Question about Taylor Expansions

[cpburris]

I was working out a problem requiring a taylor expansion of $\sqrt {1+x^2}$ (about $x=0$). I needed to go out to the 5th term in the expansion, which, while not difficult, was long and annoying as the $x^2$ necessitated chain rules and product rules when taking the derivatives and the number of terms in each derivative just keeps increasing. I was wondering if a function which is strictly dependent on $x^n$ ($x^2$ terms but no terms linear in $x$ for example), where $x$ is the small parameter, whether it is permissible to perform a change of variable, say $y=x^n$, and perform the taylor expansion with respect to $y$ (as $x$ is small, then of course $x^n$ is small). I don't see any problem with doing that, but I wanted to make sure there isn't something I am missing.

 

[fresh_42]

I don't see a problem either. You basically define a new function and consider that. But why would you want to do this? You evaluate at $x=0$ anyway which makes many terms automatically zero. And you cannot take the $y-$result and re-substitute $y=x^n$ in the polynomial. This is wrong. In the Taylor series we have terms $\left. \dfrac{df(x^n)}{dx}\right|_{x=0}=\left. \dfrac{df(y)}{dx}\right|_{x=0} = \left. \dfrac{df(y)}{dy}\right|_{y=y(0)}\cdot \left. \dfrac{dy(x)}{dx}\right|_{x=0}$ and you cannot pretend as if the second factor equals $1$. And the higher the degree, the more factors and sums you get.

It might work in your example with the many zeros, but you cannot say it in general. You can consider the new function, but if you change the variable again, the derivatives are different: differentiation and substitution do not commute.

 

[mathman]

Since $\sqrt{1+x^2}=(1+x^2)^\frac{1}{2}$, binomial expansion can be used $(1+x^2)^\frac{1}{2}=1+\frac{1}{2}x^2+\frac{(1/2)(-1/2)}{2!}x^4+⋯+\frac{(1/2)(-1/2)...(-(2k-3))/2)}{k!})x^{2k}+⋯$.