## anonymous 3 years ago What does it mean for a function to be C^infinity?

1. anonymous

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

2. anonymous

$\mathbb C^\infty$Like this?

3. 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

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

Wait, no, nevermind, ignore me.

8. anonymous

What class is this?

9. anonymous

higher linear algebra

10. anonymous

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

11. helder_edwin

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

12. anonymous

Exactly, amen

13. anonymous

$$\mathbb C^\infty$$ means something else.

14. anonymous

$C^\infty(R)->C^\infty(R)$

15. anonymous

$$C$$ or $$\mathbb C$$?

16. 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?

17. anonymous

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

18. anonymous

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

19. anonymous

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

20. anonymous

@Zarkon You too get in here.

21. anonymous

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

22. anonymous

Manifold calculus? Hard stuff to learn in linear algebra.

23. 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.

24. anonymous

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

25. anonymous

Two completely different things.