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9.1 Power Series9.1 Power Series

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What You Will Learn

• All continuous functions can be represented as a polynomial

• Polynomials are easy to integrate and differentiate

• Calculators use polynomials to calculate trig functions, logarithmic functions etc.

• Downfall of polynomial equivalent functions is that they have an infinite number of terms.

This is an example of an This is an example of an infinite seriesinfinite series..

1

1

Start with a square one Start with a square one unit by one unit:unit by one unit:

1

21

21

4

1

4+

1

81

8+

1

161

16+

1

32 1

64

1

32+

1

64+ 1+⋅⋅⋅=

This series This series convergesconverges (approaches a limiting value.) (approaches a limiting value.)

Many series do not converge:Many series do not converge:1 1 1 1 1

1 2 3 4 5+ + + + +⋅⋅⋅=∞

→

In an infinite series:In an infinite series: 1 2 31

n kk

a a a a a∞

=

+ + +⋅⋅⋅+ +⋅⋅⋅=∑

aa11, a, a22,…,… are are termsterms of the series. of the series. aann is the is the nnthth term term..

Partial sums:Partial sums: 1 1S a=

2 1 2S a a= +

3 1 2 3S a a a= + +

1

n

n kk

S a=

=∑

nnthth partial sum partial sum

If If SSnn has a limit as , then the series converges, has a limit as , then the series converges,

otherwise it otherwise it divergesdiverges..

n → ∞

→

Geometric Series:

In a In a geometric seriesgeometric series, each term is found by multiplying , each term is found by multiplying

the preceding term by the same number, the preceding term by the same number, rr..

2 3 1 1

1

n n

n

a ar ar ar ar ar∞

− −

=

+ + + +⋅⋅⋅+ +⋅⋅⋅=∑

This converges to if , and diverges if .This converges to if , and diverges if .1

a

r−1r < 1r ≥

1 1r− < < is the is the interval of convergenceinterval of convergence..

→

Geometric Series

n

n

1 - rS = a

1 - r

⎛ ⎞⎜ ⎟⎝ ⎠

Partial Sum of a Geometric Series:

Sn = a + ar + ar2 + ar3 + … + arn-1

-[r Sn = ar + ar2 + ar3 + … + arn

Sn – r Sn = a + arn

Sn (1 – r) = a (1 - rn)

Sum of Converging Series

nn

nn

1 - rlim S = a , if r < 1, then r goes to zero and

1 - r

a S =

1 - rif r > 1, the series diverges.

Note: The interval of convergence is r < 1

→∞

⎛ ⎞⎜ ⎟⎝ ⎠

Power Series Using Calculator

( )

102

x = 1

2

To calculate a partial sum of a power series on the calculator,

x

you can find the expanded form by entering:

seq x , x, 1, 10

to get 1 4 9 16 2

∑

{ }

( )( )2

5 36 49 64 81 100

and the sum by entering:

sum seq x , x, 1, 10

to get 385

Example 1:Example 1:

3 3 3 3

10 100 1000 10000+ + + +⋅⋅⋅

.3 .03 .003 .0003+ + + +⋅⋅⋅ .333...=

1

3=

310

11

10−

aa

rr

3109

10

=3

9=

1

3=

→

1 1 11

2 4 8− + − +⋅⋅⋅

11

12

⎛ ⎞− −⎜ ⎟⎝ ⎠

11

12

=+

13

2

=2

3=

aa

rr

Example 2:Example 2:

→

The partial sum of a geometric series is:The partial sum of a geometric series is:( )1

1

n

n

a rS

r

−=

−

If thenIf then1r <( )1

lim1

n

n

a r

r→∞

−

− 1

a

r=

−

0

If and we let , then:If and we let , then:1x < r x=

2 31 x x x+ + + +⋅⋅⋅ 1

1 x=

−The more terms we use, the better our approximation The more terms we use, the better our approximation (over the interval of convergence.)(over the interval of convergence.)

→

Example of a Power Series

The function y = 1

1 - x can be written as the power series:

Sn = 1 + x + x2 + x3 + ...

On your calculator, enter y1 =

1

1 - x and

y2 = sum seq xn , n, 0, 20( )( )

Then test a value: Enter y1(.5)

Enter y2(.5)

What do you notice?

A power series is in this form:

c x c c x c x c x c xnn

nn

n

= + + + +⋅⋅⋅+ +⋅⋅⋅=

∞

∑ 0 1 22

33

0

or

c x a c c x a c x a c x a c x ann

nn

n

( ) ( ) ( ) ( ) ( )− = + − + − + − +⋅⋅⋅+ − +⋅⋅⋅=

∞

∑ 0 1 22

33

0

The coefficients c0, c1, c2… are constants.

The center “a” is also a constant.

(The first series would be centered at the origin if you graphed it. The second series would be shifted left or right. “a” is the new center.)

→

Once we have a series that we know, we can find a new Once we have a series that we know, we can find a new series by doing the same thing to the left and right hand series by doing the same thing to the left and right hand sides of the equation.sides of the equation.

This is a geometric series where This is a geometric series where r=-xr=-x..

1

x

x+To find a series forTo find a series for multiply both sides by multiply both sides by xx..

2 311

1x x x

x= − + − +⋅⋅⋅

+

2 3 4

1

xx x x x

x= − + − ⋅⋅⋅

+

1

1 x+Example 3:Example 3:

→

Example 4:Example 4:

Given:Given: 2 311

1x x x

x= + + + +⋅⋅⋅

−find:find:

( )2

1

1 x−

1

1

d

dx x−( ) 11

dx

dx−

= − ( ) 21 1x

−= − − ⋅−

( )2

1

1 x=

−

So:So:( )

( )2 32

11

1

dx x x

dxx= + + + + ⋅⋅⋅

−

2 31 2 3 4x x x= + + + +⋅⋅⋅

We differentiated term by term.We differentiated term by term.→

Example 5:Example 5:

Given:Given: 2 311

1x x x

x= − + − +⋅⋅⋅

+find:find: ( )ln 1 x+

( )1ln 1

1dx x c

x= + +

+∫

2 311

1t t t

t= − + − +⋅⋅⋅

+

hmm?hmm?

→

Example 5:Example 5: 2 311

1t t t

t= − + − +⋅⋅⋅

+

( )2 3

0 0

11

1

x xdt t t t dt

t= − + − + ⋅⋅⋅

+∫ ∫

( ) 2 3 4

00

1 1 1ln 1

2 3 4

xx

t t t t t+ = − + − + ⋅⋅⋅

( ) ( ) 2 3 41 1 1ln 1 ln 1 0

2 3 4x x x x x+ − + = − + − + ⋅⋅⋅

( ) 2 3 41 1 1ln 1

2 3 4x x x x x+ = − + − + ⋅⋅⋅ 1 1x− < <

→

( ) 2 3 41 1 1ln 1

2 3 4x x x x x+ = − + − + ⋅⋅⋅ 1 1x− < <

The previous examples of infinite series approximated The previous examples of infinite series approximated simple functions such as or .simple functions such as or .1

3

1

1 x−

This series would allow us to calculate a transcendental This series would allow us to calculate a transcendental function to as much accuracy as we like using only function to as much accuracy as we like using only pencil and paper!pencil and paper!

Convergent Series

Only two kinds of series converge:

1) Geometric whose | r | < 1

2) Telescoping series

Example of a telescoping series: the middle terms cancel out

( )n 1 n = 1

n

1 1 1 = - n n + 1 n n + 1

1 1 1 1 1 1 1 = 1 - + - + - + - +...

2 2 3 3 4 4 5

= sum of 1 - last term

= 1 + lim

∞ ∞

=

→

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

∑ ∑

1 -

n + 1

= 1 + 0 = 1

∞

⎛ ⎞⎜ ⎟⎝ ⎠

Finding a series for tan-1 x

1. Find a power series that represents on (-1,1)

2. Use integration to find a power series that represents

tan-1 x.

3. Graph the first four partial sums. Do the graphs suggest convergence on the open interval (-1, 1)?

4. Do you think that the series for tan-1 x converges at x = 1?

1 - x2 + x4 - x6 + ...+ (-1)nx2n + ...

1

1+ x2∫ dx = x - x3

3 +

x5

5 -

x7

7 +...+ (-1)n x2n+1

2n+1 +...

Yes to

4

1

1 + x2( )

Yes

ex

Guess the function

Define a function f by a power series as follows:

2 3 4 nx x x xf (x) = 1 + x + + + +...+ + ...

2! 3! 4! n!

Find f ‘(x).

What function is this?