All nodes lies in (a,b)
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Let
{
x
i
}
i
=
1
l
{\displaystyle \{x_{i}\}_{i=1}^{l}\!\,}
be the nodes that lie in the interval
(
a
,
b
)
{\displaystyle (a,b)\!\,}
.
Let
q
l
(
x
)
=
∏
i
=
1
l
(
x
−
x
i
)
{\displaystyle q_{l}(x)=\prod _{i=1}^{l}(x-x_{i})\!\,}
which is a polynomial of degree
l
{\displaystyle l\!\,}
.
Let
p
n
(
x
)
=
∏
i
=
1
n
(
x
−
x
i
)
=
q
l
(
x
)
∏
i
=
1
n
−
l
(
x
−
x
i
)
{\displaystyle p_{n}(x)=\prod _{i=1}^{n}(x-x_{i})=q_{l}(x)\prod _{i=1}^{n-l}(x-x_{i})\!\,}
which is a polynomial of degree
n
>
l
{\displaystyle n>l\!\,}
.
Then
⟨
p
n
,
q
l
⟩
=
∫
a
b
q
l
2
(
x
)
∏
i
=
1
n
−
l
(
x
−
x
i
)
⏟
r
(
x
)
≠
0
{\displaystyle \langle p_{n},q_{l}\rangle =\int _{a}^{b}q_{l}^{2}(x)\underbrace {\prod _{i=1}^{n-l}(x-x_{i})} _{r(x)}\neq 0\!\,}
since
r
(
x
)
{\displaystyle r(x)\!\,}
is of one sign in the interval
(
a
,
b
)
{\displaystyle (a,b)\!\,}
since for
i
=
1
,
2
,
…
n
−
l
{\displaystyle i=1,2,\ldots n-l\!\,}
,
x
i
∉
(
a
,
b
)
.
{\displaystyle x_{i}\not \in (a,b).\!\,}
This implies
q
l
{\displaystyle q_{l}\!\,}
is of degree
n
{\displaystyle n\!\,}
since otherwise
⟨
p
n
,
q
l
⟩
=
0
{\displaystyle \langle p_{n},q_{l}\rangle =0\!\,}
from the orthogonality of
p
n
{\displaystyle p_{n}\!\,}
.
All weights positive
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