MHB What to do when "second differences" are different?

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The discussion revolves around analyzing a set of data points to determine the type of mathematical model they represent. The user calculated first and second differences but found inconsistency in the second differences. Participants clarified that if first differences are constant, a linear model is indicated, while constant second differences suggest a quadratic model. The conversation also touched on the possibility of an exponential model if the ratios of consecutive terms are constant. A question was raised about the impact of changing the first data point from -16 to +16 on the ratios.
MRF2
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Hey, I'm using data points:

X: -1; 0; 1; 2; 3
Y:-16; 4; 1; 1/4; 1/16

I solved for the first differences, and got:
-12; -3; -1/4; -3/16

I then solved for second differences, and got:
9; 11/4; 1/16

Is my math just wrong in a way I can't see, or...?
Thanks!
 
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So, if the first-level differences are the same, you've got a linear model. If the second-level differences are the same, you've got a quadratic model. What do you have if the ratios are constant?
 
Ackbach said:
So, if the first-level differences are the same, you've got a linear model. If the second-level differences are the same, you've got a quadratic model. What do you have if the ratios are constant?

Exponential model?
 
MRF2 said:
Exponential model?

I didn't find the ratios to be the same.
 
Yes, it would be exponential if the ratios are the same. Is the first data point -16 or 16?

Hmm:
4/16 = 1/4
1/4 = 1/4
(1/4)/1 = 1/4
(1/16)/(1/4) = (1/16) * (4/1) = 1/4

So, if the first data point is +16 instead of -16, would the ratios be the same?
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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