The Method of Least Squares - Elements of Scientific Computing - page 156

leads to

10˛ C 1 X
i D1

! ˇ D

1 X

t i y i : (5.17)

i D1

Here

1 X

t i D 1 C 2 C 3 CC 10 D 55 (5.18)

i D1

and

1 X

y i D 0:29 C 0:14 C 0:19 CC 0:29 D 3:12; (5.19)

i D1

so we have

10˛ C 55ˇ D 3:12: (5.20)

Next, since

1 X

@F
@ˇ

D 2 .˛ C ˇt i y i /t i ; (5.21)

i D1

the condition

@F
@ˇ

D 0 (5.22)

gives

1 X ! ˛ C 1 X

i D1

! ˇ D

X

10

t i t i y i t i : (5.23)

i D1 i D1

Here,

1 X

t i

D 1 C 2 2 C 3 2 CC 10 2 D 385

(5.24)

i D1

and

X

10

t i y i D 1 0:29 C 2 0:14 C 3 0:19 CC 10 0:29 D 20; (5.25)

i D1