# Algorithm for Fractional power calculation

9 vues (au cours des 30 derniers jours)
Life is Wonderful le 28 Fév 2022
Hi
1. fractional base is no problem with current implementation.
2. To support fractional exponents, get the n-th root for any given number b. How to implement algorithm to get a numerical approximation ?
3. Current approach is inefficient, because it loops e times
b = [-32:32]; % example input values
e = [-3:3]; % example input values but doesn't support fraction's
power_function(b,e)
p = 1;
if e < 0
e = abs(e);
multiplier = 1/b;
else
multiplier = b;
end
for k = 1:e
p(:) = p * multiplier; % n-th root for any given number
end
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### Réponses (1)

Alan Stevens le 28 Fév 2022
How about using the Newton-Raphson algorithm. Here's the basic idea:
% x^n = b
% Let f(x) = x^n - b
% dfdx(x) = n*x^(n-1)
%
% Use Newton-Raphson iteration
% x1 = initial guess
% err = 1;
% while err > tol
% x = x1 - f(x1)/dfdx(x1);
% err = abs(x - x1);
% x1 = x
% end
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Life is Wonderful le 12 Mar 2022
Modifié(e) : Life is Wonderful le 13 Mar 2022
Hi Walter,
Please see below code which is computing exp() and taylor series. At the 1 iteration final exp() value is calculated/converge. Please find the attachment
I could see with one multiplication and one addition, I could get the exp() THDN = -108dBc with x = +23 & -145 with x = -23 dBc which is very much efficient. The same works for fixed point as well . One need to be careful with division operator ( this takes lots time ).
I have added Excel spread sheet ( to replicate my below implementation ) for offline calculation. It has upto 10 fraction place of accuracy ( which pretty good for any standard as acceptance criteria). Please find the attachment.
Next step - With below code, do you have a proposal where number iteration is reduced [ I think one can optimize 90 iteration ] and If one do a loop unroll [smaller than 5 iteration ] , it can be done one max 5 steps. Any proposal/ suggestion ?
Thanks !!
n = 1e2;
x = 23.0; % for taylor series [ -22:22] for 32 bit DSP
inpval = 23.0; % for built-in exp()
% create a function and pass the args
expval = 1.0; % initialize as double
fprintf('%20s|%30s|%26s|%25s|%25s|\n--------------------+------------------------------+--------------------------+-------------------------+-------------------------+\n',"indx","expval","exp(x)",'diff','Nbits');
for i = n-1:-1:1
expval = 1 + x * expval / i;
fprintf('%20d|%30.10f|%26.15f|%25.10f|%25.10f|\n',i,expval,exp(inpval),abs(exp(inpval) - expval), log2(exp(inpval)));
end
Walter Roberson le 12 Mar 2022
Sorry, I do not know.

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