I have a tutorial question on Taylor series.
$$f(x,y,z) = e^{(\sqrt{x y}+\sqrt{x z}+\sqrt{z y})}$$
I managed to get the Taylor series at (1,1,1)
Is it possible to get the expansion at (0,0,0)? Because at (0,0,0) the derivative is 0/0 which is confusing…how do we evaluate 0/0 in Taylor series?
Thanks!
Best Answer
As copper.hat said in the comments, it is not possible to do a Taylor series in $x$, $y$ or $z$ around the point $(0,0,0)$ because the function has no derivative in that point.
What you can do in this point, however, is to do a Taylor expansion of just the exponential function, and insert the argument; this will give you an expansion in square roots: $$\begin{aligned} \mathrm e^{\sqrt{xy}+\sqrt{xz}+\sqrt{yz}} &= \sum_{n=0}^\infty \frac{1}{n!}(\sqrt{xy}+\sqrt{xz}+\sqrt{yz})^n\\ &= 1 + \sqrt{xy}+\sqrt{xz}+\sqrt{yz}\\ &\phantom{= 1} + \frac12 xy+\frac12 xz+\frac12 yz + \sqrt{xy}\sqrt{xz}+\sqrt{xy}\sqrt{yz}+\sqrt{xz}\sqrt{yz} + \dots \end{aligned}$$