MHB Proving Function Polynomial in Coordinates is Differentiable Everywhere

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A function \( f:\mathbb{R}^n\rightarrow \mathbb{R} \) that is polynomial in its coordinates can be shown to be differentiable everywhere by applying the chain rule in conjunction with properties of differentiable functions. The projections \( \pi_i \) are differentiable, and the sum and product of differentiable functions retain differentiability. By expressing \( f \) as a sum of products of differentiable functions, the chain rule can be utilized to demonstrate that compositions of these functions remain differentiable. An example illustrates this by showing that if \( g(t) = t^3 \) is differentiable and composed with a projection, the resulting function is also differentiable. Thus, the polynomial function \( f \) is differentiable everywhere in \( \mathbb{R}^n \).
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The question is:
Using the chain rule to prove that a function $f:\mathbb{R}^n\rightarrow \mathbb{R}$ which is polynomial in the coordinates is differentiable everywhere.
(The chain rule is for the use under function composition circumstances, how to apply it here to prove that the function $f$ which is polynomial in the coordinates is differentiable everywhere?)
 
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ianchenmu said:
(The chain rule is for the use under function composition circumstances, how to apply it here to prove that the function $f$ which is polynomial in the coordinates is differentiable everywhere?)

We can write:

$f:\mathbb{R}^n\to \mathbb{R},\quad f(x_1,\ldots,x_n)=\displaystyle\sum_{i_1\geq 0,\ldots,i_n\geq 0}a_{i_1,\ldots,i_n}x_1^{i_1}\ldots x_n^{i_n}$

with finitely many nonnull real coefficientes $a_{i_1,\ldots,i_n}$. Now, use the following properties:$(1)\;$ The projections $\pi_i:\mathbb{R}^n\to\mathbb{R},\quad \pi_i(x_1,\ldots,x_n)=x_i$ are differentable on $\mathbb{R}^n$.

$(2)\;$ Every constant funcion is differentiable on $\mathbb{R}^n$.

$(3)\;$ The product of two differentiable functions on $\mathbb{R}^n$ is differentiable on $\mathbb{R}^n$.

$(4)\;$ The sum of two differentiable functions on $\mathbb{R}^n$ is differentiable on $\mathbb{R}^n$.
 
Fernando Revilla said:
We can write:

$f:\mathbb{R}^n\to \mathbb{R},\quad f(x_1,\ldots,x_n)=\displaystyle\sum_{i_1\geq 0,\ldots,i_n\geq 0}a_{i_1,\ldots,i_n}x_1^{i_1}\ldots x_n^{i_n}$

with finitely many nonnull real coefficientes $a_{i_1,\ldots,i_n}$. Now, use the following properties:$(1)\;$ The projections $\pi_i:\mathbb{R}^n\to\mathbb{R},\quad \pi_i(x_1,\ldots,x_n)=x_i$ are differentable on $\mathbb{R}^n$.

$(2)\;$ Every constant funcion is differentiable on $\mathbb{R}^n$.

$(3)\;$ The product of two differentiable functions on $\mathbb{R}^n$ is differentiable on $\mathbb{R}^n$.

$(4)\;$ The sum of two differentiable functions on $\mathbb{R}^n$ is differentiable on $\mathbb{R}^n$.

But where you used the chain rule? I mean, how to use the chain rule to prove that $f$ is differentiable everywhere?
 
ianchenmu said:
But where you used the chain rule? I mean, how to use the chain rule to prove that $f$ is differentiable everywhere?

Perhaps an example will provide you the adequate outline. Consider $\pi_7(x_1,\ldots,x_n)=x_7$ and $g:\mathbb{R}\to \mathbb{R}$ given by $g(t)=t^3$. As $\pi_7$ and $g$ are differentiable, $(g\circ \pi_7)[(x_1,\ldots,x_n)]=x_7^3$ is also differentiable, etc, etc,...
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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