Prove that this function differentiable endles times

• transgalactic
In summary, this function differentiates at zero and has a Taylor Series at 0 but is not equal to it except at 0.
transgalactic
prove that this function differentiable endles times on x=0 ?? http://img502.imageshack.us/img502/6778/83126617mm0.th.gif

i was told
"once we can express the function as a power series around zero and it is differentiable at zero, we know it is infinitely differentiable"
differentiable around zero part:
$$f'(0)=\lim_{x->0}\frac{e^\frac{-1}{x^2}-0}{x-0} \\ =\lim_{x->0}\frac x{2e^{1/x^2}}$$
i know that exponentials grow faster then polinomials but the power of the exp is 1/x^2
so the denominator goes to infinity but faster then the numenator so the expression
goes to 0.
power series around zero part:
the power series of e^x
$$g(x)=e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^3)\\$$
i put -1/x^2 instead of x
$$g(x)=e^x=1+\frac{-1}{x^2}+\frac{x^{4}}{2}+\frac{-x^6}{6}+O(-x^6)\\$$

what now??

Last edited by a moderator:
The first derivative of [tex]e^{-1/x^2}[/itex] is, of course, $e^{-1/x^2}(2x^{-3})$. The second derivative, using the product rule is $e^{-1/x^2}(4x^{-6})+ e^{-1/x^2}(-6x^{-4})$.

Can you use "proof by induction" to show that every derivative is $e^{-1/x^2}$ times terms with negative powers of x? It doesn't really matter what those powers are because, as x goes to 0, the exponential will "dominate" and the derivative will go to 0 as x goes to 0 so we can take every derivative to be 0 at x= 0.

That means, by the way, that we could form its Taylor series at x= 0 (its MacLaurin series) as $0+ 0x+ 0x^2+ \cdot\cdot\cdot= 0$. But obviously f(x) is NOT identically 0. This is an example of a function that is infinitely differentiable so has Taylor series at 0 but is NOT equal to that series except at 0.

my power series is fine
i took the power series for e^t and substitutes t=-1/x^2

so i got a power series and differentiation of the first around 0

is it a solution

how to do it by induction?

1. What does it mean for a function to be differentiable endlessly?

Differentiability is a mathematical concept that refers to the smoothness of a function. A function that is differentiable endlessly, also known as infinitely differentiable, means that it can be differentiated an infinite number of times, resulting in a smooth curve without any abrupt changes or corners.

2. How can I prove that a function is differentiable endlessly?

In order to prove that a function is differentiable endlessly, you must show that it can be differentiated an infinite number of times without any discontinuities or singularities. This can be demonstrated through various mathematical techniques such as the limit definition of a derivative or using the Taylor series expansion.

3. What are the benefits of having a function that is differentiable endlessly?

A function that is differentiable endlessly has many practical applications, especially in physics and engineering. It allows for precise calculations and modeling of complex systems, as well as providing a smooth and continuous representation of real-world phenomena.

4. Can a function be differentiable endlessly at all points?

No, not all functions are differentiable endlessly. Some functions may have points of discontinuity or singularities where the function is not smooth and cannot be differentiated. This is often the case with functions that have sharp corners or cusps.

5. How does the concept of endlessly differentiable functions relate to the concept of analytic functions?

Analytic functions are a subtype of differentiable functions that are differentiable endlessly in a specific region and have a corresponding Taylor series expansion. Therefore, all analytic functions are also endlessly differentiable, but not all endlessly differentiable functions are necessarily analytic.

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