why ezplot(f,[3000,4000]) would not work

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Abdulaziz Abutunis 2016-2-28

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评论： Walter Roberson 2016-2-28

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Hi all,

I have this next function which is the result of solving for Px in my code

f=@(Px) (16*(-(((27*Px)/2 - 65625)*((27*Px)/2 + 65625))/4)^(1/2))/27;

the problem is that when I use the ezplot(f,[3000,5000]) it will plot wrong plot. However, if I use

ezplot((16*(-(((27*Px)/2 - 65625)*((27*Px)/2 + 65625))/4)^(1/2))/27,[3000,5000])

the curve will be correct. Please if you have any suggestion to solve this issue advise me.

Thanks

Aziz

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John D'Errico 2016-2-28

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No. Actually, the result that you claim does work, will fail in MATLAB.

ezplot((16*(-(((27*Px)/2 - 65625)*((27*Px)/2 + 65625))/4)^(1/2))/27,[3000,5000])
Undefined function or variable 'Px'.

I think you have things confused.

Abdulaziz Abutunis 2016-2-28

I am sorry I forgot the ' when I posted the function it should be ezplot('(16*(-(((27*Px)/2 - 65625)*((27*Px)/2 + 65625))/4)^(1/2))/27',[3000,5000])

Thank you, Aziz

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Answer 1

Walter Roberson 2016-2-28

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Your second form relies upon whatever Px happens to be in memory, and will fail if Px is not a scalar.

Possibly what you meant to post was

ezplot('(16*(-(((27*Px)/2 - 65625)*((27*Px)/2 + 65625))/4)^(1/2))/27',[3000,5000])

When I try that, the output is identical to plotting with your f anonymous function.

When I use your f anonymous function, ezplot gives a warning about f not being vectorized. You can remove that by vectorizing it:

f=@(Px) (16*(-(((27*Px)/2 - 65625).*((27*Px)/2 + 65625))/4).^(1/2))/27;

The output is the same exactly as for the non-vectorized version, and the same exactly as for the string version.

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Abdulaziz Abutunis 2016-2-28

Thank you Walter for the quick response. you are right, I forgot to add '...'. As you mentioned, the output for the vectorized and non-vectorized are the same (a straight line), however, the string version gave different curve which is correct when I checked a solved problem.

Thank you, Aziz