Find the 3rd order Taylor polynomial of $f(x,y)=\log(1+x-y)$ at $a=(0,0)$
So I'll just list out the partials I computed:
$f_x=\frac{1}{x-y+1}$, $f_y=\frac{-1}{x-y+1}$, $f_{xx}=\frac{-1}{(x-y+1)^2}$. $f_{xy}=\frac{-1}{(x-y+1)^2}$, $f_{yy}=\frac{1}{(x-y+1)^2}$, $f_{xxx}=\frac{2}{(x-y+1)^3}$.$f_{xxy}=\frac{2}{(x-y+1)^3}$,$f_{yyx}=\frac{-2}{(x-y+1)^3}$, $f_{yyy}=\frac{2}{(x-y+1)^2}$
My Taylor polynomial is then $f(x,y)=f(0,0)+f_x(0,0)(x)+f_y(0,0)(y)+\frac{1}{2}(f_{xx}(0,0)(x)^2+2f_{xy}(0,0)(xy)+f_{yy}(0,0)(y^2))+\frac{1}{6}(f_{xxx}(0,0)(x)^3+3f_{xxy}(0,0)(x)^2(y)+3f_{yyx}(0,0)(y)^2(x)+f_{yyy}(0,0)(y)^3)$
Which using my partials I found is:
$f(x,y)=0+x-y+\frac{1}{2}(-x^2-2xy+y^2)+\frac{1}{6}(2x^3+6x^2y-6xy^2+2y^3)$
Is this correct?
Best Answer
Consider $\varphi(t) = \log(1+t)$ so that $\varphi^{(k)}(t) = (-1)^{k+1}(k-1)!(1+t)^{-k}.$ Hence, $\mathbf{T}_3\varphi(t) = t-\dfrac{t^2}{2}+\dfrac{t^3}{3}.$ From general theory of Taylor polynomials is known that $$\mathbf{T}_3f(x, y) = \mathbf{T}_3\varphi(x-y) = x - y - \dfrac{x^2}{2} + xy - \dfrac{y^2}{2} + \dfrac{x^3}{3}+x^2y+xy^2+\dfrac{y^3}{3}.$$ You solution is correct (just inneficient and clumsy; try your method to find the 20th degree and compare it with the method presented here). Q.E.D.