# part8-08 by wuzhengqin

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```									Math 104 - Calculus I

Part 8
Power series, Taylor series
Series
First .. a review of what we have done so far:
1. We examined series of constants and learned that we can
say everything there is to say about geometric and
telescoping series.
2. We developed tests for convergence of series of constants.
3. We considered power series, derived formulas and other
tricks for finding them, and know them for a few functions.
4. We used the ratio tests to determine intervals on which
power series converge, and use the other tests to check
convergence at the endpoints of the intervals.
Geometric series
a/(1-r) = a + ar + ar2 + ar3 + ...
provided |r|< 1.

We often use partial fractions to detect
telescoping series, for which we can
calculate explicitly the partial sums sn
Tests for convergence for
series of constants
Fundamental divergence test (nth term must go
to zero for convergence to be possible)
Integral test
Comparison and limit comparison tests
Ratio test
Root test
Alternating series test
Power series
f(x) = a0 + a1 x + a2 x2 + a3 x3 + ...
where an = f(n)(0)/n!

We know the series for ex, sin(x), cos(x),
1/(1-x), and a few related functions.
Convergence of power series
Before we get too excited about finding series, let's make sure that,
at the very least, the series converge.
Next week, we'll deal with the question of whether they converge
to the function we expect. But for now, we'll assume that if they
converge, they converge to the function they "came from".
(Strictly speaking, this is not always true -- but it is true for a large
class of functions, which includes nearly all the ones
encountered in basic science and mathematics. This fact was not
fully appreciated until the early part of the twentieth century.)
Fortunately, most of the question of whether power series converge
is answered fairly directly by the ratio test.
Recall that...


for a series of constants  bn , we have that the
n 0
series converges (absolutely) if the the limit of the
absolute value of bbn1 is less than one, diverges if
n

the limit is greater than one, and the test is
indeterminate if the limit equals one.
To use the ratio test on power series, just leave the x
there and calculate the limit for each value of x.
This will give an inequality that x must satisfy in
order for the series to converge.
For the series for the
exponential function...
x
For e , we carry out the ratio test as follows :
( n 1 )
bn 1      x       n!      x
                      . No matter wha t x
n n 1
bn     ( n  1 )! x
x
is, we have lim            0. Since the ratio limit is
n  n  1
less than one for all x, the series converges
for all x.

Calculate the series for the function sin(x) and
determine for which x the series converges.
Here’s a more interesting example


xn                                    x ( n 1) n nx
 n - - The ratio test this time gives (n  1) x n  n  1 .
n 1

The limit of this as n tends to infinity is simply x.
The conclusionis that theseries convergesif x  1 and
diverges if x  1 (reminiscent of the geometric series).
What remains...
is to check thepoints x  1 and x  -1. For x  1, the

series becomes 1 , the harmonic series, which
n
n 1

we know to be divergent.

For x  1, the series becomes         ( 1) n
n      , the alternating
n 1

harmonic series, which weknow to be (conditionally)
convergent.
Final Conclusion

The series     xn
n    convergesif - 1  x  1
n 1

and diverges otherwise.

For which values of x does the series     ( 2 x )n
n2
converge?
n 1

A. -1 < x < 1
B. -2 < x < 2
C. 1/2 < x < 1/2
D. -2 < x < 2
E. -1/2 < x < 1/2
One more...

For which values of x does the series  nx n converge?
n 1

A. -1 < x < 1
B. -1 < x < 1
C. 1 < x < 1
D. -1 < x < 1
E. 0 < x < 1
From these examples,
...it should be apparent that power series
converge for values of x in an interval that
is centered at zero, i.e., an interval of the
form [-a, a] , (-a, a], [-a, a) or (-a, a) (where
a might be either zero or infinity). The
interval is called the interval of convergence
and the number a is called the radius of
convergence .
Let’s go back
To finding series of functions:
There are two ways:
1. Use the formula an  f n (0)/n! We've found series for e x and sin( x) :

x 2 x3 x 4         xn
e  1  x     ...  
x

2! 3! 4!     n  0 n!

x3 x5 x7          (1) n x ( 2 n 1)
sin( x)  x     ...
3! 5! 7!            (2n  1)!
We could do other series this way,but method 2 is more fun :
The other way

2. Start from known series and use algebraic and/or
analytic manipulation to get others :
Examples : Substitute x 2 for x everywhere
in the e x series to get :

x 4 x 6 x8        x (2n)
 1  x 2     ...  
2
e( x       )

2! 3! 4!     n  0 n!
Try this...

Take the derivative of the series for sin(x) to get

x2 x4 x6          
( 1) n x ( 2 n )
cos(x)  1          ...  
2! 4! 6!         n 0    (2n)!
Integrate both sides of the geometric
series from 0 to x to get:

x          x       x       x         x
        dt   1dt   tdt   t dt   t dt  ...
1                              2         3
0 1-t        0        0       0        0

x 2 x3 x 4
- ln(1 - x)  x            ...
2   3  4
Negate both sides and replace x by (-x)
everywhere to get:
2    3     4
x   x    x
ln(1  x)  x             ...
2   3    4
(This shows that the fourth series sums to ln(2)).
Start from the geometric series again...
And substitute x 2 for x everywhere it appears to get

1
 1  x 2  x 4  x 6  x 8  x10  ...
1 x2
Now integrate both sides from 0 to x to get :
x 3 x 5 x 7 x 9 x11
arctan(x)  x                      ...
3    5      7    9 11
(This shows that thefirst series we saw earlier

converges to       .)
4

How to get the other one
from previously

( n12 ) ?
n 1
Application of Series
1. Limits: Series give a good idea of the behavior
of functions in the neighborhood of 0:
sin( x)
We know for other reasons that                 lim          1
x 0    x
We could do this by series:
sin( x)         
(1) n x 2 n          x2 x4 x6
lim           lim                     lim 1         1  0  0...  1
n  0 ( 2n  1)!
x 0    x      x 0                      x 0   3! 5! 7!
This can be used on
complicated limits...
x  sin( x)
Calculate the limit: lim              (  x3 )
x 0
1 e
A.   0
B.   1/6
C.   1
D.   1/12
E.   does not exist
Application of series
(continued)
2. Approximate evaluation of integrals: Many
integrals that cannot be evaluated in closed form
(i.e., for which no elementary anti-derivative exists)
can be approximated using series (and we can even
estimate how far off the approximations are).

1
e
( x2 )
Example: Calculate                    dx to the nearest 0.001.
0
We begin by...

substituti ng - x 2 for x in the series we already know for e x ,
and integratin g it. This will give us a numerical series that
1 (  x2 )
converges to the answer : e
0         dx is approximat ed by

1        x4 x6               1       1      1

2
1 x        dx  ...  1                   ...
0         2! 3!               3 5  2! 7  3!
According to Maple...
The last series is an alternating series with
decreasing terms. We need to find the first one
that is less than 0.0005 to ensure that the error
will be less than 0.001. According to Maple:
evalf(1/(7*factorial(3))), evalf(1/(9*factorial(4))),evalf( 1/(11*factorial(5)));
.02380952381, .004629629630, .0007575757576
evalf(1/(13*factorial(6)));
.0001068376068
Keep going...
So it's enough to go out to the 5! term. We do this
as follows:
Sum((-1)^n/((2*n+1)*factorial(n)),n=0..5) = sum((-1)^n/((2*n+1)
*factorial(n)),n=0..5);
5


n 0
( 1) n
( 2 n 1) n!      31049
41580

evalf(%);

.7467291967=.7467291967
and finally...
1
So we get that  e              ( x2 )
dx  .747 to the
0
nearest thousandth.

Again, according to Maple, the actual answer
(to 10 places) is
evalf(int(exp(-x^2),x=0..1));

.74669241330
Try this...
Sum the first four nonzero terms to approximate
1
 cos(
0
x )dx

A. 0.7635
B. 0.5637
C. 0.3567
D. 0.6357
E. 0.6735

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