Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have the following equation:

eq 1:

$f=100 \left(\left(\frac{z^{c/252}}{y^{b/252}}\right)^{\frac{252}{c-b}}-1\right)-100 \left(\left(\frac{y^{b/252}}{x^{a/252}}\right)^{\frac{252}{b-a}}-1\right)$

Given [a, b, c, x, y, z], solving for f is trivial, e.g.:

eq 2:

$100 \left(\left(\frac{1.1030^{1188/252}}{1.1015^{937/252}}\right)^{\frac{252}{1188-937}}-1\right)-100 \left(\left(\frac{1.1015^{937/252}}{1.0979^{687/252}}\right)^{\frac{252}{937-687}}-1\right) = -0.283604$

The problem is solving for y given [a, b, c, x, z, f].

I tried WolframAlpha and inputing the values does give me a numerical solution for y so I know it's possible. The following for example returns y $\approx$ 1.10150000054474...

eq 3:

$100 \left(\left(\frac{1.1030^{1188/252}}{y^{937/252}}\right)^{\frac{252}{1188-937}}-1\right)-100 \left(\left(\frac{y^{937/252}}{1.0979^{687/252}}\right)^{\frac{252}{937-687}}-1\right) = -0.283604$

I've read about Newton–Raphson but I'm not sure how to implement it since I can't isolate y and neither can WolframAlpha aparently. Asking it to solve for y results in a timeout.

Any ideas on an alternative method or maybe some way to isolate the y variable on eq 1?

UPDATE Limitless' comments below seem to confirm that there is no closed-form way to solve for y, so the focus is on finding an appropriate numerical method to achieve that.


a,b,c are non-zero positive integers

x,z may be positive or negative, but not zero. Though I will accept solutions that assume they are positive only.

share|cite|improve this question
The best I can get is the following simplification: $$\frac{f}{100}=z^{\frac{c}{c-b}}y^{\frac{b}{b-c}}-x^{\frac{a}{a-b}}y^{\frac{b}{‌​b-a}}$$ – 000 Apr 17 '12 at 20:17
By making substitutions, it seems $y$ does not have a solution in terms of $a,b,c,x,z,\text{ and } \vec{f}$. See Wolfram Alpha's wisdom. – 000 Apr 17 '12 at 20:25
Well, WA did solve it though. See eq3 where I have all terms but y. It resulted in "Numerical Solution: y $\approx$ 1.10150000054474..." – indiosmo Apr 17 '12 at 20:30
Just because WA solved it does not mean it has a solution in terms of the variables. (That's akin to saying sextics have a general solution because WA can solve them. ;)) Most likely, it is a numerical method (that I do not know, as none of the obvious possibilities are coming to mind as likely. . .) that is being employed. – 000 Apr 17 '12 at 20:33
To be fair, my question does ask specifically for a numerical method to solve this, as I realize it can't be solved analytically. – indiosmo Apr 17 '12 at 20:35
up vote 0 down vote accepted

Since Newton Raphson requires the first derivative of this equation (hard to compute symbolically), here is a Regula-Falsi implementation that converges to 1.101500000544740 written in C (compile with gcc and run).

share|cite|improve this answer
You could also post a copy of that here. – 000 Apr 17 '12 at 22:04
Done. I prefer using pastebin though. – chemeng Apr 17 '12 at 22:06
I'll study the method to understand it better but I tested with several values and it works, so I'm accepting your answer. – indiosmo Apr 17 '12 at 22:17
The Regula-Falsi method is actually an alternative of the Newton-Raphson. Instead of using the first derivative it uses an approximation of the form $f'(x_n)\approx \frac{f(x_n)-f(x_{n-1})}{x_n-x_{n-1}}$ and therefore doesn't require the calculation of the first derivative. Anyway,glad i could help. – chemeng Apr 17 '12 at 22:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.