hey math whizzes, what's this thing called...

[lighthousekeeper]

What's this thing called:

X = aA + bB + cC + dD + eE

where if you have enough examples of values of sets of A, B, C, D, E, and X, you can solve for a, b, c, d, and e.

So you might have:
1.2 = a(.2) + b(.1) + c(.2) + d(.4) + e(.7)
2.4 = a(.3) + b(-.1) + c(.2) + d(.1) + e(.3)
1.8 = a(.4) + b(.1) + c(.2) + d(.3) + e(.4)
1.5 = a(.6) + b(.3) + c(.3) + d(.8) + e(.1)
2.2 = a(.8) + b(.1) + c(.2) + d(.6) + e(.2)
2.6 = a(.9) + b(.6) + c(.1) + d(.4) + e(.1)
(don't go trying to figure out this one above - i just pulled random numbers)

Is there a term for this type of equation? I was thinking polynomial equation, but I think that's a little different.

[cuervo72]

linear?

[st.cronin]

hard?

[Shkspr]

A system of linear equations. I believe using a matrix would be the preferred approach to a solution.

[lighthousekeeper]

actually what I'm really trying to answer is if I've got a set of 1000 of these equations:

1.2 = a(.2) + b(.1) + c(.2) + d(.4) + e(.7)
2.4 = a(.3) + b(-.1) + c(.2) + d(.1) + e(.3)
1.8 = a(.4) + b(.1) + c(.2) + d(.3) + e(.4)
...
1.5 = a(.6) + b(.3) + c(.3) + d(.8) + e(.1)
2.2 = a(.8) + b(.1) + c(.2) + d(.6) + e(.2)
2.6 = a(.9) + b(.6) + c(.1) + d(.4) + e(.1)

Where the actual values of a, b, c, d, and e aren't always the same value, but rather distributed along a normal distribution, how can I find out the most probable value for a, b, c, d, and e?

[ISiddiqui]

I believe using a matrix would be the preferred approach to a solution.

http://imagecache2.allposters.com/images/15/MATRIX.JPG

[Shkspr]

Oh, crap, are those Markov chains? I quit.

[lighthousekeeper]

Oh, crap, are those Markov chains? I quit.

well you gets points for even knowing what a Markov chain is.

[Mr. Wednesday]

actually what I'm really trying to answer is if I've got a set of 1000 of these equations:

1.2 = a(.2) + b(.1) + c(.2) + d(.4) + e(.7)
2.4 = a(.3) + b(-.1) + c(.2) + d(.1) + e(.3)
1.8 = a(.4) + b(.1) + c(.2) + d(.3) + e(.4)
...
1.5 = a(.6) + b(.3) + c(.3) + d(.8) + e(.1)
2.2 = a(.8) + b(.1) + c(.2) + d(.6) + e(.2)
2.6 = a(.9) + b(.6) + c(.1) + d(.4) + e(.1)

Where the actual values of a, b, c, d, and e aren't always the same value, but rather distributed along a normal distribution, how can I find out the most probable value for a, b, c, d, and e?

If you try to find the a, b, c, d, and e values to minimize the error in an overdetermined system (many more equations than there are unknowns), that's linear regression (for this problem setup).

[JonInMiddleGA]

http://www.sportsecyclopedia.com/nhl/montreal/Markovhabs.jpg

+

http://www.krmfg.com/images/original%20images/chain.jpg

[lighthousekeeper]

If you try to find the a, b, c, d, and e values to minimize the error in an overdetermined system (many more equations than there are unknowns), that's linear regression (for this problem setup).

ah that's it - thanks!

[lighthousekeeper]

fyi, here's a great site for doing linear regression calculations - did exactly what I was looking for:

http://www.wessa.net/esteq.wasp

[RPI-Fan]

It's a linear regression. VERY easy to do in Excel. Let me know if you need help.