Section 9.7 and 9.10: Taylor
Polynomials and
Approximations/Taylor and
Maclaurin Series
Power Series for Functions
We can create a Power Series (or polynomial series) that
can approximate a function around a certain value of the
function.
In order to approximate the function, we force the series
to match the function at its value through many
derivatives.
𝑥2
2
Example: 𝑇2 (𝑥) = 1 + 𝑥 + approximates 𝑓 𝑥 = 𝑒 𝑥 at
𝑥 = 0 because
𝑓 0 = 𝑇2 0 = 1
𝑓′ 0 = 𝑇2 ′(0) = 1
𝑓′′ 0 = 𝑇2 ′′(0) = 1
Taylor Series Generated by f at x=0
Also known as a Maclaurin Series
generated by 𝑓.
Let 𝑓 be a function with derivatives of all orders throughout
some open interval containing 0. Then the Taylor series
generated by 𝒇 at 𝒙 = 𝟎 is:
′′ 0
𝑛 0
𝑓
𝑓
𝑓 0 + 𝑓′ 0 ∙ 𝑥 +
𝑥2 + ⋯ +
𝑥𝑛 + ⋯
2!
𝑛!
∞
𝑓 𝑘 (0) 𝑘
𝑥
𝑘!
𝑘=0
Verifying the Maclaurin Series
Justify that the Taylor Series centered at 𝑥 = 0 matches any
function at its value through all of its derivatives at 𝑥 = 0.
′′ 0
′′′ 0
𝑛 0
𝑓
𝑓
𝑓
𝑓 0 + 𝑓′ 0 ∙ 𝑥 +
𝑥2 +
𝑥3 + ⋯ +
𝑥𝑛 + ⋯
2!
3!
𝑛!
′′ 0
′′′ 0
𝑛 0
𝑓
𝑓
𝑓
02 +
03 + ⋯ +
0𝑛 + ⋯
If 𝑥 = 0, does 𝑓 0 + 𝑓 ′ 0 ∙ 0 +
2!
3!
𝑛!
the series
equal 𝑓(0)?
If 𝑥 = 0, does
the
derivative of
the series
equal 𝑓′(0)?
= 𝑓 0 Yes!
′
Derivative: 0 + 𝑓 0 + 𝑓
′′
0 𝑥
𝑓′′′ 0
+
2!
𝑓 ′′′ 0
𝑥 = 0: 0 + 𝑓 ′ 0 + 𝑓 ′′ 0 ∙ 0 + +
= 𝑓′ 0
Yes!
2!
2
𝑥 +
𝑓𝑛 0
⋯+
(𝑛−1)!
𝑓𝑛 0
02 … +
𝑛 − 1!
𝑥 𝑛−1 + ⋯
0𝑛−1 + ⋯
Verifying the Maclaurin Series
Justify that the Taylor Series centered at 𝑥 = 0 matches any
function at its value through all of its derivatives at 𝑥 = 0.
′′ 0
′′′ 0
𝑛 0
𝑓
𝑓
𝑓
𝑓 0 + 𝑓′ 0 ∙ 𝑥 +
𝑥2 +
𝑥3 + ⋯ +
𝑥𝑛 + ⋯
2!
3!
𝑛!
If 𝑥 = 0, does
the second
derivative of
the series
equal
𝑓′′(0)?
2nd
Deriv: 0 + 0 +
𝑓 ′′
0 +
𝑓 ′′′
0 𝑥+
𝑓𝑛 0
⋯+
(𝑛−2)!
𝑓𝑛 0
𝑥 = 0: 0 + 0 + 𝑓 ′′ 0 + 𝑓 ′′′ 0 ∙ 0 + ⋯ +
(𝑛 − 2)!
𝑥 𝑛−2 + ⋯
0𝑛−2 + ⋯
= 𝑓′′ 0 Yes!
This pattern will continue. If 𝑥 = 0, then the nth derivative of the
series will equal 𝑓 𝑛 0 . The Taylor Series meets the defined
requirements.
Taylor Series v Taylor Polynomial
Taylor Polynomial: Have a finite number of terms
and are approximations of the function around a
value of 𝑥.
′′ 0
𝑛 0
𝑓
𝑓
𝑓 0 + 𝑓′ 0 ∙ 𝑥 +
𝑥2 + ⋯ +
𝑥𝑛
2!
𝑛!
Taylor Series: Have a infinite number of terms and
are equivalent to the function for values of 𝑥.
′′
𝑛
𝑓
0
𝑓
0 𝑛
′
2
𝑓 0 +𝑓 0 ∙𝑥+
𝑥 + ⋯+
𝑥 +⋯
2!
𝑛!
Maclaurin Series to Memorize
1
1−𝑥
∞
𝑛
𝑛=0 𝑥
= 1 + 𝑥 + 𝑥2 + 𝑥3 + ⋯ + 𝑥𝑛 + ⋯ =
𝑥
𝑒 =1+𝑥+
𝑥2
2!
+
𝑥3
3!
+⋯+
𝑥𝑛
𝑛!
for 𝑥 < 1
𝑛
∞ 𝑥
𝑛=0 𝑛!
+⋯=
for 𝑥 < 1
Odd degree powers and sin 𝑥 is an odd function
sin 𝑥 = 𝑥 −
𝑥3
3!
+
𝑥5
5!
− ⋯ + −1
2𝑛+1
𝑛 𝑥
+
2𝑛+1 !
⋯=
∞
𝑛=0
−1
2𝑛+1
𝑛 𝑥
2𝑛+1 !
Starts at 0 because sin 0 = 0
for all
real 𝑥
Even degree powers and cos 𝑥 is an even function
cos 𝑥 = 1 −
𝑥2
2!
+
𝑥4
4!
− ⋯ + −1
Starts at 1 because cos 0 = 1
2𝑛
𝑛 𝑥
+
2𝑛 !
⋯=
∞
𝑛=0
−1
2𝑛
𝑛 𝑥
2𝑛 !
for all
real 𝑥
We will discuss the interval of convergence later.
Example 1
Suppose 𝑔 is a function with derivatives of all orders for
real numbers. Assume 𝑔 0 = −4, 𝑔′ 0 = 2, 𝑔′′ 0 = 6,
and 𝑔′′′ 0 = −8. Write the second-degree the Taylor
polynomial for g at 𝑥 = 0, and use it to approximate 𝑔(0.5).
Notice the degree required.
Substitute 𝑥 = 0.5 into the polynomial to
approximate the function 𝑔:
Use the Formula for a second-degree
polynomial:
2
−4
+
2
∙
0.5
+
3
∙
0.5
′′ 0
𝑔
′
2
𝑔 0 +𝑔 0 ∙𝑥+
2!
6 2
−4 + 2 ∙ 𝑥 + 𝑥
2!
−4 + 2𝑥 + 3𝑥 2
𝑥
−2.25
Example 2
Find the Taylor series for 𝑓 𝑥 = (1 + 𝑥)3 at 𝑥 = 0.
Find the function and derivatives of 𝑓(𝑥):
Find the value of the function and
each derivative if 𝑥 = 0:
𝑓 𝑥 = (1 + 𝑥)3
𝑓 0 =1
𝑓′ 𝑥 = 3(1 + 𝑥)2
𝑓′ 0 = 3
1
𝑓′′ 𝑥 = 6(1 + 𝑥)
𝑓′′′ 𝑥 = 6
All future
derivatives
will be zero.
𝑓 (4) 𝑥 = 0
𝑓′′ 0 = 6
𝑓′′′ 0 = 6
𝑓′′′ 0 = 0
Substitute into the formula:
6 2 6 3
1 + 3 ∙ 𝑥 + 𝑥 + 𝑥 = 1 + 3𝑥 + 3𝑥 2 + 𝑥 3
2!
3!
Notice: A Taylor series for a polynomial function IS the polynomial function in
standard form.
Example 3
Find the Maclaurin series for 𝑓 𝑥 = 𝑥 2 𝑒 𝑥 .
Find the function and derivatives of 𝑓(𝑥):
𝑓 𝑥 = 𝑥2𝑒 𝑥
𝑓 ′ (𝑥) = 𝑥 2 𝑒 𝑥 + 2𝑥𝑒 𝑥
𝑓 ′′ (𝑥) = 𝑥 2 𝑒 𝑥 + 4𝑥𝑒 𝑥 + 2𝑒 𝑥
𝑓′′′ 𝑥 = 𝑥 2 𝑒 𝑥 + 6𝑥𝑒 𝑥 + 6𝑒 𝑥
𝑓 (4) (𝑥) = 𝑥 2 𝑒 𝑥 + 8𝑥𝑒 𝑥 + 12𝑒 𝑥
𝑓 (5) (𝑥) = 𝑥 2 𝑒 𝑥 + 10𝑥𝑒 𝑥 + 20𝑒 𝑥
𝑓
6
(𝑥) = 𝑥 2 𝑒 𝑥 + 12𝑥𝑒 𝑥 + 30𝑒 𝑥
Substitute into the formula:
Find the value of the function and
each derivative if 𝑥 = 0:
Not
enough
terms to
see a
pattern.
𝑓 0 =0
𝑓′ 0 = 0
𝑓′′ 0 = 2
𝑓′′′ 0 = 6
𝑓 (4) (0) = 12
𝑓 (5) (0) = 20
𝑓
6
(0) = 30
4
5
𝑛+2
𝑥
𝑥
2 2 6 3 12 3 20 4 30 5
𝑥
+
+ ⋯+
0 + 0 ∙ 𝑥 + 𝑥 + 𝑥 + 𝑥 + 𝑥 + 𝑥 + ⋯ = 𝑥2 + 𝑥3 +
+⋯
2! 3!
2!
3!
4!
5!
6!
𝑛!
There are several method for
generating new Taylor Series
from known ones: integrate term
by term, differentiate term by
term, using algebra, or using
substitution.
Example 3: Method 2
Find the Maclaurin series for 𝑓 𝑥 = 𝑥 2 𝑒 𝑥 .
We know the Maclaurin
series for 𝑒 𝑥 :
We know the Maclaurin
series for 𝑥 2 :
1+𝑥+
𝑥2
2!
+
𝑥3
3!
+ ⋯+
𝑥𝑛
𝑛!
∞
+⋯=
𝑛=0
𝑥𝑛
𝑛!
𝑥2
Multiply the Maclaurin series for 𝑒 𝑥 by the Maclaurin series for 𝑥 2 :
2
3
𝑛
𝑥
𝑥
𝑥
𝑥 2 (1 + 𝑥 + + + ⋯ +
+⋯)
2! 3!
𝑛!
4
5
𝑛+2
𝑥
𝑥
𝑥
2
3
𝑥 + 𝑥 + + + ⋯+
+⋯=
2! 3!
𝑛!
∞
𝑛=0
𝑥 𝑛+2
𝑛!
Example 4
Find the Maclaurin series for 𝑓 𝑥 =
Find the function and derivatives of 𝑓(𝑥):
𝑓 𝑥 =
𝑓 ′ (𝑥)
𝑓′′ 𝑥 =
𝑓 ′′′ (𝑥)
=
=
2
−𝑥
𝑒
2
3
−𝑥
−8𝑥 𝑒
2
−𝑥
− 2𝑒
+
2
2
−𝑥
12𝑥𝑒
2
𝑓 (4) (𝑥) = 16𝑥 4 𝑒 −𝑥 − 48𝑥 2 𝑒 −𝑥 + 12𝑒 −𝑥
2
Find the value of the function and
each derivative if 𝑥 = 0:
𝑓 0 =1
Not
enough 𝑓′ 0 = 0
terms to
see a 𝑓 ′′ 0 = −2
pattern.
2
−𝑥
−2𝑥𝑒
2
2
−𝑥
4𝑥 𝑒
2
5
𝑥 = −32𝑥 5 𝑒 −𝑥 + 160𝑥 3 𝑒 −𝑥 − 120𝑥𝑒 −𝑥
𝑓
6
𝑥 = 64𝑥 6 𝑒 −𝑥 − 480𝑥 4 𝑒 −𝑥 + 720𝑥 2 𝑒 −𝑥 − 120𝑒 −𝑥
2
1+0∙𝑥+
Substitute into
the formula:
2
−2 2
𝑥
2!
+
0 3
𝑥
3!
𝑓′′′ 0 = 0
𝑓 (4) 0 = 12
𝑓 (5) 0 = 0
𝑓 (6) 0 = −120
2
𝑓
2
2
−𝑥
𝑒 .
+
2
2
12 4
𝑥
4!
+
4
6
𝑥
𝑥
1 − 𝑥2 + − + ⋯ +
2! 3!
0 5
𝑥
5!
+
−120 6
𝑥
6!
+⋯
(−1)𝑛 𝑥 2𝑛
+⋯=
𝑛!
∞
𝑛=0
(−1)𝑛 𝑥 2𝑛
𝑛!
There are several method for
generating new Taylor Series
from known ones: integrate term
by term, differentiate term by
term, using algebra, or using
substitution.
Example 4: Method 2
Find the Maclaurin series for 𝑓 𝑥 =
2
−𝑥
𝑒 .
We know the Maclaurin series for 𝑒 𝑥 :
2
3
𝑛
𝑥
𝑥
𝑥
1 + 𝑥 + + + ⋯+
+⋯=
2! 3!
𝑛!
∞
𝑛=0
𝑥𝑛
𝑛!
Substitute −𝑥 2 for 𝑥 in the Maclaurin series for 𝑒 𝑥 :
2 2
2 3
2 𝑛
(−𝑥
)
(−𝑥
)
(−𝑥
)
2
1 + (−𝑥 ) +
+
+ ⋯+
+⋯
2!
3!
𝑛!
1 − 𝑥2 +
𝑥4
2!
−
𝑥6
3!
∞
𝑛 2𝑛
+ ⋯+
(−1) 𝑥
𝑛!
+⋯=
𝑛=0
(−1)𝑛 𝑥 2𝑛
𝑛!
Finding ANY Taylor Series
It should be possible to approximate functions around values other than
𝑥 = 0. How should we change the Maclaurin Series to make a Taylor
Series centered at any 𝑥 = 𝑎? Remember it must still match the function
at its value through all of its derivatives at 𝑥 = 𝑎.
Change
Change the 0 to an 𝑎:
′′
𝑛
𝑓
𝑎
𝑓
𝑎 𝑛
′
2
𝑓 𝑎 +𝑓 𝑎 ∙𝑥+
𝑥 + ⋯+
𝑥 +⋯
2!
𝑛!
the 𝑥 to
𝑥 − 𝑎.
′′′ 𝑎
𝑛 𝑎
If 𝑥 = 𝑎,
𝑓
𝑓
′
′′
2
𝑛−1 + ⋯
≠ 𝑓′ 𝑎
does the 0 + 𝑓 𝑎 + 𝑓 𝑎 ∙ 𝑎 + + 2! 𝑎 … + 𝑛 − 1! 𝑎
series equal
No…
𝑓(𝑎)?
After substituting 𝑎 for 𝑥, we need these values
to be 0 in order for the sum to equal 𝑓′(𝑎).
′′ 𝑎
𝑛 𝑎
𝑓
𝑓
𝑓 𝑎 + 𝑓 ′ 𝑎 ∙ (𝑥 − 𝑎) +
(𝑥 − 𝑎)2 + ⋯ +
(𝑥 − 𝑎)𝑛 + ⋯
2!
𝑛!
Finding ANY Taylor Series
It should be possible to approximate functions around values other than
𝑥 = 0. How should we change the Maclaurin Series to make a Taylor
Series centered at any 𝑥 = 𝑎? Remember it must still match the function
at its value through all of its derivatives at 𝑥 = 𝑎.
Change
Change the 0 to an 𝑎:
the 𝑥 to
𝑥 − 𝑎.
′′
𝑛
𝑓
𝑎
𝑓
𝑎
′
2
𝑓 𝑎 + 𝑓 𝑎 ∙ (𝑥 − 𝑎) +
(𝑥 − 𝑎) + ⋯ +
(𝑥 − 𝑎)𝑛 + ⋯
2!
𝑛!
If 𝑥 = 𝑎, Deriv: 0 + 𝑓 ′ 𝑎 + 𝑓 ′′ 𝑎 (𝑥 − 𝑎) + 𝑓′′′ 𝑎 (𝑥 − 𝑎)2 + ⋯ + 𝑓 𝑛 𝑎 (𝑥 − 𝑎)𝑛−1 + ⋯
2!
(𝑛−1)!
does the
derivative
′′′ 𝑎
𝑛 𝑎
𝑓
𝑓
of the 𝑥 = 𝑎: 0 + 𝑓 ′ 𝑎 + 𝑓 ′′ 𝑎 (𝑎 − 𝑎) +
(𝑎 − 𝑎)2 + ⋯ +
(𝑎 − 𝑎)𝑛−1 + ⋯
2!
(𝑛 − 1)!
series
equal
= 𝑓′ 𝑎
𝑓′(𝑎)?
YES!
Taylor Series Generated by f at x=a
Let 𝑓 be a function with derivatives of all orders throughout
some open interval containing 𝑎. Then the Taylor series
generated by 𝒇 at 𝒙 = 𝒂 is:
′′ 𝑎
𝑛 𝑎
𝑓
𝑓
𝑓 𝑎 + 𝑓 ′ 𝑎 ∙ (𝑥 − 𝑎) +
(𝑥 − 𝑎)2 + ⋯ +
(𝑥 − 𝑎)𝑛 + ⋯
2!
𝑛!
∞
𝑘=0
𝑓
𝑘
(𝑎)
(𝑥 − 𝑎)𝑘
𝑘!
Example
Consider 𝑓 𝑥 = 𝑒 𝑥 .
(a) Find the fourth degree Taylor Polynomial for 𝑓 𝑥 = 𝑒 𝑥
centered at 𝑐 = 1.
Find the function and derivatives of 𝑓(𝑥):
Find the value of each derivative if
𝑥 = 1:
𝑓 𝑥 = 𝑒𝑥
𝑓 1 = 𝑒1
𝑓′ 𝑥 = 𝑒 𝑥
𝑓′ 1 = 𝑒1
𝑓′′ 𝑥 = 𝑒 𝑥
𝑓′′ 1 = 𝑒1
𝑓′′′ 𝑥 = 𝑒 𝑥
𝑓′′′ 1 = 𝑒1
𝑓 (4) 𝑥 = 𝑒 𝑥
𝑓 (4) 1 = 𝑒1
Substitute into
the formula for
4 terms:
𝑒1
+
𝑒1
∙ (𝑥 − 1) +
𝑒
2
𝑒1
(𝑥
2!
−
1
𝑒
1)2 +
3!
𝑒
6
(𝑥 −
1
𝑒
1)3 +
4!
𝑒 + 𝑒(𝑥 − 1) + (𝑥 − 1)2 + (𝑥 − 1)3 +
(𝑥 − 1)4
𝑒
(𝑥
24
− 1)4
Example
Consider 𝑓 𝑥 = 𝑒 𝑥 .
(b) Approximate 𝑒1.5 and 𝑒10 with the Taylor Polynomial.
Substitute 𝑥 = 1.5, 10 into the Taylor polynomial:
𝑒 + 𝑒(1.5 − 1) +
𝑒
(1.5
2
−
𝑒
2
1) + (1.5 −
6
𝑒
3
1) + (1.5
24
− 1)4
= 4.481
𝑒 + 𝑒(10 − 1) +
𝑒
(10
2
− 1)
2
𝑒
+ (10
6
− 1)
3
𝑒
+ (10
24
− 1)4
= 1210.655
Since 𝑒1.5 ≈ 4.481 and 𝑒10 ≈ 22026.466, the Taylor polynomial is a
better approximation when it is closer to center value.
Extension
Find the Maclaurin series for 𝑓 𝑥 = 𝑒 𝑖𝑥 .
We know the Maclaurin
series for 𝑒 𝑥 :
Substitute
𝑖𝑥 for 𝑥 in
the
Maclaurin
series for
𝑒 𝑥:
𝑥2 𝑥3 𝑥4 𝑥5
𝑥𝑛
1 + 𝑥 + + + + …+
+⋯
2! 3! 4! 5!
𝑛!
𝑖𝑥 2
𝑖𝑥 3
𝑖𝑥 4
𝑖𝑥 5
1 + 𝑖𝑥 +
+
+
+
+⋯
2!
3!
4!
5!
𝑖2𝑥2 𝑖3𝑥3 𝑖4𝑥4 𝑖5𝑥5
1 + 𝑖𝑥 +
+
+
+
+⋯
2!
3!
4!
5!
−𝑥 2 −𝑖𝑥 3 𝑥 4 𝑖𝑥 5
1 + 𝑖𝑥 +
+
+ +
+⋯
2!
3!
4!
5!
−𝑥 2 𝑥 4
−𝑥 3 𝑥 5
(1 +
+ + ⋯ ) + 𝑖(𝑥 +
+ + ⋯)
2!
4!
3!
5!
cos 𝑥 + 𝑖 sin 𝑥
Extension Continued
Find the Maclaurin series for 𝑓 𝑥 = 𝑒 𝑖𝑥 .
Therefore:
Let
𝑥 = 𝜋:
𝑖𝑥
= cos 𝑥 + 𝑖 sin 𝑥
𝑖𝜋
= cos 𝜋 + 𝑖 sin 𝜋
𝑒
𝑒
𝑒
An equation relating the 5
most important numbers in
mathematics!
𝑖𝜋
𝑒
= −1 + 0
𝑖𝜋
+1=0