If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Integral Calculus

### Course: Integral Calculus>Unit 5

Lesson 16: Representing functions as power series

# Differentiating power series

Within its interval of convergence, the derivative of a power series is the sum of derivatives of individual terms: [Σf(x)]'=Σf'(x). See how this is used to find the derivative of a power series.

## Want to join the conversation?

• How could 0^0 power be 1?
• 0^0 is kind of undefined, so the only way to evaluate it is limits. You've got `lim x->0 (x^0)`, `lim x->0 (x^x)`, and `lim x->0 (0->x)`; the middle of these is probably the most important. The limits are, respectively, `1`, `undefined`, and `undefined`. Also, the right-hand limit of the middle function is `1`. Where your confusion (I think) is coming from is that the right-hand limit of the third is, in fact, `0`. Also supporting the statement `0^0=1` is a somewhat fundamental definition of exponentiation: `x^y` means start with one, and multiply it by `x` `y` times. It is easy to see that in this, `0^0=1`

Edit: After watching the video, it appears the function in question is `f(x)=k*x^0`, and this is indeed `k*1` for all `x`, including `x=0`
• Sal claims in this pre-algebra video that 0^0 is undefined, so I'm at a loss here.
• If we factor out x^2 from the expanded f(x), then it is just the Maclaurin series of sin(x).

So, f(x) = x^2*sin(x)
• why doesn't Sal add one to the lower limit of summation each time he takes the derivative of the power series, I thought that's what you had to do with series when taking its derivative?
• isn't this x^2*sin(x)? could you use that here?
• Yes, I expanded the first few terms and rewrote the function as x^2*sin(x). Thought it might be faster, but I needed to apply the product rule for the derivative a few times. However, it removes the need to worry about n (in the video it seemed convenient that we only needed to worry about n=0, but I imagine in other problems this may not be the case).
(1 vote)
• Don't we get different results using those two methods it seams to me that using the first method for the fifth derivative evaluated at 0 we get 20 . But using the second one we get 0. May somoene clarify
• How do we know when a geometric series is finite or infinite ?
(1 vote)
• |r| > 1 means it is divergent (infinite)
|r| < 1 means it is convergent (finite)
where r is the common ratio
• Why did Sal choose to expand only to the first three terms? Because we're taking the third derivative? Or because that's generally considered a reasonable approximation?
(1 vote)
• He stopped at the first three terms because he did not need any other terms to calculate the third derivative at x=0. If, say, he wanted to calculate the 10th derivative at x=0, then he would need to expand at least up to x^10, as lower terms differentiate to 0 and higher terms evaluate to 0. If, say, he wanted an arbitrary derivative at a value of x other than x=0, then he would in fact need the entire series, or he could calculate an error bound and get close enough, or he could recognize that that series is actually equal to x^2*sin(x), or one of various other methods.