Recent content by jjj888

What if the I have a large number of data points but they are all in a rough shape of a parabola, will I have the same problem with contitioning / dependency?

I know this is going to get a little beyond my grasp, but what causes the matrix to be ill-conditioned? I understand that in the end you are still trying to fit something to something else and there is no exact definition. What are these more direct methods for finding coefficients for a...

Thanks. I did discover my x sums were wrong. Although I noticed the the matrix values and the calculator values were slightly different going down the decimal line. Am I to assume the matrix would be more accurate because the calculator uses some other algo? Or did your values match exactly?

| n Ʃx Ʃx2| |a| |y| | Ʃx Ʃx2 Ʃx3| |b| |Ʃxy| | Ʃx2 Ʃx3 Ʃx4| |c| |x2y| This was the base matrix I used. From the partial derivs of the sqrared errors and such.

I wrote the matrix down wrong. Here is what I had. | 3 7 39| |a| = -4 | 7 39 233| |b| = -28 |39 233 1311| |c| = -156 a = 0.3436 b = 0.1237 c = -0.1512

I'm trying to understand the derivation of polynomial regression. Given data points: [(-1,-1),(2,-1),(6,-2)]. So a 2nd degree curve will be a concave downward parabola. My calculator produces the equation: -0.0357x2+0.0357x-0.9285. Which fits the data good. But if I try to do it manually...

So, if dx = (dx/dt)dt then, x=1/A∫txf(x)(dx/dt)dt y=1/A∫t(1/2)[f(x)2](dx/dt)dt xc = ∫t[(0.9t3+5.1t2)(-1.8t3-3.9t2+13.2t)(2.7t2+10.2t)]dt/∫t[(-1.8t3-3.9t2+13.2t)(2.7t2+10.2t)]dt = (125.59/37.26) = 3.37 yc =...

Thanks, but before I try to decifer this I'm still having notation issues, what do these symbols mean: ∫A, ∫x, ∫y, ∫t. Are they definites of that variable, or is there something else. Is there a different way to write it? Thanks

I'm having trouble understanding how xc = ∫Ax.dx.dy/∫Adx.dy is equal to x=1/A∫abxf(x)dx? Anyway here's my x derivation using x=1/A∫abxf(x)dx: x = 1/4.85∫01((0.9t3+5.1t2)(-1.8t3-3.9t2+13.2t)) = 2.656 I set it up the same way as the previous problem. I would think this is a fairly...

I did, but i wasn't too familiar with the notation. I'll take a look again.

Ok so I'm adding on to this a bit. I have: x=0.9t3+5.1t2 y=-1.8t3-3.9t2+13.2t using : x=1/A∫abxf(x)dx y=1/A∫ab(1/2)[f(x)]2dx for centroid I get: x=2.656 y=2.957 in cad Iget: x=3.371 y=3.2872 which makes more sense similarly I have: x=-0.35t3-0.9t2+8.25t+6...

Thanks, I acctually did find the answer. x = 3t y = 5t^3-9t^2+6t I got the answer using the formula below. This is the first formula I encountered when I started. x=1/A∫abxf(x)dx y=1/A∫ab(1/2)[f(x)]2dx My error was with the area. I used the area under the curve of the...

you're going to have to break that down for me a bit.

But, As I said above, when I plot the curve out in CAD the centroid cordinates are: x = 1.8 y = 0.7086 But I can't seem to get this answer. I'm fairly certain this answer is correct, it makes sense when you look at it. Very confusing at this point.

If the density is constant you would use 1. But I read this somewhere else, and they said to just integrate my original functions normally, which also doesn't give me: x=1.5 y=-1.25