50 1 Mathematical Physics
Differentiating, limn=∞(
−2(n^2 +n1))
=∞∞
Differentiating again, limn=∞(
−^22
)
=−1(=L)
∣
∣
∣
∣
1
L
∣
∣
∣
∣=
∣
∣
∣
∣
1
− 1
∣
∣
∣
∣=^1
The series (A)isI. Absolutely convergent when|Lx|<1or|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 whenxlies between−1 and+ 1
By II the series is divergent whenxis less than−1 or greater than+ 1
By III there is no test whenx=±1.
Thus the given series is said to have [− 1 ,1] as the interval of convergence.1.36 f(x)=logx;f(1)= 0
f′(x)=1
x;f′(1)= 1f′′(x)=−1
x^2;f′′(1)=− 1f′′′(x)=2
x^3;f′′′(1)= 2
Substitute in the Taylor seriesf(x)=f(a)+(x−a)
1!f′(a)+(x−a)^2
2!f′′(a)+(x−a)^3
3!f′′′(a)+···logx= 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 placingx=0, we get
f(x)=cosx,f(0)= 1
f′(x)=−sinxf′(0)= 0
f′′(x)=−cosx,f′′(0)=− 1
f′′′(x)=sinx,f′′′(0)= 0
fiv(x)=cosx,fiv(0)= 1
etc.
Substituting in (1)cosx= 1 −x^2
2!+
x^4
4!−
x^6
6!+···
The series is convergent with all the values ofx.