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davester248
 2 years ago
Best ResponseYou've already chosen the best response.0The answer will always be that number to the square root of infinity.

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.2\[\mathbb C^\infty\]Like this?

extremity
 2 years ago
Best ResponseYou've already chosen the best response.0Why 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

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.2That means it's infinite dimensional complex. Are you in a modern algebra class?

extremity
 2 years ago
Best ResponseYou've already chosen the best response.0ok whats\[C^\infty(R)\] cause thats what im getting at

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.2I'm a little hazy, but isn't that the CauchyReimmann surface?

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.2Wait, no, nevermind, ignore me.

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.2What class is this?

extremity
 2 years ago
Best ResponseYou've already chosen the best response.0higher linear algebra

extremity
 2 years ago
Best ResponseYou've already chosen the best response.0its not very specific, but its a linear map !!

helder_edwin
 2 years ago
Best ResponseYou've already chosen the best response.1\(C^\infty\) means that it can be differentiated infinitelymany times.

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.2\(\mathbb C^\infty\) means something else.

extremity
 2 years ago
Best ResponseYou've already chosen the best response.0\[C^\infty(R)>C^\infty(R)\]

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.2\(C\) or \(\mathbb C\)?

extremity
 2 years ago
Best ResponseYou've already chosen the best response.0single 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?

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.2Ah okay. \(C(R)\) means the complex conjugate of \(R\). This is the sansserif \(C\) as opposed to the shell \(\mathbb C\).

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.2At least, in particle physics. Maybe it has another meaning in higher linear algebra. Apologies. Infinity complex conjugates doesn't sound right.

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.2@TuringTest Hehe, you're better at math than me.

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.2@Zarkon You too get in here.

extremity
 2 years ago
Best ResponseYou've already chosen the best response.0All good guys ! http://sci4um.com/about19318.html So f(x)=0 is C^infinity :) im happy now ! thanks guys !

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.2Manifold calculus? Hard stuff to learn in linear algebra.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1I 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 finitedegree, I don't know about otherwise) polynomials are infinitely differentiable.

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.2Yes, @extremity found it. Why do they use the same symbol as the one you use when you learn about symmetry in physics?

badreferences
 2 years ago
Best ResponseYou've already chosen the best response.2Two completely different things.
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