Non integral exponent for taylor expansion using sage

Question

This is my function

var('h,r')
f=r^2*arccos((r-h)/r)-(r-h)*sqrt(2*r*h-h^2)

taylor(f,h,0,3)

Result:

-1/5*sqrt(2)*h^(5/2)/sqrt(r) + 4/3*sqrt(2)*h^(3/2)*sqrt(r)

I expected an expression of the form ax^3+bx^2+cx+d but I got 5/2 and 3/2 as exponents for h. Why is that?

Answer 1

This is a plot from -1 to 1 of the function for $r=3$.

Notice, like sqrt, it doesn't continue beyond zero. Indeed, the complex plot makes it clear there is a branch cut in the usual implementation, which would indicate $h=0$ is a 'bad point', presumably then also not analytic and not capable of being expanded as a (Taylor) series. Apparently in this case (I'm too lazy to check), like sqrt, it isn't even a pole or it would have a Laurent expansion.

See this math.SX question for details which I found quite interesting - that there was a name for a 'branch point' as opposed to other singularity types was not emphasized when I learned complex analysis lo these many years ago.

Answer 2

Rewrite $$\cos ^{-1}\left(\frac{r-h}{r}\right)=\cos ^{-1}(1-x)=\sqrt{2} \sqrt{x}+\frac{x^{3/2}}{6 \sqrt{2}}+\frac{3 x^{5/2}}{80 \sqrt{2}}+O\left(x^{7/2}\right)$$ where $x=\frac hr$.

On the other hand, rewrite $$\sqrt{2rh-h^2}=\sqrt{2rh}\sqrt{1-\frac{h}{2r}}=\sqrt{2rh}\sqrt{1-y}=\sqrt{2rh}\Big(1-\frac{y}{2}-\frac{y^2}{8}-\frac{y^3}{16}+O\left(y^4\right)\Big)$$ where $y=\frac {h}{2r}$.

Replace $x$ and $y$ by their definition and you will see how appear powers $3/2$ and $5/2$.