What do people mean by "Divide unity by something" in Math?

Question

I was reading a book, and on one page it said:

Let the series$$S=A+Bx+Cx^2+Dx^3+Ex^4+Fx^5+\ldots+\&c\tag1$$ Divide unity by $S$ as far as two terms of the quotient, which will be the form $p+qx$, and write the remainder in the form $\ldots$

The book also included an example:

$\ldots $$$S=1+3x+6x^2+10x^3+15x^4+21x^5+\ldots\&c\tag2$$ Then, we shall have $\frac 1S=1-3x+\ldots$ with a remainder $$3x^2+8x^3+15x^4+21x^5+\&c\tag3$$

$$\vdots$$

Questions:

1. What did the book mean by "Divide unity by $S$? Do they mean take the reciprocal of $S$, or something entirely different? I do know that unity means $1$, so we're dividing by $1$ somewhere.
2. How did the book get $\frac 1S=1-3x+\ldots\&c$?

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At this point, I am kind of baffled by what they mean. I spent almost half an hour on this, and just cannot figure it out!

Any help is appreciated!

Answer 1

Method 1:

For$|x|<1$, $$f(x)=\sum_{k=0}^\infty x^{k}=\frac {1}{1-x}.$$ $$f(1-S)=\frac {1}{S}=\sum_{k=0}^\infty (1-S)^{k}.$$ Then, $$\frac {1}{S}=1+(1-S)+(1-S)^2+(1-S)^3+\ldots$$ $$=1+(-3x-6x^2-10x^3-\ldots)+(-3x-6x^2-10x^3-\ldots)^2+(-3x-6x^2-10x^3-\ldots)^3+\ldots$$ $$=1+(\color{red}{-3x}\color{blue}{-6x^2}\color{green}{-10x^3}-\ldots)+(\color{blue}{9x^2}\color{green}{+36x^3}+60x^4+36x^4-\ldots)+(\color{green}{-27x^3}+\ldots)+\ldots$$ $$=1\color{red}{-3x}\color{blue}{+3x^2}\color{green}{-x^3}+\ldots$$

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Method 2:

Let $\frac 1 S =a_0+a_1x+a_2x^2+a_3x^3+\ldots$. $$S\cdot \frac 1 S=(1+3x+6x^2+10x^3+\ldots)(a_0+a_1x+a_2x^2+a_3x^3+\ldots)$$ $$=a_0+(3a_0+a_1)x+(6a_0+3a_1+a_2)x^2+(10a_0+6a_1+3a_2+a_3)x^3+\ldots=1$$ Therefore, \begin{array}{cccc} &~~~~a_0&&&&&& &=1 \\ &~~3a_0&+&~a_1&&&&&=0 \\ &~~6a_0&+&3a_1&+&~a_2&&&=0 \\ &10a_0&+&6a_1&+&3a_2&+&~a_3&=0 \\ \end{array} $$\ldots\ldots$$

$\therefore a_0=1, a_1=-3, a_2=3,a_3=-1, \ldots$
$$\frac 1 S=a_0+a_1x+a_2x^2+a_3x^3+\cdot\cdot\cdot=1-3x+3x^2-x^3+\ldots.$$

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Method 3:

Rewrite unity $1$ as $1+0x+0x^2+0x^3+\ldots$.


$\qquad \qquad \qquad \qquad \qquad \qquad ~1-3x+3x^2-~~x^3~+\ldots$

$~~1+3x+6x^2+10x^3+\ldots\overline{)1+0x+0x^2+~0x^3+\ldots }$

$\qquad \qquad \qquad \qquad \qquad \qquad 1+3x+6x^2+10x^3+\ldots$

$\qquad\qquad\qquad\qquad\qquad\qquad\quad\overline{-3x-6x^2-10x^3+\ldots} $

$\qquad\qquad\qquad\qquad\qquad\qquad\quad-3x-9x^2-18x^3+\ldots$

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\overline{3x^2+~~8x^3+\ldots} $

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad3x^2+~~9x^3+\ldots $

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\overline{-x^3+\ldots}$

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad-x^3+\ldots$

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\overline{\ldots} $

$$\therefore \frac 1 S=1-3x+3x^2-x^3+\ldots$$

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Choose one of the methods you like :)

Answer 2

Presumably they're thinking of dividing $1$ by $S$ with a quotient $Q(x)$ and remainder $R(x)$ such that $$ 1 = (Q(x))S + R $$ where $Q$ is a polynomial of some degree $n$, chosen so that it "cancels" the terms of degree $0, \ldots, n-1$ in the division, so $R$ is a series starting with a term in $x^n$. For $S = 1+3x+\cdots$ and for $Q(x)$ of degree $1$ (so $Q(x)=p+qx$), we can work out $Q$ by synthetic division. After two steps we have this: \begin{align} & \phantom{)\ } 1 - 3x + \cdots \\ 1+3x+6x^2+10x^3+15x^4+21x^5+\cdots \ & \overline{)\ 1 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad} \\ & \phantom{)\ } \underline{1+3x+6x^2+10x^3+15x^4+21x^5+\cdots}\\ & \phantom{)\ 1} -3x-6x^2-10x^3-15x^4-21x^5+\cdots\\ & \phantom{)\ 1} \underline{{}-3x-9x^2-18x^3-30x^4-45x^5+\cdots}\\ & \phantom{)\ 1-3x-{}} 3x^2+\phantom{1}8x^3+15x^4+24x^5+\cdots\\ \end{align}

In other words, \begin{multline} 1 = (1 - 3x)(1+3x+6x^2+10x^3+15x^4+21x^5+\cdots) + {}\\ (3x^2+8x^3+15x^4+24x^5+\cdots), \end{multline} so the remainder is $3x^2+8x^3+15x^4+24x^5+\cdots$.

I notice that the $x^5$ term in my remainder is different from what is written in the question, but that could just be a transcription error somewhere.