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