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Have been dealing with this equation for a while, but yet able to get a simple expression.

I have already done numerical solution and seems to be working just fine, but I would like to get a closed expression.

The problem is to find $x$ form the following:

$\frac{2LxB}{2Lx+\sqrt{b^2+x^2}} + 2b\sqrt{\frac{2LxB}{2Lx+\sqrt{b^2+x^2}}} = Ax^2$

What I tried:

$y^2 = \frac{2LxB}{2Lx+\sqrt{b^2+x^2}}$ So the equation above resumes to:

$y^2 + 2by = Ax^2$

Then found the solution for $y$:

$y = \sqrt{b^2+Ax^2} -b$.

Then, trying equating the above, with the definition of $y$:

$\frac{2LxB}{2Lx+\sqrt{b^2+x^2}} = (\sqrt{b^2+Ax^2} -b)^2$

However, the problem is the term inside the square root, since $A$ is present in only one of them.

Any trick or technique you may know to solve this?

Thanks folks!

kentropy
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  • what is $Lx$ ? $L \cdot x$ or $L_x$ ? – G Cab Jan 30 '20 at 19:12
  • These are all independent factors. Meaning $L\cdot x$. – kentropy Jan 30 '20 at 19:13
  • I cannot think of any viable analytic approach. Are you looking instead for an approximated solution ? – G Cab Jan 30 '20 at 21:14
  • Thanks for your replies. Well, I actually managed to get a numerical result using the Matlab solver, which worked fine for my immediate needs. But I would like to get a closed expression as it would look more elegant in my paper. Now, I'm just claiming the right x is given by the solution of such equation. But getting a closed expression would be great!. – kentropy Jan 30 '20 at 21:19
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    well, a closed formula is always attainable by just going on in isolating the square roots and squaring. The matter is that you will reach to an equation with x elevated at something not less than $6$ and containing most of the lower powers of it. So it should not be much useful. Probably you need instead an asymptotic approximation. – G Cab Jan 30 '20 at 21:39

1 Answers1

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The problem is that, beside the trivial $x=0$, you face a quintic polynomial $$-64 b^6 B^2 L^2+16 B^2 L^2 \left(L^2 \left(B-4 b^2\right)^2-4 (A+1) b^4\right)x-$$ $$8 \left(A B L^2 \left(A b^2 \left(B-4 b^2\right)+8 B \left(L^2 \left(B-4 b^2\right)+b^2\right)\right)\right)x^2+$$ $$A^2 \left(A^2 b^4+8 B L^2 \left(2 (A+2) b^2-B\right)+32 B L^4 \left(3 B-4 b^2\right)\right)x^3-$$ $$2 \left(A^3 \left(4 L^2-1\right) \left(A b^2+8 B L^2\right)\right)x^4+A^4 \left(1-4 L^2\right)^2 x^5=0$$

So, no hope for a closed form solution.

Claude Leibovici
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  • Hi. Thanks for your reply. Yes, I also got the trivial $x=0$, but of course, not useful for my needs. So did you get that huge polynomial just by expanding our all factors? (just wondering) – kentropy Jan 31 '20 at 19:39
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    @kennentropy. Do you know another way to do it ? The question is : are you looking for the first root only ? Is it small compared to $b$ ? May be, we could make some approximations of the solution. Just let me know. Cheers :-) – Claude Leibovici Feb 01 '20 at 07:12