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**soni85****Member**- Registered: 2014-02-19
- Posts: 6

hi, i have this equation

where,

I- identity matrix. (n*n)

Q-(n*n) matrix

so,

If u=2 and xp=1;

then,

(G(1) - (n*1) matrix ,depends on 'u'; It has 0.5 in u-1 and u+1 th position. (count starts from 0); P(xp) - (1*n) matrix ,depends on xp ; It has 0.5 in xp-1 and xp+1 th position.)

If xp < u then the E(xp) is constant, if xp > u then E(xp) is linearly decreasing. But is it possible to get a final simple equation with notations u , xp and n(matrix size)?

*Last edited by soni85 (2014-02-19 21:32:24)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,237

Hi;

Welcome to the forum.

What is E(xp)? Do you have an example?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,501

hi soni85

Welcome to the forum.

I- identity matrix. (n*n) AND Q-(n*n) matrix IMPLIES inverse is n*n

P-(1*n) matrix ...

G(1)-(1*n) matrix, ....

Did you mean that G is n*1 ?

Sizes: (1*n)(n*n)(1*n) ???

I assume that Q and G are fully known, and that P and X are not (other than the O.5s)

So

will result in a known matrix n*1 let's say R

The problem is then

If, say, n = 5, then these are too many variables to produce a unique solution.

And where did 'u' come from ?

I suspect I'm misunderstanding your problem. Would you please post an example with specific values to make this clearer ? Thanks.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**soni85****Member**- Registered: 2014-02-19
- Posts: 6

let say u=2 and xp=1;

then

Q will be like this,

G(1) depends on 'u'; It has 0.5 in u-1 and u+1 th position. (count starts from 0);

P(xp) depends on xp ; It has 0.5 in xp-1 and xp+1 th position.

*Last edited by soni85 (2014-02-20 00:03:22)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,501

Sorry, I'm still confused about this.

Please post the entire equation, not just a bit. And still 'u' has appeared out of nowhere ???

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**soni85****Member**- Registered: 2014-02-19
- Posts: 6

I've modified my question. Hope its clear now.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,501

Have a look at this:

http://www.mathsisfun.com/algebra/matri … lying.html

Particularly look at the section on what size of matrix may be multiplied by what size of matrix.

There is also a matrix calculator on MIF.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**soni85****Member**- Registered: 2014-02-19
- Posts: 6

sorry about that... i was confused while typing. I've changed it now.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,501

I've changed G to a **4** by 1 column matrix.

So we have

xp = 1 and u = 2, determining the position of the first 0.5 in the row or column, and

E will be a 1 by 1 matrix.

You want a formula for E that depends on u and xp. Is that right ?

if xp > u then E(xp) is linearly decreasing.

What does this mean? Surely E is determined as number whatever u and xp are ?

Bob

*Last edited by bob bundy (2014-02-20 01:06:09)*

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**soni85****Member**- Registered: 2014-02-19
- Posts: 6

Yes thats right.

If xp > u then E(xp) linearly decreases as xp increases.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,501

I am going to replace 'xp' by 'v' to make the notation easier to follow. And I will replace (I-Q)-¹ by a single matrix R

The value of v picks out the values of R from row v and from row v+2, adds them and multiplies by 0.5

The value of u picks out column u and column u+2, adds them and multiplies by 0.5.

Everything else is a zero, so just 4 elements of R are picked out.

The result is

where

is the element in the ith row and jth column of RI have no idea what to do when xp>u as you have not said the exact linear dependency rule.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**soni85****Member**- Registered: 2014-02-19
- Posts: 6

thanks for your effort. Yes finally E is based on four values. So if i change v i'll get a different E value for the same u. What i need is an another simpler equation for E, not with all these matrices. I know it from simulations that E=2(n-v) if v > u and E=2(n-u) if v < u. But is it possible to get these results from the equation given in the question? I saw a paper with similar problem. They've solved it using recurrence relations.

Title: "the random walk between a reflecting and an absorbing barrier". (I couldn't use the url)

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,501

hi soni85

I had a look at that link and couldn't see any connection with what you've been asking about. What are your simulations about?

Please go back to the beginning and describe what this is about and how you arrived at the matrix equation. There may be other members who will recognise this problem then.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,237

I saw a paper with similar problem. They've solved it using recurrence relations.

It does not look like your Q came from an absorbing chain. For one thing it is not stochastic. If you derived it from a random walk please provide the original problem.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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