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Math Help - Is it possible to solve this kind of polynomial?

  1. #1
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    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.
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  2. #2
    Super Member Bacterius's Avatar
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    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 algebra´c way do to it, you can just use a Newton approximation and solve for x
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    Quote Originally Posted by Bacterius View Post
    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 algebra´c 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.

    In the meantime, what about this idea...

    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?
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  4. #4
    Super Member Bacterius's Avatar
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    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.
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    Quote Originally Posted by rainer View Post

    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
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    Quote Originally Posted by Bacterius View Post
    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)
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  7. #7
    Super Member Bacterius's Avatar
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    But what you wrote is not correct, you can't divide the argument, only the result.
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    f(x)\div x \neq f\left(\frac{x}{x}\right)
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  9. #9
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    Quote Originally Posted by Bacterius View Post
    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!
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