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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.
\(C\) 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.
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.
Two completely different things.