1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Homework Helper

    Using induction you can show that




    If you need to prove these yourself then you can manipulate the first sum by turning it into the well known
    and if you need to prove this too... well... use geometry I suppose.

    and for the second,

    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


    User Avatar
    Homework Helper

    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)


    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,



    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


    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


    User Avatar
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook