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My Own Problem

  1. Jun 20, 2013 #1
    My friend was doing an analytical geometry problem and a shape appeared that I wanted to find the area of with my new knowledge of integrals. I found the area and i'm now working to find an equation for the nth area as the size of the shape changes for all integers. After doing the math I come to this sequence, (4/3), (32/3), (108/3). I stopped at the third solution because the math is a little time consuming and repetitive. Is there an equation that will represent the nth solution? Just hoping to get some help. Thanks.
     
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  3. Jun 20, 2013 #2

    tiny-tim

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    hi kevinnn! :smile:

    i've no idea what your area is,

    but if you divide your numbers by 4/3, you get 1, 8, 27 :wink:
     
  4. Jun 21, 2013 #3

    HallsofIvy

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    You should understand that, just because the first few terms of a sequence have a simple relation, it is not necessary that the rest of the sequence have that relation. The classical example is the "circle region" problem: Place n points around the circumference of a circle, NOT uniformly spaced, so that when you draw all lines connecting any two of those points, no more than two such lines intersect in a single point. How many sectors does that divide the interior of the circle into?

    n= 0; 1. n= 2; 2. n= 3; 4. n= 4; 8. n= 5; 16, n= 6; 32. n= 7, 63.
    The first 7 terms are [tex]2^{n-1}[/tex] but that fails for n> 6.
     
  5. Jun 22, 2013 #4
    Yes i'm aware of the fact that just because it appears that a formula for the nth term in a sequence can be found it may not always exist. Do you by any chance know if it can be shown that a nth term expression for a sequence exists or doesn't exist? Possibly mathematical induction? Any other method that a first semester calculus, going into second semester, could understand? Thanks.
     
  6. Jun 29, 2013 #5
    Got it.

    Stat plot on a graphing calculator almost always yields insight. Once you see the points as
    (1, 4/3) (2, 32/3) (3, 108/3)
    where (x,y), things get better.

    So basically, from the graph I got, I accidentally plotted the points as their inverses where (y,x) and saw that it was definitely some power function. Quick switch to (x,y) did a power regression on the calculator (although this usually only helps you get to a rough idea, this worked great) and got a pretty obvious answer that I was too lazy to see right off the bat.

    A=(4/3)(n^3)

    Should've been obvious... That matches the data points perfectly for this set, try two more calculations by hand to verify that this works moderately well, although you could calculate a point off in the distance by hand, like when n=30 to see if it still works for high n.
     
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