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Please explain to me how to convert this to closed form:

$$P_n = 1.75P_{n-1}, P_0=40.$$

And how to convert these into recursive form:

$$ \begin{align*} Y &= 5.4x-27, \\ Y &= 3800\cdot1.04^x. \end{align*} $$

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We will begin to calculate, to see what is going on.

We have $P_0=40$. Therefore, by the given recurrence, $P_1=(1.75)P_0=(1.75)(40)$.

Important: Don't plug $(1.75)(40)$ into a calculator! In fact, to prevent the temptation, I will write instead $P_1=(1.75)P_0$. We want to be always aware of the structure of our answer, and plugging numbers into a calculator tends to hide structure.

What is $P_2$? by the recurrence, it is $(1.75)P_1$. But $P_1=(1.75)P_0$, so $P_2=(1.75)(1.75)P_0$. For convenience write instead $P_2=(1.75)^2P_0$.

What is $P_3$? By the recurrence, $P_3=(1.75)P_2$. But from our expression for $P_2$, we get $P_3=(1.75)(1.75)^2P_0$, which is $(1.75)^3P_0$.

Maybe compute a bit further. Or not. It is I think clear that the same reasoning will give $P_4=(1.75)^4P_0$, and in general $P_n=(1.75)^nP_0$ for any non-negative integer $n$. So we conclude that $$P_n=((1.75)^n)(40).$$

Now let's look at your $Y=5.4x -27$. Presumably the intent is that $x$ is a non-negative integer. In order to bring out the analogies with the first problem, I will write $n$ instead of $x$. So $Y=5.4n-27$. Note that $Y$ depends on $n$. Bring out this dependence by writing instead $$Y_n=5.4n -27.$$

Note that $Y_0=-27$. What is $Y_1$? It is $5.4$ more than $Y_0$. So $Y_1=5.4+Y_0$.

What is $Y_2$? From the formula, it is $(2)(5.4)-27$. Note that this is $5.4$ more than $Y_1$. And what is $Y_3$? It is $(3)(5.4)-27$, which is $5.4$ more than $Y_2$. It is clear that $Y_4$ is $5.4$ more than $Y_3$, and that in general $Y_n$ is $5.4$ more than $Y_{n-1}$.

We reach the conclusion that $Y_n=Y_{n-1}+5.4$, with $Y_0=-27$.

We could proceed slightly more abstractly. Note that $Y_n=5.4n-27$ and that $Y_{n-1}=5.4(n-1)-27$. Subtract. After a bit of algebra, we get $Y_n-Y_{n-1}=5.4$, or equivalently $Y_n=Y_{n-1}+5.4$.

Finally, we look at the equation that you wrote as $$Y=3800(1.04)^x$$

Again, I imagine $x$ is supposed to be a non-negative integer. I will keep the $x$, though I am tempted to replace it by $n$. But definitely, "$Y$" is not good, we should write it either as $Y_x$ or $Y(x)$, to make explicit the dependence on $x$.

So we have $$Y_x=3800(1.04)^x$$ Now look at my discussion of the first question you had. What was the answer to that? And what was the question? In our current problem, the question looks like the answer to the first problem we dealt with.

Or more directly, we have from the given equation that $Y_0=3800$. Can you see that to get the "next" $Y$, you multiply the current $Y$ by $1.04$? So $Y_x=(1.04)Y_{x-1}$.

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(Just pointing out two typos, but you're missing a $ about halfway through and you have a comma pretending to be a decimal at the very end.) – El'endia Starman May 31 '11 at 22:08
@El'endia Starman: Thank you for pointing out the typos. Hope I have fixed them. I am typo-prone. This is aggravated by the fact that many of my posts are too long. – André Nicolas May 31 '11 at 23:45
Yep, you got them. – El'endia Starman Jun 1 '11 at 0:03

For the first part: write out the first $P_i$s without evaluating them, and see what you get.

For the second part: see how to $Y$ changes from $x = z$ to $x = z + 1$, and turn your findings into a "recursive form".

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