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 β 1)(x β 2) = 2x β x2 β 2x3 + x4.
Following are the graphs of P(x) against the lower
degree terms of P(x).
4. Taylor Expansions
y=2xβx2β2x3+x4
y=2x
Let P(x) = (x + 1)x(x β 1)(x β 2) = 2x β x2 β 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=2xβx2β2x3+x4y=2xβx2β2x3+x4
y=2x y=2xβx2
Let P(x) = (x + 1)x(x β 1)(x β 2) = 2x β x2 β 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=2xβx2β2x3+x4y=2xβx2β2x3+x4y=2xβx2β2x3+x4
y=2x y=2xβx2
y=2xβx2β2x3
Let P(x) = (x + 1)x(x β 1)(x β 2) = 2x β x2 β 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=2xβx2β2x3+x4y=2xβx2β2x3+x4y=2xβx2β2x3+x4
y=2x y=2xβx2
y=2xβx2β2x3
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 β 0.
Let P(x) = (x + 1)x(x β 1)(x β 2) = 2x β x2 β 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 β x2 β 2x3 + x4 as
β2(x β 1) β (x β 1)2 β 2(x β 1)3 + (x β 1)4
9. Taylor Expansions
But we may also express P(x) = 2x β x2 β 2x3 + x4 as
β2(x β 1) β (x β 1)2 β 2(x β 1)3 + (x β 1)4, i.e. as a
polynomial in (x β 1) or geometrically based at x = 1.
10. Taylor Expansions
But we may also express P(x) = 2x β x2 β 2x3 + x4 as
β2(x β 1) β (x β 1)2 β 2(x β 1)3 + (x β 1)4, i.e. as a
polynomial in (x β 1) or geometrically based at x = 1.
Following are graphs comparing
y = P(x) = β2(x β1) β (x β 1)2 β 2(x β 1)3 + (x β 1)4
and its lower degree expansions at x = 1.
11. Taylor Expansions
But we may also express P(x) = 2x β x2 β 2x3 + x4 as
β2(x β 1) β (x β 1)2 β 2(x β 1)3 + (x β 1)4, i.e. as a
polynomial in (x β 1) or geometrically based at x = 1.
y=β2(x β 1)
Following are graphs comparing
y = P(x) = β2(x β1) β (x β 1)2 β 2(x β 1)3 + (x β 1)4
and its lower degree expansions at x = 1.
12. Taylor Expansions
But we may also express P(x) = 2x β x2 β 2x3 + x4 as
β2(x β 1) β (x β 1)2 β 2(x β 1)3 + (x β 1)4, i.e. as a
polynomial in (x β 1) or geometrically based at x = 1.
y=β2(x β 1) y=β2(x β 1) β (x β 1)2
Following are graphs comparing
y = P(x) = β2(x β1) β (x β 1)2 β 2(x β 1)3 + (x β 1)4
and its lower degree expansions at x = 1.
13. Taylor Expansions
But we may also express P(x) = 2x β x2 β 2x3 + x4 as
β2(x β 1) β (x β 1)2 β 2(x β 1)3 + (x β 1)4, i.e. as a
polynomial in (x β 1) or geometrically based at x = 1.
y=β2(x β 1) y=β2(x β 1) β (x β 1)2 y=β2(x β 1) β (x β 1)2 β 2(x β 1)3
Following are graphs comparing
y = P(x) = β2(x β1) β (x β 1)2 β 2(x β 1)3 + (x β 1)4
and its lower degree expansions at x = 1.
14. Taylor Expansions
But we may also express P(x) = 2x β x2 β 2x3 + x4 as
β2(x β 1) β (x β 1)2 β 2(x β 1)3 + (x β 1)4, i.e. as a
polynomial in (x β 1) or geometrically based at x = 1.
These are the best successive estimations of P(x) for
x β 1 because the higher degree (x β 1) terms are
more negligible than the lower degree ones.
y=β2(x β 1) y=β2(x β 1) β (x β 1)2 y=β2(x β 1) β (x β 1)2 β 2(x β 1)3
Following are graphs comparing
y = P(x) = β2(x β1) β (x β 1)2 β 2(x β 1)3 + (x β 1)4
and its lower degree expansions at x = 1.
15. Taylor Expansions
But we may also express P(x) = 2x β x2 β 2x3 + x4 as
β2(x β 1) β (x β 1)2 β 2(x β 1)3 + (x β 1)4, i.e. as a
polynomial in (x β 1) or geometrically based at x = 1.
These are the best successive estimations of P(x) for
x β 1 because the higher degree (x β 1) terms are
more negligible than the lower degree ones.
We call them the Taylor polynomials at x = 1 of P(x).
y=β2(x β 1) y=β2(x β 1) β (x β 1)2 y=β2(x β 1) β (x β 1)2 β 2(x β 1)3
Following are graphs comparing
y = P(x) = β2(x β1) β (x β 1)2 β 2(x β 1)3 + (x β 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) = Ξ£k=0
n
(x β 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) = Ξ£k=0
n
(x β 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 β a)+
1!
pn(x) = Ξ£k=0
n
(x β 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 β a)
++
1!
pn(x) = Ξ£k=0
n
(x β a)k
k!
f(k)(a)
f(2)(a)(x β 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 β a)
++
1!
pn(x) = Ξ£k=0
n
(x β a)k
k!
f(k)(a)
f(2)(a)(x β a)2
+ ..2!
+ f(n)(a)(x β 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 β a)
++
1!
pn(x) = Ξ£k=0
n
(x β a)k
k!
f(k)(a)
f(2)(a)(x β a)2
+ ..2!
+ f(n)(a)(x β a)n
n!
and the Taylor series of f(x) as:
P(x) = Ξ£k=0
(x β a)k
k!
f(k)(a)β
= 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 β a)
++
1!
pn(x) = Ξ£k=0
n
(x β a)k
k!
f(k)(a)
f(2)(a)(x β a)2
+ ..2!
+ f(n)(a)(x β a)n
n!
and the Taylor series of f(x) as:
P(x) = Ξ£k=0
(x β a)k
k!
f(k)(a)β
pn(x) = Ξ£k=0
n
xk
k!
f(k)(0)
= a0 + a1x + a2x2 + a3x3 + . . anxn
are βregular lookingβ 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 β a)
++
1!
pn(x) = Ξ£k=0
n
(x β a)k
k!
f(k)(a)
f(2)(a)(x β a)2
+ ..2!
+ f(n)(a)(x β a)n
n!
and the Taylor series of f(x) as:
P(x) = Ξ£k=0
(x β a)k
k!
f(k)(a)β
pn(x) = Ξ£k=0
n
xk
k!
f(k)(0)
= a0 + a1x + a2x2 + a3x3 + . . anxn
are βregular lookingβ polynomials centered at x = 0,
and Taylor expansion are polynomials in (x β a)k
centered at x = a.
= f(a)
24. Example A. a. Find the Taylorβexpansion of
P(x) = 1 + x + x2 around x = 1.
We need the derivatives of f(x):
(1) (1)
f(1) = 1 + 1 + 1 ο f(1) = 3
Taylor Expansions
f (x) = 1 + 2x ο f (1) = 3
(2) (2)
f (x) = 2 ο f (1) = 2
(n)
f (1) = 0 for n β₯ 3. Hence,
= f(1) = 3p0(x)
f(1)(1)(x β 1)= f(1)+p1(x)
1!
= 3(x β 1)3 +
1!
f(1)(1)(x β 1)
+= f(1)+p2(x)
1!
f(2)(1)(x β 1)2
2!
= 3(x β 1)3 +
1!
+ 2(x β 1)2
2! = 3 + 3(x β 1) + 1(x β 1)2
= 1 + x + x2
(= 3x)
pn(x) = 1 + x + x2 for n β₯ 2.
25. Example A.
b. Here is a direct method for expressing
P(x) = 1 + x + x2 as a polynomial in (x β 1).
Taylor Expansions
26. Example A.
b. Here is a direct method for expressing
P(x) = 1 + x + x2 as a polynomial in (x β 1).
Taylor Expansions
We are to find a, b, and c such that
1 + x + x2 = a + b(x β 1) + c(x β 1)2.
27. Example A.
b. Here is a direct method for expressing
P(x) = 1 + x + x2 as a polynomial in (x β 1).
Taylor Expansions
We are to find a, b, and c such that
1 + x + x2 = a + b(x β 1) + c(x β 1)2.
Expanding the highest degree term to compare the
highest degree x2,
1 + x + x2 = a + b(x β 1) + cx2 β 2cx + c
28. Example A.
b. Here is a direct method for expressing
P(x) = 1 + x + x2 as a polynomial in (x β 1).
Taylor Expansions
We are to find a, b, and c such that
1 + x + x2 = a + b(x β 1) + c(x β 1)2.
Expanding the highest degree term to compare the
highest degree x2,
1 + x + x2 = a + b(x β 1) + cx2 β 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 β 1).
Taylor Expansions
We are to find a, b, and c such that
1 + x + x2 = a + b(x β 1) + c(x β 1)2.
Expanding the highest degree term to compare the
highest degree x2,
1 + x + x2 = a + b(x β 1) + cx2 β 2cx + c
we see that c = 1.
Plugging this back into the equation and simplifying
we have 1 + x = a + b(x β 1) β 2x + 1
30. Example A.
b. Here is a direct method for expressing
P(x) = 1 + x + x2 as a polynomial in (x β 1).
Taylor Expansions
We are to find a, b, and c such that
1 + x + x2 = a + b(x β 1) + c(x β 1)2.
Expanding the highest degree term to compare the
highest degree x2,
1 + x + x2 = a + b(x β 1) + cx2 β 2cx + c
we see that c = 1.
Plugging this back into the equation and simplifying
we have 1 + x = a + b(x β 1) β 2x + 1
or that 3x = a + b(x β 1)
31. Example A.
b. Here is a direct method for expressing
P(x) = 1 + x + x2 as a polynomial in (x β 1).
Taylor Expansions
We are to find a, b, and c such that
1 + x + x2 = a + b(x β 1) + c(x β 1)2.
Expanding the highest degree term to compare the
highest degree x2,
1 + x + x2 = a + b(x β 1) + cx2 β 2cx + c
we see that c = 1.
Plugging this back into the equation and simplifying
we have 1 + x = a + b(x β 1) β 2x + 1
or that 3x = a + b(x β 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 β 1).
Taylor Expansions
We are to find a, b, and c such that
1 + x + x2 = a + b(x β 1) + c(x β 1)2.
Expanding the highest degree term to compare the
highest degree x2,
1 + x + x2 = a + b(x β 1) + cx2 β 2cx + c
we see that c = 1.
Plugging this back into the equation and simplifying
we have 1 + x = a + b(x β 1) β 2x + 1
or that 3x = a + b(x β 1).
From this we see that b = 3,
and a = 3 also.
34. Example B.
Taylor Expansions
Ο
2 .
cos(x)
βsin(x)
βcos(x)
sin(x)
At x = Ο
2 , we have the
0, β1, 0, 1, 0, β1, 0, 1, β¦
0
0
β11
sequence of coefficients
Find the Taylorβexpansion of
f(x) = cos(x) at x =
35. Example B.
Taylor Expansions
Ο
2 .
cos(x)
βsin(x)
βcos(x)
sin(x)
At x = Ο
2 , we have the
0, β1, 0, 1, 0, β1, 0, 1, β¦ so the Taylor expansions is
(x β Ο/2)β +T(x) = 0
1!
(x β Ο/2)3
+ 0
3!+ 0
(x β Ο/2)5
5!
β + 0
(x β Ο/2)7
7!
..+
0
0
β11
sequence of coefficients
Find the Taylorβexpansion of
f(x) = cos(x) at x =
36. Example B.
Taylor Expansions
Ο
2 .
cos(x)
βsin(x)
βcos(x)
sin(x)
At x = Ο
2 , we have the
0, β1, 0, 1, 0, β1, 0, 1, β¦ so the Taylor expansions is
(x β Ο/2)β +T(x) = 0
1!
(x β Ο/2)3
+ 0
3!+ 0
(x β Ο/2)5
5!
β + 0
(x β Ο/2)7
7!
..+
0
0
β11
sequence of coefficients
The pattern 0, β1, 0, 1, 0, β1, 0, 1, .. may be given as
βsin(nΟ/2),
Find the Taylorβexpansion of
f(x) = cos(x) at x =
y = βsin(x)
37. Example B.
Taylor Expansions
Ο
2 .
cos(x)
βsin(x)
βcos(x)
sin(x)
At x = Ο
2 , we have the
0, β1, 0, 1, 0, β1, 0, 1, β¦ so the Taylor expansions is
(x β Ο/2)β +T(x) = 0
1!
(x β Ο/2)3
+ 0
3!+ 0
(x β Ο/2)5
5!
β + 0
(x β Ο/2)7
7!
..+
0
0
β11
sequence of coefficients
T(x)= Ξ£
βsin(nΟ/2)(x β Ο/2)2n+1
(2n + 1)!n=0
β
The pattern 0, β1, 0, 1, 0, β1, 0, 1, .. may be given as
βsin(nΟ/2), so the Taylor series
at x = Ο/2 of cos(x) is:
Find the Taylorβexpansion of
f(x) = cos(x) at x =
y = βsin(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) = xβ1, y(2) = βxβ2, y(3) = 2xβ3,
y(4) = β3*2xβ4, y(5) = 4!xβ5,
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) = xβ1, y(2) = βxβ2, y(3) = 2xβ3,
y(4) = β3*2xβ4, y(5) = 4!xβ5, we have the formula
y(n)=(β1)nβ1(n β 1)! xβn 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) = xβ1, y(2) = βxβ2, y(3) = 2xβ3,
y(4) = β3*2xβ4, y(5) = 4!xβ5, we have the formula
y(n)=(β1)nβ1(n β 1)! xβn for n = 1, 2, 3 ..
Example C. Expand Ln(x) at x = 1.
Ln(1) = 0, y(1)(1) = 1, y(2)(1) = β1, y(3)(1) = 2,
y(4)(1) = β3*2, y(5)(1) = 4!, β¦
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) = xβ1, y(2) = βxβ2, y(3) = 2xβ3,
y(4) = β3*2xβ4, y(5) = 4!xβ5, we have the formula
y(n)=(β1)nβ1(n β 1)! xβn for n = 1, 2, 3 ..
Example C. Expand Ln(x) at x = 1.
Ln(1) = 0, y(1)(1) = 1, y(2)(1) = β1, y(3)(1) = 2,
y(4)(1) = β3*2, y(5)(1) = 4!, β¦
So that the general formula is:
y(n) = (β1)nβ1(n β 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) = xβ1, y(2) = βxβ2, y(3) = 2xβ3,
y(4) = β3*2xβ4, y(5) = 4!xβ5, we have the formula
y(n)=(β1)nβ1(n β 1)! xβn for n = 1, 2, 3 ..
Example C. Expand Ln(x) at x = 1.
Ln(1) = 0, y(1)(1) = 1, y(2)(1) = β1, y(3)(1) = 2,
y(4)(1) = β3*2, y(5)(1) = 4!, β¦
and that the general formula is:
y(n) = (β1)nβ1(n β 1)! for n = 1, 2, 3..
And the Taylor series of In(x) around x = 1 is:
47. Taylor Expansions
T(x) = =Ξ£ n=1
(x β 1)n
n!
(β1)nβ1(n β 1)! β
Ξ£
n=1
(x β 1)n
n
(β1)nβ1β
(x β 1) β(x β 1)2
2 +
(x β 1)3
3
β (x β 1)4
4 +
(x β 1)5
5 β¦=
Setting x = 2 in the series, we see that the sum of
the alternating harmonic series is
= Ln(2)1 β 1
2 + 1
3 β 1
4
+ β β¦
1
5
is the Taylor series of Ln(x) at x = 1.