**Question Corner and Discussion Area**

Thank you for maintaining such an interesting and useful web page. The following problem comes from a real life application: defining a smooth calculational grid for a computational fluid dynamics program.Closed-form solutions are more of a rarity than one might think, and not as important as one might think. The equation you are trying to solve can be written asConsider the sum of N terms of a geometric series:

S=A1 +A1*R+A1*R^2 + ... +A1*R^(N-1)We know that this is equivalent to:

S=A1 * (1 -R^N) / (1 -R)We can rearrange this equation to solve for

Nto get:

N=LN((S*R+A1 -S) /A1) /LN(R).The challenge is to solve for

R. From the form of the first equation, we know that what we seek is one of the roots of a polynomial of degreeN-1. If we view the equation in physical terms we should satisfy ourselves that 1) there is a unique real solution and 2) there are reasonable restrictions that we can apply that will eliminate the unsought complex roots.The physical model that the algebra represents is:

Sis a distance to be spanned byNcomputational cells. The first cell has a width ofA1 the second a with ofA1*Rand so on. What we seek is an expansion (or contraction) factor,R, that will cause the sum ofNcell widths to be exactlyS.From this we can state the following inequalities:

From a purely intuitive view there ought to be a closed form generalized solution for

S> 0 ;there must be a distance to span. A1 > 0 ;the first cell must have width. A1 <S;we won't span the distance with the first cell. N> 1 ;we have to have at least a second cell to cover the remaining distance. R> 0 ;follows from above. R. But I have been unable to see my way to anything but an iterative numerical solution. Can you supply some insight?Thank you and best wishes, Ned Piburn

whereR^N+R^(N-1) + ... +R+ 1 -q= 0 (*)

(Incidentally, although the form *S *= *A*1 * (1 - *R*^*N*) / (1 - *R*) looks
simpler, it's not any easier to solve for *R*; because although, when
you rearrange it into a polynomial equation *R*^*N *- *q R *+ *q *- 1 = 0, it
looks nicer than (*), it is simply (*R*-1) times (*). So in finding its
roots the first thing you'd do is discover and discard the root *R*=1;
then you'd divide by (*R*-1) and be left with (*). So it's no easier to
use this form; in fact, it's mathematically less convenient because
you have the extra root *R*=1 to worry about).

Solutions by radicals of polynomials of degree <= 4 were known by the
1500's (the degree 2 case being the familiar quadratic formula). For a
couple of hundred years mathematicians tried and failed to find such a
formula for quintic (degree 5) equations; eventually it was discovered
that no such general formula exists, because associated to each such
polynomial is an abstract mathematical object called a *Galois group*,
and only when the Galois group has certain nice properties does a
solution by radicals exist. All Galois groups for polynomials of
degree <= 4 have this property, but the Galois groups for most degree
>= 5 polynomials do not. Some do, but equation (*) is not among them
(except for very special values of *q* like 0 or 1).

There are general solutions of quintic equations involving other
functions like hypergeometric functions, but from a practical point of
view that doesn't matter too much since the only way to evaluate them
is by numerical calculations anyway. And besides, that's only for *N*=5.

However, just because there isn't a "closed-form" solution doesn't
mean *R* is not expressible as a function of *N* and *q*! In fact, your
"physical arguments" can be translated into precise mathematics. The
conditions you gave, namely that *N *> 1, *R *> 0, and *S *> *A*1 (so that
*q *> 1), mean that *R* is uniquely defined as a function of *N* and *q*. The
mathematical proof of that goes like this:

Let *f*(*R*) = *R*^*N *+ · · · + *R *+ (1-*q*). This *f* is a continuous function,
with *f*(0) < 0 and *f*(*q*-1) > 0. (Reason: *f*(0) = 1 - *q *< 0 since *q *> 1,
and *f*(*q*-1) = (*q*-1)^*N *+ · · · + (*q*-1)^2 + (*q*-1) + (1-*q*) which, since
(*q*-1)^*N*, . . . , (*q*-1)^2 are all positive, is greater than
(*q*-1) + (1-*q*) = 0.)

Therefore, by the intermediate value theorem, *f*(*R*)=0 for some
value of *R* in the range 0 < *R *< *q*-1.

Also, this *R* is unique: there can be no other *R *> 0 for whichs
*f*(*R*) = 0, because *f* is an increasing function when *R *> 0
(its derivative is
*NR*^(*N*-1) + ... + 2*R *+ 1 which is positive).

Therefore, for each *N*>1 and *q*>1, there is a unique corresponding *R*>0
for which (*) holds. The equation *does* define *R* as a function
of *N* and *q*. You could even go on to describe properties of that
function; you can prove that it's continuous and differentiable, for
example.

It just happens that this function isn't representable as a
combination of the few functions we're familiar with, so to calculate
its values we need numerical approximations. But that's not an
unexpected thing; very few functions are. Even the trigonometric and
exponential functions cannot be written as combinations of the simpler
functions; that's why we give new names to them. The same thing is
true of the function representing *R* in terms of *N* and *q*; if you wanted
to use it a lot, you'd just give that function a name and a notation,
the way one did with functions like "sin" and "log". To calculate its
values you use some kind of numerical method; that's exactly what one
does when one calculates sines or logarithms.

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Last updated: April 19, 1999

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