Can We Simplify the Taylor Series Expansion for e^(f(x,y))?

[thatboi]

I know that for 1 variable, one can write $e^{f(x)} = \sum_{n = 0}^{\infty}\frac{(f(x))^n}{n!}$. In the case of 2-variables $f(x,y)$, I assume we cannot write $e^{f(x,y)} = \sum_{n = 0}^{\infty}\frac{(f(x,y))^n}{n!}$ right (because of how the Taylor series is defined for multiple variables)? Is there still a compact way of writing this expansion?

 

[andrewkirk]

For any variable $z$ it is the case that
$$e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$$
Substituting any expression for $z$ it remains valid, as long as the expression delivers a real number.
Substituting $f(x)$ for $z$ gives your first formula.
Substituting $f(x,y)$ for $z$ gives your second formula.
Both are valid.
The definition of a Taylor series for two variables is not relevant. In neither case is the formula a Taylor series for the function of $x$ or $x,y$.

 

[PeroK]

I know that for 1 variable, one can write $e^{f(x)} = \sum_{n = 0}^{\infty}\frac{(f(x))^n}{n!}$. In the case of 2-variables $f(x,y)$, I assume we cannot write $e^{f(x,y)} = \sum_{n = 0}^{\infty}\frac{(f(x,y))^n}{n!}$ right (because of how the Taylor series is defined for multiple variables)?

Those equations are perfectly valid. But, they may not represent the Taylor series for the given function. In general, it won't even be a power series in $x$:
$$e^{f(x)} = \sum_{n=0}^\infty \frac{f(x)^n}{n!} = 1 + f(x) + \frac{f(x)^2}{2!} + \dots$$Which is fine, but it's not necessarily the Taylor series for $e^{f(x)}$

In some cases, you do get the Taylor series. For example, if we let $z = x^2$, then (as above) we get the power series:
$$e^{x^2} = \sum_{n=0}^\infty \frac{x^{2n}}{n!}$$Which is the Taylor series for $e^{x^2}$.

 

[Charles Link]

Suggestion is that if $f(0,0)$ is not approximately zero, that you factor out $e^{f(0,0)}$ from the expression for the Taylor type series of $f(x,y)$.

Edit: It leaves you with $e^{f(0,0)}e^{\Delta}=e^{f(0,0)}(1+ \Delta+ \Delta^2/2+...)$, = maybe it will work...Edit 2=even $\Delta$ is complicated to second order in $\Delta x$ and $\Delta y$=I don't see an easy way to simplify it.

In any case, ignore=my calculus was rusty today...