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Complete golden ratio formula for advanced KAAL number

KAAL number is 4.4.

Presently, 

is showing reasonably good results. 

You may recall 1+1/x+1/(x^2)=2 is the golden ratio expression. 

Then I said if we raise 1 to very small number, 2 to very small number and 3 to small number,  one gets LHS of just above expression. 

I was talking of raising only the n to some power. Say raising n to q/r.

What if we raise n to q/r, x to n+s/t and start from u/v and end at w/x. Is there an algorithm for this ?

Yes. Wolfram alpha is showing results. 

This can give advanced results for KAAL number.

But it is not in my hands. A big research team is required. 

Will this idea get materialized?

No. Because it is India. 

These two expressions give hope

KAAL number expressions were made. We used 1+2/x+3/(x^2) for p and got 1/2 number positions correct with correctly placed leading 0s.

The following two results from Wolfram Alpha are promising and we may get 2/3 number positions correct with correctly placed leading 0s.

I can check only a little with calculator and Wolfram Alpha. But I shall try.

But real thing is getting a nod from Professors of Mathematics.

My dreams of p of KAAL number

The present p

My dream of expressing p

Seems okayish

Advanced KAAL number theory is a problem. 

The value of p as below gives 2/1 numbers with 0s ahead correct.

But what if we need 3/2 numbers with 0s ahead correct.

The following helps.

You may note that the seed number is 227. And 227 is related to 772 as 1000 minus 772 is 227. And we are using 1.079 and 1.79.

What I mean in nutshell is that further scope in KAAL number exists. If. only I get a chance.  Only exam qualified guys get access to big computers. Not me. 

Few more infinite series to help find the right p1 and p2

 

Currently we are using the same p for 4.4^p/x and one(x)+pzero(x). This gives us first two digits correctly placed with the trailing 0s correct.

Maybe the above will help in separating p and making them p1 and p2, 

Whether e^ix or 4.4^p/x ?

I asked Google AI the question?

The answer