Question: Least Squares Method

Now, there are two unknows, and i cant seem to get it right. 

Its "the method of the least squares", and it has gone wrong. How terrible, now ive made many books, but this book always sets me up, you may have guessed it: the Dutch Math book strikes again... 

Ive tried to use a value, it looks like i used the rights steps, but the wrong input, the solution does not give me anything that looks like it is doing the rigth thing. Checking the new formula does not give the solution it is suppose to give. Now i know they mentioned that x=1/I but that does not give the right answer either. 

Okey the question states: "A battery delivers a current I to a circuit with an external(not inside the battery) resistance Ry. The battery has an internal resistance Ri and delivers a source tension (voltage) U. These are the formulas U=(Ry+Ri)*I, of which follows: Ry+Ri=U/I. We state that x= 1/I and then we get Ry+Ri=U*x, of which follows: Ry=U*x-Ri. 

That last formula gives a linear conjunction between x and Ry. There are a couple of measurements done, of which different values of the resistance Ry and the current I is measured (look at table 5.4). The value of x is determined by the smallest squares method." 

While i did not use x instead of I, it made no difference. I was no where close to the results. I think i am misschien something.

`<,>`(`<|>`(1, 1.5, 2, 2.5, 3), `<|>`(1.31, 1.01, .81, .67, .56))

Matrix(%id = 18446745647253576510)



proc (t) options operator, arrow, function_assign; (1/2)*t end proc


1, 3/2, 2, 5/2, 3


Sum((1/2)*t, t = 2 .. 6) = 10





convert(`<,>`(1.31, 1.01, .81, .67, .56), `+`)






`<,>`(1, 1.5, 2, 2.5, 3).`<,>`(1.31, 1.01, .81, .67, .56)




proc (t) options operator, arrow, function_assign; (1/4)*t^2 end proc


1, 9/4, 4, 25/4, 9


Sum((1/4)*t^2, t = 2 .. 6) = 45/2


solve([(45/2)*a+10*b = 7.80000000000000071, 2*a+b = 4.36], [a, b])

[[a = -14.32000000, b = 33.00000000]]


"f(t):=-14.32 t+33"

proc (t) options operator, arrow, function_assign; -14.32*t+33 end proc


plot(proc (t) options operator, arrow, function_assign; -14.32*t+33 end proc)






Maybe someone knows an answer to this. 

Thank you!


The Function 


Please Wait...