(Deceptively?) Simple question about Taylor series expansions

In summary, under certain conditions, it is correct to say that the function u(x) \in L^2(-\infty,\infty) can be written as a Taylor series expansion with the form u(x-t) = u(x) - \frac{du}{dx}t + \frac 12 \frac{d^2u}{dx^2}t^2 - \cdots = \sum_{n=0}^\infty \frac{u^{(n)}(x)}{n!}(-t)^n. These conditions include the function being analytic and not being a bump function, which is a counterexample where the Taylor expansion does not converge to the function. In the case of complex
  • #1
AxiomOfChoice
533
1
Under what circumstances is it correct to say of the function [itex]u(x) \in L^2(-\infty,\infty)[/itex] that

[tex]
u(x-t) = u(x) - \frac{du}{dx}t + \frac 12 \frac{d^2u}{dx^2}t^2 - \cdots = \sum_{n=0}^\infty \frac{u^{(n)}(x)}{n!}(-t)^n.
[/tex]
 
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  • #2
We can do that if the function is analytic. I don't really know if there are specific conditions which I could give.

Some notes:
1) Saying that u is differentiable an infinite number of times is not enough. A bump function is a counterexample. http://en.wikipedia.org/wiki/Bump_function These functions have the property that theTaylor expansion doesn't converge to the function.

2) If we were talking about complex functions instead of real functions then the answer is simpler. A complex function is analytic if and only if it is differentiable. However, these functions will rarely be in [itex]L^2[/itex].
 

1. What is a Taylor series expansion?

A Taylor series expansion is a method used in mathematics to represent a function as an infinite sum of terms. These terms are calculated using derivatives of the function at a specific point, called the center of expansion. The more terms that are included in the expansion, the more accurate the approximation of the function becomes.

2. How is a Taylor series expansion calculated?

A Taylor series expansion is calculated using the formula: f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ... + f^n(a)(x-a)^n/n!, where f(a) is the value of the function at the center of expansion, f'(a) is the first derivative of the function at the center, and so on. This formula can be used to find the value of the function at any point within the radius of convergence.

3. What is the radius of convergence for a Taylor series expansion?

The radius of convergence is the distance from the center of expansion where the Taylor series expansion is valid. It is the distance between the center and the closest point where the function is not infinitely differentiable. The radius of convergence can be determined by using the ratio test or the root test.

4. What are some applications of Taylor series expansions?

Taylor series expansions have many applications in mathematics, physics, and engineering. They are commonly used to approximate functions that are difficult to integrate or differentiate. They are also used in numerical analysis to solve differential equations and in statistics to estimate the parameters of a distribution. Additionally, Taylor series expansions are used in signal processing to compress data and in computer graphics to create smooth curves and surfaces.

5. Are Taylor series expansions always accurate?

No, Taylor series expansions are only accurate within their radius of convergence. Outside of this radius, the approximation becomes increasingly inaccurate. Additionally, Taylor series expansions can only approximate smooth and continuous functions. If a function is not infinitely differentiable, the Taylor series expansion will not converge to the function. Therefore, it is important to carefully consider the radius of convergence and the properties of the function before using a Taylor series expansion to approximate it.

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