Upload
math266
View
161
Download
4
Embed Size (px)
Citation preview
Differentiation and Integration of Power Series
Differentiation and Integration of Power Series Theorem (Derivative and integral of Taylor series) :
Σk=0
∞
Differentiation and Integration of Power Series
Let f(x) = P(x) = the Taylor series = ck(x – a)k for
all x in the interval (a – R, a + R) where R is the radius of convergence, then
Theorem (Derivative and integral of Taylor series) :
Σk=0
∞
Differentiation and Integration of Power Series
Let f(x) = P(x) = the Taylor series = ck(x – a)k for
all x in the interval (a – R, a + R) where R is the radius of convergence, then
Theorem (Derivative and integral of Taylor series) :
I. the Taylor series of f '(x) is P'(x) = [ck(x – a)k]'
with radius of convergence R and f '(x) = P'(x) forall x in the interval (a – R, a + R).
Σk=0
∞
Σk=0
∞
Differentiation and Integration of Power Series
Let f(x) = P(x) = the Taylor series = ck(x – a)k for
all x in the interval (a – R, a + R) where R is the radius of convergence, then
Theorem (Derivative and integral of Taylor series) :
I. the Taylor series of f '(x) is P'(x) = [ck(x – a)k]'
with radius of convergence R and f '(x) = P'(x) forall x in the interval (a – R, a + R).
Σk=0
∞
II. the Taylor series of ∫f(x)dx is ∫P(x)dx= ∫ck(x – a)kdx
with radius of convergence R and ∫f(x)dx = ∫P(x)dx forall x in the interval (a – R, a + R).
Σk=0
∞
Σk=0
∞
Differentiation and Integration of Power Series
Let f(x) = P(x) = the Taylor series = ck(x – a)k for
all x in the interval (a – R, a + R) where R is the radius of convergence, then
Theorem (Derivative and integral of Taylor series) :
I. the Taylor series of f '(x) is P'(x) = [ck(x – a)k]'
with radius of convergence R and f '(x) = P'(x) forall x in the interval (a – R, a + R).
Σk=0
∞
II. the Taylor series of ∫f(x)dx is ∫P(x)dx= ∫ck(x – a)kdx
with radius of convergence R and ∫f(x)dx = ∫P(x)dx forall x in the interval (a – R, a + R).
Σk=0
∞
III. ∫ f(x)dx = ∫ ck(x – a)kdx with the series convergesabsolutely and (c, d) is any interval in (a – R, a + R).
Σk=0
∞
c
d
c
d
Example: Given the Mac series of
Differentiation and Integration of Power Series
Σk=0 (2k+1)!
(-1)kx2k+1∞x –
3!x3
+ 5!x5
+ .. =7!x7
–
Σk=0 (2k)!
(-1)kx2k∞
+ 4!x4
6!x6
8!x8
+ 1 – – – .. =2!x2
sin(x) =
cos(x) =
Example: Given the Mac series of
[sin(x)]' =
Differentiation and Integration of Power Series
Σk=0 (2k+1)!
(-1)kx2k+1∞x –
3!x3
+ 5!x5
+ .. =7!x7
–
Σk=0 (2k)!
(-1)kx2k∞
+ 4!x4
6!x6
8!x8
+ 1 – – – .. =2!x2
sin(x) =
cos(x) =
The derivative of the power series of sin(x) is
Example: Given the Mac series of
[sin(x)]' = [
Differentiation and Integration of Power Series
Σk=0 (2k+1)!
(-1)kx2k+1∞x –
3!x3
+ 5!x5
+ .. =7!x7
–
Σk=0 (2k)!
(-1)kx2k∞
+ 4!x4
6!x6
8!x8
+ 1 – – – .. =2!x2
sin(x) =
cos(x) =
x – 3!x3
+ 5!x5
+ .. ]' 7!x7
–
The derivative of the power series of sin(x) is
Example: Given the Mac series of
[sin(x)]' = [
Differentiation and Integration of Power Series
Σk=0 (2k+1)!
(-1)kx2k+1∞x –
3!x3
+ 5!x5
+ .. =7!x7
–
Σk=0 (2k)!
(-1)kx2k∞
+ 4!x4
6!x6
8!x8
+ 1 – – – .. =2!x2
sin(x) =
cos(x) =
x – 3!x3
+ 5!x5
+ .. ]' 7!x7
–
+ 4!x4
6!x6
+ … = 1 – – 2!x2
The derivative of the power series of sin(x) is
Example: Given the Mac series of
[sin(x)]' = [
Differentiation and Integration of Power Series
Σk=0 (2k+1)!
(-1)kx2k+1∞x –
3!x3
+ 5!x5
+ .. =7!x7
–
Σk=0 (2k)!
(-1)kx2k∞
+ 4!x4
6!x6
8!x8
+ 1 – – – .. =2!x2
sin(x) =
cos(x) =
x – 3!x3
+ 5!x5
+ .. ]' 7!x7
–
+ 4!x4
6!x6
+ … = 1 – – 2!x2
= cos(x)
The derivative of the power series of sin(x) is
Example: Given the Mac series of
[sin(x)]' = [
Differentiation and Integration of Power Series
Σk=0 (2k+1)!
(-1)kx2k+1∞x –
3!x3
+ 5!x5
+ .. =7!x7
–
Σk=0 (2k)!
(-1)kx2k∞
+ 4!x4
6!x6
8!x8
+ 1 – – – .. =2!x2
sin(x) =
cos(x) =
x – 3!x3
+ 5!x5
+ .. ]' 7!x7
–
+ 4!x4
6!x6
+ … = 1 – – 2!x2
= cos(x)
The derivative of the power series of sin(x) is
∫sin(x)dx = The antiderivative of the power series of sin(x) is
Example: Given the Mac series of
[sin(x)]' = [
Differentiation and Integration of Power Series
Σk=0 (2k+1)!
(-1)kx2k+1∞x –
3!x3
+ 5!x5
+ .. =7!x7
–
Σk=0 (2k)!
(-1)kx2k∞
+ 4!x4
6!x6
8!x8
+ 1 – – – .. =2!x2
sin(x) =
cos(x) =
x – 3!x3
+ 5!x5
+ .. ]' 7!x7
–
+ 4!x4
6!x6
+ … = 1 – – 2!x2
= cos(x)
The derivative of the power series of sin(x) is
∫sin(x)dx = ∫xdx – ∫ dx 3!x3
+ ∫ dx 5!x5
–…
The antiderivative of the power series of sin(x) is
Example: Given the Mac series of
[sin(x)]' = [
Differentiation and Integration of Power Series
Σk=0 (2k+1)!
(-1)kx2k+1∞x –
3!x3
+ 5!x5
+ .. =7!x7
–
Σk=0 (2k)!
(-1)kx2k∞
+ 4!x4
6!x6
8!x8
+ 1 – – – .. =2!x2
sin(x) =
cos(x) =
x – 3!x3
+ 5!x5
+ .. ]' 7!x7
–
+ 4!x4
6!x6
+ … = 1 – – 2!x2
= cos(x)
The derivative of the power series of sin(x) is
∫sin(x)dx = ∫xdx – ∫ dx 3!x3
+ ∫ dx 5!x5
–…
The antiderivative of the power series of sin(x) is
4!x4
6!x6= + 2!
x2– – … + c
Example: Given the Mac series of
[sin(x)]' = [
Differentiation and Integration of Power Series
Σk=0 (2k+1)!
(-1)kx2k+1∞x –
3!x3
+ 5!x5
+ .. =7!x7
–
Σk=0 (2k)!
(-1)kx2k∞
+ 4!x4
6!x6
8!x8
+ 1 – – – .. =2!x2
sin(x) =
cos(x) =
x – 3!x3
+ 5!x5
+ .. ]' 7!x7
–
+ 4!x4
6!x6
+ … = 1 – – 2!x2
= cos(x)
The derivative of the power series of sin(x) is
∫sin(x)dx = ∫xdx – ∫ dx 3!x3
+ ∫ dx 5!x5
–…
The antiderivative of the power series of sin(x) is
4!x4
6!x6= + 2!
x2= -cos(x) + c – – … + c
Example: Given the Mac series of
[sin(x)]' = [
Differentiation and Integration of Power Series
Σk=0 (2k+1)!
(-1)kx2k+1∞x –
3!x3
+ 5!x5
+ .. =7!x7
–
Σk=0 (2k)!
(-1)kx2k∞
+ 4!x4
6!x6
8!x8
+ 1 – – – .. =2!x2
sin(x) =
cos(x) =
x – 3!x3
+ 5!x5
+ .. ]' 7!x7
–
+ 4!x4
6!x6
+ … = 1 – – 2!x2
= cos(x)
The derivative of the power series of sin(x) is
∫sin(x)dx = ∫xdx – ∫ dx 3!x3
+ ∫ dx 5!x5
–…
The antiderivative of the power series of sin(x) is
4!x4
6!x6= + 2!
x2= cos(x) + c – – … + c
They all have infinite radius of convergence.
Example: Use the fact [ ]' =
Differentiation and Integration of Power Series
1 – x 1
(1 – x)2 1
to find the Mac series of .
(1 – x)2 1
Example: Use the fact [ ]' =
Differentiation and Integration of Power Series
1 – x 1
(1 – x)2 1
to find the Mac series of .
(1 – x)2 1
The Mac series of =
1 – x 1 Σ
k=0
∞xk = 1 + x + x2 + x3 + ..
Example: Use the fact [ ]' =
Differentiation and Integration of Power Series
1 – x 1
(1 – x)2 1
to find the Mac series of .
(1 – x)2 1
The Mac series of =
1 – x 1 Σ
k=0
∞xk = 1 + x + x2 + x3 + ..
Hence the Mac series of it's derivative is 1 – x
1 Σk=0
∞xk = 1 + x + x2 + x3 + .. =
Example: Use the fact [ ]' =
Differentiation and Integration of Power Series
1 – x 1
(1 – x)2 1
to find the Mac series of .
(1 – x)2 1
The Mac series of =
1 – x 1 Σ
k=0
∞xk = 1 + x + x2 + x3 + ..
Hence the Mac series of it's derivative is 1 – x
1 Σk=0
∞xk = 1 + x + x2 + x3 + .. =
(1 – x)2 1
ddx
Example: Use the fact [ ]' =
Differentiation and Integration of Power Series
1 – x 1
(1 – x)2 1
to find the Mac series of .
(1 – x)2 1
The Mac series of =
1 – x 1 Σ
k=0
∞xk = 1 + x + x2 + x3 + ..
Hence the Mac series of it's derivative is
Σk=0
xk[ ]' ∞
1 – x 1 Σ
k=0
∞xk = 1 + x + x2 + x3 + .. =
(1 – x)2 1
ddx
ddx
=
Example: Use the fact [ ]' =
Differentiation and Integration of Power Series
1 – x 1
(1 – x)2 1
to find the Mac series of .
(1 – x)2 1
The Mac series of =
1 – x 1 Σ
k=0
∞xk = 1 + x + x2 + x3 + ..
Hence the Mac series of it's derivative is
Σk=0
xk[ ]' = ∞Σk=1
kxk-1 ∞
1 – x 1 Σ
k=0
∞xk = 1 + x + x2 + x3 + .. =
(1 – x)2 1
ddx
ddx
=
Example: Use the fact [ ]' =
Differentiation and Integration of Power Series
1 – x 1
(1 – x)2 1
to find the Mac series of .
(1 – x)2 1
The Mac series of =
1 – x 1 Σ
k=0
∞xk = 1 + x + x2 + x3 + ..
Hence the Mac series of it's derivative is
Σk=0
xk[ ]' = ∞Σk=1
kxk-1 = 1 + 2x + 3x2 + 4x3 +…∞
1 – x 1 Σ
k=0
∞xk = 1 + x + x2 + x3 + .. =
(1 – x)2 1
ddx
ddx
=
Example: Use the fact [ ]' =
Differentiation and Integration of Power Series
1 – x 1
(1 – x)2 1
to find the Mac series of .
(1 – x)2 1
The Mac series of =
1 – x 1 Σ
k=0
∞xk = 1 + x + x2 + x3 + ..
Hence the Mac series of it's derivative is
Σk=0
xk
The radius of convergence of 1/(x – 1) is R = 1,
[ ]' = ∞Σk=1
kxk-1 = 1 + 2x + 3x2 + 4x3 +…∞
1 – x 1 Σ
k=0
∞xk = 1 + x + x2 + x3 + .. =
(1 – x)2 1
ddx
ddx
=
Example: Use the fact [ ]' =
Differentiation and Integration of Power Series
1 – x 1
(1 – x)2 1
to find the Mac series of .
(1 – x)2 1
The Mac series of =
1 – x 1 Σ
k=0
∞xk = 1 + x + x2 + x3 + ..
Hence the Mac series of it's derivative is
Σk=0
xk
The radius of convergence of 1/(x – 1) is R = 1,hence R =1 is the radius of convergence of
[ ]' = ∞Σk=1
kxk-1 = 1 + 2x + 3x2 + 4x3 +…∞
Σk=1
∞kxk-1,
1 – x 1 Σ
k=0
∞xk = 1 + x + x2 + x3 + .. =
(1 – x)2 1
ddx
ddx
=
Example: Use the fact [ ]' =
Differentiation and Integration of Power Series
1 – x 1
(1 – x)2 1
to find the Mac series of .
(1 – x)2 1
The Mac series of =
1 – x 1 Σ
k=0
∞xk = 1 + x + x2 + x3 + ..
Hence the Mac series of it's derivative is
Σk=0
xk
The radius of convergence of 1/(x – 1) is R = 1,hence R =1 is the radius of convergence of and 1/(1 – x)2 = for all x in (-1, 1).
[ ]' = ∞Σk=1
kxk-1 = 1 + 2x + 3x2 + 4x3 +…∞
Σk=1
∞kxk-1,
Σk=1
∞kxk-1
1 – x 1 Σ
k=0
∞xk = 1 + x + x2 + x3 + .. =
(1 – x)2 1
ddx
ddx
=
We may use the integral formula to find the power series of some non-elementary functions.
Differentiation and Integration of Power Series
We may use the integral formula to find the power series of some non-elementary functions.
Differentiation and Integration of Power Series
Example: Find the Mac series of ∫ dxx sin(x)
We may use the integral formula to find the power series of some non-elementary functions.
Differentiation and Integration of Power Series
Example: Find the Mac series of ∫ dxx sin(x)
The Mac series of is
x sin(x)
x – 3!x3
+ 5!x5
+ .. 7!x7
– ( ) / x
We may use the integral formula to find the power series of some non-elementary functions.
Differentiation and Integration of Power Series
Example: Find the Mac series of ∫ dxx sin(x)
The Mac series of is
x sin(x)
x – 3!x3
+ 5!x5
+ .. 7!x7
– ( ) / x
= 1 – 3!x2
+ 5!x4
+ .. 7!x6
–
We may use the integral formula to find the power series of some non-elementary functions.
Differentiation and Integration of Power Series
Example: Find the Mac series of ∫ dxx sin(x)
The Mac series of is
x sin(x)
x – 3!x3
+ 5!x5
+ .. 7!x7
– ( ) / x
= 1 – 3!x2
+ 5!x4
+ .. 7!x6
– Σk=0 (2k+1)!
(-1)kx2k∞=
We may use the integral formula to find the power series of some non-elementary functions.
Differentiation and Integration of Power Series
Example: Find the Mac series of ∫ dxx sin(x)
The Mac series of is
x sin(x)
x – 3!x3
+ 5!x5
+ .. 7!x7
– ( ) / x
= 1 – 3!x2
+ 5!x4
+ .. 7!x6
– Σk=0 (2k+1)!
(-1)kx2k∞=
So the Mac series of ∫ dx = x sin(x) Σ ∫
k=0 (2k+1)!(-1)kx2k∞ dx
We may use the integral formula to find the power series of some non-elementary functions.
Differentiation and Integration of Power Series
Example: Find the Mac series of ∫ dxx sin(x)
The Mac series of is
x sin(x)
x – 3!x3
+ 5!x5
+ .. 7!x7
– ( ) / x
= 1 – 3!x2
+ 5!x4
+ .. 7!x6
– Σk=0 (2k+1)!
(-1)kx2k∞=
So the Mac series of ∫ dx = x sin(x) Σ ∫
k=0 (2k+1)!(-1)kx2k∞ dx
= Σ k=0 (2k+1)(2k+1)!
(-1)kx2k+1∞
We may use the integral formula to find the power series of some non-elementary functions.
Differentiation and Integration of Power Series
Example: Find the Mac series of ∫ dxx sin(x)
The Mac series of is
x sin(x)
x – 3!x3
+ 5!x5
+ .. 7!x7
– ( ) / x
= 1 – 3!x2
+ 5!x4
+ .. 7!x6
– Σk=0 (2k+1)!
(-1)kx2k∞=
So the Mac series of ∫ dx = x sin(x) Σ ∫
k=0 (2k+1)!(-1)kx2k∞ dx
= Σ k=0 (2k+1)(2k+1)!
(-1)kx2k+1∞ = x – 3*3!x3
+ + .. – 5*5!x5
7*7!x7
Differentiation and Integration of Power Series
Example: Find ∫ e-x dx accurate to 3 decimal places. x=0
12
Differentiation and Integration of Power Series
Example: Find ∫ e-x dx accurate to 3 decimal places. x=0
1
∫ e-x dx is not an elementary function. We have to use Taylor expansion to obtain the answer we want.
2
2
Differentiation and Integration of Power Series
Example: Find ∫ e-x dx accurate to 3 decimal places.
ex =
n! xn
x=0
1
∫ e-x dx is not an elementary function. We have to use Taylor expansion to obtain the answer we want.
2
2
Σk=0
∞
Differentiation and Integration of Power Series
Example: Find ∫ e-x dx accurate to 3 decimal places.
ex =
n! xn
x=0
1
∫ e-x dx is not an elementary function. We have to use Taylor expansion to obtain the answer we want.
2
2
Σk=0
∞e-x =
n! (-x2)nΣk=0
∞2 =
Differentiation and Integration of Power Series
Example: Find ∫ e-x dx accurate to 3 decimal places.
ex =
n! xn
x=0
1
∫ e-x dx is not an elementary function. We have to use Taylor expansion to obtain the answer we want.
2
2
Σk=0
∞e-x =
n! (-x2)nΣk=0
∞2 x2 + 2!x4
3!x6
4!x8
+ = 1 – – – ..
Differentiation and Integration of Power Series
Example: Find ∫ e-x dx accurate to 3 decimal places.
ex =
n! xn
x=0
1
∫ e-x dx is not an elementary function. We have to use Taylor expansion to obtain the answer we want.
2
2
Σk=0
∞e-x =
n! (-x2)nΣk=0
∞2 x2 + 2!x4
3!x6
4!x8
+ = 1
Hence ∫ e-x dx =2
n! (-x)2n
Σ ∫ dx = k=0
∞
– – – ..
Differentiation and Integration of Power Series
Example: Find ∫ e-x dx accurate to 3 decimal places.
ex =
n! xn
x=0
1
∫ e-x dx is not an elementary function. We have to use Taylor expansion to obtain the answer we want.
2
2
Σk=0
∞e-x =
n! (-x2)nΣk=0
∞2 x2 + 2!x4
3!x6
4!x8
+ = 1
Hence ∫ e-x dx =2
n! (-x2)n
Σ ∫ dx = k=0
∞
(2n+1)n! Σk=0
∞ (-1)nx2n+1
– – – ..
Differentiation and Integration of Power Series
Example: Find ∫ e-x dx accurate to 3 decimal places.
ex =
n! xn
x=0
1
∫ e-x dx is not an elementary function. We have to use Taylor expansion to obtain the answer we want.
2
2
Σk=0
∞e-x =
n! (-x2)nΣk=0
∞2 x2 + 2!x4
3!x6
4!x8
+ = 1
Hence ∫ e-x dx =2
n! (-x2)n
Σ ∫ dx = k=0
∞
(2n+1)n! Σk=0
∞
= c + x
(-1)nx2n+1
3x3
+ 5*2!x5
+ .. 7*3!x7
– – – ..
– –
Differentiation and Integration of Power Series
Example: Find ∫ e-x dx accurate to 3 decimal places.
ex =
n! xn
x=0
1
∫ e-x dx is not an elementary function. We have to use Taylor expansion to obtain the answer we want.
2
2
Σk=0
∞e-x =
n! (-x2)nΣk=0
∞2 x2 + 2!x4
3!x6
4!x8
+ = 1
Hence ∫ e-x dx =2
n! (-x2)n
Σ ∫ dx = k=0
∞
(2n+1)n! Σk=0
∞
= c + x
(-1)nx2n+1
3x3
+ 5*2!x5
+ .. 7*3!x7
= P(x)
– – – ..
– –
Differentiation and Integration of Power Series
Example: Find ∫ e-x dx accurate to 3 decimal places.
ex =
n! xn
x=0
1
∫ e-x dx is not an elementary function. We have to use Taylor expansion to obtain the answer we want.
2
2
Σk=0
∞e-x =
n! (-x2)nΣk=0
∞2 x2 + 2!x4
3!x6
4!x8
+ = 1
Hence ∫ e-x dx =2
n! (-x2)n
Σ ∫ dx = k=0
∞
(2n+1)n! Σk=0
∞
= c + x
(-1)nx2n+1
3x3
+ 5*2!x5
+ .. 7*3!x7
= P(x)
∫ e-x dx = P(1) – P(0) x=0
12
– – – ..
– –
Therefore
Differentiation and Integration of Power Series
Example: Find ∫ e-x dx accurate to 3 decimal places.
ex =
n! xn
x=0
1
∫ e-x dx is not an elementary function. We have to use Taylor expansion to obtain the answer we want.
2
2
Σk=0
∞e-x =
n! (-x2)nΣk=0
∞2 x2 + 2!x4
3!x6
4!x8
+ = 1
Hence ∫ e-x dx =2
n! (-x2)n
Σ ∫ dx = k=0
∞
(2n+1)n! Σk=0
∞
= c + x
(-1)nx2n+1
3x3
+ 5*2!x5
+ .. 7*3!x7
= P(x)
∫ e-x dx = P(1) – P(0) x=0
1
= 1 31 + 5*2!
1 + .. 7*3!1
2
– – – ..
– –
– –
Therefore
which is a "decreasing" alternating series.
Differentiation and Integration of Power Series
∫ e-x dx x=0
1
= 1 2Henceis a convergent alternating series.
31 + 5*2!
1 + 7*3!1– – 1
9*4! – 111*5! + 13*6! …
1
Differentiation and Integration of Power Series
∫ e-x dx x=0
1
= 1 2Henceis a convergent alternating series.
From theorem of alternaing series, any "decreasing" convergent alternating series, it's sum S satisfies |S| < |a1=1st term|.
31 + 5*2!
1 + 7*3!1– – 1
9*4! – 111*5! + 13*6! …
1
Differentiation and Integration of Power Series
∫ e-x dx x=0
1
= 1 2Henceis a convergent alternating series.
From theorem of alternaing series, any "decreasing" convergent alternating series, it's sum S satisfies |S| < |a1=1st term|.
By trial and error, we find that < 0.0005. 13*6!1
31 + 5*2!
1 + 7*3!1– – 1
9*4! – 111*5! + 13*6! …
1
Differentiation and Integration of Power Series
∫ e-x dx x=0
1
= 1 2Henceis a convergent alternating series.
From theorem of alternaing series, any "decreasing" convergent alternating series, it's sum S satisfies |S| < |a1=1st term|.
By trial and error, we find that < 0.0005. 13*6!1
So the tail series 1 – .. < 0.0005 + 13*6!1
15*7!1
17*8!–
31 + 5*2!
1 + 7*3!1– – 1
9*4! – 111*5! + 13*6! …
1
Differentiation and Integration of Power Series
∫ e-x dx x=0
1
= 1 31 + 5*2!
1 + 7*3!12 – – Hence
is a convergent alternating series.
From theorem of alternaing series, any "decreasing" convergent alternating series, it's sum S satisfies |S| < |a1=1st term|.
By trial and error, we find that < 0.0005. 13*6!1
So the tail series 1 – .. < 0.0005 + 13*6!1
15*7!1
17*8!
This implies the sum of the front terms 1
31 + 5*2!
1 + 7*3!1– –
–
19*4! – 1
11*5! 0.74673
is accurate to 3 decimal places.
19*4! – 1
11*5! + 13*6! …1