Mathematical Methods for Physics and Engineering : A Comprehensive Guide

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4.2 SUMMATION OF SERIES


Integrating the RHS by parts we find


S(x)/x=x^2 expx− 2 xexpx+2expx+c,

where the value of the constant of integrationccanbefixedbytherequirementthat
S(x)/x=0atx= 0. Thus we find thatc=−2 and that the sum is given by


S(x)=x^3 expx− 2 x^2 expx+2xexpx− 2 x.

Often, however, we require the sum of a series that does not depend on a

variable. In this case, in order that we may differentiate or integrate the series,


we define a function of some variablexsuch that the value of this function is


equal to the sum of the series for some particular value ofx(usually atx=1).


Sum the series

S=1+

2


2


+


3


22


+


4


23


+···.


Let us begin by defining the function


f(x)=1+2x+3x^2 +4x^3 +···,

so that the sumS=f(1/2). Integrating this function we obtain

f(x)dx=x+x^2 +x^3 +···,


which we recognise as an infinite geometric series with first terma=xand common ratio
r=x. Therefore, from (4.4), we find that the sum of this series isx/(1−x). In other words

f(x)dx=


x
1 −x

,


so thatf(x)isgivenby


f(x)=

d
dx

( x

1 −x

)


=


1


(1−x)^2

.


The sum of the original series is thereforeS=f(1/2) = 4.


Aside from differentiation and integration, an appropriate substitution can

sometimes transform a series into a more familiar form. In particular, series with


terms that contain trigonometric functions can often be summed by the use of


complex exponentials.


Sum the series

S(θ)=1+cosθ+

cos 2θ
2!

+


cos 3θ
3!

+···.


Replacing the cosine terms with a complex exponential, we obtain


S(θ)=Re

{


1+expiθ+

exp 2iθ
2!

+


exp 3iθ
3!

+···


}


=Re

{


1+expiθ+

(expiθ)^2
2!

+


(expiθ)^3
3!

+···


}


.

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