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I'm attempting to work out a problem of the form $a = bx + cx^2 + O(x^3)$, where I need to solve for $x$. To be honest, I don't really know how to work this out. Someone suggested that it can be done recursively, by writing

$x = \frac{a}{b} -\frac{c}{b}x^2 - O(x^3)$, squaring this and inserting it into the original equation. In that way, I could solve it recursively, was the idea. I've tried working this out on paper but without any success. It's based on solving this problem: enter image description here

If someone could give me a hint as to how this is done, I'd be very grateful. Oh, and it might be relevant for the solution: both $z$ and $\lambda^3n$ are small; close to zero.

user3183724
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1 Answers1

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With $a=\frac{\lambda^3n}{2s+1},\ b=2^{-3/2}$, the left hand side in the picture reads

$$a=z-b\,z^2-{\mathcal O}(z^3),$$

resembling the quadratic equation

$$\left(a+{\mathcal O}(z^3)\right)-z+b\,z^2=0.$$

Treating ${\mathcal O}(z^3)$ like a constant, we can isolate the occurrences of $z$ up to $z^2$ and find

$$z=\frac{1}{2b}\left(1-\sqrt{1-4\,b\left(a+{\mathcal O}(z^3)\right)}\right)=a+b\,a^2-\left(1+2\,a\,b+6\,(a\,b)^2\right)\cdot{\mathcal O}(z^3)+{\mathcal O}(a^3\cdot z^3).$$

Ignoring powers higher than $z^2$, you get

$$z\approx a+b\,a^2=\frac{\lambda^3n}{2s+1}+2^{-3/2}\left(\frac{\lambda^3n}{2s+1}\right)^2.$$

Nikolaj-K
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