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I have this equation:


and I am told this:

Note that for a fixed amplitude $A$, the equation relating amplitude and $f$ can be converted to a 4th degree polynomial in $f$ by squaring both sides (to eliminate the square root), taking the reciprocal of both sides, and then subtracting $A$. Do this algebraic manipulation "by hand."

I got it down to a quadratic:

$$f^2 = \frac{F^4-F^4y^2}{y^2(F^2C^2-2F^2+1)}$$

Please help :)

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Joe, I hope you don't mind, I rolled back and then incorporated your edits into my edit. – Zev Chonoles Mar 6 '12 at 22:47
@ZevChonoles: Don't mind at all. – Joe Johnson 126 Mar 6 '12 at 22:48
Somehow, when you "got it down to a quadratic," the symbol $A$ disappeared altogether. How'd you do that? – Gerry Myerson Mar 6 '12 at 22:49

$$A=\frac{1}{\sqrt{\left(1-\left(\frac{f}{F}\right)^2\right)^2+c^2\left(\frac{f}{F}\right)^2}}$$ $$A^2 = \frac{1}{(1-(\frac{f}{F})^2)^2+c^2(\frac{f}{F})^2}$$ $$\frac{1}{A^2} = (1-(\frac{f}{F})^2)^2+c^2(\frac{f}{F})^2$$ $$\frac{1}{A^2} = 1+\frac{f^4}{F^4} - \frac{2f^2}{F^2} + \frac{c^2f^2}{F^2}$$ $$\frac{1}{A^2} = \frac{1}{F^4} f^4 + \left( \frac{c^2-2}{F^2}\right)f^2+1$$

where the last line gives a 4 degree polynomial in $f$. (Not sure where to subtract $A$!)

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Surely the "subtract $A$" was a mistake. – Gerry Myerson Mar 6 '12 at 23:10
I assume so, but perhaps I am missing some clever algebraic trick? – Aru Ray Mar 7 '12 at 3:30
I don't think you're missing anything. The problem statement is very sloppy, asking to convert an equation to a polynomial - strictly speaking, that makes no sense. It's not surprising that someone who could be that sloppy could also be sloppy about "subtract $A$". – Gerry Myerson Mar 7 '12 at 6:23

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