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Summation progressive and then regressive data.

  1. Nov 9, 2009 #1
    I just thought I would share this, I was about to ask you fine people how to do this when I realized the square root of the sum of progressive to regressive data equals the highest point.
    I.E.

    1+2+3+4+5+6+5+4+3+2+1=36
    6[tex]^{2}[/tex]=36
    and I tried this a few times and the results were the same.

    so then I began to think what if the pattern is not in intervals of 1.
    0+2+4+6+8+10+8+6+4+2+0=50
    which is 10[tex]^{2}[/tex]/2
    I JUMPED FOR JOY!
    so we can say

    the sum of all numbers that progress and then regress is equal to the point of regression squared divided by the average change.
    [tex]\frac{R^{2}_{p}}{\bar{\Delta}}[/tex]

    so... I am just working on this while I type now.
    If we make the numbers non-uniform...such as: 1+2+4+5+6+8+9+7+6+5+3+2+0 which is 1,2,1,2,1,2,1,2... instead of the normal 1,1,1,1 or 2,2,2,2,2..

    1+2+4+5+6+8+9+7+6+5+3+2+0=58

    I have been working on this for a while now...no luck..any help?
     
  2. jcsd
  3. Nov 9, 2009 #2

    jgens

    User Avatar
    Gold Member

    Neat! Before working on more complex cases, try understanding why [itex]1 + 2 + \dots + n + \dots + 2 + 1 = n^2[/itex]. This might help you figure out the answer to some of your questions. :wink:

    Hint: Try reorganizing the order that you sum the numbers and hopefully you'll find the connection.
     
  4. Nov 9, 2009 #3
    1+2+3+4+5+4+3+2+1
    The reason my method is so is because of my previous explained methods (as I have noticed you have read.)

    If you take the sequence
    [tex]\sum[/tex][tex]^{100}_{n=1}[/tex]n which is equal to 5050

    and explain it using my method
    you divide a square along it's diagonal and then find the area of one of the triangle and shade in that region.
    Then you find the midpoint of the unshaded triangles leg and the the mid-point of it's hypotenuses and do this again with the smaller triangle. Find the area of the smallest triangle.

    The area of the smallest triangle + the area of the largest triangles = the first half of the progressive to regressive summation

    1+2+3+4...+99+100.

    The other half of the summation comes in with the still unshaded regions of the square.
    shaded-unshaded= the summation of 1 to 1 less than the point of regression (I.E. 100)

    therefor the unshaded region accounts for
    +99+98+97.....+3+2+1+0


    so the shaded plus unshaded equals the sum of the progressive to regressive pattern.

    and unshaded plus shaded equals the whole area of the square which sides = the point of regression. therefor
    the summation equals the point of regression squared...


    please correct me if I am wrong :D I most surely am somewere ( WILL U HELP NOW? :D)
     
    Last edited: Nov 9, 2009
  5. Nov 11, 2009 #4
    Hi!

    First of all, you said that the sequence of the numbers in the sum 1+2+4+5+6+8+9+7+6+5+3+2+0 differed by alternating 1's and 2's. However, it differs by 1, 2, 1, 1, 2, 1, (-)2, (-)1, (-)1, (-)2, (-)1, (-)2; you have adjacent 1's in two places (4+5+6 and 7+6+5). If I understand you correctly, I think you are looking for the series:

    1+2+4+5+7+8+8+7+5+4+2+1=54

    (1, 2, 3, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5, 4, 3, 2, 1)

    This is easier to compute than your previous randomized 58. My solution to this problem would be to divide it up into cases with a maximum number n (in this case, 9 would be the maximum number rather than 8). This number n would have to fall in one of the following cases.

    Case I.

    [tex] n=3p\equiv 0 \bmod {3} [/tex]

    Case II.

    [tex] n=3p+1\equiv 1 \bmod {3} [/tex]

    Case III.

    [tex] n=3p+2\equiv 2 \bmod {3} [/tex]

    I divide it up into these 3 cases because they each have different results, as shown below. The reason I knew they would, though, is because the increments of 1 and 2 repeat every 3 (=1+2) numbers n. I will now use p rather than n to simplify the calculations.

    Case I.

    As in the example with n=9, the numbers are symmetric about n, so you could rewrite this as:

    (1) [tex]1+2+4+5+7+...(3p-2)+(3p-1)+(3p-1)+(3p-2)+...+7+5+4+2+1[/tex]

    (2) [tex]2(1+2+4+5+7+...+(3p-2)+(3p-1))[/tex]

    (3) [tex]2(1+4+7+...+(3p-2))+2(2+5+8+...+(3p-1))[/tex]

    (4) [tex]2\sum_{k=1}^{p}{(3k-2)}+2\sum_{k=1}^{p}{(3k-1)}[/tex]

    (5) [tex]2(\frac{3p(p+1)}{2}-2p)+2(\frac{3p(p+1)}{2}-p)[/tex]

    (6) [tex]2(\frac{p(3p-1)}{2})+2(\frac{p(3p+1)}{2})[/tex]

    (7) [tex]3p^2-p+3p^2+p[/tex]

    (8) [tex]6(\frac{n}{3})^2[/tex]

    (9) [tex]\frac{2n^2}{3}[/tex]

    Although there are quicker ways of solving this, I thought that this would be a good start to showing you how to do it. As you can see, it simplifies nicely. I numbered the lines of work in case you have any questions.

    Now you can solve the other two cases!
     
    Last edited: Nov 11, 2009
  6. Nov 11, 2009 #5
    Oh wow, thank you! this method is rather cumbersome compared to most that I have done, but I hope to learn from it :D
     
  7. Nov 11, 2009 #6
    Yes, it is more advanced, but it will help prepare you for even crazier ones, such as the Fibonacci Sequence! Another crazy one I managed to do was derive how many triangles there are in increasingly large triangles for a certain side length (f(n) where n was the side length of the equilateral triangle) in "IQ Triangles." For an example of what I mean, see http://www2.potsdam.edu/parksjm/Puzzle80.htm.

    It is crazier than others because of the increasing and decreasing mixed intervals. The point where it changes from increasing to decreasing is not very well defined, so that is why there need to be more than one cases, making it extremely cumbersome.
     
  8. Nov 11, 2009 #7
    what does the equal sign with a line above it mean? forgive me for my lack of knowledge xD lol
     
  9. Nov 11, 2009 #8
    Haha.. It's okay =P.

    It simply means "is equivalent to" and it is used with modulus arithmetic. It's simply syntax because if you write [tex]107=59 \bmod {3}[/tex] it looks as if you're saying [tex]107=59[/tex] which is obviously false. So, the "Ancient Math Gods" decided to write [tex] 107 \equiv 59 \bmod {3} [/tex] to clarify that. For more information on modulus arithmetic, look up Fermat's Little Theorem, which is often used in modern day cryptanalysis and cryptology (number theory).
     
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