MathCity.org Exercise 2.8 (Solutions)



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Exercise 2.8 (Solutions)Page 101

Calculus and Analytic Geometry, MATHEMATICS 12 Available online @ , Version: 3.0

Taylor Series Expansion of Function

f (x + h) = f (x) + hf (x) + h2 f (x) + h3 f (x) + ....

2!

3!

Maclaurin Series

f (x) = f (0) + xf (0) + x2 f (0) + x3 f (0) + ....

2!

3!

Question # 1 Apply the Maclaurin series expansion to prove that: (i) ln(1 + x) = x - x2 + x3 - x4 + ......

234 (ii) cos x = 1 - x2 + x4 - x6 + ......

2! 4! 6! (iii) 1 + x = 1 + x - x2 + x3 + .....

2 8 16 (iv) ex = 1 + x + x2 + x3 + .....

2! 3! (v) e2x = 1 + 2x + 4x2 + 8x3 + .....

2! 3! Solution (i) Let f (x) = ln(1 + x)

f (0) = ln(1 + 0) = 0

f (x) = d ln(1 + x) = 1

dx

1+ x

f (0) = 1 = 1 1+ 0

f (x) = d (1 + x)-1 = - (1 + x)-2

dx

f (0) = - (1 + 0)-2 = -1

f (x)

=

d dx

-(1

+

x)-2

=

+ 2(1 + x)-3

f (0) = 2(1 + 0)-3 = 2

f (iv) (x) = d 2(1 + x)-3 = - 6(1 + x)-4

dx

f (iv) (0) = - 6(1 + 0)-4

= -6

By Maclaurin series



FSc-II / Ex- 2.8 - 2

f (x) = f (0) + xf (0) + x2 f (0) + x3 f (0) + ......

2!

3!

ln(1 + x) = 0 + x(1) + x2 (-1) + x3 (2) + x4 (-6) + ......

2!

3! 4!

= x - x2 + x3 (2) - x4 (6) + ...... 21 3 21 43 21

= x - x2 + x3 - x4 + ...... 234

(ii) Let f (x) = cos x f (0) = cos(0) = 1

f (x) = d cos x = - sin x dx

f (0) = - sin(0) = 0

f (x) = d (-sin x) = - cos x

dx

f (0) = - cos(0) = -1

f (x) = d (-cos x) = + sin x

dx

f (0) = sin(0) = 0

f (iv) (x) = d sin x = cos x dx

f (iv) (x) = cos(0) = 1

f (v) (x) = d cos x = - sin x f (v) (x) = - sin(0) = 0 dx

f (vi) (0) = d (-sin x) = - cos x

dx

f (vi) (0) = - cos(0) = -1

Now by Maclaurin series

f (x) = f (0) + xf (0) + x2 f (0) + x3 f (0) + ......

2!

3!

cos x = 1 + x(0) + x2 (-1) + x3 (0) + x4 (1) + x5 (0) + x6 (-1) + ......

2!

3! 4! 5! 6!

= 1 + 0 - x2 + 0 + x4 + 0 - x6 + ......

2!

4!

6!

= 1 - x2 + x4 - x6 + ...... 2! 4! 6!

(iii) Let f (x) = 1 + x

=

(1

+

x

)

1 2

1

f (0) = (1 + 0)2

= 1

f (x) =

d

(1 +

x

)

1 2

dx

=

1

(1

+

x

)-

1 2

(1)

=

1

(1

+

x

)-

1 2

2

2

f (0) =

1

(1

+

0

)-

1 2

=

1

2

2

f ( x)

=

d dx

1 2

(1 +

x)-

1 2

=

-

1

(1

+

x)-

3 2

4

f (0)

=

-

1

(1

+

0

)

-

3 2

=

- 1

4

4

f (x)

=

-

1 4

d dx

(1

+

x

)-

3 2

=

-

1 4

-

3 2

(1

+

x)-

5 2

=

3

(1

+

x)-

5 2

8

f (0)

=

3 (1+

)0

-

5 2

=

3

8

8

Now by Maclaurin series

f (x) = f (0) + xf (0) + x2 f (0) + x3 f (0) + ......

2!

3!

1+ x

=

1+ x 1 + 2

x2 2!

-

1 4

+

x3 3 + ..... 3! 8

=

1+ x 1 + 2

x2 2

-

1 4

+

x3 6

3 + ..... 8

= 1 + x - x2 + x3 + ..... 2 8 16

(iv) Let f (x) = ex f (0) = e0 = 1

( ) f ( x) = d ex = ex f (0) = e0 = 1 dx

( ) f ( x) = d ex = ex f (0) = e0 = 1 dx

( ) f ( x) = d ex = ex f (0) = e0 = 1 dx

By Maclaurin series

f (x) = f (0) + xf (0) + x2 f (0) + x3 f (0) + ......

2!

3!

ex = 1 + x(1) + x2 (1) + x3 (1) + ..... 2! 3!

= 1 + x + x2 + x3 + ..... 2! 3!

(v) Let f (x) = e2x f (0) = e2(0) = e0 = 1

( ) f ( x) = d e2x = 2e2x dx

FSc-II / Ex- 2.8 - 3

FSc-II / Ex- 2.8 - 4

f (0) = 2e2(0) = 2(1) = 2

( ) ( ) f ( x) = 2 d e2x = 2 2e2x = 4e2x dx

f (0) = 4e2(0) = 4(1) = 4

( ) ( ) f ( x) = 4 d e2x = 4 2e2x = 8e2x dx

f (0) = 8e2(0) = 8

By Maclaurin series

f (x) = f (0) + xf (0) + x2 f (0) + x3 f (0) + ......

2!

3!

e2x = 1 + x(2) + x2 (4) + x3 (8) + ..... 2! 3!

= 1 + 2x + 4x2 + 8x3 + ..... 2! 3!

[

Question # 2

Show that

cos( x + h) = cos x - hsin x - h2 cos x + h3 sin x + ......

2

3

and evaluate cos61 .

Solution Let f (x) = cos x

f ( x) = d cos x = - sin x

dx

f ( x) = - d sin x = - cos x

dx

f ( x) = - d cos x = - (-sin x) = sin x

dx

By Taylor series

f (x + h) = f (x) + hf (x) + h2 f (x) + h3 f (x) + ....

2!

3!

cos( x + h) = cos x + h(-sin x) + h2 (-cos x) + h3 (sin x) + ......

2!

3!

cos( x + h) = cos x - hsin x - h2 cos x + h3 sin x + ......

2

3

Put x = 60 and h = 1 = = 0.01745 rad 180

cos(60 +1) = cos60 - (0.01745)sin 60 - (0.01745)2 cos60 + (0.01745)3 sin 60 + ......

2

3

FSc-II / Ex- 2.8 - 5

cos61 = 0.5 - (0.01745)(0.866) - (0.000305) (0.5) + (0.00000531) (0.866) + ......

2

6

= 0.5 - 0.0151117 - 0.000076125 + 0.000000072 + ...... = 0.484812247 0.4848

Question # 3

Show that

2x+h =

2x

1

+

(ln

2)h

+

(ln

2)2

h2 2!

+

(ln

2)3

h3 3!

+

......

Solution Let f (x) = 2x

f (x) = d 2x

dx

d ax = ax ln a dx

= 2x ln 2

( ) f ( x) = ln 2 d 2x = ln 2 2x ln 2 = (ln 2)2 2x dx

f ( x) = (ln 2)2 d 2x = (ln 2)2 2x ln 2 = (ln 2)3 2x

dx

Now by Taylor series

f (x + h) = f (x) + hf (x) + h2 f (x) + h3 f (x) + ....

2!

3!

2x+h = 2x + h 2x ln 2 + h2 (ln 2)2 2x + h3 (ln 2)3 2x + ......

2!

3!

=

2x

1 +

(ln

2)

h

+

(ln

2)2

h2 2!

+

(ln

2)3

h3 3!

+

......

Book:

Exercise 2.8, page 101 Text Book of Algebra and Trigonometry Class XII Punjab Textbook Board, Lahore.

Available online at in PDF Format (Picture format to view online). Updated: September,12,2017.

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