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I am in confidence with Taylor expansion of function $f\colon R \to R$, but I when my professor started to use higher order derivatives and multivariate Taylor expansion of $f\colon R^n \to R$ and $f\colon R^n \to R^m$ I felt lost.

Can somean explain to me from scratch multivariate Taylor?

In particular I don't understand the notation $$ f(x+h) = \sum_{k=0}^p \frac{1}{k!} f^{(k)}(x)[h,...,h] + O(h^{p+1}) $$ Why we need the k-linear form $ \frac{1}{k!} f^{(k)}(x)[h,...,h]$? This k-linear form is the derivative or the derivative is only $f^{(k)}(x)$?

I'm quite lost. Thank you.

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I think that s/he is using multi-indices. see… – Baby Dragon Mar 15 '13 at 17:06
No, it's not using multi-indices... – UNKNOWN Mar 15 '13 at 17:15

One can think about Taylor's theorem in calculus as applying in the following cases:

  1. Scalar-valued functions of a scalar variable, i.e. $f : \mathbb{R} \rightarrow \mathbb{R}$
  2. Vector-valued functions of a scalar variable, i.e. $f : \mathbb{R} \rightarrow \mathbb{R}^n$
  3. Scalar-valued functions of a vector variable, i.e. $f : \mathbb{R}^n \rightarrow \mathbb{R}$
  4. Vector-valued functions of a vector variable, i.e. $f : \mathbb{R}^n \rightarrow \mathbb{R}^m$

All of these can be derived & proven based on nothing more than integration by parts (the last one needs to be developed in a banach space & the third one is more commonly reduced to the first one which is just a shorthand for re-proving it via integration by parts) if you set things up correctly (as is done in Lang's Undergraduate, Real & Functional Analysis books) & so your main obstacle here is formalism - this is no small obstacle as we'll see below.

Now I'm not sure if your expression for Taylor's formula is map 3 or map 4, one would think it is map 3 since you used the word "linear form" which is standard parlance for maps from vector spaces into a field but you did ask about maps of the form $f : \mathbb{R}^n \rightarrow \mathbb{R}^m$ - are you sure you are differentiating these kinds of maps because they add a whole world of complexity compared to the first 3?

If you're asking about maps of the form in 3 then some intuition is given in this video & some examples & a proof are given in this video. After these you should have enough of a grasp of what's going on & if you focus on developing the formalism properly you should be able to prove it yourself in more general spaces.

If you're actually asking about map 4 then you may be used to the definition of the derivative of the last map as $f : \mathbb{R}^n \rightarrow \mathbb{R}^m$ as something like $\mathcal{f'} : \mathbb{R}^n \rightarrow \mathcal{L}(\mathbb{R}^n,\mathbb{R}^m)$ satisfying all the conditions a differentiable map does, well it's second derivative is defined similarly using a map of the form $f'': \mathbb{R}^n \rightarrow \mathcal{L}(\mathbb{R}^n,\mathcal{L}(\mathbb{R}^n,\mathbb{R}^m))$ & so on, however you see no such entity as a "k-linear form" in any of this & that's because there is a theorem which allows one to think of maps like the second derivative above in terms of multilinear maps & so one can re-cast the theory using multilinear maps which eases the development & allows for nice proofs etc... but without this being explained it might appear odd to randomly start pulling out multilinear maps.

In any case the derivative is a linear map by definition & so that is why you're coming across the word linear, but since you didn't put subscripts on the $[h,...,h]$ I'm not sure how deep I can go, because I see two possibilities here so if the above isn't enough of an explanation as it stands just let me know.

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This is a nice answer. Your link… does not seem to be working. – Baby Dragon Mar 15 '13 at 19:03
Thanks man, the page works for me so if you can't access it just type "banach isomorphism multilinear", & maybe the word "toplinear" along with it, into google & you should get it - if not I'll type it all out explicitly for you. – sponsoredwalk Mar 15 '13 at 19:17
Now it works. Go figure. – Baby Dragon Mar 15 '13 at 19:28
Upvote for linking to videos. – elaRosca Oct 12 '13 at 8:31

For 3 variables: $$f(x,y,z)=f(x_0,y_0,z_0)$$ $$+\frac{\partial f_0}{\partial x}(x-x_0)+\frac{\partial f_0}{\partial y}(y-y_0)+\frac{\partial f_0}{\partial z}(z-z_0)\quad \Rightarrow Order 1$$ $$+\frac{1}{2} \bigg(\frac{\partial^2 f_0}{\partial x^2}(x-x_0)^2+\frac{\partial^2 f_0}{\partial y^2}(y-y_0)^2+\frac{\partial^2 f_0}{\partial z^2}(z-z_0)^2+2\frac{\partial^2 f_0}{\partial x\partial y}(x-x_0)(y-y_0) $$ $$+2\frac{\partial^2 f_0}{\partial x\partial z}(x-x_0)(z-z_0)+2\frac{\partial^2 f_0}{\partial z\partial y}(z-z_0)(y-y_0)\bigg)\quad \Rightarrow Order 2$$ And it goes like this to higher orders

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What do you mean with the notation $f_0$ ? – F.F. Nov 18 '13 at 9:12
$f_0$ means "evaluated at $(x_0,y_0,z_0)$ – Occupy Gezi Apr 6 at 15:34

It is enough to understand the case $f:\ {\mathbb R}^n\to {\mathbb R}$, and we expand $f$ at $x=0$. So we are interested in polynomials of low degree $d$ in the variables $x_1$, $\ldots$, $x_n$ that approximate $f$ in the neighborhood of $0\in{\mathbb R}^n$.

The best polynomial of degree $\leq0$ for this purpose is obviously the constant function $$j^0:\quad x\mapsto j^0(x)= f(0)\ .$$ In fact we have $$f(x)=j^0(x)+ r(x),\quad \lim_{x\to0} r(x)=0\ .$$ The best approximating polynomial of degree $\leq1$ is the function $$j^1:\ x\mapsto j^1(x)=f(0)+\nabla f(0)\cdot x\ ,$$ where $\nabla f(0):=\left({\partial f\over\partial x_1},\ldots,{\partial f\over\partial x_n}\right)_0$. In fact we have $$f(x)=j^1(x)+ r(x),\quad \lim_{x\to0} {|r(x)|\over |x|}=0\ .$$ Advancing to $d=2$ we have to be aware that the dimension of the space of homogeneous polynomials of degree $2$ is ${n(n+1)\over2}$, because we have to envisage all terms of the form $x_k^2$ and also mixed terms $x_ix_j$ with $i\ne j$. It follows that $j^2(x)$ will have the already known terms of degree $0$ and $1$ and "quadratically in $n$ many" terms of degree $2$. The latter are best arranged in a matricial way, so that one has a sum $\sum_{1\leq i\leq n,\ 1\leq j\leq n}a_{ij} x_ix_j$. The coefficients $a_{ij}$ will be related to the second partials of $f$ at $0$ in a particular way established in the proof of the general theorem. Going into the details one obtains $$j^2(x)=f(0)+\nabla f(0)\cdot x+{1\over2}\sum_{1\leq i\leq n,\ 1\leq j\leq n}\left({\partial^2 f\over\partial x_i\partial x_j}\right)_0 x_ix_j\ ,$$ and one can prove that $$f(x)=j^2(x)+ r(x),\quad \lim_{x\to0} {|r(x)|\over |x|^2}=0\ .$$ The formula you displayed is a systematic and condensed way to write this all up. E.g., in degree $3$ we shall obtain a triple sum of $n^3$ monomials $x_i x_j x_k$, each of them multiplied with the appropriate third partial derivative of $f$ at the origin.

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For 2 variables I believe the following expression to be correct. If it is incorrect then I should like to know!

$$f(x,y) = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \beta_{m,n} (x-x_0)^m (y-y_0)^n$$

$$\beta_{m,n} = \frac{1}{n! m!} \left[ \frac{\partial^{m+n}}{\partial x^m \partial y^n} \ f(x,y) \right]_{(x=x_0,y=y_0)}$$

Where the square bracket notation with coordinate subscript means evaluate at.

I previously derived the following formula, however I now suspect that it is not correct. Perhaps someone better than myself at proof by induction could verify this?

$$f(x,y) = \sum_{k=0}^{\infty} \frac{1}{k!} \sum_{j=0}^{k} \alpha_{k,j} (x-x_0)^{j} (y-y_0)^{k-j}$$

$$\alpha_{k,j} = \,^{k}C_{j} \left[ \frac{\partial^{k}}{\partial x^j \partial y^{k-j}} \ f(x,y) \right]_{(x=x_0,y=y_0)}$$

In order to prove the first form, I did the following:

$$\left[ f(x,y) \right]_{(x0,y0)} = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \beta_{m,n} \cdot 0^m \cdot 0^n = \beta_{0,0}$$

Differentiating partially with respect to x we find

$$\left[ \frac{\partial f(x,y)}{\partial x} \right]_{(x0,y0)} = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \left( \beta_{m,n} \cdot m \cdot 0^{m-1} \cdot 0^n + \left(\frac{\partial \beta_{m,n}}{\partial x} \right) \cdot0^m\cdot0^n\right) = \beta_{1,0} $$ where $\tfrac{\partial \beta_{m,n}}{\partial x} = 0$.

Continuing this process for all combinations of partial derivatives with respect to both x and y, we find that in general $$\beta_{m,n}=\frac{1}{m!n!}\cdot \left[ \frac{\partial f(x,y)}{\partial x} \right]_{(x=x0,y=y0)}$$

Unless I have made a mistake that is. (In which case please add a comment so I can correct my answer.)

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I have thought of a potential mistake here, what if x or y implicitly / explicitly depend on the other? ie: x=x(y), or y=y(x) ? Will this affect the result? – user3728501 Jul 12 '15 at 17:13

This is really a comment/question, but I am making it an answer so that it shows up.

For case 3 of sponsoredwalk's March 15, '13 answer, i.e., scalar-valued functions of a vector variable, is there a "simple" formula expressing the kth order term as an operation on kth order tensors? For example, 1st order term with inner product of x and gradient is in terms of 1st order tensor (vector), 2nd order term is quadratic form with x and Hessian, which is 2nd order tensor (matrix).

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