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

 Réponse acceptée

Walter Roberson
Walter Roberson le 28 Août 2019

0 votes

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?

Plus de réponses (2)

Fawaz Hjouj
Fawaz Hjouj le 28 Août 2019
Modifié(e) : Fawaz Hjouj le 31 Août 2019

0 votes

Isnt this series convergent?

3 commentaires

Walter Roberson
Walter Roberson le 28 Août 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 le 29 Août 2019
Who is "Wally"?
Steven Lord
Steven Lord le 29 Août 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 :)

Connectez-vous pour commenter.

Fawaz Hjouj
Fawaz Hjouj le 31 Août 2019

0 votes

Sorry wrong person

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