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Convert to formula

  1. Aug 1, 2010 #1
    1. The problem statement, all variables and given/known data
    please convert the below(P) into formula using (m), or give an algorithm using (m) to get (P)

    m=2, P=0
    m=3, P=3
    m=4, P=3+5
    m=5, P=3+5+5
    m=6, P=3+5+5+7
    m=7, P=3+5+5+7+9
    m=8, P=3+5+5+7+9+9
    m=9, P=3+5+5+7+9+9+11
    m=10, P=3+5+5+7+9+9+11+13
    m=11, P=3+5+5+7+9+9+11+13+13
    m=12, P=3+5+5+7+9+9+11+13+13+15
    m=13, P=3+5+5+7+9+9+11+13+13+15+17
    m=14, P=3+5+5+7+9+9+11+13+13+15+17+17
    and so on


    2. Relevant equations



    3. The attempt at a solution
    I did not understand how to evaluate it into a formula
     
    Last edited: Aug 2, 2010
  2. jcsd
  3. Aug 2, 2010 #2

    Mentallic

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    Homework Helper

    Using induction you can show that

    [tex]3+5+7+...+(2n+1)=n(n+2)[/tex]

    and

    [tex]5+9+13+...+(4n+1)=n(2n+3)[/tex]

    If you need to prove these yourself then you can manipulate the first sum by turning it into the well known
    [tex]1+3+5+...=n^2[/tex]
    and if you need to prove this too... well... use geometry I suppose.

    and for the second,
    [tex]=4+8+12+...+4n-n[/tex]
    [tex]=4(1+2+3+...+n)-n[/tex]


    Now I couldn't quite find how to merge both formulae together to have a single function P=f(m) for all values of m>1. I'll keep working on it.
     
  4. Aug 2, 2010 #3
    yeah thanks, I got some view on the series, But I too cannot understand the mixed series. So please keep trying. I also will try on it.
     
  5. Aug 2, 2010 #4

    Mentallic

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    Ok I figured out a nice pattern and after much blood, sweat and tears, I finally solved it. There will be 3 cases to consider.

    Are you familiar with modulo arithmetic?
    For m(mod 3)=0, that is, m=3,6,9,12...

    We have the pattern:

    m=3 , P = (3) + (0)
    m=6 , P = (3+5+7) + (5)
    m=9 , P = (3+5+7+9+11) + (5+9)
    m=12, P = (3+5+7+9+11+13+15) + (5+9+13)

    etc.

    Now since in this case we don't need to consider what is happening in the other cases and we are dealing with m that are multiples of 3, let's take [tex]\frac{m}{3}=x[/tex]. So for m=12, x=4 and you can consider that as being the 4th sequence in this lone pattern.

    for x=1, we have 1 in the first bracket, and 0 in the second
    for x=2, we have 3 in the first bracket, and 1 in the second
    for x=3, we have 5 in the first bracket, and 2 in the second
    for x=2, we have 7 in the first bracket, and 3 in the second

    Can you see the pattern arising?

    for x=x, we have (2x-1) in the first bracket, and (x-1) in the second

    Now looking at the formulae I gave you earlier,

    [tex]
    3+5+7+...+(2n+1)=n(n+2)
    [/tex]
    and

    [tex]
    5+9+13+...+(4n+1)=n(2n+3)
    [/tex]

    Let the first bracket with the first formula be n=2x-1. What this means is we have to substitute n=2x-1 into the sum formula n(n-2).
    Secondly, substitute n=x-1 in the second formula.

    So we obtain [tex]P=(2x-1)(2x+1)+(x-1)(2x+1)=(3x-2)(2x+1)[/tex]

    That's the formula for m(mod 3)=0 where m/3=x.

    So let's take m=12, thus x=4.

    m=12, P=3+5+5+7+9+9+11+13+13+15=90

    P=(3.4-2)(2.4+1)=10.9=90

    It works for all others too :smile:

    Now maybe you can try apply this idea to the other two cases? Let m(mod 3)=1, i.e. m=4,7,10... and [tex]\frac{m-1}{3}=x[/tex].

    For m(mod 3)=2 let [tex]\frac{m+1}{3}=x[/tex]

    This will finally satisfy all your summations.
     
  6. Aug 2, 2010 #5
    thanks for the replies
     
  7. Aug 3, 2010 #6

    Mentallic

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    Homework Helper

    No problem. Are you able to find the answer to the other two cases?
     
  8. Aug 3, 2010 #7
    yeah I got it from other forum.as

    Let n = [m/3] ; [] mean integer part
    and k = m mod 3 ; mod mean modulus (remainder of the integer divide m/3)

    That is:
    m --> n , k
    3 --> 1, 0
    4 --> 1, 1
    5 --> 1, 2
    6 --> 2, 0
    7 --> 2, 1
    8 --> 2, 2
    9 --> 3, 0
    ...............


    Then P(n,k) = n*(6*n-1) + k*(4*n+1) - 2
     
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