Derivative theorem

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

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

Is this theorem correct?
 
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Answers and Replies

  • #2
rachmaninoff
Is this homework? You can prove it by combining the two basic identities,

[tex]\frac{d}{dx} \ln x = \frac{1}{x} \equiv x^{-1}[/tex]

[tex]\frac{d}{dx} x^{m} = m x^{m-1} \, (m \neq 0) [/tex]
 
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  • #3
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  • #4
TD
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You could prove it by induction on n.
Check for n = 1, assume valid for n = k, prove then for n = k+1.
 
  • #5
matt grime
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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
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Generic Iteration...


Is this the correct iterated integral of this function?

iterated integration theorem: (generic)
[tex](I^n f)(x) = \int_0^x (-1)^{n + 1} \frac{(n - 1)!}{t^n} dt = \ln x[/tex]
 
  • #7
Tide
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Orion's Conjecture Belted

Orion1 said:

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

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

Is this theorem correct?

No, it is not correct. It fails when n = 0.
 
  • #8
Galileo
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Orion1 said:


[tex](I^n f)(x) = \int_0^x (-1)^{n + 1} \frac{(n - 1)!}{t^n} dt = \ln x[/tex]
That integral doesn't converge.
 
  • #9
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theorem...


theorem:
[tex]\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.[/tex]
 
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  • #10
Tide
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Orion1 said:


theorem:
[tex]\frac{d^n}{dx^n} (\ln x) = (-1)^{n + 1} \frac{(|n - 1|)!}{x^n}[/tex]

Is this theorem correct when n = 0?
No, the zeroth derivative of the natural log is -- the natural log!
 
  • #11
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Then this theorem abstains at [tex]n > 0[/tex].

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

[tex]\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.[/tex]

Are there any more challenges to this theorem?
 
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  • #12
Hurkyl
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I'm worried about nonintegral values of n.
 
  • #13
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iterated integration theorem:
[tex](I^n f)(x) = \int_1^x (-1)^{n + 1} \frac{(n - 1)!}{t^n} dt = \ln x[/tex]

Is this theorem correct? Does this theorem converge?
 
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  • #14
matt grime
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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
lurflurf
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matt grime said:
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.
Orion1 seems to be attempting to make use a a result due to Cauchy that if I is the operator defined by
[tex](If)(x):=\int_0^x f(t)dt[/tex]
then if the operator is applied n time the result is
[tex](I^nf)(x)=\frac{1}{\Gamma(n)}\int_0^x (x-t)^{n-1}f(t)dt[/tex]
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
matt grime
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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
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foundation theorem...

lurflurf said:
Orion1 seems to be attempting to make use a a result due to Cauchy that if I is the operator defined by
[tex](If)(x):=\int_0^x f(t)dt[/tex]

then if the operator is applied n time the result is
[tex](I^nf)(x)=\frac{1}{\Gamma(n)}\int_0^x (x-t)^{n-1}f(t)dt[/tex]

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.

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?

[tex](I^nf)(x)=\frac{1}{\Gamma(n)}\int_0^x (x-t)^{n-1}f(t)dt[/tex]

[tex]\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.[/tex]

fractional calculus theorem: (correct symbolic method?)
[tex](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[/tex]

In response to Tide's post, [tex]n = 0[/tex] appears to be special case:
[tex](I^0f)(x) = \frac{1}{\Gamma(0)} \int_0^x \frac{f(t)}{(x-t)}dt \; \; n = 0[/tex]
 
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  • #18
benorin
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Orion1 said:

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?

[tex](I^nf)(x)=\frac{1}{\Gamma(n)}\int_0^x (x-t)^{n-1}f(t)dt[/tex]

[tex]\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.[/tex]

fractional calculus theorem: (correct symbolic method?)
[tex](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[/tex]

In response to Tide's post, [tex]n = 0[/tex] appears to be special case:
[tex](I^0f)(x) = \frac{1}{\Gamma(0)} \int_0^x \frac{f(t)}{(x-t)}dt \; \; n = 0[/tex]
Fractional Integrals are (by the given operator) only defined for n>0, atleast per the following link:

http://mathworld.wolfram.com/FractionalIntegral.html
 
  • #19
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Derivative Dogma...


Tide said:
No, it is not correct. It fails when n = 0.
derivative integer factorial theorem:
[tex]\frac{d^n}{dx^n} (\ln x) = (-1)^{n + 1} \frac{(n - 1)!}{x^n} \; \; n > 0[/tex]
benorin said:
Fractional Integrals are (by the given operator) only defined for 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!:
[tex]\frac{d^n}{dx^n} f(x) = g(x) \; \; n > 0[/tex]

However when n = 0:
[tex]\frac{d^n}{dx^n} f(x) = f(x) \; \; n = 0[/tex]

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?
 
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