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Hi ALL,
I needed the Taylor series ( about x=0) of cos[2*pi*x] for some application.
I wrote this little simple code:
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Xvalue= -10:.01:10;
Yvalue =zeros(1, length(Xvalue));
for w =1: length(Xvalue)
x0=Xvalue(w);
ss=0;
for mm = 0:100
ss=ss+ (-1)^(mm ) * ((2*pi*x0)^(2*mm)) / factorial(2*mm) ;
end
Yvalue(w)=ss;
end
figure(1)
plot(Xvalue, Yvalue)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
I got the below figure, where after x is about =6, the y values started to be NAN.
comments please

Accepted Answer
More Answers (2)
Fawaz Hjouj
on 28 Aug 2019
Edited: Fawaz Hjouj
on 31 Aug 2019
0 votes
3 Comments
Walter Roberson
on 28 Aug 2019
The value might be 0 in the limit, but you are not working in the limit, you are working with floating point numbers with finite precision and you are overflowing that finite precision.
Note the calculating the product of (2*pi*x0)^2 / (2*M) over M=0:100 would not suffer from the same overflow .
Walter Roberson
on 29 Aug 2019
Who is "Wally"?
Steven Lord
on 29 Aug 2019
Fawaz Hjouj
on 31 Aug 2019
0 votes
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