# Thread: Second derivative of infinite series gives 0^0 situation

1. ## Second derivative of infinite series gives 0^0 situation

What is the correct way to handle this problem?

Evaluate the second derivative wrt x of the summation of [(2x+1)^(n+1)]/(n!) from n=0..inf at x=-1/2

My solution was to take the derivative twice, giving:

summation of [4(n+1)(2x+1)^(n-1)]/[(n-1)!] from n=1..inf

then substituting x=-1/2 giving

summation of [4(n+1)*0^(n-1)]/[(n-1)!] from n=1..inf

Now the mistake I originally made here was to assume 0^(n-1) will cancel to zero making the solution zero, but the first term of the series creates a 0^0 situation which evidently should be treated as 1 -- resulting in the final solution being 8. I'm not very well versed on the proper way to handle 0^0 situations and was wondering if I should always treat 0^0 = 1 in problems like this?

I recognize that I could just break out the first two terms of the original series before taking the two derivatives to avoid getting to a 0^0 situation, but I wasn't sure if this was necessarily the best way to do it. Thanks.

2. Originally Posted by drumist
I recognize that I could just break out the first two terms of the original series before taking the two derivatives to avoid getting to a 0^0 situation, but I wasn't sure if this was necessarily the best way to do it. Thanks.
This is the wisest way to handle this kind of problem
I personally take a great care of the first terms of a series in order to be sure of what I am doing

3. Originally Posted by drumist
What is the correct way to handle this problem?

Evaluate the second derivative wrt x of the summation of [(2x+1)^(n+1)]/(n!) from n=0..inf at x=-1/2

My solution was to take the derivative twice, giving:

summation of [4(n+1)(2x+1)^(n-1)]/[(n-1)!] from n=1..inf

then substituting x=-1/2 giving

summation of [4(n+1)*0^(n-1)]/[(n-1)!] from n=1..inf

Now the mistake I originally made here was to assume 0^(n-1) will cancel to zero making the solution zero, but the first term of the series creates a 0^0 situation which evidently should be treated as 1 -- resulting in the final solution being 8. I'm not very well versed on the proper way to handle 0^0 situations and was wondering if I should always treat 0^0 = 1 in problems like this?

I recognize that I could just break out the first two terms of the original series before taking the two derivatives to avoid getting to a 0^0 situation, but I wasn't sure if this was necessarily the best way to do it. Thanks.
Originally Posted by running-gag
This is the wisest way to handle this kind of problem
I personally take a great care of the first terms of a series in order to be sure of what I am doing
Let's give a little stronger hint here. Your second derivative is

$y''(x) = \sum_{n = 0}^{\infty} \frac{4n(n + 1)}{n!}(2x + 1)^{n - 1}$

There is a problem with simplifying the coefficient the way you did. Compare what happens to these two "identical" expressions when n = 0:
$\frac{4n(n +1)}{n!}$ <--The unsimplified version

$\frac{4(n + 1)}{(n - 1)!}$ <--The simplification

The first expression is 0. The second expression is undefined because (-1)! is undefined. In order to correctly calculate the series at n = 0 you have to use the first expression. What does this do when you plug x = -1/2 into your series?

-Dan