How to write relationship of b and c [closed]












-1












$begingroup$


Hi I am doing a research which involves matrices, and I wish to find the relation between $b$ and $c$. The nxn matrix is given as follows:
$$
begin{bmatrix}
1/2 & -c & 1/2 & 0 & 0 & 0 & ... & 0 \
0 & 1/2 & -c & 1/2 & 0 & 0 & ... & 0 \
0 & 0 & 1/2 & -c & 1/2 & 0 & ... & 0 \
...\
0 & 0 & 0 & 0 & ... & 0 & 1/2 & -c \
0 & 0 & 0 & 0 & 0 & ... & 0 & 1/2 \
end{bmatrix}
cdot
begin{bmatrix}
b_{1} \
b_{2} \
b_{3} \
...\
b_{N-2} \
b_{N-1} \
end{bmatrix}
=
begin{bmatrix}
a_{1} \
a_{2} \
a_{3} \
...\
a_{N-1} \
a_{N} \
end{bmatrix}$$



It is crucial for me to find the relation between $b$ and $c$. Can someone help me drop some hints on how I can achieve that. Sorry the matrix is not well written. It is suppose to be in the form of $Atilde{b}=tilde{a}$.






matrices matrix-equations







share|cite|improve this question











$endgroup$



closed as off-topic by Hans Lundmark, Lee David Chung Lin, Nosrati, José Carlos Santos, max_zorn Jan 20 at 22:46


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lee David Chung Lin, Nosrati, José Carlos Santos, max_zorn

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Should your matrix be begin{bmatrix} 1/2 & -c & 1/2 & 0 & 0 & 0 & ... & 0 \ 0 & 1/2 & -c & 1/2 & 0 & 0 & ... & 0 \ 0 & 0 & 1/2 & -c & 1/2 & 0 & ... & 0 \ ...\ 0 & 0 & 0 & 0 & ... & 0& 1/2 & -c \ 0 & 0 & 0 & 0 & 0 & ... & 0 & 1/2 \ end{bmatrix}?
    $endgroup$
    – Lord Shark the Unknown
    Jan 20 at 8:39










  • $begingroup$
    Thank you for pointing that out.
    $endgroup$
    – shahrina ismail
    Jan 20 at 23:27
















-1












$begingroup$


Hi I am doing a research which involves matrices, and I wish to find the relation between $b$ and $c$. The nxn matrix is given as follows:
$$
begin{bmatrix}
1/2 & -c & 1/2 & 0 & 0 & 0 & ... & 0 \
0 & 1/2 & -c & 1/2 & 0 & 0 & ... & 0 \
0 & 0 & 1/2 & -c & 1/2 & 0 & ... & 0 \
...\
0 & 0 & 0 & 0 & ... & 0 & 1/2 & -c \
0 & 0 & 0 & 0 & 0 & ... & 0 & 1/2 \
end{bmatrix}
cdot
begin{bmatrix}
b_{1} \
b_{2} \
b_{3} \
...\
b_{N-2} \
b_{N-1} \
end{bmatrix}
=
begin{bmatrix}
a_{1} \
a_{2} \
a_{3} \
...\
a_{N-1} \
a_{N} \
end{bmatrix}$$



It is crucial for me to find the relation between $b$ and $c$. Can someone help me drop some hints on how I can achieve that. Sorry the matrix is not well written. It is suppose to be in the form of $Atilde{b}=tilde{a}$.






matrices matrix-equations







share|cite|improve this question











$endgroup$



closed as off-topic by Hans Lundmark, Lee David Chung Lin, Nosrati, José Carlos Santos, max_zorn Jan 20 at 22:46


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lee David Chung Lin, Nosrati, José Carlos Santos, max_zorn

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Should your matrix be begin{bmatrix} 1/2 & -c & 1/2 & 0 & 0 & 0 & ... & 0 \ 0 & 1/2 & -c & 1/2 & 0 & 0 & ... & 0 \ 0 & 0 & 1/2 & -c & 1/2 & 0 & ... & 0 \ ...\ 0 & 0 & 0 & 0 & ... & 0& 1/2 & -c \ 0 & 0 & 0 & 0 & 0 & ... & 0 & 1/2 \ end{bmatrix}?
    $endgroup$
    – Lord Shark the Unknown
    Jan 20 at 8:39










  • $begingroup$
    Thank you for pointing that out.
    $endgroup$
    – shahrina ismail
    Jan 20 at 23:27














-1












-1








-1





$begingroup$


Hi I am doing a research which involves matrices, and I wish to find the relation between $b$ and $c$. The nxn matrix is given as follows:
$$
begin{bmatrix}
1/2 & -c & 1/2 & 0 & 0 & 0 & ... & 0 \
0 & 1/2 & -c & 1/2 & 0 & 0 & ... & 0 \
0 & 0 & 1/2 & -c & 1/2 & 0 & ... & 0 \
...\
0 & 0 & 0 & 0 & ... & 0 & 1/2 & -c \
0 & 0 & 0 & 0 & 0 & ... & 0 & 1/2 \
end{bmatrix}
cdot
begin{bmatrix}
b_{1} \
b_{2} \
b_{3} \
...\
b_{N-2} \
b_{N-1} \
end{bmatrix}
=
begin{bmatrix}
a_{1} \
a_{2} \
a_{3} \
...\
a_{N-1} \
a_{N} \
end{bmatrix}$$



It is crucial for me to find the relation between $b$ and $c$. Can someone help me drop some hints on how I can achieve that. Sorry the matrix is not well written. It is suppose to be in the form of $Atilde{b}=tilde{a}$.






matrices matrix-equations







share|cite|improve this question











$endgroup$




Hi I am doing a research which involves matrices, and I wish to find the relation between $b$ and $c$. The nxn matrix is given as follows:
$$
begin{bmatrix}
1/2 & -c & 1/2 & 0 & 0 & 0 & ... & 0 \
0 & 1/2 & -c & 1/2 & 0 & 0 & ... & 0 \
0 & 0 & 1/2 & -c & 1/2 & 0 & ... & 0 \
...\
0 & 0 & 0 & 0 & ... & 0 & 1/2 & -c \
0 & 0 & 0 & 0 & 0 & ... & 0 & 1/2 \
end{bmatrix}
cdot
begin{bmatrix}
b_{1} \
b_{2} \
b_{3} \
...\
b_{N-2} \
b_{N-1} \
end{bmatrix}
=
begin{bmatrix}
a_{1} \
a_{2} \
a_{3} \
...\
a_{N-1} \
a_{N} \
end{bmatrix}$$



It is crucial for me to find the relation between $b$ and $c$. Can someone help me drop some hints on how I can achieve that. Sorry the matrix is not well written. It is suppose to be in the form of $Atilde{b}=tilde{a}$.






matrices matrix-equations




matrices matrix-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 20 at 23:24







shahrina ismail

















asked Jan 20 at 8:37









shahrina ismailshahrina ismail

15519




15519




closed as off-topic by Hans Lundmark, Lee David Chung Lin, Nosrati, José Carlos Santos, max_zorn Jan 20 at 22:46


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lee David Chung Lin, Nosrati, José Carlos Santos, max_zorn

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Hans Lundmark, Lee David Chung Lin, Nosrati, José Carlos Santos, max_zorn Jan 20 at 22:46


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lee David Chung Lin, Nosrati, José Carlos Santos, max_zorn

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    Should your matrix be begin{bmatrix} 1/2 & -c & 1/2 & 0 & 0 & 0 & ... & 0 \ 0 & 1/2 & -c & 1/2 & 0 & 0 & ... & 0 \ 0 & 0 & 1/2 & -c & 1/2 & 0 & ... & 0 \ ...\ 0 & 0 & 0 & 0 & ... & 0& 1/2 & -c \ 0 & 0 & 0 & 0 & 0 & ... & 0 & 1/2 \ end{bmatrix}?
    $endgroup$
    – Lord Shark the Unknown
    Jan 20 at 8:39










  • $begingroup$
    Thank you for pointing that out.
    $endgroup$
    – shahrina ismail
    Jan 20 at 23:27


















  • $begingroup$
    Should your matrix be begin{bmatrix} 1/2 & -c & 1/2 & 0 & 0 & 0 & ... & 0 \ 0 & 1/2 & -c & 1/2 & 0 & 0 & ... & 0 \ 0 & 0 & 1/2 & -c & 1/2 & 0 & ... & 0 \ ...\ 0 & 0 & 0 & 0 & ... & 0& 1/2 & -c \ 0 & 0 & 0 & 0 & 0 & ... & 0 & 1/2 \ end{bmatrix}?
    $endgroup$
    – Lord Shark the Unknown
    Jan 20 at 8:39










  • $begingroup$
    Thank you for pointing that out.
    $endgroup$
    – shahrina ismail
    Jan 20 at 23:27
















$begingroup$
Should your matrix be begin{bmatrix} 1/2 & -c & 1/2 & 0 & 0 & 0 & ... & 0 \ 0 & 1/2 & -c & 1/2 & 0 & 0 & ... & 0 \ 0 & 0 & 1/2 & -c & 1/2 & 0 & ... & 0 \ ...\ 0 & 0 & 0 & 0 & ... & 0& 1/2 & -c \ 0 & 0 & 0 & 0 & 0 & ... & 0 & 1/2 \ end{bmatrix}?
$endgroup$
– Lord Shark the Unknown
Jan 20 at 8:39




$begingroup$
Should your matrix be begin{bmatrix} 1/2 & -c & 1/2 & 0 & 0 & 0 & ... & 0 \ 0 & 1/2 & -c & 1/2 & 0 & 0 & ... & 0 \ 0 & 0 & 1/2 & -c & 1/2 & 0 & ... & 0 \ ...\ 0 & 0 & 0 & 0 & ... & 0& 1/2 & -c \ 0 & 0 & 0 & 0 & 0 & ... & 0 & 1/2 \ end{bmatrix}?
$endgroup$
– Lord Shark the Unknown
Jan 20 at 8:39












$begingroup$
Thank you for pointing that out.
$endgroup$
– shahrina ismail
Jan 20 at 23:27




$begingroup$
Thank you for pointing that out.
$endgroup$
– shahrina ismail
Jan 20 at 23:27










1 Answer
1






active

oldest

votes


















1












$begingroup$

Believing Lord Shark's version, you can just do back substitution, starting at the bottom. You get
$$frac 12b_N=a_N\b_N=2a_N\frac 12b_{N-1}-cb_N=a_{N-1}\b_{N-1}=2a_{N-1}+2cb_N\
frac 12b_{N-2}-cb_{N-1}+frac 12b_N=a_{N-2}\
b_{N-2}=2(a_{N-2}+cb_{N-1})-b_N$$
and so on up the line






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But how does this helps me get the relation between b and c, because I still have all the a_N terms in there.
    $endgroup$
    – shahrina ismail
    Jan 20 at 23:27










  • $begingroup$
    It has to depend on the $a$'s. They are part of the equation. If all the $a$s are zero, so are all the $b$'s regardless of $c$.
    $endgroup$
    – Ross Millikan
    Jan 20 at 23:29


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Believing Lord Shark's version, you can just do back substitution, starting at the bottom. You get
$$frac 12b_N=a_N\b_N=2a_N\frac 12b_{N-1}-cb_N=a_{N-1}\b_{N-1}=2a_{N-1}+2cb_N\
frac 12b_{N-2}-cb_{N-1}+frac 12b_N=a_{N-2}\
b_{N-2}=2(a_{N-2}+cb_{N-1})-b_N$$
and so on up the line






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But how does this helps me get the relation between b and c, because I still have all the a_N terms in there.
    $endgroup$
    – shahrina ismail
    Jan 20 at 23:27










  • $begingroup$
    It has to depend on the $a$'s. They are part of the equation. If all the $a$s are zero, so are all the $b$'s regardless of $c$.
    $endgroup$
    – Ross Millikan
    Jan 20 at 23:29
















1












$begingroup$

Believing Lord Shark's version, you can just do back substitution, starting at the bottom. You get
$$frac 12b_N=a_N\b_N=2a_N\frac 12b_{N-1}-cb_N=a_{N-1}\b_{N-1}=2a_{N-1}+2cb_N\
frac 12b_{N-2}-cb_{N-1}+frac 12b_N=a_{N-2}\
b_{N-2}=2(a_{N-2}+cb_{N-1})-b_N$$
and so on up the line






share|cite|improve this answer









$endgroup$













  • $begingroup$
    But how does this helps me get the relation between b and c, because I still have all the a_N terms in there.
    $endgroup$
    – shahrina ismail
    Jan 20 at 23:27










  • $begingroup$
    It has to depend on the $a$'s. They are part of the equation. If all the $a$s are zero, so are all the $b$'s regardless of $c$.
    $endgroup$
    – Ross Millikan
    Jan 20 at 23:29














1












1








1





$begingroup$

Believing Lord Shark's version, you can just do back substitution, starting at the bottom. You get
$$frac 12b_N=a_N\b_N=2a_N\frac 12b_{N-1}-cb_N=a_{N-1}\b_{N-1}=2a_{N-1}+2cb_N\
frac 12b_{N-2}-cb_{N-1}+frac 12b_N=a_{N-2}\
b_{N-2}=2(a_{N-2}+cb_{N-1})-b_N$$
and so on up the line






share|cite|improve this answer









$endgroup$



Believing Lord Shark's version, you can just do back substitution, starting at the bottom. You get
$$frac 12b_N=a_N\b_N=2a_N\frac 12b_{N-1}-cb_N=a_{N-1}\b_{N-1}=2a_{N-1}+2cb_N\
frac 12b_{N-2}-cb_{N-1}+frac 12b_N=a_{N-2}\
b_{N-2}=2(a_{N-2}+cb_{N-1})-b_N$$
and so on up the line







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 20 at 21:52









Ross MillikanRoss Millikan

298k23198371




298k23198371












  • $begingroup$
    But how does this helps me get the relation between b and c, because I still have all the a_N terms in there.
    $endgroup$
    – shahrina ismail
    Jan 20 at 23:27










  • $begingroup$
    It has to depend on the $a$'s. They are part of the equation. If all the $a$s are zero, so are all the $b$'s regardless of $c$.
    $endgroup$
    – Ross Millikan
    Jan 20 at 23:29


















  • $begingroup$
    But how does this helps me get the relation between b and c, because I still have all the a_N terms in there.
    $endgroup$
    – shahrina ismail
    Jan 20 at 23:27










  • $begingroup$
    It has to depend on the $a$'s. They are part of the equation. If all the $a$s are zero, so are all the $b$'s regardless of $c$.
    $endgroup$
    – Ross Millikan
    Jan 20 at 23:29
















$begingroup$
But how does this helps me get the relation between b and c, because I still have all the a_N terms in there.
$endgroup$
– shahrina ismail
Jan 20 at 23:27




$begingroup$
But how does this helps me get the relation between b and c, because I still have all the a_N terms in there.
$endgroup$
– shahrina ismail
Jan 20 at 23:27












$begingroup$
It has to depend on the $a$'s. They are part of the equation. If all the $a$s are zero, so are all the $b$'s regardless of $c$.
$endgroup$
– Ross Millikan
Jan 20 at 23:29




$begingroup$
It has to depend on the $a$'s. They are part of the equation. If all the $a$s are zero, so are all the $b$'s regardless of $c$.
$endgroup$
– Ross Millikan
Jan 20 at 23:29



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