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Calculus Level 3

I gave three numbers $x \ne y_1 \ne z \not\in \{-1; 0; 1\}$ to a machine.

The machine was given to do the following program:

Step 1: Calculate $\large y_{n + 1} = \dfrac{zx - x + y_n}{z^2}$ .

Step 2: Compare $\ \large y_1$ and $\large \ y_{n + 1}$ .

If $\large \ y_1 = y_{n + 1}$ then the machine will end the program and tell the result. Otherwise, the machine will continue computing, repeating the sequence.

Will the machine end the program and tell the result with the right choice of numbers?

No, it won't. Yes, it will. Cannot be determined.

1 solution

Gabriel Teran
Feb 26, 2018

We can see it from the algorithmic point of view and a little bit of probability. Here we are giving an idea of a recursive function where the input values will be x, y (1), z and n. A very important contribution is to know that the values of x, y (1) and z are different and also do not belong to the set {-1,0,1}, which can help us to discard the option: "It can not be determined" and of entries of the type z = 1 and x = -1 where it can happen that y (1) = y (n + 1).

Continuing with the explanation, the function will start looking for the value of y (n + 1), when it bumps with y (n) it will call itself to look for this value but it will trip with y (n- 1) and it will be called again to find this value and if successively until it reaches y (2) where it meets y (1) which is a value that we have and will be replaced in the function, achieving a first value that is the of y (2) and later that of y (3) ... y (n + 1) and it is not going to stop before because it is very difficult to get the value of y (1) having values like xz -x y (all exprension) / z ^ 2 affecting the value of y (k). We can take as an example the factorial function where there is only one case where n! = 1! and it is n = 1, in the others this does not happen, then the same happens here is unlikely not to say impossible that y (n + 1) = y (1) knowing that x! = y (1)! = z and ! e {-1,0,1}

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