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Find an approximately value of ln3 with Taylor

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Hanna Sundling
Hanna Sundling on 29 Nov 2019
Answered: Raunak Gupta on 4 Dec 2019
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.Skärmavbild 2019-11-29 kl. 11.59.45.png
Skärmavbild 2019-11-29 kl. 11.56.13.png

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Answers (1)

Raunak Gupta
Raunak Gupta on 4 Dec 2019
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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