What does it mean for a function to be C^infinity?

- anonymous

What does it mean for a function to be C^infinity?

- jamiebookeater

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- anonymous

The answer will always be that number to the square root of infinity.

- anonymous

\[\mathbb C^\infty\]Like this?

- anonymous

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

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## More answers

- anonymous

That means it's infinite dimensional complex. Are you in a modern algebra class?

- anonymous

ok whats\[C^\infty(R)\]
cause thats what im getting at

- anonymous

I'm a little hazy, but isn't that the Cauchy-Reimmann surface?

- anonymous

Wait, no, nevermind, ignore me.

- anonymous

What class is this?

- anonymous

higher linear algebra

- anonymous

its not very specific, but its a linear map !!

- helder_edwin

\(C^\infty\) means that it can be differentiated infinitely-many times.

- anonymous

Exactly, amen

- anonymous

\(\mathbb C^\infty\) means something else.

- anonymous

\[C^\infty(R)->C^\infty(R)\]

- anonymous

\(C\) or \(\mathbb C\)?

- anonymous

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?

- anonymous

Ah okay.
\(C(R)\) means the complex conjugate of \(R\). This is the sans-serif \(C\) as opposed to the shell \(\mathbb C\).

- anonymous

At least, in particle physics. Maybe it has another meaning in higher linear algebra. Apologies.
Infinity complex conjugates doesn't sound right.

- anonymous

@TuringTest Hehe, you're better at math than me.

- anonymous

@Zarkon You too get in here.

- anonymous

All good guys !
http://sci4um.com/about19318.html
So f(x)=0 is C^infinity :) im happy now ! thanks guys !

- anonymous

Manifold calculus? Hard stuff to learn in linear algebra.

- TuringTest

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.

- anonymous

Yes, @extremity found it. Why do they use the same symbol as the one you use when you learn about symmetry in physics?

- anonymous

Two completely different things.

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