What is the algorithm to calculate decimal power of a number ? (Like 2^1.3082)

Question

Need of algorithm

Answer 1

I don't know for sure how `pow` does it, but you can do it with a taylor series expansion.
f(x) = f(0) + x f'(0) / 1! + x² f''(0) / 2! + x³ f'''(0) / 3! + ...
Now, letting f(x) = n^x:
n^x = 1 + x log n / 1! + x² log² n / 2! + x³ log³ n / 3! + ...
So you can calculate `let logn = Math.log(n)` and then sum up `(x**i) * (logn**i) / factorial(i)` - as many of them as you need to get an accurate result.
Of course we still have exponents, but it's all whole numbers. We are also trading `pow` for `log`, but you can find a similar series for the logarithm function (the mercator series).
Maybe we can even simplify the above, to get rid of all exponents.
n^x = 1 + x log n / 1! + x² log² n / 2! + x³ log³ n / 3! + ...
n^x = 1 + x log n * ( 1 + x log n / 2 * ( 1 + x log n / 3 * ( 1 + ... ) ) )
Which means in code:
for(let i = some_value; i > 0; --i)
result = 1 + result * x * logn / i;
EDIT: I coded it up, it doesn't look too bad!
https://code.sololearn.com/WsE1NAsJHFKJ/?ref=app

Answer 2

Use this code
https://code.sololearn.com/WqkbU5cI2xc3/?ref=app

Answer 3

Depends,
For square root (x^0.5)
I think it is based in testing.
You start counting from 0 to ∞ with step of y,
if n * n == x: return n
elif n * n > x: n -= y, y *= 0.1
If n*n is greater than x, the step would be decreased.
I have an old code about it:
https://code.sololearn.com/cGLkePeW9INZ/?ref=app

Answer 4

Thank you very much Schindlabua ☺️☺️☺️☺️
I was searching for this algorithm for more than 3 months but finally you helped me.
I don't know how to say thanks to you but again thank you very much !!!!!!

Answer 5

Sorry, i misunderstood your questions.

Answer 6

But Calviղ I want the power to be a decimal value
For example:- 2^0.00002449
pow() library can do this but I want to do that without using pow() library.
How does the pow() library did it ???
🤔🤔🤔

Answer 7

Hah, no worries! You're welcome!