Introduction to Power Series

Understanding power series and their properties

🎯 Introduction to Power Series

What is a Power Series?

A power series is an infinite series of the form:

n=0cn(xa)n=c0+c1(xa)+c2(xa)2+c3(xa)3+\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

where:

  • cnc_n are coefficients (constants)
  • xx is a variable
  • aa is the center

💡 Key Idea: A power series is like a polynomial with infinitely many terms!


Power Series Centered at a=0a = 0

When a=0a = 0, we get a Maclaurin-type power series:

n=0cnxn=c0+c1x+c2x2+c3x3+\sum_{n=0}^{\infty} c_n x^n = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \cdots

This is the most common form!


Example 1: Geometric Series as Power Series

The geometric series:

n=0xn=1+x+x2+x3+\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots

This is a power series with cn=1c_n = 1 for all nn, centered at a=0a = 0.

Converges when: x<1|x| < 1

Sum: 11x\frac{1}{1-x} (when x<1|x| < 1)


Power Series as Functions

A power series defines a function:

f(x)=n=0cn(xa)nf(x) = \sum_{n=0}^{\infty} c_n (x-a)^n

The domain of ff is the set of all xx values where the series converges.


Example 2: Recognize the Function

What function does n=0xnn!\sum_{n=0}^{\infty} \frac{x^n}{n!} represent?

This is the Taylor series for exe^x!

ex=1+x+x22!+x33!+x44!+e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots

Converges for all xx (we'll prove this later).


Convergence of Power Series

For a given power series cn(xa)n\sum c_n(x-a)^n, exactly one of the following is true:

  1. Converges only at x=ax = a (trivial case)

  2. Converges for all xx (radius of convergence R=R = \infty)

  3. Converges for xa<R|x - a| < R and diverges for xa>R|x - a| > R

where RR is the radius of convergence.


The Radius of Convergence

The radius of convergence RR is:

  • R=0R = 0 if series only converges at center
  • 0<R<0 < R < \infty if series converges on an interval
  • R=R = \infty if series converges for all xx

Interval of convergence: (aR,a+R)(a - R, a + R) or [aR,a+R][a - R, a + R] or variations

(Must check endpoints separately!)


Finding Radius of Convergence

Use the Ratio Test:

1R=limncn+1cn\frac{1}{R} = \lim_{n \to \infty} \left|\frac{c_{n+1}}{c_n}\right|

Or equivalently:

R=limncncn+1R = \lim_{n \to \infty} \left|\frac{c_n}{c_{n+1}}\right|

(If the limit exists or is \infty)


Example 3: Find Radius

Find the radius of convergence for n=0xnn+1\sum_{n=0}^{\infty} \frac{x^n}{n+1}.

Use Ratio Test: Let an=xnn+1a_n = \frac{x^n}{n+1}

limnan+1an=limnxn+1n+2xnn+1\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \left|\frac{\frac{x^{n+1}}{n+2}}{\frac{x^n}{n+1}}\right|

=limnxn+1n+2= \lim_{n \to \infty} \left|x \cdot \frac{n+1}{n+2}\right|

=xlimnn+1n+2=x1=x= |x| \cdot \lim_{n \to \infty} \frac{n+1}{n+2} = |x| \cdot 1 = |x|


For convergence: x<1|x| < 1

Radius of convergence: R=1R = 1


Example 4: Power Series Centered at a=2a = 2

Find the radius of convergence for n=1(x2)nn3n\sum_{n=1}^{\infty} \frac{(x-2)^n}{n \cdot 3^n}.

Use Ratio Test: Let an=(x2)nn3na_n = \frac{(x-2)^n}{n \cdot 3^n}

limnan+1an=limn(x2)n+1(n+1)3n+1n3n(x2)n\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \left|\frac{(x-2)^{n+1}}{(n+1) \cdot 3^{n+1}} \cdot \frac{n \cdot 3^n}{(x-2)^n}\right|

=limnx23nn+1= \lim_{n \to \infty} \left|\frac{x-2}{3} \cdot \frac{n}{n+1}\right|

=x231=x23= \frac{|x-2|}{3} \cdot 1 = \frac{|x-2|}{3}


For convergence: x23<1\frac{|x-2|}{3} < 1

x2<3|x - 2| < 3

Radius of convergence: R=3R = 3

Center: a=2a = 2

Interval (before checking endpoints): (23,2+3)=(1,5)(2-3, 2+3) = (-1, 5)


Operations on Power Series

If f(x)=cnxnf(x) = \sum c_n x^n and g(x)=dnxng(x) = \sum d_n x^n both converge for x<R|x| < R:

Addition/Subtraction:

f(x)±g(x)=(cn±dn)xnf(x) \pm g(x) = \sum (c_n \pm d_n) x^n

Scalar Multiplication:

kf(x)=(kcn)xnk \cdot f(x) = \sum (k \cdot c_n) x^n

Multiplication by xmx^m:

xmf(x)=cnxn+mx^m f(x) = \sum c_n x^{n+m}


Differentiation of Power Series

Theorem: If f(x)=n=0cn(xa)nf(x) = \sum_{n=0}^{\infty} c_n (x-a)^n with radius R>0R > 0, then:

f(x)=n=1ncn(xa)n1f'(x) = \sum_{n=1}^{\infty} n \cdot c_n (x-a)^{n-1}

Differentiate term by term!

The derivative has the same radius of convergence RR.


Example 5: Differentiate Power Series

Find the derivative of f(x)=n=0xnn!f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}.

Differentiate term by term:

f(x)=n=1nxn1n!=n=1xn1(n1)!f'(x) = \sum_{n=1}^{\infty} \frac{n \cdot x^{n-1}}{n!} = \sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!}

Reindex with m=n1m = n - 1:

=m=0xmm!= \sum_{m=0}^{\infty} \frac{x^m}{m!}

Notice: f(x)=f(x)f'(x) = f(x)!

This confirms f(x)=exf(x) = e^x (since ddx[ex]=ex\frac{d}{dx}[e^x] = e^x).


Integration of Power Series

Theorem: If f(x)=n=0cn(xa)nf(x) = \sum_{n=0}^{\infty} c_n (x-a)^n with radius R>0R > 0, then:

f(x)dx=C+n=0cnn+1(xa)n+1\int f(x)\,dx = C + \sum_{n=0}^{\infty} \frac{c_n}{n+1}(x-a)^{n+1}

Integrate term by term!

The integral has the same radius of convergence RR.


Example 6: Integrate Power Series

Find n=0xndx\int \sum_{n=0}^{\infty} x^n\,dx for x<1|x| < 1.

Integrate term by term:

n=0xndx=C+n=0xn+1n+1\int \sum_{n=0}^{\infty} x^n\,dx = C + \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}

We know n=0xn=11x\sum_{n=0}^{\infty} x^n = \frac{1}{1-x} for x<1|x| < 1.

So: 11xdx=ln1x+C\int \frac{1}{1-x}\,dx = -\ln|1-x| + C

This matches: ln(1x)=n=0xn+1n+1=n=1xnn-\ln(1-x) = \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} = \sum_{n=1}^{\infty} \frac{x^n}{n}


Uniqueness of Power Series

Theorem: If cn(xa)n=dn(xa)n\sum c_n(x-a)^n = \sum d_n(x-a)^n for all xx in some interval, then:

cn=dn for all nc_n = d_n \text{ for all } n

Power series representation is unique!


Example 7: Find Coefficients

If n=0cnxn=11x\sum_{n=0}^{\infty} c_n x^n = \frac{1}{1-x} for x<1|x| < 1, find cnc_n.

We know: 11x=1+x+x2+x3+\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots

By uniqueness: cn=1c_n = 1 for all n0n \geq 0.


Substitution in Power Series

Can substitute u=f(x)u = f(x) into a power series!

Example: If n=0xn=11x\sum_{n=0}^{\infty} x^n = \frac{1}{1-x} for x<1|x| < 1:

Replace xx with x2-x^2:

n=0(x2)n=11(x2)=11+x2\sum_{n=0}^{\infty} (-x^2)^n = \frac{1}{1-(-x^2)} = \frac{1}{1+x^2}

1x2+x4x6+=11+x21 - x^2 + x^4 - x^6 + \cdots = \frac{1}{1+x^2} (for x<1|x| < 1)


⚠️ Common Mistakes

Mistake 1: Forgetting Absolute Value

WRONG: x2<3|x - 2| < 3 means x<5x < 5

RIGHT: x2<3|x - 2| < 3 means 3<x2<3-3 < x - 2 < 3, so 1<x<5-1 < x < 5


Mistake 2: Not Checking Endpoints

Radius tells you interval is (aR,a+R)(a-R, a+R), but series might converge at one or both endpoints!

Must check x=aRx = a - R and x=a+Rx = a + R separately.


Mistake 3: Wrong Index After Differentiation

When differentiating n=0cnxn\sum_{n=0}^{\infty} c_n x^n:

Get n=1ncnxn1\sum_{n=1}^{\infty} n c_n x^{n-1} (starts at n=1n=1, not n=0n=0!)

The n=0n=0 term disappears (constant → 0).


Mistake 4: Confusing Center and Radius

For cn(x2)n\sum c_n(x-2)^n with R=3R = 3:

  • Center: a=2a = 2
  • Radius: R=3R = 3
  • Interval: (1,5)(-1, 5) before checking endpoints

Not centered at 0!


📝 Practice Strategy

  1. Identify center aa: Look at (xa)n(x - a)^n
  2. Use Ratio Test: Find liman+1an\lim|\frac{a_{n+1}}{a_n}|, set <1< 1 for convergence
  3. Solve for xx: Get xa<R|x - a| < R
  4. Check endpoints: Test x=a±Rx = a \pm R separately using other tests
  5. For differentiation: Differentiate each term, start sum at n=1n=1
  6. For integration: Integrate each term, add constant
  7. Known series: Memorize 11x=xn\frac{1}{1-x} = \sum x^n and ex=xnn!e^x = \sum \frac{x^n}{n!}

📚 Practice Problems

1Problem 1medium

Question:

Find a power series representation for f(x)=11+x3f(x) = \frac{1}{1+x^3} and determine its radius of convergence.

💡 Show Solution

Step 1: Use known geometric series

We know: 11u=n=0un\frac{1}{1-u} = \sum_{n=0}^{\infty} u^n for u<1|u| < 1


Step 2: Substitute u=x3u = -x^3

11+x3=11(x3)=n=0(x3)n\frac{1}{1+x^3} = \frac{1}{1-(-x^3)} = \sum_{n=0}^{\infty} (-x^3)^n

=n=0(1)nx3n= \sum_{n=0}^{\infty} (-1)^n x^{3n}

=1x3+x6x9+x12= 1 - x^3 + x^6 - x^9 + x^{12} - \cdots


Step 3: Find radius of convergence

For geometric series, we need u<1|u| < 1:

x3<1|-x^3| < 1 x3<1|x^3| < 1 x<1|x| < 1

Radius of convergence: R=1R = 1


Answer: 11+x3=n=0(1)nx3n\frac{1}{1+x^3} = \sum_{n=0}^{\infty} (-1)^n x^{3n} for x<1|x| < 1

2Problem 2medium

Question:

Find the radius and interval of convergence for the power series:

n=0xnn+1\sum_{n=0}^{\infty} \frac{x^n}{n+1}

💡 Show Solution

Solution:

Use the Ratio Test with an=xnn+1a_n = \frac{x^n}{n+1}:

an+1an=xn+1n+2n+1xn\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{x^{n+1}}{n+2} \cdot \frac{n+1}{x^n}\right|

=xn+1n+2= |x| \cdot \frac{n+1}{n+2}

L=limnxn+1n+2=x1=xL = \lim_{n \to \infty} |x| \cdot \frac{n+1}{n+2} = |x| \cdot 1 = |x|

Series converges when L<1L < 1: x<1|x| < 1

Radius of convergence: R=1R = 1

Check endpoints:

At x=1x = 1: 1n+1\sum \frac{1}{n+1} diverges (harmonic-like)

At x=1x = -1: (1)nn+1\sum \frac{(-1)^n}{n+1} converges (alternating series test)

Interval of convergence: [1,1)[-1, 1)

3Problem 3hard

Question:

Find the power series representation for ln(1+x)\ln(1+x) by integrating the series for 11+x\frac{1}{1+x}.

💡 Show Solution

Step 1: Start with known series

11+x=11(x)=n=0(x)n=n=0(1)nxn\frac{1}{1+x} = \frac{1}{1-(-x)} = \sum_{n=0}^{\infty} (-x)^n = \sum_{n=0}^{\infty} (-1)^n x^n

for x<1|x| < 1.


Step 2: Integrate both sides

11+xdx=n=0(1)nxndx\int \frac{1}{1+x}\,dx = \int \sum_{n=0}^{\infty} (-1)^n x^n\,dx

ln1+x+C=n=0(1)nxn+1n+1\ln|1+x| + C = \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1}


Step 3: Find constant

At x=0x = 0: ln1=0\ln 1 = 0, and the series sum is 0.

So C=0C = 0.

For 1<x<1-1 < x < 1: 1+x>01 + x > 0, so 1+x=1+x|1+x| = 1+x.

ln(1+x)=n=0(1)nxn+1n+1\ln(1+x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1}


Step 4: Reindex

Let m=n+1m = n + 1:

ln(1+x)=m=1(1)m1xmm\ln(1+x) = \sum_{m=1}^{\infty} (-1)^{m-1} \frac{x^m}{m}

Or more commonly written as:

ln(1+x)=n=1(1)n1nxn\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^n

=xx22+x33x44+= x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots

for x<1|x| < 1.


Answer: ln(1+x)=n=1(1)n1nxn\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^n for x<1|x| < 1

4Problem 4expert

Question:

If f(x)=n=0x2n(2n)!f(x) = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}, find f(x)f''(x) and identify the function.

💡 Show Solution

Step 1: Write out first few terms

f(x)=1+x22!+x44!+x66!+f(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdots


Step 2: Find first derivative

f(x)=n=12nx2n1(2n)!f'(x) = \sum_{n=1}^{\infty} \frac{2n \cdot x^{2n-1}}{(2n)!}

=n=1x2n1(2n1)!= \sum_{n=1}^{\infty} \frac{x^{2n-1}}{(2n-1)!}

=x+x33!+x55!+= x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots


Step 3: Find second derivative

f(x)=n=1(2n1)x2n2(2n1)!f''(x) = \sum_{n=1}^{\infty} \frac{(2n-1) \cdot x^{2n-2}}{(2n-1)!}

=n=1x2n2(2n2)!= \sum_{n=1}^{\infty} \frac{x^{2n-2}}{(2n-2)!}

Reindex with m=n1m = n - 1:

=m=0x2m(2m)!=f(x)= \sum_{m=0}^{\infty} \frac{x^{2m}}{(2m)!} = f(x)


Step 4: Identify the function

We have f(x)=f(x)f''(x) = f(x) with f(0)=1f(0) = 1 and f(0)=0f'(0) = 0.

This is the differential equation for cosx\cos x!

But wait, f(0)=cos0=1f(0) = \cos 0 = 1

And f(0)=sin0=0f'(0) = -\sin 0 = 0

Actually, d2dx2[cosx]=cosx\frac{d^2}{dx^2}[\cos x] = -\cos x, not +cosx+\cos x.

Let me reconsider...

If g(x)=coshx=ex+ex2g(x) = \cosh x = \frac{e^x + e^{-x}}{2}, then:

coshx=n=0x2n(2n)!\cosh x = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}

And d2dx2[coshx]=coshx\frac{d^2}{dx^2}[\cosh x] = \cosh x


Answer: f(x)=f(x)f''(x) = f(x), and f(x)=coshxf(x) = \cosh x (hyperbolic cosine)

Or: f(x)=ex+ex2f(x) = \frac{e^x + e^{-x}}{2}