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Math Help - Binomial Coefficient

  1. #1
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    Binomial Coefficient

     (1/1-x)(1+x^{5}+x^{10}+x^{15}+...+x^{100})

    I need the coefficient of  x^{100}

    can anyone help?

    If you have any tips for solving these kinds of problems, it would be really appreciated.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    I Think

    Quote Originally Posted by p00ndawg View Post
     (1/1-x)(1+x^{5}+x^{10}+x^{15}+...+x^{100})

    I need the coefficient of  x^{100}

    can anyone help?

    If you have any tips for solving these kinds of problems, it would be really appreciated.
    The answer is N/A...there is no x^100 term after you expand
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    The answer is N/A...there is no x^100 term after you expand
    wat?

    it should be 21. There has to be a term, (1/1-x) = (1+ x+ x^{2}+x^{3}+x^{4}+...)
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    O....

    Quote Originally Posted by p00ndawg View Post
    wat?

    it should be 21. There has to be a term, (1/1-x) = (1++ x+ x^{2}+x^{3}+x^{4}+...)
    Do you mean the power series for \frac{1}{1-x}...you know that only applys for -1<x<1 right? you cant make a general statement unless you also state that?
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  5. #5
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    Quote Originally Posted by Mathstud28 View Post
    Do you mean the power series for \frac{1}{1-x}...you know that only applys for -1<x<1 right? you cant make a general statement unless you also state that?
    well originally the question was find the number of ways to make change for 1 dollar using up to 9 pennies and any number of nickels and dimes.

    I just figured it was implied, my bad.

    I just reduced it about as far as I can, but I dont know how to get the number of coefficients easily.

    not sure if theres an easy way to do it, but my professor did it in his head lol.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Hmm

    Quote Originally Posted by p00ndawg View Post
    well originally the question was find the number of ways to make change for 1 dollar using up to 9 pennies and any number of nickels and dimes.

    I just figured it was implied, my bad.

    I just reduced it about as far as I can, but I dont know how to get the number of coefficients easily.

    not sure if theres an easy way to do it, but my professor did it in his head lol.
    That sounds more like a _nC_r problem...yeah?
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  7. #7
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    Quote Originally Posted by Mathstud28 View Post
    That sounds more like a _nC_r problem...yeah?
    its a generating function problem, cleverly disguised as a binomial coefficient problem, cleverly disguised as an innocent, yet disturbing way of dividing up a dollar.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by p00ndawg View Post
    its a generating function problem, cleverly disguised as a binomial coefficient problem, cleverly disguised as an innocent, yet disturbing way of dividing up a dollar.
    are you sure that it is the power series for \frac{1}{1-x}?
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  9. #9
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    Quote Originally Posted by Mathstud28 View Post
    are you sure that it is the power series for \frac{1}{1-x}?
    yea, I originally had the problem like this..


    (1 + x + x^{2} +...+x^{9})(1+x^{5}+x^{10}+x^{15}+...+x^{100})(1+x  ^{10}+x^{20}+x^{30}+...+x^{100})


    I then reduced the first expression to :
    (1-x^{10}/1-x) left the second expression untouched, and reduced the third to, (1/1-x^{10}) .

    I actually have the problem worked out, but the last line where the professor got the answer left me blank, because I cant see how he got 21 so easily.
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    haha

    Quote Originally Posted by p00ndawg View Post
    yea, I originally had the problem like this..


    (1 + x + x^{2} +...+x^{9})(1+x^{5}+x^{10}+x^{15}+...+x^{100})(1+x  ^{10}+x^{20}+x^{30}+...+x^{100})


    I then reduced the first expression to :
    (1-x^{10}/1-x) left the second expression untouched, and reduced the third to, (1/1-x^{10}) .

    I actually have the problem worked out, but the last line where the professor got the answer left me blank, because I cant see how he got 21 so easily.
    Ok this may be a really stupid question...but did he show you some step that gave the right answer or did he just know the answer? if it is the latter did you ever consider this is a planned out problem?
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  11. #11
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    Quote Originally Posted by Mathstud28 View Post
    Ok this may be a really stupid question...but did he show you some step that gave the right answer or did he just know the answer? if it is the latter did you ever consider this is a planned out problem?
    He explained it, and told us the logic to what the answer was, undeniably I couldnt really write down what he was saying, but what he said made sense.

    I just cant figure it out.

    It was also a review problem for my test tomorrow but these binomial coefficient problems are kicking my ass.

    Can you help me with this problem then..

    (x^{3}+x^{4}+x^{5}+x^{6}+x^{7}...)^{3}

    I need the coefficient of x^{10}
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Oh yeah

    Quote Originally Posted by p00ndawg View Post
    He explained it, and told us the logic to what the answer was, undeniably I couldnt really write down what he was saying, but what he said made sense.

    I just cant figure it out.

    It was also a review problem for my test tomorrow but these binomial coefficient problems are kicking my ass.

    Can you help me with this problem then..

    (x^{3}+x^{4}+x^{5}+x^{6}+x^{7}...)^{3}

    I need the coefficient of x^{10}
    sure its 3
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  13. #13
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    Quote Originally Posted by Mathstud28 View Post
    sure its 3
    now you're becoming like my teacher. lol

    how???
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  14. #14
    Super Member PaulRS's Avatar
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    Quote Originally Posted by p00ndawg View Post
    It was also a review problem for my test tomorrow but these binomial coefficient problems are kicking my ass.

    Can you help me with this problem then..

    (x^{3}+x^{4}+x^{5}+x^{6}+x^{7}...)^{3}

    I need the coefficient of x^{10}
    Yep, it's 3. You always have to choose one term from 1st product, one from the second and one from the 3rd, but in order to get x^{10}, two of these must be x^{3} and the other x^4. Thus the number of permutations of (x^3)(x^3)(x^4) is your coefficient.
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  15. #15
    Super Member PaulRS's Avatar
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    Quote Originally Posted by p00ndawg View Post
     (1/1-x)(1+x^{5}+x^{10}+x^{15}+...+x^{100})

    I need the coefficient of  x^{100}

    can anyone help?

    If you have any tips for solving these kinds of problems, it would be really appreciated.
    Let: \delta(n)=0 if 5 doesn't divide n, and \delta(k)=1 if 5 does divide n

    The coefficient of x^{100} stays untouched if we change (1+x^{5}+x^{10}+x^{15}+...+x^{100}) for 1+x^{5}+x^{10}+x^{15}+...

    We have 1+x^{5}+x^{10}+x^{15}+...=\sum_{n=0}^{\infty}{\del  ta(n)\cdot{x^n}}

    Thus: (1+x+x^{2}+...)(1+x^{5}+x^{10}+x^{15}+...)=\sum_{n  =0}^{\infty}{\sum_{k=0}^n{\delta(k)}}\cdot{x^n}

    But: \sum_{k=0}^n{\delta(k)} is the number of multiples of 5 between 0 and n, thus we have <br />
\sum\limits_{k = 0}^n {\delta \left( k \right)}  = \left\lfloor {\tfrac{n}<br />
{5}} \right\rfloor  + 1<br />
where <br />
\left\lfloor x \right\rfloor <br />
is the floor function

    So the coefficient of x^{100} turns out to be <br />
\sum\limits_{k = 0}^{100} {\delta \left( k \right)}  = \left\lfloor {\tfrac{{100}}<br />
{5}} \right\rfloor  + 1 = 21<br />

    Stated in another way: For each 1,x^{5},...,x^{100}, you have only one element among 1,x^{1},x^{2},... such that the product of the pair gives x^{100}. It follows then that the coefficient of x^{100} is actually 21, which is the numbers of terms of the factor (1+x^5+...+x^{100})
    Last edited by PaulRS; April 9th 2008 at 07:29 PM.
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