Infinite Series - Finding the 10th derivative

[carlodelmundo]

Homework Statement

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?

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?

 

[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?

 

[carlodelmundo]

Man, Dick, you're very helpful!

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!

 

[Hurkyl]

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) = \, ?