Exercise 1.24

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Prove the identities

(a) $x^n - y^n = (x-y) (x^{n-1}+x^{n-2} y + \cdots +y^{n-1})$

(b) For $n$ odd, $x^n + y^n = (x+y) (x^{n-1}- x^{n-2} y + \cdots +y^{n-1})$

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
Let $R$ any commutative ring, and $x,y \in R$.

a) Let $$S = \sum_{i=0}^{n-1} x^{n-1-i} y^i.$$ Then
\begin{align*}
xS &= \sum_{i=0}^{n-1} x^{n-i} y^i = x^n + \sum_{i=1}^{n-1} x^{n-i} y^i\
yS &= \sum_{i=0}^{n-1} x^{n-1-i} y^{i+1} = \sum_{j=1}^{n} x^{n-j} y^j \qquad(j = i+1)\
&=y^n + \sum_{i=1}^{n-1} x^{n-i} y^i.
\end{align*}
So $xS - y S = x^n-y^n$,
$$x^n - y^n = (x-y) \sum_{i=0}^{n-1} x^{n-1-i} y^i = (x-y)(x^{n-1}+x^{n-2} y + \cdots +x^{n-1-i}y^i+ \cdots + y^{n-1}).$$

b) If we substitute $-y$ by $y$, we obtain
$$x^n - (-1)^n y^n = (x+ y) \sum_{i=0}^{n-1}(-1)^i x^{n-1-i} y^i .$$

If $n$ is odd,
$$x^n  + y^n = (x+ y) \sum_{i=0}^{n-1}(-1)^i x^{n-1-i} y^i  = (x+y)(x^{n-1}-x^{n-2} y + \cdots +(-1)^i x^{n-1-i}y^i+ \cdots + y^{n-1}).$$
\end{proof}

Exercise 1.24

Answers

Comments

Exercise 1.24

Answers

Comments

Add answer