I have a number sequence 0-100 that I am using in a formula. In the formula the sequenced numbers are squared and then multiplied by another number to get the answer. I want to reverse the order of the numbers in the sequence but need to figure out how to alter the formula to obtain the same basic results as prior to reversing the sequence. Seems like this should be possible but I have no idea how to do it. I'm not a math idiot but it isn't my field so simplicity in the explanation would be helpful.

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    $\begingroup$ More details would be appreciated. $\endgroup$ – Mike Pierce Nov 11 '14 at 23:45
  • $\begingroup$ Not sure what details you need. Essentially, the formula would be x squared (y) = variable number. X can be any number from 0 to 100. I am wanting to reverse x so 100 becomes 0, 99 becomes 1, etc., but after the reversal the new formula needs to be altered so the variable answer still remains the same. $\endgroup$ – Dale Nov 12 '14 at 0:24

So, the mapping $100 \to 0, 99 \to 1, 98 \to 2, \dotsc, 0 \to 100$ can be expressed as a linear function $f(x) = 100 - x$. But, if I understand you correctly, we aren't given the numbers $\{0, 1, 2, \dotsc, 100\}$, but instead given some constant mulitple of the squares of these numbers, so we have $\{a0^2, a1^2, \dotsc, a100^2\}$ for some constant $a$. So in our formula we have to undo this by dividing by $a$ and taking the square root. So our final formula would be $$ f(x) = 100 - \sqrt{\frac{x}{a}} $$

  • $\begingroup$ Thanks so much for trying to help me out - a wouldn't be a constant though. It could be any number. Let me give an example. Before reversing the sequencing one possibility would be (432) 99squared = 4,234,032. Or (10) 98.7squared = 97,416.9. It is the 4,234,032 and 97,416.9 that I still want to be able to derive after reversing the numbers. I am not sure if this changes your approach or not. $\endgroup$ – Dale Nov 12 '14 at 2:33
  • $\begingroup$ Then it just sounds like you need to un-reverse the numbers before you square them and multiply by the thing. So instead of taking $432*(99)^2$, take $432*(100-99)^2$. $\endgroup$ – Mike Pierce Nov 12 '14 at 2:54

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