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Consider the following recursion: $$C_{i+1} = a \sum_{j=1}^iC_j + b$$ where $a$ and $b$ are constants.

  1. Can series-element $C_i$ be expressed in terms of only its index $i$, $a$ and $b$?
  2. In case $C_1$ = $b$, does the answer change? simplified expression?

Thanks much.

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    Sounds like homework. Please format your question better. For example Ci+1 probably is supposed to be an element of a series and would be better written down as $C_{i+1}$. Are i and j really supposed to be different? I don't think so. What values is i supposed to take? Positive integers, $i \geq 0$? I also suspect you mean recursion instead of regression, which would make this not even a a statistical question. Please clear up your question.2012-05-09
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    Ignore the question about the difference between i and j. The pertinent question is what range of j you sum over. Unfortunately I could not edit the comment anymore. @Ken: I think your edit should have $C_{i+1}$ on the left hand and not $C_{i}+1$, but until the OP states his intentions I am hesistant to edit myself.2012-05-09
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    Forgot to add brackets around it, thanks.2012-05-09
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    Really Erik essentially did the whole thing for you. The final algebraic step yields Ci+1-Ci=aCi or Ci+1 = (a+1) Ci2012-05-09
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    Assume $a=0$, $C_1=0$, $b \ne 0$. $C_2=b$, not $(a+1)C_1$, which equals $0$.2012-05-09
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    To finish what Erik said you get Ci+1=(a+1)^(i-1) C2 = (a+1)^(i-1) (aC1+b)2012-05-09
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    With over a 1000 reputation points across the network, I would expect you to be familiar with the "Edit" function. Please edit your other answer to this question instead of this one.2012-05-09

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First, this is not really a statistical question. You seem to confuse regression and recursion. I assume your formula should be

$$ C_{i+1} = a \sum_{j=0}^{i}{C_j} + b $$

and i takes on integer values with $ i\geq 0$. Let's rewrite this as

\begin{equation} C_{i+1} = a \sum_{j=0}^{i-1}{C_j} + aC_i +b \end{equation}

Note that

$$ C_i = a \sum_{j=0}^{i-1}{C_j} + b. $$

Subtract b from both sides of that equation and use the result to eliminate the sum in the formula above for $C_{i+1}$. That's all the help I'm willing to give, unless you show that you worked on this on your own. But be very careful when arriving at $i=1$.

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    It's not homework, just a formula I need a better expression of for a certain application I'm writing at work. There have passed quite a lot of years since I graduated Uni, so I'm rusty on this. Thanks for all the help!2012-05-09
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    The question was not clear until the range of summation was put in. Also the equation could be stochastic. It could represent a times series with unknown parameters a and b. But apparently Erik guessed right and all you wanted was to know how the i=1st element in the series could be expressed recursively as a function of the ith element alone.2012-05-09
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    Almost... I need to know whether the $i$'th element could be expressed in terms of $i$ only.2012-05-09
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    @Erik, thanks for you tip: $$C_{i+1} = (a + 1) C_i$$ and hence $$C_{i+1} = (a + 1)^i C_1$$, right?2012-05-09
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    Define only in terms of i. Do you really want a formula only contain i and literal numbers? Or do you just want to eliminate the recursive part of the equation?2012-05-09
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    Your first equation is correct... The second is a bit tricky and depends on your range of i. Do notice that the way we wrote the formula for $C_{i+1}$ assumes that $C_i$ can be written as given by the recursion formula. This is not true if i+1 is the second possible value. Assuming you start with i=1 and a given start value $C_1$ you would get $C_2=aC_1+b$. Only afterwards the formula holds true, with $C_3=(a+1)C_2$.2012-05-09
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    Thanks. Can $C_i$ be generally expressed in terms of only $i$, $a$ and $b$?2012-05-09
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    Do the results differ when you start with a different value for $C_1$? If yes, then no.2012-05-09