5
$\begingroup$

Prove that $\sqrt{8}=1+\dfrac34+\dfrac{3\cdot5}{4\cdot8}+\dfrac{3\cdot5\cdot7}{4\cdot8\cdot12}+\ldots$

My work:
$\sqrt8=\bigg(1-\dfrac12\bigg)^{-\frac32}$

Now, I suppose there is some "binomial expansion with rational co-efficients" or something similar for this, which I do not know. Please help.

N.B.: Any other solution is also acceptable, I do not have any restriction regarding the solution.

$\endgroup$
  • $\begingroup$ what's $4.8.12$? $\endgroup$ – Alex Feb 19 '14 at 10:37
  • 2
    $\begingroup$ Look at the binomial series. $\endgroup$ – J.R. Feb 19 '14 at 10:37
  • $\begingroup$ @Alex 4*8*12=384 $\endgroup$ – Hawk Feb 19 '14 at 10:38
  • $\begingroup$ @TooOldForMath I could not follow the series, can you please state the rule which it follows at each term? $\endgroup$ – Hawk Feb 19 '14 at 10:41
  • $\begingroup$ @Hawk: I explained it in an answer below. $\endgroup$ – J.R. Feb 19 '14 at 10:54
5
$\begingroup$

The binomial series is "just" the Taylor series of $(1+x)^{\alpha}$ at $x=0$.

Start deriving $f(x)=(1+x)^{\alpha}$ by $x$ and you get $\alpha(1+x)^{\alpha-1}$, then $\alpha(\alpha-1)(1+x)^{\alpha-2}$ for the first derivative etc. The $n$th derivative is

$$\frac{d^n}{dx^n} (1+x)^{\alpha} = \alpha(\alpha-1)\cdots(\alpha-n+1)(1+x)^{\alpha-n}$$

So the Taylor series is

$$\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n= \sum_{n=0}^\infty \frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-n+1)}{n!} x^n$$

If $\alpha$ is a non-negative integer this series becomes a finite sum and the $n$th coefficient is simply ${\alpha\choose n}$, so it makes sense to call it a "generalized" binomial coefficient and denote it by the same symbol, also if $\alpha$ is not a non-negative integer.

Now notice that the Taylor series has convergence radius $1$ and that $(1+x)^{\alpha}$ is real-analytic, therefore

$$(1+x)^{\alpha} = \sum_{n=0}^\infty {\alpha\choose n} x^n$$

for all $|x|<1$, in particular $x=-\frac{1}{2}$ and $\alpha=-3/2$. Now notice that $${-3/2\choose n}=\frac{(-1)^n}{n!2^n} 3\cdot 5\cdots (2n+1)$$

Then,

$$\begin{align} \sqrt{8} &=(1-1/2)^{-3/2} = \sum_{n=0}^\infty \frac{(-1)^n}{n!2^n} 3\cdot 5\cdots (2n+1) \frac{1}{(-2)^n} = \sum_{n=0}^\infty \frac{3\cdot 5\cdots (2n+1)}{n! 4^n}\\ &= 1 + \frac{3}{4} + \frac{3\cdot 5}{4\cdot 8} + \frac{3\cdot 5\cdot 7}{4\cdot 8\cdot 12} + \cdots \end{align} $$

$\endgroup$
  • $\begingroup$ I understand it now! $\endgroup$ – Hawk Feb 19 '14 at 11:27
3
$\begingroup$

It can be shown that $\displaystyle (1 + x)^{\alpha} = \sum_{n = 0}^{\infty} \binom{\alpha}{n}x^n$ when $|x| < 1$. This is not really precalculus though...

EDIT: I'll add a proof for the sum:

Let $s(x)$ be the sum of the series. Then

$\begin{align} (1 + x)s'(x) &= (1 + x)\sum_{n = 0}^{\infty} n\binom{\alpha}{n}x^{n-1} \\ &= \sum_{n = 0}^{\infty} (n+1)\binom{\alpha}{n+1}x^n + \sum_{n = 0}^{\infty}n\binom{\alpha}{n}x^n \\ &= \sum_{n = 0}^{\infty} \left[(n+1)\binom{\alpha}{n+1} + n\binom{\alpha}{n}\right]x^n \\ &= \alpha \sum_{n = 0}^{\infty} \binom{\alpha}{n}x^n = \alpha \; s(x) \end{align}$

Furthermore, let $f(x) := (1+x)^{-\alpha} s(x)$. Then

$\displaystyle f'(x) = -\alpha(1+x)^{-\alpha - 1} s(x) + (1+x)^{-\alpha} s'(x) = -(1+x)^{-\alpha} s'(x) + (1+x)^{-\alpha} s'(x) = 0$

which implies that $f(x)$ is costant. Since $f(0) = 1$, we have $f(x) = 1$ and therefore $(1+x)^{-\alpha} s(x) = 1$. Multiply both sides by $(1+x)^{\alpha}$ and we are done.

$\endgroup$
  • $\begingroup$ What is $\alpha \choose n$ when $\alpha$ is fraction? $\endgroup$ – Hawk Feb 19 '14 at 10:41
  • 2
    $\begingroup$ It is called generalized binomial coefficient en.wikipedia.org/wiki/… $\endgroup$ – dani_s Feb 19 '14 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.