MATH 100 - FUNDAMENTALS OF
MATHEMATICS - Taylor and
Maclaurin Series
Question Bank - Set 2
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 express f(x) in terms
of a known series expansion and then determine the coefficients in the series.
Step 1: We know that the geometric series expansion for 1
1−xis given by:
1
1−x=
∞
X
n=0
xn
Step 2: To find the Maclaurin series for f(x), we need to express it in the
form P∞
n=0 anxn. By comparing with the geometric series expansion, we have:
f(x) = 1
1−x=
∞
X
n=0
xn
Step 3: Now, we need to find the coefficients an. From the above equation,
it is clear that an= 1 for all n.
Step 4: Therefore, the Maclaurin series for f(x) = 1
1−xis:
f(x) =
∞
X
n=0
xn= 1 + x+x2+x3+. . .
Question 2
Question
Find the Maclaurin series representation for the function f(x) = e2xsin(3x).
Solution
To find the Maclaurin series representation for the function f(x) = e2xsin(3x),
we will first compute the derivatives of f(x) at x= 0 and then write out the
Maclaurin series using the formula for the Maclaurin series representation.
Step 1: Find the derivatives of f(x)
f(x) = e2xsin(3x)
f′(x) = (2e2x)(sin(3x)) + e2x(3 cos(3x))
f′′(x) = (22e2x)(sin(3x)) + 2(2e2x)(3 cos(3x)) + (e2x)(−32sin(3x))
= 4e2xsin(3x) + 12e2xcos(3x)−9e2xsin(3x)
= 5e2xsin(3x) + 12e2xcos(3x)
f′′′(x) = 5(2e2x)(sin(3x)) + 5(12e2x)(cos(3x))
+ 12(2e2x)(cos(3x)) −12(12e2x)(sin(3x))
= 10e2xsin(3x) + 60e2xcos(3x) + 24e2xcos(3x)−144e2xsin(3x)
=−134e2xsin(3x) + 84e2xcos(3x)
Step 2: Write out the Maclaurin series Using the derivatives of f(x)
evaluated at x= 0, we can write the Maclaurin series representation of f(x):
f(x) = f(0) + f′(0)
1! x+f′′(0)
2! x2+f′′′(0)
3! x3+· · ·
Plugging in the values we found:
f(x) = 1 + 5
1!x+12
2! x2+84
3! x3+· · ·
Therefore, the Maclaurin series representation for f(x) = e2xsin(3x) is 1 +
5x+ 6x2+ 14x3+· · · .
Question 3
Question
Find the Maclaurin series for f(x) = ln(1 + x) and determine the interval of
convergence.
2
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 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(3)(x) = d
dx −1
(1 + x)2=2
(1 + x)3
.
.
.
The nth derivative of f(x) is given by
f(n)(x)=(−1)n−1(n−1)!
(1 + x)n
Step 2: Evaluate the derivatives at x= 0
f(0) = ln(1) = 0
f′(0) = 1
1= 1
f′′(0) = −1
12=−1
f(3)(0) = 2
13= 2
.
.
.
f(n)(0) = (−1)n−1(n−1)!
1= (−1)n−1(n−1)!
Step 3: Construct the Maclaurin series The Maclaurin series for f(x) =
ln(1 + x) is given by the formula
f(x) =
∞
X
n=0
f(n)(0)
n!xn=
∞
X
n=0
(−1)n−1(n−1)!
n!xn=
∞
X
n=0
(−1)n−1xn
n
Step 4: Determine the interval of convergence To find the interval of
convergence, we will use the ratio test. Let an= (−1)n−11
n. Then,
lim
n→∞
an+1
an
= lim
n→∞
(−1)n1
n+1
(−1)n−11
n
= lim
n→∞
n
n+ 1
= 1
Since the ratio test gives 1, the interval of convergence is the open interval
(−1,1).
3
Question 4
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 first find
the derivatives of f(x) at x= 0 to obtain the coefficients of the Maclaurin series.
Step 1: Find f′(x)and f′′ (x).
f(x) = 1
1 + x2
f′(x) = d
dx 1
1 + x2
=−2x
(1 + x2)2
f′′(x) = d
dx −2x
(1 + x2)2
=−2(1 + x2)2−2x(2)(1 + x2)(2x)
(1 + x2)4
=−2(1 + x2)(1 −2x2)
(1 + x2)4
=−2(1 −2x2+x2)
(1 + x2)3
=−2(1 −3x2)
(1 + x2)3
Step 2: Find f′(0) and f′′ (0).Evaluate f′(x) and f′′ (x) at x= 0:
f′(0) = −2(0)
(1 + 0)2= 0
f′′(0) = −2(1 −0)
(1 + 0)3=−2
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! +· · ·
Therefore, the Maclaurin series representation for f(x) = 1
1+x2is:
1
1 + x2= 1 −2x2+2x4
2! −2(3x2)x4
3! +· · ·
4
Question 5
Question
Find the Maclaurin series for the function f(x) = ln(1 + x) and determine its
radius of convergence.
Solution
Given f(x) = ln(1 + x), we can find its Maclaurin series representation by
differentiating this function and evaluating at x= 0.
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(n)(x)=(−1)n−1(n−1)!
(1 + x)n
Step 2: Evaluate f(n)(0).
f(n)(0) = (−1)n−1(n−1)!
1n= (−1)n−1(n−1)!
Step 3: Write out the Maclaurin series using the formula f(x) = f(0) +
f′(0)x+f′′ (0)
2! x2+· · · +f(n)(0)
n!xn.
f(x) = ln(1 + x) =
∞
X
n=0
(−1)n−1xn
n
Step 4: Determine the radius of convergence using the ratio test.
lim
n→∞
an+1
an
= lim
n→∞
(−1)nxn+1
n+1
(−1)n−1xn
n
= lim
n→∞
x
n+ 1
= 0
Since the limit is less than 1 for all x, the radius of convergence is R=∞.
5
Question 6
Question
Find the Maclaurin series for f(x) = ln(1 + x) and determine its radius of
convergence.
Solution
Step 1: Find the derivatives of f(x) and evaluate them at x= 0 to obtain the
coefficients of the Maclaurin series.
f(x) = ln(1 + x)
f′(x) = 1
1 + x⇒f′(0) = 1
f′′(x) = −1
(1 + x)2⇒f′′ (0) = −1
f′′′(x) = 2
(1 + x)3⇒f′′′ (0) = 2
f(4)(x) = −6
(1 + x)4⇒f(4)(0) = −6
.
.
.
Step 2: Write out the Maclaurin series by using the coefficients obtained.
ln(1 + x) = 1 −x+1
2x2−1
3x3+1
4x4−. . . =
∞
X
n=1
(−1)n−1xn
n
Step 3: Determine the radius of convergence using the ratio test.
R= lim
n→∞
(−1)nxn+1
n+1
(−1)n−1xn
n
= lim
n→∞ |x| · n
n+ 1 =|x|<1
So, the radius of convergence is R= 1.
Question 7
Question
Find the Maclaurin series representation for the function f(x) = 1
1+x2.
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we will first express f(x) as a
geometric series and then find the Maclaurin series by expanding it.
6
Step 1: Find the Maclaurin series for 11 + x2
We know that the geometric series is given by:
1
1−r= 1 + r+r2+r3+. . . for |r|<1.
Thus, we can express 1
1+x2as a geometric series:
1
1 + x2=1
1−(−x2)= 1 + (−x2)+(−x2)2+ (−x2)3+. . .
= 1 −x2+x4−x6+. . . =
∞
X
n=0
(−1)nx2n.
Step 2: Determine the Maclaurin series of f(x)
Since the Maclaurin series representation of 1
1+x2is P∞
n=0(−1)nx2n, we can
directly write the Maclaurin series of f(x) as:
f(x) =
∞
X
n=0
(−1)nx2n.
Therefore, the Maclaurin series representation for the function f(x) = 1
1+x2is
P∞
n=0(−1)nx2n.
Question 8
Question
Find the Maclaurin series for f(x) = excos x.
Solution
To find the Maclaurin series for f(x) = excos x, we will first express exand cos x
as power series expansions around x= 0, and then multiply them together.
Step 1: Find the power series expansion for exThe power series
expansion for exis given by:
ex=
∞
X
n=0
xn
n!
So,
ex= 1 + x+x2
2! +x3
3! +x4
4! +. . .
Step 2: Find the power series expansion for cos xThe power series
expansion for cos xis given by:
cos x=
∞
X
n=0
(−1)nx2n
(2n)!
7
So,
cos x= 1 −x2
2! +x4
4! −x6
6! +. . .
Step 3: Multiply the series Multiplying the power series expansions for
exand cos x, we get the Maclaurin series for excos x:
f(x) = excos x
= (1 + x+x2
2! +x3
3! +x4
4! +. . .)(1 −x2
2! +x4
4! −x6
6! +. . .)
= 1 + (x−x3
2! )+(x2
2! −x4
2! ·2!) + . . .
So, the Maclaurin series for f(x) = excos xis
1+(x−x3
2) + x2
2−x4
12 +. . .
Question 9
Question
Find the Maclaurin series representation for 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−xand substitute −x2in place of x.
Step 1: Find the Maclaurin series of 11−x.Starting with the geo-
metric series formula, 1
1−x= 1 + x+x2+x3+. . ., we have:
1
1−x2= 1 + x2+x4+x6+. . .
Step 2: Find the Maclaurin series of 11−x2.Substitute −x2in place
of xin the series representation of 1
1−x:
1
1 + x2= 1 −x2+x4−x6+. . .
Therefore, the Maclaurin series representation for f(x) = 1
1−x2is 1 −x2+
x4−x6+. . ..
Question 10
Question
Find the Maclaurin series for the function f(x) = ln(1 + x).
8
Solution
To find the Maclaurin series for f(x) = ln(1 + x), we can start by finding deriva-
tives of f(x) and then evaluating them at x= 0.
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
Step 2: Find f(n)(0) for each derivative.
f(0) = ln(1) = 0
f′(0) = 1
1= 1
f′′(0) = −1
12=−1
f′′′(0) = 2
13= 2
Step 3: Express the Maclaurin series in sigma notation. The Maclau-
rin series for f(x) = ln(1 + x) is given by:
f(x) =
∞
X
n=0
f(n)(0)
n!xn
Step 4: Substitute the values of f(n)(0) into the Maclaurin series
formula.
f(x) = 0 + 1x−1
2!x2+2
3!x3−. . .
f(x) = x−x2
2+x3
3−. . .
Therefore, the Maclaurin series for f(x) = ln(1 + x) is f(x) = x−x2
2+x3
3−
. . ..
Question 11
Question
Find the Maclaurin series for the function f(x) = sin2(x).
9
Solution
To find the Maclaurin series for f(x) = sin2(x), we first need to express sin2(x)
in terms of well-known Maclaurin series for elementary functions.
Step 1: Express sin2(x)using trigonometric identities We know that
sin2(x) = 1
2(1 −cos(2x)).
Step 2: Find the Maclaurin series for 12(1 −cos(2x)) We start by
finding the Maclaurin series for 1 and for cos(2x) separately. - Maclaurin series
for 1: 1 + 0x+ 0x2+ 0x3+. . . = 1 - Maclaurin series for cos(2x): cos(0) −
2 sin(0)x+0x2+(−2)2
2! cos(0)x2+(−2)3
3! sin(0)x3+. . . = 1−2x2+22
2! x2−23
3! x3+. . . =
1−2x2+ 2x2−8
3x3+. . . = 1 −8
3x3+. . .
Thus, the Maclaurin series for sin2(x) is f(x) = 1
2(1 −cos(2x)) = 1
2(1 −1 +
8
3x2−. . .) = 4
3x2−. . ..
Question 12
Question
Find the Maclaurin series for f(x) = 1
1+x2.
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we can begin by expressing f(x)
as a geometric series. Recall that the geometric series formula is:
1
1−a= 1 + a+a2+a3+. . .
where |a|<1.
Thus, we rewrite f(x) as:
1
1 + x2=1
1−(−x2)= 1 + (−x2)+(−x2)2+ (−x2)3+. . .
Now, we simplify the terms:
1−x2+x4−x6+. . .
Therefore, the Maclaurin series for f(x) = 1
1+x2is:
1−x2+x4−x6+. . . =
∞
X
n=0
(−1)nx2n
Question 13
Question
Find the Maclaurin series for f(x) = 1
x−3.
10
Solution
To find the Maclaurin series for f(x) = 1
x−3, we first need to express f(x) in
terms of a power series. Since we are looking for a Maclaurin series (centered
at x= 0), we will use the formula for the geometric series: 1
1−r=P∞
n=0 rnfor
|r|<1.
Step 1: Rewrite the given function in terms of a power series.
f(x) = 1
x−3=1
−3·1
1−x
3
=−1
3
∞
X
n=0 x
3n
=−1
3
∞
X
n=0
xn
3n=−
∞
X
n=0
xn
3n+1
Step 2: Determine the coefficients of the Maclaurin series. The Maclaurin
series for f(x) will be in the form:
f(x) =
∞
X
n=0
anxn
Comparing this with −1
3P∞
n=0
xn
3n+1 , we see that an=−1
3n+1 .
Step 3: Write down the Maclaurin series. Thus, the Maclaurin series for
f(x) = 1
x−3is:
f(x) = −
∞
X
n=0
xn
3n+1 =−x
3+x2
32+x3
33+. . .
Question 14
Question
Find the Maclaurin series expansion for f(x) = e2xcos(3x).
Solution
To find the Maclaurin series expansion for f(x) = e2xcos(3x), we will first find
the Maclaurin series expansions for e2xand cos(3x), and then multiply these
series together.
Step 1: Find the Maclaurin series expansion for e2x.The Maclaurin
series expansion for exis given by:
ex=
∞
X
n=0
xn
n!
Replacing xwith 2x, we have:
11
e2x=
∞
X
n=0
(2x)n
n!
e2x=
∞
X
n=0
2nxn
n!
So, the Maclaurin series expansion for e2xis P∞
n=0 2nxn
n!.
Step 2: Find the Maclaurin series expansion for cos(3x).The Maclau-
rin series expansion for cos(x) is given by:
cos(x) =
∞
X
n=0
(−1)nx2n
(2n)!
Replacing xwith 3x, we have:
cos(3x) =
∞
X
n=0
(−1)n(3x)2n
(2n)!
cos(3x) =
∞
X
n=0
(−1)n32nx2n
(2n)!
So, the Maclaurin series expansion for cos(3x) is P∞
n=0
(−1)n32nx2n
(2n)! .
Step 3: Multiply the two series together. To get the Maclaurin series
expansion for f(x) = e2xcos(3x), we multiply the Maclaurin series for e2xby
the Maclaurin series for cos(3x).
Multiplying the two series together term by term, we obtain:
f(x) = ∞
X
n=0
2nxn
n!! ∞
X
n=0
(−1)n32nx2n
(2n)! !
f(x) =
∞
X
n=0
n
X
k=0
2k
k!·(−1)n−k32(n−k)xn
(n−k)! ·(2(n−k))!
After simplifying, we get the Maclaurin series expansion for f(x) = e2xcos(3x).
Question 15
Question
Find the Maclaurin series representation for f(x) = 1
1+x2.
12
Solution
To find the Maclaurin series representation for f(x) = 1
1+x2, we can use the
formula for the Maclaurin series of 1
1−xand substitute x2for x.
Step 1: Find the Maclaurin series of 11−x
The Maclaurin series of 1
1−xis given by:
1
1−x= 1 + x+x2+x3+. . . =
∞
X
n=0
xn
Step 2: Substitute x2for x
Substitute x2for xin the Maclaurin series of 1
1−xto get the Maclaurin series
representation of 1
1+x2:
1
1 + x2= 1 + x2+x4+x6+. . . =
∞
X
n=0
x2n
Therefore, the Maclaurin series representation for f(x) = 1
1+x2is P∞
n=0 x2n.
Question 16
Question
Find the Maclaurin series for the function f(x) = e2xcos(3x).
Solution
To find the Maclaurin series for the given function, we will first express f(x) as
a power series, and then determine its coefficients.
Step 1: Express f(x)as a power series. We know that the Maclaurin
series for exis given by:
ex=
∞
X
n=0
xn
n!
And the Maclaurin series for cos xis:
cos x=
∞
X
n=0
(−1)nx2n
(2n)!
Therefore, the Maclaurin series for f(x) = e2xcos(3x) can be expressed as:
f(x) = ∞
X
n=0
(2x)n
n!! ∞
X
n=0
(−1)n(3x)2n
(2n)! !
Multiplying these series together will give us the desired power series for
f(x).
13
Step 2: Determine the coefficients. After multiplying the series to-
gether, we can simplify the resulting expression by collecting like terms and
finding the coefficients of xn.
The Maclaurin series for f(x) will be in the form:
f(x) =
∞
X
n=0
anxn
Therefore, we need to determine the coefficients anby finding the prod-
uct of the corresponding terms in the two series. This is a somewhat tedious
calculation, but by distributing and rearranging terms, we can determine the
coefficients of the power series for f(x).
The final expression for the Maclaurin series of f(x) = e2xcos(3x) will be
the sum of these terms with the determined coefficients.
Thus, the Maclaurin series for f(x) = e2xcos(3x) can be found by multi-
plying the Maclaurin series for e2xand cos(3x), and carefully simplifying the
resulting expression to determine the coefficients an.
Question 17
Question
Find the Maclaurin series for the function f(x) = sin2(x) up to the term involv-
ing x4.
Solution
To find the Maclaurin series for f(x) = sin2(x), we use the identity sin2(x) =
1−cos(2x)
2and then expand the cosine term as a Maclaurin series.
Step 1: Find the Maclaurin series for cos(2x).The Maclaurin series
for cos(2x) is given by
cos(2x)=1−(2x)2
2! +(2x)4
4! −(2x)6
6! +· · ·
Simplifying, we have
cos(2x)=1−2x2+2x4
3! −2x6
5! +· · ·
Step 2: Use the identity sin2(x) = 1−cos(2x)2.Substitute the Maclaurin
series for cos(2x) into the identity:
sin2(x) = 1−(1 −2x2+2x4
3! −2x6
5! +· · · )
2
Simplify to get
sin2(x) = x2−x4
3! +x6
5! − · · ·
14
Step 3: Determine the Maclaurin series for f(x) = sin2(x).Now, we
have the Maclaurin series for f(x) = sin2(x) up to the term involving x4:
f(x) = x2−x4
3! +x6
5! − · · ·
Therefore, the Maclaurin series for f(x) = sin2(x) up to the term involving
x4is x2−x4
6.
Question 18
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 to determine the coefficients of the series.
Step 1: Find f′(x)
f(x) = 1
1 + x= (1 + x)−1
Using the power rule, we have:
f′(x) = (−1)(1 + x)−2=−(1 + x)−2
Evaluate f′(0):
f′(0) = −1
Step 2: Find f′′ (x) Differentiate f′(x):
f′′(x) = −(−2)(1 + x)−3= 2(1 + x)−3
Evaluate f′′(0):
f′′(0) = 2
Step 3: Find f′′′ (x) Differentiate f′′(x):
f′′′(x) = 3(−2)(1 + x)−4=−6(1 + x)−4
Evaluate f′′′(0):
f′′′(0) = −6
Step 4: Generalize the pattern From the calculated derivatives at x= 0,
we observe the pattern:
f(n)(0) = (−1)n·n!
15
Step 5: Write the Maclaurin series The Maclaurin series for f(x) = 1
1+x
is:
f(x) = f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+· · ·
= 1 −x+x2−x3+· · ·
=
∞
X
n=0
(−1)nxn
Question 19
Question
Find the Maclaurin series for the function f(x) = 1
1+x.
Solution
To find the Maclaurin series for the function f(x) = 1
1+x, we will first express
f(x) as a geometric series, and then find the general term of the Maclaurin
series.
Step 1: Express f(x)as a geometric series The function f(x) = 1
1+x
can be expressed as a geometric series using the formula for a geometric series:
1
1 + x=1
1−(−x)=
∞
X
n=0
(−x)n.
Step 2: Find the general term of the Maclaurin series In the geo-
metric series, the nth term is (−x)n. To express this as the nth term of the
Maclaurin series, we need to find the corresponding term of the Maclaurin series,
which is f(n)(0)
n!xn.
Let’s find the nth derivative of f(x) and evaluate it at x= 0:
f(x) = 1
1 + x
f′(x) = −1
(1 + x)2
f′′(x) = 2
(1 + x)3
f′′′(x) = −6
(1 + x)4
.
.
.
From this pattern, we can see that the nth derivative of f(x) evaluated at x= 0
is:
f(n)(0) = (−1)nn!.
16
Therefore, the nth term of the Maclaurin series for f(x) is:
f(n)(0)
n!xn=(−1)nn!
n!xn= (−1)nxn.
Step 3: Write the Maclaurin series Thus, the Maclaurin series for
f(x) = 1
1+xis:
∞
X
n=0
(−1)nxn.
Question 20
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 then evaluate them at x= 0 to determine the coefficients in the
Maclaurin 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
(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
We can observe a pattern emerging with the derivatives.
Step 2: Evaluate 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
17
Step 3: Find 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+ 2x2−6x3+· · ·
Step 4: Determine the interval of convergence Since the function
f(x) = 1
1+xis in the form of a geometric series, the Maclaurin series converges
for |x|<1. Therefore, the interval of convergence is (−1,1).
Question 21
Question
Find the Maclaurin series for f(x) = sin(2x).
Solution
To find the Maclaurin series for f(x) = sin(2x), we will first find the derivatives
of f(x) at x= 0 to determine the coefficients of the series.
Step 1: Find f′(x)
f(x) = sin(2x)
f′(x) = 2 cos(2x)
Step 2: Find f′′ (x)
f′′(x) = −4 sin(2x)
Step 3: Find f′′′ (x)
f′′′(x) = −8 cos(2x)
Step 4: Find f(4)(x)
f(4)(x) = 16 sin(2x)
Step 5: Find f(5)(x)
f(5)(x) = 32 cos(2x)
Step 6: Find f(6)(x)
f(6)(x) = −64 sin(2x)
From the above calculations, we can see that the derivatives of f(x) = sin(2x)
at x= 0 alternate between ±2nwhere nis the power of xin the series expansion.
18
Step 7: Write the Maclaurin series So, the Maclaurin series for f(x) =
sin(2x) is given by:
sin(2x)=2x−23x3
3! +25x5
5! −27x7
7! +· · ·
Thus, the Maclaurin series for f(x) = sin(2x) is 2
∞
X
n=0
(−1)n22n+1x2n+1
(2n+ 1)! .
Question 22
Question
Find the Maclaurin series representation for f(x) = 1
1+x2.
Solution
To find the Maclaurin series for f(x), we first need to find the derivatives of
f(x) and evaluate them at x= 0 to find the coefficients of the series.
Step 1: Find the first few derivatives of f(x).
f(x) = 1
1 + x2
f′(x) = −(1 + x2)−2·2x=−2x(1 + x2)−2
f′′(x) = −2(1 + x2)−2−2x·(−2)(1 + x2)−3·2x= 2(3x2−1)(1 + x2)−3
Step 2: Find the Maclaurin series for f(x).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+· · ·
Now, let’s evaluate f(0), f′(0), and f′′(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
Thus, the Maclaurin series for f(x) is:
1−2
2!x2+1
4!x4−2
6!x6+· · ·
Question 23
Question
Find the Maclaurin series representation for f(x) = 1
x2+4 .
19
Solution
To find the Maclaurin series representation for f(x) = 1
x2+4 , we will first express
it in the form of a geometric series.
Step 1: Express f(x)as a geometric series We note that 1
x2+4 =
1
4·1
1+(x
2)2.
Since x
2= 1, we can rewrite 1
1+(x
2)2as an infinite geometric series:
1
1−r= 1 + r+r2+r3+. . . , for |r|<1
Here, r=−x
22=−x2
4, so we have:
1
x2+ 4 =1
4·1
1−x
22=1
4
∞
X
n=0 x
22n
Step 2: Find the Maclaurin series representation To find the Maclau-
rin series representation, we substitute a= 0 into the series from Step 1:
1
x2+ 4 =1
4
∞
X
n=0 x
22n
=1
4
∞
X
n=0
x2n
22n=1
4
∞
X
n=0
x2n
4n
Therefore, the Maclaurin series representation for f(x) = 1
x2+4 is 1
4P∞
n=0
x2n
4n.
Question 24
Question
Find the Maclaurin series for the function f(x) = e2xcos(3x).
Solution
To find the Maclaurin series for f(x) = e2xcos(3x), we can use the formula for
the Maclaurin series of exand cos(x), then multiply the two series together.
Step 1: Find the Maclaurin series for e2x.
The Maclaurin series for exis given by:
ex=
∞
X
n=0
xn
n!
Substitute 2xfor x:
e2x=
∞
X
n=0
(2x)n
n!=
∞
X
n=0
2nxn
n!
Step 2: Find the Maclaurin series for cos(3x).
20
The Maclaurin series for cos(x) is given by:
cos(x) =
∞
X
n=0
(−1)nx2n
(2n)!
Substitute 3xfor x:
cos(3x) =
∞
X
n=0
(−1)n(3x)2n
(2n)! =
∞
X
n=0
(−1)n32nx2n
(2n)!
Step 3: Multiply the two series together.
The Maclaurin series for f(x) = e2xcos(3x) is given by multiplying the two
individual Maclaurin series together:
f(x) = e2xcos(3x) = ∞
X
n=0
2nxn
n!! ∞
X
n=0
(−1)n32nx2n
(2n)! !
Now, multiply the two series term by term and simplify to obtain the Maclau-
rin series for f(x).
Question 25
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) and then evaluate those derivatives at x= 0.
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
.
.
.
21
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)x2
2! +f′′′(0)x3
3! +· · · =
∞
X
n=0
f(n)(0)xn
n!
Substitute the values we found in Step 2 into the series to get the Maclaurin
series for f(x):
f(x) = 1 −x+x2−x3+· · · =
∞
X
n=0
(−1)nxn
Question 26
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 first find
its derivatives and evaluate them at x= 0 to determine the coefficients of the
series.
Step 1: Find the derivatives of f(x).
f(x) = 1
1 + x2= (1 + x2)−1
f′(x) = −(1 + x2)−2·2x=−2x(1 + x2)−2
f′′(x) = −2(1 + x2)−2−2x·(−2)(1 + x2)−3·2x= 4x2(1 + x2)−3
f′′′(x) = 4(1 + x2)−3+ 4x2·(−3)(1 + x2)−4·2x=−24x3(1 + x2)−4
.
.
.
22
Step 2: Evaluate the derivatives at x= 0.
f(0) = 1
f′(0) = 0
f′′(0) = 4
f′′′(0) = 0
.
.
.
Step 3: Write the Maclaurin series. Since f(0) = 1, f′(0) = 0, f′′(0) =
4, f′′′(0) = 0, and so on, the Maclaurin series for f(x) is:
f(x) = 1 + 0 ·x+4
2!x2+ 0 ·x3+24
3! x4+. . . = 1 + 2x2+4
3x4+. . .
Therefore, the Maclaurin series representation for f(x) = 1
1+x2is 1 + 2x2+
4
3x4+. . ..
Question 27
Question
Find the Maclaurin series representation for the function f(x) = 1
1+x.
Solution
To find the Maclaurin series for f(x) = 1
1+x, we can use the formula for the
geometric series:
1
1−u=
∞
X
n=0
un
where |u|<1.
Step 1: Find the term uIn this case, u=−x. Since | − x|=|x|<1 for
−1<x<1, the geometric series formula holds.
Step 2: Substitute uinto the formula
1
1−(−x)=
∞
X
n=0
(−x)n
Step 3: Simplify
1
1 + x=
∞
X
n=0
(−1)nxn
Therefore, the Maclaurin series representation for f(x) = 1
1+xis P∞
n=0(−1)nxn.
23
Question 28
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 will start by finding its deriva-
tives and evaluating them at x= 0 to obtain the coefficients of the series.
Step 1: Find the derivatives
f(x) = 1
1 + x
f′(x) = d
dx 1
1 + x=−1
(1 + x)2
f′′(x) = d2
dx21
1 + x=2
(1 + x)3
f′′′(x) = d3
dx31
1 + x=−6
(1 + x)4
.
.
.
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 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:
f(x)=1−x+2
2!x2−6
3!x3+· · ·
Simplify:
f(x) = 1 −x+x2−x3+· · ·
f(x) =
∞
X
n=0
(−1)nxn
24
Step 4: Determine the interval of convergence To find the interval of
convergence, we will use the ratio test:
lim
n→∞
(−1)n+1xn+1
(−1)nxn
=|x|lim
n→∞ |x|
=|x|
The series converges if |x|<1. Therefore, the interval of convergence is (−1,1).
Question 29
Question
Find the Maclaurin series for f(x) = 1
1+x.
Solution
To find the Maclaurin series for f(x) = 1
1+x, we first find its derivatives and
evaluate them at x= 0 to find the coefficients of the series.
Step 1: Find the derivatives
f(x) = 1
1 + x
f′(x) = −1
(1 + x)2
f′′(x) = 2
(1 + x)3
f′′′(x) = −6
(1 + x)4
.
.
.
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 The Maclaurin series for f(x) = 1
1+x
is:
f(x) =
∞
X
n=0
f(n)(0)
n!xn= 1 −x+x2−x3+. . .
25
Question 30
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), evaluate them at x= 0 to find the coefficients, and then express the
series using sigma notation.
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
f(4)(x) = −6
(1 + x)4
.
.
.
Step 2: Evaluate the derivatives at x= 0 to find the coefficients
for the Maclaurin series
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
f(4)(0) = −6
(1 + 0)4=−6
.
.
.
Step 3: Express the Maclaurin series in sigma notation The Maclau-
rin series for f(x) = ln(1 + x) is
ln(1 + x) =
∞
X
n=1
(−1)n−1xn
n
26
Question 2
Question
Find the Maclaurin series representation for the function f(x) = e2xsin(3x).
Solution
To find the Maclaurin series representation for the function f(x) = e2xsin(3x),
we will first compute the derivatives of f(x) at x= 0 and then write out the
Maclaurin series using the formula for the Maclaurin series representation.
Step 1: Find the derivatives of f(x)
f(x) = e2xsin(3x)
f′(x) = (2e2x)(sin(3x)) + e2x(3 cos(3x))
f′′(x) = (22e2x)(sin(3x)) + 2(2e2x)(3 cos(3x)) + (e2x)(−32sin(3x))
= 4e2xsin(3x) + 12e2xcos(3x)−9e2xsin(3x)
= 5e2xsin(3x) + 12e2xcos(3x)
f′′′(x) = 5(2e2x)(sin(3x)) + 5(12e2x)(cos(3x))
+ 12(2e2x)(cos(3x)) −12(12e2x)(sin(3x))
= 10e2xsin(3x) + 60e2xcos(3x) + 24e2xcos(3x)−144e2xsin(3x)
=−134e2xsin(3x) + 84e2xcos(3x)
Step 2: Write out the Maclaurin series Using the derivatives of f(x)
evaluated at x= 0, we can write the Maclaurin series representation of f(x):
f(x) = f(0) + f′(0)
1! x+f′′(0)
2! x2+f′′′(0)
3! x3+· · ·
Plugging in the values we found:
f(x) = 1 + 5
1!x+12
2! x2+84
3! x3+· · ·
Therefore, the Maclaurin series representation for f(x) = e2xsin(3x) is 1 +
5x+ 6x2+ 14x3+· · · .
Question 3
Question
Find the Maclaurin series for f(x) = ln(1 + x) and determine the interval of
convergence.
2
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 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(3)(x) = d
dx −1
(1 + x)2=2
(1 + x)3
.
.
.
The nth derivative of f(x) is given by
f(n)(x)=(−1)n−1(n−1)!
(1 + x)n
Step 2: Evaluate the derivatives at x= 0
f(0) = ln(1) = 0
f′(0) = 1
1= 1
f′′(0) = −1
12=−1
f(3)(0) = 2
13= 2
.
.
.
f(n)(0) = (−1)n−1(n−1)!
1= (−1)n−1(n−1)!
Step 3: Construct the Maclaurin series The Maclaurin series for f(x) =
ln(1 + x) is given by the formula
f(x) =
∞
X
n=0
f(n)(0)
n!xn=
∞
X
n=0
(−1)n−1(n−1)!
n!xn=
∞
X
n=0
(−1)n−1xn
n
Step 4: Determine the interval of convergence To find the interval of
convergence, we will use the ratio test. Let an= (−1)n−11
n. Then,
lim
n→∞
an+1
an
= lim
n→∞
(−1)n1
n+1
(−1)n−11
n
= lim
n→∞
n
n+ 1
= 1
Since the ratio test gives 1, the interval of convergence is the open interval
(−1,1).
3
Question 4
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 first find
the derivatives of f(x) at x= 0 to obtain the coefficients of the Maclaurin series.
Step 1: Find f′(x)and f′′ (x).
f(x) = 1
1 + x2
f′(x) = d
dx 1
1 + x2
=−2x
(1 + x2)2
f′′(x) = d
dx −2x
(1 + x2)2
=−2(1 + x2)2−2x(2)(1 + x2)(2x)
(1 + x2)4
=−2(1 + x2)(1 −2x2)
(1 + x2)4
=−2(1 −2x2+x2)
(1 + x2)3
=−2(1 −3x2)
(1 + x2)3
Step 2: Find f′(0) and f′′ (0).Evaluate f′(x) and f′′ (x) at x= 0:
f′(0) = −2(0)
(1 + 0)2= 0
f′′(0) = −2(1 −0)
(1 + 0)3=−2
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! +· · ·
Therefore, the Maclaurin series representation for f(x) = 1
1+x2is:
1
1 + x2= 1 −2x2+2x4
2! −2(3x2)x4
3! +· · ·
4
Question 5
Question
Find the Maclaurin series for the function f(x) = ln(1 + x) and determine its
radius of convergence.
Solution
Given f(x) = ln(1 + x), we can find its Maclaurin series representation by
differentiating this function and evaluating at x= 0.
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(n)(x)=(−1)n−1(n−1)!
(1 + x)n
Step 2: Evaluate f(n)(0).
f(n)(0) = (−1)n−1(n−1)!
1n= (−1)n−1(n−1)!
Step 3: Write out the Maclaurin series using the formula f(x) = f(0) +
f′(0)x+f′′ (0)
2! x2+· · · +f(n)(0)
n!xn.
f(x) = ln(1 + x) =
∞
X
n=0
(−1)n−1xn
n
Step 4: Determine the radius of convergence using the ratio test.
lim
n→∞
an+1
an
= lim
n→∞
(−1)nxn+1
n+1
(−1)n−1xn
n
= lim
n→∞
x
n+ 1
= 0
Since the limit is less than 1 for all x, the radius of convergence is R=∞.
5
Question 6
Question
Find the Maclaurin series for f(x) = ln(1 + x) and determine its radius of
convergence.
Solution
Step 1: Find the derivatives of f(x) and evaluate them at x= 0 to obtain the
coefficients of the Maclaurin series.
f(x) = ln(1 + x)
f′(x) = 1
1 + x⇒f′(0) = 1
f′′(x) = −1
(1 + x)2⇒f′′ (0) = −1
f′′′(x) = 2
(1 + x)3⇒f′′′ (0) = 2
f(4)(x) = −6
(1 + x)4⇒f(4)(0) = −6
.
.
.
Step 2: Write out the Maclaurin series by using the coefficients obtained.
ln(1 + x) = 1 −x+1
2x2−1
3x3+1
4x4−. . . =
∞
X
n=1
(−1)n−1xn
n
Step 3: Determine the radius of convergence using the ratio test.
R= lim
n→∞
(−1)nxn+1
n+1
(−1)n−1xn
n
= lim
n→∞ |x| · n
n+ 1 =|x|<1
So, the radius of convergence is R= 1.
Question 7
Question
Find the Maclaurin series representation for the function f(x) = 1
1+x2.
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we will first express f(x) as a
geometric series and then find the Maclaurin series by expanding it.
6
Step 1: Find the Maclaurin series for 11 + x2
We know that the geometric series is given by:
1
1−r= 1 + r+r2+r3+. . . for |r|<1.
Thus, we can express 1
1+x2as a geometric series:
1
1 + x2=1
1−(−x2)= 1 + (−x2)+(−x2)2+ (−x2)3+. . .
= 1 −x2+x4−x6+. . . =
∞
X
n=0
(−1)nx2n.
Step 2: Determine the Maclaurin series of f(x)
Since the Maclaurin series representation of 1
1+x2is P∞
n=0(−1)nx2n, we can
directly write the Maclaurin series of f(x) as:
f(x) =
∞
X
n=0
(−1)nx2n.
Therefore, the Maclaurin series representation for the function f(x) = 1
1+x2is
P∞
n=0(−1)nx2n.
Question 8
Question
Find the Maclaurin series for f(x) = excos x.
Solution
To find the Maclaurin series for f(x) = excos x, we will first express exand cos x
as power series expansions around x= 0, and then multiply them together.
Step 1: Find the power series expansion for exThe power series
expansion for exis given by:
ex=
∞
X
n=0
xn
n!
So,
ex= 1 + x+x2
2! +x3
3! +x4
4! +. . .
Step 2: Find the power series expansion for cos xThe power series
expansion for cos xis given by:
cos x=
∞
X
n=0
(−1)nx2n
(2n)!
7
So,
cos x= 1 −x2
2! +x4
4! −x6
6! +. . .
Step 3: Multiply the series Multiplying the power series expansions for
exand cos x, we get the Maclaurin series for excos x:
f(x) = excos x
= (1 + x+x2
2! +x3
3! +x4
4! +. . .)(1 −x2
2! +x4
4! −x6
6! +. . .)
= 1 + (x−x3
2! )+(x2
2! −x4
2! ·2!) + . . .
So, the Maclaurin series for f(x) = excos xis
1+(x−x3
2) + x2
2−x4
12 +. . .
Question 9
Question
Find the Maclaurin series representation for 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−xand substitute −x2in place of x.
Step 1: Find the Maclaurin series of 11−x.Starting with the geo-
metric series formula, 1
1−x= 1 + x+x2+x3+. . ., we have:
1
1−x2= 1 + x2+x4+x6+. . .
Step 2: Find the Maclaurin series of 11−x2.Substitute −x2in place
of xin the series representation of 1
1−x:
1
1 + x2= 1 −x2+x4−x6+. . .
Therefore, the Maclaurin series representation for f(x) = 1
1−x2is 1 −x2+
x4−x6+. . ..
Question 10
Question
Find the Maclaurin series for the function f(x) = ln(1 + x).
8
Solution
To find the Maclaurin series for f(x) = ln(1 + x), we can start by finding deriva-
tives of f(x) and then evaluating them at x= 0.
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
Step 2: Find f(n)(0) for each derivative.
f(0) = ln(1) = 0
f′(0) = 1
1= 1
f′′(0) = −1
12=−1
f′′′(0) = 2
13= 2
Step 3: Express the Maclaurin series in sigma notation. The Maclau-
rin series for f(x) = ln(1 + x) is given by:
f(x) =
∞
X
n=0
f(n)(0)
n!xn
Step 4: Substitute the values of f(n)(0) into the Maclaurin series
formula.
f(x) = 0 + 1x−1
2!x2+2
3!x3−. . .
f(x) = x−x2
2+x3
3−. . .
Therefore, the Maclaurin series for f(x) = ln(1 + x) is f(x) = x−x2
2+x3
3−
. . ..
Question 11
Question
Find the Maclaurin series for the function f(x) = sin2(x).
9
Solution
To find the Maclaurin series for f(x) = sin2(x), we first need to express sin2(x)
in terms of well-known Maclaurin series for elementary functions.
Step 1: Express sin2(x)using trigonometric identities We know that
sin2(x) = 1
2(1 −cos(2x)).
Step 2: Find the Maclaurin series for 12(1 −cos(2x)) We start by
finding the Maclaurin series for 1 and for cos(2x) separately. - Maclaurin series
for 1: 1 + 0x+ 0x2+ 0x3+. . . = 1 - Maclaurin series for cos(2x): cos(0) −
2 sin(0)x+0x2+(−2)2
2! cos(0)x2+(−2)3
3! sin(0)x3+. . . = 1−2x2+22
2! x2−23
3! x3+. . . =
1−2x2+ 2x2−8
3x3+. . . = 1 −8
3x3+. . .
Thus, the Maclaurin series for sin2(x) is f(x) = 1
2(1 −cos(2x)) = 1
2(1 −1 +
8
3x2−. . .) = 4
3x2−. . ..
Question 12
Question
Find the Maclaurin series for f(x) = 1
1+x2.
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we can begin by expressing f(x)
as a geometric series. Recall that the geometric series formula is:
1
1−a= 1 + a+a2+a3+. . .
where |a|<1.
Thus, we rewrite f(x) as:
1
1 + x2=1
1−(−x2)= 1 + (−x2)+(−x2)2+ (−x2)3+. . .
Now, we simplify the terms:
1−x2+x4−x6+. . .
Therefore, the Maclaurin series for f(x) = 1
1+x2is:
1−x2+x4−x6+. . . =
∞
X
n=0
(−1)nx2n
Question 13
Question
Find the Maclaurin series for f(x) = 1
x−3.
10
Solution
To find the Maclaurin series for f(x) = 1
x−3, we first need to express f(x) in
terms of a power series. Since we are looking for a Maclaurin series (centered
at x= 0), we will use the formula for the geometric series: 1
1−r=P∞
n=0 rnfor
|r|<1.
Step 1: Rewrite the given function in terms of a power series.
f(x) = 1
x−3=1
−3·1
1−x
3
=−1
3
∞
X
n=0 x
3n
=−1
3
∞
X
n=0
xn
3n=−
∞
X
n=0
xn
3n+1
Step 2: Determine the coefficients of the Maclaurin series. The Maclaurin
series for f(x) will be in the form:
f(x) =
∞
X
n=0
anxn
Comparing this with −1
3P∞
n=0
xn
3n+1 , we see that an=−1
3n+1 .
Step 3: Write down the Maclaurin series. Thus, the Maclaurin series for
f(x) = 1
x−3is:
f(x) = −
∞
X
n=0
xn
3n+1 =−x
3+x2
32+x3
33+. . .
Question 14
Question
Find the Maclaurin series expansion for f(x) = e2xcos(3x).
Solution
To find the Maclaurin series expansion for f(x) = e2xcos(3x), we will first find
the Maclaurin series expansions for e2xand cos(3x), and then multiply these
series together.
Step 1: Find the Maclaurin series expansion for e2x.The Maclaurin
series expansion for exis given by:
ex=
∞
X
n=0
xn
n!
Replacing xwith 2x, we have:
11
e2x=
∞
X
n=0
(2x)n
n!
e2x=
∞
X
n=0
2nxn
n!
So, the Maclaurin series expansion for e2xis P∞
n=0 2nxn
n!.
Step 2: Find the Maclaurin series expansion for cos(3x).The Maclau-
rin series expansion for cos(x) is given by:
cos(x) =
∞
X
n=0
(−1)nx2n
(2n)!
Replacing xwith 3x, we have:
cos(3x) =
∞
X
n=0
(−1)n(3x)2n
(2n)!
cos(3x) =
∞
X
n=0
(−1)n32nx2n
(2n)!
So, the Maclaurin series expansion for cos(3x) is P∞
n=0
(−1)n32nx2n
(2n)! .
Step 3: Multiply the two series together. To get the Maclaurin series
expansion for f(x) = e2xcos(3x), we multiply the Maclaurin series for e2xby
the Maclaurin series for cos(3x).
Multiplying the two series together term by term, we obtain:
f(x) = ∞
X
n=0
2nxn
n!! ∞
X
n=0
(−1)n32nx2n
(2n)! !
f(x) =
∞
X
n=0
n
X
k=0
2k
k!·(−1)n−k32(n−k)xn
(n−k)! ·(2(n−k))!
After simplifying, we get the Maclaurin series expansion for f(x) = e2xcos(3x).
Question 15
Question
Find the Maclaurin series representation for f(x) = 1
1+x2.
12
Solution
To find the Maclaurin series representation for f(x) = 1
1+x2, we can use the
formula for the Maclaurin series of 1
1−xand substitute x2for x.
Step 1: Find the Maclaurin series of 11−x
The Maclaurin series of 1
1−xis given by:
1
1−x= 1 + x+x2+x3+. . . =
∞
X
n=0
xn
Step 2: Substitute x2for x
Substitute x2for xin the Maclaurin series of 1
1−xto get the Maclaurin series
representation of 1
1+x2:
1
1 + x2= 1 + x2+x4+x6+. . . =
∞
X
n=0
x2n
Therefore, the Maclaurin series representation for f(x) = 1
1+x2is P∞
n=0 x2n.
Question 16
Question
Find the Maclaurin series for the function f(x) = e2xcos(3x).
Solution
To find the Maclaurin series for the given function, we will first express f(x) as
a power series, and then determine its coefficients.
Step 1: Express f(x)as a power series. We know that the Maclaurin
series for exis given by:
ex=
∞
X
n=0
xn
n!
And the Maclaurin series for cos xis:
cos x=
∞
X
n=0
(−1)nx2n
(2n)!
Therefore, the Maclaurin series for f(x) = e2xcos(3x) can be expressed as:
f(x) = ∞
X
n=0
(2x)n
n!! ∞
X
n=0
(−1)n(3x)2n
(2n)! !
Multiplying these series together will give us the desired power series for
f(x).
13
Step 2: Determine the coefficients. After multiplying the series to-
gether, we can simplify the resulting expression by collecting like terms and
finding the coefficients of xn.
The Maclaurin series for f(x) will be in the form:
f(x) =
∞
X
n=0
anxn
Therefore, we need to determine the coefficients anby finding the prod-
uct of the corresponding terms in the two series. This is a somewhat tedious
calculation, but by distributing and rearranging terms, we can determine the
coefficients of the power series for f(x).
The final expression for the Maclaurin series of f(x) = e2xcos(3x) will be
the sum of these terms with the determined coefficients.
Thus, the Maclaurin series for f(x) = e2xcos(3x) can be found by multi-
plying the Maclaurin series for e2xand cos(3x), and carefully simplifying the
resulting expression to determine the coefficients an.
Question 17
Question
Find the Maclaurin series for the function f(x) = sin2(x) up to the term involv-
ing x4.
Solution
To find the Maclaurin series for f(x) = sin2(x), we use the identity sin2(x) =
1−cos(2x)
2and then expand the cosine term as a Maclaurin series.
Step 1: Find the Maclaurin series for cos(2x).The Maclaurin series
for cos(2x) is given by
cos(2x)=1−(2x)2
2! +(2x)4
4! −(2x)6
6! +· · ·
Simplifying, we have
cos(2x)=1−2x2+2x4
3! −2x6
5! +· · ·
Step 2: Use the identity sin2(x) = 1−cos(2x)2.Substitute the Maclaurin
series for cos(2x) into the identity:
sin2(x) = 1−(1 −2x2+2x4
3! −2x6
5! +· · · )
2
Simplify to get
sin2(x) = x2−x4
3! +x6
5! − · · ·
14
Step 3: Determine the Maclaurin series for f(x) = sin2(x).Now, we
have the Maclaurin series for f(x) = sin2(x) up to the term involving x4:
f(x) = x2−x4
3! +x6
5! − · · ·
Therefore, the Maclaurin series for f(x) = sin2(x) up to the term involving
x4is x2−x4
6.
Question 18
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 to determine the coefficients of the series.
Step 1: Find f′(x)
f(x) = 1
1 + x= (1 + x)−1
Using the power rule, we have:
f′(x) = (−1)(1 + x)−2=−(1 + x)−2
Evaluate f′(0):
f′(0) = −1
Step 2: Find f′′ (x) Differentiate f′(x):
f′′(x) = −(−2)(1 + x)−3= 2(1 + x)−3
Evaluate f′′(0):
f′′(0) = 2
Step 3: Find f′′′ (x) Differentiate f′′(x):
f′′′(x) = 3(−2)(1 + x)−4=−6(1 + x)−4
Evaluate f′′′(0):
f′′′(0) = −6
Step 4: Generalize the pattern From the calculated derivatives at x= 0,
we observe the pattern:
f(n)(0) = (−1)n·n!
15
Step 5: Write the Maclaurin series The Maclaurin series for f(x) = 1
1+x
is:
f(x) = f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+· · ·
= 1 −x+x2−x3+· · ·
=
∞
X
n=0
(−1)nxn
Question 19
Question
Find the Maclaurin series for the function f(x) = 1
1+x.
Solution
To find the Maclaurin series for the function f(x) = 1
1+x, we will first express
f(x) as a geometric series, and then find the general term of the Maclaurin
series.
Step 1: Express f(x)as a geometric series The function f(x) = 1
1+x
can be expressed as a geometric series using the formula for a geometric series:
1
1 + x=1
1−(−x)=
∞
X
n=0
(−x)n.
Step 2: Find the general term of the Maclaurin series In the geo-
metric series, the nth term is (−x)n. To express this as the nth term of the
Maclaurin series, we need to find the corresponding term of the Maclaurin series,
which is f(n)(0)
n!xn.
Let’s find the nth derivative of f(x) and evaluate it at x= 0:
f(x) = 1
1 + x
f′(x) = −1
(1 + x)2
f′′(x) = 2
(1 + x)3
f′′′(x) = −6
(1 + x)4
.
.
.
From this pattern, we can see that the nth derivative of f(x) evaluated at x= 0
is:
f(n)(0) = (−1)nn!.
16
Therefore, the nth term of the Maclaurin series for f(x) is:
f(n)(0)
n!xn=(−1)nn!
n!xn= (−1)nxn.
Step 3: Write the Maclaurin series Thus, the Maclaurin series for
f(x) = 1
1+xis:
∞
X
n=0
(−1)nxn.
Question 20
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 then evaluate them at x= 0 to determine the coefficients in the
Maclaurin 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
(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
We can observe a pattern emerging with the derivatives.
Step 2: Evaluate 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
17
Step 3: Find 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+ 2x2−6x3+· · ·
Step 4: Determine the interval of convergence Since the function
f(x) = 1
1+xis in the form of a geometric series, the Maclaurin series converges
for |x|<1. Therefore, the interval of convergence is (−1,1).
Question 21
Question
Find the Maclaurin series for f(x) = sin(2x).
Solution
To find the Maclaurin series for f(x) = sin(2x), we will first find the derivatives
of f(x) at x= 0 to determine the coefficients of the series.
Step 1: Find f′(x)
f(x) = sin(2x)
f′(x) = 2 cos(2x)
Step 2: Find f′′ (x)
f′′(x) = −4 sin(2x)
Step 3: Find f′′′ (x)
f′′′(x) = −8 cos(2x)
Step 4: Find f(4)(x)
f(4)(x) = 16 sin(2x)
Step 5: Find f(5)(x)
f(5)(x) = 32 cos(2x)
Step 6: Find f(6)(x)
f(6)(x) = −64 sin(2x)
From the above calculations, we can see that the derivatives of f(x) = sin(2x)
at x= 0 alternate between ±2nwhere nis the power of xin the series expansion.
18
Step 7: Write the Maclaurin series So, the Maclaurin series for f(x) =
sin(2x) is given by:
sin(2x)=2x−23x3
3! +25x5
5! −27x7
7! +· · ·
Thus, the Maclaurin series for f(x) = sin(2x) is 2
∞
X
n=0
(−1)n22n+1x2n+1
(2n+ 1)! .
Question 22
Question
Find the Maclaurin series representation for f(x) = 1
1+x2.
Solution
To find the Maclaurin series for f(x), we first need to find the derivatives of
f(x) and evaluate them at x= 0 to find the coefficients of the series.
Step 1: Find the first few derivatives of f(x).
f(x) = 1
1 + x2
f′(x) = −(1 + x2)−2·2x=−2x(1 + x2)−2
f′′(x) = −2(1 + x2)−2−2x·(−2)(1 + x2)−3·2x= 2(3x2−1)(1 + x2)−3
Step 2: Find the Maclaurin series for f(x).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+· · ·
Now, let’s evaluate f(0), f′(0), and f′′(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
Thus, the Maclaurin series for f(x) is:
1−2
2!x2+1
4!x4−2
6!x6+· · ·
Question 23
Question
Find the Maclaurin series representation for f(x) = 1
x2+4 .
19
Solution
To find the Maclaurin series representation for f(x) = 1
x2+4 , we will first express
it in the form of a geometric series.
Step 1: Express f(x)as a geometric series We note that 1
x2+4 =
1
4·1
1+(x
2)2.
Since x
2= 1, we can rewrite 1
1+(x
2)2as an infinite geometric series:
1
1−r= 1 + r+r2+r3+. . . , for |r|<1
Here, r=−x
22=−x2
4, so we have:
1
x2+ 4 =1
4·1
1−x
22=1
4
∞
X
n=0 x
22n
Step 2: Find the Maclaurin series representation To find the Maclau-
rin series representation, we substitute a= 0 into the series from Step 1:
1
x2+ 4 =1
4
∞
X
n=0 x
22n
=1
4
∞
X
n=0
x2n
22n=1
4
∞
X
n=0
x2n
4n
Therefore, the Maclaurin series representation for f(x) = 1
x2+4 is 1
4P∞
n=0
x2n
4n.
Question 24
Question
Find the Maclaurin series for the function f(x) = e2xcos(3x).
Solution
To find the Maclaurin series for f(x) = e2xcos(3x), we can use the formula for
the Maclaurin series of exand cos(x), then multiply the two series together.
Step 1: Find the Maclaurin series for e2x.
The Maclaurin series for exis given by:
ex=
∞
X
n=0
xn
n!
Substitute 2xfor x:
e2x=
∞
X
n=0
(2x)n
n!=
∞
X
n=0
2nxn
n!
Step 2: Find the Maclaurin series for cos(3x).
20
The Maclaurin series for cos(x) is given by:
cos(x) =
∞
X
n=0
(−1)nx2n
(2n)!
Substitute 3xfor x:
cos(3x) =
∞
X
n=0
(−1)n(3x)2n
(2n)! =
∞
X
n=0
(−1)n32nx2n
(2n)!
Step 3: Multiply the two series together.
The Maclaurin series for f(x) = e2xcos(3x) is given by multiplying the two
individual Maclaurin series together:
f(x) = e2xcos(3x) = ∞
X
n=0
2nxn
n!! ∞
X
n=0
(−1)n32nx2n
(2n)! !
Now, multiply the two series term by term and simplify to obtain the Maclau-
rin series for f(x).
Question 25
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) and then evaluate those derivatives at x= 0.
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
.
.
.
21
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)x2
2! +f′′′(0)x3
3! +· · · =
∞
X
n=0
f(n)(0)xn
n!
Substitute the values we found in Step 2 into the series to get the Maclaurin
series for f(x):
f(x) = 1 −x+x2−x3+· · · =
∞
X
n=0
(−1)nxn
Question 26
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 first find
its derivatives and evaluate them at x= 0 to determine the coefficients of the
series.
Step 1: Find the derivatives of f(x).
f(x) = 1
1 + x2= (1 + x2)−1
f′(x) = −(1 + x2)−2·2x=−2x(1 + x2)−2
f′′(x) = −2(1 + x2)−2−2x·(−2)(1 + x2)−3·2x= 4x2(1 + x2)−3
f′′′(x) = 4(1 + x2)−3+ 4x2·(−3)(1 + x2)−4·2x=−24x3(1 + x2)−4
.
.
.
22
Step 2: Evaluate the derivatives at x= 0.
f(0) = 1
f′(0) = 0
f′′(0) = 4
f′′′(0) = 0
.
.
.
Step 3: Write the Maclaurin series. Since f(0) = 1, f′(0) = 0, f′′(0) =
4, f′′′(0) = 0, and so on, the Maclaurin series for f(x) is:
f(x) = 1 + 0 ·x+4
2!x2+ 0 ·x3+24
3! x4+. . . = 1 + 2x2+4
3x4+. . .
Therefore, the Maclaurin series representation for f(x) = 1
1+x2is 1 + 2x2+
4
3x4+. . ..
Question 27
Question
Find the Maclaurin series representation for the function f(x) = 1
1+x.
Solution
To find the Maclaurin series for f(x) = 1
1+x, we can use the formula for the
geometric series:
1
1−u=
∞
X
n=0
un
where |u|<1.
Step 1: Find the term uIn this case, u=−x. Since | − x|=|x|<1 for
−1<x<1, the geometric series formula holds.
Step 2: Substitute uinto the formula
1
1−(−x)=
∞
X
n=0
(−x)n
Step 3: Simplify
1
1 + x=
∞
X
n=0
(−1)nxn
Therefore, the Maclaurin series representation for f(x) = 1
1+xis P∞
n=0(−1)nxn.
23
Question 28
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 will start by finding its deriva-
tives and evaluating them at x= 0 to obtain the coefficients of the series.
Step 1: Find the derivatives
f(x) = 1
1 + x
f′(x) = d
dx 1
1 + x=−1
(1 + x)2
f′′(x) = d2
dx21
1 + x=2
(1 + x)3
f′′′(x) = d3
dx31
1 + x=−6
(1 + x)4
.
.
.
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 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:
f(x)=1−x+2
2!x2−6
3!x3+· · ·
Simplify:
f(x) = 1 −x+x2−x3+· · ·
f(x) =
∞
X
n=0
(−1)nxn
24
Step 4: Determine the interval of convergence To find the interval of
convergence, we will use the ratio test:
lim
n→∞
(−1)n+1xn+1
(−1)nxn
=|x|lim
n→∞ |x|
=|x|
The series converges if |x|<1. Therefore, the interval of convergence is (−1,1).
Question 29
Question
Find the Maclaurin series for f(x) = 1
1+x.
Solution
To find the Maclaurin series for f(x) = 1
1+x, we first find its derivatives and
evaluate them at x= 0 to find the coefficients of the series.
Step 1: Find the derivatives
f(x) = 1
1 + x
f′(x) = −1
(1 + x)2
f′′(x) = 2
(1 + x)3
f′′′(x) = −6
(1 + x)4
.
.
.
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 The Maclaurin series for f(x) = 1
1+x
is:
f(x) =
∞
X
n=0
f(n)(0)
n!xn= 1 −x+x2−x3+. . .
25
Question 30
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), evaluate them at x= 0 to find the coefficients, and then express the
series using sigma notation.
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
f(4)(x) = −6
(1 + x)4
.
.
.
Step 2: Evaluate the derivatives at x= 0 to find the coefficients
for the Maclaurin series
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
f(4)(0) = −6
(1 + 0)4=−6
.
.
.
Step 3: Express the Maclaurin series in sigma notation The Maclau-
rin series for f(x) = ln(1 + x) is
ln(1 + x) =
∞
X
n=1
(−1)n−1xn
n
26
Question 2
Question
Find the Maclaurin series representation for the function f(x) = e2xsin(3x).
Solution
To find the Maclaurin series representation for the function f(x) = e2xsin(3x),
we will first compute the derivatives of f(x) at x= 0 and then write out the
Maclaurin series using the formula for the Maclaurin series representation.
Step 1: Find the derivatives of f(x)
f(x) = e2xsin(3x)
f′(x) = (2e2x)(sin(3x)) + e2x(3 cos(3x))
f′′(x) = (22e2x)(sin(3x)) + 2(2e2x)(3 cos(3x)) + (e2x)(−32sin(3x))
= 4e2xsin(3x) + 12e2xcos(3x)−9e2xsin(3x)
= 5e2xsin(3x) + 12e2xcos(3x)
f′′′(x) = 5(2e2x)(sin(3x)) + 5(12e2x)(cos(3x))
+ 12(2e2x)(cos(3x)) −12(12e2x)(sin(3x))
= 10e2xsin(3x) + 60e2xcos(3x) + 24e2xcos(3x)−144e2xsin(3x)
=−134e2xsin(3x) + 84e2xcos(3x)
Step 2: Write out the Maclaurin series Using the derivatives of f(x)
evaluated at x= 0, we can write the Maclaurin series representation of f(x):
f(x) = f(0) + f′(0)
1! x+f′′(0)
2! x2+f′′′(0)
3! x3+· · ·
Plugging in the values we found:
f(x) = 1 + 5
1!x+12
2! x2+84
3! x3+· · ·
Therefore, the Maclaurin series representation for f(x) = e2xsin(3x) is 1 +
5x+ 6x2+ 14x3+· · · .
Question 3
Question
Find the Maclaurin series for f(x) = ln(1 + x) and determine the interval of
convergence.
2
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 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(3)(x) = d
dx −1
(1 + x)2=2
(1 + x)3
.
.
.
The nth derivative of f(x) is given by
f(n)(x)=(−1)n−1(n−1)!
(1 + x)n
Step 2: Evaluate the derivatives at x= 0
f(0) = ln(1) = 0
f′(0) = 1
1= 1
f′′(0) = −1
12=−1
f(3)(0) = 2
13= 2
.
.
.
f(n)(0) = (−1)n−1(n−1)!
1= (−1)n−1(n−1)!
Step 3: Construct the Maclaurin series The Maclaurin series for f(x) =
ln(1 + x) is given by the formula
f(x) =
∞
X
n=0
f(n)(0)
n!xn=
∞
X
n=0
(−1)n−1(n−1)!
n!xn=
∞
X
n=0
(−1)n−1xn
n
Step 4: Determine the interval of convergence To find the interval of
convergence, we will use the ratio test. Let an= (−1)n−11
n. Then,
lim
n→∞
an+1
an
= lim
n→∞
(−1)n1
n+1
(−1)n−11
n
= lim
n→∞
n
n+ 1
= 1
Since the ratio test gives 1, the interval of convergence is the open interval
(−1,1).
3
Question 4
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 first find
the derivatives of f(x) at x= 0 to obtain the coefficients of the Maclaurin series.
Step 1: Find f′(x)and f′′ (x).
f(x) = 1
1 + x2
f′(x) = d
dx 1
1 + x2
=−2x
(1 + x2)2
f′′(x) = d
dx −2x
(1 + x2)2
=−2(1 + x2)2−2x(2)(1 + x2)(2x)
(1 + x2)4
=−2(1 + x2)(1 −2x2)
(1 + x2)4
=−2(1 −2x2+x2)
(1 + x2)3
=−2(1 −3x2)
(1 + x2)3
Step 2: Find f′(0) and f′′ (0).Evaluate f′(x) and f′′ (x) at x= 0:
f′(0) = −2(0)
(1 + 0)2= 0
f′′(0) = −2(1 −0)
(1 + 0)3=−2
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! +· · ·
Therefore, the Maclaurin series representation for f(x) = 1
1+x2is:
1
1 + x2= 1 −2x2+2x4
2! −2(3x2)x4
3! +· · ·
4
Question 5
Question
Find the Maclaurin series for the function f(x) = ln(1 + x) and determine its
radius of convergence.
Solution
Given f(x) = ln(1 + x), we can find its Maclaurin series representation by
differentiating this function and evaluating at x= 0.
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(n)(x)=(−1)n−1(n−1)!
(1 + x)n
Step 2: Evaluate f(n)(0).
f(n)(0) = (−1)n−1(n−1)!
1n= (−1)n−1(n−1)!
Step 3: Write out the Maclaurin series using the formula f(x) = f(0) +
f′(0)x+f′′ (0)
2! x2+· · · +f(n)(0)
n!xn.
f(x) = ln(1 + x) =
∞
X
n=0
(−1)n−1xn
n
Step 4: Determine the radius of convergence using the ratio test.
lim
n→∞
an+1
an
= lim
n→∞
(−1)nxn+1
n+1
(−1)n−1xn
n
= lim
n→∞
x
n+ 1
= 0
Since the limit is less than 1 for all x, the radius of convergence is R=∞.
5
Question 6
Question
Find the Maclaurin series for f(x) = ln(1 + x) and determine its radius of
convergence.
Solution
Step 1: Find the derivatives of f(x) and evaluate them at x= 0 to obtain the
coefficients of the Maclaurin series.
f(x) = ln(1 + x)
f′(x) = 1
1 + x⇒f′(0) = 1
f′′(x) = −1
(1 + x)2⇒f′′ (0) = −1
f′′′(x) = 2
(1 + x)3⇒f′′′ (0) = 2
f(4)(x) = −6
(1 + x)4⇒f(4)(0) = −6
.
.
.
Step 2: Write out the Maclaurin series by using the coefficients obtained.
ln(1 + x) = 1 −x+1
2x2−1
3x3+1
4x4−. . . =
∞
X
n=1
(−1)n−1xn
n
Step 3: Determine the radius of convergence using the ratio test.
R= lim
n→∞
(−1)nxn+1
n+1
(−1)n−1xn
n
= lim
n→∞ |x| · n
n+ 1 =|x|<1
So, the radius of convergence is R= 1.
Question 7
Question
Find the Maclaurin series representation for the function f(x) = 1
1+x2.
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we will first express f(x) as a
geometric series and then find the Maclaurin series by expanding it.
6
Step 1: Find the Maclaurin series for 11 + x2
We know that the geometric series is given by:
1
1−r= 1 + r+r2+r3+. . . for |r|<1.
Thus, we can express 1
1+x2as a geometric series:
1
1 + x2=1
1−(−x2)= 1 + (−x2)+(−x2)2+ (−x2)3+. . .
= 1 −x2+x4−x6+. . . =
∞
X
n=0
(−1)nx2n.
Step 2: Determine the Maclaurin series of f(x)
Since the Maclaurin series representation of 1
1+x2is P∞
n=0(−1)nx2n, we can
directly write the Maclaurin series of f(x) as:
f(x) =
∞
X
n=0
(−1)nx2n.
Therefore, the Maclaurin series representation for the function f(x) = 1
1+x2is
P∞
n=0(−1)nx2n.
Question 8
Question
Find the Maclaurin series for f(x) = excos x.
Solution
To find the Maclaurin series for f(x) = excos x, we will first express exand cos x
as power series expansions around x= 0, and then multiply them together.
Step 1: Find the power series expansion for exThe power series
expansion for exis given by:
ex=
∞
X
n=0
xn
n!
So,
ex= 1 + x+x2
2! +x3
3! +x4
4! +. . .
Step 2: Find the power series expansion for cos xThe power series
expansion for cos xis given by:
cos x=
∞
X
n=0
(−1)nx2n
(2n)!
7
So,
cos x= 1 −x2
2! +x4
4! −x6
6! +. . .
Step 3: Multiply the series Multiplying the power series expansions for
exand cos x, we get the Maclaurin series for excos x:
f(x) = excos x
= (1 + x+x2
2! +x3
3! +x4
4! +. . .)(1 −x2
2! +x4
4! −x6
6! +. . .)
= 1 + (x−x3
2! )+(x2
2! −x4
2! ·2!) + . . .
So, the Maclaurin series for f(x) = excos xis
1+(x−x3
2) + x2
2−x4
12 +. . .
Question 9
Question
Find the Maclaurin series representation for 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−xand substitute −x2in place of x.
Step 1: Find the Maclaurin series of 11−x.Starting with the geo-
metric series formula, 1
1−x= 1 + x+x2+x3+. . ., we have:
1
1−x2= 1 + x2+x4+x6+. . .
Step 2: Find the Maclaurin series of 11−x2.Substitute −x2in place
of xin the series representation of 1
1−x:
1
1 + x2= 1 −x2+x4−x6+. . .
Therefore, the Maclaurin series representation for f(x) = 1
1−x2is 1 −x2+
x4−x6+. . ..
Question 10
Question
Find the Maclaurin series for the function f(x) = ln(1 + x).
8
Solution
To find the Maclaurin series for f(x) = ln(1 + x), we can start by finding deriva-
tives of f(x) and then evaluating them at x= 0.
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
Step 2: Find f(n)(0) for each derivative.
f(0) = ln(1) = 0
f′(0) = 1
1= 1
f′′(0) = −1
12=−1
f′′′(0) = 2
13= 2
Step 3: Express the Maclaurin series in sigma notation. The Maclau-
rin series for f(x) = ln(1 + x) is given by:
f(x) =
∞
X
n=0
f(n)(0)
n!xn
Step 4: Substitute the values of f(n)(0) into the Maclaurin series
formula.
f(x) = 0 + 1x−1
2!x2+2
3!x3−. . .
f(x) = x−x2
2+x3
3−. . .
Therefore, the Maclaurin series for f(x) = ln(1 + x) is f(x) = x−x2
2+x3
3−
. . ..
Question 11
Question
Find the Maclaurin series for the function f(x) = sin2(x).
9
Solution
To find the Maclaurin series for f(x) = sin2(x), we first need to express sin2(x)
in terms of well-known Maclaurin series for elementary functions.
Step 1: Express sin2(x)using trigonometric identities We know that
sin2(x) = 1
2(1 −cos(2x)).
Step 2: Find the Maclaurin series for 12(1 −cos(2x)) We start by
finding the Maclaurin series for 1 and for cos(2x) separately. - Maclaurin series
for 1: 1 + 0x+ 0x2+ 0x3+. . . = 1 - Maclaurin series for cos(2x): cos(0) −
2 sin(0)x+0x2+(−2)2
2! cos(0)x2+(−2)3
3! sin(0)x3+. . . = 1−2x2+22
2! x2−23
3! x3+. . . =
1−2x2+ 2x2−8
3x3+. . . = 1 −8
3x3+. . .
Thus, the Maclaurin series for sin2(x) is f(x) = 1
2(1 −cos(2x)) = 1
2(1 −1 +
8
3x2−. . .) = 4
3x2−. . ..
Question 12
Question
Find the Maclaurin series for f(x) = 1
1+x2.
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we can begin by expressing f(x)
as a geometric series. Recall that the geometric series formula is:
1
1−a= 1 + a+a2+a3+. . .
where |a|<1.
Thus, we rewrite f(x) as:
1
1 + x2=1
1−(−x2)= 1 + (−x2)+(−x2)2+ (−x2)3+. . .
Now, we simplify the terms:
1−x2+x4−x6+. . .
Therefore, the Maclaurin series for f(x) = 1
1+x2is:
1−x2+x4−x6+. . . =
∞
X
n=0
(−1)nx2n
Question 13
Question
Find the Maclaurin series for f(x) = 1
x−3.
10
Solution
To find the Maclaurin series for f(x) = 1
x−3, we first need to express f(x) in
terms of a power series. Since we are looking for a Maclaurin series (centered
at x= 0), we will use the formula for the geometric series: 1
1−r=P∞
n=0 rnfor
|r|<1.
Step 1: Rewrite the given function in terms of a power series.
f(x) = 1
x−3=1
−3·1
1−x
3
=−1
3
∞
X
n=0 x
3n
=−1
3
∞
X
n=0
xn
3n=−
∞
X
n=0
xn
3n+1
Step 2: Determine the coefficients of the Maclaurin series. The Maclaurin
series for f(x) will be in the form:
f(x) =
∞
X
n=0
anxn
Comparing this with −1
3P∞
n=0
xn
3n+1 , we see that an=−1
3n+1 .
Step 3: Write down the Maclaurin series. Thus, the Maclaurin series for
f(x) = 1
x−3is:
f(x) = −
∞
X
n=0
xn
3n+1 =−x
3+x2
32+x3
33+. . .
Question 14
Question
Find the Maclaurin series expansion for f(x) = e2xcos(3x).
Solution
To find the Maclaurin series expansion for f(x) = e2xcos(3x), we will first find
the Maclaurin series expansions for e2xand cos(3x), and then multiply these
series together.
Step 1: Find the Maclaurin series expansion for e2x.The Maclaurin
series expansion for exis given by:
ex=
∞
X
n=0
xn
n!
Replacing xwith 2x, we have:
11
e2x=
∞
X
n=0
(2x)n
n!
e2x=
∞
X
n=0
2nxn
n!
So, the Maclaurin series expansion for e2xis P∞
n=0 2nxn
n!.
Step 2: Find the Maclaurin series expansion for cos(3x).The Maclau-
rin series expansion for cos(x) is given by:
cos(x) =
∞
X
n=0
(−1)nx2n
(2n)!
Replacing xwith 3x, we have:
cos(3x) =
∞
X
n=0
(−1)n(3x)2n
(2n)!
cos(3x) =
∞
X
n=0
(−1)n32nx2n
(2n)!
So, the Maclaurin series expansion for cos(3x) is P∞
n=0
(−1)n32nx2n
(2n)! .
Step 3: Multiply the two series together. To get the Maclaurin series
expansion for f(x) = e2xcos(3x), we multiply the Maclaurin series for e2xby
the Maclaurin series for cos(3x).
Multiplying the two series together term by term, we obtain:
f(x) = ∞
X
n=0
2nxn
n!! ∞
X
n=0
(−1)n32nx2n
(2n)! !
f(x) =
∞
X
n=0
n
X
k=0
2k
k!·(−1)n−k32(n−k)xn
(n−k)! ·(2(n−k))!
After simplifying, we get the Maclaurin series expansion for f(x) = e2xcos(3x).
Question 15
Question
Find the Maclaurin series representation for f(x) = 1
1+x2.
12
Solution
To find the Maclaurin series representation for f(x) = 1
1+x2, we can use the
formula for the Maclaurin series of 1
1−xand substitute x2for x.
Step 1: Find the Maclaurin series of 11−x
The Maclaurin series of 1
1−xis given by:
1
1−x= 1 + x+x2+x3+. . . =
∞
X
n=0
xn
Step 2: Substitute x2for x
Substitute x2for xin the Maclaurin series of 1
1−xto get the Maclaurin series
representation of 1
1+x2:
1
1 + x2= 1 + x2+x4+x6+. . . =
∞
X
n=0
x2n
Therefore, the Maclaurin series representation for f(x) = 1
1+x2is P∞
n=0 x2n.
Question 16
Question
Find the Maclaurin series for the function f(x) = e2xcos(3x).
Solution
To find the Maclaurin series for the given function, we will first express f(x) as
a power series, and then determine its coefficients.
Step 1: Express f(x)as a power series. We know that the Maclaurin
series for exis given by:
ex=
∞
X
n=0
xn
n!
And the Maclaurin series for cos xis:
cos x=
∞
X
n=0
(−1)nx2n
(2n)!
Therefore, the Maclaurin series for f(x) = e2xcos(3x) can be expressed as:
f(x) = ∞
X
n=0
(2x)n
n!! ∞
X
n=0
(−1)n(3x)2n
(2n)! !
Multiplying these series together will give us the desired power series for
f(x).
13
Step 2: Determine the coefficients. After multiplying the series to-
gether, we can simplify the resulting expression by collecting like terms and
finding the coefficients of xn.
The Maclaurin series for f(x) will be in the form:
f(x) =
∞
X
n=0
anxn
Therefore, we need to determine the coefficients anby finding the prod-
uct of the corresponding terms in the two series. This is a somewhat tedious
calculation, but by distributing and rearranging terms, we can determine the
coefficients of the power series for f(x).
The final expression for the Maclaurin series of f(x) = e2xcos(3x) will be
the sum of these terms with the determined coefficients.
Thus, the Maclaurin series for f(x) = e2xcos(3x) can be found by multi-
plying the Maclaurin series for e2xand cos(3x), and carefully simplifying the
resulting expression to determine the coefficients an.
Question 17
Question
Find the Maclaurin series for the function f(x) = sin2(x) up to the term involv-
ing x4.
Solution
To find the Maclaurin series for f(x) = sin2(x), we use the identity sin2(x) =
1−cos(2x)
2and then expand the cosine term as a Maclaurin series.
Step 1: Find the Maclaurin series for cos(2x).The Maclaurin series
for cos(2x) is given by
cos(2x)=1−(2x)2
2! +(2x)4
4! −(2x)6
6! +· · ·
Simplifying, we have
cos(2x)=1−2x2+2x4
3! −2x6
5! +· · ·
Step 2: Use the identity sin2(x) = 1−cos(2x)2.Substitute the Maclaurin
series for cos(2x) into the identity:
sin2(x) = 1−(1 −2x2+2x4
3! −2x6
5! +· · · )
2
Simplify to get
sin2(x) = x2−x4
3! +x6
5! − · · ·
14
Step 3: Determine the Maclaurin series for f(x) = sin2(x).Now, we
have the Maclaurin series for f(x) = sin2(x) up to the term involving x4:
f(x) = x2−x4
3! +x6
5! − · · ·
Therefore, the Maclaurin series for f(x) = sin2(x) up to the term involving
x4is x2−x4
6.
Question 18
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 to determine the coefficients of the series.
Step 1: Find f′(x)
f(x) = 1
1 + x= (1 + x)−1
Using the power rule, we have:
f′(x) = (−1)(1 + x)−2=−(1 + x)−2
Evaluate f′(0):
f′(0) = −1
Step 2: Find f′′ (x) Differentiate f′(x):
f′′(x) = −(−2)(1 + x)−3= 2(1 + x)−3
Evaluate f′′(0):
f′′(0) = 2
Step 3: Find f′′′ (x) Differentiate f′′(x):
f′′′(x) = 3(−2)(1 + x)−4=−6(1 + x)−4
Evaluate f′′′(0):
f′′′(0) = −6
Step 4: Generalize the pattern From the calculated derivatives at x= 0,
we observe the pattern:
f(n)(0) = (−1)n·n!
15
Step 5: Write the Maclaurin series The Maclaurin series for f(x) = 1
1+x
is:
f(x) = f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+· · ·
= 1 −x+x2−x3+· · ·
=
∞
X
n=0
(−1)nxn
Question 19
Question
Find the Maclaurin series for the function f(x) = 1
1+x.
Solution
To find the Maclaurin series for the function f(x) = 1
1+x, we will first express
f(x) as a geometric series, and then find the general term of the Maclaurin
series.
Step 1: Express f(x)as a geometric series The function f(x) = 1
1+x
can be expressed as a geometric series using the formula for a geometric series:
1
1 + x=1
1−(−x)=
∞
X
n=0
(−x)n.
Step 2: Find the general term of the Maclaurin series In the geo-
metric series, the nth term is (−x)n. To express this as the nth term of the
Maclaurin series, we need to find the corresponding term of the Maclaurin series,
which is f(n)(0)
n!xn.
Let’s find the nth derivative of f(x) and evaluate it at x= 0:
f(x) = 1
1 + x
f′(x) = −1
(1 + x)2
f′′(x) = 2
(1 + x)3
f′′′(x) = −6
(1 + x)4
.
.
.
From this pattern, we can see that the nth derivative of f(x) evaluated at x= 0
is:
f(n)(0) = (−1)nn!.
16
Therefore, the nth term of the Maclaurin series for f(x) is:
f(n)(0)
n!xn=(−1)nn!
n!xn= (−1)nxn.
Step 3: Write the Maclaurin series Thus, the Maclaurin series for
f(x) = 1
1+xis:
∞
X
n=0
(−1)nxn.
Question 20
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 then evaluate them at x= 0 to determine the coefficients in the
Maclaurin 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
(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
We can observe a pattern emerging with the derivatives.
Step 2: Evaluate 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
17
Step 3: Find 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+ 2x2−6x3+· · ·
Step 4: Determine the interval of convergence Since the function
f(x) = 1
1+xis in the form of a geometric series, the Maclaurin series converges
for |x|<1. Therefore, the interval of convergence is (−1,1).
Question 21
Question
Find the Maclaurin series for f(x) = sin(2x).
Solution
To find the Maclaurin series for f(x) = sin(2x), we will first find the derivatives
of f(x) at x= 0 to determine the coefficients of the series.
Step 1: Find f′(x)
f(x) = sin(2x)
f′(x) = 2 cos(2x)
Step 2: Find f′′ (x)
f′′(x) = −4 sin(2x)
Step 3: Find f′′′ (x)
f′′′(x) = −8 cos(2x)
Step 4: Find f(4)(x)
f(4)(x) = 16 sin(2x)
Step 5: Find f(5)(x)
f(5)(x) = 32 cos(2x)
Step 6: Find f(6)(x)
f(6)(x) = −64 sin(2x)
From the above calculations, we can see that the derivatives of f(x) = sin(2x)
at x= 0 alternate between ±2nwhere nis the power of xin the series expansion.
18
Step 7: Write the Maclaurin series So, the Maclaurin series for f(x) =
sin(2x) is given by:
sin(2x)=2x−23x3
3! +25x5
5! −27x7
7! +· · ·
Thus, the Maclaurin series for f(x) = sin(2x) is 2
∞
X
n=0
(−1)n22n+1x2n+1
(2n+ 1)! .
Question 22
Question
Find the Maclaurin series representation for f(x) = 1
1+x2.
Solution
To find the Maclaurin series for f(x), we first need to find the derivatives of
f(x) and evaluate them at x= 0 to find the coefficients of the series.
Step 1: Find the first few derivatives of f(x).
f(x) = 1
1 + x2
f′(x) = −(1 + x2)−2·2x=−2x(1 + x2)−2
f′′(x) = −2(1 + x2)−2−2x·(−2)(1 + x2)−3·2x= 2(3x2−1)(1 + x2)−3
Step 2: Find the Maclaurin series for f(x).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+· · ·
Now, let’s evaluate f(0), f′(0), and f′′(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
Thus, the Maclaurin series for f(x) is:
1−2
2!x2+1
4!x4−2
6!x6+· · ·
Question 23
Question
Find the Maclaurin series representation for f(x) = 1
x2+4 .
19
Solution
To find the Maclaurin series representation for f(x) = 1
x2+4 , we will first express
it in the form of a geometric series.
Step 1: Express f(x)as a geometric series We note that 1
x2+4 =
1
4·1
1+(x
2)2.
Since x
2= 1, we can rewrite 1
1+(x
2)2as an infinite geometric series:
1
1−r= 1 + r+r2+r3+. . . , for |r|<1
Here, r=−x
22=−x2
4, so we have:
1
x2+ 4 =1
4·1
1−x
22=1
4
∞
X
n=0 x
22n
Step 2: Find the Maclaurin series representation To find the Maclau-
rin series representation, we substitute a= 0 into the series from Step 1:
1
x2+ 4 =1
4
∞
X
n=0 x
22n
=1
4
∞
X
n=0
x2n
22n=1
4
∞
X
n=0
x2n
4n
Therefore, the Maclaurin series representation for f(x) = 1
x2+4 is 1
4P∞
n=0
x2n
4n.
Question 24
Question
Find the Maclaurin series for the function f(x) = e2xcos(3x).
Solution
To find the Maclaurin series for f(x) = e2xcos(3x), we can use the formula for
the Maclaurin series of exand cos(x), then multiply the two series together.
Step 1: Find the Maclaurin series for e2x.
The Maclaurin series for exis given by:
ex=
∞
X
n=0
xn
n!
Substitute 2xfor x:
e2x=
∞
X
n=0
(2x)n
n!=
∞
X
n=0
2nxn
n!
Step 2: Find the Maclaurin series for cos(3x).
20
The Maclaurin series for cos(x) is given by:
cos(x) =
∞
X
n=0
(−1)nx2n
(2n)!
Substitute 3xfor x:
cos(3x) =
∞
X
n=0
(−1)n(3x)2n
(2n)! =
∞
X
n=0
(−1)n32nx2n
(2n)!
Step 3: Multiply the two series together.
The Maclaurin series for f(x) = e2xcos(3x) is given by multiplying the two
individual Maclaurin series together:
f(x) = e2xcos(3x) = ∞
X
n=0
2nxn
n!! ∞
X
n=0
(−1)n32nx2n
(2n)! !
Now, multiply the two series term by term and simplify to obtain the Maclau-
rin series for f(x).
Question 25
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) and then evaluate those derivatives at x= 0.
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
.
.
.
21
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)x2
2! +f′′′(0)x3
3! +· · · =
∞
X
n=0
f(n)(0)xn
n!
Substitute the values we found in Step 2 into the series to get the Maclaurin
series for f(x):
f(x) = 1 −x+x2−x3+· · · =
∞
X
n=0
(−1)nxn
Question 26
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 first find
its derivatives and evaluate them at x= 0 to determine the coefficients of the
series.
Step 1: Find the derivatives of f(x).
f(x) = 1
1 + x2= (1 + x2)−1
f′(x) = −(1 + x2)−2·2x=−2x(1 + x2)−2
f′′(x) = −2(1 + x2)−2−2x·(−2)(1 + x2)−3·2x= 4x2(1 + x2)−3
f′′′(x) = 4(1 + x2)−3+ 4x2·(−3)(1 + x2)−4·2x=−24x3(1 + x2)−4
.
.
.
22
Step 2: Evaluate the derivatives at x= 0.
f(0) = 1
f′(0) = 0
f′′(0) = 4
f′′′(0) = 0
.
.
.
Step 3: Write the Maclaurin series. Since f(0) = 1, f′(0) = 0, f′′(0) =
4, f′′′(0) = 0, and so on, the Maclaurin series for f(x) is:
f(x) = 1 + 0 ·x+4
2!x2+ 0 ·x3+24
3! x4+. . . = 1 + 2x2+4
3x4+. . .
Therefore, the Maclaurin series representation for f(x) = 1
1+x2is 1 + 2x2+
4
3x4+. . ..
Question 27
Question
Find the Maclaurin series representation for the function f(x) = 1
1+x.
Solution
To find the Maclaurin series for f(x) = 1
1+x, we can use the formula for the
geometric series:
1
1−u=
∞
X
n=0
un
where |u|<1.
Step 1: Find the term uIn this case, u=−x. Since | − x|=|x|<1 for
−1<x<1, the geometric series formula holds.
Step 2: Substitute uinto the formula
1
1−(−x)=
∞
X
n=0
(−x)n
Step 3: Simplify
1
1 + x=
∞
X
n=0
(−1)nxn
Therefore, the Maclaurin series representation for f(x) = 1
1+xis P∞
n=0(−1)nxn.
23
Question 28
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 will start by finding its deriva-
tives and evaluating them at x= 0 to obtain the coefficients of the series.
Step 1: Find the derivatives
f(x) = 1
1 + x
f′(x) = d
dx 1
1 + x=−1
(1 + x)2
f′′(x) = d2
dx21
1 + x=2
(1 + x)3
f′′′(x) = d3
dx31
1 + x=−6
(1 + x)4
.
.
.
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 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:
f(x)=1−x+2
2!x2−6
3!x3+· · ·
Simplify:
f(x) = 1 −x+x2−x3+· · ·
f(x) =
∞
X
n=0
(−1)nxn
24
Step 4: Determine the interval of convergence To find the interval of
convergence, we will use the ratio test:
lim
n→∞
(−1)n+1xn+1
(−1)nxn
=|x|lim
n→∞ |x|
=|x|
The series converges if |x|<1. Therefore, the interval of convergence is (−1,1).
Question 29
Question
Find the Maclaurin series for f(x) = 1
1+x.
Solution
To find the Maclaurin series for f(x) = 1
1+x, we first find its derivatives and
evaluate them at x= 0 to find the coefficients of the series.
Step 1: Find the derivatives
f(x) = 1
1 + x
f′(x) = −1
(1 + x)2
f′′(x) = 2
(1 + x)3
f′′′(x) = −6
(1 + x)4
.
.
.
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 The Maclaurin series for f(x) = 1
1+x
is:
f(x) =
∞
X
n=0
f(n)(0)
n!xn= 1 −x+x2−x3+. . .
25
Question 30
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), evaluate them at x= 0 to find the coefficients, and then express the
series using sigma notation.
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
f(4)(x) = −6
(1 + x)4
.
.
.
Step 2: Evaluate the derivatives at x= 0 to find the coefficients
for the Maclaurin series
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
f(4)(0) = −6
(1 + 0)4=−6
.
.
.
Step 3: Express the Maclaurin series in sigma notation The Maclau-
rin series for f(x) = ln(1 + x) is
ln(1 + x) =
∞
X
n=1
(−1)n−1xn
n
26
Step 4: Determine the radius of convergence To find the interval of
convergence, we will use the ratio test. Let an=(−1)n−1
n.
lim
n→∞
an+1
an
= lim
n→∞
−1
n+ 1 ·n
1
= lim
n→∞
n
n+ 1 = 1
Since the limit is equal to 1, the radius of convergence is 1. Therefore, the
interval of convergence is (−1,1].
27
