Third order Taylor polynomial of $f:(0, \infty) \times \mathbb{R} \to \mathbb{R}$ given by $f(x,y) = x^y$

Question

Let $f:(0, \infty) \times \mathbb{R} \to \mathbb{R}$ be given by $f(x,y) = x^y$.

How can one determine the Taylor polynomial of the third order of the function $f(x,y)$ in the points $(1,0)$?

I have

$$f_x = yx^{y-1}$$

$$f_{xx} = (y-1)yx^{y-2}$$

$$f_{xxx} = (y-2)(y-1)yx^{y-3}$$

$$f_y = x^y \ln(x)$$

$$f_{yy} = x^y\ln^2(x)$$

$$f_{yyy} = x^y\ln^3(x)$$

Then we do

$$f(1,0) = 1^0 = 1$$ $$f_x(1,0) = 0\cdot 1^{0-1}0 $$ $$f_{xx}(1,0) = (0-1)\cdot0\cdot1^{0-2} = 0$$ $$f_{xxx}(1,0) = (0-2)\cdot(0-1)\cdot0\cdot1^{0-3} = 0$$

$$f_y(1,0) = 1^0\cdot \ln(1) = 0$$ $$f_{yy}(1,0) = 1^0 \cdot \ln^2(1) = 0$$ $$f_{yyy}(1,0)= 1^0 \cdot \ln^3(1) = 0$$

Since we have only zeros expect for $f(1,0)$, can I just say that

$$T_{(1,0),3}(x,y) = 1$$

or is that wrong?

Answer 1

I think that you forgot to take into account the cross derivatives.

In my opinion, it should be something like $$x+x \log (x)(y-1)+ \frac{1}{2}x \log ^2(x)(y-1)^2 +\frac{1}{6} x\log ^3(x)(y-1)^3 +\cdots$$

Checking with $x=0.1$ and $y=0.9$, we get $0.125880$ while the exact value should be $0.125893$.

Edit

As @Christian Blatter pointed out, I misread the point around which has to be done the expansion and what I wrote was for $(0,1)$. Doing it around $(1,0)$ it becomes $$1+(x-1) y+(x-1)^2 \left(-\frac{y}{2}+\frac{y^2}{2}\right)+(x-1)^3 \left(\frac{y}{3}-\frac{y^2}{2}+\frac{y^3}{6}\right)+\cdots $$ A small grid of values (all miising values are strictly equal to $1$) $$\left( \begin{array}{cccc} x & y & \text{approximation} & \text{exact} \\ 0.8 & -0.2 & 1.0455040 & 1.0456396 \\ 0.8 & -0.1 & 1.0225080 & 1.0225652 \\ 0.8 & +0.1 & 0.9779720 & 0.9779328 \\ 0.8 & +0.2 & 0.9564160 & 0.9563525 \\ 0.9 & -0.2 & 1.0212880 & 1.0212957 \\ 0.9 & -0.1 & 1.0105885 & 1.0105918 \\ 0.9 & +0.1 & 0.9895215 & 0.9895193 \\ 0.9 & +0.2 & 0.9791520 & 0.9791484 \\ 1.1 & -0.2 & 0.9811120 & 0.9811185 \\ 1.1 & -0.1 & 0.9905115 & 0.9905143 \\ 1.1 & +0.1 & 1.0095785 & 1.0095766 \\ 1.1 & +0.2 & 1.0192480 & 1.0192449 \\ 1.2 & -0.2 & 0.9640960 & 0.9641925 \\ 1.2 & -0.1 & 0.9818920 & 0.9819330 \\ 1.2 & +0.1 & 1.0184280 & 1.0183994 \\ 1.2 & +0.2 & 1.0371840 & 1.0371373 \end{array} \right)$$

Answer 2

Since we want the third order Taylor polynomial of $f(x,y):=x^y$ at the point $(1,0)$ we consider the auxiliary function $$g(t):=f\bigl(1+t(x-1),t y\bigr)$$ and compute its third order Taylor polynomial with respect to $t$ at $t=0$. This gives $$T_0^3g(t)=g(0)+ g'(0) t +{g''(0)\over2} t^2+{g'''(0)\over6}t^3\ .\tag{1}$$ Wecompute $$g(0)=1,\quad g'(0)=0,\quad g''(0)=2(x-1)y,\quad g'''(0)=-3(x-1)^2 y\ .$$ Putting $t=1$ at the end in $(1)$ gives all terms of total degree $\leq3$ in $(x-1)$ and $y$, and leads to $$T_{(1,0)}^3f(x,y)=1+(x-1)y-{1\over2}(x-1)^2 y\ .$$ When $x=1.1$ and $y=-0.1$ we obtain $0.9905$, while $1.1^{-0.1}=0.9905142582$.