Here's the question you clicked on:
extremity
What does it mean for a function to be C^infinity?
The answer will always be that number to the square root of infinity.
\[\mathbb C^\infty\]Like this?
Why are all polynomials C^infinity? I mean, if we consider the polynomial f(x)=x^2...how does that satisfy the definition. ps. badreferences u are correct
That means it's infinite dimensional complex. Are you in a modern algebra class?
ok whats\[C^\infty(R)\] cause thats what im getting at
I'm a little hazy, but isn't that the Cauchy-Reimmann surface?
Wait, no, nevermind, ignore me.
What class is this?
higher linear algebra
its not very specific, but its a linear map !!
\(C^\infty\) means that it can be differentiated infinitely-many times.
\(\mathbb C^\infty\) means something else.
\[C^\infty(R)->C^\infty(R)\]
\(C\) or \(\mathbb C\)?
single C. I see. Now, wolfram says that all polynomials are C^infty. Can someone explain to be how f(x) = ax^2 +bx+c is C^infty? considering that it is a polynomial?
Ah okay. \(C(R)\) means the complex conjugate of \(R\). This is the sans-serif \(C\) as opposed to the shell \(\mathbb C\).
At least, in particle physics. Maybe it has another meaning in higher linear algebra. Apologies. Infinity complex conjugates doesn't sound right.
@TuringTest Hehe, you're better at math than me.
@Zarkon You too get in here.
All good guys ! http://sci4um.com/about19318.html So f(x)=0 is C^infinity :) im happy now ! thanks guys !
Manifold calculus? Hard stuff to learn in linear algebra.
I think it's what @helder_edwin said. Any polynomial can be differentiated to 0 by taking n+1 derivatives where n is the order of the polynomial. The derivative of 0 is 0, which can, of course, be differentiated ad infinitum. Hence all (at least finite-degree, I don't know about otherwise) polynomials are infinitely differentiable.
Yes, @extremity found it. Why do they use the same symbol as the one you use when you learn about symmetry in physics?
Two completely different things.