Eureka !!

This is a big one.

Remember my pi by 2 expression. Sigma for n =2 to infinity of 2^n/comb(2n,n) gave exactly pi by 2.

When I tried my rational number series with plus sign I got 2^(m-1) approximately. And the accuracy improved for higher values of n.

The expressions are

((n+1)^m/(n+2)^m+(n-1)^m/n^m)/(((n+1)/(n+2)+(n-1)/n)^m) and ((n)^m/(n+1)^m+(n-1)^m/n^m)/(((n)/(n+1)+(n-1)/n)^m). These expressions are almost 1/(2^(m-1)). And the exactness improved as n increased.

So if I mix the above two different ideas then it will be a great thing. Because of 2^n and 2^(m-1).

I shall be working on this.

(Many people may be confused with (n-1),n,(n+1) and (n+2) or (n-1),n and (n+1). There is nothing here really. 4 consecutive numbers (the way we count with our four fingers) and 3 consecutive numbers (the way we would have counted if we had three fingers))

(What I see is a brilliant pattern in them. And of course the approximation part. Which I claim to exist in 4 consecutive numbers and not in 3 consecutive numbers)

(To prove that there can be no general approximation formula for 3 consecutive numbers will be difficult for me. To me it is as difficult as Fermat's last theorem. a^n+b^n=c^n only for n = 1 and 2. Not for n>=3. But Sir Andrew Wiles proved the Fermat's last theorem. And he won so much praise and accolades from the whole world !!)

(Basel problem was posed in 1650 and first solved by Euler in 1734. The proof on impossibility of solving the general quintic equation was given by Abel and the problem was with humanity for 250 years. Fermat's last theorem was posed in 1637. And solved by Sir Andrew Wiles in 1994)

("There can be no general approximation formula for 3 consecutive numbers". Can Indian mathematicians solve this one ? )

((1(+/-)x)^n is approximately 1(+/-)nx for small values of x is damn good and involves three numbers 1,x and n. But three consecutive numbers? Interested?)

(Also whether the quintic expressions formed by ((n+1)^m/(n+2)^m+(n-1)^m/n^m)/(((n+1)/(n+2)-(n-1)/n)^m) ; ((n)^m/(n+1)^m+(n-1)^m/n^m)/(((n)/(n+1)-(n-1)/n)^m; 

((n+1)^m/(n+2)^m-(n-1)^m/n^m)/(((n+1)/(n+2)-(n-1)/n)^m) ; ((n)^m/(n+1)^m-(n-1)^m/n^m)/(((n)/(n+1)-(n-1)/n)^m are easily solvable or not ?)(Answer - Wolfram Alpha is able to solve these type of quintic equations. I have checked.)

(So only one poser remain. Can there be a general approximation formula involving three consecutive numbers?)

This finishes.

Finally

Finally my statements are as below :-

1.

((n+1)^m/(n+2)^m-(n-1)^m/n^m)/(((n+1)/(n+2)-(n-1)/n)^m)=((n(n+1))^m-((n-1)(n+2))^m)/2^m.

It is always a positive integer for positive integer m and n.

2.

((n+1)^m/(n+2)^m-(n-1)^m/n^m)/(((n+1)/(n+2)-(n-1)/n)^m)=0

and for m equal to 2 gives -1.618033 . . . (the golden ratio)

3.

((n+1)^m/(n+2)^m-(n-1)^m/n^m)/(((n+1)/(n+2)-(n-1)/n)^m)+42 and for m equal to 2 is a fairly good prime number generator.

4.

((n+1)^m/(n+2)^m-(n-1)^m/n^m)/(((n+1)/(n+2)-(n-1)/n)^m) can be approximated to a/((n+1)^2)-b/((n+1)^3)+c/((n+1)^4)-d/((n+1)^5).

Where a,b,c and d are in Arithmetic Progression. First term is 2m. Common ratio is 2m(m-2).

5.

((n+1)^m/(n+2)^m-(n-1)^m/n^m)/(((n+1)/(n+2)-(n-1)/n)^m) traverses across number series, squared series, triangle series in a beautiful way.(Difficult to explain in words but can be demonstrated)

6.

I believe this to be the law of four fingers. Four fingers play a crucial role of approximation in human thought. And also accuracy.

7.

((n)^m/(n+1)^m-(n-1)^m/n^m)/(((n)/(n+1)-(n-1)/n)^m)=((n)^2m-((n-1)(n+2))^m)

It is always a positive integer for positive integer value of m and n.

8.

((n)^m/(n+1)^m-(n-1)^m/n^m)/(((n)/(n+1)-(n-1)/n)^m) traverses across number series, squared number series, triangle series in a beautiful way. ( Difficult to explain in words but can be demonstrated)

9.

I believe statement 7 is law of three fingers. Only accuracy. No approximations (As far as I tried, there were no general statement on approximations (the way there was in four fingers)).

10.

((n+1)^m/(n+2)^m+(n-1)^m/n^m)/(((n+1)/(n+2)-(n-1)/n)^m) also give positive integer and so does ((n)^m/(n+1)^m+(n-1)^m/n^m)/(((n)/(n+1)-(n-1)/n)^m) for positive integer value of m and n. 

Cats and dogs have 1+4 toes in their front feet. And 1+3 in their hind feet.

Monkeys, apes and humans have 1+4 and 1+4.

Birds 1+3.

Yesterday, I saw the video of a simpler solution to quadratic equations by Po-Shen-Loh of Carnegie Mellon University. I liked it so much. It was much easy than the traditional formula. Po-Shen-Loh had submitted this paper in 2019.

My good wishes to all my blog readers !!

What I mean is . . .

My favorite equations (and/or) I too have made it

My favorite expressions are down below. They are in the order I like. I do not have formal mathematics knowledge except "Engineering Mathematics". I learnt from Wikipedia all by myself. I had no books or guru to teach me.
The order and the equation:-
1. Niels Henrik Abel
2. Leonhard Euler
3. Carl Friedrich Gauss

You may recall that I too managed to make a set of equations. My rational series. Though they are hardly of great level but still there are two points that I wish to say:-
1. The polynomials are of two types. One involving (n+1)/(n+2) - (n-1)/n. And the other is n/(n+1) -(n-1)/n.
2. Apart from polynomials they can be roughly represented as geometric progression too till finite number of terms.
Here is the link to Rational Number Series and Polynomials.

If I can develop polynomials and geometric progression till finite terms reading only Wikipedia. Imagine what I can do if I get a formal training and chance to read books by great authors !! I would also like to remind that I have done "rough work" on Gauss too and have obtained approximate results. Then there are my #PythaShastri mantras.

Polynomials from the Rational Number Series

The rational number series gives exquisitely beautiful polynomials.
My mental prowess is less in making polynomials. I can make only the simple ones.So, I tried Wolfram Alpha. And it delivered the results.
I am attaching the snapshots of the polynomials. I am attaching 12 snapshots. Well, 3 of them gives golden ratio. n^2+n-1 terms.

India China tussle

India China tussle at Ladakh over border issue is very painful to read.
As a blogger I feel it is my duty to write my opinions on it.
Border means different thing to different people. When the British divided it was the British idea that mattered and most of the people obeyed.
But it is not British on the Ladakh side. India had rightly made Ladakh the union territory earlier.
But now there is a dispute.
India is the place where Gautama Buddha was born. I hope China respects the Buddhist philosophies and accepts peace. Hope the Chinese are "repelled" by India's respect for Buddha.

Two things I want to say

Firstly, I have obtained an equation for the Golden Ratio. This is indeed a big feat. This gives the golden ratio value accurately (unlike my previous blog post on golden ratio with agm logic).
Well, the equation that gives the golden ratio phi is the solution for n in
(((n+1)^2)/((n+2)^2)-((n-1)^2)/(n^2))/(((n+1)/(n+2)-(n-1)/n)^2)=0. This is from the rational number series. Actually the equation can be simplified to n^2+n-1=0.
Thus the golden ratio value as -1.6180339887. . . . . . . . .  is obtained.
Secondly, a prime number generator as effective as the Euler prime number generating polynomial is now made from the Rational number series.
The equation is
(((n+1)^2)/((n+2)^2)-((n-1)^2)/(n^2))/(((n+1)/(n+2)-(n-1)/n)^2)+42 generates prime numbers.
I have tested for 1 to 400. As much as 67% (268) of the results are prime.
Basically Euler prime generating formula was k^2-k+41. Mine is equivalent to k^2+k+41.
But both the above results can be obtained from the Rational number series. i.e the series
(((n+1)^2)/((n+2)^2)-((n-1)^2)/(n^2))/(((n+1)/(n+2)-(n-1)/n)^2).(equivalent to n^2+n-1)
So, the results are original.I feel privileged and feel honored to have got two of the finest results of mathematics. The Golden Ratio. And the Euler Prime Generating Polynomial.
Have great day!! Hope we all get freedom from the Covid-19 pandemic soon !!