Market etiquette or good manners in a minefield - page 35
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Let me correct you. Not "distances", but "squares of distances".
If anything, it's not "squares of distance" but "squares of variance".
:)
"only it wasn't the lottery, it was a game of preference, and you didn't win, you lost."
This is what we got:
On the abscissa axis is the data vector, on the ordinate axis is the forecast
I can't figure out how to add more than 3 vectors to a graph. I have Matcad 14, it has "Primary Y-axis", "Secondary Y-axis", and "X-axis".
In the figure, the secondary Y-axis is Z(right) - a straight line using the method of least squares. By the way I had to divide this vector by 10^5 to see it on the graph.
to PapaYozh
I don't get it, dearie, is what you're writing so important that you can't keep silent? Well, if you have nothing else to say?
I can't figure out how to add more than 3 vectors to a graph. I have Matcad 14, it has "Primary Y-axis", "Secondary Y-axis", and "X-axis".
Comma separated!
The format is as follows: the ordinate axis is y0[i], y1[i], y2[i]. Abscissa axis - x0[i],x[i],x[i]. Or (if samples by X coincide):Ordinate axis - y0[i],y1[i],y2[i]. Abscissa axis is x[i].
There is a mistake about 10^5. You need to search around. And from the plot you can see that the line does not visually go through the area of maximal density. Maybe I have an error in the code, but I seem to accurately derived the coefficients for a linear regression, and built on it what I drew above. Just in case, print coefficients once more or search in Inet.
Maybe I have an error in the code, but I seem to have accurately derived the coefficients for the linear regression myself, and used it to build what I drew above. Just in case, print the coefficients again, or look them up on the Internet.
No, I made a mistake! When I counted the output of a neuron, I did so: W*D, i.e. simply multiplied vectors... -:)
Now I'm rewriting it.
Does that sound about right?
The slope should be in the other direction I think. upwards is right.
You can also go the other way... I'm just messing with the ISC.
Now it looks like it!
Where is the promised training cloud?
Put a scale grid on the graph and tell me what your tangent of the angle of inclination for the straight line is.
You can go to another one... I'm just messing around with the ISC.
Any straight line can be inscribed in such a circle, at any angle to the horizontal.