0
$\begingroup$

How does one deduce the approximation of $\sqrt{7}$ to be $\frac{10837}{4096}$ by taking $x = \frac{1}{64}$ in the expansion of $\sqrt{1-x}$?

How should you approach such a question? I assume the first step would be to expand $\sqrt{1-x}$ which can only be done through binomial theorem (afaik).

That gives $\sqrt{1-x} = 1 - \frac{1}{2}(x) - \frac{1}{8}(x^2)$ + ... and so on.

How do you continue? I can't seem to figure out how taking $x = \frac{1}{64}$ accomplishes anything.

$\endgroup$
9
  • 2
    $\begingroup$ Well, $$\sqrt7=\frac83\sqrt{1-\frac1{64}}=\frac83\left(1-\frac12\frac1{64}-\frac18\left(\frac1{64}\right)^2+\ldots\right)$$ hence $$\sqrt7\approx\frac{8\cdot(8\cdot(64)^2-4\cdot64-1)}{3\cdot8\cdot(64)^2}=\cdots$$ $\endgroup$
    – Did
    Sep 29, 2018 at 9:14
  • $\begingroup$ Forgive my ignorance but, how did you get $\sqrt7 = \frac{8}{3}\sqrt{1-\frac{1}{64}}$? $\endgroup$
    – eclmist
    Sep 29, 2018 at 9:18
  • 1
    $\begingroup$ @gimusi. They don't give Field medals here ? I am really disappointed to hear that. Cheers. $\endgroup$ Sep 29, 2018 at 9:33
  • 2
    $\begingroup$ @gimusi The friendliness in this stack exchange is refreshing. Thanks! $\endgroup$
    – eclmist
    Sep 29, 2018 at 9:40
  • 1
    $\begingroup$ @Sam. You well noticed this key aspect of the site ! $\endgroup$ Sep 29, 2018 at 9:42

1 Answer 1

3
$\begingroup$

HINT

Use for example

$$\sqrt{7}=\sqrt{9\cdot \frac79}=3\sqrt{\frac79}=3\sqrt{1-\frac29}$$

or according to the other hint use that

$$\sqrt{1-\frac1{64}}=\sqrt{\frac{63}{64}}=\frac38\sqrt 7$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .