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extremity

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

  • one year ago
  • one year ago

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  1. davester248
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    The answer will always be that number to the square root of infinity.

    • one year ago
  2. badreferences
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    \[\mathbb C^\infty\]Like this?

    • one year ago
  3. extremity
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    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

    • one year ago
  4. badreferences
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    That means it's infinite dimensional complex. Are you in a modern algebra class?

    • one year ago
  5. extremity
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    ok whats\[C^\infty(R)\] cause thats what im getting at

    • one year ago
  6. badreferences
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    I'm a little hazy, but isn't that the Cauchy-Reimmann surface?

    • one year ago
  7. badreferences
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    Wait, no, nevermind, ignore me.

    • one year ago
  8. badreferences
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    What class is this?

    • one year ago
  9. extremity
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    higher linear algebra

    • one year ago
  10. extremity
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    its not very specific, but its a linear map !!

    • one year ago
  11. helder_edwin
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    \(C^\infty\) means that it can be differentiated infinitely-many times.

    • one year ago
  12. davester248
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    Exactly, amen

    • one year ago
  13. badreferences
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    \(\mathbb C^\infty\) means something else.

    • one year ago
  14. extremity
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    \[C^\infty(R)->C^\infty(R)\]

    • one year ago
  15. badreferences
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    \(C\) or \(\mathbb C\)?

    • one year ago
  16. extremity
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    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?

    • one year ago
  17. badreferences
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    Ah okay. \(C(R)\) means the complex conjugate of \(R\). This is the sans-serif \(C\) as opposed to the shell \(\mathbb C\).

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

    • one year ago
  19. badreferences
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    @TuringTest Hehe, you're better at math than me.

    • one year ago
  20. badreferences
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    @Zarkon You too get in here.

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

    • one year ago
  22. badreferences
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    Manifold calculus? Hard stuff to learn in linear algebra.

    • one year ago
  23. TuringTest
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    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.

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

    • one year ago
  25. badreferences
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    Two completely different things.

    • one year ago
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