Approximation of non-analytic function

Question

I have a function which is of the form \begin{equation} f(x) = \frac{1 - x^{1/2} + x - x^{3/2} + \ldots}{1+x^{1/2} - x + x^{3/2} - \ldots}. \end{equation} Intuitively, I would assume that for small $x$, it holds \begin{equation} f(x) \approx \frac{1-x^{1/2}}{1+x^{1/2}} \end{equation} and then, furthermore, \begin{equation} f(x) \approx 1 - a x^{1/2} + \ldots \end{equation} where $a$ is some factor. My question is: How can I determine $a$ and the range of $x$ for which this approximation is valid? Obviously, I cannot use a Taylor approximation since $f$ is not analytic and the derivative diverges in the origin.

Let me point out that I am not so much interested in the specific example above, which I have just invented. Much rather, I would like to know what is the general theory and methods behind this type of fractional functions.

Answer 1

Another way to treat this particular sort of problem. It is an analytic function of, say, $z$, where $z=x^{1/2}$.

Answer 2

Note, that the numerator, is $$ \sum_{k=0}^\infty (-x^{1/2})^k = \frac{1}{1 + x^{1/2}}, \quad |x| < 1 $$ and the denominator equals $$ 2 -\sum_{k=0}^\infty (-x^{1/2})^k = 2 - \frac 1{1 + x^{1/2}} $$ Hence \begin{align*} f(x) &= \frac{\frac 1{1 + x^{1/2}}}{\frac{2(1 + x^{1/2}) - 1}{1 + x^{1/2}}}\\ &= \frac 1{2(1 + x^{1/2}) - 1}\\ &= \frac 1{1 + 2x^{1/2}}\\ &= \sum_{k=0}^\infty (-2x^{1/2})^k\\ &= 1 - 2x^{1/2} + 4x \mp \cdots \end{align*} That is, $a = 2$ seems like a good choice.

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Addendum: If we write $f$ as $g(\sqrt x)$ with $$ g(y) = \frac{1 - y + y^2 \mp}{1 + y - y^2 \pm } $$ then we can Taylor expand $g$ nicely, giving $$ g(y) = 1 - 2y + 4y^2 \mp $$ and hence $$ f(x) = 1 - 2x^{1/2} + 4x \mp $$ and the range of convergence can be obtained by that of $g$.

Answer 3

I would suggest $$\frac{1-\sqrt{x}}{1+\sqrt{x}}\approx 1-2\sqrt{x}$$

because of $(1+\sqrt{x})(1-2\sqrt{x})=1-\sqrt{x}-2x$