1.6 The geometric progression and the binomial theorem

Analysis of functions of a single Page 1 / 1

The binomial theorem is introduced, the existences of nth roots of real numbers is explored, the binomial coefficient is defined, and a theorem providing a formula for the sum of a geometric progression is included.

There are two special algebraic identities that hold in $R$ (in fact in any field $F$ whatsoever) that we emphasize. They are bothproved by mathematical induction. The first is the formula for the sum of a geometric progression.

Geometric progression

Let $x$ be a real number, and let $n$ be a natural number. Then,

If $x \neq 1,$ then
$\sum_{j = 0}^{n} x^{j} = \frac{1 - x^{n + 1}}{1 - x} .$
If $x = 1,$ then
$\sum_{j = 0}^{n} x^{j} = n + 1 .$

The second claim is clear, since there are $n + 1$ summands and each is equal to 1.

We prove the first claim by induction. Thus, if $n = 1,$ then the assertion is true, since

\sum_{j = 0}^{1} x^{j} = x^{0} + x^{1} = 1 + x = (1 + x) \frac{1 - x}{1 - x} = \frac{1 - x^{2}}{1 - x} .

Now, supposing that the assertion is true for the natural number $k,$ i.e., that

\sum_{j = 0}^{k} x^{j} = \frac{1 - x^{k + 1}}{1 - x},

let us show that the assertion holds for the natural number $k + 1 .$ Thus

\begin{matrix} \sum_{j = 0}^{k + 1} x^{j} & = & \sum_{j = 0}^{k} x^{j} + x^{k + 1} \\ = & \frac{1 - x^{k + 1}}{1 - x} + x^{k + 1} \\ = & \frac{1 - x^{k + 1} + x^{k + 1} - x^{k + 2}}{1 - x} \\ = & \frac{1 - x^{k + 1 + 1}}{1 - x}, \end{matrix}

which completes the proof.

The second algebraic formula we wish to emphasize is the Binomial Theorem. Before stating it, we must introduce some useful notation.

Let $n$ be a natural number. As earlier in this chapter, we define $n!$ as follows:

n! = n \times (n - 1) \times (n - 2) \times ... \times 2 \times 1 .

For later notational convenience, we also define $0!$ to be 1.

If $k$ is any integer for which $0 \leq k \leq n,$ we define the binomial coefficient $(\binom{n}{k})$ by

(\binom{n}{k}) = \frac{n!}{k! (n - k)!} = \frac{n \times (n - 1) \times (n - 2) \times ... \times (n - k + 1)}{k!} .

Prove that $(\binom{n}{0}) = 1, (\binom{n}{1}) = n$ and $(\binom{n}{n}) = 1 .$
Prove that
$(\binom{n}{k}) \leq \frac{2 n^{k}}{2^{k}}$
for all natural numbers $n$ and all integers $0 \leq k \leq n .$
Prove that
$(\binom{n + 1}{k}) = (\binom{n}{k}) + (\binom{n}{k - 1})$
for all natural numbers $n$ and all integers $1 \leq k \leq n .$

If $x, y \in R$ and $n$ is a natural number, then

{(x + y)}^{n} = \sum_{k = 0}^{n} (\binom{n}{k}) x^{k} y^{n - k} .

We shall prove this theorem by induction. If $n = 1,$ then the assertion is true, for ${(x + y)}^{1} = x + y$ and

\sum_{k = 0}^{1} (\binom{1}{k}) x^{k} y^{1 - k} = (\binom{1}{0}) x^{0} y^{1} + (\binom{1}{1}) x^{1} y^{0} = x + y .

Now, assume that the assertion holds for the natural number $j;$ i.e.,

{(x + y)}^{j} = \sum_{k = 0}^{j} (\binom{j}{k}) x^{k} y^{j - k},

and let us prove that the assertion holds for the natural number $j + 1 .$ We will make use of part (c) of [link] . We have that

\begin{matrix} {(x + y)}^{j + 1} & = & (x + y) {(x + y)}^{j} \\ = & (x + y) \sum_{k = 0}^{j} (\binom{j}{k}) x^{k} y^{j - k} \\ = & x \sum_{k = 0}^{j} (\binom{j}{k}) x^{k} y^{j - k} + y \sum_{k = 0}^{j} (\binom{j}{k}) x^{k} y^{j - k} \\ = & \sum_{k = 0}^{j} (\binom{j}{k}) x^{k + 1} y^{j - k} + \sum_{k = 0}^{j} (\binom{j}{k}) x^{k} y^{j + 1 - k} \\ = & \sum_{k = 0}^{j - 1} (\binom{j}{k}) x^{k + 1} y^{j - k} + (\binom{j}{j}) x^{j + 1} y^{0} \\ + \sum_{k = 1}^{j} (\binom{j}{k}) x^{k} y^{j + 1 - k} + (\binom{j}{0}) x^{0} y^{j + 1} \\ = & x^{j + 1} + \sum_{k = 1}^{j} (\binom{j}{k - 1}) x^{k} y^{j + 1 - k} \\ + \sum_{k = 1}^{j} (\binom{j}{k}) x^{k} y^{j + 1 - k} + y^{j + 1} \\ = & x^{j + 1} + \sum_{k = 1}^{j} ((\binom{j}{k - 1}) + (\binom{j}{k})) x^{k} y^{j + 1 - k} + y^{j + 1} \\ = & x^{j + 1} + \sum_{k = 1}^{j} (\binom{j + 1}{k}) x^{k} y^{j + 1 - k} + y^{j + 1} \\ = & (\binom{j + 1}{j + 1}) x^{j + 1} y^{0} + \sum_{k = 1}^{j} (\binom{j + 1}{k}) x^{k} y^{j + 1 - k} + (\binom{j + 1}{0}) x^{0} y^{j + 1} \\ = & \sum_{k = 0}^{j + 1} (\binom{j + 1}{k}) x^{k} y^{j + 1 - k}, \end{matrix}

which shows that the assertion of the theorem holds for the natural number $j + 1 .$ This completes the proof.

The next exercise is valid in any ordered field, but, since we are mainly interested in the order field $R,$ we state everything in terms of that field.

If $x$ and $y$ are positive real numbers, and if $n$ and $k$ are natural numbers with $k \leq n,$ show that ${(x + y)}^{n} \geq (\binom{n}{k}) x^{k} y^{n - k} .$
For any positive real number $x$ and natural number $n,$ show that ${(1 + x)}^{n} \geq 1 + n x .$
For any real number $x > - 1$ and natural number $n,$ prove that ${(1 + x)}^{n} \geq 1 + n x .$ HINT: Do not try to use the binomial theorem as in part (b); it won't work because the terms are not all positive; prove this directly by induction.

There is one more important algebraic identity, which again can be proved by induction.It is actually just a corollary of the geometric progression formula.

If $x, y \in R$ and $n$ is a natural number, then

x^{n} - y^{n} = (x - y) (\sum_{j = 0}^{n - 1} x^{j} y^{n - 1 - j} .

If $n = 1$ the theorem is clear. Suppose it holds for a natural number $k,$ and let us prove the identity for the natural number $k + 1 .$ We have

\begin{matrix} x^{k + 1} - y^{k + 1} & = & x^{k + 1} - x^{k} y + x^{k} y - y^{k + 1} \\ = & (x - y) x^{k} + y (x^{k} - y^{k}) \\ = & (x - y) x^{k} + y (x - y) (\sum_{j = 0}^{k - 1} x^{j} y^{k - 1 - j}) \\ = & (x - y) x^{k} + (x - y) (\sum_{j = 0}^{k - 1} x^{j} y^{k - j} \\ = & (x - y) (x^{k} y^{k - k} + \sum_{j = 0}^{k - 1} x^{j} y^{k - j}) \\ = & (x - y) (\sum_{j = 0}^{k} x^{j} y^{k - j}) \end{matrix}

, which shows that the assertion holds for the natural number $k + 1 .$ So, by induction, the theorem is proved.

Let $x$ and $y$ be real numbers.

Let $n$ be an odd natural number; i.e., $n = 2 k + 1$ for some natural number $k .$ Show that
$x^{n} + y^{n} = (x + y) (\sum_{j = 0}^{n - 1} {(- 1)}^{j} x^{j} y^{n - 1 - j} .$
HINT: Write $x^{n} + y^{n} = x^{n} - {(- y)}^{n} .$
Show that $x^{2} + y^{2}$ can not be factored into a product of the form $(a x + b y) (c x + d y)$ for any choices of real numbers $a, b, c,$ and $d .$

Using the Binomial Theorem together with the preceding theorem, we may now investigate the existence of $n$ th roots of real numbers. This next theorem is definitely not valid in any ordered field, forit again depends on the completeness property.

Let $n$ be a natural number and let $x$ be a positive real number. Then there exists a unique positive real number $y$ such that $y^{n} = x;$ i.e., $x$ has a unique positive $n$ th root.

Note first that if $0 \leq t < s,$ then $t^{n} < s^{n} .$ (To see this, argue by induction, and use part (e) of [link] .) Using this, we mimic the proof of [link] . Thus, let $S$ be the set of all positive real numbers $t$ for which $t^{n} \leq x .$ Then $S$ is nonempty and bounded above. Indeed, if $x \geq 1,$ then $1 \in S,$ while if $x < 1,$ then $x$ itself is in $S .$ Therefore, $S$ is nonempty. Also, using part (b) of [link] , we see that $1 + (x / n)$ is an upper bound for $S .$ For, if $t > 1 + x / n,$ then

t^{n} > {(1 + (x / n))}^{n} \geq 1 + n (x / n) > x .

Now let $y = sup S,$ and let us show that $y^{n} = x .$ We rule out the other two possibilities. First, if $y^{n} > x,$ let $ϵ$ be the positive number $y^{n} - x,$ and define $ϵ^{'}$ to be the positive number $ϵ / (n y^{n - 1}) .$ Then, using [link] , choose $t \in S$ so that $y - ϵ^{'} < t \leq y .$ ( [link] is where the completeness of the ordered field $R$ is crucial.) We have

\begin{matrix} ϵ & = & y^{n} - x \\ = & y^{n} - t^{n} + t^{n} - x \\ \leq & y^{n} - t^{n} \\ = & (y - t) (\sum_{j = 0}^{n - 1} y^{j} t^{n - 1 - j}) \\ \leq & (y - t) (\sum_{j = 0}^{n - 1} y^{j} y^{n - 1 - j}) \\ = & (y - t) (\sum_{j = 0}^{n - 1} y^{n - 1} \\ < & ϵ^{'} n y^{n - 1} \\ = & ϵ, \end{matrix}

and this is a contradiction. Therefore, $y^{n}$ is not greater than $x .$

Now, if $y^{n} < x,$ let $ϵ$ be the positive number $x - y^{n},$ and choose a $δ > 0$ such that $δ < 1$ and $δ < ϵ / {(y + 1)}^{n} .$ Then, using the Binomial Theorem, we have that

\begin{matrix} {(y + δ)}^{n} & = & \sum_{k = 0}^{n} (\binom{n}{k}) y^{k} δ^{n - k} \\ = & y^{n} + \sum_{k = 0}^{n - 1} (\binom{n}{k}) y^{k} δ^{n - k} \\ = & y^{n} + δ \sum_{k = 0}^{n - 1} (\binom{n}{k}) y^{k} δ^{n - 1 - k} \\ < & y^{n} + δ \sum_{k = 0}^{n} (\binom{n}{k}) y^{k} 1^{n - k} \\ = & y^{n} + δ {(y + 1)}^{n} \\ = & x - ϵ + δ {(y + 1)}^{n} \\ < & x - ϵ + ϵ \\ = & x, \end{matrix}

implying that $y + δ \in S .$ But this is a contradiction, since $y = sup S .$ Therefore, $y^{n}$ is not less than $x,$ and so $y^{n} = x .$

We have shown the existence of a positive $n$ th root of $x .$ To see the uniqueness, suppose $y$ and $y^{'}$ are two positive $n$ th roots of $x .$ Then

\begin{matrix} 0 & = & y^{n} - {y^{'}}^{n} \\ = & (y - y^{'}) (\sum_{j = 0}^{n - 1} y^{j} {y^{'}}^{n - j - 1}, \end{matrix}

which implies that either $y - y^{'} = 0$ or $\sum_{j = 0}^{n - 1} y^{j} {y^{'}}^{n - j - 1} = 0 .$ Since this latter sum consists of positive terms, it cannot be 0, whence $y = y^{'} .$ This shows that there is but one positive $n$ th root of $x,$ and the theorem is proved.

Show that if $n = 2 k$ is an even natural number, then every positive real number has exactly two distinct $n$ th roots.
If $n = 2 k + 1$ is an odd natural number, show that every real number has exactly one $n$ th root.
If $n$ is a natural number greater than 1, prove that there is no rational number whose $n$ th power equals 2, i.e., the $n$ th root of 2 is not a rational number.

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Source: OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1

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