Which Function Gives a Better Fit?

Let's begin by plotting the two fitting curves and the data on the same set of axes.

> p0:=plot(data1,style=point, color=green):
p1:=plot(polyFunc(x),x=1790..1850, color=red):
p2:=plot(expFunc(x),x=1790..1850, color=blue):
with(plots):
display([p0,p1,p2]);

How do we decide which fit is better? We are going to use the sum of the squares of the errors as a measure of the goodness of fit over the time period 1790-1850. That is, suppose we start with data points ( ), ( ), ..., ( ) and we fit the data with the function . Then we will define the total squared error over the period to be the sum of the individual squared errors, namely, . Moreover, one fitting function will be better than another if its total squared error is smaller. Squaring the errors does not allow the signed-errors to cancel one another in the summation; sqaring also has the effect of giving more weight to bigger errors than smaller ones.

Using the sum of squared errors as the criterion, let's display the errors in tabular form and compare them.

> polyerror:=[seq([data1[i][1],abs(polyFunc(1790+10*(i-1))-data1[i][2]),abs(polyFunc(1790+10*(i-1))-data1[i][2])/data1[i][2]*100],i=1..7)]:
toterrorpoly:=add((polyFunc(1790+10*(i-1))-data1[i][2])^2,i=1..7):

with(math3):
headers:=["year","absolute error","% error"]:
printtable(polyerror,"Cubic Polynomial Fit", headers);
TotalSqrError:=toterrorpoly;

                                   Cubic Polynomial Fit

                     year          absolute error            % error

                     -----------------------------------------------

                     1790           46.300000               1.178417

                     1800          107.400000               2.027563

                     1810           26.100000                .361296

                     1820           49.200000                .511541

                     1830           77.700000                .602279

                     1840          146.900000                .858061

                     1850           55.300000                .237737

> experror:=[seq([data1[i][1],abs(expFunc(1790+10*(i-1))-data1[i][2]),abs(expFunc(1790+10*(i-1))-data1[i][2])/data1[i][2]*100],i=1..7)]:
toterrorexp:=add((expFunc(1790+10*(i-1))-data1[i][2])^2,i=1..7):

with(math3):
headers:=["year","absolute error","% error"]:
printtable(experror,"Exponential Fit", headers);
TotalSqrError:=toterrorexp;

                                     Exponential Fit

                   year            absolute error              % error

                   ---------------------------------------------------

                   1790               26.037493                .662700

                   1800               15.360584                .289987

                   1810               88.498676               1.225065

                   1820               33.677210                .350148

                   1830               27.448430                .212762

                   1840              171.605470               1.002368

                   1850               35.115430                .150963

Which fitting function is better? Discuss.