I am reading an article from 1936 with something that looks like an easy way to solve Riccati equations with variable coefficients as nice polynomials.The link is : http://www.jstor.org/stable/2301168?seq=2 . The problem is that at page 474 the author talks about some power expansion of square root of an even degree polynomial and even though he gives two examples I don't really understand what is he really doing. Could someone explain me or at least give me a hint on this?
So, you have a polynomial P(x) which is of even degree but not necessarily in a nice form (you can't make a square ) . Now you take the square root of this polynomial and you try to approximate it with a decreasing power expansion.This is what I don't get, how do you do that (and what results let you do that). For example,
$$
P(x) = x^{6}+8x^{4}-x^{3}+7x-2,
$$
The expansion(or in his words"the polynomial part of the expansion in series of descending powers of x") of the $\sqrt{P(x)}$ is given in the article as $x^{3}+7x-\frac{1}{2}$. Does anyone know how did he got there? Thank you.
Best Answer
By "descending powers of $x$", it means an expansion of the form
$$ \sum_{i=n}^{\infty} a_ix^{-i} $$ where $n$ can be any positive or negative integer (if the polynomial was of odd degree, then it would require $\sum_{i=n}^{\infty} a_ix^{\frac{1}{2}-i}$). So if we look at the polynomial again, you have
$$ \sqrt{P(x)} = \sqrt{x^6+8x^4−x^3+7x−2} = x^3\sqrt{1+8x^{-2}-x^{-3}+7x^{-5}-2x^{-6}} $$ We can now perform a taylor expansion on $X=8x^{-2}-x^{-3}+7x^{-5}-2x^{-6}$, so we have $\sqrt{1+X}$.
$$ \sqrt{1+X} \approx 1+\frac{X}{2}+O(X^2) $$ This gives
$$ \sqrt{P(x)} \approx x^3+4x-\frac{1}{2} + O(x^{-1}) $$ Note that I believe that either you've mis-copied the approximation given in the article, or the article had a typo. It should not be $x^3+7x-\frac{1}{2}$, as the square of that is
$$ (x^3+7x-\frac{1}{2})^2 = x^6 + 14x^4-x^3+49x^2-7x+\frac{1}{4} $$ which is clearly not a good approximation to $P(x)$. By comparison
$$ (x^3+4x-\frac{1}{2})^2 = x^6 + 8x^4-x^3+16x^2-4x+\frac{1}{4} $$ which agrees with $P(x)$ in the first three terms, and thus is a good approximation for large $x$.