User:Amitavakala

I profusely thank Wikipedia for donating this space to me.
Also I need to state that all the articles written here were done in the midst of extreme mental pressure and anxiety. Some mentally challenged people from other part of the world in collaboration with their counterpart in my land harassed me, maligned me, tried to make me insignificant in all possible ways on the roads, holiday tours, inside office, restaurants (while eating!), days after days months after months, years after years, relentlessly. On the top of that they claim that they made me famous @#! when the truth is that they tried to paint my reputation as most heinous man in the world.Not to mention given access to my personal data, information, private life by hacking my systems to countless people and thereby further aggravating the problem 100 times.

The base of natural logarithm : e
I found the derivation of "e" (the slope method) in the following link http://numbers.computation.free.fr/Constants/E/e.html

I representated the same derivation (slope method) here

y = ex ,

As we know d(ex)/dx = ex, I mean that is the property of e

That means, (e(x+dx) - ex)/dx = ex

edx - 1= dx e = (1 + dx)(1/dx)

If we take n = dx, then in this equation n tends to 0 and 1/n tends to infinity

so e = (1 + n)(1/n) when n tends to 0

It is noted that as we take smaller and smaller value of n, the approximate value of e goes up and up, that indicates this method approaches e from the bottom.

The above equation has been proved by differentiation (slope) method.

The value of e can be approximated by integration method also. For this we need to refer the graph y = ex ( I will try to put the graph later)

now, consider the area confined by 4 points, (x,0),(x+dx,0),(x,ex),(x+dx,e(x+dx)), This should be equal to the area confined by the curve y = ex between two points, x and x+dx, on x axis.

So, (e(x+dx) - ex)(dx/2) + exdx = e(x + dx/2) - ex

(edx-1)(1-dx/2)=dx

e=(1 + dx/(1-dx/2))(1/dx)

or, e = ((2 + dx)/(2-dx))(1/dx)

If we take n = dx, then in this equation n tends to 0 and 1/n tends to infinity

So, e = ((2 + n)/(2 - n))(1/n) when n tends to 0

It is noted, that second (area) method approaches e from the top. It also approaches e faster than first (slope) method.

But why the first method approaches e from the bottom whereas the second method approaches e from the top? The simple answer is for y = ex, dy/dx = ex (+ve),and also d2y/dx2 = ex (+ve), That means, (e(x+dx) - ex)/dx will be always greater than ex for any conceivable value of dx, so e will be always greater than (1 + dx)(1/dx).

For exactly the same reason the second method reaches e from the top.

Next question, why the second method is faster? In the second (area) method edx = 1 + dx/(1 - dx/2). If we neglect the dx/2 in the denominator, we will ultimately reach the first derivation. So simply the error term is less in the second method compared to first method, that is why it approaches e faster.

PI (Π)
Calculating the value of PI is little roundabout.

The formula is: when θ tends to 0 (here θ is in degree), SINθ(180/θ)<π<TANθ(180/θ), being said that, how to calculate SINθ and TANθ for very small value of θ. One easy approach is the use of the formulas TANθ = √(1/TAN22θ + 1) - (1/TAN2θ) and SINθ = TANθ/√(1+TAN2θ). We can start with θ= 45 degree, for which TANθ = 1 and SINθ =1/√2, for which 2.828<π<4. At 14th step we get θ= 0.00549 degree, and 3.1415926372<π<3.1415926516. Remember Π is all about circle, it relates perimeter of a circle to its area. Let us investigate the relation. If the length of the perimeter of a circle is P its area is A, and radius is R, then consider an infinitesimally small length dP on the perimeter P. For this the area subtended by the arc dP is ½(dP×R). If we integrate this area over total perimiter then we get area of the circle.

So ∫0P½(dP×R) = ½(P×R)=A Again consider the same area, if we increase the radius by dR, then the we will have a new circle with radius R+dR=R1, for the same arc dP on first circle we will have an arc dP1 on second circle, and from the property of similar triangle we will find dP1/dP = R1/R, that means ∫dP1/∫dP = R1/R, or P1/P = R1/R, and that means PαR or P = C×R and A = ½(C×R×R) =½C×R2. We just took C as 2Π and derived the value of Π at the beginning.

Circle
One of the important properties of a circle is that it has minimum boundary length with same area among all possible shapes. We can derive that with little imagination. Suppose in the beginning, an irregular area in a flat surface consists of tiny little frictionless particles with infinitesimally small areas. The flat surface is perfectly frictionless and particles are free to move in that flat surface in any direction. The particles also possess the property of gravity. That means they attract each other by gravitational force. There is no other force acting on these particles which has got a component in the flat surface. As the particles attract to each other they will try to squeeze inside the area. This is also true for the particles at the boundary. As particles at the boundary will try to squeeze inside, this will ensure at the end that minimum possible number of particles remain at the boundary, which will create minimum possible length of the boundary. Now at the end all the particles at the boundary will experience same resultant force inward, which means to the direction perpendicular to the direction of the tangent to the boundary at that point. So that no points at the boundary will be able to squeeze more into inside than the adjacent points. If resultant force has a direction which is in the direction of tangent (either side), that will mean that particle will have movement which will have a component in that direction. We already assumed that particles are stable at the end so there will not be any movements for any particles. This indicates resultant force on any particle at the boundary will be in the direction inwards which is perpendicular to the direction of the tangent at the point to the boundary line. This indicates the areas at both side of this perpendicular line have to be same and symmetrical to each other. As this is true for all the points at the boundary, this indicates that the total area has got infinite rotational symmetry. This indicates that the area is a circle. As this is true for particles in two dimension, this is equally applicable for particles in three dimension. Because of the same very reason, among all possible shapes, sphere has the least surface area with same volume. Most celestial objects are spherical in nature because of the same reason.

e and PI (Π)
We know that ex has got the unique property that no matter how many times we differentiate or integrate, it delivers the same value. Is there any other function which behaves in the same way? Elementary users of mathematics(like me) know that sine and cos functions behave “almost “ similar way.

Every double differentiation or integration results same value with a negative sign. Every quadruple differentiation or integration results same value with same sign. Point here to note is that sine and cos functions are related to vectors, whereas ex is a scalar. Keeping this in mind we will try to construct two functions which are same but will be represented in two different forms.

y = eM θ = Asinθ+Bcos θ

We need to find the value of A,B, and M.

Now Asinθ+Bcos θ actually represents a vector magnitude wise. If A is a vector in Y direction and B is a vector in X direction then Asinθ+Bcos θ will represent a vector whose direction will be θ from X axis towards Y axis. The magnitude as well as direction will change if θ changes. This vector will have a magnitude B in X direction when θ=0, and magnitude A in Y direction when θ= Π/2.

Now,

dy/dx gives dy/dx = Acos θ-Bsin θ = M eM θ

d2y/dx 2 = -Asinθ -Bcos θ = M 2eM θ= -y = - eM θ

Which strangely(but also obviously) gives M 2 =-1 or M=√(-1)=i (imaginary number).

Further,

Putting θ=0, in eM θ = Asinθ+Bcosθ, we get B=1, and θ=0, in Acos θ-Bsin θ = M eM θ gives A=M=i

So finally the equation looks like y= cosθ + isinθ= eiθ

Apperance of i is obvious as it gives much needed representation of direction in the equation.

A quick look at the final equation tells us that it is a unit circle in the X,Y plane where the positive X direction is represented by +1, positive Y direction is represented by +i,negative X direction is represented by -1, negative Y direction is represented by -i. Everytime we differentiate the function it just rotates the unit vector by 90 degree (Π/2 radian) anticlockwise, and does not change the magnitude. Integratioin rotates it clockwise.

Putting θ= Π/2, we get i = eiΠ/2

Putting θ= Π, we get -1 = eiΠ

or, eiΠ + 1 =0

Let us consider the equation R = eθ , where R =√(x2+y2)

∫√((Rdθ)2 + dr2) gives the length of the arc. So finally ∫√((Rdθ)2 + dr2)= √2∫Rdθ= √2R

Integrating between R1 and R2 gives the length of the arc √2(R2 – R1)

Let us investigate for area also. Integrating the area of the curve of R = eθ, we get

(1/2)∫(R+dR)Rdθ in this case =1/2∫dR(R+dR)= 1/2∫RdR = 1/2∫e2θdθ = (1/4)R2

So the area of the curve bounded by R1 and R2 is 1/4(R2 - R1). In general (1/2)∫(R2dθ gives the area under the curve for any R = f(θ). Considering this, if R = eθ/4Π, integration (1/2)∫(R2dθ = (1/2)∫eθ/2Πdθ = Πeθ/2Π = ΠR2. So integrating between R1 and R2 gives the area ΠR22 - ΠR12

which is the difference of the areas of the circles of radius R2 and R1. Another observation, Y = SIN(X) gives the sine curve along X axis with max 1 and min -1, R = SIN(θ) from θ = 0 to θ = Π (in 1st and 2nd quadrant)gives a circle of radius 1/2, and center at Y = 1/2, X = 0. R = COS(θ) (in 1st and 4th quadrant) gives a circle of radius 1/2, and center at X = 1/2, Y = 0.

Involute and Cycloid
Where I stopped last time, the curves I drawn of the equations reminded me two popular curves used for gear design, one is involute, whose definition I could readily remember, another is cycloid, whose definition I forgot after so many years.

So I immediately tried to construct the equation of involute, which I never tried before. Finally the equation looks like r = R√(1 + (θ - α)2 ) and  β = (θ - α) - ARCTAN(θ - α) R = base circle radius, α = angle at which "unwind" begins, from X axis, θ = total angle from origin (x) axis up to which the "string" is un-winded. β = angle of the point (which is having a 'radius' r at that moment) from the beginning of the "un-wind". So at any point of time the loci has a radius r and angle β + α from X axis. This is when we consider unwinding in +ve direction. Similar equations can be constructed considering unwinding in -ve direction. With these equations I was able to construct a involute gear teeth (without the undercut). The only constraint is that while designing we will have β (and α, which is a constant) only for our calculation, but r can not be expressed directly as a function of β (I could not do it), but for every β we can readily calculate θ, by which r can be expressed. So the design is possible. Also the area under the curve can be calculated easily by using the formula (1/2)∫(R2dθ. This may be useful while considering mass and cost of material.       --- Next was the curve cycloid, whose definition and illustration is available in wikipedia itself. And again I tried to construct the equation. Considering a circle of radius R and rolled angle θ, x = R(θ - sinθ) y = R(1 - cosθ), so at the starting θ = 0, when the point touched the ground and after 360 deg rotation clockwise, the poin will again touch the ground. Also the area under the curve is ∫ydx = R2∫(1 - cosθ)2dθ. Again the logic follows in earlier way - for each θ, we know the value of x and y. After the integration is performed the eq looks like R2 {(3/2)θ - 2SINθ +(1/4)SIN2θ)}. Integrating from start to end the area is 3ΠR2

When I finished the "Tautochrone curve problem" explained later, I again looked back to the definition of cycloid. So what if we role the circle over another circle, what will be the shape? There can be different combinations where the rotating point will come back to same starting point after only one rotation (outer radius/inner radius is a whole number).

1) both the circles have same radius r = R√(5-4cosθ)  http://cid-8f009c5ac338df79.photos.live.com/self.aspx/curves/cycloido.jpg

2) the center circle has very small radius.

http://cid-8f009c5ac338df79.photos.live.com/self.aspx/curves/cyccloid0.jpg --- Up to this far everything was smooth, but only then I stumbled on two famous problems associated with cycloid. One is Brachistochrone curve, another is Tautochrone curve, to know the problem of the first one inadvertently I also saw the solution. But then only I came to know one need to have vast knowledge in calculus to solve it independently. Nevertheless Tautochrone curve was left where I could try to sharpen my skill. ---

Tautochrone curve problem
Tautochrone curve problem - In the inverted cycloid from anywhere if we leave a ball it will take same time to reach to the bottom of the cycloid (at the mid), due to gravity, no friction encountered. I first checked, how much time it will take if we leave the ball from the top. Going by definition, dt = ds/v, where v = √2gy, ds = (√2Ry)dθ, so finally t =∫dt = ∫ds/v = ∫(√(R/g))dθ= (√(R/g))Π -- (1). This was very good as variable from numerator and denominator cancelled out each other. But the real problem starts if we leave the ball from anywhere in between. Let us drop the ball at θ= Π/2, then t =∫dt = ∫ds/v = (√(R/g))∫ΠΠ/2√{(1- COSθ)/-COSθ}dθ-- (2).

Suppose the ball is left at an angle α, where 0<α<Π. then ∫ds/v = (√(R/g))∫Πα√{(1- COSθ)/(COSα- COSθ}dθ-- (3).

Initially I tried to solve these problems by binomial expansion. And I observed at the end that those did not give accurate results for the entire range. Later I came across the work of Weierstrass, where he mentioned about pointwise continuous and uniformly continuous functions. I realized that both of the series fall under pointtwise continuous and not uniformly continuous. In these cases integration of the series will no give desired result. So I removed the entire paragraph now.

t =∫dt = ∫ds/v = ∫(√(R/g))(√(y/(y-a)))dθ,so when a = 0 numerator and denominator cancell out and integration gives(√(R/g))Π. Now "a" has some value, now I changed dθ to dy so dθ = dy/sinθ, the eq becomes

∫(√(R/g))(√(y/(y-a)))dy/(Rsinθ) = ∫(√(1/Rg))(√(y/(y-a))/(√ (2(y/R) - (y/R)2 ))dy, after few steps of rearrangement

∫(√(R/g))dy/√((a/2 - R))2 - (y - a/2 -R)2) - now it falls in standard form so the integration gives

√(R/g) arcsin ((y - R - a/2)/(a/2 - R)), no matter what the value of "a" is, integration from "a" to 2R (bottom) will always give √(R/g)((3/2)Π - Π/2) = √(R/g)Π. - (4).

If I rearrange the result of approach (4) and convert it to function of θ the final integration looks like √(R/g) arcsin ((1 + 2cosθ - cosα)/(1+ cosα)). Now if I look back to (3) and put back this eq. --- (√(R/g))∫Πα√{(1- COSθ)/(COSα- COSθ}dθ = √(R/g) arcsin ((1 + 2cosθ - cosα)/(1+ cosα)). It tells me that there was a way to arrange the eq. in(3) to put it in standard form, but it was not cospicuous to me (and I wil not dig back now).

But certainly dig back is required for (2). ∫Π/20√{(1+ SINθ)/SINθ}dθ = ∫101/√{(1- x)x}dx = arcsin (1/2 - x)/(1/2) = Π --- A few lines manipulation and I missed it earlier!.

Calculus of Variations
Brachistochrone Problem falls under Calculus of Variations, which is a pretty interesting subject to delve into. Initial few basics are available in the websites( wiki, mathworld, etc.). So I will jump straight to my point. Let us consider an area which is bounded by a curve (towards +ve Y axis) and X axis. So the curve begins at x1 ends at x2 in this case. Also throughout the range y has only one (+ve) value (solution) for any x value. Also y'= dy/dx should exist throughout the range (continuous). Now

1)F=∫ydx gives the area bounded by the curve and X axis between x1 and x2. Also

2)f=√(1 + y'2)dx gives the length of the curve bounded by x1 and x2.

So the question is what is maximum possible area bounded by the curve when curve length is constant, or what is the minimum length of the curve when area is constant? Now both F and f has +ve values in this entire range and are continuous.

Now entering into 'Calculus of Variations' we can see the function we need to vary is √(1 + y'2)/y. We need to find a minimum here.

Now using Euler-Lagrange differential equation and Beltrami identity consequtively we see that

√(1 + y'2)/y - y'(y'/(y√(1 + y'2)) = C

or, 1/(y√(1 + y'2)) = C

or, dy/dx = √(1 - C2y2)/y

or, dy(y/√(1 - C2y2))=dx/C, Integrating both sides

-√(1 - C2y2)/C) = x + a

or, (x + a)2 + y2 = 1/C2

This is the equation of a Circle whose center lies on X axis. Now obvious question comes, why it did not give the general equation of circle (x + a)2 + (y + b)2 = 1/C2. $

Let me justify the situation. Here F and f are two different functionals but they must have the same y=Φ(x). Here the the "true evolution" of y in F between x1 and x2 is constrained by "true evolution" of y in f. Now we want conditional extremisation of one functional relative to other (if that exist at all). Extremisation of the ratio of the two functionals should serve in this case.

Let us consider these two different functionals seperately. 1)F=∫ydx clearly indicates as we maximise or minimise y, F maximises or minimises. 2)f=√(1 + y'2)dx, here again going by the same Euler-Lagrange differential equation and Beltrami identity consequtively we get

√(1 + y'2) - y'2/((√(1 + y'2))=C, or, y'= √((1/c)2 -1)=m, indicates a straight line. Two different results, as they are disconnected.

Now let us try this method in polar coordinate. Here θ is equivalent to x, r is equivalent to y, and r'= dr/dθ is equivalent to y'. So F = (1/2)∫r2dθ, f = ∫√(r'2 + r2)dθ. So here the function needs to be varied is √(r'2 + r2)/r2. Again going by the same method we get √(r'2 + r2)/r2 - r'2/(r2(√(r'2 + r2))= C

or, 1/√(r'2 + r2) = C

or, (r'2 + r2) = 1/C2 By looking at this, three possible solutions come into mind.

1) r = 1/C 2) r = (1/C)sin(θ + a) 3) r = (1/C)cos(θ + a). And all of them represent circles in the domain of +ve r.

$ Four, five months after I wrote this, I was going through a book of "Calculus of Variations" (which I ordered after writing this to have a better idea on this subject) I found the same problem posed in a different way. In the book the length of the arc is assumed fixed. So it uses lagrange multiplier, as the lengh of the curve is fixed it should give one unique solution for one fixed length. So a +ve value of 'b' is possible ( although -ve is not) and also indeed derived in the book. As in my problem I did not fix the length of the arc, so it should give (and has given) only one unique solution for any (x1,0) and (x2,0), thus b = 0 in my case.

To summarise 1) In my case optmises A/P when none are fixed 2) The "book" fixed the P and optimises A. Where in 2) the existense of +ve b will make the curve bounded by X axis (between (x1,0) and (x2,0)) is a circular arc of radius R with length of P less than ΠR. Only thing to note here is that for this fix P the A is maximum (bounded by arc and X axis), although A/P is still not maximum between (x1,0) and (x2,0) (check it yourself), it will be maximum only when for the circular arc b = 0, or P = ΠR, as in my case 1).

---
[The following section is the direct result of reading the book "Calculus Gallery" by William Dunham. All the famous Mathematician's name and there specific works I came to know from this book only. Being said that I have to mention that all the derivations and procedures are purely by me, and you will not find those derivations in this book. Any resemblance to any other material elsewhere will be pure coincidence]

(In the following discussion I concentrated only between 0 and 1)

1) What is the closest rational number to 1/2? Can we think of ? Okay forget it, we can not. Can we think rational numbers close to 1/2? Sure we can, 2/3, 5/7, 3/5 and lot other. Which one is closer here ? of course 3/5. Is 3/5 the closest? Okay we can have 1/2(3/5+1/2) = 11/20, which is more close to 1/2. Is 11/20 closest to 1/2? We can further have 1/2(11/20+1/2)= 21/40, and like this we can have 41/80, 81/160, 161/320, 321/640, 641/1280, 1281/2560 and so on. We have started with 3/5, we could have started with 5/7, or 2/3 or any other rational number between 0 to 1 and execute the same procedure.

2) What is the closest rational number to (√2)-1. Okay here (√2)-1 = 0.41421356....

So I can start with 0.4= 2/5 here, next I can have 41/100, next 414/1000 = 207/500 and it will continue. Again from 2/5 to 41/100 or from 41/100 to 207/500 we can apply the same procedure in 1)

By seeing these phenomenon it seems clear that for any real number we can have a neighborhood (a very small range which contains that real number), where we will find rationals in the form of Pn/Qn where Qn → ∞. Basically for 1) we can say

For all the rationals X in 0 to 1 we can have a neighborhood of X where for every ε > 0, we can find a δ > 0, such that if 1/Qn < δ, |Pn/Qn - X| < ε,(Weierstrassian Language). Except possibly sometimes occurrence of only one very finite Qn (in the form of Pn/Qn) in the deleted neighborhood of X.

This means there are rational numbers in whose deleted neighborhood we can find numbers like 1/2 or 2/3. This brought ourselves to the very famous function called

'''"Ruler function" f(x) = 1/Q if x=P/Q in lowest terms, f(x) = 0 if x is irrational. The function states that f(x) is continuous at irrationals yet discontinuous at each rationals.''' Which should be pretty obvious by this time.

Let us consider Liouville’s famous inequation i.e. |p/q – x0| >= 1/(Aqn ), where x0 is irrational. Now for a given x0 like (√2)-1 in 2) what makes |p/q – x0| become smaller and smaller; as discussed in 2) it is q which has to be bigger and bigger. Also if we see the right side of Liouville’s inequation, we find that for a given x0, A and n are both fixed, so to make left side smaller q has to be bigger.

Now see this from the other way, what if we fix the rational p/q and make the irrational (√2)-1 closer to p/q in a sequential process described in 1). What we can see here is that both q and n are fixed here so to make left side smaller, A has to bigger. Indeed I verified this statement by performing the calculations.

As p/q is any rational between 0 and 1, we see that for any rational number we can have a neighborhood (a very small range which contains that rational number), where we will find irrationals for which A → ∞.

Liouville first proved the existence of Transcendental numbers. He used the same |p/q – x0| >= 1/(Aqn ) to prove that. I will call that number as "L". Transcendental numbers are the numbers which do not satisfy polynomial equations of the form

a1xn + a2xn-1+ --- + anxundefined +an+1 = 0, where a1 to an+1 are integers, and n is a whole number greater than 0. And those which satisfy, are called algebraic. Only few days back I came to know that equations greater than order 4 can not be managed to formulize its roots. While examining the characteristics of polynomial equations I found certain characteristics which although look pretty obvious, I thought it would be good for me if I prove those and try to draw some conclusions. So here it goes.

Henceforth I will call a number irrational if it is Algebraic.

1) If a number 'x' is algebraic and 'a' is any integer then 'x + a' is also algebraic (and vice versa). Consider the equation

a1(x +a)n + a2(x +a)n-1+ --- + an(x+a)undefined +an+1 = 0 we can expand each (x +a)m, and finally rearrange the coefficients so that at the end it transforms into another polynomial equation of x of same order but different set of coefficients (all integers). Here I proved if (x+a) is algebraic then x is algebraic)

2) If a number 'x' is algebraic and p/q is any rational, then (p/q)x is also algebraic (and vice versa). Only reconstruction is required here (multiply by qn all the terms). Here I proved if (p/q)x is algebraic then x is algebraic)

3) If we combine both 1) and 2) then it can be proved that if x is any algebraic and p/q is any rational then x + p/q is also algebraic. (This can be proved without taking help of 1) and 2).

4) If x is any algebraic then any real root of x (i.e. x(1/m) )is also algebraic (imaginary roots are excluded in this topic). They satisfy the exactly similar equation only the order of the equation changes to mn.

5) If x is any algebraic then 1/x is also algebraic. Let us say x is algebraic. To know the polynomial equation it satisfies we assign a variable against it like X = x, then we start a combination of powering (by a whole number) and rearrangement to derive the required polynomial equation. The very nature of polynomial equation suggest that the solutions must contain a combination of roots in the form of 1/m where m is whole number. Also in the similar way we can write X =1/x, or Xx = 1, then if we apply same procedure on Xx as we have applied on x, we derive the required polynomial equation for 1/x. Obviously order of the equation remain same.

Whatever I discussed above from 1) to 5) in some way defines the boundary of Algebraic numbers. So it also defines the boundary of transcendentals. We can say

1) for any transcendental T, T ± p/q is transcendental

2) T×(p/q) is transcendental

3) Tm is transcendental ( m is a whole number)

4) 1/T is is also transcendental

Of course these do not anyway prove the existence of Transcendental numbers, which was first proved by Liouville, then in more generic form by Cantor. What I understood is that Cantor proved that density of Transcendentals is infinitely more than that of Algebraic (which I believe is the final statement about Transcendentals). He used set theory to prove that. Can we have any other way to show that? Let me try

Now let us go back to 5) statement of Algebraic. "The very nature of polynomial equation suggest that the solutions must contain a combination of roots in the form of 1/m where m is whole number." Where m ≤ n (order of polynomial, and is a whole number). Obviously m can not be greater than n. "To know the polynomial equation it satisfies we assign a variable against it like X = x, then we start a combination of powering (by a whole number) and rearrangement to derive the required polynomial equation."

All first order polynomial equations have rational (p/q) solutions. Ax - B = 0 has only one solution B/A. What if the order of the equation (which is 1) is less than 1? Suppose the order of the equation is 1/2. so write it again Ax1/2 - B = 0, If we think for a few seconds we can find out that we can easily square the Ax1/2 term and the final equation is A2x - B2 = 0, thus again it falls into first order equation. Suppose the order of the equation is 2/3. so write it again Ax2/3 - B =0, A3x2 = B3, so it is a second order equation.

Going in this line we can see that if the order of the equation is any rational less than 1, we can convert them to suitable polynomial equations, which finally give algebraic solutions. Great, now what if the order of the equation is any irrational less than 1?

Now we see that we can not convert them to suitable polynomial equations which give algebraic solution. Suppose order of equation is 1/√2, then Ax1/√2 - B = 0, now if we want to remove power 1/√2 from x, we introduce power √2 in the coefficients. Which again does not satisfy the requirement of polynomial equation which gives algebraic solutions. So if we solve the above equation, what it will give? You are right, it is definitely transcendental! and how many irrationals we have between 0 and 1? infinite numbers of course. So infinite numbers of transcendental equations (which deliver infinite number of transcendental numbers) exist against only one first order equation (which delivers only one algebraic number). The same is applicable for higher order polynomial equations also. Second order equations are a1x2 + a2x1+ a3 = 0, if we fix a1, a2, a3, then we can create transcendental equations (infinite numbers) for which highest power of the equation is between 1 and 2 (irrationals). And for most of the equations we can not even express the root in the structured power form. Let me give an example. What is the root of x√3 + x√2 = 1, Here the highest power i.e. √3 is between 1 and 2.

Definitely it is a transcendental equation, and we can not find a structured expression for the solution. But a real positive solution definitely exist (look at the equation, and it definitely tells that). So I solve it with solver (in excel) and the solution is 0.64263122504568.......... Let me explain it more as it really calls for. The equivalent 2nd order polynomial equation for this is x2 + x1 = 1; now for equivalent transcendental equation, we can choose the first power (here √3) in infinite number of ways which will fall between 1 and 2, and for each of these first power, the second power (here √2) must be less than first power, and can be anything between 2 and 0. !!!!!!!!!

Add to this is, for transcendentals here I restricted it to two powered terms i.e. x√3 and x√2 to make it equivalent with the polynomial equation. Let √3 = m1, √2 = m2, then we can add infinite number of powered terms at the tail where m1>m2>m3>........ !*!*!*!*!*!*

Add to this is that we can now use the transcendentals we found in this way to the power to find new transcendentals (the proof will be exactly same as earlier).

It is really mind boggling for me.

A part of what I explained and prooved above is a proof of A. O. Gelfond's statement (1934) "If a is any algebriac number other than 0 or 1 and if b is an irrational algebraic then ab must be transcendental", but along with this it also encompasses far more transcendental numbers by the use of transcendental equations (as in x√3 + x√2 = 1). At the time of this writing A. O. Gelfond's proof is still unknown to me.

A little more discussion is needed regarding this, as it is evident now that against every algebraic number we have infinite numbers of transcendental numbers. Let me denote 'length' of a algebraic point is La and length of a transcendental point is Lt (La = Lt), then length of infinite number transcendental points related to a single algebraic point is ΣLt,

so ΣLt/La = ∞ (inifinity).

→(ΣLt + La)/La = ∞ (inifinity).

→ Σ(ΣLt + La)/ΣLa = ∞ (inifinity).

Between 0 and 1 Σ(ΣLt + La) constitutes the real line. As Σ(ΣLt + La) = 1 between 0 and 1, so ΣLa ≈ 0, and ΣLt ≈ 1

This is actually another proof of "measure of all algebraic numbers is zero" which was proved by Henri Lebesgue in 1904. This helped to show that Dirichlet's function is integrable.

Dirichlet's function Ø(x) = c, if x is rational, and Ø(x) = d, if x is irrational (including transcendentals). Naturally; set of rational numbers is a subset of set of algebriac numbers. As the length(or measure) of algebriac numbers is zero (ΣLa ≈ 0), so the length(or measure) of rational numbers is also zero (ΣLr ≈ 0). Now consider the following facts.

∫dx between two numbers 'x1' and 'x2' on real line has two parts

1) integration over transcendental points (non denumerable) and

2) integration over algebraic points (denumerable).

For integration over algebraic points, ∫dxa = ΣLa ≈ 0, and integration over transcendental points, ∫dxt = ΣLt ≈ b - a (as I proved earlier).

So ∫dx over x1,x2 = ∫dxa + ∫dxt = ∫dxt So it follows that ∫f(x)dx over x1,x2 = ∫f(x)dxa + ∫f(x)dxt = ∫f(x)dxt . So any function which is continuous over transcendental (non denumerable) points is integrable function. Dirichlet's function which is continuous over irrational (transcendetal + algebraic irrational) (non denumerable) points is automatically continuous over transcendental, so integrable.

THE END
I met a fine Mathematician today, sits nearby me in my office, opined that I should refrain from writing these type of silly and childish thing. I was really embarrassed to realize that this section is watched by so many eminent mathematicians in my office to whom the above writings are silly and childish *?!@#$%^&. I have no other option to stop writing here to save my time and to save their time also. After all they took so much pain going through all the materials and declare that these are childish. I have no words appropriate enough to thank those who went through these materials and "other things" and to those whose technological marvel allowed them to do so.

Minor observations for documentation purpose
Section A) Fermat's theorem on sums of two squares- states that an odd prime p is expressible as p = x2+y2 only if p = 4k+1 (k=1,2,3). For example, the primes 5, 13, 17, 29, 37 can be expressed as 5 = 12+22, 13 = 32+22

If p = 4k-1 then they cannot be expressed as sum of 2 suqares.

Brahmagupta–Fibonacci identity implies that the product of two sums each of two squares is itself a sum of two squares. In other words, the set of all sums of two squares is closed under multiplication

(a2+b2)(c2+d 2)=((ac+bd)2+ (ad-bc)2) =((ac-bd)2+ (ad+bc)2)

for ex. 5*13= (12+22)*(22+32)= (12+82)=(72+42) but Brahmagupta–Fibonacci identity can not be applied when both pairs are same i.e. instead of 5*13 if we use 5*5= (12+22)*(12+22) but we can still manipulate the equations to get the solution.

(a2+b2)(a2+b 2)= a4+ 2a2*b2+ b 4= (a2-b2)2 + (2ab)2

for ex. 5*5 = (12+22)2 = (32+42) That means every number which can be expressed as the sum of 2 squares, when squared will also give a pair of squares, so every prime of Fermat of the form 4K+1, when squared will also give a pair of squares. This has a recursive trend.

Sum of any two squares can be written as A= a2+ (a+x)2=2a(a+x) +(x2-1) +1. Now when the sum is odd then two cases arises, a is odd and x is odd, or a is even and x is odd. Looking at 1st part 2a(a+x) +(x2-1) and considering both the cases we see each of 2a(a+x) and (x2-1) are multiple of 4. So we can write a2+ (a+x)2=2a(a+x) +(x2-1) +1 =4k +1. As we can write it as 4k+1, so we can not write it as 4k-1.

Conclusion

1) An odd number A, which can be expressed as sum of two squares CAN NOT be expressed as 4k-1. Reverse is also true here, i.e. any odd number (including primes) which can be expressed as 4k- 1 can not be expressed as sum of two squares.

2) An odd number A, which can be expressed as sum of two squares CAN be always expressed as 4k+1. Now do all the "A"s always lead to all the "4k+1". The above statement only tells that set of "A"s is a subset of set of "4k+1". For ex. 4.2+1=9 can not be expressed as sum of two suqares. So what type of "4k+1"s do not lead to "A". Some observations are important

a) Any power of 4k+1 or (4k1+1)(4k2+1) can be expressed as some 4K +1.

b) Any even power of 4k-1 or (4k1-1)(4k2-1)can be expressed as some 4K +1. And CAN NOT be expressed as sum of two squares(refer earlier section).

c) (4k+1)*(4k-1) can be expressed as some 4K -1. So can not be expressed as 4K+1. So CAN NOT be expressed as sum of two squares

Now in a) if we already know that "4k+1" has 2 suqares then its any power will also have a pair of squares (refer earlier section). If "4k+1" comes from process b) then it can not be expressed as some of two suqares and neither of its power.(refer earlier section). For (4k1+1)(4k2+1) if both (4k1+1) and (4k2+1) have pair squares or one of them has pair squares and other is a square itself ONLY then the product can be expressed as sum of suqares.

Section B)

While examining the nature of Primes I encountered a series which I later found out that it was already noticed by Legendre. But I guess I found one more step which is not mentioned in Wiki (where I found Legendre's note) and if so it is worth mentioning here.

My effort started with finding number of primes between "a" and "na" where "n" is a whole number. If we start detecting primes from the begining then we will see that every prime is responsible for the composites between "na" and "a". We need to only consider primes which are less than or equal to square root of "na". The prime numbers are 2,3,5,7,11, ........... For 2 we will have effect ~(1/2)x2, for 3 we will have effect ~(1/3)x3,for 5 we will have effect ~(1/5)x5 and so on. So if we consider the numbers between 40 and 80 (includes 80), then we need to consider only 2,3,5,7. Now we need to calculate x2, x3, x5, x7. For x7 the calculation will be

x7 = (1 -1/2 -1/3 - 1/5 + (1/2)(1/3) + (1/2)(1/5) + (1/3)(1/5) - (1/2)(1/3)(1/5)). This is what I guess was mentioned by Legendre. But you see if you want to do it for 11 then you may bite your pen inbetween, and after certain number you will need additional support even to find out the all the terms. I ran out of steam after 11, but still could not reject the idea as this was the only almost exact calculation I have at my hand.

The shortcut I will tell later, but let us see the actual process. x7 = 4/15, x5 = 1/3,x3 = 1/2, x2 =1.

So the number of composites between 40 and 80 will be ~ 40((1/2)(1) + (1/3)(1/2) + (1/5)(1/3) + (1/7)(4/15))

The shortcut x7 = (1- 1/2)(1 - 1/3)(1- 1/5), x11 = (1- 1/2)(1 - 1/3)(1- 1/5)(1- 1/7),x13 = (1- 1/2)(1 - 1/3)(1- 1/5)(1- 1/7)(1- 1/11) and so on. Squeeze it more, x11 = x7(1-1/7), x13 = x11(1-1/11) and so on. This we can manage with a hand calculator. With the help of this shortcut and knowledge of primes below 100, I could calculate all the xi upto x97 and this allowed to calculate the number of primes below 10000, which is (1227) real close to actual number (1229), by taking starting of range from 0. M= number of primes less than N In general 1 - Σ (1/ai)xi can also be written as (1- 1/2)(1 - 1/3)(1- 1/5)...(1- 1/ap), and when ap tends to infinity this leads to well known function (inverse of it), which was first discovered by Euler. What I know so far is that Euler's approach was different to reach this function.

It is easy to see that the formula says for any Number N=a2 to consider numbers less than or equal to a, so for any prime number a1, we can consider numbers from a12 + 1 (= N1 +1) to N2-1 (N2 is a22 where a2 is next prime). Now as N2 is a composite, this can be extended to N2. This makes the the equations slightly shifted in above table and new table looks like as below. Needless to say that numbers below 10000 (1527), which is mentioned earlier, is calculated in this way, The numbers are not astrnomical but most probably the accuracy is, which startled me.

We can also conclude from this, that between any two N1 and immediate next N2, the primes are distributed "somewhat" evenly, i.e. if there are x number of primes in this range and we devide this range in 2 halfs then all the "x" can not be in any one half alone.Now we know the number of primes between a12 and a1 is almost a12(1-Σ (1/ai)xi), (neglecting number 1 here). Also the number of primes between a22 and a1 is almost a22(1-Σ (1/ai)xi).

Now if there are really x numbers of primes between a2 and a, where a is any whole number, we can roughly substitute (1-Σ (1/ai)xi) by x/a2, now if we apply this to a2 + 2a, we get x + 2x/a, so we can say there are roughly 2x/a numbers of prime exists between a2 and (a+1)2. I just checked for 100 and 121: Number of primes between 10= a and 100 is 21 = x, so 2x/a = 4.2, and number of primes between 100 and 121 is 5. The number of primes between 625 and 25=a is 105= x, so 2x/a = 8.4, the number of primes between 625 and 676 (262) is 8. Number of primes between 2500 and 50=a is 352=x, so 2x/a = 14, number of primes between 2500 and 512 is 11. Number of primes between 10000 and 100= a is 1229 - 25 = 1204, so 2x/a = 24, number of primes between 10000 and 1012= 10201 is 23.

observations continued
Theorem - Given first N number of primes Pi(i = 1,,,N), and a set "A" contains numbers from 1 to ∏Pi (product of all Pi,∏ indicates product), the number of coprimes of Pi in set A is ∏(Pi - 1) including the number 1 and the coprimes in this set will distributed symmetrically from the middle number of the set.

Prove and further-

For Σ(1/ai)xi different primes and their composites associated are like this, (when we consider primes (1/ai) up to 7 (a) for counting primes).

ZT = -2 -3 -5 -7 6 10 14 15 21 35 -30 -42 -70 -105 210. For each of this term, the error term associated for any given number N will be the remainder divided by the number itself. When the error term becomes 1, we subtract 1 from it to make it 0 (as 1 goes to the actual part) For ex. for 2 the error term can be ½ or 0. For 3 error terms can be, 1/3, 2/3, 0. When we take the given number 0, the sum of the error terms will be 0. When we take the given number 1, then the sum of the error terms will be sum of the inverse of all these numbers. If we proceed like this we will find that again at 210 the sum of the error terms will be 0. It can be easily seen that at 210/2 = 105 also the sum of the error terms will be 0. We do not need to see beyond 210 as every term will repeat afterwards. If a given number N is composite then it will have error term 0 at least for one Z (at least for one prime). If N has only 2 prime factors then we will have zeros (or ones) at 3 Zs, (two –ves for prime Zs and one +ve for the composite). In general If N has only m number of prime factors then we will have 2m -1 number of zeros (ones), and number of negative zeros (ones) will be always 1 more than the number of positive zeros (ones).

If a given number N is prime or coprime to (2,3,5,,,,a) then it will not have error term 0 for any Z. Now two consecutive numbers are always co prime. That means that we will not get error term zero at the same Z for any two consecutive numbers. So for any 2 consecutive numbers N1 and N2 if both are composites, then for N2 we will have zeros (error) at some Zs where N1 has non zero (errors)  and vice versa.

We start from N=1 and then go to N = 2, suppose sum of errors at 1 is E1, then E2 = 2*E1 + 1, ( 1 comes from the fact that number of negative zeros will be one more than number of positive zeros)  or E2 – E1 = E1 + 1. E3 = 3*E1 + 2, or E3 – E2 = E1 + 1, it will continue like this unless we encounter a prime, as prime contributes no zeros at any Z, suppose x is prime then E(x-1) = (x-1)* E1 + (x-2). So E(x) = x*E1 + (x-2), or E(x) – E(x-1) = E1, again E(x+1) – E(x) = E1 + 1. When we see at 210 also the error terms will be zero for all ZT. and if we move backwards we will see error will reduce exactly in the same way it increased from 0.

So if P is a coprime then ∏Pi - P will also be a coprime. So if P is not a coprime then error at P + error at (∏Pi - P) is zero, otherwise the sum will be equal to 1. The data was hacked. I did a mistake in my assumptions in some proof, that can only be detected by a mathematician. The writing was there only in my laptop, I did not publish it. I realized the mistake next day but did not bother to correct it. After some days a girl tried to inform me that mistake by some gestures. And actually I was waiting for that, so I got it. Now for a seasoned mathematician a clue is good enough and it seems they got the whole files (working excel file!!!!!). This is so unfair and low!!!!!
 * I have to publish this as this one and more was hacked, I still have feeling that data is still hacked without internet but this is a very crazy feeling.***

ok i will take a break from this now
Euler's formula - If G is a connected plane graph with n vertices e edges and f faces (enclosed areas), then n - e + f = 2 (including outer area, i.e. outside the closed region), or n - e + f = 1 (excluding outer area, i.e. outside the closed region)

Now what I understood from plane graph is that it is a graph with enclosed areas inside it with no crossing edges, such that from any point inside the graph you can go to any other point choosing suitable lines already there. At first I will assume all the connecting lines (edges) are straight lines. Given such a plane graph, we can see if internal areas are not all triangles, then we can keep on adding lines (edges) between points to break the polygons in smaller triangles till we reach a situation where there are no internal areas other than triangles. Now we see the map consists of all small internal triangles. We start with any such one triangle and observe that it has 3 vertices, 3 edges and one inside area, (n=3,e=3, f=1) which satisfies the formula, now we add another triangle adjacent to it, we observe we added only 2 edges, 1 vertices and 1 inside area so it becomes (n=4,e=5, f=2), and this also satisfies the formula. Next there is a possibility that we can add one extra edge that will connect two existing vertices and so also will create a one extra area, here also $n-e+f =2$ as one extra edge and one extra area cancels out each other. Now we see these are the only two process of adding area. Now we can keep on adding triangles one after another (and observe that this always satisfies the formula) until we use up all the points, and we observe that even at the final stages it satisfies the formula. Now we observe the we have number of vertices equal to the number of vertices originally given. Now as we assumed at the beginning that internal areas were not all triangles initially; so we have extra edges and areas now. So I will now remove the additional edges one by one. When we start this and remove one edge, we observe that we reduced only one face (enclosed area), (like removing a diagonal from a rectangle makes 2 areas into 1 area); an edge separates exactly two adjacent areas. In this operation also we see the formula holds true. We continue this until we reach the original graph given.

Note that in above argument if the lines are curved also, it holds true as long as only one line connects to two single points. Now if there are additional curved lines (joining two points or even the same single point), we replace all original areas which are partly or fully covered by curved lines by triangles. Now consider this set of area, as we see that this set also consists of triangles only so satisfies the formula (alongwith the other set) by the logic mentioned in the last paragraph. Now we will change these triangles to their original shapes. To do that we may add or remove points also we may change shape of the lines. When we remove a point from a triangle we remove one point plus one edge (two adjoining sides become one, as the main purpose of removing the point is it was sitting on a curved line, if the curve line contains more point, we can add them one by one and observe that each addition also creates an additional edge. In a nutshell in this operation if we add, we add same number of points and edges, and if we deduct, we deduct same number of points and edges. So essentially n - e remains same in this process, so also n - e + f remains constant and thus holds the formula. Now the above two processes are very adaptive and mixable, so much so that the triangulation technique can be applied to a graph from the very beginning, which doesn't contain any straight line at all!

Now maximum number of straight lines we can create with N points is NC2. Where each line contains exactly 2 points. Now as pointed out in earlier proof that with such points we can always make a graph which contains only triangles (in this case there will be crossing lines). We start an operation by which we start collapsing two sides of triangles on the third side, keeping one triangle fixed. Such moves destroy lines which are formed by collapsing points, of the two sides/lines, with the points on the ,third side/line, they fall.

Now we repeat the operation until we destroy all the triangles except the fixed one. Now we see we transferred all the remaining points on the sides of this fixed triangle. In this process we make sure that we collapse only the original sides (not the sides generated by the transfer process). We also note that no two original sides are col-linear.

Now possibly all the three lines of this fixed triangle contain more than 2 points, as all N points are now on these 3 lines. Here we can check no such combination of these points in these 3 lines can produce less than N -1, 2point lines.

Suppose a,b,c are the number of transferred points lie on the three sides respectively. So the number of lines containing only two points will be = a+b+c+ab+bc+ca (where a,b,c >0). Now maximum number lines possible when the points are evenly distributed over 3 sides. Minimum number possible when one side contains the largest and other two sides contain 1 point each. In this we get 3(N-4) lines, equalling this with N gives N= 6, means exactly 1 point in each side and 6 lines with only 2 points. Any other arrangement on these three line will increase the number of 2_point lines more than total number of points. When one side doesn't contain any points, to get min 2 point lines we keep only one point in one side and rest in other side. By this we get 2N -6, 2 point lines, and observe that for N = 5 we get only 4 2point lines (N-1). Now if all the points lie in one side only we get exactly N-1 2 point lines. So given N points in a plane not all in a straight line, we get minimum N-1 2 point lines (contain only 2 points).

Unfortunately the reduction can not be done where lines are perpendicular to each other. This point I missed before. Points on a perfectly square grid is non reducible. Now this is frustrating. Now I can prove a perfect square or rectangular grid has more 2 point lines than number of points but a complete proof will be still elusive. Now is it really true? So far I did not see a figure which has less than N - 1 "2 point lines". I can not get rid of that unless I see less than n-1 2 point lines.

Ok leave the 2 point line. Let us consider only lines. Given a set of N points in a plane, we added one more point and claimed that it did not create any additional line. That is only possible when this new point is radially connected to those all N points (what the heck!) or we can say it formed lines which contain more than two points including itself. That means it sits on the "center" of these lines (like center of a square) and no such points are there to which it connects directly (that would mean it created additional lines with those points). Now if the situation is not like this, we can safely assume any one point addition also will create at least one extra line.

Ok morning first thing came to mind is that multiple radial points possible. In case multiple radial lines how to approach. Let us assume 3 radial points A,B,C. We start with A and proceed towards B. As B to be radial point there must be one point in AB; also there has to be one (two, as per the definition of radial line) additional point (a1) which is not in AB. So we get at least two points before B. Now we proceed towards C (which is not in AB). We see C forms radial lines at least with A and B, which means there are at least two points in AC and in BC (now a1 can be in AC). We can apply this logic further. The thought is; in the increasing sequence from one radial point to next radial point we will always find at least 2 points which are not radial points. And each non radial point will add at least one new line as discussed.

Next question, what is the maximum number of radial points possible in a set of N points? Ok not many, we can have infinite number points with one radial point,with 2 radial point we can still have infinite number of points but the restriction becomes stronger, with 3 radial points also we can have infinite number of points also but the restriction becomes more stronger. With 4 radial points no construction is possible. Actually possible, somehow I forgot the picture I drew earlier. Typically what I followed is minimum point requirement. With two points I could draw lot if lines but I chose two to form a triangle ABC with radial points at base A,B. Immediately it became apparent that the vertex could hold the third radial point. So that makes a equilateral triangle with lines from vertexes to mid of opposite base creates the centroid which automatically becomes the fourth radial point. Are there multiple lines possible with this. I need to check again. But in that simplistic method I tried to pit fifth radial point, it gave me endless loops.

When considering these few radial points first, observing that they have at least N lines for N points we can always say any addition of a point will surely generate at least one extra line. BTW what is that extra line, it is obviously a two point line in that particular addition. So if we stop after that, we can always say the present configuration has at least one two point line.

Fact- More than two radial points can not stay in the same line. It creates a non ending sequence (a point needs one more point). Without a picture difficult to explain here. This problem can be solved if we take out the third point from the line and place it somewhere else. But then it becomes very restrictive although infinite number of points possible. Later I found out that it is possible with lited number of points.

With this somehow I observed that one fact I terribly missed earlier is the definitions I give for radial points also hold the clue for number of two point lines.

Configuration of N points with no radial points → we have N points which have at least one two point line. So least number of two point lines is N/2.

Configuration of N points with one radial points → we have N - 1 points which have at least one two point line. So least number of two point lines is (N-1)/2.

Configuration of N points with two radial points → we have N - 2 points which have at least one two point line. So least number of two point lines is (N-2)/2.

Configuration of N points with three radial points → we have N - 3 points which have at least one two point line. So least number of two point lines is (N-3)/2.

I saw today a square can have five radial points with vertexes and center. So a triangle with center can gave four radial points, a square can have five radial points, so higher polygons also can have radial points like this?

On a different note assuming N points has at least one 2 point line. We can start removing points from these two point lines one by one. Removing a point from a two point line is actually destroying a line. So if we continue from N, N-1, to N- (N-3) we destroyed N-3 lines, and we have still three more lines. So the total lines in the set was originally at least N. Also if after N-(N-X) stage if we find all the X-1 points are lying in a single line, then also we have X lines in that situation. So again initial total least number of lines were N-X + X =N lines.

Now this sounds similar to one existing proof, which says if there are at least N lines for N points, then there will be at least N +1 lines for N +1 points, which gave me impression that every additional point adds one additional line, which I saw is sometimes not possible. So the above explanation, which is easier for me to understand. In fact if number of lines m are more than number of points n then current proofs say that we can add additional m -n points which may not add any line. Now try to go little further. Consider the last stage after N - ( N -X). Where all X -1 points are lying in one line( (X-1) >=2). This has X lines and looks like a triangle with no additional points on the two sides.Now consider the point we just deleted to make this. This point at least had two 2point lines as it has to seat any where except the line on which X-1 points are residing. This tells us any N points where at most N-2 points are lying in a straight line, it has at least N+1 lines. A good example would be a rhombus with 4 vertex PT and one center point has 5 pt and 6 lines. You can't get less than this.

Extending L.M. Kellys argument in 3 dimension, it can be shown that if there are N points not all in the same plane, there is at least one plane which consists exactly 3 points. Based on this and using my previous arguments it can be shown that if there are N points not all in the same plane, there are at least N planes.

if there are N points with at most N-2 points on the same plane, there are at least N+1 planes.

This should extend to higher dimension also.

In Mth dimensional space with at least M +1 points, not all in the same (M-1)th dimensional 'face', there will be at least one (M-1)th dimensional 'face' which will contain exactly M points.

And there will be at least M +1 (M-1)th dimensional 'face'.

'' Maximum number of 3 point lines with given N points is (N2 -1)*(3/8). Here all the lines are either 3 point or 2 point lines. So we get minimum number of 2 point line (N-1)(N-3)/8. This formula is valid for N>=7. So for N =7, it gives minimum 3, 2 point lines. ''

Sylvester problem
Eli Goodman and Ricky Pollack combinatorial model can be used very easily to prove Sylvester problem. First assume that one dimensional projection starts with 123...n and after performing 180 deg rotation of direction we get n..321. There is an axis of symmetry and every point has a symmetric point (for even number of points) with which it changes position, for ex. for 1234, 1 exchanges its position with 4; 2 exchanges its position with 3 etc. For odd number of points the middle point will remain in the same position after 180 deg rotation. Assuming at some point of time 1 will start moving towards right with reversing substrings of at least 3 points (for ex. 123 to 321 but no 12 to 21) the same is true for n. And the substrings are all in increasing sequence which got decreasing sequence after reversal. At one point 1 and n will cross each other and as they both can not reach the end point together at least one of them will see decreasing substrings in front of it, which can not be reversed again.

Rolle's mean value theorem
A different proof of Rolle's mean value theorem - suppose f(x) = xn + a1xn + ......+ an; has roots m,n (n > m) in a region, where no other roots lies between m and n. In this scenario we can reformulate the eqn as f(x) = (x-m)(x-n){xn-2 + a1xn-3 + ......+ an-2}; differentiating this one we get the function f'(x)= (2x - m - n){xn-2 + a1xn-3 + ......+ an-2} + (x-m)(x-n){(n-2)xn-3 + a1(n-3)xn-4 + ......+ an-3}; after simplification.

now at x = m, f'(x) = (m-n){xn-2 + a1xn-3 + ......+ an-2} and at x = n f'(x) = (n-m){xn-2 + a1xn-3 + ......+ an-2}; now if we look at the function {xn-2 + a1xn-3 + ......+ an-2} we see that this function has same sign at x = m and at x = n, otherwise if this function has different signs at at x = m and at x = n, that means it has a zero between m and n, (a function changes sign only when it crosses x axis, i.e. y = 0, even this has a proof by Cauchy) and that would mean that f(x) has a zero (root) between m and n, which we initially assumed not true. Now going back equation of f'(x), we see f'(x) has different sign at x = m and x = n, as (m-n) is -ve and (n-m) is +ve, so f'(x) must have a zero (root) between m and n, (it must cross x axis in order to change sign).

This proves the theorem. Now instead of f(m)= f(n) = 0, we can have f(m)= f(n) = a, then also above procedure is applicable, we just need to shift the x axis by an amount a. In this way the shape of the curve does not change only the equation reformulates. This can be extended to Lagranges Mean value theorem where we need a combination of rotation of both axis and shifting of x axis.

In a more general and short way if f(x)has two roots m and n in a region where no other roots lies in between m and n we can write f(x) = (x-m)(x-n)g(x). so f'(x)= (2x - m - n)g(x) + (x-m)(x-n)g'(x); after simplification.

now at x = m, f'(x) = (m-n)g(x) and at x =, f'(x) = (n-m)g(x);

g(x)has same sign at x = m and at x = n, otherwise if this function has different signs at at x = m and at x = n, that means it has a zero between m and n, (a function changes sign only when it crosses x axis, i.e. y = 0, even this has a proof by Cauchy) and that would mean that f(x) has a zero (root) between m and n, which we initially assumed not true. Now going back equation of f'(x), we see f'(x) has different sign at x = m and x = n, as (m-n) is -ve and (n-m) is +ve, so f'(x) must have a zero (root) between m and n, (it must cross x axis in order to change sign).