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10.15 Representing Functions as Power Series

3 min readjune 18, 2024

10.15 Representing Functions as Power Series

Congrats congrats! You made it to the last AP Calculus BC topic! In this key topic, you’ll be deriving power series using different techniques you’ve learned throughout this course.

🟦Power Series

A power series, similar to what you’ve learned about Taylor polynomials, are a representation of a function using an infinite series of polynomials. It is generally expressed by the following below, where n is a non-negative integer, ana_n is a sequence of real numbers, and r is a real number.

n=0an(xr)\displaystyle\sum_{n=0}^{∞}{a_n(x-r)}

For the AP Calculus BC exam, memorizing these three frequently appearing power series can be a lifesaver: exe^x, sin(x)\sin(x), and cos(x)\cos(x). In a lot of cases, you will be able to use the original series to find the power series of a transformed version of one of these functions.

ex=n=0xnn!=1+x+x22!+x33!++xnn!e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots+\frac{x^n}{n!}
cos(x)=n=0(1)nx2n(2n)!=1x22!+x44!x66!++(1)nx2n(2n)!\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots+\frac{(-1)^nx^{2n}}{(2n)!}
sin(x)=n=0(1)nx2n+1(2n+1)!=xx33!+x55!x77!++(1)nx2n+1(2n+1)!\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots+\frac{(-1)^nx^{2n+1}}{(2n+1)!}

✏️ Power Series Example 1

Find the power series representation for x2exx^2e^x. Include the first 4 nonzero terms and the general term.

We know that the power series of exe^x is 1+x+x22!+x33!+...+xnn!+...1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}+.... So, since the function is just exe^x being multiplied by x2x^2, we can just multiply the power series of exe^x by x2x^2:

x2ex=x2[1+x+x22!+x33!+...+xnn!+...]x^2e^x=x^2[1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}+...]
=x2+x3+x42!+x53!+...+xn+2n!+...=x^2+x^3+\frac{x^4}{2!}+\frac{x^5}{3!}+...+\frac{x^{n+2}}{n!}+...

Great job! Lets look at one more example 🤗

✏️ Power Series Example 2

If h(x)h(x) is the power series centered at x=0x=0 of cos(x)\cos(x), what is h(x)h’(x)? Include the the first 3 nonzero terms and the general term.

We’re given that h(x)=1x22!+x44!+...+(1)nx2n(2n)!+...h(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}+...+\frac{(-1)^nx^{2n}}{(2n)!}+...To find h(x)h’(x), simply have to take the derivative of the series:

h(x)=x+x33!x55!+...+(1)nx2n1(2n1)!+...h'(x)=-x+\frac{x^3}{3!}-\frac{x^5}{5!}+...+\frac{(-1)^nx^{2n-1}}{(2n-1)!}+...

A cool thing to recognize is that this series is equivalent to the power series of sin(x)\sin(x), but negative; this proves that the derivative of cos(x)\cos(x) is sin(x)-\sin(x).


🔷Practice FRQ 2022 #6

The following free-response question (FRQ) is Question 6 from the 2022 AP Calculus BC examination administered by College Board. All credit to College Board.

The function f is defined by the power series f(x)=xx33+x55x77+...+(1)nx2n+12n+1+...f(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+...+\frac{(-1)^nx^{2n+1}}{2n+1}+... for all real numbers x for which the series converges.

c) Write the first four nonzero terms and the general term for an infinite series that represents f′(x).

For this problem, all you have to do is take the derivative, similar to our example 2:

f(x)=1x2+x4x6+...+(1)nx2n+...f'(x)=1-x^2+x^4-x^6+...+(-1)^nx^{2n}+...

Awesome work! 🎉


👏 Congrats! You’re done.

Congratulations, you’re done with this unit, and as a result, you’ve also reached the end of AP Calculus BC! Now, you have all the tools you need to ace that AP test this May!

As you start to review, be sure to check out our Study Tools unit for more resources and exam information.

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10.15 Representing Functions as Power Series

3 min readjune 18, 2024

10.15 Representing Functions as Power Series

Congrats congrats! You made it to the last AP Calculus BC topic! In this key topic, you’ll be deriving power series using different techniques you’ve learned throughout this course.

🟦Power Series

A power series, similar to what you’ve learned about Taylor polynomials, are a representation of a function using an infinite series of polynomials. It is generally expressed by the following below, where n is a non-negative integer, ana_n is a sequence of real numbers, and r is a real number.

n=0an(xr)\displaystyle\sum_{n=0}^{∞}{a_n(x-r)}

For the AP Calculus BC exam, memorizing these three frequently appearing power series can be a lifesaver: exe^x, sin(x)\sin(x), and cos(x)\cos(x). In a lot of cases, you will be able to use the original series to find the power series of a transformed version of one of these functions.

ex=n=0xnn!=1+x+x22!+x33!++xnn!e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots+\frac{x^n}{n!}
cos(x)=n=0(1)nx2n(2n)!=1x22!+x44!x66!++(1)nx2n(2n)!\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots+\frac{(-1)^nx^{2n}}{(2n)!}
sin(x)=n=0(1)nx2n+1(2n+1)!=xx33!+x55!x77!++(1)nx2n+1(2n+1)!\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots+\frac{(-1)^nx^{2n+1}}{(2n+1)!}

✏️ Power Series Example 1

Find the power series representation for x2exx^2e^x. Include the first 4 nonzero terms and the general term.

We know that the power series of exe^x is 1+x+x22!+x33!+...+xnn!+...1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}+.... So, since the function is just exe^x being multiplied by x2x^2, we can just multiply the power series of exe^x by x2x^2:

x2ex=x2[1+x+x22!+x33!+...+xnn!+...]x^2e^x=x^2[1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}+...]
=x2+x3+x42!+x53!+...+xn+2n!+...=x^2+x^3+\frac{x^4}{2!}+\frac{x^5}{3!}+...+\frac{x^{n+2}}{n!}+...

Great job! Lets look at one more example 🤗

✏️ Power Series Example 2

If h(x)h(x) is the power series centered at x=0x=0 of cos(x)\cos(x), what is h(x)h’(x)? Include the the first 3 nonzero terms and the general term.

We’re given that h(x)=1x22!+x44!+...+(1)nx2n(2n)!+...h(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}+...+\frac{(-1)^nx^{2n}}{(2n)!}+...To find h(x)h’(x), simply have to take the derivative of the series:

h(x)=x+x33!x55!+...+(1)nx2n1(2n1)!+...h'(x)=-x+\frac{x^3}{3!}-\frac{x^5}{5!}+...+\frac{(-1)^nx^{2n-1}}{(2n-1)!}+...

A cool thing to recognize is that this series is equivalent to the power series of sin(x)\sin(x), but negative; this proves that the derivative of cos(x)\cos(x) is sin(x)-\sin(x).


🔷Practice FRQ 2022 #6

The following free-response question (FRQ) is Question 6 from the 2022 AP Calculus BC examination administered by College Board. All credit to College Board.

The function f is defined by the power series f(x)=xx33+x55x77+...+(1)nx2n+12n+1+...f(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+...+\frac{(-1)^nx^{2n+1}}{2n+1}+... for all real numbers x for which the series converges.

c) Write the first four nonzero terms and the general term for an infinite series that represents f′(x).

For this problem, all you have to do is take the derivative, similar to our example 2:

f(x)=1x2+x4x6+...+(1)nx2n+...f'(x)=1-x^2+x^4-x^6+...+(-1)^nx^{2n}+...

Awesome work! 🎉


👏 Congrats! You’re done.

Congratulations, you’re done with this unit, and as a result, you’ve also reached the end of AP Calculus BC! Now, you have all the tools you need to ace that AP test this May!

As you start to review, be sure to check out our Study Tools unit for more resources and exam information.