-1
$\begingroup$

So you have the following experiment: Experiment

Could you find a function or sum for me to be able to calculate it for each step?

I have thought about something like this (b and a are the pipettes, x and y are the containers): b and a are the pipettes, x and y are the containers

Im not a math guy so this is propably wrong but it would be nice if you helped me out.

Thanks in advance for any responses!

  • 0
    Pst, to...redeem myself, does solving the recursive equation count as part of your question? – 2017-01-15
  • 0
    Not sure if it will help me, but I´d appreciate it. Also, it doesn´t matter that you misinterpreted my qustion. Mistakes happen to everyone. The good intention is what counts! – 2017-01-15

2 Answers 2

0

Let's go together: For $x_0$ and $y_0$ which are initial values of $A$ and $B$, respectively. in first exchange we get \begin{cases} x_1=x_0+\dfrac{1}{10}y_0-\dfrac{1}{10}x_0=\dfrac{9}{10}x_0+\dfrac{1}{10}y_0,\\ y_1=y_0+\dfrac{1}{10}x_0-\dfrac{1}{10}y_0=\dfrac{9}{10}y_0+\dfrac{1}{10}x_0. \end{cases} next \begin{cases} x_2=\dfrac{9}{10}x_1+\dfrac{1}{10}y_1,\\ y_2=\dfrac{9}{10}y_1+\dfrac{1}{10}x_1. \end{cases} so we have following two sequences simultaneously: \begin{cases} x_{n+1}=\dfrac{9}{10}x_n+\dfrac{1}{10}y_n,\\ y_{n+1}=\dfrac{9}{10}y_n+\dfrac{1}{10}x_n. \end{cases}

  • 0
    @Nico Check it for some steps. – 2017-01-14
  • 0
    I have a question. Can one rewrite the eqation to be able to input it into a calculator and get a solution for a specific n value or a graph? – 2017-01-15
  • 0
    I have not studied anything about this type of equations but maybe some software solve this type of equations or ask about _sequences simultaneously_ in **math.stackexchange** with sequences and series tag. – 2017-01-15
  • 0
    Thanks for the quick response. I´ll ask about that! – 2017-01-15
0

We start with $1$ unit of water. After one drop, we have $100-10=90\%$ water left:

$$t_1=0.9$$

And again:

$$t_2=0.9t_1=0.9^2$$

And again...

$$t_n=0.9^n$$

  • 0
    No offense but It seems like you didin´t really read my Post. Check MyGlasses Answer, its right. – 2017-01-15
  • 0
    @Nico I'm sorry, but the above is what I interpreted to be the answer to what I thought was the question :-( – 2017-01-15