On the construction of orthogonal polynomials
1
$begingroup$
In the following proof, argument goes on based on considering $C_n$ to be nonzero then it finishes the proof for $C_n=0$ :
Also if we set $C_n=0$ in Eq. (6.10) then must $m=0,1,2,...,n- 2$ in Eq. (6.12). In either case the proof fails. Am I right? If so, how to finish the proof, that is showing $Delta_n ne 0$?
real-analysis ordinary-differential-equations proof-explanation orthogonality orthogonal-polynomials
asked Dec 20 '18 at 9:23
72D72D
531116
$endgroup$
add a comment |
1
$begingroup$
In the following proof, argument goes on based on considering $C_n$ to be nonzero then it finishes the proof for $C_n=0$ :
Also if we set $C_n=0$ in Eq. (6.10) then must $m=0,1,2,...,n- 2$ in Eq. (6.12). In either case the proof fails. Am I right? If so, how to finish the proof, that is showing $Delta_n ne 0$?
real-analysis ordinary-differential-equations proof-explanation orthogonality orthogonal-polynomials
asked Dec 20 '18 at 9:23
72D72D
531116
$endgroup$
$begingroup$
Why did you exclude $m=n-1$?
$endgroup$
– metamorphy
Dec 20 '18 at 9:34
$begingroup$
@metamorphy, when $C_n ne 0$ (implicitly $C_{n+1} = 0$), $m=0,...,n-1$ so when $C_n = 0$ and $C_{n-1} ne 0$ so the argument reduces by $1$ less i.e.: $m=0,...,(n-1)-1$.
$endgroup$
– 72D
Dec 20 '18 at 9:38
add a comment |
1
1
1
$begingroup$
In the following proof, argument goes on based on considering $C_n$ to be nonzero then it finishes the proof for $C_n=0$ :
Also if we set $C_n=0$ in Eq. (6.10) then must $m=0,1,2,...,n- 2$ in Eq. (6.12). In either case the proof fails. Am I right? If so, how to finish the proof, that is showing $Delta_n ne 0$?
real-analysis ordinary-differential-equations proof-explanation orthogonality orthogonal-polynomials
asked Dec 20 '18 at 9:23
72D72D
531116
$endgroup$
In the following proof, argument goes on based on considering $C_n$ to be nonzero then it finishes the proof for $C_n=0$ :
Also if we set $C_n=0$ in Eq. (6.10) then must $m=0,1,2,...,n- 2$ in Eq. (6.12). In either case the proof fails. Am I right? If so, how to finish the proof, that is showing $Delta_n ne 0$?
real-analysis ordinary-differential-equations proof-explanation orthogonality orthogonal-polynomials
real-analysis ordinary-differential-equations proof-explanation orthogonality orthogonal-polynomials
asked Dec 20 '18 at 9:23
72D72D
531116
asked Dec 20 '18 at 9:23
72D72D
531116
asked Dec 20 '18 at 9:23
72D72D
531116
asked Dec 20 '18 at 9:23
72D72D
531116
asked Dec 20 '18 at 9:23
72D72D
531116
531116
$begingroup$
Why did you exclude $m=n-1$?
$endgroup$
– metamorphy
Dec 20 '18 at 9:34
$begingroup$
@metamorphy, when $C_n ne 0$ (implicitly $C_{n+1} = 0$), $m=0,...,n-1$ so when $C_n = 0$ and $C_{n-1} ne 0$ so the argument reduces by $1$ less i.e.: $m=0,...,(n-1)-1$.
$endgroup$
– 72D
Dec 20 '18 at 9:38
add a comment |
$begingroup$
Why did you exclude $m=n-1$?
$endgroup$
– metamorphy
Dec 20 '18 at 9:34
$begingroup$
@metamorphy, when $C_n ne 0$ (implicitly $C_{n+1} = 0$), $m=0,...,n-1$ so when $C_n = 0$ and $C_{n-1} ne 0$ so the argument reduces by $1$ less i.e.: $m=0,...,(n-1)-1$.
$endgroup$
– 72D
Dec 20 '18 at 9:38
$begingroup$
Why did you exclude $m=n-1$?
$endgroup$
– metamorphy
Dec 20 '18 at 9:34
$begingroup$
Why did you exclude $m=n-1$?
$endgroup$
– metamorphy
Dec 20 '18 at 9:34
$begingroup$
@metamorphy, when $C_n ne 0$ (implicitly $C_{n+1} = 0$), $m=0,...,n-1$ so when $C_n = 0$ and $C_{n-1} ne 0$ so the argument reduces by $1$ less i.e.: $m=0,...,(n-1)-1$.
$endgroup$
– 72D
Dec 20 '18 at 9:38
$begingroup$
@metamorphy, when $C_n ne 0$ (implicitly $C_{n+1} = 0$), $m=0,...,n-1$ so when $C_n = 0$ and $C_{n-1} ne 0$ so the argument reduces by $1$ less i.e.: $m=0,...,(n-1)-1$.
$endgroup$
– 72D
Dec 20 '18 at 9:38
add a comment |
1 Answer
1
active
oldest
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$begingroup$
You exclude $m=n-1$ for no reason. The proof proceeds by showing that $(6.10)$ has solutions $C_0,ldots,C_n$ with $C_nneq 0$. This amounts to proving $Delta_nneq 0$. If we assume the contrary, the same system with $C_n=0$ (consisting of $n$ equations, not $n-1$) has nontrivial solutions $C_0,ldots,C_{n-1}$. This corresponds to $(6.12)$ with the allowed values of $m$ exactly as specified there.
answered Dec 20 '18 at 9:47
metamorphymetamorphy
3,6821621
$endgroup$
$begingroup$
We want to prove $Delta_nneq 0$ so we suppose the contrary $Delta_n = 0$. This is OK. However, $C_n ne 0$ was required from beginning and this assumption can't be changed otherwise it will be other argument from the beginning so (6.10) can't be used. For new argument I supposed $n-2$ the last $m$; for old one $C_n ne 0$.
$endgroup$
– 72D
Dec 20 '18 at 9:52
1
$begingroup$
Maybe you get it this way: if $Delta_n=0$ then the system $$begin{align}M_0 C_0&+M_1 C_1+ldots+M_{n-1}C_{n-1}&=&color{red}{0}\M_1 C_0&+M_2 C_1+ldots+M_{n}C_{n-1}&=&color{red}{0}\ ldots\M_{n-1} C_0&+M_n C_1+ldots+M_{2n-2}C_{n-1}&=&color{red}{0}end{align}$$ has solutions in $C_0,ldots,C_{n-1}$ not all zero.
$endgroup$
– metamorphy
Dec 20 '18 at 10:02
$begingroup$
No I say that $C_n=0$ is not allowed even if $Δ_n=0$. When forming (6.10) we assumed $C_n ne 0$.
$endgroup$
– 72D
Dec 20 '18 at 10:09
1
$begingroup$
Forget your $C_n$! Do you agree with the statement in my last comment as written? (If you don't, tell me why.)
$endgroup$
– metamorphy
Dec 20 '18 at 10:11
$begingroup$
Yes, I totally agree with all of your last comment. Let me re-read the proof again...
$endgroup$
– 72D
Dec 20 '18 at 10:23
|
show 2 more comments
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$begingroup$
You exclude $m=n-1$ for no reason. The proof proceeds by showing that $(6.10)$ has solutions $C_0,ldots,C_n$ with $C_nneq 0$. This amounts to proving $Delta_nneq 0$. If we assume the contrary, the same system with $C_n=0$ (consisting of $n$ equations, not $n-1$) has nontrivial solutions $C_0,ldots,C_{n-1}$. This corresponds to $(6.12)$ with the allowed values of $m$ exactly as specified there.
answered Dec 20 '18 at 9:47
metamorphymetamorphy
3,6821621
$endgroup$
$begingroup$
We want to prove $Delta_nneq 0$ so we suppose the contrary $Delta_n = 0$. This is OK. However, $C_n ne 0$ was required from beginning and this assumption can't be changed otherwise it will be other argument from the beginning so (6.10) can't be used. For new argument I supposed $n-2$ the last $m$; for old one $C_n ne 0$.
$endgroup$
– 72D
Dec 20 '18 at 9:52
1
$begingroup$
Maybe you get it this way: if $Delta_n=0$ then the system $$begin{align}M_0 C_0&+M_1 C_1+ldots+M_{n-1}C_{n-1}&=&color{red}{0}\M_1 C_0&+M_2 C_1+ldots+M_{n}C_{n-1}&=&color{red}{0}\ ldots\M_{n-1} C_0&+M_n C_1+ldots+M_{2n-2}C_{n-1}&=&color{red}{0}end{align}$$ has solutions in $C_0,ldots,C_{n-1}$ not all zero.
$endgroup$
– metamorphy
Dec 20 '18 at 10:02
$begingroup$
No I say that $C_n=0$ is not allowed even if $Δ_n=0$. When forming (6.10) we assumed $C_n ne 0$.
$endgroup$
– 72D
Dec 20 '18 at 10:09
1
$begingroup$
Forget your $C_n$! Do you agree with the statement in my last comment as written? (If you don't, tell me why.)
$endgroup$
– metamorphy
Dec 20 '18 at 10:11
$begingroup$
Yes, I totally agree with all of your last comment. Let me re-read the proof again...
$endgroup$
– 72D
Dec 20 '18 at 10:23
|
show 2 more comments
1
$begingroup$
You exclude $m=n-1$ for no reason. The proof proceeds by showing that $(6.10)$ has solutions $C_0,ldots,C_n$ with $C_nneq 0$. This amounts to proving $Delta_nneq 0$. If we assume the contrary, the same system with $C_n=0$ (consisting of $n$ equations, not $n-1$) has nontrivial solutions $C_0,ldots,C_{n-1}$. This corresponds to $(6.12)$ with the allowed values of $m$ exactly as specified there.
answered Dec 20 '18 at 9:47
metamorphymetamorphy
3,6821621
$endgroup$
$begingroup$
We want to prove $Delta_nneq 0$ so we suppose the contrary $Delta_n = 0$. This is OK. However, $C_n ne 0$ was required from beginning and this assumption can't be changed otherwise it will be other argument from the beginning so (6.10) can't be used. For new argument I supposed $n-2$ the last $m$; for old one $C_n ne 0$.
$endgroup$
– 72D
Dec 20 '18 at 9:52
1
$begingroup$
Maybe you get it this way: if $Delta_n=0$ then the system $$begin{align}M_0 C_0&+M_1 C_1+ldots+M_{n-1}C_{n-1}&=&color{red}{0}\M_1 C_0&+M_2 C_1+ldots+M_{n}C_{n-1}&=&color{red}{0}\ ldots\M_{n-1} C_0&+M_n C_1+ldots+M_{2n-2}C_{n-1}&=&color{red}{0}end{align}$$ has solutions in $C_0,ldots,C_{n-1}$ not all zero.
$endgroup$
– metamorphy
Dec 20 '18 at 10:02
$begingroup$
No I say that $C_n=0$ is not allowed even if $Δ_n=0$. When forming (6.10) we assumed $C_n ne 0$.
$endgroup$
– 72D
Dec 20 '18 at 10:09
1
$begingroup$
Forget your $C_n$! Do you agree with the statement in my last comment as written? (If you don't, tell me why.)
$endgroup$
– metamorphy
Dec 20 '18 at 10:11
$begingroup$
Yes, I totally agree with all of your last comment. Let me re-read the proof again...
$endgroup$
– 72D
Dec 20 '18 at 10:23
|
show 2 more comments
1
1
1
$begingroup$
You exclude $m=n-1$ for no reason. The proof proceeds by showing that $(6.10)$ has solutions $C_0,ldots,C_n$ with $C_nneq 0$. This amounts to proving $Delta_nneq 0$. If we assume the contrary, the same system with $C_n=0$ (consisting of $n$ equations, not $n-1$) has nontrivial solutions $C_0,ldots,C_{n-1}$. This corresponds to $(6.12)$ with the allowed values of $m$ exactly as specified there.
answered Dec 20 '18 at 9:47
metamorphymetamorphy
3,6821621
$endgroup$
You exclude $m=n-1$ for no reason. The proof proceeds by showing that $(6.10)$ has solutions $C_0,ldots,C_n$ with $C_nneq 0$. This amounts to proving $Delta_nneq 0$. If we assume the contrary, the same system with $C_n=0$ (consisting of $n$ equations, not $n-1$) has nontrivial solutions $C_0,ldots,C_{n-1}$. This corresponds to $(6.12)$ with the allowed values of $m$ exactly as specified there.
answered Dec 20 '18 at 9:47
metamorphymetamorphy
3,6821621
answered Dec 20 '18 at 9:47
metamorphymetamorphy
3,6821621
answered Dec 20 '18 at 9:47
metamorphymetamorphy
3,6821621
answered Dec 20 '18 at 9:47
metamorphymetamorphy
3,6821621
3,6821621
$begingroup$
We want to prove $Delta_nneq 0$ so we suppose the contrary $Delta_n = 0$. This is OK. However, $C_n ne 0$ was required from beginning and this assumption can't be changed otherwise it will be other argument from the beginning so (6.10) can't be used. For new argument I supposed $n-2$ the last $m$; for old one $C_n ne 0$.
$endgroup$
– 72D
Dec 20 '18 at 9:52
1
$begingroup$
Maybe you get it this way: if $Delta_n=0$ then the system $$begin{align}M_0 C_0&+M_1 C_1+ldots+M_{n-1}C_{n-1}&=&color{red}{0}\M_1 C_0&+M_2 C_1+ldots+M_{n}C_{n-1}&=&color{red}{0}\ ldots\M_{n-1} C_0&+M_n C_1+ldots+M_{2n-2}C_{n-1}&=&color{red}{0}end{align}$$ has solutions in $C_0,ldots,C_{n-1}$ not all zero.
$endgroup$
– metamorphy
Dec 20 '18 at 10:02
$begingroup$
No I say that $C_n=0$ is not allowed even if $Δ_n=0$. When forming (6.10) we assumed $C_n ne 0$.
$endgroup$
– 72D
Dec 20 '18 at 10:09
1
$begingroup$
Forget your $C_n$! Do you agree with the statement in my last comment as written? (If you don't, tell me why.)
$endgroup$
– metamorphy
Dec 20 '18 at 10:11
$begingroup$
Yes, I totally agree with all of your last comment. Let me re-read the proof again...
$endgroup$
– 72D
Dec 20 '18 at 10:23
|
show 2 more comments
$begingroup$
We want to prove $Delta_nneq 0$ so we suppose the contrary $Delta_n = 0$. This is OK. However, $C_n ne 0$ was required from beginning and this assumption can't be changed otherwise it will be other argument from the beginning so (6.10) can't be used. For new argument I supposed $n-2$ the last $m$; for old one $C_n ne 0$.
$endgroup$
– 72D
Dec 20 '18 at 9:52
1
$begingroup$
Maybe you get it this way: if $Delta_n=0$ then the system $$begin{align}M_0 C_0&+M_1 C_1+ldots+M_{n-1}C_{n-1}&=&color{red}{0}\M_1 C_0&+M_2 C_1+ldots+M_{n}C_{n-1}&=&color{red}{0}\ ldots\M_{n-1} C_0&+M_n C_1+ldots+M_{2n-2}C_{n-1}&=&color{red}{0}end{align}$$ has solutions in $C_0,ldots,C_{n-1}$ not all zero.
$endgroup$
– metamorphy
Dec 20 '18 at 10:02
$begingroup$
No I say that $C_n=0$ is not allowed even if $Δ_n=0$. When forming (6.10) we assumed $C_n ne 0$.
$endgroup$
– 72D
Dec 20 '18 at 10:09
1
$begingroup$
Forget your $C_n$! Do you agree with the statement in my last comment as written? (If you don't, tell me why.)
$endgroup$
– metamorphy
Dec 20 '18 at 10:11
$begingroup$
Yes, I totally agree with all of your last comment. Let me re-read the proof again...
$endgroup$
– 72D
Dec 20 '18 at 10:23
$begingroup$
We want to prove $Delta_nneq 0$ so we suppose the contrary $Delta_n = 0$. This is OK. However, $C_n ne 0$ was required from beginning and this assumption can't be changed otherwise it will be other argument from the beginning so (6.10) can't be used. For new argument I supposed $n-2$ the last $m$; for old one $C_n ne 0$.
$endgroup$
– 72D
Dec 20 '18 at 9:52
$begingroup$
We want to prove $Delta_nneq 0$ so we suppose the contrary $Delta_n = 0$. This is OK. However, $C_n ne 0$ was required from beginning and this assumption can't be changed otherwise it will be other argument from the beginning so (6.10) can't be used. For new argument I supposed $n-2$ the last $m$; for old one $C_n ne 0$.
$endgroup$
– 72D
Dec 20 '18 at 9:52
1
1
$begingroup$
Maybe you get it this way: if $Delta_n=0$ then the system $$begin{align}M_0 C_0&+M_1 C_1+ldots+M_{n-1}C_{n-1}&=&color{red}{0}\M_1 C_0&+M_2 C_1+ldots+M_{n}C_{n-1}&=&color{red}{0}\ ldots\M_{n-1} C_0&+M_n C_1+ldots+M_{2n-2}C_{n-1}&=&color{red}{0}end{align}$$ has solutions in $C_0,ldots,C_{n-1}$ not all zero.
$endgroup$
– metamorphy
Dec 20 '18 at 10:02
$begingroup$
Maybe you get it this way: if $Delta_n=0$ then the system $$begin{align}M_0 C_0&+M_1 C_1+ldots+M_{n-1}C_{n-1}&=&color{red}{0}\M_1 C_0&+M_2 C_1+ldots+M_{n}C_{n-1}&=&color{red}{0}\ ldots\M_{n-1} C_0&+M_n C_1+ldots+M_{2n-2}C_{n-1}&=&color{red}{0}end{align}$$ has solutions in $C_0,ldots,C_{n-1}$ not all zero.
$endgroup$
– metamorphy
Dec 20 '18 at 10:02
$begingroup$
No I say that $C_n=0$ is not allowed even if $Δ_n=0$. When forming (6.10) we assumed $C_n ne 0$.
$endgroup$
– 72D
Dec 20 '18 at 10:09
$begingroup$
No I say that $C_n=0$ is not allowed even if $Δ_n=0$. When forming (6.10) we assumed $C_n ne 0$.
$endgroup$
– 72D
Dec 20 '18 at 10:09
1
1
$begingroup$
Forget your $C_n$! Do you agree with the statement in my last comment as written? (If you don't, tell me why.)
$endgroup$
– metamorphy
Dec 20 '18 at 10:11
$begingroup$
Forget your $C_n$! Do you agree with the statement in my last comment as written? (If you don't, tell me why.)
$endgroup$
– metamorphy
Dec 20 '18 at 10:11
$begingroup$
Yes, I totally agree with all of your last comment. Let me re-read the proof again...
$endgroup$
– 72D
Dec 20 '18 at 10:23
$begingroup$
Yes, I totally agree with all of your last comment. Let me re-read the proof again...
$endgroup$
– 72D
Dec 20 '18 at 10:23
|
show 2 more comments
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$begingroup$
Why did you exclude $m=n-1$?
$endgroup$
– metamorphy
Dec 20 '18 at 9:34
$begingroup$
@metamorphy, when $C_n ne 0$ (implicitly $C_{n+1} = 0$), $m=0,...,n-1$ so when $C_n = 0$ and $C_{n-1} ne 0$ so the argument reduces by $1$ less i.e.: $m=0,...,(n-1)-1$.
$endgroup$
– 72D
Dec 20 '18 at 9:38