E_Bezier

Normally the quadratic Bezier through three points A,B,C is:

I found an alternative that has some nice properties I'm calling the E_Bezier also through points A,B,C:

Both the Bezier and the E_Bezier are defined over the interval t=0..1, and equal A at t=0 and C at t=1, but the E_Bezier has the property that at t=1/2, it equals exactly B, which is not the case for the Bezier curve...

The acceleration of the E_Bezier curve is exactly double the acceleration of the Bezier curve...

I called it the E_Bezier because it spends "E" qual time between points A and B and points B and C...

**I might continue with an update for higher order than quadratic E_Bezier**

Balance scale series

Trying to get the least amount of elements in a series of natural numbers that can be combined by adding or subtracting unique elements of the series to get as many consecutive natural numbers as possible starting with 1. I came up with:

1,2,7, 21, 52
 For example
13=21-7-1,
14=21-7,
15=21-7+1, etc... Every natural number less than 84 can be reached by adding or subtracting these first 5 numbers...

The pattern I found is that x-sum(previous) = sum(previous)+1, solve for x, because the previous were able to reach every number up to sum(previous), and you need an x so that subtracting all of those numbers yields every number up to x...

This isn't in the Encyclopedia of Integer sequences yet, but I don't know if it's the best solution...
The name comes from the fact that on a balancing scale, you could weigh any natural number weight by putting the additions on one side of the balance and the subtractions and the object being weighed on the other...

Faster method than Newton-Raphson

This will be considering a replacement of the Newton-Raphson method for better performance changing:

to:

First recall that the Newton-Raphson method started as a generalization of the Babylonian method for finding square roots... I'll start the same way by showing this method speeds up finding square roots...

Starting with a reasonably close approximation to the square root of N such as the closest perfect square, one can first use this formula where x is the estimate.

For example with an N of 61, we can chose x to be 7 because 7^2=49 is close to 61 and then:

Now recursively use the formula again:

Which is very close to the square root of 61 compare:

After 2 iterations it's already accurate to 9 decimal places! Two iterations of the Newton Raphson method with the same initial conditions only correctly estimates 3. In fact it is better than 3 iterations of Newton-Raphson by a couple of decimal places...There are 7 arithmetical steps per iteration of this method and 5 for Newton-Raphson, so two iterations of this method take 14 arithmetic steps and get a better result than when Newton-Raphson takes 15 arithmetical steps, and Newton-Raphson would need 20 arithmetical steps to improve upon two iterations of this method... I think the difference becomes more marked when extreme accuracy is needed or when just the amount of accuracy that this method gives over Newton-Raphson is needed... And of course when very many of these calculations are needed...

So for another example if f(x) is 2*e^(x)-3 starting with a guess of .5 and doing two iterations we get:

That is very close to the actual root of:

Three iterations of Newton's method only correctly gets 5 decimal places right for more arithmetic...

Possible commutative pairing function

My conjecture is that given a,b positive integers:

is a positive integer unique to the pair a,b except for exchanging a and b...

I tried proving it a few different ways but really I don't know why it seems to work. I looked at many examples where it holds and usually there are 3,6,9, or 12 solutions over the integers but except for a,b and b,a one of the two integers is always negative...  Maybe the proof is simple, if 12 is the most solutions there can be and you know two solutions will both be negative integers, and 8 will have 1 negative and 1 positive integer, that only leaves two to be positive, but I wasn't able to get to that... Maybe also there's an argument from the symmetries that have to exist, I don't know...
For 3,4 the solutions are:

Distribution decomposition with normal curves

The formula I found to decompose a distribution into normal components, similar to how the Fourier transform decomposes a wave into sine and cosine components is:

Where n is the number of points to be interpolated... So if we know the value of the distribution say at x=1, 2, and 3 we evaluate the above with n=3 and those values for x1, x2, x3:

Now we can evaluate the above at each of x=1,2,3, with our y values of let's say .9, .7, and .8 respectively, and solve:

Now we can plot the solution:

Preference Wheel on the complex plane

This example is for giving a person the task to rank their 3 favorite choices out of 10 possible choices labeled alphabetically... The data is as follows:

The above data means the first person chose their favorite to be choice e, their second favorite to be choice d, and their third favorite to be choice b... And the second person chose their favorite to be choice i, and their second favorite to be choice g, and their third choice was b... Etc...

So I thought to make a preference chart where a person's 3rd favorite always comes halfway between their first and second favorite on a circle... First I solved these equations:

In complex numbers on the unit circle this is saying the variable in the right hand side of each equation is the point on the circle halfway between the two variables on the left hand side around the circle... And the last equation a=1 is to disallow rotational symmetry we set one variable to equal the real axis...

There were quite a few solutions, but most were trivial that all the variables equal 1 or 0 or all real number solutions, the two interesting ones are mirror images across the real axis, one of them is:

Which plotted on the unit circle is:

And see that it encodes the information perfectly for example that f is halfway around the circle between j and e, when two variables are equal the halfway point is either equal to either or on the opposite side of the circle..

So once plotted the question is can we extrapolate that someone who listed, for example, a and e as their first two favorites, which was not information given, will usually like h as a third choice... I'll have to find a real dataset to statistically determine whether that is the case or not...

Multidimensional projections on 2d plane

For 3 dimensions you can imagine that A,B,C are your 3 dimensional axes turned in such a way that looking at them they appear like this:

Define a point P on the plane to be P=a*A+b*B+c*C, with a,b,c values from the sliders...

Then we can look at all the ways to fix 2 of the sliders at 1 or 0 and move the 3rd, and tracing the paths they make gives:

Which is the "shadow" or projection of the 3d cube on the plane...

Now what if we have four orthogonal axes, so that their projection on the 2d plane looked like this:

We can define P to be P=a*A+b*B+c*C+d*D

There are a lot of ways to hold all but one of the sliders fixed at 1 or 0 and move the other one, doing so traces this onto the plane:

This is the projection on the plane of a 4 dimensional cube!