Polynomial ExpansionDate: 02/06/2003 at 14:53:59 From: Khursheed Subject: Polynomial Expansion What's the general formula for things like (a+b+c)^2 (a+b+c+d)^3 (a+b+c+d+e)^4 (...n+1 terms...)^nth power Date: 02/06/2003 at 18:31:16 From: Doctor Greenie Subject: Re: Polynomial Expansion Hello, Khursheed - I am not familiar with a general formula for this type of expansion; I think such a formula would be extremely cumbersome. It is relatively easy to determine the coefficient of any given term in such an expansion. For a detailed discussion of the process, see my answer to a question on the following page in the Dr. Math archives: Coefficients in a Trinomial Expansion http://mathforum.org/library/drmath/view/51601.html Using the ideas presented there on your example (a+b+c+d+e)^4 we could say, for example, that the coefficient of the a^2cd term is 4! ------------ = 12 (2!)(1!)(1!) There is a way to check the answers we get for the coefficients of the individual terms. If we replace each of the variables a, b, c, d, and e with "1," then the expression becomes (5)^4 = 625. This means that the sum of all the coefficients of (a+b+c+d+e)^4 should be 625. Let's see if we can "account" for this sum of all the coefficients.... (1) terms like a^4 or c^4 (one variable; to the 4th power) The number of terms like this is '5 choose 1' = C(5,1) = 5. The number of ways to choose the same variable from all four factors of (a+b+c+d+e) (in other words, the number of distinct ways of arranging, for example, the letters bbbb) is 4! -- = 1 4! The sum of the coefficients of the terms of this type is 5*1=5. (2) terms like b^3.c or a.e^3 (two variables; one to the 3rd power and the other to the first power) The number of terms like this is P(5,2) = 20. (This is a permutation rather than a combination because - for example - the b^3.c and c^3.b terms are different terms.) The number of ways to choose three of one variable and one of another from the four factors (in other words, the number of distinct ways of arranging, for example, the letters bbbc, is 4! -------- = 4 (3!)(1!) The sum of the coefficients of the terms of this type is 20*4=80. (3) terms like b^2.c^2 or a^2.e^2 (two variables; each to the 2nd power) The number of terms like this is C(5,2) = 10. (This is a combination because - for example - the b^2.c^2 and c^2.b^2 terms are the same term.) The number of ways to choose two of one variable and two of another from the four factors (in other words, the number of distinct ways of arranging, for example, the letters bbcc, is 4! -------- = 6 (2!)(2!) The sum of the coefficients of the terms of this type is 10*6=60. (4) terms like b^2.c.d or a^2.c.e (three variables; one to the 2nd power and the other two to the first power) The number of terms like this is 30. This number is a mixture of a combination and a permutation. You can think of it in either of two ways: (a) choose the one variable which is squared and then choose the two of the remaining 4 which are to the first power - this gives the number C(5,1)*C(4,2) = 5*6 = 30; or (b) choose two of the five variables that are to the first power and then choose one of the remaining three that is squared - this gives the number C(5,2)*C (3,1) = 10*3 = 30. The number of ways to choose two of one variable and one each of two others from the four factors (in other words, the number of distinct ways of arranging, for example, the letters bbcd, is 4! ------------ = 12 (2!)(1!)(1!) The sum of the coefficients of the terms of this type is 30*12=360. (5) terms like abcd (four different variables; each to the first power) The number of terms like this is C(5,4) = 5 The number of ways to choose four different variables from the four factors (in other words, the number of distinct ways of arranging, for example, the letters abcd, is 4! ---------------- = 24 (1!)(1!)(1!)(1!) The sum of the coefficients of the terms of this type is 5*24=120. So the sum of all the coefficients of all the different-type terms is 5 + 80 + 60 + 360 + 120 = 625 as it should be. Note that it IS possible to write out the whole expansion of (a+b+c+d+e)^4 using the preceding type of analysis. For this example, the (abbreviated) expansion would be (a+b+c+d+e)^4 = 1(a^4+b^4+c^4+d^4+e^4) + 4(a^3b+a^3c+a^3d+a^3e+...+c^3e+d^3e) [20 terms like this...] + 6(a^2b^2+a^2c^2+a^2d^2+...+c^2e^2+d^2e^2) [10 terms like this...] + 12(a^2bc+a^2bd+...+d^2be+d^2ce) [30 terms like this...] + 24(abcd+abce+abde+acde+bcde) I hope all this helps. Please write back if you have any further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/