# Taylor Series Expansions for sin(x)

17 views (last 30 days)
Spaceman on 22 Mar 2024
Commented: Spaceman on 8 Apr 2024
Given: sin(x)=(-1)^i*x^(2*i=1)/(2i+1)!...or... ...
Find: Code that calculates the Taylor series for sin(pi/3) using the equation above
Issue: It just seems like my code isn't approximating right, anything I am overlooking here?
My Solution: I also am unsure of how to compare the CORRECT summation results to realsin which happens to be 0.866
x=pi/3;
sum1=0;
N=10;
for i=0:N
a=(2*i+1);
sum1=(-1).^i.*(x.^(2i+1))./factorial(a);
fprintf('This is the estimate of sin(pi/3): %10e \n',sum1)
% diff=sum1-sin(pi/3);
end
% Could also use the input function to have the user enter a number>10, factorial is accurate up to N<=21
% summation here is always negative every other value?
% realsin=pi/3=0.8660
% Need a way to display how close my estimate was, difference between,
% howclose=diff(sum1, realsin)
% fprintf('Estimate was ... off of actual value: %0.10e', howclose)

Voss on 22 Mar 2024
2i should be 2*i.
And you need to accumulate the sum; as it is now sum1 is only the current term and it's never added to anything.
Voss on 8 Apr 2024
You're welcome!
Spaceman on 8 Apr 2024
You're the best, Voss!

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