2
$\begingroup$

So, in attempting to compute the condition number for the 2-norm of a matrix, I have stumbled upon a problem i can't resolve.

I have the formula

$$ \frac{1-\cos\left(\frac{n}{n+1} \pi\right)}{1-\cos\left(\frac{1}{n+1} \pi\right)}. $$

I arrived at this because the eigenvalues for my matrix are $2\left(1-\cos\left(\frac{p}{n+1} \pi\right)\right)$ for $p=1,..,n$.

The problem comes from the fact that I need to show that this formula somehow ends up being equal to $$\frac{1}{\tan^2\left(\frac{1}{2(n+1)}\pi\right)} $$

I've arrived at

$$ \frac{1-\cos\left(\pi\frac{n}{n+1}\right)}{2\sin^2\left(\frac{\pi}{2(n+1)}\right)} $$

but can't seem to get any further, mostly due to the $n$ in the numerator of the cosine.

Any help would be appreciated.

$\endgroup$

1 Answer 1

2
$\begingroup$

You have the right idea in exploiting the identity $$1 - \cos\theta \;=\; 2 \sin^2 \frac{\theta}{2}$$ to re-write the denominator. Using that on the numerator gives $$1 - \cos\left( \pi \frac{n}{n+1} \right) = 2\sin^2\frac{\pi n}{2(n+1)} \tag{$\star$}$$ Now, simply observe that the arguments of the sines combine very conveniently: $$\frac{\pi n}{2(n+1)} + \frac{\pi}{2(n+1)}= \frac{\pi(n+1)}{2(n+1)} = \frac{\pi}{2}$$ so that $(\star)$ becomes $$2\sin^2\left( \frac{\pi}{2} - \frac{\pi}{2(n+1)} \right) = 2\cos^2\left(\frac{\pi}{2(n+1)}\right)$$ since $\sin(\pi/2-\theta) = \cos\theta$. The result follows. $\square$

$\endgroup$
1
  • 1
    $\begingroup$ thanks a ton! I wasn't aware of that last identity. $\endgroup$
    – Sertii
    Oct 6, 2015 at 8:43

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.