please help me with this mathematical problem

2015 10 8
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更新时间：2015 10 8
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i have two values x and y where x=0 10 20 30 40 50 60 and y=78.54 76.19 73.65 70.85 67.72 64.26 60.19 respectively. from the above values I have to find the next values like 70 80 90 100 110... and there respective y values. please help me out.

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回答（1 个）

Walter Roberson
Walter Roberson 2015-10-8

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The next number in the series is always 19.
Given any finite list of x and y values, there are an infinite number of functions that those y values exactly at each of the x values. At any x value not given in the list, there are then an infinite number of possibilities for y, and there is no "correct" way of determining what the corresponding y "really" is.
Therefore one might as well pick any arbitrary number as being the "real" next number in the sequence.
Consider that, after all,
polyfit([0 10 20 30 40 50 60 70 80 90], [78.54 76.19 73.65 70.85 67.72 64.26 60.19 19 19 19], 9)
should be able to come up with a perfect fit, to within the limits of floating point accuracy. Any sequence extended by any arbitrary value can still mathematically be fit by some function.
polyfit([0 10 20 30 40 50 60 70 80 90], [78.54 76.19 73.65 70.85 67.72 64.26 60.19 42 42 42], 9)
should also be able to come up with a perfect fit to within floating point accuracy
So generate any value you like and you will be right.

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blessy joy
blessy joy 2015-10-8
how is the next term is 19
Thorsten
Thorsten 2015-10-8
19 was just taken as example by Walter to demonstrate that any values are "correct" in the sense that there is some function that would generate these values.
So to solve your question in a meaningful way, you have to make additional assumptions on the function that underlies your data, e.g., it has to be a linear function. Without these assumptions your question can be answered by any y values, as Walter points out.
John D'Errico
John D'Errico 2015-10-8
Oh, come on! We all know that 42 is the right answer! 19 is just silly.
In all seriousness, you need to be careful when you use polynomial extrapolation, or ANY class of extrapolant.
A quick check shows that a 4th order polynomial fit seems realistic. Not overkill, and it extrapolate out as far as 100 in a reasonable way. Thus...
p = polyfit(x,y,4);
I'll let you figure out how to use the model. (hint: help polyfit)
Image Analyst
Image Analyst 2015-10-8
Maybe you could make an argument for 1, on account of Benford's law
https://en.wikipedia.org/wiki/Benford's_law
Bonus points if you come up with a MATLAB routine to mine random web sites to empirically come up with Benford's Law distribution.
The submissions in the File Exchange don't do that. <http://www.mathworks.com/matlabcentral/fileexchange/?utf8=%E2%9C%93&term=benford>

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