anonymous
  • anonymous
What does it mean for a function to be C^infinity?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
The answer will always be that number to the square root of infinity.
anonymous
  • anonymous
\[\mathbb C^\infty\]Like this?
anonymous
  • 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
  • anonymous
That means it's infinite dimensional complex. Are you in a modern algebra class?
anonymous
  • anonymous
ok whats\[C^\infty(R)\] cause thats what im getting at
anonymous
  • anonymous
I'm a little hazy, but isn't that the Cauchy-Reimmann surface?
anonymous
  • anonymous
Wait, no, nevermind, ignore me.
anonymous
  • anonymous
What class is this?
anonymous
  • anonymous
higher linear algebra
anonymous
  • anonymous
its not very specific, but its a linear map !!
helder_edwin
  • helder_edwin
\(C^\infty\) means that it can be differentiated infinitely-many times.
anonymous
  • anonymous
Exactly, amen
anonymous
  • anonymous
\(\mathbb C^\infty\) means something else.
anonymous
  • anonymous
\[C^\infty(R)->C^\infty(R)\]
anonymous
  • anonymous
\(C\) or \(\mathbb C\)?
anonymous
  • 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
  • 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
  • anonymous
At least, in particle physics. Maybe it has another meaning in higher linear algebra. Apologies. Infinity complex conjugates doesn't sound right.
anonymous
  • anonymous
@TuringTest Hehe, you're better at math than me.
anonymous
  • anonymous
@Zarkon You too get in here.
anonymous
  • anonymous
All good guys ! http://sci4um.com/about19318.html So f(x)=0 is C^infinity :) im happy now ! thanks guys !
anonymous
  • anonymous
Manifold calculus? Hard stuff to learn in linear algebra.
TuringTest
  • 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
  • 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
  • anonymous
Two completely different things.

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