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Integration and series

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data

    f9jn0p.jpg

    2. Relevant equations
    Gamma(n+1) = n(gamma(n))


    3. The attempt at a solution
    I have searched around trying to figure out how to do this. So far I have come out with:

    Gamma(n) = (n-1)(n-2)...
    Gamma(n+1)=n(n-1)(n-2)...2(1) otherwise known as Gamma(n+1)=n!

    Problem is... I don't fully understand what I did because I followed a like example. Can someone please help me out with this. Also I have not even attempted the second part, so ignore that for now.
     
  2. jcsd
  3. Sep 13, 2009 #2
    First part.

    [tex]\Gamma(n+1) = n\Gamma(n)[/tex]

    [tex]\Gamma(\frac{5}{2})=\Gamma(1+\frac{3}{2})=\frac{3}{2}\Gamma(1+\frac{1}{2})=\frac{3}{2}\frac{1}{2}\Gamma({\frac{1}{2})=\frac{3\sqrt{\pi}}{4}[/tex]



    This part is for the proof.


    Assume true for k

    [tex]\Gamma(k+\frac{1}{2})=\frac{1.3.5...(2k-1)\sqrt{\pi}}{2^k}[/tex]

    Then prove for k+1

    [tex]\Gamma(k+1+\frac{1}{2})=(k+\frac{1}{2})\left(\frac{1.3.5...(2k-1)\sqrt{\pi}}{2^k}\right)[/tex]


    Note the recursion relation, then,


    [tex]\Gamma(k+1+\frac{1}{2})=\frac{1}{2}(2(k+1)-1)\frac{1.3.5...(2k-1)\sqrt{\pi}}{2^k}=\frac{1.3.5...(2k-1)(2(k+1)-1)\sqrt{\pi}}{2^{k+1}}[/tex]


    Complete the proof by proving for k=1 and hence all positive integers.
     
    Last edited: Sep 13, 2009
  4. Sep 13, 2009 #3
    A little confused because I think you combined both parts of that. Both of those have nothing to do with eachother, but otherwise you helped me with the first one.
     
  5. Sep 13, 2009 #4
    I just worked out the value of 5/2 for the first one, and for the second I used the relationship we worked out at the beginning and proof by induction to answer the last part of it. I've separated them now too.
     
  6. Sep 14, 2009 #5
    Thank you for the help, but I still don't see the proof from (1.3.5...(2n-1))/2^n == (2n)!/((4^n)(n!))

    It's fine. I'm going to move on and get help from a classmate when the time comes. Thanks for the help though!
     
  7. Sep 14, 2009 #6
    I'll clarify:

    The question asks you to prove that

    [tex] \Gamma(n+\frac{1}{2}) = \frac{1.3.5.7...(2n-1)}{2^n}\sqrt{\pi}[/tex]

    We can sort of see this in the 5/2 example, all the numerators are odd numbers and the denominator is always 2, so the odd numbers will multiply successively n-times on the top and 2 will be of the power n on the bottom, and always at the end is the square root of pi. To prove this first "assume" that it is true for any number say, k.

    [tex]\Gamma(k+\frac{1}{2}) = \frac{1.3.5.7...(2k-1)}{2^k}\sqrt{\pi}[/tex]

    We assume this is true, and now we try to prove it for k+1, we want the next odd number to appear on the numerator and for another 2 to be multiplied in the denominator (or half on the numerator) to make the right hand side look like this:


    [tex]\frac{1.3.5.7...(2k-1)(2(k+1)-1)}{2^{k+1}}\sqrt{\pi}[/tex]


    Once this happens all we have to do is show that the equation holds for n=1 and then we get it true for n=1,n=1+1 etc.

    From the relationship we established,

    [tex] \Gamma(n+1)=n\Gamma(n) [/tex]

    We can add 1 quite nicely:

    Remember that we have assumed that

    [tex]\Gamma[k+\frac{1}{2}] = \frac{1.3.5.7...(2k-1)}{2^k}\sqrt{\pi}[/tex]

    So for k+1:

    [tex]\Gamma(k+1+\frac{1}{2}) = (k+\frac{1}{2})\Gamma(k+\frac{1}{2}) [/tex]

    (k+1/2 is n here.)

    [tex]\Gamma(k+1+\frac{1}{2}) = (k+\frac{1}{2})\frac{1.3.5.7...(2k-1)}{2^k}\sqrt{\pi} [/tex]

    Now all we need to do is rearrange it

    [tex](k+\frac{1}{2})\frac{1.3.5.7...(2k-1)}{2^k}\sqrt{\pi} = \frac{1}{2}(2k+1)\frac{1.3.5.7...(2k-1)}{2^k}\sqrt{\pi} [/tex]


    Notice that we have a half and a 2k+1, which if you look is actually (2(k+1)-1), our next odd number.

    [tex]\frac{1}{2}(2k+1)\frac{1.3.5.7...(2k-1)}{2^k}\sqrt{\pi} = \frac{1.3.5.7...(2k-1)(2(k+1)-1)}{2^{k+1}}\sqrt{\pi}[/tex]

    So if it's true for k it is true for k+1, if you prove it for k=1 you have proved it then for all positive n.

    [tex] \Gamma(1+\frac{1}{2}) = \frac{\sqrt{\pi}}{2} = \frac{1.\sqrt{\pi}}{2^1} [/tex]

    [tex] \Rightarrow \Gamma(n+\frac{1}{2}) = \frac{1.3.5.7...(2n-1)}{2^n}\sqrt{\pi}[/tex]

    For positive n.


    Edit: The final part of the proof would just be manipulating the expression to match the RHS.
     
    Last edited: Sep 14, 2009
  8. Sep 14, 2009 #7
    Thanks, I understood it much better from that.
     
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