Solve an Algebraic Equation: A+B=C

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The discussion focuses on finding a non-time-series algebraic equation for the relationship between variables A, B, and C, where A + B always equals 10. The user provides specific data points for A and B, along with corresponding values for C, and seeks an equation that accurately represents these relationships. Suggestions for potential equations include variations involving the differences and products of A and B. A proposed solution involves a symmetric function around x=5, leading to the equation f(x) = 0.08x^4, which is tested against the given values. The goal is to derive a polynomial equation that fits the provided data points without incorporating time as a variable.
seasnake
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Can anyone figure this out... searching for an algebraic equation (not a student)

I'm trying to unravel some equations in a spreadsheet and as such I need to find a non-time-series, algebraic equation for the following data expressed for finding C:

A + B always equals 10 (i.e. A1 + B1 = 10; A2 + B2 = 10; ...)

when:

A0 = 0... B0 = 10 ... C0 = 10

A1 = 1 ... B1 = 9 ... C1 = 4

A2 = 2 ... B2 = 8 ... C2 = 2

A3 = 3 ... B3 = 7 ... C3 = 0.83 repeating 3

A4 = 4 ... B4 = 6 ... C4 = 0.2

A5 = 5 ... B5 = 5 ... C5 = 0

A6 = 6 ... B6 = 4 ... C6 = 0.2

A7 = 7 ... B7 = 3 ... C7 = 0.83 repeating 3

A8 = 8 ... B8 = 2 ... C8 = 2

A9 = 9 ... B9 = 1 ... C9 = 4

A10 = 10 ... B10 = 0 ... C10 = 10

an eqational attempts would be something along the lines of:

(A - B) / (A + B) = C

(A - B) / (A times B) = C

(A - B)^2 / (A + B) = C

Note: C3 = C7 = 0.83 = 5 / 6 (serves as a decent clue as it displays both a numerator and a denominator)

once again, I am not seeking a time series equation...
 
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What's a 'not time series equation'? Do you want an exact polynomial through those points or what?
 


seasnake said:
I'm trying to unravel some equations in a spreadsheet and as such I need to find a non-time-series, algebraic equation for the following data expressed for finding C:

A + B always equals 10 (i.e. A1 + B1 = 10; A2 + B2 = 10; ...)

when:

A0 = 0... B0 = 10 ... C0 = 10

A1 = 1 ... B1 = 9 ... C1 = 4

A2 = 2 ... B2 = 8 ... C2 = 2

A3 = 3 ... B3 = 7 ... C3 = 0.83 repeating 3

A4 = 4 ... B4 = 6 ... C4 = 0.2

A5 = 5 ... B5 = 5 ... C5 = 0

A6 = 6 ... B6 = 4 ... C6 = 0.2

A7 = 7 ... B7 = 3 ... C7 = 0.83 repeating 3

A8 = 8 ... B8 = 2 ... C8 = 2

A9 = 9 ... B9 = 1 ... C9 = 4

A10 = 10 ... B10 = 0 ... C10 = 10

an eqational attempts would be something along the lines of:

(A - B) / (A + B) = C

(A - B) / (A times B) = C

(A - B)^2 / (A + B) = C

Note: C3 = C7 = 0.83 = 5 / 6 (serves as a decent clue as it displays both a numerator and a denominator)

once again, I am not seeking a time series equation...
You are seeking an equation. "Time" is not a mathematical concept so "time series" is not relevant here.

Since x+ y= 10, y is fixed once you know x so you can just look for a function of x giving those values. I notice this is symmertric about x= 5 so I would look for f(x)= a(x-5)2n Taking f(0)= a(5)2n= 10 and f(1)= a(4)2n= 4 we can divide the first equation by the second and get
\frac{5^{2n}}{4^{2n}}= \left(\frac{5}{4}\right)^{2n}= \frac{10}{4}= \frac{5}{2}[/itex]<br /> Taking log of both sides, 2n log(5/4)= log(5/2) or 2n(0.096910)=0.39794 so 2n= 4.106. Let&#039;s just call that &quot;4&quot;. Then, from f(0)= a(5)<sup>2n</sup>= a(5<sup>4</sup>)= 125a= 10, a= 10/125= 0.08.<br /> <br /> See if f(x,y)= 0.08 x<sup>4</sup> fits the other values.
 
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