In the solution given how they got the circled term .
I could not understand that as it contain infinite series
Asked
Active
Viewed 145 times
0
-
See http://math.stackexchange.com/questions/2125682/to-prove-tana-2-tan2a-4-tan4a-8-cot8a-cota – lab bhattacharjee Mar 29 '17 at 10:33
-
This term is zero. – Mar 29 '17 at 10:41
-
Possible duplicate of To prove $ \tan(A) + 2 \tan(2A) +4\tan4A + 8 \cot8A =\cot(A) $ – skyking Mar 29 '17 at 10:41
-
@skyking no it is not similiar to that – Koolman Mar 29 '17 at 10:44
-
@labbhattacharjee that question does not contain infinite series. – Koolman Mar 29 '17 at 10:45
-
@Koolman, Did u get the answer? – lab bhattacharjee Mar 29 '17 at 11:04
-
@labbhattacharjee now I am more confused . Totally blank – Koolman Mar 29 '17 at 11:07
1 Answers
2
We use the identity $\tan x = \cot x - 2\cot 2x$. That is $2^{-k}\tan 2^{-k}x = 2^{-k}\cot x - 2^{-(k-1)}\cot 2^{-(k-1)}x$ so terms will cancel. To put it more formally:
Let $S_n = \sum_1^n 2^{-k} \tan 2^{-k}x$. Now assume that $S_n = -\cot x + 2^{-1}\cot 2^{-n}x$ then we have:
$$S_{n+1} = S_n + 2^{-(n+1)}\tan 2^{-(n+1)}x = -\cot x + 2^{-1}\cot 2^{-n}x + 2^{-(n+1)}\left(\cot 2^{-(n+1)}x - 2\cot 2^{-n}x\right)\\ = -\cot x + 2^{-(n+1)}\cot 2^{-n(+1)}x $$
Also we see that.
$S_1 = {1\over 2}\tan{x\over 2} = {1\over2}\left(\cot {x\over 2} - 2\cot x\right) = -\cot x+2^{-1}\cot 2^{-1}x$$
This by induction shows that $S_n = -\cot x + 2^{-1}\cot 2^{-n}x$ for all $n\in\mathbb N$.
The only part where infinity is involved now is to realize that $\sum_1^\infty 2^{-k} \tan 2^{-k}x$ is indeed $\lim S_n$. If you're only require it to converge conditionally it surely is since $S_n$ is the partial sums. If you require absolute convergence you can observe that $0<2^{-k}\tan x/2^k < 2^{-k}\tan x/2 < 2^{-k}$ in the interval, which is enough to have uniform and absolute convergence.
skyking
- 16,654