Results 1 to 10 of 10

Math Help - Please find coefficient of x^4

  1. #1
    Member
    Joined
    Oct 2007
    Posts
    134

    Please find coefficient of x^4

    (x+2y)^10
    My answer: cooficient of X^4=210* 3^6=153090

    This cannot be right! Please help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    13440x^4 y^6 is what MathCad gives.
    {{10} \choose {4}} ( 2^6 )
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member angel.white's Avatar
    Joined
    Oct 2007
    Posts
    723
    Awards
    1
    Quote Originally Posted by oceanmd View Post
    (x+2y)^10
    My answer: cooficient of X^4=210* 3^6=153090

    This cannot be right! Please help
    It should be 13,440 by the Binomial theorem. Do you understand what Plato said?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    Quote Originally Posted by Plato View Post
    13440x^4 y^6 is what MathCad gives.
    {{10} \choose {4}} ( 2^6 )
    Basically, this stems from Pascal's triangle / Binomial theorem.

    Coefficient of some term x^my^n for (ax + by)^{m+n} is the following:

    {{m+n} \choose {m}} ( a^m b^n )
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2007
    Posts
    134
    I still do not understand why it is

     {10\choose 4} x^4 (2y)^6, doesn't it have to be  {10\choose 6} x^4 (2y)^6. The formula that my textbook gives is
     {n\choose n-j} x^j a^(n-j),

    so in my example n=10, j=4

    Please help to understand.

    Thank you
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Quote Originally Posted by oceanmd View Post
    Please help to understand.
    That is easy: {{N} \choose {k}} = {{N} \choose {N - k}}
    {{10} \choose {4}} = {{10} \choose {10 - 4}}
    \frac{{N!}}{{\left( {K!} \right)\left( {N - K} \right)!}} = \frac{{N!}}{{\left[ {N - K} \right]!\left( {N - \left( {N - K} \right)} \right)!}}<br />
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2007
    Posts
    134
    Thank you for your reply. So, whether it's 4 or 6 does not matter. They are both correct. And the answer is 210 for either one. Correct?

    Thank you
    Follow Math Help Forum on Facebook and Google+

  8. #8
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    Quote Originally Posted by oceanmd View Post
    Thank you for your reply. So, whether it's 4 or 6 does not matter. They are both correct. And the answer is 210 for either one. Correct?

    Thank you

    \frac{10!}{6!4!}=\frac{10!}{4!6!}

    {{10}\choose{4}}={{10}\choose{6}}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Oct 2007
    Posts
    134
    Thank you, colby
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member angel.white's Avatar
    Joined
    Oct 2007
    Posts
    723
    Awards
    1
    Quote Originally Posted by oceanmd View Post
    Thank you for your reply. So, whether it's 4 or 6 does not matter. They are both correct. And the answer is 210 for either one. Correct?

    Thank you
    Correct, but I personally think it is best to use the iteration you are on. For example, 0, then 1 then 2 then 3 then 4 then 5, 6, 7, 8, 9, 10

    This helps you keep track of where you are, you could technically do
    10, 9, 8, 7, 6, 5, 6, 7, 8, 9, 10 or 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0 or 0, 1, 2, 3, 4, 5, 4, 3, 2, 1, 0

    Or heck, any combination in between, such as 10, 1, 8, 3, 6, 5, 6, 7, 2, 1, 10

    But with all those terms, organization will be helpful, and doing it in order will probably keep you from making errors.

    Also, don't forget to multiply that combination of (n, n-k) by the coefficient of x to the power of n- k, and y to the power of k.

    so
    {{10} \choose {4}} (1x)^4(2y)^6


    {{10} \choose {4}} (1^4 x^4) (2^6 y^6)

    {{10} \choose {4}} 2^6 x^4 y^6

    (210) (64) x^4 y^6

    13440 x^4 y^6

    On a side note, {{10} \choose {6}} ={{10} \choose {4}} is the reason why pascal's triangle is symmetrical. (ie: coefficients of 1, 2, 1 and 1, 3, 3, 1 etc)
    -------------


    So, for the purposes of visualizing what's going on, the total breakdown will be:
    {{10} \choose {0}} (x)^{10}(2y)^0 + {{10} \choose {1}} (x)^9(2y)^1 + {{10} \choose {2}} (x)^8(2y)^2 + {{10} \choose {3}} (x)^7(2y)^3 + {{10} \choose {4}} (x)^6(2y)^4  + {{10} \choose {5}} (x)^5(2y)^5 + {{10} \choose {6}} (x)^4(2y)^6 + {{10} \choose {7}} (x)^3(2y)^7 + {{10} \choose {8}} (x)^2(2y)^8 + {{10} \choose {9}} (x)^1(2y)^9 + {{10} \choose {10}} (x)^0(2y)^{10}

    Notice how easy it is to find any given term, because the combinations are sequential you can just find {{10}\choose{6}} to find our term. This would be a bit harrier if we randomly alternated between our 2 options, you could easily lose track of where you are in the middle, and would have to count from the beginning. That is why I recommend you count sequentially from 0 to n in your combinations.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. how to find the leading coefficient
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 1st 2009, 08:12 AM
  2. Find coefficient of x^5
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: December 7th 2007, 04:45 AM
  3. please find coefficient
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: December 7th 2007, 12:05 AM
  4. Please find coefficient
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 6th 2007, 08:52 PM
  5. Find the coefficient (Precalc II)
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: August 27th 2007, 03:22 PM

Search Tags


/mathhelpforum @mathhelpforum