>	data:=[[1,2], [2,2], [5,6], [7,9], [8,10]];

 

(1)

 

>

 

>	plot(data,style=point);

 

How far away is y=x+1.2 from this data?   need some idea of distance(line, data) 

bad idea: sum of differences in y values 

>	f:=x->x+1.2;

 

(2)

 

>

f(3);

 

(3)

 

>

g:=x+1.2;

 

(4)

 

>

g(3);

 

(5)

 

can reference elements in a list using [] 

>

data;

 

(6)

 

>

data[3];

 

(7)

 

>	data[3][1];

 

(8)

 

>	data[3][2];

 

(9)

 

>	f(data[3][1]);

 

(10)

 

>	f(data[3][1])-data[3][2];

 

(11)

 

>	seq( f(data[i][1])-data[i][2], i=1..5);

 

(12)

 

>	sum( f(data[i][1])-data[i][2], i=1..5);

 

(13)

 

>	h:=x->0.9*x+1.2;

 

(14)

 

>	sum( h(data[i][1])-data[i][2], i=1..5); seq( h(data[i][1])-data[i][2], i=1..5);

 

 

	(15)

 

>

 

Bad, gives 0 when fit isn't perfect. 

next try, consider sum(absvalue) 

>	sum( abs(h(data[i][1])-data[i][2]), i=1..5);

 

(16)

 

>	sum( abs(f(data[i][1])-data[i][2]), i=1..5);

 

(17)

 

>

 

want to write a function that gives me abs difference between  f and a point.  This is  |f(p_x) - p_y| 

>	err := (f, p) -> abs( f(p[1]) - p[2]);

 

(18)

 

>	err( h, [5,6]);

 

(19)

 

>	err( h, data[2]);

 

(20)

 

Note that garbage in gives a garbage answer. 

>

err(3,7);

 

(21)

 

now distance is the sum of the errors 

>	nops(data);

 

(22)

 

>	nops( {1, 2,3,4,5,6,7});

 

(23)

 

>

nops(f);

 

(24)

 

>	sum( abs(h(data[i][1])-data[i][2]), i=1..nops(data));

 

(25)

 

>	sum( err(h,data[i]), i=1..nops(data));

 

(26)

 

>	dist:= h-> sum( err(h,data[i]), i=1..nops(data));

 

(27)

 

>

dist(h);

 

(28)

 

>

dist(f);

 

(29)

 

>	dist(exp);evalf(%);

 

 

	(30)

 

>	dist(x->3*x+2);

 

(31)

 

>

 

>

 

Try distance for various lines of slope 1, varying intercept. 

You should think about how to do this, and we'll go over it next week.