1.3 Solutions 49
h=−f(v)
f′(v);f(v)=f(− 2 .1)f′(v)=dy
dx|v;f′(v)= 10. 23h=−0. 039
10. 23
=− 0. 0038
To a first approximation the root is− 2. 1 − 0 .0038123 or− 2 .1038123. As a
second approximation, assume the root to be
a=− 2. 1038123 +h,
Putv 1 =− 2. 1038123
h 1 =−f(v 1 )/f′(v 1 )=− 0. 000814 / 6. 967 =− 0. 0001168
The second approximation, therefore, givesa=− 2 .1039291.
The third and higher approximations can be made in this fashion. The first
approximation will be usually good enough in practice.1.34 y(x)=x
(^2) exp(−x (^2) )(1)
Turning points are determined from the location of maxima and minima. Dif-
ferentiating (1) and setting dy/dx= 0
dy/dx= 2 x(1−x^2 )exp(−x^2 )= 0
x= 0 ,+ 1 ,−1. These are the turning points.
We can now find whether the turning points are maxima or minima.
dy
dx
=2(x−x^3 )e−x
2
y
′′
=2(2x^4 − 5 x^2 +1)e−x
2
Forx= 0 ,d
(^2) y
dx^2 =+^2 →minimum
Forx=+ 1 ,,d
(^2) y
dx^2 =−^4 e
− (^1) →maximum
Forx=− 1 ,,d
(^2) y
dx^2 =−^4 e
− (^1) →maximum
y(x)=x^2 e−x
2
is an even function becausey(−x)=+y(x)
1.3.6 Series............................................
1.35 The given series isx−
x^2
22+
x^3
32−
x^4
42+··· (A)
The series formed by the coefficients is1 −1
22
+
1
32
−
1
42
+··· (B)
lim
n=∞(
an+ 1
an)
=lim
n=∞[
−
n^2
(n+1)^2]
=
∞
∞
Apply L’Hospital rule.