Answer to Question #104542 in Calculus for Michael james

Question

Solve the Taylor's expression
F(x,y)=e^xcosy. (π,0)

Answer 1

We find the partial derivatives of $f(x,y)= e^x cos(y)$

The partial derivatives of with respect to 'x' are $f{_x}= e^x cos(y)$

$f{_{xx}}= e^xcos(y)$

$f{_{xxx}}=e^x cos(y)......$

$\therefore f({{\pi},{0}})=f{_{x}}({\pi},0)=f{_{xx}}({\pi},0)....= e^{\pi}cos(0) = e^{\pi}$

The partial derivatives of $f(x,y)$

with respect to y are $f_{{y}}=e^x(-sin(y))$

$f_{{yy}}=e^x(-cos(y))$

$f_{{yyy}}=e^x(sin(y))....$

$f_{{y}}(\pi,0)=0=f_{{yyy}}(\pi,0)=f_{{yyyyy}}(\pi,0)....$

And $f_{{yy}}=-1$

$f_{{yyyy}}=1$ ....

The partial derivatives of $\frac{\partial 2f}{\partial x\partial y}= 0$

Substituting these values in Taylor's

series we get

$f(x,y) = e^{{\pi}}+(x-\pi)e^{{\pi}}(1)+(x-\pi)^2e^{{\pi}}+0+y^2(-e^{{\pi}})+....$