Solving equation with parts of a sum and a sum

**fysikbengt** · June 16th 2011, 11:38 AM

Is it possible to solve any of these equations and thus find $\Delta x_{n}$ or $\sum_{0}^{n}\Delta x_{n}$ ?

Define

$\Delta x_{0}=V\frac{l}{v_{0}}$

and

$\Delta x_{1}=V\frac{l-\Delta x_{0}}{v_{0}+V}$

and

$\Delta x_{n+1}=V\frac{l-\sum_{0}^{n}\Delta x_{n}}{v_{0}+nV}$ .

Since $V\ll v_{0}$ this can be simplified(?) to

$\Delta x_{n+1}-\Delta x_{n}=\frac{V \Delta x_{n}}{v_{0}+nV}$ .

Maybe this euation is easier to solve. I cannot get any further.

**ojones** · June 16th 2011, 05:58 PM

Recurrence relations can sometimes be solved using a $Z$ transform.

**Sudharaka** · June 16th 2011, 06:13 PM

Originally Posted by fysikbengt

Is it possible to solve any of these equations and thus find $\Delta x_{n}$ or $\sum_{0}^{n}\Delta x_{n}$ ?

Define

$\Delta x_{0}=V\frac{l}{v_{0}}$

and

$\Delta x_{1}=V\frac{l-\Delta x_{0}}{v_{0}+V}$

and

$\Delta x_{n+1}=V\frac{l-\sum_{0}^{n}\Delta x_{n}}{v_{0}+nV}$ .

Since $V\ll v_{0}$ this can be simplified(?) to

$\Delta x_{n+1}-\Delta x_{n}=\frac{V \Delta x_{n}}{v_{0}+nV}$ .

Maybe this euation is easier to solve. I cannot get any further.

Dear fysikbengt,

$\Delta x_{n+1}=V\left(\frac{l-\displaystyle\sum_{0}^{n}\Delta x_{n}}{v_{0}+nV}\right)$

Therefore, $\Delta x_{n}=V\left(\frac{l-\displaystyle\sum_{0}^{n-1}\Delta x_{n}}{v_{0}+(n-1)V}\right)$

$\Delta x_{n+1}-\Delta x_{n}=V\left(\frac{l-\displaystyle\sum_{0}^{n}\Delta x_{n}}{v_{0}+nV}\right)-V\left(\frac{l-\displaystyle\sum_{0}^{n-1}\Delta x_{n}}{v_{0}+(n-1)V}\right)$

Since, $V<<v_0~;~$ $v_0+nV\approx{v_0+(n-1)V}$

$\Delta x_{n+1}-\Delta x_{n}=V\left(\frac{l-\displaystyle\sum_{0}^{n}\Delta x_{n}}{v_{0}+nV}\right)-V\left(\frac{l-\displaystyle\sum_{0}^{n-1}\Delta x_{n}}{v_{0}+nV}\right)=-\left(\frac{V \Delta x_{n}}{v_{0}+nV}\right)$

So I think you should have a the minus sign for your last expression.

**fysikbengt** · June 17th 2011, 11:07 AM

Originally Posted by Sudharaka

Dear fysikbengt,

$\Delta x_{n+1}=V\left(\frac{l-\displaystyle\sum_{0}^{n}\Delta x_{n}}{v_{0}+nV}\right)$

Therefore, $\Delta x_{n}=V\left(\frac{l-\displaystyle\sum_{0}^{n-1}\Delta x_{n}}{v_{0}+(n-1)V}\right)$

$\Delta x_{n+1}-\Delta x_{n}=V\left(\frac{l-\displaystyle\sum_{0}^{n}\Delta x_{n}}{v_{0}+nV}\right)-V\left(\frac{l-\displaystyle\sum_{0}^{n-1}\Delta x_{n}}{v_{0}+(n-1)V}\right)$

Since, $V<<v_0~;~$ $v_0+nV\approx{v_0+(n-1)V}$

$\Delta x_{n+1}-\Delta x_{n}=V\left(\frac{l-\displaystyle\sum_{0}^{n}\Delta x_{n}}{v_{0}+nV}\right)-V\left(\frac{l-\displaystyle\sum_{0}^{n-1}\Delta x_{n}}{v_{0}+nV}\right)=-\left(\frac{V \Delta x_{n}}{v_{0}+nV}\right)$

So I think you should have a the minus sign for your last expression.

Yes, of course the relation is convergent. The minus sign is there in my notes, it is a typo. Thanks for noticing.

**fysikbengt** · June 17th 2011, 11:15 AM

Originally Posted by ojones

Recurrence relations can sometimes be solved using a $Z$ transform.

I tried to study $Z$ transforms and realised it is a full field not easily mastered. I have even forgot most of what I have read about Fourier transforms. So, I might just as well give up. This is not about laziness I hope.

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