7/28/2019 Solved Problems Maclaurin's
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Q 1 : Using the Maclaurins series or otherwise establish the series for thefollowing functions :
(i) sec2 x (ii) log (1 +x+x2 +x3)
(i) sec2x
Let y= sec2x
Integrating w.r.t. xboth sides, we get
ydx = sec2xdx+ c= tanx+ c
y dx =x+3
1x3 +
15
2x5+ + c
Now differentiating w.r.t. xwe get
y = 1 +x2 +3
2x4+
(ii) log (1 + x+x2 +x3)
Let y = log (1 +x+x2 +x3)
x
x n
1
1
=xn1 +xn2 +xn3 + + 1
We getx
x
1
1 4=x3 +x2 + x +1 = 1 +x+x2 +x3
log (1 +x+x2 +x3) = log
x
x
1
1 4
= log (1 x4) log (1x)
7/28/2019 Solved Problems Maclaurin's
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2
=
....
2
84 x
x
....
432
432 xxxx
y =x+2
2x+
3
3x+x4
1
4
1+
y =x+2
2x
+3
3x
4
3x4+
Q 2 : Using the Machlaurins series prove the following :
(i) Ifx= y2
1y2 +
3
1y3
4
1y3
4
1y4+ then
y =x+!2
2x
+!3
3x
+ and viceversa.
(ii) tan1
cos1
sin
x
x=xsin +
2
2x
sin 2 +3
3x
sin 3.
(i) Given thatx= y2
1y2 +
3
1y3
4
1y3
4
1y4+
Observe R.H.S. its log (1 + y)
x = log (1 + y)
Which is also, 1 + y= ex
or 1 + y = 1 +!1
x+
!2
2x+
!3
3x+ (series ofexon RHS)
y =!1
x+
!2
2x
+!3
3x
+
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(ii) Let y = tan1
cos1
sin
x
x
tan y =
cos1
sin
x
x
Now by Eulers formula for tan y
tan y =)(
)(iyiy
iyiy
eei
ee
)(
)(iyiy
iyiy
eei
ee
=
cos1
sin
x
x
or1
12
2
iy
iy
e
e=
cos1
sin
x
ix
Using componendo and dividendo
i.eNumeratoratorDeno
NumeratoratorDeno
min
minon both sides, we get
1
2iye=
sincos1
sincos1
ixx
ixx
e2iy =
i
i
xe
xe
1
1 ( ei = cos i sin )
or 2iy = log
i
i
xe
xe
1
1
= log (1 xe-i ) log (1 xei)
=
....
32
3322
iii exexxe
....
3
3
2
322
eiii xexxe
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Taking common coefficients ofx,x2,x3
=x(ei ei)+2
2x(e2i e-2i)+
3
3x(e3ie-3i)+
2iy = 2isin x+ 2isin 2
2x+ 2isin 3
3
3x+
(using the formula 2i sin n = ei ein)
y =xsin +2
2x
sin 2 +3
3x
sin 3
Q 3 : Expand the following functions into a series with a minimum of threenonzero terms :
(i) tan1xa
xa
Let y = tan1xa
xa
Putx= a tan or tan1
a
x
Now, y = tan1)/1(
)/1(
axa
axa
= tan1
tan4/tan1
tan4/tan
1
4tan
= tan1 tan (/4 + ) = /4 +
=4
+ tan1a
x
y =4
+
ax
31
3
a
x +51
5
a
x +
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(ii) tan1x
x1
Let y = tan1x
x 112
tan1x
x 11 2
Putx= tan or = tan1 x
y = tan1
tan
tan1 2= tan1
tan
1sec
= tan1
cos
sin
1cos
1
= tan1
sin
cos1
= tan1
2cos
2sin2
22sin2
= tan 1
2tan
=2
=
2
1tan1x
y =2
1tan1x=
2
1
....
53
53 xxx
(iii) tan12
1
2
x
x
Let y = tan1
21
2
x
x
tan1
21
2
x
x
Putx= tan or = tan1 x
y = tan1
2tan1
tan2= tan1 (tan 2)
= 2
y = 2 tan1x
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y = 2
.....
53
53 xxx
(iv) cos1
1
1
xx
xx
Let y = cos1
1
1
xx
xx
= cos1
xx
xx
1
1
= cos1
1
12
2
x
x (1)
= cos1
2
2
1
1
x
x
Putx= tan or = tan1x
y = cos1
2
2
tan1
tan1
= cos1 (cos2) = cos1 (cos (2))
= 2 = 2 tan1x
y = 2
.....
53
53 xxx
(v) cos1 (tanh logx)
Let y = cos1 tanh logx
Trick! Since
tanh =
ee
ee=
1
12
2
e
e
and elogA = A,
we get y = cos1
1
1log2
log2
x
x
e
e= cos1
1
12log
2log
x
x
e
e
7/28/2019 Solved Problems Maclaurin's
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= cos1
1
12
2
x
x
= 2 tan1x
y = 2
.....
53
53 xxx
(vi) tan1pxq
qxp
Let y = tan1
pxq
qxp
Tip 1
= tan1)./1(
)/(
xqpq
xqpq
= tan1
xq
p
xq
p
.1
Tip 2
Since tan1 A tan1B = tan1AB
BA
1
We get, y = tan1q
p tan1x Tip 3
y = tan1q
p
.....
53
53 xxx Tip 4
7/28/2019 Solved Problems Maclaurin's
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Q 4 : Find the series expansions of the following function :
tan122
22
11
11
xx
xx
Let y = tan122
22
11
11
xx
xx
Rationalizing denominator we get,
y = tan1
2222
2222
)1()1(
)11)(11(
xx
xxxx
= tan1
22
222
11)1()1(
xxxx
= tan1
2
2222
2
112)1()1(
x
xxxx
= tan1
2
4
2
122
x
x= tan1
2
411
x
x
Putx2 = sin or = sin1 (x2)
y = tan1
sin
sin11 2
= tan1
sin
cos1= tan1
2/cos2/sin2
2/cos2 2
= tan1
2cot = tan1 tan
22
y =2
2
=
2
2
1sin1 (x2)
Using series for sin-1
y =2
2
1
.....
7642
531
542
31
32
114106
2 xxxx
7/28/2019 Solved Problems Maclaurin's
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Q 6 : Expand the function in an ascending powers ofxto a minimum of
3 nonzero terms xsin1 .
Solution:
Let y = xsin1
y =2
cos2
sin22
cos2
sin 22xxxx
y =2
cos2
sin22
cos2
sin 22xxxx
or y = sin2x + cos
2x
Now using the series for sin xand cosxwe get
y =
....
25
1
2!3
1
2
53xxx
+
....
2!4
1
2!2
11
42xx
Collecting the coefficients we get
= 1 +2
x
8
2x
48
3x+
384
4x
+3840
5x
Q 7 : Expand the function in an ascending powers ofxto a minimum of 3 nonzero terms excosx.
Solution.:
Using the series for ex where x has to be treated as xcosx
y = 1 +!1
xcosx+
!2
2x
cos2x+!3
3x
cos3x+
= 1 +!1
x
.....
!4!21
42 xx
7/28/2019 Solved Problems Maclaurin's
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+!2
1x2
242
.....!4!2
1
xx
+!3
1x3
342
.....!4!2
1
xx
+
= 1 +x2
3x+
24
5x
+2
2x
2
4x+
+6
3x+
= 1 +x+2
3x+x3
6
1
2
1
= 1 +x+2
2x
2
1x3 +
Q 8 : Expand in powers ofx, exsinx.
Solution :
Letxsinx= y
....!4!3!2
1432
sin yyy
yee yxx
.....)sin(!3
1)sin(
!2
1sin1 32sin xxxxxxe xx
........!3
2....!3!2
.....!5!3
13
33253
x
xxx
xxxxx
........)(!3
....)!3(!3
2
!2....
!5!31 3
3
2
642
253
x
xxxx
xxxxx
....!3!3!2120!3
166464
2 xxxxx
x
....120
1
3
11 642 xxx
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Q 9: Expand in powers ofx,1xe
x.
Hence, prove that ....720
1
12
11
1
1
2
42
xxe
ex
x
x
.
Solution :
We have
1....!3!2
11 32
xx
x
x
e
xx
....!5!4!3!2
5432
xxxx
x
x
....!5!4!3!2
1
1432 xxxx
1432
....!5!4!3!2
1
xxxx
232432
....!4!3!2
....!5!4!3!2
1
xxxxxxx
........!2....!3
2
!2
43
xxx
....243664
....1202462
14432432
xxxxxxxx
....16
....88
443
yxx
....720122
142
xxx
Now,
1
21
21
21
21
1
2 xx
x
x
x
e
x
e
ex
e
ex
12
xe
xx
....72012
1....720122
121
1
2
4242 xxxxxx
e
exx
x
7/28/2019 Solved Problems Maclaurin's
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7/28/2019 Solved Problems Maclaurin's
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Q 11 : Expand log (1 +x+x2 +x3) up tox8.
Solution :
log (1 +x+ 2x+x3) = log (1 +x)(1 +x2)
x
xxxxxx
1
)1)(1)(1(log)1log(
232
)1log()1log(1
1log 4
4
xxx
x
...
8765432...
2
876543284 xxxxxxxx
xx
....8
3
7654
3
32
8765432
xxxxxxx
x
Q 12 : Prove that ....32
)1log(32
32 xxxxxx .
Solution :
log(1 x+x2x3) = log (1 x) (1 +x2)
= log (1 x)+ log (1 +x2)
....2
....32
42
32
x
xxx
x
....32
32
xx
x