2
$\begingroup$

While determining fourier series expansion of following function.

f(x) = $\sqrt{1-\cos x} $

here $f(-x) = f(x)$ so it is even. But

the above function can also be written as $f(x) = \sqrt 2\sin(x/2)$

here $f(-x) = -f(x)$ so by that logic it is odd

Can anyone explain where I am wrong. Thanks in advance

$\endgroup$
5
  • 4
    $\begingroup$ Yes: $\sqrt{A^2}=\lvert A\rvert$. $\endgroup$
    – Bernard
    Aug 19, 2016 at 13:31
  • $\begingroup$ @ raj Please check my edit/formatting. To take your notation literally, we have $\sqrt 2\sin(x/2)$. But did you mean \sqrt{2\sin(x/2)} = $\sqrt{2\sin(x/2)}$? $\endgroup$
    – amWhy
    Aug 19, 2016 at 13:38
  • $\begingroup$ No the original notation is correct, and now i understand where i went wrong thank you $\endgroup$
    – raj kapoor
    Aug 19, 2016 at 13:40
  • $\begingroup$ $cos(x)$ is even, so? $\endgroup$
    – tired
    Aug 19, 2016 at 13:43
  • $\begingroup$ we can also write $ \sqrt{1-\cos x} $ as $\sqrt 2\sin(x/2)$ and sin is odd function because of which I was confused $\endgroup$
    – raj kapoor
    Aug 19, 2016 at 13:49

1 Answer 1

Reset to default
2
$\begingroup$

As said @Bernard in a comment, $\forall A\in\Bbb R,\sqrt{A^2}=|A|$.

Hence $$f(x)=\sqrt{2}\sqrt{\sin^2(x/2)}=\sqrt{2}|\sin(x/2)|$$ and you can check that $f$ is even.

$\endgroup$
10
  • $\begingroup$ $\sqrt{A^2}\,\,$ $\endgroup$
    – user307169
    Aug 19, 2016 at 13:41
  • $\begingroup$ Thank you I understood where i went wrong $\endgroup$
    – raj kapoor
    Aug 19, 2016 at 13:51
  • $\begingroup$ Why $\sqrt(A^2)=|A|$?@paf $\endgroup$
    – Sathasivam K
    Aug 19, 2016 at 14:09
  • $\begingroup$ If $A<0$, then $\sqrt{A^2}=-A$ (e.g. $\sqrt{(-3)^2}=\sqrt{9}=3$). $\endgroup$
    – paf
    Aug 19, 2016 at 14:12
  • 1
    $\begingroup$ Square root is the inverse of squaring only on positive numbers. $\endgroup$
    – paf
    Aug 19, 2016 at 14:39

You must log in to answer this question.

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