MATH 100 - FUNDAMENTALS OF
MATHEMATICS - Taylor and
Maclaurin Series
Question Bank - Set 5
Liberty University
Question 1
Question
Find the Maclaurin series for the function f(x) = 1
1+x.
Solution
To find the Maclaurin series for f(x) = 1
1+x, we will first find the derivatives of
f(x) at x= 0, and then apply the formula for the Maclaurin series.
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 derivatives at x= 0
f(0) = 1
f′(0) = −1
f′′(0) = 2
f′′′(0) = −6
.
.
.
Step 3: Write the Maclaurin series The Maclaurin series for a function
f(x) is given by
f(x) = f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+. . . =
∞
X
n=0
f(n)(0)
n!xn.
Thus, for f(x) = 1
1+x, the Maclaurin series is
∞
X
n=0
f(n)(0)
n!xn= 1 −x+x2−x3+. . . =
∞
X
n=0
(−1)nxn.
Question 2
Question
Find the Maclaurin series for the function f(x) = ln(1 + x).
Solution
To find the Maclaurin series for the function f(x) = ln(1 + x), we will first
find the derivatives of f(x) and evaluate them at x= 0 in order to find the
coefficients of the Maclaurin series.
Step 1: Find the 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
and so on...
2
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
and so on...
Step 3: Write out the Maclaurin series Since the Maclaurin series of a
function f(x) centered at x= 0 is given by
f(x) = f(0) + f′(0)x+f′′ (0)x2
2! +f′′′(0)x3
3! +f(4)(0)x4
4! +. . .
we have the Maclaurin series for f(x) = ln(1 + x) as
ln(1 + x) = x−x2
2+x3
3−x4
4+. . .
Question 3
Question
Find the Maclaurin series for f(x) = 1
1+x.
Solution
To find the Maclaurin series for f(x) = 1
1+x, we start by recalling the geometric
series formula:
1
1−r= 1 + r+r2+r3+· · ·
where |r|<1.
We can rewrite f(x) as:
f(x) = 1
1 + x= 1 −x+x2−x3+· · ·
Step 1: Let’s match this with the form of the geometric series:
1
1−r= 1 + r+r2+r3+· · ·
3
To match the above form, we need to express f(x) as:
f(x)=1−x+x2−x3+· · · = 1 + (−x)+(−x)2+ (−x)3+· · ·
Step 2: Now our function matches the required form of the geometric series,
we can express f(x) as the sum of an infinite geometric series with first term
a= 1 and common ratio r=−x:
f(x) = 1 −x+x2−x3+· · · =
∞
X
n=0
(−1)nxn
Step 3: Therefore, the Maclaurin series for f(x) = 1
1+xis:
∞
X
n=0
(−1)nxn
Question 4
Question
Find the Maclaurin series for the function f(x) = 1
x2+4 .
Solution
To find the Maclaurin series for f(x), we will first express f(x) as a geometric
series by expanding 1
x2+4 as a sum of terms involving xk.
Step 1: Find the expression for f(x)as a geometric series. We
recognize that 1
x2+4 can be expressed in the form of 1
1−uwhere u=−x2
4.
Therefore, we have:
f(x) = 1
x2+ 4 =1
4·1
1−(−x2/4) =1
4
∞
X
n=0 −x2
4n
Step 2: Write out the Maclaurin series for f(x).Expanding the
geometric series, we get:
f(x) = 1
4
∞
X
n=0 −x2
4n
=1
4
∞
X
n=0
(−1)n·x2n
4n
=1
41−x2
4+x4
42−x6
43+. . .
Therefore, the Maclaurin series for f(x) = 1
x2+4 is:
f(x) = 1
41−x2
4+x4
42−x6
43+. . .
4
Question 5
Question
Find the Maclaurin series for f(x) = sin2(x).
Solution
To find the Maclaurin series for f(x) = sin2(x), we can first rewrite f(x) in terms
of trigonometric identities. Recall that sin(2x) = 2 sin(x) cos(x). Therefore, we
have:
f(x) = 1
2−1
2cos(2x)
Now, we need to find the Maclaurin series for cos(2x), which we can do by
finding the Maclaurin series for cos(x) and then substituting 2xfor x. The
Maclaurin series for cos(x) is:
cos(x)=1−x2
2! +x4
4! −x6
6! +· · · =
∞
X
n=0
(−1)nx2n
(2n)!
Now, substituting xwith 2xgives the Maclaurin series for cos(2x):
cos(2x)=1−(2x)2
2! +(2x)4
4! −(2x)6
6! +· · · =
∞
X
n=0
(−1)n(2x)2n
(2n)!
Substitute this series back into the expression for f(x) to get the Maclaurin
series for f(x) = sin2(x):
f(x) = 1
2−1
21−(2x)2
2! +(2x)4
4! −(2x)6
6! +· · ·
Simplify this expression to get the Maclaurin series for f(x) = sin2(x):
f(x) = 1
2−1
2+x2
2! −x4
4! +x6
6! − · · · =
∞
X
n=0
(−1)nx2n+2
(2n+ 2)!
Therefore, the Maclaurin series for f(x) = sin2(x) is:
f(x) =
∞
X
n=0
(−1)nx2n+2
(2n+ 2)!
Question 6
Question
Find the Maclaurin series representation for the function f(x) = 1
1+x2.
5
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we can recall the geometric series
formula: 1
1−r= 1 + r+r2+r3+. . . , for |r|<1.
Step 1: Find the Maclaurin series for 1
1−x.
The function 1
1−xhas a known Maclaurin series:
1
1−x= 1 + x+x2+x3+. . . =
∞
X
n=0
xn
Step 2: Find the Maclaurin series for 1
1+x2.
Using the fact that 1
1−x= 1 + x+x2+. . ., we can substitute −x2for x:
1
1+(−x2)= 1 −x2+x4−x6+. . . =
∞
X
n=0
(−1)nx2n
Therefore, the Maclaurin series representation for f(x) = 1
1+x2is:
f(x) =
∞
X
n=0
(−1)nx2n= 1 −x2+x4−x6+. . .
Question 7
Question
Find the Maclaurin series for the function f(x) = ln(1 + x).
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. The Maclaurin series will be the Taylor
series centered at x= 0.
Step 1: Find the 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
.
.
.
6
Step 2: Evaluate the derivatives at x= 0
f(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 the Maclaurin series The Maclaurin series for ln(1 + x)
is given by:
ln(1 + x) = f(x) = f(0) + f′(0)
1! x+f′′(0)
2! x2+f′′′(0)
3! x3+· · ·
Substitute the values we found:
ln(1 + x) = 0 + 1 ·x−1·x2
2+ 2 ·x3
6+· · ·
Therefore, the Maclaurin series for ln(1 + x) is:
ln(1 + x) = x−x2
2+x3
3−x4
4+· · ·
Question 8
Question
Find the Maclaurin series for ln(1 + x) by differentiating its Taylor series.
Solution
To find the Maclaurin series for ln(1 + x), we will first find the Taylor series for
ln(1 + x) centered at a= 0 and then differentiate it term by term.
Step 1: Find the Taylor series for ln(1 + x)centered at a= 0 The
Taylor series for ln(1 + x) centered at a= 0 is given by:
ln(1 + x) =
∞
X
n=1
(−1)n−1xn
n
Step 2: Differentiate the Taylor series term by term Differentiating
the series term by term, we get:
d
dx ln(1 + x) = d
dx
∞
X
n=1
(−1)n−1xn
n
7
=
∞
X
n=1
d
dx (−1)n−1xn
n
=
∞
X
n=1
(−1)n−1xn−1
Therefore, the Maclaurin series for ln(1 + x) is:
ln(1 + x) =
∞
X
n=1
(−1)n−1xn
n
Question 9
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 its
derivatives and evaluating them at x= 0.
f(x) = ln(1 + x)
f′(x) = 1
1 + x(differentiate using chain rule)
f′′(x) = −1
(1 + x)2(differentiate the derivative of f)
f(3)(x) = 2
(1 + x)3(differentiate f′′(x))
f(4)(x) = −6
(1 + x)4(differentiate f(3)(x))
.
.
.
Now, let’s evaluate each derivative at x= 0 to find the coefficients of the
8
Maclaurin series for f(x).
f(0) = ln(1 + 0) = ln(1) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f(3)(0) = 2
(1 + 0)3= 2
f(4)(0) = −6
(1 + 0)4=−6
.
.
.
Therefore, the Maclaurin series for f(x) = ln(1 + x) is
ln(1 + x) = x−x2
2+x3
3−x4
4+· · · =
∞
X
n=1
(−1)n−1xn
n.
Question 10
Question
Find the Maclaurin series for the function f(x) = e2xsin(3x).
Solution
To find the Maclaurin series of f(x), we will use the formula for the Maclaurin
series of a product of two functions. The Maclaurin series for e2xand sin(3x)
are known, so we will use these series to find the series for f(x).
Step 1: Find the Maclaurin series for e2xThe Maclaurin series for e2x
is:
e2x=
∞
X
n=0
(2x)n
n!=
∞
X
n=0
2nxn
n!
Step 2: Find the Maclaurin series for sin(3x) The Maclaurin series for
sin(3x) is:
sin(3x) =
∞
X
n=0
(−1)n(3x)2n+1
(2n+ 1)! =
∞
X
n=0
(−1)n32n+1x2n+1
(2n+ 1)!
Step 3: Find the Maclaurin series for f(x) = e2xsin(3x) To find the
series for the product f(x) = e2xsin(3x), we multiply the series for e2xand
9
sin(3x):
f(x) = e2xsin(3x)
= ∞
X
n=0
2nxn
n!!· ∞
X
n=0
(−1)n32n+1x2n+1
(2n+ 1)! !
=
∞
X
n=0
n
X
k=0
2kxk
k!·(−1)n−k32(n−k)+1x2(n−k)+1
(2(n−k) + 1)!
=
∞
X
n=0
n
X
k=0
2k(−1)n−k32(n−k)+1xk+2(n−k)+1
k!(2(n−k) + 1)!
Simplifying further, we find:
f(x) =
∞
X
n=0
n
X
k=0
2k(−1)n−k32(n−k)+1x2n+1
k!(2(n−k) + 1)!
Therefore, the Maclaurin series for f(x) = e2xsin(3x) is given by:
∞
X
n=0
n
X
k=0
2k(−1)n−k32(n−k)+1x2n+1
k!(2(n−k) + 1)!
Question 11
Question
Find the Maclaurin series for the function 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) and then evaluate them at x= 0 to find the coefficients of the Maclaurin
series.
Step 1: Find the first few derivatives of f(x)
f(x) = (1 + x2)−1
f′(x) = −1(1 + x2)−2·2x=−2x(1 + x2)−2
f′′(x) = d
dx(−2x(1 + x2)−2) = −2(1 + x2)−2−2(−2x)(−2)(1 + x2)−3= 2(1 + x2)−3(2x2−1)
f′′′(x) = d
dx 2(1 + x2)−3(2x2−1)= 6(1 + x2)−4(2x2−1) −4(1 + x2)−3(4x)
= 2(1 + x2)−4(12x2−6−16x2) = 2(1 + x2)−4(12x2−6−16x2)
=−8x(1 + x2)−4
10
Step 2: Evaluate the derivatives at x= 0
f(0) = 1
f′(0) = 0
f′′(0) = 2
f′′′(0) = 0
Since the odd derivatives evaluated at x= 0 are all zero, the Maclaurin series
will only involve even powers of x. Thus, the Maclaurin series for f(x) = 1
1+x2
is:
f(x) = 1 + 2x2+8x4
3! +. . .
Question 12
Question
Find the Maclaurin series for the function f(x) = sin x
x.
Solution
To find the Maclaurin series for f(x) = sin x
x, we can first find the Maclaurin
series for sin xand then divide by x.
Step 1: Find the Maclaurin series for sin xThe Maclaurin series for
sin xis given by:
sin x=x−x3
3! +x5
5! −x7
7! +. . . =
∞
X
n=0
(−1)nx2n+1
(2n+ 1)!
Step 2: Divide by xto find the Maclaurin series for f(x) = sin xx
Dividing sin xby xgives:
sin x
x= 1 −x2
3! +x4
5! −x6
7! +. . . =
∞
X
n=0
(−1)nx2n
(2n+ 1)!
Therefore, the Maclaurin series for f(x) = sin x
xis P∞
n=0
(−1)nx2n
(2n+1)! .
Question 13
Question
Let f(x) = 1
1+x2. Find the first four nonzero terms of the Maclaurin series for
f(x).
11
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we recall 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 nth derivative of f(x) evaluated at x= 0.
Step 1: Find the derivatives of f(x).
Let’s start by finding the derivatives of f(x) = 1
1+x2:
f(x) = 1
1 + x2
f′(x) = −2x
(1 + x2)2
f′′(x) = 2(3x2−1)
(1 + x2)3
f′′′(x) = −4x(5x4−10x2+ 3)
(1 + x2)4
f(4)(x) = 4(35x4−45x2+ 6)
(1 + x2)5
Step 2: Evaluate the derivatives at x= 0.
We need to evaluate f(n)(0) for n= 0,1,2,3,4:
f(0)(0) = f(0) = 1
f(1)(0) = −2
f(2)(0) = 6
f(3)(0) = 0
f(4)(0) = 24
Step 3: Write the Maclaurin series.
Using the formula for the Maclaurin series, we have:
f(x)=1−2x+ 3x2−6x3+ 24x4+. . .
Therefore, the first four nonzero terms of the Maclaurin series for f(x) =
1
1+x2are 1 −2x+ 3x2−6x3+ 24x4.
Question 14
Question
Find the Maclaurin series for the function f(x) = 1
1+xand determine its interval
of convergence.
12
Solution
To find the Maclaurin series for f(x) = 1
1+x, we will use the fact that 1
1+x=
1−x+x2−x3+x4−. . . which is a geometric series where the common ratio
is −xand the first term is 1.
Step 1: Find the Maclaurin series using the formula for a geometric series.
f(x) = 1 −x+x2−x3+x4−. . . =
∞
X
n=0
(−1)nxn
Step 2: Determine the interval of convergence using the ratio test. Let’s
apply the ratio test:
lim
n→∞
(−1)n+1xn+1
(−1)nxn
= lim
n→∞ |x|=|x|
The series converges if |x|<1. Therefore, the interval of convergence is
(−1,1).
Question 15
Question
Find the Maclaurin series for the function f(x) = e−2x2.
Solution
To find the Maclaurin series for f(x), we will first find the Maclaurin series for
exand then substitute −2x2in place of x.
Step 1: Find the Maclaurin series for exThe Maclaurin series for ex
is:
ex=
∞
X
n=0
xn
n!
Step 2: Substitute −2x2for xTo find the Maclaurin series for e−2x2, we
substitute −2x2for xin the Maclaurin series for ex:
e−2x2=
∞
X
n=0
(−2x2)n
n!=
∞
X
n=0
(−2)nx2n
n!
Step 3: Simplify the series We can simplify the series further by sepa-
rating out the even powers of x:
e−2x2=
∞
X
n=0
(−1)n2n
n!x2n
Therefore, the Maclaurin series for f(x) = e−2x2is:
f(x) =
∞
X
n=0
(−1)n2n
n!x2n
13
Question 16
Question
Find the Maclaurin series for f(x) = ln(1 + x), and determine the interval of
convergence.
Solution
To find the Maclaurin series for f(x) = ln(1 + x), we’ll start by finding the
derivatives of f(x) and evaluating them at x= 0 to find the coefficients of the
series.
Step 1: Find the derivatives of f(x)
f(x) = ln(1 + x)⇒f′(x) = 1
1 + x⇒f′′ (x) = −1
(1 + x)2
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
Step 3: Write out the Maclaurin series for f(x) Since the Maclaurin series
is given by
f(x) = f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+. . .
we have
ln(1 + x) = 0 + 1x−1
2x2+1
3x3−. . .
Thus, the Maclaurin series for f(x) = ln(1 + x) is
ln(1 + x) = x−1
2x2+1
3x3−. . .
Step 4: Determine the interval of convergence The Maclaurin series for
ln(1 + x) converges for −1< x ≤1 (i.e., |x|<1) as it is based on the Taylor
series of a function around x= 0 which converges within the radius of conver-
gence. Thus, the interval of convergence is −1< x ≤1.
Question 17
Question
Find the Maclaurin series for the function f(x) = e2xcos(x).
14
Solution
To find the Maclaurin series for the function f(x) = e2xcos(x), we will utilize
the properties of Maclaurin series for e2xand cos(x).
Step 1: Find the Maclaurin series for e2x.The Maclaurin series for ex
is given by:
ex= 1 + x+x2
2! +x3
3! +x4
4! +. . . =
∞
X
n=0
xn
n!
Substitute 2xfor xin the series:
e2x= 1 + 2x+(2x)2
2! +(2x)3
3! +(2x)4
4! +. . . =
∞
X
n=0
(2x)n
n!
Simplyfying:
e2x= 1 + 2x+ 2x2+4x3
3+4x4
6+. . . =
∞
X
n=0
2nxn
n!
Step 2: Find the Maclaurin series for cos(x).The Maclaurin series for
cos(x) is given by:
cos(x)=1−x2
2! +x4
4! −x6
6! +. . . =
∞
X
n=0
(−1)nx2n
(2n)!
Substitute xfor xin the series:
cos(x)=1−x2
2! +x4
4! −x6
6! +. . . =
∞
X
n=0
(−1)nx2n
(2n)!
Step 3: Find the Maclaurin series for f(x) = e2xcos(x).Since f(x) =
e2xcos(x), the Maclaurin series for f(x) will be the product of the Maclaurin
series for e2xand cos(x):
f(x) = e2xcos(x) = ∞
X
n=0
2nxn
n!! ∞
X
n=0
(−1)nx2n
(2n)!!
Expanding the product, we get:
f(x) =
∞
X
n=0
n
X
k=0
2k(−1)n−kxn
k!(n−k)!
Therefore, the Maclaurin series for f(x) = e2xcos(x) is:
f(x) =
∞
X
n=0
n
X
k=0
2k(−1)n−kxn
k!(n−k)!
15
Question 18
Question
Find the Maclaurin series for f(x) = (1 + x)−1/2.
Solution
To find the Maclaurin series for f(x) = (1 + x)−1/2, we can start by finding
the derivatives of f(x) at x= 0 and writing out the general formula for the
Maclaurin series.
Step 1: Find the derivatives of f(x).
f(x) = (1 + x)−1/2
f′(x) = −1
2(1 + x)−3/2
f′′(x) = −1
2· −3
2(1 + x)−5/2=3
4(1 + x)−5/2
f′′′(x) = 3
4· −5
2(1 + x)−7/2=−15
8(1 + x)−7/2
.
.
.
Step 2: Determine the pattern and write the general formula for the Maclau-
rin series. From the derivatives, we can see that the nth derivative of f(x) eval-
uated at x= 0 is (−1)n(2n−1)!!
2n(1 + x)−(2n+1)/2. Therefore, the Maclaurin series
for f(x) is given by
f(x) =
∞
X
n=0
(−1)n(2n−1)!!
2nxn
Step 3: Express the Maclaurin series in terms of f(x). Substitute x=−x2
into the Maclaurin series formula to get:
(1 + x)−1/2=
∞
X
n=0
(−1)n(2n−1)!!
2n(−x2)n
Therefore, the Maclaurin series for f(x) = (1 + x)−1/2is
f(x) =
∞
X
n=0
(2n−1)!!
2nx2n
Question 19
Question
Find the Maclaurin series for the function f(x) = excos(x).
16
Solution
To find the Maclaurin series for f(x) = excos(x), we will first find the Maclaurin
series for exand cos(x) separately, and then multiply them together.
Step 1: Find the Maclaurin series for exThe Maclaurin series for ex
is given by:
ex=
∞
X
n=0
xn
n!
Step 2: Find the Maclaurin series for cos(x) The Maclaurin series for
cos(x) is given by:
cos(x) =
∞
X
n=0
(−1)nx2n
(2n)!
Step 3: Multiply the series together To find the Maclaurin series for
f(x) = excos(x), we multiply the series for exand cos(x) together term by
term:
f(x) = excos(x)
= ∞
X
n=0
xn
n!! ∞
X
n=0
(−1)nx2n
(2n)! !
=
∞
X
n=0 n
X
k=0
1
k!(n−k)!
(−1)n−k
(2(n−k))!!xn
Therefore, the Maclaurin series for f(x) = excos(x) is:
f(x) =
∞
X
n=0
cnxn
where
cn=
n
X
k=0
1
k!(n−k)!
(−1)n−k
(2(n−k))!
Question 20
Question
Find the Maclaurin series for the function f(x) = 1
1+x2.
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we will first find the derivative of
f(x) and evaluate it at x= 0 to find the coefficients of the series.
17
Step 1: Find the derivatives of f(x)
f(x) = 1
1 + x2
f′(x) = −2x
(1 + x2)2
f′′(x) = 2(3x2−1)
(1 + x2)3
f′′′(x) = −24x(1 −5x2)
(1 + x2)4
Step 2: Evaluate the derivatives at x= 0
f(0) = 1
f′(0) = 0
f′′(0) = 2
f′′′(0) = 0
Step 3: Write the Maclaurin series using the coefficients The Maclau-
rin series for f(x) is given by
f(x) = 1 + 0 ·x+2
2!x2+ 0 ·x3+24
4! x4+. . .
Simplifying this series, we get
f(x) = 1 + x2+6
5x4+. . .
Question 21
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 find the derivatives of
f(x) and evaluate them at x= 0 to determine the coefficients of the series.
Step 1: Find the first few derivatives of f(x).
f(x) = 1
1 + x
f′(x) = d
dx 1
1 + x=−(1 + x)−2=−1
(1 + x)2
f′′(x) = d
dx −1
(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
18
Step 2: Evaluate the derivatives at x= 0.
f(0) = 1
f′(0) = −1
f′′(0) = 2
f′′′(0) = −6
Step 3: Write the Maclaurin series for f(x).The Maclaurin series for
a function f(x) is given by
f(x) = f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+· · ·
Substitute the evaluated derivatives at x= 0 into the series:
f(x)=1−x+2
2!x2−6
3!x3+· · ·
Simplifying, we get the Maclaurin series for f(x) = 1
1+x:
f(x) = 1 −x+x2−x3+· · · =
∞
X
n=0
(−1)nxn
Question 22
Question
Find the Maclaurin series for f(x) = 1
1+x2and determine the interval of con-
vergence.
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we first need to express it as a
geometric series. Recall that the geometric series formula is:
1
1−r=
∞
X
n=0
rn
for |r|<1.
Step 1: Find the geometric series form We can rewrite f(x) = 1
1+x2
as:
f(x) = 1
1−(−x2)=
∞
X
n=0
(−x2)n=
∞
X
n=0
(−1)nx2n
Step 2: Determine the interval of convergence The interval of conver-
gence of a power series can be found using the ratio test. Let’s apply the ratio
test to the series P∞
n=0(−1)nx2n:
lim
n→∞
an+1
an
= lim
n→∞
(−1)n+1x2(n+1)
(−1)nx2n
= lim
n→∞ |x2|
19
The series converges if |x2|<1, which simplifies to −1< x < 1. Therefore,
the interval of convergence is (−1,1).
Thus, the Maclaurin series for f(x) = 1
1+x2is
∞
X
n=0
(−1)nx2n= 1 −x2+x4−x6+. . . (for −1<x<1)
Question 23
Question
Find the Maclaurin series for 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 then evaluate them at x= 0.
Step 1: Find the derivatives of f(x)
f(x) = ln(1 + x)
f′(x) = 1
1 + x(by the chain rule)
f′′(x) = −1
(1 + x)2(by the chain rule again)
f′′′(x) = 2
(1 + x)3(by the chain rule again)
f(4)(x) = −6
(1 + x)4(by the chain rule again)
Step 2: Evaluate derivatives at x= 0
f(0) = ln(1) = 0
f′(0) = 1
1= 1
f′′(0) = −1
12=−1
f′′′(0) = 2
13= 2
f(4)(0) = −6
14=−6
Therefore, the Maclaurin series for f(x) = ln(1 + x) is:
ln(1 + x) = 0 + x−1
2x2+1
3x3−1
4x4+. . .
20
Step 3: Determine the interval of convergence To find the interval of
convergence, we will use the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
1
n+1
1
n
= lim
n→∞
n
n+ 1
= 1 ⇒radius of convergence = R= 1
Thus, the interval of convergence is (−1,1].
Question 24
Question
Find the Maclaurin series for the function f(x) = 1
1+x2by differentiating the
function term-by-term.
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we first need to know the Maclau-
rin series for 1
1−x, which is
1
1−x= 1 + x+x2+x3+· · · =
∞
X
n=0
xn.
Next, we substitute −x2in place of xin the series:
1
1 + x2= 1 −x2+x4−x6+· · · =
∞
X
n=0
(−1)nx2n.
Therefore, the Maclaurin series for the function f(x) = 1
1+x2is P∞
n=0(−1)nx2n.
Question 25
Question
Find the Maclaurin series for the function f(x) = sin2(x).
Solution
To find the Maclaurin series for f(x) = sin2(x), we will first find the Maclaurin
series for sin(x) and then square it.
Step 1: Find the Maclaurin series for sin(x).The Maclaurin series for
sin(x) is given by:
sin(x) = x−x3
3! +x5
5! −x7
7! +· · · =
∞
X
n=0
(−1)nx2n+1
(2n+ 1)!
21
Step 2: Square the Maclaurin series for sin(x).Squaring the Maclaurin
series for sin(x), we get:
(sin(x))2= ∞
X
n=0
(−1)nx2n+1
(2n+ 1)!!2
Expanding the squared series gives:
(sin(x))2=x2−2
3!x4+2
5!x6−2
7!x8+· · ·
Therefore, the Maclaurin series for f(x) = sin2(x) is:
f(x) = x2−2
3!x4+2
5!x6−2
7!x8+· · ·
Question 26
Question
Find the Maclaurin series for f(x) = 1
1+x+x2and determine the interval of
convergence.
Solution
To find the Maclaurin series for f(x) = 1
1+x+x2, we will begin by expressing
f(x) as a geometric series.
Step 1: Find the geometric series representation Since 1
1+x+x2=
1
(x+1/2)2−1/4, we rewrite f(x) as
1
1 + x+x2=1
(x+ 1/2)2−1/4=1
(x+ 1/2−1/2)(x+ 1/2+1/2) =A
x+ 1/2−1/2+B
x+ 1/2+1/2,
to find the constants Aand B.
Step 2: Find the values of Aand BWe rewrite 1
1+x+x2using partial
fraction decomposition to get:
1
1 + x+x2=A
x+B
x+ 1.
Multiplying through by x(x+ 1) gives
1 = A(x+ 1) + Bx.
Substituting x= 0 in the equation above:
1 = A(0 + 1) =⇒A= 1.
22
Substituting x=−1 in the equation above:
1 = B(−1) =⇒B=−1.
Step 3: Write out the Maclaurin series Our function f(x) can be
rewritten as
f(x) = 1
1 + x+x2=1
x−1
x+ 1.
Thus, the Maclaurin series for f(x) is
f(x)=1−x+x2−x3+x4−x5+. . . =
∞
X
n=0
(−1)nxn.
Step 4: Determine the interval of convergence To find the interval
of convergence for the series, note that the series converges for −1< x < 1.
Therefore, the interval of convergence is (−1,1).
Question 27
Question
Find the Maclaurin series for f(x) = 1
1+xby directly differentiating the geomet-
ric series representation of f(x).
Solution
To find the Maclaurin series for f(x) = 1
1+x, we will directly differentiate the
geometric series representation of f(x).
The geometric series representation of 1
1+xis:
1
1 + x= 1 −x+x2−x3+x4−. . . =
∞
X
n=0
(−1)nxn
Step 1: Differentiate the geometric series term by term
f′(x) = d
dx ∞
X
n=0
(−1)nxn!
Differentiating the series term by term:
f′(x) =
∞
X
n=0
d
dx((−1)nxn)
f′(x) =
∞
X
n=0
(−1)n·nxn−1
23
Step 2: Find the second derivative of the series
f′′(x) = d
dx ∞
X
n=0
(−1)n·nxn−1!
Differentiating the series term by term:
f′′(x) =
∞
X
n=0
d
dx((−1)n·nxn−1)
f′′(x) =
∞
X
n=0
(−1)n·n(n−1)xn−2
Step 3: Generalize the derivative pattern By observing the derivatives,
we can see that the nth derivative of f(x) evaluated at x= 0 is
f(n)(0) = (−1)n·n!
Step 4: Write the Maclaurin series for f(x) The Maclaurin series for
f(x) = 1
1+xis:
f(x) = f(0) + f′(0)
1! x+f′′(0)
2! x2+f′′′(0)
3! x3+. . .
= 1 −x+x2−x3+x4−. . . =
∞
X
n=0
(−1)nxn
Therefore, the Maclaurin series for f(x) = 1
1+xis P∞
n=0(−1)nxn.
Question 28
Question
Find the Maclaurin series for the function f(x) = 1
1+x.
Solution
To find the Maclaurin series for f(x) = 1
1+x, we will expand the function in a
power series centered at x= 0.
24
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
(1 + x)3
f′′′(x) = d
dx 21
(1 + x)3
=−61
(1 + x)4
Step 2: Find the Maclaurin series. The Maclaurin series expansion of
f(x) is given by
f(x) = f(0) + f′(0)
1! x+f′′(0)
2! x2+f′′′(0)
3! x3+. . .
Substitute the derivatives of f(x) we found into the formula:
f(0) = 1
1+0 = 1
f′(0) = −1
(1 + 0)2=−1
f′′(0) = 2 1
(1 + 0)3= 2
f′′′(0) = −61
(1 + 0)4=−6
Therefore, the Maclaurin series for f(x) = 1
1+xis:
1−x+ 2x2−6x3+. . .
Question 29
Question
Find the Maclaurin series for the function f(x) = sinx
2.
25
Solution
To find the Maclaurin series for f(x) = sinx
2, we will first find the Maclaurin
series for sin(x) and then replace xwith x
2.
Step 1: Find the Maclaurin series for sin(x). The Maclaurin series for sin(x)
is given by:
sin(x) = x−x3
3! +x5
5! −x7
7! +· · · =
∞
X
n=0
(−1)nx2n+1
(2n+ 1)!
Step 2: Replace xwith x
2in the Maclaurin series for sin(x). Substitute
x=x
2:
f(x) = sin x
2=
∞
X
n=0
(−1)nx
22n+1
(2n+ 1)!
=
∞
X
n=0
(−1)nx2n+1
22n+1(2n+ 1)!
=
∞
X
n=0
(−1)nx2n+1
22n+1(2n+ 1)!
Therefore, the Maclaurin series for f(x) = sinx
2is:
f(x) =
∞
X
n=0
(−1)nx2n+1
22n+1(2n+ 1)!
Question 30
Question
Find the Maclaurin series representation for the function f(x) = sin2(x).
Solution
To find the Maclaurin series representation for f(x) = sin2(x), we first need to
express f(x) in terms of a known Maclaurin series. Recall the trigonometric
identity sin2(x) = 1−cos(2x)
2.
Step 1: Find the Maclaurin series for cos(2x). We know that the Maclaurin
series for cos(x) is given by:
cos(x)=1−x2
2! +x4
4! −x6
6! +. . . =
∞
X
n=0
(−1)nx2n
(2n)!
Substitute 2xfor xto get the Maclaurin series for cos(2x):
cos(2x)=1−(2x)2
2! +(2x)4
4! −(2x)6
6! +. . . =
∞
X
n=0
(−1)n(2x)2n
(2n)!
26
Step 3: Write the Maclaurin series The Maclaurin series for a function
f(x) is given by
f(x) = f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+. . . =
∞
X
n=0
f(n)(0)
n!xn.
Thus, for f(x) = 1
1+x, the Maclaurin series is
∞
X
n=0
f(n)(0)
n!xn= 1 −x+x2−x3+. . . =
∞
X
n=0
(−1)nxn.
Question 2
Question
Find the Maclaurin series for the function f(x) = ln(1 + x).
Solution
To find the Maclaurin series for the function f(x) = ln(1 + x), we will first
find the derivatives of f(x) and evaluate them at x= 0 in order to find the
coefficients of the Maclaurin series.
Step 1: Find the 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
and so on...
2
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
and so on...
Step 3: Write out the Maclaurin series Since the Maclaurin series of a
function f(x) centered at x= 0 is given by
f(x) = f(0) + f′(0)x+f′′ (0)x2
2! +f′′′(0)x3
3! +f(4)(0)x4
4! +. . .
we have the Maclaurin series for f(x) = ln(1 + x) as
ln(1 + x) = x−x2
2+x3
3−x4
4+. . .
Question 3
Question
Find the Maclaurin series for f(x) = 1
1+x.
Solution
To find the Maclaurin series for f(x) = 1
1+x, we start by recalling the geometric
series formula:
1
1−r= 1 + r+r2+r3+· · ·
where |r|<1.
We can rewrite f(x) as:
f(x) = 1
1 + x= 1 −x+x2−x3+· · ·
Step 1: Let’s match this with the form of the geometric series:
1
1−r= 1 + r+r2+r3+· · ·
3
To match the above form, we need to express f(x) as:
f(x)=1−x+x2−x3+· · · = 1 + (−x)+(−x)2+ (−x)3+· · ·
Step 2: Now our function matches the required form of the geometric series,
we can express f(x) as the sum of an infinite geometric series with first term
a= 1 and common ratio r=−x:
f(x) = 1 −x+x2−x3+· · · =
∞
X
n=0
(−1)nxn
Step 3: Therefore, the Maclaurin series for f(x) = 1
1+xis:
∞
X
n=0
(−1)nxn
Question 4
Question
Find the Maclaurin series for the function f(x) = 1
x2+4 .
Solution
To find the Maclaurin series for f(x), we will first express f(x) as a geometric
series by expanding 1
x2+4 as a sum of terms involving xk.
Step 1: Find the expression for f(x)as a geometric series. We
recognize that 1
x2+4 can be expressed in the form of 1
1−uwhere u=−x2
4.
Therefore, we have:
f(x) = 1
x2+ 4 =1
4·1
1−(−x2/4) =1
4
∞
X
n=0 −x2
4n
Step 2: Write out the Maclaurin series for f(x).Expanding the
geometric series, we get:
f(x) = 1
4
∞
X
n=0 −x2
4n
=1
4
∞
X
n=0
(−1)n·x2n
4n
=1
41−x2
4+x4
42−x6
43+. . .
Therefore, the Maclaurin series for f(x) = 1
x2+4 is:
f(x) = 1
41−x2
4+x4
42−x6
43+. . .
4
Question 5
Question
Find the Maclaurin series for f(x) = sin2(x).
Solution
To find the Maclaurin series for f(x) = sin2(x), we can first rewrite f(x) in terms
of trigonometric identities. Recall that sin(2x) = 2 sin(x) cos(x). Therefore, we
have:
f(x) = 1
2−1
2cos(2x)
Now, we need to find the Maclaurin series for cos(2x), which we can do by
finding the Maclaurin series for cos(x) and then substituting 2xfor x. The
Maclaurin series for cos(x) is:
cos(x)=1−x2
2! +x4
4! −x6
6! +· · · =
∞
X
n=0
(−1)nx2n
(2n)!
Now, substituting xwith 2xgives the Maclaurin series for cos(2x):
cos(2x)=1−(2x)2
2! +(2x)4
4! −(2x)6
6! +· · · =
∞
X
n=0
(−1)n(2x)2n
(2n)!
Substitute this series back into the expression for f(x) to get the Maclaurin
series for f(x) = sin2(x):
f(x) = 1
2−1
21−(2x)2
2! +(2x)4
4! −(2x)6
6! +· · ·
Simplify this expression to get the Maclaurin series for f(x) = sin2(x):
f(x) = 1
2−1
2+x2
2! −x4
4! +x6
6! − · · · =
∞
X
n=0
(−1)nx2n+2
(2n+ 2)!
Therefore, the Maclaurin series for f(x) = sin2(x) is:
f(x) =
∞
X
n=0
(−1)nx2n+2
(2n+ 2)!
Question 6
Question
Find the Maclaurin series representation for the function f(x) = 1
1+x2.
5
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we can recall the geometric series
formula: 1
1−r= 1 + r+r2+r3+. . . , for |r|<1.
Step 1: Find the Maclaurin series for 1
1−x.
The function 1
1−xhas a known Maclaurin series:
1
1−x= 1 + x+x2+x3+. . . =
∞
X
n=0
xn
Step 2: Find the Maclaurin series for 1
1+x2.
Using the fact that 1
1−x= 1 + x+x2+. . ., we can substitute −x2for x:
1
1+(−x2)= 1 −x2+x4−x6+. . . =
∞
X
n=0
(−1)nx2n
Therefore, the Maclaurin series representation for f(x) = 1
1+x2is:
f(x) =
∞
X
n=0
(−1)nx2n= 1 −x2+x4−x6+. . .
Question 7
Question
Find the Maclaurin series for the function f(x) = ln(1 + x).
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. The Maclaurin series will be the Taylor
series centered at x= 0.
Step 1: Find the 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
.
.
.
6
Step 2: Evaluate the derivatives at x= 0
f(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 the Maclaurin series The Maclaurin series for ln(1 + x)
is given by:
ln(1 + x) = f(x) = f(0) + f′(0)
1! x+f′′(0)
2! x2+f′′′(0)
3! x3+· · ·
Substitute the values we found:
ln(1 + x) = 0 + 1 ·x−1·x2
2+ 2 ·x3
6+· · ·
Therefore, the Maclaurin series for ln(1 + x) is:
ln(1 + x) = x−x2
2+x3
3−x4
4+· · ·
Question 8
Question
Find the Maclaurin series for ln(1 + x) by differentiating its Taylor series.
Solution
To find the Maclaurin series for ln(1 + x), we will first find the Taylor series for
ln(1 + x) centered at a= 0 and then differentiate it term by term.
Step 1: Find the Taylor series for ln(1 + x)centered at a= 0 The
Taylor series for ln(1 + x) centered at a= 0 is given by:
ln(1 + x) =
∞
X
n=1
(−1)n−1xn
n
Step 2: Differentiate the Taylor series term by term Differentiating
the series term by term, we get:
d
dx ln(1 + x) = d
dx
∞
X
n=1
(−1)n−1xn
n
7
=
∞
X
n=1
d
dx (−1)n−1xn
n
=
∞
X
n=1
(−1)n−1xn−1
Therefore, the Maclaurin series for ln(1 + x) is:
ln(1 + x) =
∞
X
n=1
(−1)n−1xn
n
Question 9
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 its
derivatives and evaluating them at x= 0.
f(x) = ln(1 + x)
f′(x) = 1
1 + x(differentiate using chain rule)
f′′(x) = −1
(1 + x)2(differentiate the derivative of f)
f(3)(x) = 2
(1 + x)3(differentiate f′′(x))
f(4)(x) = −6
(1 + x)4(differentiate f(3)(x))
.
.
.
Now, let’s evaluate each derivative at x= 0 to find the coefficients of the
8
Maclaurin series for f(x).
f(0) = ln(1 + 0) = ln(1) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f(3)(0) = 2
(1 + 0)3= 2
f(4)(0) = −6
(1 + 0)4=−6
.
.
.
Therefore, the Maclaurin series for f(x) = ln(1 + x) is
ln(1 + x) = x−x2
2+x3
3−x4
4+· · · =
∞
X
n=1
(−1)n−1xn
n.
Question 10
Question
Find the Maclaurin series for the function f(x) = e2xsin(3x).
Solution
To find the Maclaurin series of f(x), we will use the formula for the Maclaurin
series of a product of two functions. The Maclaurin series for e2xand sin(3x)
are known, so we will use these series to find the series for f(x).
Step 1: Find the Maclaurin series for e2xThe Maclaurin series for e2x
is:
e2x=
∞
X
n=0
(2x)n
n!=
∞
X
n=0
2nxn
n!
Step 2: Find the Maclaurin series for sin(3x) The Maclaurin series for
sin(3x) is:
sin(3x) =
∞
X
n=0
(−1)n(3x)2n+1
(2n+ 1)! =
∞
X
n=0
(−1)n32n+1x2n+1
(2n+ 1)!
Step 3: Find the Maclaurin series for f(x) = e2xsin(3x) To find the
series for the product f(x) = e2xsin(3x), we multiply the series for e2xand
9
sin(3x):
f(x) = e2xsin(3x)
= ∞
X
n=0
2nxn
n!!· ∞
X
n=0
(−1)n32n+1x2n+1
(2n+ 1)! !
=
∞
X
n=0
n
X
k=0
2kxk
k!·(−1)n−k32(n−k)+1x2(n−k)+1
(2(n−k) + 1)!
=
∞
X
n=0
n
X
k=0
2k(−1)n−k32(n−k)+1xk+2(n−k)+1
k!(2(n−k) + 1)!
Simplifying further, we find:
f(x) =
∞
X
n=0
n
X
k=0
2k(−1)n−k32(n−k)+1x2n+1
k!(2(n−k) + 1)!
Therefore, the Maclaurin series for f(x) = e2xsin(3x) is given by:
∞
X
n=0
n
X
k=0
2k(−1)n−k32(n−k)+1x2n+1
k!(2(n−k) + 1)!
Question 11
Question
Find the Maclaurin series for the function 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) and then evaluate them at x= 0 to find the coefficients of the Maclaurin
series.
Step 1: Find the first few derivatives of f(x)
f(x) = (1 + x2)−1
f′(x) = −1(1 + x2)−2·2x=−2x(1 + x2)−2
f′′(x) = d
dx(−2x(1 + x2)−2) = −2(1 + x2)−2−2(−2x)(−2)(1 + x2)−3= 2(1 + x2)−3(2x2−1)
f′′′(x) = d
dx 2(1 + x2)−3(2x2−1)= 6(1 + x2)−4(2x2−1) −4(1 + x2)−3(4x)
= 2(1 + x2)−4(12x2−6−16x2) = 2(1 + x2)−4(12x2−6−16x2)
=−8x(1 + x2)−4
10
Step 2: Evaluate the derivatives at x= 0
f(0) = 1
f′(0) = 0
f′′(0) = 2
f′′′(0) = 0
Since the odd derivatives evaluated at x= 0 are all zero, the Maclaurin series
will only involve even powers of x. Thus, the Maclaurin series for f(x) = 1
1+x2
is:
f(x) = 1 + 2x2+8x4
3! +. . .
Question 12
Question
Find the Maclaurin series for the function f(x) = sin x
x.
Solution
To find the Maclaurin series for f(x) = sin x
x, we can first find the Maclaurin
series for sin xand then divide by x.
Step 1: Find the Maclaurin series for sin xThe Maclaurin series for
sin xis given by:
sin x=x−x3
3! +x5
5! −x7
7! +. . . =
∞
X
n=0
(−1)nx2n+1
(2n+ 1)!
Step 2: Divide by xto find the Maclaurin series for f(x) = sin xx
Dividing sin xby xgives:
sin x
x= 1 −x2
3! +x4
5! −x6
7! +. . . =
∞
X
n=0
(−1)nx2n
(2n+ 1)!
Therefore, the Maclaurin series for f(x) = sin x
xis P∞
n=0
(−1)nx2n
(2n+1)! .
Question 13
Question
Let f(x) = 1
1+x2. Find the first four nonzero terms of the Maclaurin series for
f(x).
11
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we recall 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 nth derivative of f(x) evaluated at x= 0.
Step 1: Find the derivatives of f(x).
Let’s start by finding the derivatives of f(x) = 1
1+x2:
f(x) = 1
1 + x2
f′(x) = −2x
(1 + x2)2
f′′(x) = 2(3x2−1)
(1 + x2)3
f′′′(x) = −4x(5x4−10x2+ 3)
(1 + x2)4
f(4)(x) = 4(35x4−45x2+ 6)
(1 + x2)5
Step 2: Evaluate the derivatives at x= 0.
We need to evaluate f(n)(0) for n= 0,1,2,3,4:
f(0)(0) = f(0) = 1
f(1)(0) = −2
f(2)(0) = 6
f(3)(0) = 0
f(4)(0) = 24
Step 3: Write the Maclaurin series.
Using the formula for the Maclaurin series, we have:
f(x)=1−2x+ 3x2−6x3+ 24x4+. . .
Therefore, the first four nonzero terms of the Maclaurin series for f(x) =
1
1+x2are 1 −2x+ 3x2−6x3+ 24x4.
Question 14
Question
Find the Maclaurin series for the function f(x) = 1
1+xand determine its interval
of convergence.
12
Solution
To find the Maclaurin series for f(x) = 1
1+x, we will use the fact that 1
1+x=
1−x+x2−x3+x4−. . . which is a geometric series where the common ratio
is −xand the first term is 1.
Step 1: Find the Maclaurin series using the formula for a geometric series.
f(x) = 1 −x+x2−x3+x4−. . . =
∞
X
n=0
(−1)nxn
Step 2: Determine the interval of convergence using the ratio test. Let’s
apply the ratio test:
lim
n→∞
(−1)n+1xn+1
(−1)nxn
= lim
n→∞ |x|=|x|
The series converges if |x|<1. Therefore, the interval of convergence is
(−1,1).
Question 15
Question
Find the Maclaurin series for the function f(x) = e−2x2.
Solution
To find the Maclaurin series for f(x), we will first find the Maclaurin series for
exand then substitute −2x2in place of x.
Step 1: Find the Maclaurin series for exThe Maclaurin series for ex
is:
ex=
∞
X
n=0
xn
n!
Step 2: Substitute −2x2for xTo find the Maclaurin series for e−2x2, we
substitute −2x2for xin the Maclaurin series for ex:
e−2x2=
∞
X
n=0
(−2x2)n
n!=
∞
X
n=0
(−2)nx2n
n!
Step 3: Simplify the series We can simplify the series further by sepa-
rating out the even powers of x:
e−2x2=
∞
X
n=0
(−1)n2n
n!x2n
Therefore, the Maclaurin series for f(x) = e−2x2is:
f(x) =
∞
X
n=0
(−1)n2n
n!x2n
13
Question 16
Question
Find the Maclaurin series for f(x) = ln(1 + x), and determine the interval of
convergence.
Solution
To find the Maclaurin series for f(x) = ln(1 + x), we’ll start by finding the
derivatives of f(x) and evaluating them at x= 0 to find the coefficients of the
series.
Step 1: Find the derivatives of f(x)
f(x) = ln(1 + x)⇒f′(x) = 1
1 + x⇒f′′ (x) = −1
(1 + x)2
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
Step 3: Write out the Maclaurin series for f(x) Since the Maclaurin series
is given by
f(x) = f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+. . .
we have
ln(1 + x) = 0 + 1x−1
2x2+1
3x3−. . .
Thus, the Maclaurin series for f(x) = ln(1 + x) is
ln(1 + x) = x−1
2x2+1
3x3−. . .
Step 4: Determine the interval of convergence The Maclaurin series for
ln(1 + x) converges for −1< x ≤1 (i.e., |x|<1) as it is based on the Taylor
series of a function around x= 0 which converges within the radius of conver-
gence. Thus, the interval of convergence is −1< x ≤1.
Question 17
Question
Find the Maclaurin series for the function f(x) = e2xcos(x).
14
Solution
To find the Maclaurin series for the function f(x) = e2xcos(x), we will utilize
the properties of Maclaurin series for e2xand cos(x).
Step 1: Find the Maclaurin series for e2x.The Maclaurin series for ex
is given by:
ex= 1 + x+x2
2! +x3
3! +x4
4! +. . . =
∞
X
n=0
xn
n!
Substitute 2xfor xin the series:
e2x= 1 + 2x+(2x)2
2! +(2x)3
3! +(2x)4
4! +. . . =
∞
X
n=0
(2x)n
n!
Simplyfying:
e2x= 1 + 2x+ 2x2+4x3
3+4x4
6+. . . =
∞
X
n=0
2nxn
n!
Step 2: Find the Maclaurin series for cos(x).The Maclaurin series for
cos(x) is given by:
cos(x)=1−x2
2! +x4
4! −x6
6! +. . . =
∞
X
n=0
(−1)nx2n
(2n)!
Substitute xfor xin the series:
cos(x)=1−x2
2! +x4
4! −x6
6! +. . . =
∞
X
n=0
(−1)nx2n
(2n)!
Step 3: Find the Maclaurin series for f(x) = e2xcos(x).Since f(x) =
e2xcos(x), the Maclaurin series for f(x) will be the product of the Maclaurin
series for e2xand cos(x):
f(x) = e2xcos(x) = ∞
X
n=0
2nxn
n!! ∞
X
n=0
(−1)nx2n
(2n)!!
Expanding the product, we get:
f(x) =
∞
X
n=0
n
X
k=0
2k(−1)n−kxn
k!(n−k)!
Therefore, the Maclaurin series for f(x) = e2xcos(x) is:
f(x) =
∞
X
n=0
n
X
k=0
2k(−1)n−kxn
k!(n−k)!
15
Question 18
Question
Find the Maclaurin series for f(x) = (1 + x)−1/2.
Solution
To find the Maclaurin series for f(x) = (1 + x)−1/2, we can start by finding
the derivatives of f(x) at x= 0 and writing out the general formula for the
Maclaurin series.
Step 1: Find the derivatives of f(x).
f(x) = (1 + x)−1/2
f′(x) = −1
2(1 + x)−3/2
f′′(x) = −1
2· −3
2(1 + x)−5/2=3
4(1 + x)−5/2
f′′′(x) = 3
4· −5
2(1 + x)−7/2=−15
8(1 + x)−7/2
.
.
.
Step 2: Determine the pattern and write the general formula for the Maclau-
rin series. From the derivatives, we can see that the nth derivative of f(x) eval-
uated at x= 0 is (−1)n(2n−1)!!
2n(1 + x)−(2n+1)/2. Therefore, the Maclaurin series
for f(x) is given by
f(x) =
∞
X
n=0
(−1)n(2n−1)!!
2nxn
Step 3: Express the Maclaurin series in terms of f(x). Substitute x=−x2
into the Maclaurin series formula to get:
(1 + x)−1/2=
∞
X
n=0
(−1)n(2n−1)!!
2n(−x2)n
Therefore, the Maclaurin series for f(x) = (1 + x)−1/2is
f(x) =
∞
X
n=0
(2n−1)!!
2nx2n
Question 19
Question
Find the Maclaurin series for the function f(x) = excos(x).
16
Solution
To find the Maclaurin series for f(x) = excos(x), we will first find the Maclaurin
series for exand cos(x) separately, and then multiply them together.
Step 1: Find the Maclaurin series for exThe Maclaurin series for ex
is given by:
ex=
∞
X
n=0
xn
n!
Step 2: Find the Maclaurin series for cos(x) The Maclaurin series for
cos(x) is given by:
cos(x) =
∞
X
n=0
(−1)nx2n
(2n)!
Step 3: Multiply the series together To find the Maclaurin series for
f(x) = excos(x), we multiply the series for exand cos(x) together term by
term:
f(x) = excos(x)
= ∞
X
n=0
xn
n!! ∞
X
n=0
(−1)nx2n
(2n)! !
=
∞
X
n=0 n
X
k=0
1
k!(n−k)!
(−1)n−k
(2(n−k))!!xn
Therefore, the Maclaurin series for f(x) = excos(x) is:
f(x) =
∞
X
n=0
cnxn
where
cn=
n
X
k=0
1
k!(n−k)!
(−1)n−k
(2(n−k))!
Question 20
Question
Find the Maclaurin series for the function f(x) = 1
1+x2.
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we will first find the derivative of
f(x) and evaluate it at x= 0 to find the coefficients of the series.
17
Step 1: Find the derivatives of f(x)
f(x) = 1
1 + x2
f′(x) = −2x
(1 + x2)2
f′′(x) = 2(3x2−1)
(1 + x2)3
f′′′(x) = −24x(1 −5x2)
(1 + x2)4
Step 2: Evaluate the derivatives at x= 0
f(0) = 1
f′(0) = 0
f′′(0) = 2
f′′′(0) = 0
Step 3: Write the Maclaurin series using the coefficients The Maclau-
rin series for f(x) is given by
f(x) = 1 + 0 ·x+2
2!x2+ 0 ·x3+24
4! x4+. . .
Simplifying this series, we get
f(x) = 1 + x2+6
5x4+. . .
Question 21
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 find the derivatives of
f(x) and evaluate them at x= 0 to determine the coefficients of the series.
Step 1: Find the first few derivatives of f(x).
f(x) = 1
1 + x
f′(x) = d
dx 1
1 + x=−(1 + x)−2=−1
(1 + x)2
f′′(x) = d
dx −1
(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
18
Step 2: Evaluate the derivatives at x= 0.
f(0) = 1
f′(0) = −1
f′′(0) = 2
f′′′(0) = −6
Step 3: Write the Maclaurin series for f(x).The Maclaurin series for
a function f(x) is given by
f(x) = f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+· · ·
Substitute the evaluated derivatives at x= 0 into the series:
f(x)=1−x+2
2!x2−6
3!x3+· · ·
Simplifying, we get the Maclaurin series for f(x) = 1
1+x:
f(x) = 1 −x+x2−x3+· · · =
∞
X
n=0
(−1)nxn
Question 22
Question
Find the Maclaurin series for f(x) = 1
1+x2and determine the interval of con-
vergence.
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we first need to express it as a
geometric series. Recall that the geometric series formula is:
1
1−r=
∞
X
n=0
rn
for |r|<1.
Step 1: Find the geometric series form We can rewrite f(x) = 1
1+x2
as:
f(x) = 1
1−(−x2)=
∞
X
n=0
(−x2)n=
∞
X
n=0
(−1)nx2n
Step 2: Determine the interval of convergence The interval of conver-
gence of a power series can be found using the ratio test. Let’s apply the ratio
test to the series P∞
n=0(−1)nx2n:
lim
n→∞
an+1
an
= lim
n→∞
(−1)n+1x2(n+1)
(−1)nx2n
= lim
n→∞ |x2|
19
The series converges if |x2|<1, which simplifies to −1< x < 1. Therefore,
the interval of convergence is (−1,1).
Thus, the Maclaurin series for f(x) = 1
1+x2is
∞
X
n=0
(−1)nx2n= 1 −x2+x4−x6+. . . (for −1<x<1)
Question 23
Question
Find the Maclaurin series for 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 then evaluate them at x= 0.
Step 1: Find the derivatives of f(x)
f(x) = ln(1 + x)
f′(x) = 1
1 + x(by the chain rule)
f′′(x) = −1
(1 + x)2(by the chain rule again)
f′′′(x) = 2
(1 + x)3(by the chain rule again)
f(4)(x) = −6
(1 + x)4(by the chain rule again)
Step 2: Evaluate derivatives at x= 0
f(0) = ln(1) = 0
f′(0) = 1
1= 1
f′′(0) = −1
12=−1
f′′′(0) = 2
13= 2
f(4)(0) = −6
14=−6
Therefore, the Maclaurin series for f(x) = ln(1 + x) is:
ln(1 + x) = 0 + x−1
2x2+1
3x3−1
4x4+. . .
20
Step 3: Determine the interval of convergence To find the interval of
convergence, we will use the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
1
n+1
1
n
= lim
n→∞
n
n+ 1
= 1 ⇒radius of convergence = R= 1
Thus, the interval of convergence is (−1,1].
Question 24
Question
Find the Maclaurin series for the function f(x) = 1
1+x2by differentiating the
function term-by-term.
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we first need to know the Maclau-
rin series for 1
1−x, which is
1
1−x= 1 + x+x2+x3+· · · =
∞
X
n=0
xn.
Next, we substitute −x2in place of xin the series:
1
1 + x2= 1 −x2+x4−x6+· · · =
∞
X
n=0
(−1)nx2n.
Therefore, the Maclaurin series for the function f(x) = 1
1+x2is P∞
n=0(−1)nx2n.
Question 25
Question
Find the Maclaurin series for the function f(x) = sin2(x).
Solution
To find the Maclaurin series for f(x) = sin2(x), we will first find the Maclaurin
series for sin(x) and then square it.
Step 1: Find the Maclaurin series for sin(x).The Maclaurin series for
sin(x) is given by:
sin(x) = x−x3
3! +x5
5! −x7
7! +· · · =
∞
X
n=0
(−1)nx2n+1
(2n+ 1)!
21
Step 2: Square the Maclaurin series for sin(x).Squaring the Maclaurin
series for sin(x), we get:
(sin(x))2= ∞
X
n=0
(−1)nx2n+1
(2n+ 1)!!2
Expanding the squared series gives:
(sin(x))2=x2−2
3!x4+2
5!x6−2
7!x8+· · ·
Therefore, the Maclaurin series for f(x) = sin2(x) is:
f(x) = x2−2
3!x4+2
5!x6−2
7!x8+· · ·
Question 26
Question
Find the Maclaurin series for f(x) = 1
1+x+x2and determine the interval of
convergence.
Solution
To find the Maclaurin series for f(x) = 1
1+x+x2, we will begin by expressing
f(x) as a geometric series.
Step 1: Find the geometric series representation Since 1
1+x+x2=
1
(x+1/2)2−1/4, we rewrite f(x) as
1
1 + x+x2=1
(x+ 1/2)2−1/4=1
(x+ 1/2−1/2)(x+ 1/2+1/2) =A
x+ 1/2−1/2+B
x+ 1/2+1/2,
to find the constants Aand B.
Step 2: Find the values of Aand BWe rewrite 1
1+x+x2using partial
fraction decomposition to get:
1
1 + x+x2=A
x+B
x+ 1.
Multiplying through by x(x+ 1) gives
1 = A(x+ 1) + Bx.
Substituting x= 0 in the equation above:
1 = A(0 + 1) =⇒A= 1.
22
Substituting x=−1 in the equation above:
1 = B(−1) =⇒B=−1.
Step 3: Write out the Maclaurin series Our function f(x) can be
rewritten as
f(x) = 1
1 + x+x2=1
x−1
x+ 1.
Thus, the Maclaurin series for f(x) is
f(x)=1−x+x2−x3+x4−x5+. . . =
∞
X
n=0
(−1)nxn.
Step 4: Determine the interval of convergence To find the interval
of convergence for the series, note that the series converges for −1< x < 1.
Therefore, the interval of convergence is (−1,1).
Question 27
Question
Find the Maclaurin series for f(x) = 1
1+xby directly differentiating the geomet-
ric series representation of f(x).
Solution
To find the Maclaurin series for f(x) = 1
1+x, we will directly differentiate the
geometric series representation of f(x).
The geometric series representation of 1
1+xis:
1
1 + x= 1 −x+x2−x3+x4−. . . =
∞
X
n=0
(−1)nxn
Step 1: Differentiate the geometric series term by term
f′(x) = d
dx ∞
X
n=0
(−1)nxn!
Differentiating the series term by term:
f′(x) =
∞
X
n=0
d
dx((−1)nxn)
f′(x) =
∞
X
n=0
(−1)n·nxn−1
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Step 2: Find the second derivative of the series
f′′(x) = d
dx ∞
X
n=0
(−1)n·nxn−1!
Differentiating the series term by term:
f′′(x) =
∞
X
n=0
d
dx((−1)n·nxn−1)
f′′(x) =
∞
X
n=0
(−1)n·n(n−1)xn−2
Step 3: Generalize the derivative pattern By observing the derivatives,
we can see that the nth derivative of f(x) evaluated at x= 0 is
f(n)(0) = (−1)n·n!
Step 4: Write the Maclaurin series for f(x) The Maclaurin series for
f(x) = 1
1+xis:
f(x) = f(0) + f′(0)
1! x+f′′(0)
2! x2+f′′′(0)
3! x3+. . .
= 1 −x+x2−x3+x4−. . . =
∞
X
n=0
(−1)nxn
Therefore, the Maclaurin series for f(x) = 1
1+xis P∞
n=0(−1)nxn.
Question 28
Question
Find the Maclaurin series for the function f(x) = 1
1+x.
Solution
To find the Maclaurin series for f(x) = 1
1+x, we will expand the function in a
power series centered at x= 0.
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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
(1 + x)3
f′′′(x) = d
dx 21
(1 + x)3
=−61
(1 + x)4
Step 2: Find the Maclaurin series. The Maclaurin series expansion of
f(x) is given by
f(x) = f(0) + f′(0)
1! x+f′′(0)
2! x2+f′′′(0)
3! x3+. . .
Substitute the derivatives of f(x) we found into the formula:
f(0) = 1
1+0 = 1
f′(0) = −1
(1 + 0)2=−1
f′′(0) = 2 1
(1 + 0)3= 2
f′′′(0) = −61
(1 + 0)4=−6
Therefore, the Maclaurin series for f(x) = 1
1+xis:
1−x+ 2x2−6x3+. . .
Question 29
Question
Find the Maclaurin series for the function f(x) = sinx
2.
25
Solution
To find the Maclaurin series for f(x) = sinx
2, we will first find the Maclaurin
series for sin(x) and then replace xwith x
2.
Step 1: Find the Maclaurin series for sin(x). The Maclaurin series for sin(x)
is given by:
sin(x) = x−x3
3! +x5
5! −x7
7! +· · · =
∞
X
n=0
(−1)nx2n+1
(2n+ 1)!
Step 2: Replace xwith x
2in the Maclaurin series for sin(x). Substitute
x=x
2:
f(x) = sin x
2=
∞
X
n=0
(−1)nx
22n+1
(2n+ 1)!
=
∞
X
n=0
(−1)nx2n+1
22n+1(2n+ 1)!
=
∞
X
n=0
(−1)nx2n+1
22n+1(2n+ 1)!
Therefore, the Maclaurin series for f(x) = sinx
2is:
f(x) =
∞
X
n=0
(−1)nx2n+1
22n+1(2n+ 1)!
Question 30
Question
Find the Maclaurin series representation for the function f(x) = sin2(x).
Solution
To find the Maclaurin series representation for f(x) = sin2(x), we first need to
express f(x) in terms of a known Maclaurin series. Recall the trigonometric
identity sin2(x) = 1−cos(2x)
2.
Step 1: Find the Maclaurin series for cos(2x). We know that the Maclaurin
series for cos(x) is given by:
cos(x)=1−x2
2! +x4
4! −x6
6! +. . . =
∞
X
n=0
(−1)nx2n
(2n)!
Substitute 2xfor xto get the Maclaurin series for cos(2x):
cos(2x)=1−(2x)2
2! +(2x)4
4! −(2x)6
6! +. . . =
∞
X
n=0
(−1)n(2x)2n
(2n)!
26
Simplify this series to get:
cos(2x)=1−2x2+4x4
3! −8x6
6! +. . . =
∞
X
n=0
(−1)n22nx2n
(2n)!
Step 2: Rewrite sin2(x) in terms of the Maclaurin series for cos(2x). Using
the trigonometric identity, we have:
sin2(x) = 1−cos(2x)
2=1
2−1
2cos(2x)
Substitute the Maclaurin series for cos(2x) into the expression above:
sin2(x) = 1
2−1
21−2x2+4x4
3! −8x6
6! +. . .
Simplify to get:
sin2(x) = 1
2−1
2+x2−2x4
3! +4x6
6! −. . .
sin2(x) = x2−x4
3+x6
45 −. . .
Therefore, the Maclaurin series for f(x) = sin2(x) is:
f(x) = x2−x4
3+x6
45 −. . .
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