0
$\begingroup$

How does $\sqrt{\frac{2(cos(x)-1)}{cos(x)}} \approx x$?

At first I thought it was an example of binomial approximation but I was unable to approximate is to just $x$. I could only get $$\left(\frac{2cos(x)}{cos(x)}-\frac{2}{cos(x)}\right)^{\frac{1}{2}}$$ $$\left(2-\frac{2}{cos(x)}\right)^{\frac{1}{2}} \approx 2-\frac{1}{cos(x)}$$

Am I using the binomial approximation theorem correctly? Or is there another way of approximating this value?

$\endgroup$
  • $\begingroup$ $\cos(x)= 1- x^2/2 + \cdots$ $\endgroup$ – gammatester Oct 29 '17 at 20:11
  • $\begingroup$ Is it in a neighbourhood of $0$? Then the radicand is negative. $\endgroup$ – Bernard Oct 29 '17 at 20:11
  • 1
    $\begingroup$ I think you have a sign problem. For $x$ near $0$ your expression is imaginary. Perhaps you meant $1-\cos x$ in the numerator? $\endgroup$ – lulu Oct 29 '17 at 20:14
  • $\begingroup$ @lulu unfortunatly no, it is $cos(x)-1$ $\endgroup$ – Niamh O'SS Oct 29 '17 at 20:17
  • 1
    $\begingroup$ No, really, what you wrote down is not correct. Try it for small positive values of $x$. With the sign change I proposed you get $|x|$ for small $x$. $\endgroup$ – lulu Oct 29 '17 at 20:25
2
$\begingroup$

Near $0$?

Well, first of all you have a sign error. For small $x\neq 0$ we have $0<\cos x<1$ so the expression inside the radical is clearly negative. Accordingly, I believe you are trying to show that $$\sqrt {2\,\frac {1-\cos x}{\cos x}}\approx |x|$$ for $x$ near $0$. At least, this has the advantage of being true.

To see it, note that $$\cos x \approx 1 -\frac {x^2}2\implies \frac 1{\cos x}\approx1+\frac {x^2}2$$ Here the second approximation follows from the first by means of first stage of the Geometric series $\frac 1{1-z}\approx 1+z$.

It follows that the modified expression is approximately $$\sqrt {2\times \left(1+\frac {x^2}2\right)\times \frac {x^2}2}\approx \sqrt {x^2+\frac {x^4}2}\approx |x|$$

$\endgroup$
  • $\begingroup$ The jump from$$\cos x\approx\dots$$to$$\frac1{\cos x}\approx\dots$$is probably not immediately clear. $\endgroup$ – Simply Beautiful Art Oct 29 '17 at 21:13
  • $\begingroup$ @SimplyBeautifulArt Thanks, I'll add a sentence. $\endgroup$ – lulu Oct 29 '17 at 21:17

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.