Graphics Reference
In-Depth Information
Table 9.1.
Pascal's triangle
i
n
0
1
2
3
4
5
6
0
1
1
1
1
2
1
2
1
3
1
3
3
1
4
1
4
6
4
1
5
1
5
10
10
5
1
6
1
6
15
20
15
6
1
(
x
+
a
)
0
=1
(
x
+
a
)
1
=1
x
+1
a
(
x
+
a
)
2
=1
x
2
+2
ax
+1
a
2
(
x
+
a
)
3
=1
x
3
+3
ax
2
+3
a
2
x
+1
a
3
(
x
+
a
)
4
=1
x
4
+4
ax
3
+6
a
2
x
2
+4
a
3
x
+1
a
4
which reveals Pascal's triangle as coe
cients of the polynomial terms.
Pascal, however, recognized other qualities in the numbers, in that they
described the odds governing combinations. For example, to determine the
probability of any girl-boy combination in a family of 6 children, we sum the
numbers in the 6th row of Pascal's triangle:
1+6+15+20+15+6+1=64
.
The number (1) at the start and end of the 6th row represent the chances of
getting 6 boys or 6 girls, i.e. 1 in 64. The next number (6) represents the next
most likely combination: 5 boys and 1 girl, or 5 girls and 1 boy, i.e. 6 in 64.
The centre number (20) applies to 3 boys and 3 girls, for which the chances
are 20 in 64.
term in (9.5) is nothing more than a generator for Pascal's
triangle. The powers of
t
and (1
Thus the
n
i
t
) in (9.5) appear as shown in Table 9.2
for different values of
n
and
i
. When the two sets of results are combined,
we get the complete Bernstein polynomial terms shown in Table 9.3. One
very important property of these terms is that they sum to unity, which is an
important feature of any interpolant.
As the sum of (1
−
−
t
)and
t
is 1,
t
)+
t
)
n
= 1
((1
−
(9.7)
which is why we can use the binomial expansion of (1
−
t
)and
t
as interpolants.
When
n
= 2 we obtain the quadratic form
t
)
2
2
t
(1
t
)
t
2
(1
−
−
(9.8)
