please help me with this mathematical problem

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blessy joy
blessy joy 2015 年 10 月 8 日
コメント済み: Image Analyst 2015 年 10 月 8 日
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.

回答 (1 件)

Walter Roberson
Walter Roberson 2015 年 10 月 8 日
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.
  5 件のコメント
Walter Roberson
Walter Roberson 2015 年 10 月 8 日

19 has been mathematically established as being correct. See http://www.whydomath.org/Reading_Room_Material/ian_stewart/9505.html

Image Analyst
Image Analyst 2015 年 10 月 8 日
Maybe you could make an argument for 1, on account of 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.

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