find finverse of cumtrapz()
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Aishwarya Radhakrishnan
on 22 Sep 2019
Commented: Star Strider
on 22 Sep 2019
Hi,
I have to invert the function that calculates cumtrapz of an equation as follows:
mu = 0;
var = 1;
%x = -9:0.01:9;
syms x
f(x) = cumtrapz((2*pi*(v))^(-0.5)* exp(-((x-mu).^2)/(2*(v))))
i = finverse(f)
fplot(i)
But this gives me :
f(x) =
0
Warning: Unable to find functional inverse.
> In symengine
In sym/privBinaryOp (line 1032)
In sym/finverse (line 40)
In inverse_cdf (line 7)
i(x) =
Empty sym: 0-by-1
However the cumtrapz function does work:
mu = 0;
var = 1;
x = -9:0.01:9;
y = (2*pi*(v))^(-0.5)* exp(-((x-mu).^2)/(2*(v)));
i = cumtrapz(y);
plot(i)
what can i do to invert this function
0 Comments
Accepted Answer
Star Strider
on 22 Sep 2019
Edited: Star Strider
on 22 Sep 2019
Since ‘f’:
mu = 0;
v = 1;
syms x
f(x) = int((2*pi*(v))^(-0.5)* exp(-((x-mu).^2)/(2*(v))))
evaluates to:
f(x) =
(7186705221432913*2^(1/2)*pi^(1/2)*erf((2^(1/2)*x)/2))/36028797018963968
3 Comments
Star Strider
on 22 Sep 2019
If you want to use cumtrapz, the interp1 function is likely the best option:
mu = 0;
v = 1;
x = -7:0.01:7;
fx = cumtrapz(1/sqrt(2*pi*(v)) * exp(-((x-mu).^2)/(2*(v))));
y = [5 25 50 75 95];
fi = interp1(fx, x, y, 'linear','extrap')
figure
plot(x, fx)
hold on
plot(fi, y, '+')
hold off
grid
This plots ‘+’ markers at the appropriate values of the ‘y’ vector. I call the inverse ‘fi’. I had to restrict your original ‘x’ vector because with the original vector, the ‘fx’ points were not unique, as interp1 defines that.
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