complex function inverse plot
显示 更早的评论
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.
2 个评论
syms x
y=(5:0.01:6);
y=1./(log(x)-0.1);
g = finverse(y);
plot(y,g);
For this you need symbolic tool box, I would like you to read this link: https://de.mathworks.com/matlabcentral/answers/286749-find-inverse-of-a-function
fima v
2020-2-26
采纳的回答
KSSV
2020-2-26
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)
更多回答（1 个）
John D'Errico
2020-2-26
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.
类别
在 帮助中心 和 File Exchange 中查找有关 Assumptions 的更多信息
Translated by 
