Transcedental roots
- Tolaso J Kos
- Administrator
- Posts: 867
- Joined: Sat Nov 07, 2015 6:12 pm
- Location: Larisa
- Contact:
Transcedental roots
Prove that the non zero roots of the equation $\tan x =x $ are transcedental.
Imagination is much more important than knowledge.
Re: Transcedental roots
We have that:
$$\tan x = \frac{e^{ix}- e^{-ix}}{i \left ( e^{ix} +e^{-ix} \right )}= -i \frac{e^{2ix}- 1}{e^{2ix}+1} \Rightarrow e^{2ix} = \frac{1+i\tan x}{1-i \tan x}$$
If $x$ is an algebraic number, then $2ix$ is also algebraic hence from Lindemann's theorem $e^{2ix}$ is transcedental (for $x\neq 0$). Hence $\tan x$ must also be transcedental.
$$\tan x = \frac{e^{ix}- e^{-ix}}{i \left ( e^{ix} +e^{-ix} \right )}= -i \frac{e^{2ix}- 1}{e^{2ix}+1} \Rightarrow e^{2ix} = \frac{1+i\tan x}{1-i \tan x}$$
If $x$ is an algebraic number, then $2ix$ is also algebraic hence from Lindemann's theorem $e^{2ix}$ is transcedental (for $x\neq 0$). Hence $\tan x$ must also be transcedental.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Create an account or sign in to join the discussion
You need to be a member in order to post a reply
Create an account
Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute
Sign in
Who is online
Users browsing this forum: No registered users and 19 guests