Mixing
Solve the following system of equations: $[x[i-1]-(2-h^2)x[i]+x[i+1]=0, x[1]=1, x[50]-x[49]=h$, where...
$begingroup$
This is my attempt at the solution using MuPAD:
$x[1]:=1; h:=0.1; x[50]-x[49]=h$
eqs:= $[x[i-1]-(2-h^2)x[i]+x[i+1]=0; i=2,...,49]$
vars:=$[x[i]; i = 1,...,50]$
numeric::linsolve(eqs,vars);
The problem is that $x[50]$ and $x[49]$ are not defined yet. What is the correct approach here? Do I solve the equation generally and then solve for every term $x[1,...,50]$ individually afterwards? Presumably the general solution will be an equation for each term $x[2,...,50]$. How do I pass the variables $h$ and $x[50]-x[49]$ to the general solution?
computer-algebra-systems
edited Feb 28 '18 at 18:44
Lorenzo B.
1,8402520
asked Feb 28 '18 at 18:18
JensJens
13
$endgroup$
add a comment |
$begingroup$
This is my attempt at the solution using MuPAD:
$x[1]:=1; h:=0.1; x[50]-x[49]=h$
eqs:= $[x[i-1]-(2-h^2)x[i]+x[i+1]=0; i=2,...,49]$
vars:=$[x[i]; i = 1,...,50]$
numeric::linsolve(eqs,vars);
The problem is that $x[50]$ and $x[49]$ are not defined yet. What is the correct approach here? Do I solve the equation generally and then solve for every term $x[1,...,50]$ individually afterwards? Presumably the general solution will be an equation for each term $x[2,...,50]$. How do I pass the variables $h$ and $x[50]-x[49]$ to the general solution?
computer-algebra-systems
edited Feb 28 '18 at 18:44
Lorenzo B.
1,8402520
asked Feb 28 '18 at 18:18
JensJens
13
$endgroup$
add a comment |
$begingroup$
This is my attempt at the solution using MuPAD:
$x[1]:=1; h:=0.1; x[50]-x[49]=h$
eqs:= $[x[i-1]-(2-h^2)x[i]+x[i+1]=0; i=2,...,49]$
vars:=$[x[i]; i = 1,...,50]$
numeric::linsolve(eqs,vars);
The problem is that $x[50]$ and $x[49]$ are not defined yet. What is the correct approach here? Do I solve the equation generally and then solve for every term $x[1,...,50]$ individually afterwards? Presumably the general solution will be an equation for each term $x[2,...,50]$. How do I pass the variables $h$ and $x[50]-x[49]$ to the general solution?
computer-algebra-systems
edited Feb 28 '18 at 18:44
Lorenzo B.
1,8402520
asked Feb 28 '18 at 18:18
JensJens
13
$endgroup$
This is my attempt at the solution using MuPAD:
$x[1]:=1; h:=0.1; x[50]-x[49]=h$
eqs:= $[x[i-1]-(2-h^2)x[i]+x[i+1]=0; i=2,...,49]$
vars:=$[x[i]; i = 1,...,50]$
numeric::linsolve(eqs,vars);
The problem is that $x[50]$ and $x[49]$ are not defined yet. What is the correct approach here? Do I solve the equation generally and then solve for every term $x[1,...,50]$ individually afterwards? Presumably the general solution will be an equation for each term $x[2,...,50]$. How do I pass the variables $h$ and $x[50]-x[49]$ to the general solution?
computer-algebra-systems
computer-algebra-systems
edited Feb 28 '18 at 18:44
Lorenzo B.
1,8402520
asked Feb 28 '18 at 18:18
JensJens
13
edited Feb 28 '18 at 18:44
Lorenzo B.
1,8402520
asked Feb 28 '18 at 18:18
JensJens
13
edited Feb 28 '18 at 18:44
Lorenzo B.
1,8402520
edited Feb 28 '18 at 18:44
Lorenzo B.
1,8402520
edited Feb 28 '18 at 18:44
Lorenzo B.
1,8402520
1,8402520
asked Feb 28 '18 at 18:18
JensJens
13
asked Feb 28 '18 at 18:18
JensJens
13
asked Feb 28 '18 at 18:18
JensJens
13
13
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
For a problem of this size you can simply generate the system of equations and solve it.
Here's the Mathematica code to do it:
h = 1/10;
eql[0] = Table[x[i - 1] - (2 - h^2) x[i] + x[i + 1] == 0, {i, 2, 49}];
eql[1] = {x[50] - x[49] == h, x[1] == 1}
Solve[eql[0]~Join~eql[1], Table[x[i], {i, 1, 50}], Reals]
answered Jan 12 at 19:23
ablmfablmf
2,54342452
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For a problem of this size you can simply generate the system of equations and solve it.
Here's the Mathematica code to do it:
h = 1/10;
eql[0] = Table[x[i - 1] - (2 - h^2) x[i] + x[i + 1] == 0, {i, 2, 49}];
eql[1] = {x[50] - x[49] == h, x[1] == 1}
Solve[eql[0]~Join~eql[1], Table[x[i], {i, 1, 50}], Reals]
answered Jan 12 at 19:23
ablmfablmf
2,54342452
$endgroup$
add a comment |
$begingroup$
For a problem of this size you can simply generate the system of equations and solve it.
Here's the Mathematica code to do it:
h = 1/10;
eql[0] = Table[x[i - 1] - (2 - h^2) x[i] + x[i + 1] == 0, {i, 2, 49}];
eql[1] = {x[50] - x[49] == h, x[1] == 1}
Solve[eql[0]~Join~eql[1], Table[x[i], {i, 1, 50}], Reals]
answered Jan 12 at 19:23
ablmfablmf
2,54342452
$endgroup$
add a comment |
$begingroup$
For a problem of this size you can simply generate the system of equations and solve it.
Here's the Mathematica code to do it:
h = 1/10;
eql[0] = Table[x[i - 1] - (2 - h^2) x[i] + x[i + 1] == 0, {i, 2, 49}];
eql[1] = {x[50] - x[49] == h, x[1] == 1}
Solve[eql[0]~Join~eql[1], Table[x[i], {i, 1, 50}], Reals]
answered Jan 12 at 19:23
ablmfablmf
2,54342452
$endgroup$
For a problem of this size you can simply generate the system of equations and solve it.
Here's the Mathematica code to do it:
h = 1/10;
eql[0] = Table[x[i - 1] - (2 - h^2) x[i] + x[i + 1] == 0, {i, 2, 49}];
eql[1] = {x[50] - x[49] == h, x[1] == 1}
Solve[eql[0]~Join~eql[1], Table[x[i], {i, 1, 50}], Reals]
answered Jan 12 at 19:23
ablmfablmf
2,54342452
answered Jan 12 at 19:23
ablmfablmf
2,54342452
answered Jan 12 at 19:23
ablmfablmf
2,54342452
answered Jan 12 at 19:23
ablmfablmf
2,54342452
2,54342452
add a comment |
add a comment |
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