1000 Solved Problems in Modern Physics

50 1 Mathematical Physics

Differentiating, limn=∞

(

−2(n^2 +n1)

)

=∞∞

Differentiating again, limn=∞

(

−^22

)

=−1(=L)

∣

∣

∣

∣

1

L

∣

∣

∣

∣=

∣

∣

∣

∣

1

− 1

∣

∣

∣

∣=^1

The series (A)is

I. Absolutely convergent when|Lx|<1or|x|>

∣

∣L^1

∣

∣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)= 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)+···

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.