something
    That's what I was thinking, but I wasn't really thinking
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    (Original post by bogstandardname)
    Is the supposed to be a natural log?
    Yup! (we could, if we wished here, use the Gamma function)
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    (Original post by Smaug123)
    Yep, it is - the stuff done in Part IB is pretty basic, but I just can't get the Euler-Lagrange equations to stick in my head, and they take me about two minutes to derive (during which time I could have solved a question if I knew the E-L equations). I'm resorting to Anki to learn them by heart :P
    Hahaha #CambridgeMathsLAD :lol:

    No idea what any of that stuff is bro!

    Anki?
    (Original post by Hasufel)
    Yup! (we could, if we wished here, use the Gamma function)
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    Do you mean the incomplete* Gamma function? (as said a few posts up)
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    (Original post by Jkn)
    <spoilered>Oo is that that 'Variational Principles' stuff? </spoilered>
    The highlight of the course is when we prove that in the problem of finding the function that minimises the distance between two points, a straight line extremises that distance, and it is a local minimum. I don't think we ever proved that it was a global minimum.
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    (Original post by Jkn)
    Hahaha #CambridgeMathsLAD :lol:

    No idea what any of that stuff is bro!

    Anki?
    Euler-Lagrange is the differential equation which you use to extremise a functional (in the same way as you use the differential equation \dfrac{df}{dx} = 0 to extremise a function).
    Seriously, IB Variational Principles contains about five theorems, and the rest of it is examples :P
    Anki is a spaced-repetition program (like Memrise, if you ever used that) - it's designed for optimal rote-learning of small facts. I've used it to rote-learn a proof of the Sylow theorems, for example (by splitting the three theorems up into 51 short linked facts).
    I hasten to add that I'm not one of these people that finds Cambridge maths basic in general - I struggle just as much as anyone else from the look of your preparation, the first year will essentially be entirely stuff you already know (possibly excepting groups)…
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    (Original post by Jkn)
    Hahaha #CambridgeMathsLAD :lol:

    No idea what any of that stuff is bro!

    Anki?

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    Do you mean the incomplete* Gamma function? (as said a few posts up)
    Yes, indeed.
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    (Original post by Hasufel)
    PROBLEM 277

    Derive (pardon the pun) a series representation, excluding the constant of integration, for:

    \displaystyle \int x^{n}(In(x))^{m}dx

    (Hint: one way is to incorporate derivatives into your answer)
    Not at all sure on this one, I had to look up a few properties and what not, which I don't think I can prove but:
    As that's apparently \ln(x)
    Solution 277:
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    I = \displaystyle\int x^n(\ln(x))^m \ dx
    Let:
    u=-\ln(x) \Rightarrow dx=-e^{-u}du

\Rightarrow I = -\displaystyle\int e^{-(n+1)u}(-u)^m \ du = (-1)^{m+1}\displaystyle\int e^{-(n+1)u}u^m \ du
    Let:
    t=(n+1)u \Rightarrow dt=(n+1)du

\Rightarrow I = \left(\dfrac{-1}{n+1} \right)^{m+1}\displaystyle\int e^{-t}t^m \ dt

\Rightarrow I = -\left(\dfrac{-1}{n+1} \right)^{m+1}\Gamma (m+1,t)
    Since:
    \Gamma (m+1,t)= \Gamma (m+1) -\gamma(m+1,t)

\Rightarrow I = \left( \dfrac{-1}{n+1} \right)^{m+1} \left( t^{m+1} \displaystyle\sum_{k=0}^{\infty}  (-1)^k \dfrac{t^k}{k!(m+k+1)}-\Gamma(m+1) \right)
    Subbing in for t yields: (please excuse blunders )
    t=\dfrac{-1}{\ln(x)}

\Rightarrow I = \left( \dfrac{-1}{n+1} \right)^{m+1} \left( \dfrac{-\ln(x)}{n+1} \right)^{m+1}  \displaystyle\sum_{k=0}^{\infty}  (-1)^k  \dfrac{\left(\dfrac{\ln(x)}{n+1} \right)^k}{k!(m+k+1)}-\Gamma(m+1)\right)

\Rightarrow I=\left(\dfrac{-1}{n+1}\right)^{2(m+1)}(\ln(x))^  {m+1}  \displaystyle\sum_{k=0}^{\infty}  (-1)^{2k}  \dfrac{\ln(x)^k}{(n+1)^kk!(m+k+1  )} - \left( \dfrac{-1}{n+1} \right)^{m+1}\Gamma(m+1)


    Where m is a positive integer:
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    I believe that:
    I = \displaystyle\sum_{k=0}^m \left( \dfrac{(-1)^k m!\ln(x)^{m-k}x^{n+1}}{(m-k)!(n+1)^{k+1}} \right)
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    Solution 285

    Firstly  \dfrac{d}{dx}(x^n) = nx^{n-1}

    Repeating,  \dfrac{d^a}{dx^a}(x^n) = \dfrac{n!}{(n-a)!}x^{n-a}

    Generalise for any real a with gamma function;  \dfrac{d^a}{dx^a}(x^n) = \dfrac{\Gamma(n+1)}{\Gamma(n-a+1)}x^{n-a}

    Set  a = \frac{1}{2} and we get  H(x^n) = \dfrac{\Gamma(n+1)}{\Gamma(n+ \frac{1}{2})}x^{n- \frac{1}{2}}

    Repeating H twice gives;

     H^2(x^n) = H(\dfrac{\Gamma(n+1)}{\Gamma(n+ \frac{1}{2})}x^{n- \frac{1}{2}}) 



\ \\



= \dfrac{\Gamma(n+1)}{\Gamma(n+ \frac{1}{2})} H(x^{n-\frac{1}{2}})



\ \\



= \dfrac{\Gamma(n+1)}{\Gamma(n+  \frac{1}{2})}\dfrac{\Gamma(n+ \frac{1}{2})}{\Gamma(n)}x^{n-1}

    =nx^{n-1}

    And so we find  H^2(x^n) = D(x^n)
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    Problem 287 */***

    Prove that \displaystyle 2(ab+bc+ca)^4=a^4(b-c)^4+b^4(c-a)^4+c^4(a-b)^4 when a+b+c=0.

    Another problem that came from an interesting place and has a rather insane generalisation. As usual, the tall tale will follow a correct proof!
    (Original post by Smaug123)
    Euler-Lagrange is the differential equation which you use to extremise a functional (in the same way as you use the differential equation \dfrac{df}{dx} = 0 to extremise a function).
    Seriously, IB Variational Principles contains about five theorems, and the rest of it is examples :P
    Anki is a spaced-repetition program (like Memrise, if you ever used that) - it's designed for optimal rote-learning of small facts. I've used it to rote-learn a proof of the Sylow theorems, for example (by splitting the three theorems up into 51 short linked facts).
    I hasten to add that I'm not one of these people that finds Cambridge maths basic in general - I struggle just as much as anyone else from the look of your preparation, the first year will essentially be entirely stuff you already know (possibly excepting groups)…
    Sounds a bit brutal, I may take a peak later in the summer.

    Oh right, yeah I've used similar things before. They tended to help with boring slogs (spanish vocal, stats definitions, physics definitions, etc..), I used several iPhone apps.

    Hmm, I doubt that! I still have {groups}, matrices, a lot of vector stuff, applications of s. relativity to solving problems, all of probability, etc.. from 1A that I haven't really scratched the surface on! My fear is that this would mean I have to spend my whole first year focusing on the areas of maths I disliked so much that I didn't touch on it over the summer! ..like I doubt I will have much fun with integration over the next year or two! :lol:
    (Original post by joostan)
    I = -\left(\dfrac{-1}{n+1} \right)^{m+1}\Gamma (m+1,t)
    Nice stuff man.

    Ah, that's what I got!
    I = \displaystyle\sum_{k=0}^m \left( \dfrac{(-1)^k m!\ln(x)^{m-k}x^{n+1}}{(m-k)!(n+1)^{k+1}} \right)
    Looks similar to the series I derived using IBP => recurrence relation (functional equation) => solve. Probably the same but my notebooks/ridiculous piles of paper are not beside me atm! :lol:
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    (Original post by Jkn)

    Nice stuff man.

    Ah, that's what I got!

    Looks similar to the series I derived using IBP => recurrence relation (functional equation) => solve. Probably the same but my notebooks/ridiculous piles of paper are not beside me atm! :lol:
    Cheers
    Yeah I did it by IBP at first (hence deriving them is a positive integer series) but then I realised it was for all values, so along came the gamma :cool:
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    (Original post by FireGarden)
    Generalise for any real a with gamma function;  \dfrac{d^a}{dx^a}(x^n) = \dfrac{\Gamma(n+1)}{\Gamma(n-a+1)}x^{n-a}
    Nice stuff, welcome back!

    BUT you are only half-done (I am trying to force people to provide elementary solutions, sorry! ) as the following issues arise: if the gamma function is the solution to the 'smooth curve passion though the factorials' problem, then why can it be applied here also? I there any guarantee that the fractional derivate (something fixed) is going to se the same function that was only created to pass through the certain integer points? (why must the fractional derivative be smooth anyway?) Also, the gamma function is not the only such function that passes through these points (as I recently found out!) etc.. (you get the idea).

    Anyway, I conclude that it is non-obvious (so we require a proper mathematical derivation). Enjoy

    Edit: Oh and the form given for the half derivative can be simplified by using Gamma function properties (which you may use without proof as the property has been derive before on this thread).
    (Original post by joostan)
    Cheers
    Yeah I did it by IBP at first (hence deriving them is a positive integer series) but then I realised it was for all values, so along came the gamma :cool:
    It seemed odd that it could be stated in such a simple form, especially when he asked for a series solution ;0
 
 
 
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