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 Maths question I need answered pretty much now 
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Post Maths question I need answered pretty much now • Posted: Tue Apr 21, 2009 4:46 pm
Alright, I've been given a table of values, time, and the percentage corresponding to the time. (The percentage being y(t), our function).

I had to rearrange y = 100/(1 + e^(c1 +c2t)), to isolate c1 + c2t, which I believe I did, rearranging it to c1 + c2t = ln(100/y + 1)

Now, I have to find a 10x2 matrix, A, and column vector (read: 10x1 matrix), b, such that A multiplied by (c1 c2)' = b, where the elements of b are functions of y. :S (NOTE: (c1 c2)', for those who don't know, means turn the thing vertically. It's just easy horizontal notation of vectors. So, it's a 2x1 matrix, with c1 being above and c2 below.)

Trouble is, I haven't a clue how to do that. Help?

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Post Re: Maths question I need answered pretty much now • Posted: Tue Apr 21, 2009 4:50 pm
Well first, isn't it suppose to be c1 + c2t = ln(100/y - 1) ?

Other than that, I don't think I can help that much. Almost failed this course when I took it...

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Post Re: Maths question I need answered pretty much now • Posted: Tue Apr 21, 2009 5:03 pm
Smack. I can't remember, sorry Yaz. -_- I do recognize something here, though. hmmm.

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Post Re: Maths question I need answered pretty much now • Posted: Tue Apr 21, 2009 5:50 pm
Table:

t y
0 1
1 5
2 10
3 50
4 61
5 75
6 85
7 95
8 99
9 99.5

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Post Re: Maths question I need answered pretty much now • Posted: Tue Apr 21, 2009 6:06 pm
So, c is column, correct? I have to ask, because you're like in Australia, and we have a tendency to use different terms. -_-

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Post Re: Maths question I need answered pretty much now • Posted: Tue Apr 21, 2009 6:09 pm
...c? Oh, you mean c1 and c2! No, those are just letters. Like x and y. The point of this assignment is to figure out what those letters equal. To do that, I have to come up with a 10x2 matrix, etc.

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Post Re: Maths question I need answered pretty much now • Posted: Tue Apr 21, 2009 6:12 pm
Hmmm. I thought that you could put the columns as there own matrices (X) is one matrix, (Y) is another, and put those matrices into the equation...Time to google it. :P

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Post Re: Maths question I need answered pretty much now • Posted: Tue Apr 21, 2009 6:17 pm
Wait, so are you having problems setting up the matrix or how to solve for b?

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Post Re: Maths question I need answered pretty much now • Posted: Tue Apr 21, 2009 6:20 pm
...Both? I just need the matrix. We're solving for (c1 c2)'.

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Post Re: Maths question I need answered pretty much now • Posted: Tue Apr 21, 2009 6:25 pm
Well assuming you plug in the numbers, wouldn't your matrix be:

[A]

1 0
1 1
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9

With [b]

4.615120517
3.044522438
2.397895273
1.098612289
0.970530501
0.84729786
0.777704569
0.719122667
0.698184975
0.695656592

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Post Re: Maths question I need answered pretty much now • Posted: Tue Apr 21, 2009 6:28 pm
That's what I was thinking, but how do you get the other matrix? Multiply the new one by the old one? (A.k.a. Multiply Mat. A x Mat. B=Mat C.) I thought you needed a new matrix too (as in, a new 10x2, matrix) :P

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Post Re: Maths question I need answered pretty much now • Posted: Tue Apr 21, 2009 6:31 pm
Oh, he needs to use the matlab function to do that. I can't remember what it is, something like [A]*[b] or the inverse of something. But I think those are the two matrices anyway.

I hated this math so much btw.

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Post Re: Maths question I need answered pretty much now • Posted: Tue Apr 21, 2009 10:55 pm
Yeah, that was A, and b was comprised of the answers to that "ln(100/y - 1)" expression. What's more, I was getting hung up on a phrase which, if I had've followed my gut instinct, would've gotten it.

See, what I was getting caught up on was "functions of y". Not y(t), but f(y)...i.e. the equation they had me find (c1 + c2t). Also, every instinct told me to make the second column of A the times in the table (i.e. 0, 1, 2, etc.). Thus, if I had realised that b were the values of f(y), I would easily have realised the first column of A was 1s.

Now, the problem is, finding the reduced row echelon form of [A|b] would be pointless, as I had far too many values. What I needed was to turn A into a nice square matrix. My friend, somehow, realised that multiplying A by its transpose (A') would give me a nice 2x2 matrix (A'*A is 2x10 * 10x2). This gave me some matrix...c. Then I multiplied b by A', and it gave me a 2x1 matrix, d (A'b = d). Now, I had cx = d. Finding the reduced row echelon of [c|d] then proceeded to give me my values for c1 and c2.

So simple, in hindsight. Don't you agree?

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