1000 Solved Problems in Modern Physics

(Romina) #1

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|>


∣^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)= 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.
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