color graphic of a function in matlab
1 次查看(过去 30 天)
显示 更早的评论
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/542061/image.png)
hi i need to write this in matlab and make a graphic that looks like this
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/542066/image.jpeg)
i need to put the function in a code and make this graphic i dont know how to do either
0 个评论
采纳的回答
Walter Roberson
2021-3-7
编辑:Walter Roberson
2021-3-7
Example:
xvec = linspace(0,10*pi, 100);
yvec = linspace(0,5*pi, 50);
[Xg, Yg] = meshgrid(xvec, yvec);
Zg = sin(Xg) .* tanh(Yg);
h = pcolor(Xg, Yg, Zg); set(h, 'edgecolor', 'none')
You will need to figure out how to effectively do the infinite summation.
7 个评论
Walter Roberson
2021-3-12
if a is infinity and b is pi then the denominator is n*sinh(n*a*pi/pi) which is n*sinh(n*infinity). For all positive n that is n*sinh(infinity) and positive times infinity is positive infinity. So each term of the denominator is positive infinity.
When the denominator is infinity the results of the division are 0 except possibly if the numerator is infinity. Can the numerator be infinity? If y is positive then as n approaches infinity the sinh approaches infinity. But if x is positive then x*n approaches infinity and sin(infinity) is undefined, being anywhere in the range -1 to +1. You would need careful consideration of the possibility of +/-infinity divided by infinity. I suspect that the reasoning might be that the constant infinity in the denominator "approaches" infinity faster than the numerator that is not constant infinity, so I suspect that you always get 0, but I am not certain.
So for a infinity and b positive I believe that the sum is either 0 or (because of n infinite) undefined.
Walter Roberson
2021-3-12
Maple says that with the denominiator being infinite that the terms are all 0 including at n = infinity.
更多回答(0 个)
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Graphics Performance 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!