x31 power series

Power Series

A power series P(x) is a "polynomial" in x of infinitely many terms.

Power Series

A power series P(x) is a "polynomial" in x of infinitely many terms. That is,

Σn=0

∞anxn

= a0 + a1x + a2x2 + a3x3 + a4x4 ….

Power Series

P(x) =


Σn=0

∞anxn

= a0 + a1x + a2x2 + a3x3 + a4x4 ….

Note that P(0) = a0.

Power Series

P(x) =


Σn=0

∞anxn

= a0 + a1x + a2x2 + a3x3 + a4x4 ….

The basic question for a power series is to determine the value(s) of x where it converges or diverges.


Power Series

P(x) =


Σn=0

∞anxn

= a0 + a1x + a2x2 + a3x3 + a4x4 ….

Example: A. Let P(x) = 1 + x + x2 + x3 + x4 …



Power Series

P(x) =


Σn=0

∞anxn

= a0 + a1x + a2x2 + a3x3 + a4x4 ….

Example: A. Let P(x) = 1 + x + x2 + x3 + x4 …From the theorem of geometric series, given x = r,I. P(r) = 1 + r + r2 + r3 + r4 + … converges if | r | < 1.



Power Series

P(x) =


Σn=0

∞anxn

= a0 + a1x + a2x2 + a3x3 + a4x4 ….

Example: A. Let P(x) = 1 + x + x2 + x3 + x4 …From the theorem of geometric series, given x = r,I. P(r) = 1 + r + r2 + r3 + r4 + … converges if | r | < 1.II. P(r) diverges if | r | > 1.



Power Series

P(x) =


Σn=0

∞anxn

= a0 + a1x + a2x2 + a3x3 + a4x4 ….

Example: A. Let P(x) = 1 + x + x2 + x3 + x4 …From the theorem of geometric series, given x = r,I. P(r) = 1 + r + r2 + r3 + r4 + … converges if | r | < 1.II. P(r) diverges if | r | > 1.



Power Series

P(x) =

0-1 converges (absolutely) 1 divergesdiverges

P(x) = Σ xn

Example: B. Let P(x) = 1 + Power Series

x x2+ 2 + x3

3 + x4

4 +..

Example: B. Let P(x) = 1 +

Use the ceiling theorem:

Power Series x x2

+ 2 + x3

3 + x4

4 +..

P(x) = 1 + |x| |x|2+ 2 + 3 + 4 +..|x|3 |x|4

1 +|x| + |x2| + |x3| + |x4| +.. …vI vI vI vI vI



Power Series x x2

+ 2 + x3

3 + x4

4 +..

P(x) = 1 + |x| |x|2+ 2 + 3 + 4 +..|x|3 |x|4

1 +|x| + |x2| + |x3| + |x4| +.. …vI vI vI vI vI

and 1 +|x| + |x2| + |x3| + |x4| +… converges for | x | < 1,

we conclude that



Power Series x x2

+ 2 + x3

3 + x4

4 +..

P(x) = 1 + |x| |x|2+ 2 + 3 + 4 +..|x|3 |x|4

1 +|x| + |x2| + |x3| + |x4| +.. …vI vI vI vI vI

and 1 +|x| + |x2| + |x3| + |x4| +… converges for | x | < 1,

we conclude that

P(x) = 1 + |x||x|2

+ 2 + 3 + 4 +..|x|3 |x|4

also converges absolutely for |x| < 1.

Power Series For x = ±1:

P(1) = 1 + is basically the harmonic series, so it diverges.

Power Series

1 1+ 2 + 13 + 1

4 +..For x = ±1:


Power Series

1 1+ 2 + 13 + 1

4 +..

P(-1) = 1 – is basically the alternaing harmonic series. It converges conditionally.

1 1 –2+ 13 + 1

4 – ..

For x = ±1:


Power Series

1 1+ 2 + 13 + 1

4 +..

For x = | r | > 1,


1 1 –2+ 13 + 1

4 – ..

For x = ±1:


Power Series

1 1+ 2 + 13 + 1

4 +..

For x = | r | > 1, the general terms of P(r)


1 1 –2+ 13 + 1

4 – ..

For x = ±1:

rn

n ∞ as n ∞


Power Series

1 1+ 2 + 13 + 1

4 +..



1 1 –2+ 13 + 1

4 – ..

For x = ±1:

rn

n ∞ as n ∞ (verify this by L'Hopital's rule).


Power Series

1 1+ 2 + 13 + 1

4 +..


P(r) = 1 + r r2+ 2 + 3 + 4 +.. diverges.r3 r4


1 1 –2+ 13 + 1

4 – ..

For x = ±1:

rn

n ∞ as n ∞ (verify this by L'Hopital's rule).Hence


Power Series

1 1+ 2 + 13 + 1

4 +..


P(r) = 1 + r r2+ 2 + 3 + 4 +.. diverges.r3 r4


1 1 –2+ 13 + 1

4 – ..

For x = ±1:

rn


We summarize the result using the following picture:


Power Series

1 1+ 2 + 13 + 1

4 +..


P(r) = 1 + r r2+ 2 + 3 + 4 +.. diverges.r3 r4


1 1 –2+ 13 + 1

4 – ..

For x = ±1:

rn



0-1 converges (absolutely) 1 divergesdiverges

P(x) = Σ xn

n


Power Series

1 1+ 2 + 13 + 1

4 +..


P(r) = 1 + r r2+ 2 + 3 + 4 +.. diverges.r3 r4


1 1 –2+ 13 + 1

4 – ..

For x = ±1:

rn



0-1 converges (absolutely) 1

converges conditionally diverges

divergesdiverges

P(x) = Σ xn

n

Theorem: Given a power series , one of the following must be true:

Power Series Σn=1

P(x)∞

Theorem: Given a power series , one of the following must be true: I. P(x) converges at x = 0, diverges everywhere else.

Power Series Σn=1

P(x)∞


Power Series Σn=1

P(x)∞

0

converges at 0 divergesdiverges


Power Series Σn=1

P(x)∞

II. P(x) converges (absolutely) for | x | < R for some positive R,

0



Power Series Σn=1

P(x)∞

II. P(x) converges (absolutely) for | x | < R for some positive R, diverges for | x | > R

0



Power Series Σn=1

P(x)∞

II. P(x) converges (absolutely) for | x | < R for some positive R, diverges for | x | > R and at x = ±R, the series may converge or diverges.

0



Power Series Σn=1

P(x)∞

II. P(x) converges (absolutely) for | x | < R for some positive R, diverges for | x | > R and at x = ±R, the series may converge or diverges. R is called the radius of convergence.

0



Power Series Σn=1

P(x)∞


0


0

R converges (absolutely) R

? ?

divergesdiverges


Power Series Σn=1

P(x)∞


0


0

R converges (absolutely) R

? ?

divergesdiverges

III. P(x) converges everywhere.

0convergesconverges

Power Series We may calculate the the radius of convergence using the ratio test or the root test.

Power Series

Example: C. Discuss the convergence of P(x) =Σ

n=0

∞ xn

n! .

We may calculate the the radius of convergence using the ratio test or the root test.

Power Series

Since the coefficients are factorials, use the ratio test:


n=0

∞ xn

n! .


Power Series

Since the coefficients are factorials, use the ratio test:Σn=0

∞anxn

= a0 + a1x + a2x2 + a3x3 ... converges P(x) = for x if |an+1xn+1|* |anxn|

1 < 1 as n ∞.


n=0

∞ xn

n! .


Power Series


∞anxn


1 < 1 as n ∞.

|an+1xn+1|* |anxn|

1 =xn+1

(n+1)!* n!xn


n=0

∞ xn

n! .


Power Series


∞anxn


1 < 1 as n ∞.

|an+1xn+1|* |anxn|

1 =xn+1

(n+1)!* n!xn =

xn+1


n=0

∞ xn

n! .


Power Series


∞anxn


1 < 1 as n ∞.

|an+1xn+1|* |anxn|

1 =xn+1

(n+1)!* n!xn =

xn+1


n=0

∞ xn

n! .


As n ∞, for any real number x, xn+1 0 < 1.

Power Series


∞anxn


1 < 1 as n ∞.

|an+1xn+1|* |anxn|

1 =xn+1

(n+1)!* n!xn =

xn+1


n=0

∞ xn

n! .


As n ∞, for any real number x, xn+1 0 < 1.

Hence the series converegs for all numbers x.

Example: D. Find the radius of convergence of P(x) =

Power Series

Σn=0

∞ xn

2n


Power Series

Σn=0

∞ xn

2n

By the root test, for any real number x,

Σn=0

∞anxn

= a0 + a1x + a2x2 + a3x3 + a4x4 …. P(x) = converges if lim |anxn| < 1, diverges if its > 1. n

n∞


Power Series

Σn=0

∞ xn

2n


Σn=0

∞anxn

= a0 + a1x + a2x2 + a3x3 + a4x4 …. P(x) = converges if lim |anxn| < 1, diverges if its > 1.

anxn = xn

2n,

n

n∞


Power Series

Σn=0

∞ xn

2n


Σn=0

∞anxn


anxn = xn

2n, so the series converges if

n

n

n∞

lim |xn/2n| n∞


Power Series

Σn=0

∞ xn

2n


Σn=0

∞anxn


anxn = xn


n

n

n∞

lim |xn/2n| = |x/2| < 1 n∞


Power Series

Σn=0

∞ xn

2n


Σn=0

∞anxn


anxn = xn


n

n

n∞

lim |xn/2n| = |x/2| < 1 or | x | < 2.n∞


Power Series

Σn=0

∞ xn

2n


Σn=0

∞anxn


anxn = xn

2n,

So the radius of convergence R = 2.

so the series converges if

n

n

n∞

lim |xn/2n| = |x/2| < 1 or | x | < 2.n∞


Power Series

Σn=0

∞ xn

2n


Σn=0

∞anxn


anxn = xn

2n,

So the radius of convergence R = 2.

so the series converges if

n

n

n∞

lim |xn/2n| = |x/2| < 1 or | x | < 2.n∞

Therefore the series converges for | x | < 2 and diverges | x | > 2.

Power Series E. Discuss the convergence at x = ±2 then summarize the convergence of P(x).


P(2) = Σn=0

∞ 2n

2n


P(2) = Σn=0

∞ 2n

2n = 1 + 1 + 1 + … so it diverges at x = 2.


P(2) = Σn=0

∞ 2n


P(-2) = Σn=0

∞ (-2)n

2n


P(2) = Σn=0

∞ 2n


P(-2) = Σn=0

∞ (-2)n

2n = 1 – 1 + 1 – 1.. so it diverges at x = -2.


P(2) = Σn=0

∞ 2n


P(-2) = Σn=0

∞ (-2)n


So P(x) converges for | x | < 2, diverges for | x | > 2.


P(2) = Σn=0

∞ 2n


P(-2) = Σn=0

∞ (-2)n



A power series in (x – a) is defined to be Σn=0

∞an(x – a)n.


P(2) = Σn=0

∞ 2n


P(-2) = Σn=0

∞ (-2)n




∞an(x – a)n.

We call this the power series centered at x = a.


P(2) = Σn=0

∞ 2n


P(-2) = Σn=0

∞ (-2)n




∞an(x – a)n.

The power series

Σn=0

∞anxn

We call this the power series centered at x = a.

is the special case of

Σn=0

∞an(x – a)n with a = 0.

Theorem: Given a power series , one of the following must be true:

Power Series Σn=0

∞an(x – a)n

Theorem: Given a power series , one of the following must be true: I. It converges at x = a, diverges everywhere else.

Power Series

a

converges at a divergesdiverges

Σn=0

∞an(x – a)n


Power Series

II. It converges (absolutely) for | x – a | < R for some positive R, diverges for | x – a | > R and at x = a±R, the series may converge or diverges. (a – R, a + R) is called the interval of convergence.

a


a

a – R converges (absolutely) a + R

? ?

divergesdiverges

Σn=0

∞an(x – a)n


Power Series

II. It converges (absolutely) for | x – a | < R for some positive R, diverges for | x – a | > R and at x = a±R, the series may converge or diverges. (a – R, a + R) is called the interval of convergence.

a


a

a – R converges (absolutely) a + R

? ?

divergesdiverges

III. It converges everywhere.

aconergesconerges

Σn=0

∞an(x – a)n

Example: F. Discuss the convergence of P(x) =

Power Series

Σn=1

∞ Ln(n)(x – 3)n

n


Power Series

Σn=1

∞ Ln(n)(x – 3)n

n

Use the ratio test,


Power Series

Σn=1

∞ Ln(n)(x – 3)n

n

Use the ratio test,

|an+1 (x –a)n+1|* |an(x – a)n|

1


Power Series

Σn=1

∞ Ln(n)(x – 3)n

n

Use the ratio test,

|an+1 (x –a)n+1|* |an(x – a)n|

1

= Ln(n+1)(x – 3)n+1

n+1 * Ln(n)(x – 3)nn


Power Series

Σn=1

∞ Ln(n)(x – 3)n

n

Use the ratio test,

|an+1 (x –a)n+1|* |an(x – a)n|

1

= Ln(n+1)(x – 3)n+1

n+1 * Ln(n)(x – 3)nn

= n * Ln(n+1) (x – 3)(n+1) * Ln(n)


Power Series

Σn=1

∞ Ln(n)(x – 3)n

n

Use the ratio test,

|an+1 (x –a)n+1|* |an(x – a)n|

1

= Ln(n+1)(x – 3)n+1

n+1 * Ln(n)(x – 3)nn

= n * Ln(n+1) (x – 3)(n+1) * Ln(n)

, as n ∞


Power Series

Σn=1

∞ Ln(n)(x – 3)n

n

Use the ratio test,

|an+1 (x –a)n+1|* |an(x – a)n|

1

= Ln(n+1)(x – 3)n+1

n+1 * Ln(n)(x – 3)nn

= n * Ln(n+1) (x – 3)(n+1) * Ln(n)

, as n ∞

1 1


Power Series

Σn=1

∞ Ln(n)(x – 3)n

n

Use the ratio test,

|an+1 (x –a)n+1|* |an(x – a)n|

1

= Ln(n+1)(x – 3)n+1

n+1 * Ln(n)(x – 3)nn

= n * Ln(n+1) (x – 3)(n+1) * Ln(n)

, as n ∞ we get |x – 3|

1 1


Power Series

Σn=1

∞ Ln(n)(x – 3)n

n

Use the ratio test,

|an+1 (x –a)n+1|* |an(x – a)n|

1

= Ln(n+1)(x – 3)n+1

n+1 * Ln(n)(x – 3)nn

= n * Ln(n+1) (x – 3)(n+1) * Ln(n)

, as n ∞ we get |x – 3|

1 1

The series converges if |x – 3| < 1

Power Series For the end points of the interval:

P(4) =

Power Series

Σn=1

∞ Ln(n)(4 – 3)n

n

For the end points of the interval:

P(4) =

Power Series

Σn=1

∞ Ln(n)(4 – 3)n

n = Σn=1

∞ Ln(n)n


P(4) =

Power Series

Σn=1

∞ Ln(n)(4 – 3)n

n = Σn=1

∞ Ln(n)n > Σ

n=2

∞ 1n = ∞


P(4) =

Power Series

Σn=1

∞ Ln(n)(4 – 3)n

n

Σn=1

∞ Ln(n)(2 – 3)n

n

= Σn=1

∞ Ln(n)n > Σ

n=2

∞ 1n = ∞

P(2) =


P(4) =

Power Series

Σn=1

∞ Ln(n)(4 – 3)n

n

Σn=1

∞ Ln(n)(2 – 3)n

n

= Σn=1

∞ Ln(n)n > Σ

n=2

∞ 1n = ∞

P(2) =


= Σn=1

∞ (-1)nLn(n)n

P(4) =

Power Series

Σn=1

∞ Ln(n)(4 – 3)n

n

Σn=1

∞ Ln(n)(2 – 3)n

n

= Σn=1

∞ Ln(n)n

is an alternating series with term 0,

> Σn=2

∞ 1n = ∞

P(2) =


= Σn=1

∞ (-1)nLn(n)n

P(4) =

Power Series

Σn=1

∞ Ln(n)(4 – 3)n

n

Σn=1

∞ Ln(n)(2 – 3)n

n

= Σn=1

∞ Ln(n)n

is an alternating series with term 0, and we check that Ln(n)/(n+1) > Ln(n+1)/(n+1) so it converges. However it converges conditionally.

> Σn=2

∞ 1n = ∞

P(2) =


= Σn=1

∞ (-1)nLn(n)n

Hence we have the following graph for convergence:

3

2 converges (absolutely) 4

converges conditionally

divergesdiverges

x31 power series

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