Leibniz's rule and differentiability of a function.

In summary, for α,β>=1, the function f is continuous at 0, differentiable at 0, continuously differentiable at 0, and twice differentiable at 0. The sum Ʃ(k=0,n) keκx can be found using Leibniz's rule.
  • #1
peripatein
880
0
Hi,

Homework Statement



(I) The following function is defined for α,β>0:
f(x) = { xβsin(1/xα), x≠0;
{ 0, x=0

I was asked for the values of α,β for which f(x) would be continuous at 0, differentiable at 0, continuously differentiable at 0, and twice differentiable at 0.

(II) I was asked to find the sum Ʃ(k=0,n) keκx

(III) Using Leibniz's rule, I am trying to evaluate (exsinx)(n).

Homework Equations


The Attempt at a Solution



(I) Would it be correct to write that f(x) is continuous at x=0 for any Natural α,β>=1?
I tried to equate the limits of the first derivatives at x=0 and got that there are no values of α,β for which f(x) is differentiable at 0. Is that correct? There also no values of α,β for which f(x) is continuously differentiable nor for which the second derivative would exist at x=0.
Are these statements correct?

(II) I first found the sum of the geometric progression Ʃ(k=0,n) eκx, which is (ex(n+1) - 1)/(ex - 1).
Is this result now to be multiplied by Ʃ(k=0,n) k, i.e. the arithmetic progression whose first term is 1 and last enx, with d=1?

(III) I know the final answer, namely 2n/2exsin(x+π/4), but do not quite understand how it was derived. Could someone please explain it to me? PS. I am not allowed to apply Euler's formula.
 
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  • #2
peripatein said:
Hi,

Homework Statement



(I) The following function is defined for α,β>0:
f(x) = { xβsin(1/xα), x≠0;
{ 0, x=0

I was asked for the values of α,β for which f(x) would be continuous at 0, differentiable at 0, continuously differentiable at 0, and twice differentiable at 0.

The Attempt at a Solution



(I) Would it be correct to write that f(x) is continuous at x=0 for any Natural α,β>=1?
I tried to equate the limits of the first derivatives at x=0 and got that there are no values of α,β for which f(x) is differentiable at 0. Is that correct? There also no values of α,β for which f(x) is continuously differentiable nor for which the second derivative would exist at x=0.
Are these statements correct?

Assuming [itex]f: [0,\infty) \to \mathbb{R}[/itex] and with one-sided limits at 0:

[itex]f(x)[/itex] is continuous at 0 for any [itex]\beta > 0[/itex], because [itex]|f(x) - 0| \leq |x^{\beta}| \to 0[/itex] as [itex]x \to 0[/itex].

[itex]f(x)[/itex] is differentiable at 0 with [itex]f'(0) = 0[/itex] for any [itex]\beta > 1[/itex], since [itex]f(x)/x = x^{\beta - 1}\sin(x^{-\alpha}) \to 0[/itex] as [itex]x \to 0[/itex] by the above.

For the next two, I need the result that if
[tex]
g(x) = \left\{\begin{array}{r@{\quad}l}
0 & x = 0 \\
x^\gamma \cos(x^{-\alpha}) & x \neq 0 \end{array}
\right.
[/tex]
then [itex]g(x)[/itex] is continuous at 0 for [itex]\gamma > 0[/itex], because [itex]|g(x)| \leq |x^\gamma| \to 0[/itex] as [itex]x \to 0[/itex].

[itex]f(x)[/itex] is continuously differentiable at 0 with [itex]f'(0) = 0[/itex] provided [itex]\beta > 1[/itex] and [itex]\beta > \alpha + 1[/itex], since for [itex]x \neq 0[/itex],
[itex]f'(x) = \beta x^{\beta - 1}\sin(x^{-\alpha}) - \alpha x^{\beta - \alpha - 1}\cos(x^{-\alpha})[/itex].

[itex]f(x)[/itex] is twice differentiable at 0 with [itex]f''(0) = 0[/itex] provided [itex]\beta > 2[/itex] and [itex]\beta > \alpha + 2[/itex], since then [itex]f'(x)/x = \beta x^{\beta - 2}\sin(x^{-\alpha}) - \alpha x^{\beta - \alpha - 2}\cos(x^{-\alpha}) \to 0[/itex] as [itex]x \to 0[/itex] by the above.

If you want to consider [itex]x < 0[/itex] as well then these results certainly hold for integer [itex]\alpha[/itex] and [itex]\beta[/itex].

(II) I was asked to find the sum Ʃ(k=0,n) keκx

(II) I first found the sum of the geometric progression Ʃ(k=0,n) eκx, which is (ex(n+1) - 1)/(ex - 1).
Is this result now to be multiplied by Ʃ(k=0,n) k, i.e. the arithmetic progression whose first term is 1 and last enx, with d=1?

No. The easiest way to do this sum is
[tex]
\sum_{k=0}^n ke^{kx} = \sum_{k=0}^n \frac{\mathrm{d}}{\mathrm{d}x} e^{kx}
= \frac{\mathrm{d}}{\mathrm{d}x} \sum_{k=0}^n e^{kx}
[/tex]
by linearity of the derivative.

(III) Using Leibniz's rule, I am trying to evaluate (exsinx)(n).

(III) I know the final answer, namely 2n/2exsin(x+π/4), but do not quite understand how it was derived. Could someone please explain it to me? PS. I am not allowed to apply Euler's formula.

I think that result is wrong: it doesn't hold when [itex]n = 2[/itex].

Starting with [itex]n = 1[/itex]:
[tex](e^x \sin x)' = e^x (\sin x + \cos x) = \sqrt 2 e^x \sin (x + (\pi/4))[/tex]
since if
[tex]R \sin (x + \alpha) = R \sin x \cos \alpha + R \cos x \sin \alpha = \sin x + \cos x[/tex]
then [itex]R = \sqrt 2[/itex] and [itex]\tan \alpha = 1[/itex] so that [itex]\alpha = \pi/4[/itex].
Then for [itex]n = 2[/itex]:
[tex](e^x \sin x)'' = (\sqrt 2 e^x \sin (x + (\pi/4)))'
= 2e^x \sin (x + (\pi/2))[/tex]
This suggests that the actual result is
[tex](e^x \sin x)^{(n)} = 2^{n/2} e^x \sin(x + (n\pi/4))
[/tex]
which can be proven by induction.
 
Last edited:
  • #3
Pasmith, in your answer to my first question, for f(x) to be differentiable at x=0, why ought not alpha to be less than beta - 1?
 
  • #4
peripatein said:
Pasmith, in your answer to my first question, for f(x) to be differentiable at x=0, why ought not alpha to be less than beta - 1?

Because
[tex]f'(0) = \lim_{x \to 0^{+}} \frac{f(x) -f(0)}{x} = \lim_{x \to 0^{+}} x^{\beta - 1}\sin(x^{-\alpha})[/tex]
which exists for all [itex]\alpha> 0[/itex] if [itex]\beta > 1[/itex] and doesn't exist for any [itex]\alpha > 0[/itex] if [itex]\beta \leq 1[/itex]. The condition on [itex]\alpha[/itex] is only required for the derivative to be continuous at 0.
 
  • #5
Do we demand f'(0)=0 and f''(0)=0, so that the derivative is continuous, as every differentiable function must also be continuous? Is that why?
 
  • #6
peripatein said:
Do we demand f'(0)=0 and f''(0)=0, so that the derivative is continuous, as every differentiable function must also be continuous? Is that why?

The only way for f'(0) to exist at all is if f'(0) = 0. Derivatives are not necessarily continuous, and existence of f'(0) is necessary but not sufficient for f'(x) to be continuous at 0; the extra condition on [itex]\alpha[/itex] is required. It is certainly necessary that f'(x) be continuous at 0 for f to be twice differentiable at 0, but for
[tex]
f''(0) = \lim_{x \to 0} \frac{f'(x) - f'(0)}{x} = \lim_{x \to 0} \frac{f'(x)}{x}
[/tex]
to exist a stronger condition is required, and that results in f''(0) = 0.
 

1. What is Leibniz's rule?

Leibniz's rule, also known as the generalized product rule, is a formula for computing the derivative of a function that is the product of two other functions. It is often used in multivariable calculus and can be written as d/dx (f(x)g(x)) = f'(x)g(x) + f(x)g'(x).

2. How is Leibniz's rule different from the product rule?

The product rule is a specific case of Leibniz's rule, where the two functions being multiplied are simply x and y. Leibniz's rule is a more general form that can be applied to a wider variety of functions.

3. What is the importance of differentiability in a function?

Differentiability is an important property of a function because it allows us to calculate its derivative, which gives us information about the rate of change of the function. This information is useful in many areas of math and science, such as optimization problems and modeling real-world phenomena.

4. Can every function be differentiated using Leibniz's rule?

No, not every function can be differentiated using Leibniz's rule. The functions being multiplied must themselves be differentiable in order for Leibniz's rule to be applicable. Additionally, the function being differentiated must also be differentiable at the point where the rule is being applied.

5. How is Leibniz's rule related to the chain rule?

The chain rule is a general rule for finding the derivative of a composite function. Leibniz's rule can be seen as a special case of the chain rule, where the composite function is a product of two simpler functions. In this way, Leibniz's rule can be thought of as an extension of the chain rule.

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