Proving That f is Analytic with All Orders of Derivatives

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In summary, the conversation discusses the concept of proving something using derivatives and the concern about taking the derivative of an inequality. It is mentioned that this may not always be justified, and Taylor's theorem is suggested as a possible solution. The conversation ends with the thought that solving differential inequalities may not be as easy as it seems.
  • #1
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f is analytic and has all orders of derivatives.

I'm trying to prove something here but doing this makes me uneasy... is there anything wrong with saying:

If [tex]|f(z)| \leq |z^{n}|[/tex] then,

[tex]|f'(z)| \leq |nz^{n-1}|[/tex] so

[tex]|f''(z)| \leq |n(n-1)z^{n-2}|[/tex]
.
.
.

[tex]|f^{n}(z)| \leq n!z^{0}[/tex]

[tex]|f^{n}(z)| \leq n![/tex]

[tex]|f^{n+1}(z)| \leq 0[/tex]

[tex]|f^{n+1}(z)| = 0[/tex]

I can find no support for taking the derivative of the inequality like this in my calc book.
 
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  • #2
Taking the derivative of both sides of the inequality is not justified. For example, consider f(x) = sin(e^x) for x > 1. |f(x)| < x^2 but it is not that case that |f'(x)| = |e^x cos(e^x)| < 2x (for example, when x=ln(2*pi)). However, the fact that youre comparing an analytic function to a polynomial in z makes me think that Taylor's theorem may help you.
 
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  • #3
Thanks. IT just seemed way too easy.

How do they solve differential inequalities then, I wonder.
 

1. What is the definition of an analytic function?

An analytic function is a mathematical function that can be expressed as a power series in a neighborhood of every point in its domain. This means that at every point in its domain, the function can be represented as an infinite sum of terms, each of which is a constant multiple of a power of the difference between the point and a fixed reference point.

2. How do you prove that a function is analytic with all orders of derivatives?

To prove that a function f is analytic with all orders of derivatives, you must show that the function can be represented as a power series at every point in its domain. This can be done by taking the limit of the Taylor series expansion of the function as the number of terms approaches infinity. If the limit exists and is equal to the function, then f is analytic with all orders of derivatives.

3. What is the significance of proving that a function is analytic with all orders of derivatives?

The significance of proving that a function is analytic with all orders of derivatives is that it guarantees the function's smoothness and regularity. This means that the function has derivatives of all orders, and these derivatives are continuous and follow a specific pattern. Analytic functions are also important in complex analysis, as they have many useful properties and play a crucial role in solving mathematical problems.

4. Can a function be analytic with all orders of derivatives at some points but not others?

Yes, it is possible for a function to be analytic with all orders of derivatives at some points but not others. This is because the definition of an analytic function requires it to be represented as a power series in a neighborhood of every point in its domain. So, if the function fails to have a power series representation at a certain point, it is not analytic at that point. However, it may still be analytic at other points in its domain.

5. Are all continuous functions analytic with all orders of derivatives?

No, not all continuous functions are analytic with all orders of derivatives. While continuous functions are guaranteed to have all first-order derivatives, they may not have derivatives of higher orders. To be analytic with all orders of derivatives, a function must have derivatives of all orders and follow a specific pattern. This is a stricter requirement than just being continuous.

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