TITLE: Pade approximant for the function $\sqrt{1+x}$
QUESTION [5 upvotes]: I'm doing the followiwng exercise:

The objective is to obtain an approximation for the square root of any given number using the expression
$$\sqrt{1+x}=f(x)\cdot\sqrt{1+g(x)}$$
where g(x) is an infinitesimal. If we choose $f(x)$ as an approximation of $\sqrt{1+x}$, then we can calculate $g(x)$:
$$g(x)=\frac{1+x}{f^2(x)}-1$$
$f(x)$ can be chosen as a rational function $p(x)/q(x)$, such that $p$ and $q$ have the same degree and it's Mclaurin series is equal to the Mclaurin series of the function $\sqrt{1+x}$ until some degree. Find a rational function $f(x):=p(x)/q(x)$, quotient of two linear polynomials, such that the McLaurin series of $p(x)-\sqrt{1+x}\cdot q(x)$ have the three first terms equal to $0$.

How can I do this? Has something to be with the Pade approximant?
Any hint would be really appreciated. Thanks for your time.

REPLY [2 votes]: You can work this out without MacLaurin, starting from
$$p^2(x)\approx(1+x)q^2(x).$$
Then with
$$p(x)=(1+ax)^2,q(x)=(1+bx)^2,$$
we have
$$1+2ax+a^2x^2\approx1+(2b+1)x+(b^2+2b)x^2+b^2x^3,$$
and we identify
$$2a=2b+1,\\a^2=b^2+2b$$
which gives the solution
$$\sqrt{1+x}\approx\frac{1+\frac34x}{1+\frac14x}.$$