# What is linear interpolation?

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.

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 $y=2$
• $y=2$ 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 2 '13 at 23:42
• 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 3 '13 at 0:15