Here is a nasty trig induction question i dont even pretend to know how to begin. I think it is a little out of the scope of my course, but would like to try it, just in case.
thanks jacs
Add $\displaystyle tan^{-1}(\frac{1}{2(n+1)^{2}})$ to both sides
That would be next in your series.
$\displaystyle tan^{-1}(\frac{1}{2\cdot{1^{2}}})+tan^{-1}(\frac{1}{2\cdot{2^{2}}})+$$\displaystyle ...........+tan^{-1}(\frac{1}{2n^{2}})+tan^{-1}(\frac{1}{2(n+1)^{2}})$$\displaystyle =\frac{\pi}{4}-tan^{-1}(\frac{1}{2n+1})+tan^{-1}(\frac{1}{2(n+1)^{2}})$
Now, your mission is to show that $\displaystyle \frac{\pi}{4}-tan^{-1}(\frac{1}{2n+1})+tan^{-1}(\frac{1}{2(n+1)^{2}})$$\displaystyle =\frac{\pi}{4}-tan^{-1}(\frac{1}{2(n+1)+1})=\frac{\pi}{4}-tan^{1}(\frac{1}{2n+3})$
It does, but can you show how?.
I am afraid i have no idea how to proceed at all. This question i think is a little beyond what were are supposed to be able to do at our course level. It was presented as a challenge question, but even with your start, i don't know how to proceed.
jacs
For $\displaystyle n=1$
The left hand side is,
$\displaystyle \tan^{-1} \frac{1}{2\cdot 1^2}=\tan^{-1} \frac{1}{2}=\frac{\pi}{4}$
The right hand side is,
$\displaystyle \frac{\pi}{4}-\tan^{-1} \frac{1}{2(1)+1}=\frac{\pi}{4}-\mbox{ some non-zero number}$
No wonder you cannot prove it!
Hey PH, I believe you made a slight error.Originally Posted by ThePerfectHacker
$\displaystyle tan^{-1}(1)=\frac{\pi}{4}$
Left side:
$\displaystyle tan^{-1}(\frac{1}{2})\neq\frac{\pi}{4}$
It equals 0.463647609001
The right side:
$\displaystyle \frac{\pi}{4}-tan^{-1}(\frac{1}{3})=0.46347609001$
It is true for n=1
He has proven you wrong - he just has not proven that:Originally Posted by ThePerfectHacker
$\displaystyle
\arctan(1/2)=\frac{\pi}{4}-\arctan(1/3)
$
which follows immediately from taking tan of both sides and the trig identity
for the tan of a difference of two angles.
RonL
Yes, Cap'N. That method proves the general result, also.
$\displaystyle tan(u-v)=\frac{tan(u)-tan(v)}{1+tan(u)tan(v)}$
Skipping ahead a lot of algebra:
$\displaystyle \frac{\frac{1}{2(n+1)^{2}}-\frac{1}{2n+1}}{1+\frac{1}{2(n+1)^{2}}\cdot\frac{1 }{2n+1}}=\frac{-1}{2n+3}$
$\displaystyle tan(\frac{-1}{2n+3})=-tan(\frac{1}{2n+3})$