I had posted earlier that "Becoming Extroverted"(my blog) ideas result in sigma n!/(n^n). That is why this term occurs in faster pi expressions of Ramanujan and Chudnovsky brothers.
I am working on root 3 into sigma n!/(n^n).
The above results in 3.14 in just three iterations.
The equation has a scope and am further developing it. I am also trying to use "Becoming Extroverted" ideas in above to try and get further places of pi.
The screenshot of the last expression:
I am working on root 3 into sigma n!/(n^n).
The above results in 3.14 in just three iterations.
The equation has a scope and am further developing it. I am also trying to use "Becoming Extroverted" ideas in above to try and get further places of pi.
The screenshot of the last expression:
My final submission
1/pi to 10 places; the maximum my calculator can handle:
Divide 2.01........ by root 10. Divide again by 2.0031343434. You get 1 by pi.
I believe the last
16 comments:
I am trying to use 5! in root 3 into sigma n!/(n^n). There is some progress as further iterations gave 3 more places. But beyond that I am not sure.
So, the result is 3 places of pi each for 2 iterations. And the end.
I read in the internet that Chudnovsky brothers algorithm gives 4 places of pi for each iteration and it keeps going. So that is where the world is.
Another thing is the continued fraction 2/(3+1/(5+2/(7+3/(9+4/(11+5 . . . .
The above continued fraction is interesting and gives pi/5 (wait a second !!) to 99.79 % accuracy.
I don't think it is perfect but then pi/5 for odd and even numbers do not exist for continued fractions. Maybe, some day someone will make it.
1/pi = 1/(3+1/(7-1/(11+1/(15-1/(19+1 . . .
It is Madhava of Sangamagrama formula tried in continued fractions.
It is a new beginning. Ages old formula of Madhava of Sangamagrama. Reapplied. And it seems correct.
I work on intuitions in constants, especially after "measuring behaviour" resulted in a pi expressions.
I thought about "Becoming Extroverted" also in pi lines. "Becoming Extroverted" was more of equal to.
A person feels he is n!. Great combination. But the other is n^n. Too great an extrovert !!
So I wondered as n!/n^n.
So one thing is n!/(n^n) or n!/((n!)^n).
Another thing is what can one do when somebody is b^n. My intuitions say one has to try a.n+b.
These may be mad. But that's how I think.
So, sigma {[dn!/(en^n)][a.n+c/b^n] is extroverted response equation. A pi formula !!
And that is what is Ramanujan pi formulas are all about.
You need to work-out or take guidance from Ramanujan materials to get a,b,c,d,e. And then you get pi.
See Ramanujan Sato series in Wikipedia.
I feel proud that I am involving in mathematics of British Raj times.
Sigma ((n!)/((n!)^n))(comb(2n,n)) works a little.
The above by 1.77 gives 3.14159 (six places ).
I have put all the extroverted ideas together namely
1. (n!)/((n!^n). Jealous
2. comb(2n,n). Doubling. Measuring behaviour.
3. (3^(0.5))/((5!)^n). Five fingers and phase difference.
I made
((n!)(2n!)(3^(0.5)))/(((n!)^(n+2))(120^n)).
The result is 2.0289717.
Nothing great. Maybe multiplying and/or dividing by rational or irrational numbers may give results of pi or 1/pi.
I tried some more
1. (n!)/((n!)^n). Jealous.
2. comb(2n,n). Doubling.
3. (3^(0.5n)). Root terms.
4. e^(11^0.5). Numerator of 5th term in the series of e^(11^0.5); 11 is heegner number.(121)
5. 60n. Denominator of 1st term in the series for 1/11.(60 is 5 ×12)
6. 1. An addition to say 60n is lagging 121^n.
((n!)(2n!)(3^(0.5n))(60n+1))/(((n!)^(n+2))(121^n))
The result is 3.75876688697.
No great shakes.
I shall finish my poser.
1. (n!)/((n!)^n)
2. comb(2n,n)
3. 144. 1/11 series.
4. Root 2.
5. 121. e ^(root 11)
11 is heegner number.
So sigma from 0 to 100 of
((n!)(2n!)(144n+(2^(0.5)))/(((n!)^(n+2))(121^n))
results in 5.2418538001.
Dividing above by 1.6685338865 is pi.
1.6685338865 seems to be rational. Maybe 100/59.
This finishes.
Among the deceased mathematicians from India; I like
Ramanujan
Madhava of Sangamagrama
Harishchandra
Harishchandra was unlucky not to get Abel prize. He seemed great.
I don't like much Aryabhata, Bhaskara. Very simple work have they done.
Someone says in internet that they have discovered zero. What's this?
I like Dr.Vashist Narayan Singh too. But he doesn't seem to be of complex numbers. Maybe something big.
I have pasted the screenshot for the last but one comment.
Now
10/6(1+11/10000+20/1000000+33/100000000+60/10000000000) is very nearly 1.6685338865.
It's great!! Yoo hoo !!
Maybe a very quick pi expressions is here. One step away.
This means jealousy, combination concepts are correct. Pi god says so.
Initially it was like finding a needle in a haystack 50 square meters. Now it has been reduced to 1 square meter.
And the result error is quite symmetric. Oooooooh wow!!
The result error is symmetrical but not possibly in mathematical terms.
I think pi as a quick expressions is still far.
I tried the following:
((n!)(2n!)(120n))/((n!^(n))(121^n))
My logic was:
1.(n!)/((n!)^n). Jealousy due to extrovertedness.
2. comb(2n,n). Measuring behavior.
3. (5!)n or 120n. Fingers of hand.
4. 121^n. Heegner number squared.
The above results divided by (root 10 into 2.00313434343434) is 1 by pi to 10 places. The maximum a calculator can display.
Only checking needed is the recurring part in 2.0031343434 using large computers.
I am pasting the screenshot above.
I believe the last
1. (n!)/((n!)^n). Jealous.
2. comb(2n,n). Measuring behavior.
3. 163. Heegner number.
4. 10,5,2. Hands.
8 places of 1/pi obtained.
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