anonymous
  • anonymous
sum of (sin 2kx/11 + i cos2kx/11) from k=1 to k=10 where i=square root of negative one
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\[\sum_{k=1}^{10} [(\sin 2k \pi/11) +i(\cos 2k \pi/11) ]\]
anonymous
  • anonymous
what do you want to find here? ^_^
anonymous
  • anonymous
the value of this summation=)

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anonymous
  • anonymous
Well. First take 1/11 as a common factor and cut the summation to two parts.
anonymous
  • anonymous
sorry common factor??
anonymous
  • anonymous
You will see that the sum of (sin2k pi) is always zero for any value for k=1,2,3,..
anonymous
  • anonymous
Yeah.
anonymous
  • anonymous
sin (2k pi) is always zero for k=1,2,3,..*
anonymous
  • anonymous
yes.. so the sine part comes out to be zero. what about the cosine part?
anonymous
  • anonymous
on the contract cos (2k pi) is one for k=1,2,..
anonymous
  • anonymous
So the summation of cos (2k pi) is from k=1 to k=10 is equal to 10. right?
anonymous
  • anonymous
Are you following? :)
anonymous
  • anonymous
well the answer is supposed to be -i
anonymous
  • anonymous
\[{1 \over 11}\sum_{k=1}^{10}\sin(2k \pi)+{i \over 11}\sum_{k=1}^{10}\cos (2k \pi)=0+{i \over 11}(10)\]
anonymous
  • anonymous
Are you sure?
anonymous
  • anonymous
wait let me check again.
anonymous
  • anonymous
Is 11 part of the angle?
anonymous
  • anonymous
1/11 is not multiplied to sin but its multiplied to its angle
anonymous
  • anonymous
isn't cos(20 pi) = -1? which will give you -i/11?
anonymous
  • anonymous
LOL
anonymous
  • anonymous
what x_x
anonymous
  • anonymous
and 1/11 = not part of the angle
anonymous
  • anonymous
You should have put that in parenthesis >.<
anonymous
  • anonymous
why didn't you put it in the angle =_=...
anonymous
  • anonymous
Well, you got a solution for "a" problem :P
anonymous
  • anonymous
hey anwar, aren't you supposed to find the summation of both sin and cos from 1 - 10?
anonymous
  • anonymous
sorry about that guys
anonymous
  • anonymous
Yeah sstarica
anonymous
  • anonymous
It's OK.. But, I am not sure if I have enough time to do the problem. Just let me think about it quickly.
anonymous
  • anonymous
okay anwar
anonymous
  • anonymous
is it possible to solve using euler's representation of complex no.?
anonymous
  • anonymous
like this : \[[\sum_{k=1}^{10}\sin(\frac{2 \pi}{11}) + i \cos(\frac{2\pi}{11}) ] + [\sum_{k=1}^{10}\sin(\frac{2(10)\pi}{11}) + i \cos(\frac{2(10) \pi}{11})]\] ? .-. right
anonymous
  • anonymous
That's what I am thinking about right now.
anonymous
  • anonymous
lol
anonymous
  • anonymous
yeah all through 2,3,4 to 10
anonymous
  • anonymous
@sstarica: you just substitute for k=1,2,3,.. and add them up.
anonymous
  • anonymous
no need, there's a faster way to compute the sum without going through the 1 2 3 stages :)
anonymous
  • anonymous
take first term and add it with last term
anonymous
  • anonymous
and whats that?
anonymous
  • anonymous
something like that?
anonymous
  • anonymous
I don't know lol
anonymous
  • anonymous
Well yeah.. You just need to see a pattern.
anonymous
  • anonymous
lol..
anonymous
  • anonymous
my prof did that last time, he took the first term and added it up with the last term, then got the answer, or it's prolly one of my hallucinations? I just remembered he did something like that
anonymous
  • anonymous
z_z oh nvm me proceed.
anonymous
  • anonymous
\[i \sum_{k=1}^{10} [\cos(2k \pi/11) -i \sin(2k \pi/11)]\] =\[i \sum_{k=1}^{10} (e ^{-2k \pi/11})\]
anonymous
  • anonymous
i dont know how to proceed any further
anonymous
  • anonymous
that's right, but there is an i in the power.
anonymous
  • anonymous
there is?
anonymous
  • anonymous
There should be :)
anonymous
  • anonymous
o.o
anonymous
  • anonymous
what?!
anonymous
  • anonymous
yea there should be.. otherwise its not a correct representation of the complex number
anonymous
  • anonymous
o..o
anonymous
  • anonymous
\[i \sum_{k=1}^{10} (e ^{-2k \pi i/11})\] should look like this
anonymous
  • anonymous
=.=
anonymous
  • anonymous
hey star u workin on it?
anonymous
  • anonymous
I gtg now.. I'll BRB
anonymous
  • anonymous
bye ^_^
anonymous
  • anonymous
bye
anonymous
  • anonymous
I am back, but you're gone. :(
anonymous
  • anonymous
I think I am always getting the idea.
anonymous
  • anonymous
not always, almost :)
anonymous
  • anonymous
giv it a try..=)
anonymous
  • anonymous
sin(2 pi/11)+sin(20 pi/11)=?
anonymous
  • anonymous
Hello, anyone there?
anonymous
  • anonymous
sin x+sin y=2sin((x+y)/2)+....., right?
anonymous
  • anonymous
I have no idea lol
anonymous
  • anonymous
I'm trying to fix the DAMN PRINTER! >_<
anonymous
  • anonymous
Apply it to sin(2 pi/11)+sin(20 pi/11)
anonymous
  • anonymous
excuse my language ^^"
anonymous
  • anonymous
sin((2 * pi) / 11) + sin((20 * pi) / 11) = 0
anonymous
  • anonymous
It will give you 0. Same with sin(4 pi/11)+sin(18 pi/11), and so on.
anonymous
  • anonymous
yup
anonymous
  • anonymous
Do you see that the "sin" part of the summation will be 0?
anonymous
  • anonymous
agreed... wat abt the cosine part
anonymous
  • anonymous
trying it :).. just give me few minutes.
anonymous
  • anonymous
1,2,3,4,5,6,7,8,9........60
anonymous
  • anonymous
yup sure
anonymous
  • anonymous
what are you saying starica?
anonymous
  • anonymous
I'm counting ^_^
anonymous
  • anonymous
3 mins have passed
anonymous
  • anonymous
its still "few"
anonymous
  • anonymous
lol yeah~
anonymous
  • anonymous
=) arent you trying?
anonymous
  • anonymous
Don't count -.-
anonymous
  • anonymous
cosx+cosy=2cos((x+y)/2)cos((x-y)/2)
anonymous
  • anonymous
I'm too tired to try and I love counting lol =P
anonymous
  • anonymous
it looks simple though u_u
anonymous
  • anonymous
well you should know..:P
anonymous
  • anonymous
If I try, I'll figure it out, but I'm tired :( ...
anonymous
  • anonymous
why dont you go take a nap for a change..
anonymous
  • anonymous
I'm in campus, and I can't sleep during the day
anonymous
  • anonymous
r u an undergrad student?
anonymous
  • anonymous
mhm, a first year ^_^
anonymous
  • anonymous
gr8.. ehat are you majoring in?
anonymous
  • anonymous
*what
anonymous
  • anonymous
IT , computer engineering hopefully.
anonymous
  • anonymous
I am tired :)
anonymous
  • anonymous
You said you want to go to chemical engineering, right?
anonymous
  • anonymous
isnt workin out anwar?:-(
anonymous
  • anonymous
Not really. My friend is distracting me :P .. He wants me to go with him get lunch :)
anonymous
  • anonymous
lol.. anyways i'd better leave now.. thanx for your effort anwar and sstarica..=)
anonymous
  • anonymous
No problem. If I get the idea, I will just post the solution here.
anonymous
  • anonymous
okay buddy.. bye
anonymous
  • anonymous
Bye!!
anonymous
  • anonymous
not really, I'm staying in IT

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