C^infty on U: Question & Counter Example

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In summary, C^infty on U refers to a class of functions that are infinitely differentiable on a given open set U in a mathematical space. This means that the functions have derivatives of all orders on U. A function is infinitely differentiable if it has derivatives of all orders at every point in its domain, making it smooth and well-defined at every point. C^infty on U is significant in mathematics, particularly in fields such as analysis and differential geometry, as it allows for the study of smooth and continuous functions. An example of a function that is not infinitely differentiable on U is f(x) = |x| on the open set (-1,1). C^infty on U functions can be represented by their
  • #1
Palindrom
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Suppose h and h o f are both C^infty on f(U) and U correspondingly, with U an open set. Assume now that f(U) is open. Is f C^infty on U? Does anyone have a counter example?

I've edited the question so that it would actually make sense.
 
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  • #2
trivially false. take h constant.
 
  • #3
How very stupid of me. Thank you.
 

FAQ: C^infty on U: Question & Counter Example

What is C^infty on U?

C^infty on U refers to a class of functions that are infinitely differentiable on a given open set U in a mathematical space. This means that the functions have derivatives of all orders on U.

What does it mean for a function to be infinitely differentiable?

A function is infinitely differentiable if it has derivatives of all orders at every point in its domain. This means that the function is smooth and has a well-defined slope at every point.

What is the significance of C^infty on U in mathematics?

C^infty on U is an important concept in mathematical analysis and differential geometry. It allows for the study of smooth and continuous functions that have derivatives of all orders, which are essential in many fields such as physics, engineering, and economics.

Can you provide an example of a function that is not infinitely differentiable on U?

Yes, the function f(x) = |x| is not infinitely differentiable on the open set U = (-1,1). This is because the function is not differentiable at x = 0, as it has a sharp corner at that point.

How is C^infty on U related to Taylor series?

C^infty on U functions can be represented by their Taylor series, which is an infinite sum of terms that approximate the function at a given point. This series is useful in approximating the value of a function and its derivatives at any point within its domain.

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