VedekPako Posted June 10, 2009 Share Posted June 10, 2009 May algebra isn't that good, but I have a problem I would like to solve. If Mars' tectonics lasted only 800 million years and Venus' plate tectonics lasted for about three billion years and Earth's plate tectonics should shut down in about 2 billion years from now, then, I should be able to estimate when plate tectonics ceases on worlds, given their mass to Earth. 0.3 = 0.8 0.4 = A 0.5 = B 0.6 = C 0.7 = D 0.8 = E 0.9 = 3 1.0 = 6 1.1 = F 1.2 = G 1.3 = H Etc. How can I find the values of A though H and beyond? If I know that 0.3 is 0.8 and 1.0 is 6, then I should be able to solve this, but I can't figure it out. Link to comment Share on other sites More sharing options...
the tree Posted June 10, 2009 Share Posted June 10, 2009 No, you shouldn't be able to solve it. A couple of data points is not enough information to know anything about a function - or even if a well behaved function is relevant. Link to comment Share on other sites More sharing options...
Bignose Posted June 11, 2009 Share Posted June 11, 2009 0.3 = 0.8 0.4 = A 0.5 = B 0.6 = C 0.7 = D 0.8 = E 0.9 = 3 1.0 = 6 1.1 = F 1.2 = G 1.3 = H Be careful with your notation here, because writing 1.0 = 6 is essentially meaningless. Because it is simply a false statement. You meant something like f(1.0)=6. And you are trying to discover what f(x) is given f(1.0)=6, f(0.9)=3, f(0.3)=0.8. Then, as tree said, being limited to only three points, extrapolation or interpolation to other points is a very iffy proposition. I.e. you can find the best-fit line to those three points, but if the phenomena is not linear, then using the line is going to give you erroneous results. These things are usually best worked on from attempting to determine the pathology of the function in the first place and then doing the fitting. It usually doesn't work the other way around. It really won't work very well with only three points. Just as an example, there is a second order polynomial that will go through all three of those points. But, again, that certainly doesn't mean that the phenomena is truly a second order polynomial. And, a second order polynomial will give you some farcically bad results -- it will turn back at some point and that turning point is not a physically possible point. It fits your data, but doesn't mean anything. Link to comment Share on other sites More sharing options...
the tree Posted June 11, 2009 Share Posted June 11, 2009 That was a much better explanation. Also, probably worth noting, this is has nothing to do with algebra. Link to comment Share on other sites More sharing options...
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