Wize AP Calculus (BC) Textbook > Power Series

Functions and Power Series

Popular Courses

AP Exam Prep

0:00 / 0:00

Representing Functions as Power Series
We can rewrite certain functions as power series.

Recall
Geometric Series: ∑n=0∞arn=a1−r{\displaystyle \sum_{n=0}^\infty}ar^n=\frac{a}{1-r}n=0∑∞​arn=1−ra​ when ∣r∣<1|r|<1∣r∣<1

Rewriting Functions as Power Series
We can rewrite 11−u(x)\frac{1}{1-u\left(x\right)}1−u(x)1​ as ∑n=0∞[u(x)]n{\displaystyle \sum_{n=0}^\infty}
\left[u\left(x\right)\right]^nn=0∑∞​[u(x)]nwhen ∣u(x)∣<1|u(x)|<1∣u(x)∣<1

*Sometimes we'll need to use partial fraction decomposition to rewrite our function into geometric series form

Practice: Power Series Representation
Find a power series representation of f(x)=1x2+4f(x)=\frac{1}{x^2+4}f(x)=x2+41​ and determine the radius of convergece.

f(x)=∑n=0∞(−x24)nf(x)=
{\displaystyle \sum_{n=0}^\infty}
\left(-\frac{x^2}{4}\right)^nf(x)=n=0∑∞​(−4x2​)n
R=1R=1R=1
Interval of convergence: (−2,2)(-2,2)(−2,2)

f(x)=14∑n=0∞(−x24)nf(x)=\frac{1}{4}
{\displaystyle \sum_{n=0}^\infty}
\left(-\frac{x^2}{4}\right)^nf(x)=41​n=0∑∞​(−4x2​)n 
R=2R=2R=2
Interval of convergence: (−2,2)(-2,2)(−2,2)

f(x)=14∑n=0∞(x24)nf(x)=\frac{1}{4}
{\displaystyle \sum_{n=0}^\infty}
\left(\frac{x^2}{4}\right)^nf(x)=41​n=0∑∞​(4x2​)n
R=2R=2R=2
Interval of convergence: (−2,2)(-2,2)(−2,2)

f(x)=4∑n=0∞(−x24)nf(x)=4
{\displaystyle \sum_{n=0}^\infty}
\left(-\frac{x^2}{4}\right)^nf(x)=4n=0∑∞​(−4x2​)n
R=4R=4R=4
Interval of convergence: (−2,2](-2,2](−2,2]

f(x)=4∑n=0∞(x24)nf(x)=4
{\displaystyle \sum_{n=0}^\infty}
\left(\frac{x^2}{4}\right)^nf(x)=4n=0∑∞​(4x2​)n
R=4R=4R=4
Interval of convergence: [−2,2)[-2,2)[−2,2)

I don't know

Practice: Power Series Representation
Find a power series represetnation of f(x)=2x−11−x2f(x)=\frac{2x-1}{1-x^2}f(x)=1−x22x−1​ and determine the radius of convergece.

f(x)=∑n=0∞[12+3(−1)n2]xnf(x)={\displaystyle \sum_{n=0}^\infty}\left[\frac{1}{2}+\frac{3(-1)^n}{2}\right]x^nf(x)=n=0∑∞​[21​+23(−1)n​]xn; Radius of convergence is R=12R=\frac{1}{2}R=21​

f(x)=∑n=0∞[12−3(−1)n2]xnf(x)={\displaystyle \sum_{n=0}^\infty}\left[\frac{1}{2}-\frac{3(-1)^n}{2}\right]x^nf(x)=n=0∑∞​[21​−23(−1)n​]xn; Radius of convergence is R=1R=1R=1

f(x)=∑n=0∞[3(−1)n2]xnf(x)={\displaystyle \sum_{n=0}^\infty}\left[\frac{3(-1)^n}{2}\right]x^nf(x)=n=0∑∞​[23(−1)n​]xn; Radius of convergence is R=1R=1R=1

f(x)=∑n=0∞[32−3(−1)n2]xnf(x)={\displaystyle \sum_{n=0}^\infty}\left[\frac{3}{2}-\frac{3(-1)^n}{2}\right]x^nf(x)=n=0∑∞​[23​−23(−1)n​]xn; Radius of convergence is R=12R=\frac{1}{2}R=21​

f(x)=∑n=0∞[32+3(−1)n2]xnf(x)={\displaystyle \sum_{n=0}^\infty}\left[\frac{3}{2}+\frac{3(-1)^n}{2}\right]x^nf(x)=n=0∑∞​[23​+23(−1)n​]xn; Radius of convergence is R=1R=1R=1

I don't know

Practice Question
Find a power series representation of f(x)=7x2+x−12f\left(x\right)=\frac{7}{x^2+x-12}f(x)=x2+x−127​ and determine the radius of convergence.

f(x)=∑n=0∞[−13n+1−(−1)n4n+1]xnf(x)=
{\displaystyle \sum_{n=0}^\infty}
\left[-\frac{1}{3^{n+1}}
-

\frac{(-1)^n}{4^{n+1}}\right]x^nf(x)=n=0∑∞​[−3n+11​−4n+1(−1)n​]xn
R=3R=3R=3
Interval of convergence: [−3,3][-3,3][−3,3]

f(x)=∑n=0∞[−13n+1−(−1)n4n+1]xnf(x)=
{\displaystyle \sum_{n=0}^\infty}
\left[-\frac{1}{3^{n+1}}
-

\frac{(-1)^n}{4^{n+1}}\right]x^nf(x)=n=0∑∞​[−3n+11​−4n+1(−1)n​]xn
R=3R=3R=3
Interval of convergence: (−3,3)(-3,3)(−3,3)

f(x)=∑n=0∞[−13n+1−(−1)n4n+1]xnf(x)=
{\displaystyle \sum_{n=0}^\infty}
\left[-\frac{1}{3^{n+1}}
-

\frac{(-1)^n}{4^{n+1}}\right]x^nf(x)=n=0∑∞​[−3n+11​−4n+1(−1)n​]xn
R=4R=4R=4
Interval of convergence: [−4,4][-4,4][−4,4]

f(x)=∑n=0∞[−13n+1−(−1)n4n+1]xnf(x)=
{\displaystyle \sum_{n=0}^\infty}
\left[-\frac{1}{3^{n+1}}
-

\frac{(-1)^n}{4^{n+1}}\right]x^nf(x)=n=0∑∞​[−3n+11​−4n+1(−1)n​]xn
R=0R=0R=0
x=0x=0x=0

f(x)=∑n=0∞[−13n+1−(−1)n4n+1]xnf(x)=
{\displaystyle \sum_{n=0}^\infty}
\left[-\frac{1}{3^{n+1}}
-

\frac{(-1)^n}{4^{n+1}}\right]x^nf(x)=n=0∑∞​[−3n+11​−4n+1(−1)n​]xn
R=4R=4R=4
Interval of convergence: (−4,4)(-4,4)(−4,4)

I don't know

0:00 / 0:00

Differentiation and Integration of Power Series
If the radius of convergence of a power seires is R>0R>0R>0, we can differentiate and integrate the terms in the power series
f(x)=∑n=0∞cn(x−a)n=c0+c1(x−a)+c2(x−a)2+c3(x−a)3+...f\left(x\right)=
{\displaystyle \sum_{n=0}^\infty}
c_n\left(x-a\right)^n=c_0+c_1\left(x-a\right)+c_2\left(x-a\right)^2+c_3\left(x-a\right)^3+...f(x)=n=0∑∞​cn​(x−a)n=c0​+c1​(x−a)+c2​(x−a)2+c3​(x−a)3+...

Differentiate
[∑n=0∞cn(x−a)n]′=∑n=1∞n cn(x−a)n−1\left[{\displaystyle \sum_{n=0}^\infty}
c_n\left(x-a\right)^n\right]'

=
{\displaystyle \sum_{n=1}^\infty}
n\ c_n\left(x-a\right)^{n-1}[n=0∑∞​cn​(x−a)n]′=n=1∑∞​n cn​(x−a)n−1

Integrate
∫∑n=0∞cn(x−a)ndx=C+∑n=0∞cnn+1(x−a)n+1{\displaystyle \int}{\displaystyle \sum_{n=0}^\infty}c_n\left(x-a\right)^ndx=

C+{\displaystyle \sum_{n=0}^\infty}\frac{c_n}{n+1}\left(x-a\right)^{n+1}∫n=0∑∞​cn​(x−a)ndx=C+n=0∑∞​n+1cn​​(x−a)n+1

*The radius of convergence will remain the same R.

When will we need to do this?
If f(x)f(x)f(x) can't be rewritten in geometric series form, but f′(x)f'(x)f′(x) can.
1. Differentiate the function
2. Write out the power series representation using the new geometric series forms
3. Integrate this new power series to get the power series for f(x)f(x)f(x)


If f(x)f(x)f(x) can't be rewritten in geometric series form, but ∫f(x) dx{\displaystyle \int}f(x) \ dx∫f(x) dx can.
1. Integrate the function
2. Write out the power series representation using the new geometric series forms
3. Differentiate this new power series to get the power series for f(x)f(x)f(x)

Practice: Power Series Representation
Find a power series represetnation of f(x)=1(1−x)3f(x)=\frac{1}{\left(1-x\right)^3}f(x)=(1−x)31​ and determine the radius of convergence.

f(x)=12∑n=2∞(n−1)xn−2f(x)=\frac{1}{2}
{\displaystyle \sum_{n=2}^\infty}(n-1)x^{n-2}f(x)=21​n=2∑∞​(n−1)xn−2
R=1R=1R=1

f(x)=∑n=2∞(n)xn−2f(x)=
{\displaystyle \sum_{n=2}^\infty}(n)x^{n-2}f(x)=n=2∑∞​(n)xn−2
R=0R=0R=0

f(x)=12∑n=2∞(n)(n−1)xn−2f(x)=\frac{1}{2}
{\displaystyle \sum_{n=2}^\infty}(n)(n-1)x^{n-2}f(x)=21​n=2∑∞​(n)(n−1)xn−2
R=1R=1R=1

f(x)=2∑n=2∞(n)(n−1)xn+2f(x)=2
{\displaystyle \sum_{n=2}^\infty}(n)(n-1)x^{n+2}f(x)=2n=2∑∞​(n)(n−1)xn+2
R=1R=1R=1

f(x)=2∑n=2∞(n−1)xn+2f(x)=2
{\displaystyle \sum_{n=2}^\infty}(n-1)x^{n+2}f(x)=2n=2∑∞​(n−1)xn+2
R=2R=2R=2

I don't know

Practice: Power Series Representation
Find a power series represenetation of f(x)=ln⁡(1+x2)f\left(x\right)=\ln\left(1+x^2\right)f(x)=ln(1+x2) and determine the radius of convergene.

f(x)=C+∑n=0∞(−1)nx2n+2nf(x)=C+
{\displaystyle\sum_{n=0}^\infty}
(-1)^n 
\frac{x^{2n+2}}{n}f(x)=C+n=0∑∞​(−1)nnx2n+2​
R=0R=0R=0

f(x)=C+∑n=0∞x2n+2nf(x)=C+
{\displaystyle\sum_{n=0}^\infty}

\frac{x^{2n+2}}{n}f(x)=C+n=0∑∞​nx2n+2​
R=1R=1R=1

f(x)=C+∑n=0∞x2n+2n+1f(x)=C+
{\displaystyle\sum_{n=0}^\infty}

\frac{x^{2n+2}}{n+1}f(x)=C+n=0∑∞​n+1x2n+2​
R=1R=1R=1

f(x)=C+∑n=0∞(−1)nx2n+2n+1f(x)=C+
{\displaystyle\sum_{n=0}^\infty}
(-1)^n 
\frac{x^{2n+2}}{n+1}f(x)=C+n=0∑∞​(−1)nn+1x2n+2​
R=1R=1R=1

f(x)=C+∑n=0∞(−1)nx2n+2n+1f(x)=C+
{\displaystyle\sum_{n=0}^\infty}
(-1)^n 
\frac{x^{2n+2}}{n+1}f(x)=C+n=0∑∞​(−1)nn+1x2n+2​
R=0R=0R=0

I don't know

Practice: Power Series Representation 
Find a power series represetnation of f(x)=x arctan⁡x2f(x)=x\ \arctan x^2f(x)=x arctanx2 and determine the radius of convergece.

f(x)=x+∑n=0∞[(2(−1)nx4n+3(4n+3))(14n+2+1)]f(x)=x+
{\displaystyle \sum_{n=0}^\infty}
\left[
\left(2(-1)^n\frac{x^{4n+3}}{(4n+3)}\right)
\left(
\frac{1}{4n+2}+1
\right)
\right]f(x)=x+n=0∑∞​[(2(−1)n(4n+3)x4n+3​)(4n+21​+1)]
R=1R=1R=1

f(x)=x+∑n=0∞[(2(−1)nx4n+3(4n+3))]f(x)=x
+
{\displaystyle \sum_{n=0}^\infty}
\left[
\left(2(-1)^n\frac{x^{4n+3}}{(4n+3)}\right)

\right]f(x)=x+n=0∑∞​[(2(−1)n(4n+3)x4n+3​)]
R=1R=1R=1

f(x)=C+2x+∑n=0∞[((−1)nx4n+5(4n+5))(14n+2+1)]f(x)=C+2x
+
{\displaystyle \sum_{n=0}^\infty}
\left[
\left((-1)^n\frac{x^{4n+5}}{(4n+5)}\right)
\left(
\frac{1}{4n+2}+1
\right)
\right]f(x)=C+2x+n=0∑∞​[((−1)n(4n+5)x4n+5​)(4n+21​+1)]
R=0R=0R=0

f(x)=D+Cx+∑n=0∞[(2(−1)nx4n+3(4n+3))]f(x)=D+Cx
+
{\displaystyle \sum_{n=0}^\infty}
\left[
\left(2(-1)^n\frac{x^{4n+3}}{(4n+3)}\right)

\right]f(x)=D+Cx+n=0∑∞​[(2(−1)n(4n+3)x4n+3​)]
R=0R=0R=0

f(x)=D+Cx+∑n=0∞[(2(−1)nx4n+3(4n+3))(14n+2+1)]f(x)=D+Cx
+
{\displaystyle \sum_{n=0}^\infty}
\left[
\left(2(-1)^n\frac{x^{4n+3}}{(4n+3)}\right)
\left(
\frac{1}{4n+2}+1
\right)
\right]f(x)=D+Cx+n=0∑∞​[(2(−1)n(4n+3)x4n+3​)(4n+21​+1)]
R=1R=1R=1

I don't know