Taylor series for $\sqrt{1+\sqrt{x}}$ around $x=0$

Question

I want a Taylor series expansion of $\sqrt{1+\sqrt{x}}$ around $x=0$. I have two doubts here and these are as follows:

1. I see that the first and higher order derivatives of the above function blows up at $x=0$ and so I should not be able to find its Taylor expansion. But mathematica gives me the finite answer. How?
2. Let's say I want a series expansion upto 1st order in $x$. I guess mathematica does not put $x=0$ whenever it blows up. Taylor series of any function $f(x)$ around $x=0$ is given as: $$ f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+... $$ Here I see that each terms containing derivatives contribute something finite to the terms of order $x^{1/2}$ and $x$. How would I combine all these contribution in order to get the final coefficients of terms of order $x^{1/2}$ and $x$.

Answer 1

Do you seek a Taylor series in $x$, or one in $\sqrt{x}$? The former is impossible because (i) the first derivative diverges at $x=0$ (as would be expected from the small-$x$ approximation $1+\frac{x}{2\sqrt{x}}$) and (ii) the Taylor series in $\sqrt{x}$ naturally introduces half-odd-integer powers of $x$. Just as $\sqrt{1+y}\approx 1+\frac{y}{2}-\frac{y^2}{8}$, $\sqrt{1+\sqrt{x}}\approx 1+\frac{\sqrt{x}}{2}-\frac{x}{8}$. This begins a Taylor series, just one in $\sqrt{x}$, not in $x$.

Answer 2

I don't know the formal term for this, but this is what you are looking for:

Let $u^2 = x$

then: $\sqrt{1 + \sqrt{x}} = \sqrt{1+ u} $

This has taylor series: that can be computed (it's also a standard one so open to memorization too). The terms are:

$$1 + \frac{1}{2}u - \frac{1}{2!}\frac{1}{2^2}u^2 + \frac{1}{3!}\frac{3}{2^3} u^3 ... $$

So recall $x = \sqrt{u}$ (which side + or - you take is a subtle question and worth a good amount of thought). To yield:

$$f(x) = 1 + \frac{1}{2}\sqrt{x} - \frac{1}{2!}\frac{1}{2^2}x + \frac{1}{3!}\frac{3}{2^3} x\sqrt{x} - \frac{1}{4!}\frac{3\times 4}{2^4}x^2... $$

This is not a taylor series or even a traditional power series but it is the type of object you are looking for.

Answer 3

If you want a series in terms of $t=x^{1/2}$ and $x$, it is enough to consider a series in terms of $x^{1/2}$, as $x=t^2$ is just the square.

Now you can expand with the generalized binomal theorem

$$\sqrt{1+\sqrt{t^2}}=\sqrt{1+t}=1+\frac t2-\frac{t^2}8+\frac{x^3}{16}-\frac{5x^4}{128}\cdots \\=1+\frac{x^{1/2}}2-\frac{x}8+\frac{x^{3/2}}{16}-\frac{5x^2}{128}\cdots$$

This is not a Taylor series.