Interpolation and Extrapolation

Date: 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/