Square root of a Taylor expansion

Question

This is probably a silly question:

If the Taylor expansion at $t=1$ is given by $$ f(t)=c^2(t-1)^2+o(t-1)^2 $$ ($c$ is a constant)

how can I then approximate $$ \sqrt{f(t)}? $$

Answer 1

First notice that \begin{align} f(t) &= c^2 (1-t)^2 + \mathcal{O}( (1-t)^2) \\ &= c^2 (1-t)^2 \, \left(1 + \frac{1}{c^2} \, \mathcal{O}(1) \right) \end{align} A third term would be required in the expansion to obtain something like $$f(t) = c^2 (1-t)^2 + a_{1} (1-t)^4 + \mathcal{O}((1-t)^6)$$ such that $$f(t) = c^2 (1-t)^2 \, \left(1 + \frac{a_{1}}{c^2} \, (1-t)^2 + \mathcal{O}((1-t)^4) \right).$$

Now the square root follows: \begin{align} \sqrt{f(t)} &= c (1-t) \, \sqrt{1 + \frac{a_{1}}{c^2} (1-t)^2 + \mathcal{O}((1-t)^4)} \\ &= c (1-t) \, \left[ 1 + \frac{1}{2} \, \left(\frac{a_{1}}{c^2} (1-t)^2 + \mathcal{O}((1-t)^4) \right) - \frac{1}{8} \, \left( \frac{a_{1}}{c^2} (1-t)^2 + \mathcal{O}((1-t)^4) \right)^2 + \cdots \right] \\ &= c (1-t) \, \left( 1 + \frac{a_{1}}{2 \, c^2} \, (1-t)^2 + \mathcal{O}((1-t)^4) \right) \\ &= c (1-t) + \frac{a_{1}}{2 c} \, (1-t)^3 + \mathcal{O}((1-t)^5). \end{align}

Note: Use was made of $$\sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \cdots$$

Answer 2

There are many options and I'm not quite sure about what you mean by "approximate" $\sqrt{f(t)}$.

Assuming that you want to work on the Taylor expansion of, say, $$g(t)=\sqrt{f(t)}$$ at $t=1$, maybe the simplest way is to follow Taylor's formula, which leads us to calculate (and decide whether they exist, in fact) the values $$g(1),g'(1),g''(1),g'''(1),\ldots$$

The obvious detail is that $g(1)=\sqrt{f(1)}=\sqrt 0=0$. But even the existence of $g'(1)$ is not clear, since $f(1)=0$ and the function $x\mapsto\sqrt{x}$ is not differentiable at $t=0$. It might be the case that $g'(t)$ exists or not, so we have to use the definition of derivative.

The derivative $g'(1)$ exists iff the following limit exists and is finite $$\lim_{t\to1}\frac{g(t)-g(1)}{t-1},$$ that is $$\lim_{t\to 1}\frac{\sqrt{c^2(t-1)^2+o(t-1)^2}-0}{t-1}=\lim_{t\to1}|t-1|\cdot\frac{\sqrt{c^2+\frac{o(t-1)^2}{(t-1)^2}}}{t-1},$$ but by definition of "little $o$", it is $$\lim_{t\to1} \frac{o(t-1)^2}{(t-1)^2}=0,$$ and $\lim_{t\to1}\frac{|t-1|}{t-1}$ does not exist, since it is $1$ for $t\to1^+$ and $-1$ for $t\to1^-$ which gives $$g'(1^+)= \sqrt{c^2+0}=|c|$$ and $$g'(1^-)= -|c|$$.

So $g'(1)$ only exists for $c=0$ and is $0$, since the limit is taken for the product of $\frac{|t-1|}{t-1}$ (which is bounded) and $\sqrt{\frac{o(t-1)^2}{(t-1)^2}}$, which goes to $0$.

But with the information given, in that case you can only say that $$g(t)=\sqrt{o(t-1)^2}=o(|t-1|)$$.

And when $c\neq0$, you can only use Taylor for order $0$, that is $$g(t)=0+o(1)=o(1).$$

But, as I've said, I don't quite understand the whole idea of you question, so if you have something more specific in mind... just let me know.