COS(2pix)
古いコメントを表示
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

採用された回答
その他の回答 (2 件)
Fawaz Hjouj
2019 年 8 月 28 日
編集済み: Fawaz Hjouj
2019 年 8 月 31 日
0 投票
3 件のコメント
Walter Roberson
2019 年 8 月 28 日
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
2019 年 8 月 29 日
Who is "Wally"?
Steven Lord
2019 年 8 月 29 日
カテゴリ
ヘルプ センター および File Exchange で 2-D and 3-D Plots についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!