A Result

These results are interesting. 

Books needed

Justification -

I wrote the area of a perfect and hence simple right angled triangle is zero. And I also wrote that I need this statement for proofs. Well what I was talking of was about analysis. Hence a book on analysis.

I wrote n should be treated as pi and 2n as pi squared. What I was talking of was about infinity. Hence a book on infinity by the great man himself. Euler.

The three consecutive numbers and four consecutive number theory has given me hope in a subject I dreaded. Algebra. I want to see how much does this concept relevant. Hence books on Algebra by great men themselves. Euler. Maclaurin.

Notebooks of Ramanujan, Disquisitiones Arithmeticae and a book on Niels Henrik Abel are mathematical biographies that I wish to possess.

Is it valid?

Made it !!

At last !!

I wanted to put four variables and make an equation like Niels Henrik Abel.

I have made it. Here it is. It is exact. ( I have checked only for 2 values, though.)

I shall check on this more. And blog if there is an error. 

Have a nice day, dear blog readers !!

An exact expression

I was trying hard to get an exact equation. There were many approximate results with my four consecutive numbers and three consecutive numbers theory.
I made this one. This is an exact equation. Put any values for n and a. And bingo !! You can verify that the results hold exactly.(Still if there is any discrepancy, do let me know)

Check out these two expressions

The two expressions are approximately equal. Not exactly equal. But approximately. Till 7 places of decimals or more. Also the number of terms in LHS and RHS in both are odd number of terms. Not even number. In case the number of terms are even number then it gives some other value.
Again, many readers of my blog may be confused. I am sorry for that. Perhaps I need to write a paper on this. I shall think about it.

In total I made three approximate expressions. I am putting all the three expressions here for the help of the reader.

Have a nice day, my dear readers.

At present . . .

I watched the movie Khoj(2017) recently. I liked it immensely. It is a Bengali/Hindi movie. It is a detective story. The locales were superb. The music was excellent. And if you start watching the movie, then you cannot help but watch till the end. This movie was recommended by Youtube logic. It has English subtitles.
I am not in a position to hear radio as my travel time is highly reduced. As compared to my Delhi life earlier. Therefore the pleasure of hearing new songs and using the Shazam app for the song name is no longer there. I feel terribly sad. This is a big sacrifice I had to make as I am a big songs lover. Internet and the search logic is hardly an answer to this. Because the pleasure of listening to a fantastic new song is only there in FM radio. This is how I feel. I really miss those listening to "Suno na sangemarmar" or "Khamoshiyan" in FM radio in Delhi roads. Dukh hondha hai, pape !!(Punjabi)
It is great that India has come up with COVAXIN, a vaccine for coronavirus. It is a great feat by our scientists and medical doctors. The doctors and the medicos have been working so hard for the last few months. I am sure this vaccine will work and give freedom from this disease.
My good wishes to all of my blog readers !!

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 !!