Infinite Series - Finding the 10th derivative

1. Apr 11, 2009

carlodelmundo

1. The problem statement, all variables and given/known data

Let T(x) = $$\sum^{\infty}_{k=0}$$ $$\frac{1}{2^k}$$ $$\frac{(x-3)}{k!}$$k be the Taylor series for a function f. What is the value of f10(3), the tenth derivative of f at x = 3?

3. The attempt at a solution

I have a very small idea of actually starting this problem. Can I just derive the infinite series by taking derivatives of $$\frac{1}{2^k}$$ $$\frac{(x-3)}{k!}$$k 10 times?

Basically, what is the first step in this problem?

2. Apr 11, 2009

Dick

I'm sure you mean (1/2^k)*(x-3)^k/k!. Yes, think about taking the derivative 10 times and then putting x=3. Only one term will survive. What is it? Practice by thinking about x^k. If you take 10 derivatives of that and then put x=0 you will almost always get 0 unless k has a particular value. What is it?

3. Apr 11, 2009

carlodelmundo

Thanks I got (1/2)^10 or 9.765 * 10^-4.

You're right about that one term. When k = 10, then its 0^0, or 1. Thanks for the tips! Much appreciated!

4. Apr 11, 2009

Hurkyl

Staff Emeritus
I just want to emphasize the significance of the problem.

It's too easy to let your thinking run in only one direction -- you compute derivatives so that you may find the Taylor series to your function.

But the other direction is also important -- sometimes we want to find the derivative, and it turns out that we can compute the Taylor series easily. So, we compute the Taylor series so that we may find a derivative!

For example, suppose that we have

$$f(x) = \frac{1}{1-x}$$

and we want to know the eighteenth derivative:

$$f^{(18)}(0) = \, ?$$