# 1983 BC 5 part C

1. Jun 5, 2008

### keithk

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 $$f(x) = \sum^{\infty}_{n=0}a_{n}x^n$$ find the value of f'(1)

$$a_{0} = 1$$ and $$a_{n} = (7/n)a_{n-1}$$

2. Relevant equations
None maybe?

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

This was as far as I was able to get. Mathematica tells me that the answer is 7e^7.
I know that $$\sum^{\infty}_{n=1}n/(n)! = e$$ and that $$\sum^{\infty}_{n=0}7^n/(n)! = e^7$$ (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. Jun 6, 2008

### keithk

Ok I think I got it.
If $$f(x) = \sum^{\infty}_{n=0}a_{n}x^n$$ find the value of f'(1)
$$a_{0} = 1$$ and $$a_{n} = (7/n)a_{n-1}$$

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

Sorry if that is a bit hard to follow.

Last edited: Jun 6, 2008
3. Jun 6, 2008

### keithk

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:
$$f'(1) = 7/0! + 7^2/1! + 7^3/2! ... \sum^{\infty}_{n=1}\frac{7^n }{(n-1)!} = 7(1 + 7 + 7^2/2! + 7^3/3!...)$$

4. Jun 6, 2008

### Defennder

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

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

This matches the expression for $$\frac{f'(1)}{7} = 1 + 7 + \ ... \ \frac{7^{n-1}}{(n-1)!}$$, except that it ends with n-1 rather n. However since both are infinite series, they are one and the same.