# Derivative theorem

1. Aug 16, 2005

### Orion1

Any calculus researchers interested in disproving this theorem with some simple integers?

Orion1 derivative integer factorial theorem:
$$\frac{d^n}{dx^n} (\ln x) = (-1)^{n + 1} \frac{(n - 1)!}{x^n}$$

Is this theorem correct?

Last edited: Aug 16, 2005
2. Aug 16, 2005

### rachmaninoff

Is this homework? You can prove it by combining the two basic identities,

$$\frac{d}{dx} \ln x = \frac{1}{x} \equiv x^{-1}$$

$$\frac{d}{dx} x^{m} = m x^{m-1} \, (m \neq 0)$$

Last edited by a moderator: Aug 16, 2005
3. Aug 18, 2005

### Orion1

Expired Post...

Expired Post...

Last edited: Aug 19, 2005
4. Aug 18, 2005

### TD

You could prove it by induction on n.
Check for n = 1, assume valid for n = k, prove then for n = k+1.

5. Aug 18, 2005

### matt grime

They prove it because they do, as any one who knows about talyor series for instance could tell you. Again, do not name things after yourself, that is not allowed in mathematics, not least when it is this elementary and well known.

6. Aug 23, 2005

### Orion1

Generic Iteration...

Is this the correct iterated integral of this function?

iterated integration theorem: (generic)
$$(I^n f)(x) = \int_0^x (-1)^{n + 1} \frac{(n - 1)!}{t^n} dt = \ln x$$

7. Aug 23, 2005

### Tide

Orion's Conjecture Belted

No, it is not correct. It fails when n = 0.

8. Aug 24, 2005

### Galileo

That integral doesn't converge.

9. Aug 25, 2005

### Orion1

theorem...

theorem:
$$\frac{d^n}{dx^n} (\ln x) = \left\{ \begin{array}{rcl} \ln x & \text{if} & n = 0 \\ (-1)^{n + 1} \frac{(n - 1)!}{x^n} & \text{if} & n > 0 \end{array} \right.$$

Last edited: Aug 26, 2005
10. Aug 26, 2005

### Tide

No, the zeroth derivative of the natural log is -- the natural log!

11. Aug 26, 2005

### Orion1

Then this theorem abstains at $$n > 0$$.

theorem:
$$\frac{d^n}{dx^n} (\ln x) = (-1)^{n + 1} \frac{(n - 1)!}{x^n} \; \; n > 0$$

$$\frac{d^n}{dx^n} (\ln x) = \left\{ \begin{array}{rcl} \ln x & \text{if} & n = 0 \\ (-1)^{n + 1} \frac{(n - 1)!}{x^n} & \text{if} & n > 0 \end{array} \right.$$

Are there any more challenges to this theorem?

Last edited: Aug 26, 2005
12. Aug 26, 2005

### Hurkyl

Staff Emeritus
I'm worried about nonintegral values of n.

13. Aug 26, 2005

### Orion1

iterated integration theorem:
$$(I^n f)(x) = \int_1^x (-1)^{n + 1} \frac{(n - 1)!}{t^n} dt = \ln x$$

Is this theorem correct? Does this theorem converge?

Last edited: Aug 26, 2005
14. Aug 26, 2005

### matt grime

no, that doesn't work since the integral you wrote down evaluates to

(-1)^n(n-2)!x^{-n+1} +(-1)^n(n-2)!

it isn't an "iterated" integral, it is just an integral.

and theorems do not converge. integrals can converge, sequences can converge.

15. Aug 26, 2005

### lurflurf

Orion1 seems to be attempting to make use a a result due to Cauchy that if I is the operator defined by
$$(If)(x):=\int_0^x f(t)dt$$
then if the operator is applied n time the result is
$$(I^nf)(x)=\frac{1}{\Gamma(n)}\int_0^x (x-t)^{n-1}f(t)dt$$
This result is often used as the foundation for the so called fractional calcus when operators of nonintegral order are considered. Since the right hand side is well defined for n not an integer this is taken as a definition of the left hant side when n is not an integer.

16. Aug 26, 2005

### matt grime

i had guessed what he was *trying* to do, but we can't be sure, and i was pointing out what he *had* done

17. Sep 2, 2005

### Orion1

foundation theorem...

That is correct, I was uncertain how the symbolic method described such a theorem, it certainly is not listed in my calculus book.

As I understand this theorem, the integer, or non-integer must be defined 'prior' to performing the actual integration for each successive iteration.

I suppose the next question, does this theorem obey the fractional calculus foundation theorem?

$$(I^nf)(x)=\frac{1}{\Gamma(n)}\int_0^x (x-t)^{n-1}f(t)dt$$

$$\frac{d^n}{dx^n} (\ln x) = \left\{ \begin{array}{rcl} \ln x & \text{if} & n = 0 \\ (-1)^{n + 1} \frac{(n - 1)!}{x^n} & \text{if} & n > 0 \end{array} \right.$$

fractional calculus theorem: (correct symbolic method?)
$$(I^nf)(x) = \frac{1}{\Gamma(n)} \int_0^x (x-t)^{n-1}f(t)dt = \frac{1}{\Gamma(n)} \int_0^x (-1)^{n + 1} \frac{(n - 1)!}{t^n} dt = \ln x \; \; n > 0$$

In response to Tide's post, $$n = 0$$ appears to be special case:
$$(I^0f)(x) = \frac{1}{\Gamma(0)} \int_0^x \frac{f(t)}{(x-t)}dt \; \; n = 0$$

Last edited: Sep 2, 2005
18. Oct 9, 2005

### benorin

Fractional Integrals are (by the given operator) only defined for n>0, atleast per the following link:

http://mathworld.wolfram.com/FractionalIntegral.html

19. Oct 12, 2005

### Orion1

Derivative Dogma...

derivative integer factorial theorem:
$$\frac{d^n}{dx^n} (\ln x) = (-1)^{n + 1} \frac{(n - 1)!}{x^n} \; \; n > 0$$
Then the original theorem is correct, because like Fractional Integrals, Fractional Derivatives are only defined for n > 0?

I disagree that n = 0 derivatives are actually derivatives at all!:
$$\frac{d^n}{dx^n} f(x) = g(x) \; \; n > 0$$

However when n = 0:
$$\frac{d^n}{dx^n} f(x) = f(x) \; \; n = 0$$

It just seems that any derivative theorem would always fail at n = 0, because n = 0 is not a derivative!

Can anyone actually state a 'Fractional Calculus' derivative theorem that does not fail at n = 0?

Last edited: Oct 12, 2005