My Math Forum  

Go Back   My Math Forum > Math Forums > Math

Math General Math Forum - For general math related discussion and news


Thanks Tree5Thanks
Closed Thread
 
LinkBack Thread Tools Display Modes
February 7th, 2019, 06:36 AM   #1
Newbie
 
Joined: Feb 2019
From: 48/3 Maniya Housing Society Khalid Bin Waleed Road Karachi Pakistan

Posts: 14
Thanks: 0

Math Focus: Foundations, Philosophy
On the New Calculus

I would like for my idea on taking a derivative using a new technique evaluated:

I have made the post more concise and precise here. All prior messages can be disregarded (particularly the partial definition of a straight line).

Some theory (below) before the definitions. This is to convey how I conceptualized the beginning of what I believe is a "New Calculus". I thought about how an arc of a circle can be defined in a topological sense. This led me to the definition of a line, and eventually, a circle, square, rectangle, and equilateral triangle. Therefore, I want to convey the initial part, as it is paramount to the understanding of what follows in terms of definitions.

A circle is formed by very short lines ( not horizontal or vertical, but "angulated") on the circumference, as the limit of those lines, “delta x,” go to zero. So, the circumference of a circle is the sum of these small lines that tend to go to zero in the limit. Importantly, though, these lines are not horizontal or vertical, thereby not providing epsilon as the limit (as they go to zero in the limit).

Suppose that there are two points on a circle's circumference. The curvature between the points comprises of small lines “delta x”, where the limit of each delta x goes to zero.

If the distance between the curvature is formed by “angulated” distances (very small), then the curvature (or the arc would never form). It would be in a sense chiseled shaped object. As delta x goes to zero (x being the distance between two points on the circumference of a circle), the sum of these lines forms the circumference of a circle.

So, the circumference of a circle is a sum of pieces (or distances) X, where the limit of delta X goes to 0.

It is summation of delta x where the limit of delta x goes to zero. Or X the curvature is the sum of pieces delta x as the limit of each delta x goes to zero

The summation of delta x as x goes to zero is the circumference of the circle

From a point X on the circumference of a circle, the distance between point X and point Y (on the circumference) is the summation of delta x's as the limit of each delta x goes to zero.

A straight line (whether vertical or horizontal ) defined upon a certain space is formed of adjoining distances, delta x, as the limit of each distance delta x goes to zero resulting in an error e. The summation of (0 to n) of these e's= ne. Therefore, the length of a line over a certain space can be defined as ne.

A square can be defined by its diagonal, let's call it s.
This is because all sides are equal and the diagonal determines the length of the sides.

For a square which can be defined by its diagonal:

Sum of delta x as the limit of delta x goes to zero.

(The diagonal)^2= (Summation (0 to n) as the limit of each delta x goes to zero)^2. + (summation (o to n) as the limit of each delta x goes to zero)^2

Thus the diagonal= Square Root( (summation (o to n) of e)^2+ Square Root (Summation (o to n) of e)^2

diagonal= square root( (ne)^2+(ne)^2)

Thus the diagonal = sqrt(2*(n *e)^2)
Thus the diagonal= sqrt(2)*en

A circle's Circumference= 2 pi r
Circumference= 2 pi ((Summation (0 to n)(lim as delta r goes to zero)
Circumference= 2 pi (Summation (0 to n) of e)
Circumference= 2 pi (ne)

The above definition works because even if r is not a horizontal line, it should give the same result as a horizontal line because all the lines from the center of the circle to the cirumference comprise of the r, and thus should provide an equivalent result. This is a necessary and sufficient truth, that is, by definition it follows that all the "rs" should provide the same result.

Moreover,

definition of (r)=circumference/2pi= definition of a circumference/2pi=2pi(ne))/2pi=ne

definition of an area of a circle:

Area of a circle= pi (r^2)
definition of the area of a circle= pi (summation (n=0 to n=n)(limit of delta x as it goes to zero)^2
=pi(en)^2

A rectangle’s diagonal can be defined by:

Diagonal ^2= ( Summation (n=0 to n=n)(limit of delta x as it goes to zero)) ^2+(summation (j=o to j=k) (limit of delta y as it goes to zero)^2

Diagonal^2= (Summation from 0 to n of e)^2+(summation from j=0 to j=k of e)^2
Diagonal^2= (ne)^2+ (ke)^2
Diagonal^2= n^2(e^2) + (k^2)e^2
Diagonal^2= e^2(n^2+k^2)
Diagonal= e* sqrt(n^2+k^2)

where n is not equal to k

Equilateral Triangle

Altitude of an Equilateral triangle =(1/2) * √3 * a

where a is the length of the side. Let's call the altitude h

a= 2h/Sqrt(3)

a=2/sqrt(3) *{(sum (o to n))(limit as delta h goes to 0)

a=2/Sqrt(3) *en

perimeter of an equilateral triangle = 3a
definition of perimeter= (6/sqrt(3)) *en

Isoceles Triangle: Let the 2 identical sides be 'b'. Let's call the height of the triangle as h. Let's call the base (the third side) 'a'.

Then h = sqrt (b^2-1/4(a^2))

h^2=b^2-(1/4)(a^2)

b^2= h^2+ (1/4)(a^2)

b=Sqrt (h^2+(1/4)(a^2))

definition of b= sqrt((en)^2+1/4(en)^2)

The above follows because a is a horizontal line and h is a vertical line.

definition of b = Sqrt(5/4(en^2))
definition of b = Sqrt(5/4) *en
definition of b = sqrt(5)/2 *en

Also,

definition of b^2= h^2+(1/4(a^2)

= (en)^2+(1/4(en)^2)
= 5/4(en)^2

Therefore, definition of b =
b = sqrt(5/4) en
b =sqrt(5)/2*en

definition of h = en
definition of a = en

perimeter = a+2b
definition of the perimeter = definition (of a)+ definition (of 2b)
definition of the perimeter = en+2 ((Sqrt(5)/2)en)
definition of the perimeter = en+sqrt(5)en

surface area of a cube = 6* (l^2) where l is the length of the side

definition of a surface area of the cube = 6* (en)^2
definition of a length of the side of a cube = en

Note: the e, epsilon,is either horizontal or vertical when defining limits.

Special right triangle 45-45-90 with two legs equal inscribed at the origin: (en)^2+(en)^2= hypotenuse^2
hypotenuse^2 = 2(en)^2
Hypotenuse = Sqrt (2) en

Therefore, the definition of the line y=x is Sqrt(2)en
This is because the line y=x cuts the origin on the Cartesian plane in two pieces of 45 degree angles and is represented by the hypotenuse in the above theory.

Similarly y = -x has the same definition. This is because space can exist, but the contrary doesn't hold true. Non-existence is not a finite quantity.

Inscribing a circle with center at origin will not produce definitions of lines, such as y = 2x, as a circle has tangible boundaries, whereas the special right triangle (or its diagonal in this instance) on the positive side of the Cartesian plane can stretch as long as possible, even infinitely longer. In other words, the line y = x would have bounds within the circumference of a circle. It wouldn't define it appropriately.

Among the transformations, translation doesn't change the definition of an object. Therefore, where the shape is placed on the Cartesian please doesn't change its definition.
This is easily understood with the basic example:

derivative of y = nx + a
dy/dx = n

Therefore, the position of the object along translations doesn't change the derivative (whether it is a positions sideways or b positions high or low). Likewise, for the definition of the derivative y = nx + a would not be different since y = x + a has the same definition as y = x. Therefore, the translations of objects wouldn't change their definitions.

Furthermore, let's equate one unit of the line with one unit on the Cartesian Plane. Then, a line defined on the plane becomes intuitively more accessible.

For instance, the definition of a horizontal line defined upon a space (0 to 10 on the Cartesian Plane for the sake of assumption) would be 10e (that is, n = 10).

This is not a necessary and sufficient truth as we can also define 2 units on the Cartesian plane and equate it with each of n pieces of a line.

However, equating one unit on the Cartesian Plane with one e on the line seems to be a reasonable template to further develop these ideas.

Once we accept the assumption that one unit on the Cartesian plane corresponds with one unit of the line en, then half a unit of the line (or, say, 9.5) will correspond with .5e. This is because the piece e is proportional to its distance on a piece of a line (this is because the line is either vertical or horizontal in this case).

N.B Definition of A = def (B+C)

Def A = def (B) + def (C)

Similarly,

Definition of A = def (B - C)
Def (A) = Def (B) - Def(C)

Def(A) = Def (B + C)
Def(A)/a = Def (B + C)/a where a is any positive number

Def f derivative of x = (def(f(x)+ delta x) -def (f(x))/ def (delta x)
Where limit of delta x goes to zero.

Suppose f(x) = x

Definition of f`(x) = lim as delta x goes to zero (def (f(X) + delta x)) - def f(x))/def (delta x)

def (f(X)) = sqrt(2)en
def (delta x) = e
Def(f(x) + delta x) = Sqrt(2) en+e
DEf (f(x) + delta x) - def (f(x))= Sqrt(2)en + e - Sqrt(2)(en)
DEf (f(x) + delta x) - def (f(x))/def(deltaX) = (Sqrt(2)en+e-Sqrt(2)(en))/e
= e((Sqrt(2n) + 1 - Sqrt(2n))/e
= 1

Note that when we substitute Sqrt(2) en as the definition, we have already taken the limit to zero of f(x) or f(x + delta x)

for the definition of the derivative of y = 2x

def 2(x+deltax)-def (2x)/def (delta x)
2 def(x+delta x)- def (2x)/def(delta x)
I have mentioned in the above theory how to find the definition of the line y = 2x.

Since
Def (A) = Def(aB) where a is an integer
Then Def(A) = a(Def B)

Therefore it follows,
def y = def (2x)
def y = 2( Def x)

def y = 2*Sqrt(2) en
= 2^(3/2) en
def (f(X)) = 2^(3/2) en
Def(f(x)+delta x) = 2(f(X) + delta x) = 2(Sqrt(2) en + e)
Def (delta x) = e

Therefore, definition of a derivative = (2(Sqrt(2)en + e)) - 2^(3/2)en)/e
= ((2^3/2)en + 2e_ - 2^(3/2)en)/e
= 2e/e = 2

For the definition of the derivative of y = nx
definition of nx = n Sqrt(2) en
DEf(f(x)) = definition of nx = n Sqrt(2)e(n1)
where n1 is the subscript to distinguish from the other n
Def (F(X+Delta X) = n(Sqrt 2*e(n1)+e)
Def (Delta X) = e

Therefore, the definition of the derivative is
n((Sqrt(2)*e(n1) + e) - n Sqrt(2)*e(n1)/e
n(Sqrt(2)* e (n1) + e - Sqrt (2)*e(n1)/e
Then n(e)/e = n

Therefore, the definition of the derivative nx = n

N.B The limits of delta x approaching zero are embedded in the definitions.

On the Philosophy of Mathematics:

Please further note that the “New Calculus” shows that the laws of Mathematics are universal, that despite the differences in this calculus, we get equations which are workable and mean something. These laws of Mathematics exist in nature, just like the laws of physics. We decode these laws based upon logic,too, but the very birth of these ideas is mostly intuitive. Even then, so, we “discover” abstract Mathematics which exists in space, truthful, but we can only grasp this through our basic intuitions at first, and “then” justify it through logic. But something should not emerge correct based on something that is incorrect, however, since these are the laws of Mathematics (which in its truest form exist in nature), they can be approached somewhat correctly from wrong foundations. This is the proof for the very existence of these laws, that despite the flaws in the Old Calculus, we could get correct results a lot of the times. This is because we accessed that part of Mathematics in nature that we could have approached from those incorrect foundations. If there weren't these laws, then the Old Calculus would never have come to be and all the Mathematical foundations would have to be 100 percent correct for Mathematics to work (even) slightly.
For example def derivative = def (ax) = a def(x) where a is an integer.
However, how do we come to accept this assumption?

We do trial and error with equations and realize that this works. Then we accept this assumption based upon the evidence that it leads to solid conclusions. This is based on intuition and discovery rather than being a self-evident truth. Thus, we discover Mathematics rather than create it ourselves. And, since we discover Mathematics, it exists with its laws in nature just like the laws of physics.

For instance, E (aX + b) = aE(x) + b follows from integrands which have an axiom that integral of (a* f(x)) with limits from b to c is the equivalent to (a*(integral of f(x) with limits from b to c)). This is one of the foundations of calculus, but the rule has its justification in discovery rather than it being self-evident (that is, even if it is axiomatic, it is not self-evident).

If we couldn't see 2 + 2 = 4 in this universe, then it would be impossible to know what 2 + 2 would amount to; the only way to know it is 4 in this universe is to add two objects with two other objects. Once, then, we know it is 4, we can go further and claim 1 and 3 equal 4 using the logic: that is, half of two is one and 1.5 of 2 is 3. Therefore, 1 and 3 add up to 4. Logic follows from experience. In this case the discovery is that someone saw 2 objects and 2 other objects, and realized 2 times 2 is 4. Then we could realize what 4 amounts to. And, then we realized that 3 is a so and so fraction of 4, and 1 is so and so fraction of 2 (or 4).

In other words, these numbers exist in nature and are (were) a matter of discovery rather than being based on the notion that their properties are self-evident.
We can imagine a universe without any objects; in this universe, there wouldn't be any laws of Mathematics since the concept of a number wouldn't exist. There would be no logic. But since the universe exists with objects in space, there is logic and Mathematics. Mathematics is conditional on the existence of something, a finite quantity.
In other words there will not be the notion of divisibility in a universe with no objects. Since objects exist in space, there is Mathematics. Thus, the existence of Mathematics is conditional on the existence of the finite–and “hence” limit of delta x going to zero is something finite and not the non-existence of space, that is zero– and thus, discovery. It is a discovery because there are multiple objects in space from which we infer that, say, 2 + 2 = 4
Laws of mathematics vanish when we consider a universe without an object, and as the objects in the universe are created (discovered), so are the laws of Mathematics.

e(n+5) - e(n) = 5e
e(n+a) - e(n) = a*e

This is because there are a spaces between e(n + a) and en, with each of those spaces going in the limit to zero and yielding e.

For the definition of derivative of x^2 = def f(x + delta x) - def (f(x))/e
therefore the definition of ((x+deltax)^2 - def (x^2))/e
Therefore, (def (x^2) + def(2*x*delta x) + def (delata x)^2 - def(X^2))/e

Therefore, (def(2*x*delta x) + def (delta x)^2)/e

def (Delta x)^2 is the definition of a parabola.

Like other objects that have been defined before, the definition of the parabola would remain the same regardless of where it is situated on the Cartesian plane. So, we can consider the definition of a parabola which is inscribed at the origin, and replace it in the equation above.
(x-h)^2 = 4p(y-k)
focus is (h,k + p)
directrix y = k - p
vertex is (0,0)
(defy = en)
def (k-p) = en

focus on (0,0)
k + p = 0
k = -p

x^2 = 4p(y - k)
x^2 = 4 p(y + p)
= 4py + 4p^2
y = 0
x^2 = 4p^2

def (y) = en
y - k = -p
k - y = p
defk - en = defp

x^2 = 4p(y - k)
x^2 = 4p(k - p - k)
x^2 = 4p(-p)
x^2 = -4p^2
def (x^2) = -def (4p^2)
Since p is a point on a line, the definition should be e
def (X^2) = -4e^2

Which means for any (p,q) the limit as (p,q) goes to zero is e.

Therefore, (def(2*x*delta x) + def (delta x)^2)/e
Therefore, ((def(2*x*delta x) - 4e^2)/e
Moreover, def (2x*deltax)/e - 4e

N.B: When we solve limits that can be evaluated using conjugates, how do we know which result is correct?

For instance,

A typical example is: Evaluate lim as x goes to 4 of ((x^1/2)-2)/(x-4)

Right here when we look at the limits, it comes out as 0/0 (when using the Newtonian Calculus). Then we use the conjugates and the result of the limit is 1/4. However, how do we come to accept the second conclusion and disregard the first conclusion? Because it gives, we think, an answer that is correct. However, from a logical point of view both the conclusions have merit (when thinking using the Newtonian Calculus).

This is why I am making the argument that there are problems with the old calculus and the justifications aren't terse. It is by trial and error that we realize what works and what doesn't. What is accepted, then, becomes part of the books. Calculus is based on incorrect foundations and has a lot of logical inconsistency. I encourage people to work at what I am calling the "New Calculus".

def y = x + a Where a is an integer is def y = x since y=x+a is a translation y = x and the definitions don't change under translations.

As long as a line is bounded within space, it is a partial line. Definition of say y = x requires the definition of the complete line.


The basic idea behind my method is that as space gets lesser and lesser, it converges to a finite point e. That is, as the limit goes to zero, it results in epsilon. Then if we take equal spaces on the Cartesian plane, each results in the same limit e. A line defined on a space, therefore, results in e + e + e... + en where n is the subscript. This results in the definition of line en. The rest just follows from this idea. We take definitions of objects, and use the Newtonian method of the derivative to find similar results using the definitions.
This may have benefits beyond what we may conceive at this point. You may look at the derivative of nx to check that it cannot be a fluke. The derivative of x may have happened by chance, but nx adds veracity to the idea.

Note when we define lines like y = x, they stretch infinite. Here, though, we can define a partial line. Like y = 5 between so and so, that is, y = 5 between x = 2 and x = 3.

In this manner, we can develop derivatives of objects not possible before, but it is a starting point, from where we can further develop this new Mathematics. There has to be some benefit to this method.

Last edited by skipjack; February 7th, 2019 at 03:17 PM.
omeradmani is offline  
 
February 9th, 2019, 11:45 PM   #2
Newbie
 
Joined: Feb 2019
From: 48/3 Maniya Housing Society Khalid Bin Waleed Road Karachi Pakistan

Posts: 14
Thanks: 0

Math Focus: Foundations, Philosophy
Can anyone attempt to find the definition of x*delta x in the way other definitions have been taken?
omeradmani is offline  
February 10th, 2019, 12:05 AM   #3
Global Moderator
 
Joined: Dec 2006

Posts: 20,966
Thanks: 2216

Can you boil all that down to just one short paragraph? At present, you seem to be rambling incoherently all over the place, with no useful content at all.
skipjack is offline  
February 10th, 2019, 12:41 AM   #4
Newbie
 
Joined: Feb 2019
From: 48/3 Maniya Housing Society Khalid Bin Waleed Road Karachi Pakistan

Posts: 14
Thanks: 0

Math Focus: Foundations, Philosophy
If you can’t understand it, it is not my fault.

This is the summary:

The basic idea behind my method is that as space gets lesser and lesser, it converges to a finite point e. That is, as the limit goes to zero (of delta x), it results in epsilon. Then if we take equal spaces on the Cartesian plane, each results in the same limit e. A line defined on a space, therefore, results in e + e + e... + en where n is the subscript. This results in the definition of line en. The rest just follows from this idea. We take definitions of objects, and use the Newtonian method of the derivative to find similar results using the definitions.

It is useful. I have taken the derivative of x and nx using a different technique.

Last edited by skipjack; February 10th, 2019 at 01:19 AM.
omeradmani is offline  
February 10th, 2019, 01:19 AM   #5
Global Moderator
 
Joined: Dec 2006

Posts: 20,966
Thanks: 2216

Why is that useful?
Thanks from topsquark
skipjack is offline  
February 10th, 2019, 04:01 AM   #6
Math Team
 
topsquark's Avatar
 
Joined: May 2013
From: The Astral plane

Posts: 2,266
Thanks: 934

Math Focus: Wibbly wobbly timey-wimey stuff.
Can you please do me a favor and demonstrate your method of taking the derivative of the function $\displaystyle y = 3x^2 - 2x + 4$?

-Dan
topsquark is online now  
February 10th, 2019, 04:11 AM   #7
Newbie
 
Joined: Feb 2019
From: 48/3 Maniya Housing Society Khalid Bin Waleed Road Karachi Pakistan

Posts: 14
Thanks: 0

Math Focus: Foundations, Philosophy
Dear Topsquark,

Please read it carefully. X^2 is a work in progress.
omeradmani is offline  
February 10th, 2019, 04:39 AM   #8
Senior Member
 
Joined: Dec 2015
From: somewhere

Posts: 634
Thanks: 91

Since it gives the same result , what is it used for ?
Thanks from omeradmani
idontknow is offline  
February 10th, 2019, 07:25 AM   #9
Newbie
 
Joined: Feb 2019
From: 48/3 Maniya Housing Society Khalid Bin Waleed Road Karachi Pakistan

Posts: 14
Thanks: 0

Math Focus: Foundations, Philosophy
Idontknow,

Why did you delete your post?

And, it may give different results if developed further.
omeradmani is offline  
February 10th, 2019, 08:09 AM   #10
Math Team
 
topsquark's Avatar
 
Joined: May 2013
From: The Astral plane

Posts: 2,266
Thanks: 934

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
Originally Posted by omeradmani View Post
Dear Topsquark,

Please read it carefully. X^2 is a work in progress.
Then what good is your method? I'd suggest waiting to introduce your work until you could at least do simple calculations or ask for help in working it out.

-Dan
Thanks from SDK
topsquark is online now  
Closed Thread

  My Math Forum > Math Forums > Math

Tags
calculus, derivative, foundations, new mathematics



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Calculus Made Easy or A first course in calculus Happy Math Books 0 December 13th, 2014 03:20 AM
Is is it possible to learn pre-calculus and calculus quickly? KO1337 Academic Guidance 2 May 22nd, 2014 04:41 AM
Which Calculus:A First Course in Calculus VS Thomas Calculus pingpong Math Books 0 August 13th, 2013 03:25 PM
Calculus help Bird Calculus 3 February 19th, 2012 11:52 AM
What pre-calculus needed for this calculus class? tdod Calculus 3 December 14th, 2011 11:24 AM





Copyright © 2019 My Math Forum. All rights reserved.