07.2 Boundary between decisions

Content and key concepts

On the previous page, one neuron by step.
Let's change the weights and see how the dividing line moves. Zero weights can remove the whole normal boundary.
He collects inputs, adds a shift, and applies a non-linear rule. All parts of this calculation are already available in school arithmetic.
Для изменения рисунка включитеJavaScript; расчёт и теория доступны далее.
Weight sets the norm to the border
For classification by z, the boundary is defined by the equation wTx+b=0. If wx0 is hyperplane. For its two points, the difference v satisfies wTv=0: the vector w is perpendicular to all directions inside the border.
At w=[1,-1], b=5, on the plane, you get x2=x1+.5. The point [1.2] is on the negative side because z=5. The number z is not equal to the geometric distance without regulation:
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d(x,\mathcal H)=\frac{|w^Tx+b|}{\|w\|_2},\qquad w\ne0.
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For the selected point, the distance .5/G2. The multiplication w and b by one positive number preserves the boundary and the threshold classes, but scales the score. Therefore, the sigmoid probability at the same geometric boundary may become more pronounced.
Experiment.
Set w2=0: the boundary is vertical. Then cancel both weights. Score becomes constant: at b ± 0, one class will be selected everywhere; at b=0, all points will be on the threshold. You cannot draw one normal dividing line in the latter case.
Debug.
The formula x2=-(w1x1+b)/w2 requires the processing of a zero denominator. The lab is building a border crossing with the frame and separately explains the event. It cannot be assumed that the "surer classified" point is simply because on the automatically scalable drawing it is further from the line.
There is no border z = 0 at b = 0; there is no space at the door.
Cool function and saturation. How does this question continue the result?
Python examples of Chapter 07
Academic base