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Saturday, 9 August 2014

CHAPTER 9 - Introduction to Rational Index

In the previous section, we discussed the expansion of {(x + y)^n}, where n  is a natural number. We’ll extend that discussion to a more general scenario now. In particular, we’ll consider the expansion of {(1 + x)^n}, where n is a rational number and |x|\,\, < \,\,1. Note that any binomial of the form {(a + b)^n} can be reduced to this form.:
{(a + b)^n} = {\left\{ {a\left( {1 + \dfrac{b}{a}} \right)} \right\}^n} (we are assuming |a| > |b| )
 = {a^n}{\left( {1 + \dfrac{b}{a}} \right)^n}
 = {a^n}{(1 + x)^n} where |x|<1
The general binomial theorem states that
\begin{array}{l}  {(1 + x)^n} = 1 + nx + \dfrac{{n(n - 1)}}{{2!}}{x^2} + \dfrac{{n(n - 1)(n - 2)}}{{3!}}\;{x^3} + \ldots \;\\  \\  \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots  + \dfrac{{n(n - 1)(n - 2)\ldots (n - r + 1)}}{{r!}}{x^r} + \ldots \infty  \end{array}
That is, there are an infinite number of terms in the expansion with the general term given by
{T_{r + 1}} = \dfrac{{n(n - 1)(n - 2)\ldots (n - r + 1)}}{{r!}}{x^r}
For an approximate proof of this expansion, we proceed as follows: assuming that the expansion contains an infinite number of terms, we have:
{\left( {1 + x} \right)^n} = {a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} +\ldots + {a_n}{x^n} +\ldots \infty
Putting x = 0  gives {a_o} = 1. Now differentiating once gives
n{\left( {1 + x} \right)^{n - 1}} = {a_1} + 2{a_2}x + 3{a_3}{x^2} + \ldots \infty
Putting x = 0 gives {a_1} = n.
Proceeding in this way, we find that the {r^{th}} coefficient is given by
{a_n} = \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\ldots \left( {n - r + 1} \right)}}{{r!}}
Note that if n is a natural number, then this expansion reduces to the expansion obtained earlier, because {T_{r + 1}} becomes ^n{C_r}{x^r}, and the expansion terminates for r > n. For the general {T_{r + 1}}, we obviously cannot use ^n{C_r} since that is defined only for natural n.
One very important point that we are emphasizing again is that the general expansion holds only for |x|\; < 1.
Let us denote the genral binomial coefficient by {V_r}. Thus, we have
{V_r} = \dfrac{{n(n - 1)(n - 2) \ldots (n - r + 1)}}{{r!}}
and {(1 + x)^n} = \sum\limits_{r = 0}^\infty  {} {V_r}{x^r}
Let us discuss some particularly interesting expansions. In all cases, |x|\; < \;1:
(1) {(1 + x)^{ - 1}}
Since n =  - 1, we see that
{V_r} = \dfrac{{( - 1)(( - 1) - 1)(( - 1) - 2) \ldots (( - 1) - r + 1)}}{{r!}}
 = {( - 1)^r}
so that the expansion is {(1 + x)^{ - 1}} = 1 - x + {x^2} - {x^3} + \ldots  \infty
(2) {(1 - x)^{ - 1}}
Again {V_r} = {( - 1)^r} and thus {(1 - x)^{ - 1}} = 1 + x + {x^2} + {x^3} +\ldots \infty
(3) {(1 + x)^{ - 2}}
We have n =  - 2;
{V_r} = \dfrac{{( - 2)(( - 2) - 1)(( - 2) - 2) \ldots  (( - 2) - r + 1)}}{{r!}}
 = \dfrac{{{{( - 1)}^r}\;(r + 1)!}}{{r!}}
 = {( - 1)^r} \cdot \;(r + 1)
Thus,
{(1 + x)^{ - 2}} = 1 - 2x + 3{x^2} - 4{x^3} + \ldots \infty
(4) {(1 - x)^{ - 2}}
Again {V_r} = {( - 1)^r} \cdot (r + 1) so that
{(1 - x)^{ - 2}} = 1 + 2x + 3{x^2} + 4{x^3} +\ldots \infty
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