Saturday, 9 August 2014

CHAPTER 5 - Worked Out Examples 3

     Example: 5  

How will you expand the multinomial expression {\left( {{x_1} + {x_2} +\ldots + {x_m}} \right)^n}?
Solution: 5

We will approach this problem using combinatorics. Note that a general term of the expansion would be of the form (without the coefficient)
x_1^{{n_1}}x_2^{{n_2}}x_3^{n{  _3}}\ldots  x_m^{{n_m}}\ldots(1)
where the various powers must always sum to n (why?).
{n_1} + {n_2} + {n_3} + \ldots  + {n_m} = n
Now, to evaluate the coefficient of the term in (1) , we consider the multinomial expression in expanded form:
\underbrace {\left( {{x_1} + {x_2} + \ldots + {x_m}} \right)\left( {{x_1} + {x_2} +\ldots + {x_m}} \right) \ldots \left( {{x_1} + {x_2} +\ldots {x_m}} \right)}_{n\,\,{\rm{times}}}
To generate the term in (1) , we must get {x_1} from {n_1} terms, {x_2} from {n_1} terms and so on. Let us find the number of ways in which this can be done.
First select those {n_1} multinomials that will contribute {x_1} : this can be done in {}^n{C_{{n_1}}} ways. Now, from the remaining \left( {n - {n_1}} \right) multinomials, select those {n_2} multinomials that will contribute {x_2} : this can be done in {}^{(n - {n_1})}{C_{{n_2}}} ways. Continuing this process, we see that the number of ways to get {x_1} from {n_1} {x_2} from {n_2} \ldots  and so on, that is, the number of times the term in (1)  will be generated in the expansion is
^n{C_{{n_1}}} \times {\,^{\left( {n - {n_1}} \right)}}{C_{{n_2}}} \times {\,^{\left( {n - {n_1} - {n_2}} \right)}}{C_{{n_3}}} \times\ldots
 = \dfrac{{n!}}{{{n_1}!\left( {n - {n_1}} \right)!}}\,\, \times \,\,\dfrac{{\left( {n - {n_1}} \right)!}}{{{n_2}!\left( {n - {n_1} - {n_2}} \right)!}}\,\, \times \,\,\dfrac{{\left( {n - {n_1} - {n_2}} \right)!}}{{{n_3}!\left( {n - {n_1} - {n_2} - {n_3}} \right)!}}\,\, \times \,\,\ldots
 = \,\,\dfrac{{n!}}{{{n_1}!\,\,{n_2}!\ldots {n_m}!}}\,
This is what is known as the general multinomial coefficient. The multinomial expansion can now be written compactly as
{\left( {{x_1} + {x_2} + \ldots  + {x_m}} \right)^n} = \sum {\dfrac{{n!}}{{{n_1}!{n_2}! \ldots {n_m}!}}} \,\,x_1^{{n_1}}x_2^{{n_2}}\ldots \,x_m^{{n_m}}
where the summation is carried out over all possible combinations of the {n_i}‘s such that \sum {{n_i} = n} . For example, in {\left( {{x_1} + {x_2} + {x_3}} \right)^4}, let us consider some terms in the expansion:
{x_1}^2{x_2}{x_3}\dfrac{{4!}}{{2!1!1!}} = 12
{x_1}^3{x_2}\dfrac{{4!}}{{3!1!}} = 4
{\left( {{x_1} + {x_2} + {x_3}} \right)^4}{x_2}^4\dfrac{{4!}}{{4!}} = 1
{x_1}{x_2}{x_3}^2\dfrac{{4!}}{{1!1!2!}} = 12
     Example: 6     

Find the coefficient of {x^4} in the expansion of {\left( {1 + x - \dfrac{2}{{{x^2}}}} \right)^{10}}.
Solution: 6

From the previous example, the general term in the expansion will be
\dfrac{{10!}}{{{n_1}!\;{n_2}!\;{n_3}!}}\;{(1)^{{n_1}}}{(x)^{{n_{^2}}}}{\left( {\dfrac{{ - 2}}{{{x^2}}}} \right)^{{n_{^3}}}}
 = \dfrac{{10}}{{{n_1}!\;{n_2}!\;{n_3}!}}\;{x^{{n_2} - 2{n_3}}}{( - 2)^{{n_3}}}
where {n_1} + {n_2} + {n_3} must be 10.
Now, {x^4} is generated whenever {n_2} - 2{n_3} = 4. The possible values of the triplet ({n_1},\;{n_2},\;{n_3}) can now simply be listed out:
({n_1},\;{n_2},\;{n_3}) \equiv (6,\;4,\;0),\;(3,\;6,\;1),\;(0,\;8,\;2)
Thus, the (total) coefficient of {x^4} is
\dfrac{{10!}}{{6!\;4!\;0!}}{( - 2)^0} + \dfrac{{10!}}{{3!\;6!\;1!}}{( - 2)^1} + \dfrac{{10!}}{{0!\;8!\;2!}}{( - 2)^2}
 - 1290 (verify)

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