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I am solving non linear recursive relations and I stumbled upon this: $$x(n)*x(n+1)=3,\ x(0)=3$$ If you start calculating the values after $0$ you will notice that a pattern emerges: $$x(0) = 3$$ $$x(1) = 1$$ $$x(2) = 3$$ $$x(3) = 1$$ $$...$$ From this I concluded that $x(n) = (-1)^n + 2$

BUT! If you try to solve the problem more conservatively you will convert the relation to a linear one: $log_{3} (x(n)) + log_{3}(x(n+1)) = 1$ $Set\ g(n)=log_{3}(x(n)),\ g(0) = log_{3}(x(0))=1$

$g(n+1) + g(n) = 1 \Rightarrow g(n)=\frac{1}{2}*(-1)^n + \frac{1}{2} \Rightarrow x(n) = 3^{\frac{1}{2}*((-1)^n + 1)} $

The fact that $3^{\frac{1}{2}*((-1)^n + 1)} = (-1)^n + 2$ can also be confirmed by graphing the two functions. My question is: is there a way to go from the one to the other instead of just proving that they are equal?

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You're not going to be able to manipulate one into the other because they're not the same function for non-integer input. For example, let $n = 1/2$ then

$$ (-1)^{1/2} + 2 = 2 + i $$

and

$$ 3^{\frac{(-1)^{1/2} + 1}2} = e^{\frac{1 + i}{2}\log 3} = 3^{1/2}[\cos(\tfrac{1}2\log 3) + i\sin(\tfrac{1}2\log 3)] \approx 1.47724 + 0.904296 i $$

If you don't know a lot about complex numbers you're going to just have to take my word for now.

Here's a simpler example:

$$ \cos(n \pi) = (-1)^n $$

for integer values of $n$. But $\cos(\frac12\pi) = 0$ and $(-1)^{1/2} = i$.

So if you have two functions that are equal for integer inputs but are unequal for non-integer inputs. At some point in "simplifying" you will want to use the fact that your inputs are integers. Here the property is that

$$ (-1)^n = \begin{cases} 1 & n \text{ is even} \\ -1 & n \text{ is odd} \end{cases}. $$

And because $(-1)^n$ is in some sense defined based on whether or not $n$ is even or odd, in order to simplify, you're going to want to separate your computation into two cases: when $n$ is even and when $n$ is odd.

Here when you separate the two cases it is easy to see that for even integers, both functions are equal to $3$ and for odd integers, equal to $1$. So they represent the same function on the integers.

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  • $\begingroup$ Thank you! Would graphing the functions and using the z axis as the complex plane reveal the trig function? am I only seeing the points are which cos goes through the RxR plane when I graph it as a real function? $\endgroup$ – Mario Drouga Jan 13 at 5:05
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    $\begingroup$ @Mario Possibly. Personally I find it easier to plot the real part and complex parts separately just in the xy-plane. For instance $(-1)^x = e^{x \log(-1)} = e^{\pi x i} = \cos(\pi x) + i \sin(\pi x)$. Converting $3^{\frac{(-1)^x + 1}2}$ to a real and imaginary part will be a lot messier. $\endgroup$ – Trevor Gunn Jan 13 at 5:29
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This is of the same type of situation as saying that $2$ and $1 + 1$ are the same. They are different expressions for the same value.

In your case, let $f\left(n\right) = \left(-1\right)^n + 2$, for $n \ge 0 \text{ and } n \in N$. For even values of $n$, the value is $1 + 2 = 3$ and for odd values of $n$, the value is $-1 + 3 = 1$.

Similarly, let $h\left(n\right) = 3^{\frac{\left(-1\right)^n + 1}{2}}$. For $n$ being even here, the value of $\left(-1\right)^n = 1$, so the function becomes $3^{\frac{\left(1 + 1 \right)}{2}} = 3^1 = 3$. With odd values of $n$, $\left(-1\right)^n = -1$, so the function becomes $3^{\frac{\left(-1 + 1\right)}{2}} = 3^0 = 1$.

As you can see, the $2$ functions always have the same value for all non-negative values of $n$. They are just different ways to express the set of results. However, I prefer $f\left(n\right) = \left(-1\right)^n + 2$ as, to me at least, it's simpler to understand and use.

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  • $\begingroup$ You've proven that both functions are equal. Can you manipulate the form of function h(n) to get function f(n)? $\endgroup$ – Mario Drouga Jan 13 at 4:13
  • $\begingroup$ @MarioDrouga I'm not quite sure what you mean by "manipulate". How would you, for example, propose to "manipulate" $1 + 1$ to show it's the same as $2$? $\endgroup$ – John Omielan Jan 13 at 4:17
  • $\begingroup$ What if you haven't arrived to the solution of f(n)? Would you be able to simplify g(n) enough to get you to f(n)? $\endgroup$ – Mario Drouga Jan 13 at 4:19
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    $\begingroup$ @MarioDrouga The most basic of "simplifications" are the numeric values themselves, just like $2$ is usually consider simpler than $1 + 1$. With the set of values you got, there are very many different ways to represent them, with you just using $2$. In general, don't worry about determining more than $1$ way to specify results, much less figuring out how to go directly from one to another. Instead, you would normally just try to find a relatively simple & straightforward way to represent your results & use this, with the understanding that it's not the one & only unique way to do it. $\endgroup$ – John Omielan Jan 13 at 4:22

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