Find an approximately value of ln3 with Taylor
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I am going to find the approximately value of ln3 by putting x = 1/3 and sum all the terms with a bigger abs-value than 1e-8. Anyone who know how I should continue my code?
I got the taylorserie of lnx for x = 1 as help in the task.
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Raunak Gupta
2019 年 12 月 4 日
Hi,
You may find below code useful. I am using ln(1-x) expansion of Taylor Series as it is more intuitive to code. It is not be very different from ln(1+x).
tol = 1e-8;
s = 0;
x = 2/3;
count = 1;
while 1
update = (x.^count)./count;
s = s - update;
count = count + 1;
if ~(abs(update) > tol)
return;
end
end
% s is the finalvalue of ln(1/3)
% reversing the sign of s will give ln(3)
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