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Section 9.8 and 9.9: Power Series

Section 9.8 and 9.9: Power Series

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Section 9.8 and 9.9: Power Series. A Polynomial Series. Consider the Polynomial Series:. Investigate the partial sums of the sequence and compare the results to on a graph. Window: and . A Polynomial Series. Consider the Polynomial Series:. - PowerPoint PPT Presentation

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Page 1: Section 9.8 and 9.9: Power Series

Section 9.8 and 9.9: Power Series

Page 2: Section 9.8 and 9.9: Power Series

A Polynomial SeriesConsider the Polynomial Series:

Investigate the partial sums of the sequence and compare the results to on a graph.

Window: and

Page 3: Section 9.8 and 9.9: Power Series

A Polynomial SeriesConsider the Polynomial Series:

Investigate the partial sums of the sequence and compare the results to on a graph.

Window: and

Page 4: Section 9.8 and 9.9: Power Series

A Polynomial SeriesConsider the Polynomial Series:

Investigate the partial sums of the sequence and compare the results to on a graph.

Window: and

Page 5: Section 9.8 and 9.9: Power Series

A Polynomial SeriesConsider the Polynomial Series:

Investigate the partial sums of the sequence and compare the results to on a graph.

Window: and

Page 6: Section 9.8 and 9.9: Power Series

A Polynomial SeriesConsider the Polynomial Series:

Investigate the partial sums of the sequence and compare the results to on a graph.

Window: and

Page 7: Section 9.8 and 9.9: Power Series

A Polynomial SeriesConsider the Polynomial Series:

Investigate the partial sums of the sequence and compare the results to on a graph.

Window: and

Page 8: Section 9.8 and 9.9: Power Series

A Polynomial SeriesConsider the Polynomial Series:

Window: and

On , , , , , ,… converges to . The polynomial series is a good approximation of on .

The sequence of polynomials

converges to a rational expression.

Why?

Page 9: Section 9.8 and 9.9: Power Series

A Polynomial SeriesConsider the Polynomial Series:

The series is a geometric series.∙ 𝒙

The constant ratio is .

The first term is 1.

∙ 𝒙 ∙ 𝒙

Therefore the sum is when .

So: when .

This is the only interval the equation

is true. (Same as the

graphs.)

Page 10: Section 9.8 and 9.9: Power Series

Power Series Centered at x=0An expression of the form:

is a power series centered at .

The interval of convergence is the domain (values for ) for which the series converges.

(A series of powers of x)

Page 11: Section 9.8 and 9.9: Power Series

Example 1Find a power series to represent and give its interval of convergence.

11 1ar x The expression is the sum

of a Geometric Series: 1

1 x

The initial term is .

The constant ratio is .

Generate the Series: 1 x 2x 3x ... 11 n 1nx ...

∙−𝒙∙−𝒙 ∙−𝒙Since the series is Geometric, the

series converges when : 1x 1,1

So the interval of convergence is:

Page 12: Section 9.8 and 9.9: Power Series

Example 2Find a power series to represent and give its interval of convergence.

21

1 1ar x The expression is the sum

of a Geometric Series: 21

1 x

The same as the last

series except it

has .Generate the

Series with the previous series

by replacing the ’s with :

1 2x 22x 32x ... 11 n 12 nx

...

Since the series is Geometric, the series converges when :

2 1x 1,1

So the interval of convergence is:

1−𝑥2+𝑥4−𝑥6+…+(−1)𝑛−1𝑥2𝑛−2+…

Page 13: Section 9.8 and 9.9: Power Series

Example 3Find a power series to represent and give its interval of convergence.

11 2ar x The expression is the sum

of a Geometric Series: 1

1 1x

The initial term is .

The constant ratio is .Generate the Series:

1 1x 21x 31x ... 11 nx ...∙ 𝒙−𝟏 ∙ 𝒙−𝟏 ∙ 𝒙−𝟏

Since the series is Geometric, the series converges when : 1 1x 0,2

So the interval of convergence is no longer centered at 0:

Page 14: Section 9.8 and 9.9: Power Series

Power Series Centered at x=aAn expression of the form:

is a power series centered at .

The interval of convergence is the domain (values for ) for which the series converges.

(A series of powers of x-a)

Page 15: Section 9.8 and 9.9: Power Series

Theorem 1: Term-by-Term Differentiation

If

converges for , then the series

obtained by differentiating the series for term by term, converges for and represents on that interval.

If the series for converges for all , then so does the series for .

Not so Straightforward.

Page 16: Section 9.8 and 9.9: Power Series

Theorem 1: Term-by-Term Differentiation

That theorem can be confusing. What it says is that if…

• A power series can be differentiated term by term to form a new series.

• The new series will converge to the derivative of the function represented by the original series.

• The new series will at least converge on the same interval as the original series.

This gives a way to generate new connections between functions and series.

Can it converge at a larger interval? Does it include the endpoints?

We will not prove this. It is assumed to be true.

Page 17: Section 9.8 and 9.9: Power Series

ExampleFind a power series to represent .

21 11 1

ddx x x

Notice:

We already know:for

To find the power series,

we differentiate both sides of the equation

piece by piece:

𝑑𝑑𝑥 ( 1

1−𝑥 )= 𝑑𝑑𝑥 (1+𝑥+𝑥2+𝑥3+…+𝑥𝑛+…   )

21

1 x012x 23x 1... ...nnx

The last theorem guarantees this series will AT LEAST converge on the same interval as :

The series is no longer Geometric. It could converge on a larger interval.

Page 18: Section 9.8 and 9.9: Power Series

Example: Check the EndpointsFind a power series to represent .

21

1 x12x 23x 1... ...nnx

Let’s check the endpoint of the interval of convergence to see if the resulting series converges.

If :

1 2 1 23 1 24 1 25 1 26 1 ... 1 2 3 4 5 6 ...

We know: For at least

Graph the partial sums:

Each Successive term in the sequence of

partial sums is outside the two previous terms

in this sequence .

The Series Diverges

Page 19: Section 9.8 and 9.9: Power Series

Example: Check the EndpointsFind a power series to represent .

21

1 x12x 23x 1... ...nnx

The endpoint is not in the interval of convergence. Now check the other endpoint .

If :

1 2 1 23 1 24 1 25 1 26 1 ...1 2 3 4 5 6 ... ...n

We know: For at least

Notice:

By the nth Term Test, the Series Diverges.

Page 20: Section 9.8 and 9.9: Power Series

Example: ConclusionFind a power series to represent .

12x 23x 1... ...nnx

The following equation is true for :

A Power Series to represent that rational expression is:

The interval of convergence is

21

1 x12x 23x 1... ...nnx

Page 21: Section 9.8 and 9.9: Power Series

Theorem 2: Term-by-Term Integration

If

converges for , then the series

obtained by integrating the series for term by term, converges for and represents on that interval.

If the series for converges for all , then so does the series for the integral.Not so Straightforward.

Page 22: Section 9.8 and 9.9: Power Series

Theorem 1: Term-by-Term IntegrationThat theorem can also be confusing.

What it says is that if…• A power series can be integrated term by term to form a

new series.• The new series will converge to the integral of the

function represented by the original series.• The new series will at least converge on the same

interval as the original series.

This gives a way to generate new connections between functions and series.

Can it converge at a larger interval? Does it include the endpoints?

We will not prove this. It is assumed to be true.

Page 23: Section 9.8 and 9.9: Power Series

Example 1Find a power series to represent .

11 ln 1x dx x Notice:

We already know:for To find

the power series,

we integrate

both sides of

the equation piece by piece:

( 11+𝑥 )𝑑𝑥= (1−𝑥+𝑥2− 𝑥3+…+(−1)𝑛−1𝑥𝑛+…)𝑑𝑥

ln 1 x C x 2

2x 3

3x 4

4x

1( 1)... ...n nxn

Solve for C.

Page 24: Section 9.8 and 9.9: Power Series

Example 1: Solve for CWhat happens to the after you integrate both side of the power series?

( 11+𝑥 )𝑑𝑥= (1−𝑥+𝑥2− 𝑥3+…+(−1)𝑛−1𝑥𝑛+…)𝑑𝑥

x 2

2x

3

3x 4

4x

1( 1)... ...n nxn

ln 1 x C Let .

12 3 4 ( 1) 00 0 02 3 4ln 1 0 0 ... ...

n n

nC

0C The is equal to 0. Now go back to the problem.

Page 25: Section 9.8 and 9.9: Power Series

Example 1: Check the EndpointsThe series below converges for at least :

ln 1 x x 2

2x

3

3x 4

4x

1( 1)... ...n nxn

The series is no longer Geometric, the interval of convergence could have changed. Let’s check the endpoint of the interval of

convergence to see if the resulting series converges.

1 212

313

414

1( 1) 1... ...nn

n

1 1 1 12 3 41 ... ...n This series is

the opposite of the

Divergent Harmonic Series.

1 1 1 12 3 41 ... ...n

Therefore, does not result in a

convergent series and is not in the

interval of convergence.

Page 26: Section 9.8 and 9.9: Power Series

Example 1: Check the Endpoints

The endpoint is not in the interval of convergence. Now check the other endpoint .

12 3 4 1 11 1 12 3 41 ... ...

n n

n

111 1 1

2 3 41 ... ...n

n

This is the Alternating Harmonic

Series.

We know the Alternate Harmonic Series converges. So the series above also converges for . Thus, is in the interval of convergence.

The series below converges for at least :

ln 1 x x 2

2x

3

3x 4

4x

1( 1)... ...n nxn

Page 27: Section 9.8 and 9.9: Power Series

Example 1: ConclusionFind a power series to represent .

The following equation is true for :

A Power Series to represent that natural log expression is:

The interval of convergence is x 2

2x

3

3x 4

4x

1( 1)... ...n nxn

ln 1 x x 2

2x

3

3x 4

4x

1( 1)... ...n nxn

And since the equation above holds for , we now know the value for the Alternating Harmonic Series:

111 1 12 3 4ln 2 1 ... ...

n

n

Page 28: Section 9.8 and 9.9: Power Series

The Alternating Harmonic Series

The Alternating Harmonic Series converges to :

0 12

1?S

ln𝟐≈𝟎 .𝟔𝟗𝟑𝟏𝟒𝟕

Page 29: Section 9.8 and 9.9: Power Series

Example 2Find a power series to represent .

21

1arctan

xdx x

Notice:

We already know: for

To find the power series, we integrate both sides of the equation piece by

piece:

( 11+𝑥2 )𝑑𝑥= (1−𝑥2+𝑥4−𝑥6+…+(−1)𝑛−1𝑥2𝑛−2+… )𝑑𝑥

arctan x C x 3

3x 5

5x 7

7x

1 2 1( 1)2 1... ...n nxn

Solve for C.

Page 30: Section 9.8 and 9.9: Power Series

Example 2: Solve for CWhat happens to the after you integrate both side of the power series?

( 11+𝑥2 )𝑑𝑥= (1−𝑥2+𝑥4−𝑥6+…+(−1)𝑛−1𝑥2𝑛−2+… )𝑑𝑥

x 3

3x

5

5x 7

7x

1 2 1( 1)2 1... ...n nxn

arctan x C

Let .1 2 13 5 7 ( 1) 00 0 0

3 5 7 2 1arctan 0 0 ... ...n n

nC

0C The is equal to 0. Now go back to the problem.

Page 31: Section 9.8 and 9.9: Power Series

Example 2Find a power series to represent .

21

1arctan

xdx x

Notice:

We already know: for

To find the power

series, we integrate

both sides of the equation

piece by piece:

( 11+𝑥2 )𝑑𝑥= (1−𝑥2+𝑥4−𝑥6+…+(−1)𝑛−1𝑥2𝑛−2+… )𝑑𝑥

arctan x x 3

3x 5

5x 7

7x

1 2 1( 1)2 1... ...n nxn

The last theorem guarantees this series will AT LEAST converge on the

same interval as :

Page 32: Section 9.8 and 9.9: Power Series

Example 2: Check the EndpointsThe series below converges for at least :

arctan x x 3

3x

5

5x 7

7x

1 2 1( 1)2 1... ...n nxn

The series is no longer Geometric, the interval of convergence could have changed. Let’s check the endpoint of the interval of

convergence to see if the resulting series converges.

1 313

515

717

2 11( 1) 12 1... ...

nn

n

( 1)1 1 13 5 7 2 11 ... ...

n

n

Graph the partial sums:

Each Successive term in the sequence of

partial sums is inside the two previous terms

in this sequence.

The Series Converges and is in the interval of

convergence

Page 33: Section 9.8 and 9.9: Power Series

Example 2: Check the EndpointsThe series below converges for at least :

arctan x x 3

3x

5

5x 7

7x

1 2 1( 1)2 1... ...n nxn

The endpoint IS in the interval of convergence. Now check the other endpoint .

1 313 51

5 717 2 11( 1) 1

2 1... ...nn

n

1( 1)1 1 13 5 7 2 11 ... ...

n

n

This is the opposite of the alternating sequence from the other endpoint.

Thus, the series converges and is in the interval of convergence.

Page 34: Section 9.8 and 9.9: Power Series

Example: ConclusionFind a power series to represent .

The following equation is true for :

A Power Series to represent that inverse tangent expression is:

The interval of convergence is

And since the equation above holds for , we now know the following:

x 3

3x

5

5x 7

7x

1 2 1( 1)2 1... ...n nxn

arctan x x 3

3x

5

5x 7

7x

1 2 1( 1)2 1... ...n nxn

111 1 14 3 5 7 2 1arctan1 1 ... ...

n

n

Page 35: Section 9.8 and 9.9: Power Series

Approximating Pi with a Power Series

From the previous example we know: 111 1 1

4 3 5 7 2 11 ... ...n

n

Notice what happens when we multiply by 4.

14 14 4 43 5 7 2 14 ... ...

n

n

With a Power Series, we have found a way to approximate Pi.

Test the series by finding partial sums to confirm the result.

Page 36: Section 9.8 and 9.9: Power Series

Closure

Interval of Convergence is a challenging topic. In later sections we will learn actual tests to determine if a series converges.

The biggest thing we should be concerned with is our ability to make new series from

another series. And the possibility the series will be better than the original (i.e. it will

converge for more values of x).