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10.15 - Representing Functions as Power Series

Quick Reference: The Essential Series

From the previous lesson, these need to be memorized:
ex=∑n=0∞xnn!=1+x+x22!+x33!+⋯e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdotssin⁡(x)=∑n=0∞(−1)nx2n+1(2n+1)!=x−x33!+x55!−x77!+⋯\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!} + \cdotscos⁡(x)=∑n=0∞(−1)nx2n(2n)!=1−x22!+x44!−x66!+⋯\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!} + \cdots11+x=∑n=0∞(−1)nxn=1−x+x2−x3+⋯\frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n = 1 - x + x^2 - x^3 + \cdots

Differentiating Power Series

Power series can be differentiated term by term within their interval of convergence.
If f(x)=∑n=0∞anxn, then f′(x)=∑n=0∞ddx[anxn]\text{If } f(x) = \sum_{n=0}^{\infty} a_n x^n, \text{ then } f'(x) = \sum_{n=0}^{\infty} \frac{d}{dx}[a_n x^n]
There are two methods, and both give the same result. Pick whichever is more convenient.

Method 1: Differentiate Term by Term

Strategy: write out several terms of the series, differentiate each one individually, then find the pattern.

Example 1: Derivative of an Exponential Series

Given: f(x)=∑n=0∞x3nn!f(x) = \sum_{n=0}^{\infty} \frac{x^{3n}}{n!}
Step 1: Write out the first few terms
f(x)=1+x3+x62!+x93!+x124!+⋯f(x) = 1 + x^3 + \frac{x^6}{2!} + \frac{x^9}{3!} + \frac{x^{12}}{4!} + \cdots
Step 2: Differentiate each term
ddx[1]=0\frac{d}{dx}[1] = 0ddx[x3]=3x2\frac{d}{dx}[x^3] = 3x^2ddx[x62!]=6x52!=3x51!\frac{d}{dx}\left[\frac{x^6}{2!}\right] = \frac{6x^5}{2!} = \frac{3x^5}{1!}ddx[x93!]=9x83!=3x82!\frac{d}{dx}\left[\frac{x^9}{3!}\right] = \frac{9x^8}{3!} = \frac{3x^8}{2!}ddx[x124!]=12x114!=3x113!\frac{d}{dx}\left[\frac{x^{12}}{4!}\right] = \frac{12x^{11}}{4!} = \frac{3x^{11}}{3!}
Step 3: Write the derivative series
f′(x)=3x2+3x51!+3x82!+3x113!+⋯f'(x) = 3x^2 + \frac{3x^5}{1!} + \frac{3x^8}{2!} + \frac{3x^{11}}{3!} + \cdots

Method 2: Differentiate the Summation Directly

Strategy: apply the derivative to the general term in summation notation, then adjust the starting index if you need to.

Example 1 Revisited: Using Direct Differentiation

Given: f(x)=∑n=0∞x3nn!f(x) = \sum_{n=0}^{\infty} \frac{x^{3n}}{n!}
Step 1: Differentiate the general term
f′(x)=∑n=0∞ddx[x3nn!]=∑n=0∞3n⋅x3n−1n!f'(x) = \sum_{n=0}^{\infty} \frac{d}{dx}\left[\frac{x^{3n}}{n!}\right] = \sum_{n=0}^{\infty} \frac{3n \cdot x^{3n-1}}{n!}
Step 2: Adjust the starting index
When n=0n = 0, the term becomes 3(0)⋅x−10!=0\frac{3(0) \cdot x^{-1}}{0!} = 0. Since the first term is zero, we start at n=1n = 1:
f′(x)=∑n=1∞3n⋅x3n−1n!f'(x) = \sum_{n=1}^{\infty} \frac{3n \cdot x^{3n-1}}{n!}
Step 3: Simplify by factoring
The nn in the numerator cancels with one factor in n!n!:
nn!=nn⋅(n−1)!=1(n−1)!\frac{n}{n!} = \frac{n}{n \cdot (n-1)!} = \frac{1}{(n-1)!}
Therefore:
f′(x)=∑n=1∞3x3n−1(n−1)!f'(x) = \sum_{n=1}^{\infty} \frac{3x^{3n-1}}{(n-1)!}
Step 4: Verify this matches Method 1
Plugging in n=1,2,3,…n = 1, 2, 3, \ldots:
  • n=1n = 1: 3x20!=3x2\frac{3x^{2}}{0!} = 3x^2 ✓
  • n=2n = 2: 3x51!=3x5\frac{3x^{5}}{1!} = 3x^5 ✓
  • n=3n = 3: 3x82!\frac{3x^{8}}{2!} ✓
That matches the term-by-term result.

Important: Adjusting the Starting Index

Key principle: after differentiating, if the first term (at n=0n = 0) equals zero, change the starting index to n=1n = 1.
Why this matters: - Starting at n=0n = 0 would include a meaningless zero term - Starting at n=1n = 1 makes the series cleaner - Both methods have to arrive at the same pattern
Common situation:
∑n=0∞n⋅(expression)→∑n=1∞n⋅(expression)\sum_{n=0}^{\infty} n \cdot (\text{expression}) \quad \rightarrow \quad \sum_{n=1}^{\infty} n \cdot (\text{expression})
The n=0n = 0 term vanishes because of the factor of nn out front.

Integrating Power Series

Power series can be integrated term by term within their interval of convergence.
If f(x)=∑n=0∞anxn, then ∫f(x) dx=∑n=0∞∫anxn dx\text{If } f(x) = \sum_{n=0}^{\infty} a_n x^n, \text{ then } \int f(x) \, dx = \sum_{n=0}^{\infty} \int a_n x^n \, dx
The big advantage: some integrals that are difficult or flat-out impossible to evaluate directly become manageable once you represent the function as a power series.

Example 2: Integrating a Sine Function

Evaluate ∫0xsin⁡(t5) dt\int_0^x \sin(t^5) \, dt.
The challenge: the standard techniques, like substitution and integration by parts, don’t work here.
The solution: use the power series representation.
Step 1: Write the Maclaurin series for sin⁡(t)\sin(t)
sin⁡(t)=t−t33!+t55!−t77!+⋯\sin(t) = t - \frac{t^3}{3!} + \frac{t^5}{5!} - \frac{t^7}{7!} + \cdots
Step 2: Substitute t5t^5 for tt
sin⁡(t5)=t5−(t5)33!+(t5)55!−(t5)77!+⋯\sin(t^5) = t^5 - \frac{(t^5)^3}{3!} + \frac{(t^5)^5}{5!} - \frac{(t^5)^7}{7!} + \cdots=t5−t153!+t255!−t357!+⋯= t^5 - \frac{t^{15}}{3!} + \frac{t^{25}}{5!} - \frac{t^{35}}{7!} + \cdots
Step 3: Integrate term by term
∫0xsin⁡(t5) dt=∫0x(t5−t153!+t255!−t357!+⋯ )dt\int_0^x \sin(t^5) \, dt = \int_0^x \left(t^5 - \frac{t^{15}}{3!} + \frac{t^{25}}{5!} - \frac{t^{35}}{7!} + \cdots\right) dt=[t66−t1616⋅3!+t2626⋅5!−t3636⋅7!+⋯ ]0x= \left[\frac{t^6}{6} - \frac{t^{16}}{16 \cdot 3!} + \frac{t^{26}}{26 \cdot 5!} - \frac{t^{36}}{36 \cdot 7!} + \cdots\right]_0^x
Step 4: Evaluate at the bounds
When t=0t = 0, every term vanishes. When t=xt = x:
=x66−x1616⋅3!+x2626⋅5!−x3636⋅7!+⋯= \frac{x^6}{6} - \frac{x^{16}}{16 \cdot 3!} + \frac{x^{26}}{26 \cdot 5!} - \frac{x^{36}}{36 \cdot 7!} + \cdots
Simplified answer:
∫0xsin⁡(t5) dt=x66−x1616⋅6+x2626⋅120−x3636⋅5040+⋯\int_0^x \sin(t^5) \, dt = \frac{x^6}{6} - \frac{x^{16}}{16 \cdot 6} + \frac{x^{26}}{26 \cdot 120} - \frac{x^{36}}{36 \cdot 5040} + \cdots
Note: usually you leave the denominators unfactored, like 16⋅3!16 \cdot 3! rather than 96, unless the problem specifically asks you to simplify.

Example 3: Another Integration Using Series

Evaluate ∫0xe−t2 dt\int_0^x e^{-t^2} \, dt.
Step 1: Start with the series for ete^t
et=1+t+t22!+t33!+t44!+⋯e^t = 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \frac{t^4}{4!} + \cdots
Step 2: Substitute −t2-t^2 for tt
e−t2=1+(−t2)+(−t2)22!+(−t2)33!+(−t2)44!+⋯e^{-t^2} = 1 + (-t^2) + \frac{(-t^2)^2}{2!} + \frac{(-t^2)^3}{3!} + \frac{(-t^2)^4}{4!} + \cdots=1−t2+t42!−t63!+t84!−⋯= 1 - t^2 + \frac{t^4}{2!} - \frac{t^6}{3!} + \frac{t^8}{4!} - \cdots
Step 3: Integrate term by term
∫0xe−t2 dt=∫0x(1−t2+t42!−t63!+t84!−⋯ )dt\int_0^x e^{-t^2} \, dt = \int_0^x \left(1 - t^2 + \frac{t^4}{2!} - \frac{t^6}{3!} + \frac{t^8}{4!} - \cdots\right) dt=[t−t33+t55⋅2!−t77⋅3!+t99⋅4!−⋯ ]0x= \left[t - \frac{t^3}{3} + \frac{t^5}{5 \cdot 2!} - \frac{t^7}{7 \cdot 3!} + \frac{t^9}{9 \cdot 4!} - \cdots\right]_0^x
Step 4: Evaluate
=x−x33+x510−x742+x9216−⋯= x - \frac{x^3}{3} + \frac{x^5}{10} - \frac{x^7}{42} + \frac{x^9}{216} - \cdots

General Patterns

Differentiation Pattern

ddx[∑n=0∞anxn]=∑n=1∞n⋅anxn−1\frac{d}{dx}\left[\sum_{n=0}^{\infty} a_n x^n\right] = \sum_{n=1}^{\infty} n \cdot a_n x^{n-1}
Index shift: it often changes from n=0n = 0 to n=1n = 1.

Integration Pattern

∫[∑n=0∞anxn]dx=∑n=0∞anxn+1n+1+C\int \left[\sum_{n=0}^{\infty} a_n x^n\right] dx = \sum_{n=0}^{\infty} \frac{a_n x^{n+1}}{n+1} + C
Index stays the same: it usually stays at n=0n = 0.

Building Series from the Geometric Series

The geometric series is a ready-made template:
a1−r=∑n=0∞arn=a+ar+ar2+⋯when ∣r∣<1\frac{a}{1 - r} = \sum_{n=0}^{\infty} a r^n = a + ar + ar^2 + \cdots \quad \text{when } |r| < 1
If a little algebra turns a function into first term1−ratio\frac{\text{first term}}{1 - \text{ratio}}, you can write its power series right away, and ∣ratio∣<1|\text{ratio}| < 1 gives the interval of convergence. Combine this with substitution, differentiation, and integration to reach many more functions.

Example 4: Algebra, Then the Geometric Series

Find a power series for f(x)=32−xf(x) = \dfrac{3}{2 - x} and its interval of convergence.
Step 1: Get a 1 in front of the denominator. Factor out the 2:
32−x=32⋅11−x2\frac{3}{2 - x} = \frac{3}{2} \cdot \frac{1}{1 - \frac{x}{2}}
Step 2: Use the geometric series with a=32a = \frac{3}{2} and r=x2r = \frac{x}{2}:
32−x=∑n=0∞32(x2)n=∑n=0∞3xn2n+1=32+3x4+3x28+3x316+⋯\frac{3}{2 - x} = \sum_{n=0}^{\infty} \frac{3}{2}\left(\frac{x}{2}\right)^n = \sum_{n=0}^{\infty} \frac{3x^n}{2^{n+1}} = \frac{3}{2} + \frac{3x}{4} + \frac{3x^2}{8} + \frac{3x^3}{16} + \cdots
Step 3: Find the interval. The series converges only when ∣x2∣<1\left|\frac{x}{2}\right| < 1, so −2<x<2-2 < x < 2. At x=±2x = \pm 2 the ratio is ±1\pm 1 and the terms don’t go to 0, so both endpoints diverge. The interval is (−2,2)(-2, 2).

Example 5: Integrate a Geometric Series

Find a power series for ln⁡(1+x)\ln(1 + x) and its interval of convergence.
Step 1: Start from a known series. Since ddxln⁡(1+x)=11+x\frac{d}{dx}\ln(1 + x) = \frac{1}{1 + x}, use
11+t=1−t+t2−t3+⋯for ∣t∣<1\frac{1}{1 + t} = 1 - t + t^2 - t^3 + \cdots \quad \text{for } |t| < 1
Step 2: Integrate term by term from 00 to xx. Because ln⁡(1+0)=0\ln(1 + 0) = 0, no extra constant shows up:
ln⁡(1+x)=∫0x11+t dt=x−x22+x33−x44+⋯=∑n=0∞(−1)nxn+1n+1\ln(1 + x) = \int_0^x \frac{1}{1 + t}\,dt = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1}
Step 3: Check the endpoints. Integrating keeps the radius at R=1R = 1, but the endpoints can change. At x=1x = 1 the series is 1−12+13−⋯1 - \frac{1}{2} + \frac{1}{3} - \cdots, the alternating harmonic series, which converges (that’s where its sum ln⁡2\ln 2 comes from). At x=−1x = -1 it’s −(1+12+13+⋯ )-\left(1 + \frac{1}{2} + \frac{1}{3} + \cdots\right), which diverges. So the interval is (−1,1](-1, 1], one more endpoint than the series for 11+x\frac{1}{1 + x} had.

When to Use a Series Representation

Use power series integration when:
  1. Standard integration techniques fail
  1. The integrand involves compositions like sin⁡(x2)\sin(x^2), e−x2e^{-x^2}, or cos⁡(x3)\cos(x^3)
  1. You need a polynomial approximation of an integral
  1. You’re working with special functions, like the error function or Fresnel integrals
Examples of “impossible” integrals that series make possible: - ∫e−x2 dx\int e^{-x^2} \, dx, which shows up in probability with the normal distribution - ∫sin⁡(x)x dx\int \frac{\sin(x)}{x} \, dx, the sine integral function - ∫1ln⁡(x) dx\int \frac{1}{\ln(x)} \, dx, the logarithmic integral

Interval of Convergence After Operations

Important property: differentiation and integration preserve the interval of convergence, except possibly at the endpoints.
Example: - If ∑anxn\sum a_n x^n converges on (−R,R)(-R, R) - Then ∑nanxn−1\sum n a_n x^{n-1} converges on (−R,R)(-R, R) - And ∑anxn+1n+1\sum \frac{a_n x^{n+1}}{n+1} converges on (−R,R)(-R, R)
Endpoint behavior may change, so the endpoints need to be checked separately after differentiating or integrating.

Practical Workflow

For differentiation problems:
  1. Identify the parent function or series
  1. Choose your method: term-by-term or direct
  1. Apply the power rule to each term, or to the general term
  1. Check whether the starting index needs adjusting
  1. Simplify if you can, by factoring out common terms
For integration problems:
  1. Recognize the integrand as a modified standard function
  1. Write the appropriate series
  1. Integrate term by term, using the power rule in reverse
  1. Apply the bounds if it’s a definite integral
  1. Leave it in series form, and don’t try to find a closed form

Common Simplifications

Factorial Manipulation

nn!=1(n−1)!\frac{n}{n!} = \frac{1}{(n-1)!}

Exponent Manipulation

(xa)b=xab(x^a)^b = x^{ab}
Power to a power means you multiply the exponents.

After Integration

xn+1n+1\frac{x^{n+1}}{n+1}
Keep the factorials separate. Write x1616⋅3!\frac{x^{16}}{16 \cdot 3!} rather than x1696\frac{x^{16}}{96}.

Multiple Choice Strategy

When you’re working series problems on a timed test:
  1. Write out the first 3 or 4 terms of the series
  1. Compare with the answer choices early
  1. Match patterns before fully simplifying
  1. Eliminate answers with the wrong signs or exponents
  1. Don’t oversimplify. Answers are often left with factorials in them

Key Takeaways

  • Term-by-term differentiation: ddx[∑anxn]=∑ddx[anxn]\frac{d}{dx}\left[\sum a_n x^n\right] = \sum \frac{d}{dx}[a_n x^n]
  • Term-by-term integration: ∫[∑anxn]dx=∑∫anxn dx\int \left[\sum a_n x^n\right] dx = \sum \int a_n x^n \, dx
  • Two methods for derivatives: term-by-term, or direct summation differentiation
  • Adjust the starting index when the first term becomes zero, which is common after differentiating
  • Integration makes hard problems tractable. ∫0xsin⁡(t5) dt\int_0^x \sin(t^5) \, dt becomes manageable
  • The interval of convergence is preserved, except possibly at the endpoints
  • Leave the denominators factored. Write 26⋅5!26 \cdot 5! rather than multiplying it out
  • The power rule still applies. Use standard calculus rules on each term
  • Simplify nn!=1(n−1)!\frac{n}{n!} = \frac{1}{(n-1)!} whenever you can
  • Both methods have to agree. Term-by-term and direct differentiation give identical results
  • Series enable “impossible” integrals. ∫e−x2dx\int e^{-x^2} dx has no elementary antiderivative
  • Memorize the four essential series so you can quickly generate derivatives and integrals
  • Geometric template: rewrite the function as a1−r\frac{a}{1 - r} to get ∑arn\sum ar^n right away, and ∣r∣<1|r| < 1 gives the interval
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