Taylor series for 2 dimensions second order terms

Question

We were learning about Taylor series in our introductory lecture to Mathematics for Chemistry and were required to understand applying Taylor series to 2 dimensional problems, where $x$ and $t$ are both inputs, not $t$ being the output of $x$.

I could understand the latter using the following with reasonable explanation:

$$f(x)\approx f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+ \frac{f'''(x_0)}{3!}(x-x_0)^3+\cdots.$$

But the following didn't make so much sense.

$$f(x,t)=f(a,b)+f_x(a,b)\cdot (x-a)+f_t(a,b)\cdot (t-b)+\frac{1}{2}f_{xx}(a,b)\cdot (x-a)^2+$$ $$\frac{1}{2}f_{xt}(a,b)\cdot (x-a)(t-b)+f_{tx}(a,b)\cdot (x-a)(t-b)+ \frac{1}{2}f_{tt}(a,b)\cdot (t-b)^2+\cdots$$

The terms

$$f_x(a,b)\cdot (x-a)+f_t(a,b)\cdot (t-b)$$

made sense as each "axes of input" contributed a certain amount to the graph and can be assessed independently by keeping one variable constant.

However, the terms

$$\frac{1}{2}f_{xt}(a,b)\cdot (x-a)(t-b)+f_{tx}(a,b)\cdot (x-a)(t-b)+ \frac{1}{2}f_{tt}(a,b)\cdot (t-b)^2+\cdots$$

made less sense.

I understand using the 1-dimensional Taylor series that this can be written as

$$\frac{1}{2}\left[{}f_{xt}(a,b)\cdot (x-a)(t-b)+2f_{tx}(a,b)\cdot (x-a)(t-b)+f_{tt}(a,b)\cdot (t-b)^2\right]+\cdots$$

where the $\frac{1}{2}$ comes from the factorial as above.

I can't seem to understand from any material how these terms above can be found. More specifically, I can't see why we need a $2f_{tx}$ term in this expansion. I know that this set of terms looks similar enough to a $(x+y)^2$ expansion i.e. $x^2 + 2xy + y^2$ but I don't feel that helped me very much.

Why is there a $2f_{tx}$ term? How can these terms above be found?

Answer 1

There seems to be a mistake in the formula you've been given. The coefficient of the $f_{tx}$ term should be the same as the coefficient of the other second-order terms. Under the usual hypotheses $f_{tx}=f_{xt}$ and one usually sees the formula written with only one of these terms, and a coefficient of $2$. Perhaps that's where the confusion came from.

These slides may help answer your questions about how the terms in the formula arise.

Answer 2

You can understand better, I thin, using the notation of differential operators: rewrite the 2-varianles formula as \begin{align} f(x,t)&=f(a,b)+\frac{\partial f }{\partial x}\bigg|_{(a,b)}\cdot (x-a)+\frac{\partial f }{\partial x}\bigg|_{(a,b)} (t-b)\\ &\phantom{=}+\frac{1}{2}\frac{\partial^2 f }{\partial x^2}(a,b)\cdot (x-a)^2+\frac{\partial^2 f }{\partial x\partial t}(a,b)\cdot (x-a)(t-b)+ \frac{1}{2}\frac{\partial^2 f }{\partial t^2}(a,b)\cdot (t-b)^2+\cdots \\ &=f(a,b)+\frac{\partial f }{\partial x}\bigg|_{(a,b)}(x-a)+\frac{\partial f }{\partial x}\bigg|_{(a,b)}(t-b)+\frac{1}{2}\left(\frac{\partial f }{\partial x}\bigg|_{(a,b)} (x-a)+\frac{\partial f }{\partial x}\bigg|_{(a,b)} (t-b)\right)^{\mkern-5mu(2)}+\cdots , \end{align} where the symbolic square is defined as \begin{multline}\left(\frac{\partial f }{\partial x}\bigg|_{(a,b)} (x-a)+\frac{\partial f }{\partial x}\bigg|_{(a,b)} (t-b)\right)^{\mkern-5mu(2)}= \\\frac{\partial^2 f}{\partial x^2}\bigg|_{(a,b)}(x-a)^2+2\frac{\partial^2 f }{\partial x\partial t}\bigg|_{(a,b)}(x-a)(t-b)+\frac{\partial^2 f}{\partial t^2}\bigg|_{(a,b)}(t-b)^2 \end{multline} You can guess the 3rd order term in Taylor's formula would be \begin{align}&\frac1{3!}\left(\frac{\partial f }{\partial x}\bigg|_{(a,b)} (x-a)+\frac{\partial f }{\partial x}\bigg|_{(a,b)} (t-b)\right)^{\mkern-5mu(3)}= \\&\frac1{3!}\left(\frac{\partial^3 f}{\partial x^3}\bigg|_{(a,b)}(x-a)^3+3\frac{\partial^3 f }{\partial x^2\partial t}\bigg|_{(a,b)}(x-a)^2(t-b)+3\frac{\partial^3 f }{\partial x\partial t^2}\bigg|_{(a,b)}(x-a)(t-b)^2+\frac{\partial^3 f}{\partial t^3}\bigg|_{(a,b)}(t-b)^3\right) \end{align}