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1983 BC 5 part C

  • Thread starter keithk
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  • #1
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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.

Homework Statement


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]


Homework Equations


None maybe?


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:

Answers and Replies

  • #2
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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.
 
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  • #3
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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]
 
  • #4
Defennder
Homework Helper
2,591
5
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.
 

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