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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Walter Roberson
Walter Roberson 2019년 8월 28일

0 개 추천

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?

추가 답변 (2개)

Fawaz Hjouj
Fawaz Hjouj 2019년 8월 28일
편집: Fawaz Hjouj 2019년 8월 31일

0 개 추천

Isnt this series convergent?

댓글 수: 3

Walter Roberson
Walter Roberson 2019년 8월 28일
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 2019년 8월 29일
Who is "Wally"?
Steven Lord
Steven Lord 2019년 8월 29일
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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Fawaz Hjouj
Fawaz Hjouj 2019년 8월 31일

0 개 추천

Sorry wrong person

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