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I Need An Equation

  1. Nov 8, 2003 #1
    I have the following pattern and i need an equation that if x = 1 then y will equal the fist number of it, if x = 2, then y will equal the 2nd number.

    The pattern is:
    2, 4, 16, 256, 65536

    Hint: It keeps squaring itself.

    Any ideas?
     
  2. jcsd
  3. Nov 8, 2003 #2

    NateTG

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    Homework Helper

    Try writing all of them as powers of 2:
    21,22,24,28,216

    It should be easy from there...
     
  4. Nov 8, 2003 #3
    But this doesnt help me much becasue it doesnt find my numbers i need in an equation...
     
  5. Nov 8, 2003 #4
    f(x) = 2^2x
     
    Last edited: Nov 8, 2003
  6. Nov 8, 2003 #5
    I had already thought of this but it does not work...Here is teh pattern it gives you.
    4,16,64,256
    This is the pattern I want...
    2, 4, 16, 256, 65536

    There is not supposed to be 64 and it also skips 2
     
  7. Nov 8, 2003 #6
    No, that gives 22, 24, 26, ...

    I don't think we should be giving explicit formulas anyway, since it sounds like homework.
     
  8. Nov 8, 2003 #7
    Qwerty, take NateTG's suggestion and first find a formula for just the exponents 1, 2, 4, 8, 16, ... from that you can derive a formula for the actual sequence.
     
  9. Nov 9, 2003 #8
    it turns out to be quite a function with an even nicer derivative
     
  10. Nov 9, 2003 #9
    Acually no, it's not for homework....its for a project im working on(programming).

    And I don't need your help anymore, I found the answer to my question myself.
    Answer: 2^(2^x)
     
  11. Nov 9, 2003 #10
    if you first put x=1 (as you requested) to this equation you get 2^(2^1)=4 which is not the first number 2 ofcourse.
    but if you put first x=0 then this is the equation you were looking for this pattern.
     
  12. Nov 9, 2003 #11
    Yes this is an error in my first post, 0 is supposed to be the first number subsituted into the equation.
     
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