MacLaurin Expansions

1. Maclaurin Expansions
Given an input x, a polynomial P(x) is an algebraic
formula in the sense that only arithmetic operations +,
–, *, / are needed to obtain the output P(x).

2. Maclaurin Expansions
Given an input x, a polynomial P(x) is an algebraic
formula in the sense that only arithmetic operations +,
–, *, / are needed to obtain the output P(x).
This is not the case for sin(x) or In(x).

3. Maclaurin Expansions
Given an input x, a polynomial P(x) is an algebraic
formula in the sense that only arithmetic operations +,
–, *, / are needed to obtain the output P(x).
This is not the case for sin(x) or In(x). For example,
given x = 2, there is no obvious way to calculate In(2).

4. Maclaurin Expansions
Given an input x, a polynomial P(x) is an algebraic
formula in the sense that only arithmetic operations +,
–, *, / are needed to obtain the output P(x).
This is not the case for sin(x) or In(x). For example,
given x = 2, there is no obvious way to calculate In(2).
But we may produce polynomials, i.e. sequences of
arithmetic steps based on the input x to estimate In(x).

5. Maclaurin Expansions
Given an input x, a polynomial P(x) is an algebraic
formula in the sense that only arithmetic operations +,
–, *, / are needed to obtain the output P(x).
This is not the case for sin(x) or In(x). For example,
given x = 2, there is no obvious way to calculate In(2).
But we may produce polynomials, i.e. sequences of
arithmetic steps based on the input x to estimate In(x).
These polynomials are called the Taylor polynomials,
or Maclaurin polynomials (if centered at 0) of In(x).

6. Maclaurin Expansions
Given an input x, a polynomial P(x) is an algebraic
formula in the sense that only arithmetic operations +,
–, *, / are needed to obtain the output P(x).
This is not the case for sin(x) or In(x). For example,
given x = 2, there is no obvious way to calculate In(2).
But we may produce polynomials, i.e. sequences of
arithmetic steps based on the input x to estimate In(x).
These polynomials are called the Taylor polynomials,
or Maclaurin polynomials (if centered at 0) of In(x).
The construction of the Maclaurin polynomials
is based on the observation that
the lower degree terms of a polynomial
P(x) = a0 + a1x + a2x2 + a3x3 + . . anxn
give the best approximations of P(x) around x = 0.

7. Maclaurin Expansions
Let P(x) = (x + 1)x(x – 1)(x – 2) = 2x – x2 – 2x3 + x4.

8. Maclaurin Expansions
Let P(x) = (x + 1)x(x – 1)(x – 2) = 2x – x2 – 2x3 + x4.
Following are the graphs of P(x) against the lower
degree terms of P(x).

9. Maclaurin Expansions
Let P(x) = (x + 1)x(x – 1)(x – 2) = 2x – x2 – 2x3 + x4.
Following are the graphs of P(x) against the lower
degree terms of P(x).
y=2x–x2–2x3+x4
y=2x

10. Maclaurin Expansions
Let P(x) = (x + 1)x(x – 1)(x – 2) = 2x – x2 – 2x3 + x4.
Following are the graphs of P(x) against the lower
degree terms of P(x).
y=2x–x2–2x3+x4y=2x–x2–2x3+x4
y=2x y=2x–x2

11. Maclaurin Expansions
y=2x–x2–2x3+x4
Let P(x) = (x + 1)x(x – 1)(x – 2) = 2x – x2 – 2x3 + x4.
Following are the graphs of P(x) against the lower
degree terms of P(x).
y=2x–x2–2x3+x4y=2x–x2–2x3+x4
y=2x y=2x–x2
y=2x–x2–2x3

12. Maclaurin Expansions
y=2x–x2–2x3+x4
Let P(x) = (x + 1)x(x – 1)(x – 2) = 2x – x2 – 2x3 + x4.
Following are the graphs of P(x) against the lower
degree terms of P(x).
y=2x–x2–2x3+x4y=2x–x2–2x3+x4
y=2x y=2x–x2
y=2x–x2–2x3
The closer and closer approximation of P(x) by its
lower degree terms comes as no surprise because
the higher degree terms are more negligible than the
lower degree terms for x’s that are near 0.

13. Maclaurin Expansions
y=2x–x2–2x3+x4
Let P(x) = (x + 1)x(x – 1)(x – 2) = 2x – x2 – 2x3 + x4.
Following are the graphs of P(x) against the lower
degree terms of P(x).
y=2x–x2–2x3+x4y=2x–x2–2x3+x4
y=2x y=2x–x2
y=2x–x2–2x3
The closer and closer approximation of P(x) by its
lower degree terms comes as no surprise because
the higher degree terms are more negligible than the
lower degree terms for x’s that are near 0.
On the other hand, terms of a polynomial P(x) may be
calculated using the derivatives of P(x).

14. Maclaurin Expansions
y=2x–x2–2x3+x4
Let P(x) = (x + 1)x(x – 1)(x – 2) = 2x – x2 – 2x3 + x4.
Following are the graphs of P(x) against the lower
degree terms of P(x).
y=2x–x2–2x3+x4y=2x–x2–2x3+x4
y=2x y=2x–x2
y=2x–x2–2x3
The closer and closer approximation of P(x) by its
lower degree terms comes as no surprise because
the higher degree terms are more negligible than the
lower degree terms for x’s that are near 0.
On the other hand, terms of a polynomial P(x) may be
calculated using the derivatives of P(x).
This calculation leads us to the Maclaurin polynomials
of differentiable functions in general.

15. Given a function f(x) that is infinitely differentiable at
x = 0, we define the polynomial
=pn(x)
Maclaurin Expansions

16. Given a function f(x) that is infinitely differentiable at
x = 0, we define the polynomial
= f(0)pn(x)
f(0)(0)
0!
Maclaurin Expansions

17. Given a function f(x) that is infinitely differentiable at
x = 0, we define the polynomial
f '(0)x= f(0)+pn(x)
f(0)(0)
0!
f(1)(0)
1!
Maclaurin Expansions

18. Given a function f(x) that is infinitely differentiable at
x = 0, we define the polynomial
f '(0)x
f(2)(0)
+
2!
= f(0)+ x2pn(x)
f(0)(0)
0!
f(1)(0)
1!
Maclaurin Expansions

19. Given a function f(x) that is infinitely differentiable at
x = 0, we define the polynomial
f '(0)x
f(2)(0)
+
2!
= f(0)+ x2pn(x)
f(0)(0)
0!
f(3)(0)
+
3! x3..
f(1)(0)
1!
Maclaurin Expansions

20. Given a function f(x) that is infinitely differentiable at
x = 0, we define the polynomial
f '(0)x
f(2)(0)
+
2!
= f(0)+ x2pn(x)
f(0)(0)
0!
f(3)(0)
+
3! x3..
f(n)(0)
+
n! xn
f(1)(0)
1!
Maclaurin Expansions

21. Given a function f(x) that is infinitely differentiable at
x = 0, we define the polynomial
f '(0)x
f(2)(0)
+
2!
= f(0)+ x2pn(x)
f(0)(0)
0!
f(3)(0)
+
3! x3..
f(n)(0)
+
n! xn
f(1)(0)
1!
or pn(x) = Σk=0
n
xk
k!
f(k)(0)
Maclaurin Expansions

22. Given a function f(x) that is infinitely differentiable at
x = 0, we define the polynomial
f '(0)x
f(2)(0)
+
2!
= f(0)+ x2pn(x)
f(0)(0)
0!
f(3)(0)
+
3! x3..
f(n)(0)
+
n! xn
f(1)(0)
1!
or pn(x) = Σk=0
n
xk
k!
f(k)(0)
This is called the n'th (degree) Maclaurin polynomial
(Mac-poly) of f(x).
Maclaurin Expansions

23. Given a function f(x) that is infinitely differentiable at
x = 0, we define the polynomial
f '(0)x
f(2)(0)
+
2!
= f(0)+ x2pn(x)
f(0)(0)
0!
f(3)(0)
+
3! x3..
f(n)(0)
+
n! xn
f(1)(0)
1!
or pn(x) = Σk=0
n
xk
k!
f(k)(0)
This is called the n'th (degree) Maclaurin polynomial
(Mac-poly) of f(x).
If n = ∞, we have the Maclaurin series (Mac-series):
P(x) =Σk=0
xk.k!
f(k)(0)∞
Maclaurin Expansions
They are referred to as the Mac-expansions of f(x).

24. Maclaurin Expansions
The derivatives of order 0, 1, 2, . ., n, of pn(x) at x = 0
are the same as the derivatives of f(x).

25. = f(0)pn(0)
In other words,
Maclaurin Expansions
The derivatives of order 0, 1, 2, . ., n, of pn(x) at x = 0
are the same as the derivatives of f(x).

26. f (0)
= f(0)
+ #x + #x2 + ..#xn-1|x=0
pn(0)
In other words,
=pn(0)
(1) (1)
Maclaurin Expansions
The derivatives of order 0, 1, 2, . ., n, of pn(x) at x = 0
are the same as the derivatives of f(x).

27. f (0)
= f(0)
+ #x + #x2 + ..#xn-1|x=0 = f (0)
pn(0)
In other words,
=pn(0)
(1) (1) (1)
Maclaurin Expansions
The derivatives of order 0, 1, 2, . ., n, of pn(x) at x = 0
are the same as the derivatives of f(x).

28. f (0)
= f(0)
+ #x + #x2 + ..#xn-1|x=0 = f (0)
pn(0)
In other words,
=pn(0)
(1)
f (0)+ #x + #x2 + ..#xn-2|x=0=pn(0)
(2) (2)
(1) (1)
Maclaurin Expansions
The derivatives of order 0, 1, 2, . ., n, of pn(x) at x = 0
are the same as the derivatives of f(x).

29. f (0)
= f(0)
+ #x + #x2 + ..#xn-1|x=0 = f (0)
pn(0)
In other words,
=pn(0)
(1)
f (0)+ #x + #x2 + ..#xn-2|x=0 = f (0)=pn(0)
(2) (2) (2)
(1) (1)
Maclaurin Expansions
The derivatives of order 0, 1, 2, . ., n, of pn(x) at x = 0
are the same as the derivatives of f(x).

30. f (0)
= f(0)
+ #x + #x2 + ..#xn-1|x=0 = f (0)
pn(0)
In other words,
=pn(0)
(1)
f (0)+ #x + #x2 + ..#xn-2|x=0 = f (0)=pn(0)
(2) (2) (2)
(1) (1)
and so on, up to pn(0) = f (0).(n) (n)
Maclaurin Expansions
The derivatives of order 0, 1, 2, . ., n, of pn(x) at x = 0
are the same as the derivatives of f(x).

31. f (0)
= f(0)
+ #x + #x2 + ..#xn-1|x=0 = f (0)
pn(0)
In other words,
=pn(0)
(1)
f (0)+ #x + #x2 + ..#xn-2|x=0 = f (0)=pn(0)
(2) (2) (2)
(1) (1)
and so on, up to pn(0) = f (0).(n) (n)
Maclaurin Expansions
The derivatives of order 0, 1, 2, . ., n, of pn(x) at x = 0
are the same as the derivatives of f(x).
In fact, pn(x) is the only polynomial with degree ≤ n
whose derivatives of order = 0, 1, 2,. . n, at x = 0
are the same as those of f(x).

32. f (0)
= f(0)
+ #x + #x2 + ..#xn-1|x=0 = f (0)
pn(0)
In other words,
=pn(0)
(1)
f (0)+ #x + #x2 + ..#xn-2|x=0 = f (0)=pn(0)
(2) (2) (2)
(1) (1)
and so on, up to pn(0) = f (0).(n) (n)
In fact, pn(x) is the only polynomial with degree ≤ n
whose derivatives of order = 0, 1, 2,. . n, at x = 0
are the same as those of f(x).
Likewise the Mac-series is the only power series that
has all its derivatives agree with those of f(x) at x = 0.
Maclaurin Expansions
The derivatives of order 0, 1, 2, . ., n, of pn(x) at x = 0
are the same as the derivatives of f(x).

33. The following are some examples of the Mac-polys
and Mac-series of basic functions calculated via the
definition.
Maclaurin Expansions

34. The following are some examples of the Mac-polys
and Mac-series of basic functions calculated via the
definition. Later, we will use these expansions to
calculate the expansions of other functions (instead of
using the definition).
Maclaurin Expansions

35. The following are some examples of the Mac-polys
and Mac-series of basic functions calculated via the
definition. Later, we will use these expansions to
calculate the expansions of other functions (instead of
using the definition).
Example A. Find the Mac-expansions of
f(x) = 1 + x + x2 + x3 + x4 around x = 0.
Maclaurin Expansions

36. The following are some examples of the Mac-polys
and Mac-series of basic functions calculated via the
definition. Later, we will use these expansions to
calculate the expansions of other functions (instead of
using the definition).
Example A. Find the Mac-expansions of
f(x) = 1 + x + x2 + x3 + x4 around x = 0.
We need the derivatives of f(x):
Maclaurin Expansions

37. The following are some examples of the Mac-polys
and Mac-series of basic functions calculated via the
definition. Later, we will use these expansions to
calculate the expansions of other functions (instead of
using the definition).
Example A. Find the Mac-expansions of
f(x) = 1 + x + x2 + x3 + x4 around x = 0.
We need the derivatives of f(x):
f (x) = 1 + 2x + 3x2 + 4x3(1)
Maclaurin Expansions

38. The following are some examples of the Mac-polys
and Mac-series of basic functions calculated via the
definition. Later, we will use these expansions to
calculate the expansions of other functions (instead of
using the definition).
Example A. Find the Mac-expansions of
f(x) = 1 + x + x2 + x3 + x4 around x = 0.
We need the derivatives of f(x):
f (x) = 1 + 2x + 3x2 + 4x3  f (0) = 1.(1) (1)
Maclaurin Expansions

39. The following are some examples of the Mac-polys
and Mac-series of basic functions calculated via the
definition. Later, we will use these expansions to
calculate the expansions of other functions (instead of
using the definition).
Example A. Find the Mac-expansions of
f(x) = 1 + x + x2 + x3 + x4 around x = 0.
We need the derivatives of f(x):
f (x) = 1 + 2x + 3x2 + 4x3  f (0) = 1.(1) (1)
f (x) = 2 + 3*2x + 4*3x2(2)
Maclaurin Expansions

40. The following are some examples of the Mac-polys
and Mac-series of basic functions calculated via the
definition. Later, we will use these expansions to
calculate the expansions of other functions (instead of
using the definition).
Example A. Find the Mac-expansions of
f(x) = 1 + x + x2 + x3 + x4 around x = 0.
We need the derivatives of f(x):
f (x) = 1 + 2x + 3x2 + 4x3  f (0) = 1.(1) (1)
f (x) = 2 + 3*2x + 4*3x2  f (0) = 2!
(2) (2)
Maclaurin Expansions

41. The following are some examples of the Mac-polys
and Mac-series of basic functions calculated via the
definition. Later, we will use these expansions to
calculate the expansions of other functions (instead of
using the definition).
Example A. Find the Mac-expansions of
f(x) = 1 + x + x2 + x3 + x4 around x = 0.
We need the derivatives of f(x):
f (x) = 1 + 2x + 3x2 + 4x3  f (0) = 1.(1) (1)
f (x) = 2 + 3*2x + 4*3x2  f (0) = 2!
(2) (2)
f (x) = 3*2 + 4*3*2x
(3)
Maclaurin Expansions

42. The following are some examples of the Mac-polys
and Mac-series of basic functions calculated via the
definition. Later, we will use these expansions to
calculate the expansions of other functions (instead of
using the definition).
Example A. Find the Mac-expansions of
f(x) = 1 + x + x2 + x3 + x4 around x = 0.
We need the derivatives of f(x):
f (x) = 1 + 2x + 3x2 + 4x3  f (0) = 1.(1) (1)
f (x) = 2 + 3*2x + 4*3x2  f (0) = 2!
(2) (2)
f (x) = 3*2 + 4*3*2x  f (0) = 3!
(3) (3)
Maclaurin Expansions

43. The following are some examples of the Mac-polys
and Mac-series of basic functions calculated via the
definition. Later, we will use these expansions to
calculate the expansions of other functions (instead of
using the definition).
Example A. Find the Mac-expansions of
f(x) = 1 + x + x2 + x3 + x4 around x = 0.
We need the derivatives of f(x):
f (x) = 1 + 2x + 3x2 + 4x3  f (0) = 1.(1) (1)
f (x) = 2 + 3*2x + 4*3x2  f (0) = 2!
(2) (2)
f (x) = 3*2 + 4*3*2x  f (0) = 3!
(3) (3)
f (x) = 4*3*2
(4)
Maclaurin Expansions

44. The following are some examples of the Mac-polys
and Mac-series of basic functions calculated via the
definition. Later, we will use these expansions to
calculate the expansions of other functions (instead of
using the definition).
Example A. Find the Mac-expansions of
f(x) = 1 + x + x2 + x3 + x4 around x = 0.
We need the derivatives of f(x):
f (x) = 1 + 2x + 3x2 + 4x3  f (0) = 1.(1) (1)
f (x) = 2 + 3*2x + 4*3x2  f (0) = 2!
(2) (2)
f (x) = 3*2 + 4*3*2x  f (0) = 3!
(3) (3)
f (x) = 4*3*2  f (0) = 4!
(4) (4)
Maclaurin Expansions

45. The following are some examples of the Mac-polys
and Mac-series of basic functions calculated via the
definition. Later, we will use these expansions to
calculate the expansions of other functions (instead of
using the definition).
Example A. Find the Mac-expansions of
f(x) = 1 + x + x2 + x3 + x4 around x = 0.
We need the derivatives of f(x):
f (x) = 1 + 2x + 3x2 + 4x3  f (0) = 1.(1) (1)
f (x) = 2 + 3*2x + 4*3x2  f (0) = 2!
(2) (2)
f (x) = 3*2 + 4*3*2x  f (0) = 3!
(3) (3)
f (x) = 4*3*2  f (0) = 4!
(4) (4)
f (x) = 0 for n > 5
(n)
Maclaurin Expansions

46. Hence p0(x) = f(0) = 1
Maclaurin Expansions

47. Hence p0(x) = f(0) = 1
f '(0)x= f(0)+p1(x)
Maclaurin Expansions

48. Hence p0(x) = f(0) = 1
f '(0)x = 1 + 1x= f(0)+p1(x)
Maclaurin Expansions

49. Hence p0(x) = f(0) = 1
f '(0)x = 1 + 1x= f(0)+p1(x)
f(2)(0)
+
2! x2f '(0)x= f(0)+p2(x)
Maclaurin Expansions

50. Hence p0(x) = f(0) = 1
f '(0)x = 1 + 1x= f(0)+p1(x)
f(2)(0)
+
2! x2f '(0)x= f(0)+p2(x) = 1 + 1x + 2!
2!
x2
Maclaurin Expansions

51. Hence p0(x) = f(0) = 1
f '(0)x = 1 + 1x= f(0)+p1(x)
f(2)(0)
+
2! x2f '(0)x= f(0)+p2(x) = 1 + 1x + 2!
2!
x2
= 1 + x + x2
Maclaurin Expansions

52. Hence p0(x) = f(0) = 1
f '(0)x = 1 + 1x= f(0)+p1(x)
f(2)(0)
+
2! x2f '(0)x= f(0)+p2(x) = 1 + 1x + 2!
2!
x2
= 1 + x + x2
f(2)(0)
+
2! x2 f(3)(0)
+
3! x3f '(0)x= f(0)+p3(x)
Maclaurin Expansions

53. Hence p0(x) = f(0) = 1
f '(0)x = 1 + 1x= f(0)+p1(x)
f(2)(0)
+
2! x2f '(0)x= f(0)+p2(x) = 1 + 1x + 2!
2!
x2
= 1 + x + x2
f(2)(0)
+
2! x2 f(3)(0)
+
3! x3f '(0)x= f(0)+p3(x)
+ x2
+ x31x= 1 + 2!
2!
3!
3!
Maclaurin Expansions

54. Hence p0(x) = f(0) = 1
f '(0)x = 1 + 1x= f(0)+p1(x)
f(2)(0)
+
2! x2f '(0)x= f(0)+p2(x) = 1 + 1x + 2!
2!
x2
= 1 + x + x2
f(2)(0)
+
2! x2 f(3)(0)
+
3! x3f '(0)x= f(0)+p3(x)
+ x2
+ x31x= 1 + 2!
2!
3!
3!
= 1 + x + x2 + x3
Maclaurin Expansions

55. Hence p0(x) = f(0) = 1
f '(0)x = 1 + 1x= f(0)+p1(x)
=p4(x)
f(2)(0)
+
2! x2f '(0)x= f(0)+p2(x) = 1 + 1x + 2!
2!
x2
= 1 + x + x2
f(2)(0)
+
2! x2 f(3)(0)
+
3! x3f '(0)x= f(0)+p3(x)
+ x2
+ x31x= 1 + 2!
2!
3!
3!
= 1 + x + x2 + x3
+ x2
+ x31x1+ 2!
2!
3!
3!
+ x44!
4!
Maclaurin Expansions

56. Hence p0(x) = f(0) = 1
f '(0)x = 1 + 1x= f(0)+p1(x)
=p4(x)
f(2)(0)
+
2! x2f '(0)x= f(0)+p2(x) = 1 + 1x + 2!
2!
x2
= 1 + x + x2
f(2)(0)
+
2! x2 f(3)(0)
+
3! x3f '(0)x= f(0)+p3(x)
+ x2
+ x31x= 1 + 2!
2!
3!
3!
= 1 + x + x2 + x3
+ x2
+ x31x1+ 2!
2!
3!
3!
+ x44!
4!
= 1 + x + x2 + x3 + x4
Maclaurin Expansions

57. Hence p0(x) = f(0) = 1
f '(0)x = 1 + 1x= f(0)+p1(x)
=p4(x)
f(2)(0)
+
2! x2f '(0)x= f(0)+p2(x) = 1 + 1x + 2!
2!
x2
= 1 + x + x2
f(2)(0)
+
2! x2 f(3)(0)
+
3! x3f '(0)x= f(0)+p3(x)
+ x2
+ x31x= 1 + 2!
2!
3!
3!
= 1 + x + x2 + x3
+ x2
+ x31x1+ 2!
2!
3!
3!
+ x44!
4!
= 1 + x + x2 + x3 + x4
For n > 5, pn(x) = 1 + x + x2 + x3 + x4 = f(x)
Maclaurin Expansions

58. In general, if P(x) = a0 + a1x + a2x2 + .. +akxk
is a polynomial of degree k,
Maclaurin Expansions

59. In general, if P(x) = a0 + a1x + a2x2 + .. +akxk
is a polynomial of degree k, then
p0(x) = a0
Maclaurin Expansions

60. In general, if P(x) = a0 + a1x + a2x2 + .. +akxk
is a polynomial of degree k, then
p0(x) = a0
p1(x) = a0 + a1x
Maclaurin Expansions

61. In general, if P(x) = a0 + a1x + a2x2 + .. +akxk
is a polynomial of degree k, then
p0(x) = a0
p1(x) = a0 + a1x
p2(x) = a0 + a1x + a2x2
Maclaurin Expansions

62. In general, if P(x) = a0 + a1x + a2x2 + .. +akxk
is a polynomial of degree k, then
p0(x) = a0
p1(x) = a0 + a1x
p2(x) = a0 + a1x + a2x2
.
.
Maclaurin Expansions

63. In general, if P(x) = a0 + a1x + a2x2 + .. +akxk
is a polynomial of degree k, then
p0(x) = a0
p1(x) = a0 + a1x
p2(x) = a0 + a1x + a2x2
.
.
pk(x) = a0 + a1x + a2x2.. + akxk = P(x)
Maclaurin Expansions

64. In general, if P(x) = a0 + a1x + a2x2 + .. +akxk
is a polynomial of degree k, then
p0(x) = a0
p1(x) = a0 + a1x
p2(x) = a0 + a1x + a2x2
.
.
pk(x) = a0 + a1x + a2x2.. + akxk = P(x)
and for all n > k, pn(x) = a0 + a1x + a2x2.. + akxk = P(x).
Maclaurin Expansions

65. In general, if P(x) = a0 + a1x + a2x2 + .. +akxk
is a polynomial of degree k, then
Fact. The Mac-polynomials of degree k or larger
of a polynomial P(x) of degree k, is P(x) itself.
p0(x) = a0
p1(x) = a0 + a1x
p2(x) = a0 + a1x + a2x2
.
.
pk(x) = a0 + a1x + a2x2.. + akxk = P(x)
and for all n > k, pn(x) = a0 + a1x + a2x2.. + akxk = P(x).
Maclaurin Expansions

66. In general, if P(x) = a0 + a1x + a2x2 + .. +akxk
is a polynomial of degree k, then
Fact. The Mac-polynomials of degree k or larger
of a polynomial P(x) of degree k, is P(x) itself.
For an infinitely differentiable function such as
f(x) = ex, we can compute its Mac-expansions in the
same manner.
p0(x) = a0
p1(x) = a0 + a1x
p2(x) = a0 + a1x + a2x2
.
.
pk(x) = a0 + a1x + a2x2.. + akxk = P(x)
and for all n > k, pn(x) = a0 + a1x + a2x2.. + akxk = P(x).
Maclaurin Expansions

67. Maclaurin Expansions
Example B. Find the Mac-expansions of f(x) = ex
around x = 0.

68. Maclaurin Expansions
Example B. Find the Mac-expansions of f(x) = ex
around x = 0.
We need the derivatives of f(x):
f (x) = ex  f (0) = 1 for all n.(n) (n)

69. f(1)(0)x
f(2)(0)
+
2!
= f(0)+ x2pn(x) f(3)(0)
+
3! x3..
f(n)(0)
+
n! xn
Therefore the n'th Mac-polynomial of ex is
Maclaurin Expansions
Example B. Find the Mac-expansions of f(x) = ex
around x = 0.
We need the derivatives of f(x):
f (x) = ex  f (0) = 1 for all n.(n) (n)

70. f(1)(0)x
f(2)(0)
+
2!
= f(0)+ x2pn(x) f(3)(0)
+
3! x3..
f(n)(0)
+
n! xn
Therefore the n'th Mac-polynomial of ex is
x +
2!
= 1 +
x2
+ .. ++
3!
x3
n!
xn
Maclaurin Expansions
Example B. Find the Mac-expansions of f(x) = ex
around x = 0.
We need the derivatives of f(x):
f (x) = ex  f (0) = 1 for all n.(n) (n)

71. f(1)(0)x
f(2)(0)
+
2!
= f(0)+ x2pn(x) f(3)(0)
+
3! x3..
f(n)(0)
+
n! xn
P(x) = Σk=0 k! .
xk
Therefore the n'th Mac-polynomial of ex is
x +
2!
= 1 +
x2
+ .. ++
3!
x3
n!
xn
The Mac-series of ex is
∞
x +
2!
1 +
x2
+ .. ++
3!
x3
n! ..
xn
=
Maclaurin Expansions
Example B. Find the Mac-expansions of f(x) = ex
around x = 0.
We need the derivatives of f(x):
f (x) = ex  f (0) = 1 for all n.(n) (n)

72. f(1)(0)x
f(2)(0)
+
2!
= f(0)+ x2pn(x) f(3)(0)
+
3! x3..
f(n)(0)
+
n! xn
P(x) = Σk=0 k! .
xk
Therefore the n'th Mac-polynomial of ex is
x +
2!
= 1 +
x2
+ .. ++
3!
x3
n!
xn
The Mac-series of ex is
∞
x +
2!
1 +
x2
+ .. ++
3!
x3
n! ..
xn
=
Maclaurin Expansions
Example B. Find the Mac-expansions of f(x) = ex
around x = 0.
We need the derivatives of f(x):
f (x) = ex  f (0) = 1 for all n.(n) (n)
Here are some of the graphic comparisons of ex
to its Mac-polys.

73. y = ex
y=x+1
The graphs of Mac-polys for y = ex
Maclaurin Expansions

74. y = ex
y=x+1
y=x2/2+x+1
Maclaurin Expansions
The graphs of Mac-polys for y = ex

75. y = ex
y=x+1
y=x2/2+x+1
y=x3/6+x2/2+x+1
Maclaurin Expansions
The graphs of Mac-polys for y = ex

76. Example C.
Find the Mac-expansions of f(x) = sin(x).
Maclaurin Expansions

77. Example C.
Find the Mac-expansions of f(x) = sin(x).
To find the derivatives of all orders of sin(x)
we arrange them in a circle.
Maclaurin Expansions
sin(x)
cos(x)
–sin(x)
–cos(x)

78. Example C.
Find the Mac-expansions of f(x) = sin(x).
To find the derivatives of all orders of sin(x)
we arrange them in a circle.
Maclaurin Expansions
sin(x)
cos(x)
–sin(x)
–cos(x)
derivative:
0th,

79. Example C.
Find the Mac-expansions of f(x) = sin(x).
To find the derivatives of all orders of sin(x)
we arrange them in a circle.
Maclaurin Expansions
sin(x)
cos(x)
–sin(x)
–cos(x)
derivative:
0th,
derivative:
1st,

80. Example C.
Find the Mac-expansions of f(x) = sin(x).
To find the derivatives of all orders of sin(x)
we arrange them in a circle.
Maclaurin Expansions
sin(x)
cos(x)
–sin(x)
–cos(x)
derivative:
0th,
derivative:
1st,
derivative:
2nd,

81. Example C.
Find the Mac-expansions of f(x) = sin(x).
To find the derivatives of all orders of sin(x)
we arrange them in a circle.
Maclaurin Expansions
sin(x)
cos(x)
–sin(x)
–cos(x)
derivative:
0th,
derivative:
1st,
derivative:
2nd,
derivative:
3rd,

82. Example C.
Find the Mac-expansions of f(x) = sin(x).
To find the derivatives of all orders of sin(x)
we arrange them in a circle.
Maclaurin Expansions
sin(x)
cos(x)
–sin(x)
–cos(x)
derivative:
0th, 4th, 8th, ..
derivative:
1st,
derivative:
2nd,
derivative:
3rd,

83. Example C.
Find the Mac-expansions of f(x) = sin(x).
To find the derivatives of all orders of sin(x)
we arrange them in a circle.
Maclaurin Expansions
sin(x)
cos(x)
–sin(x)
–cos(x)
derivative:
0th, 4th, 8th, ..
derivative:
1st, 5th, 9th, ..
derivative:
2nd,..
derivative:
3rd,..

84. Example C.
Find the Mac-expansions of f(x) = sin(x).
To find the derivatives of all orders of sin(x)
we arrange them in a circle.
Maclaurin Expansions
sin(x)
cos(x)
–sin(x)
–cos(x)
derivative:
0th, 4th, 8th, ..
derivative:
1st, 5th, 9th, ..
derivative:
2nd, 6th, 10th, ..
derivative:
3rd, 7th, 11th, ..

85. So at x = 0,
the derivatives are:
0
1
0
–1
derivative:
0th, 4th, 8th, ..
derivative:
1st, 5th, 9th, ..
derivative:
2nd, 6th, 10th, ..
derivative:
3rd, 7th, 11th, ..
Maclaurin Expansions

86. So at x = 0,
the derivatives are:
0
1
0
–1
derivative:
0th, 4th, 8th, ..
derivative:
1st, 5th, 9th, ..
derivative:
2nd, 6th, 10th, ..
derivative:
3rd, 7th, 11th, ..
Maclaurin Expansions
1x
0
+
2!
= 0 + x2
P(x)
–1
+
3!
x3
Setting 0, 1, 0, –1, 0, 1, 0, –1, ..
for f(n)(0) in the expansion:
0
+
4!
x4 1
+
5!
x5
+
6!
x60
+
7!
x7..–1

87. So at x = 0,
the derivatives are:
0
1
0
–1
derivative:
0th, 4th, 8th, ..
derivative:
1st, 5th, 9th, ..
derivative:
2nd, 6th, 10th, ..
derivative:
3rd, 7th, 11th, ..
Maclaurin Expansions
1x
0
+
2!
= 0 + x2
P(x)
–1
+
3!
x3
Setting 0, 1, 0, –1, 0, 1, 0, –1, ..
for f(n)(0) in the expansion:
0
+
4!
x4 1
+
5!
x5
+
6!
x60
+
7!
x7..–1
1x=or P(x) –
3!
x3
+ 5!
x5
+..
7!
x7
–

88. So at x = 0,
the derivatives are:
0
1
0
–1
derivative:
0th, 4th, 8th, ..
derivative:
1st, 5th, 9th, ..
derivative:
2nd, 6th, 10th, ..
derivative:
3rd, 7th, 11th, ..
Maclaurin Expansions
1x
0
+
2!
= 0 + x2
P(x)
–1
+
3!
x3
Setting 0, 1, 0, –1, 0, 1, 0, –1, ..
for f(n)(0) in the expansion:
0
+
4!
x4 1
+
5!
x5
+
6!
x60
+
7!
x7..–1
1x=or P(x) –
3!
x3
+ 5!
x5
+..
7!
x7
–
Writing {1, –3, 5, –7, ..} as {(–1)k(2k + 1)},

89. So at x = 0,
the derivatives are:
0
1
0
–1
derivative:
0th, 4th, 8th, ..
derivative:
1st, 5th, 9th, ..
derivative:
2nd, 6th, 10th, ..
derivative:
3rd, 7th, 11th, ..
Maclaurin Expansions
1x
0
+
2!
= 0 + x2
P(x)
–1
+
3!
x3
Σk=0 (2k+1)!
(–1)kx2k+1
as the Mac-series of sin(x).
∞
Setting 0, 1, 0, –1, 0, 1, 0, –1, ..
for f(n)(0) in the expansion:
0
+
4!
x4 1
+
5!
x5
+
6!
x60
+
7!
x7..–1
1x=or P(x) –
3!
x3
+ 5!
x5
+..
7!
x7
–
Writing {1, –3, 5, –7, ..} as {(–1)k(2k + 1)}, we have
=P(x) 1x –
3!
x3
+ 5!
x5
+.. =
7!
x7
–

90. Maclaurin Expansions
Source:
Wikipedia

91. Maclaurin Expansions
Example D. Find the Mac-expansions of f(x) = (1 – x)–1.

92. Again, let's obtain the pattern of the derivatives first.
Maclaurin Expansions
Example D. Find the Mac-expansions of f(x) = (1 – x)–1.

93. Again, let's obtain the pattern of the derivatives first.
f(x) = (1 – x)–1 so at x = 0, f(0) = 1
Maclaurin Expansions
Example D. Find the Mac-expansions of f(x) = (1 – x)–1.

94. Again, let's obtain the pattern of the derivatives first.
f(x) = (1 – x)–1 so at x = 0, f(0) = 1
f (x) = (1 – x)–2 so at x = 0, f (x) = 1(1) (1)
Maclaurin Expansions
Example D. Find the Mac-expansions of f(x) = (1 – x)–1.

95. Again, let's obtain the pattern of the derivatives first.
f(x) = (1 – x)–1 so at x = 0, f(0) = 1
f (x) = (1 – x)–2 so at x = 0, f (x) = 1(1) (1)
f (x) = 2(1 – x)–3 so at x = 0, f (x) = 2!(2) (2)
Maclaurin Expansions
Example D. Find the Mac-expansions of f(x) = (1 – x)–1.

96. Again, let's obtain the pattern of the derivatives first.
f(x) = (1 – x)–1 so at x = 0, f(0) = 1
f (x) = (1 – x)–2 so at x = 0, f (x) = 1(1) (1)
f (x) = 2(1 – x)–3 so at x = 0, f (x) = 2!(2) (2)
f (x) = 3*2(1 – x)–4 so at x = 0, f (x) = 3!(3) (3)
Maclaurin Expansions
Example D. Find the Mac-expansions of f(x) = (1 – x)–1.

97. Again, let's obtain the pattern of the derivatives first.
f(x) = (1 – x)–1 so at x = 0, f(0) = 1
f (x) = (1 – x)–2 so at x = 0, f (x) = 1(1) (1)
f (x) = 2(1 – x)–3 so at x = 0, f (x) = 2!(2) (2)
f (x) = 3*2(1 – x)–4 so at x = 0, f (x) = 3!(3) (3)
In general, f (x) = n!(n)
Maclaurin Expansions
Example D. Find the Mac-expansions of f(x) = (1 – x)–1.

98. Again, let's obtain the pattern of the derivatives first.
f(x) = (1 – x)–1 so at x = 0, f(0) = 1
f (x) = (1 – x)–2 so at x = 0, f (x) = 1(1) (1)
f (x) = 2(1 – x)–3 so at x = 0, f (x) = 2!(2) (2)
f (x) = 3*2(1 – x)–4 so at x = 0, f (x) = 3!(3) (3)
In general, f (x) = n!(n)
Therefore, P(x) = 1x +
2!
1+ x2
+
3!
x3
+
4!
x4
+ … or2! 3! 4!
Maclaurin Expansions
Example D. Find the Mac-expansions of f(x) = (1 – x)–1.

99. Again, let's obtain the pattern of the derivatives first.
f(x) = (1 – x)–1 so at x = 0, f(0) = 1
f (x) = (1 – x)–2 so at x = 0, f (x) = 1(1) (1)
f (x) = 2(1 – x)–3 so at x = 0, f (x) = 2!(2) (2)
f (x) = 3*2(1 – x)–4 so at x = 0, f (x) = 3!(3) (3)
In general, f (x) = n!(n)
Therefore, P(x) = 1x +
2!
1+ x2
+
3!
x3
+
4!
x4
+ … or2! 3! 4!
P(x) = 1 + x + x2 + x3 + x4 .. is the Mac-series of 1 – x .
1
Maclaurin Expansions
Example D. Find the Mac-expansions of f(x) = (1 – x)–1.

100. Summary of the Mac-series
I. For polynomials P, a Mac-poly of degree k consists of the
first k-terms of the polynomial P. Mac-series of polynomials
are the polynomials themselves.
II. For ex, its Σ
k=0
k! .
xk∞
x + 2!1 +
x2
+ .. ++ 3!
x3
n! ..
xn
=
Σ
k=0 (2k+1)!
(-1)kx2k+1∞
x –
3!
x3
+ 5!
x5
+ .. =7!
x7
–III. For sin(x), its
IV. For cos(x), its Σ
k=0 (2k)!
(-1)kx2k∞
+ 4!
x4
6!
x6
8!
x8
+1 – – – .. =2!
x2
V. For , its
(1 – x )
1
1 + x + x2 + x3 + x4 .. = Σ
k=0
∞
xk
Computation Techniques for Maclaurin Expansions
VI. For Ln(1 + x), its + 3
x3
4
x4
5
x5
+x – –2
x2
.. Σ
k=1 k .
(-1)k+1xk∞
=