# Thread: Solving a finite power series for x

1. ## Solving a finite power series for x

A few days ago, EcoDTR posted a question about a number series to which Soroban made a reply including this statement:
If the problem was actrually: 1 + x + x^2 + x^3 + ... = 5

. . we can solve for x . . . otherwise, forget it!
Can you elaborate a little bit on how to solve for x in the case of a finite power series? I.E. 1 + x + x^2 + x^3 + ... + x^24 + x^25 + x^26 = y?

For y=244, I know the answer is approximately 1.1413211694 through trial and error using my computer to run out the summation as I plugged in numbers. Now I'd like to know the math behind why that is the answer or how I can calculate x given a known y. Any pointers?

-=*>Chuck<*=-

2. Originally Posted by chuckdcc
A few days ago, EcoDTR posted a question about a number series to which Soroban made a reply including this statement:

Can you elaborate a little bit on how to solve for x in the case of a finite power series? I.E. 1 + x + x^2 + x^3 + ... + x^24 + x^25 + x^26 = y?
This is a "geometric series"
If $\displaystyle x\not = 1$
Thus, its sum is,
$\displaystyle \frac{1-x^{26}}{1-x}=y$
Thus,
$\displaystyle 1-x^{26}=y-xy$
Thus,
$\displaystyle x^{26}-xy+(y-1)=0$
This is a 26th degree polynomial equation.
In general they cannot be solved when degree is 5 or above.

3. Originally Posted by ThePerfectHacker
This is a "geometric series"
If $\displaystyle x\not = 1$
Thus, its sum is,
$\displaystyle \frac{1-x^{26}}{1-x}=y$
Thus,
$\displaystyle 1-x^{26}=y-xy$
Thus,
$\displaystyle x^{26}-xy+(y-1)=0$
This is a 26th degree polynomial equation.
In general they cannot be solved when degree is 5 or above.

If $\displaystyle y>1$ Descartes rule of signs tells us that this has either 0 or 2 positive roots. But $\displaystyle x=1$ is always a root of $\displaystyle x^{26}-xy+(y-1)=0$ (which may not be a solution for technical reasons), which guarantees that there is at least one other positive root, and this is the solution we seek (it may also be 1).

It may be necessary to find it numerically but it does exist.

If $\displaystyle 0<y<1$, there are no solutions.

If $\displaystyle y=1$, then $\displaystyle x=0$ is the solution.

I'll think about non-positive $\displaystyle y$ later.

RonL

4. Originally Posted by ThePerfectHacker
This is a "geometric series"
If $\displaystyle x\not = 1$
Thus, its sum is,
$\displaystyle \frac{1-x^{26}}{1-x}=y$
Thus,
$\displaystyle 1-x^{26}=y-xy$
Thus,
$\displaystyle x^{26}-xy+(y-1)=0$
This is a 26th degree polynomial equation.
In general they cannot be solved when degree is 5 or above.
I think that the power in the sum function should be 27 instead of 26... $\displaystyle \frac{1-x^{27}}{1-x}=y$ But no matter, because you also state that degrees higher that 5 generally can't be solved. So my answer is: Can't do it.

Interestingly, I arrived at my question in trying to fit a geometric series between two known y values and used various substitutions to arrive at $\displaystyle y=\frac{x^{27}-1}{x-1}$ which can be reduced to $\displaystyle y=\sum_{n=0}^{26}{x^n}$ by factoring the $\displaystyle x-1$ out of the numerator to get rid of the denominator. I managed to find a declaration of this equality in my old Calculus textbox, but no mention of the ability or inability to solve for x.

I ended up writing a simple program to work out the first 256 values of x for the integer values of y from 1 to 256, but it seemed like a clunky way to get the numbers I was after. I would much rather have identified a workable formula than do lookups in a chart. When I saw the post by Soroban, I recognized it as related to my series and was hoping there might have been a formula I had missed.

Thank you both for your time,
-=*>Chuck<*=-