Binomial Expressions

$(1-r)^{-1}=1+r+r^2+r^3+\cdots$

$\sum_{i=0}^{\infty} r^i \quad|r|<1$

$(1+r)^{-1}=1-r+r^2-r^3$

$\sum_{i=0}^{\infty}(-r)^i \ \ |r|<1 .$

$\text { What is }(0.8)^{-1}$

Solution:

$(0.8)^{-1}=\left(\frac{8}{10}\right)^{-1}=\left(\frac{10}{8}\right)$

$=1.25$

Can we get the same answer by expanding its infinite sum?

We know $(0.8)^{-1}=(1-0.2)^{-1}$

$(1-r)^{-1} \ \ |r| < 1$

$\because|0.2| < 1 .$

By the series expansion.

$\underbrace{1+0.2^2}_{=1.2}+ \underbrace{(0.2^2+0.2^3+ \cdots)} _{0.05 ?}$

But the answer is $1.25$

$0.2^2+0.2^3+0.2^4+\cdots \cdot$
$= 0.2^2 (\underbrace{1+0.2+0.2^2+\cdots}_{\text {G.P. Series. }})$
$= 0.2^2 \cdot \frac{1-r^n}{1-r}$
$= lim \ n \rightarrow \infty \ 0.2^2 \times \left(\frac{1-0.2^n}{1-0.2}\right)$

$0.2^2 \times \frac{1}{1-0.2}$

$=0.2^2 \times \frac{1}{0.8}=\frac{0.04}{0.8}$

$=\frac{1}{20}=0.05$

$\therefore$ The whole sum is

$1.2+0.05$

$= 1.25$

$(1+0.2)^{-1}$

Solution:

We knows the answer.

$=(1 \cdot 2)^{-1}=\left(\frac{12}{10}\right)^{-1}$

$=\frac{10}{12}=\frac{5}{6}$

$=0.8333$

$(1+r)^{-1}=1-r+r^2-r^3+ \cdots$
$\therefore (1+0.2)^{-1}$
$\underbrace{1-0.2}_{=0.8} + \underbrace{(0.2)^2-(0.2)^3+\cdots} _{=0.0333 \cdots}$

$(0.2)^2(1-0.2 + 0.2^2 - \ldots)$
$\text { common ratio }, r=-0.2$
$\ =0.04\left(\frac{1}{1+0.2}\right)$

$0.04 \times \left(\frac{1}{1.2}\right)$

$= \frac{4}{120}=\frac{1}{30}$

$= 0.0333 .$

$\therefore$ The answer is $0.8333 …$

Is it true for only $-1$ ?

can we have similar expansion say for $-2,-3$?

Or say some rational number say $\frac{1}{2}, \frac{2}{3}$ ?

We have studied Binomial Theorem for positive integer $n$.

$(1+x)^n=1+{}^nC_1 x+{}^nC_2 x^2 +\cdots+{}^nC_{n-1} x^{n-1} + x^n$

a. Number of terms in finite.

b. We could use combination ${}^n C_r$ for $r=0,1, \cdot\cdot\cdot n.$

The problem with negative index is

i) The number of terms is infinite.

ii) We cannot use ${}^{-n}C_r$ as This is not defined.

Let us first see what is $(1-r)^{-2} .|r|<1$

Our aim is to find out the coefficients of the expansion.

Let us assume

$(1-r)^{-2}=a_0+a_1 r+a_2 r^2+\cdots$

We knows that $(1-r)^{-2}=(1-r)^{-1} \times (1-r)^{-1}$

$\left(1+r+r^2+\cdots \cdot\right) \times \left(1+r+r^2+\cdots\right)$

$\left(1+r+r^2+\cdots\right) \times \left(1+r+r^2+\cdots\right)$

coeff of $r^0=1 \quad \therefore a_0=1$

Coeff of $r=(1+1)=2 \quad \therefore a_1=2$

coeff of $r^2=(1+1+1)=3 \quad \therefore a_2=3$

coeff of $r^3=4 \quad \therefore a_3=4$

In ganeral coeff of $r^k={k+1}$

$\therefore (1-r)^{-2}$

$=1+2 r+3 r^2+4 r^3+ \cdots$

From here we can easily see that:

$(1+r)^{-2}=(1-(-r))^{-2}$

$=1-2 r+3 r^2-4 r^3+\cdots$

$\text { Expand }(1+r)^{-2} \text { as }$

$(1+r)^{-1} \times (1+r)^{-1}$

what in $(1-r)^{-3}?$

Solution: Again as before we assume

$(1-r)^{-3}=b_0+b_1 r+b_2 r^2+b_3 r^3 + \cdot \cdot \cdot$

And we try to find the values of $b_0, b_1, b_2 \ldots$. etc.

We will write

$(1-r)^{-3}=(1-r)^{-2} \times (1-r)^{-1}$

$(1-r)^{-3}=b_0+b_1 r+b_2 r^2+\cdots \ (1-r)^{-2} \times (1-r)^{-1}= \ \left(1+2 r+3 r^2+4 r^3+\cdots\right) \ \ \times \left(1+ r+r^2+\cdots\right) \ \text { coeff of } r^0=1 \quad \therefore b_0=1 \ \text { coess of } r=(1+2) \therefore b_1=3 \ \text { coesf of } r^2=(1+2+3)=6 \quad \therefore b_2=6 \ \text { coesf } of r^3=(1+2+3+4)=b_3=10$

$\therefore(1-r)^{-3}$

$=1+3 r+6 r^2+10 r^3+\ldots .$

The general term: coeff of $r^k$ in.

$\text { 1. }(k+1)+1 \cdot k+1 \cdot(k-1)+\cdots+1 \cdot 1$

$=1 \cdot(1+2+\cdots+k+1)$

$=\frac{(k+1)(k+2)}{2}=b_k= \ \text{Coeff of} \ r^k$

$\begin{aligned} \therefore & (1-r)^{-3} \\ & =\sum_{k=0}^{\infty} \frac{(k+1)(k+2)}{2} r^k \end{aligned}$

Obviously, it is not possible to compute these terms for arbitrarily large $n.$ We have the formula:

For positive integral $n$. we have coefficient of

$r^k={}^n C_k \text {. }$ not valid for $-n$. we write ${}^nC_r$ as

$\frac{n !}{r !(n-r) !}=\frac{r(n-1) \cdots(n-r-1)}{r !}$

Let us denote by $c(-n, k)$ the term $\frac{(-n)(-n-1)(-n-2) \cdots(-n-k+1)}{k !}$

We shall see that $c(-n, k)$ in the coeff of $r^k$ in the expension of $(1+x)^{-x} \quad|x|<1$

Consider

$(1-r)^{-2}$ coeff of $-r=\frac{-2}{1!}=-2$

coeff of $(-r)^2=\frac{-\dot{2}(-3)}{2!}=3$

$\operatorname{coeff}(-r)^3=\frac{-2(-3 !(-4)}{3 !}=-4$

$\therefore$ The expansion is:

$1+(-2)(-r)+3(-r)^2+(-4)(-r)^3 + \cdot \cdot$

$=1+2 r+3 r^2+4 r^3 + \cdot \cdot$

Let us verify for $(1+r)^{-3}$ coeff $r=\frac{-3}{1!}=-3$

coeff $r^2=\frac{-3(-4)}{2 !}=6$

coeff $r^3=\frac{-3(-4)(-5)}{3 !}=-10$

$(1+r)^{-3}$

$1-3 r+6 r^2-10 r^3+\cdots$

what is $(1-x)^{\frac{1}{2}}$ or $(1-x)^{-\frac{1}{2}}$

Consider $(1-x)^{-\frac{1}{2}}$

we know that

$(1-x)^{-1}=(1-x)^{-\frac{1}{2}} \times (1-x)^{-\frac{1}{2}} |x|<1$

$1+x+x^2+x^3+\ldots$

Let us assume

$(1-x)^{-\frac{1}{2}}=a_0+a_1 x+a_2 x^2$

$(1-x)^{-\frac{1}{2}} \times (1-x)^{-\frac{1}{2}}$

$=\left(a_0+a_1 x+a_2 x^2+a_3 x^3\right)$

$\quad \times \left(a_0+a_1 x+a_2 x^2+a_3 x^3\right) = \ {1+x+x^2+x^3+\cdots}$

${\therefore a_0^2= 1 \therefore a_0 = \pm1}$

But we take the positive value

$\therefore G_0=1$

$\text { coeff of } x=G_0 G_1+G_1 G_0$

$=2\cdot G_0 \cdot G_1 =2 G_1=1$

$\operatorname{from}(1-x)^{-1}$

$2 G_1=1 \quad \therefore G_1=\frac{1}{2}$

$\text { coeff } x^2=G_0 G_2+G_1^2+G_2 G_0 =1$

$2 G_0 G_2+G_1^2=1$

$2 G_0 G_2=1-\frac{1}{4}=\frac{3}{4}$

$\therefore 2G_2=\frac{3}{4} \quad \therefore G_2=\frac{3}{8}$

$\therefore$ we get that

$(1-x)^{-\frac{1}{2}}=1+\frac{1}{2} x+\frac{3}{8} x^2+ \cdot \cdot \cdot$

We shall see whether we get the same

by using the expansion the way we did for negative integer

For negative integer we have seen that coeff of $x^k$ in

$(1-x)^{-n}=\frac{(-n)(-n-1)–(-n-k+1)}{k !}$

Let us apply the same for $\frac{p}{a}$, in particular here for $-\frac{1}{2}$

By using similar expansion we get:

$\quad(1-x)^{-\frac{1}{2}}=$

$1-\left(-\frac{1}{2}\right) x+\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)}{2 !}(x)^2$

$=1+\frac{1}{2} x+\frac{3}{8} x^2+\cdots$

$G_0=1$

$G_1=\frac{1}{2}$

$G_2=\frac {3}{8}$

This we get the same answer