Inverse of log formula
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Lisa
el 1 de Mayo de 2017
Respondida: Image Analyst
el 1 de Mayo de 2017
Hi, I have the following formula:
y = -log(x / mean(x))
y and x are vector’s.
However, in my case I only know y, but not x. Therefore, I would like to solve the inverse of the formula for x. I am aware of that exp is the inverse of log:
x=-exp(y)
but I am not sure how to consider the term mean(x)?
Thank you in advance.
Lisa
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Respuesta aceptada
Star Strider
el 1 de Mayo de 2017
You cannot determine ‘mean(x)’ unless you have access to the original ‘x’ vector. (If you had ‘x’, you would not need to do the inversion.) You can only recover the normalised ratio ‘x/mean(x)’.
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Image Analyst
el 1 de Mayo de 2017
Lisa, is this what you are after - to find the x for a y that is a specified value? You can do it numerically like this, where I find the x where y = 1:
% Define a range of x values.
x = sort(rand(1, 500), 'ascend');
% Compute the y values over that range.
fprintf('mean(x) = %f\n', mean(x));
y = -log(x / mean(x));
% Plot curve.
plot(x, y, 'b-', 'LineWidth', 2);
grid on;
xlabel('x', 'FontSize', 20);
ylabel('y', 'FontSize', 20);
% Find the x closest to y = 1 (for example).
[minValue, index] = min(abs(y-1))
% Get x and y values
x1 = x(index);
y1 = y(index);
fprintf('Closest point: y = %f at x = %f\n', y1, x1);
% Draw lines
hold on;
line([0, x1], [y1, y1], 'Color', 'r', 'LineWidth', 2);
line([x1, x1], [0, y1], 'Color', 'r', 'LineWidth', 2);
ax = gca;
ax.XAxisLocation = 'origin';
In the command window:
mean(x) = 0.514679
minValue =
0.0017406261319084
index =
93
Closest point: y = 1.001741 at x = 0.189011
And the plot of the results:
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