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?).
i.e.,
 ${n_1} + {n_2} + {n_3} + \ldots + {n_m} = n$
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:
 Multinomial Term Coefficient ${x_1}^2{x_2}{x_3}$ $\dfrac{{4!}}{{2!1!1!}} = 12$ ${x_1}^3{x_2}$ $\dfrac{{4!}}{{3!1!}} = 4$ ${\left( {{x_1} + {x_2} + {x_3}} \right)^4}$ ${x_2}^4$ $\dfrac{{4!}}{{4!}} = 1$ ${x_1}{x_2}{x_3}^2$ $\dfrac{{4!}}{{1!1!2!}} = 12$ $\vdots$
 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)