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Binomial expansion

  1. Nov 4, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the coefficient of x^n in the expansion of ( 1 + x/1! + x^2/2! + x^3/3! + ......... + x^n/n! )^2 .

    2. Relevant equations


    3. The attempt at a solution
    At first glance, this looks like the polynomial form of e^x, but the expansion of e^x goes to infinity, so any use of that seems ruled out. Furthermore, I can't try expanding this or writing it in terms of the rth term and equating the power of x to n and doing something with that. Basically, I don't know how to begin.
     
  2. jcsd
  3. Nov 4, 2014 #2

    Dick

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    Ask yourself whether the terms that are missing from the expansion of e^x would make any difference to the coefficient of the x^n term.
     
  4. Nov 4, 2014 #3

    PeroK

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    Why not work it out for n = 2, n = 3 and possibility n =4 and see if you can work out what's happening.
     
  5. Nov 4, 2014 #4
    I tried for n=2, I got 2. However, looking at it, I don't see any pattern that might be of much use for figuring out x^n.
     
  6. Nov 4, 2014 #5

    Dick

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    Try it for the other ones and compare with the coefficient of x^n in (e^x)^2.
     
  7. Nov 4, 2014 #6

    Ray Vickson

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    Think of putting two copies of your polynomial side-by-side; these are the two factors in the squared expression. The coefficient of x^k in the product consists of a sum of coefficient products: 1= 1/0! from factor 1 and 1/k! from factor 2, or 1/1! from factor 1 and 1/(k-1)! from factor 2, or 1/2! from factor 1 and 1/(k-2)! from factor 2, or .... or 1/k! from factor 1 and 1/0! from factor 2. You want the sum of all the two-by-two products of these numbers.
     
  8. Nov 4, 2014 #7
    If we write the squaring multiplication with one of the terms reversed, it's may make it easier to see the components of the x^n coefficient in the answer.

    ##
    \begin{matrix}
    ( & 1 & + &x/1! & + &x^2/2! & + & \dots & + & x^{n-2}/(n-2)! & + &x^{n-1}/(n-1)! & + & x^n/n! & ) & \times \\
    ( & x^n/n! & + & x^{n-1}/(n-1)! & + & x^{n-2}/(n-2)! & + & \dots & + & x^2/2! & + & x/1! & + & 1 & ) &
    \end{matrix}
    ##
     
  9. Nov 4, 2014 #8

    Dick

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    Sure, you can write down the series. But it's also not terribly hard just to write down the number that it must sum to.
     
  10. Nov 6, 2014 #9
    So I want to sum the following-
    ( 1/0! * 1/n! ) + ( 1/1! * 1/(n-1)! ) + ...... + ( 1/n! * 1/0! )
    How do I sum that up to get the answer 2^n/n! ?
     
  11. Nov 6, 2014 #10

    Dick

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    What powers of x are in the difference between your product and the power series expansion of e^(2x)?
     
  12. Nov 6, 2014 #11
    Does 1/(k!(n-k)!) look familiar to you at all? How about n!/(k!(n-k)!) ?
     
  13. Nov 6, 2014 #12
    .
    Yes of course. So I multiply and divide by n!, and I get ( C0+C1+...Cn )/n! Which is just 2^n/n!

    Thank you everyone :D
     
  14. Nov 6, 2014 #13

    Dick

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    Yes, but you didn't even have to sum the series. Your product is identical to the power series for e^(2x) up to order n. The nth term of that is 2^n/n!.
     
  15. Nov 6, 2014 #14
    That's pretty neat. I haven't formally learnt power series and stuff, so I didn't know that.
     
  16. Nov 6, 2014 #15
    I just googled operations on power series, and that's definitely a much better way of doing this problem.
     
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