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

 Risposta accettata

Walter Roberson
Walter Roberson il 28 Ago 2019

0 voti

for mm = 0:100
ss=ss+ (-1)^(mm ) * ((2*pi*x0)^(2*mm)) / factorial(2*mm) ;
What is (2*pi*6)^(2*100) ?
What is factorial(2*100) ?
What is the ratio of those two?

Più risposte (2)

Fawaz Hjouj
Fawaz Hjouj il 28 Ago 2019
Modificato: Fawaz Hjouj il 31 Ago 2019

0 voti

Isnt this series convergent?

3 Commenti

Walter Roberson
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
Walter Roberson il 29 Ago 2019
Who is "Wally"?
Steven Lord
Steven Lord il 29 Ago 2019
According to Wikipedia "Wally or Wallie is a given name, and a nickname for Wallace which ultimately means 'Wales' and Walter."
I don't recognize many of the real people listed on that Wikipedia page; I recognized many more from the Fictional characters section :)

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