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I am having trouble grokking why it is, assuming that the function is analytic everywhere (and many other assumptions that I am, no doubt, naively assuming), that this is true:


I am familiar with the one-variabled Taylor series, and intuitively feel why the 'linear' multivariable terms should be as they are.

In short, I ask for a proof of this equality. If possible, it would be nice to have an answer free of unnecessary compaction of notation (such as table of partial derivatives).

As a auxiliary question, I see a direct analogy with the first 2 terms $f(x,y)=f(x_0,y_0)+[f'_x(x_0,y_0)(x-x_0)+f'_y(x_0,y_0)(y-y_0)]$ and the total differential $f(x,y)-f(x_0,y_0)=\Delta f(x,y)=f'_x(x_0,y_0)\Delta x+f'_y(x_0,y_0)\Delta y$.

When $\Delta x $ and $\Delta y $ are not infinitesimally small, can I use the third term in the Taylor multivariable series to get closer to the real total differential?

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3 Answers 3

up vote 2 down vote accepted

Let $\phi(\boldsymbol{r})$ be a scalar field, and $\boldsymbol{a} \cdot \nabla \phi$ gives the directional derivative of $\phi$ in the direction of $a$. That is,

$$\boldsymbol{a} \cdot \nabla \phi(\boldsymbol{r}) = \lim_{t\to 0} \frac{\phi(\boldsymbol{r} + \boldsymbol{a} t) - \phi(\boldsymbol{r})}{t}$$

Now let's consider $\Phi(t) = \phi(\boldsymbol{r}_0 + \boldsymbol{a}t)$ for some finite $t$. Now, let's expand this in powers of $t$. This is a one-dimensional Taylor series.

$$\Phi(t) = \Phi(0) + \Phi'(0)t + \frac{1}{2!} \Phi''(0) t^2 + \ldots$$

To substitute back in $\Phi(t) = \phi(\boldsymbol{r}_0+\boldsymbol{a}t)$, we must compute derivatives of $\Phi$ in terms of $\phi$. Again, we resort to the basic definition of the derivative.

$$\Phi'(0) = \lim_{t\to 0} \frac{\phi(\boldsymbol{r}_0+\boldsymbol{a}t) - \phi(\boldsymbol{r}_0)}{t} = \boldsymbol{a} \cdot \nabla \phi(\boldsymbol{r})\Big|_{\boldsymbol{r}=\boldsymbol{r}_0}$$

And similarly for higher derivatives. This enables us to write,

$$\phi(\boldsymbol{r}_0+\boldsymbol{a}t) = \phi(\boldsymbol{r}_0) + [\boldsymbol{a} \cdot \nabla \phi(\boldsymbol{r})] \Big|_{\boldsymbol{r}=\boldsymbol{r}_0} t + \frac{1}{2!} [\boldsymbol{a} \cdot \nabla][\boldsymbol{a} \cdot \nabla]\phi(\boldsymbol{r}) \Big|_{\boldsymbol{r}=\boldsymbol{r}_0} t^2 + \ldots$$

It is not difficult to show that this form reproduces the form of the original question. Take $t=1$ and let $\boldsymbol{a} = (x-x_0, y-y_0)$ and $\boldsymbol{r}_0 = (x_0, y_0)$. Thus, we have built multivariate Taylor series from the well-established case of a single variable, just by use of the directional derivative.

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Concise and understandable- excellent answer, thank you. –  Alyosha Oct 27 '12 at 18:50

Let $u \in \mathbb{R}^m, \, h \in \mathbb{R}^m, \, t \in \mathbb{R},$ and $F(t)=f(u+th).$ Suppose that $F$ can be expanded into Taylor's series $$F(t)=\sum\limits_{n=0}^{\infty}{\frac{1}{n!}}F^{(n)}(0)t^n.\tag{*}$$ Taylor's expansion for $f$ can be obtained from $({}^{*})$ by differentiating $f$ and then put $t=1$.

For the case $n=2$ $$f(u+h)=\sum\limits_{n=0}^{\infty}{{\frac{1}{n!}}d^{n}f(u)},$$ where $u=(x, \, y)\quad h=(dx,\, dy),$ $$d^{n}f(u)=\sum\limits_{k=0}^{n}{\binom{n}{k}}\frac{\partial^n{f}}{\partial{x}^k {}\partial{y}^{n-k}}dx^kdy^{n-k}.$$

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Isn't that just an explanation for the single-variable expansion, or is a generalisation for the m-variable expansion? If so, why is $F''(t)=f''_{xx}(x,y) (x)^2+f''_{yx}(x,y)(x)(y)+f''_{yy}(x,y)(y)^2$? Most possibly a mundane thing, but this is what I'm having trouble with. I'm happy with your definition as a generalisation, but not how to apply it. –  Alyosha Oct 26 '12 at 18:28
But why is the last equality correct? –  Alyosha Oct 27 '12 at 10:10

Intuitively its quite clear: the multivariable analog of the first derivative is the gradient, which is exactly the second term evaluated at Ro. The second derivitave generalizes to the Hessian, which is best represented in matrix form, and that is your second term... The trick in deriving this is to define an s=R-R0 so g(s) = f(R-Ro) and use the chain rule/mean value theorem as usual.

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