- #1
mufq15
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I need to find an example of such a function. I know that x^2sin(1/x) is differentiable but not C^1, but I'm having trouble extending this to C^2.
Twice differentiable but not C^2
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SUMMARY
The discussion centers on the exploration of functions that are twice differentiable but not C². The function f(x) = x⁴sin(1/x) is presented as an example, demonstrating that while it is differentiable, its second derivative is not continuous due to the oscillatory nature of sin(1/x). The conversation highlights the relationship between the differentiability of a function and its anti-derivative, emphasizing that the anti-derivative is always at least as differentiable as the original function.
PREREQUISITES- Understanding of differentiability and continuity in calculus
- Familiarity with the properties of trigonometric functions, specifically sine
- Knowledge of limits and their role in defining derivatives
- Basic skills in finding derivatives and anti-derivatives of functions
- Study the properties of C¹ and C² functions in detail
- Learn about the implications of differentiability on the continuity of derivatives
- Explore examples of functions that are differentiable but not C²
- Investigate the behavior of oscillatory functions and their derivatives
Mathematicians, calculus students, and educators seeking to deepen their understanding of differentiability, particularly in the context of functions that exhibit complex behavior like oscillations.
- #2
arildno
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Really?
What about an anti-derivative to the function you've posted?
What about an anti-derivative to the function you've posted?
- #3
mufq15
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Um, I understand that that would work, but I don't think I know how to take an antiderivative of that (or am I just being silly?) Are you suggesting I just write it as the integral of that?
- #4
arildno
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Yes, you won't be able to write your anti-derivative in terms of elementary functions, however the really important insight is that the anti-derivative of a function is always differentiable to a greater or equal extent as your original function. For functions finitely differentiable, the strict inequality holds, for infinitely differentiable functions, the "equality" holds.
Another choice should readily suggest itself by considering WHAT IS IT THAT MAKES YOUR FUNCTION ONCE DIFFERENTIABLE?
The answer is that the power of the x multiplied with the sine is big enough to kill off the crazy behaviour of the sine function!
(The amplitude of the function becomes so small that the pathological oscillation of the function becomes "irrelevant")
Thus, we may ask:
What if we make the power of the amplitude even bigger?
Might not this make also the derivative of our function not only continuous, but also differentiable?
Consider the function:
f(x)=x^{4}\sin(\frac{1}{x}), x\neq{0}, f(0)=0
Now, the DERIVATIVE of this function is readily found out to be:
f'(x)=4x^{3}\sin(\frac{1}{x})-x^{2}\cos(\frac{1}{x}), x\neq{0}, f'(0)=0
Now, see if you manage to differentiate this function everywhere, that is find the second derivative of f.
Another choice should readily suggest itself by considering WHAT IS IT THAT MAKES YOUR FUNCTION ONCE DIFFERENTIABLE?
The answer is that the power of the x multiplied with the sine is big enough to kill off the crazy behaviour of the sine function!
(The amplitude of the function becomes so small that the pathological oscillation of the function becomes "irrelevant")
Thus, we may ask:
What if we make the power of the amplitude even bigger?
Might not this make also the derivative of our function not only continuous, but also differentiable?
Consider the function:
f(x)=x^{4}\sin(\frac{1}{x}), x\neq{0}, f(0)=0
Now, the DERIVATIVE of this function is readily found out to be:
f'(x)=4x^{3}\sin(\frac{1}{x})-x^{2}\cos(\frac{1}{x}), x\neq{0}, f'(0)=0
Now, see if you manage to differentiate this function everywhere, that is find the second derivative of f.
- #5
mufq15
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Oh, okay. The sin(1/x) in the second derivative makes it not continuous. Thanks for your help.
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