What are upper indices $\partial x_1^{\alpha_1}$ used for in multivariate differentiability?

Question

What are upper indices $\partial x_1^{\alpha_1}$ used for in multivariate differentiability?

Because the definition writes:

$$\frac{\partial^{\alpha} f}{\partial x_1^{\alpha_1}...\partial x_n^{\alpha_n}}$$

$\alpha = \alpha_1 + ... + \alpha_n \leq k$

$\alpha_i$ are non-neg. integers.

I wonder when does the $\lt$ actually hold?

Or why aren't $\alpha_i=1$ always?

Answer 1

It is called multi-index notation. Usually, it is introduced before the Taylor expansion to make the notation easier on the eyes.

You may find this useful:

https://en.wikipedia.org/wiki/Multi-index_notation

Answer 2

Multi-indices seem to be a major issue today.

Maybe a simple example can help to illustrate the definition: let, e.g, for $n=3$ and $\alpha = (1,0,2)$. Then $$\frac{\partial^\alpha f}{\partial x_1^{\alpha_1}\partial x_2^{\alpha_2}\partial x_3^{\alpha_3}} = \frac{\partial^3 f}{\partial x_1\partial^2 x_3}$$

Also, $|\alpha| = \sum_i \alpha_i = 3$

But also $\alpha =(0,1,0) $ is a valid multi index, for which you simply get

$$ \frac{\partial^\alpha f}{\partial x_1^{\alpha_1}\partial x_2^{\alpha_2}\partial x_3^{\alpha_3}} = \frac{\partial f}{\partial x_2}$$ and in this case $|\alpha| = 1$.

Answer 3

While technically it might be possible to use partial derivative notation for functions of one variable, in practically every case we actually have in mind a function of more than one variable, like so:

$$ y = f(x_1, x_2, x_3,\ldots, x_n). $$

For such a function, as you know, it is possible to write partial derivatives of orders higher than the first derivative, for example, $$\frac{\partial^2 f}{\partial x_1^2},\quad \frac{\partial^2 f}{\partial x_3^2},\quad \frac{\partial^3 f}{\partial x_1^3},\quad \text{or}\quad \frac{\partial^4 f}{\partial x_n^4}.$$

You can also have higher-order derivatives with respect to more than one of the variables; for example, if $n = 6$ we can write $$\frac{\partial^4 f}{\partial x_1^2\partial x_3\partial x_6}.$$

If we allow the convention that $\partial x_i^1$ means the same thing as $\partial x_i$ in the "denominator" of a partial derivative, and that $\partial x_i^0$ means the same thing as writing nothing at all in the "denominator" (except what is written there using other notation), then $$\frac{\partial^4 f}{\partial x_1^2\partial x_3\partial x_6} = \frac{\partial^4 f}{\partial x_1^2\partial x_2^0\partial x_3^1 \partial x_4^0\partial x_5^0\partial x_6^1},$$ which is exactly what your formula would be if $n = 6,$ $\alpha_1 = 2,$ $\alpha_3 = \alpha_6 = 1,$ and $\alpha_2 = \alpha_4 = \alpha_5 = 0.$ Note that in this case $\alpha = 4 = 2 + 0 + 1 + 0 + 0 + 1 = \alpha_1 + \cdots + \alpha_6.$

So that's all the formula is. It's just a generic way of writing a partial derivative of whatever order you happen to require over a function of whatever number of variables you happen to be using, in which you might be differentiating with respect to each variable zero, one, or more times, differentiating over each variable as often as you need.

If we note that the fourth-order derivative in this example is just another way of writing a compound application of partial derivatives to the function, like this, $$\frac{\partial^4 f}{\partial x_1^2\partial x_3\partial x_6} = \frac{\partial}{\partial x_1}\frac{\partial}{\partial x_1} \frac{\partial}{\partial x_3}\frac{\partial}{\partial x_6}f, $$ then it should be obvious why the index in the "numerator" (written $\alpha$ in your formula) has to be the sum of the indices in the "denominator."

The reason why $\alpha \leq k$ must relate to some way in which $k$ is used earlier in the definition, so to make sense of that or to say when it is $<$ rather than $=$ we would have to look at the whole definition.