MHB Proving sin⁻¹(ix): 2nπ ± i log (√1+x²+x)

  • Thread starter Thread starter Suvadip
  • Start date Start date
  • Tags Tags
    Inverse
Suvadip
Messages
68
Reaction score
0
I need to prove
$$sin^{-1}(ix)=2n\pi\pm i log (\sqrt{1+x^2}+x)$$

I can prove $$sin^{-1}(ix)=2n\pi+ i log (\sqrt{1+x^2}+x)$$

How to prove the other part. Please help
 
Physics news on Phys.org
suvadip said:
I need to prove
$$sin^{-1}(ix)=2n\pi\pm i log (\sqrt{1+x^2}+x)$$

I can prove $$sin^{-1}(ix)=2n\pi+ i log (\sqrt{1+x^2}+x)$$

How to prove the other part. Please help

You have to find the z for which is...

$\displaystyle \sin z = \frac{e^{i\ z}-e^{- i\ z}}{2\ i} = i\ x$ (1)

Setting in (1) $\displaystyle e^{i\ z}=y$ You arrive to the equation...$\displaystyle y^{2} + 2\ x\ y -1 =0 $ (2)... which is solved for $\displaystyle y= - x \pm \sqrt{1+x^{2}}$ so that is... $\displaystyle \sin^{-1} (i\ x) = 2\ \pi\ i\ n - i\ \ln (- x \pm \sqrt{1+x^{2}}) = 2\ \pi\ i\ n + i\ \ln (x \pm \sqrt{1+x^{2}})$ (3)

Kind regards

$\chi$ $\sigma$
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
792
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K