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Numerical Integration 39
(3.1.9) z (n)
k := 2(2n + α + 1)ˆy (n)
k
−
1
6(2n + α + 1)
5
4(1 − ˆy (n)
−
k
)2
1
1 − ˆy (n)
k
− 1 + 3α 2
Let us permute the values of z (n)
k , 1 ≤ k ≤ n, in such a way they are in increasing order.
After this rearrangement, we denote the new sequence by ˆz (n)
k , 1 ≤ k ≤ n. Finally, we
have
(3.1.10) ξ (n)
k ≈ ˆz (n)
k
, 1 ≤ k ≤ n.
By virtue of theorem 3.1.2, one also gets limn→+∞ ξ (n)
n = +∞, limn→+∞ ξ (n)
1 = 0.
Moreover, the largest zero tends to infinity like 4n, while the smallest zero behaves like
π 2
4n . We plot in figure 3.1.3 the Laguerre zeroes for α = 0 and n = 10 in the interval
]0,40[.
Figure 3.1.3 - Laguerre zeroes for α = 0 and n = 10.
Zeroes in the Hermite case - It is sufficient to recall the relations (1.7.9)
and (1.7.10).
Therefore, when n is even, positive zeroes are approximated by
ˆz (n/2)
k , 1 ≤ k ≤ n
2 ,
where these quantities are obtained with the procedure described above with α = −1 2 .
When n is odd, positive zeroes are approximated by
these terms are evaluated for α = 1
2 . Moreover ξ(n)
(n+1)/2
ˆz ((n−1)/2)
k
, 1 ≤ k ≤ n−1
2 , where
= 0. For all n, the zeroes are
symmetrically distributed around x = 0. A plot in the interval ]−6,6[ is given in figure
3.1.4 for n = 15.
Figure 3.1.4 - Hermite zeroes for n = 15.
.