# definition of the second derivative

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3 hours ago, 113 said:

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}

f′′(x)=f(x+2dx)2f(x+dx)+f(x)(dx)2

Sadly you have completely misunderstood my question and points.

Perhaps this was my fault for not being detailed enough.

You have used the arithmetic operation " + " (amongst others)

My consideration concerns whether this is indeed legitimate and, if so, in what system of algebra?

Consider this:

If I work with the set of integers the axioms guarantee that use of the three operations, addition, subtraction and multiplication will result in another member of that set (ie another integer).

eg 10+7 = 17;  10-7 = 3;  10 * 7 = 70

all work fine and are well defined.

But 10/7 is not defined, and further, reciprocals do not exist in this algebra.

Equally the sum (10 + 10/7) is not defined in this system

Now you have the sum (x + 2dx) which begs two questions

How is multiplication defined in your working?

In other words which system of algebra (arithmetic) are you working?

This is necessary since you cannot be using the standard axioms of arithmetic.

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52 minutes ago, studiot said:

Now you have the sum (x + 2dx) which begs two questions

Do you think the sum (x + 2dx) is using some different system of algebra than, for example,  the sum (x + dx) ?

Quote

How is multiplication defined in your working?

In other words which system of algebra (arithmetic) are you working?

This is necessary since you cannot be using the standard axioms of arithmetic.

I did not invent my own system of arithmetic. I am currently learning what John L. Bell has written in his book. I think you should ask the same question about what axioms of arithmetic are used in an infinitesimal approach: dx is nilsquare infinitesimal, meaning (dx)² = 0  is true, but dx=0 need not be true at the same time.  So how is multiplication defined here? What axioms of arithmetic are being used? Maybe they are to be found is John L.Bell's book, he writes: "As we show in this book, within smooth infinitesimal analysis the basic calculus and differential geometry can be developed along traditional ‘infinitesimal’ lines – with full rigour – using straightforward calculations with infinitesimals in place of the limit concept. And in the 1970s startling new developments in the mathematical discipline of category theory led to the creation of smooth infinitesimal analysis, a rigorous axiomatic theory of nilsquare and nonpunctiform infinitesimals."

Edited by 113
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28 minutes ago, 113 said:

Do you think the sum (x + 2dx) is using some different system of algebra than, for example,  the sum (x + dx) ?

I don't know since you haven't told me.

But whether you use (x+dx) or (x+2x) or (x+anything that is not a real number) is irrelevant until you justify what arithmetic you are using that permits this and defines what the result is.

Do you consider (x+dx) to be a real number or what?

28 minutes ago, 113 said:

I did not invent my own system of arithmetic. I am currently learning what John L. Bell has written in his book. I think you should ask the same question about what axioms of arithmetic are used in an infinitesimal approach: dx is nilsquare infinitesimal, meaning (dx)² = 0  is true, but dx=0 need not be true at the same time.  So how is multiplication defined here? What axioms of arithmetic are being used? Maybe they are to be found is John L.Bell's book, he writes: "As we show in this book, within smooth infinitesimal analysis the basic calculus and differential geometry can be developed along traditional ‘infinitesimal’ lines – with full rigour – using straightforward calculations with infinitesimals in place of the limit concept. And in the 1970s startling new developments in the mathematical discipline of category theory led to the creation of smooth infinitesimal analysis, a rigorous axiomatic theory of nilsquare and nonpunctiform infinitesimals."

You are still missing the points of the questions I am asking.

Since it is your proposition it is for you to state clearly the system of algebra (and its rules) in which you are working.

SIA has problems of its own. Here is an extract from a recent paper.

Mathematicians have struggled with the philosophic problem of dealing with this for centuries, but no one has yet come up with a watertight answer, or a better one that the limit process which has many other uses to boot.

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5 hours ago, studiot said:

I don't know since you haven't told me.

But whether you use (x+dx) or (x+2x) or (x+anything that is not a real number) is irrelevant until you justify what arithmetic you are using that permits this and defines what the result is.

Do you consider (x+dx) to be a real number or what?

I don't know. Maybe the answer can be found in the book I am studying.

John L. Bell is defining the ‘derivative’ of an arbitrary given function f : R → R.

For fixed x in R, define the function g:  Δ → R by g(ε) = f(x + ε) so that

f(x + ε) = f(x) + εf'(x)

is the fundamental equation of the differential calculus in S for arbitrary x in R and ε in Δ.( Δ may be considered an infinitesimal neighbourhood or microneigbourhood of 0).

Also he is stating Microaffiness Axiom:

For any map f:Δ → R there exist unique a, b  ϵ R such that f(ε) = a + bε

for all ε ϵ Δ

He draws a conclusion: Our single most important underlying assumption will be: in S, all curves determined by functions from R to R satisfy the Principle of Microstraightness. The Principle of Microaffineness may be construed as asserting that, in S, the microneighbourhood Δ can be subjected only to translations and rotations, i.e. behaves as if it were an infinitesimal ‘rigid rod’. Δ may also be thought of as a generic tangent vector because Microaffineness entails that it can be ‘brought into coincidence’ with the tangent to any curve at any point on it. Since we will shortly show that Δ does not reduce to a single point, it will be, so to speak, ‘large enough’ to have a slope but ‘too small’ to bend.

Quote

You are still missing the points of the questions I am asking. Since it is your proposition it is for you to state clearly the system of algebra (and its rules) in which you are working.SIA has problems of its own. Here is an extract from a recent paper.Mathematicians have struggled with the philosophic problem of dealing with this for centuries, but no one has yet come up with a watertight answer, or a better one that the limit process which has many other uses to boot.

I don't have all the answers to your questions. I am only studying the subject. Don't expect me to have all the answers if no-one else has been able to find them. I am

looking for them in the books. I am not developing my own system of algebra.

Edited by 113
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4 hours ago, 113 said:

I don't know. Maybe the answer can be found in the book I am studying.

Well Bell is aware of the problem, as were Lawvere, Bishop and the authors of my extract, Hellman and Shapiro.

As far as I am aware all use a scheme (with variations) that runs as follows

Postulate an 'extended' set  R (usually extended with infinitesimals of some sort)

Postulate a suitable set of axioms to deduce the properties that are inheritable in R alone

Show that the usual rules of calculus work in the extended set

Use the axioms to transfer this to claculus in R.

Unfortunately the notation and language employed is not universal - even the word Language is given a special meaning
So this means that a new language needs to be learned to appreciate each one.

Here are some references.

Non Standard analysis  - Martin Davis (book)

Classical continuum without Points -pdf

Bell of course you already have.

Quote

He draws a conclusion: Our single most important underlying assumption will be: in S, all curves determined by functions from R to R satisfy the Principle of Microstraightness. The Principle of Microaffineness may be construed as asserting that, in S, the microneighbourhood Δ can be subjected only to translations and rotations, i.e. behaves as if it were an infinitesimal ‘rigid rod’. Δ may also be thought of as a generic tangent vector because Microaffineness entails that it can be ‘brought into coincidence’ with the tangent to any curve at any point on it. Since we will shortly show that Δ does not reduce to a single point, it will be, so to speak, ‘large enough’ to have a slope but ‘too small’ to bend.

This is remarkably similar to what I say to people about the application of Vector Spaces to physical problems.

That it is not a matter of a simple map from one set to another, but that a third set needs to be involved.

Perhaps this will be the way forward.

Finally you shoud beware of the use of the following symbols

Δx; δx; dx and D(x)

These are used somewhat inconsistently to my way of thinking

Edited by studiot
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• 4 weeks later...
On 9/2/2019 at 12:10 AM, studiot said:

Well Bell is aware of the problem, as were Lawvere, Bishop and the authors of my extract, Hellman and Shapiro.

As far as I am aware all use a scheme (with variations) that runs as follows

Postulate an 'extended' set  R (usually extended with infinitesimals of some sort)

Postulate a suitable set of axioms to deduce the properties that are inheritable in R alone

Show that the usual rules of calculus work in the extended set

Use the axioms to transfer this to claculus in R.

I am beginning to suspect that calculus is not based on real numbers. Look at the definition of the derivative:

$\frac{dy}{dx} = \lim_{h\to\ 0}\frac{f(x+h) - f(x)}{h}$

where h is finite.

What is dy/dx? An infinitesimal ratio? A ratio of two infinitesimals dy and dx ? It seems to me that we are not dealing with real numbers anymore if dy and dx

are not real numbers.

Edited by 113
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dy/dx is a symbol for the derivative - it is not a ratio.

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• 2 weeks later...
On 9/2/2019 at 12:10 AM, studiot said:

Finally you shoud beware of the use of the following symbols

Δx; δx; dx and D(x)

These are used somewhat inconsistently to my way of thinking

The limit defintion of derivative in my previous post contains only the symbols h (corresponding to Δx) and dx. There is no δx.

It seems to me that the introduction of "differential calculus" gives rise to symbol δx. Then there seems to appear two representations:

f'(x) = dy/dx

f'(x) = δy/δx

I think it is possible the usage of δy/δx was chosen to escape the problem arising in real number calculus, the problem of 0/0.

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1 hour ago, 113 said:

The limit defintion of derivative in my previous post contains only the symbols h (corresponding to Δx) and dx. There is no δx.

It seems to me that the introduction of "differential calculus" gives rise to symbol δx. Then there seems to appear two representations:

f'(x) = dy/dx

f'(x) = δy/δx

I think it is possible the usage of δy/δx was chosen to escape the problem arising in real number calculus, the problem of 0/0.

I stand by what I have said.

On 9/1/2019 at 10:10 PM, studiot said:

Finally you shoud beware of the use of the following symbols

Δx; δx; dx and D(x)

These are used somewhat inconsistently to my way of thinking

The use of δx may have started with Leibnitz, but it has been used extensively in many other roles since.

This often clouds his view of δy/δx as a limit as two infinitesimals simultaneously tend to zero.

Edited by studiot
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On 9/1/2019 at 2:18 AM, 113 said:

f(x)=f(x+dx)f(x)dx

f(x+dx)=f(x+2dx)f(x+dx)dx

f′′(x)=df(x)dx =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)=f(x+2dx)2f(x+dx)+f(x)(dx)2

To be rigorous use$\Delta x$,  not $dx$ in the expressions and take limits as $\Delta x \to 0$

.

Edited by mathematic
typo for latex
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• 4 months later...

Well, you can't define or calculate the derivative of a function using only the four arithmetic operations.  I would have thought you would have learned that in whatever course you were introduced to the derivative.  You have to have the concept of a "limit" as well:  $$\frac{df}{dx}(a)= \lim_{h\to 0} \frac{f(a+ h)- f(a)}{h}$$.

Edited by Country Boy
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• 2 months later...

This is what people want you to say:

$f''\left(x\right)=\frac{f\left(x+2\varepsilon\right)+f\left(x\right)-2f\left(x+\varepsilon\right)}{\varepsilon^{2}}+o\left(\varepsilon\right)$

Where,

$\lim_{\varepsilon\rightarrow0}o\left(\varepsilon\right)=0$

Can't be false because it's an identity as long as f(x) is twice differentiable at x. Utter the words and fall on your knees.

Sorry, were it not for CoVid-19 I would be sleeping like a baby.

Edited by joigus
forgot a symbol
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The expression is familiar to me because I've seen it in lattice problems in physics.

Here's a test run of your expression with,

$f\left(x\right)=\sin x$

It should give,

$f''\left(x\right)=-\sin x$

Let's see:

$\frac{\sin\left(x+2\varepsilon\right)+\sin x-2\sin\left(x+\varepsilon\right)}{\varepsilon^{2}}$

Using,

$\sin\left(x+2\varepsilon\right)=\sin x\cos\left(2\varepsilon\right)+\sin\left(2\varepsilon\right)\cos x$

$\sin\left(x+\varepsilon\right)=\sin x\cos\varepsilon+\sin\varepsilon\cos x$

$\cos\left(2\varepsilon\right)=\cos^{2}\varepsilon-\sin^{2}\varepsilon$

We get, after simplifications, to,

$-\sin x-\varepsilon\cos x+\frac{\varepsilon^{2}}{4}\sin x$

Works for powers of x, works for sines and cosines, therefore exponentials and products of them. Works quite well I think when things are smooth.

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