I was working with one dataset and found the curve to be sigmoidal. i have fitted the curve and got the equation A2+((A1-A2)/1+exp((x-x0)/dx)) where:
x0 : Mid point of the curve
dx : slope of the curve
I need to find the slope and midpoint in order to give generalized equation. any suggestions?
You should be able to simplify the modeling of the sigmoid with a function of the following form:
The source includes code in R showing how to fit your data to the sigmoid curve, which you can adapt to whatever language you're writing in. The source also notes the following form:
Which you can adapt the linked R code to solve for. The nice thing about the general functions here will be that you can solve for the derivative from them. Also, you should note that the midpoint of the sigmoid is just where the derivative of dx (or dx^2) is 0 (where it goes from neg to pos or vice versa).
Assuming your equation is a misprint of
A2+(A1-A2)/(1+exp((x-x0)/dx))
then your graph does not reflect zero residual, since in your graph the upper shoulder is sharper than the lower shoulder.
Likely the problem is your starting values. Try using the native R function SSfpl, as in
nls(y ~ SSfpl(x,A2,A1,x0,dx))
Related
I am trying to implement polynomial regression using the least squares method. There was a problem while plotting the 3rd graph, it is not displayed.
I think it's about the implementation of the formula y=ax+b.
But in my case, in first I got experimental data values using inline functions polyfit and polyval.
x=0:0.1:5;
y=3*x+2;
y1=y+randn(size(y));
k=1;#Polynom
X1=0:0.01:10
B=polyfit(x,y1,k);
Y1=polyval(B,X1);
And after all, I am already using a linear model to solve the polynomial regression using the method of least squares.
Y2=Y1'*x+B'; -----this problem formula
subplot(3,2,3);
plot(x,Y1,'-b',X1,y1,'LineWidth');
title('y1=ax+b');
xlabel('x');
ylabel('y');
grid on;
As a result, no graph is drawn.
check size of the vector: x and Y1 are not same length, same for X1 and y1.
You probably want to plot as:
plot(x,y1,'-b',X1,Y1,'LineWidth', 1);
I start this thread asking for your help in Excel.
The main goal is to determine the coordinates of the intersection point P=(x,y) between two curves (curve A, curve B) modeled by points.
The curves are non-linear and each defining point is determined using complex equations (equations are dependent by a lot of parameters chosen by user, as well as user will choose the number of points which will define the accuracy of the curves). That is to say that each curve (curve A and curve B) is always changing in the plane XY (Z coordinate is always zero, we are working on the XY plane) according to the input parameters and the number of the defining points is also depending by the user choice.
My first attempt was to determine the intersection point through the trend equations of each curve (I used the LINEST function to determine the coefficients of the polynomial equation) and by solving the solution putting them into a system. The problem is that Excel is not interpolating very well the curves because they are too wide, then the intersection point (the solution of the system) is very far from the real solution.
Then, what I want to do is to shorten the ranges of points to be able to find two defining trend equations for the curves, cutting away the portion of curves where cannot exist the intersection.
Today, in order to find the solution, I plot the curves on Siemens NX cad using multi-segment splines with order 3 and then I can easily find the coordinates of the intersection point. Please notice that I am using the multi-segment splines to be more precise with the approximation of the functions curve A and curve B.
Since I want to avoid the CAD tool and stay always on Excel, is there a way to select a shorter range of the defining points close to the intersection point in order to better approximate curve A and curve B with trend equations (Linest function with 4 points and 3rd order spline) and then find the solution?
I attach a picture to give you an example of Curve A and Curve B on the plane:
https://postimg.cc/MfnKYqtk
At the following link you can find the Excel file with the coordinate points and the curve plot:
https://www.mediafire.com/file/jqph8jrnin0i7g1/intersection.xlsx/file
I hope to solve this problem with your help, thank you in advance!
kalo86
Your question gave me some days of thinking and research.
With the help of https://pomax.github.io/bezierinfo/
§ 27 - Intersections (Line-line intersections)
and
§ 28 - Curve/curve intersection
your problem can be solved in Excel.
About the mystery of Excel smoothed lines you find details here:
https://blog.splitwise.com/2012/01/31/mystery-solved-the-secret-of-excel-curved-line-interpolation/
The author of this fit is Dr. Brian T. Murphy, PhD, PE from www.xlrotor.com. You find details here:
https://www.xlrotor.com/index.php/our-company/about-dr-murphy
https://www.xlrotor.com/index.php/knowledge-center/files
=>see Smooth_curve_bezier_example_file.xls
https://www.xlrotor.com/smooth_curve_bezier_example_file.zip
These knitted together you get the following results for the intersection of your given curves:
for the straight line intersection:
(x = -1,02914127711195 / y = 23,2340949174492)
for the smooth line intersection:
(x = -1,02947493047196 / y = 23,2370611219553)
For a full automation of your task you would need to add more details regarding the needed accuracy and what details you need for further processing (and this is actually not the scope of this website ;-).
Intersection of the straight lines:
Intersection of the smoothed lines:
comparison charts:
solution,
Thank you very much for the anwer, you perfectly centered my goal.
Your solution (for the smoothed lines) is very very close to what I determine in Siemens NX.
I'm going to read the documentation at the provided link https://pomax.github.io/bezierinfo/ in order to better understand the math behind this argument.
Then, to resume my request, you have been able to find the coordinates (x,y) of the intersection point between two curves without passing through an advanced CAD system with a very good precision.
I am starting to study now, best regards!
kalo86
Is there a simple equation which given the area of the shaded part and the mean, gives you the corresponding sigma for a normal distribuion?
P.S the shaded part corresponds to the area under the section of the Gaussian curve which lies on the negative x-axis. In my application this will correspond to the cross over probability.
Thanks
Do I understand correctly that you mean area left of x=0?
Area left of zero is simply \Phi((0 - \mu)/\sigma) where \mu is the mean of the distribution (1) and \sigma is the variance (what you are looking for). \Phi() is the normal cdf. You can easily (sort of) solve it for \sigma:
In case of normals \Phi((0 - \mu)/\sigma) = a is equivalent to \Phi(1/\sigma) = 1 - a (a is the area under the curve).
You cannot invert \Phi() easily but software will just do it. In R the inverse is qnorm() and \sigma will be 1/qnorm(1-a).
Yesterday, I posted a question about general concept of SVM Primal Form Implementation:
Support Vector Machine Primal Form Implementation
and "lejlot" helped me out to understand that what I am solving is a QP problem.
But I still don't understand how my objective function can be expressed as QP problem
(http://en.wikipedia.org/wiki/Support_vector_machine#Primal_form)
Also I don't understand how QP and Quasi-Newton method are related
All I know is Quasi-Newton method will SOLVE my QP problem which supposedly formulated from
my objective function (which I don't see the connection)
Can anyone walk me through this please??
For SVM's, the goal is to find a classifier. This problem can be expressed in terms of a function that you are trying to minimize.
Let's first consider the Newton iteration. Newton iteration is a numerical method to find a solution to a problem of the form f(x) = 0.
Instead of solving it analytically we can solve it numerically by the follwing iteration:
x^k+1 = x^k - DF(x)^-1 * F(x)
Here x^k+1 is the k+1th iterate, DF(x)^-1 is the inverse of the Jacobian of F(x) and x is the kth x in the iteration.
This update runs as long as we make progress in terms of step size (delta x) or if our function value approaches 0 to a good degree. The termination criteria can be chosen accordingly.
Now consider solving the problem f'(x)=0. If we formulate the Newton iteration for that, we get
x^k+1 = x - HF(x)^-1 * DF(x)
Where HF(x)^-1 is the inverse of the Hessian matrix and DF(x) the gradient of the function F. Note that we are talking about n-dimensional Analysis and can not just take the quotient. We have to take the inverse of the matrix.
Now we are facing some problems: In each step, we have to calculate the Hessian matrix for the updated x, which is very inefficient. We also have to solve a system of linear equations, namely y = HF(x)^-1 * DF(x) or HF(x)*y = DF(x).
So instead of computing the Hessian in every iteration, we start off with an initial guess of the Hessian (maybe the identity matrix) and perform rank one updates after each iterate. For the exact formulas have a look here.
So how does this link to SVM's?
When you look at the function you are trying to minimize, you can formulate a primal problem, which you can the reformulate as a Dual Lagrangian problem which is convex and can be solved numerically. It is all well documented in the article so I will not try to express the formulas in a less good quality.
But the idea is the following: If you have a dual problem, you can solve it numerically. There are multiple solvers available. In the link you posted, they recommend coordinate descent, which solves the optimization problem for one coordinate at a time. Or you can use subgradient descent. Another method is to use L-BFGS. It is really well explained in this paper.
Another popular algorithm for solving problems like that is ADMM (alternating direction method of multipliers). In order to use ADMM you would have to reformulate the given problem into an equal problem that would give the same solution, but has the correct format for ADMM. For that I suggest reading Boyds script on ADMM.
In general: First, understand the function you are trying to minimize and then choose the numerical method that is most suited. In this case, subgradient descent and coordinate descent are most suited, as stated in the Wikipedia link.
I apologise for the newbishness of this question in advance but I am stuck. I am trying to solve this question,
I can do parts i)-1v) but I am stuck on v. I know to calculate the margin y, you do
y=2/||W||
and I know that W is the normal to the hyperplane, I just don't know how to calculate it. Is this always
W=[1;1] ?
Similarly, the bias, W^T * x + b = 0
how do I find the value x from the data points? Thank you for your help.
Consider building an SVM over the (very little) data set shown in Picture for an example like this, the maximum margin weight vector will be parallel to the shortest line connecting points of the two classes, that is, the line between and , giving a weight vector of . The optimal decision surface is orthogonal to that line and intersects it at the halfway point. Therefore, it passes through . So, the SVM decision boundary is:
Working algebraically, with the standard constraint that , we seek to minimize . This happens when this constraint is satisfied with equality by the two support vectors. Further we know that the solution is for some . So we have that:
Therefore a=2/5 and b=-11/5, and . So the optimal hyperplane is given by
and b= -11/5 .
The margin boundary is
This answer can be confirmed geometrically by examining picture.