MATH 100 - FUNDAMENTALS OF
MATHEMATICS - Taylor and
Maclaurin Series
Question Bank - Set 1
Liberty University
Question 1
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 express sin2(x) in
terms of elementary functions and then find its Maclaurin series by expanding
it as a Taylor series.
Step 1: Express sin2(x)as a combination of elementary functions
We know that sin2(x) = 1
2(1 −cos(2x)).
Step 2: Find the Maclaurin series for 1and cos(2x) The Maclaurin
series for 1 is 1 + 0x+ 0x2+ 0x3+. . . = 1.
The Maclaurin series for cos(2x) is 1 −1
2(2x)2+1
4! (2x)4−1
6! (2x)6+. . .
Simplifying, we get cos(2x) = 1 −2x2+2x4
3! −2x6
5! +. . .
Step 3: Multiply by 12and subtract to find the Maclaurin series
for sin2(x) Multiplying by 1
2and subtracting, we get:
sin2(x) = 1
2(1 −cos(2x)) = 1
2+x2−x4
3! +x6
5! −. . .
Thus, the Maclaurin series for f(x) = sin2(x) is 1
2+x2−x4
3! +x6
5! −. . ..
Question 2
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 use the geometric series
formula.
Step 1: Find the Maclaurin series for 11−x.The Maclaurin series
for 1
1−xis given by:
1
1−x=
∞
X
n=0
xn
for |x|<1.
Step 2: Substitute x2for x.Substitute x2for xto find the Maclaurin
series for f(x) = 1
1+x2:
1
1 + x2=
∞
X
n=0
(x2)n=
∞
X
n=0
x2n
for |x2|<1.
Step 3: Write the Maclaurin series for f(x).Since the Maclaurin series
for f(x) = 1
1+x2is P∞
n=0 x2n, the Maclaurin series for f(x) = 1
1+x2is:
f(x) =
∞
X
n=0
x2n
for |x2|<1.
Question 3
Question
Find the Maclaurin series of the function f(x) = ln(1 + x).
Solution
To find the Maclaurin series of f(x) = ln(1 + x), we will first find the derivatives
of f(x) and then evaluate them at x= 0 to determine the coefficients of the
series.
2
Step 1: Find the first few derivatives of f(x) = ln(1 + 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
.
.
.
Step 2: Evaluate the derivatives at x= 0.
f(0) = ln(1 + 0) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f′′′(0) = 2
(1 + 0)3= 2
f(4)(0) = −6
(1 + 0)4=−6
.
.
.
Step 3: Write down the Maclaurin series using the coefficients found in Step
2. The Maclaurin series for ln(1 + x) is:
ln(1 + x) = x−x2
2+x3
3−x4
4+· · ·
Therefore, the Maclaurin series of f(x) = ln(1 + x) is f(x) = P∞
n=1
(−1)n−1xn
n.
Question 4
Question
Find the Maclaurin series for 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.
3
Step 1: Find the Maclaurin series for sin(x).The Maclaurin series for
sin(x) is:
sin(x) = x−x3
3! +x5
5! −x7
7! +. . . =
∞
X
n=0
(−1)nx2n+1
(2n+ 1)!
Step 2: Square the Maclaurin series for sin(x).Squaring the Maclaurin
series for sin(x) gives us the Maclaurin series for sin2(x):
sin2(x) = ∞
X
n=0
(−1)nx2n+1
(2n+ 1)!!2
=
∞
X
n=0 (−1)nx2n+1
(2n+ 1)!·(−1)mx2m+1
(2m+ 1)!
=
∞
X
n=0
∞
X
m=0
(−1)n+mx2n+1x2m+1
(2n+ 1)!(2m+ 1)!
=
∞
X
n=0
∞
X
m=0
(−1)n+mx2n+2m+2
(2n+ 1)!(2m+ 1)!
Now, we can simplify the inner sum to get the Maclaurin series for sin2(x):
sin2(x) =
∞
X
n=0
(−1)n
(2n+ 1)! ·
n
X
m=0
(−1)m
(2(m−n) + 1)!x2n+2m+2
Question 5
Question
Find the Maclaurin series of the function f(x) = 1
x+2 .
Solution
To find the Maclaurin series of f(x), we will first find the Maclaurin series of
1
1+xby using the formula for the Maclaurin series of 1
1−xand substituting x
with −x.
Step 1: Find the Maclaurin series of 1
1+x. The Maclaurin series for 1
1−xis
given by:
∞
X
n=0
xn
Therefore, the Maclaurin series for 1
1+xis:
∞
X
n=0
(−1)nxn
4
Step 2: Find the Maclaurin series of f(x) = 1
x+2 . We can express f(x)
as 1
x+2 =1
2·1
1+ x
2. Since the Maclaurin series for 1
1+xis P∞
n=0(−1)nxn, the
Maclaurin series for 1
1+ x
2is:
∞
X
n=0 −1
2n
xn=
∞
X
n=0
(−1)n
2nxn
Multiplying by 1
2, we get the Maclaurin series for f(x):
∞
X
n=0
(−1)n
2n+1 xn
Question 6
Question
Find the Maclaurin series for f(x) = sin2(x).
Solution
To find the Maclaurin series for f(x) = sin2(x), we first need to find the Maclau-
rin series for sin2(x).
sin2(x) = (sin(x))2=eix −e−ix
2i2
=1
4e2ix −2eixe−ix +e−2ix
=1
4e2ix −2 + e−2ix.
Now, we can find the Maclaurin series for f(x) = 1
4e2ix −2 + e−2ix.
=1
41+2ix +(2ix)2
2! +(2ix)3
3! +· · · − 2+1−2ix +(−2ix)2
2! −(2ix)3
3! +· · ·
=1
42 + (2ix)2
2! +(2ix)4
4! +· · · − 2
=1
4−4x2+16x4
4! −64x6
6! +· · ·
=−x2+x4
6−x6
45 +· · · .
Therefore, the Maclaurin series for f(x) = sin2(x) is −x2+x4
6−x6
45 +· · · .
5
Question 7
Question
Find the Maclaurin series for the function f(x) = excos(x).
Solution
To find the Maclaurin series for f(x) = excos(x), we will use the Maclaurin
series for exand cos(x) and then multiply them together.
Step 1: Find the Maclaurin series for exThe Maclaurin series for ex
is:
ex=
∞
X
n=0
xn
n!
Step 2: Find the Maclaurin series for cos(x) The Maclaurin series for
cos(x) is:
cos(x) =
∞
X
n=0
(−1)nx2n
(2n)!
Step 3: Multiply the two series together Now, to find the Maclaurin
series for f(x) = excos(x), we multiply the Maclaurin series for exand cos(x):
f(x) = excos(x) = ∞
X
n=0
xn
n!! ∞
X
n=0
(−1)nx2n
(2n)! !
Step 4: Simplify the series To find the Maclaurin series for f(x), we
multiply the series term by term:
f(x) =
∞
X
n=0 n
X
k=0
(−1)kxn+k
k!(n−k)!(2n)!!
Therefore, the Maclaurin series for f(x) = excos(x) is:
∞
X
n=0 n
X
k=0
(−1)kxn+k
k!(n−k)!(2n)!!
Question 8
Question
Find the Maclaurin series for f(x) = 1
1−x2.
6
Solution
To find the Maclaurin series for f(x) = 1
1−x2, we will first find the derivatives
of f(x) and use them to find the coefficients in the series expansion.
Step 1: Find the derivatives of f(x).
f(x) = 1
1−x2
f′(x) = d
dx 1
1−x2=2x
(1 −x2)2= 2x(1 + x2)
f′′(x) = d
dx 2x(1 + x2)= 2(1 + x2)+4x2= 2(1 + 3x2)
f′′′(x) = d
dx 2(1 + 3x2)= 6x
Step 2: Find the Maclaurin series. The Maclaurin series expansion 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 find the values of f(0), f′(0), f′′(0), and f′′′(0):
f(0) = 1
1−0= 1
f′(0) = 2(0)(1 + 02) = 0
f′′(0) = 2(1) = 2
f′′′(0) = 0
Therefore, the Maclaurin series for f(x) is:
f(x) = 1 + 2
2!x2+0+. . . = 1 + x2
Question 9
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 then evaluate them at x= 0 to find the coefficients of the series.
7
Step 1: Find the derivatives of f(x).
f(x) = 1
1 + x
f′(x) = −(1 + x)−2=−(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: Evaluate the derivatives at x= 0.
f(0) = 1
f′(0) = −1
f′′(0) = 2
f′′′(0) = −6
.
.
.
Step 3: Write out the Maclaurin series. The Maclaurin series for f(x)
is given by
f(x) = f(0) + f′(0)x+f′′ (0)x2
2! +f′′′(0)x3
3! +· · ·
Substitute the values we calculated above:
f(x) = 1 −x+x2−x3+· · · =
∞
X
n=0
(−1)nxn
Therefore, the Maclaurin series for f(x) = 1
1+xis P∞
n=0(−1)nxn.
Question 10
Question
Find the Maclaurin series for e2x.
Solution
To find the Maclaurin series for e2x, we can use the fact that the Maclaurin
series for exis P∞
n=0
xn
n!and substitute 2xin place of x.
Step 1: Substitute 2xfor xin the Maclaurin series for ex:
e2x=
∞
X
n=0
(2x)n
n!
8
Step 2: Simplify the expression by expanding (2x)n:
e2x=
∞
X
n=0
2nxn
n!
Step 3: Rewrite the series in sigma notation:
e2x= 1 + 2x+22x2
2! +23x3
3! +· · ·
Step 4: Simplify the terms in the series:
e2x= 1 + 2x+ 2x2+4x3
3! +· · ·
Step 5: Therefore, the Maclaurin series for e2xis 1 + 2x+ 2x2+4x3
3! +· · · .
Question 11
Question
Find the Maclaurin series representation for the function f(x) = ln(1 + x) and
determine the interval of convergence.
Solution
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: Find the Maclaurin series. We can express the Maclaurin series for
ln(1 + x) as:
f(x) =
∞
X
n=0
f(n)(0)
n!xn=f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+f(4)(0)
4! x4+. . .
Substitute the derivatives at x= 0:
9
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
Therefore, the Maclaurin series is:
ln(1 + x) = x−x2
2+x3
3−x4
4+. . .
Step 3: Determine the interval of convergence. The interval of convergence
of a Maclaurin series is the set of all xvalues for which the series converges.
By using the Ratio Test, we find the interval of convergence to be −1< x ≤1.
This means the Maclaurin series for ln(1 + x) converges for −1< x ≤1.
Question 12
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 then evaluate them at x= 0 to get the coefficients of the series.
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 differentiating the derivative of f(x))
f′′′(x) = 2
(1 + x)3(by differentiating f′′(x))
f(4)(x) = −6
(1 + x)4(by differentiating f′′′(x))
.
.
.
The nth derivative of f(x) can be written as
f(n)(x)=(−1)n−1(n−1)!
(1 + x)n
10
Step 2: Evaluate the derivatives at x= 0 Now, we evaluate the deriva-
tives at x= 0 to get the coefficients of the Maclaurin series:
f(0) = ln(1 + 0) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f′′′(0) = 2
(1 + 0)3= 2
f(4)(0) = −6
(1 + 0)4=−6
.
.
.
Step 3: Write down the Maclaurin series The Maclaurin series for
ln(1 + x) is
ln(1 + x) =
∞
X
n=1
(−1)n−1xn
n=x−x2
2+x3
3−x4
4+· · ·
Question 13
Question
Find the Maclaurin series for the function f(x) = ln(1 + x) by differentiating
the series for f′(x) = 1
1+xterm by term.
Solution
To find the Maclaurin series for f(x) = ln(1 + x), we will differentiate the
Maclaurin series for f′(x) = 1
1+xterm by term.
Step 1: Find the Maclaurin series for f′(x) = 1
1+x: The Maclaurin series for
1
1+xis:
1
1 + x= 1 −x+x2−x3+· · · =
∞
X
n=0
(−1)nxn
Step 2: Differentiate the series term by term to find the series for f(x) =
ln(1 + x): Differentiating the series term by term, we get:
f(x) = Z1
1 + xdx =Z∞
X
n=0
(−1)nxndx
=
∞
X
n=0
(−1)nZxndx =
∞
X
n=0
(−1)n
n+ 1 xn+1 +C
11
=
∞
X
n=0
(−1)n
n+ 1 xn+1 +C
Therefore, the Maclaurin series for f(x) = ln(1 + x) is:
ln(1 + x) =
∞
X
n=0
(−1)n
n+ 1 xn+1 +C
Question 14
Question
Find the Maclaurin series expansion for f(x) = ln(1 + x).
Solution
To find the Maclaurin series expansion for f(x) = ln(1 + x), we start by finding
the derivatives of f(x) at x= 0 and then use the formula for the Maclaurin
series expansion.
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: Find the Maclaurin series expansion The Maclaurin series
expansion for f(x) is given by
∞
X
n=0
f(n)(0)
n!xn=
∞
X
n=0
f(n)(0)
n!xn
12
By evaluating the derivatives at x= 0, we get
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
.
.
.
Therefore, the Maclaurin series expansion 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 15
Question
Find the Maclaurin series representation for ln(1 + x).
Solution
To find the Maclaurin series representation for ln(1 + x), we will first find the
derivatives of ln(1 + x) and then evaluate them at x= 0.
Step 1: Find the derivatives of ln(1 + 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
.
.
.
13
Step 2: Evaluate the derivatives at x= 0.
f(0) = ln(1 + 0) = ln(1) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f′′′(0) = 2
(1 + 0)3= 2
f(4)(0) = −6
(1 + 0)4=−6
.
.
.
Step 3: Write the Maclaurin series. Since f(0) = 0, the Maclaurin series for
ln(1 + x) is:
ln(1 + x) = x−x2
2+x3
3−x4
4+x5
5− · · · =
∞
X
n=1
(−1)n−1xn
n
Question 16
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) at x= 0.
14
Step 1: Find the derivatives of f(x)
f(x) = 1
1 + x2
f′(x) = −2x
(1 + x2)2
f′′(x) = −2(1 + x2)2−4x2(1 + x2)
(1 + x2)4
=2x2−2
(1 + x2)3
f′′′(x) = −8x(1 + x2)3−(2x2−2)(3(1 + x2)22x)
(1 + x2)6
=−8x−6x3
(1 + x2)5
f(4)(x) = −6(1 + x2)5−(−8x−6x3)(5(1 + x2)42x)
(1 + x2)8
=−6−24x2+ 30x2+ 24x4
(1 + x2)7
=24x4−6
(1 + x2)7
Step 2: Find f(n)(0) To find the Maclaurin series for f(x), we need to
evaluate the derivatives f(n)(0).
f′(0) = 0; f′′(0) = −2; f′′′ (0) = 0; f(4)(0) = −6
Step 3: Write the Maclaurin series The Maclaurin series representation
of f(x) = 1
1+x2is given by the formula:
f(x) = f(0) + f′(0)x+f′′ (0)x2
2! +f′′′(0)x3
3! +f(4)(0)x4
4! +. . .
Substitute the values of the derivatives at x= 0:
f(x)=1−2x2−6x4
4!
Therefore, the Maclaurin series for f(x) = 1
1+x2is 1 −2x2−3x4
4+. . . .
Question 17
Question
Find the Maclaurin series for f(x) = 1
1+x2.
15
Solution
Step 1: Find the derivatives of f(x) and evaluate them at x= 0 to find the
coefficients of the Maclaurin series.
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(3x2−1)
(1 + x2)3
f(3)(x) = d
dx 2(3x2−1)
(1 + x2)3=−24x(3x2−2)
(1 + x2)4
f(4)(x) = d
dx −24x(3x2−2)
(1 + x2)4=24(12x3−5x)
(1 + x2)5
Step 2: Evaluate the derivatives at x= 0.
f(0) = 1
f′(0) = 0
f′′(0) = −2
f(3)(0) = 0
f(4)(0) = 24
Step 3: Write out the Maclaurin series. Since f(0) = 1 and all odd deriva-
tives evaluated at x= 0 are 0, the Maclaurin series for f(x) is:
f(x)=1−2x2+ 24x4+. . .
Question 18
Question
Find the Maclaurin series for the function f(x) = ln(1 + x) and determine the
radius 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 to determine the coefficients of the
series.
16
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
Step 2: Find the Maclaurin series expansion. 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) evaluated at x= 0 into the series expan-
sion:
f(x) = ln(1 + x) = 0 + x−x2
2+x3
3−x4
4+· · ·
Therefore, the Maclaurin series for f(x) is
ln(1 + x) = x−x2
2+x3
3−x4
4+· · ·
Step 3: Determine the radius of convergence. In general, the radius
of convergence Rof a power series P∞
n=0 cnxnis given by
R= lim
n→∞
cn
cn+1
In this case, cn=1
n+1 , so
R= lim
n→∞
1
n+1
1
n+2
= lim
n→∞
n+ 2
n+ 1
= 1
Therefore, the radius of convergence of the Maclaurin series for ln(1 + x) is
R= 1.
Question 19
Question
Find the Maclaurin series for f(x) = e−x2.
17
Solution
To find the Maclaurin series for f(x) = e−x2, we can use the fact that the
Maclaurin series for exis P∞
n=0
xn
n!.
Step 1: Find the Maclaurin series for e−x2by substituting −x2for xin the
Maclaurin series for ex.
∞
X
n=0
(−x2)n
n!=
∞
X
n=0
(−1)n·x2n
n!
Step 2: Simplify the expression by splitting the sum into its even and odd
terms. ∞
X
n=0
(−1)n·x2n
n!=
∞
X
n=0
(−1)n·x2n
(2n)! +
∞
X
n=0
(−1)n·x2n
(2n+ 1)!
Step 3: Recognize that the first series is the Maclaurin series for cos(x) and
the second series is the Maclaurin series for sin(x). Thus, the Maclaurin series
for f(x) = e−x2is:
e−x2= cos(x) + sin(x)
e−x2=
∞
X
n=0
(−1)n·x2n
(2n)! +
∞
X
n=0
(−1)n·x2n
(2n+ 1)!
Question 20
Question
Find the Maclaurin series representation for the function f(x) = ln(1 + x).
Solution
To find the Maclaurin series representation of f(x) = ln(1 + x), we will first find
the derivatives of f(x) at x= 0, which will give us 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
f′′′(x) = 2
(1 + x)3
f(4)(x) = −6
(1 + x)4
.
.
.
18
Step 2: Evaluate the derivatives at x= 0.
f(0) = ln(1 + 0) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f′′′(0) = 2
(1 + 0)3= 2
f(4)(0) = −6
(1 + 0)4=−6
.
.
.
Step 3: Write down the Maclaurin series. The Maclaurin series rep-
resentation of ln(1 + x) is
ln(1 + x) = x−x2
2+x3
3−x4
4+· · · + (−1)n−1xn
n+· · ·
Therefore, the Maclaurin series representation for f(x) = ln(1 + x) is
ln(1 + x) = x−x2
2+x3
3−x4
4+· · ·
Question 21
Question
Find the Maclaurin series for ln(1 + x) up to the term in x5.
Solution
To find the Maclaurin series for ln(1 + x), we’ll start by finding the derivatives
of ln(1 + x) and evaluating them at x= 0. Then, we’ll use these derivatives to
build the Maclaurin series by using the formula for the Taylor series.
19
Step 1: Find the derivatives of ln(1 + 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) = −3·2
(1 + x)4
f(5)(x) = 4·3·2
(1 + x)5=24
(1 + x)5
Step 2: Evaluate derivatives at x= 0
f(0) = ln(1 + 0) = ln(1) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f′′′(0) = 2
(1 + 0)3= 2
f(4)(0) = −3·2
(1 + 0)4=−6
f(5)(0) = 24
(1 + 0)5= 24
Step 3: Build the Maclaurin series The Maclaurin series for a function
f(x) centered at x= 0 is given by:
f(x) = f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+f(4)(0)
4! x4+f(5)(0)
5! x5+. . .
Substitute the values we found earlier:
ln(1 + x) = 0 + 1 ·x−1
2·x2+2
6·x3−6
24 ·x4+24
120 ·x5+. . .
Therefore, the Maclaurin series for ln(1 + x) up to the term in x5is:
ln(1 + x) = x−1
2x2+1
3x3−1
4x4+1
5x5+. . .
Question 22
Question
Find the Maclaurin series for the function f(x) = 1
1+x2.
20
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we can use the geometric series
formula:
1
1−(−x2)=
∞
X
n=0
(−x2)n
Step 1: Rewrite the geometric series formula with the right substitution.
We rewrite the formula as:
1
1 + x2=
∞
X
n=0
(−1)nx2n
Step 2: Identify the Maclaurin series.
By comparing the formula above to the Maclaurin series representation:
1
1 + x=
∞
X
n=0
(−1)nxn
we have the Maclaurin series for f(x):
1
1 + x2=
∞
X
n=0
(−1)nx2n
Therefore, the Maclaurin series for f(x) = 1
1+x2is P∞
n=0(−1)nx2n.
Question 23
Question
Find the Maclaurin series for f(x) = 1
1+x2, and determine the interval of con-
vergence.
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we will use the geometric series
formula:
1
1−u=
∞
X
n=0
un
where |u|<1.
Step 1: Find the Maclaurin series Let u=−x2. Then, we have:
1
1 + x2=1
1−(−x2)=
∞
X
n=0
(−x2)n
21
Simplifying the series gives:
1
1 + x2=
∞
X
n=0
(−1)nx2n
Therefore, the Maclaurin series for f(x) = 1
1+x2is P∞
n=0(−1)nx2n.
Step 2: Determine the interval of convergence To find the interval of
convergence, we will use the ratio test. Let an= (−1)nx2n.
The ratio test states:
lim
n→∞
an+1
an
=L
If L < 1, then the series converges absolutely. If L > 1, then the series diverges.
If L= 1, the test is inconclusive.
Calculating the limit:
lim
n→∞
(−1)n+1x2(n+1)
(−1)nx2n
= lim
n→∞ |x2|=|x2|
For the series to converge, |x2|<1, which implies −1<x<1. Therefore,
the interval of convergence is (−1,1).
Question 24
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 Maclaurin
series for sin(x) and then replace xwith 2x.
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 2x.
f(x) = sin(2x) =
∞
X
n=0
(−1)n(2x)2n+1
(2n+ 1)! =
∞
X
n=0
(−1)n22n+1x2n+1
(2n+ 1)! =
∞
X
n=0
(−1)n22n+1x2n+1
(2n+ 1)!
Therefore, the Maclaurin series for f(x) = sin(2x) is P∞
n=0
(−1)n22n+1x2n+1
(2n+1)! .
22
Question 25
Question
Find the Maclaurin series for the function f(x) = ln(1 + x) and determine the
interval of convergence.
Solution
To find the Maclaurin series for f(x) = ln(1 + x), we will use the formula for
the Maclaurin series of ln(1 + x), which is given by:
ln(1 + x) =
∞
X
n=1
(−1)n+1xn
n
Step 1: Find the Maclaurin series We can see that the Maclaurin series
for ln(1 + x) is:
ln(1 + x) =
∞
X
n=1
(−1)n+1xn
n
Step 2: Determine the interval of convergence To find the interval of
convergence, we use the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
(−1)n+2xn+1
n+1
(−1)n+1xn
n
= lim
n→∞
n|x|
n+ 1
=|x|
The series converges for |x|<1. Therefore, the interval of convergence is
(−1,1].
Question 26
Question
Find the Maclaurin series for the function f(x) = cos(x) sin(x).
Solution
To find the Maclaurin series for f(x) = cos(x) sin(x), we will first find the
Maclaurin series for cos(x) and sin(x), and then multiply them together.
Step 1: Find the Maclaurin series for cos(x) The Maclaurin series for
cos(x) is given by:
cos(x) =
∞
X
n=0
(−1)nx2n
(2n)!
23
Step 2: Find the Maclaurin series for sin(x) The Maclaurin series for
sin(x) is given by:
sin(x) =
∞
X
n=0
(−1)nx2n+1
(2n+ 1)!
Step 3: Multiply the two series together Multiplying the Maclaurin
series for cos(x) and sin(x) together, we get:
f(x) = cos(x) sin(x) = ∞
X
n=0
(−1)nx2n
(2n)!! ∞
X
n=0
(−1)nx2n+1
(2n+ 1)!!
Expanding this product will give us the Maclaurin series for f(x) = cos(x) sin(x).
Question 27
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 can start by recalling the
Maclaurin series for ln(1 + x):
ln(1 + x) = x−x2
2+x3
3−x4
4+. . . =
∞
X
n=1
(−1)n−1xn
n
Using the fact that ln(1 −x) = −ln(1 + x), we get:
ln(1 −x) = −
∞
X
n=1
(−1)n−1xn
n
Step 1: Reorder the terms inside the summation:
ln(1 −x) =
∞
X
n=1
(−1)nxn
n
Step 2: Rewrite the series with nstarting from 0:
ln(1 −x) =
∞
X
n=0
(−1)n+1 xn
n+ 1
Step 3: This is the Maclaurin series for f(x) = ln(1 −x):
ln(1 −x) =
∞
X
n=0
(−1)n+1 xn
n+ 1
24
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 first find the derivatives of
f(x) at x= 0 and then use the formula for the Maclaurin series.
Step 1: Find the derivatives of f(x).
f(x) = 1
1 + x
f′(x) = d
dx 1
1 + x=−1
(1 + x)2
f′′(x) = d2
dx2 −1
(1 + x)2!=2
(1 + x)3
f′′′(x) = d3
dx3 2
(1 + x)3!=−6
(1 + x)4
.
.
.
It can be seen that the nth derivative of f(x) evaluated at x= 0 is of the form
(−1)nn!.
Step 2: Write the Maclaurin series for f(x).The Maclaurin series for
a function 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) at x= 0 into the formula:
f(x) = 1 −x+x2−x3+· · ·
=
∞
X
n=0
(−1)nxn(Maclaurin series for 1
1 + x)
Therefore, the Maclaurin series for f(x) = 1
1+xis P∞
n=0(−1)nxn.
Question 29
Question
Find the Maclaurin series for the function f(x) = excos(x).
25
Solution
To find the Maclaurin series of f(x) = excos(x), we can use the Maclaurin series
for exand cos(x) and then multiply the resulting series.
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 for exand cos(x) Multiplying the series for
exand cos(x), we get:
f(x) = excos(x) = ∞
X
n=0
xn
n!! ∞
X
n=0
(−1)nx2n
(2n)!!
Step 4: Simplify the expression By multiplying the series together and
simplifying, we get the Maclaurin series for f(x) = excos(x).
Therefore, the Maclaurin series for f(x) = excos(x) is [final Maclaurin se-
ries].
Question 30
Question
Find the Maclaurin series for f(x) = sin2(x).
Solution
To find the Maclaurin series for f(x) = sin2(x), we can start by using the known
Maclaurin series for sin(x).
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: Find the Maclaurin series for f(x) = sin2(x) Since f(x) =
sin2(x), we can write:
f(x) = (sin(x))2
f(x)=(
∞
X
n=0
(−1)nx2n+1
(2n+ 1)!)2
26
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we will use the geometric series
formula.
Step 1: Find the Maclaurin series for 11−x.The Maclaurin series
for 1
1−xis given by:
1
1−x=
∞
X
n=0
xn
for |x|<1.
Step 2: Substitute x2for x.Substitute x2for xto find the Maclaurin
series for f(x) = 1
1+x2:
1
1 + x2=
∞
X
n=0
(x2)n=
∞
X
n=0
x2n
for |x2|<1.
Step 3: Write the Maclaurin series for f(x).Since the Maclaurin series
for f(x) = 1
1+x2is P∞
n=0 x2n, the Maclaurin series for f(x) = 1
1+x2is:
f(x) =
∞
X
n=0
x2n
for |x2|<1.
Question 3
Question
Find the Maclaurin series of the function f(x) = ln(1 + x).
Solution
To find the Maclaurin series of f(x) = ln(1 + x), we will first find the derivatives
of f(x) and then evaluate them at x= 0 to determine the coefficients of the
series.
2
Step 1: Find the first few derivatives of f(x) = ln(1 + 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
.
.
.
Step 2: Evaluate the derivatives at x= 0.
f(0) = ln(1 + 0) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f′′′(0) = 2
(1 + 0)3= 2
f(4)(0) = −6
(1 + 0)4=−6
.
.
.
Step 3: Write down the Maclaurin series using the coefficients found in Step
2. The Maclaurin series for ln(1 + x) is:
ln(1 + x) = x−x2
2+x3
3−x4
4+· · ·
Therefore, the Maclaurin series of f(x) = ln(1 + x) is f(x) = P∞
n=1
(−1)n−1xn
n.
Question 4
Question
Find the Maclaurin series for 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.
3
Step 1: Find the Maclaurin series for sin(x).The Maclaurin series for
sin(x) is:
sin(x) = x−x3
3! +x5
5! −x7
7! +. . . =
∞
X
n=0
(−1)nx2n+1
(2n+ 1)!
Step 2: Square the Maclaurin series for sin(x).Squaring the Maclaurin
series for sin(x) gives us the Maclaurin series for sin2(x):
sin2(x) = ∞
X
n=0
(−1)nx2n+1
(2n+ 1)!!2
=
∞
X
n=0 (−1)nx2n+1
(2n+ 1)!·(−1)mx2m+1
(2m+ 1)!
=
∞
X
n=0
∞
X
m=0
(−1)n+mx2n+1x2m+1
(2n+ 1)!(2m+ 1)!
=
∞
X
n=0
∞
X
m=0
(−1)n+mx2n+2m+2
(2n+ 1)!(2m+ 1)!
Now, we can simplify the inner sum to get the Maclaurin series for sin2(x):
sin2(x) =
∞
X
n=0
(−1)n
(2n+ 1)! ·
n
X
m=0
(−1)m
(2(m−n) + 1)!x2n+2m+2
Question 5
Question
Find the Maclaurin series of the function f(x) = 1
x+2 .
Solution
To find the Maclaurin series of f(x), we will first find the Maclaurin series of
1
1+xby using the formula for the Maclaurin series of 1
1−xand substituting x
with −x.
Step 1: Find the Maclaurin series of 1
1+x. The Maclaurin series for 1
1−xis
given by:
∞
X
n=0
xn
Therefore, the Maclaurin series for 1
1+xis:
∞
X
n=0
(−1)nxn
4
Step 2: Find the Maclaurin series of f(x) = 1
x+2 . We can express f(x)
as 1
x+2 =1
2·1
1+ x
2. Since the Maclaurin series for 1
1+xis P∞
n=0(−1)nxn, the
Maclaurin series for 1
1+ x
2is:
∞
X
n=0 −1
2n
xn=
∞
X
n=0
(−1)n
2nxn
Multiplying by 1
2, we get the Maclaurin series for f(x):
∞
X
n=0
(−1)n
2n+1 xn
Question 6
Question
Find the Maclaurin series for f(x) = sin2(x).
Solution
To find the Maclaurin series for f(x) = sin2(x), we first need to find the Maclau-
rin series for sin2(x).
sin2(x) = (sin(x))2=eix −e−ix
2i2
=1
4e2ix −2eixe−ix +e−2ix
=1
4e2ix −2 + e−2ix.
Now, we can find the Maclaurin series for f(x) = 1
4e2ix −2 + e−2ix.
=1
41+2ix +(2ix)2
2! +(2ix)3
3! +· · · − 2+1−2ix +(−2ix)2
2! −(2ix)3
3! +· · ·
=1
42 + (2ix)2
2! +(2ix)4
4! +· · · − 2
=1
4−4x2+16x4
4! −64x6
6! +· · ·
=−x2+x4
6−x6
45 +· · · .
Therefore, the Maclaurin series for f(x) = sin2(x) is −x2+x4
6−x6
45 +· · · .
5
Question 7
Question
Find the Maclaurin series for the function f(x) = excos(x).
Solution
To find the Maclaurin series for f(x) = excos(x), we will use the Maclaurin
series for exand cos(x) and then multiply them together.
Step 1: Find the Maclaurin series for exThe Maclaurin series for ex
is:
ex=
∞
X
n=0
xn
n!
Step 2: Find the Maclaurin series for cos(x) The Maclaurin series for
cos(x) is:
cos(x) =
∞
X
n=0
(−1)nx2n
(2n)!
Step 3: Multiply the two series together Now, to find the Maclaurin
series for f(x) = excos(x), we multiply the Maclaurin series for exand cos(x):
f(x) = excos(x) = ∞
X
n=0
xn
n!! ∞
X
n=0
(−1)nx2n
(2n)! !
Step 4: Simplify the series To find the Maclaurin series for f(x), we
multiply the series term by term:
f(x) =
∞
X
n=0 n
X
k=0
(−1)kxn+k
k!(n−k)!(2n)!!
Therefore, the Maclaurin series for f(x) = excos(x) is:
∞
X
n=0 n
X
k=0
(−1)kxn+k
k!(n−k)!(2n)!!
Question 8
Question
Find the Maclaurin series for f(x) = 1
1−x2.
6
Solution
To find the Maclaurin series for f(x) = 1
1−x2, we will first find the derivatives
of f(x) and use them to find the coefficients in the series expansion.
Step 1: Find the derivatives of f(x).
f(x) = 1
1−x2
f′(x) = d
dx 1
1−x2=2x
(1 −x2)2= 2x(1 + x2)
f′′(x) = d
dx 2x(1 + x2)= 2(1 + x2)+4x2= 2(1 + 3x2)
f′′′(x) = d
dx 2(1 + 3x2)= 6x
Step 2: Find the Maclaurin series. The Maclaurin series expansion 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 find the values of f(0), f′(0), f′′(0), and f′′′(0):
f(0) = 1
1−0= 1
f′(0) = 2(0)(1 + 02) = 0
f′′(0) = 2(1) = 2
f′′′(0) = 0
Therefore, the Maclaurin series for f(x) is:
f(x) = 1 + 2
2!x2+0+. . . = 1 + x2
Question 9
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 then evaluate them at x= 0 to find the coefficients of the series.
7
Step 1: Find the derivatives of f(x).
f(x) = 1
1 + x
f′(x) = −(1 + x)−2=−(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: Evaluate the derivatives at x= 0.
f(0) = 1
f′(0) = −1
f′′(0) = 2
f′′′(0) = −6
.
.
.
Step 3: Write out the Maclaurin series. The Maclaurin series for f(x)
is given by
f(x) = f(0) + f′(0)x+f′′ (0)x2
2! +f′′′(0)x3
3! +· · ·
Substitute the values we calculated above:
f(x) = 1 −x+x2−x3+· · · =
∞
X
n=0
(−1)nxn
Therefore, the Maclaurin series for f(x) = 1
1+xis P∞
n=0(−1)nxn.
Question 10
Question
Find the Maclaurin series for e2x.
Solution
To find the Maclaurin series for e2x, we can use the fact that the Maclaurin
series for exis P∞
n=0
xn
n!and substitute 2xin place of x.
Step 1: Substitute 2xfor xin the Maclaurin series for ex:
e2x=
∞
X
n=0
(2x)n
n!
8
Step 2: Simplify the expression by expanding (2x)n:
e2x=
∞
X
n=0
2nxn
n!
Step 3: Rewrite the series in sigma notation:
e2x= 1 + 2x+22x2
2! +23x3
3! +· · ·
Step 4: Simplify the terms in the series:
e2x= 1 + 2x+ 2x2+4x3
3! +· · ·
Step 5: Therefore, the Maclaurin series for e2xis 1 + 2x+ 2x2+4x3
3! +· · · .
Question 11
Question
Find the Maclaurin series representation for the function f(x) = ln(1 + x) and
determine the interval of convergence.
Solution
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: Find the Maclaurin series. We can express the Maclaurin series for
ln(1 + x) as:
f(x) =
∞
X
n=0
f(n)(0)
n!xn=f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+f(4)(0)
4! x4+. . .
Substitute the derivatives at x= 0:
9
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
Therefore, the Maclaurin series is:
ln(1 + x) = x−x2
2+x3
3−x4
4+. . .
Step 3: Determine the interval of convergence. The interval of convergence
of a Maclaurin series is the set of all xvalues for which the series converges.
By using the Ratio Test, we find the interval of convergence to be −1< x ≤1.
This means the Maclaurin series for ln(1 + x) converges for −1< x ≤1.
Question 12
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 then evaluate them at x= 0 to get the coefficients of the series.
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 differentiating the derivative of f(x))
f′′′(x) = 2
(1 + x)3(by differentiating f′′(x))
f(4)(x) = −6
(1 + x)4(by differentiating f′′′(x))
.
.
.
The nth derivative of f(x) can be written as
f(n)(x)=(−1)n−1(n−1)!
(1 + x)n
10
Step 2: Evaluate the derivatives at x= 0 Now, we evaluate the deriva-
tives at x= 0 to get the coefficients of the Maclaurin series:
f(0) = ln(1 + 0) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f′′′(0) = 2
(1 + 0)3= 2
f(4)(0) = −6
(1 + 0)4=−6
.
.
.
Step 3: Write down the Maclaurin series The Maclaurin series for
ln(1 + x) is
ln(1 + x) =
∞
X
n=1
(−1)n−1xn
n=x−x2
2+x3
3−x4
4+· · ·
Question 13
Question
Find the Maclaurin series for the function f(x) = ln(1 + x) by differentiating
the series for f′(x) = 1
1+xterm by term.
Solution
To find the Maclaurin series for f(x) = ln(1 + x), we will differentiate the
Maclaurin series for f′(x) = 1
1+xterm by term.
Step 1: Find the Maclaurin series for f′(x) = 1
1+x: The Maclaurin series for
1
1+xis:
1
1 + x= 1 −x+x2−x3+· · · =
∞
X
n=0
(−1)nxn
Step 2: Differentiate the series term by term to find the series for f(x) =
ln(1 + x): Differentiating the series term by term, we get:
f(x) = Z1
1 + xdx =Z∞
X
n=0
(−1)nxndx
=
∞
X
n=0
(−1)nZxndx =
∞
X
n=0
(−1)n
n+ 1 xn+1 +C
11
=
∞
X
n=0
(−1)n
n+ 1 xn+1 +C
Therefore, the Maclaurin series for f(x) = ln(1 + x) is:
ln(1 + x) =
∞
X
n=0
(−1)n
n+ 1 xn+1 +C
Question 14
Question
Find the Maclaurin series expansion for f(x) = ln(1 + x).
Solution
To find the Maclaurin series expansion for f(x) = ln(1 + x), we start by finding
the derivatives of f(x) at x= 0 and then use the formula for the Maclaurin
series expansion.
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: Find the Maclaurin series expansion The Maclaurin series
expansion for f(x) is given by
∞
X
n=0
f(n)(0)
n!xn=
∞
X
n=0
f(n)(0)
n!xn
12
By evaluating the derivatives at x= 0, we get
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
.
.
.
Therefore, the Maclaurin series expansion 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 15
Question
Find the Maclaurin series representation for ln(1 + x).
Solution
To find the Maclaurin series representation for ln(1 + x), we will first find the
derivatives of ln(1 + x) and then evaluate them at x= 0.
Step 1: Find the derivatives of ln(1 + 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
.
.
.
13
Step 2: Evaluate the derivatives at x= 0.
f(0) = ln(1 + 0) = ln(1) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f′′′(0) = 2
(1 + 0)3= 2
f(4)(0) = −6
(1 + 0)4=−6
.
.
.
Step 3: Write the Maclaurin series. Since f(0) = 0, the Maclaurin series for
ln(1 + x) is:
ln(1 + x) = x−x2
2+x3
3−x4
4+x5
5− · · · =
∞
X
n=1
(−1)n−1xn
n
Question 16
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) at x= 0.
14
Step 1: Find the derivatives of f(x)
f(x) = 1
1 + x2
f′(x) = −2x
(1 + x2)2
f′′(x) = −2(1 + x2)2−4x2(1 + x2)
(1 + x2)4
=2x2−2
(1 + x2)3
f′′′(x) = −8x(1 + x2)3−(2x2−2)(3(1 + x2)22x)
(1 + x2)6
=−8x−6x3
(1 + x2)5
f(4)(x) = −6(1 + x2)5−(−8x−6x3)(5(1 + x2)42x)
(1 + x2)8
=−6−24x2+ 30x2+ 24x4
(1 + x2)7
=24x4−6
(1 + x2)7
Step 2: Find f(n)(0) To find the Maclaurin series for f(x), we need to
evaluate the derivatives f(n)(0).
f′(0) = 0; f′′(0) = −2; f′′′ (0) = 0; f(4)(0) = −6
Step 3: Write the Maclaurin series The Maclaurin series representation
of f(x) = 1
1+x2is given by the formula:
f(x) = f(0) + f′(0)x+f′′ (0)x2
2! +f′′′(0)x3
3! +f(4)(0)x4
4! +. . .
Substitute the values of the derivatives at x= 0:
f(x)=1−2x2−6x4
4!
Therefore, the Maclaurin series for f(x) = 1
1+x2is 1 −2x2−3x4
4+. . . .
Question 17
Question
Find the Maclaurin series for f(x) = 1
1+x2.
15
Solution
Step 1: Find the derivatives of f(x) and evaluate them at x= 0 to find the
coefficients of the Maclaurin series.
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(3x2−1)
(1 + x2)3
f(3)(x) = d
dx 2(3x2−1)
(1 + x2)3=−24x(3x2−2)
(1 + x2)4
f(4)(x) = d
dx −24x(3x2−2)
(1 + x2)4=24(12x3−5x)
(1 + x2)5
Step 2: Evaluate the derivatives at x= 0.
f(0) = 1
f′(0) = 0
f′′(0) = −2
f(3)(0) = 0
f(4)(0) = 24
Step 3: Write out the Maclaurin series. Since f(0) = 1 and all odd deriva-
tives evaluated at x= 0 are 0, the Maclaurin series for f(x) is:
f(x)=1−2x2+ 24x4+. . .
Question 18
Question
Find the Maclaurin series for the function f(x) = ln(1 + x) and determine the
radius 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 to determine the coefficients of the
series.
16
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
Step 2: Find the Maclaurin series expansion. 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) evaluated at x= 0 into the series expan-
sion:
f(x) = ln(1 + x) = 0 + x−x2
2+x3
3−x4
4+· · ·
Therefore, the Maclaurin series for f(x) is
ln(1 + x) = x−x2
2+x3
3−x4
4+· · ·
Step 3: Determine the radius of convergence. In general, the radius
of convergence Rof a power series P∞
n=0 cnxnis given by
R= lim
n→∞
cn
cn+1
In this case, cn=1
n+1 , so
R= lim
n→∞
1
n+1
1
n+2
= lim
n→∞
n+ 2
n+ 1
= 1
Therefore, the radius of convergence of the Maclaurin series for ln(1 + x) is
R= 1.
Question 19
Question
Find the Maclaurin series for f(x) = e−x2.
17
Solution
To find the Maclaurin series for f(x) = e−x2, we can use the fact that the
Maclaurin series for exis P∞
n=0
xn
n!.
Step 1: Find the Maclaurin series for e−x2by substituting −x2for xin the
Maclaurin series for ex.
∞
X
n=0
(−x2)n
n!=
∞
X
n=0
(−1)n·x2n
n!
Step 2: Simplify the expression by splitting the sum into its even and odd
terms. ∞
X
n=0
(−1)n·x2n
n!=
∞
X
n=0
(−1)n·x2n
(2n)! +
∞
X
n=0
(−1)n·x2n
(2n+ 1)!
Step 3: Recognize that the first series is the Maclaurin series for cos(x) and
the second series is the Maclaurin series for sin(x). Thus, the Maclaurin series
for f(x) = e−x2is:
e−x2= cos(x) + sin(x)
e−x2=
∞
X
n=0
(−1)n·x2n
(2n)! +
∞
X
n=0
(−1)n·x2n
(2n+ 1)!
Question 20
Question
Find the Maclaurin series representation for the function f(x) = ln(1 + x).
Solution
To find the Maclaurin series representation of f(x) = ln(1 + x), we will first find
the derivatives of f(x) at x= 0, which will give us 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
f′′′(x) = 2
(1 + x)3
f(4)(x) = −6
(1 + x)4
.
.
.
18
Step 2: Evaluate the derivatives at x= 0.
f(0) = ln(1 + 0) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f′′′(0) = 2
(1 + 0)3= 2
f(4)(0) = −6
(1 + 0)4=−6
.
.
.
Step 3: Write down the Maclaurin series. The Maclaurin series rep-
resentation of ln(1 + x) is
ln(1 + x) = x−x2
2+x3
3−x4
4+· · · + (−1)n−1xn
n+· · ·
Therefore, the Maclaurin series representation for f(x) = ln(1 + x) is
ln(1 + x) = x−x2
2+x3
3−x4
4+· · ·
Question 21
Question
Find the Maclaurin series for ln(1 + x) up to the term in x5.
Solution
To find the Maclaurin series for ln(1 + x), we’ll start by finding the derivatives
of ln(1 + x) and evaluating them at x= 0. Then, we’ll use these derivatives to
build the Maclaurin series by using the formula for the Taylor series.
19
Step 1: Find the derivatives of ln(1 + 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) = −3·2
(1 + x)4
f(5)(x) = 4·3·2
(1 + x)5=24
(1 + x)5
Step 2: Evaluate derivatives at x= 0
f(0) = ln(1 + 0) = ln(1) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f′′′(0) = 2
(1 + 0)3= 2
f(4)(0) = −3·2
(1 + 0)4=−6
f(5)(0) = 24
(1 + 0)5= 24
Step 3: Build the Maclaurin series The Maclaurin series for a function
f(x) centered at x= 0 is given by:
f(x) = f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+f(4)(0)
4! x4+f(5)(0)
5! x5+. . .
Substitute the values we found earlier:
ln(1 + x) = 0 + 1 ·x−1
2·x2+2
6·x3−6
24 ·x4+24
120 ·x5+. . .
Therefore, the Maclaurin series for ln(1 + x) up to the term in x5is:
ln(1 + x) = x−1
2x2+1
3x3−1
4x4+1
5x5+. . .
Question 22
Question
Find the Maclaurin series for the function f(x) = 1
1+x2.
20
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we can use the geometric series
formula:
1
1−(−x2)=
∞
X
n=0
(−x2)n
Step 1: Rewrite the geometric series formula with the right substitution.
We rewrite the formula as:
1
1 + x2=
∞
X
n=0
(−1)nx2n
Step 2: Identify the Maclaurin series.
By comparing the formula above to the Maclaurin series representation:
1
1 + x=
∞
X
n=0
(−1)nxn
we have the Maclaurin series for f(x):
1
1 + x2=
∞
X
n=0
(−1)nx2n
Therefore, the Maclaurin series for f(x) = 1
1+x2is P∞
n=0(−1)nx2n.
Question 23
Question
Find the Maclaurin series for f(x) = 1
1+x2, and determine the interval of con-
vergence.
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we will use the geometric series
formula:
1
1−u=
∞
X
n=0
un
where |u|<1.
Step 1: Find the Maclaurin series Let u=−x2. Then, we have:
1
1 + x2=1
1−(−x2)=
∞
X
n=0
(−x2)n
21
Simplifying the series gives:
1
1 + x2=
∞
X
n=0
(−1)nx2n
Therefore, the Maclaurin series for f(x) = 1
1+x2is P∞
n=0(−1)nx2n.
Step 2: Determine the interval of convergence To find the interval of
convergence, we will use the ratio test. Let an= (−1)nx2n.
The ratio test states:
lim
n→∞
an+1
an
=L
If L < 1, then the series converges absolutely. If L > 1, then the series diverges.
If L= 1, the test is inconclusive.
Calculating the limit:
lim
n→∞
(−1)n+1x2(n+1)
(−1)nx2n
= lim
n→∞ |x2|=|x2|
For the series to converge, |x2|<1, which implies −1<x<1. Therefore,
the interval of convergence is (−1,1).
Question 24
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 Maclaurin
series for sin(x) and then replace xwith 2x.
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 2x.
f(x) = sin(2x) =
∞
X
n=0
(−1)n(2x)2n+1
(2n+ 1)! =
∞
X
n=0
(−1)n22n+1x2n+1
(2n+ 1)! =
∞
X
n=0
(−1)n22n+1x2n+1
(2n+ 1)!
Therefore, the Maclaurin series for f(x) = sin(2x) is P∞
n=0
(−1)n22n+1x2n+1
(2n+1)! .
22
Question 25
Question
Find the Maclaurin series for the function f(x) = ln(1 + x) and determine the
interval of convergence.
Solution
To find the Maclaurin series for f(x) = ln(1 + x), we will use the formula for
the Maclaurin series of ln(1 + x), which is given by:
ln(1 + x) =
∞
X
n=1
(−1)n+1xn
n
Step 1: Find the Maclaurin series We can see that the Maclaurin series
for ln(1 + x) is:
ln(1 + x) =
∞
X
n=1
(−1)n+1xn
n
Step 2: Determine the interval of convergence To find the interval of
convergence, we use the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
(−1)n+2xn+1
n+1
(−1)n+1xn
n
= lim
n→∞
n|x|
n+ 1
=|x|
The series converges for |x|<1. Therefore, the interval of convergence is
(−1,1].
Question 26
Question
Find the Maclaurin series for the function f(x) = cos(x) sin(x).
Solution
To find the Maclaurin series for f(x) = cos(x) sin(x), we will first find the
Maclaurin series for cos(x) and sin(x), and then multiply them together.
Step 1: Find the Maclaurin series for cos(x) The Maclaurin series for
cos(x) is given by:
cos(x) =
∞
X
n=0
(−1)nx2n
(2n)!
23
Step 2: Find the Maclaurin series for sin(x) The Maclaurin series for
sin(x) is given by:
sin(x) =
∞
X
n=0
(−1)nx2n+1
(2n+ 1)!
Step 3: Multiply the two series together Multiplying the Maclaurin
series for cos(x) and sin(x) together, we get:
f(x) = cos(x) sin(x) = ∞
X
n=0
(−1)nx2n
(2n)!! ∞
X
n=0
(−1)nx2n+1
(2n+ 1)!!
Expanding this product will give us the Maclaurin series for f(x) = cos(x) sin(x).
Question 27
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 can start by recalling the
Maclaurin series for ln(1 + x):
ln(1 + x) = x−x2
2+x3
3−x4
4+. . . =
∞
X
n=1
(−1)n−1xn
n
Using the fact that ln(1 −x) = −ln(1 + x), we get:
ln(1 −x) = −
∞
X
n=1
(−1)n−1xn
n
Step 1: Reorder the terms inside the summation:
ln(1 −x) =
∞
X
n=1
(−1)nxn
n
Step 2: Rewrite the series with nstarting from 0:
ln(1 −x) =
∞
X
n=0
(−1)n+1 xn
n+ 1
Step 3: This is the Maclaurin series for f(x) = ln(1 −x):
ln(1 −x) =
∞
X
n=0
(−1)n+1 xn
n+ 1
24
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 first find the derivatives of
f(x) at x= 0 and then use the formula for the Maclaurin series.
Step 1: Find the derivatives of f(x).
f(x) = 1
1 + x
f′(x) = d
dx 1
1 + x=−1
(1 + x)2
f′′(x) = d2
dx2 −1
(1 + x)2!=2
(1 + x)3
f′′′(x) = d3
dx3 2
(1 + x)3!=−6
(1 + x)4
.
.
.
It can be seen that the nth derivative of f(x) evaluated at x= 0 is of the form
(−1)nn!.
Step 2: Write the Maclaurin series for f(x).The Maclaurin series for
a function 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) at x= 0 into the formula:
f(x) = 1 −x+x2−x3+· · ·
=
∞
X
n=0
(−1)nxn(Maclaurin series for 1
1 + x)
Therefore, the Maclaurin series for f(x) = 1
1+xis P∞
n=0(−1)nxn.
Question 29
Question
Find the Maclaurin series for the function f(x) = excos(x).
25
Solution
To find the Maclaurin series of f(x) = excos(x), we can use the Maclaurin series
for exand cos(x) and then multiply the resulting series.
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 for exand cos(x) Multiplying the series for
exand cos(x), we get:
f(x) = excos(x) = ∞
X
n=0
xn
n!! ∞
X
n=0
(−1)nx2n
(2n)!!
Step 4: Simplify the expression By multiplying the series together and
simplifying, we get the Maclaurin series for f(x) = excos(x).
Therefore, the Maclaurin series for f(x) = excos(x) is [final Maclaurin se-
ries].
Question 30
Question
Find the Maclaurin series for f(x) = sin2(x).
Solution
To find the Maclaurin series for f(x) = sin2(x), we can start by using the known
Maclaurin series for sin(x).
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: Find the Maclaurin series for f(x) = sin2(x) Since f(x) =
sin2(x), we can write:
f(x) = (sin(x))2
f(x)=(
∞
X
n=0
(−1)nx2n+1
(2n+ 1)!)2
26
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we will use the geometric series
formula.
Step 1: Find the Maclaurin series for 11−x.The Maclaurin series
for 1
1−xis given by:
1
1−x=
∞
X
n=0
xn
for |x|<1.
Step 2: Substitute x2for x.Substitute x2for xto find the Maclaurin
series for f(x) = 1
1+x2:
1
1 + x2=
∞
X
n=0
(x2)n=
∞
X
n=0
x2n
for |x2|<1.
Step 3: Write the Maclaurin series for f(x).Since the Maclaurin series
for f(x) = 1
1+x2is P∞
n=0 x2n, the Maclaurin series for f(x) = 1
1+x2is:
f(x) =
∞
X
n=0
x2n
for |x2|<1.
Question 3
Question
Find the Maclaurin series of the function f(x) = ln(1 + x).
Solution
To find the Maclaurin series of f(x) = ln(1 + x), we will first find the derivatives
of f(x) and then evaluate them at x= 0 to determine the coefficients of the
series.
2
Step 1: Find the first few derivatives of f(x) = ln(1 + 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
.
.
.
Step 2: Evaluate the derivatives at x= 0.
f(0) = ln(1 + 0) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f′′′(0) = 2
(1 + 0)3= 2
f(4)(0) = −6
(1 + 0)4=−6
.
.
.
Step 3: Write down the Maclaurin series using the coefficients found in Step
2. The Maclaurin series for ln(1 + x) is:
ln(1 + x) = x−x2
2+x3
3−x4
4+· · ·
Therefore, the Maclaurin series of f(x) = ln(1 + x) is f(x) = P∞
n=1
(−1)n−1xn
n.
Question 4
Question
Find the Maclaurin series for 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.
3
Step 1: Find the Maclaurin series for sin(x).The Maclaurin series for
sin(x) is:
sin(x) = x−x3
3! +x5
5! −x7
7! +. . . =
∞
X
n=0
(−1)nx2n+1
(2n+ 1)!
Step 2: Square the Maclaurin series for sin(x).Squaring the Maclaurin
series for sin(x) gives us the Maclaurin series for sin2(x):
sin2(x) = ∞
X
n=0
(−1)nx2n+1
(2n+ 1)!!2
=
∞
X
n=0 (−1)nx2n+1
(2n+ 1)!·(−1)mx2m+1
(2m+ 1)!
=
∞
X
n=0
∞
X
m=0
(−1)n+mx2n+1x2m+1
(2n+ 1)!(2m+ 1)!
=
∞
X
n=0
∞
X
m=0
(−1)n+mx2n+2m+2
(2n+ 1)!(2m+ 1)!
Now, we can simplify the inner sum to get the Maclaurin series for sin2(x):
sin2(x) =
∞
X
n=0
(−1)n
(2n+ 1)! ·
n
X
m=0
(−1)m
(2(m−n) + 1)!x2n+2m+2
Question 5
Question
Find the Maclaurin series of the function f(x) = 1
x+2 .
Solution
To find the Maclaurin series of f(x), we will first find the Maclaurin series of
1
1+xby using the formula for the Maclaurin series of 1
1−xand substituting x
with −x.
Step 1: Find the Maclaurin series of 1
1+x. The Maclaurin series for 1
1−xis
given by:
∞
X
n=0
xn
Therefore, the Maclaurin series for 1
1+xis:
∞
X
n=0
(−1)nxn
4
Step 2: Find the Maclaurin series of f(x) = 1
x+2 . We can express f(x)
as 1
x+2 =1
2·1
1+ x
2. Since the Maclaurin series for 1
1+xis P∞
n=0(−1)nxn, the
Maclaurin series for 1
1+ x
2is:
∞
X
n=0 −1
2n
xn=
∞
X
n=0
(−1)n
2nxn
Multiplying by 1
2, we get the Maclaurin series for f(x):
∞
X
n=0
(−1)n
2n+1 xn
Question 6
Question
Find the Maclaurin series for f(x) = sin2(x).
Solution
To find the Maclaurin series for f(x) = sin2(x), we first need to find the Maclau-
rin series for sin2(x).
sin2(x) = (sin(x))2=eix −e−ix
2i2
=1
4e2ix −2eixe−ix +e−2ix
=1
4e2ix −2 + e−2ix.
Now, we can find the Maclaurin series for f(x) = 1
4e2ix −2 + e−2ix.
=1
41+2ix +(2ix)2
2! +(2ix)3
3! +· · · − 2+1−2ix +(−2ix)2
2! −(2ix)3
3! +· · ·
=1
42 + (2ix)2
2! +(2ix)4
4! +· · · − 2
=1
4−4x2+16x4
4! −64x6
6! +· · ·
=−x2+x4
6−x6
45 +· · · .
Therefore, the Maclaurin series for f(x) = sin2(x) is −x2+x4
6−x6
45 +· · · .
5
Question 7
Question
Find the Maclaurin series for the function f(x) = excos(x).
Solution
To find the Maclaurin series for f(x) = excos(x), we will use the Maclaurin
series for exand cos(x) and then multiply them together.
Step 1: Find the Maclaurin series for exThe Maclaurin series for ex
is:
ex=
∞
X
n=0
xn
n!
Step 2: Find the Maclaurin series for cos(x) The Maclaurin series for
cos(x) is:
cos(x) =
∞
X
n=0
(−1)nx2n
(2n)!
Step 3: Multiply the two series together Now, to find the Maclaurin
series for f(x) = excos(x), we multiply the Maclaurin series for exand cos(x):
f(x) = excos(x) = ∞
X
n=0
xn
n!! ∞
X
n=0
(−1)nx2n
(2n)! !
Step 4: Simplify the series To find the Maclaurin series for f(x), we
multiply the series term by term:
f(x) =
∞
X
n=0 n
X
k=0
(−1)kxn+k
k!(n−k)!(2n)!!
Therefore, the Maclaurin series for f(x) = excos(x) is:
∞
X
n=0 n
X
k=0
(−1)kxn+k
k!(n−k)!(2n)!!
Question 8
Question
Find the Maclaurin series for f(x) = 1
1−x2.
6
Solution
To find the Maclaurin series for f(x) = 1
1−x2, we will first find the derivatives
of f(x) and use them to find the coefficients in the series expansion.
Step 1: Find the derivatives of f(x).
f(x) = 1
1−x2
f′(x) = d
dx 1
1−x2=2x
(1 −x2)2= 2x(1 + x2)
f′′(x) = d
dx 2x(1 + x2)= 2(1 + x2)+4x2= 2(1 + 3x2)
f′′′(x) = d
dx 2(1 + 3x2)= 6x
Step 2: Find the Maclaurin series. The Maclaurin series expansion 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 find the values of f(0), f′(0), f′′(0), and f′′′(0):
f(0) = 1
1−0= 1
f′(0) = 2(0)(1 + 02) = 0
f′′(0) = 2(1) = 2
f′′′(0) = 0
Therefore, the Maclaurin series for f(x) is:
f(x) = 1 + 2
2!x2+0+. . . = 1 + x2
Question 9
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 then evaluate them at x= 0 to find the coefficients of the series.
7
Step 1: Find the derivatives of f(x).
f(x) = 1
1 + x
f′(x) = −(1 + x)−2=−(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: Evaluate the derivatives at x= 0.
f(0) = 1
f′(0) = −1
f′′(0) = 2
f′′′(0) = −6
.
.
.
Step 3: Write out the Maclaurin series. The Maclaurin series for f(x)
is given by
f(x) = f(0) + f′(0)x+f′′ (0)x2
2! +f′′′(0)x3
3! +· · ·
Substitute the values we calculated above:
f(x) = 1 −x+x2−x3+· · · =
∞
X
n=0
(−1)nxn
Therefore, the Maclaurin series for f(x) = 1
1+xis P∞
n=0(−1)nxn.
Question 10
Question
Find the Maclaurin series for e2x.
Solution
To find the Maclaurin series for e2x, we can use the fact that the Maclaurin
series for exis P∞
n=0
xn
n!and substitute 2xin place of x.
Step 1: Substitute 2xfor xin the Maclaurin series for ex:
e2x=
∞
X
n=0
(2x)n
n!
8
Step 2: Simplify the expression by expanding (2x)n:
e2x=
∞
X
n=0
2nxn
n!
Step 3: Rewrite the series in sigma notation:
e2x= 1 + 2x+22x2
2! +23x3
3! +· · ·
Step 4: Simplify the terms in the series:
e2x= 1 + 2x+ 2x2+4x3
3! +· · ·
Step 5: Therefore, the Maclaurin series for e2xis 1 + 2x+ 2x2+4x3
3! +· · · .
Question 11
Question
Find the Maclaurin series representation for the function f(x) = ln(1 + x) and
determine the interval of convergence.
Solution
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: Find the Maclaurin series. We can express the Maclaurin series for
ln(1 + x) as:
f(x) =
∞
X
n=0
f(n)(0)
n!xn=f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+f(4)(0)
4! x4+. . .
Substitute the derivatives at x= 0:
9
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
Therefore, the Maclaurin series is:
ln(1 + x) = x−x2
2+x3
3−x4
4+. . .
Step 3: Determine the interval of convergence. The interval of convergence
of a Maclaurin series is the set of all xvalues for which the series converges.
By using the Ratio Test, we find the interval of convergence to be −1< x ≤1.
This means the Maclaurin series for ln(1 + x) converges for −1< x ≤1.
Question 12
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 then evaluate them at x= 0 to get the coefficients of the series.
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 differentiating the derivative of f(x))
f′′′(x) = 2
(1 + x)3(by differentiating f′′(x))
f(4)(x) = −6
(1 + x)4(by differentiating f′′′(x))
.
.
.
The nth derivative of f(x) can be written as
f(n)(x)=(−1)n−1(n−1)!
(1 + x)n
10
Step 2: Evaluate the derivatives at x= 0 Now, we evaluate the deriva-
tives at x= 0 to get the coefficients of the Maclaurin series:
f(0) = ln(1 + 0) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f′′′(0) = 2
(1 + 0)3= 2
f(4)(0) = −6
(1 + 0)4=−6
.
.
.
Step 3: Write down the Maclaurin series The Maclaurin series for
ln(1 + x) is
ln(1 + x) =
∞
X
n=1
(−1)n−1xn
n=x−x2
2+x3
3−x4
4+· · ·
Question 13
Question
Find the Maclaurin series for the function f(x) = ln(1 + x) by differentiating
the series for f′(x) = 1
1+xterm by term.
Solution
To find the Maclaurin series for f(x) = ln(1 + x), we will differentiate the
Maclaurin series for f′(x) = 1
1+xterm by term.
Step 1: Find the Maclaurin series for f′(x) = 1
1+x: The Maclaurin series for
1
1+xis:
1
1 + x= 1 −x+x2−x3+· · · =
∞
X
n=0
(−1)nxn
Step 2: Differentiate the series term by term to find the series for f(x) =
ln(1 + x): Differentiating the series term by term, we get:
f(x) = Z1
1 + xdx =Z∞
X
n=0
(−1)nxndx
=
∞
X
n=0
(−1)nZxndx =
∞
X
n=0
(−1)n
n+ 1 xn+1 +C
11
=
∞
X
n=0
(−1)n
n+ 1 xn+1 +C
Therefore, the Maclaurin series for f(x) = ln(1 + x) is:
ln(1 + x) =
∞
X
n=0
(−1)n
n+ 1 xn+1 +C
Question 14
Question
Find the Maclaurin series expansion for f(x) = ln(1 + x).
Solution
To find the Maclaurin series expansion for f(x) = ln(1 + x), we start by finding
the derivatives of f(x) at x= 0 and then use the formula for the Maclaurin
series expansion.
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: Find the Maclaurin series expansion The Maclaurin series
expansion for f(x) is given by
∞
X
n=0
f(n)(0)
n!xn=
∞
X
n=0
f(n)(0)
n!xn
12
By evaluating the derivatives at x= 0, we get
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
.
.
.
Therefore, the Maclaurin series expansion 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 15
Question
Find the Maclaurin series representation for ln(1 + x).
Solution
To find the Maclaurin series representation for ln(1 + x), we will first find the
derivatives of ln(1 + x) and then evaluate them at x= 0.
Step 1: Find the derivatives of ln(1 + 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
.
.
.
13
Step 2: Evaluate the derivatives at x= 0.
f(0) = ln(1 + 0) = ln(1) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f′′′(0) = 2
(1 + 0)3= 2
f(4)(0) = −6
(1 + 0)4=−6
.
.
.
Step 3: Write the Maclaurin series. Since f(0) = 0, the Maclaurin series for
ln(1 + x) is:
ln(1 + x) = x−x2
2+x3
3−x4
4+x5
5− · · · =
∞
X
n=1
(−1)n−1xn
n
Question 16
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) at x= 0.
14
Step 1: Find the derivatives of f(x)
f(x) = 1
1 + x2
f′(x) = −2x
(1 + x2)2
f′′(x) = −2(1 + x2)2−4x2(1 + x2)
(1 + x2)4
=2x2−2
(1 + x2)3
f′′′(x) = −8x(1 + x2)3−(2x2−2)(3(1 + x2)22x)
(1 + x2)6
=−8x−6x3
(1 + x2)5
f(4)(x) = −6(1 + x2)5−(−8x−6x3)(5(1 + x2)42x)
(1 + x2)8
=−6−24x2+ 30x2+ 24x4
(1 + x2)7
=24x4−6
(1 + x2)7
Step 2: Find f(n)(0) To find the Maclaurin series for f(x), we need to
evaluate the derivatives f(n)(0).
f′(0) = 0; f′′(0) = −2; f′′′ (0) = 0; f(4)(0) = −6
Step 3: Write the Maclaurin series The Maclaurin series representation
of f(x) = 1
1+x2is given by the formula:
f(x) = f(0) + f′(0)x+f′′ (0)x2
2! +f′′′(0)x3
3! +f(4)(0)x4
4! +. . .
Substitute the values of the derivatives at x= 0:
f(x)=1−2x2−6x4
4!
Therefore, the Maclaurin series for f(x) = 1
1+x2is 1 −2x2−3x4
4+. . . .
Question 17
Question
Find the Maclaurin series for f(x) = 1
1+x2.
15
Solution
Step 1: Find the derivatives of f(x) and evaluate them at x= 0 to find the
coefficients of the Maclaurin series.
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(3x2−1)
(1 + x2)3
f(3)(x) = d
dx 2(3x2−1)
(1 + x2)3=−24x(3x2−2)
(1 + x2)4
f(4)(x) = d
dx −24x(3x2−2)
(1 + x2)4=24(12x3−5x)
(1 + x2)5
Step 2: Evaluate the derivatives at x= 0.
f(0) = 1
f′(0) = 0
f′′(0) = −2
f(3)(0) = 0
f(4)(0) = 24
Step 3: Write out the Maclaurin series. Since f(0) = 1 and all odd deriva-
tives evaluated at x= 0 are 0, the Maclaurin series for f(x) is:
f(x)=1−2x2+ 24x4+. . .
Question 18
Question
Find the Maclaurin series for the function f(x) = ln(1 + x) and determine the
radius 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 to determine the coefficients of the
series.
16
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
Step 2: Find the Maclaurin series expansion. 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) evaluated at x= 0 into the series expan-
sion:
f(x) = ln(1 + x) = 0 + x−x2
2+x3
3−x4
4+· · ·
Therefore, the Maclaurin series for f(x) is
ln(1 + x) = x−x2
2+x3
3−x4
4+· · ·
Step 3: Determine the radius of convergence. In general, the radius
of convergence Rof a power series P∞
n=0 cnxnis given by
R= lim
n→∞
cn
cn+1
In this case, cn=1
n+1 , so
R= lim
n→∞
1
n+1
1
n+2
= lim
n→∞
n+ 2
n+ 1
= 1
Therefore, the radius of convergence of the Maclaurin series for ln(1 + x) is
R= 1.
Question 19
Question
Find the Maclaurin series for f(x) = e−x2.
17
Solution
To find the Maclaurin series for f(x) = e−x2, we can use the fact that the
Maclaurin series for exis P∞
n=0
xn
n!.
Step 1: Find the Maclaurin series for e−x2by substituting −x2for xin the
Maclaurin series for ex.
∞
X
n=0
(−x2)n
n!=
∞
X
n=0
(−1)n·x2n
n!
Step 2: Simplify the expression by splitting the sum into its even and odd
terms. ∞
X
n=0
(−1)n·x2n
n!=
∞
X
n=0
(−1)n·x2n
(2n)! +
∞
X
n=0
(−1)n·x2n
(2n+ 1)!
Step 3: Recognize that the first series is the Maclaurin series for cos(x) and
the second series is the Maclaurin series for sin(x). Thus, the Maclaurin series
for f(x) = e−x2is:
e−x2= cos(x) + sin(x)
e−x2=
∞
X
n=0
(−1)n·x2n
(2n)! +
∞
X
n=0
(−1)n·x2n
(2n+ 1)!
Question 20
Question
Find the Maclaurin series representation for the function f(x) = ln(1 + x).
Solution
To find the Maclaurin series representation of f(x) = ln(1 + x), we will first find
the derivatives of f(x) at x= 0, which will give us 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
f′′′(x) = 2
(1 + x)3
f(4)(x) = −6
(1 + x)4
.
.
.
18
Step 2: Evaluate the derivatives at x= 0.
f(0) = ln(1 + 0) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f′′′(0) = 2
(1 + 0)3= 2
f(4)(0) = −6
(1 + 0)4=−6
.
.
.
Step 3: Write down the Maclaurin series. The Maclaurin series rep-
resentation of ln(1 + x) is
ln(1 + x) = x−x2
2+x3
3−x4
4+· · · + (−1)n−1xn
n+· · ·
Therefore, the Maclaurin series representation for f(x) = ln(1 + x) is
ln(1 + x) = x−x2
2+x3
3−x4
4+· · ·
Question 21
Question
Find the Maclaurin series for ln(1 + x) up to the term in x5.
Solution
To find the Maclaurin series for ln(1 + x), we’ll start by finding the derivatives
of ln(1 + x) and evaluating them at x= 0. Then, we’ll use these derivatives to
build the Maclaurin series by using the formula for the Taylor series.
19
Step 1: Find the derivatives of ln(1 + 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) = −3·2
(1 + x)4
f(5)(x) = 4·3·2
(1 + x)5=24
(1 + x)5
Step 2: Evaluate derivatives at x= 0
f(0) = ln(1 + 0) = ln(1) = 0
f′(0) = 1
1+0 = 1
f′′(0) = −1
(1 + 0)2=−1
f′′′(0) = 2
(1 + 0)3= 2
f(4)(0) = −3·2
(1 + 0)4=−6
f(5)(0) = 24
(1 + 0)5= 24
Step 3: Build the Maclaurin series The Maclaurin series for a function
f(x) centered at x= 0 is given by:
f(x) = f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+f(4)(0)
4! x4+f(5)(0)
5! x5+. . .
Substitute the values we found earlier:
ln(1 + x) = 0 + 1 ·x−1
2·x2+2
6·x3−6
24 ·x4+24
120 ·x5+. . .
Therefore, the Maclaurin series for ln(1 + x) up to the term in x5is:
ln(1 + x) = x−1
2x2+1
3x3−1
4x4+1
5x5+. . .
Question 22
Question
Find the Maclaurin series for the function f(x) = 1
1+x2.
20
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we can use the geometric series
formula:
1
1−(−x2)=
∞
X
n=0
(−x2)n
Step 1: Rewrite the geometric series formula with the right substitution.
We rewrite the formula as:
1
1 + x2=
∞
X
n=0
(−1)nx2n
Step 2: Identify the Maclaurin series.
By comparing the formula above to the Maclaurin series representation:
1
1 + x=
∞
X
n=0
(−1)nxn
we have the Maclaurin series for f(x):
1
1 + x2=
∞
X
n=0
(−1)nx2n
Therefore, the Maclaurin series for f(x) = 1
1+x2is P∞
n=0(−1)nx2n.
Question 23
Question
Find the Maclaurin series for f(x) = 1
1+x2, and determine the interval of con-
vergence.
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we will use the geometric series
formula:
1
1−u=
∞
X
n=0
un
where |u|<1.
Step 1: Find the Maclaurin series Let u=−x2. Then, we have:
1
1 + x2=1
1−(−x2)=
∞
X
n=0
(−x2)n
21
Simplifying the series gives:
1
1 + x2=
∞
X
n=0
(−1)nx2n
Therefore, the Maclaurin series for f(x) = 1
1+x2is P∞
n=0(−1)nx2n.
Step 2: Determine the interval of convergence To find the interval of
convergence, we will use the ratio test. Let an= (−1)nx2n.
The ratio test states:
lim
n→∞
an+1
an
=L
If L < 1, then the series converges absolutely. If L > 1, then the series diverges.
If L= 1, the test is inconclusive.
Calculating the limit:
lim
n→∞
(−1)n+1x2(n+1)
(−1)nx2n
= lim
n→∞ |x2|=|x2|
For the series to converge, |x2|<1, which implies −1<x<1. Therefore,
the interval of convergence is (−1,1).
Question 24
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 Maclaurin
series for sin(x) and then replace xwith 2x.
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 2x.
f(x) = sin(2x) =
∞
X
n=0
(−1)n(2x)2n+1
(2n+ 1)! =
∞
X
n=0
(−1)n22n+1x2n+1
(2n+ 1)! =
∞
X
n=0
(−1)n22n+1x2n+1
(2n+ 1)!
Therefore, the Maclaurin series for f(x) = sin(2x) is P∞
n=0
(−1)n22n+1x2n+1
(2n+1)! .
22
Question 25
Question
Find the Maclaurin series for the function f(x) = ln(1 + x) and determine the
interval of convergence.
Solution
To find the Maclaurin series for f(x) = ln(1 + x), we will use the formula for
the Maclaurin series of ln(1 + x), which is given by:
ln(1 + x) =
∞
X
n=1
(−1)n+1xn
n
Step 1: Find the Maclaurin series We can see that the Maclaurin series
for ln(1 + x) is:
ln(1 + x) =
∞
X
n=1
(−1)n+1xn
n
Step 2: Determine the interval of convergence To find the interval of
convergence, we use the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
(−1)n+2xn+1
n+1
(−1)n+1xn
n
= lim
n→∞
n|x|
n+ 1
=|x|
The series converges for |x|<1. Therefore, the interval of convergence is
(−1,1].
Question 26
Question
Find the Maclaurin series for the function f(x) = cos(x) sin(x).
Solution
To find the Maclaurin series for f(x) = cos(x) sin(x), we will first find the
Maclaurin series for cos(x) and sin(x), and then multiply them together.
Step 1: Find the Maclaurin series for cos(x) The Maclaurin series for
cos(x) is given by:
cos(x) =
∞
X
n=0
(−1)nx2n
(2n)!
23
Step 2: Find the Maclaurin series for sin(x) The Maclaurin series for
sin(x) is given by:
sin(x) =
∞
X
n=0
(−1)nx2n+1
(2n+ 1)!
Step 3: Multiply the two series together Multiplying the Maclaurin
series for cos(x) and sin(x) together, we get:
f(x) = cos(x) sin(x) = ∞
X
n=0
(−1)nx2n
(2n)!! ∞
X
n=0
(−1)nx2n+1
(2n+ 1)!!
Expanding this product will give us the Maclaurin series for f(x) = cos(x) sin(x).
Question 27
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 can start by recalling the
Maclaurin series for ln(1 + x):
ln(1 + x) = x−x2
2+x3
3−x4
4+. . . =
∞
X
n=1
(−1)n−1xn
n
Using the fact that ln(1 −x) = −ln(1 + x), we get:
ln(1 −x) = −
∞
X
n=1
(−1)n−1xn
n
Step 1: Reorder the terms inside the summation:
ln(1 −x) =
∞
X
n=1
(−1)nxn
n
Step 2: Rewrite the series with nstarting from 0:
ln(1 −x) =
∞
X
n=0
(−1)n+1 xn
n+ 1
Step 3: This is the Maclaurin series for f(x) = ln(1 −x):
ln(1 −x) =
∞
X
n=0
(−1)n+1 xn
n+ 1
24
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 first find the derivatives of
f(x) at x= 0 and then use the formula for the Maclaurin series.
Step 1: Find the derivatives of f(x).
f(x) = 1
1 + x
f′(x) = d
dx 1
1 + x=−1
(1 + x)2
f′′(x) = d2
dx2 −1
(1 + x)2!=2
(1 + x)3
f′′′(x) = d3
dx3 2
(1 + x)3!=−6
(1 + x)4
.
.
.
It can be seen that the nth derivative of f(x) evaluated at x= 0 is of the form
(−1)nn!.
Step 2: Write the Maclaurin series for f(x).The Maclaurin series for
a function 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) at x= 0 into the formula:
f(x) = 1 −x+x2−x3+· · ·
=
∞
X
n=0
(−1)nxn(Maclaurin series for 1
1 + x)
Therefore, the Maclaurin series for f(x) = 1
1+xis P∞
n=0(−1)nxn.
Question 29
Question
Find the Maclaurin series for the function f(x) = excos(x).
25
Solution
To find the Maclaurin series of f(x) = excos(x), we can use the Maclaurin series
for exand cos(x) and then multiply the resulting series.
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 for exand cos(x) Multiplying the series for
exand cos(x), we get:
f(x) = excos(x) = ∞
X
n=0
xn
n!! ∞
X
n=0
(−1)nx2n
(2n)!!
Step 4: Simplify the expression By multiplying the series together and
simplifying, we get the Maclaurin series for f(x) = excos(x).
Therefore, the Maclaurin series for f(x) = excos(x) is [final Maclaurin se-
ries].
Question 30
Question
Find the Maclaurin series for f(x) = sin2(x).
Solution
To find the Maclaurin series for f(x) = sin2(x), we can start by using the known
Maclaurin series for sin(x).
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: Find the Maclaurin series for f(x) = sin2(x) Since f(x) =
sin2(x), we can write:
f(x) = (sin(x))2
f(x)=(
∞
X
n=0
(−1)nx2n+1
(2n+ 1)!)2
26
Expanding the square, we get:
f(x) = ∞
X
n=0
(−1)nx2n+1
(2n+ 1)!! ∞
X
m=0
(−1)mx2m+1
(2m+ 1)!
] Now, we multiply the series and simplify:
f(x) =
∞
X
n=0
∞
X
m=0
(−1)n+mx2n+1x2m+1
(2n+ 1)!(2m+ 1)!
f(x) =
∞
X
n=0
∞
X
m=0
(−1)n+mx2n+2m+2
(2n+ 1)!(2m+ 1)!
Now, we combine the nand mterms into a single index k=n+m:
f(x) =
∞
X
k=0
k
X
n=0
(−1)n+k−nx2k+2
(2n+ 1)!(2(k−n) + 1)!
f(x) =
∞
X
k=0
(−1)kx2k+2
(2k+ 1)!
k
X
n=0 k
n
Finally, we simplify the expression to obtain the Maclaurin series for f(x) =
sin2(x).
27
