Taylor and Maclaurin Series

Representing functions as infinite series

๐ŸŽฏ 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)+fโ€ฒ(a)(xโˆ’a)+fโ€ฒโ€ฒ(a)2!(xโˆ’a)2+โ‹ฏ+f(n)(a)n!(xโˆ’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

=โˆ‘k=0nf(k)(a)k!(xโˆ’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 nโ†’โˆžn \to \infty, we get the Taylor series:

f(x)=โˆ‘n=0โˆžf(n)(a)n!(xโˆ’a)nf(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n

=f(a)+fโ€ฒ(a)(xโˆ’a)+fโ€ฒโ€ฒ(a)2!(xโˆ’a)2+fโ€ฒโ€ฒโ€ฒ(a)3!(xโˆ’a)3+โ‹ฏ= f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots

๐Ÿ’ก 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)=โˆ‘n=0โˆžf(n)(0)n!xnf(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n

=f(0)+fโ€ฒ(0)x+fโ€ฒโ€ฒ(0)2!x2+fโ€ฒโ€ฒโ€ฒ(0)3!x3+โ‹ฏ= 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 fโ€ฒ(x),fโ€ฒโ€ฒ(x),fโ€ฒโ€ฒโ€ฒ(x),โ€ฆf'(x), f''(x), f'''(x), \ldots

Step 2: Evaluate at center: f(a),fโ€ฒ(a),fโ€ฒโ€ฒ(a),โ€ฆf(a), f'(a), f''(a), \ldots

Step 3: Look for pattern in derivatives

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

Step 5: Sum from n=0n = 0 to โˆž\infty


Example 1: Maclaurin Series for exe^x

Step 1: Find derivatives

f(x)=ex,fโ€ฒ(x)=ex,fโ€ฒโ€ฒ(x)=ex,โ€ฆf(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 fโ€ฒ(0)=1f'(0) = 1 fโ€ฒโ€ฒ(0)=1f''(0) = 1 f(n)(0)=1f^{(n)}(0) = 1 for all nn


Step 3: Write Taylor series

ex=โˆ‘n=0โˆž1n!xn=1+x+x22!+x33!+x44!+โ‹ฏe^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=โˆžR = \infty)


Example 2: Maclaurin Series for sinโกx\sin x

Step 1: Find derivatives

f(x)=sinโกxf(x) = \sin x fโ€ฒ(x)=cosโกxf'(x) = \cos x fโ€ฒโ€ฒ(x)=โˆ’sinโกxf''(x) = -\sin x fโ€ฒโ€ฒโ€ฒ(x)=โˆ’cosโกxf'''(x) = -\cos x f(4)(x)=sinโกxf^{(4)}(x) = \sin x (pattern repeats!)


Step 2: Evaluate at x=0x = 0

f(0)=sinโก0=0f(0) = \sin 0 = 0 fโ€ฒ(0)=cosโก0=1f'(0) = \cos 0 = 1 fโ€ฒโ€ฒ(0)=โˆ’sinโก0=0f''(0) = -\sin 0 = 0 fโ€ฒโ€ฒโ€ฒ(0)=โˆ’cosโก0=โˆ’1f'''(0) = -\cos 0 = -1 f(4)(0)=sinโก0=0f^{(4)}(0) = \sin 0 = 0 f(5)(0)=cosโก0=1f^{(5)}(0) = \cos 0 = 1

Pattern: 0,1,0,โˆ’1,0,1,0,โˆ’1,โ€ฆ0, 1, 0, -1, 0, 1, 0, -1, \ldots


Step 3: Only odd powers survive!

sinโกx=xโˆ’x33!+x55!โˆ’x77!+โ‹ฏ\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots

=โˆ‘n=0โˆž(โˆ’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 cosโกx\cos x

Pattern in derivatives at x=0x = 0:

f(0)=1,fโ€ฒ(0)=0,fโ€ฒโ€ฒ(0)=โˆ’1,fโ€ฒโ€ฒโ€ฒ(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,โˆ’1,0,1,0,โˆ’1,0,โ€ฆ1, 0, -1, 0, 1, 0, -1, 0, \ldots


Only even powers!

cosโกx=1โˆ’x22!+x44!โˆ’x66!+โ‹ฏ\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots

=โˆ‘n=0โˆž(โˆ’1)n(2n)!x2n= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n}

Converges for all xx!


Example 4: Taylor Series for lnโกx\ln x centered at a=1a = 1

Step 1: Find derivatives

f(x)=lnโกxf(x) = \ln x fโ€ฒ(x)=1x=xโˆ’1f'(x) = \frac{1}{x} = x^{-1} fโ€ฒโ€ฒ(x)=โˆ’xโˆ’2f''(x) = -x^{-2} fโ€ฒโ€ฒโ€ฒ(x)=2xโˆ’3f'''(x) = 2x^{-3} f(4)(x)=โˆ’6xโˆ’4=โˆ’3!xโˆ’4f^{(4)}(x) = -6x^{-4} = -3!x^{-4} f(n)(x)=(โˆ’1)nโˆ’1(nโˆ’1)!xnf^{(n)}(x) = \frac{(-1)^{n-1}(n-1)!}{x^n} for nโ‰ฅ1n \geq 1


Step 2: Evaluate at x=1x = 1

f(1)=lnโก1=0f(1) = \ln 1 = 0 fโ€ฒ(1)=1f'(1) = 1 fโ€ฒโ€ฒ(1)=โˆ’1f''(1) = -1 fโ€ฒโ€ฒโ€ฒ(1)=2f'''(1) = 2 f(n)(1)=(โˆ’1)nโˆ’1(nโˆ’1)!f^{(n)}(1) = (-1)^{n-1}(n-1)! for nโ‰ฅ1n \geq 1


Step 3: Write Taylor series

lnโกx=โˆ‘n=1โˆž(โˆ’1)nโˆ’1(nโˆ’1)!n!(xโˆ’1)n\ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(n-1)!}{n!}(x-1)^n

=โˆ‘n=1โˆž(โˆ’1)nโˆ’1n(xโˆ’1)n= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}(x-1)^n

=(xโˆ’1)โˆ’(xโˆ’1)22+(xโˆ’1)33โˆ’(xโˆ’1)44+โ‹ฏ= (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \cdots

Converges for 0<xโ‰ค20 < x \leq 2 (interval: (0,2](0, 2])


Relationship Between sinโกx\sin x and cosโกx\cos x

Notice: ddx[sinโกx]=cosโกx\frac{d}{dx}[\sin x] = \cos x

Differentiate the series for sinโกx\sin x:

ddx[xโˆ’x33!+x55!โˆ’โ‹ฏโ€‰]\frac{d}{dx}\left[x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\right]

=1โˆ’3x23!+5x45!โˆ’โ‹ฏ= 1 - \frac{3x^2}{3!} + \frac{5x^4}{5!} - \cdots

=1โˆ’x22!+x44!โˆ’โ‹ฏ=cosโกx= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots = \cos x โœ“

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 eโˆ’x2e^{-x^2}

Start with: ex=โˆ‘n=0โˆžxnn!e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}

Substitute xโ†’โˆ’x2x \to -x^2:

eโˆ’x2=โˆ‘n=0โˆž(โˆ’x2)nn!=โˆ‘n=0โˆž(โˆ’1)nx2nn!e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}

=1โˆ’x2+x42!โˆ’x63!+x84!โˆ’โ‹ฏ= 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 xsinโกxx \sin x

Start with: sinโกx=โˆ‘n=0โˆž(โˆ’1)nx2n+1(2n+1)!\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}

Multiply by xx:

xsinโกx=xโ‹…โˆ‘n=0โˆž(โˆ’1)nx2n+1(2n+1)!x \sin x = x \cdot \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}

=โˆ‘n=0โˆž(โˆ’1)nx2n+2(2n+1)!= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+2}}{(2n+1)!}

=x2โˆ’x43!+x65!โˆ’x87!+โ‹ฏ= x^2 - \frac{x^4}{3!} + \frac{x^6}{5!} - \frac{x^8}{7!} + \cdots


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

From Example 5: eโˆ’x2=โˆ‘n=0โˆž(โˆ’1)nx2nn!e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}

Integrate term by term:

โˆซeโˆ’x2dx=C+โˆ‘n=0โˆž(โˆ’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+xโˆ’x31!โ‹…3+x52!โ‹…5โˆ’x73!โ‹…7+โ‹ฏ= 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 aโ‰ 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=โˆ‘n=0โˆže2n!(xโˆ’2)ne^x = \sum_{n=0}^{\infty} \frac{e^2}{n!}(x-2)^n

=e2[1+(xโˆ’2)+(xโˆ’2)22!+(xโˆ’2)33!+โ‹ฏโ€‰]= e^2 \left[1 + (x-2) + \frac{(x-2)^2}{2!} + \frac{(x-2)^3}{3!} + \cdots\right]

=e2โˆ‘n=0โˆž(xโˆ’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:

limโกnโ†’โˆžRn(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,sinโกx,cosโกx,lnโก(1+x)e^x, \sin x, \cos x, \ln(1+x)), this happens on their interval of convergence.


โš ๏ธ Common Mistakes

Mistake 1: Wrong Factorial in General Term

For sinโกx=xโˆ’x33!+x55!โˆ’โ‹ฏ\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots:

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

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


Mistake 2: Wrong Starting Index

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

WRONG: Sum starts at n=1n = 1

RIGHT: โˆ‘n=0โˆž(โˆ’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 lnโกx\ln x centered at a=1a = 1:

WRONG: lnโกx=โˆ‘(โˆ’1)nโˆ’1xnn\ln x = \sum \frac{(-1)^{n-1} x^n}{n}

RIGHT: lnโกx=โˆ‘(โˆ’1)nโˆ’1(xโˆ’1)nn\ln x = \sum \frac{(-1)^{n-1} (x-1)^n}{n} (must use (xโˆ’1)n(x-1)^n!)


Mistake 4: Computing All Derivatives Unnecessarily

To find series for e3xe^{3x}:

WRONG: Compute fโ€ฒ(x),fโ€ฒโ€ฒ(x),fโ€ฒโ€ฒโ€ฒ(x),โ€ฆf'(x), f''(x), f'''(x), \ldots

RIGHT: Use ex=โˆ‘xnn!e^x = \sum \frac{x^n}{n!}, substitute xโ†’3xx \to 3x:

e3x=โˆ‘n=0โˆž(3x)nn!=โˆ‘n=0โˆž3nxnn!e^{3x} = \sum_{n=0}^{\infty} \frac{(3x)^n}{n!} = \sum_{n=0}^{\infty} \frac{3^n x^n}{n!}


๐Ÿ“ Practice Strategy

  1. Memorize basic series: ex,sinโกx,cosโกx,11โˆ’x,lnโก(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: sinโกx\sin x has only odd powers, cosโกx\cos x only even
  8. Write general term: Essential for summation notation

๐Ÿ“š Practice Problems

1Problem 1medium

โ“ Question:

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

๐Ÿ’ก Show Solution

Method 1: Using derivatives

f(x)=(1โˆ’x)โˆ’1f(x) = (1-x)^{-1} fโ€ฒ(x)=(1โˆ’x)โˆ’2f'(x) = (1-x)^{-2} fโ€ฒโ€ฒ(x)=2(1โˆ’x)โˆ’3f''(x) = 2(1-x)^{-3} fโ€ฒโ€ฒโ€ฒ(x)=6(1โˆ’x)โˆ’4=3!(1โˆ’x)โˆ’4f'''(x) = 6(1-x)^{-4} = 3!(1-x)^{-4} f(n)(x)=n!(1โˆ’x)โˆ’(n+1)f^{(n)}(x) = n!(1-x)^{-(n+1)}

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

Maclaurin series: 11โˆ’x=โˆ‘n=0โˆžn!n!xn=โˆ‘n=0โˆžxn\frac{1}{1-x} = \sum_{n=0}^{\infty} \frac{n!}{n!}x^n = \sum_{n=0}^{\infty} x^n

=1+x+x2+x3+x4+โ‹ฏ= 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.

11โˆ’x=โˆ‘n=0โˆžxn\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n


Interval of convergence:

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

โˆฃxโˆฃ<1|x| < 1

At x=1x = 1: โˆ‘1\sum 1 diverges

At x=โˆ’1x = -1: โˆ‘(โˆ’1)n\sum (-1)^n diverges

Interval: (โˆ’1,1)(-1, 1)


Answer: 11โˆ’x=โˆ‘n=0โˆžxn\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n for โˆฃxโˆฃ<1|x| < 1

2Problem 2hard

โ“ Question:

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

๐Ÿ’ก Show Solution

Solution:

Method 1: Using the definition

Maclaurin series: f(x)=โˆ‘n=0โˆžf(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
  • fโ€ฒ(x)=2e2xf'(x) = 2e^{2x}, so fโ€ฒ(0)=2f'(0) = 2
  • fโ€ฒโ€ฒ(x)=4e2xf''(x) = 4e^{2x}, so fโ€ฒโ€ฒ(0)=4f''(0) = 4
  • fโ€ฒโ€ฒโ€ฒ(x)=8e2xf'''(x) = 8e^{2x}, so fโ€ฒโ€ฒโ€ฒ(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=โˆ‘n=0โˆžxnn!=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

โ“ Question:

Use the Maclaurin series for exe^x to find the Maclaurin series for f(x)=e2xcosโกxf(x) = e^{2x} \cos x.

๐Ÿ’ก Show Solution

Step 1: Write known series

ex=โˆ‘n=0โˆžxnn!=1+x+x22!+x33!+โ‹ฏe^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots

cosโกx=โˆ‘n=0โˆž(โˆ’1)nx2n(2n)!=1โˆ’x22!+x44!โˆ’โ‹ฏ\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 xโ†’2xx \to 2x in series for exe^x:

e2x=โˆ‘n=0โˆž(2x)nn!=โˆ‘n=0โˆž2nxnn!e^{2x} = \sum_{n=0}^{\infty} \frac{(2x)^n}{n!} = \sum_{n=0}^{\infty} \frac{2^n x^n}{n!}

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

=1+2x+2x2+4x33+โ‹ฏ= 1 + 2x + 2x^2 + \frac{4x^3}{3} + \cdots


Step 3: Multiply series

e2xcosโกx=(1+2x+2x2+4x33+โ‹ฏโ€‰)(1โˆ’x22+x424โˆ’โ‹ฏโ€‰)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: 1โ‹…1=11 \cdot 1 = 1

xx term: 2xโ‹…1=2x2x \cdot 1 = 2x

x2x^2 term: 2x2โ‹…1+1โ‹…(โˆ’x22)=2x2โˆ’x22=3x222x^2 \cdot 1 + 1 \cdot (-\frac{x^2}{2}) = 2x^2 - \frac{x^2}{2} = \frac{3x^2}{2}

x3x^3 term: 4x33โ‹…1+2xโ‹…(โˆ’x22)=4x33โˆ’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) 1โ‹…x424+2x2โ‹…(โˆ’x22)+(higherย orderย terms)1 \cdot \frac{x^4}{24} + 2x^2 \cdot (-\frac{x^2}{2}) + (\text{higher order terms}) =x424โˆ’x4+โ‹ฏ= \frac{x^4}{24} - x^4 + \cdots

(This gets tedious!)


First few terms:

e2xcosโกx=1+2x+3x22+x33+โ‹ฏe^{2x} \cos x = 1 + 2x + \frac{3x^2}{2} + \frac{x^3}{3} + \cdots


Answer: e2xcosโกx=1+2x+3x22+x33+โ‹ฏe^{2x}\cos x = 1 + 2x + \frac{3x^2}{2} + \frac{x^3}{3} + \cdots

(Usually only first 3-4 terms are requested)

4Problem 4medium

โ“ Question:

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

๐Ÿ’ก Show Solution

Step 1: Rewrite in useful form

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

Let u=xโˆ’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: 11โˆ’r=โˆ‘n=0โˆžrn\frac{1}{1-r} = \sum_{n=0}^{\infty} r^n for โˆฃrโˆฃ<1|r| < 1

Replace rr with โˆ’u-u:

11+u=11โˆ’(โˆ’u)=โˆ‘n=0โˆž(โˆ’u)n=โˆ‘n=0โˆž(โˆ’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=xโˆ’1u = x-1

1x=โˆ‘n=0โˆž(โˆ’1)n(xโˆ’1)n\frac{1}{x} = \sum_{n=0}^{\infty} (-1)^n (x-1)^n

=1โˆ’(xโˆ’1)+(xโˆ’1)2โˆ’(xโˆ’1)3+โ‹ฏ= 1 - (x-1) + (x-1)^2 - (x-1)^3 + \cdots


Step 4: Find interval of convergence

Need โˆฃuโˆฃ<1|u| < 1: โˆฃxโˆ’1โˆฃ<1|x-1| < 1 โˆ’1<xโˆ’1<1-1 < x - 1 < 1 0<x<20 < x < 2


Answer: 1x=โˆ‘n=0โˆž(โˆ’1)n(xโˆ’1)n\frac{1}{x} = \sum_{n=0}^{\infty} (-1)^n(x-1)^n for 0<x<20 < x < 2

5Problem 5hard

โ“ Question:

Find the first 4 nonzero terms of the Maclaurin series for f(x) = cos(x).

๐Ÿ’ก Show Solution

Step 1: Maclaurin series formula: f(x) = ฮฃ[fโฝโฟโพ(0)/n!]ยทxโฟ

Step 2: Find derivatives at x = 0: f(x) = cos(x) โ†’ f(0) = 1 f'(x) = -sin(x) โ†’ f'(0) = 0 f''(x) = -cos(x) โ†’ f''(0) = -1 f'''(x) = sin(x) โ†’ f'''(0) = 0 fโฝโดโพ(x) = cos(x) โ†’ fโฝโดโพ(0) = 1 fโฝโตโพ(x) = -sin(x) โ†’ fโฝโตโพ(0) = 0 fโฝโถโพ(x) = -cos(x) โ†’ fโฝโถโพ(0) = -1 fโฝโทโพ(x) = sin(x) โ†’ fโฝโทโพ(0) = 0 fโฝโธโพ(x) = cos(x) โ†’ fโฝโธโพ(0) = 1

Step 3: Write first 4 nonzero terms: cos(x) = f(0) + f''(0)xยฒ/2! + fโฝโดโพ(0)xโด/4! + fโฝโถโพ(0)xโถ/6! + fโฝโธโพ(0)xโธ/8! + ... = 1 + (-1)xยฒ/2 + (1)xโด/24 + (-1)xโถ/720 + (1)xโธ/40320 + ... = 1 - xยฒ/2 + xโด/24 - xโถ/720 + ...

Step 4: Pattern: cos(x) = ฮฃ[(-1)โฟยทx^(2n)]/(2n)! from n=0 to โˆž

Answer: 1 - xยฒ/2 + xโด/24 - xโถ/720 (first 4 nonzero terms)