Take the 2-minute tour ×
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.

I have this equation:

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

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 :)

share|improve this question
    
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
3  
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

1 Answer 1

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

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

share|improve this answer
    
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
1  
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

Your Answer

 
discard

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.