Mathematics for Computer Science

14.1. The Value of an Annuity 407

differentiating equation 14.2 leads to:

d

dx

(^) n 1
X
iD 0
xi

!

D

d

dx



1 xn

1 x



: (14.11)

The left-hand side of equation 14.11 is simply

nX 1

iD 0

d

dx

.xi/D

nX 1

iD 0

ixi^1 :

The right-hand side of equation 14.11 is

nxn^1 .1x/.1/.1xn/

.1x/^2

D

nxn^1 CnxnC 1 xn

.1x/^2

D

1 nxn^1 C.n1/xn

.1x/^2

:

Hence, equation 14.11 means that

nX 1

iD 0

ixi^1 D

1 nxn^1 C.n1/xn

.1x/^2

:

Incidentally, Problem 14.2 shows how the perturbation method could also be ap-
plied to derive this formula.
Often, differentiating or integrating messes up the exponent ofxin every term.
In this case, we now have a formula for a sum of the form

P

ixi^1 , but we want a
formula for the series

P

ixi. The solution is simple: multiply byx. This gives:

nX 1

iD 1

ixiD

xnxnC.n1/xnC^1

.1x/^2

(14.12)

and we have the desired closed-form expression for our sum^3. It’s a little compli-
cated looking, but it’s easier to work with than the sum.
Notice that ifjxj< 1, then this series converges to a finite value even if there are
infinitely many terms. Taking the limit of equation 14.12 asntends infinity gives
the following theorem:

(^3) Since we could easily have made a mistake in the calculation, it is always a good idea to go back
and validate a formula obtained this way with a proof by induction.