Page 24 - Foundations Of Differential Calculus

1. On Finite Differences 7
4
Example 3. Find the differences of all orders of x .
4
4
I
Let y = x . Since y =(x + ω) , we have
3 2 2 3 4
∆y =4ωx +6ω x +4ω x + ω ,
which is the first difference. Then, from what we have already found,
3 2 2 3 4
∆4ωx =12ω x +12ω x +4ω ,
2 2 3 4
∆6ω x =12ω x +6ω ,
3 4
∆4ω x =4ω ,
4
∆ω =0.
When these are combined, we have the second difference
2 2 3 4
∆∆y =12ω x +24ω x +14ω .
Furthermore, since
2 2
3
4
∆12ω x =24ω x +12ω ,
3 4
∆24ω x =24ω ,
4
∆14ω =0,
we obtain the third difference
3 3 4
∆ y =24ω x +36ω .
Finally, we have the fourth difference
4 4
∆ y =24ω ,
and since this is constant, all differences of higher order vanish.
n
Example 4. Find the differences of all orders of x .
II
I
n
n
n
n
Let y = x . Since y =(x + ω) ,y =(x +2ω) ,y III =(x +3ω) ,... ,
the expanded powers are as follows:
n
y = x ,
I n n n−1 n (n − 1) 2 n−2 n (n − 1) (n − 2) 3 n−3
y = x + ωx + ω x + ω x
1 1 · 2 1 · 2 · 3
+ ··· ,
II n n n−1 n (n − 1) 2 n−2 n (n − 1) (n − 2) 3 n−3
y = x + 2ωx + 4ω x + 8ω x
1 1 · 2 1 · 2 · 3
+ ··· ,
III n n n−1 n (n − 1) 2 n−2 n (n − 1) (n − 2) 3 n−3
y = x + 3ωx + 9ω x + 27ω x
1 1 · 2 1 · 2 · 3
+ ··· ,
IV n n n−1 n (n − 1) 2 n−2 n (n − 1) (n − 2) 3 n−3
y = x + 4ωx + 16ω x + 64ω x
1 1 · 2 1 · 2 · 3
+ ··· .