Interpolation and ExtrapolationDate: 08/10/99 at 21:40:18 From: Ramon Santiago Subject: Interpolation I can't remember the proper way of solving interpolation problems. It's been a long time since I finished my degree, and very few times that I've needed to interpolate something. I want to find the step-by-step procedure for finding the missing values of X and Y based on existing given values as shown below: 26 -> 1800 28 -> X 32 -> 2400 44 -> Y I hope somebody there in your group can help me. Thanks Ramon Date: 08/11/99 at 03:04:47 From: Doctor Floor Subject: Re: Interpolation Dear Ramon, Thanks for writing to Ask Dr. Math. Let me first explain a little bit about the term "interpolation." "Interpolation" means that you have to calculate a value between ("inter" is Latin for between) two given values. There are several types of interpolation, but in this case I suppose you want to use "linear interpolation," which is based on the fact that the values are increasing or decreasing linearly. "Extrapolation" is the term you should use when you have to calculate a value before or beyond the given values ("extra" is Latin for outside). That's what you should do to find Y. Let's get to your question: 26 -> 1800 28 -> X 32 -> 2400 44 -> Y We see that if the left value increases from 26 to 32, which is +6, then the right value increases from 1800 to 2400, which is +600. If we assume linear behavior, then we must conclude that if the left value increases by 1, the right increases by 100. And we find: Interpolating: When the left value increases from 26 to 28, which is +2, the right value increases by 200. So X = 1800+200 = 2000. Extrapolating: When the left value increases from 32 to 44, which is +12, the right value increases by 1200. So Y = 2400+1200 = 3600. If you need more help, just write us back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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