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anonymous

  • 5 years ago

nc0 + nc3 + nc6 ......Please help

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  1. anonymous
    • 5 years ago
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    it goes on until nCn

  2. anonymous
    • 5 years ago
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    expand (1+ x)^n bionomialy then put x = the cube roots of unity you'll get 3 equations which on adding and subtracting will allow you to create the series you want

  3. anonymous
    • 5 years ago
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    p.s. what courses are you studying?

  4. anonymous
    • 5 years ago
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    ive just graduated from year 12

  5. anonymous
    • 5 years ago
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    neat.....which college/university/institute are you trying for

  6. anonymous
    • 5 years ago
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    i think ill get into one of the iits

  7. anonymous
    • 5 years ago
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    no problem

  8. anonymous
    • 5 years ago
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    so r u indian?

  9. anonymous
    • 5 years ago
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    wt are you doing??

  10. anonymous
    • 5 years ago
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    i dont see an answer

  11. anonymous
    • 5 years ago
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    hes right...that druvidfae is bloody bright

  12. anonymous
    • 5 years ago
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    z^3 = 1 , how does that help ?

  13. anonymous
    • 5 years ago
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    u put in the cube roots of unity one by one and add all three equations...the required series comes out

  14. anonymous
    • 5 years ago
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    because all other binomial coefficients like nc1 and nc2 and so on have a (1 + w + w^2) multiplied

  15. anonymous
    • 5 years ago
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    get it??

  16. anonymous
    • 5 years ago
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    ok so ( 1 + z^1/3 ) ( 1 + z^1/3)( 1 + z ^1/3)

  17. anonymous
    • 5 years ago
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    but roots of unity are not real

  18. anonymous
    • 5 years ago
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    yes..first put in 1, then w and then w^2..u dont have to use only real nos ;)

  19. anonymous
    • 5 years ago
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    one sec, im a bit rusty here, ( 1 + x ) ^m = 1 + mx + m(m-1)/2 x^2 + ..

  20. anonymous
    • 5 years ago
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    same way if i wanted nC0 + nC5 + nC10... i would have put in the fifth roots of unity..

  21. anonymous
    • 5 years ago
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    sorry had to leave for a moment

  22. anonymous
    • 5 years ago
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    hey druvidfae wtr u studying>?

  23. anonymous
    • 5 years ago
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    but n is not given?

  24. anonymous
    • 5 years ago
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    the answer is in terms of n

  25. anonymous
    • 5 years ago
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    yes you would the problem...is that you would have lrage number of equations also in other situations only substitutions for a prime no. will work.

  26. anonymous
    • 5 years ago
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    Though I'm doing my degree in computer science I am interested in all branches of science and mathematics ... particular abstract maths(algebra,m-th,etc)

  27. anonymous
    • 5 years ago
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    oh wait, so youre using the fact that ( 1 + w + w^2 ) = ...

  28. anonymous
    • 5 years ago
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    where w^3 = 1

  29. anonymous
    • 5 years ago
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    = 0

  30. anonymous
    • 5 years ago
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    1 + n + n(n-1)/2 *1 +.... 1 + nw + n(n-1)/2 *w + ... 1 + nw^2 + (n)(n-1)/2 w^2 + ...

  31. anonymous
    • 5 years ago
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    woops

  32. anonymous
    • 5 years ago
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    Though I'm doing my degree in computer science I am interested in all branches of science and mathematics ... particular abstract maths(algebra,m-th,etc)

  33. anonymous
    • 5 years ago
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    since the roots repeat themselves modularly i.e.after every 3 powers. the sequence effectively has only three variable inserting subsequent values allows the offset to change thus allowing us to affect every third element symmetrically

  34. anonymous
    • 5 years ago
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    1 + n + n(n-1)/2 *1 +.... 1 + nw + n(n-1)/2 *w^2 + ... 1 + nw^2 + (n)(n-1)/2 w^4 + ..

  35. anonymous
    • 5 years ago
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    ok, and w^4 = w , that sort of thing

  36. anonymous
    • 5 years ago
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    whats the left side of this expression?

  37. anonymous
    • 5 years ago
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    do you have an equation ?

  38. anonymous
    • 5 years ago
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    i get 3 + 0 + 0 + ...

  39. anonymous
    • 5 years ago
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    its similar to interference in physics .Waves with a particular offset (phi) affect the the superposition in a certain way (standing waves, packets ,etc) here we are changing the wavelength (period) rather than the offset to match our sequence.once we've got our wavelength we continue to change the offset to get the shape we need.

  40. anonymous
    • 5 years ago
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    canto:yes... the principle can be extended to other problems as well.

  41. anonymous
    • 5 years ago
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    whoa whoa , slow down there buddy

  42. anonymous
    • 5 years ago
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    first thing is first, youre using the identity ( 1 + 1) ^n = Sum nCk , from k=0 to n ?

  43. anonymous
    • 5 years ago
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    you guys leave out too many steps.

  44. anonymous
    • 5 years ago
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    youre doing ( 1 + w)^n, (1+1)^n, and ( 1 + w^2) ^n, adding them

  45. anonymous
    • 5 years ago
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    for this question we don't need the 2^n identity

  46. anonymous
    • 5 years ago
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    right

  47. anonymous
    • 5 years ago
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    yes

  48. anonymous
    • 5 years ago
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    ( 1 + w)^n + (1+1)^n, + ( 1 + w^2) ^n = 3 + n( 1 + w + w^2) + nC2 ( 1 + w^2 + w^4 ) + ...

  49. anonymous
    • 5 years ago
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    ( 1 + w)^n + (1+1)^n + ( 1 + w^2) ^n = 3 + n( 1 + w + w^2) + nC2 ( 1 + w^2 + w^4 ) + nC3 ( 1 + w^3 + w^8 ) + ... nCn ( 1 + w^n + w^(2^n))

  50. anonymous
    • 5 years ago
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    IS THAT WHAT YOU ARE DOING ????

  51. anonymous
    • 5 years ago
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    and then stuff vanishes on the right side, but the left side i dont know what it is

  52. anonymous
    • 5 years ago
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    the left side is a closed formula?

  53. anonymous
    • 5 years ago
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    i would like you guys to actually solve it, lol

  54. anonymous
    • 5 years ago
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    im still learning this stuff, sorry

  55. anonymous
    • 5 years ago
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    i messed up

  56. anonymous
    • 5 years ago
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    ( 1 + w)^n + (1+1)^n + ( 1 + w^2) ^n = 3 + n( 1 + w + w^2) + nC2 ( 1 + w^2 + w^4 ) + nC3 ( 1 + w^3 + w^6 ) + ... nCn ( 1 + w^n + w^(2n))

  57. anonymous
    • 5 years ago
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    yes...as for the left side notice that the w and w^2 factors have imaginary parts which are inverse of one another expanding this using euler's form will reveal that the imaginary parts get canceled out.

  58. anonymous
    • 5 years ago
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    ok , im a little worried about the right side though

  59. anonymous
    • 5 years ago
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    = 3 + 0 + 0 + nC3 + 0 + 0

  60. anonymous
    • 5 years ago
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    why all non-3 get reduced to 0 . that is the sequence we want

  61. anonymous
    • 5 years ago
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    ( 1 + w)^n + (1+1)^n + ( 1 + w^2) ^n = 3 + 3*n( 1 + w + w^2) + 3*nC2 ( 1 + w^2 + w^4 ) + 3*nC3 ( 1 + w^3 + w^6 ) + ... 3*nCn ( 1 + w^n + w^(2n))

  62. anonymous
    • 5 years ago
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    so the final answer is [2^n + (1+w)^n + (1+w^2)^n ]/3 = nC0 + n C 3 + ... n C n

  63. anonymous
    • 5 years ago
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    although n is not necessarily a multiple of 3

  64. anonymous
    • 5 years ago
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    why all non-3 get reduced to 0 . that is the sequence we want

  65. anonymous
    • 5 years ago
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    im just checking , do you agree with my closed form formula ?

  66. anonymous
    • 5 years ago
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    also we have to assume that n | 3

  67. anonymous
    • 5 years ago
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    errr, 3 | n

  68. anonymous
    • 5 years ago
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    yeah the l.h.s. reduces to something like 3^n (note there are probably some more factors i'm missing)

  69. anonymous
    • 5 years ago
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    hmmm, no that doesnt make sense, it should be less than 2^n

  70. anonymous
    • 5 years ago
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    since nC0 + nC1 + ... n Cn = 2^n

  71. anonymous
    • 5 years ago
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    oh you mean before dividing by 3 ?

  72. anonymous
    • 5 years ago
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    if n is not a multiple of the then the last 1 or 2 factors do not appear in the final series........since series is up to n.

  73. anonymous
    • 5 years ago
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    we have deletion of terms

  74. anonymous
    • 5 years ago
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    oh right

  75. anonymous
    • 5 years ago
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    then you would zeroes as the last 1 or two terms ?

  76. anonymous
    • 5 years ago
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    ok im going to test this , offhand i dont know the third root of unity of 1

  77. anonymous
    • 5 years ago
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    e^ ( i * pi/3 ) ?

  78. anonymous
    • 5 years ago
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    no... \[w =(1+ i \sqrt{3})/2 ; w^2 = (1-i \sqrt{3})/2\] expand both bionomially notice that odd powers cancel and the remaining terms are of the form 3^n

  79. anonymous
    • 5 years ago
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    1 = e^(2*Pi *i) , so 1^(1/3) = e^(2pi/3 *i)

  80. anonymous
    • 5 years ago
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    yes your forms are also correct

  81. anonymous
    • 5 years ago
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    sorry, i have to start from fundamental principles, i have horrible memory

  82. anonymous
    • 5 years ago
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    1 = e^( i*[ 2*Pi + 2pi *n ) , so 1^(1/3) = e^(i * [2pi/3 +2pi/3*n)

  83. anonymous
    • 5 years ago
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    so 1^1/3 = e ^ ( i * (2pi/3 , 4pi/3

  84. anonymous
    • 5 years ago
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    oh its easier if we start with theta = 0

  85. anonymous
    • 5 years ago
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    1 = e^( i*[ 0+ 2pi *n ) , so 1^(1/3) = e^(i * [2pi/3*n)

  86. anonymous
    • 5 years ago
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    so we have z^(1/3) = { 1 , e^(i*2pi/3) , e^ ( i*4pi/3)

  87. anonymous
    • 5 years ago
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    i assume you have it memorized down well :) so i agree with that

  88. anonymous
    • 5 years ago
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    so w and w^2 are conjugates, interesting

  89. anonymous
    • 5 years ago
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    but it might be more succinct calculator wise to use exponential polar form

  90. anonymous
    • 5 years ago
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    [2^n + (1+e^(i*2pi/3))^n + (1+e^(i*4pi/3))^n ]/3 = nC0 + n C 3 + ... n C n

  91. anonymous
    • 5 years ago
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    now test, let n = 8

  92. anonymous
    • 5 years ago
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    sorry my early values have a slight correction the real part should be negative.

  93. anonymous
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    let n = 8 , so we have [2^8 + (1+e^(i*2pi/3))^8 + (1+e^(i*4pi/3))^8 ]/3 = 8C0 + 8 C 3 + ... 8C8 ?

  94. anonymous
    • 5 years ago
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    or last term should be 8 C 6

  95. anonymous
    • 5 years ago
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    does your calculator allow complex calculations......i.e. calculations involving iota based calculations.

  96. anonymous
    • 5 years ago
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    yes

  97. anonymous
    • 5 years ago
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    i get 85 for left side 85 + 3.3666666666666666666666E -13i

  98. anonymous
    • 5 years ago
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    the last term should be\[(n & 6)\]

  99. anonymous
    • 5 years ago
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    yessssssssssss

  100. anonymous
    • 5 years ago
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    it works

  101. anonymous
    • 5 years ago
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    *nc6

  102. anonymous
    • 5 years ago
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    if n = 8

  103. anonymous
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    8C0 + 8C 3 + 8 C 6 = 85

  104. anonymous
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    so the imaginary part vanishes there

  105. anonymous
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    haha thats funny, my calcualto has 3.33333333336 E -13 i

  106. anonymous
    • 5 years ago
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    so that vanishes , its like 3.3 *10^-13 * i

  107. anonymous
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    for the left side its 85 + 3.3666666666666666666666 E -13i E means 10^

  108. anonymous
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    which calculator do you have ??

  109. anonymous
    • 5 years ago
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    why didnt him actually solve it, thats annoying

  110. anonymous
    • 5 years ago
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    TI 84

  111. anonymous
    • 5 years ago
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    its easy to use, and fast to learn , the TI 89 is more powerful, but harder to use

  112. anonymous
    • 5 years ago
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    im pissed, maple cant solve tan (105)

  113. anonymous
    • 5 years ago
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    its an exact angle

  114. anonymous
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    TI 89 has multiple color codes, so light blue, light green for functions

  115. anonymous
    • 5 years ago
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    my calculator:casio fx82 and fx 100 my numerical software :octave

  116. anonymous
    • 5 years ago
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    it should be nC0 + nC3 + ... n C n divisible by 3

  117. anonymous
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    or something like n C k where k mod 3 is not equal to zero

  118. anonymous
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    not necessarily (due to the additional 2^n factor)

  119. anonymous
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    sum nC k , where nC k = 0 when k mod 3 ! = 0, and otherwise it is normal

  120. anonymous
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    i mean for the right hand side

  121. anonymous
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    yeah on the r.h.s.

  122. anonymous
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    great, i feel like i solved it. but you take credit for starting it

  123. anonymous
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    but we didnt simplify it all the way

  124. anonymous
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    LHS we have 2^n / 3 + `

  125. anonymous
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    oh right, youre real part should be negative

  126. anonymous
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    2pi/3 angle is in quadrant 2

  127. anonymous
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    also x^n = 1 gives you the pentagonal shapes

  128. anonymous
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    or z^n = 1 , i should say

  129. anonymous
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    northeast quadrant

  130. anonymous
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    northwest quadrant

  131. anonymous
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    starting from (1,0) angle 0 , going counterclockwise

  132. anonymous
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    ahhh.. memory so fickle art thou

  133. anonymous
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    quandrant 1 is 0 < theta < 90 q2 90 < theta < 180 , etc

  134. anonymous
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    q3 180 < theta < 270 , et

  135. anonymous
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    oh every nice, the 1-1 cancels the inside of the binomial

  136. anonymous
    • 5 years ago
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    2^n / 3 + ( i * sqrt 3 / 2 )^n + (-i sqrt 3 / 2 ) ^n = n C0 + n C 3 + ..

  137. anonymous
    • 5 years ago
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    notice that when you have n mod 3 = 0, then i^n + (-i)^n both go to zero

  138. anonymous
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    no i take that back

  139. anonymous
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    i^n + (-i)^n is interesting

  140. anonymous
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    it has the sequence for n = 1,2,3, 0 , -2, 0 , 2, 0, -2 , ...

  141. anonymous
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    oh thats because i^n +(- i) ^n = i^n ( 1 + (-1)^n)

  142. anonymous
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    i mesed up , i didnt see the 1 + (-1/2 + i sqrt 3 / 2 ) . so its

  143. anonymous
    • 5 years ago
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    we have 2^n/3 + 1/3* ( 1- i sqrt 3 / 2 ) ^n + ( 1 + i sqrt 3 / 2 ) ^n

  144. anonymous
    • 5 years ago
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    2^n*1/3 + 1/3* ( 1- i sqrt 3 / 2 ) ^n + ( 1 + i sqrt 3 / 2 ) ^n

  145. anonymous
    • 5 years ago
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    and we can use an identity ( a + bi ) ^n + ( a - bi)^n

  146. anonymous
    • 5 years ago
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    or z^n + z*^n = ...

  147. anonymous
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    where z* is the conjugate of z

  148. anonymous
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    i hope im not boring you

  149. anonymous
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    no definately not

  150. anonymous
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    i might need some help here

  151. anonymous
    • 5 years ago
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    i did the case for i^n + (-i)^n

  152. anonymous
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    sure...and what came up

  153. anonymous
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    i^n ( 1 - (-1)^n ) , well a sequence

  154. anonymous
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    for n = 1 ,2,3 we have 0,-2,0,2,0,-2 ...

  155. anonymous
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    lets try the case a + bi ^n + ( a -bi)^n

  156. anonymous
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    yes that's what we're supposed to get remember with i the gap between terms is 2

  157. anonymous
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    oh the (a + bi)^n + ( a - bi)^n seems to diverge

  158. anonymous
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    i bet you by looking at the binomial form, we can simplify , by looking at the series

  159. anonymous
    • 5 years ago
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    so its easier if we divide out the a ,

  160. anonymous
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    so ( 1 + ki ) ^n + ( 1 - ki)^n ,

  161. anonymous
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    oh very nice we get

  162. anonymous
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    1 + ki + (ki)^2 + k^3 *i^3 + ... 1 -ki + (ki^2) - k^3i^3 + ...

  163. anonymous
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    = 1 + (ki)^2 + (ki)^4 + ...

  164. anonymous
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    and we know 1 + x^2 + x^4 + ... = sum x^(2n)

  165. anonymous
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    so we get sum (ki)^(2n) for n=0...

  166. anonymous
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    that is equal to 1 / ( 1- (ki)^2 )

  167. anonymous
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    ( 1 + ki ) ^n + ( 1 - ki)^n , = 1 / ( 1 - (ki)^2 ) ?

  168. anonymous
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    only if |ki|< 1

  169. anonymous
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    nevermind, this wont help us , lol

  170. anonymous
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    also i messed up, it should be ( 1 + ki ) ^n + ( 1 - ki)^n = 2 + 2(ki)^2 + 2 (ki)^4 + ...

  171. anonymous
    • 5 years ago
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    oh crap, i left out the binomial coefficients

  172. anonymous
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    1 + nC1*ki + nC2(ki)^2 +nC3 k^3 *i^3 + ... 1 - nC0* ki + nC2* (ki^2) -nC3 k^3i^3 + ...

  173. anonymous
    • 5 years ago
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    actually the power series of complex variables shows some diverging oscillation type things

  174. anonymous
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    oh its an alternating series

  175. anonymous
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    (( 1 + ki ) ^n + ( 1 - ki)^n) /2 = 1 - nC2 k^2 + nC4 k^4 - nC6 k^6 + ...

  176. anonymous
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    yay it works

  177. anonymous
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    i test it on k = 3, n = 4

  178. anonymous
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    i dont know if this is .... useful formula. can we simplify the right side?

  179. anonymous
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    might be a well known series

  180. anonymous
    • 5 years ago
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    so what were you saying earlier, how can we simplify this result, the conjugates w, and w^2 appear

  181. anonymous
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    the final answer is then [ 2^n + ( 1/2 - i sqrt (3 )/ 2 ) ^n + ( 1/2+ i sqrt (3) / 2 ) ^n ] / 3

  182. anonymous
    • 5 years ago
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    yes

  183. anonymous
    • 5 years ago
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    but if we can find a nice simplification of ( a + bi)^n + ( a - bi)^n, that would be nice

  184. anonymous
    • 5 years ago
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    and this argument works for any prime p , because ( 1 + w + w^2 + ... w^p-1) = 0

  185. anonymous
    • 5 years ago
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    sorry but i dont see where the application is for this type of problem,

  186. anonymous
    • 5 years ago
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    but its very pretty nonetheless

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