Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 1983 BC 5 part C

  1. Jun 5, 2008 #1
    Sum of a Series - 1983 BC 5 part C

    I don't have the actual problem (this is for a friend) but this is what I could gather from what she was saying.

    1. The problem statement, all variables and given/known data
    If [tex]f(x) = \sum^{\infty}_{n=0}a_{n}x^n[/tex] find the value of f'(1)

    [tex]a_{0} = 1[/tex] and [tex]a_{n} = (7/n)a_{n-1}[/tex]


    2. Relevant equations
    None maybe?


    3. The attempt at a solution
    Ok so after differentiating,
    [tex]f(x) = \sum^{\infty}_{n=1}n a_{n}x^{n-1}[/tex]
    Writing out the terms and subbing 1 for x got me to,
    [tex]f(x) = \sum^{\infty}_{n=1}7^n/(n-1)![/tex]
    or
    [tex]f(x) = \sum^{\infty}_{n=1}n 7^n/(n)![/tex]

    This was as far as I was able to get. Mathematica tells me that the answer is 7e^7.
    I know that [tex]\sum^{\infty}_{n=1}n/(n)! = e[/tex] and that [tex] \sum^{\infty}_{n=0}7^n/(n)! = e^7[/tex] (which doesn't really help because we're
    starting at 1). But with both parts in there I'm not sure what to do
     
    Last edited: Jun 5, 2008
  2. jcsd
  3. Jun 6, 2008 #2
    Ok I think I got it.
    If [tex]f(x) = \sum^{\infty}_{n=0}a_{n}x^n[/tex] find the value of f'(1)
    [tex]a_{0} = 1[/tex] and [tex]a_{n} = (7/n)a_{n-1}[/tex]

    [tex]f(x) = 1/0! + 7x/1! + 7^2x^2/2! + 7^3x^3/3!... \sum^{\infty}_{n=0}\frac{7^nx^n}{n!}[/tex]
    [tex]f'(x) = 0 + 7/0! + 7^2x/1! + 7^3x^2/2! ... \sum^{\infty}_{n=1}\frac{7^n x^{n-1}}{(n-1)!} = \sum^{\infty}_{n=1}\frac{n7^nx^{n-1}}{n!}[/tex]
    [tex]f'(1) = 7/0! + 7^2/1! + 7^3/2! ... \sum^{\infty}_{n=1}\frac{n7^n }{n!} = 7(1 + 7 + 7^2/2! + 7^3/3!...)[/tex]
    [tex]e^x = 1 + x + x^2/2! + x^3/3! +x^4/4!... \sum^{\infty}_{n=0}\frac{x^n }{n!}[/tex]
    [tex]e^7 = 1 + 7 + 7^2/2! + 7^3/3! + 7^4/4!... \sum^{\infty}_{n=0}\frac{7^n }{n!}[/tex]
    Recall that:
    [tex]f'(1) = 7/0! + 7^2/1! + 7^3/2! ... \sum^{\infty}_{n=1}\frac{n7^n }{n!} = 7(1 + 7 + 7^2/2! + 7^3/3!...)[/tex]
    and compare to e^7:
    Which gives:
    7e^7

    Sorry if that is a bit hard to follow.
     
    Last edited: Jun 6, 2008
  4. Jun 6, 2008 #3
    Ohh I can't edit it anymore...I think it is correct but I wanted to change the forms of some stuff.

    I like this better, for example:
    [tex]f'(1) = 7/0! + 7^2/1! + 7^3/2! ... \sum^{\infty}_{n=1}\frac{7^n }{(n-1)!} = 7(1 + 7 + 7^2/2! + 7^3/3!...)[/tex]
     
  5. Jun 6, 2008 #4

    Defennder

    User Avatar
    Homework Helper

    I got the same answer as you, 7e^7.

    By the power series: [tex]e^7 = 1 + 7 + \frac{7^2}{2!} + \frac{7^3}{3!} + \frac{7^4}{4!} + \ ... \ \frac{7^n}{n!} [/tex].

    This matches the expression for [tex]\frac{f'(1)}{7} = 1 + 7 + \ ... \ \frac{7^{n-1}}{(n-1)!} [/tex], except that it ends with n-1 rather n. However since both are infinite series, they are one and the same.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook