# Is it possible to solve this kind of polynomial?

• Jul 23rd 2010, 01:23 AM
rainer
Is it possible to solve this kind of polynomial?
Hello,

Is there any way to solve polynomial equations of the form:

$ax^{1+u}+bx+cx^{1-u}+d=0$

?

I really hope so.

PS- it probably doesn't matter, but for the particular problem I have in mind, $u=\frac{i n\pi}{z}$, where "i" is the imaginary unit.
• Jul 23rd 2010, 01:38 AM
Bacterius
Well that's not a polynomial then, since it has non-integer exponents. I guess it could *perhaps* be boiled down to a cubic, but I'm not sure if that is still valid if we have a complex number as an exponent.

Anyway, if you can't solve it, it should remain continuous and derivable regardless of what "u" is, so if you can't find an algebraïc way do to it, you can just use a Newton approximation and solve for x :)
• Jul 23rd 2010, 02:02 AM
rainer
Quote:

Originally Posted by Bacterius
Well that's not a polynomial then, since it has non-integer exponents. I guess it could *perhaps* be boiled down to a cubic, but I'm not sure if that is still valid if we have a complex number as an exponent.

Anyway, if you can't solve it, it should remain continuous and derivable regardless of what "u" is, so if you can't find an algebraïc way do to it, you can just use a Newton approximation and solve for x :)

Newton approximation eh? I'll have to look that up.

Say d is actually d(x), that is to say, d is a function of x...

$ax^{1+u}+bx+cx^{1-u}+d(x)=0$

(we know that d(1) is defined)

Now say I divide the argument of d(x) by x. Since I have to do this to all x's, wouldn't this then yield...

$ax^{u}+b+cx^{-u}+d\left(\frac{x}{x}\right)=0$

Then multiplying through by $x^u$...

$ax^{2u}+x^u(b+d(1))+c=0$

Would that work?
• Jul 23rd 2010, 02:09 AM
Bacterius
Nice try but no, that doesn't work, $\frac{f(x)}{x} \neq f(1)$ except if $f(x) = x$ (which is obviously useless)
When you divide the polynomial like you did here, you don't divide the argument given to the function, you divide its result.
• Jul 23rd 2010, 02:10 AM
pickslides
Quote:

Originally Posted by rainer

Now say I divide the argument of d(x) by x. Since I have to do this to all x's, wouldn't this then yield...

$ax^{u}+b+cx^{-u}+d\left(\frac{x}{x}\right)=0$

No that would give $ax^{u}+b+cx^{-u}+\frac{d(x)}{x}=0$
• Jul 23rd 2010, 02:17 AM
rainer
Quote:

Originally Posted by Bacterius
Nice try but no, that doesn't work, $\frac{f(x)}{x} \neq f(1)$ except if $f(x) = x$ (which is obviously useless)
When you divide the polynomial like you did here, you don't divide the argument given to the function, you divide its result.

But I'm not doing $\frac{f(x)}{x} \neq f(1)$

I'm doing $f\left(\frac{x}{x}\right) = f(1)$
• Jul 23rd 2010, 02:27 AM
Bacterius
But what you wrote is not correct, you can't divide the argument, only the result.
• Jul 23rd 2010, 02:37 AM
pickslides
$f(x)\div x \neq f\left(\frac{x}{x}\right)$
• Jul 23rd 2010, 02:41 AM
rainer
Quote:

Originally Posted by Bacterius
But what you wrote is not correct, you can't divide the argument, only the result.

Ohhhh yeah, now I get it. Newton's method it is then!