Transcript Calculus for the Natural Sciences

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...

bn
for a series of constants 
, we have that the
n 0
series converges (absolutely) if the the limit of the
absolute value of bb is less than one, diverges if
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.
n 1
n
x
For the series for the
exponential function...
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.
Your turn...
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
- - T heratio test thistimegives

.

n
(n  1) x
n 1
n 1 n
T helimit of thisas n tendsto infinityis simply x.

T heconclusionis that theseries convergesif x  1 and
divergesif x  1 (reminiscent of thegeometricseries).
What remains...
is to check thepointsx  1 and x  -1. For x  1, the

series becomes 1n , theharmonicseries, which
n 1
we know to be divergent.

For x  1, theseries becomes 
n 1
( 1) n
n
, thealternating
harmonicseries, which we know to be (conditionally)
convergent.
Final Conclusion

T heseries 
n 1
xn
n
convergesif - 1  x  1
and divergesotherwise.
OK, your turn...

For which values of x does theseries 
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
( 2 x )n
n2
converge?
One more...

For whichvalues of x does theseries  nxn 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:
T hereare two ways :
1. Use theformulaan  f n (0)/n! We've found series for e x and sin(x) :

x 2 x3 x 4
xn
e  1  x     ...  
2! 3! 4!
n  0 n!
x
x3 x5 x7
(1) n x ( 2 n 1)
sin(x)  x     ...
3! 5! 7!
(2n  1)!
We could do otherseries this way, but method2 is morefun :
The other way
2. Start from known series and use algebraic and/or
analyticmanipulation toget others:
Examples: Substitute x 2 for x everywhere
in thee x series to get :
e( x
2
)
4
6
8
(2n)

x
x
x
x
 1  x 2     ...  
2! 3! 4!
n  0 n!
Try this...
Take the derivative of the series for sin(x) to get
x
x
x
(1) x
cos(x) 1  
  ...  
2! 4! 6!
(2n)!
n 0
2
4
6

n
( 2n)
Integrate both sides of the geometric
series from 0 to x to get:

x
1
0 1-t
x
x
x
0
0
0
x
dt   1dt   tdt   t dt   t dt  ...
2
x 2 x3 x 4
- ln(1- x)  x 
 
 ...
2
3
4
0
3
Negate both sides and replace x by (-x)
everywhere to get:
2
3
4
x
x
x
ln(1  x)  x 
 
 ...
2
3
4
(T hisshows that thefourthseries sums to ln(2)).
Start from the geometric series again...
And substitute x 2 for x everywhere it appears to get
1
2
4
6
8
10

1

x

x

x

x

x
 ...
2
1 x
Now int egrat ebot h sides from 0 t o x t oget :
x 3 x 5 x 7 x 9 x11
arct an(x)  x  



 ...
3
5
7
9 11
(T hisshows t hat t hefirst series we saw earlier

convergest o .)
4
A challenge to think about...
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:
We know for other reasons that
sin( x)
lim
1
x 0
x
We could do this by series:

sin(x)
(1) n x 2n
x 2 x 4 x6
lim
 lim 
 lim 1     1  0  0...  1
x 0
x 0
x0
x
3! 5! 7!
n 0 (2n  1)!
This can be used on
complicated limits...
Calculate the limit: lim
x 0
A.
B.
C.
D.
E.
0
1/6
1
1/12
does not exist
x  sin(x)
1 e
(  x3 )
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
Example: Calculate
e
0
( x2 )
dx to the nearest 0.001.
We begin by...
substituting - 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
dx is approximat ed by
0
4
6
1
x
x
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
dx  .747 to the
So we get that  e
0
nearest thousandth.
(  x2 )
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
A. 0.7635
B. 0.5637
C. 0.3567
D. 0.6357
E. 0.6735
x )dx
Application of series
(cont.)
3. Differential equations : Another important
application of series is to find "solutions" to
differential equations (when other methods
fail).
Earlier, we found the series for e x based on
the differential equation it satisfies (y ' = y ),
together with the initial condition y(0)=1.
Airy functions
The equation y '' + xy = 0 has no "elementary"
solution (the solutions of this equation are called
"Airy functions", named after a famous British
Royal Astronomer). But we can find series for
the solution that satisfies y(0)=1, y '(0)=0, as
follows:
From the initial conditions, we can assume that
the series for y(x) begins:
y( x)  1  a2 x  a3 x  a4 x  ... an x  ...
2
3
4
n
Airy (continued)
Then the series for yseries:
2
3


y ( x)  2a2  6a3 x  12a4 x  20a5 x  ...
 (n  1)(n  2)an 2 x  ...
n
So...
y  xy  2a2  (6a3  1) x  12a4 x 2  (20a5  a2 ) x 3  (30a6  a3 ) x 4  ...
 ((n  1)(n  2)an  2  an 1 ) x n  ...
We work our way from left to right -- since y’’+ xy is supposed
to be zero, every coefficient must be zero.
Therefore:
1
a2  0, a3   16 , a4  a5  0, a6  180
, etc.
So the series for y(x) begins as
follows...
y ( x)  1 
x3
6

x6
180
 ...
(we actuallyknow that t henext
9
nonzerotermis the x term.)
What is the denominator of the
9
x term of this series?
A. 720
B. 1080
C. 1440
D. 4440
E. 12960
Application of series (cont.)
4. Algebraic equations depending on a parameter ("regular
perturbations").
Consider the equation x 2  ax  4  0 . We could calculate
the solutions of this equation using the quadratic formula, but it
will be instructive to think of the two solutions (x) as functions
of the parameter a .
We will find the power series of one of these functions.
Keep in mind that we are thinking of x as a function of a (and not
the other way around)!
Write x = f(a).
First, if a=0, there are two roots, x = 2 and x = -2. We'll calculate
the series for the larger root, so f(0)=2.
Work it out...
As we did for thedifferential equation,we will assume that wecan write
f(x)  2  b1a  b2 a 2  b3a 3  ...
and calculatethefirst few coefficients. Remember,x  f(a).
T heequation x 2  ax  4  0 becomes(we' ll supress the... for thetimebeing) :
(2  b1a  b2 a 2  b3a 3 ) 2  a(2  b1a  b2 a 2  b3a 3 )  4  0
Because we don't haveall of the termscontainingpowers of a higher than a 3 .
Right to left
Multiplying out and collectingtermsby power of a yields:
(2b1b2  4b3  b2 )a 3  (b1  b1 2  4b2 )a 2  (4b1  2)a  0
Workfromright toleft thistimeand get that:
b1   12 , thenb2  161 , and finallyb3  0. T herefore
f (a)  2  
a
2
a 4 term)
a2
16
 ... (where we know thenext termis the
To see how good the approximation is, we
plot the "actual" solution (in blue) and our
approximation (in red) on the same graph
Graph
And….
What is the beginning of the series for the
other root?
A max/min problem
During World War II, the blood of thousands of recruits had to be
tested for various diseases. The incidence of the diseases was
quite low (less than one per hundred). Assume that the
probability that a random person has the disease is p (for some p
between 0 and 1, but much closer to 0 than to 1). Let q = 1 - p be
the probability that the person does not have the disease.
Because the tests were costly, someone had the bright idea to pool
several (x) samples at a time. If the pooled sample tested
negative, then all x samples were negative. But if the pooled
sample tested positive, then each individual sample must be
tested again.
max/min
T heprobability thatno - onein thex samples testspositiveis q x .
T hisis theprobability thatonly one test willbe required for thex recruits.
T heprobability thatsomeonein thex samples testspositiveis 1  q x .
T hisis theprobability thatx  1 testswill be required for thex recruits.
max/min
So, theaveragenumber of testsperformedper recruit will be
A( x) 
qx
x

( 1-q x )(x 1 )
x
 1  q x  1x .
Obviously,theArmy wanted to minimizeA, given q.
T hederivativeof A(x) with respect tox is
 q x ln (q) x12 Weneed to set thisequal to zero and solvefor x,
in order tominimizeA.
Rewrite
We'll rewrit e t hisequat ion a lit t le,replacingq by 1 - p :
1
ln(1  p ) x  ( 1-p)
x  0
2
Remember,we want t osolvefor xas a funct ionof p.
For p  0, t heequat ionseems t o become0 x 2  1,
which means t hat lim x  .
p 0 
Undaunted...
we try expanding the terms in powers of p. We know that
ln(1-p) = -p + ... and
1
(1 p ) x
 1  ...
If we substitute this much into the equation, we get:
 px  1  0
2
The solution of this is
the limit!).
x
1
p
(which is at least consistent with
What we have here is an example of an asymptotic series , or, a
series of powers other than whole numbers.
We can calculate more coefficients by expanding the equation in
higher powers of p.