complex function inverse plot
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fima v
el 26 de Feb. de 2020
Respondida: KSSV
el 26 de Feb. de 2020
Hello, i have a plot a function
y=1/(ln(x)-0.1)
how can i plot X as a function of Y ,when y=[5:0.01:6]
Thanks.
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KSSV
el 26 de Feb. de 2020
On solving it manually....we have:
y = 1/(log(x)-0.1) ;
x = exp(1/y+0.1) ;
To plot:
y=[5:0.01:6] ;
x = exp(1./y+0.1) ;
plot(x,y)
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Más respuestas (1)
John D'Errico
el 26 de Feb. de 2020
Note that the natural log function is NOT written as ln in MATLAB, but just as log.
Next, in general, it may be impossible to plot your relationship, because there may be infinitely many disjoint regions over which your plot would need to be generated. For example, suppose you wanted to plot the regions where something as simple as sin(1./x) is between .5 and 1?
If a solution exists, where the function has a simple inverse, and if you have the symbolic toolbox, then finverse can help you.
syms X
Y = 1/(log(X) - 0.1)
fplot(finverse(Y),[5,6]);
xlabel 'Y'
ylabel 'X'
grid on
Or, if you want to use plot, you could do it like this:
syms X
Y = 1/(log(X) - 0.1);
y = 5:.01:6;
xfun = matlabFunction(finverse(Y));
plot(y,xfun(y),'-');
xlabel 'Y'
ylabel 'X'
grid on
If you don't have the symbolic toolbox, then you would need to use a loop, and then use fzero to compute the inverse.
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