Understanding linear transformation

[schniefen]

TL;DR

Understanding the linear transformation of $F(\mathbf{u})(t)=\mathbf{u}^{(n)}(t)+a_1\mathbf{u}^{(n-1)}(t)+...+a_n\mathbf{u}(t)$.

How can the function $F(\mathbf{u})(t)=\mathbf{u}^{(n)}(t)+a_1\mathbf{u}^{(n-1)}(t)+...+a_n\mathbf{u}(t)$, where $\mathbf{u}\in U=C^n(\mathbf{R})$ (i.e. the space of all $n$ times continuously differentiable functions on $\mathbf{R}$) be a linear transformation (from $U$) to $V=C(\mathbf{R})$? In $F$, isn't the term $\mathbf{u}^{(n)}(t)$ not eliminable and so one always gets a vector in $C^n(\mathbf{R})$?

 

[fresh_42]

What do you mean by eliminable? $u^{n}$ is only continuous after $n$ differentiations and so is $F(u)$.

 

[schniefen]

What do you mean by eliminable? $u^{n}$ is only continuous after $n$ differentiations and so is $F(u)$.

Doesn't $(\mathbf{u}^{(n)}(t),\mathbf{u}^{(n-1)}(t),...,\mathbf{u}(t))$ represent a basis for $U=C^n(\mathbf{R})$, and so in order to transform a vector from $U$ to $V=C(\mathbf{R})$ (with basis $\mathbf{u}(t)$), one would have to be able to cancel the other basis vectors by choosing appropriate coefficients in $F$?

 

[fresh_42]

What is the dimension of $C^n(\mathbb{R})\,?$ Compare it with the dimension of the space of all polynomials of a certain degree and lower: What is $\dim_\mathbb{R} \{\,a_nx^n+\ldots+a_1x+a_0\,|\,a_i\in \mathbb{R}\,\}\,?$

 

[schniefen]

Is $\dim C^n(\mathbb{R})=\dim_\mathbb{R} \{\,a_nx^n+\ldots+a_1x+a_0\,|\,a_i\in \mathbb{R}\,\}\,=n$?

 

[fresh_42]

We have $n+1$ independent directions $a_i$, which is finite. But regardless the plus one, do we have $\sin (t)$ in the vector space of polynomials? It is certainly in $C^n(\mathbb{R})$.

 

[schniefen]

What is the basis for $C^n(\mathbb{R})$?

 

[fresh_42]

I don't know one. Maybe Legendre polynomials, but I'm not even sure whether if it's countable or not. I suspect it isn't.

 

[schniefen]

But doesn't $\dim C^n(\mathbb{R})=n+1$ imply a basis of $n+1$ vectors?

 

[fresh_42]

$\dim C^n(\mathbb{R})=n+1$ implies anything, because it is wrong. We have $P^k(\mathbb{R}) \subseteq C^n(\mathbb{R})$ for all $k,n \in \mathbb{N}$, where $P^k$ is the algebra of (real) polynomials of degree at most $k$, and we have $\sin \notin P^k$ for any $k$.

This shows, that $\dim C^n(\mathbb{R})$ is at least countably infinite and that $P^n(\mathbb{R})$ is a proper subset.

 

[Stephen Tashi]

In $F$, isn't the term $\mathbf{u}^{(n)}(t)$ not eliminable and so one always gets a vector in $C^n(\mathbf{R})$?

A vector in $C^n({R})$ is also vector in $C({R})$.

I think what you mean to ask about is a situation where $u(t)$ is $n$ times continuously differentiable, but not $n+1$ times continuously differentiable. You are worried that the $u^n(t)$ term in $F(u)$ might cause $F'(u) = u^{(n+1)} + ...$ not to exist. I think there is a distinction between the notation $C({R})$ and $C^1({R})$. For $F(u)$ to be in $C({R})$ , it need not be differentiable. It only has to be continuous. The function $u^{(n)}$ is continuous even though it may not be differentiable.