Solving equation with parts of a sum and a sum

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

Define

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

and

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

and

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

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

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

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

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

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Originally Posted by fysikbengt

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

Define

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

and

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

and

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

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

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

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

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Dear fysikbengt,

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

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

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

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

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

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

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Originally Posted by Sudharaka

Dear fysikbengt,

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

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

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

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

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

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

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

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Originally Posted by ojones

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

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I tried to study $\displaystyle Z$ 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.