Taylor Expansions

1. Taylor Expansions

2. 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
Contribute the most for the approximations of P(x)
around x = 0. So the best 1 term approximation is
a0. The best 2-term approximation is a0 + a1x and so
on.
Taylor Expansions

3. 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 contribute the
most for the approximations of P(x) around x = 0.
Taylor 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).

4. Taylor Expansions
y=2x–x2–2x3+x4
y=2x
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).
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 contribute the
most for the approximations of P(x) around x = 0.

5. Taylor Expansions
y=2x–x2–2x3+x4y=2x–x2–2x3+x4
y=2x y=2x–x2
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).
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 contribute the
most for the approximations of P(x) around x = 0.

6. Taylor Expansions
y=2x–x2–2x3+x4y=2x–x2–2x3+x4y=2x–x2–2x3+x4
y=2x y=2x–x2
y=2x–x2–2x3
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).
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 contribute the
most for the approximations of P(x) around x = 0.

7. Taylor Expansions
y=2x–x2–2x3+x4y=2x–x2–2x3+x4y=2x–x2–2x3+x4
y=2x y=2x–x2
y=2x–x2–2x3
This is so because the higher degree terms in
P(x) = a0 + a1x + a2x2 + a3x3 + . . anxn are more
negligible than the lower degree terms for x ≈ 0.
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).
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 contribute the
most for the approximations of P(x) around x = 0.

8. Taylor Expansions
But we may also express P(x) = 2x – x2 – 2x3 + x4 as
–2(x – 1) – (x – 1)2 – 2(x – 1)3 + (x – 1)4

9. Taylor Expansions
But we may also express P(x) = 2x – x2 – 2x3 + x4 as
–2(x – 1) – (x – 1)2 – 2(x – 1)3 + (x – 1)4, i.e. as a
polynomial in (x – 1) or geometrically based at x = 1.

10. Taylor Expansions
But we may also express P(x) = 2x – x2 – 2x3 + x4 as
–2(x – 1) – (x – 1)2 – 2(x – 1)3 + (x – 1)4, i.e. as a
polynomial in (x – 1) or geometrically based at x = 1.
Following are graphs comparing
y = P(x) = –2(x –1) – (x – 1)2 – 2(x – 1)3 + (x – 1)4
and its lower degree expansions at x = 1.

11. Taylor Expansions
But we may also express P(x) = 2x – x2 – 2x3 + x4 as
–2(x – 1) – (x – 1)2 – 2(x – 1)3 + (x – 1)4, i.e. as a
polynomial in (x – 1) or geometrically based at x = 1.
y=–2(x – 1)
Following are graphs comparing
y = P(x) = –2(x –1) – (x – 1)2 – 2(x – 1)3 + (x – 1)4
and its lower degree expansions at x = 1.

12. Taylor Expansions
But we may also express P(x) = 2x – x2 – 2x3 + x4 as
–2(x – 1) – (x – 1)2 – 2(x – 1)3 + (x – 1)4, i.e. as a
polynomial in (x – 1) or geometrically based at x = 1.
y=–2(x – 1) y=–2(x – 1) – (x – 1)2
Following are graphs comparing
y = P(x) = –2(x –1) – (x – 1)2 – 2(x – 1)3 + (x – 1)4
and its lower degree expansions at x = 1.

13. Taylor Expansions
But we may also express P(x) = 2x – x2 – 2x3 + x4 as
–2(x – 1) – (x – 1)2 – 2(x – 1)3 + (x – 1)4, i.e. as a
polynomial in (x – 1) or geometrically based at x = 1.
y=–2(x – 1) y=–2(x – 1) – (x – 1)2 y=–2(x – 1) – (x – 1)2 – 2(x – 1)3
Following are graphs comparing
y = P(x) = –2(x –1) – (x – 1)2 – 2(x – 1)3 + (x – 1)4
and its lower degree expansions at x = 1.

14. Taylor Expansions
But we may also express P(x) = 2x – x2 – 2x3 + x4 as
–2(x – 1) – (x – 1)2 – 2(x – 1)3 + (x – 1)4, i.e. as a
polynomial in (x – 1) or geometrically based at x = 1.
These are the best successive estimations of P(x) for
x ≈ 1 because the higher degree (x – 1) terms are
more negligible than the lower degree ones.
y=–2(x – 1) y=–2(x – 1) – (x – 1)2 y=–2(x – 1) – (x – 1)2 – 2(x – 1)3
Following are graphs comparing
y = P(x) = –2(x –1) – (x – 1)2 – 2(x – 1)3 + (x – 1)4
and its lower degree expansions at x = 1.

15. Taylor Expansions
But we may also express P(x) = 2x – x2 – 2x3 + x4 as
–2(x – 1) – (x – 1)2 – 2(x – 1)3 + (x – 1)4, i.e. as a
polynomial in (x – 1) or geometrically based at x = 1.
These are the best successive estimations of P(x) for
x ≈ 1 because the higher degree (x – 1) terms are
more negligible than the lower degree ones.
We call them the Taylor polynomials at x = 1 of P(x).
y=–2(x – 1) y=–2(x – 1) – (x – 1)2 y=–2(x – 1) – (x – 1)2 – 2(x – 1)3
Following are graphs comparing
y = P(x) = –2(x –1) – (x – 1)2 – 2(x – 1)3 + (x – 1)4
and its lower degree expansions at x = 1.

16. Taylor Expansions
We define the n'th Taylor polynomial of f(x) at x = a as:
pn(x) = Σk=0
n
(x – a)k
k!
f(k)(a)

17. Taylor Expansions
We define the n'th Taylor polynomial of f(x) at x = a as:
pn(x) = Σk=0
n
(x – a)k
k!
f(k)(a)
= f(a)

18. Taylor Expansions
We define the n'th Taylor polynomial of f(x) at x = a as:
f(1)(a)(x – a)+
1!
pn(x) = Σk=0
n
(x – a)k
k!
f(k)(a)
= f(a)

19. Taylor Expansions
We define the n'th Taylor polynomial of f(x) at x = a as:
f(1)(a)(x – a)
++
1!
pn(x) = Σk=0
n
(x – a)k
k!
f(k)(a)
f(2)(a)(x – a)2
2!
= f(a)

20. Taylor Expansions
We define the n'th Taylor polynomial of f(x) at x = a as:
f(1)(a)(x – a)
++
1!
pn(x) = Σk=0
n
(x – a)k
k!
f(k)(a)
f(2)(a)(x – a)2
+ ..2!
+ f(n)(a)(x – a)n
n!
= f(a)

21. Taylor Expansions
We define the n'th Taylor polynomial of f(x) at x = a as:
f(1)(a)(x – a)
++
1!
pn(x) = Σk=0
n
(x – a)k
k!
f(k)(a)
f(2)(a)(x – a)2
+ ..2!
+ f(n)(a)(x – a)n
n!
and the Taylor series of f(x) as:
P(x) = Σk=0
(x – a)k
k!
f(k)(a)∞
= f(a)

22. Taylor Expansions
The Maclaurin expansions of f(x)
We define the n'th Taylor polynomial of f(x) at x = a as:
f(1)(a)(x – a)
++
1!
pn(x) = Σk=0
n
(x – a)k
k!
f(k)(a)
f(2)(a)(x – a)2
+ ..2!
+ f(n)(a)(x – a)n
n!
and the Taylor series of f(x) as:
P(x) = Σk=0
(x – a)k
k!
f(k)(a)∞
pn(x) = Σk=0
n
xk
k!
f(k)(0)
= a0 + a1x + a2x2 + a3x3 + . . anxn
are “regular looking” polynomials centered at x = 0,
= f(a)

23. Taylor Expansions
The Maclaurin expansions of f(x)
We define the n'th Taylor polynomial of f(x) at x = a as:
f(1)(a)(x – a)
++
1!
pn(x) = Σk=0
n
(x – a)k
k!
f(k)(a)
f(2)(a)(x – a)2
+ ..2!
+ f(n)(a)(x – a)n
n!
and the Taylor series of f(x) as:
P(x) = Σk=0
(x – a)k
k!
f(k)(a)∞
pn(x) = Σk=0
n
xk
k!
f(k)(0)
= a0 + a1x + a2x2 + a3x3 + . . anxn
are “regular looking” polynomials centered at x = 0,
and Taylor expansion are polynomials in (x – a)k
centered at x = a.
= f(a)

24. Example A. a. Find the Taylor–expansion of
P(x) = 1 + x + x2 around x = 1.
We need the derivatives of f(x):
(1) (1)
f(1) = 1 + 1 + 1  f(1) = 3
Taylor Expansions
f (x) = 1 + 2x  f (1) = 3
(2) (2)
f (x) = 2  f (1) = 2
(n)
f (1) = 0 for n ≥ 3. Hence,
= f(1) = 3p0(x)
f(1)(1)(x – 1)= f(1)+p1(x)
1!
= 3(x – 1)3 +
1!
f(1)(1)(x – 1)
+= f(1)+p2(x)
1!
f(2)(1)(x – 1)2
2!
= 3(x – 1)3 +
1!
+ 2(x – 1)2
2! = 3 + 3(x – 1) + 1(x – 1)2
= 1 + x + x2
(= 3x)
pn(x) = 1 + x + x2 for n ≥ 2.

25. Example A.
b. Here is a direct method for expressing
P(x) = 1 + x + x2 as a polynomial in (x – 1).
Taylor Expansions

26. Example A.
b. Here is a direct method for expressing
P(x) = 1 + x + x2 as a polynomial in (x – 1).
Taylor Expansions
We are to find a, b, and c such that
1 + x + x2 = a + b(x – 1) + c(x – 1)2.

27. Example A.
b. Here is a direct method for expressing
P(x) = 1 + x + x2 as a polynomial in (x – 1).
Taylor Expansions
We are to find a, b, and c such that
1 + x + x2 = a + b(x – 1) + c(x – 1)2.
Expanding the highest degree term to compare the
highest degree x2,
1 + x + x2 = a + b(x – 1) + cx2 – 2cx + c

28. Example A.
b. Here is a direct method for expressing
P(x) = 1 + x + x2 as a polynomial in (x – 1).
Taylor Expansions
We are to find a, b, and c such that
1 + x + x2 = a + b(x – 1) + c(x – 1)2.
Expanding the highest degree term to compare the
highest degree x2,
1 + x + x2 = a + b(x – 1) + cx2 – 2cx + c
we see that c = 1.

29. Example A.
b. Here is a direct method for expressing
P(x) = 1 + x + x2 as a polynomial in (x – 1).
Taylor Expansions
We are to find a, b, and c such that
1 + x + x2 = a + b(x – 1) + c(x – 1)2.
Expanding the highest degree term to compare the
highest degree x2,
1 + x + x2 = a + b(x – 1) + cx2 – 2cx + c
we see that c = 1.
Plugging this back into the equation and simplifying
we have 1 + x = a + b(x – 1) – 2x + 1

30. Example A.
b. Here is a direct method for expressing
P(x) = 1 + x + x2 as a polynomial in (x – 1).
Taylor Expansions
We are to find a, b, and c such that
1 + x + x2 = a + b(x – 1) + c(x – 1)2.
Expanding the highest degree term to compare the
highest degree x2,
1 + x + x2 = a + b(x – 1) + cx2 – 2cx + c
we see that c = 1.
Plugging this back into the equation and simplifying
we have 1 + x = a + b(x – 1) – 2x + 1
or that 3x = a + b(x – 1)

31. Example A.
b. Here is a direct method for expressing
P(x) = 1 + x + x2 as a polynomial in (x – 1).
Taylor Expansions
We are to find a, b, and c such that
1 + x + x2 = a + b(x – 1) + c(x – 1)2.
Expanding the highest degree term to compare the
highest degree x2,
1 + x + x2 = a + b(x – 1) + cx2 – 2cx + c
we see that c = 1.
Plugging this back into the equation and simplifying
we have 1 + x = a + b(x – 1) – 2x + 1
or that 3x = a + b(x – 1).
From this we see that b = 3.

32. Example A.
b. Here is a direct method for expressing
P(x) = 1 + x + x2 as a polynomial in (x – 1).
Taylor Expansions
We are to find a, b, and c such that
1 + x + x2 = a + b(x – 1) + c(x – 1)2.
Expanding the highest degree term to compare the
highest degree x2,
1 + x + x2 = a + b(x – 1) + cx2 – 2cx + c
we see that c = 1.
Plugging this back into the equation and simplifying
we have 1 + x = a + b(x – 1) – 2x + 1
or that 3x = a + b(x – 1).
From this we see that b = 3,
and a = 3 also.

33. Example B.
Taylor Expansions
π
2 .
Find the Taylor–expansion of
f(x) = cos(x) at x =

34. Example B.
Taylor Expansions
π
2 .
cos(x)
–sin(x)
–cos(x)
sin(x)
At x = π
2 , we have the
0, –1, 0, 1, 0, –1, 0, 1, …
0
0
–11
sequence of coefficients
Find the Taylor–expansion of
f(x) = cos(x) at x =

35. Example B.
Taylor Expansions
π
2 .
cos(x)
–sin(x)
–cos(x)
sin(x)
At x = π
2 , we have the
0, –1, 0, 1, 0, –1, 0, 1, … so the Taylor expansions is
(x – π/2)– +T(x) = 0
1!
(x – π/2)3
+ 0
3!+ 0
(x – π/2)5
5!
– + 0
(x – π/2)7
7!
..+
0
0
–11
sequence of coefficients
Find the Taylor–expansion of
f(x) = cos(x) at x =

36. Example B.
Taylor Expansions
π
2 .
cos(x)
–sin(x)
–cos(x)
sin(x)
At x = π
2 , we have the
0, –1, 0, 1, 0, –1, 0, 1, … so the Taylor expansions is
(x – π/2)– +T(x) = 0
1!
(x – π/2)3
+ 0
3!+ 0
(x – π/2)5
5!
– + 0
(x – π/2)7
7!
..+
0
0
–11
sequence of coefficients
The pattern 0, –1, 0, 1, 0, –1, 0, 1, .. may be given as
–sin(nπ/2),
Find the Taylor–expansion of
f(x) = cos(x) at x =
y = –sin(x)

37. Example B.
Taylor Expansions
π
2 .
cos(x)
–sin(x)
–cos(x)
sin(x)
At x = π
2 , we have the
0, –1, 0, 1, 0, –1, 0, 1, … so the Taylor expansions is
(x – π/2)– +T(x) = 0
1!
(x – π/2)3
+ 0
3!+ 0
(x – π/2)5
5!
– + 0
(x – π/2)7
7!
..+
0
0
–11
sequence of coefficients
T(x)= Σ
–sin(nπ/2)(x – π/2)2n+1
(2n + 1)!n=0
∞
The pattern 0, –1, 0, 1, 0, –1, 0, 1, .. may be given as
–sin(nπ/2), so the Taylor series
at x = π/2 of cos(x) is:
Find the Taylor–expansion of
f(x) = cos(x) at x =
y = –sin(x)

38. We can't expand Ln(x) at x = 0 but we can expand it
at x = 1.
Taylor Expansions
Example C. Expand Ln(x) at x = 1.

39. We can't expand Ln(x) at x = 0 but we can expand it
at x = 1.
Taylor Expansions
From the pattern y(1) = x–1, y(2) = –x–2, y(3) = 2x–3,
y(4) = –3*2x–4, y(5) = 4!x–5,
Example C. Expand Ln(x) at x = 1.

40. We can't expand Ln(x) at x = 0 but we can expand it
at x = 1.
Taylor Expansions
From the pattern y(1) = x–1, y(2) = –x–2, y(3) = 2x–3,
y(4) = –3*2x–4, y(5) = 4!x–5, we have the formula
y(n)=(–1)n–1(n – 1)! x–n for n = 1, 2, 3 ..
Example C. Expand Ln(x) at x = 1.

41. We can't expand Ln(x) at x = 0 but we can expand it
at x = 1.
Taylor Expansions
At x = 1, we have the sequence of coefficients:
From the pattern y(1) = x–1, y(2) = –x–2, y(3) = 2x–3,
y(4) = –3*2x–4, y(5) = 4!x–5, we have the formula
y(n)=(–1)n–1(n – 1)! x–n for n = 1, 2, 3 ..
Example C. Expand Ln(x) at x = 1.
Ln(1) = 0, y(1)(1) = 1, y(2)(1) = –1, y(3)(1) = 2,
y(4)(1) = –3*2, y(5)(1) = 4!, …

42. We can't expand Ln(x) at x = 0 but we can expand it
at x = 1.
Taylor Expansions
At x = 1, we have the sequence of coefficients:
From the pattern y(1) = x–1, y(2) = –x–2, y(3) = 2x–3,
y(4) = –3*2x–4, y(5) = 4!x–5, we have the formula
y(n)=(–1)n–1(n – 1)! x–n for n = 1, 2, 3 ..
Example C. Expand Ln(x) at x = 1.
Ln(1) = 0, y(1)(1) = 1, y(2)(1) = –1, y(3)(1) = 2,
y(4)(1) = –3*2, y(5)(1) = 4!, …
So that the general formula is:
y(n) = (–1)n–1(n – 1)! for n = 1, 2, 3..

43. We can't expand Ln(x) at x = 0 but we can expand it
at x = 1.
Taylor Expansions
At x = 1, we have the sequence of coefficients:
From the pattern y(1) = x–1, y(2) = –x–2, y(3) = 2x–3,
y(4) = –3*2x–4, y(5) = 4!x–5, we have the formula
y(n)=(–1)n–1(n – 1)! x–n for n = 1, 2, 3 ..
Example C. Expand Ln(x) at x = 1.
Ln(1) = 0, y(1)(1) = 1, y(2)(1) = –1, y(3)(1) = 2,
y(4)(1) = –3*2, y(5)(1) = 4!, …
and that the general formula is:
y(n) = (–1)n–1(n – 1)! for n = 1, 2, 3..
And the Taylor series of In(x) around x = 1 is:

44. Taylor Expansions
T(x) = Σ
(x – 1)n
n!
(–1)n–1(n – 1)!
n=1
∞

45. Taylor Expansions
T(x) = =Σ n=1
(x – 1)n
n!
(–1)n–1(n – 1)! ∞
Σ
n=1
(x – 1)n
n
(–1)n–1∞

46. Taylor Expansions
T(x) = =Σ n=1
(x – 1)n
n!
(–1)n–1(n – 1)! ∞
Σ
n=1
(x – 1)n
n
(–1)n–1∞
(x – 1) –(x – 1)2
2 +
(x – 1)3
3
– (x – 1)4
4 +
(x – 1)5
5 …=

47. Taylor Expansions
T(x) = =Σ n=1
(x – 1)n
n!
(–1)n–1(n – 1)! ∞
Σ
n=1
(x – 1)n
n
(–1)n–1∞
(x – 1) –(x – 1)2
2 +
(x – 1)3
3
– (x – 1)4
4 +
(x – 1)5
5 …=
Setting x = 2 in the series, we see that the sum of
the alternating harmonic series is
= Ln(2)1 – 1
2 + 1
3 – 1
4
+ – …
1
5
is the Taylor series of Ln(x) at x = 1.