Page 36 - Foundations Of Differential Calculus

1. On Finite Differences 19
In a similar way we obtain
6 1 5 1
x
3 4
4
5
5
Σx = − x + ωx − ω x
6ω 2 12 12
and
6 x 7 1 6 1 5 1 3 3 1 5
Σx = − x + ωx − ω x − ω x.
7ω 2 2 6 42
Later we will show an easier method to obtain these expressions.
28. If the given difference is for a polynomial function of x, then its sum
(or the function of which it is the difference) can easily be found with these
formulas. Since the difference is made up of different powers of x,wefind
the sum of each term and then collect all of these terms.
2
Example 1. Find the function whose difference is ax + bx + c.
We find the sum of each term by means of the formulas found above:
3 2
2 ax ax aωx
Σax = − + ,
3ω 2 6
bx 2 bx
Σbx = − ,
2ω 2
cx
Σc = .
ω
When we collect these sums we obtain
3 2
2 ax aω − b 2 aω − 3bω +6c
Σ ax + bx + c = − x + x + C,
3ω 2ω 6ω
2
which is the desired function, whose difference is ax + bx + c.
4 2 2 4
Example 2. Find the function whose difference is x − 2ω x + ω .
Following the same method we obtain

4 1 5 1 4 ωx 3 ω 3
Σx = x − x + − x,
5ω 2 3 30
3
2 2 2ω 3 2 2 ω
−Σ2ω x = − x + ω x − x,
3 3
4 3
+Σω = ω x,
so that the desired function is
1 5 1 4 1 3 2 2 19 3
x − x − ωx + ω x + ω x + C.
5ω 2 3 30
As a check, if instead of x we put x+ω and from this expression we subtract
4 2 2 4
the one we have found, the given difference x −2ω x +ω is what remains.