2
$\begingroup$

Let $f(z,w)$ be an analytic function in two variables where $w=w(z)$ is dependent on $z$ ($z$ is the independent variable). Then $f(z,w)$ has a power series expansion centered at $(z_0,w(z_0))$

$f(z,w)=\displaystyle\sum_{k,n=0}^\infty a_{k,n}(z-z_0)^k(w-w(z_0))^n$.

I've seen a general Taylor expansion for two independent variables using partial derivatives, i.e.

$f(x,y)=f(x_0,y_0)+[f_x(x_0,y_0)(x-x_0)+f_y(y-y_0)]+\frac{1}{2!}[f_{xx}(x-x_0)^2+2f_{xy}(x-x_0)(y-y_0)+f_{yy}(y-y_0)^2]+...$

Can I interpret this to give the power series where one variable is dependent of the other? What would the partial derivatives be?

$\endgroup$

1 Answer 1

0
$\begingroup$

In my opinion it does not make sense searching: How to transform an analytic function $f(x,y)$ of two variables to an analytic function $f(x,g(x))$ of one variable? E.g., take $f(x,y)= x+y$.

On the other hand, one can always ask whether an equation of type $f(x,y)=0$ implicitly defines a function $y=y(x)$ or $x=x(y)$. The theorem on implicit functions states: If the partial derivative $f_y(x_0,y_0) \neq 0$ then a neighbourhoud and a function $y=g(x)$ exist, such that $f(x,y)=0$ in the neighbourhoud iff $y=g(x)$. In this case $\frac {dg}{dx}(x)=-\frac{f_x(x,g(x))}{f_y(x,g(x))}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.