Question

sum of (sin 2kx/11 + i cos2kx/11) from k=1 to k=10
where i=square root of negative one

Answer 1

\[\sum_{k=1}^{10} [(\sin 2k \pi/11) +i(\cos 2k \pi/11) ]\]

Answer 2

what do you want to find here? ^_^

Answer 3

the value of this summation=)

Answer 4

Well. First take 1/11 as a common factor and cut the summation to two parts.

Answer 5

sorry common factor??

Answer 6

You will see that the sum of (sin2k pi) is always zero for any value for k=1,2,3,..

Answer 7

Yeah.

Answer 8

sin (2k pi) is always zero for k=1,2,3,..*

Answer 9

yes.. so the sine part comes out to be zero. what about the cosine part?

Answer 10

on the contract cos (2k pi) is one for k=1,2,..

Answer 11

So the summation of cos (2k pi) is from k=1 to k=10 is equal to 10. right?

Answer 12

Are you following? :)

Answer 13

well the answer is supposed to be -i

Answer 14

\[{1 \over 11}\sum_{k=1}^{10}\sin(2k \pi)+{i \over 11}\sum_{k=1}^{10}\cos (2k \pi)=0+{i \over 11}(10)\]

Answer 15

Are you sure?

Answer 16

wait let me check again.

Answer 17

Is 11 part of the angle?

Answer 18

1/11 is not multiplied to sin but its multiplied to its angle

Answer 19

isn't cos(20 pi) = -1? which will give you -i/11?

Answer 20

LOL

Answer 21

what x_x

Answer 22

and 1/11 = not part of the angle

Answer 23

You should have put that in parenthesis >.<

Answer 24

why didn't you put it in the angle =_=...

Answer 25

Well, you got a solution for "a" problem :P

Answer 26

hey anwar, aren't you supposed to find the summation of both sin and cos from 1 - 10?

Answer 27

sorry about that guys

Answer 28

Yeah sstarica

Answer 29

It's OK.. But, I am not sure if I have enough time to do the problem. Just let me think about it quickly.

Answer 30

okay anwar

Answer 31

is it possible to solve using euler's representation of complex no.?

Answer 32

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

Answer 33

That's what I am thinking about right now.

Answer 34

lol

Answer 35

yeah all through 2,3,4 to 10

Answer 36

@sstarica: you just substitute for k=1,2,3,.. and add them up.

Answer 37

no need, there's a faster way to compute the sum without going through the 1 2 3 stages :)

Answer 38

take first term and add it with last term

Answer 39

and whats that?

Answer 40

something like that?

Answer 41

I don't know lol

Answer 42

Well yeah.. You just need to see a pattern.

Answer 43

lol..

Answer 44

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

Answer 45

z_z oh nvm me proceed.

Answer 46

\[i \sum_{k=1}^{10} [\cos(2k \pi/11) -i \sin(2k \pi/11)]\]
=\[i \sum_{k=1}^{10} (e ^{-2k \pi/11})\]

Answer 47

i dont know how to proceed any further

Answer 48

that's right, but there is an i in the power.

Answer 49

there is?

Answer 50

There should be :)

Answer 51

o.o

Answer 52

what?!

Answer 53

yea there should be.. otherwise its not a correct representation of the complex number

Answer 54

o..o

Answer 55

\[i \sum_{k=1}^{10} (e ^{-2k \pi i/11})\]
should look like this

Answer 56

=.=

Answer 57

hey star u workin on it?

Answer 58

I gtg now.. I'll BRB

Answer 59

bye ^_^

Answer 60

bye

Answer 61

I am back, but you're gone. :(

Answer 62

I think I am always getting the idea.

Answer 63

not always, almost :)

Answer 64

giv it a try..=)

Answer 65

sin(2 pi/11)+sin(20 pi/11)=?

Answer 66

Hello, anyone there?

Answer 67

sin x+sin y=2sin((x+y)/2)+....., right?

Answer 68

I have no idea lol

Answer 69

I'm trying to fix the DAMN PRINTER! >_<

Answer 70

Apply it to sin(2 pi/11)+sin(20 pi/11)

Answer 71

excuse my language ^^"

Answer 72

sin((2 * pi) / 11) + sin((20 * pi) / 11) = 0

Answer 73

It will give you 0. Same with sin(4 pi/11)+sin(18 pi/11), and so on.

Answer 74

yup

Answer 75

Do you see that the "sin" part of the summation will be 0?

Answer 76

agreed... wat abt the cosine part

Answer 77

trying it :).. just give me few minutes.

Answer 78

1,2,3,4,5,6,7,8,9........60

Answer 79

yup sure

Answer 80

what are you saying starica?

Answer 81

I'm counting ^_^

Answer 82

3 mins have passed

Answer 83

its still "few"

Answer 84

lol yeah~

Answer 85

=)
arent you trying?

Answer 86

Don't count -.-

Answer 87

cosx+cosy=2cos((x+y)/2)cos((x-y)/2)

Answer 88

I'm too tired to try and I love counting lol =P

Answer 89

it looks simple though u_u

Answer 90

well you should know..:P

Answer 91

If I try, I'll figure it out, but I'm tired :( ...

Answer 92

why dont you go take a nap for a change..

Answer 93

I'm in campus, and I can't sleep during the day

Answer 94

r u an undergrad student?

Answer 95

mhm, a first year ^_^

Answer 96

gr8.. ehat are you majoring in?

Answer 97

*what

Answer 98

IT , computer engineering hopefully.