Taylor Series
I. Quiz
II. Homework
III. Taylor Series
Recall that the taylor polynomial of degree n for a differentiable function f(x) centered at x = c is
sum from 0 to n of 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.
Definition: The Taylor Series for f(x) centered at x = c is
If c = 0 we call this series the Mclaurin Series for f(x).
Recall that the error of the nth degree Taylor Polynomial is given by
R = f(n+1)(z)/(n+1)! (z - c)n+1
Hence if lim R = 0 then the Taylor Series converges.
Example:
Find the McLaurin Series expansion for
f(x) = cos(x)
We compute:
f(0) = 1, f'(0) = 0, f''(0) = -1, f(3)(0) = 0, f(4)(0) = 1, f(5)(0) = 0, f(6)(0) = -1,...
Hence we have the series
1 - x2 /2+ x4/4! - x6/6! + x8/8! + ...
We see that the series is
sum from 0 to infinity of (-1)nx2n/(2n)!
Exercises Find the Taylor series expansion for
A) sin(x) centered at x = pi/2
B) sinh(x) centered at x = 0
IV. Statistics
The Standard Normal Distribution function is defined by
f(x) = 1/sqrt(2 pi) e(-x2/2)
We define the probability as follows:
P(a < x < b) = int from a to b of f(x) dx
Use McLaurin series to approximate the probability of getting a B in this class if the average is 70 and the standard deviation is 10.
A B corresponds to between 1 and 2 standard deviations from the mean, hence we need to compute
We can calculate the first many terms on the calculator to get an approximate value.
VI) Limits
In the first quarter you learned a proof that lim as x -> 0 (1 - cosx)/x = 0
In the second quarter you used L'Hopitals rule. Now we will do it a third way:
We have
cosx = sum (-1)nx2n/(2n)! = 1 - x2/2 + x3/6 - ...
Hence 1 - cos x = x2/2 - x3/6 + ...
and (1 - cos x)/x = x/2 - x2/6 + ...
When x = 0 we have 0.
Exercise: Prove L'Hopital's Rule using power series.
VII. Addition and Subtraction of Power Series.
Theorem: Suppose that we have two functions and their power series representations
f(x) = sum anxn and
g(x) = sum bnxn
Then
f(x) + g(x) = sum (an + bn)xn
f(x) - g(x) = sum (an - bn)xn
Example: We have that the power series representation of ln(1 - x) + 1/(1 - x) is
sum -xn+ 1/(n + 1) + sum xn
sum -xn+ 1/(n + 1) + xn
= (-x + 1) + (-x2/2 + x) + (-x3/3 + x2) + (-x4/4 + x3)
= 1 + 1/2 x2 + 2/3 x3 + 3/4 x4 + ...
Exercise: Find the power Series Representation for arctan x + arctanh x.
VIII. Multiplication of Power Series
Suppose we have two power series
f(x) = sum anxn and g(x) = sum bnxn
What is f(x)g(x)?
Consider the following example
Let f(x) = ex/(1 - x) = ex( 1/(1 - x)) = (sum xn/n!)(sum xn)
= (1 + x + x2/2 + x3/6 + x4/24 + ...)(1 + x + x2 + x3 + x4+ ...)
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 = 65/24We can continue this process indefinitely, or better yet use a computer to generate the terms.
IX. Division of Power Series:
Suppose we want to find the power series representation of arctanx/ex
= (sum (-1)nx2n+1/(2n + 1))/(sum xn/n!)
= (x - x3/3 - x5/5+ x7/7 -...) /(1 + x + x2/2 + x3/6 + x4/24 + ...) = c0 + c1x + c2x2 + ...
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.
Manipulate series interactively