Taylor Series

Taylor Series

Recall that the Taylor polynomial of degree n for a differentiable function f(x) centered 
at x = c is

        \(  \displaystyle\sum_{k=0}^{n} \frac{f^{(k)}(c)}{k!}(x-c)^k  \)

If we let n approach infinity, we arrive at the Taylor Series for f(x) centered at x = c.

If c = 0 we call this series the Mclaurin Series for f(x).  Recall that the error of the n^th degree Taylor Polynomial is given by

                    f ⁽ⁿ⁺¹⁾(z)
        R =                        (z - c)ⁿ⁺¹  
                      (n + 1)!

Hence if 

        \(  \lim\limits_{n\to \infty} R = 0   \)

then the Taylor Series converges.

Example

Find the McLaurin Series expansion for

        f(x) = cos(x)

Solution

We construct the following table.

Hence we have the series

          x²        x⁴        x⁶        x⁸   
1  -           +         -         +           - ...
          2!        4!        6!        8!

Notice that the series only contains even powers of x and even factorials.  Even numbers can be represented by 2n.  Also notice that this is an alternating series, hence the McLaurin series is

        \(  \lim\limits_{n\to \infty} \frac{(-1)^n x^{2n}}{(2n)!}    \)

Exercises  Find the Taylor series expansion for

1. sin(x) centered at \( x = \frac{\pi}{2}\)

2. sinh(x) centered at x = 0

Statistics

The Standard Normal Distribution function is defined by

We define the probability as follows:

Example:

Use McLaurin series and the fact that

        \(  \displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!} \)

to approximate the probability of getting a "B" in this class if the average is 70 and the standard deviation is 10 and the instructor grades on a "curve".  A "B" corresponds to between 1 and 2 standard deviations from the mean, hence we need to compute

       \( P(1 < x < 2) = \int_1^2 \frac{1}{\sqrt{2\pi}}e^-\frac{x^2}{2}dx = \frac{1}{\sqrt{2\pi}} \int_1^2 \displaystyle\sum_{n=0}^{\infty} \frac{(-\frac{x^2}{2})^n}{n!}dx  \)

       \(   = \frac{1}{\sqrt{2\pi}}  \displaystyle\sum_{n=0}^{\infty}\int_1^2 \frac{(-\frac{x^2}{2})^n}{n!}dx  = \frac{1}{\sqrt{2\pi}}  \displaystyle\sum_{n=0}^{\infty}\int_1^2 \frac{(-1)^n x^{2n}}{2^n n!}dx  \)

       \(  = [ \frac{1}{\sqrt{2\pi}}  \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2^n (2n+1) n!}]_1^2 = \frac{1}{\sqrt{2\pi}}  \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n+1}}{2^n (2n+1) n!} - \frac{1}{\sqrt{2\pi}}  \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n }{2^n (2n+1) n!} \)

We can calculate the first many terms on the calculator to get an approximate value of 

         0.76

Limits

In the first quarter you learned a proof that 


        \(  \lim\limits_{x\to 0} \frac{1 - cos(x)}{x}    \)

In the second quarter you used L'Hopitals rule.  Now we will do it a third way:  We have
 

       \( cos(x)  =  \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2} + \frac{x^4}{24} - ... \)

Hence 

                                 x²         x⁴
        1  -  cos x  =            -           + ...
                                 2          24

Now divide both sides by x to get

         1  -  cos x            x          x³
                            =           -           + ...
                x                 2          24


When x = 0, the right hand side becomes zero, hence so does the left hand side.
 

Exercise

Prove L'Hopital's Rule using power series. 
 

Addition and Subtraction of Power Series

Example:  

We have that the power series representation of 

                              1
        ln(1 - x) +               
                           1 - x

 is

       \(   \displaystyle\sum_{n=0}^{\infty} \frac{-x^{n+1}}{n+1} + \displaystyle\sum_{n=0}^{\infty} x^n = \displaystyle\sum_{n=0}^{\infty} (\frac{-x^{n+1}}{n+1} + x^n)   \)
       \( = (-x+1) +(-\frac{x^2}{2} + x) +(-\frac{x^3}{3} + x^2) +(-\frac{x^4}{4} + x^3 + ... \)

       \( = 1 +\frac{1}{2}x^2 + \frac{2}{3}x^3 + \frac{3}{4}x^4 + ... \)

Exercise

Find the power Series Representation for 

        arctan x + arctanh x
 

Multiplication of Power Series

Suppose we have two power series

               \( f(x) =  \displaystyle\sum_{n=0}^{\infty} a_n x^n  \)


and 

         \( g(x) =  \displaystyle\sum_{n=0}^{\infty} b_n x^n  \)

What is the power series for

        f(x)g(x)

Consider the following example.  Let 

         \( \frac{e^x}{1-x} = e^x \frac{1}{1-x} =  (\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!})(\displaystyle\sum_{n=0}^{\infty} x^n)  \)

       \( =    ( 1 + x +\frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + ...)(1 + x + x^2 + x^3 + x^4 + ...)\) 

 
We can multiply these series as though they were finite series.  We collect the coefficients:

• The constant term is 1.

• The first degree term is 1 + 1 = 2.

• The second degree term is 1 + 1 + 1/2 = 5/2.

• The third degree term is 1 + 1 + 1/2 + 1/6 = 8/3

• The fourth degree term is 1 + 1 + 1/2 + 1/6 + 1/24 = 65/24

We can continue this process indefinitely, or better yet use a computer to generate the terms.

The series is

                      5             8              65
        1 + x +       x²   +        x³  +           x⁴  + ...
                      2             3              24     
 

Division of Power Series

Suppose we want to find the power series representation of 

        

        \( \frac{arctan(x)}{e^x} = \frac{  \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}}{\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!}}  \)  

       \(  = \frac{x - \frac{x^3}{3} +  \frac{x^5}{5} -  \frac{x^7}{7} + ... }{1 + x + \frac{x^2}{2} +  \frac{x^3}{6} +  \frac{x^4}{24} + ... } = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + ... \)  

We multiply by the denominator and equate coefficients:

        (c₀ + c₁x + c₂x² + ...)(1 + x + x²/2 + x³/6 + x⁴/24 + ...) = (x - x³/3 +  x⁵/5- x⁷/7 +...)

• The constant coefficient gives us  c₀ = 0.

• The first degree term gives us c₀ + c₁ = 1. Hence c₁ = 1.

• The second degree term gives us 1 + c₂ =  0. Hence c₂ = -1.

• The third degree term gives us 1/2 - 1 + c₃ = -1/3.  Hence c₃ = 1 - 1/2 - 1/3 = 1/6.

and so on.  

The series is

                       1
        x - x² +       x³ + ...
                       6

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