Previously
we mentioned the results for the sum of the natural numbers, where n is
the number in the series and Sn is the sum of these n:
n
1
2
3
4
5
S_{n}
1
3
6
10
15
Δ^{1}
2
3
4
5
Δ^{2}
1
1
1
We noted the series is based on a quadratic because the second
difference, Δ^{2}, is a constant. So S_{n}
is of the form a_{0}n^{2}+a_{1}n+a_{2}.
We note that the constant, a2, is 0 because the sum of 0 terms is 0, S_{0}=0.
Because Δ^{n}=a_{0}n!,
when the steps are 1, we know:
Δ^{2}=a_{0}2!, so:
a_{0}2!=1, and a_{0}=1/2!=1/2
As the polynomial is a quadratic, and we would have needed 3
simultaneous equations to solve for the 3 unknowns, but we have already
identified 2 of them, we can find the remaining coefficent, a_{1},
from an equation with one unknown, for example:
S_{1}=1=(1/2)n^{2}+a_{1}n
As n is 1, we have 1=(1/2)+a_{1
}And so the equation is:
(a_{1}n^{2}+a_{2}n)/2,
or
n/2(n+1)
Generalizing
Polynomials Of Degree 2 With Algebra
Consider the equation:
f(x)=a_{0}x^{2}+a_{1}x+a_{2
}[1.01]
x
Δ^{0}
or f(x)
Δ^{1
}or Δ
Δ^{2
}
0
a_{2}
1
a_{0}+a_{1}+a_{2}
a_{0}+a_{1}
2
4a_{0}+2a_{1}+a_{2}
3a_{0}+a_{1}
2a_{0}
3
9a_{0}+3a_{1}+a_{2}
5a_{0}+a_{1}
2a_{0}
With the previous example, we can write the coefficients down
immediately. So:
Δ^{2 }=1=2a_{0},
so a_{0}=1/2
Δ^{ }=1=a_{0}+a_{0},
so a_{1}=1/2
And a_{0}=0.
So, writing n for x, the equation is:
S_{n}=(1/2)n^{2}+(1/2)n_{
}
We can note that:
f(0)=a_{2}, and generally, f(0)=a_{n}
because this is the constant term, independent of x [1.02]
Δf(1)=a_{0}+a_{1}, and
generally:
[1.03]
because this is just replacing the x's with 1, leaving the coefficients.
Δf(0), which is f(1)-f(0), is a_{0}+a_{1}
and generally:
[1.04]
because Δf(0)=f(1)-f(0), f(0)=a_{n}, and
f(1) is the sum of the
coefficients (by substituting 1 for x in a general polynomial).
Example
x
Δ
Δ^{2}
Δ^{3}
0
5
1
12
7
2
29
17
10
3
56
27
10
0
Because we find that Δ^{2}
is a constant, 10, we conclude the polynomial is one of degree 2.
f(x)=a_{0}x^{2}+a_{1}x+a_{2
}
Using our results for a polynomial of degree 2, we note:
a_{2}=5
2!a_{0}=10, so a_{0}=5
Δf(0)=a_{0}+a_{1}=7=5+a_{1}
so, a_{1}=2
The underlying equation is therefore:
5x2+2x+5
Generalizing
Polynomials Of Degree 3 With Algebra
We can do the same algebra we did before with a degree 2 polynomial
with a degree 3 (n=3):
f(x)=a_{0}x^{3}+a_{1}x^{2}+a_{2}x+a_{3}
x
Δ^{0}
or f(x)
Δ^{1
}or Δ
Δ^{2
}
Δ^{3
}
0
a_{3}
1
a_{0}+a_{1}+a_{2+}a_{3}
a_{0}+a_{1}+a_{2}
2
8a_{0}+4a_{1}+2a_{2}+a_{3}
7a_{0}+3a_{1}+a_{2}
6a_{0}+2a_{1}
3
27a_{0}+9a_{1}+3a_{2}+a_{3}
19a_{0}+5a_{1}+a_{2}
12a_{0}+2a_{1}
6a_{0}
4
64a_{0}+16a_{1}+4a_{2}+a_{3}
37a_{0}+7a_{1}+a_{2}
18a_{0}+2a_{1}
6a_{0}
We note that what we have previously concluded applies here too (as it
will to any polynomial) That is:
f(0)=a_{n}, which is a_{3}
here.
f(1) is the sum of all the coefficients, a_{0}+a_{1}+a_{2+}a_{3}
Δf(0)=the sum of the coefficients, except the last a_{0}+a_{1}+a_{2}
The constant term is n!a_{0}
Tantalisingly, Δ^{2
}=6a_{0}+2a_{1}
and Δ^{3
}=6a_{0} which suggests we
see 3! appearing in the second difference. This appears to be the case
when n=2. However, as we shall see later, it is not that simple.
Example
Δ
Δ^{2}
Δ^{3}
0
5
1
17
12
2
47
30
18
3
107
60
30
12
4
209
102
42
12
At the third difference we obtain a constant so this is a polynomial
with n=3
f(x)=a_{0}x^{3}+a_{1}x^{2}+a_{2}x+a_{3}
We note a_{3}=5
a_{0}=12/3!, so a_{0}=2
From the general table above, we note that Δ^{2
}f(0)=6a_{0}+2a_{1}
So 18=12+2a_{1}
So, a_{1}=3
Δf(0)=sum of coefficient=a_{0}+a_{1}+a_{2}=12,
2+3+a_{2}=12
So a_{2}=7
The equation is therefore:
f(x)=2x^{3}+3x^{2}+7x+5
Generalizing
Polynomials Of Degree 4 With Algebra
We can do the same algebra we did before with a degree 3 polynomial
with a degree 4 (n=4):
f(x)=a_{0}x^{4}+a_{1}x^{3}+a_{2}x^{2}+a_{3}x+a_{4}
x
Δ^{0}
or f(x)
Δ^{1
}or Δ
Δ^{2
}
Δ^{3
}
Δ^{4
}
0
a_{4}
1
a_{0}+a_{1}+a_{2+}a_{3+}a_{4}
a_{0}+a_{1}+a_{2}+a_{3}
2
16a_{0}+8a_{1}+2a_{2}+2a_{3}+a_{4}
15a_{0}+7a_{1}+3a_{2}+a_{3}
14a_{0}+6a_{1}+2a_{2}
3
81a_{0}+27a_{1}+9a_{2}+3a_{3}+a_{4}
65a_{0}+19a_{1}+5a_{2}+a_{3}
50a_{0}+12a_{1}+2a_{2}
36a_{0}+6a_{1}
4
256a_{0}+64a_{1}+16a_{2}+4a_{3}+a_{4}
175a_{0}+37a_{1}+7a_{2}+a_{3}
110a_{0}+18a_{1}+2a_{2}
60a_{0}+6a_{1}
24a_{0}
5
625a_{0}+125a_{1}+25a_{2}+5a_{3}+a_{4}
369a_{0}+61a_{1}+9a_{2}+a_{3}
194a_{0}+24a_{1}+2a_{2}
84a_{0}+6a_{1}
24a_{0}
Looking at Δ^{3
} we note that n! does not appear in a
simple way, but we also note that 6a_{1} looks like
(n-1)!a_{1} and so do the other Δ^{n-1
}f(0) -- in the tables above. It seems there are patterns emerging, but
we will not tax the algebra further and wait until we look at factorial
polynomials for clarification. In fact, factorial polynomials help us
greatly with difference equations, and regular polynomials can be
converted to factorial polynomials. (We finally discover how to compute these values on this page).
Generalizing
Polynomials Of Degree 5 With Algebra
We can do the same algebra we did before with a degree 4 polynomial
with a degree 5 (n=5):
f(x)=a_{0}x^{54}+a_{1}x^{4}+a_{2}x^{3}+a_{3}x^{2}+a_{4}x+a_{5}
x
Δ^{0}
or f(x)
Δ^{1
}or Δ
Δ^{2
}
Δ^{3
}
Δ^{4
}
Δ^{5
}
0
a_{5}
1
a_{0}+a_{1}+a_{2+}a_{3+}a_{4}+a_{5}
a_{0}+a_{1}+a_{2}+a_{3}+a_{4}
2
32a_{0}+16a_{1}+8a_{2}+4a_{3}+2a_{4}+a_{5}
31a_{0}+15a_{1}+7a_{2}+3a_{3}+a_{4}
30a_{0}+14a_{1}+6a_{2}+2a_{3}
3
243a_{0}+81a_{1}+27a_{2}+9a_{3}+3a_{4}+a_{5}
211a_{0}+65a_{1}+19a_{2}+5a_{3}+a_{4}
180a_{0}+50a_{1}+12a_{2}+2a_{3}
150a_{0}+36a_{1}+6a_{2}
4
1024a_{0}+256a_{1}+64a_{2}+16a_{3}+4a_{4}+a_{5}
781a_{0}+175a_{1}+37a_{2}+7a_{3}+a_{4}
570a_{0}+110a_{1}+18a_{2}+2a_{3}
390a_{0}+60a_{1}+6a_{2}
240a_{0}+24a_{1}
_{}
5
3125a_{0}+625a_{1}+125a_{2}+25a_{3}+5a_{4}+a_{5}
2101a_{0}+369a_{1}+61a_{2}+9a_{3}+a_{4}
1320a_{0}+194a_{1}+24a_{2}+2a_{3}
750a_{0}+84a_{1}+6a_{2}
360a_{0}+24a_{1}
120a_{0}
6
7776a_{0}+1296a_{1}+216a_{2}+36a_{3}+6a_{4}+a_{5}
4651a_{0}+671a_{1}+91a_{2}+11a_{3}+a_{4}
2550a_{0}+302a_{1}+30a_{2}+2a_{3}
1230a_{0}+108a_{1}+6a_{2}
480a_{0}+24a_{1}
120a_{0}
Again, patterns seem to be suggested, but they aren't
particularly clear yet. One approach we can take is to transform the
polynomials into another form that might yield their patterns more
clearly. We will look at factorial polynomials next to try to
understand difference behaviour better.
Example
Finally, on this page, we will look at an example of a polynomial degree 5
Δ^{0}
Δ1
Δ2
Δ3
Δ4
Δ5
0
11
1
22
11
2
95
73
62
3
452
357
284
222
4
1723
1271
914
630
408
5
5186
3463
2192
1278
648
240
6
13007
7821
4358
2166
888
240
This is, or appears to be a polynomial of degree 5:
f(x)=a_{0}x^{54}+a_{1}x^{4}+a_{2}x^{3}+a_{3}x^{2}+a_{4}x+a_{5} a_{5}=11 (f(0) ) a_{0}=240/5!=2 Δ^{4}f(0)=408=240a_{0}+24a_{1}^{ } So, 24a_{1}=408-2*240=-72 a_{1}=-3 Δ^{3}f(0)=150a_{0}+36a_{1}+6a_{2} Substituting known values, 150*2+36(-3)_{}+6a_{2}=222 a_{2}=5 Δ^{2}f(0)=30a_{0}+14a_{1}+6a_{2}+2a_{3} Substituting known values: 30a_{0}+14a_{1}+6a_{2}+2a_{3}=62=30*2+14(-3)_{}+6*5_{}+2a_{3} a_{3}=7 _{} Δf(0)=a_{0}+a_{1}+a_{2}+a_{3}+a_{4}
Substituting known values, 2-3+5+7+a_{4}=11 So a_{4}=0 So the equation is:
f(x)=2x^{54}-3x^{4}+5x^{3}+7x^{2}+11_{}
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