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Polynomial Expansion

Date: 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/ 
Associated Topics:
High School Permutations and Combinations
High School Polynomials

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