I have made a formula for 2n factorial. I have called it the devil formula because 3.3 can be 3 or 3.33333 . Similarly 0.066 can be 0.06 or 0.0666. And so can 1.3 be 1.33 or 1.3333. This formula can be adjusted and near value of 2n factorial be reached. However this formula works from 100 to above.
The formula is
Hope you have a fine day !!
You may be aware that I made 2n! formula of the type ((((n+1)(n+2))/2)^(n)). I have made it very approximate.
The formula for (2n)! is
The Wolfram Alpha results to confirm that the formula is now very approximate.
Have a nice weekend!!
(Today on September 8th, 2024)
If you are doing a programming in computer, then you may use the following. The results will be more accurate.
I have been working on ((((n+1)(n+2))/2)^n) since April. 5 months. I derived the formula from Measuring Behavior expression or what I term it to be equivalent to Wallis pi. I was trying to get 0.5 factorial. While doing that I discovered the above pattern for (2n)!. So a journey of 5 months of intense application ended yesterday night.
Of course, James Stirling formula for n! is far more superior and has far more fidelity.
But the expression (((n+1)(n+2))/2) is the sum of first (n+1) numbers. And that is the reason why I persisted with this formula. Because sum of (n+1) numbers to the power n is related to 2n!. I found that interesting!!
I was not sure whether two of my approximations statements were true or not.
They were
(a/b)^k
the next numbers
Math GPT has done analysis and found out that the statements are true
(a/b)^k
the next numbers
I am feeling satisfied. Have a nice day !!
I have modified my earlier formula and it now fetches result in the calculator range to 1 percent. Given below is the screenshot of the formula
And the Wolfram Alpha screenshot to confirm 1 percent error claim.
Have a nice day !!
I hereby submit my approximate 2n factorial formula
It calculates 2n factorial till 2n is 0.1 million or 1 lakh.
Any more improvements, if any will be done to this formula only. And that is what I have thought about.
Have a nice day !!
You may be aware that I made a formula for (2n)!. It was (2n) factorial was approximately equal to (((n+1)(n+2))/2)^n. Here is the link
It was a crude formula and the result was not accurate.
So I posted a query in Wolfram Alpha to plot (((n+1)(n+2))/2)^n divided by (2n)!. The resultant graph seems like a half binomial distribution graph. Half bell graph.
As the graph is known to us the formula (((n+1)(n+2))/2)^2 can be made more accurate.
If you thought factorials of decimals and integers can be found out only by integral of e or integral of log functions, think again !!
Here is a formula which uses summation till infinity and still finds factorials of decimals and integers. How about that ?
This formula is very liked by me also because I did an experiment to derive it.
(I had done this formula earlier and have made a blog post. In this blog post I have reframed the expression in some other manner)
