Solved Problems Maclaurin's

7/28/2019 Solved Problems Maclaurin's

1/13

1

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)


2/13

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

+


3/13

3

(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


4/13

4

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 +


5/13

5

(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


6/13

6

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


7/13

7

= 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


8/13

8

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


9/13

9

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


10/13

10

+!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


11/13

11

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


12/13


13/13

13

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

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Solved Problems Maclaurin's