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Taylor and Maclaurin Series

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Talor and Maclaurin Series
Recall that we can rewrite functions of the form f(x)=11−u(x)f\left(x\right)=\frac{1}{1-u\left(x\right)}f(x)=1−u(x)1​ as power series f(x)=∑n=0∞[u(x)]nf(x)={\displaystyle \sum_{n=0}^\infty}
\left[u\left(x\right)\right]^nf(x)=n=0∑∞​[u(x)]n.
It turns out that there's a whole group of special functions that can also be rewritten as power series.

Taylor Series
Let f(x)f\left(x\right)f(x) be a function whose derivatives are defined on an interval containing the value x=ax=ax=a. 
The Taylor Series of f(x)f\left(x\right)f(x) about x=ax=ax=a can be written as f(x)=∑n=0∞f(n)(a)n!(x−a)nf(x)=\sum\limits^\infty_{n=0}\frac{f^{(n)}(a)}{n!}(x-a)^nf(x)=n=0∑∞​n!f(n)(a)​(x−a)n
Expanding: f(x)=f(a)+f′(a)(x−a)+f′′(a)2!(x−a)2+f′′′(a)3!(x−a)3....f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3....f(x)=f(a)+f′(a)(x−a)+2!f′′(a)​(x−a)2+3!f′′′(a)​(x−a)3....

Identifying Clue
If the question asks you to find the Taylor Series Representation of a function centered at x=a
*Usually it'll say "using the definition..."

Steps
1. Find the first few derivatives of the function 

2. Substitute x=ax=ax=a into all these derivatives--These are now your coefficient!
*We need to spot a pattern between these numbers

3. Write out the power series representation using the formula

If the question also asks for the Radius or Interval of Convergence --> use the ratio or root test!


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Taylor Polynomial of Order N
If we only take the first N terms of a Taylor series (i.e. the Nth partial sum), this is called the Taylor Polynomial of order N:
PN(x)=∑n=0Nf(n)(a)n!(x−a)n.P_N(x)=\sum\limits^N_{n=0}\frac{f^{(n)}(a)}{n!}(x-a)^n.PN​(x)=n=0∑N​n!f(n)(a)​(x−a)n.
It can be computed if the function f(x)f(x)f(x) has at least N derivatives on an interval containing x=ax=ax=a.


Maclaurin Series  
The specific case where a=0a=0a=0 is called the  Maclaurin Series of f(x)f(x)f(x): 
 f(x)=∑n=0∞f(n)(0)n!xn=f(0)+f′(0)x+f′′(0)2!x2+f′′′(0)3!x3....f(x)=\sum\limits^\infty_{n=0}\frac{f^{(n)}(0)}{n!}x^n=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3....f(x)=n=0∑∞​n!f(n)(0)​xn=f(0)+f′(0)x+2!f′′(0)​x2+3!f′′′(0)​x3....

Practice Question
Find the Taylor series representation for f(x)=11+2xf(x)=\frac{1}{1+2x}f(x)=1+2x1​ about c=−2c=-2c=−2. Then, state the radius of convergence.
  

f(x)=−∑n=0∞2n3n+1(x+2)n\displaystyle f(x)=-{\displaystyle \sum_{n=0}^\infty}
\frac{2^n}{3^{n+1}}(x+2)^nf(x)=−n=0∑∞​3n+12n​(x+2)n
Radius of convergence is R=12R=\frac{1}{2}R=21​

f(x)=−∑n=0∞2n3n+1(x+2)n\displaystyle f(x)=-{\displaystyle \sum_{n=0}^\infty}
\frac{2^n}{3^{n+1}}(x+2)^nf(x)=−n=0∑∞​3n+12n​(x+2)n
Radius of convergence is R=32R=\frac{3}{2}R=23​

f(x)=∑n=0∞2n3n(x+2)n\displaystyle f(x)={\displaystyle \sum_{n=0}^\infty}
\frac{2^n}{3^{n}}(x+2)^nf(x)=n=0∑∞​3n2n​(x+2)n
Radius of convergence is R=12R=\frac{1}{2}R=21​

f(x)=∑n=0∞2n3n(x+2)n\displaystyle f(x)={\displaystyle \sum_{n=0}^\infty}
\frac{2^n}{3^{n}}(x+2)^nf(x)=n=0∑∞​3n2n​(x+2)n
Radius of convergence is R=32R=\frac{3}{2}R=23​

f(x)=−∑n=0∞2n+13n+1(x+2)n\displaystyle f(x)=-{\displaystyle \sum_{n=0}^\infty}
\frac{2^n+1}{3^{n+1}}(x+2)^nf(x)=−n=0∑∞​3n+12n+1​(x+2)n
Radius of convergence is R=12R=\frac{1}{2}R=21​

I don't know

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Common/Known Maclaurin Series
Here is a list of Maclaurin series representations of common functions



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Identifying Clues
The question asks you to find the Maclaurin series of a function and doesn't say anything about "using the definition"
The function can be a slight modification of a function in the above table
The function can be a product or quotient of function types in the above table
You are asked to find what a series converges to
The given series is not a power series but its convergence cannot be found using our regular series knowledge
Recognize the given series as one of the ones in this table and determine the corresponding x value needed

Practice Question
Express xcos⁡2xx\cos2xxcos2x  as a power series. For what xxx is the series valid? 
  

∑n=0∞4n(2n)!x2n+1\displaystyle\sum\limits^\infty_{n=0}\frac{4^n}{(2n)!}x^{2n+1}n=0∑∞​(2n)!4n​x2n+1
Valid for x=0x=0x=0

∑n=0∞4n(2n)!x2n+1\displaystyle\sum\limits^\infty_{n=0}\frac{4^n}{(2n)!}x^{2n+1}n=0∑∞​(2n)!4n​x2n+1
Valid for x∈(−∞,∞)x\in(-\infty,\infty)x∈(−∞,∞)

∑n=0∞4n(2n)!x2n\displaystyle\sum\limits^\infty_{n=0}\frac{4^n}{(2n)!}x^{2n}n=0∑∞​(2n)!4n​x2n
Valid for x∈(−∞,∞)x\in(-\infty,\infty)x∈(−∞,∞)

∑n=0∞(−4)n(2n)!x2n+1\displaystyle\sum\limits^\infty_{n=0}\frac{(-4)^n}{(2n)!}x^{2n+1}n=0∑∞​(2n)!(−4)n​x2n+1
Valid for x=0x=0x=0

∑n=0∞(−4)n(2n)!x2n+1\displaystyle\sum\limits^\infty_{n=0}\frac{(-4)^n}{(2n)!}x^{2n+1}n=0∑∞​(2n)!(−4)n​x2n+1
Valid for x∈(−∞,∞)x\in(-\infty,\infty)x∈(−∞,∞)

I don't know

Practice Question
Express x3ln⁡(1−x24)x^3\ln(1-\frac{x^2}{4})x3ln(1−4x2​)   as a power series. Then, state the first 3 non-zero terms and the radius of convergence. 

The power series: ∑n=1∞−14nnx2n+3\displaystyle\sum\limits^\infty_{n=1}\frac{-1}{4^nn}x^{2n+3}n=1∑∞​4nn−1​x2n+3
The first 3 non-zero terms: −14x5−132x7−1192x9+...\displaystyle-\frac{1}{4}x^5-\frac{1}{32}x^7-\frac{1}{192}x^9+...−41​x5−321​x7−1921​x9+...
The radius of convergence is R=2R=2R=2

The power series: ∑n=1∞(−1)n4nnx2n+3\displaystyle\sum\limits^\infty_{n=1}\frac{(-1)^n}{4^nn}x^{2n+3}n=1∑∞​4nn(−1)n​x2n+3
The first 3 non-zero terms: −14x5+132x7−1192x9+...\displaystyle-\frac{1}{4}x^5+\frac{1}{32}x^7-\frac{1}{192}x^9+...−41​x5+321​x7−1921​x9+...
The radius of convergence is R=2R=2R=2

The power series: ∑n=1∞(−1)n4nx2n+3\displaystyle\sum\limits^\infty_{n=1}\frac{(-1)^n}{4^n}x^{2n+3}n=1∑∞​4n(−1)n​x2n+3
The first 3 non-zero terms: −14x5−116x7−164x9+...\displaystyle-\frac{1}{4}x^5-\frac{1}{16}x^7-\frac{1}{64}x^9+...−41​x5−161​x7−641​x9+...
The radius of convergence is R=2R=2R=2

The power series: ∑n=1∞−14nnx2n+1\displaystyle\sum\limits^\infty_{n=1}\frac{-1}{4^nn}x^{2n+1}n=1∑∞​4nn−1​x2n+1
The first 3 non-zero terms: −14x3−132x5−1192x7+...\displaystyle-\frac{1}{4}x^3-\frac{1}{32}x^5-\frac{1}{192}x^7+...−41​x3−321​x5−1921​x7+...
The radius of convergence is R=2R=2R=2

I don't know

Practice Question
Find the Maclaurin series for  f(x)=∫ex2  ⁣dxf(x)={\displaystyle\int}e^{x^2}\de{x}f(x)=∫ex2 dx.
  

C+∑n=0∞1(2n+1)n!x2n+1\displaystyle C+\sum\limits^\infty_{n=0}\frac{1}{(2n+1)n!}x^{2n+1}C+n=0∑∞​(2n+1)n!1​x2n+1

C−∑n=0∞1(2n+1)n!x2n+1\displaystyle C-\sum\limits^\infty_{n=0}\frac{1}{(2n+1)n!}x^{2n+1}C−n=0∑∞​(2n+1)n!1​x2n+1

C−∑n=0∞1(2n−1)n!x2n−1C-\displaystyle \sum\limits^\infty_{n=0}\frac{1}{(2n-1)n!}x^{2n-1}C−n=0∑∞​(2n−1)n!1​x2n−1

∑n=0∞1(2n)n!x2n\displaystyle \sum\limits^\infty_{n=0}\frac{1}{(2n)n!}x^{2n}n=0∑∞​(2n)n!1​x2n

I don't know

Practice Question
Evaluate the series 1−13+15−17+19−111+...1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+...1−31​+51​−71​+91​−111​+....