[TIB] Re: help!! ti-89
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[TIB] Re: help!! ti-89
Tavis Segura wrote:
> Yikes!!! Forgive me if my rewrite offends you; all that indirection was
> driving me nuts. All I did was replace the indirection/index expressions
> with matrices, which would be a very good thing to learn if you are
> going to be working with data arranged in this manner. I haven't used
> the solver often, so I need to look it up if I am going to have an
> answer this weekend.
>
> For lack of a comment character, I'm using a double slash // in its
> place.. just replace them with the comment symbol (c) when you try to
> use it on your calc..
FYI, on the TI-89 and 92+ the comment character is the copyright symbol,
[DIAMOND][)]. I'd just use // or # here though, since the copyright
symbol is a pain to put in, and everyone recognizes // and #.
>
> Tavis
>
>>> Thank you all for answering my previous questions
>>> In the following programm I use the FDM (finite difference method) to
>>> calculate the voltage in a net of a particular shape.
>>> I don't know how to solve the system of linear equations
>>> If you any idea please tell me!
>>> Here is the code
>
> fdm()
> Prgm
> ClrIO
> Request "Mx=",mx,0
> Request "Nx=",nx,0
> Request "My=",my,0
> Request "Ny=",ny,0
> Request "internal V",v1,0
> Request "external V",v2,0
> expr(v1)»v1
> expr(v2)»v2
> expr(mx)»mx
> expr(nx)»nx
> expr(my)»my
> expr(ny)»ny
>
> // Note newMat(a,b) creates a matrix
> // with a rows and b columns.
> newMat(nx,ny)»f
> newMat(nx,my-1)»eqa
> newMat(mx-1,ny)»eqb
>
> // In place of #("f"&string(i)&"_"&string(j)), I used f[i,j].
> // Much easier to read, isn't it?
> // I don't know if I cleaned up everything, so you might want
> // to double-check it.
>
> For i,1,nx
> // Since j is always 1, there's no need to keep storing to j
> // Just put it in the index
> v1»f[i,1]
> EndFor
> For j,1,ny
> v1»f[1,j]
> EndFor
> For i,mx,nx
> v2»f[i,my]
> EndFor
> For j,my,ny
> v1»f[mx,j]
> EndFor
> For j,1,my
> f[nx-1,j]»f[nx+1,j]
> EndFor
> For i,1,mx
> f[i,ny-1]»f[i,ny+1]
> EndFor
> For i,2,nx
> For j,2,my-1
> (f[i+1,j]+f[i-1,j]+f[i,j+1]+f[i,j-1])/4=f[i,j]»eqa[i,j]
> Disp eqa[i,j]:Pause
> EndFor
> EndFor
> For i,2,mx-1
> For j,my,ny
> (f[i+1,j]+f[i-1,j]+f[i,j+1]+f[i,j-1])/4=f[i,j]»eqb[i,j]
> Disp eqb[i,j]:Pause
> EndFor
> EndFor
> eqa[2,2]»bool1
> For i,2,nx
> For j,2,my-1
> bool1 and eqa[i,j]»bool1
> EndFor
> EndFor
> eqb[2,my]»bool2
> For i,2,mx-1
> For j,my,ny
> bool2 and eqb[i,j]»bool2
> EndFor
> EndFor
> solve(bool1 and bool2,???????????)»sol
> disp sol
> pause
> EndPrgm
>
>>> The difficult part is with the question marks.
>>> Please note tha nx>mx and ny>my
>
>
> I know I haven't answered your question, but keep in mind, there is a
> lot of theory behind this, and I don't exactly know the system you just
> built, and don't use solve very often. I'll try to help you out, but I
> need to study this program more closely first.
>
>
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>
>
--
Robert Mohr
mohr.42@osu.edu
mohrr@cis.ohio-state.edu
References: