Expanding $left(sum_r=1, rneq i^mT_rright)left(sum_s=1, sneq i^mT_s^-1right).$

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I'm pretty sure that this has been asked before but I can't find it anywhere. All my search results are clouded by GCSE revision on expanding brackets.

The Problem:

Fix $minBbb N$ and $iinoverline1, m$. Expand and simplify $$left(sum_r=1,\ rneq i^mT_rright)left(sum_s=1,\ sneq i^mT_s^-1right)$$ for non-zero complex numbers $(T_j)_jinoverline1, m$.

My Attempt:

We have

$$beginalign
left(sum_r=1,\ rneq i^mT_rright)left(sum_s=1,\ sneq i^mT_s^-1right)&=1+T_1T_2^-1+dots +T_1T_i-1^-1+0+T_1T_i+1^-1+dots +T_1T_m^-1 \
&+T_2T_1^-1+1+dots +T_2T_i-1^-1+0+T_2T_i+1^-1+dots +T_2T_m^-1 \
&+ \
&vdots \
&+T_i-1T_1^-1+dots +T_i-1T_i-2^-1+1+0+T_i-1T_i+1^-1+dots +T_i-1T_m^-1 \
&+T_i+1T_1^-1+dots +T_i+1T_i-1^-1+0+1+T_i+1T_i+2^-1+dots +T_i
+1T_m^-1 \
&+ \
&vdots \
&+T_mT_1^-1+dots +T_mT_i-1^-1+0+T_mT_i+1^-1+dots +1 \
&=(m-1)+X,
endalign$$

but I don't know what that $X$ should be.

I expect to see binomial coefficients in there.

Added Complication:

I would prefer not to relabel. This problem arose in my research, where it's important to keep track on the subscripts.

Please help :)

sequences-and-series

share|cite|improve this question

edited Jul 14 at 21:47

asked Jul 14 at 17:09

Shaun

7,41392972

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why the downvote?
  – Shaun
  Jul 14 at 19:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You may as well assume you have a sequence of $m-1$ nonzero complex numbers (just omit the $i$th one) and then forget about the extra restriction on the indices. Then write the product as $sum_r sum_s T_r/T_s$ and proceed from there.
  – Yakov Shklarov
  Jul 14 at 21:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @YakovShklarov Ah, see, that first thing - relabeling and omitting the $i$th term - won't work for me, because in what I'm doing it's important to keep track of the indices; I should have mentioned that in the above. I'm sorry. I'll add it now.
  – Shaun
  Jul 14 at 21:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @YakovShklarov Nevermind, I see what to do now.
  – Shaun
  Jul 14 at 21:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Not the indices, the subscripts.
  – Shaun
  Jul 14 at 22:54
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

I'm pretty sure that this has been asked before but I can't find it anywhere. All my search results are clouded by GCSE revision on expanding brackets.

The Problem:

Fix $minBbb N$ and $iinoverline1, m$. Expand and simplify $$left(sum_r=1,\ rneq i^mT_rright)left(sum_s=1,\ sneq i^mT_s^-1right)$$ for non-zero complex numbers $(T_j)_jinoverline1, m$.

My Attempt:

We have

$$beginalign
left(sum_r=1,\ rneq i^mT_rright)left(sum_s=1,\ sneq i^mT_s^-1right)&=1+T_1T_2^-1+dots +T_1T_i-1^-1+0+T_1T_i+1^-1+dots +T_1T_m^-1 \
&+T_2T_1^-1+1+dots +T_2T_i-1^-1+0+T_2T_i+1^-1+dots +T_2T_m^-1 \
&+ \
&vdots \
&+T_i-1T_1^-1+dots +T_i-1T_i-2^-1+1+0+T_i-1T_i+1^-1+dots +T_i-1T_m^-1 \
&+T_i+1T_1^-1+dots +T_i+1T_i-1^-1+0+1+T_i+1T_i+2^-1+dots +T_i
+1T_m^-1 \
&+ \
&vdots \
&+T_mT_1^-1+dots +T_mT_i-1^-1+0+T_mT_i+1^-1+dots +1 \
&=(m-1)+X,
endalign$$

but I don't know what that $X$ should be.

I expect to see binomial coefficients in there.

Added Complication:

I would prefer not to relabel. This problem arose in my research, where it's important to keep track on the subscripts.

Please help :)

sequences-and-series

share|cite|improve this question

edited Jul 14 at 21:47

asked Jul 14 at 17:09

Shaun

7,41392972

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why the downvote?
  – Shaun
  Jul 14 at 19:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You may as well assume you have a sequence of $m-1$ nonzero complex numbers (just omit the $i$th one) and then forget about the extra restriction on the indices. Then write the product as $sum_r sum_s T_r/T_s$ and proceed from there.
  – Yakov Shklarov
  Jul 14 at 21:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @YakovShklarov Ah, see, that first thing - relabeling and omitting the $i$th term - won't work for me, because in what I'm doing it's important to keep track of the indices; I should have mentioned that in the above. I'm sorry. I'll add it now.
  – Shaun
  Jul 14 at 21:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @YakovShklarov Nevermind, I see what to do now.
  – Shaun
  Jul 14 at 21:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Not the indices, the subscripts.
  – Shaun
  Jul 14 at 22:54
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I'm pretty sure that this has been asked before but I can't find it anywhere. All my search results are clouded by GCSE revision on expanding brackets.

The Problem:

Fix $minBbb N$ and $iinoverline1, m$. Expand and simplify $$left(sum_r=1,\ rneq i^mT_rright)left(sum_s=1,\ sneq i^mT_s^-1right)$$ for non-zero complex numbers $(T_j)_jinoverline1, m$.

My Attempt:

We have

$$beginalign
left(sum_r=1,\ rneq i^mT_rright)left(sum_s=1,\ sneq i^mT_s^-1right)&=1+T_1T_2^-1+dots +T_1T_i-1^-1+0+T_1T_i+1^-1+dots +T_1T_m^-1 \
&+T_2T_1^-1+1+dots +T_2T_i-1^-1+0+T_2T_i+1^-1+dots +T_2T_m^-1 \
&+ \
&vdots \
&+T_i-1T_1^-1+dots +T_i-1T_i-2^-1+1+0+T_i-1T_i+1^-1+dots +T_i-1T_m^-1 \
&+T_i+1T_1^-1+dots +T_i+1T_i-1^-1+0+1+T_i+1T_i+2^-1+dots +T_i
+1T_m^-1 \
&+ \
&vdots \
&+T_mT_1^-1+dots +T_mT_i-1^-1+0+T_mT_i+1^-1+dots +1 \
&=(m-1)+X,
endalign$$

but I don't know what that $X$ should be.

I expect to see binomial coefficients in there.

Added Complication:

I would prefer not to relabel. This problem arose in my research, where it's important to keep track on the subscripts.

Please help :)

sequences-and-series

share|cite|improve this question

edited Jul 14 at 21:47

asked Jul 14 at 17:09

Shaun

7,41392972

I'm pretty sure that this has been asked before but I can't find it anywhere. All my search results are clouded by GCSE revision on expanding brackets.

The Problem:

Fix $minBbb N$ and $iinoverline1, m$. Expand and simplify $$left(sum_r=1,\ rneq i^mT_rright)left(sum_s=1,\ sneq i^mT_s^-1right)$$ for non-zero complex numbers $(T_j)_jinoverline1, m$.

My Attempt:

We have

$$beginalign
left(sum_r=1,\ rneq i^mT_rright)left(sum_s=1,\ sneq i^mT_s^-1right)&=1+T_1T_2^-1+dots +T_1T_i-1^-1+0+T_1T_i+1^-1+dots +T_1T_m^-1 \
&+T_2T_1^-1+1+dots +T_2T_i-1^-1+0+T_2T_i+1^-1+dots +T_2T_m^-1 \
&+ \
&vdots \
&+T_i-1T_1^-1+dots +T_i-1T_i-2^-1+1+0+T_i-1T_i+1^-1+dots +T_i-1T_m^-1 \
&+T_i+1T_1^-1+dots +T_i+1T_i-1^-1+0+1+T_i+1T_i+2^-1+dots +T_i
+1T_m^-1 \
&+ \
&vdots \
&+T_mT_1^-1+dots +T_mT_i-1^-1+0+T_mT_i+1^-1+dots +1 \
&=(m-1)+X,
endalign$$

but I don't know what that $X$ should be.

I expect to see binomial coefficients in there.

Added Complication:

I would prefer not to relabel. This problem arose in my research, where it's important to keep track on the subscripts.

Please help :)

sequences-and-series

share|cite|improve this question

edited Jul 14 at 21:47

asked Jul 14 at 17:09

Shaun

7,41392972

share|cite|improve this question

share|cite|improve this question

edited Jul 14 at 21:47

edited Jul 14 at 21:47

edited Jul 14 at 21:47

asked Jul 14 at 17:09

Shaun

7,41392972

asked Jul 14 at 17:09

Shaun

7,41392972

asked Jul 14 at 17:09

Shaun

7,41392972

7,41392972

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why the downvote?
  – Shaun
  Jul 14 at 19:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You may as well assume you have a sequence of $m-1$ nonzero complex numbers (just omit the $i$th one) and then forget about the extra restriction on the indices. Then write the product as $sum_r sum_s T_r/T_s$ and proceed from there.
  – Yakov Shklarov
  Jul 14 at 21:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @YakovShklarov Ah, see, that first thing - relabeling and omitting the $i$th term - won't work for me, because in what I'm doing it's important to keep track of the indices; I should have mentioned that in the above. I'm sorry. I'll add it now.
  – Shaun
  Jul 14 at 21:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @YakovShklarov Nevermind, I see what to do now.
  – Shaun
  Jul 14 at 21:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Not the indices, the subscripts.
  – Shaun
  Jul 14 at 22:54
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why the downvote?
  – Shaun
  Jul 14 at 19:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You may as well assume you have a sequence of $m-1$ nonzero complex numbers (just omit the $i$th one) and then forget about the extra restriction on the indices. Then write the product as $sum_r sum_s T_r/T_s$ and proceed from there.
  – Yakov Shklarov
  Jul 14 at 21:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @YakovShklarov Ah, see, that first thing - relabeling and omitting the $i$th term - won't work for me, because in what I'm doing it's important to keep track of the indices; I should have mentioned that in the above. I'm sorry. I'll add it now.
  – Shaun
  Jul 14 at 21:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @YakovShklarov Nevermind, I see what to do now.
  – Shaun
  Jul 14 at 21:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Not the indices, the subscripts.
  – Shaun
  Jul 14 at 22:54
  

  
  
  
  

Why the downvote?
– Shaun
Jul 14 at 19:16

Why the downvote?
– Shaun
Jul 14 at 19:16

1

1

You may as well assume you have a sequence of $m-1$ nonzero complex numbers (just omit the $i$th one) and then forget about the extra restriction on the indices. Then write the product as $sum_r sum_s T_r/T_s$ and proceed from there.
– Yakov Shklarov
Jul 14 at 21:38

You may as well assume you have a sequence of $m-1$ nonzero complex numbers (just omit the $i$th one) and then forget about the extra restriction on the indices. Then write the product as $sum_r sum_s T_r/T_s$ and proceed from there.
– Yakov Shklarov
Jul 14 at 21:38

@YakovShklarov Ah, see, that first thing - relabeling and omitting the $i$th term - won't work for me, because in what I'm doing it's important to keep track of the indices; I should have mentioned that in the above. I'm sorry. I'll add it now.
– Shaun
Jul 14 at 21:44

@YakovShklarov Ah, see, that first thing - relabeling and omitting the $i$th term - won't work for me, because in what I'm doing it's important to keep track of the indices; I should have mentioned that in the above. I'm sorry. I'll add it now.
– Shaun
Jul 14 at 21:44

@YakovShklarov Nevermind, I see what to do now.
– Shaun
Jul 14 at 21:50

@YakovShklarov Nevermind, I see what to do now.
– Shaun
Jul 14 at 21:50

Not the indices, the subscripts.
– Shaun
Jul 14 at 22:54

Not the indices, the subscripts.
– Shaun
Jul 14 at 22:54

add a comment | 

1 Answer
1

active

oldest

votes

up vote
0
down vote



accepted

It's much simpler than I thought:

$$X=sum_r=1,\ rneq i^msum_s=1,\ sneq i,\ sneq r^mfracT_rT_s.$$

share|cite|improve this answer

answered Jul 14 at 21:56

Shaun

7,41392972

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

It's much simpler than I thought:

$$X=sum_r=1,\ rneq i^msum_s=1,\ sneq i,\ sneq r^mfracT_rT_s.$$

share|cite|improve this answer

answered Jul 14 at 21:56

Shaun

7,41392972

add a comment | 

up vote
0
down vote



accepted

It's much simpler than I thought:

$$X=sum_r=1,\ rneq i^msum_s=1,\ sneq i,\ sneq r^mfracT_rT_s.$$

share|cite|improve this answer

answered Jul 14 at 21:56

Shaun

7,41392972

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

It's much simpler than I thought:

$$X=sum_r=1,\ rneq i^msum_s=1,\ sneq i,\ sneq r^mfracT_rT_s.$$

share|cite|improve this answer

answered Jul 14 at 21:56

Shaun

7,41392972

It's much simpler than I thought:

$$X=sum_r=1,\ rneq i^msum_s=1,\ sneq i,\ sneq r^mfracT_rT_s.$$

share|cite|improve this answer

answered Jul 14 at 21:56

Shaun

7,41392972

share|cite|improve this answer

share|cite|improve this answer

answered Jul 14 at 21:56

Shaun

7,41392972

answered Jul 14 at 21:56

Shaun

7,41392972

answered Jul 14 at 21:56

Shaun

7,41392972

7,41392972

add a comment | 

add a comment | 

 

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