# What is the complete Partial Derivatives approximation formula to calculate the value of a function?

We know the approximation formula using partial derivatives to calculate the value of a function with some variables $$x$$ and $$y$$, which can be extended to any number of variables to be

$$$$f(x, y) \approx \frac{\partial f(x, y)}{\partial x} \Delta x + \frac{\partial f(x, y)}{\partial y} \Delta y$$$$ What are the higher order terms involving second and $$n^{th}$$ order derivatives if you want the approximation to be better?

I recently learnt the formula and I don't know who derived it first or have any leads on what papers to read to get the full formula or proof so any help is appreciated.

Edit:

Assuming it is the Taylor's theorem, is the second order term for this approximation

$$$$\frac{1}{2!} (\frac{\partial^2 f(x, y)}{\partial x^2} {\Delta x}^2 + \frac{\partial^2 f(x, y)}{\partial y^2} {\Delta y}^2 + \frac{2\partial^2 f(x,y)}{\partial x \partial y}\Delta x \Delta y )$$$$ and there was a constant for $$0^{th}$$ order term $$f(x_{1}, y_{1})$$ where you plug in the value of $$x$$ and $$y$$ which I am unsure of, so the approximation up until the second order term should be

$$$$f(x, y) \approx f(x_{1}, y_{1}) + \frac{\partial f(x, y)}{\partial x} \Delta x + \frac{\partial f(x, y)}{\partial y} \Delta y + \frac{1}{2!} (\frac{\partial^2 f(x, y)}{\partial x^2} {\Delta x}^2 + \frac{\partial^2 f(x, y)}{\partial y^2} {\Delta y}^2 \\ + \frac{2\partial^2 f(x,y)}{\partial x \partial y}\Delta x \Delta y )$$$$

where $${\Delta x} = x - x_{1}, \Delta y = y - y_{1}$$ and you plug in $$x = x_{1}, y = y_{1}$$ after partial differentiation except in $$\Delta x, \Delta y$$ and the $$0^{th}$$ order term is a bit unclear, so I'm unsure if there's a mistake understanding the realization of $$x$$ and $$y$$ in there somewhere, or what the notation means when they define it in 2 variables/dimensions here with respect to $$f(\alpha)$$, perhaps they meant $$f(x_{1},y_{1})$$ as the $$0^{th}$$ order term? So we can use it to calculate properties from the equation at some point $$x_{1}$$, $$y_{1}$$ on $$f(x,y)$$, like closeness to some tangent plane or asymptotic analysis to get approximation with predetermined error value with higher order terms.

• See en.wikipedia.org/wiki/Taylor%27s_theorem under the "Generalizations" section. – Rahul Apr 16 at 10:52
• Thanks, there is a section for partial derivatives in the Generalizations section! But it says it is in one variable which is a bit confusing, but I assume it means it extends to all variables with partial derivatives but I see some terms which partial differentiate with 'x' then with 'y'? – devssh Apr 16 at 11:09
• It might be more illuminating if you absorb the last term into the parenthesized expression as $2{\partial^2f \over \partial x\partial y}\Delta x\Delta y$. – amd Apr 16 at 18:36

With the generalized Taylor series : $$f(x,y)= \sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\left(\frac{\partial^{j+k} f}{\partial x^j\partial y^k}\right)_{(x_1,y_1)} \frac{(x-x_1)^j(y-y_1)^k}{j! k!}$$ See the section "Taylor series in several variables" in : https://en.wikipedia.org/wiki/Taylor_series#Generalization

The first term with $$(j=0,k=0)$$ is $$f(x_1,y_1)$$.

The term with $$(j=1,k=0)$$ is $$\left(\frac{\partial f}{\partial x}\right)_{(x_1,y_1)} \frac{(x-x_1)}{1}$$.

The term with $$(j=0,k=1)$$ is $$\left(\frac{\partial f}{\partial y}\right)_{(x_1,y_1)} \frac{(y-y_1)}{1}$$.

The term with $$(j=1,k=1)$$ is $$\left(\frac{\partial^{2} f}{\partial x\partial y}\right)_{(x_1,y_1)} \frac{(x-x_1)(y-y_1)}{1}$$

The term with $$(j=2,k=0)$$ is $$\left(\frac{\partial^2 f}{\partial x^2}\right)_{(x_1,y_1)} \frac{(x-x_1)^2}{2}$$.

The term with $$(j=0,k=2)$$ is $$\left(\frac{\partial^2 f}{\partial y^2}\right)_{(x_1,y_1)} \frac{(y-y_1)^2}{2}$$.

and so on …

For example :

The term with $$(j=3,k=2)$$ is $$\left(\frac{\partial^5 f}{\partial x^3\partial y^2}\right)_{(x_1,y_1)} \frac{(x-x_1)^3(y-y_1)^2}{12}$$.

For an approximate, takes only a limited number of terms $$j\leq J$$ and $$k\leq K$$ instead of $$\infty$$.

• Thanks for validating that. I wasn't clear on the j!k! denominator as well but now it seems to check out – devssh Apr 16 at 14:04