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#1 Help Me ! » complex » 2009-10-02 07:26:52
- mathkeep
- Replies: 0
.
#2 Re: Help Me ! » integrate » 2009-08-25 01:35:35
[hide]
Nevermind this I made a mistake on the first line
anyways atleast u tried..
#3 Re: Help Me ! » integrate » 2009-08-24 23:42:06
hi Ricky!
heyee..i have got the solution for the integral..
but the hint u provided isnt working for the integral
i converted the above expression into the form.. ...
n then i applied the form
after then i put tan(x)=t
n then putting all values..ultimately gave me
after that m confused what to do next..
#4 Re: Help Me ! » integrate » 2009-08-24 14:46:14
Can I ask what textbook this came out of ?
actually it came randomly to my mind..i was solving the integral of
n i thought to find the integral that i asked in post #1
#5 Re: Help Me ! » integrate » 2009-08-24 11:16:27
yep i need the steps..
#6 Help Me ! » integrate » 2009-08-24 06:43:49
- mathkeep
- Replies: 8
Integrate :
#7 Re: Help Me ! » Limit » 2009-08-05 23:41:52
Unfortunately soroban, the question was to prove that it converges to 0, not just to show convergence.
hmm..well Sorobon ...Ricky is right..anyways...that may help me in some other question 
:D
#8 Re: Help Me ! » Limit » 2009-08-05 23:36:20
I think a much easier proof would be to just note that
Note that this is just using the fact that:
And then making this swap for each n, n-1, n-2, ... except for 1.
This shows that
yeah a nice proof..thanks Ricky..
#9 Re: Help Me ! » Limit » 2009-08-05 08:48:41
hi Riad Zaidan!
thanks for the proof..got convinced from that easily..
n juriguen thanks for the link too
#10 Help Me ! » Limit » 2009-08-05 07:01:27
- mathkeep
- Replies: 12
here is a question of limit...
actually my limit is coming out to be 0...i thought in this way...
write the limit as :
so the further ratios will make 1 smaller and smaller....thus making the fraction almost zero...
but i want a rigorous proof..
#11 Re: Help Me ! » solve it! » 2009-08-02 02:23:37
hmm..any more replies???
i also think we are done with all the solutions...
#12 Re: Help Me ! » solve it! » 2009-08-02 02:20:02
Hi mathkeep;
Yes, only when n is even. Juriguen has that indicated in post #4.
ok..i did not see that ...thanks for telling ..i think we have got all the solutions now..from post #4...
what do u people say??
#13 Re: Help Me ! » solve it! » 2009-08-02 02:14:32
Hi Juriguen;
Your idea was a lot better than mine so no apology necessary.
Hi mathkeep;
Also for n>3 and odd, x takes imaginary values. For n=3 we have x = 2 π m+(2.35619449... + .881373587... i) m ∈ Z. The complex coefficients do not have a simple form.
nice work but i want real solutions only...
#14 Re: Help Me ! » solve it! » 2009-08-02 02:13:28
Indeed, it would be something like:
But this is not so easy any more!
Jose
yeas u r right now...but as u said this not so easy anymore....
n i have another problem:
when n is odd x={2k(pi)+1}....
will always be =-1 n will always be =0...i mean that can be the solution only if n is even.#15 Re: Help Me ! » solve it! » 2009-08-02 01:53:00
hi juriguen, also m not able to get it..why are u taking n to be a even when the solution is
....cuzat x= ... and ... i mean it doesnt matter wheher n is even or odd...correct me if m wrong
#16 Re: Help Me ! » solve it! » 2009-08-02 01:38:08
hi juriguen!
yeah...ur solution seems to be correct but i could not understand this step:
i mean was that an assumption...or what?
#17 Re: Help Me ! » solve it! » 2009-08-02 01:27:10
hi bobbym..yeah..you r right but what are the other solutions..i was able to make it out..that
is always a solution..n am looking for the method of solving this question too
#18 Help Me ! » solve it! » 2009-08-02 01:03:29
- mathkeep
- Replies: 19
solve the equation:
where n is a given positive integer
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