Main content

## 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)

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Finding function from power series by integrating

AP.CALC:

LIM‑8 (EU)

, LIM‑8.D (LO)

, LIM‑8.D.6 (EK)

, LIM‑8.G (LO)

, LIM‑8.G.1 (EK)

When a power series S₁ is an antiderivative of a geometric series S₂, we can find the function represented by S₁ by integrating the expression for S₂.

## Want to join the conversation?

- Isn't the answer ln |1-2x| -2?(5 votes)
- No, g'(0)= - 2, but g(0)=0, when we substitute 0 to the second given function. This way c=0 and g(x)=ln |1-2x|(3 votes)

- I would really love to see some context for this massive unit....It's all good to understand how to calculate the interval of convergence for geometric series and power series and their derivatives and integral functions......but what is it good for?(2 votes)
- Power series (Taylor, MacLaurin, geometric, in general) are useful for estimation. While that doesn't sound too exciting, it is apparently extremely useful in analysis, physics, and engineering, for models and differential equations that are difficult to evaluate analytically, but can be approximated to arbitrary precision numerically (because power series are polynomials, which are easy to deal with).

https://www.ams.org/journals/mcom/1966-20-093/S0025-5718-1966-0187402-4/S0025-5718-1966-0187402-4.pdf

https://math.stackexchange.com/questions/766798/what-are-power-series-used-for-a-reference-request

https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80(4 votes)

- Why do we substitute x with 0 to find C?(1 vote)
- We need to find C in order to determine what the function actually is. We know that the equation holds for all x, so we can plug in any x we like to get an equation where we can solve for C. We choose x=0 because it makes almost everything vanish, so solving for C is easier.(4 votes)

- isn't the other function the infinite series -2x/n*(1-2x)?(2 votes)
- I've had this problem in partial fraction expansion before. Suppose all capital letters are constants, and the expression:
*x + A = Cx^3 + 4*

Is true for all x. Then apparently we can sub a convenient x value, like x = 0, and get that A = 4? I don't get the logic of this; A = 4 makes both sides equal when x = 0, but why does it work for all other x as well? I would appreciate any help!(2 votes) - for example, if i substitute x=3 both sides, i didn't find same values for the answer(2 votes)
- Is what we found only true when -1/2<x<1/2 ?(1 vote)

## Video transcript

- [Instructor] We know that
for x in the open interval from negative 1/2 to 1/2, that negative two over one minus two x is equal to this series. And it says using this
fact, find the function that corresponds to the following series. And like always, pause this video and see if you can work through it. All right, so the first
thing you might say is well how do we know that this
expression is equal to this series? And you might recognize this
series as a geometric series where the first term is negative two and then the common ratio
to get each successive term, we're multiplying by two x,
we're multiplying by two x. And so for a geometric series like this, the sum is gonna be the first term, negative two over one
minus the common ratio, which is exactly what we have over there. So that's why we know that. And this right over here, this gives us our radius of convergence, the x values for which this
thing will actually converge. But now that we feel good
about this first statement, let's try to answer their actual question. So we wanna find the function that corresponds to the following series. So my instinct is to say, well
how does this series relate to the series they gave us? Let's see, this first
term is a negative two, this first term is a negative two x, this is a negative four x, this
is a negative two x squared. So the thing that might
jump out at you is that this second series they gave
us is the antiderivative of this first one. Or we could say this first
series is a derivative of the second one. What's the derivative of
negative two x with respect to x? Well, it's negative two. What's the derivative of
negative two x squared with respect to x? Well, it's negative four
x and so on and so forth. And so if we were to
call this, let me call, let me call this thing right over here, actually, let me just say
that this is equal to g of x. Well, the way I can take
this right hand side and get g of x is by
taking the antiderivative, taking the indefinite integral. And so what we can do, we can take the indefinite integral of both sides, dx. So dx. On the right-hand side
I'm gonna get g of x. And on the left-hand
side, I am going to get, actually let me write it this way. So on the left-hand side,
I'll just rewrite it, I'm gonna get the indefinite integral of, I'll write it as negative two dx over one minus two x. And that's just another way of writing what I have on the left-hand side here. This is equal to g of x, right? If I take the antiderivative of this or an antiderivative of this, is this g of x, where the
constant would've been zero. So I'll say is equal to g of x. And so the key here is well,
what's the indefinite integral of this stuff, and you
might immediately recognize what I have here in the bottom, I have its derivative here on the top. If I consider this to be u. If I say u is equal to one minus two x, this is just u substitution, then du is equal to the derivative of this with respect to x which
is negative two dx. And so I have the du right over here. So let me rewrite all of this. I can rewrite this as
the integral of du over u is equal to g of x, and this we can rewrite as the natural log of the absolute value of u plus c is equal to g of x. And then we can undo our u substitution, and so for u I'll undo,
I will substitute back the one minus two x. So I can write the natural log of the absolute value of one
minus two x is equal to plus c, I don't wanna for get that, plus c is equal to g of x. And so the next thing we wanna
do is well, what is our c? And the easiest way to figure out c is let's substitute zero for x. And so let's think
about this a little bit. So if I put a zero
here, if I put x equals, actually let me just write it again. So I have the natural
log, if I say x is zero, this is gonna be the natural log of the absolute value of one plus c is equal to g of zero. So what's g of zero? G of zero, everyone of these
terms is equal to zero. G of zero is equal to zero. So is equal to zero. Well, natural log of one is just zero. So zero plus c is equal to
zero, c is equal to zero. So there you have it. I just took the
antiderivative of both sides of this equation right over here. I figured out, when I
substituted zero for x, then okay, my constant
here is going to be zero, and I get that, I get that this series, this g of x is equal to the natural log of the absolute
value of one minus two x, is equal to the natural
log of the absolute value of one minus two x. Now, if we keep this restriction that we're in this open interval, then this is always going to be positive. And we wouldn't have to
write the absolute value, but we could be safe by
writing the absolute value. But there you go. Using the fact above,
we found the function that corresponds to this following series. And I guess you could
say that the trick of it was recognizing that the series up here is a derivative
of the series down here. And so this is going to be the derivative of the following series, and so we just took the
antiderivative of both sides, or the indefinite integral of both sides.