MATH 100 - FUNDAMENTALS OF
MATHEMATICS - Taylor and
Maclaurin Series
Question Bank - Set 3
Liberty University
Question 1
Question
Find the Maclaurin series for the function f(x) = 1
1+x.
Solution
To find the Maclaurin series for f(x) = 1
1+x, we will first find the derivatives of
f(x) 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 + x
f′(x) = −(1 + x)−2=−1
(1 + x)2
f′′(x) = 2(1 + x)−3=2
(1 + x)3
f′′′(x) = −6(1 + x)−4=−6
(1 + x)4
.
.
.
Step 2: 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
.
.
.
Step 3: Write out the Maclaurin series. Since 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! +· · · ,
the Maclaurin series for f(x) = 1
1+xis
1−x+ 2x2−6x3+· · · .
Question 2
Question
Find the Maclaurin series for f(x) = e2xcos(x).
Solution
To find the Maclaurin series for f(x) = e2xcos(x), we will need to find the
derivatives of f(x) at x= 0 and use these derivatives to form the Maclaurin
series.
Step 1: Find f(0) and f′(0).
f(x) = e2xcos(x)
f(0) = e2·0cos(0) = 1
To find f′(x), we will use the product rule:
f′(x)=(e2x·cos(x))′=e2x(−sin(x)) + cos(x)(2e2x)=2e2xcos(x)−e2xsin(x)
f′(0) = 2e2·0cos(0) −e2·0sin(0) = 2 ·1·1−1·0=2
Step 2: Find f′′ (0) and f′′′ (0).
f′′(x) = (2e2xcos(x)−e2xsin(x))′= 2e2x(−sin(x))−e2xcos(x)+2e2xcos(x)−e2xsin(x)
2
f′′(0) = 2 ·1·0−1·1+2·1·1−1·0=3
To find f′′′ (x), we differentiate f′′(x):
f′′′(x) = (2e2x(−sin(x)) −e2xcos(x)+2e2xcos(x)−e2xsin(x))′
f′′′(x) = 2e2x(−cos(x))−e2x(−sin(x))+2e2x(−sin(x))−e2xcos(x)+2e2xsin(x)−e2xcos(x)
f′′′(0) = 2 ·1·(−1) −1·0+2·1·0−1·1+2·1·0−1·1 = −5
Step 3: Write the Maclaurin series for f(x).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 known values of f(0), f′(0), f′′(0), and f′′′(0):
f(x) = 1 + 2x+3
2x2−5
6x3+. . .
Therefore, the Maclaurin series for f(x) = e2xcos(x) is 1 + 2x+3
2x2−5
6x3+. . ..
Question 3
Question
Find the Taylor series representation for f(x) = excentered at a= 0.
Solution
To find the Taylor series representation of f(x) = excentered at a= 0, we will
use the formula for a Taylor series:
f(x) = f(a) + f′(a)
1! (x−a) + f′′ (a)
2! (x−a)2+f′′′(a)
3! (x−a)3+· · ·
Step 1: Find f(n)(x)for all n.
f(x) = ex
f′(x) = ex
f′′(x) = ex
.
.
.
f(n)(x) = exfor all n
3
Step 2: Evaluate f(n)(a)for all n.
f(0) = e0= 1
f′(0) = e0= 1
f′′(0) = e0= 1
.
.
.
f(n)(0) = e0= 1 for all n
Step 3: Plug in the values into the formula for the Taylor series.
Since f(n)(0) = 1 for all n, the Taylor series becomes:
f(x) = 1 + 1
1!x+1
2!x2+1
3!x3+· · · =
∞
X
n=0
xn
n!
Therefore, the Taylor series representation for f(x) = excentered at a= 0
is P∞
n=0
xn
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 can start by expressing
sin2(x) in terms of well-known trigonometric functions. Recall the trigonometric
identity: sin2(x) = 1
2−1
2cos(2x).
Step 1: Apply the trigonometric identity.
f(x) = sin2(x) = 1
2−1
2cos(2x)
Step 2: Find the Maclaurin series for 1
2. The Maclaurin series for 1
2is
simply 1
2.
Step 3: Find the Maclaurin series for −1
2cos(2x). To find the Maclaurin
series for −1
2cos(2x), we can start with the Maclaurin series for cos(x): cos(x) =
1−x2
2! +x4
4! −x6
6! +. . .. Now, substitute xwith 2xin the series to get the Maclaurin
series for cos(2x).
cos(2x) = 1 −(2x)2
2! +(2x)4
4! −(2x)6
6! +. . .
Simplify this expression.
cos(2x)=1−2x2+4x4
3! −8x6
6! +. . .
4
Now, multiply by −1
2to get the Maclaurin series for −1
2cos(2x).
−1
2cos(2x) = −1
2+x2−2x4
3! +4x6
6! −. . .
Step 4: Add the Maclaurin series for 1
2and −1
2cos(2x) to get the Maclaurin
series for f(x) = sin2(x).
f(x) = 1
2−1
2cos(2x) = 1
2+−1
2+x2−2x4
3! +4x6
6! −. . .
Now, combine like terms.
f(x) = x2−2x4
3! +4x6
6! −. . .
Therefore, the Maclaurin series for f(x) = sin2(x) is x2−2x4
3! +4x6
6! −. . ..
Question 5
Question
Find the Maclaurin series representation for f(x) = 1
1+x.
Solution
To find the Maclaurin series for f(x), we will use the fact that 1
1+x= 1 −x+
x2−x3+x4−. . . for |x|<1.
Step 1: Find the formula for the Maclaurin series of f(x).
Recall that the Maclaurin series for f(x) is given by:
f(x) = f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+. . .
Step 2: Find the derivatives of f(x) at x= 0.
We have f(x) = 1
1+x, so:
f(0) = 1
1+0 = 1
f′(x) = d
dx 1
1 + x=−1
(1 + x)2
f′(0) = −1
(1 + 0)2=−1
f′′(x) = d
dx −1
(1 + x)2=2
(1 + x)3
5
f′′(0) = 2
(1 + 0)3= 2
The higher-order derivatives will follow a similar pattern.
Step 3: Plug f(0), f′(0), f′′ (0) into the formula for the Maclaurin series.
The Maclaurin series representation for f(x) is:
f(x)=1−x+x2−x3+x4−. . .
Question 6
Question
Find the Maclaurin series representation for the function f(x) = 1
(1+x)2.
Solution
To find the Maclaurin series representation for f(x) = 1
(1+x)2, we will first find
the derivatives of f(x) and then evaluate them at x= 0 to get the coefficients
of the Maclaurin series.
Step 1: Find the first few derivatives of f(x).
f(x) = 1
(1 + x)2
f′(x) = d
dx 1
(1 + x)2=−2(1 + x)−3
f′′(x) = d
dx −2(1 + x)−3=−6(1 + x)−4
f′′′(x) = d
dx −6(1 + x)−4= 24(1 + x)−5
Step 2: Evaluate the derivatives at x= 0.
f(0) = 1
f′(0) = −2
f′′(0) = −6
f′′′(0) = 24
Step 3: Write out the Maclaurin series using the coefficients found
above. Since the Maclaurin series for a function can be expressed as:
f(x) =
∞
X
n=0
f(n)(0)
n!xn
The Maclaurin series for f(x) is:
f(x)=1−2x−6x2+ 24x3+. . .
6
Question 7
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 its derivatives and
then evaluate them at x= 0.
Step 1: Find the first few derivatives of f(x).
f(x) = (1 −x)−1
f′(x)=(−1)(1 −x)−2
f′′(x)=(−1)(−2)(1 −x)−3
f′′′(x)=(−1)(−2)(−3)(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 down the general term for the Maclaurin series. The
general term for the Maclaurin series is given by:
f(n)(0) ·xn
n!
Step 4: Write down the Maclaurin series for f(x).Putting it all
together, the Maclaurin series for f(x) = 1
1−xis:
1−x+x2−x3+x4−. . . =
∞
X
n=0
(−1)nxn
Question 8
Question
Find the Maclaurin series representation for the function f(x) = ln(1 + x).
7
Solution
To find the Maclaurin series representation for f(x) = ln(1 + x), we will first
need to find its derivatives at x= 0 to obtain the coefficients of the series.
Step 1: Find the first few derivatives of f(x).
f(x) = ln(1 + x)
f′(x) = 1
1 + x,
f′′(x) = d
dx 1
1 + x=−1
(1 + x)2,
f′′′(x) = d
dx −1
(1 + x)2=2
(1 + x)3,
f(4)(x) = d
dx 2
(1 + x)3=−6
(1 + x)4.
Step 2: Find f(n)(0) for all n.
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 the Maclaurin series. The Maclaurin series representa-
tion of f(x) = ln(1 + x) is given by:
f(x) =
∞
X
n=0
f(n)(0)
n!xn=x−x2
2+x3
3−x4
4+· · ·
Question 9
Question
Find the Maclaurin series for the function f(x) = sin(2x).
Solution
To find the Maclaurin series for f(x) = sin(2x), we will first express sin(2x) in
terms of its known Taylor series expansion and then determine its Maclaurin
series.
8
Step 1: Derive the Maclaurin series for sin(2x)using 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)n
(2n+ 1)!x2n+1.
By substituting 2xin place of x, we have:
sin(2x)=2x−(2x)3
3! +(2x)5
5! −(2x)7
7! +· · · .
Step 2: Simplify the above series to obtain the Maclaurin series
for sin(2x).Simplifying the expression above, we get:
sin(2x)=2x−8x3
6+32x5
120 −128x7
5040 +· · ·
= 2x−4x3
3+2x5
15 −4x7
315 +· · ·
=
∞
X
n=0
(−1)n22n+1x2n+1
(2n+ 1)! .
Therefore, the Maclaurin series for f(x) = sin(2x) is:
f(x)=2x−4x3
3! +2x5
5! −4x7
7! +· · · =
∞
X
n=0
(−1)n22n+1x2n+1
(2n+ 1)! .
Question 10
Question
Find the Maclaurin series representation of the function f(x) = 1
1−x+x2.
Solution
To find the Maclaurin series of f(x), we will first express f(x) as a geometric
series.
Step 1: Find the geometric series form of f(x).Consider the geometric
series formula: 1
1−r= 1 + r+r2+r3+. . .. We can express f(x) in terms of a
geometric series:
f(x) = 1
1−x+x2=1
(1 −x)2−(x2)=1
(1 −x)2(1 + x).
Splitting the expression, we can rewrite f(x) as:
f(x) = A
1−x+B
(1 −x)2+C
1 + x
9
for some constants A,B, and C.
Multiplying through by the denominators and solving for A,B, and C, we
find:
A= 1, B = 2, C = 1.
Therefore, we have:
f(x) = 1
1−x+2
(1 −x)2+1
1 + x= 1 + x+x2+ 2x3+ 3x4+. . . .
Step 2: Write the Maclaurin series of f(x).Since f(x) has the repre-
sentation 1 + x+x2+ 2x3+ 3x4+. . ., the Maclaurin series for f(x) is:
f(x) = 1 + x+x2+ 2x3+ 3x4+. . . .
Therefore, the Maclaurin series representation of f(x) = 1
1−x+x2is 1 + x+
x2+ 2x3+ 3x4+. . ..
Question 11
Question
Find the Maclaurin series for the function f(x) = 1
1+x+x2.
Solution
To find the Maclaurin series for f(x) = 1
1+x+x2, we will first express f(x) as a
geometric series and then find the Maclaurin series representation.
Step 1: Find the derivative of f(x)
f(x) = 1
1 + x+x2= (1 + x+x2)−1
f′(x) = −1(1 + x+x2)−2(1 + 2x) = −1+2x
(1 + x+x2)2
Step 2: Find the second derivative of f(x)
f′(x) = −1+2x
(1 + x+x2)2
f′′(x) = d
dx −1+2x
(1 + x+x2)2
=2−2(1 + 2x)(1 + 2x)
(1 + x+x2)3=−2−4x
(1 + x+x2)3
Step 3: Find the third derivative of f(x)
f′′(x) = −2−4x
(1 + x+x2)3
10
f′′′(x) = d
dx −2−4x
(1 + x+x2)3
=−4−4(1 + 2x)(−4−4x)
(1 + x+x2)4=16 + 24x+ 8x2
(1 + x+x2)4
Step 4: Find the fourth derivative of f(x)
f′′′(x) = 16 + 24x+ 8x2
(1 + x+x2)4
f(4)(x) = d
dx 16 + 24x+ 8x2
(1 + x+x2)4
=48 + 56x+ 16x2
(1 + x+x2)5
Step 5: Generalize the pattern After computing several derivatives, we
see a pattern emerging where the n-th derivative of f(x) is given by:
f(n)(x) = Pn(x)
(1 + x+x2)n+1
where Pn(x) is a polynomial of degree n.
Step 6: Write 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+· · ·
Step 7: Evaluate the derivatives at x= 0 Evaluating the derivatives at
x= 0, we find:
f(0) = 1
1+0+02= 1
f′(0) = −1
f′′(0) = −1
1
f′′′(0) = 1
1
f(4)(0) = 0
1
Step 8: Substitute into the Maclaurin series formula Therefore, the
Maclaurin series for f(x) = 1
1+x+x2is:
1−x−x2+x3+· · ·
11
Question 12
Question
Find the Maclaurin series for the function f(x) = 1
1+x2.
Solution
To find the Maclaurin series for the function f(x) = 1
1+x2, we will first find the
derivatives of f(x) and evaluate them at x= 0 to find the coefficients of the
series.
Step 1: Find the derivatives of f(x)
f(x) = 1
1 + x2
f′(x) = −2x
(1 + x2)2
f′′(x) = −2(1 + x2)2−2x(2)(1 + x2)(2x)
(1 + x2)4
=−2−2x2−8x2
(1 + x2)3=−2−10x2
(1 + x2)3
The pattern for the derivatives of f(x) becomes apparent:
f′(x) = −2x
(1 + x2)2
f′′(x) = −2(1 + x2)−2x(2)(x)
(1 + x2)3=−2−6x2
(1 + x2)3
f′′′(x) = −2(1 + x2)−6x2(2x)
(1 + x2)4=−2−14x2
(1 + x2)4
.
.
.
Step 2: Evaluate the derivatives at x= 0 At x= 0:
f(0) = 1
f′(0) = 0
f′′(0) = −2
f′′′(0) = 0
f′′′′(0) = −2
.
.
.
Step 3: Write the Maclaurin series for f(x) Since f(0) = 1, f′(0) = 0,
f′′(0) = −2, f′′′ (0) = 0, f′′′′ (0) = −2, and so on, the Maclaurin series for f(x)
is:
f(x)=1−2x2+ 2x4−2x6+. . . =
∞
X
n=0
(−1)nx2n
12
Question 13
Question
Find the Maclaurin series for f(x) = sin(x) cos2(x).
Solution
To find the Maclaurin series for f(x) = sin(x) cos2(x), we will first expand sin(x)
and cos(x) into their Maclaurin series and then multiply them together.
Step 1: Find the Maclaurin series for sin(x).
sin(x) = x−x3
3! +x5
5! −x7
7! +. . . =
∞
X
n=0
(−1)nx2n+1
(2n+ 1)!
Step 2: Find the Maclaurin series for cos(x).
cos(x)=1−x2
2! +x4
4! −x6
6! +. . . =
∞
X
n=0
(−1)nx2n
(2n)!
Step 3: Substitute the Maclaurin series for sin(x) and cos(x) into f(x).
f(x) = (
∞
X
n=0
(−1)nx2n+1
(2n+ 1)!)(
∞
X
n=0
(−1)nx2n
(2n)!)(
∞
X
n=0
(−1)nx2n
(2n)!)
Step 4: Simplify the expression by expanding the product.
f(x) = sin(x) cos2(x)=(x−x3
3! +x5
5! −. . .)(1−x2
2! +x4
4! −. . .)(1−x2
2! +x4
4! −. . .)
Step 5: Find the Maclaurin series for f(x) by multiplying out the series
and simplifying.
f(x) = x−x3
2−x5
10 +x7
14 +. . . =
∞
X
n=0 (−1)n2−(n+ 1)
(2n+ 1)! x2n+1
Question 14
Question
Find the Maclaurin series for the function f(x) = 1
2−xand determine the interval
of convergence.
13
Solution
To find the Maclaurin series for f(x) = 1
2−x, we will first find the derivatives of
f(x) at x= 0 to obtain the coefficients of the series.
Step 1: Find the first few derivatives of f(x)
f(x) = 1
2−x
f′(x) = d
dx 1
2−x=1
(2 −x)2
f′′(x) = d
dx 1
(2 −x)2=2
(2 −x)3
f′′′(x) = d
dx 2
(2 −x)3=2·3
(2 −x)4=6
(2 −x)4
.
.
.
Step 2: Find the coefficients of the Maclaurin series The Maclaurin
series for a function f(x) is given by f(x) = P∞
n=0
f(n)(0)
n!xn. Therefore, the
Maclaurin series for f(x) = 1
2−xis:
f(x) =
∞
X
n=0
f(n)(0)
n!xn= 1 + 2x+ 4x2+ 8x3+. . .
Step 3: Determine the interval of convergence The interval of conver-
gence for a Maclaurin series can be determined using the ratio test. The ratio
test states that if
L= lim
n→∞
an+1
an
then the series converges for |x|< L.
Applying the ratio test to the Maclaurin series of f(x), we have:
L= lim
n→∞
2n+1
(n+ 1)! ·n!
2n·xn+1
xn
= lim
n→∞
2
n+ 1|x|= 0
Since L= 0, the interval of convergence is (−∞,∞), which means the Maclaurin
series for f(x) converges for all real numbers x.
Therefore, the Maclaurin series for f(x) = 1
2−xis 1 + 2x+ 4x2+ 8x3+. . .
with an interval of convergence of (−∞,∞).
Question 15
Question
Find the Maclaurin series for f(x) = sinx2.
14
Solution
To find the Maclaurin series for f(x) = sinx2, we will first find the Maclaurin
series for sin(x) and then substitute x2in place of x.
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: Substitute x2in place of x.Substitute x2in place of xin the
Maclaurin series for sin(x):
sinx2= (x2)−(x2)3
3! +(x2)5
5! −(x2)7
7! +· · · =
∞
X
n=0
(−1)n(x2)2n+1
(2n+ 1)!
Step 3: Simplify the series. Simplify the series by expanding (x2)2n+1:
sinx2=x2−x6
3! +x10
5! −x14
7! +· · · =
∞
X
n=0
(−1)nx4n+2
(2n+ 1)!
Therefore, the Maclaurin series for f(x) = sinx2is P∞
n=0(−1)nx4n+2
(2n+1)! .
Question 16
Question
Find the Maclaurin series for the function f(x) = sinx
2.
Solution
To find the Maclaurin series for f(x) = sinx
2, we can start by finding the
derivatives of f(x) at x= 0.
Step 1: Find the first few derivatives of f(x)at x= 0.
f(x) = sinx
2
f′(x) = 1
2cosx
2
f′′(x) = −1
4sinx
2
f′′′(x) = −1
8cosx
2
f(4)(x) = 1
16 sinx
2
15
Step 2: Evaluate the derivatives at x= 0.
f(0) = sin(0) = 0
f′(0) = 1
2cos(0) = 1
2
f′′(0) = −1
4sin(0) = 0
f′′′(0) = −1
8cos(0) = −1
8
f(4)(0) = 1
16 sin(0) = 0
Step 3: Write out the Maclaurin series using the derivatives eval-
uated at x= 0.The Maclaurin series for f(x) = sinx
2is:
f(x) = f(0) + f′(0)x+f′′ (0)x2
2! +f′′′(0)x3
3! +· · ·
Substitute the evaluated derivatives into the series:
f(x) = 0 + 1
2x+ 0 ·x2
2! −1
8·x3
3! + 0 ·x4
4! +· · ·
Simplifying, we get:
sinx
2=1
2x−1
48x3+· · ·
Therefore, the Maclaurin series for f(x) = sinx
2is 1
2x−1
48 x3+· · ·
Question 17
Question
Find the Maclaurin series for f(x) = ln(1 + x).
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):
ln(1 + x) =
∞
X
n=1
(−1)n+1 xn
n.
Step 1: Determine the derivative of f(x).
The derivative of ln(1 + x) is f′(x) = 1
1+x.
Step 2: Find the Maclaurin series of f(x).
16
Since we know that ln(1 + x) is equal to its own Maclaurin series, we have:
ln(1 + x) =
∞
X
n=1
(−1)n+1 xn
n=x−x2
2+x3
3−x4
4+. . .
Therefore, the Maclaurin series for f(x) = ln(1 + x) is
∞
X
n=1
(−1)n+1 xn
n.
Question 18
Question
Determine the Maclaurin series for f(x) = 1
1+xand find the interval of conver-
gence.
Solution
Step 1: The Maclaurin series for f(x) = 1
1+xis given by the geometric series
formula:
f(x) =
∞
X
n=0
(−1)nxn
for −1<x<1.
Step 2: Let’s verify the interval of convergence by using the ratio test. The
ratio test states that if limn→∞
an+1
an
=L, then the series converges absolutely:
lim
n→∞
(−1)n+1xn+1
(−1)nxn
= lim
n→∞
x
1
=|x|<1
Therefore, the interval of convergence is −1<x<1.
Step 3: Thus, the Maclaurin series for f(x) = 1
1+xis:
f(x) =
∞
X
n=0
(−1)nxn,−1<x<1
Question 19
Question
Find the Maclaurin series for the function f(x) = 1
1+x2.
17
Solution
To find the Maclaurin series for the function f(x) = 1
1+x2, we will first find the
derivatives of f(x) at x= 0 in order to obtain 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)(2x)(1 + x2)
(1 + x2)4
=−2(1 + x2)[1 + x2+ 2x2]
(1 + x2)4
=−2(1 + 3x2+x4)
(1 + x2)3
f(3)(x) = −2(1 + x2)3−(−2)(1 + 3x2+x4)(2x)
(1 + x2)6
=−2(1 + x2)3+ 4(1 + 3x2+x4)x
(1 + x2)6
=−2(1 + 3x2+ 3x4+x6)+4x+ 12x3+ 4x5
(1 + x2)6
=−2−6x2−6x4−2x6+ 4x+ 12x3+ 4x5
(1 + x2)6
Step 2: Evaluate the derivatives at x= 0.
f(0) = 1
f′(0) = 0
f′′(0) = −2
f(3)(0) = 0
Step 3: Write the Maclaurin series. Since f(0) = 1, f′(0) = 0, f′′(0) =
−2, and f(3)(0) = 0, the Maclaurin series for f(x) = 1
1+x2is:
f(x)=1−2x2+ 0x4+ 0x6+· · · = 1 −2x2
Question 20
Question
Find the Maclaurin series for the function f(x) = 1
1+x2.
18
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we will first find the derivatives of
f(x) at x= 0 and then use these derivatives to construct the Maclaurin series.
Step 1: Find the derivatives of f(x).
f(x) = 1
1 + x2
f′(x) = −(1 + x2)−2·2x=−2x
(1 + x2)2
f′′(x) = −2(1 + x2)−2·1+4x2(1 + x2)−3=2(3x2−1)
(1 + x2)3
f′′′(x) = 2(3x2−1)(1 + x2)−3·2x+ 6x(1 + x2)−3−3(2x)(1 + x2)−4=−12x
(1 + x2)4
Step 2: Evaluate the derivatives at x= 0.
f(0) = 1
f′(0) = 0
f′′(0) = −2
f′′′(0) = 0
Step 3: Construct the Maclaurin series using the derivatives at
x= 0.The Maclaurin series for a function f(x) is given by:
f(x) = f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+· · ·
Substitute the derivative values at x= 0 into the Maclaurin series:
f(x)=1−2x2+0
3!x3+· · ·
f(x)=1−2x2
Therefore, the Maclaurin series for f(x) = 1
1+x2is 1 −2x2.
Question 21
Question
Find the Maclaurin series for the function f(x) = 1
1+x.
19
Solution
To find the Maclaurin series for f(x) = 1
1+x, we will first find the derivatives of
f(x) and then evaluate them at x= 0 to find the coefficients of the series.
Step 1: Find the derivatives of f(x).
f(x) = 1
1 + x
f′(x) = d
dx 1
1 + x=−1
(1 + x)2
f′′(x) = d
dx −1
(1 + x)2=2
(1 + x)3
f′′′(x) = d
dx 2
(1 + x)3=−6
(1 + x)4
.
.
.
f(n)(x) = (−1)n·n!
(1 + x)n+1
Step 2: Evaluate derivatives at x= 0.
f(0) = 1
f′(0) = −1
f′′(0) = 2
f′′′(0) = −6
.
.
.
f(n)(0) = (−1)n·n!
Step 3: Write the Maclaurin series. The Maclaurin series for f(x) is
given by:
f(x) =
∞
X
n=0
f(n)(0)
n!xn=
∞
X
n=0
(−1)nxn
Therefore, the Maclaurin series for f(x) = 1
1+xis P∞
n=0(−1)nxn.
Question 22
Question
Find the Maclaurin series for f(x) = 1
x2(1+x).
Solution
To find the Maclaurin series for f(x) = 1
x2(1+x), we can start by expressing f(x)
as a sum of simpler functions with known Maclaurin series.
20
Step 1: Factorize the denominator.
f(x) = 1
x2(1 + x)=A
x+B
x2+C
1 + x
where A,B, and Care constants to be determined.
Step 2: Expand and simplify.
1 = A(1 + x) + Bx(1 + x) + Cx2
Step 3: Determine the constants A,Band Cby equating coefficients of
like terms.
A= 1
B+A= 0
C+B= 0
Solving the system of equations gives A= 1, B=−1, C= 1.
Step 4: Substitute back into the original expression to obtain f(x).
f(x) = 1
x−1
x2+1
1 + x
Step 5: Find the Maclaurin series for each term:
1
x= 1 −x+x2−x3+. . . =
∞
X
n=0
(−1)nxn
1
x2= 1 −2x+ 3x2−4x3+. . . =
∞
X
n=0
(−1)n(n+ 1)xn
1
1 + x= 1 −x+x2−x3+. . . =
∞
X
n=0
(−1)nxn
Step 6: Combine the Maclaurin series for each term to get the Maclaurin
series for f(x).
f(x) =
∞
X
n=0
(−1)nxn−
∞
X
n=0
(−1)n(n+ 1)xn+
∞
X
n=0
(−1)nxn
f(x) =
∞
X
n=0
[(−1)n−(−1)n(n+ 1) + (−1)n]xn
f(x) =
∞
X
n=0
(−1)n[1 −(n+ 1) + 1]xn
f(x) =
∞
X
n=0
(−1)n(2 −n)xn
21
Question 23
Question
Determine the Maclaurin series for the function f(x) = 1
1+x2.
Solution
To find the Maclaurin series for the function f(x) = 1
1+x2, we will first find the
Maclaurin series for the function g(x) = 1
1−xand then substitute x2in place of
x.
Step 1: Find Maclaurin series for g(x) = 11−x.The Maclaurin series
for 1
1−xis a geometric series given by:
1
1−x=
∞
X
n=0
xn
Step 2: Substitute x2into the series. Substitute x2in place of xto get
the Maclaurin series for 1
1+x2:
f(x) = 1
1 + x2=
∞
X
n=0
(x2)n=
∞
X
n=0
x2n
Therefore, the Maclaurin series for f(x) = 1
1+x2is P∞
n=0 x2n.
Question 24
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 express f(x)
in terms of a known Maclaurin series. We’ll start by expressing sin2(x) as a
product of trigonometric functions.
Step 1: Express sin2(x) as a product of trigonometric functions. Since
sin2(x) = 1
2−1
2cos(2x), we can rewrite it as 1
2−1
2cos(2x).
Step 2: 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! +· · · =
∞
X
n=0
(−1)n(2x)2n
(2n)!
22
Step 3: Substitute the Maclaurin series for cos(2x) into 1
2−1
2cos(2x). Sub-
stitute the Maclaurin series for cos(2x) into 1
2−1
2cos(2x) to obtain:
sin2(x) = 1
2−1
21−(2x)2
2! +(2x)4
4! − · · ·
Step 4: Simplify the expression. sin2(x) = 1
2−1
2+x2
2! −x4
4! +· · ·
sin2(x) = x2
2! −x4
4! +· · ·
Therefore, the Maclaurin series for f(x) = sin2(x) is P∞
n=1(−1)n−1x2n
(2n)! .
Question 25
Question
Find the Maclaurin series for the function f(x) = sin2(x).
Solution
To find the Maclaurin series for f(x) = sin2(x), we will first express sin2(x) in
terms of trigonometric functions and then find the Maclaurin series for sin2(x).
Step 1: Express f(x)in terms of trigonometric functions We know
that sin2(x) = 1
2(1 −cos(2x)).
Step 2: Find the Maclaurin series for f(x) Finding the Maclaurin series
for f(x) = 1
2(1−cos(2x)) involves finding the Maclaurin series for 1 and cos(2x)
individually and then combining them.
Step 3: Find the Maclaurin series for 1 The Maclaurin series for 1 is
simply 1 + 0x+ 0x2+ 0x3+. . . = 1.
Step 4: Find the Maclaurin series for cos(2x) The Maclaurin series for
cos(2x) can be found using the Maclaurin series for cos(x):
cos(x)=1−x2
2! +x4
4! −x6
6! +. . .
Substitute 2xfor x:
cos(2x) = 1 −(2x)2
2! +(2x)4
4! −(2x)6
6! +. . .
Simplify:
cos(2x) = 1 −2x2+8x4
24 −32x6
720 +. . .
cos(2x)=1−2x2+2x4
3−4x6
45 +. . .
Step 5: Combine the Maclaurin series Since f(x) = 1
2(1 −cos(2x)),
the Maclaurin series for f(x) is:
f(x) = 1
2−x2+x4
3−2x6
45 +. . .
Therefore, the Maclaurin series for f(x) = sin2(x) is 1
2−x2+x4
3−2x6
45 +. . ..
23
Question 26
Question
Find the Maclaurin series for the function f(x) = 1
1+x2.
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we will first find the derivatives of
f(x) and evaluate them at x= 0 to find the coefficients of the Maclaurin series.
Step 1: Find the derivatives of f(x)
f(x) = 1
1 + x2
f′(x) = −2x
(1 + x2)2
f′′(x) = −2(1 + x2)2−8x2(1 + x2)
(1 + x2)4
=−2−6x2−6x4
(1 + x2)3
f′′′(x) = 6x(1 −5x2)
(1 + x2)4
Step 2: Evaluate the derivatives at x= 0
f(0) = 1
f′(0) = 0
f′′(0) = −2
f′′′(0) = 0
Step 3: Write the Maclaurin series Since the derivative f′′(x) evaluated
at x= 0 is −2, and all higher derivatives at 0 are 0, the Maclaurin series for
f(x) is:
f(x)=1−2x2+ 0x3+ 0x4+. . . = 1 −2x2
Therefore, the Maclaurin series for f(x) = 1
1+x2is 1 −2x2.
Question 27
Question
Find the Maclaurin series for the function f(x) = 1
1+x2.
24
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we’ll first find the derivatives of
f(x) and then express them as a series centered at x= 0.
Step 1: Find the derivatives of f(x).
f(x) = 1
1 + x2
f′(x) = −2x
(1 + x2)2
f′′(x) = 2(3x2−1)
(1 + x2)3
f′′′(x) = 2(1 −10x2+ 3x4)
(1 + x2)4
Step 2: Observe a pattern in the derivatives. We can notice a pattern in
the derivatives:
f(x) : 1
f′(x) : −2x
f′′(x) : 2(3x2−1) = 6x2−2
f′′′(x) : 2(1 −10x2+ 3x4) = 2 −20x2+ 6x4
Step 3: Generalize the pattern. From the pattern above, we can make a
conjecture for the n-th derivative of f(x):
f(n)(x)=(−1)n·Pn(x)
(1 + x2)n+1
where Pn(x) is a polynomial of degree n.
Step 4: Find the Maclaurin series. The Maclaurin series for f(x) = 1
1+x2is
given by:
f(x) =
∞
X
n=0
(−1)n·Pn(0)
(n+ 1)!xn
Therefore, the Maclaurin series for f(x) = 1
1+x2is:
∞
X
n=0
(−1)n·Pn(0)
(n+ 1)!xn
Question 28
Question
Find the Maclaurin series for f(x) = 1
2−x.
25
Solution
To find the Maclaurin series for f(x) = 1
2−x, we will start by finding the deriva-
tives of f(x) at x= 0.
Step 1: Find the first derivative
f′(x) = d
dx 1
2−x
Using the quotient rule, we have:
f′(x) = 2
(2 −x)2
Step 2: Find the second derivative
f′′(x) = d
dx 2
(2 −x)2
Applying the chain rule and the power rule, we get:
f′′(x) = 4
(2 −x)3
Step 3: Find the third derivative
f′′′(x) = d
dx 4
(2 −x)3
Following the same steps as above, we obtain:
f′′′(x) = 12
(2 −x)4
Step 4: General pattern By taking derivatives, we can observe a general
pattern emerging:
f(n)(x) = n!·2
(2 −x)n+1
Step 5: Express f(x)as a Maclaurin series To express f(x) as a Maclau-
rin series, we use the formula:
f(x) = f(0) + f′(0)
1! x+f′′(0)
2! x2+f′′′(0)
3! x3+· · ·
Substitute the derivatives we found earlier:
f(x) = 1 + x+ 2x2+ 6x3+· · ·
Therefore, the Maclaurin series for f(x) = 1
2−xis 1 + x+ 2x2+ 6x3+· · · .
26
Question 29
Question
Find the Maclaurin series representation for f(x) = 1
x+1 .
Solution
To find the Maclaurin series representation for f(x) = 1
x+1 , we will begin by
finding the derivatives of f(x) at x= 0 in order to construct the Maclaurin
series.
Step 1: Find f′(x)
f(x) = 1
x+ 1
f′(x) = d
dx 1
x+ 1
=−1
(x+ 1)2
Step 2: Find f′′ (x)
f′(x) = −1
(x+ 1)2
f′′(x) = d
dx −1
(x+ 1)2
=2
(x+ 1)3
Step 3: Find f′′′ (x)
f′′(x) = 2
(x+ 1)3
f′′′(x) = d
dx 2
(x+ 1)3
=−6
(x+ 1)4
Step 4: Generalize the pattern After finding the first few derivatives of
f(x), we notice a pattern where the nth derivative is of the form:
f(n)(x)=(−1)n·n!·(x+ 1)−(n+1)
Step 5: Write the Maclaurin series The Maclaurin series for f(x) is
given by:
f(x) = f(0) + f′(0)
1! x+f′′(0)
2! x2+f′′′(0)
3! x3+· · ·
Substitute f(0), f′(0), f′′(0), etc. into the above formula to get the Maclau-
rin series representation for f(x) = 1
x+1 :
27
f(x)=1−x+x2−x3+x4− · · · =
∞
X
n=0
(−1)n·xn
Question 30
Question
Find the Maclaurin series for the function f(x) = ln(1 + x) and determine its
interval of convergence.
Solution
To find the Maclaurin series for f(x) = ln(1 + x), we can start by finding the
derivative of f(x) and evaluating it at x= 0 to determine the coefficients of the
series.
Step 1: Find higher-order 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: Evaluate 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
.
.
.
Step 3: Write the Maclaurin series The Maclaurin series for f(x) =
ln(1 + x) is:
f(x) =
∞
X
n=0
f(n)(0)
n!xn=x−x2
2+x3
3−x4
4+· · · =
∞
X
n=1
(−1)n−1xn
n
28
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
.
.
.
Step 3: Write out the Maclaurin series. Since 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! +· · · ,
the Maclaurin series for f(x) = 1
1+xis
1−x+ 2x2−6x3+· · · .
Question 2
Question
Find the Maclaurin series for f(x) = e2xcos(x).
Solution
To find the Maclaurin series for f(x) = e2xcos(x), we will need to find the
derivatives of f(x) at x= 0 and use these derivatives to form the Maclaurin
series.
Step 1: Find f(0) and f′(0).
f(x) = e2xcos(x)
f(0) = e2·0cos(0) = 1
To find f′(x), we will use the product rule:
f′(x)=(e2x·cos(x))′=e2x(−sin(x)) + cos(x)(2e2x)=2e2xcos(x)−e2xsin(x)
f′(0) = 2e2·0cos(0) −e2·0sin(0) = 2 ·1·1−1·0=2
Step 2: Find f′′ (0) and f′′′ (0).
f′′(x) = (2e2xcos(x)−e2xsin(x))′= 2e2x(−sin(x))−e2xcos(x)+2e2xcos(x)−e2xsin(x)
2
f′′(0) = 2 ·1·0−1·1+2·1·1−1·0=3
To find f′′′ (x), we differentiate f′′(x):
f′′′(x) = (2e2x(−sin(x)) −e2xcos(x)+2e2xcos(x)−e2xsin(x))′
f′′′(x) = 2e2x(−cos(x))−e2x(−sin(x))+2e2x(−sin(x))−e2xcos(x)+2e2xsin(x)−e2xcos(x)
f′′′(0) = 2 ·1·(−1) −1·0+2·1·0−1·1+2·1·0−1·1 = −5
Step 3: Write the Maclaurin series for f(x).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 known values of f(0), f′(0), f′′(0), and f′′′(0):
f(x) = 1 + 2x+3
2x2−5
6x3+. . .
Therefore, the Maclaurin series for f(x) = e2xcos(x) is 1 + 2x+3
2x2−5
6x3+. . ..
Question 3
Question
Find the Taylor series representation for f(x) = excentered at a= 0.
Solution
To find the Taylor series representation of f(x) = excentered at a= 0, we will
use the formula for a Taylor series:
f(x) = f(a) + f′(a)
1! (x−a) + f′′ (a)
2! (x−a)2+f′′′(a)
3! (x−a)3+· · ·
Step 1: Find f(n)(x)for all n.
f(x) = ex
f′(x) = ex
f′′(x) = ex
.
.
.
f(n)(x) = exfor all n
3
Step 2: Evaluate f(n)(a)for all n.
f(0) = e0= 1
f′(0) = e0= 1
f′′(0) = e0= 1
.
.
.
f(n)(0) = e0= 1 for all n
Step 3: Plug in the values into the formula for the Taylor series.
Since f(n)(0) = 1 for all n, the Taylor series becomes:
f(x) = 1 + 1
1!x+1
2!x2+1
3!x3+· · · =
∞
X
n=0
xn
n!
Therefore, the Taylor series representation for f(x) = excentered at a= 0
is P∞
n=0
xn
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 can start by expressing
sin2(x) in terms of well-known trigonometric functions. Recall the trigonometric
identity: sin2(x) = 1
2−1
2cos(2x).
Step 1: Apply the trigonometric identity.
f(x) = sin2(x) = 1
2−1
2cos(2x)
Step 2: Find the Maclaurin series for 1
2. The Maclaurin series for 1
2is
simply 1
2.
Step 3: Find the Maclaurin series for −1
2cos(2x). To find the Maclaurin
series for −1
2cos(2x), we can start with the Maclaurin series for cos(x): cos(x) =
1−x2
2! +x4
4! −x6
6! +. . .. Now, substitute xwith 2xin the series to get the Maclaurin
series for cos(2x).
cos(2x) = 1 −(2x)2
2! +(2x)4
4! −(2x)6
6! +. . .
Simplify this expression.
cos(2x)=1−2x2+4x4
3! −8x6
6! +. . .
4
Now, multiply by −1
2to get the Maclaurin series for −1
2cos(2x).
−1
2cos(2x) = −1
2+x2−2x4
3! +4x6
6! −. . .
Step 4: Add the Maclaurin series for 1
2and −1
2cos(2x) to get the Maclaurin
series for f(x) = sin2(x).
f(x) = 1
2−1
2cos(2x) = 1
2+−1
2+x2−2x4
3! +4x6
6! −. . .
Now, combine like terms.
f(x) = x2−2x4
3! +4x6
6! −. . .
Therefore, the Maclaurin series for f(x) = sin2(x) is x2−2x4
3! +4x6
6! −. . ..
Question 5
Question
Find the Maclaurin series representation for f(x) = 1
1+x.
Solution
To find the Maclaurin series for f(x), we will use the fact that 1
1+x= 1 −x+
x2−x3+x4−. . . for |x|<1.
Step 1: Find the formula for the Maclaurin series of f(x).
Recall that the Maclaurin series for f(x) is given by:
f(x) = f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+. . .
Step 2: Find the derivatives of f(x) at x= 0.
We have f(x) = 1
1+x, so:
f(0) = 1
1+0 = 1
f′(x) = d
dx 1
1 + x=−1
(1 + x)2
f′(0) = −1
(1 + 0)2=−1
f′′(x) = d
dx −1
(1 + x)2=2
(1 + x)3
5
f′′(0) = 2
(1 + 0)3= 2
The higher-order derivatives will follow a similar pattern.
Step 3: Plug f(0), f′(0), f′′ (0) into the formula for the Maclaurin series.
The Maclaurin series representation for f(x) is:
f(x)=1−x+x2−x3+x4−. . .
Question 6
Question
Find the Maclaurin series representation for the function f(x) = 1
(1+x)2.
Solution
To find the Maclaurin series representation for f(x) = 1
(1+x)2, we will first find
the derivatives of f(x) and then evaluate them at x= 0 to get the coefficients
of the Maclaurin series.
Step 1: Find the first few derivatives of f(x).
f(x) = 1
(1 + x)2
f′(x) = d
dx 1
(1 + x)2=−2(1 + x)−3
f′′(x) = d
dx −2(1 + x)−3=−6(1 + x)−4
f′′′(x) = d
dx −6(1 + x)−4= 24(1 + x)−5
Step 2: Evaluate the derivatives at x= 0.
f(0) = 1
f′(0) = −2
f′′(0) = −6
f′′′(0) = 24
Step 3: Write out the Maclaurin series using the coefficients found
above. Since the Maclaurin series for a function can be expressed as:
f(x) =
∞
X
n=0
f(n)(0)
n!xn
The Maclaurin series for f(x) is:
f(x)=1−2x−6x2+ 24x3+. . .
6
Question 7
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 its derivatives and
then evaluate them at x= 0.
Step 1: Find the first few derivatives of f(x).
f(x) = (1 −x)−1
f′(x)=(−1)(1 −x)−2
f′′(x)=(−1)(−2)(1 −x)−3
f′′′(x)=(−1)(−2)(−3)(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 down the general term for the Maclaurin series. The
general term for the Maclaurin series is given by:
f(n)(0) ·xn
n!
Step 4: Write down the Maclaurin series for f(x).Putting it all
together, the Maclaurin series for f(x) = 1
1−xis:
1−x+x2−x3+x4−. . . =
∞
X
n=0
(−1)nxn
Question 8
Question
Find the Maclaurin series representation for the function f(x) = ln(1 + x).
7
Solution
To find the Maclaurin series representation for f(x) = ln(1 + x), we will first
need to find its derivatives at x= 0 to obtain the coefficients of the series.
Step 1: Find the first few derivatives of f(x).
f(x) = ln(1 + x)
f′(x) = 1
1 + x,
f′′(x) = d
dx 1
1 + x=−1
(1 + x)2,
f′′′(x) = d
dx −1
(1 + x)2=2
(1 + x)3,
f(4)(x) = d
dx 2
(1 + x)3=−6
(1 + x)4.
Step 2: Find f(n)(0) for all n.
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 the Maclaurin series. The Maclaurin series representa-
tion of f(x) = ln(1 + x) is given by:
f(x) =
∞
X
n=0
f(n)(0)
n!xn=x−x2
2+x3
3−x4
4+· · ·
Question 9
Question
Find the Maclaurin series for the function f(x) = sin(2x).
Solution
To find the Maclaurin series for f(x) = sin(2x), we will first express sin(2x) in
terms of its known Taylor series expansion and then determine its Maclaurin
series.
8
Step 1: Derive the Maclaurin series for sin(2x)using 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)n
(2n+ 1)!x2n+1.
By substituting 2xin place of x, we have:
sin(2x)=2x−(2x)3
3! +(2x)5
5! −(2x)7
7! +· · · .
Step 2: Simplify the above series to obtain the Maclaurin series
for sin(2x).Simplifying the expression above, we get:
sin(2x)=2x−8x3
6+32x5
120 −128x7
5040 +· · ·
= 2x−4x3
3+2x5
15 −4x7
315 +· · ·
=
∞
X
n=0
(−1)n22n+1x2n+1
(2n+ 1)! .
Therefore, the Maclaurin series for f(x) = sin(2x) is:
f(x)=2x−4x3
3! +2x5
5! −4x7
7! +· · · =
∞
X
n=0
(−1)n22n+1x2n+1
(2n+ 1)! .
Question 10
Question
Find the Maclaurin series representation of the function f(x) = 1
1−x+x2.
Solution
To find the Maclaurin series of f(x), we will first express f(x) as a geometric
series.
Step 1: Find the geometric series form of f(x).Consider the geometric
series formula: 1
1−r= 1 + r+r2+r3+. . .. We can express f(x) in terms of a
geometric series:
f(x) = 1
1−x+x2=1
(1 −x)2−(x2)=1
(1 −x)2(1 + x).
Splitting the expression, we can rewrite f(x) as:
f(x) = A
1−x+B
(1 −x)2+C
1 + x
9
for some constants A,B, and C.
Multiplying through by the denominators and solving for A,B, and C, we
find:
A= 1, B = 2, C = 1.
Therefore, we have:
f(x) = 1
1−x+2
(1 −x)2+1
1 + x= 1 + x+x2+ 2x3+ 3x4+. . . .
Step 2: Write the Maclaurin series of f(x).Since f(x) has the repre-
sentation 1 + x+x2+ 2x3+ 3x4+. . ., the Maclaurin series for f(x) is:
f(x) = 1 + x+x2+ 2x3+ 3x4+. . . .
Therefore, the Maclaurin series representation of f(x) = 1
1−x+x2is 1 + x+
x2+ 2x3+ 3x4+. . ..
Question 11
Question
Find the Maclaurin series for the function f(x) = 1
1+x+x2.
Solution
To find the Maclaurin series for f(x) = 1
1+x+x2, we will first express f(x) as a
geometric series and then find the Maclaurin series representation.
Step 1: Find the derivative of f(x)
f(x) = 1
1 + x+x2= (1 + x+x2)−1
f′(x) = −1(1 + x+x2)−2(1 + 2x) = −1+2x
(1 + x+x2)2
Step 2: Find the second derivative of f(x)
f′(x) = −1+2x
(1 + x+x2)2
f′′(x) = d
dx −1+2x
(1 + x+x2)2
=2−2(1 + 2x)(1 + 2x)
(1 + x+x2)3=−2−4x
(1 + x+x2)3
Step 3: Find the third derivative of f(x)
f′′(x) = −2−4x
(1 + x+x2)3
10
f′′′(x) = d
dx −2−4x
(1 + x+x2)3
=−4−4(1 + 2x)(−4−4x)
(1 + x+x2)4=16 + 24x+ 8x2
(1 + x+x2)4
Step 4: Find the fourth derivative of f(x)
f′′′(x) = 16 + 24x+ 8x2
(1 + x+x2)4
f(4)(x) = d
dx 16 + 24x+ 8x2
(1 + x+x2)4
=48 + 56x+ 16x2
(1 + x+x2)5
Step 5: Generalize the pattern After computing several derivatives, we
see a pattern emerging where the n-th derivative of f(x) is given by:
f(n)(x) = Pn(x)
(1 + x+x2)n+1
where Pn(x) is a polynomial of degree n.
Step 6: Write 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+· · ·
Step 7: Evaluate the derivatives at x= 0 Evaluating the derivatives at
x= 0, we find:
f(0) = 1
1+0+02= 1
f′(0) = −1
f′′(0) = −1
1
f′′′(0) = 1
1
f(4)(0) = 0
1
Step 8: Substitute into the Maclaurin series formula Therefore, the
Maclaurin series for f(x) = 1
1+x+x2is:
1−x−x2+x3+· · ·
11
Question 12
Question
Find the Maclaurin series for the function f(x) = 1
1+x2.
Solution
To find the Maclaurin series for the function f(x) = 1
1+x2, we will first find the
derivatives of f(x) and evaluate them at x= 0 to find the coefficients of the
series.
Step 1: Find the derivatives of f(x)
f(x) = 1
1 + x2
f′(x) = −2x
(1 + x2)2
f′′(x) = −2(1 + x2)2−2x(2)(1 + x2)(2x)
(1 + x2)4
=−2−2x2−8x2
(1 + x2)3=−2−10x2
(1 + x2)3
The pattern for the derivatives of f(x) becomes apparent:
f′(x) = −2x
(1 + x2)2
f′′(x) = −2(1 + x2)−2x(2)(x)
(1 + x2)3=−2−6x2
(1 + x2)3
f′′′(x) = −2(1 + x2)−6x2(2x)
(1 + x2)4=−2−14x2
(1 + x2)4
.
.
.
Step 2: Evaluate the derivatives at x= 0 At x= 0:
f(0) = 1
f′(0) = 0
f′′(0) = −2
f′′′(0) = 0
f′′′′(0) = −2
.
.
.
Step 3: Write the Maclaurin series for f(x) Since f(0) = 1, f′(0) = 0,
f′′(0) = −2, f′′′ (0) = 0, f′′′′ (0) = −2, and so on, the Maclaurin series for f(x)
is:
f(x)=1−2x2+ 2x4−2x6+. . . =
∞
X
n=0
(−1)nx2n
12
Question 13
Question
Find the Maclaurin series for f(x) = sin(x) cos2(x).
Solution
To find the Maclaurin series for f(x) = sin(x) cos2(x), we will first expand sin(x)
and cos(x) into their Maclaurin series and then multiply them together.
Step 1: Find the Maclaurin series for sin(x).
sin(x) = x−x3
3! +x5
5! −x7
7! +. . . =
∞
X
n=0
(−1)nx2n+1
(2n+ 1)!
Step 2: Find the Maclaurin series for cos(x).
cos(x)=1−x2
2! +x4
4! −x6
6! +. . . =
∞
X
n=0
(−1)nx2n
(2n)!
Step 3: Substitute the Maclaurin series for sin(x) and cos(x) into f(x).
f(x) = (
∞
X
n=0
(−1)nx2n+1
(2n+ 1)!)(
∞
X
n=0
(−1)nx2n
(2n)!)(
∞
X
n=0
(−1)nx2n
(2n)!)
Step 4: Simplify the expression by expanding the product.
f(x) = sin(x) cos2(x)=(x−x3
3! +x5
5! −. . .)(1−x2
2! +x4
4! −. . .)(1−x2
2! +x4
4! −. . .)
Step 5: Find the Maclaurin series for f(x) by multiplying out the series
and simplifying.
f(x) = x−x3
2−x5
10 +x7
14 +. . . =
∞
X
n=0 (−1)n2−(n+ 1)
(2n+ 1)! x2n+1
Question 14
Question
Find the Maclaurin series for the function f(x) = 1
2−xand determine the interval
of convergence.
13
Solution
To find the Maclaurin series for f(x) = 1
2−x, we will first find the derivatives of
f(x) at x= 0 to obtain the coefficients of the series.
Step 1: Find the first few derivatives of f(x)
f(x) = 1
2−x
f′(x) = d
dx 1
2−x=1
(2 −x)2
f′′(x) = d
dx 1
(2 −x)2=2
(2 −x)3
f′′′(x) = d
dx 2
(2 −x)3=2·3
(2 −x)4=6
(2 −x)4
.
.
.
Step 2: Find the coefficients of the Maclaurin series The Maclaurin
series for a function f(x) is given by f(x) = P∞
n=0
f(n)(0)
n!xn. Therefore, the
Maclaurin series for f(x) = 1
2−xis:
f(x) =
∞
X
n=0
f(n)(0)
n!xn= 1 + 2x+ 4x2+ 8x3+. . .
Step 3: Determine the interval of convergence The interval of conver-
gence for a Maclaurin series can be determined using the ratio test. The ratio
test states that if
L= lim
n→∞
an+1
an
then the series converges for |x|< L.
Applying the ratio test to the Maclaurin series of f(x), we have:
L= lim
n→∞
2n+1
(n+ 1)! ·n!
2n·xn+1
xn
= lim
n→∞
2
n+ 1|x|= 0
Since L= 0, the interval of convergence is (−∞,∞), which means the Maclaurin
series for f(x) converges for all real numbers x.
Therefore, the Maclaurin series for f(x) = 1
2−xis 1 + 2x+ 4x2+ 8x3+. . .
with an interval of convergence of (−∞,∞).
Question 15
Question
Find the Maclaurin series for f(x) = sinx2.
14
Solution
To find the Maclaurin series for f(x) = sinx2, we will first find the Maclaurin
series for sin(x) and then substitute x2in place of x.
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: Substitute x2in place of x.Substitute x2in place of xin the
Maclaurin series for sin(x):
sinx2= (x2)−(x2)3
3! +(x2)5
5! −(x2)7
7! +· · · =
∞
X
n=0
(−1)n(x2)2n+1
(2n+ 1)!
Step 3: Simplify the series. Simplify the series by expanding (x2)2n+1:
sinx2=x2−x6
3! +x10
5! −x14
7! +· · · =
∞
X
n=0
(−1)nx4n+2
(2n+ 1)!
Therefore, the Maclaurin series for f(x) = sinx2is P∞
n=0(−1)nx4n+2
(2n+1)! .
Question 16
Question
Find the Maclaurin series for the function f(x) = sinx
2.
Solution
To find the Maclaurin series for f(x) = sinx
2, we can start by finding the
derivatives of f(x) at x= 0.
Step 1: Find the first few derivatives of f(x)at x= 0.
f(x) = sinx
2
f′(x) = 1
2cosx
2
f′′(x) = −1
4sinx
2
f′′′(x) = −1
8cosx
2
f(4)(x) = 1
16 sinx
2
15
Step 2: Evaluate the derivatives at x= 0.
f(0) = sin(0) = 0
f′(0) = 1
2cos(0) = 1
2
f′′(0) = −1
4sin(0) = 0
f′′′(0) = −1
8cos(0) = −1
8
f(4)(0) = 1
16 sin(0) = 0
Step 3: Write out the Maclaurin series using the derivatives eval-
uated at x= 0.The Maclaurin series for f(x) = sinx
2is:
f(x) = f(0) + f′(0)x+f′′ (0)x2
2! +f′′′(0)x3
3! +· · ·
Substitute the evaluated derivatives into the series:
f(x) = 0 + 1
2x+ 0 ·x2
2! −1
8·x3
3! + 0 ·x4
4! +· · ·
Simplifying, we get:
sinx
2=1
2x−1
48x3+· · ·
Therefore, the Maclaurin series for f(x) = sinx
2is 1
2x−1
48 x3+· · ·
Question 17
Question
Find the Maclaurin series for f(x) = ln(1 + x).
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):
ln(1 + x) =
∞
X
n=1
(−1)n+1 xn
n.
Step 1: Determine the derivative of f(x).
The derivative of ln(1 + x) is f′(x) = 1
1+x.
Step 2: Find the Maclaurin series of f(x).
16
Since we know that ln(1 + x) is equal to its own Maclaurin series, we have:
ln(1 + x) =
∞
X
n=1
(−1)n+1 xn
n=x−x2
2+x3
3−x4
4+. . .
Therefore, the Maclaurin series for f(x) = ln(1 + x) is
∞
X
n=1
(−1)n+1 xn
n.
Question 18
Question
Determine the Maclaurin series for f(x) = 1
1+xand find the interval of conver-
gence.
Solution
Step 1: The Maclaurin series for f(x) = 1
1+xis given by the geometric series
formula:
f(x) =
∞
X
n=0
(−1)nxn
for −1<x<1.
Step 2: Let’s verify the interval of convergence by using the ratio test. The
ratio test states that if limn→∞
an+1
an
=L, then the series converges absolutely:
lim
n→∞
(−1)n+1xn+1
(−1)nxn
= lim
n→∞
x
1
=|x|<1
Therefore, the interval of convergence is −1<x<1.
Step 3: Thus, the Maclaurin series for f(x) = 1
1+xis:
f(x) =
∞
X
n=0
(−1)nxn,−1<x<1
Question 19
Question
Find the Maclaurin series for the function f(x) = 1
1+x2.
17
Solution
To find the Maclaurin series for the function f(x) = 1
1+x2, we will first find the
derivatives of f(x) at x= 0 in order to obtain 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)(2x)(1 + x2)
(1 + x2)4
=−2(1 + x2)[1 + x2+ 2x2]
(1 + x2)4
=−2(1 + 3x2+x4)
(1 + x2)3
f(3)(x) = −2(1 + x2)3−(−2)(1 + 3x2+x4)(2x)
(1 + x2)6
=−2(1 + x2)3+ 4(1 + 3x2+x4)x
(1 + x2)6
=−2(1 + 3x2+ 3x4+x6)+4x+ 12x3+ 4x5
(1 + x2)6
=−2−6x2−6x4−2x6+ 4x+ 12x3+ 4x5
(1 + x2)6
Step 2: Evaluate the derivatives at x= 0.
f(0) = 1
f′(0) = 0
f′′(0) = −2
f(3)(0) = 0
Step 3: Write the Maclaurin series. Since f(0) = 1, f′(0) = 0, f′′(0) =
−2, and f(3)(0) = 0, the Maclaurin series for f(x) = 1
1+x2is:
f(x)=1−2x2+ 0x4+ 0x6+· · · = 1 −2x2
Question 20
Question
Find the Maclaurin series for the function f(x) = 1
1+x2.
18
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we will first find the derivatives of
f(x) at x= 0 and then use these derivatives to construct the Maclaurin series.
Step 1: Find the derivatives of f(x).
f(x) = 1
1 + x2
f′(x) = −(1 + x2)−2·2x=−2x
(1 + x2)2
f′′(x) = −2(1 + x2)−2·1+4x2(1 + x2)−3=2(3x2−1)
(1 + x2)3
f′′′(x) = 2(3x2−1)(1 + x2)−3·2x+ 6x(1 + x2)−3−3(2x)(1 + x2)−4=−12x
(1 + x2)4
Step 2: Evaluate the derivatives at x= 0.
f(0) = 1
f′(0) = 0
f′′(0) = −2
f′′′(0) = 0
Step 3: Construct the Maclaurin series using the derivatives at
x= 0.The Maclaurin series for a function f(x) is given by:
f(x) = f(0) + f′(0)x+f′′ (0)
2! x2+f′′′(0)
3! x3+· · ·
Substitute the derivative values at x= 0 into the Maclaurin series:
f(x)=1−2x2+0
3!x3+· · ·
f(x)=1−2x2
Therefore, the Maclaurin series for f(x) = 1
1+x2is 1 −2x2.
Question 21
Question
Find the Maclaurin series for the function f(x) = 1
1+x.
19
Solution
To find the Maclaurin series for f(x) = 1
1+x, we will first find the derivatives of
f(x) and then evaluate them at x= 0 to find the coefficients of the series.
Step 1: Find the derivatives of f(x).
f(x) = 1
1 + x
f′(x) = d
dx 1
1 + x=−1
(1 + x)2
f′′(x) = d
dx −1
(1 + x)2=2
(1 + x)3
f′′′(x) = d
dx 2
(1 + x)3=−6
(1 + x)4
.
.
.
f(n)(x) = (−1)n·n!
(1 + x)n+1
Step 2: Evaluate derivatives at x= 0.
f(0) = 1
f′(0) = −1
f′′(0) = 2
f′′′(0) = −6
.
.
.
f(n)(0) = (−1)n·n!
Step 3: Write the Maclaurin series. The Maclaurin series for f(x) is
given by:
f(x) =
∞
X
n=0
f(n)(0)
n!xn=
∞
X
n=0
(−1)nxn
Therefore, the Maclaurin series for f(x) = 1
1+xis P∞
n=0(−1)nxn.
Question 22
Question
Find the Maclaurin series for f(x) = 1
x2(1+x).
Solution
To find the Maclaurin series for f(x) = 1
x2(1+x), we can start by expressing f(x)
as a sum of simpler functions with known Maclaurin series.
20
Step 1: Factorize the denominator.
f(x) = 1
x2(1 + x)=A
x+B
x2+C
1 + x
where A,B, and Care constants to be determined.
Step 2: Expand and simplify.
1 = A(1 + x) + Bx(1 + x) + Cx2
Step 3: Determine the constants A,Band Cby equating coefficients of
like terms.
A= 1
B+A= 0
C+B= 0
Solving the system of equations gives A= 1, B=−1, C= 1.
Step 4: Substitute back into the original expression to obtain f(x).
f(x) = 1
x−1
x2+1
1 + x
Step 5: Find the Maclaurin series for each term:
1
x= 1 −x+x2−x3+. . . =
∞
X
n=0
(−1)nxn
1
x2= 1 −2x+ 3x2−4x3+. . . =
∞
X
n=0
(−1)n(n+ 1)xn
1
1 + x= 1 −x+x2−x3+. . . =
∞
X
n=0
(−1)nxn
Step 6: Combine the Maclaurin series for each term to get the Maclaurin
series for f(x).
f(x) =
∞
X
n=0
(−1)nxn−
∞
X
n=0
(−1)n(n+ 1)xn+
∞
X
n=0
(−1)nxn
f(x) =
∞
X
n=0
[(−1)n−(−1)n(n+ 1) + (−1)n]xn
f(x) =
∞
X
n=0
(−1)n[1 −(n+ 1) + 1]xn
f(x) =
∞
X
n=0
(−1)n(2 −n)xn
21
Question 23
Question
Determine the Maclaurin series for the function f(x) = 1
1+x2.
Solution
To find the Maclaurin series for the function f(x) = 1
1+x2, we will first find the
Maclaurin series for the function g(x) = 1
1−xand then substitute x2in place of
x.
Step 1: Find Maclaurin series for g(x) = 11−x.The Maclaurin series
for 1
1−xis a geometric series given by:
1
1−x=
∞
X
n=0
xn
Step 2: Substitute x2into the series. Substitute x2in place of xto get
the Maclaurin series for 1
1+x2:
f(x) = 1
1 + x2=
∞
X
n=0
(x2)n=
∞
X
n=0
x2n
Therefore, the Maclaurin series for f(x) = 1
1+x2is P∞
n=0 x2n.
Question 24
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 express f(x)
in terms of a known Maclaurin series. We’ll start by expressing sin2(x) as a
product of trigonometric functions.
Step 1: Express sin2(x) as a product of trigonometric functions. Since
sin2(x) = 1
2−1
2cos(2x), we can rewrite it as 1
2−1
2cos(2x).
Step 2: 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! +· · · =
∞
X
n=0
(−1)n(2x)2n
(2n)!
22
Step 3: Substitute the Maclaurin series for cos(2x) into 1
2−1
2cos(2x). Sub-
stitute the Maclaurin series for cos(2x) into 1
2−1
2cos(2x) to obtain:
sin2(x) = 1
2−1
21−(2x)2
2! +(2x)4
4! − · · ·
Step 4: Simplify the expression. sin2(x) = 1
2−1
2+x2
2! −x4
4! +· · ·
sin2(x) = x2
2! −x4
4! +· · ·
Therefore, the Maclaurin series for f(x) = sin2(x) is P∞
n=1(−1)n−1x2n
(2n)! .
Question 25
Question
Find the Maclaurin series for the function f(x) = sin2(x).
Solution
To find the Maclaurin series for f(x) = sin2(x), we will first express sin2(x) in
terms of trigonometric functions and then find the Maclaurin series for sin2(x).
Step 1: Express f(x)in terms of trigonometric functions We know
that sin2(x) = 1
2(1 −cos(2x)).
Step 2: Find the Maclaurin series for f(x) Finding the Maclaurin series
for f(x) = 1
2(1−cos(2x)) involves finding the Maclaurin series for 1 and cos(2x)
individually and then combining them.
Step 3: Find the Maclaurin series for 1 The Maclaurin series for 1 is
simply 1 + 0x+ 0x2+ 0x3+. . . = 1.
Step 4: Find the Maclaurin series for cos(2x) The Maclaurin series for
cos(2x) can be found using the Maclaurin series for cos(x):
cos(x)=1−x2
2! +x4
4! −x6
6! +. . .
Substitute 2xfor x:
cos(2x) = 1 −(2x)2
2! +(2x)4
4! −(2x)6
6! +. . .
Simplify:
cos(2x) = 1 −2x2+8x4
24 −32x6
720 +. . .
cos(2x)=1−2x2+2x4
3−4x6
45 +. . .
Step 5: Combine the Maclaurin series Since f(x) = 1
2(1 −cos(2x)),
the Maclaurin series for f(x) is:
f(x) = 1
2−x2+x4
3−2x6
45 +. . .
Therefore, the Maclaurin series for f(x) = sin2(x) is 1
2−x2+x4
3−2x6
45 +. . ..
23
Question 26
Question
Find the Maclaurin series for the function f(x) = 1
1+x2.
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we will first find the derivatives of
f(x) and evaluate them at x= 0 to find the coefficients of the Maclaurin series.
Step 1: Find the derivatives of f(x)
f(x) = 1
1 + x2
f′(x) = −2x
(1 + x2)2
f′′(x) = −2(1 + x2)2−8x2(1 + x2)
(1 + x2)4
=−2−6x2−6x4
(1 + x2)3
f′′′(x) = 6x(1 −5x2)
(1 + x2)4
Step 2: Evaluate the derivatives at x= 0
f(0) = 1
f′(0) = 0
f′′(0) = −2
f′′′(0) = 0
Step 3: Write the Maclaurin series Since the derivative f′′(x) evaluated
at x= 0 is −2, and all higher derivatives at 0 are 0, the Maclaurin series for
f(x) is:
f(x)=1−2x2+ 0x3+ 0x4+. . . = 1 −2x2
Therefore, the Maclaurin series for f(x) = 1
1+x2is 1 −2x2.
Question 27
Question
Find the Maclaurin series for the function f(x) = 1
1+x2.
24
Solution
To find the Maclaurin series for f(x) = 1
1+x2, we’ll first find the derivatives of
f(x) and then express them as a series centered at x= 0.
Step 1: Find the derivatives of f(x).
f(x) = 1
1 + x2
f′(x) = −2x
(1 + x2)2
f′′(x) = 2(3x2−1)
(1 + x2)3
f′′′(x) = 2(1 −10x2+ 3x4)
(1 + x2)4
Step 2: Observe a pattern in the derivatives. We can notice a pattern in
the derivatives:
f(x) : 1
f′(x) : −2x
f′′(x) : 2(3x2−1) = 6x2−2
f′′′(x) : 2(1 −10x2+ 3x4) = 2 −20x2+ 6x4
Step 3: Generalize the pattern. From the pattern above, we can make a
conjecture for the n-th derivative of f(x):
f(n)(x)=(−1)n·Pn(x)
(1 + x2)n+1
where Pn(x) is a polynomial of degree n.
Step 4: Find the Maclaurin series. The Maclaurin series for f(x) = 1
1+x2is
given by:
f(x) =
∞
X
n=0
(−1)n·Pn(0)
(n+ 1)!xn
Therefore, the Maclaurin series for f(x) = 1
1+x2is:
∞
X
n=0
(−1)n·Pn(0)
(n+ 1)!xn
Question 28
Question
Find the Maclaurin series for f(x) = 1
2−x.
25
Solution
To find the Maclaurin series for f(x) = 1
2−x, we will start by finding the deriva-
tives of f(x) at x= 0.
Step 1: Find the first derivative
f′(x) = d
dx 1
2−x
Using the quotient rule, we have:
f′(x) = 2
(2 −x)2
Step 2: Find the second derivative
f′′(x) = d
dx 2
(2 −x)2
Applying the chain rule and the power rule, we get:
f′′(x) = 4
(2 −x)3
Step 3: Find the third derivative
f′′′(x) = d
dx 4
(2 −x)3
Following the same steps as above, we obtain:
f′′′(x) = 12
(2 −x)4
Step 4: General pattern By taking derivatives, we can observe a general
pattern emerging:
f(n)(x) = n!·2
(2 −x)n+1
Step 5: Express f(x)as a Maclaurin series To express f(x) as a Maclau-
rin series, we use the formula:
f(x) = f(0) + f′(0)
1! x+f′′(0)
2! x2+f′′′(0)
3! x3+· · ·
Substitute the derivatives we found earlier:
f(x) = 1 + x+ 2x2+ 6x3+· · ·
Therefore, the Maclaurin series for f(x) = 1
2−xis 1 + x+ 2x2+ 6x3+· · · .
26
Question 29
Question
Find the Maclaurin series representation for f(x) = 1
x+1 .
Solution
To find the Maclaurin series representation for f(x) = 1
x+1 , we will begin by
finding the derivatives of f(x) at x= 0 in order to construct the Maclaurin
series.
Step 1: Find f′(x)
f(x) = 1
x+ 1
f′(x) = d
dx 1
x+ 1
=−1
(x+ 1)2
Step 2: Find f′′ (x)
f′(x) = −1
(x+ 1)2
f′′(x) = d
dx −1
(x+ 1)2
=2
(x+ 1)3
Step 3: Find f′′′ (x)
f′′(x) = 2
(x+ 1)3
f′′′(x) = d
dx 2
(x+ 1)3
=−6
(x+ 1)4
Step 4: Generalize the pattern After finding the first few derivatives of
f(x), we notice a pattern where the nth derivative is of the form:
f(n)(x)=(−1)n·n!·(x+ 1)−(n+1)
Step 5: Write the Maclaurin series The Maclaurin series for f(x) is
given by:
f(x) = f(0) + f′(0)
1! x+f′′(0)
2! x2+f′′′(0)
3! x3+· · ·
Substitute f(0), f′(0), f′′(0), etc. into the above formula to get the Maclau-
rin series representation for f(x) = 1
x+1 :
27
f(x)=1−x+x2−x3+x4− · · · =
∞
X
n=0
(−1)n·xn
Question 30
Question
Find the Maclaurin series for the function f(x) = ln(1 + x) and determine its
interval of convergence.
Solution
To find the Maclaurin series for f(x) = ln(1 + x), we can start by finding the
derivative of f(x) and evaluating it at x= 0 to determine the coefficients of the
series.
Step 1: Find higher-order 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: Evaluate 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
.
.
.
Step 3: Write the Maclaurin series The Maclaurin series for f(x) =
ln(1 + x) is:
f(x) =
∞
X
n=0
f(n)(0)
n!xn=x−x2
2+x3
3−x4
4+· · · =
∞
X
n=1
(−1)n−1xn
n
28
Step 4: Determine interval of convergence Using the ratio test, we
can find the interval of convergence of the series:
lim
n→∞
(−1)nxn+1
n+1
(−1)n−1xn
n
=|x|lim
n→∞
n
n+ 1 <1
This simplifies to |x|<1, so the interval of convergence is (−1,1].
29
