Taylor and Maclaurin Series

Representing functions as infinite series

🎯 Taylor and Maclaurin Series

Taylor Polynomials Review

For a function ff with derivatives at x=ax = a:

Taylor polynomial of degree nn:

Pn(x)=f(a)+f′(a)(x−a)+f′′(a)2!(x−a)2+⋯+f(n)(a)n!(x−a)nP_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n

=∑k=0nf(k)(a)k!(x−a)k= \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k

This is a finite polynomial that approximates ff near x=ax = a.

Taylor Series (Infinite!)

If we let n→∞n \to \infty, we get the Taylor series:

f(x)=∑n=0∞f(n)(a)n!(x−a)nf(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n

=f(a)+f′(a)(x−a)+f′′(a)2!(x−a)2+f′′′(a)3!(x−a)3+⋯= f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots

💡 Key Idea: Taylor series is the "best" power series representation of ff centered at aa!

Maclaurin Series (Special Case)

When a=0a = 0, we get a Maclaurin series:

f(x)=∑n=0∞f(n)(0)n!xnf(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n

=f(0)+f′(0)x+f′′(0)2!x2+f′′′(0)3!x3+⋯= f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots

Most common Taylor series are Maclaurin series!

Finding Taylor Series - The Recipe

Step 1: Find derivatives f′(x),f′′(x),f′′′(x),…f'(x), f''(x), f'''(x), \ldots

Step 2: Evaluate at center: f(a),f′(a),f′′(a),…f(a), f'(a), f''(a), \ldots

Step 3: Look for pattern in derivatives

Step 4: Write general term f(n)(a)n!(x−a)n\frac{f^{(n)}(a)}{n!}(x-a)^n

Step 5: Sum from n=0n = 0 to ∞\infty

Example 1: Maclaurin Series for exe^x

Step 1: Find derivatives

f(x)=ex,f′(x)=ex,f′′(x)=ex,…f(x) = e^x, \quad f'(x) = e^x, \quad f''(x) = e^x, \quad \ldots

All derivatives are exe^x!

Step 2: Evaluate at a=0a = 0

f(0)=e0=1f(0) = e^0 = 1 f′(0)=1f'(0) = 1 f′′(0)=1f''(0) = 1 f(n)(0)=1f^{(n)}(0) = 1 for all nn

Step 3: Write Taylor series

ex=∑n=0∞1n!xn=1+x+x22!+x33!+x44!+⋯e^x = \sum_{n=0}^{\infty} \frac{1}{n!}x^n = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots

Converges for all xx! (radius R=∞R = \infty)

Example 2: Maclaurin Series for sin⁡x\sin x

Step 1: Find derivatives

f(x)=sin⁡xf(x) = \sin x f′(x)=cos⁡xf'(x) = \cos x f′′(x)=−sin⁡xf''(x) = -\sin x f′′′(x)=−cos⁡xf'''(x) = -\cos x f(4)(x)=sin⁡xf^{(4)}(x) = \sin x (pattern repeats!)

Step 2: Evaluate at x=0x = 0

f(0)=sin⁡0=0f(0) = \sin 0 = 0 f′(0)=cos⁡0=1f'(0) = \cos 0 = 1 f′′(0)=−sin⁡0=0f''(0) = -\sin 0 = 0 f′′′(0)=−cos⁡0=−1f'''(0) = -\cos 0 = -1 f(4)(0)=sin⁡0=0f^{(4)}(0) = \sin 0 = 0 f(5)(0)=cos⁡0=1f^{(5)}(0) = \cos 0 = 1

Pattern: 0,1,0,−1,0,1,0,−1,…0, 1, 0, -1, 0, 1, 0, -1, \ldots

Step 3: Only odd powers survive!

sin⁡x=x−x33!+x55!−x77!+⋯\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots

=∑n=0∞(−1)n(2n+1)!x2n+1= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1}

Converges for all xx!

Example 3: Maclaurin Series for cos⁡x\cos x

Pattern in derivatives at x=0x = 0:

f(0)=1,f′(0)=0,f′′(0)=−1,f′′′(0)=0,f(4)(0)=1f(0) = 1, \quad f'(0) = 0, \quad f''(0) = -1, \quad f'''(0) = 0, \quad f^{(4)}(0) = 1

Pattern: 1,0,−1,0,1,0,−1,0,…1, 0, -1, 0, 1, 0, -1, 0, \ldots

Only even powers!

cos⁡x=1−x22!+x44!−x66!+⋯\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots

=∑n=0∞(−1)n(2n)!x2n= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n}

Converges for all xx!

Example 4: Taylor Series for ln⁡x\ln x centered at a=1a = 1

Step 1: Find derivatives

f(x)=ln⁡xf(x) = \ln x f′(x)=1x=x−1f'(x) = \frac{1}{x} = x^{-1} f′′(x)=−x−2f''(x) = -x^{-2} f′′′(x)=2x−3f'''(x) = 2x^{-3} f(4)(x)=−6x−4=−3!x−4f^{(4)}(x) = -6x^{-4} = -3!x^{-4} f(n)(x)=(−1)n−1(n−1)!xnf^{(n)}(x) = \frac{(-1)^{n-1}(n-1)!}{x^n} for n≥1n \geq 1

Step 2: Evaluate at x=1x = 1

f(1)=ln⁡1=0f(1) = \ln 1 = 0 f′(1)=1f'(1) = 1 f′′(1)=−1f''(1) = -1 f′′′(1)=2f'''(1) = 2 f(n)(1)=(−1)n−1(n−1)!f^{(n)}(1) = (-1)^{n-1}(n-1)! for n≥1n \geq 1

Step 3: Write Taylor series

ln⁡x=∑n=1∞(−1)n−1(n−1)!n!(x−1)n\ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(n-1)!}{n!}(x-1)^n

=∑n=1∞(−1)n−1n(x−1)n= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}(x-1)^n

=(x−1)−(x−1)22+(x−1)33−(x−1)44+⋯= (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \cdots

Converges for 0<x≤20 < x \leq 2 (interval: (0,2](0, 2])

Relationship Between sin⁡x\sin x and cos⁡x\cos x

Notice: ddx[sin⁡x]=cos⁡x\frac{d}{dx}[\sin x] = \cos x

Differentiate the series for sin⁡x\sin x:

ddx[x−x33!+x55!−⋯ ]\frac{d}{dx}\left[x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\right]

=1−3x23!+5x45!−⋯= 1 - \frac{3x^2}{3!} + \frac{5x^4}{5!} - \cdots

=1−x22!+x44!−⋯=cos⁡x= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots = \cos x ✓

Series confirm calculus relationships!

Using Known Series (Smart Way!)

Instead of computing all derivatives, use:

  • Substitution
  • Differentiation/Integration
  • Algebraic manipulation

Example 5: Find Series for e−x2e^{-x^2}

Start with: ex=∑n=0∞xnn!e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}

Substitute x→−x2x \to -x^2:

e−x2=∑n=0∞(−x2)nn!=∑n=0∞(−1)nx2nn!e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}

=1−x2+x42!−x63!+x84!−⋯= 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \frac{x^8}{4!} - \cdots

Much easier than computing derivatives!

Example 6: Find Series for xsin⁡xx \sin x

Start with: sin⁡x=∑n=0∞(−1)nx2n+1(2n+1)!\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}

Multiply by xx:

xsin⁡x=x⋅∑n=0∞(−1)nx2n+1(2n+1)!x \sin x = x \cdot \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}

=∑n=0∞(−1)nx2n+2(2n+1)!= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+2}}{(2n+1)!}

=x2−x43!+x65!−x87!+⋯= x^2 - \frac{x^4}{3!} + \frac{x^6}{5!} - \frac{x^8}{7!} + \cdots

Example 7: Find Series for ∫e−x2dx\int e^{-x^2} dx

From Example 5: e−x2=∑n=0∞(−1)nx2nn!e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}

Integrate term by term:

∫e−x2dx=C+∑n=0∞(−1)nx2n+1n!(2n+1)\int e^{-x^2} dx = C + \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{n!(2n+1)}

=C+x−x31!⋅3+x52!⋅5−x73!⋅7+⋯= C + x - \frac{x^3}{1! \cdot 3} + \frac{x^5}{2! \cdot 5} - \frac{x^7}{3! \cdot 7} + \cdots

Note: This integral has no elementary antiderivative, but we can express it as a series!

Taylor Series Centered at a≠0a \neq 0

Example: Find Taylor series for exe^x centered at a=2a = 2.

All derivatives of exe^x are exe^x, so:

f(n)(2)=e2f^{(n)}(2) = e^2 for all nn

Taylor series:

ex=∑n=0∞e2n!(x−2)ne^x = \sum_{n=0}^{\infty} \frac{e^2}{n!}(x-2)^n

=e2[1+(x−2)+(x−2)22!+(x−2)33!+⋯ ]= e^2 \left[1 + (x-2) + \frac{(x-2)^2}{2!} + \frac{(x-2)^3}{3!} + \cdots\right]

=e2∑n=0∞(x−2)nn!= e^2 \sum_{n=0}^{\infty} \frac{(x-2)^n}{n!}

When Does Taylor Series Equal the Function?

The Taylor series equals f(x)f(x) when:

lim⁡n→∞Rn(x)=0\lim_{n \to \infty} R_n(x) = 0

where Rn(x)R_n(x) is the remainder (error) after nn terms.

For most common functions (like ex,sin⁡x,cos⁡x,ln⁡(1+x)e^x, \sin x, \cos x, \ln(1+x)), this happens on their interval of convergence.

⚠️ Common Mistakes

Mistake 1: Wrong Factorial in General Term

For sin⁡x=x−x33!+x55!−⋯\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots:

WRONG: General term is (−1)nxnn!\frac{(-1)^n x^n}{n!}

RIGHT: General term is (−1)nx2n+1(2n+1)!\frac{(-1)^n x^{2n+1}}{(2n+1)!} (only odd powers!)

Mistake 2: Wrong Starting Index

For sin⁡x\sin x, first nonzero term is xx (when n=0n = 0).

WRONG: Sum starts at n=1n = 1

RIGHT: ∑n=0∞(−1)nx2n+1(2n+1)!\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} (starts at n=0n = 0)

Mistake 3: Forgetting Center for Taylor Series

For ln⁡x\ln x centered at a=1a = 1:

WRONG: ln⁡x=∑(−1)n−1xnn\ln x = \sum \frac{(-1)^{n-1} x^n}{n}

RIGHT: ln⁡x=∑(−1)n−1(x−1)nn\ln x = \sum \frac{(-1)^{n-1} (x-1)^n}{n} (must use (x−1)n(x-1)^n!)

Mistake 4: Computing All Derivatives Unnecessarily

To find series for e3xe^{3x}:

WRONG: Compute f′(x),f′′(x),f′′′(x),…f'(x), f''(x), f'''(x), \ldots

RIGHT: Use ex=∑xnn!e^x = \sum \frac{x^n}{n!}, substitute x→3xx \to 3x:

e3x=∑n=0∞(3x)nn!=∑n=0∞3nxnn!e^{3x} = \sum_{n=0}^{\infty} \frac{(3x)^n}{n!} = \sum_{n=0}^{\infty} \frac{3^n x^n}{n!}

📝 Practice Strategy

  1. Memorize basic series: ex,sin⁡x,cos⁡x,11−x,ln⁡(1+x)e^x, \sin x, \cos x, \frac{1}{1-x}, \ln(1+x)
  2. Use substitution: Replace xx with f(x)f(x) in known series
  3. Multiply/divide by powers: For xkf(x)x^k f(x), just multiply series by xkx^k
  4. Differentiate/integrate: Term by term when needed
  5. For derivatives at aa: Look for patterns to avoid computing all
  6. Check first few terms: Make sure they match Taylor polynomial
  7. Odd/even functions: sin⁡x\sin x has only odd powers, cos⁡x\cos x only even
  8. Write general term: Essential for summation notation

📚 Practice Problems

1Problem 1medium

❓ Question:

Find the Maclaurin series for f(x)=11−xf(x) = \frac{1}{1-x} and state the interval of convergence.

💡 Show Solution

Method 1: Using derivatives

f(x)=(1−x)−1f(x) = (1-x)^{-1} f′(x)=(1−x)−2f'(x) = (1-x)^{-2} f′′(x)=2(1−x)−3f''(x) = 2(1-x)^{-3} f′′′(x)=6(1−x)−4=3!(1−x)−4f'''(x) = 6(1-x)^{-4} = 3!(1-x)^{-4} f(n)(x)=n!(1−x)−(n+1)f^{(n)}(x) = n!(1-x)^{-(n+1)}

At x=0x = 0: f(n)(0)=n!f^{(n)}(0) = n!

Maclaurin series: 11−x=∑n=0∞n!n!xn=∑n=0∞xn\frac{1}{1-x} = \sum_{n=0}^{\infty} \frac{n!}{n!}x^n = \sum_{n=0}^{\infty} x^n

=1+x+x2+x3+x4+⋯= 1 + x + x^2 + x^3 + x^4 + \cdots

Method 2: Geometric series (faster!)

This is just the geometric series with ratio r=xr = x.

11−x=∑n=0∞xn\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n

Interval of convergence:

Geometric series converges when ∣r∣<1|r| < 1:

∣x∣<1|x| < 1

At x=1x = 1: ∑1\sum 1 diverges

At x=−1x = -1: ∑(−1)n\sum (-1)^n diverges

Interval: (−1,1)(-1, 1)

Answer: 11−x=∑n=0∞xn\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n for ∣x∣<1|x| < 1

2Problem 2hard

❓ Question:

Find the first four nonzero terms of the Maclaurin series for f(x)=e2xf(x) = e^{2x}.

💡 Show Solution

Solution:

Method 1: Using the definition

Maclaurin series: f(x)=∑n=0∞f(n)(0)n!xnf(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n

Find derivatives:

  • f(x)=e2xf(x) = e^{2x}, so f(0)=1f(0) = 1
  • f′(x)=2e2xf'(x) = 2e^{2x}, so f′(0)=2f'(0) = 2
  • f′′(x)=4e2xf''(x) = 4e^{2x}, so f′′(0)=4f''(0) = 4
  • f′′′(x)=8e2xf'''(x) = 8e^{2x}, so f′′′(0)=8f'''(0) = 8
  • f(4)(x)=16e2xf^{(4)}(x) = 16e^{2x}, so f(4)(0)=16f^{(4)}(0) = 16

Series: e2x=1+2x1!+4x22!+8x33!+16x44!+...e^{2x} = 1 + \frac{2x}{1!} + \frac{4x^2}{2!} + \frac{8x^3}{3!} + \frac{16x^4}{4!} + ...

=1+2x+2x2+4x33+2x43+...= 1 + 2x + 2x^2 + \frac{4x^3}{3} + \frac{2x^4}{3} + ...

Method 2: Using known series

We know: 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!} + ...

Substitute 2x2x for xx:

e2x=1+2x+(2x)22!+(2x)33!+(2x)44!+...e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \frac{(2x)^4}{4!} + ...

First four terms: 1+2x+2x2+4x331 + 2x + 2x^2 + \frac{4x^3}{3}

3Problem 3hard

❓ Question:

Use the Maclaurin series for exe^x to find the Maclaurin series for f(x)=e2xcos⁡xf(x) = e^{2x} \cos x.

💡 Show Solution

Step 1: Write known series

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!} + \cdots

cos⁡x=∑n=0∞(−1)nx2n(2n)!=1−x22!+x44!−⋯\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots

Step 2: Find e2xe^{2x}

Substitute x→2xx \to 2x in series for exe^x:

e2x=∑n=0∞(2x)nn!=∑n=0∞2nxnn!e^{2x} = \sum_{n=0}^{\infty} \frac{(2x)^n}{n!} = \sum_{n=0}^{\infty} \frac{2^n x^n}{n!}

=1+2x+4x22!+8x33!+⋯= 1 + 2x + \frac{4x^2}{2!} + \frac{8x^3}{3!} + \cdots

=1+2x+2x2+4x33+⋯= 1 + 2x + 2x^2 + \frac{4x^3}{3} + \cdots

Step 3: Multiply series

e2xcos⁡x=(1+2x+2x2+4x33+⋯ )(1−x22+x424−⋯ )e^{2x} \cos x = \left(1 + 2x + 2x^2 + \frac{4x^3}{3} + \cdots\right)\left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots\right)

Multiply term by term (collect like powers):

Constant term: 1⋅1=11 \cdot 1 = 1

xx term: 2x⋅1=2x2x \cdot 1 = 2x

x2x^2 term: 2x2⋅1+1⋅(−x22)=2x2−x22=3x222x^2 \cdot 1 + 1 \cdot (-\frac{x^2}{2}) = 2x^2 - \frac{x^2}{2} = \frac{3x^2}{2}

x3x^3 term: 4x33⋅1+2x⋅(−x22)=4x33−x3=x33\frac{4x^3}{3} \cdot 1 + 2x \cdot (-\frac{x^2}{2}) = \frac{4x^3}{3} - x^3 = \frac{x^3}{3}

x4x^4 term: (from multiple products) 1⋅x424+2x2⋅(−x22)+(higher order terms)1 \cdot \frac{x^4}{24} + 2x^2 \cdot (-\frac{x^2}{2}) + (\text{higher order terms}) =x424−x4+⋯= \frac{x^4}{24} - x^4 + \cdots

(This gets tedious!)

First few terms:

e2xcos⁡x=1+2x+3x22+x33+⋯e^{2x} \cos x = 1 + 2x + \frac{3x^2}{2} + \frac{x^3}{3} + \cdots

Answer: e2xcos⁡x=1+2x+3x22+x33+⋯e^{2x}\cos x = 1 + 2x + \frac{3x^2}{2} + \frac{x^3}{3} + \cdots

(Usually only first 3-4 terms are requested)

4Problem 4medium

❓ Question:

Find the Taylor series for f(x)=1xf(x) = \frac{1}{x} centered at a=1a = 1.

💡 Show Solution

Step 1: Rewrite in useful form

f(x)=1x=11+(x−1)f(x) = \frac{1}{x} = \frac{1}{1 + (x-1)}

Let u=x−1u = x - 1, so x=1+ux = 1 + u:

f(x)=11+uf(x) = \frac{1}{1+u}

Step 2: Use geometric series

We know: 11−r=∑n=0∞rn\frac{1}{1-r} = \sum_{n=0}^{\infty} r^n for ∣r∣<1|r| < 1

Replace rr with −u-u:

11+u=11−(−u)=∑n=0∞(−u)n=∑n=0∞(−1)nun\frac{1}{1+u} = \frac{1}{1-(-u)} = \sum_{n=0}^{\infty} (-u)^n = \sum_{n=0}^{\infty} (-1)^n u^n

Step 3: Substitute back u=x−1u = x-1

1x=∑n=0∞(−1)n(x−1)n\frac{1}{x} = \sum_{n=0}^{\infty} (-1)^n (x-1)^n

=1−(x−1)+(x−1)2−(x−1)3+⋯= 1 - (x-1) + (x-1)^2 - (x-1)^3 + \cdots

Step 4: Find interval of convergence

Need ∣u∣<1|u| < 1: ∣x−1∣<1|x-1| < 1 −1<x−1<1-1 < x - 1 < 1 0<x<20 < x < 2

Answer: 1x=∑n=0∞(−1)n(x−1)n\frac{1}{x} = \sum_{n=0}^{\infty} (-1)^n(x-1)^n for 0<x<20 < x < 2

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