Gravity (split)

[Strange]

I have given you that formula. If you experiment with it in the 2 modes I have said. You will get the formulas mentioned. Namely

circumference Cir=2*Pi*r=Pi*D,

Circle's area Pi*(r^2),

sphere's area 4*Pi*(r^2)=Pi*(D^2),

sphere's volume (4/3)*Pi*(r^3).

All those familiar formulas, of circular geometry that were developed (derived) with calculus.

There is a natural system within, & the key to calculus.

 

The catch is those are calculus results. That's a calculus summary then.

 

And you now need to show that these results are incorrect.

 

Or stop claiming that calculus gives the wrong results.

[Capiert]

And you now need to show that these results are incorrect.

I did not say those results are not correct.

That is the problem.

They look to me like they are derived from another formula,

which you do not use.

Namely a derivative

y'=(x^exp)-((x-1)^exp)

if the original formula is

y=x^exp.

exp=exponent.

 

If so then

C=(my derivative)-(your standard derivative).

 

Or stop claiming that calculus gives the wrong results.

?

[Strange]

They look to me like they are derived from another formula,

 

Please show this derivation.

[imatfaal]

...

Namely a derivative

y'=(x^exp)-((x-1)^exp)

if the original formula is

y=x^exp.

exp=exponent.

 

Can you talk me through the above?

 

[latex]y = x^z[/latex]

 

and you say that

 

[latex]\frac{dy}{dx} = (x^z) - ((x-1)^z)[/latex]

 

ok so call z=2 and look at when the graph crosses the y axis

 

graphically we know that the slope of the curve at that point is horizontal (draw it if you do not believe me) ie zero but your formula would have the slope as -1 (which is 45 degree slope from top left to bottom right )

[Capiert]

Before I do, 1 note. Your standard deriviative gives only the most significant term.

The other terms (=rest of the details) are not included in your standard (=approximation).

 

The method is finite element calculus, starting with a circle.

Using what we know (cirles area), I chop it up into (N=)4 pieces, =3rings+core.

(Caution: I like to wrongly say 4 rings, for simplicity sake, calling the core: a ring without a hole.)

Each element can be labled.

 

The nth Dimension formula uses other syntax.

Similarities are r=x, n=exp.

It's irritating, not to confuse them.

I've attempted to avoid e in the past, respecting confusion with Euler's.

Instead I used n for the exponent, but need that for element number.

Today I used exp for the previous mail. Last minute decision.

 

Ok so far?

[swansont]

Before I do, 1 note. Your standard deriviative gives only the most significant term.

The other terms (=rest of the details) are not included in your standard (=approximation).

 

Ok so far?

No. You've been asked for examples. Give examples.

 

Such as x^2

 

Is the derivative not 2x? Enlighten us.

 

Use another example if you wish, but give an example.

[ajb]

Capiert -- try wikipedia first.

[Capiert]

Can you talk me through the above?

I will try.

 

[latex]y = x^z[/latex]

 

and you say that

 

[latex]\frac{dy}{dx} = (x^z) - ((x-1)^z)[/latex]

No, that looks only partly correct. The dx looks like too much. Other than that it looks like you have transscribed correctly. I think dx=x/N. d=1/N is the basis.

[latex]{dy}= (x^z) - ((x-1)^z)[/latex]

Then I would expect

[latex]\frac{dy}{dx} = ((x^z) - (x-1)^z)*\frac{N}{x}[/latex]

 

ok so call z=2

ok

and look at when the graph crosses the y axis

I've never plotted the beast (before). Ok. x=0.5 (y'=0)

But your suggestion looks like a good idea.

 

graphically we know that the slope of the curve at that point is horizontal

?

(draw it if you do not believe me)

?

I'm sorry, visually it does not quite look exactly horizontal to me when I magnify in excel.

I get a slight indication that the curve (y=x^2) is not completely horizontal, at x=0.5

& that (=my) derivative has a slope of m=2, & an offset b=-1.

y'=2*x-1.

i.e. zero

?

but your formula would have the slope as -1

(which is 45 degree slope from top left

to bottom right )

?

I'm sorry, that's not what I get.

It looks more like 60 degrees to me.

Edited May 31, 2016 by Capiert

[swansont]

I will try.

No, that looks only partly correct. The dx looks like too much. Other than that it looks like you have transscribed correctly. I think dx=x/N. d=1/N is the basis.

.

What, pray tell, is N?

[Capiert]

What, pray tell, is N?

N is the number of elements that we divide the thing into.

If 1=100%, then we can e.g. divide into N slices, if it were a loaf of bread.

Infinitesimal calculus makes dx very small without saying exactly how small.

Here we have a way to set the thickness, &/or resolution.

Please show this derivation.

Exacta (prep)

 

Sun 2016 05 22 05:56 PS Wi 11.3 C blue

 

Ok, let' see how we got that (nth Dimension formula).

To be thorough we should go thru every step,

till we have enough info to make the formula (to test).

I'll try to make it painless.

But caution!

I am going to use the same (syntax) symbol

n#nth

(they) are NOT the same!

 

To start with:

 

Take a point

 

(Sorry I don't know an easy, no cost, drawing program,

with raster, editable with formula numbers.)

 

draw a circle

with radius

r=1 [cm]. (But we'll drop units to make it easy.)

 

The line thru the center (point) is the diameter D=2*r.

 

We notice the circumference Cir=Pi*D

is (factor) Pi=3.14 larger than the line (D).

 

We're told the circle's area is

A=Pi*(r^2) (from school).

A1=Pi*1

 

Doubling the radius (r2=2), we get

A2=Pi*(r2^2)

A2=Pi*4

 

The area_ring is

AR2=A2-A1=Pi*(4-1)=Pi*3

 

Let's continue 2 more times, for r3=3 & r4=4.

 

A3=Pi*(r3^2)=Pi*9

A4=Pi*(r4^2)=Pi*16.

 

Their rings(' area) are

AR3=A3-A2=Pi*(9-4)=Pi*5

AR4=A4-A3=Pi*(16-9)=Pi*7.

 

What do we notice?

The areas are Pi*(1, 4, 9, 16), depends on r^2

The ring areas are Pi*(1, 3, 5, 7), +2 added (increments).

 

For those who don't know Pi*(1, 4, 9, 16)=Pi*1, Pi*4, Pi*9, Pi*16.

..06:40

 

If we divide those by Pi it's

(1, 4, 9, 16) circle areas

(1, 3, 5, 7) ring areas

 

(1st ring has no hole, so it's not really a ring, but for simplicity (to avoid distraction).

Tell me what to call it.)

 

..06:50

 

The number of rings we had (there) were N=4.

Each ring had its own number n=1, 2, 3, 4.

 

(You call them "elements".)

 

So we know the name of each element.

AR(1, 2, 3, 4), ring's_area. (=Area of ring number n).

 

Each ring_area is

ARn=An-A(n-1)

the nth circle's_area minus the previous (circle's area) n-1.

i.e. the existing number's (circle_area) n, minus the previous n-1 (circle's_area).

 

..07:15

 

What constitutes (=makes up) that? The areas formula, of each ring (=element).

(n is which ring's number)

Thus, any ring's_area is

ARn=Pi*(rn^2)-Pi*((r(n-1))^2), grouping Pi out

ARn=Pi*((rn^2)-((r(n-1))^2)), grouping r out (not really very legal, but we try)

ARn=Pi*(r^2)*((n^2)-((n-1)^2)),

we see the ring's area is determined by a difference term (n^2)-((n-1)^2)

(multiplied by the original (maximum) area, using (maximum) radius r).

 

That is the differential! (Reduced to syntax.) (A part* of what Newton searched (for).)

 

(Is it all there? I doubt it. Why should everything you guys have done be without error.

The Royal society wants skeptics, because science is in a mess, invented by students.

Each different with their own ways.

A professor is still learning. They don't know everything. What for a better student than that?

Nobody can know everything. & what you know?

You got it from somebody else, mostly. There is no consistency.)

 

It's a factor, multipled to the area.

Because that is where it came from (the radius squared).

n^2-(n-1)^2

That is an area calculation, using a squaring, ^2.

That factor now defines the operation (activity)

about subtracting 2 (similar) squared terms (with exponents).

Meaning the original area formula remains original! To obtain an element.

 

We all know a(n infinitesimally) very small element

(limit going to zero thick, or extremely "thin")

is described by a differential. (?)

In terms, (our example was N=4, but now N~oo is almost infinite)

for the 1st element n=1, & the number of elements N=oo (limit).

The thickness (width) is 1/N=1/oo.

 

The integral:

 

Goal: n/N into the integrating formula.

oo=infinity

 

1 (=100%) can be divided into any number N of parts, (not just 100). Each called 1% or 1 cent.

or

1 can be divided into any number N of parts, (not just 4).

 

The number of parts, is not the same as their size.

However, the number N influences their size, inversely (proportionally).

Large N, small size. Small N, large size.

As long as we have an equality (equivalence), we have an exact (accurate) answer.

It does not matter what value N is. It could even be infinity, if possible.

That is amazing accuracy. Predictability.

That means "no" error!

 

The method (integration) is

to multiply the same element (width)

by the number of elements N,

to get the area.

 

The element width is the differential (or difference).

..10:01

 

 

 

ARn/Pi*(r^2)=((n^2)-(n-1)^2))

 

 

I hope that helps a bit, & it's ok for you, otherwise typos got me again.

Edited May 31, 2016 by Capiert

[Strange]

 

 

That means "no" error!

 

So why do you keep saying it doesn't work?

[Capiert]

Capiert -- try wikipedia first.

Oh je! We'll never get done if we do that.

How about later?

So why do you keep saying it doesn't work?

Because the details

(=lesser terms, not the most significant term)

are missing from the approximation (standard derivative, not mine)

(if I'm allowed to say that so).

 

I'm a bit reluctant to call it mine (egoism),

but I don't know how to distinguish the 2 otherwise, publically.

I'd rather be egoistic at/in other things.

Edited May 31, 2016 by Capiert

[swansont]

Are you aware that A=Pi*r^2 is the result you get from calculus?

[Strange]

Because the details

(=lesser terms, not the most significant term)

are missing from the approximation (standard derivative, not mine)

 

 

Are they? Why does it give the right answer then?

 

You keep claiming there are errors, but have still failed to show an example.

[Capiert]

Are you aware that A=Pi*r^2 is the result you get from calculus?

Yes, that is what I meant by catch in post 50.

Are they?

According to that (=my calculations, of nth D formula, & (my) derivative)

those lesser terms seem missing. I cannot find them in your solutions,

when they exist in mine.

Why does it give the right answer then?

That's the puzzle. Riddle if you will.

How are you guys getting the right answer?

Or are you getting the right answer?

Are the circular answers complete?

I addressed a different problem yesterday, (on the COW string)

but nobody answered it.

How do we calculate the volume of a sphere?

Nobody presented the math (of shells) to orient me.

Is this forum here to ignore my questions,

although I answer yours?

If you do it correctly from my perspective*,

then I will see no error.

But if that simple calculation is wrong*,

then I will see an error.

The error is the lack of accuracy, in the approximation.

I must evaluate your method.

It is a reasonable arguement goal.

 

You keep claiming there are errors, but have still failed to show an example.

Please give me the math example I requested.

I will take it from there if worthy.

Edited June 1, 2016 by Capiert

[Strange]

The error is the lack of accuracy, in the approximation.

 

 

You keep claiming this. But have provided no evidence it is true.

[elfmotat]

Is this forum here to ignore my questions,

although I answer yours?

 

If you keep making claims without evidence, yes.

 

Please give me the math example I requested.

I will take it from there if worthy.

 

I don't really understand what you're asking for. The nearest thought to my mind would be the following example:

 

[math]y = x^2[/math]

 

If we introduce a displacement [math]\Delta x[/math], then we have:

 

[math]\Delta y = (x+\Delta x)^2 - x^2 = 2x \Delta x + (\Delta x)^2[/math]

 

or:

 

[math]\frac{\Delta y}{\Delta x} = 2x + \Delta x[/math]

 

This is well-defined for all [math]\Delta x \neq 0[/math]. But, the whole point of calculus is the concept of the limit. If we take the limit as [math]\Delta x \to 0[/math], then:

 

[math]\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} 2x + \Delta x = 2x[/math]

 

We simply define:

 

[math]\frac{dy}{dy} := \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = 2x[/math].

 

Is there something wrong with what I just did? If so, why/how? That's calculus, at its most basic. Alternatively (and more hand-wavy), we can introduce an infinitesimal displacement [math]\Delta x = \epsilon[/math] such that [math]\epsilon^2 = 0[/math]:

 

[math]\Delta y = (x+\epsilon)^2 - x^2 = 2x \epsilon + \epsilon^2 = 2x \epsilon[/math]

 

or:

 

[math]\frac{\Delta y}{\Delta x} = 2x[/math].

 

Is this what you're taking issue with? The introduction of a number such that its square is zero? Both of these methods can be made much more rigorous and well-defined. There is nothing logically inconsistent here, unless you'd care to point out the flaw?

Edited June 1, 2016 by elfmotat

[imatfaal]

I will try.

 

No, that looks only partly correct. The dx looks like too much. Other than that it looks like you have transscribed correctly. I think dx=x/N. d=1/N is the basis.

[latex]{dy}= (x^z) - ((x-1)^z)[/latex]

Then I would expect

[latex]\frac{dy}{dx} = ((x^z) - (x-1)^z)*\frac{N}{x}[/latex]

 

ok

I've never plotted the beast (before). Ok. x=0.5 (y'=0)

But your suggestion looks like a good idea.

 

?

?

I'm sorry, visually it does not quite look exactly horizontal to me when I magnify in excel.

I get a slight indication that the curve (y=x^2) is not completely horizontal, at x=0.5

& that (=my) derivative has a slope of m=2, & an offset b=-1.

y'=2*x-1.

?

?

I'm sorry, that's not what I get.

It looks more like 60 degrees to me.

 

 

https://docs.google.com/spreadsheets/d/1fHskcfO5nJy13S8EH1hN4QXd13eIcX6Xkfq4O8U7W1E/edit?usp=sharing

 

Here is a google spreadsheet. The curve is y=x^2. You can change the increment of x in cell B1 ( 0 is fixed and you get 20 numbers either side). It is initially set at an increment of 1. That looks pretty horizontal to me at the y axis. Feel free to zoom in by changing the increment to anything you like (the graph will not change shape - is that a clue?)

 

That you do not know immediately that the slope of 45 degrees is a slope of 1 or -1 and must MUST MUST correspond with y = x or y = -x is crucial. If you move one space left to right and one space up to down then the slope MUST be -1 and the angle MUST be 45 degrees. This is very simple and not even pre-calc.

 

BTW - you have never plotted y=x^2, it doesn't look as if you know where the y-axis is, don't understand basis graphs, and you have no idea of current knowledge/notation; yet you have the bare-faced temerity to claim that calculus is wrong; that is so arrogant and blinkered to transcend silliness and become bizarre!

[Phi for All]

!

Moderator Note

The OP has failed to support the assertions made in accordance with the rules of the Speculation section. Specifically, continuous claims of errors were never supported by evidence or examples. It's clear that many gaps in mathematics knowledge are hampering successful discussion here.

 

Thread closed. You are not welcome to bring this subject up again until you've studied it quite a bit better.