50
1 Mathematical Physics
Differentiating, lim n=∞
−
2n
2(n+1)
=
∞
∞
Differentiating again, lim n=∞
−
2
2
= −1(= L)
1
L
=
1
−1
= 1
The series (A) is
I. Absolutely convergent when |Lx| < 1 or |x| >
1
L
i.e. −
1
L
< x <
+
1
L
II. Divergent when |Lx| > 1, or |x| >
1
L
III. No test when |Lx| = 1, or |x| =
1
L
.
By I the series is absolutely convergent when x lies between −1 and +1
By II the series is divergent when x is less than −1 or greater than +1
By III there is no test when x = ±1.
Thus the given series is said to have [−1, 1] as the interval of convergence.
1.36 f (x) = log x; f (1) = 0
f
(x) =
1
x
; f
(1) = 1
f
(x) = −
1
x 2 ; f
(1) = −1
f
(x) =
2
x 3 ; f
(1) = 2
Substitute in the Taylor series
f (x) = f (a) +
(x − a)
1!
f
(a) +
(x − a)
2
2!
f
(a) +
(x − a)
3
3!
f
(a) + · · ·
log x = 0 + (x − 1) −
1
2
(x − 1)
2
+
1
3
(x − 1)
2
− · · ·
1.37 Use the Maclaurin’s series
f (x) = f (0) +
x
1!
f
(0) +
x
2
2!
f
(0) +
x
3
3!
f
(0) + · · ·
(1)
Differentiating first and then placing x = 0, we get
f (x) = cos x, f (0) = 1
f
(x) = − sin x f
(0) = 0
f
(x) = − cos x, f
(0) = −1
f
(x) = sin x, f
(0) = 0
f
iv (x) = cos x, f
iv (0) = 1
etc.
Substituting in (1)
cos x = 1 −
x
2
2!
+
x
4
4!
−
x
6
6!
+ · · ·
The series is convergent with all the values of x.