(PS4) Power Series Calculus#

In this lesson we are going to see:

  • how to differentiate and integrate a power series.

  • how to use differentiation or integration to help express a given function as a power series.

Review Videos#

Calculus of a Power Series#

Differentiation and Integration

The function defined by a power series with radius of convergence \(R>0\) is continuous, differentiable, and integrable on the interval \((a-R,a+R)\).

Additionally, we can differentiate and integrate the power series term-by-term.

Specifically, if we have the function:

\[\begin{split} \begin{aligned} f(x) & = \sum_{n=0}^{\infty}c_n (x-a)^n \\ & \\ & = c_0 + c_1 (x-a) + c_2 (x-a)^2 + c_3 (x-a)^3 +\cdots \end{aligned} \end{split}\]

Then we have the derivative:

\[\begin{split} \begin{aligned} f'(x) & = \sum_{n=1}^{\infty}\dfrac{d}{dx} \bigg[c_n (x-a)^n\bigg] \\ & \\ & = c_1 + 2c_2 (x-a) + 3c_3 (x-a)^2 + \cdots \end{aligned} \end{split}\]

And antiderivative:

\[\begin{split} \begin{aligned} \int f(x)\; dx & = C+ \sum_{n=0}^{\infty} \int \bigg[c_n (x-a)^n\bigg] \; dx \\ & \\ & = C+ c_0(x-a) + \dfrac{c_1}{2}(x-a)^2 + \dfrac{c_2}{3} (x-a)^3 + \cdots \end{aligned} \end{split}\]

Note:

  • Radius of Convergence stays the same after differentiation or integration.

  • Interval of Convergence may change at the endpoints.

Example 1#

Differentiate the following function:

\[ f(x)=\sum_{n=0}^{\infty} \dfrac{x^n}{(2n)!} \]

Try for yourself and then click to see the steps.

When we differentiate a power series, we differentiate term by term (just like regular polynomials). To help visualize this process, let’s write out the first few terms of this series:

\[\begin{split} \begin{aligned} f(x)=\sum_{n=0}^{\infty} \dfrac{x^n}{(2n)!} & = \dfrac{x^0}{(2\cdot0)!} + \dfrac{x^1}{(2\cdot1)!} + \dfrac{x^2}{(2\cdot2)!}+ \dfrac{x^3}{(2\cdot3)!}+ \dfrac{x^4}{(2\cdot4)!}+\cdots \\ & \\ & = 1 + \dfrac{x}{2!} + \dfrac{x^2}{4!}+ \dfrac{x^3}{6!}+ \dfrac{x^4}{8!}+\cdots \end{aligned} \end{split}\]

To differentiate this power series we then calculate the derivative of each individual term:

\[\begin{split} \begin{aligned} f'(x)& =\sum_{n=0}^{\infty} \dfrac{d}{dx}\left[ \dfrac{x^n}{(2n)!} \right] = \bigg[1\bigg]'+ \bigg[\dfrac{x}{2!}\bigg]' + \bigg[\dfrac{x^2}{4!}\bigg]'+ \bigg[\dfrac{x^3}{6!}\bigg]'+ \bigg[\dfrac{x^4}{8!}\bigg]'+\cdots \\ & \\ & =\sum_{n=1}^{\infty} \dfrac{n\,x^{n-1}}{(2n)!} \qquad = 0+ \dfrac{1}{2!} + \dfrac{2x}{4!}+ \dfrac{3x^2}{6!}+ \dfrac{4x^3}{8!}+\cdots \end{aligned} \end{split}\]

Notice that since the first term of the series was a constant, its derivative went to \(0\). For this reason we changed the starting index value from \(n=0\) in the original power series to \(n=1\) in our derivative.

We previously found the radius of convergence for the original power series for \(f(x)\) to be \(R=\infty\).

Since differentiating a power series does not change the radius of convergence, that means the radius for this brand new series is also \(R=\infty\).

\[\begin{split} \begin{aligned} f(x)=\sum_{n=0}^{\infty} \dfrac{x^n}{(2n)!} & \xrightarrow{\qquad \text{ROC} \qquad } R=\infty\\ & \\ f'(x) = \sum_{n=1}^{\infty} \dfrac{n\,x^{n-1}}{(2n)!} & \xrightarrow{\qquad \text{ROC} \qquad } R=\infty \end{aligned} \end{split}\]

More Power Series Representations#

Sometimes we can find a power series representation for a function by first differentiating or integrating to get our prototype form.

Our Prototype

\[ \dfrac{1}{1-x} = 1+x+x^2+x^3+\cdots = \sum_{n=0}^{\infty} x^n \qquad \text{for } |x|<1 \]

  • Interval of Convergence: \((-1,1)\)

  • Radius of Convergence: \(R=1\)

If we use this prototype to help us find a power series, then we can use the inequality \(|r|<1\) to help us find the Radius of Convergence as well.

Example 2#

Find a power series representation for the following function centered at \(a=0\). Find the Radius of Convergence as well.

\[ f(x)=\dfrac{1}{(1-2x)^2} \]

Try for yourself and then click to see the steps.

Our strategy for finding a power series representation is to first get the function to match up with our prototype \(\tfrac{1}{1-x}\). Currently, the exponent of 2 around the entire denominator is preventing this. So before we get to the power series we are instead going to try using some calculus first.

We are going to try either integrating or differentiating the function \(f(x)\) and see which gets us closer to the form of our prototype.

\[ f(x) \xrightarrow[\text{Integrate?}]{\text{Differentiate?}} \]

If we differentiate \(f(x)\) we get:

\[ f'(x) = \dfrac{4}{(1-2x)^3} \]

which is taking us in the wrong direction (the exponent in the denominator is now bigger). If we integrate the function instead we get something much more closely related to our prototype \(\tfrac{1}{1-x}\):

\[ \int f(x)\; dx = \int \dfrac{1}{(1-2x)^2}\; dx = \dfrac{1}{2}\cdot \dfrac{1}{1-2x} \]

That looks better.

We’re now ready to find the power series representation, not for our original function but rather its antiderivative.

\[ \dfrac{1}{(1-2x)^2} \xrightarrow{\text{Integrate}}\dfrac{1}{2}\cdot \dfrac{1}{1-2x} \]

Using our prototype formula \(\displaystyle \dfrac{1}{1-x}=\sum_{n=0}^{\infty} x^n\) with \(r=2x\) we get power series:

\[ \dfrac{1}{2}\cdot \dfrac{1}{1-2x} = \dfrac{1}{2}\cdot \sum_{n=0}^{\infty} (2x)^n \]

This is pretty good for the power series, but we usually want to get it into the form \(\sum c_k x^k\) where we separate all of the constant terms from the powers of \(x\).

\[ \dfrac{1}{2}\cdot \dfrac{1}{1-2x} = \dfrac{1}{2}\cdot \sum_{n=0}^{\infty} (2x)^n = \dfrac{1}{2}\cdot \sum_{n=0}^{\infty} 2^n x^n = \sum_{n=0}^{\infty} 2^{n-1} x^n \]

This is better formatting.

Now that we have the power series for the antiderivative, the question is: How do we get back to the original function? To effectively “undo” the integral we did at the beginning, we differentiate:

\[ \dfrac{1}{(1-2x)^2} \xrightarrow{\text{Integrate}}\dfrac{1}{2}\cdot \dfrac{1}{1-2x}\xrightarrow{\text{Power Series}} \sum_{n=0}^{\infty} 2^{n-1} x^n \xrightarrow{\text{Differentiate}} \sum_{n=1}^{\infty} 2^{n-1} n x^{n-1} \]

There we have the power series for our original function.

\[ \dfrac{1}{(1-2x)^2} = \sum_{n=1}^{\infty} 2^{n-1} n x^{n-1} \]

Comments:

  • Notice that when we differentiated, we just used our usual power rule:

\[ \dfrac{d}{dx}\bigg[ 2^{n-1} x^n \bigg] = 2^{n-1} \cdot \dfrac{d}{dx}\bigg[ x^n \bigg] = 2^{n-1} \cdot n x^{n-1} \]

  • When \(n=0\), the (first) term in the series before we differentiate is \(2^{-1}x^0 = \tfrac{1}{2}\) which is a constant. When we do actually differentiate, the derivative process is going to effectively cancel off this \(n=0\) term. So we can change the starting index to \(n=1\).

Finally, to get the radius of convergence we start with the \(r\)-value that we plugged into our prototype formula back when we first got a power series. In this example we used, \(r=2x\).

Since we used our prototype formula, we know that we started things off by dealing with a geometric series. This means we can use the inequality \(|r|<1\) to help us find the Radius of Convergence and the first-pass Interval of Convergence.

\[ |r|<1 \iff |2x|<1 \iff 2|x|<1 \iff |x| < \dfrac{1}{2} \]

Once we have simplified the inequality down to where we have \(|x|\) or \(|x-a|\) all by itself on the left side, the number on the other side of the inequality is the ROC:

\[ R=\dfrac{1}{2} \]

Since differentiating a power series does not change the radius of convergence, this is the radius of convergence for our answer.

Example 3#

Find a power series representation for the following function. Find the Radius of Convergence as well.

\[ f(x)=\ln(2+x) \]

Try for yourself and then click to see the steps.

Our strategy for finding a power series representation is to first get the function to match up with our prototype \(\tfrac{1}{1-x}\). Currently, we don’t even have a fraction, so before we get to the power series we try some calculus first.

We are going to try either integrating or differentiating the function \(f(x)\) and see which gets us closer to the form of our prototype.

\[ f(x) \xrightarrow[\text{Integrate?}]{\text{Differentiate?}} \]

If we differentiate \(f(x)\) we get:

\[ f'(x) = \dfrac{1}{2+x}\big[ 2+x\big]' = \dfrac{1}{2+x} \]

which looks pretty close to our prototype \(\tfrac{1}{1-x}\). If we tried integrating we would get:

\[ \int f(x)\; dx = \int \ln(2+x) \; dx = (2+x)\ln(2+x) -(2+x) \]

Which definitely looks like it is taking us in the wrong direction.

We’re now ready to find the power series representation, not for our original function but rather its derivative.

\[ \ln(2+x)\xrightarrow{\text{Differentiate}} \dfrac{1}{2+x} \]

We’re almost ready to use our prototype formula \(\displaystyle \dfrac{1}{1-x}=\sum_{n=0}^{\infty} x^n\) but before we do, we need to reformat it a bit to find the \(r\)-value to plug in. We start by factoring off the \(2\) in the denominator:

\[ \dfrac{1}{2+x} = \dfrac{1}{2+\tfrac{2x}{2}} = \dfrac{1}{2}\cdot \dfrac{1}{1+\tfrac{x}{2}} \]

Next we introduce our double negative to get the subtraction sign we need:

\[ \dfrac{1}{2+x} = \dfrac{1}{2}\cdot \dfrac{1}{1+\tfrac{x}{2}} = \dfrac{1}{2}\cdot \dfrac{1}{1- \left(-\tfrac{x}{2}\right)} \]

From here we can plug \(r=-\dfrac{x}{2}\) into our prototype formula to get the power series:

\[ \dfrac{1}{2+x} =\dfrac{1}{2}\cdot \dfrac{1}{1- \left(-\tfrac{x}{2}\right)} = \dfrac{1}{2}\cdot \sum_{n=0}^{\infty} \left( -\dfrac{x}{2}\right)^n \]

This is pretty good for the power series, but we usually want to get it into the form \(\sum c_k x^k\) where we separate all of the constant terms from the powers of \(x\).

\[ \dfrac{1}{2+x} = \dfrac{1}{2}\cdot \sum_{n=0}^{\infty} \left( -\dfrac{x}{2}\right)^n = \dfrac{1}{2}\cdot \sum_{n=0}^{\infty} \dfrac{(-1)^n}{2^n} x^n = \sum_{n=0}^{\infty} \dfrac{(-1)^n}{2^{n+1}} x^n \]

This is better formatting.

Now that we have the power series for the derivative, the question is: How do we get back to the original function? To effectively “undo” the derivative we did at the beginning, we integrate:

\[ \ln(2+x)\xrightarrow{\text{Differentiate}} \dfrac{1}{2+x} \xrightarrow{\text{Power Series}} \sum_{n=0}^{\infty} \dfrac{(-1)^n}{2^{n+1}} x^n \xrightarrow{\text{Integration}} C+ \sum_{n=0}^{\infty} \dfrac{(-1)^n}{2^{n+1}}\dfrac{1}{n+1} x^{n+1} \]

There we have the power series for our original function.

\[ \ln(2+x) = C+ \sum_{n=0}^{\infty} \dfrac{(-1)^n}{2^{n+1}}\dfrac{1}{n+1} x^{n+1} \]

Comments:

  • Notice that when we integrated, we just used our usual power rule:

\[ \int \dfrac{(-1)^n}{2^{n+1}} x^n \; dx = \dfrac{(-1)^n}{2^{n+1}}\cdot \int x^n \; dx = \dfrac{(-1)^n}{2^{n+1}}\cdot \dfrac{1}{n+1}x^{n+1} \]

  • Since our last step here was integration, we needed to also include \(+C\).

We’re not quite done yet. There is some ambiguity in our power series function with that unknown constant \(+C\), an ambiguity that the original function \(\ln (2+x)\) does not have. This means we need to figure out the value for \(C\) that makes this equation true:

\[ \ln(2+x) = C+ \sum_{n=0}^{\infty} \dfrac{(-1)^n}{2^{n+1}}\dfrac{1}{n+1} x^{n+1} \]

Since we want these two different formulas to be equal (and represent the same function), they need to agree at all \(x\)-values. In particular, they need to agree at the center point of our power series \(a=0\).

Plugging \(a=0\) into both sides gives us:

\[ \ln(2+0) = C+ \sum_{n=0}^{\infty} \dfrac{(-1)^n}{2^{n+1}}\dfrac{1}{n+1} 0^{n+1} \]

In the series on the right, every term has a power of 0, which all end of canceling out, giving us:

\[ \ln(2) = C + 0 \]

We can then use this to update our power series:

\[ \ln(2+x) = \ln(2) + \sum_{n=0}^{\infty} \dfrac{(-1)^n}{2^{n+1}}\dfrac{1}{n+1} x^{n+1} \]

Finally, to get the radius of convergence we start with the \(r\)-value that we plugged into our prototype formula back when we first got a power series. In this example we used, \(r=-\dfrac{x}{2}\).

Since we used our prototype formula, we know that we started things off by dealing with a geometric series. This means we can use the inequality \(|r|<1\) to help us find the Radius of Convergence and the first-pass Interval of Convergence.

\[ |r|<1 \iff \left|-\dfrac{x}{2}\right|<1 \iff \dfrac{|x|}{2}<1 \iff |x| < 2 \]

Once we have simplified the inequality down to where we have \(|x|\) or \(|x-a|\) all by itself on the left side, the number on the other side of the inequality is the ROC:

\[ R=2 \]

Since integrating a power series does not change the radius of convergence, this is the radius of convergence for our answer.

Example 4#

Evaluate the following integral using power series:

\[ \int_0^{1/2}\dfrac{x}{1+x^5}\; dx \]

Try for yourself and then click to see the steps.

To start us off in solving this problem, we are going to first find the power series representation for the function we are trying to integrate.

We first notice that the function is close to the format of our prototype: \(\tfrac{1}{1-x}\). We just need to do a little work to get it there completely:

Reformat:

\[ \dfrac{x}{1+x^5} = x\cdot \dfrac{1}{1+x^5} = x\cdot \dfrac{1}{1-\big(-x^5\big)} \]

We first separated off the numerator, so that we had a \(1\) on top, and then used our standard double negative trick in the denominator.

We’re now ready to apply our prototype formula with \(r=-x^5\)

\[ \text{Prototype: } \dfrac{1}{1-x}=\sum_{n=0}^{\infty} x^n \]

Plugging this in we get:

\[ \dfrac{x}{1+x^5} = x\cdot \dfrac{1}{1-\big(-x^5\big)} = x\cdot \sum_{n=0}^{\infty} (-x^5)^n \]

Reformat: Before we get to the integration, we want to reformat this power series to get it into the form: \(\sum c_k x^k\) where we separate all of the constants and \(x\)-terms.

\[\begin{split} \begin{aligned} \dfrac{x}{1+x^5} & = x\cdot \sum_{n=0}^{\infty} (-x^5)^n \\ & \\ & = x\cdot \sum_{n=0}^{\infty} (-1)^n (x^5)^n \\ & \\ & = x\cdot \sum_{n=0}^{\infty} (-1)^n x^{5n} \\ & \\ & = \sum_{n=0}^{\infty} (-1)^n x\cdot x^{5n} \\ & \\ & = \sum_{n=0}^{\infty} (-1)^n x^{5n+1} \\ \end{aligned} \end{split}\]

And there we go, we have a power series ready to be integrated.

We’re now ready to find the antiderivative of our power series, and we’ll do this by just finding the antiderivative of each individual term.

\[\begin{split} \begin{aligned} \int_0^{1/2}\dfrac{x}{1+x^5}\; dx & = \int_0^{1/2}\sum_{n=0}^{\infty} (-1)^n x^{5n+1}\; dx \\ & \\ & = \left[\sum_{n=0}^{\infty} (-1)^n \tfrac{1}{5n+2}x^{5n+2} \right]_{x=0}^{x=1/2}\\ & \\ \end{aligned} \end{split}\]

In the first step here, we just replaced the function with its power series representation. In the second step we integrated each term of the power series using regular power rule. (This is why it was helpful to get the power series into the correct format \(\sum c_k x^k\) first.)