how to prove two propositions in summation formula
$begingroup$
This question is from a book writing by Wu-Yi Hsiang which is not a homework assignment.
It is about summation formula for a polynomial $f(x)$, $deg(f(x))=l$，we set
$g_k(x)=frac{1}{k!}prodlimits_{i=0}^{k-1}(x-i)$, in which k means the power of polynomial.
Clearly, $g_k(i)=0, 0leqslant i leqslant k-1$, $g_k(k)=1$
Define,
$$sumlimits_{i=0}^{n-1}g_k(i)=S_{g_k}(n)$$.
as summation formula
$$bigtriangleup f(x)=f(x+1)-f(x)$$
as difference operator
prove two propositions
proposition 1：
$$S_{g_k}(x)=g_{k+1}(x)=frac{1}{(k+1)!}prodlimits_{i=0}^{k}(x-i)$$
While author just use this is following prove , maybe it is trivial for author, I can not understand how to get this formula
proposition 2：
$$f(x)=sumlimits_{k=0}^lc_kg_k(x)$$
which
$$c_k=bigtriangleup^k f(0)$$
Author said the $f(x)$ can be uniquely determined by the combination of $g_k(x)$，I want to prove this but failed. Then he proves how to get coefficients and I understand that.
Thanks.
linear-algebra
asked Dec 29 '18 at 11:46
vnjeypvsvnjeypvs
11
$endgroup$
add a comment |
$begingroup$
This question is from a book writing by Wu-Yi Hsiang which is not a homework assignment.
It is about summation formula for a polynomial $f(x)$, $deg(f(x))=l$，we set
$g_k(x)=frac{1}{k!}prodlimits_{i=0}^{k-1}(x-i)$, in which k means the power of polynomial.
Clearly, $g_k(i)=0, 0leqslant i leqslant k-1$, $g_k(k)=1$
Define,
$$sumlimits_{i=0}^{n-1}g_k(i)=S_{g_k}(n)$$.
as summation formula
$$bigtriangleup f(x)=f(x+1)-f(x)$$
as difference operator
prove two propositions
proposition 1：
$$S_{g_k}(x)=g_{k+1}(x)=frac{1}{(k+1)!}prodlimits_{i=0}^{k}(x-i)$$
While author just use this is following prove , maybe it is trivial for author, I can not understand how to get this formula
proposition 2：
$$f(x)=sumlimits_{k=0}^lc_kg_k(x)$$
which
$$c_k=bigtriangleup^k f(0)$$
Author said the $f(x)$ can be uniquely determined by the combination of $g_k(x)$，I want to prove this but failed. Then he proves how to get coefficients and I understand that.
Thanks.
linear-algebra
asked Dec 29 '18 at 11:46
vnjeypvsvnjeypvs
11
$endgroup$
$begingroup$
I already know the power of sum of the k power polynomial is k+1
$endgroup$
– vnjeypvs
Dec 29 '18 at 11:48
1
$begingroup$
Nicely typeset, but what was the question again? Are you just asking us to prove some claim? The one about $g_{k+1}(x)$? Or the formula for $c_k$? Where is this question from? It might even be a homework assignment, and that presses people's buttons. I think we need a bit of context to understand exactly what the question is. It may help, if you take a look at the guide for new askers.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 20:13
add a comment |
$begingroup$
This question is from a book writing by Wu-Yi Hsiang which is not a homework assignment.
It is about summation formula for a polynomial $f(x)$, $deg(f(x))=l$，we set
$g_k(x)=frac{1}{k!}prodlimits_{i=0}^{k-1}(x-i)$, in which k means the power of polynomial.
Clearly, $g_k(i)=0, 0leqslant i leqslant k-1$, $g_k(k)=1$
Define,
$$sumlimits_{i=0}^{n-1}g_k(i)=S_{g_k}(n)$$.
as summation formula
$$bigtriangleup f(x)=f(x+1)-f(x)$$
as difference operator
prove two propositions
proposition 1：
$$S_{g_k}(x)=g_{k+1}(x)=frac{1}{(k+1)!}prodlimits_{i=0}^{k}(x-i)$$
While author just use this is following prove , maybe it is trivial for author, I can not understand how to get this formula
proposition 2：
$$f(x)=sumlimits_{k=0}^lc_kg_k(x)$$
which
$$c_k=bigtriangleup^k f(0)$$
Author said the $f(x)$ can be uniquely determined by the combination of $g_k(x)$，I want to prove this but failed. Then he proves how to get coefficients and I understand that.
Thanks.
linear-algebra
asked Dec 29 '18 at 11:46
vnjeypvsvnjeypvs
11
$endgroup$
This question is from a book writing by Wu-Yi Hsiang which is not a homework assignment.
It is about summation formula for a polynomial $f(x)$, $deg(f(x))=l$，we set
$g_k(x)=frac{1}{k!}prodlimits_{i=0}^{k-1}(x-i)$, in which k means the power of polynomial.
Clearly, $g_k(i)=0, 0leqslant i leqslant k-1$, $g_k(k)=1$
Define,
$$sumlimits_{i=0}^{n-1}g_k(i)=S_{g_k}(n)$$.
as summation formula
$$bigtriangleup f(x)=f(x+1)-f(x)$$
as difference operator
prove two propositions
proposition 1：
$$S_{g_k}(x)=g_{k+1}(x)=frac{1}{(k+1)!}prodlimits_{i=0}^{k}(x-i)$$
While author just use this is following prove , maybe it is trivial for author, I can not understand how to get this formula
proposition 2：
$$f(x)=sumlimits_{k=0}^lc_kg_k(x)$$
which
$$c_k=bigtriangleup^k f(0)$$
Author said the $f(x)$ can be uniquely determined by the combination of $g_k(x)$，I want to prove this but failed. Then he proves how to get coefficients and I understand that.
Thanks.
linear-algebra
linear-algebra
asked Dec 29 '18 at 11:46
vnjeypvsvnjeypvs
11
asked Dec 29 '18 at 11:46
vnjeypvsvnjeypvs
11
edited Dec 30 '18 at 1:13
vnjeypvs
asked Dec 29 '18 at 11:46
vnjeypvsvnjeypvs
11
asked Dec 29 '18 at 11:46
vnjeypvsvnjeypvs
11
asked Dec 29 '18 at 11:46
vnjeypvsvnjeypvs
11
11
$begingroup$
I already know the power of sum of the k power polynomial is k+1
$endgroup$
– vnjeypvs
Dec 29 '18 at 11:48
1
$begingroup$
Nicely typeset, but what was the question again? Are you just asking us to prove some claim? The one about $g_{k+1}(x)$? Or the formula for $c_k$? Where is this question from? It might even be a homework assignment, and that presses people's buttons. I think we need a bit of context to understand exactly what the question is. It may help, if you take a look at the guide for new askers.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 20:13
add a comment |
$begingroup$
I already know the power of sum of the k power polynomial is k+1
$endgroup$
– vnjeypvs
Dec 29 '18 at 11:48
1
$begingroup$
Nicely typeset, but what was the question again? Are you just asking us to prove some claim? The one about $g_{k+1}(x)$? Or the formula for $c_k$? Where is this question from? It might even be a homework assignment, and that presses people's buttons. I think we need a bit of context to understand exactly what the question is. It may help, if you take a look at the guide for new askers.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 20:13
$begingroup$
I already know the power of sum of the k power polynomial is k+1
$endgroup$
– vnjeypvs
Dec 29 '18 at 11:48
$begingroup$
I already know the power of sum of the k power polynomial is k+1
$endgroup$
– vnjeypvs
Dec 29 '18 at 11:48
1
1
$begingroup$
Nicely typeset, but what was the question again? Are you just asking us to prove some claim? The one about $g_{k+1}(x)$? Or the formula for $c_k$? Where is this question from? It might even be a homework assignment, and that presses people's buttons. I think we need a bit of context to understand exactly what the question is. It may help, if you take a look at the guide for new askers.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 20:13
$begingroup$
Nicely typeset, but what was the question again? Are you just asking us to prove some claim? The one about $g_{k+1}(x)$? Or the formula for $c_k$? Where is this question from? It might even be a homework assignment, and that presses people's buttons. I think we need a bit of context to understand exactly what the question is. It may help, if you take a look at the guide for new askers.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 20:13
add a comment |
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$begingroup$
I already know the power of sum of the k power polynomial is k+1
$endgroup$
– vnjeypvs
Dec 29 '18 at 11:48
1
$begingroup$
Nicely typeset, but what was the question again? Are you just asking us to prove some claim? The one about $g_{k+1}(x)$? Or the formula for $c_k$? Where is this question from? It might even be a homework assignment, and that presses people's buttons. I think we need a bit of context to understand exactly what the question is. It may help, if you take a look at the guide for new askers.
$endgroup$
– Jyrki Lahtonen
Dec 29 '18 at 20:13