Efficiency of a system

1) Efficiency of a system = equality in a system
2) Equality can be in space dimension( like symmetry, copies) or in time dimension (like greater ability to resist change).
3) Different options arise because of increase of inequalities or inefficiencies in the original system.
4) Efficiency= (Ability*Frequency)/(number of options)
5) Number of options in the universe is continuously increasing as the universe is continuously degrading( as the second law of thermodynamics state) but the rate of degradation is continuously decreasing as the number of options is growing. It is approaching 0.
6) As the total rate of degradation is very low, for extra degradation in one system there must be improvement in other system(s). Life utilizes this principle.
7) Rate of degradation decreases with each unit degradation while rate of improvement increases with each unit improvement. Due to this a fair changing system should improve at a rate of [ (d^2)(options)\dt ] * efficiency/(no. of options)^2 unless changes are totally non discrete.
8) There can be no fair changing system when the change is even slightly discrete, there will always be an element of randomness in discretely changing systems. Randomness will increase as discreteness increases or rate of change increases.

How the brain works

Serotonin - dopamine exchange theory

Economics/business style model

 

Last conscious thought you had was analyzed brain's reward mechanism. Based on its analysis it was given a certain amounts of dopamine and/or pain.

The value of a thought = (quantity of dopamine) - (quantity of pain). This value must be paid for by the next thought if it wants to replace the current thought (and be analysed by the reward area). But this payment must be done in serotonin depending on the exchange rate ( which in turn depends on genes and current circumstances). The amount of serotonin paid, goes into the bank account ( or should we say, memory account) of the previous thought.

Now after making the payment the current thought gets the chance to be analyzed by the reward area. After analysis the current thought gets its value in exchange of which it gets certain amount of serotonin. If the amount of serotonin gained is more than the amount invested it goes back to the memory bank with a profit, else a loss. The thoughts which have a high amount of serotonin in their memory accounts can buy their way into the consciousness more frequently.

Like begets like, our consciousness adapts to produce more of which comes more frequently. If the total amount of profits exceeds losses (all thoughts combined) the total number of thoughts in the mind increases, but to do this brain must consume extra energy and space, but both are limited in the body, so we cannot be much more happier than sad. In depressed people the memory area is smaller than in healthy people. Antidepressants which increase the amount of serotonin in brain also cause an increase in memory area of the brain. In children, when the brain can grow in size happiness can exceed pain.

According to theory pain must be soon be eliminated from the brain as painful thoughts which make losses will soon run out of serotonin and lose their power to enter consciousness. But this is not true in practice because of following mechanism. When a thought releases high amount of pain or involves a lot of brain area while releasing pain the brain realises that dwelling on this particular problem is beneficial for survival. In this case GABA is released which inhibits other thoughts from replacing the painful thought, only a small leeway is left where the positive thoughts can replace parts of the painful thoughts and thus take us towards the solution of the problem. Endorphins which decrease perception of pain do so by blocking GABA (in the central nervous system). Practices like meditation decrease pain by not letting GABA do its work. Whenever GABA makes the mind empty the meditator  returns back to object of concentration.

A new and very simple technique to approximate pi

Pi

Let the radius of a circle be R. We take a quarter of that circle.quarter circleNow a right angled triangle ABC can be drawn starting from center of the circle with AB (its height) extending along the left edge of the quarter circle, it’s base BC starting from B and touching the circumference of the quarter circle at C and AC being the hypotenuse of the circle also joins center of the circle to its circumference, in other words it is the radius of the circle

pi

 

 

 

 

When AB = radius of the circle then BC will be zero as AC is always equal to the radius of the circle, when AB is ‘h’ less than R then BC will be equal to √((R)^2-(R-h)^2) ,or √(2Rh-h^2). Now if this quarter circle is converted into a rectangle with length R, it’s breath(b) will be the average of  √(2Rh-h^2) as h increases from 0 to R . Area of the quarter circle may be written as R*b, we know it’s also equal to pi(R)^2/4 ,therefore b=piR/4 is the average of  √(2Rh-h^2) .The area of the quarter circle is also equal to the integral of  √(2Rh-h^2)dh as h moves from 0 to R. Now if we take the integral of (2Rh-h^2)dh instead of √(2Rh-h^2)dh we will get R(R^2)-(R^3)/3 which is the integral of squares of side  √(2Rh-h^2) as h moves from 0 to R, so can we approximate the average square to have a side of b? We will check that later, for now let us do it, the average square has an area of b^2(dh) so the total area of the squares is b^2(number of squares)dh, or b^2dh(R/dh)=R(b^2) but the area of the figure is also = R(h^2)-(h^3)/3,

Therefore we have:-

R(b^2)= R(R^2)-(R^3)/3

We know that b=piR/4

R(piR/4)^2=2(R^3)/3

Or, (pi^2)/16 =2/3

Or, (pi^2)=32/3

This gives us the value of pi to be 3.266, so what did we do wrong..well we approximated the value of average square to be b^2. b^2 is the square of integral of '√(2Rh-h^2)dh divided by R', while average square is the integral of (2Rh-h^2)dh divided by R so we are substracting the square of integral of (√x/R) from integral of (x/R) where x=2Rh-h^2. So what will be the difference? The integral of x/r is (x)^2/2R(dx/dt) if dx/dt is constant, if it isn’t it will turn into an infinite series and we don’t want that so instead of taking the value of dx/dt at h=R we take the average value of dx/dt. Dx/dt = d(2Rh-h^2) or, 2R-2h which changes from 2R to 0 , so the average value of dx/dt will be R. Therefore the integral of x/R is is (x)^2/2(R)^2 and the integral of √(x)/R is 2x√(x)/3R(dx/dt) or 2x√(x)/3(R^2) and that squared is 4(x^3)/9(R^4) ,so we have :- (x^2)/2(R^2)-4(x)^3/9(R^4) ,the value of x is (2Rh-h^2) at h=R it becomes (R)^2, so we have:-

((R^2)^2)/2(R)^2 - 4((R^2)^3)/9(R)^4

=(R^2)(1/2 – 4/9)

=(R^2)(1/18)

This (multiplied by dh) is the difference in the area of average square from b^2, therefore the difference in area of the whole figure will be R*(R^2)(1/18) or, (R^3)/18.

So we have  :-  R(b^2) +(R^3)/18= R(R^2)-(R^3)/3

Or, R(piR/4)^2 + (R^3)/18= R(R^2) - (R^3)/3

Or, (pi)^2/16 = 2/3 – 1/18

Or, (pi)^2 =11*16/18

Or, (pi)^2 =176/18

Or, (pi)^2=9.7777777778

Or, pi = 3.127

So what did we do wrong now, well we integrated x^2 and √x for dx instead of dh but that was a fine approximation as you can see.

A Theory for Finding the Best Move in Chess at Any Given Position

Theory 1

Any player's chances of winning in chess can be approximated to two major factors, the amount of squares he controls and the amount of time he has. Amount of squares controlled is simply the number of squares in which he can bring one of his pieces but  the opponent cannot because of impossibility of the feat or because of obvious loss of material and the amount of time can be said to be the amount of difference between moves required by the opponent to accomplish his best attack and the moves required by the player to thwart that attack. For example opponent is threatening to kill your bishop in 3 moves with that being his best attack and it can be defended in 1 move then you have time of 2 moves to accomplish your own plans. Now your score at any given moment is (number of squares you control)*(amount of time you have) - (" " opponent controls)*(" " time opponent has). Your goal is to make a move that increases your score by maximum amount. 1 pawn has a value of 3 squares and can be at most times traded for control of 3 extra squares.

Theory 2

Okay so theory 1 a very basic theory but it had some flaws, it only counts the total squares not the squares of individual pieces, so some pieces may be hyperactive others under active which is not a desirable situation, so a better approach would be to take into account ∆S/2S. Where ∆S is the increase in number of squares available to a particular piece and S the number of squares it initially had (if ∆S is much more than 1 please take the average value of S instead of its initial value in the denominator). If one move increases the squares of more than 1 piece add them up separately. Another flaw in theory 1 was that time can vary absurdly, so we have no standard value of a square and no correlation between the value of a piece and value of a square, if I say 3 squares is worth 1 pawn, you may say when time doubles -'now any 1.5 squares is worth a pawn' or any 1 square is worth a pawn if time has tripled which is absurd. Let us take another approach. 1 threat may require 1 move to be solved, in that case opponent solves that threat instead of increasing S, so a threat that requires 1 move to be solved, or in other words wasting your opponent's 1 move is worth average ∆S/2S, but then in one move opponent may also cause some material damage. Average game lasts 40 moves and excluding the king there are 39 points on the board. But then average game may not be played till checkmate or without mistakes so let's take a conservative figure, say 100 moves. Then 39 points require 100 moves to be extinguished so 1 move is worth 0.4 pawns + average ∆S/2s gained (which can be approximated to 1/4). So a threat that requires 1 move to be solved is worth 2/3 pawns provided you do also increase the average number of squares while giving the threat. A more rigorous approach will be to calculate the decrease in advantage provided by opponent's best move(∆bm). For example opponent was threatening to kill your pawn, you in turn threatened opponent's queen, now the opponent saves his queen instead of killing your pawn so ∆bm is 1. So the score change can be approximated to be 'forced win +  ∆S/2s + ∆bm '(or "∆S/2s +2(opp's move wasted)/3" if you want to calculate lesser). Now will the move with highest score increase be the best move? It may be but there is a big catch, 1 best move may be better than a number of good moves. So a move that does not increase or even decreases the score but prepares for a second move that will increase the score by a higher amount than 2 alternate best moves combined is probably better than the alternate moves, so you must look for combination of moves that increase the score by maximum amount but that again would increase our work.

Theory 2++

Forced win + ∆S/2S + ∆bm(or 2∆(opp's moves)/3) , is a good theory and if practiced rigorously would easily bring your rating to 1800. But it's not the end of of the story, it explains the importance of mobility, tempo, and piece activity but fails to incorporate 'piece coordination'-a very important principle of chess. It basically means your pieces must defend each other, attack together (i.e. if one piece becomes a martyr while vanquishing a greater foe his teammate must revenge him) and work together towards decreasing the possibilities and eventually annihilating a piece. Let's go step by step more the defense the better. Each piece with 1 extra defender than the number of attackers is worth 0.2 points more. For 2 extra defenders 0.282, and for 3 extra defenders 2root(3) and so on(also don't forget to take the value of the 1/2S of the square where you are defending a piece). Next pieces working together to attack the opponent accomplish more. Now if number of attackers is more than the number of defenders, you get a win so we won't account for that. Now if 2 pieces are working together to reduce 2 square available to an opponent's pieces the total value of squares reduced is 2*root(2+1)/2S where S is the initial number of squares that piece had. Also don't forget to add 2*root(2+1)/2S1 and 2*root(2+1)/2S2  for two of your own pieces. The general algorithm is (no. of sqs changed)*root(no. of pieces involved)/2(total no. of sqs). So the total score change in a move is Forced win + ∆Sroot(pieces involved)/2S + ∆bm(or 2∆(opp's moves)/3) +0.2∆root(extra defense), your goal is to find a combination of moves that increase the score by maximum amount.

Why We Don't Use Pythagoras Theorem for Finding Tangential Velocity or Centripetal Acceleration

If we use pythagoras theorem for finding tangential velocity or centripetal acceleration that is, we

√(R^2 + (vsinx)^2) - R

or,(√(R^2 + (vsinx)^2) - R )* (√(R^2 + (vsinx)^2) + R)/(√(R^2 + (vsinx)^2) + R)

or, (R^2 + (vsinx)^2) - R^2/(√(R^2 + (vsinx)^2) + R)

or,(vsinx)^2/(√(R^2 + (vsinx)^2) + R)

or, approximately : (vsinx)^2/2R

But we know centripetal force is (vsinx)^2/R , so what did we do wrong over here? Is there some law which states we must use differentiation and not pythagoras theorem? Well there is a problem in taking what we got as centripetal acceleration. How do we know it's acceleration? Sure tangential velocity is acceleration but then how do we know it's velocity and not distance travelled? For time =1 unit we got displacement or tangential velocity whatever it is to be (vsinx)^2/2R. Now let's put t=2 seconds and find the total of this change, if the result gets multiplied by 2 then it's proportional to t and we can say it's tangential velocity but if it gets multiplied by 4 then it's proportional to t^2 and we may say it's distance traveled and in case it's neither proportional to t nor t^2 we may say we did some mistake somewhere. At t=1 total change was (vsinx)^2/2R and cosx increased by v(sinx)^2/R and sinx changed by an extremely small amount and can still be approximated as 1. After t=1 and before t=2 change will be vcosx +(√(R^2 + (vsinx)^2) - R ) = (vsinx)^2/2R +(vsinx)^2/R . Total change = change at t=1 and change at t=2 or,  =(vsinx)^2/2R + (vsinx)^2/2R +(vsinx)^2/R =4(vsinx)^2/2R

So as you can see what we get from pythagoras theorem is proportional to t^2 and hence is displacement not velocity.

Displacement(S)= u^2 + a(t^2)/2

at t=1, S= a(t^2)/2

= Acceleration/2

Therefore Acceleration = (vsinx)^2/R

And if we want displacement to be zero we must ensure acceleration is zero. Hence gravitational acceleration will be equal to (vsinx)^2/R for our familiar circular or elliptical orbit.