can you write a vector function?
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I am looking to express a parametrized vector function and draw the surface..
2 个评论
Adam Danz
2019-6-23
How can we help you with that? Where are you stuck?
Thomas Starai
2019-6-23
采纳的回答
Star Strider
2019-6-24
You pretty much wrote it yourself:
x = @(u,v) [2-v.*sin(u/2).*sin(u); 2-v.*sin(u/2).*cos(u); v.*cos(u/2)];
See the documentation sections on Anonymous Functions (link) and Vectorization (link) for an extended discussion of both. The documentation for the ndgrid (link) function will also be helpful.
4 个评论
Thomas Starai
2019-6-24
Star Strider
2019-6-24
You could, however that is going to be slower and less efficient than using the anonymous function implementation.
The Symbolic Math Toolbox is best for derivations and such that are done once, then using the matlabFunction function, implemented iin base MATLAB, taking advantage of its computational efficiencies.
Thomas Starai
2019-6-24
Star Strider
2019-6-24
The @ operator denotes a function handle, and is necessary if you want to define it as an anonymous function. You could define it as a function file, however that is usually reserved for more complicated functions. See Function Basics for details.
You don’t have to define it as a function if you don’t want to — it will simply execute as a statement instead, although you could not then use it as a function later in your code — however in your original Question you specified that it is a function. See What Is a Function Handle? for an extended discussion.
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Ali Zain
2020-1-6
0 个投票
y= function(t)=[0.2582;
-0.3631 cos(0.21910*(t);
0.3572 cos(0.4158)(t);
0.3473*cos(0.6180t);
0.3336*cos(0.8135t);
0.3162*cos(t);
0.2954*cos(1.1756t);
0.2714*cos(1.3383t);
0.2443*cos(1.4863t);
0.2146*cos(1.6180t);
0.1826*cos(1.7321t);
0.1485*cos(1.8271t);
0.1128*cos(1.902t);
-0.0759*cos(1.9563t);
0.0382*cos(1.9890t)]
why it is giving error...
1 个评论
Star Strider
2020-1-6
Except for the second line, there is no multiplication operator.
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