62whichever you feel like :)
341To be honest, I have all of Titu's books, but I have never encountered this problem! Just shows how carefully I have read them :D.
Its a beautiful problem and so I am sorely tempted to post my solution (not able to effectively hide the latexed portions)
Let z = \cos \frac{2 \pi}{3^n+1} + i \sin \frac{2 \pi}{3^n+1}.
Then z ^{3^n+1} = 1
Also 2 \cos \frac {2 \pi . 3^k}{3^n+1} = z^{3^k}+ \frac{1}{z^{3^k}}
Hence the given expression can be written as
= \left(1 + z^3 + \frac{1}{z^3} \right) \left(1+z^9 + \frac{1}{z^9} \right)...\left(1+z^{3^n} + \frac{1}{z^{3^n}} \right)
= \frac{\left(1 + z^3 + z^6 \right) \left(1+z^9 + z^{18} \right)...\left(1+z^{3^n} + \left(z^{3^n} \right)^2 \right)}{z^{3+9+...+3^n}}
The trick is to multiply this expression by (1-z3) and the numerator will collapse domino fashion so that expression becomes
\frac{1-z^{3^n+1}}{(1-z^3)z^{3 \left(\frac{3^n-1}{2} \right)}}
Since z^{3^n+1} =1 \Rightarrow z^{3^n} = \frac{1}{z}
z^{3 \left(\frac{3^n-1}{2} \right)} = \frac{z^{3 \left(\frac{3^n+1}{2} \right)}}{z^3} = \frac{ \left(z^{ \left(\frac{3^n+1}{2} \right)}\right)^3}{z^3}
But z^{ \left(\frac{3^n+1}{2} \right)} = -1
Hence the expression simplifies to \frac{-z^3 \left(1 - \frac{1}{z^3} \right)}{1-z^3} = \boxed{1}
Hats off to the person who posed this one
1can u please tell how that idea struck ur mind that "if i mutiply the numerator with 1-z^3 the numerator collapses to single term" if one dedicates 8 hours a day for math,,,wil he be able to concieve such ideas after 5 months?????
62
Lokesh Verma
·2009-09-29 22:49:13
@xyz.. you will be able to solve these... you need to practice some more of this kind.. that is the trick..
I Loved one phrase.. an opening one from Problem solving strategies.. which says "Problem solving can be learned only by solving problems"
341
Hari Shankar
·2009-09-29 22:56:07
Thats the kind of question I wish every student asks - what motivated this approach? I would recommend that you guys go after those who post solutions and ask why on earth did you do that?
In fact just before posting the solution I noticed a post http://www.targetiit.com/iit-jee-forum/posts/calculate-the-product-for-n-2-11506.html which you will realize is on very similar lines.
I will try to retrace the line of thought in my mind. When I first saw the problem, from previous problems I knew that when you see 2 cos θ one of the ways of writing it is z + 1/z where z = cos θ + i sin θ
This led me up a wrong alley at first, because the next step usually is to look for a polynomial in z+1/z
Then I retrieved one of the 1st thoughtsI had when I saw the problem that I need something like (1+x+x2) because it will give me (1-x3) which will then multiply into the next term etc. - reason being that the variable in a bracket is the cube of the argument in the previous bracket [this approach has a name - wishful thinking - and is an amazing aid in problem solving]
That way of writing was of course staring me in my face and the rest was easy.
The key is as you noted, practice. There is a book called 'The Art and Craft of Problem Solving' by Paul Zeitz which discusses different strategies to improve one's problem-solving abilities. They mention trying out jumble-puzzles as an important method. Reason is that it involves recognizing something you already know that is placed in an unfamiliar context, much like math problems.
62
Lokesh Verma
·2009-09-29 23:03:43
@Prophet sir.. awesome :) [1]
this one was from Mathematical Olympiad Challenges...
Also, the problem has a telescopic product method...
I gave it for that :)
341
Hari Shankar
·2009-09-29 23:04:33
nishant sir: my guess is that this is from Complex Nrs A-Z. Could u help me locate the problem
62
Lokesh Verma
·2009-09-29 23:13:43
It is exercise 1.9 last problem....
341
Hari Shankar
·2009-09-29 23:21:42
Ya located it. Actually their solution is more awesome
62
Lokesh Verma
·2009-09-29 23:27:45
But I like yours more..
Not because it is more complex.. but bcos i have an intrinsic bias for solving trigo questions by complex :D
62
Lokesh Verma
·2009-11-11 00:28:00
Can someone think of the alternate solution given in the book? That is nice too!
