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I am learning about linear interpolation however, we were not taught how to formally solve a problem using linear interpolation. A practice problem involving is the following:

Find how long it will take for a certain sum of money to double itself at compound annual interest of 20%. Solve this problem. Then solve this problem by first finding $(1.2)^3$ and $(1.2)^4$ and then by linear interpolation, approximating $x$ such that $(1.2)^x=2$. Show the result so obtained is equal to the Mesopotamian solution expressed sexagesimally as 3;47,13,20.

Stefan4024
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Kyla
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1 Answers1

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In linear interpolation, you have two points on the curve and draw a straight line through them. You use that line to approximate the value at other points of the curve. Your two points are $(3,y_1)$ and $(4,y_2)$, where you are supposed to calculate $y_1,y_2$. Then you use the two-point form of the line $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$ to find the $x$ where $y=2$

Ross Millikan
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  • Is y=2 the answer to my specific problem or is it a generalization in order to find the answer to the problem – Kyla Oct 02 '13 at 23:40
  • $y=2$ is what you are looking for because that represents doubling of the original principal. Your $y$ values are the ratio of the current value to the original. – Ross Millikan Oct 02 '13 at 23:42
  • Okay, I see now where the 3 and 4 come into play with the x-values. Now why does it want to ask to find (1.2)^3 and (1.2)^4 first though, what are we finding with those numbers? – Kyla Oct 02 '13 at 23:49
  • You are finding two points on the curve that bracket the point of interest. You won't have $y_1,y_2$ without that. The curve is an exponential. I suggest you graph it over $[1,1.5]$, plot these points, draw the straight line through them, then graph it over $[1.2,1.3]$, plot the points and the straight line. It will help you see what is happening. The straight line is an approximation to the curve. – Ross Millikan Oct 03 '13 at 00:15
  • I have graphed the points and the curve to see it – Kyla Oct 03 '13 at 00:30
  • The point is that the straight line is what your interpolation will follow. It is pretty close, but not exact. Suppose you have a function that is very expensive to calculate, but you need values at lots of points. You might choose to calculate it at a number of points, tabulate them, and use linear interpolation between. Of course, having more points makes the interpolation more accurate, but it takes a bigger table. That is the point of the exercise. – Ross Millikan Oct 03 '13 at 03:21