How can I Plot a primitive function?

Question

Nicolas Kennis 2016-10-16

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https://ww2.mathworks.cn/matlabcentral/answers/307515-how-can-i-plot-a-primitive-function

评论： Nicolas Kennis 2016-12-19

I'd like to use ezplot to draw a function. This is my code:

ray=0.11;
m=3.5;
C=2*10^(-12);
DS=140;
CONST=1/((DS^m)*C*pi^(m/2));
syms x
F= @(x) CONST*((x.^(-m/2))./((1.6906*(x/ray).^5-4.6813*(x/ray).^4+5.4253*(x/ray).^3-2.8213*(x/ray).^2+0.8716*(x/ray)+0.4251).^m));
syms x
z=int(F,x,0,0.04);
syms x
ezplot(z,[0,0.04])

I always find an error while I want to plot this function... Why? This is the error :

Error using inlineeval (line 14)
Error in inline expression ==> int(2000./(x.^(7./4).*((2179.*x)./275 - (28213.*x.^2)./121 + (5425300.*x.^3)./1331 - (468130000.*x.^4)./14641 +
(16906000000.*x.^5)./161051 + 4251./10000).^(7./2)), x, 0, 1./25)
 Undefined function 'int' for input arguments of type 'double'.
Error in inline/feval (line 33)
        INLINE_OUT_ = inlineeval(INLINE_INPUTS_, INLINE_OBJ_.inputExpr, INLINE_OBJ_.expr);
Error in ezplotfeval (line 51)
    z = feval(f,x(1));
Error in ezplot>ezplot1 (line 468)
    [y, f, loopflag] = ezplotfeval(f, x);
Error in ezplot (line 144)
                [hp, cax] = ezplot1(cax, f{1}, vars, labels, args{:});
Error in sym/ezplot (line 76)
      h = ezplot(fhandle(f),varargin{:});

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Answer 1

Walter Roberson 2016-10-19

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在 MATLAB Online 中打开

xvals = linspace(0, 0.04, 1000);
y = F(xvals);
z = cumtrapz(y);
plot(xvals, z);

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Walter Roberson 2016-10-24

You cannot get an exact plot. No display can position points with infinite precision. You cannot position a point at x = 1, y = Pi exactly. You can position a point at x = 1, y = some approximation of Pi.

There is no closed form formula for your function, Numeric integration is your only choice. Numeric integration is always approximate. The better numeric integration routines are adaptive, but they all come down to trapezoidal rule or similar applied with respect to the y values at x positions they sampled at.

If you want more precision, sample more finely. Sample at irregular intervals if you want. logspace() instead of linspace().

Your formula for F has only x as its free variable, and in your original post you defined z=int(F,x,0,0.04); which is definite integration of a function with respect to a single variable . The result is going to be a specific numeric value. But then you want to plot that with respect to x.

Your more recent posting uses indefinite integration for z, and asks to plot the result, but it does not define either a left or right boundary, so there cannot be any solution.

Your formula for F approaches infinity quickly as x approaches 0 from the right. The exact value of the integral is going to depend strongly on where you start integrating -- and in turn that means that the plot is going to depend a lot on where you start integrating. It is not the case that there is "thin" infinity were the values might be large but very narrow in such a way that the integral stays finite. Each additional decimal place closer to 0 gives an F value about 56 times larger (10^(7/4) is the ratio)

Nicolas Kennis 2016-10-25

编辑：Nicolas Kennis 2016-10-25

在 MATLAB Online 中打开

Walter, here you'll find the code I'm using:

 ray   = 0.11;
m     = 3.5;
C     = 2*10^(-12);
DS    = 140;
CONST = 1/((DS^m)*C*pi^(m/2));
F     = @(x) CONST*((x.^(-m/2))./((1.6906*(x/ray).^5-4.6813*(x/ray).^4+5.4253*(x/ray).^3- ...
             2.8213*(x/ray).^2+0.8716*(x/ray)+0.4251).^m));
F2     = @(x) CONST*((x.^(-m/2))./((2.3652*(x/ray).^5-7.1218*(x/ray).^4+8.5106*(x/ray).^3- ...
             4.6314*(x/ray).^2+1.5432*(x/ray)-0.0088).^m));
xvals = linspace(0.0007, 0.003, 1000);
xvals2 = linspace(0.0007, 0.003, 1000);
y = F(xvals);
y2 = F2(xvals2);
z = cumtrapz(y);
z2 = cumtrapz(y2);
plot(xvals, z,'color','r');hold on;
plot(xvals2, z2,'color','b');
view(-90,90);
set(gca,'ydir','reverse')
xlabel('Profondeur de fissure (m)');
ylabel('Nombre de cycles (N)');
set(gca,'YScale','log');
grid on

It gives me a result. When I tried to put logspace() instead of linspace(), I've have a very big difference. Is it normal? It seems to be good with linspace() but not with logspace()...

Nicolas Kennis 2016-12-19

编辑：Nicolas Kennis 2016-12-19

在 MATLAB Online 中打开

Hello, I do evolved in my function and I've a new code. I've an error but I don't know why? Can you help me? My goal is tho find this integral! I have this error : Error using * , Inner matrix dimensions must agree.

 ray   = 0.11;
m     = 3.5;
C     = 2*10^(-12);
DS    = 140;
CONST = 1/(C*pi^(m/2));
F     = @(x) CONST*((x.^(-m/2))./(((452092178*x.^3-10493093*x.^2+78093.85*x-42.08).^m).*((1.6906*(x/ray).^5-4.6813*(x/ray).^4+5.4253*(x/ray).^3-2.8213*(x/ray).^2+0.8716*(x/ray)+0.4251).^m)));
W=integral(F,0.002,0.01);
disp(['intégrale au point A = : ' num2str(W)]);

Nicolas Kennis 2016-12-19

OK I found the problem... ;-)

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Answer 2

Mischa Kim 2016-10-19

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https://ww2.mathworks.cn/matlabcentral/answers/307515-how-can-i-plot-a-primitive-function#answer_239665

在 MATLAB Online 中打开

Nicolas, I would assume that there is no closed-form, symbolic solution for your integral. So instead use numeric integration and the integral function

 ray   = 0.11;
 m     = 3.5;
 C     = 2*10^(-12);
 DS    = 140;
 CONST = 1/((DS^m)*C*pi^(m/2));
 F     = @(x) CONST*((x.^(-m/2))./((1.6906*(x/ray).^5-4.6813*(x/ray).^4+5.4253*(x/ray).^3- ...
             2.8213*(x/ray).^2+0.8716*(x/ray)+0.4251).^m));
 z     = integral(F,0.01,0.04)

Also, because of the first term in F and the lower limit of x the integral will be infinite. I changed it to 0.01 for a finite result.

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Nicolas Kennis 2016-10-19

编辑：Nicolas Kennis 2016-10-19

Kim, Thanks for your answer! I really appreciate.

So, it means that there is no solution to plot the primitive of F (from 0.0007 to 0.05 for example)?

Thanks in advance

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How can I Plot a primitive function?

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