Friday, July 25, 2025

Calculating a^b from infinite series

How to calculate a^b from infinite series?

I have used (arithmetic progression)/(geometric progression). An experiment that I did. The (arithmetic progression)/(geometric progression) is n/(a^n). You may recall that I multiplied n/(a^n) with (x^n)/(n!) and got gamma function earlier. 

Now, a^b.

11^(1/5) and 11^7. 


This could be useful in mathematical software calculations and engineering calculations. Provided the terms converge faster.



I think the expression converges very fast.  10 or so terms are good enough for 5 decimal places of accuracy.

And 4 decimal places in just 5 iteration 


2 comments:

Kirtivasan Ganesan said...

My best answer to what is the need for 4^(4/5) and all for calculating 4^(1/5).
It is in the concept. The dimensions of an object that we perceive as we go nearer to it changes as per (arithmetic progression)/(geometric progression).
However measurements do not change.

Kirtivasan Ganesan said...

We can treat at higher values of n, say 11^(318/319) as 11