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 PresentationTRANSCRIPT
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
A Polynomial SeriesConsider the Polynomial Series:
Investigate the partial sums of the sequence and compare the results to on a graph.
Window: and
A Polynomial SeriesConsider the Polynomial Series:
Investigate the partial sums of the sequence and compare the results to on a graph.
Window: and
A Polynomial SeriesConsider the Polynomial Series:
Investigate the partial sums of the sequence and compare the results to on a graph.
Window: and
A Polynomial SeriesConsider the Polynomial Series:
Investigate the partial sums of the sequence and compare the results to on a graph.
Window: and
A Polynomial SeriesConsider the Polynomial Series:
Investigate the partial sums of the sequence and compare the results to on a graph.
Window: and
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?
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.)
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)
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:
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+…
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:
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)
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.
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.
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.
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
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.
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
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.
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.
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.
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.
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.
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
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
The Alternating Harmonic Series
The Alternating Harmonic Series converges to :
0 12
1?S
ln𝟐≈𝟎 .𝟔𝟗𝟑𝟏𝟒𝟕
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.
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.
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 :
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
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.
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
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.
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).