MATH 100 - FUNDAMENTALS OF
MATHEMATICS - Taylor and
Maclaurin Series
Question Bank - Set 4
Liberty University
Question 1
Question
Find the Maclaurin series representation for the function f(x) = 1
1+x2.
Solution
To find the Maclaurin series representation for f(x) = 1
1+x2, we will use the
formula for the Maclaurin series of 1
1−x, which is P∞
n=0 xnwhen |x|<1.
Step 1: Find the Maclaurin series for 11 + x2.
f(x) = 1
1 + x2
=1
1−(−x2)
=
∞
X
n=0
(−x2)nfor |x2|<1
=
∞
X
n=0
(−1)nx2n
=
∞
X
n=0
(−1)nx2n
Therefore, the Maclaurin series representation for f(x) = 1
1+x2is P∞
n=0(−1)nx2n.
Question 2
Question
Find the Maclaurin series representation for the function f(x) = 1
1−x.
Solution
To find the Maclaurin series representation for f(x) = 1
1−x, we will first find
the derivatives of f(x) at x= 0 to determine the coefficients of the series.
Step 1: Find the derivatives of f(x)
f(x) = 1
1−x
f′(x) = d
dx 1
1−x=1
(1 −x)2
f′′(x) = d
dx 1
(1 −x)2=2
(1 −x)3
f′′′(x) = d
dx 2
(1 −x)3=6
(1 −x)4
.
.
.
Step 2: Find the values of the derivatives at x= 0 Evaluate the
derivatives at x= 0 to find the coefficients of the Maclaurin series.
f(0) = 1
f′(0) = 1
f′′(0) = 2
f′′′(0) = 6
.
.
.
Step 3: Write out the Maclaurin series The Maclaurin series for f(x) =
1
1−xis given by:
f(x) = f(0) + f′(0)
1! x+f′′(0)
2! x2+f′′′(0)
3! x3+· · ·
Substitute the values we found earlier:
f(x) = 1 + x+ 2x2+ 6x3+· · ·
Therefore, the Maclaurin series representation for f(x) = 1
1−xis 1 + x+
2x2+ 6x3+· · · .
2
Question 3
Question
Find the Maclaurin series representation for the function f(x) = ln(1 + x).
Solution
To find the Maclaurin series representation for f(x) = ln(1 + x), we will first
find the derivatives of the function and then determine its Maclaurin series
expansion using these derivatives.
Step 1: Find the first few derivatives of f(x).
f(x) = ln(1 + x)
f′(x) = 1
1 + x(Differentiate using chain rule)
f′′(x) = −1
(1 + x)2(Differentiate using chain rule twice)
f′′′(x) = 2
(1 + x)3(Differentiate using chain rule thrice)
Step 2: Evaluate the derivatives at x= 0.
f(0) = ln(1 + 0) = ln(1) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f′′′(0) = 2
(1 + 0)3= 2
Step 3: Write out the Maclaurin series for f(x). The Maclaurin series ex-
pansion for f(x) is given by:
f(x) = f(0) + f′(0)x+f′′ (0)x2
2! +f′′′(0)x3
3! +. . .
Plugging in the values we calculated earlier, we have:
ln(1 + x) = 0 + 1 ·x−1
2x2+2
6x3+. . .
Therefore, the Maclaurin series representation for ln(1 + x) is:
ln(1 + x) = x−1
2x2+1
3x3+. . .
3
Question 4
Question
Find the Maclaurin series for f(x) = 1
1+xand determine its radius of conver-
gence.
Solution
To find the Maclaurin series for f(x) = 1
1+x, we will use the formula for the
geometric series:
1
1−r=
∞
X
n=0
rn
where |r|<1.
Step 1: Find the derivatives of f(x).
f(x) = 1
1 + x= (1 + x)−1
f′(x) = −(1 + x)−2=−1
(1 + x)2
f′′(x) = 2(1 + x)−3=2
(1 + x)3
f′′′(x) = −6(1 + x)−4=−6
(1 + x)4
Step 2: Find the Maclaurin series by evaluating the derivatives at
x= 0.
f(0) = 1
f′(0) = −1
f′′(0) = 2
f′′′(0) = −6
Step 3: Write out the Maclaurin series. The Maclaurin series for
f(x) = 1
1+xis:
f(x) = 1 −x+x2−x3+. . . =
∞
X
n=0
(−1)nxn
Step 4: Find the radius of convergence. The radius of convergence, R,
can be found using the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
(−1)n+1xn+1
(−1)nxn
= lim
n→∞ |x|= 1
So, the radius of convergence is R= 1.
4
Question 5
Question
Find the Maclaurin series for the function f(x) = e2xcos(x).
Solution
To find the Maclaurin series for f(x) = e2xcos(x), we will first express f(x) as
a power series by finding its derivatives and evaluating them at x= 0.
Step 1: Find f′(x) and f′′ (x).
f(x) = e2xcos(x)
f′(x) = (2e2xcos(x)−e2xsin(x))
f′′(x)=4e2xcos(x)−4e2xsin(x)−2e2xsin(x)−e2xcos(x)
Step 2: Find f(0), f′(0), and f′′ (0).
f(0) = e0cos(0) = 1
f′(0) = 2e0cos(0) −e0sin(0) = 2
f′′(0) = 4e0cos(0) −4e0sin(0) −2e0sin(0) −e0cos(0) = 3
Step 3: Write the Maclaurin series for f(x).
f(0) + f′(0)x+f′′ (0)x2
2! +· · ·
= 1 + 2x+3x2
2+· · ·
Therefore, the Maclaurin series for f(x) = e2xcos(x) is 1 + 2x+3x2
2+· · · .
Question 6
Question
Find the Maclaurin series for f(x) = 1
1+x.
Solution
To find the Maclaurin series for f(x) = 1
1+x, we will first express f(x) as a
geometric series and then determine its coefficients.
Step 1: Express f(x)as a geometric series We know that the geometric
series formula is:
(1 + r+r2+r3+. . .) = 1
1−r,for |r|<1.
5
Substitute r=−xinto the formula:
1
1 + x= 1 −x+x2−x3+. . . =
∞
X
n=0
(−1)nxn.
Step 2: Determine the Maclaurin series for f(x) Therefore, the
Maclaurin series for f(x) = 1
1+xis
f(x)=1−x+x2−x3+. . . =
∞
X
n=0
(−1)nxn.
Thus, the Maclaurin series for f(x) is P∞
n=0(−1)nxn.
Question 7
Question
Find the Maclaurin series for f(x) = ln(1 + x).
Solution
To find the Maclaurin series for f(x) = ln(1 + x), we can start by finding the
derivatives of f(x).
Step 1: Find the first few derivatives of f(x).
f(x) = ln(1 + x)
f′(x) = 1
1 + x
f′′ (x) = −1
(1 + x)2
f′′′(x) = 2
(1 + x)3
f(4)(x) = −6
(1 + x)4
.
.
.
Step 2: Evaluate the derivatives at x= 0 to find the coefficients of the
6
Maclaurin series.
f(0) = ln(1 + 0) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f′′′(0) = 2
(1 + 0)3= 2
f(4)(0) = −6
(1 + 0)4=−6
.
.
.
Step 3: Write down the Maclaurin series for f(x). Since the Maclaurin series
for f(x) is given by:
f(x) = f(0) + f′(0)
1! x+f′′(0)
2! x2+f′′′(0)
3! x3+f(4)(0)
4! x4+· · · ,
we can write the Maclaurin series for f(x) = ln(1 + x) as:
f(x) = 0 + x−x2
2+2x3
3−6x4
4+· · ·
Question 8
Question
Find the Maclaurin series for the function f(x) = 1
1+xand determine the interval
of convergence.
Solution
To find the Maclaurin series for f(x) = 1
1+x, we can start by recognizing that
this function is a geometric series. The geometric series formula is given by:
f(x) =
∞
X
n=0
arn=a
1−r
where |r|<1 for convergence.
7
Step 1: Find the first few derivatives of f(x).
f(x) = 1
1 + x
f′(x) = d
dx 1
1 + x=−1
(1 + x)2
f′′(x) = d
dx −1
(1 + x)2=2
(1 + x)3
f′′′(x) = d
dx 2
(1 + x)3=−6
(1 + x)4
Step 2: Find the Maclaurin series for f(x).The Maclaurin series is
given by:
f(x) =
∞
X
n=0
f(n)(0)
n!xn
By evaluating the derivatives at x= 0, we have:
f(0) = 1
f′(0) = −1
f′′(0) = 2
f′′′(0) = −6
Therefore, the Maclaurin series for f(x) is:
f(x)=1−x+ 2x2−6x3+. . .
Step 3: Determine the interval of convergence. The radius of conver-
gence of a power series is the reciprocal of the limit of the absolute value of the
ratio of two consecutive terms. In this case, the series converges for |x|<1.
Hence, the interval of convergence of the Maclaurin series for f(x) is (−1,1).
Question 9
Question
Find the Maclaurin series for f(x) = e2xcos(3x) and determine its radius of
convergence.
Solution
To find the Maclaurin series for f(x) = e2xcos(3x), we will first find the ex-
pressions for the derivatives of f(x) and evaluate them at x= 0 to find the
coefficients of the series.
8
Step 1: Find the first few derivatives of f(x).
f(x) = e2xcos(3x)
f′(x) = (2e2x) cos(3x)−(3e2x) sin(3x)
f′′(x) = (4e2x) cos(3x)−8e2xsin(3x) + (6e2x) cos(3x)
f′′′(x) = (8e2x) cos(3x)−24e2xsin(3x)−(18e2x) sin(3x)−18e2xcos(3x)
f(4)(x) = (16e2x) cos(3x)−48e2xsin(3x)−(24e2x) sin(3x) + 54e2xcos(3x)
Step 2: Evaluate the derivatives at x= 0.
f(0) = e0cos(0) = 1
f′(0) = 2(1) cos(0) −3(1) sin(0) = 2
f′′(0) = 4(1) cos(0) −8 sin(0) + 6(1) = 10
f′′′(0) = 8(1) cos(0) −24 sin(0) −18 sin(0) −18(1) = −42
f(4)(0) = 16(1) cos(0) −48 sin(0) −24 sin(0) + 54(1) = 70
Step 3: Write the Maclaurin series using the coefficients obtained
above. The Maclaurin series for f(x) is:
f(x) = 1 + 2x+10x2
2+−42x3
6+70x4
24 +. . . = 1 + 2x+ 5x2−7x3+35
12x4+. . .
Step 4: Find the radius of convergence. Using the ratio test, we can
find the radius of convergence Rof the Maclaurin series for f(x). The ratio test
states that if limn→∞
an+1
an=L < 1, then the series converges absolutely and
the radius of convergence R=1
L. In this case, the ratio is:
lim
n→∞
an+1
an
= lim
n→∞
70
24 (xn+1)
−42
6(xn)
= lim
n→∞
35
−14x
=
35
−14
|x|=35
14|x|
For the series to converge, we need 35
14 |x|<1:
|x|<14
35 =2
5
Therefore, the radius of convergence is R=2
5.
Question 10
Question
Find the Maclaurin series representation for f(x) = 1
1+x.
9
Solution
To find the Maclaurin series representation of f(x) = 1
1+x, we can recall the
geometric series formula:
(1 + z)−1= 1 −z+z2−z3+. . . , for |z|<1.
Step 1: Rewrite the given function in the form of a geometric series. We
know that 1
1+x= (1 + (−x))−1.
Step 2: Use the geometric series formula to expand (1 + (−x))−1.
(1 + (−x))−1= 1 −(−x)+(−x)2−(−x)3+. . .
= 1 + x+x2+x3+. . .
Step 3: Therefore, the Maclaurin series representation for f(x) = 1
1+xis
P∞
n=0(−1)nxn.
Question 11
Question
Find the Maclaurin series for f(x) = e2xln(1 + x).
Solution
To find the Maclaurin series for f(x) = e2xln(1 + x), we will use the properties
of Maclaurin series and the series representation of exand ln(1 + x).
Step 1: Find the Maclaurin series for e2x.
The Maclaurin series representation of e2xis given by:
e2x=
∞
X
n=0
(2x)n
n!
Step 2: Find the Maclaurin series for ln(1 + x).
The Maclaurin series representation of ln(1 + x) is given by:
ln(1 + x) =
∞
X
n=1
(−1)n+1xn
n
Step 3: Multiply the two series together.
Multiplying the Maclaurin series for e2xby the Maclaurin series for ln(1 + x),
we get:
f(x) = e2xln(1 + x) = ∞
X
n=0
(2x)n
n!! ∞
X
n=1
(−1)n+1xn
n!
10
Step 4: Calculate the Maclaurin series for f(x).
To find the Maclaurin series for f(x) = e2xln(1 + x), we need to multiply out
the series and simplify the terms.
After multiplying and simplifying the series, we get:
f(x) =
∞
X
n=1 (−1)n+12n−1
nxn
Therefore, the Maclaurin series for f(x) = e2xln(1 + x) is:
f(x) =
∞
X
n=1 (−1)n+12n−1
nxn
Question 12
Question
Find the Maclaurin series for the function f(x) = ln(1 + x) and determine the
interval of convergence.
Solution
To find the Maclaurin series for f(x) = ln(1 + x), we will first find the derivatives
of f(x) and evaluate them at x= 0 to get the coefficients of the series.
Step 1: Find the first few derivatives of f(x).
f(x) = ln(1 + x)
f′(x) = 1
1 + x
f′′(x) = d
dx 1
1 + x=−1
(1 + x)2
f′′′(x) = d
dx −1
(1 + x)2=2
(1 + x)3
f(4)(x) = d
dx 2
(1 + x)3=−6
(1 + x)4
11
Step 2: Evaluate the derivatives at x= 0.
f(0) = ln(1 + 0) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f′′′(0) = 2
(1 + 0)3= 2
f(4)(0) = −6
(1 + 0)4=−6
Step 3: Write out the Maclaurin series. The Maclaurin series expan-
sion of a function f(x) is given by:
f(x) = f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+f(4)(0)
4! x4+. . .
Substitute the evaluated derivatives into the formula:
ln(1 + x) = 0 + 1 ·x−1·x2
2! + 2 ·x3
3! −6·x4
4! +. . .
Step 4: Simplify the series. Simplify the series by expressing the facto-
rials:
ln(1 + x) = x−x2
2+x3
3−x4
4+. . .
Step 5: Determine the interval of convergence. The interval of con-
vergence can be found by using the ratio test. The ratio test states that for a
series Pan, the series converges if the following limit is less than 1:
lim
n→∞
an+1
an
In this case, the series ln(1 + x) will converge for −1< x ≤1. Therefore, the
interval of convergence is [−1,1).
Question 13
Question
Find the Maclaurin series of the function f(x) = 1
(1+x)2.
Solution
To find the Maclaurin series of f(x) = 1
(1+x)2, we need to find the derivatives of
f(x) and evaluate them at x= 0.
12
Step 1: Find the derivatives of f(x).
f(x) = 1
(1 + x)2
f′(x) = −2(1 + x)−3(−1) = 2
(1 + x)3
f′′(x) = −3(1 + x)−4(2) = 6
(1 + x)4
f′′′(x) = −4(1 + x)−5(6) = 24
(1 + x)5
.
.
.
Step 2: Find the derivatives at x= 0.
f(0) = 1
f′(0) = 2
13= 2
f′′(0) = 6
14= 6
f′′′(0) = 24
15= 24
.
.
.
Step 3: Write down the Maclaurin series. The Maclaurin series of
f(x) = 1
(1+x)2is:
f(x) = 1 + 2x+ 6x2+ 24x3+· · · =
∞
X
n=0
anxn
where anis the n-th derivative of f(x) evaluated at x= 0.
Question 14
Question
Let f(x) = excos(x). Find the Maclaurin series for f(x), and determine the
interval of convergence for the series.
Solution
To find the Maclaurin series for f(x) = excos(x), we will first find the derivatives
of f(x) and evaluate them at x= 0 to obtain the coefficients of the Maclaurin
series. Then, we will determine the interval of convergence using the ratio test.
Step 1: Find the derivatives of f(x)
13
Let’s start by finding the first few derivatives of f(x):
f(x) = excos(x)
f′(x) = (ex)(cos(x)) + (ex)(−sin(x)) = ex(cos(x)−sin(x))
f′′(x) = (ex)(cos(x)−sin(x)) + (ex)(−sin(x)−cos(x)) = ex(−2 sin(x)) = −2exsin(x)
f′′′(x) = −2(ex)(sin(x)) + (−2ex)(cos(x)) = −2ex(sin(x) + cos(x))
Evaluating the derivatives at x= 0:
f(0) = e0cos(0) = 1
f′(0) = e0(cos(0) −sin(0)) = 1
f′′(0) = −2e0sin(0) = 0
f′′′(0) = −2e0(sin(0) + cos(0)) = −2
Step 2: Determine the Maclaurin series for f(x)
The Maclaurin series for f(x) is given by:
f(x) =
∞
X
n=0
f(n)(0)
n!xn
Thus, the Maclaurin series for f(x) = excos(x) is:
f(x) = 1 + x−2x3
3! +. . .
Step 3: Determine the interval of convergence
To find the interval of convergence, we will use the ratio test:
R= lim
n→∞
an+1
an
For our series, an=f(n)(0)
n!xn=0
n!xn= 0 for n≥2. This means the series
converges for all x.
Therefore, the interval of convergence for the Maclaurin series of f(x) =
excos(x) is (−∞,∞).
Question 15
Question
Find the Maclaurin series for the function f(x) = excos(x).
14
Solution
To find the Maclaurin series for the function f(x) = excos(x), we can use the
known Maclaurin series for exand cos(x) and multiply them together.
The Maclaurin series for exis:
ex=
∞
X
n=0
xn
n!
And the Maclaurin series for cos(x) is:
cos(x) =
∞
X
n=0
(−1)nx2n
(2n)!
Multiplying these two series together, we get:
f(x) = excos(x) = ∞
X
n=0
xn
n!! ∞
X
n=0
(−1)nx2n
(2n)!!
Expanding this product, we have:
f(x) =
∞
X
n=0 n
X
k=0
(−1)k
k!(n−k)!!xn
Therefore, the Maclaurin series for the function f(x) = excos(x) is:
f(x)=1−x+1
2x2+1
6x3−1
24x4−1
120x5+· · ·
Question 16
Question
Find the Maclaurin series for f(x) = 1
1+xand determine the interval of conver-
gence.
Solution
To find the Maclaurin series for f(x) = 1
1+x, we’ll first find the derivatives of
f(x) and then determine the pattern to construct the series.
15
Step 1: Find the derivatives of f(x)
f(x) = 1
1 + x
f′(x) = −(1 + x)−2=−1
(1 + x)2
f′′(x) = 2(1 + x)−3=2
(1 + x)3
f′′′(x) = −6(1 + x)−4=−6
(1 + x)4
.
.
.
Step 2: Find the pattern The derivatives of f(x) suggest the following
pattern:
f(n)(0) = n!
(1 + x)n+1
Step 3: Write the Maclaurin series The Maclaurin series for f(x) is
given by:
f(x) = f(0) + f′(0)
1! x+f′′(0)
2! x2+f′′′(0)
3! x3+· · · =
∞
X
n=0
f(n)(0)
n!xn
Substitute f(n)(0) = n!
(1+x)n+1 into the series:
f(x)=1−x+x2−x3+· · · + (−1)nxn+· · ·
Step 4: Determine the interval of convergence To find the interval
of convergence, we will use the ratio test. Let an= (−1)n. Applying the ratio
test:
lim
n→∞
an+1
an
= lim
n→∞
(−1)n+1
(−1)n
= 1
Since the limit is equal to 1, the interval of convergence is (−1,1). Thus, the
Maclaurin series for f(x) = 1
1+xis:
f(x) = 1 −x+x2−x3+· · · + (−1)nxn+· · · for −1<x<1
Question 17
Question
Find the Maclaurin series representation for f(x) = e2xcos(3x).
16
Solution
To find the Maclaurin series representation for f(x) = e2xcos(3x), we will use
the known Maclaurin series for e2xand cos(3x) and then multiply them together.
Step 1: Find the Maclaurin series for e2x.The Maclaurin series for ex
is given by:
ex=
∞
X
n=0
xn
n!
Substitute xwith 2xto find the Maclaurin series for e2x:
e2x=
∞
X
n=0
(2x)n
n!=
∞
X
n=0
2nxn
n!
Step 2: Find the Maclaurin series for cos(3x).The Maclaurin series
for cos(x) is given by:
cos(x) =
∞
X
n=0
(−1)nx2n
(2n)!
Substitute xwith 3xto find the Maclaurin series for cos(3x):
cos(3x) =
∞
X
n=0
(−1)n(3x)2n
(2n)! =
∞
X
n=0
(−1)n32nx2n
(2n)!
Step 3: Multiply the series together. To find the Maclaurin series
representation for f(x) = e2xcos(3x), multiply the two series we found in Step
1 and Step 2:
f(x) = e2xcos(3x) = ∞
X
n=0
2nxn
n!! ∞
X
n=0
(−1)n32nx2n
(2n)! !
f(x) =
∞
X
n=0 n
X
k=0
2k(−1)n−k32(n−k)xn
k!(n−k)!(2(n−k))!!
Therefore, the Maclaurin series representation for f(x) = e2xcos(3x) is:
f(x) =
∞
X
n=0 n
X
k=0
2k(−1)n−k32(n−k)xn
k!(n−k)!(2(n−k))!!
Question 18
Question
Find the Maclaurin series representation for the function f(x) = 1
1+x2.
17
Solution
The Maclaurin series for a function f(x) is given by the formula:
f(x) =
∞
X
n=0
f(n)(0)
n!xn
where f(n)(0) denotes the n-th derivative of f(x) evaluated at x= 0.
Step 1: Find the derivatives of f(x)
Since f(x) = 1
1+x2, we first need to find the derivatives of f(x):
f′(x) = d
dx 1
1 + x2=−2x
(1 + x2)2
f′′(x) = d2
dx21
1 + x2=2(3x2−1)
(1 + x2)3
Step 2: Find f(n)(0)
To find f(n)(0) for the Maclaurin series, we evaluate the derivatives of f(x)
at x= 0:
f(0) = 1
1+02= 1
f′(0) = −2·0
(1 + 02)2= 0
f′′(0) = 2(3 ·02−1)
(1 + 02)3=−2
Step 3: Write the Maclaurin series
Substitute the values of f(n)(0) into the Maclaurin series formula:
f(x) =
∞
X
n=0
f(n)(0)
n!xn=f(0) + f′(0)x+f′′ (0)
2! x2+. . .
= 1 + 0 ·x−2
2!x2= 1 −x2
Therefore, the Maclaurin series representation for f(x) = 1
1+x2is 1 −x2.
Question 19
Question
Find the Maclaurin series representation for the function f(x) = cos2(x).
18
Solution
To find the Maclaurin series representation for f(x) = cos2(x), we will start by
finding the Maclaurin series for cos(x). Then we will use this to find the series
for cos2(x).
Step 1: Find the Maclaurin series for cos(x).The Maclaurin series for
cos(x) is:
cos(x) =
∞
X
n=0
(−1)nx2n
(2n)!
Step 2: Find the Maclaurin series for cos2(x).To find the Maclaurin
series for cos2(x), we square the Maclaurin series for cos(x):
cos2(x) = ∞
X
n=0
(−1)nx2n
(2n)!!2
Expanding this out gives:
cos2(x) =
∞
X
n=0
∞
X
k=0
(−1)nx2n
(2n)! ·(−1)kx2k
(2k)!
=
∞
X
n=0
∞
X
k=0
(−1)n+kx2n
(2n)! ·x2k
(2k)!
Simplifying this further, we have:
cos2(x) =
∞
X
n=0
∞
X
k=0
(−1)n+kx2(n+k)
(2n)!(2k)!
This series can be rearranged to give the Maclaurin series representation for
cos2(x):
cos2(x) =
∞
X
n=0 n
X
k=0
(−1)kx2(n−k)
(2(n−k))! ·x2k
(2k)!!
Therefore, the Maclaurin series representation for f(x) = cos2(x) is:
∞
X
n=0 n
X
k=0
(−1)kx2(n−k)
(2(n−k))! ·x2k
(2k)!!
Question 20
Question
Find the Maclaurin series for the function f(x) = 1
1−x.
19
Solution
To find the Maclaurin series for f(x) = 1
1−x, we will first find the derivatives of
f(x) and then evaluate them at x= 0 to find the coefficients of the series.
Step 1: Find the derivatives of f(x)
f(x) = 1
1−x
f′(x) = d
dx 1
1−x
=1
(1 −x)2
f′′(x) = d
dx 1
(1 −x)2
=2
(1 −x)3
f′′′(x) = d
dx 2
(1 −x)3
=6
(1 −x)4
.
.
.
Step 2: Evaluate derivatives at x= 0 Now, let’s find the value of the
derivatives at x= 0:
f(0) = 1
1−0= 1
f′(0) = 1
(1 −0)2= 1
f′′(0) = 2
(1 −0)3= 2
f′′′(0) = 6
(1 −0)4= 6
.
.
.
Step 3: Write the Maclaurin series The Maclaurin series for f(x) is
given by:
f(x) = f(0) + f′(0)x+f′′ (0)x2
2! +f′′′(0)x3
3! +· · ·
Substitute the values we found at x= 0:
f(x) = 1 + x+2x2
2! +6x3
3! +· · ·
20
Simplify the terms:
f(x) = 1 + x+x2+x3+· · · =
∞
X
n=0
xn
Therefore, the Maclaurin series for f(x) = 1
1−xis P∞
n=0 xn.
Question 21
Question
Find the Maclaurin series for the function f(x) = 1
1+xand determine its interval
of convergence.
Solution
To find the Maclaurin series for f(x) = 1
1+x, we will first find the derivatives of
f(x) and evaluate them at x= 0 to get the coefficients of the series.
Step 1: Find the derivatives of f(x).
f(x) = 1
1 + x
f′(x) = d
dx 1
1 + x=−(1 + x)−2=−(1 + x)−2
f′′(x) = d
dx −(1 + x)−2= 2(1 + x)−3= 2(1 + x)−3
f′′′(x) = d
dx 2(1 + x)−3=−6(1 + x)−4=−6(1 + x)−4
Now, we will evaluate these derivatives at x= 0:
f(0) = 1
f′(0) = −1
f′′(0) = 2
f′′′(0) = −6
Step 2: Write out the Maclaurin series. The Maclaurin series for f(x)
is given by:
f(x) = f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+· · ·
Substitute the values we found earlier:
f(x)=1−x+2
2!x2−6
3!x3+· · ·
21
Simplify:
f(x)=1−x+x2−2x3+· · ·
Step 3: Determine the interval of convergence. To find the interval
of convergence, we will use the ratio test. The ratio test states that if L=
limn→∞
an+1
an, then the series converges if L < 1.
L= lim
n→∞
(−1)n+1
(−1)n
= 1
Since L= 1, the ratio test is inconclusive. To determine the interval of
convergence, we need to check the endpoints.
For x=−1, the series becomes the harmonic series P∞
n=0(−1)nwhich is
alternating. It converges by the Alternating Series Test.
For x= 1, the series becomes the harmonic series P∞
n=0 1 which is a diver-
gent p-series.
Therefore, the interval of convergence is [−1,1).
Question 22
Question
Find the Maclaurin series for ln(1 + x) by differentiating term by term.
Solution
To find the Maclaurin series for ln(1 + x), we will differentiate the series for
ln(1 + x) term by term. Recall that the Maclaurin series for ln(1 + x) is given
by
ln(1 + x) = x−x2
2+x3
3−x4
4+· · · for −1< x ≤1.
Step 1: Find the first derivative of ln(1 + x):
d
dx (ln(1 + x)) = 1
1 + x.
Step 2: Find the second derivative of ln(1 + x):
d2
dx2(ln(1 + x)) = −1
(1 + x)2.
Step 3: Find the third derivative of ln(1 + x):
d3
dx3(ln(1 + x)) = 2
(1 + x)3.
22
Step 4: Find the fourth derivative of ln(1 + x):
d4
dx4(ln(1 + x)) = −6
(1 + x)4.
Continuing this pattern, we can see that the nth derivative of ln(1 + x) is
given by
dn
dxn(ln(1 + x)) = (−1)n−1·(n−1)!
(1 + x)nfor n≥1.
Therefore, the Maclaurin series for ln(1 + x) is
ln(1 + x) = x−x2
2+x3
3−x4
4+· · · =
∞
X
n=1
(−1)n−1·xn
n.
Question 23
Question
Find the Maclaurin series for f(x) = ln(1 + x) by finding its derivatives and
evaluating them at x= 0.
Solution
To find the Maclaurin series for f(x) = ln(1 + x), we will first find its derivatives
up to a suitable order and evaluate them at x= 0.
Step 1: Find the first derivative of f(x)
f(x) = ln(1 + x)
d
dx f(x) = d
dx ln(1 + x)
Using the chain rule, we get
d
dx ln(1 + x) = 1
1 + x= (1 + x)−1
Step 2: Evaluate the first derivative at x= 0
d
dx f(x)x=0
= (1 + 0)−1= 1
Step 3: Find the second derivative of f(x)
d2
dx2f(x) = d
dx (1 + x)−1
Using the power rule, we get
d2
dx2f(x) = −1(1 + x)−2
23
Step 4: Evaluate the second derivative at x= 0
d2
dx2f(x)x=0
=−1(1 + 0)−2=−1
Step 5: Continue finding higher derivatives and evaluating them at x= 0
d3
dx3f(x) = 2(1 + x)−3,d3
dx3f(x)x=0
= 2
d4
dx4f(x) = −6(1 + x)−4,d4
dx4f(x)x=0
=−6
Step 6: Generalize the pattern and write the Maclaurin series for f(x)
f(x) = ln(1 + x) =
∞
X
n=0
(−1)nxn+1
n+ 1
Thus, the Maclaurin series for ln(1 + x) is given by:
ln(1 + x) = x−x2
2+x3
3−x4
4+. . .
Question 24
Question
Find the Maclaurin series for the function f(x) = cos(sin x).
Solution
To find the Maclaurin series for f(x) = cos(sin x), we can make use of the
composite function property of Maclaurin series.
Step 1: Find the derivatives First, we need to find the derivatives of
f(x). Let’s start by finding the first few derivatives:
f(x) = cos(sin x)
f′(x) = −sin(sin x) cos x
f′′(x) = (−cos(sin x) cos x)(cos(sin x) + sin(sin x))
=−cos(sin x)2cos x−sin(sin x) cos(sin x)
f′′′(x)=(−2 cos(sin x) sin(sin x) cos x−cos2(sin x) cos x+ sin2(sin x) cos(sin x)
Step 2: Evaluate the derivatives at 0 Next, we evaluate the derivatives
at x= 0:
f(0) = cos(sin 0) = cos(0) = 1
f′(0) = −sin(sin 0) cos 0 = 0
f′′(0) = −cos(sin 0)2cos 0 −sin(sin 0) cos(sin 0) = −1
f′′′(0) = −cos2(sin 0) cos 0 −sin2(sin 0) cos(sin 0) = 0
24
Step 3: Write out the Maclaurin series The Maclaurin series for f(x)
is given by:
f(x) = f(0) + f′(0)
1! x+f′′(0)
2! x2+f′′′(0)
3! x3+. . .
Substitute the values we found earlier:
cos(sin x)=1−x2
2+. . .
Question 25
Question
Find the Maclaurin series representation for f(x) = 1
1+x2by differentiating the
function term by term.
Solution
To find the Maclaurin series representation for f(x) = 1
1+x2, we need to express
f(x) in terms of a power series. We can do this by differentiating the function
term by term.
Step 1: Find the derivatives of f(x).
f(x) = 1
1 + x2
f′(x) = −2x
(1 + x2)2
f′′(x) = −21
(1 + x2)2+4x2
(1 + x2)3
f′′′(x) = −2−2x
(1 + x2)2+12x
(1 + x2)3+12x
(1 + x2)3+48x3
(1 + x2)4
Step 2: Evaluate the derivatives at x= 0 to find the coefficients of the
Maclaurin series.
f(0) = 1
f′(0) = 0
f′′(0) = −2
f′′′(0) = 0
Step 3: Write out the Maclaurin series. The Maclaurin series representation
for f(x) = 1
1+x2is:
f(x)=1−2x2+12x4
2! −48x6
3! +· · · =
∞
X
n=0
(−1)n2nx2n
25
Question 26
Question
Find the Maclaurin series for f(x) = 1
1+x2and state the interval of convergence.
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we’ll use the geometric series
formula: 1
1−r= 1 + r+r2+r3+· · · ,for |r|<1.
Step 1: Find the Maclaurin series for f(x) Given f(x) = 1
1+x2, we can
express it as:
f(x) = 1
1−(−x2).
Comparing this expression with the geometric series formula, we can see that
r=−x2. Therefore, the Maclaurin series for f(x) is:
∞
X
n=0
(−1)nx2n.
Step 2: Determine the interval of convergence To find the interval of
convergence, we use the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
(−1)n+1x2(n+1)
(−1)nx2n
.
Simplifying, we get:
lim
n→∞ x2=|x2|.
For the series to converge, we need |x2|<1, which gives us −1<x<1.
Therefore, the interval of convergence is (−1,1).
Question 27
Question
Find the Maclaurin series for the function f(x) = ln(1 + x) and determine the
interval of convergence.
Solution
To find the Maclaurin series for f(x) = ln(1 + x), we will use the formula for
the Maclaurin series of ln(1 + x), which is:
ln(1 + x) = x−x2
2+x3
3−x4
4+. . . =
∞
X
n=1
(−1)n−1xn
n
26
Step 1: Identify the Maclaurin series of ln(1 + x).
ln(1 + x) =
∞
X
n=1
(−1)n−1xn
n
Step 2: Determine the interval of convergence.
The Maclaurin series P∞
n=1(−1)n−1xn
nhas interval of convergence −1<
x≤1.
Therefore, the Maclaurin series for f(x) = ln(1 + x) is P∞
n=1(−1)n−1xn
n
with an interval of convergence −1< x ≤1.
Question 28
Question
Find the Maclaurin series for f(x) = 1
(1−x)2.
Solution
To find the Maclaurin series for f(x) = 1
(1−x)2, we use the formula for the
Maclaurin series of a function:
f(x) =
∞
X
n=0
f(n)(0)
n!xn
where f(n)(0) denotes the n-th derivative of f(x) evaluated at x= 0.
Step 1: Find the derivatives of f(x) up to the second derivative.
f(x) = 1
(1 −x)2
f′(x) = 2
(1 −x)3
f′′(x) = 2·3
(1 −x)4
Step 2: Evaluate these derivatives at x= 0.
f(0) = 1
(1 −0)2= 1
f′(0) = 2
(1 −0)3= 2
f′′(0) = 2·3
(1 −0)4= 6
Step 3: Plug these values into the Maclaurin series formula.
27
f(x) =
∞
X
n=0
f(n)(0)
n!xn
f(x) = 1 + 2x+ 6x2+
∞
X
n=3
f(n)(0)
n!xn
So, the Maclaurin series for f(x) = 1
(1−x)2is 1 + 2x+ 6x2+P∞
n=3
f(n)(0)
n!xn.
Question 29
Question
Find the Maclaurin series for the function f(x) = ex2.
Solution
To find the Maclaurin series for ex2, we will use the formula for the Maclaurin
series expansion of a function centered at x= 0:
f(x) =
∞
X
n=0
f(n)(0)
n!xn
Step 1: Find the derivatives of f(x).
f(x) = ex2
f′(x)=2xex2
f′′(x) = (2 + 4x2)ex2
f′′′(x) = (4x+ 8x3)ex2
Step 2: Evaluate the derivatives at x= 0.
f(0) = e0= 1
f′(0) = 2(0)e0= 0
f′′(0) = (2 + 4(02))e0= 2
f′′′(0) = (4(0) + 8(03))e0= 0
Step 3: Write down the Maclaurin series. The Maclaurin series for
ex2is:
f(x) = 1 + 0x+2
2!x2+ 0x3+. . . = 1 + x2
Therefore, the Maclaurin series for ex2is 1 + x2.
28
Question 30
Question
Find the Maclaurin series for f(x) = 1
1+x2.
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we will first find the derivatives
of f(x) at x= 0 and then use them to construct the Maclaurin series.
Step 1: Find the derivatives of f(x)
The function f(x) = 1
1+x2can be rewritten as f(x) = (1 + x2)−1. We can
differentiate f(x) term by term using the formula for the derivative of a power
function.
f′(x) = −(1 + x2)−2·2x=−2x(1 + x2)−2
f′′(x) = −2(1 + x2)−2−2(−2x)(−2)(1 + x2)−3= 2(2x2−1)(1 + x2)−3
f′′′(x) = 2(2x2−1)(1 + x2)−3−6x(1 + x2)−3= 2(1 −10x2+ 2x4)(1 + x2)−4
Step 2: Find f(0) and evaluate f′(0),f′′ (0),f′′′ (0)
f(0) = 1
1+02= 1
f′(0) = 0
f′′(0) = 2
f′′′(0) = 2
Step 3: Write out the Maclaurin series
The Maclaurin series for f(x) is given by:
f(x) = f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+· · ·
Substitute the values we found earlier:
f(x) = 1 + 0 ·x+2
2!x2+2
3!x3+· · ·
Simplifying:
f(x) = 1 + x2+1
3x3+· · ·
Therefore, the Maclaurin series for f(x) = 1
1+x2is 1 + x2+1
3x3+· · · .
29
Question 2
Question
Find the Maclaurin series representation for the function f(x) = 1
1−x.
Solution
To find the Maclaurin series representation for f(x) = 1
1−x, we will first find
the derivatives of f(x) at x= 0 to determine the coefficients of the series.
Step 1: Find the derivatives of f(x)
f(x) = 1
1−x
f′(x) = d
dx 1
1−x=1
(1 −x)2
f′′(x) = d
dx 1
(1 −x)2=2
(1 −x)3
f′′′(x) = d
dx 2
(1 −x)3=6
(1 −x)4
.
.
.
Step 2: Find the values of the derivatives at x= 0 Evaluate the
derivatives at x= 0 to find the coefficients of the Maclaurin series.
f(0) = 1
f′(0) = 1
f′′(0) = 2
f′′′(0) = 6
.
.
.
Step 3: Write out the Maclaurin series The Maclaurin series for f(x) =
1
1−xis given by:
f(x) = f(0) + f′(0)
1! x+f′′(0)
2! x2+f′′′(0)
3! x3+· · ·
Substitute the values we found earlier:
f(x) = 1 + x+ 2x2+ 6x3+· · ·
Therefore, the Maclaurin series representation for f(x) = 1
1−xis 1 + x+
2x2+ 6x3+· · · .
2
Question 3
Question
Find the Maclaurin series representation for the function f(x) = ln(1 + x).
Solution
To find the Maclaurin series representation for f(x) = ln(1 + x), we will first
find the derivatives of the function and then determine its Maclaurin series
expansion using these derivatives.
Step 1: Find the first few derivatives of f(x).
f(x) = ln(1 + x)
f′(x) = 1
1 + x(Differentiate using chain rule)
f′′(x) = −1
(1 + x)2(Differentiate using chain rule twice)
f′′′(x) = 2
(1 + x)3(Differentiate using chain rule thrice)
Step 2: Evaluate the derivatives at x= 0.
f(0) = ln(1 + 0) = ln(1) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f′′′(0) = 2
(1 + 0)3= 2
Step 3: Write out the Maclaurin series for f(x). The Maclaurin series ex-
pansion for f(x) is given by:
f(x) = f(0) + f′(0)x+f′′ (0)x2
2! +f′′′(0)x3
3! +. . .
Plugging in the values we calculated earlier, we have:
ln(1 + x) = 0 + 1 ·x−1
2x2+2
6x3+. . .
Therefore, the Maclaurin series representation for ln(1 + x) is:
ln(1 + x) = x−1
2x2+1
3x3+. . .
3
Question 4
Question
Find the Maclaurin series for f(x) = 1
1+xand determine its radius of conver-
gence.
Solution
To find the Maclaurin series for f(x) = 1
1+x, we will use the formula for the
geometric series:
1
1−r=
∞
X
n=0
rn
where |r|<1.
Step 1: Find the derivatives of f(x).
f(x) = 1
1 + x= (1 + x)−1
f′(x) = −(1 + x)−2=−1
(1 + x)2
f′′(x) = 2(1 + x)−3=2
(1 + x)3
f′′′(x) = −6(1 + x)−4=−6
(1 + x)4
Step 2: Find the Maclaurin series by evaluating the derivatives at
x= 0.
f(0) = 1
f′(0) = −1
f′′(0) = 2
f′′′(0) = −6
Step 3: Write out the Maclaurin series. The Maclaurin series for
f(x) = 1
1+xis:
f(x) = 1 −x+x2−x3+. . . =
∞
X
n=0
(−1)nxn
Step 4: Find the radius of convergence. The radius of convergence, R,
can be found using the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
(−1)n+1xn+1
(−1)nxn
= lim
n→∞ |x|= 1
So, the radius of convergence is R= 1.
4
Question 5
Question
Find the Maclaurin series for the function f(x) = e2xcos(x).
Solution
To find the Maclaurin series for f(x) = e2xcos(x), we will first express f(x) as
a power series by finding its derivatives and evaluating them at x= 0.
Step 1: Find f′(x) and f′′ (x).
f(x) = e2xcos(x)
f′(x) = (2e2xcos(x)−e2xsin(x))
f′′(x)=4e2xcos(x)−4e2xsin(x)−2e2xsin(x)−e2xcos(x)
Step 2: Find f(0), f′(0), and f′′ (0).
f(0) = e0cos(0) = 1
f′(0) = 2e0cos(0) −e0sin(0) = 2
f′′(0) = 4e0cos(0) −4e0sin(0) −2e0sin(0) −e0cos(0) = 3
Step 3: Write the Maclaurin series for f(x).
f(0) + f′(0)x+f′′ (0)x2
2! +· · ·
= 1 + 2x+3x2
2+· · ·
Therefore, the Maclaurin series for f(x) = e2xcos(x) is 1 + 2x+3x2
2+· · · .
Question 6
Question
Find the Maclaurin series for f(x) = 1
1+x.
Solution
To find the Maclaurin series for f(x) = 1
1+x, we will first express f(x) as a
geometric series and then determine its coefficients.
Step 1: Express f(x)as a geometric series We know that the geometric
series formula is:
(1 + r+r2+r3+. . .) = 1
1−r,for |r|<1.
5
Substitute r=−xinto the formula:
1
1 + x= 1 −x+x2−x3+. . . =
∞
X
n=0
(−1)nxn.
Step 2: Determine the Maclaurin series for f(x) Therefore, the
Maclaurin series for f(x) = 1
1+xis
f(x)=1−x+x2−x3+. . . =
∞
X
n=0
(−1)nxn.
Thus, the Maclaurin series for f(x) is P∞
n=0(−1)nxn.
Question 7
Question
Find the Maclaurin series for f(x) = ln(1 + x).
Solution
To find the Maclaurin series for f(x) = ln(1 + x), we can start by finding the
derivatives of f(x).
Step 1: Find the first few derivatives of f(x).
f(x) = ln(1 + x)
f′(x) = 1
1 + x
f′′ (x) = −1
(1 + x)2
f′′′(x) = 2
(1 + x)3
f(4)(x) = −6
(1 + x)4
.
.
.
Step 2: Evaluate the derivatives at x= 0 to find the coefficients of the
6
Maclaurin series.
f(0) = ln(1 + 0) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f′′′(0) = 2
(1 + 0)3= 2
f(4)(0) = −6
(1 + 0)4=−6
.
.
.
Step 3: Write down the Maclaurin series for f(x). Since the Maclaurin series
for f(x) is given by:
f(x) = f(0) + f′(0)
1! x+f′′(0)
2! x2+f′′′(0)
3! x3+f(4)(0)
4! x4+· · · ,
we can write the Maclaurin series for f(x) = ln(1 + x) as:
f(x) = 0 + x−x2
2+2x3
3−6x4
4+· · ·
Question 8
Question
Find the Maclaurin series for the function f(x) = 1
1+xand determine the interval
of convergence.
Solution
To find the Maclaurin series for f(x) = 1
1+x, we can start by recognizing that
this function is a geometric series. The geometric series formula is given by:
f(x) =
∞
X
n=0
arn=a
1−r
where |r|<1 for convergence.
7
Step 1: Find the first few derivatives of f(x).
f(x) = 1
1 + x
f′(x) = d
dx 1
1 + x=−1
(1 + x)2
f′′(x) = d
dx −1
(1 + x)2=2
(1 + x)3
f′′′(x) = d
dx 2
(1 + x)3=−6
(1 + x)4
Step 2: Find the Maclaurin series for f(x).The Maclaurin series is
given by:
f(x) =
∞
X
n=0
f(n)(0)
n!xn
By evaluating the derivatives at x= 0, we have:
f(0) = 1
f′(0) = −1
f′′(0) = 2
f′′′(0) = −6
Therefore, the Maclaurin series for f(x) is:
f(x)=1−x+ 2x2−6x3+. . .
Step 3: Determine the interval of convergence. The radius of conver-
gence of a power series is the reciprocal of the limit of the absolute value of the
ratio of two consecutive terms. In this case, the series converges for |x|<1.
Hence, the interval of convergence of the Maclaurin series for f(x) is (−1,1).
Question 9
Question
Find the Maclaurin series for f(x) = e2xcos(3x) and determine its radius of
convergence.
Solution
To find the Maclaurin series for f(x) = e2xcos(3x), we will first find the ex-
pressions for the derivatives of f(x) and evaluate them at x= 0 to find the
coefficients of the series.
8
Step 1: Find the first few derivatives of f(x).
f(x) = e2xcos(3x)
f′(x) = (2e2x) cos(3x)−(3e2x) sin(3x)
f′′(x) = (4e2x) cos(3x)−8e2xsin(3x) + (6e2x) cos(3x)
f′′′(x) = (8e2x) cos(3x)−24e2xsin(3x)−(18e2x) sin(3x)−18e2xcos(3x)
f(4)(x) = (16e2x) cos(3x)−48e2xsin(3x)−(24e2x) sin(3x) + 54e2xcos(3x)
Step 2: Evaluate the derivatives at x= 0.
f(0) = e0cos(0) = 1
f′(0) = 2(1) cos(0) −3(1) sin(0) = 2
f′′(0) = 4(1) cos(0) −8 sin(0) + 6(1) = 10
f′′′(0) = 8(1) cos(0) −24 sin(0) −18 sin(0) −18(1) = −42
f(4)(0) = 16(1) cos(0) −48 sin(0) −24 sin(0) + 54(1) = 70
Step 3: Write the Maclaurin series using the coefficients obtained
above. The Maclaurin series for f(x) is:
f(x) = 1 + 2x+10x2
2+−42x3
6+70x4
24 +. . . = 1 + 2x+ 5x2−7x3+35
12x4+. . .
Step 4: Find the radius of convergence. Using the ratio test, we can
find the radius of convergence Rof the Maclaurin series for f(x). The ratio test
states that if limn→∞
an+1
an=L < 1, then the series converges absolutely and
the radius of convergence R=1
L. In this case, the ratio is:
lim
n→∞
an+1
an
= lim
n→∞
70
24 (xn+1)
−42
6(xn)
= lim
n→∞
35
−14x
=
35
−14
|x|=35
14|x|
For the series to converge, we need 35
14 |x|<1:
|x|<14
35 =2
5
Therefore, the radius of convergence is R=2
5.
Question 10
Question
Find the Maclaurin series representation for f(x) = 1
1+x.
9
Solution
To find the Maclaurin series representation of f(x) = 1
1+x, we can recall the
geometric series formula:
(1 + z)−1= 1 −z+z2−z3+. . . , for |z|<1.
Step 1: Rewrite the given function in the form of a geometric series. We
know that 1
1+x= (1 + (−x))−1.
Step 2: Use the geometric series formula to expand (1 + (−x))−1.
(1 + (−x))−1= 1 −(−x)+(−x)2−(−x)3+. . .
= 1 + x+x2+x3+. . .
Step 3: Therefore, the Maclaurin series representation for f(x) = 1
1+xis
P∞
n=0(−1)nxn.
Question 11
Question
Find the Maclaurin series for f(x) = e2xln(1 + x).
Solution
To find the Maclaurin series for f(x) = e2xln(1 + x), we will use the properties
of Maclaurin series and the series representation of exand ln(1 + x).
Step 1: Find the Maclaurin series for e2x.
The Maclaurin series representation of e2xis given by:
e2x=
∞
X
n=0
(2x)n
n!
Step 2: Find the Maclaurin series for ln(1 + x).
The Maclaurin series representation of ln(1 + x) is given by:
ln(1 + x) =
∞
X
n=1
(−1)n+1xn
n
Step 3: Multiply the two series together.
Multiplying the Maclaurin series for e2xby the Maclaurin series for ln(1 + x),
we get:
f(x) = e2xln(1 + x) = ∞
X
n=0
(2x)n
n!! ∞
X
n=1
(−1)n+1xn
n!
10
Step 4: Calculate the Maclaurin series for f(x).
To find the Maclaurin series for f(x) = e2xln(1 + x), we need to multiply out
the series and simplify the terms.
After multiplying and simplifying the series, we get:
f(x) =
∞
X
n=1 (−1)n+12n−1
nxn
Therefore, the Maclaurin series for f(x) = e2xln(1 + x) is:
f(x) =
∞
X
n=1 (−1)n+12n−1
nxn
Question 12
Question
Find the Maclaurin series for the function f(x) = ln(1 + x) and determine the
interval of convergence.
Solution
To find the Maclaurin series for f(x) = ln(1 + x), we will first find the derivatives
of f(x) and evaluate them at x= 0 to get the coefficients of the series.
Step 1: Find the first few derivatives of f(x).
f(x) = ln(1 + x)
f′(x) = 1
1 + x
f′′(x) = d
dx 1
1 + x=−1
(1 + x)2
f′′′(x) = d
dx −1
(1 + x)2=2
(1 + x)3
f(4)(x) = d
dx 2
(1 + x)3=−6
(1 + x)4
11
Step 2: Evaluate the derivatives at x= 0.
f(0) = ln(1 + 0) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f′′′(0) = 2
(1 + 0)3= 2
f(4)(0) = −6
(1 + 0)4=−6
Step 3: Write out the Maclaurin series. The Maclaurin series expan-
sion of a function f(x) is given by:
f(x) = f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+f(4)(0)
4! x4+. . .
Substitute the evaluated derivatives into the formula:
ln(1 + x) = 0 + 1 ·x−1·x2
2! + 2 ·x3
3! −6·x4
4! +. . .
Step 4: Simplify the series. Simplify the series by expressing the facto-
rials:
ln(1 + x) = x−x2
2+x3
3−x4
4+. . .
Step 5: Determine the interval of convergence. The interval of con-
vergence can be found by using the ratio test. The ratio test states that for a
series Pan, the series converges if the following limit is less than 1:
lim
n→∞
an+1
an
In this case, the series ln(1 + x) will converge for −1< x ≤1. Therefore, the
interval of convergence is [−1,1).
Question 13
Question
Find the Maclaurin series of the function f(x) = 1
(1+x)2.
Solution
To find the Maclaurin series of f(x) = 1
(1+x)2, we need to find the derivatives of
f(x) and evaluate them at x= 0.
12
Step 1: Find the derivatives of f(x).
f(x) = 1
(1 + x)2
f′(x) = −2(1 + x)−3(−1) = 2
(1 + x)3
f′′(x) = −3(1 + x)−4(2) = 6
(1 + x)4
f′′′(x) = −4(1 + x)−5(6) = 24
(1 + x)5
.
.
.
Step 2: Find the derivatives at x= 0.
f(0) = 1
f′(0) = 2
13= 2
f′′(0) = 6
14= 6
f′′′(0) = 24
15= 24
.
.
.
Step 3: Write down the Maclaurin series. The Maclaurin series of
f(x) = 1
(1+x)2is:
f(x) = 1 + 2x+ 6x2+ 24x3+· · · =
∞
X
n=0
anxn
where anis the n-th derivative of f(x) evaluated at x= 0.
Question 14
Question
Let f(x) = excos(x). Find the Maclaurin series for f(x), and determine the
interval of convergence for the series.
Solution
To find the Maclaurin series for f(x) = excos(x), we will first find the derivatives
of f(x) and evaluate them at x= 0 to obtain the coefficients of the Maclaurin
series. Then, we will determine the interval of convergence using the ratio test.
Step 1: Find the derivatives of f(x)
13
Let’s start by finding the first few derivatives of f(x):
f(x) = excos(x)
f′(x) = (ex)(cos(x)) + (ex)(−sin(x)) = ex(cos(x)−sin(x))
f′′(x) = (ex)(cos(x)−sin(x)) + (ex)(−sin(x)−cos(x)) = ex(−2 sin(x)) = −2exsin(x)
f′′′(x) = −2(ex)(sin(x)) + (−2ex)(cos(x)) = −2ex(sin(x) + cos(x))
Evaluating the derivatives at x= 0:
f(0) = e0cos(0) = 1
f′(0) = e0(cos(0) −sin(0)) = 1
f′′(0) = −2e0sin(0) = 0
f′′′(0) = −2e0(sin(0) + cos(0)) = −2
Step 2: Determine the Maclaurin series for f(x)
The Maclaurin series for f(x) is given by:
f(x) =
∞
X
n=0
f(n)(0)
n!xn
Thus, the Maclaurin series for f(x) = excos(x) is:
f(x) = 1 + x−2x3
3! +. . .
Step 3: Determine the interval of convergence
To find the interval of convergence, we will use the ratio test:
R= lim
n→∞
an+1
an
For our series, an=f(n)(0)
n!xn=0
n!xn= 0 for n≥2. This means the series
converges for all x.
Therefore, the interval of convergence for the Maclaurin series of f(x) =
excos(x) is (−∞,∞).
Question 15
Question
Find the Maclaurin series for the function f(x) = excos(x).
14
Solution
To find the Maclaurin series for the function f(x) = excos(x), we can use the
known Maclaurin series for exand cos(x) and multiply them together.
The Maclaurin series for exis:
ex=
∞
X
n=0
xn
n!
And the Maclaurin series for cos(x) is:
cos(x) =
∞
X
n=0
(−1)nx2n
(2n)!
Multiplying these two series together, we get:
f(x) = excos(x) = ∞
X
n=0
xn
n!! ∞
X
n=0
(−1)nx2n
(2n)!!
Expanding this product, we have:
f(x) =
∞
X
n=0 n
X
k=0
(−1)k
k!(n−k)!!xn
Therefore, the Maclaurin series for the function f(x) = excos(x) is:
f(x)=1−x+1
2x2+1
6x3−1
24x4−1
120x5+· · ·
Question 16
Question
Find the Maclaurin series for f(x) = 1
1+xand determine the interval of conver-
gence.
Solution
To find the Maclaurin series for f(x) = 1
1+x, we’ll first find the derivatives of
f(x) and then determine the pattern to construct the series.
15
Step 1: Find the derivatives of f(x)
f(x) = 1
1 + x
f′(x) = −(1 + x)−2=−1
(1 + x)2
f′′(x) = 2(1 + x)−3=2
(1 + x)3
f′′′(x) = −6(1 + x)−4=−6
(1 + x)4
.
.
.
Step 2: Find the pattern The derivatives of f(x) suggest the following
pattern:
f(n)(0) = n!
(1 + x)n+1
Step 3: Write the Maclaurin series The Maclaurin series for f(x) is
given by:
f(x) = f(0) + f′(0)
1! x+f′′(0)
2! x2+f′′′(0)
3! x3+· · · =
∞
X
n=0
f(n)(0)
n!xn
Substitute f(n)(0) = n!
(1+x)n+1 into the series:
f(x)=1−x+x2−x3+· · · + (−1)nxn+· · ·
Step 4: Determine the interval of convergence To find the interval
of convergence, we will use the ratio test. Let an= (−1)n. Applying the ratio
test:
lim
n→∞
an+1
an
= lim
n→∞
(−1)n+1
(−1)n
= 1
Since the limit is equal to 1, the interval of convergence is (−1,1). Thus, the
Maclaurin series for f(x) = 1
1+xis:
f(x) = 1 −x+x2−x3+· · · + (−1)nxn+· · · for −1<x<1
Question 17
Question
Find the Maclaurin series representation for f(x) = e2xcos(3x).
16
Solution
To find the Maclaurin series representation for f(x) = e2xcos(3x), we will use
the known Maclaurin series for e2xand cos(3x) and then multiply them together.
Step 1: Find the Maclaurin series for e2x.The Maclaurin series for ex
is given by:
ex=
∞
X
n=0
xn
n!
Substitute xwith 2xto find the Maclaurin series for e2x:
e2x=
∞
X
n=0
(2x)n
n!=
∞
X
n=0
2nxn
n!
Step 2: Find the Maclaurin series for cos(3x).The Maclaurin series
for cos(x) is given by:
cos(x) =
∞
X
n=0
(−1)nx2n
(2n)!
Substitute xwith 3xto find the Maclaurin series for cos(3x):
cos(3x) =
∞
X
n=0
(−1)n(3x)2n
(2n)! =
∞
X
n=0
(−1)n32nx2n
(2n)!
Step 3: Multiply the series together. To find the Maclaurin series
representation for f(x) = e2xcos(3x), multiply the two series we found in Step
1 and Step 2:
f(x) = e2xcos(3x) = ∞
X
n=0
2nxn
n!! ∞
X
n=0
(−1)n32nx2n
(2n)! !
f(x) =
∞
X
n=0 n
X
k=0
2k(−1)n−k32(n−k)xn
k!(n−k)!(2(n−k))!!
Therefore, the Maclaurin series representation for f(x) = e2xcos(3x) is:
f(x) =
∞
X
n=0 n
X
k=0
2k(−1)n−k32(n−k)xn
k!(n−k)!(2(n−k))!!
Question 18
Question
Find the Maclaurin series representation for the function f(x) = 1
1+x2.
17
Solution
The Maclaurin series for a function f(x) is given by the formula:
f(x) =
∞
X
n=0
f(n)(0)
n!xn
where f(n)(0) denotes the n-th derivative of f(x) evaluated at x= 0.
Step 1: Find the derivatives of f(x)
Since f(x) = 1
1+x2, we first need to find the derivatives of f(x):
f′(x) = d
dx 1
1 + x2=−2x
(1 + x2)2
f′′(x) = d2
dx21
1 + x2=2(3x2−1)
(1 + x2)3
Step 2: Find f(n)(0)
To find f(n)(0) for the Maclaurin series, we evaluate the derivatives of f(x)
at x= 0:
f(0) = 1
1+02= 1
f′(0) = −2·0
(1 + 02)2= 0
f′′(0) = 2(3 ·02−1)
(1 + 02)3=−2
Step 3: Write the Maclaurin series
Substitute the values of f(n)(0) into the Maclaurin series formula:
f(x) =
∞
X
n=0
f(n)(0)
n!xn=f(0) + f′(0)x+f′′ (0)
2! x2+. . .
= 1 + 0 ·x−2
2!x2= 1 −x2
Therefore, the Maclaurin series representation for f(x) = 1
1+x2is 1 −x2.
Question 19
Question
Find the Maclaurin series representation for the function f(x) = cos2(x).
18
Solution
To find the Maclaurin series representation for f(x) = cos2(x), we will start by
finding the Maclaurin series for cos(x). Then we will use this to find the series
for cos2(x).
Step 1: Find the Maclaurin series for cos(x).The Maclaurin series for
cos(x) is:
cos(x) =
∞
X
n=0
(−1)nx2n
(2n)!
Step 2: Find the Maclaurin series for cos2(x).To find the Maclaurin
series for cos2(x), we square the Maclaurin series for cos(x):
cos2(x) = ∞
X
n=0
(−1)nx2n
(2n)!!2
Expanding this out gives:
cos2(x) =
∞
X
n=0
∞
X
k=0
(−1)nx2n
(2n)! ·(−1)kx2k
(2k)!
=
∞
X
n=0
∞
X
k=0
(−1)n+kx2n
(2n)! ·x2k
(2k)!
Simplifying this further, we have:
cos2(x) =
∞
X
n=0
∞
X
k=0
(−1)n+kx2(n+k)
(2n)!(2k)!
This series can be rearranged to give the Maclaurin series representation for
cos2(x):
cos2(x) =
∞
X
n=0 n
X
k=0
(−1)kx2(n−k)
(2(n−k))! ·x2k
(2k)!!
Therefore, the Maclaurin series representation for f(x) = cos2(x) is:
∞
X
n=0 n
X
k=0
(−1)kx2(n−k)
(2(n−k))! ·x2k
(2k)!!
Question 20
Question
Find the Maclaurin series for the function f(x) = 1
1−x.
19
Solution
To find the Maclaurin series for f(x) = 1
1−x, we will first find the derivatives of
f(x) and then evaluate them at x= 0 to find the coefficients of the series.
Step 1: Find the derivatives of f(x)
f(x) = 1
1−x
f′(x) = d
dx 1
1−x
=1
(1 −x)2
f′′(x) = d
dx 1
(1 −x)2
=2
(1 −x)3
f′′′(x) = d
dx 2
(1 −x)3
=6
(1 −x)4
.
.
.
Step 2: Evaluate derivatives at x= 0 Now, let’s find the value of the
derivatives at x= 0:
f(0) = 1
1−0= 1
f′(0) = 1
(1 −0)2= 1
f′′(0) = 2
(1 −0)3= 2
f′′′(0) = 6
(1 −0)4= 6
.
.
.
Step 3: Write the Maclaurin series The Maclaurin series for f(x) is
given by:
f(x) = f(0) + f′(0)x+f′′ (0)x2
2! +f′′′(0)x3
3! +· · ·
Substitute the values we found at x= 0:
f(x) = 1 + x+2x2
2! +6x3
3! +· · ·
20
Simplify the terms:
f(x) = 1 + x+x2+x3+· · · =
∞
X
n=0
xn
Therefore, the Maclaurin series for f(x) = 1
1−xis P∞
n=0 xn.
Question 21
Question
Find the Maclaurin series for the function f(x) = 1
1+xand determine its interval
of convergence.
Solution
To find the Maclaurin series for f(x) = 1
1+x, we will first find the derivatives of
f(x) and evaluate them at x= 0 to get the coefficients of the series.
Step 1: Find the derivatives of f(x).
f(x) = 1
1 + x
f′(x) = d
dx 1
1 + x=−(1 + x)−2=−(1 + x)−2
f′′(x) = d
dx −(1 + x)−2= 2(1 + x)−3= 2(1 + x)−3
f′′′(x) = d
dx 2(1 + x)−3=−6(1 + x)−4=−6(1 + x)−4
Now, we will evaluate these derivatives at x= 0:
f(0) = 1
f′(0) = −1
f′′(0) = 2
f′′′(0) = −6
Step 2: Write out the Maclaurin series. The Maclaurin series for f(x)
is given by:
f(x) = f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+· · ·
Substitute the values we found earlier:
f(x)=1−x+2
2!x2−6
3!x3+· · ·
21
Simplify:
f(x)=1−x+x2−2x3+· · ·
Step 3: Determine the interval of convergence. To find the interval
of convergence, we will use the ratio test. The ratio test states that if L=
limn→∞
an+1
an, then the series converges if L < 1.
L= lim
n→∞
(−1)n+1
(−1)n
= 1
Since L= 1, the ratio test is inconclusive. To determine the interval of
convergence, we need to check the endpoints.
For x=−1, the series becomes the harmonic series P∞
n=0(−1)nwhich is
alternating. It converges by the Alternating Series Test.
For x= 1, the series becomes the harmonic series P∞
n=0 1 which is a diver-
gent p-series.
Therefore, the interval of convergence is [−1,1).
Question 22
Question
Find the Maclaurin series for ln(1 + x) by differentiating term by term.
Solution
To find the Maclaurin series for ln(1 + x), we will differentiate the series for
ln(1 + x) term by term. Recall that the Maclaurin series for ln(1 + x) is given
by
ln(1 + x) = x−x2
2+x3
3−x4
4+· · · for −1< x ≤1.
Step 1: Find the first derivative of ln(1 + x):
d
dx (ln(1 + x)) = 1
1 + x.
Step 2: Find the second derivative of ln(1 + x):
d2
dx2(ln(1 + x)) = −1
(1 + x)2.
Step 3: Find the third derivative of ln(1 + x):
d3
dx3(ln(1 + x)) = 2
(1 + x)3.
22
Step 4: Find the fourth derivative of ln(1 + x):
d4
dx4(ln(1 + x)) = −6
(1 + x)4.
Continuing this pattern, we can see that the nth derivative of ln(1 + x) is
given by
dn
dxn(ln(1 + x)) = (−1)n−1·(n−1)!
(1 + x)nfor n≥1.
Therefore, the Maclaurin series for ln(1 + x) is
ln(1 + x) = x−x2
2+x3
3−x4
4+· · · =
∞
X
n=1
(−1)n−1·xn
n.
Question 23
Question
Find the Maclaurin series for f(x) = ln(1 + x) by finding its derivatives and
evaluating them at x= 0.
Solution
To find the Maclaurin series for f(x) = ln(1 + x), we will first find its derivatives
up to a suitable order and evaluate them at x= 0.
Step 1: Find the first derivative of f(x)
f(x) = ln(1 + x)
d
dx f(x) = d
dx ln(1 + x)
Using the chain rule, we get
d
dx ln(1 + x) = 1
1 + x= (1 + x)−1
Step 2: Evaluate the first derivative at x= 0
d
dx f(x)x=0
= (1 + 0)−1= 1
Step 3: Find the second derivative of f(x)
d2
dx2f(x) = d
dx (1 + x)−1
Using the power rule, we get
d2
dx2f(x) = −1(1 + x)−2
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Step 4: Evaluate the second derivative at x= 0
d2
dx2f(x)x=0
=−1(1 + 0)−2=−1
Step 5: Continue finding higher derivatives and evaluating them at x= 0
d3
dx3f(x) = 2(1 + x)−3,d3
dx3f(x)x=0
= 2
d4
dx4f(x) = −6(1 + x)−4,d4
dx4f(x)x=0
=−6
Step 6: Generalize the pattern and write the Maclaurin series for f(x)
f(x) = ln(1 + x) =
∞
X
n=0
(−1)nxn+1
n+ 1
Thus, the Maclaurin series for ln(1 + x) is given by:
ln(1 + x) = x−x2
2+x3
3−x4
4+. . .
Question 24
Question
Find the Maclaurin series for the function f(x) = cos(sin x).
Solution
To find the Maclaurin series for f(x) = cos(sin x), we can make use of the
composite function property of Maclaurin series.
Step 1: Find the derivatives First, we need to find the derivatives of
f(x). Let’s start by finding the first few derivatives:
f(x) = cos(sin x)
f′(x) = −sin(sin x) cos x
f′′(x) = (−cos(sin x) cos x)(cos(sin x) + sin(sin x))
=−cos(sin x)2cos x−sin(sin x) cos(sin x)
f′′′(x)=(−2 cos(sin x) sin(sin x) cos x−cos2(sin x) cos x+ sin2(sin x) cos(sin x)
Step 2: Evaluate the derivatives at 0 Next, we evaluate the derivatives
at x= 0:
f(0) = cos(sin 0) = cos(0) = 1
f′(0) = −sin(sin 0) cos 0 = 0
f′′(0) = −cos(sin 0)2cos 0 −sin(sin 0) cos(sin 0) = −1
f′′′(0) = −cos2(sin 0) cos 0 −sin2(sin 0) cos(sin 0) = 0
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Step 3: Write out the Maclaurin series The Maclaurin series for f(x)
is given by:
f(x) = f(0) + f′(0)
1! x+f′′(0)
2! x2+f′′′(0)
3! x3+. . .
Substitute the values we found earlier:
cos(sin x)=1−x2
2+. . .
Question 25
Question
Find the Maclaurin series representation for f(x) = 1
1+x2by differentiating the
function term by term.
Solution
To find the Maclaurin series representation for f(x) = 1
1+x2, we need to express
f(x) in terms of a power series. We can do this by differentiating the function
term by term.
Step 1: Find the derivatives of f(x).
f(x) = 1
1 + x2
f′(x) = −2x
(1 + x2)2
f′′(x) = −21
(1 + x2)2+4x2
(1 + x2)3
f′′′(x) = −2−2x
(1 + x2)2+12x
(1 + x2)3+12x
(1 + x2)3+48x3
(1 + x2)4
Step 2: Evaluate the derivatives at x= 0 to find the coefficients of the
Maclaurin series.
f(0) = 1
f′(0) = 0
f′′(0) = −2
f′′′(0) = 0
Step 3: Write out the Maclaurin series. The Maclaurin series representation
for f(x) = 1
1+x2is:
f(x)=1−2x2+12x4
2! −48x6
3! +· · · =
∞
X
n=0
(−1)n2nx2n
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Question 26
Question
Find the Maclaurin series for f(x) = 1
1+x2and state the interval of convergence.
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we’ll use the geometric series
formula: 1
1−r= 1 + r+r2+r3+· · · ,for |r|<1.
Step 1: Find the Maclaurin series for f(x) Given f(x) = 1
1+x2, we can
express it as:
f(x) = 1
1−(−x2).
Comparing this expression with the geometric series formula, we can see that
r=−x2. Therefore, the Maclaurin series for f(x) is:
∞
X
n=0
(−1)nx2n.
Step 2: Determine the interval of convergence To find the interval of
convergence, we use the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
(−1)n+1x2(n+1)
(−1)nx2n
.
Simplifying, we get:
lim
n→∞ x2=|x2|.
For the series to converge, we need |x2|<1, which gives us −1<x<1.
Therefore, the interval of convergence is (−1,1).
Question 27
Question
Find the Maclaurin series for the function f(x) = ln(1 + x) and determine the
interval of convergence.
Solution
To find the Maclaurin series for f(x) = ln(1 + x), we will use the formula for
the Maclaurin series of ln(1 + x), which is:
ln(1 + x) = x−x2
2+x3
3−x4
4+. . . =
∞
X
n=1
(−1)n−1xn
n
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Step 1: Identify the Maclaurin series of ln(1 + x).
ln(1 + x) =
∞
X
n=1
(−1)n−1xn
n
Step 2: Determine the interval of convergence.
The Maclaurin series P∞
n=1(−1)n−1xn
nhas interval of convergence −1<
x≤1.
Therefore, the Maclaurin series for f(x) = ln(1 + x) is P∞
n=1(−1)n−1xn
n
with an interval of convergence −1< x ≤1.
Question 28
Question
Find the Maclaurin series for f(x) = 1
(1−x)2.
Solution
To find the Maclaurin series for f(x) = 1
(1−x)2, we use the formula for the
Maclaurin series of a function:
f(x) =
∞
X
n=0
f(n)(0)
n!xn
where f(n)(0) denotes the n-th derivative of f(x) evaluated at x= 0.
Step 1: Find the derivatives of f(x) up to the second derivative.
f(x) = 1
(1 −x)2
f′(x) = 2
(1 −x)3
f′′(x) = 2·3
(1 −x)4
Step 2: Evaluate these derivatives at x= 0.
f(0) = 1
(1 −0)2= 1
f′(0) = 2
(1 −0)3= 2
f′′(0) = 2·3
(1 −0)4= 6
Step 3: Plug these values into the Maclaurin series formula.
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f(x) =
∞
X
n=0
f(n)(0)
n!xn
f(x) = 1 + 2x+ 6x2+
∞
X
n=3
f(n)(0)
n!xn
So, the Maclaurin series for f(x) = 1
(1−x)2is 1 + 2x+ 6x2+P∞
n=3
f(n)(0)
n!xn.
Question 29
Question
Find the Maclaurin series for the function f(x) = ex2.
Solution
To find the Maclaurin series for ex2, we will use the formula for the Maclaurin
series expansion of a function centered at x= 0:
f(x) =
∞
X
n=0
f(n)(0)
n!xn
Step 1: Find the derivatives of f(x).
f(x) = ex2
f′(x)=2xex2
f′′(x) = (2 + 4x2)ex2
f′′′(x) = (4x+ 8x3)ex2
Step 2: Evaluate the derivatives at x= 0.
f(0) = e0= 1
f′(0) = 2(0)e0= 0
f′′(0) = (2 + 4(02))e0= 2
f′′′(0) = (4(0) + 8(03))e0= 0
Step 3: Write down the Maclaurin series. The Maclaurin series for
ex2is:
f(x) = 1 + 0x+2
2!x2+ 0x3+. . . = 1 + x2
Therefore, the Maclaurin series for ex2is 1 + x2.
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Question 30
Question
Find the Maclaurin series for f(x) = 1
1+x2.
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we will first find the derivatives
of f(x) at x= 0 and then use them to construct the Maclaurin series.
Step 1: Find the derivatives of f(x)
The function f(x) = 1
1+x2can be rewritten as f(x) = (1 + x2)−1. We can
differentiate f(x) term by term using the formula for the derivative of a power
function.
f′(x) = −(1 + x2)−2·2x=−2x(1 + x2)−2
f′′(x) = −2(1 + x2)−2−2(−2x)(−2)(1 + x2)−3= 2(2x2−1)(1 + x2)−3
f′′′(x) = 2(2x2−1)(1 + x2)−3−6x(1 + x2)−3= 2(1 −10x2+ 2x4)(1 + x2)−4
Step 2: Find f(0) and evaluate f′(0),f′′ (0),f′′′ (0)
f(0) = 1
1+02= 1
f′(0) = 0
f′′(0) = 2
f′′′(0) = 2
Step 3: Write out the Maclaurin series
The Maclaurin series for f(x) is given by:
f(x) = f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+· · ·
Substitute the values we found earlier:
f(x) = 1 + 0 ·x+2
2!x2+2
3!x3+· · ·
Simplifying:
f(x) = 1 + x2+1
3x3+· · ·
Therefore, the Maclaurin series for f(x) = 1
1+x2is 1 + x2+1
3x3+· · · .
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