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I've got a small recursive formula:

$X_k = (X_{k-1} * 1.4) - 230$

How can I convert this to a implicit formula?

Best regards

Jeel Shah
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Lukas
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2 Answers2

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The $-230$ term is the problem, so let's define $y_k=X_k+n$ and substitute it in, then choose $n$ to get rid of the constant term. We have $$y_k-n=1.4(y_{k-1}-n)-230\\y_k=1.4y_{k-1}-1.4n+n-230\\y_k=1.4y_{k-1}-230-0.4n$$

so if we set $n=-\frac {230}{0.4}=-575$ the constants will cancel. We have $y_k=1.4^ky_0$ or $$X_k=1.4^k(X_0-575)+575$$

Ross Millikan
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I think you are asking for an explicit expression for $x_k$. The generic equation is

$$x_k - a x_{k-1} = b$$

The solution is a combination of a homogeoneous solution $x_k^{(H)}$ and an inhomogeneous solution $x_k^{(I)}$. $x_k^{(H)}$ in this case is

$$x_k^{(H)} = C a^k$$

where $C$ is a constant determined by initial conditions. $x_k^{(I)}$ on the other hand may be chosen to be a linear function whose difference is constant:

$$x_k^{(I)} = A k +B$$

$$A k +B - a (A (k-1) + B) = A (1-a) k + B(1-a) + A a = b$$

from which we see that $A=0$ and $B = b/(1-a)$. Therefore, the general solution to the equation is

$$x_k = C a^k + \frac{b}{1-a}$$

In your case, $a=1.4$ and $b=-230$.

Ron Gordon
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  • Thank's, helped me a lot. But isn't there something missing in the general solution? The formular from Ross Millikan is a little bit different, but works how it should. – Lukas Feb 18 '13 at 16:33
  • Our results are equivalent. What do you think is missing? Ross's is in terms of an initial value, whereas I left mine open to whatever condition is needed. – Ron Gordon Feb 18 '13 at 16:55
  • I thought this is the diference. SO C is (initialvalue - the b / (1 - a) stuff? Thank's for your answer anyway! It helped me a lot! – Lukas Feb 19 '13 at 09:08