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## Integral Calculus

### Course: Integral Calculus > Unit 5

Lesson 16: Representing functions as power series- Integrating power series
- Differentiating power series
- Integrate & differentiate power series
- Finding function from power series by integrating
- Integrals & derivatives of functions with known power series
- Interval of convergence for derivative and integral
- Converting explicit series terms to summation notation
- Converting explicit series terms to summation notation (n ≥ 2)

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# 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?(61 votes)
- 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`

(4 votes)

- Sal claims in this pre-algebra video that 0^0 is undefined, so I'm at a loss here.

https://www.khanacademy.org/math/pre-algebra/pre-algebra-exponents-radicals/pre-algebra-exponents/v/powers-of-zero(11 votes)- If this helps: try taking the following limt

lim f'''(x) , which indeed turns out to be 6

x->0

also check this out : https://www.quora.com/What-is-0-0-the-zeroth-power-of-zero-1(7 votes)

- 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)(10 votes) - 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?(10 votes)
- I guess he overlooked that in the video-making process.

Source: http://www.sosmath.com/diffeq/series/series02/series02.html(1 vote)

- isn't this x^2*sin(x)? could you use that here?(8 votes)
- 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(4 votes)
- 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(4 votes)

- 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.(3 votes)

- In the 3rd derivative, could (2n+3)(2n+2)(2n+1)/(2n+1)! be simplified to (2n+3)(2n+2)/(2n)!, since for any whole value of n, (2n+1) will equal (2n+1)!/(2n)!(2 votes)
- Yes, but this is not necessary, as we already have 0^2n term, and the rest doesn't really matter.(1 vote)

- sorry there might be a chain rule mistake at 6 minutes(1 vote)

## Video transcript

- [Instructor] We're told here that f(x) is equal to this infinite series, and we need to figure out what is the third derivative of f,
evaluated at x equals zero. And like always, pause this
video and see if you can work it out on your own
before we do it together. Alright, so there's two
ways to approach this. One is we could just take the derivative of this expression while
it's in sigma notation. The other way you do it is we could just expand out f(x) and take
the derivative three times, and see if we get an answer
that, I guess, makes sense. Let me do it the second way first. Let me just expand it out. F(x) is equal to, let's see,
when n is equal to zero, this is negative one to the
zero, which is just one, times x, times x to the zero plus three, so it's gonna be x to the third over two times zero, so that's
zero plus one factorial, so that's just going to be over one. Then the next term,
when n is equal to one, well now it's gonna be
negative one to the one, so now we're just gonna
have a negative out front. Negative, and it's gonna be
two times one plus three, so that's gonna be x to the fifth power over two times one plus
one, so it's gonna be two plus one is three factorial. So it's gonna be x to the fifth over six. And then when x is equal to two, this is going to be positive again, and it's gonna be x to the seventh power over five factorial. Is that right? Yeah. Five factorial, and five factorial... Actually, let me just write
that out as five factorial. Five factorial would be,
what, it would be 120. It'd be five times four
times six, so it'd be 120. But we could just keep going minus plus, and it goes on and on and on forever. Well now let's just take the derivatives. F'(x) is going to be equal to, we're still applying the power rule here, it's going to be three x squared minus 5/6x to the fourth, plus seven over five factorial x to the sixth, I'm just applying the power rule, minus plus, we just keep going
on and on and on forever. The second derivative, f''(x) is going to be equal to, apply
the power rule again. It's going to be six x to the
first minus four times five over six, I'll just write
that as 20/6x to the third, plus six times seven, so
it's 42 over five factorial x to the fifth, and we're
just gonna keep on going, minus plus, keep going,
or alternate between minus something then plus something,
on and on and on forever. Then we get to the third derivative. The third derivative is equal to, let's see, the derivative of six x is six, and then we have minus
20 times three is 60/6, which of course is 10, x squared, plus five times 42 is what,
210 over five factorial times x to the fourth power, minus plus over and over and over again, and then we just evaluate this at zero. F'''(0), well, when x is equal to zero, all of these terms with
xes are gonna go to zero, and you're just gonna be
left with this six here. So f'', the third derivative evaluated at zero is just equal to six. Now another way that we
could've tackled this is just kept it in this sigma notation. We could've said that f'(x) is equal to the infinite sum, and
actually, let me line them up. So this is where we
did f'(x) expanded out, but we could've said
f'(x) is equal to the sum from n equals zero to infinity, and you take the derivative here, you're gonna get, and
you're taking the derivative with respect to x, so for that purpose, you assume everything else is, well, the n is just gonna tell us, is gonna tell us how we
change from term to term, so if we take the derivative
with respect to x here, use the power rule, bring the
two n plus three out front, so it's gonna be negative one to the n times two n plus three,
times x to the decrement, the exponent, two n plus two over two n plus one factorial. And then if you wanna take
the second derivative, and this is the same thing as this. If you take the second derivative, f''(x), well now we're taking the sum from zero to infinity of negative one to the n. Let me move over to the right a little bit so we have some space. And now, we take this exponent out front, so you're gonna have two n plus three times two n plus two,
all of that's going to be over two n plus one factorial,
and this is gonna be times x to the two n plus one. All I'm doing every time,
it seems really complicated, I'm just taking the exponent out front, multiplying it out front,
and then decrementing it. So two n plus two minus
one is two n plus one. And then if I wanna take
the third derivative, the third derivative is
the sum where n equals zero to infinity, negative one to the n. We take this, bring it, multiply it, so we're gonna have two n plus three times two n plus two,
times two n plus one, all of that over two n plus one factorial, and then that is going to be times x to the two n power. Now let's now evaluate this
thing when x is equal to zero. F'''(0) is gonna be the
sum from n equals zero to infinity of negative
one to the nth power. This is gonna be interesting. We're gonna have all of this business, two n plus three times two n plus two times two n plus one, all of that over two n plus one factorial, times zero to the two n power. You might be tempted to say, well, hey, if zero to these different powers, maybe everything's gonna be zero, but remember, we're
starting at n equals zero, so for any n that's not equal to zero, this zero to that power
is just gonna be zero and that term's gonna disappear, kinda like what we saw
when we expanded it out. And so the only term that matters here is when n is equal to zero. So this is just going to be equal to, because for n equals one,
two, three, four, five, all the way to infinity,
this thing is gonna dominate. It's just gonna multiply,
it's gonna be zero. You're just gonna zero everything out. And so this is just gonna
reduce to the first term, when n equals zero,
and when n equals zero, it's gonna be negative one to the zero. This is gonna be, which is just one. Let me just write that as one. Times, this is three times two times one over one factorial, and
then times zero to the zero, which is equal to one. So this is equal to one,
and so this is equal to six. Either way, I think
the first way we did it was a little bit more straightforward, a little bit more
intuitive, closer to what you might have seen
before, but it's important to realize that we did
the same thing both times, we just kept it in the sigma notation on this time to the right. This technique is useful
because you'll see it a lot in math, where you might
wanna do things a little bit more of a general way,
and so it might be helpful to take the derivatives while you stay in that sigma notation.