36 differentiation and integration of power series
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fferentiation and Integration of Power Serie
Transcript of 36 differentiation and integration of power series
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 converges
absolutely 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)n
Σk=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)n
Σk=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)n
Σk=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)n
Σk=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)n
Σk=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)n
Σk=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)n
Σk=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)n
Σk=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 2Hence
is 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 2Hence
is 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 2Hence
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
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 2Hence
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!–
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
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