Prove that $sum_{j=0}^n sum_{k=0}^j p_k q_{j-k} r_{n-j} = sum_{k=0}^n sum_{j=k}^n p_k q_{j-k} r_{n-j}$ ...
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Prove that $sum_{j=0}^n sum_{k=0}^j p_k q_{j-k} r_{n-j} = sum_{k=0}^n sum_{j=k}^n p_k q_{j-k} r_{n-j}$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Prove that $sum_{kge 1} k/(k+1)! = 1$how to prove that $sum_{k=1}^{m}k!k=(m+1)!-1$ without induction?How to prove that $sum_{k=0}^n binom{2n+1}{2k+1} = 4^{n}$Prove that $1.49<sum_{k=1}^{99}frac{1}{k^2}<1.99$Looking for a clever simplification of $sum_{j=2}^{T_L}(j-1-O_L)(2j-n-1)+sum_{i=T^U}^{n-1}(O^U-n+i)(2i-n-1)$What is the asymptotic behaviour of $sum_{p_kleq x}kp_k$, where $p_k$ is the kth prime number?How to prove that $sum_{i=j}^nn-i = sum_{i=1}^{n-j}i$?Prove that $ sum_{x=1}^ infty frac{1}{x!} = e -1 $Convergence of $sum_{k=1}^infty p_k^{-z}$, $p$ is prime.Tools for finding bounds on power series
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I encounter this problem when proving that $Bbb R[[X]] := (Bbb R^Bbb N,+,cdot)$ is actually a formal power series ring over $Bbb R$.
Let $(p_n mid n in Bbb N), (q_n mid n in Bbb N), (r_n mid n in Bbb N)$ be sequences in $Bbb R$. Prove that $$sum_{j=0}^n sum_{k=0}^j p_k q_{j-k} r_{n-j} = sum_{k=0}^n sum_{j=k}^n p_k q_{j-k} r_{n-j}$$
I have tried to manipulate the indices $j,k,n$ but to no avail. Please leave me some hints. Thank you so much!
summation
asked Mar 26 at 10:18
Le Anh DungLe Anh Dung
1,2571621
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add a comment |
$begingroup$
I encounter this problem when proving that $Bbb R[[X]] := (Bbb R^Bbb N,+,cdot)$ is actually a formal power series ring over $Bbb R$.
Let $(p_n mid n in Bbb N), (q_n mid n in Bbb N), (r_n mid n in Bbb N)$ be sequences in $Bbb R$. Prove that $$sum_{j=0}^n sum_{k=0}^j p_k q_{j-k} r_{n-j} = sum_{k=0}^n sum_{j=k}^n p_k q_{j-k} r_{n-j}$$
I have tried to manipulate the indices $j,k,n$ but to no avail. Please leave me some hints. Thank you so much!
summation
asked Mar 26 at 10:18
Le Anh DungLe Anh Dung
1,2571621
$endgroup$
add a comment |
$begingroup$
I encounter this problem when proving that $Bbb R[[X]] := (Bbb R^Bbb N,+,cdot)$ is actually a formal power series ring over $Bbb R$.
Let $(p_n mid n in Bbb N), (q_n mid n in Bbb N), (r_n mid n in Bbb N)$ be sequences in $Bbb R$. Prove that $$sum_{j=0}^n sum_{k=0}^j p_k q_{j-k} r_{n-j} = sum_{k=0}^n sum_{j=k}^n p_k q_{j-k} r_{n-j}$$
I have tried to manipulate the indices $j,k,n$ but to no avail. Please leave me some hints. Thank you so much!
summation
asked Mar 26 at 10:18
Le Anh DungLe Anh Dung
1,2571621
$endgroup$
I encounter this problem when proving that $Bbb R[[X]] := (Bbb R^Bbb N,+,cdot)$ is actually a formal power series ring over $Bbb R$.
Let $(p_n mid n in Bbb N), (q_n mid n in Bbb N), (r_n mid n in Bbb N)$ be sequences in $Bbb R$. Prove that $$sum_{j=0}^n sum_{k=0}^j p_k q_{j-k} r_{n-j} = sum_{k=0}^n sum_{j=k}^n p_k q_{j-k} r_{n-j}$$
I have tried to manipulate the indices $j,k,n$ but to no avail. Please leave me some hints. Thank you so much!
summation
summation
asked Mar 26 at 10:18
Le Anh DungLe Anh Dung
1,2571621
asked Mar 26 at 10:18
Le Anh DungLe Anh Dung
1,2571621
asked Mar 26 at 10:18
Le Anh DungLe Anh Dung
1,2571621
asked Mar 26 at 10:18
Le Anh DungLe Anh Dung
1,2571621
asked Mar 26 at 10:18
Le Anh DungLe Anh Dung
1,2571621
1,2571621
add a comment |
add a comment |
2 Answers
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You are adding the same numbers in a different order The condition $0 leq k leq j leq n$ is same as $j geq k $ , $j leq n$ and $0leq k$ so the terms on the two sides are the same.
answered Mar 26 at 10:23
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
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add a comment |
$begingroup$
All you have to prove is that the sets of indices the sums run over are the same. In fact, they are both equal to
$$S{(j, k)| 0leq j,lleq nland kleq j}$$
That is, you can prove that in both sums, the index pair $(j,k)$ appears in the sum if and only if $(j, k)in S$.
answered Mar 26 at 10:22
5xum5xum
92.9k395162
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
You are adding the same numbers in a different order The condition $0 leq k leq j leq n$ is same as $j geq k $ , $j leq n$ and $0leq k$ so the terms on the two sides are the same.
answered Mar 26 at 10:23
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
$endgroup$
add a comment |
$begingroup$
You are adding the same numbers in a different order The condition $0 leq k leq j leq n$ is same as $j geq k $ , $j leq n$ and $0leq k$ so the terms on the two sides are the same.
answered Mar 26 at 10:23
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
$endgroup$
add a comment |
$begingroup$
You are adding the same numbers in a different order The condition $0 leq k leq j leq n$ is same as $j geq k $ , $j leq n$ and $0leq k$ so the terms on the two sides are the same.
answered Mar 26 at 10:23
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
$endgroup$
You are adding the same numbers in a different order The condition $0 leq k leq j leq n$ is same as $j geq k $ , $j leq n$ and $0leq k$ so the terms on the two sides are the same.
answered Mar 26 at 10:23
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
answered Mar 26 at 10:23
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
answered Mar 26 at 10:23
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
answered Mar 26 at 10:23
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
76.6k53471
add a comment |
add a comment |
$begingroup$
All you have to prove is that the sets of indices the sums run over are the same. In fact, they are both equal to
$$S{(j, k)| 0leq j,lleq nland kleq j}$$
That is, you can prove that in both sums, the index pair $(j,k)$ appears in the sum if and only if $(j, k)in S$.
answered Mar 26 at 10:22
5xum5xum
92.9k395162
$endgroup$
add a comment |
$begingroup$
All you have to prove is that the sets of indices the sums run over are the same. In fact, they are both equal to
$$S{(j, k)| 0leq j,lleq nland kleq j}$$
That is, you can prove that in both sums, the index pair $(j,k)$ appears in the sum if and only if $(j, k)in S$.
answered Mar 26 at 10:22
5xum5xum
92.9k395162
$endgroup$
add a comment |
$begingroup$
All you have to prove is that the sets of indices the sums run over are the same. In fact, they are both equal to
$$S{(j, k)| 0leq j,lleq nland kleq j}$$
That is, you can prove that in both sums, the index pair $(j,k)$ appears in the sum if and only if $(j, k)in S$.
answered Mar 26 at 10:22
5xum5xum
92.9k395162
$endgroup$
All you have to prove is that the sets of indices the sums run over are the same. In fact, they are both equal to
$$S{(j, k)| 0leq j,lleq nland kleq j}$$
That is, you can prove that in both sums, the index pair $(j,k)$ appears in the sum if and only if $(j, k)in S$.
answered Mar 26 at 10:22
5xum5xum
92.9k395162
answered Mar 26 at 10:22
5xum5xum
92.9k395162
answered Mar 26 at 10:22
5xum5xum
92.9k395162
answered Mar 26 at 10:22
5xum5xum
92.9k395162
92.9k395162
add a comment |
add a comment |
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