1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $y=2$

$\endgroup$
  • $\begingroup$ Is y=2 the answer to my specific problem or is it a generalization in order to find the answer to the problem $\endgroup$ – Kyla Oct 2 '13 at 23:40
  • $\begingroup$ $y=2$ 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. $\endgroup$ – Ross Millikan Oct 2 '13 at 23:42
  • $\begingroup$ 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? $\endgroup$ – Kyla Oct 2 '13 at 23:49
  • $\begingroup$ 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. $\endgroup$ – Ross Millikan Oct 3 '13 at 0:15
  • $\begingroup$ I have graphed the points and the curve to see it $\endgroup$ – Kyla Oct 3 '13 at 0:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.