definition of the second derivative

[113]

Is it possible to define the second derivative of f(x) in this way:

\[ f''(x) = \frac{f(x+2dx) -2(f+dx) + f(x)}{(dx)^2} \]

I am using a finite difference approximation called "Second order forward" from the link, I use dx instead of h:

https://en.wikipedia.org/wiki/Finite_difference#Higher-order_differences

Edited May 23, 2019 by 113

[studiot]

I wouldn't recommend it since the dx² in

[math]\frac{{{d^2}y}}{{d{x^2}}}[/math]

means something slightly different than h² , ie not the same as (dx)² would be.

 

Edited May 23, 2019 by studiot

[mathematic]

You can use it. f'(x)=(f(x+dx)-f(x))/dx.  f'''(x)=(f'{x+dx)-f'(x))/dx is exactly the expression  you gave.

[studiot]

It would help if you explained what you are trying to do.

Are you trying to obtain an estimate of the derivative from the finite differences?

You realise that you need multiple points to obtain the second and higher difference, you don't square the first differences?

For numerical differentiation it is normal to use

forward differences at the beginning of a table (where you have tabulated values below you)
backward differences at the end of the table (where you have tabulated values above you)
central differences in the middle of the table where you have tabulated values (both above and below you)

 

Edited May 23, 2019 by studiot

[uncool]

51 minutes ago, mathematic said:

You can use it. f'(x)=(f(x+dx)-f(x))/dx.  f'''(x)=(f'{x+dx)-f'(x))/dx is exactly the expression  you gave.

This isn't exactly true; the actual definition involves lots of "there exists" and "for all"s. If the function is nice enough, then yes, this "pseudo-proof" tells you that the second derivative must be equal to the expression given; however, the second derivative does not necessarily exist.

[mathematic]

22 hours ago, uncool said:

This isn't exactly true; the actual definition involves lots of "there exists" and "for all"s. If the function is nice enough, then yes, this "pseudo-proof" tells you that the second derivative must be equal to the expression given; however, the second derivative does not necessarily exist.

I have assumed all these niceties are in place, so I answered the question as posed.

[uncool]

"As posed", the question does not assume the "niceties". It's asking about using this as a definition; when talking about definitions, "niceties" cannot be assumed unless explicitly stated.

[113]

On 5/24/2019 at 12:37 AM, studiot said:

I wouldn't recommend it since the dx² in

d2ydx2

means something slightly different than h² , ie not the same as (dx)² would be.

 

dx is nilsquare infinitesimal, meaning (dx)² = 0  is true, but dx=0 need not be true at the same time.

https://en.wikipedia.org/wiki/Smooth_infinitesimal_analysis

A problem seems to arise because there appears to be a division by zero in that case.

[mathematic]

dx etc. are symbols used for derivatives and integrals.  They are not supposed to have numeric values.

[113]

On 5/28/2019 at 11:46 PM, mathematic said:

dx etc. are symbols used for derivatives and integrals.  They are not supposed to have numeric values.

What's wrong with division by zero ?

[mathematic]

16 hours ago, 113 said:

What's wrong with division by zero ?

Because it is nonsense.

[113]

On 8/8/2019 at 12:38 AM, mathematic said:

Because it is nonsense.

it is possible to use division by zero:

https://en.wikipedia.org/wiki/Signed_zero

[Ghideon]

26 minutes ago, 113 said:

it is possible to use division by zero:

https://en.wikipedia.org/wiki/Signed_zero

That is the IEEE 754 standard for floating-point arithmetic. It is intended to simplify some computing problems. It does not make it possible to divide by zero. As far as I know it is to have programs behave in a well defined way in certain situations.

 

Edited August 25, 2019 by Ghideon

[studiot]

On 5/23/2019 at 10:03 PM, 113 said:

I am using a finite difference approximation called "Second order forward" from the link, I use dx instead of h:

https://en.wikipedia.org/wiki/Finite_difference#Higher-order_differences

 

On 5/23/2019 at 11:05 PM, studiot said:

It would help if you explained what you are trying to do.

It would help if you answered this question.

For instance are you attempting a numerical solution of a differential equation, looking for end point matching curvatures in a finite difference mesh, or trying to calculate over a finite element mesh?

I wonder if you are mixing up the finite or discrete calculus with the analytical calculus?

The definition in post 1 is missing the Limit that is used to avoid the division by zero in analytical calculus.

But It is an extremely poor way to undertake finite differences.
That may be due to the reluctance of some Americans to acknowledge and use the capital Greek delta for the differences.

 

Edited August 25, 2019 by studiot

[113]

1 hour ago, studiot said:

 

It would help if you answered this question.

 

Let's choose an example  \[ f(x) = x^2 \]

using the definition in my first post, obtain the second derivative of f(x)

Edited August 25, 2019 by 113

[studiot]

24 minutes ago, 113 said:

Let's choose an example  f(x) = x^2

using the definition in my first post, obtain the second derivative of f(x)

This is a purely analytical example, so why are you wanting to use a definition from an article entitled

Quote

I am using a finite difference approximation called "Second order forward" from the link, I use dx instead of h:

https://en.wikipedia.org/wiki/Finite_difference#Higher-order_differences

 

Do you understand the distinction between the finite difference calculus and the analytical calculus?

[113]

18 minutes ago, studiot said:

This is a purely analytical example, so why are you wanting to use a definition from an article entitled

just to see how useful an infinitesimal approach is

[studiot]

47 minutes ago, 113 said:

just to see how useful an infinitesimal approach is

 

Finite differences are not infinitesimals.

 

Further have you considered the second analytic derivative in your example, which is a constant ?

[113]

15 minutes ago, studiot said:

 

Finite differences are not infinitesimals.

In my first post dx is an infinitesimal

15 minutes ago, studiot said:

 

Further have you considered the second analytic derivative in your example, which is a constant ?

yes, I have obtained f''(x) = 2 using the definition in my first post

[studiot]

1 hour ago, studiot said:

Do you understand the distinction between the finite difference calculus and the analytical calculus?

Please answer this question directly.
Otherwise I can see no point continuing the conversation.

[113]

10 minutes ago, studiot said:

Please answer this question directly.
Otherwise I can see no point continuing the conversation.

I am not here to talk about those subjects. There are already enough books about them available. In the beginning, in my second post, I told that I am dealing

with an infinitesimal approach: dx is nilsquare infinitesimal, meaning (dx)² = 0  is true, but dx=0 need not be true at the same time.

[113]

There is a book available, even for free download,  A Primer of Infinitesimal Analysis by John L.Bell. It is possibly what I am looking for. The book says that: "A remarkable recent development in mathematics is the refounding, on a rigorous basis, of the idea of infinitesimal quantity, a notion which, before being supplanted in the nineteenth century by the limit concept, played a seminal role within the calculus and mathematical analysis."-direct quote

Also an interesting note from the book: "A final remark: The theory of infinitesimals presented here should not be confused with that known as nonstandard analysis, invented by Abraham Robinson in the 1960s. The infinitesimals figuring in his formulation are ‘invertible’ (arising, in fact, as the ‘reciprocals’ of infinitely large quantities), while those with which we shall be concerned, being nilpotent, cannot possess inverses." -direct quote

Edited August 29, 2019 by 113

[113]

On 8/29/2019 at 7:32 PM, 113 said:

There is a book available, even for free download,  A Primer of Infinitesimal Analysis by John L.Bell. It is possibly what I am looking for. The book says that: "A remarkable recent development in mathematics is the refounding, on a rigorous basis, of the idea of infinitesimal quantity, a notion which, before being supplanted in the nineteenth century by the limit concept, played a seminal role within the calculus and mathematical analysis."-direct quote

Also an interesting note from the book: "A final remark: The theory of infinitesimals presented here should not be confused with that known as nonstandard analysis, invented by Abraham Robinson in the 1960s. The infinitesimals figuring in his formulation are ‘invertible’ (arising, in fact, as the ‘reciprocals’ of infinitely large quantities), while those with which we shall be concerned, being nilpotent, cannot possess inverses." -direct quote

The question is:  why can't John L. Bell's nilpotent infinitesimals possess inverses?

[studiot]

On 8/25/2019 at 8:49 PM, 113 said:

I am not here to talk about those subjects. There are already enough books about them available. In the beginning, in my second post, I told that I am dealing

with an infinitesimal approach: dx is nilsquare infinitesimal, meaning (dx)² = 0  is true, but dx=0 need not be true at the same time.

 

I was in two minds about continuing this discussion.

However although you belatedly (second post) announced that you are considering dx as an 'infinitesimal', you still haven't clarified first opening post.

What type  of algebra are you using that allows you to write

f(x + 2dx) ?

What makes you think that this exists?

Can you point to a theorem that supports this view?

[113]

8 hours ago, studiot said:

 

I was in two minds about continuing this discussion.

However although you belatedly (second post) announced that you are considering dx as an 'infinitesimal', you still haven't clarified first opening post.

What type  of algebra are you using that allows you to write

f(x + 2dx) ?

\[ f'(x) = \frac{f(x+dx) - f(x)}{dx}\]

\[ f'(x + dx) = \frac{f(x + 2dx) - f(x + dx)}{dx}\]

\[ f''(x) = \frac{df'(x)}{dx}\ = \frac{f'(x+dx) - f'(x)}{dx}\]

from which, after a calculation (I skip writing this lengthy LaTeX code now, you may try it yourself), it is possible to get the result in my first post, the definition of second derivative:

\[ f''(x) = \frac{f(x+2dx) - 2f(x + dx) + f(x)}{(dx)^2}\]

 

 

Edited September 1, 2019 by 113