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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

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Più risposte (2)
Fawaz Hjouj
il 28 Ago 2019
Modificato: Fawaz Hjouj
il 31 Ago 2019
0 voti
3 Commenti
Walter Roberson
il 28 Ago 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
il 29 Ago 2019
Who is "Wally"?
Steven Lord
il 29 Ago 2019
Fawaz Hjouj
il 31 Ago 2019
0 voti
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