The document discusses Taylor series approximations. It defines Taylor polynomials and provides the Taylor remainder theorem, which expresses the remainder term using derivatives of the original function. It then gives examples of using Taylor polynomials to approximate the values of exponential and trigonometric functions to a given decimal place accuracy by choosing an appropriate number of terms in the polynomial.

1. Approximations Using Taylor Expansions

2. We have the Taylor's Remainder Theorem:
Approximations Using Taylor Expansions

3. We have the Taylor's Remainder Theorem:
Approximations Using Taylor Expansions
Let f(x) be an infinitely differentiable function over
some open interval that contains [a, b]

4. We have the Taylor's Remainder Theorem:
Approximations Using Taylor Expansions
Let f(x) be an infinitely differentiable function over
some open interval that contains [a, b]
a
( )[ ]
b
f(x) is infinitely differentiable in here

5. We have the Taylor's Remainder Theorem:
Approximations Using Taylor Expansions
Let f(x) be an infinitely differentiable function over
some open interval that contains [a, b] and pn(x) be
the n'th Taylor-poly expanded at a,
a
( )[ ]
b
f(x) is infinitely differentiable in here

6. We have the Taylor's Remainder Theorem:
Approximations Using Taylor Expansions
Let f(x) be an infinitely differentiable function over
some open interval that contains [a, b] and pn(x) be
the n'th Taylor-poly expanded at a,
then there exists a "c" between a and b
a
( )[ ]
bc
f(x) is infinitely differentiable in here

7. We have the Taylor's Remainder Theorem:
Approximations Using Taylor Expansions
Let f(x) be an infinitely differentiable function over
some open interval that contains [a, b] and pn(x) be
the n'th Taylor-poly expanded at a,
then there exists a "c" between a and b such that
f(b) = pn(b) + (b – a)n+1
(n + 1)!
f(n+1)(c)
a
( )[ ]
bc
f(x) is infinitely differentiable in here

8. We have the Taylor's Remainder Theorem:
Approximations Using Taylor Expansions
Let f(x) be an infinitely differentiable function over
some open interval that contains [a, b] and pn(x) be
the n'th Taylor-poly expanded at a,
then there exists a "c" between a and b such that
f(b) = pn(b) + (b – a)n+1
(n + 1)!
f(n+1)(c)
a
( )[ ]
bc
f(x) is infinitely differentiable in here
We use the remainder formula to control the error
when we use Taylor polynomials to approximate f(b).

9. An approximation is said to be accurate to n decimal
places if the error is less than 0.5 x 10-n.
Approximations Using Taylor Expansions

10. An approximation is said to be accurate to n decimal
places if the error is less than 0.5 x 10-n. Therefore,
1 decimal place of accuracy  error < 0.05
Approximations Using Taylor Expansions

11. An approximation is said to be accurate to n decimal
places if the error is less than 0.5 x 10-n. Therefore,
1 decimal place of accuracy  error < 0.05
Approximations Using Taylor Expansions
2 decimal places of accuracy  error < 0.005

12. An approximation is said to be accurate to n decimal
places if the error is less than 0.5 x 10-n. Therefore,
1 decimal place of accuracy  error < 0.05
Approximations Using Taylor Expansions
2 decimal places of accuracy  error < 0.005
3 decimal places of accuracy  error < 0.0005

13. An approximation is said to be accurate to n decimal
places if the error is less than 0.5 x 10-n. Therefore,
1 decimal place of accuracy  error < 0.05
Approximations Using Taylor Expansions
2 decimal places of accuracy  error < 0.005
3 decimal places of accuracy  error < 0.0005
Example A. The Mac-polynomial of ex is
x + 2!1 +
x2
+ .. ++ 3!
x3
n!
xn
pn(x) =
Find an n such that pn(1) approximates e1 with an
accuracy to 4 decimal places.

14. An approximation is said to be accurate to n decimal
places if the error is less than 0.5 x 10-n. Therefore,
1 decimal place of accuracy  error < 0.05
Approximations Using Taylor Expansions
2 decimal places of accuracy  error < 0.005
3 decimal places of accuracy  error < 0.0005
Example A. The Mac-polynomial of ex is
x + 2!1 +
x2
+ .. ++ 3!
x3
n!
xn
pn(x) =
Find an n such that pn(1) approximates e1 with an
accuracy to 4 decimal places.
f(n+1)(c)
(n+1)!
bn+1 isThe error term Rn(b) = ec
(n+1)!
1n+1
for some c between 0 and b = 1.

15. Approximations Using Taylor Expansions
Assuming that we know that e < 3, then the error
ec
(n+1)!Rn(1) = <
3
(n+1)!

16. Approximations Using Taylor Expansions
Assuming that we know that e < 3, then the error
ec
(n+1)!Rn(1) = <
3
(n+1)!
For n large enough such that
we will have pn(1) accurate to 4 decimal places.
0.00005 > 3
(n+1)! > Rn(1)

17. Approximations Using Taylor Expansions
Assuming that we know that e < 3, then the error
ec
(n+1)!Rn(1) = <
3
(n+1)!
For n large enough such that
we will have pn(1) accurate to 4 decimal places.
0.00005 > 3
(n+1)!
By trying different values for n (trial and error),
we find that 0.00005 > 3/9!
> Rn(1)

18. Approximations Using Taylor Expansions
Assuming that we know that e < 3, then the error
ec
(n+1)!Rn(1) = <
3
(n+1)!
For n large enough such that
we will have pn(1) accurate to 4 decimal places.
0.00005 > 3
(n+1)!
By trying different values for n (trial and error),
we find that 0.00005 > 3/9! or that (n+1) = 9,
hence n = 8 is large enough for the requires
accuracy.
> Rn(1)

19. Approximations Using Taylor Expansions
Assuming that we know that e < 3, then the error
ec
(n+1)!Rn(1) = <
3
(n+1)!
For n large enough such that
we will have pn(1) accurate to 4 decimal places.
0.00005 > 3
(n+1)!
By trying different values for n (trial and error),
we find that 0.00005 > 3/9! or that (n+1) = 9,
hence n = 8 is large enough for the requires
accuracy.
Therefore p8(1) = 1 + 1 + 1/2! + 1/3! + .. + 1/8!
 2.7183  e approximates e accurately to
at least 4 places.
> Rn(1)

20. Approximations Using Taylor Expansions
Example B. Approximate cos(3o) to an accuracy of 3
decimal places.

21. Approximations Using Taylor Expansions
Example B. Approximate cos(3o) to an accuracy of 3
decimal places.
Set b = 3o as π/60.

22. Approximations Using Taylor Expansions
Example B. Approximate cos(3o) to an accuracy of 3
decimal places.
1 –pn(x) =
Expanding around 0, the Mac-poly of cos(x) is
+ 4!
x4
6!
x6
8!
x8
+–2!
x2
..
Set b = 3o as π/60.

23. Approximations Using Taylor Expansions
Example B. Approximate cos(3o) to an accuracy of 3
decimal places.
1 –pn(x) =
with the error term Rn( ) =
(n+1)!
Expanding around 0, the Mac-poly of cos(x) is
+ 4!
x4
6!
x6
8!
x8
+–2!
x2
..
Set b = 3o as π/60.
π
60
( )n+1,cos(n+1)(c)π
60
,and at b = π/60,

24. Approximations Using Taylor Expansions
Example B. Approximate cos(3o) to an accuracy of 3
decimal places.
1 –pn(x) =
with the error term Rn( ) =
(n+1)!
Expanding around 0, the Mac-poly of cos(x) is
+ 4!
x4
6!
x6
8!
x8
+–2!
x2
..
Set b = 3o as π/60.
π
60
( )n+1,
where c is some number between 0 and π/60.
cos(n+1)(c)π
60
,and at b = π/60,

25. Approximations Using Taylor Expansions
Example B. Approximate cos(3o) to an accuracy of 3
decimal places.
1 –pn(x) =
with the error term Rn( ) =
(n+1)!
Expanding around 0, the Mac-poly of cos(x) is
+ 4!
x4
6!
x6
8!
x8
+–2!
x2
..
Set b = 3o as π/60.
π
60
( )n+1,
where c is some number between 0 and π/60.
|cosn+1(c)| < 1,
cos(n+1)(c)π
60
,and at b = π/60,

26. Approximations Using Taylor Expansions
Example B. Approximate cos(3o) to an accuracy of 3
decimal places.
1 –pn(x) =
with the error term Rn( ) =
(n+1)!
Expanding around 0, the Mac-poly of cos(x) is
+ 4!
x4
6!
x6
8!
x8
+–2!
x2
..
Set b = 3o as π/60.
π
60
( )n+1,
where c is some number between 0 and π/60.
|cosn+1(c)| < 1, hence |Rn( )| <
cos(n+1)(c)π
60
,and at b = π/60,
π
60 (n+1)!
(π/60)n+1
.

27. Approximations Using Taylor Expansions
Example B. Approximate cos(3o) to an accuracy of 3
decimal places.
1 –pn(x) =
with the error term Rn( ) =
(n+1)!
Expanding around 0, the Mac-poly of cos(x) is
+ 4!
x4
6!
x6
8!
x8
+–2!
x2
..
Set b = 3o as π/60.
π
60
( )n+1,
where c is some number between 0 and π/60.
|cosn+1(c)| < 1, hence |Rn( )| <
cos(n+1)(c)π
60
,and at b = π/60,
π
60 (n+1)!
(π/60)n+1
.
We want 0.0005 > (n+1)!
(π/60)n+1
.

28. Approximations Using Taylor Expansions
Example B. Approximate cos(3o) to an accuracy of 3
decimal places.
1 –pn(x) =
with the error term Rn( ) =
(n+1)!
Expanding around 0, the Mac-poly of cos(x) is
+ 4!
x4
6!
x6
8!
x8
+–2!
x2
..
Set b = 3o as π/60.
π
60
( )n+1,
where c is some number between 0 and π/60.
|cosn+1(c)| < 1, hence |Rn( )| <
cos(n+1)(c)π
60
,and at b = π/60,
π
60 (n+1)!
(π/60)n+1
.
We want 0.0005 > (n+1)!
(π/60)n+1
.
By trial and error n = 2 is sufficient.

29. Approximations Using Taylor Expansions
Example B. Approximate cos(3o) to an accuracy of 3
decimal places.
1 –pn(x) =
with the error term Rn( ) =
(n+1)!
Expanding around 0, the Mac-poly of cos(x) is
+ 4!
x4
6!
x6
8!
x8
+–2!
x2
..
Set b = 3o as π/60.
π
60
( )n+1,
where c is some number between 0 and π/60.
|cosn+1(c)| < 1, hence |Rn( )| <
cos(n+1)(c)π
60
,and at b = π/60,
π
60 (n+1)!
(π/60)n+1
.
We want 0.0005 > (n+1)!
(π/60)n+1
.
By trial and error n = 2 is sufficient. So p2( )π/60
= 1 – (π/60)2/2

30. Approximations Using Taylor Expansions
Example B. Approximate cos(3o) to an accuracy of 3
decimal places.
1 –pn(x) =
with the error term Rn( ) =
(n+1)!
Expanding around 0, the Mac-poly of cos(x) is
+ 4!
x4
6!
x6
8!
x8
+–2!
x2
..
Set b = 3o as π/60.
π
60
( )n+1,
where c is some number between 0 and π/60.
|cosn+1(c)| < 1, hence |Rn( )| <
cos(n+1)(c)π
60
,and at b = π/60,
π
60 (n+1)!
(π/60)n+1
.
We want 0.0005 > (n+1)!
(π/60)n+1
.
By trial and error n = 2 is sufficient. So p2( )π/60
= 1 – (π/60)2/2  0.998629 gives us the desired result.

31. Differentiation and Integration of Power Series

32. Example C. 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
–
Σ
k=0 (2k)!
(–1)kx2k
+ 4!
x4
6!
x6
8!
x8
+1 – – – .. =2!
x2
sin(x) =
cos(x) =

33. Example C. 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
–
Σ
k=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

34. Example C. 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
–
Σ
k=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

35. Example C. 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
–
Σ
k=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
+ ∫ dx5!
x5
–…
The antiderivative of the power series of sin(x) is

36. Example C. 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
–
Σ
k=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
+ ∫ dx5!
x5
–…
The antiderivative of the power series of sin(x) is
4!
x4
6!
x6
= +
2!
x2
– – … + c

37. Example C. 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
–
Σ
k=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
+ ∫ dx5!
x5
–…
The antiderivative of the power series of sin(x) is
4!
x4
6!
x6
= +
2!
x2
= –cos(x) + constant– – … + c

38. Example C. 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
–
Σ
k=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
+ ∫ dx5!
x5
–…
The antiderivative of the power series of sin(x) is
4!
x4
6!
x6
= +
2!
x2
= –cos(x) + constant– – … + c
Both have infinite radius of convergence.

39. Differentiation and Integration of Power Series
Theorem (Derivative and integral of Taylor series) :

40. 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) :

41. 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) for
all x in the interval (a – R, a + R).
Σk=0
∞

42. 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) for
all 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 for
all x in the interval (a – R, a + R).
Σk=0
∞

43. 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) for
all x in the interval (a – R, a + R).
Σk=0
∞
III. ∫ f(x)dx = ∫ ck(x – a)kdx with the series converging
absolutely and (c, d) is any interval in (a – R, a + R).
Σk=0
∞
c
d
c
d
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 for
all x in the interval (a – R, a + R).
Σk=0
∞

44. The function y = ∫ sin(x)/x dx is non–elementary.
Differentiation and Integration of Power Series
x
sin(x)y =
(0,1)

45. The function y = ∫ sin(x)/x dx is non–elementary.
Hence to graph y = ∫ sin(x)/x dx we may estimate its
function value using its Mac-expansion.
Differentiation and Integration of Power Series
x
sin(x)y =
(0,1)

46. The function y = ∫ sin(x)/x dx is non–elementary.
Hence to graph y = ∫ sin(x)/x dx we may estimate its
function value using its Mac-expansion.
Differentiation and Integration of Power Series
Example D. Find the Mac series of ∫ dx. Graph it.x
sin(x)
x
sin(x)y =
(0,1)

47. The function y = ∫ sin(x)/x dx is non–elementary.
Hence to graph y = ∫ sin(x)/x dx we may estimate its
function value using its Mac-expansion.
Differentiation and Integration of Power Series
Example D. Find the Mac series of ∫ dx. Graph it.x
sin(x)
The Mac series of isx
sin(x)
x –
3!
x3
+ 5!
x5
+ ..7!
x7
–( ) / x
x
sin(x)y =
(0,1)

48. The function y = ∫ sin(x)/x dx is non–elementary.
Hence to graph y = ∫ sin(x)/x dx we may estimate its
function value using its Mac-expansion.
Differentiation and Integration of Power Series
Example D. Find the Mac series of ∫ dx. Graph it.x
sin(x)
The Mac series of isx
sin(x)
x –
3!
x3
+ 5!
x5
+ ..7!
x7
–( ) / x
= 1 –
3!
x2
+ 5!
x4
+ ..7!
x6
– Σk=0
=
x
sin(x)y =
(0,1)
(2k+1)!
(–1)kx2k

49. The function y = ∫ sin(x)/x dx is non–elementary.
Hence to graph y = ∫ sin(x)/x dx we may estimate its
function value using its Mac-expansion.
Differentiation and Integration of Power Series
Example D. Find the Mac series of ∫ dx. Graph it.x
sin(x)
The Mac series of isx
sin(x)
x –
3!
x3
+ 5!
x5
+ ..7!
x7
–( ) / x
= 1 –
3!
x2
+ 5!
x4
+ ..7!
x6
– Σk=0
=
So the Mac series of ∫ dx =x
sin(x) Σ ∫k=0 (2k+1)!
(–1)kx2k
dx
x
sin(x)y =
(0,1)
(2k+1)!
(–1)kx2k

50. The function y = ∫ sin(x)/x dx is non–elementary.
Hence to graph y = ∫ sin(x)/x dx we may estimate its
function value using its Mac-expansion.
Differentiation and Integration of Power Series
Example D. Find the Mac series of ∫ dx. Graph it.x
sin(x)
The Mac series of isx
sin(x)
x –
3!
x3
+ 5!
x5
+ ..7!
x7
–( ) / x
= 1 –
3!
x2
+ 5!
x4
+ ..7!
x6
– Σk=0
=
So the Mac series of ∫ dx =x
sin(x) Σ ∫k=0 (2k+1)!
(–1)kx2k
dx
= Σk=0 (2k+1)(2k+1)!
(–1)kx2k+1
x
sin(x)y =
(0,1)
(2k+1)!
(–1)kx2k

51. The function y = ∫ sin(x)/x dx is non–elementary.
Hence to graph y = ∫ sin(x)/x dx we may estimate its
function value using its Mac-expansion.
Differentiation and Integration of Power Series
Example D. Find the Mac series of ∫ dx. Graph it.x
sin(x)
The Mac series of isx
sin(x)
x –
3!
x3
+ 5!
x5
+ ..7!
x7
–( ) / x
= 1 –
3!
x2
+ 5!
x4
+ ..7!
x6
– Σk=0
=
So the Mac series of ∫ dx =x
sin(x) Σ ∫k=0 (2k+1)!
(–1)kx2k
dx
= Σk=0 (2k+1)(2k+1)!
(–1)kx2k+1
= x –
3*3!
x3
+ –
5*5!
x5
7*7!
x7
x
sin(x)y =
(0,1)
sin(x)
(2k+1)!
(–1)kx2k
+ ..

52. The function y = ∫ sin(x)/x dx is non–elementary.
Hence to graph y = ∫ sin(x)/x dx we may estimate its
function value using its Mac-expansion.
Differentiation and Integration of Power Series
Example D. Find the Mac series of ∫ dx. Graph it.x
sin(x)
The Mac series of isx
sin(x)
x –
3!
x3
+ 5!
x5
+ ..7!
x7
–( ) / x
= 1 –
3!
x2
+ 5!
x4
+ ..7!
x6
– Σk=0
=
So the Mac series of ∫ dx =x
sin(x) Σ ∫k=0 (2k+1)!
(–1)kx2k
dx
= Σk=0 (2k+1)(2k+1)!
(–1)kx2k+1
= x –
3*3!
x3
+ –
5*5!
x5
7*7!
x7
x
sin(x)y =
(0,1)
x
sin(x)y = ∫ dx
(2k+1)!
(–1)kx2k
+ ..
Here is its graph.

53. Differentiation and Integration of Power Series
Example E. Find ∫ e–x dx accurate to 3 decimal places.x=0
1
2

54. Differentiation and Integration of Power Series
Example E. Find ∫ e–x dx accurate to 3 decimal places.x=0
1
∫ e–x dx is not an elementary function.
2
2

55. Differentiation and Integration of Power Series
Example E. Find ∫ e–x dx accurate to 3 decimal places.x=0
1
∫ e–x dx is not an elementary function. We have to
use Mac-expansion to obtain the answer we want.
2
2

56. Differentiation and Integration of Power Series
Example E. 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 Mac-expansion to obtain the answer we want.
2
2
Σk=0
e–x = n!
(–x2)n
Σk=0
2

57. Differentiation and Integration of Power Series
Example E. 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 Mac-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 – – – ..

58. Differentiation and Integration of Power Series
Example E. 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 Mac-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
– – – ..

59. Differentiation and Integration of Power Series
Example E. 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 Mac-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
– – – ..

60. Differentiation and Integration of Power Series
Example E. 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 Mac-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
3
x3
+
5*2!
x5
+ ..7*3!
x7
= P(x)
– – – ..
– –

61. Differentiation and Integration of Power Series
Example E. 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 Mac-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
3
x3
+
5*2!
x5
+ ..7*3!
x7
= P(x)
∫ e–x dx = P(1) – P(0)x=0
1
2
– – – ..
– –
Therefore

62. Differentiation and Integration of Power Series
Example E. 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 Mac-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
3
x3
+
5*2!
x5
+ ..7*3!
x7
= P(x)
∫ e–x dx = P(1) – P(0)x=0
1
= 1 3
1
+ 5*2!
1
+ ..
7*3!
1
2
– – – ..
– –
– –
Therefore
which is a "decreasing" alternating series.

63. Differentiation and Integration of Power Series
∫ e–x dxx=0
1
= 1 3
1
+ 5*2!
1
+7*3!
12
– –Hence
is a convergent alternating series.
1
9*4!
–
1
11*5!
+
13*6! …
1

64. Differentiation and Integration of Power Series
∫ e–x dxx=0
1
= 1 3
1
+ 5*2!
1
+7*3!
12
– –Hence
is a convergent alternating series.
Furthermore the sum of a "decreasing" alternating
series S satisfies that |S| < |a1=1st term|.
1
9*4!
–
1
11*5!
+
13*6! …
1

65. Differentiation and Integration of Power Series
∫ e–x dxx=0
1
= 1 3
1
+ 5*2!
1
+7*3!
12
– –Hence
is a convergent alternating series.
Furthermore the sum of a "decreasing" alternating
series S satisfies that |S| < |a1=1st term|.
To estimate the sum to the 3rd decimal place means
the error must be < 0.0005.
1
9*4!
–
1
11*5!
+
13*6! …
1

66. Differentiation and Integration of Power Series
∫ e–x dxx=0
1
= 1 3
1
+ 5*2!
1
+7*3!
12
– –Hence
is a convergent alternating series.
Furthermore the sum of a "decreasing" alternating
series S satisfies that |S| < |a1=1st term|.
To estimate the sum to the 3rd decimal place means
the error must be < 0.0005.
By trial and error, we find that < 0.0005.13*6!
1
1
9*4!
–
1
11*5!
+
13*6! …
1

67. Differentiation and Integration of Power Series
∫ e–x dxx=0
1
= 1 3
1
+ 5*2!
1
+7*3!
12
– –Hence
is a convergent alternating series.
Furthermore the sum of a "decreasing" alternating
series S satisfies that |S| < |a1=1st term|.
To estimate the sum to the 3rd decimal place means
the error must be < 0.0005.
By trial and error, we find that < 0.0005.13*6!
1
1
9*4!
–
1
11*5!
+
13*6! …
1
So by the tail-series theorem,
1
– .. < 0.0005+
13*6!
1
15*7!
1
17*8!
–

68. Differentiation and Integration of Power Series
∫ e–x dxx=0
1
= 1 3
1
+ 5*2!
1
+7*3!
12
– –Hence
is a convergent alternating series.
Furthermore the sum of a "decreasing" alternating
series S satisfies that |S| < |a1=1st term|.
To estimate the sum to the 3rd decimal place means
the error must be < 0.0005.
By trial and error, we find that < 0.0005.13*6!
1
1
9*4!
–
1
11*5!
+
13*6! …
1
So by the tail-series theorem,
1
– .. < 0.0005+
13*6!
1
15*7!
1
17*8!
–
This means the sum of the front terms
1 3
1
+ 5*2!
1
+7*3!
1
– –
1
9*4!
–
1
11*5!
 0.74673
is accurate to 3 decimal places.