Expanding $(1 - x + 2y)^3$ in powers of $x-1$ and $y-2$ with a Taylor series

Question

I would like to do this.

I observe that I can write $$f(x,y) = (1 - x + 2y)^3 = (2(y-2) - (x-1) + 4)^3.$$ It's easy to do this via algebra directly.

However, I'm asked to do it by computing the Taylor series. Is it correct to say that I could expand the function $$f(a,b) = (2a - b + 4)^3$$ (where $a = y-2$ and $b = x-1$) around the point $(2,1)$ to the third-order term like so: $$f(a,b) = f(2,1) + f_a(2,1)a + f_b(2,1)b + \frac{1}{2}f_{aa}(2,1)a^2 + \frac{1}{2}f_{bb}(2,1)b^2 + f_{ab}(2,1)ab + \cdots$$ (where I've neglected to write the third order term for simplicity's sake)?

Or am I missing the point of the question / is my approach incorrect?

Answer 1

I don't understand what you're doing, you seem to be using $f$ with two different meanings.

The function is a third degree polynomial, therefore the fourth order partials are constantly zero and the Taylor series of $f$ of order $3$ (or above) coincides with $f$ (i.e. the remainder is $0$).

Thus, for all $(x,y)\in \mathbb R^2$, $f(x,y)=f((1,2)+(x-1, y-2))=(-(x-1)+2(y-2)+4)^3$ and it's easy to get the taylor expansion with powers of $(x-1)$ and $(y-2)$.

If you must use the formula for Taylor series, in the notation at the bottom of the linked section set $\mathrm x=(\mathrm a+(x-1, y-2))$ where $\mathrm a=(1,2)$ to get

$$f(x,y)=f(1,2)+\dfrac{\partial f}{\partial x}(1,2)(x-1)+\dfrac{\partial f}{\partial y}(1,2)(y-2)+\dfrac 1{2!}\left[\dfrac{\partial ^2f}{\partial x^2}(1,2)(x-1)^2+2\dfrac{\partial ^2f}{\partial y\partial x}(1,2)(x-1)(y-2)+\dfrac{\partial ^2f}{\partial y^2}(1,2)(y-2)^2\right]+R_2(x-1, y-2),$$ where $R_2$ is a function such that $\lim \limits_{(x,y)\to (0,0)}\left[\dfrac{R_2(x,y)}{\Vert (x,y)\Vert^2}\right]=0$.

Answer 2

Since the given function $f$ is a polynomial of degree $3$ in $x$ and $y\>$ its Taylor series of order $3$ set up at any point $(a,b)$ gives the exact value of $f$ for all increments $(X,Y)$. Now the third order Taylor expansion in general looks like $$\eqalign{f(a+X,b+Y)&=f(a,b)+(f_x X+f_y Y)+{1\over2}(f_{xx}X^2+2f_{xy}XY+f_{yy}Y^2)\cr &\qquad+{1\over6}(f_{xxx}X^3+3f_{xxy}X^2Y+3f_{xyy}XY^2+f_{yyy}Y^3)+ R\ ,\cr}\tag{1}$$ where the partial derivatives of $f$ have to be taken at $(a,b)$, whence are constants in $(1)$, and $R=0$ in the case at hand.

To solve your problem (the hard way!) you have to put $$(a,b):=(1,2),\quad X:=x-1,\quad Y:=y-2$$ in the above formula. You will then obtain $$f(x,y)=f(1,2)+f_x(1,2) (x-1)+f_y(1,2) (y-2)+{1\over2}\bigl(f_{xx}(1,2)(x-1)^2+2f_{xy}(1,2)(x-1)(y-2)+f_{yy}(1,2)(y-2)^2\bigr)+\ldots\ ,$$ where I have omitted the third degree terms.

Answer 3

Denoting $s=2a-b+4$ for conciseness, the successive partial derivatives are: $$s^3$$ $$\color{blue}3.2.s^2,\color{blue}3.\bar1.s^2$$ $$\color{blue}{3.2}.2.2.s,\color{blue}{3.2}.2.\bar1.s,\color{blue}{3.2}.\bar1.\bar1.s$$ $$\color{blue}{3.2.1}.2.2.2,\color{blue}{3.2.1}.2.2.\bar1,\color{blue}{3.2.1}.2.\bar1.\bar1,\color{blue}{3.2.1}.\bar1.\bar1.\bar1$$ Evaluated at $(0,0)$, $s=4$: $$64$$ $$96,-48$$ $$96,-48,24$$ $$48,-24,12,-6$$ Hence the Taylor development $$64+96a-48b+\frac12(96a^2-2.48ab+24b^2)+\frac16(48a^3-3.24a^2b+3.12ab^2-6b^3).$$ $$=64+96a-48b+48a^2-48ab+12b^2+8a^3-12a^2b+6ab^2-b^3,$$ as you can obtain by direct expansion of $s^3$.