Proving kth Derivative of f:R->R is 0 at 0

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In summary, f(x) is defined to be 0 if x is less than or equal to 0 and e^-1/x if x is greater than 0. It can be shown that f is infinitely differentiable and all its derivatives at 0 are equal to 0. This is proven by taking the first and second derivatives and applying L'Hopital's rule. For the kth derivative, we can use the fact that after differentiating e^-1/x multiple times, we will only end up with e^-1/x multiplied by some type of function of x. By showing that this function tends to 0 as x approaches 0, it can be concluded that all derivatives of f at 0 are equal to
  • #1
Nusc
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Define f:R->R by fx = 0 if x<=0 and e^-1/x if x>0

Show that f is c^inf and that all the derivatives of f at 0 vanish; that is, f^(k)0=0 for every k.


After taking first and secondderivatives we just apply l'hostpials rule to show that the derivatives vanish.

My problem is for the kth derivative.

We know after taking the derivative we get some polynomail say p(x) so

f^k(x) = p(x) * e^-1/x


What do I do now?
 
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  • #2
After you differentiate e^-{1/x} some number of times you only ever end up with e^-{1/x}*(some type of function of x), work out what 'type' of function I mean, and then show that for any function of that type the limit is zero as x tends to zero.
 
  • #3
I assume that by that type of function you mean the polynomial I spoke of. Do I need to contruct a lemma?
 
  • #4
You need to prove something. What you call it is up to you.
 
  • #5
Show that If g: R -> R is analytic in a nbhrd of 0 then gx = sum g^k0 x^k/k! for all x in the symmetric int. with center 0. then f, as defined above, cannot be analytic at 0

Well it is assumed that all derivatives vanish at 0 therefore no power series centered at 0 converges to f on a nbhrd of 0

Is that enough justifcation?
 

1. What is the significance of proving the kth derivative of a function is 0 at 0?

The significance of this proof is that it shows that the function has a horizontal tangent line at the point (0,0). This means that the function is very flat at this point and does not have any steep changes in slope. It can also indicate that the function is continuous and differentiable at this point.

2. How is the kth derivative of a function at 0 calculated?

The kth derivative of a function at 0 is calculated by taking the limit of the kth derivative of the function as x approaches 0. This can be done using the definition of a derivative or by using differentiation rules for higher order derivatives.

3. What does it mean for a function to have a horizontal tangent line?

A horizontal tangent line means that the slope of the function at that point is 0. In other words, the function is not changing at that point and remains constant. This can also be interpreted as the function having a flat spot or a point of inflection at that point.

4. Can a function have a horizontal tangent line at a point without having a derivative of 0 at that point?

Yes, it is possible for a function to have a horizontal tangent line at a point without having a derivative of 0 at that point. This can occur if the function has a sharp or discontinuous change in slope at that point. In this case, the function would not be differentiable at that point.

5. Why is it important to prove the kth derivative of a function is 0 at 0?

Proving the kth derivative of a function is 0 at 0 is important because it can help us understand the behavior of the function at this point. It can also provide information about the smoothness and continuity of the function. Additionally, this proof can be used in further mathematical calculations and to make predictions about the behavior of the function near this point.

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