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Can this be a two part function? Or does it have to be one function?
I do not understand what that means.
As far as I am concerned anything that will generate the answer.
If you're looking for the solutions in parentheses then you are looking for integer division. If you use the floor function then you will obtain what you are looking for. Have you seen the floor function before?
The floor function generates a rounded down number?
Yes, it basically truncates (gets rid of) the decimal.
I am attempting to determine if lets say there were 1072 cycles... what would b equal?
I see what you are saying. Hm.
I am not interested so much in the answer as to what B is as to how to go about solving this problem of generating the formula to determine any answer for B.
I can think of a recursive formula...
I would greatly appreciate that if you could show me how you generated it.
Since the result of each cycle is the input into the next one, the function can be defined recursively.
I am not certain how to compute that formula.
That is the problem with it -- it is recursive so there is no straightforward way to solve it. You would need to know the value of every cycle prior to the last cycle.
Is there a way to put that formula into a spreadsheet?
|dw:1356828418136:dw| This corrected version of the recursion yields correct results. I am not sure how to insert something like this into a spreadsheet. However, a programming routine could help. Unfortunately for larger and larger cycles, more recursion results in longer processing and more memory demand.
Here Ba is the (rounded off) value of B on the ath cycle. So, B_10 would give you the result of the 10th cycle.
I have to run but I will be back shortly. Can you show me a step by step solution for that formula? I am not certain how the mechanics of that work.
Let's say you want to know the 4th cycle. B4 = [H3 + B3 + 1]/10 H3 = H2 + B2 B3 = [H2 + B2 + 1]/10 H2 = H1 + B1 B3 = [H1 + B1 + B2 + 1]/10 H1 = H0 + B0 = 100 + 10 = 110 B0 = 10 Thus H2 = 110 + 10 = 120 B1 = [H0 + B0 + 1]/10 = 11 B2 = [H1 + B1 + 1]/10 = 12 So far we have H2 and B2, so we can solve for B3 B3 = [H2 + B2 + 1]/10 = 13 And H3 = H2 + B2 = 120 + 12 = 132 So for B4... B4 = [H3 + B3 + 1]/10 B4 = [(132 + 13)/10] = 14 It's a lengthy process and a computer program would be advised to calculate any further values this way.