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Clarification on alternative expression of sum

How to find the general solution of $(1+x^2)y''+2xy'-2y=0$. How to express by means of elementary functions?Multiplying two summations together exactly.Expressing $zinmathbb{C}[[w]]$ as a power series in $yinmathbb{C}[[z]]$.Expression by Power Series Expansion and Binomial ExpansionChanging summation in a power seriesExplicit expression of a given power seriesPairwise sum and divideCauchy -Schwarz for double summationI know my series converges; how do I get a closed-form expression for it?Evaluating Sum of $dfrac{i}{(-x)^i}$

1

0

$begingroup$

I am not familiar with alternative expression of sum shown below,

$ d_k=sum_{i+j+l=k}a_ib_jc_l $

How it does work?

for $k = 4 $ then,

$d_{4} = sum_{i+j+l=4}a_ib_jc_l = ...$

How do I express it in standard summation format namely $sum_{n=0}^{infty}$
This is from the formula to calculate the product of three summations (power series) or $sum_{i=0}^{infty} a_i x^i cdot sum_{j=0}^{infty} b_j x^j cdot sum_{l=0}^{infty} c_l x^l = sum_{k=0}^{infty} d_kx^k$

summation power-series

share|cite|improve this question

edited Mar 13 at 15:04

Aschoolar

asked Mar 12 at 22:03

AschoolarAschoolar

2131210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can't you solve $i+j+l=4$ ?
  $endgroup$
  – Yves Daoust
  Mar 12 at 22:07
  

  
  
  
  

add a comment | 

1

0

$begingroup$

I am not familiar with alternative expression of sum shown below,

$ d_k=sum_{i+j+l=k}a_ib_jc_l $

How it does work?

for $k = 4 $ then,

$d_{4} = sum_{i+j+l=4}a_ib_jc_l = ...$

How do I express it in standard summation format namely $sum_{n=0}^{infty}$
This is from the formula to calculate the product of three summations (power series) or $sum_{i=0}^{infty} a_i x^i cdot sum_{j=0}^{infty} b_j x^j cdot sum_{l=0}^{infty} c_l x^l = sum_{k=0}^{infty} d_kx^k$

summation power-series

share|cite|improve this question

edited Mar 13 at 15:04

Aschoolar

asked Mar 12 at 22:03

AschoolarAschoolar

2131210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can't you solve $i+j+l=4$ ?
  $endgroup$
  – Yves Daoust
  Mar 12 at 22:07
  

  
  
  
  

add a comment | 

$begingroup$

I am not familiar with alternative expression of sum shown below,

$ d_k=sum_{i+j+l=k}a_ib_jc_l $

How it does work?

for $k = 4 $ then,

$d_{4} = sum_{i+j+l=4}a_ib_jc_l = ...$

How do I express it in standard summation format namely $sum_{n=0}^{infty}$
This is from the formula to calculate the product of three summations (power series) or $sum_{i=0}^{infty} a_i x^i cdot sum_{j=0}^{infty} b_j x^j cdot sum_{l=0}^{infty} c_l x^l = sum_{k=0}^{infty} d_kx^k$

summation power-series

share|cite|improve this question

edited Mar 13 at 15:04

Aschoolar

asked Mar 12 at 22:03

AschoolarAschoolar

2131210

$endgroup$

share|cite|improve this question

edited Mar 13 at 15:04

Aschoolar

asked Mar 12 at 22:03

AschoolarAschoolar

2131210

share|cite|improve this question

edited Mar 13 at 15:04

Aschoolar

asked Mar 12 at 22:03

AschoolarAschoolar

2131210

asked Mar 12 at 22:03

AschoolarAschoolar

2131210

asked Mar 12 at 22:03

AschoolarAschoolar

2131210

3 Answers
3

active

oldest

votes

1

$begingroup$

First, it is assumed that the indices are non-negative.

Second step, it means that the sum is to be taken for the triples $(i,j,l)$ for which $i+j+l=k |0 le i,j,l$. These are the "weak" compositions of $k$ into three parts

So for e.g. $k=4$ you have to sum $a_0b_0c_4+cdots+a_0b_1c_3+cdots$

share|cite|improve this answer

answered Mar 12 at 22:18

G CabG Cab

20.4k31341

$endgroup$

add a comment | 

1

$begingroup$

You can get
explicit indices like this
(assuming that
the lower index of summation
is $0$):

$begin{array}\
d_k
&=sum_{i+j+l=k}a_ib_jc_l\
&=sum_{i=0}^{k}
a_i sum_{j=0}^{k-i}b_jc_{k-i-j}\
end{array}
$

If the summation
is over an inequality,
you get one more level
of summation:

$begin{array}\
e_k
&=sum_{0 le i+j+l le k}a_ib_jc_l\
&=sum_{i=0}^{k}
a_i sum_{j=0}^{k-i}b_jsum_{l=0}^{k-i-j}c_{l}\
end{array}
$

Work out for yourself
what these are
if the lower index of summation
is $1$ instead of $0$.

share|cite|improve this answer

answered Mar 12 at 22:56

marty cohenmarty cohen

74.4k549129

$endgroup$

add a comment | 

0

$begingroup$

The representation of $d_k$ as triple sum $sum_{i+j+l=k}a_ib_jc_l$ can be derived by applying Cauchy Series multiplication twice.

We obtain
begin{align*}
left(sum_{i=0}^inftyright.&left. a_i x^iright)left(sum_{j=0}^infty b_j x^jright)left(sum_{j=0}^infty b_j x^jright)\
&=left(sum_{n=0}^inftyleft(sum_{{i+j=n}atop{i,jgeq 0}}a_ib_jright)x^nright)left(sum_{j=0}^infty b_j x^jright)tag{1}\
&=sum_{k=0}^inftyleft(sum_{{n+l=k}atop{n,lgeq 0}}left(sum_{{i+j=n}atop{i,jgeq 0}}right)c_lright)x^k\
&=sum_{k=0}^inftyleft(color{blue}{sum_{{i+j+l=k}atop{i,j,l}}a_ib_jc_l}right)x^ktag{2}\
end{align*}

From (1) we also obtain

begin{align*}
left(sum_{n=0}^inftyright.&left.left(sum_{{i+j=n}atop{i,jgeq 0}}a_ib_jright)x^nright)left(sum_{j=0}^infty b_j x^jright)\
&=left(sum_{n=0}^inftyleft(sum_{i=0}^na_ib_{n-i}right)x^nright)left(sum_{l=0}^infty c_lx^lright)\
&=sum_{k=0}^inftyleft(sum_{{n+l=k}atop{n,lgeq 0}}left(sum_{i=0}^na_ib_{n-i}right)c_lright)x^l\
&=sum_{k=0}^inftyleft(color{blue}{sum_{n=0}^ksum_{i=0}^na_ib_{n-i}c_{k-n}}right)x^ktag{3}
end{align*}
Comparing coefficients of equal powers of $x$ in (2) and (3) shows equality of the sums.

share|cite|improve this answer

edited Mar 15 at 7:08

answered Mar 14 at 22:08

Markus ScheuerMarkus Scheuer

62.7k460150

$endgroup$

add a comment | 

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1

$begingroup$

First, it is assumed that the indices are non-negative.

Second step, it means that the sum is to be taken for the triples $(i,j,l)$ for which $i+j+l=k |0 le i,j,l$. These are the "weak" compositions of $k$ into three parts

So for e.g. $k=4$ you have to sum $a_0b_0c_4+cdots+a_0b_1c_3+cdots$

share|cite|improve this answer

answered Mar 12 at 22:18

G CabG Cab

20.4k31341

$endgroup$

add a comment | 

1

$begingroup$

First, it is assumed that the indices are non-negative.

Second step, it means that the sum is to be taken for the triples $(i,j,l)$ for which $i+j+l=k |0 le i,j,l$. These are the "weak" compositions of $k$ into three parts

So for e.g. $k=4$ you have to sum $a_0b_0c_4+cdots+a_0b_1c_3+cdots$

share|cite|improve this answer

answered Mar 12 at 22:18

G CabG Cab

20.4k31341

$endgroup$

add a comment | 

$begingroup$

First, it is assumed that the indices are non-negative.

Second step, it means that the sum is to be taken for the triples $(i,j,l)$ for which $i+j+l=k |0 le i,j,l$. These are the "weak" compositions of $k$ into three parts

So for e.g. $k=4$ you have to sum $a_0b_0c_4+cdots+a_0b_1c_3+cdots$

share|cite|improve this answer

answered Mar 12 at 22:18

G CabG Cab

20.4k31341

$endgroup$

share|cite|improve this answer

answered Mar 12 at 22:18

G CabG Cab

20.4k31341

answered Mar 12 at 22:18

G CabG Cab

20.4k31341

answered Mar 12 at 22:18

G CabG Cab

20.4k31341

1

$begingroup$

You can get
explicit indices like this
(assuming that
the lower index of summation
is $0$):

$begin{array}\
d_k
&=sum_{i+j+l=k}a_ib_jc_l\
&=sum_{i=0}^{k}
a_i sum_{j=0}^{k-i}b_jc_{k-i-j}\
end{array}
$

If the summation
is over an inequality,
you get one more level
of summation:

$begin{array}\
e_k
&=sum_{0 le i+j+l le k}a_ib_jc_l\
&=sum_{i=0}^{k}
a_i sum_{j=0}^{k-i}b_jsum_{l=0}^{k-i-j}c_{l}\
end{array}
$

Work out for yourself
what these are
if the lower index of summation
is $1$ instead of $0$.

share|cite|improve this answer

answered Mar 12 at 22:56

marty cohenmarty cohen

74.4k549129

$endgroup$

add a comment | 

1

$begingroup$

You can get
explicit indices like this
(assuming that
the lower index of summation
is $0$):

$begin{array}\
d_k
&=sum_{i+j+l=k}a_ib_jc_l\
&=sum_{i=0}^{k}
a_i sum_{j=0}^{k-i}b_jc_{k-i-j}\
end{array}
$

If the summation
is over an inequality,
you get one more level
of summation:

$begin{array}\
e_k
&=sum_{0 le i+j+l le k}a_ib_jc_l\
&=sum_{i=0}^{k}
a_i sum_{j=0}^{k-i}b_jsum_{l=0}^{k-i-j}c_{l}\
end{array}
$

Work out for yourself
what these are
if the lower index of summation
is $1$ instead of $0$.

share|cite|improve this answer

answered Mar 12 at 22:56

marty cohenmarty cohen

74.4k549129

$endgroup$

add a comment | 

$begingroup$

You can get
explicit indices like this
(assuming that
the lower index of summation
is $0$):

$begin{array}\
d_k
&=sum_{i+j+l=k}a_ib_jc_l\
&=sum_{i=0}^{k}
a_i sum_{j=0}^{k-i}b_jc_{k-i-j}\
end{array}
$

If the summation
is over an inequality,
you get one more level
of summation:

$begin{array}\
e_k
&=sum_{0 le i+j+l le k}a_ib_jc_l\
&=sum_{i=0}^{k}
a_i sum_{j=0}^{k-i}b_jsum_{l=0}^{k-i-j}c_{l}\
end{array}
$

Work out for yourself
what these are
if the lower index of summation
is $1$ instead of $0$.

share|cite|improve this answer

answered Mar 12 at 22:56

marty cohenmarty cohen

74.4k549129

$endgroup$

share|cite|improve this answer

answered Mar 12 at 22:56

marty cohenmarty cohen

74.4k549129

answered Mar 12 at 22:56

marty cohenmarty cohen

74.4k549129

answered Mar 12 at 22:56

marty cohenmarty cohen

74.4k549129

0

$begingroup$

The representation of $d_k$ as triple sum $sum_{i+j+l=k}a_ib_jc_l$ can be derived by applying Cauchy Series multiplication twice.

We obtain
begin{align*}
left(sum_{i=0}^inftyright.&left. a_i x^iright)left(sum_{j=0}^infty b_j x^jright)left(sum_{j=0}^infty b_j x^jright)\
&=left(sum_{n=0}^inftyleft(sum_{{i+j=n}atop{i,jgeq 0}}a_ib_jright)x^nright)left(sum_{j=0}^infty b_j x^jright)tag{1}\
&=sum_{k=0}^inftyleft(sum_{{n+l=k}atop{n,lgeq 0}}left(sum_{{i+j=n}atop{i,jgeq 0}}right)c_lright)x^k\
&=sum_{k=0}^inftyleft(color{blue}{sum_{{i+j+l=k}atop{i,j,l}}a_ib_jc_l}right)x^ktag{2}\
end{align*}

From (1) we also obtain

begin{align*}
left(sum_{n=0}^inftyright.&left.left(sum_{{i+j=n}atop{i,jgeq 0}}a_ib_jright)x^nright)left(sum_{j=0}^infty b_j x^jright)\
&=left(sum_{n=0}^inftyleft(sum_{i=0}^na_ib_{n-i}right)x^nright)left(sum_{l=0}^infty c_lx^lright)\
&=sum_{k=0}^inftyleft(sum_{{n+l=k}atop{n,lgeq 0}}left(sum_{i=0}^na_ib_{n-i}right)c_lright)x^l\
&=sum_{k=0}^inftyleft(color{blue}{sum_{n=0}^ksum_{i=0}^na_ib_{n-i}c_{k-n}}right)x^ktag{3}
end{align*}
Comparing coefficients of equal powers of $x$ in (2) and (3) shows equality of the sums.

share|cite|improve this answer

edited Mar 15 at 7:08

answered Mar 14 at 22:08

Markus ScheuerMarkus Scheuer

62.7k460150

$endgroup$

add a comment | 

0

$begingroup$

The representation of $d_k$ as triple sum $sum_{i+j+l=k}a_ib_jc_l$ can be derived by applying Cauchy Series multiplication twice.

We obtain
begin{align*}
left(sum_{i=0}^inftyright.&left. a_i x^iright)left(sum_{j=0}^infty b_j x^jright)left(sum_{j=0}^infty b_j x^jright)\
&=left(sum_{n=0}^inftyleft(sum_{{i+j=n}atop{i,jgeq 0}}a_ib_jright)x^nright)left(sum_{j=0}^infty b_j x^jright)tag{1}\
&=sum_{k=0}^inftyleft(sum_{{n+l=k}atop{n,lgeq 0}}left(sum_{{i+j=n}atop{i,jgeq 0}}right)c_lright)x^k\
&=sum_{k=0}^inftyleft(color{blue}{sum_{{i+j+l=k}atop{i,j,l}}a_ib_jc_l}right)x^ktag{2}\
end{align*}

From (1) we also obtain

begin{align*}
left(sum_{n=0}^inftyright.&left.left(sum_{{i+j=n}atop{i,jgeq 0}}a_ib_jright)x^nright)left(sum_{j=0}^infty b_j x^jright)\
&=left(sum_{n=0}^inftyleft(sum_{i=0}^na_ib_{n-i}right)x^nright)left(sum_{l=0}^infty c_lx^lright)\
&=sum_{k=0}^inftyleft(sum_{{n+l=k}atop{n,lgeq 0}}left(sum_{i=0}^na_ib_{n-i}right)c_lright)x^l\
&=sum_{k=0}^inftyleft(color{blue}{sum_{n=0}^ksum_{i=0}^na_ib_{n-i}c_{k-n}}right)x^ktag{3}
end{align*}
Comparing coefficients of equal powers of $x$ in (2) and (3) shows equality of the sums.

share|cite|improve this answer

edited Mar 15 at 7:08

answered Mar 14 at 22:08

Markus ScheuerMarkus Scheuer

62.7k460150

$endgroup$

add a comment | 

$begingroup$

The representation of $d_k$ as triple sum $sum_{i+j+l=k}a_ib_jc_l$ can be derived by applying Cauchy Series multiplication twice.

We obtain
begin{align*}
left(sum_{i=0}^inftyright.&left. a_i x^iright)left(sum_{j=0}^infty b_j x^jright)left(sum_{j=0}^infty b_j x^jright)\
&=left(sum_{n=0}^inftyleft(sum_{{i+j=n}atop{i,jgeq 0}}a_ib_jright)x^nright)left(sum_{j=0}^infty b_j x^jright)tag{1}\
&=sum_{k=0}^inftyleft(sum_{{n+l=k}atop{n,lgeq 0}}left(sum_{{i+j=n}atop{i,jgeq 0}}right)c_lright)x^k\
&=sum_{k=0}^inftyleft(color{blue}{sum_{{i+j+l=k}atop{i,j,l}}a_ib_jc_l}right)x^ktag{2}\
end{align*}

From (1) we also obtain

begin{align*}
left(sum_{n=0}^inftyright.&left.left(sum_{{i+j=n}atop{i,jgeq 0}}a_ib_jright)x^nright)left(sum_{j=0}^infty b_j x^jright)\
&=left(sum_{n=0}^inftyleft(sum_{i=0}^na_ib_{n-i}right)x^nright)left(sum_{l=0}^infty c_lx^lright)\
&=sum_{k=0}^inftyleft(sum_{{n+l=k}atop{n,lgeq 0}}left(sum_{i=0}^na_ib_{n-i}right)c_lright)x^l\
&=sum_{k=0}^inftyleft(color{blue}{sum_{n=0}^ksum_{i=0}^na_ib_{n-i}c_{k-n}}right)x^ktag{3}
end{align*}
Comparing coefficients of equal powers of $x$ in (2) and (3) shows equality of the sums.

share|cite|improve this answer

edited Mar 15 at 7:08

answered Mar 14 at 22:08

Markus ScheuerMarkus Scheuer

62.7k460150

$endgroup$

share|cite|improve this answer

edited Mar 15 at 7:08

answered Mar 14 at 22:08

Markus ScheuerMarkus Scheuer

62.7k460150

answered Mar 14 at 22:08

Markus ScheuerMarkus Scheuer

62.7k460150

answered Mar 14 at 22:08

Markus ScheuerMarkus Scheuer

62.7k460150