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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