proving

1. 1. Prove 0! = 1 in two ways
a. Algebraic approach
If n! is defined as the product of all positive integers from 1 to n, then:
1! = 1 ⋅ 1 = 1
2! = 1 ⋅ 2 = 2
3! = 1 ⋅ 2 ⋅ 3 = 6
4! = 1 ⋅ 2 ⋅ 3 ⋅ 4 = 24
...
n ! = 1 ⋅ 2 ⋅ 3 ⋅ ... ⋅ ( n − 2 ) ⋅ ( n − 1) ⋅ n
Logically, n ! can also be expressed as n ! = n ⋅ ( n − 1) ! .
Therefore, at n = 1 , using n ! = n ⋅ ( n − 1) !
1! = 1 ⋅ ( 1 − 1) !
1! = 1 ⋅ 0!
Since we have shown above that 1! = 1 , we substitute it. Now we have,
1 = 1 ⋅ 0!
which simplifies to 1 = 0! or 0! = 1
b. Higher function (gamma function)
The gamma function Γ ( n ) is defined by the following integral:
∞
Γ ( n ) = ∫ e− x x n −1dx
0
We derive an expression for Γ ( n + 1) as a function of Γ ( n )
∞
Γ ( n + 1) = ∫ e− x x n +1−1dx
0
∞
= ∫ e− x x n dx
0
We use integration by parts to solve this integral
∞
 − xn 
∞ ∞
∫ e x dx =  x  + n ∫ e − x x n −1dx
−x n
 e 0
0 0
At infinity, Bernoulli’s rule states that
− xn − n!⋅ x 0
= lim =0
lim
x →∞ e x ex
x →∞
∞
 − xn 
So the first term,  x  , evaluates to zero, which leaves
 e 0
∞
Γ ( n + 1) = n ∫ e − x x n −1dx
0
= nΓn = n !
Using this formula derive a pattern

2. Γ ( 3) = Γ ( 2 + 1) = 2Γ ( 2 ) = 2 ⋅ 1! = 2! = 2
Γ ( 4 ) = Γ ( 3 + 1) = 3Γ ( 3) = 3 ⋅ 2! = 3! = 6
Γ ( n + 1) = nΓ ( n ) = n ⋅ ( n − 1) ! = n !
If we use n=1
Γ ( 1 + 1) = Γ ( 2 ) = 1Γ ( 1)
1! = 1 ⋅ ( 1 − 1) !
= 1 ⋅ ( 0!)
0! = 1
2
n 
n
n
n∑ xi 2 −  ∑ xi 
∑( x − x)
2
i
2. Prove the variance  i =1 
= i =1
s2 = i =1
n ( n − 1)
n −1
Proof by definition:
n
∑( x − x)
2
i
s2 = i =1
n −1
n
∑( x − 2 xxi + x 2 )
2
i
= i =1
n −1
n n n
∑x − 2 x ∑ xi + ∑ x 2
2
i
= i =1 i =1 i =1
n −1
n n
∑ xi 2 − 2 x ∑ xi + nx 2
= i =1 i =1
n −1
n
∑x i
By replacing x with we’ll have,
i =1
n
2
  n 
n
 ∑ xi  ∑ xi
 n  
n
∑   ∑ xi  + n  i =1
xi 2 − 2  i =1 
n   i =1  n 
i =1
   
   
=
n −1
2
 n  n  n 
 ∑ xi  ∑ xi  n  ∑ xi 
n
∑ xi 2 − 2  i =1  i =1  +  i =n12 
n
= i =1
n −1

3. 2
 n  n   n 
 ∑ xi  ∑ xi   ∑ xi 
n
∑ xi − 2  i =1  i =1  +  i =1n 
2
n
= i =1
n −1
2
 n  n   n 
n
n∑ xi − 2  ∑ xi  ∑ xi  +  ∑ xi 
2
 i =1  i =1   i =1 
= i =1
n ( n − 1)
2 2
n  n 
n
n∑ xi 2 − 2  ∑ xi  +  ∑ xi 
 i =1   i =1 
= i =1
n ( n − 1)
2
n 
n
n∑ xi −  ∑ xi  2
We now have 2  i =1 
s = i =1
n ( n − 1)
∑  ∑ fd ' 
fx x − x'
3. Prove that x = = x' + i  where d ' = is unit deviation and x ' is assumed
n
n i
 
mean
x − x'
From the given d ' = we can derive the value of x
i
 ' x − x' 
d = i
i

id ' = x − x '
By adding x ' to both sides of the equality we’ll have
x = id ' + x '
We substitute the value of x to the equation x = ∑
fx
n
∑ f ( id + x )
' '
x=
n
i ∑ f d ' + ∑ fx '
=
n
i ∑ f d ∑ fx '
'
x= +
n n
We substitute the definition ∑ = 1 to x = ∑
i f d ' ∑ fx '
f
+
n n n
i∑ f d ∑ f '
'
x= + x
n n

4. i∑ f d '
+ ( 1) x '
=
n
i∑ f d '
= + x'
n
i∑ f d '
We now have x = x ' +
n
4. Derive the equation of linear regression y = a + bx .
∑ y ∑ x − ∑ x∑ xy 2
a=
n∑ x − ( ∑ x )
2
2
∑ xy − ∑ x∑ y
b=
n∑ x − ( ∑ x )
2
2
y = ∑ a + bx
(1)
∑ y = an + b∑ x
x ( y = a + bx )
(2)
xy = ax + bx 2
∑ xy = a∑ x + b∑ x 2
( ∑ xy = a∑ x + b∑ x ) n
( ∑ y = an + b∑ x ) ∑ x 2
2
(3)
( ∑ xy = a∑ x + b∑ x ) ∑ x ( ∑ y = an + b∑ x ) ∑ x
2
∑ x ∑ y = ∑ x ( an ) + ∑ x ( b∑ x ) n ∑ xy = an∑ x + bn∑ x
2 2 2 2
(4)
- ∑ x ∑ xy = ∑ x ( a ∑ x ) + ∑ x ( b∑ x ) ∑ x∑ y = an∑ x + b∑ x∑ x
2
-
∑ x ∑ y − ∑ x∑ xy = an∑ x − a ( ∑ x ) n∑ xy − ∑ x∑ y = bn∑ x − b ( ∑ x )
2 2
2 2 2
n∑ xy − ∑ x∑ y b ( n∑ x − ( ∑ x ) )
∑ x ∑ y − ∑ x∑ xy = a ( n∑ x − a ∑ x∑ x )
2
2
2
2
=
(5)
n∑ x − a ( ∑ x ) n∑ x − a ( ∑ x )
2 2
n∑ x − ( ∑ x ) n∑ x − ( ∑ x )
2 2
2 2 2 2
n∑ xy − ∑ x ∑ y
∑ x ∑ y − ∑ x∑ xy
2
b=
a=
n∑ x 2 − ( ∑ x )
n∑ x − a ( ∑ x )
2
2
2