# ( ) ∑ ∑ ∑

```Homework 5 Solutions
Ch. 8 2,4,5,10,18,23,24
8.2
∑ x ∑ y − ∑ x ∑ xy = 20 ⋅ 24 − 0 ⋅ 22 = 6
4 ⋅ 20 − 0
N ∑ x − (∑ x )
N ∑ xy − ∑ x ∑ y 4 ⋅ 22 − 0 ⋅ 24
=
= 1 .1
B=
80
N ∑ x − (∑ x )
2
A=
2
2
2
2
2
Therefore, the equation for the line is
y = 1.1x + 6.0
8.4
From equation 8.8,
AN + B ∑ x = ∑ y
Divide by N:
∑ x = ∑ y or
A+ B
A + Bx = y
N
N
Therefore, the point (x, y ) lies on the best-fit line.
8.5
For a line through the origin, A = 0, so χ 2 = ∑
Minimize this with respect to B:
Therefore,
∑x y
i
i
( yi − Bxi )2
σ y2
∂χ 2 − 2
=
∑ xi ( yi − Bxi ) = 0
∂B σ y2
− B ∑ xi2 = 0 or B =
∑ xy
∑x
2
8.10
For the weighted fit, the weights are inversely proportional to the uncertainies squared:
wi = (1 / 0.5 2 ,1 / 0.5 2 ,1 / 12 ) = (4,4,1)
Using these weights, calculate A and B:
∑ wx 2 ∑ wy − ∑ wx ∑ wxy = 29 ⋅ 22 − 15 ⋅ 38 = 1.89
A=
2
9 ⋅ 29 − 15 2
∑ w∑ wx 2 − (∑ wx )
B=
∑ w∑ wxy − ∑ wx ∑ wy = 9 ⋅ 38 − 15 ⋅ 22 = 0.33
36
∑ w∑ wx − (∑ wx )
2
2
Now calculate A and B for unweighted data points:
∑ x 2 ∑ y − ∑ x ∑ xy = 14 ⋅ 7 − 6 ⋅14 = 2.33
A=
2
3 ⋅14 − 6 2
N ∑ x 2 − (∑ x )
B=
N ∑ xy − ∑ x ∑ y
N ∑ x 2 − (∑ x )
2
=
3 ⋅14 − 6 ⋅ 7
= 0.00
6
8.18
We have B =
∑ xy , where ∆ does not depend on y.
∆
we need its derivatives with respect to the y’s:
∂B xi
=
∂y i ∆
∑ (x σ )
=
2
Therefore, σ
2
B
i y
2
∆
∑x
(∑ x )
2
=σ
2
y
2 2
=
σ y2
∑x
2
To calculate the uncertainty in B,
σB =
σy
∑x
2
8.23
We have χ 2 = ∑ ( Af i + Bg i − yi ) / σ y2 , where f i ≡ f (xi ) and g i ≡ g (xi )
2
Minimize this with respect to A and B:
∂χ 2
∂χ 2
∝ ∑ f i (Af i + Bg i − yi ) = 0
∝ ∑ g i ( Af i + Bg i − yi ) = 0
∂B
∂A
Simplify this a bit:
A∑ f 2 + B ∑ fg = ∑ fy
A∑ fg + B ∑ g 2 = ∑ gy
These are the equations given.
8.24
We will use the equations from the problem above. First, calculate the function at the
various values of x and use this to calculate the coefficients in the equations. The results
are:
∑ f 2 = 2.22 ∑ g 2 = 2.78 ∑ fg = 0.00 ∑ fy = 6.48 ∑ gy = 14.63
The equations become
2.22 A = 6.48
2.78 B = 14.63
Therefore,
A = 2.92
B = 5.26
The fit is rather poor – it is likely the frequency is not correct.
```