# Thread: Expansion of taylor's formula in 3 variables?

1. ## Expansion of taylor's formula in 3 variables?

I am having difficulties figuring out how to write the full expansion of taylor's formula in 3 variables (x,y,t), for degrees 2, 3 and even higher degree terms.

It is for a master thesis project in computer vision.

The remainder term(s) is not necessary, but would be nice to have as well.

This is taylor's formula expansion for f(x) containing up to first degree terms only (with remainder $\displaystyle R_1$):

$\displaystyle f(b)=f(a)+f_x(a)(b-a) + R_1$

And, if we exchange some variables: $\displaystyle b=x+\Delta x$ and $\displaystyle a=x$, we have:

$\displaystyle f(x + \Delta x)=f(x)+f_x(x) \Delta x + R_1$

Using this notation instead, this is my taylor's formula expansion for f(x,y,t) up to first degree terms (partial derivatives in x, y, and t):

$\displaystyle f(x + \Delta x, y + \Delta y, t + \Delta t)=$
$\displaystyle f(x,y,t) + f_x(x,y,t) \Delta x + f_y(x,y,t) \Delta y + f_t(x,y,t) \Delta t + \text{remainder terms...}$

Now I need the expansion in all three variables, for higher degrees (second, third, perhaps four) How to do this? I have found functions for this, but these are using multi-index vectors for indexing, and unfortunately I have been unable to wrap my head around such notation.

As a side note, I currently expect mixed/crossed partials to be equal for function f(x,y,t) (which I do not know analytically).

Thanks,
Intel4004

2. Let $\displaystyle S$ be a non-trivial open convex subset of $\displaystyle \mathbb{R}^n$. Then suppose $\displaystyle f: S \mapsto \mathbb{R}$ is a $\displaystyle \mathcal{C}^k$ function.
Then,
$\displaystyle f(\bold{a}+\bold{h}) = \sum_{|\alpha|\leq k} \frac{\partial ^{\alpha} f(\bold{a})}{\alpha !}\bold{h}^{\alpha} + R_k(\bold{h})$.
Here we are using multi-index notation

3. Originally Posted by ThePerfectHacker
Let $\displaystyle S$ be a non-trivial open convex subset of $\displaystyle \mathbb{R}^n$. Then suppose $\displaystyle f: S \mapsto \mathbb{R}$ is a $\displaystyle \mathcal{C}^k$ function.
Then,
$\displaystyle f(\bold{a}+\bold{h}) = \sum_{|\alpha|\leq k} \frac{\partial ^{\alpha} f(\bold{a})}{\alpha !}\bold{h}^{\alpha} + R_k(\bold{h})$.
Here we are using multi-index notation
Exactly! That is the formula that I have tried to understand. Actually, if you could tell me how the summation is supposed to iterate with $\displaystyle \alpha$, I believe I could go a long way. That is the only thing that stops me.

As an iterator value in the summation, one uses the length/size $\displaystyle |\alpha|=(\alpha_1 + \cdots + \alpha_n)$ ...right? But what is the limit of the sum? What terms doees the sum produce?

Thanks.

And, btw, in any event it applies to where you are, merry xmas

Intel4004

4. The formula is familiar to the last one in this PDF:

http://www.math.washington.edu/~solomyak/TEACH/425/05/taylor.pdf

So as I understand, $\displaystyle \alpha_n$ in $\displaystyle \alpha$, is an integer, describing how many times we differentiate with respect to variable number $\displaystyle n$?

And since $\displaystyle f$ is a $\displaystyle \mathcal{C}^k$ function, it is k differentiable, right? Does that mean that all crossed partials (up to k'th order) can be regarded equal?

And, btw, what do you mean?

Like this?:

$\displaystyle f(\bold{a}+\bold{h}) = \sum_{|\alpha|\leq k} \left ( \frac{\partial ^{\alpha} f(\bold{a})}{\alpha !}\bold{h}^{\alpha} \right ) + R_k(\bold{h})$

or:

$\displaystyle f(\bold{a}+\bold{h}) = \sum_{|\alpha|\leq k} \left ( \frac{\partial ^{\alpha} f(\bold{a})}{\alpha !}\bold{h}^{\alpha} + R_k(\bold{h}) \right )$

5. The one I wrote is the correct one.

Suppose $\displaystyle f: \mathbb{R}^2\mapsto \mathbb{R}$ is $\displaystyle \mathcal{C}^3$ on a open convex set.
Then,
$\displaystyle f+\frac{f_x}{1!}+\frac{f_y}{1!}+\frac{f_{xx}}{2!}+ \frac{2f_{xy}}{2!}+\frac{f_{yy}}{2!}+\frac{f_{xxx} }{1!}+\frac{3f_{xyy}}{3!}+\frac{3f_{xxy}}{3!}+\fra c{f_{yyy}}{3!}$

The reason why we divide $\displaystyle f_{xxy}$ by $\displaystyle 3!$ is because of the three partial derivatives but there is a $\displaystyle 3$ in the numerator is because we can permute $\displaystyle f_{xxy}$ as $\displaystyle f_{xyx}, f_{yxx}$ and all of them are the same.

That was the Taylor's theorem for 2 variables. The same situation for 3 or more but it gets messier. The general pattern is the multi-index formula.

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# taylor expansion in three variables

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