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ParthKohli

  • one year ago

Continuing what I left off yesterday.

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  1. ParthKohli
    • one year ago
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    @ganeshie8 urgent...

  2. ParthKohli
    • one year ago
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    \[x_1 + x_2 + x_3 = 20\]where\[-3 \le x_1 , x_2 , x_3 \]

  3. ParthKohli
    • one year ago
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    That reduces to the coefficient of \(x^{20}\) in\[(x^{-3} + x^{-2} + \cdots + x^{20})^3\]

  4. ParthKohli
    • one year ago
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    How do I calculate the above?

  5. ParthKohli
    • one year ago
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    @dan815 use elementary techniques please

  6. Compassionate
    • one year ago
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    When it comes to these, it really depends on what exactly you want to do with it. There's a few different ways I can play around. Be more precise

  7. ParthKohli
    • one year ago
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    \[=\left( x^{-3}\cdot\dfrac{1-x^{30}}{1-x}\right)^3\]

  8. ParthKohli
    • one year ago
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    \[= x^{-9}(1-x^{30})^3(1-x)^{-3 }\]

  9. ParthKohli
    • one year ago
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    \[x^{-9} (1 - x^{90} - 3x^{30} + 3x^{60})(1-x)^{-3}\]

  10. ParthKohli
    • one year ago
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    \[= (x^{-9} - x^{81}-3x^{21} + 3x^{51} )(1-x)^{-3}\]

  11. ParthKohli
    • one year ago
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    now the coefficient of \(x^r\) in \((1 - x)^{-n}\) is\[\binom{n+r-1}{r-1}\]

  12. ganeshie8
    • one year ago
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    you could also simply use stars and bars

  13. ganeshie8
    • one year ago
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    \[x_1 + x_2 + x_3 = 20\]where\[-3 \le x_1 , x_2 , x_3 \] is same as solving \[(y_1+3) + (y_2+3) + (y_3+3) = 20\]where\[0 \le y_1 , y_2 , y_3 \]

  14. ParthKohli
    • one year ago
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    hmm, yeah. is it \(y_1 + 3\) or \(y_1 - 3\)?

  15. ganeshie8
    • one year ago
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    \(y_i=x_i+3\) so yeah it should be \(y_i-3\)

  16. ParthKohli
    • one year ago
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    ok, that's nice.\[\binom{n-1}{k-1}\]

  17. dan815
    • one year ago
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    watcha doing?

  18. dan815
    • one year ago
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    u want int solutions?

  19. ParthKohli
    • one year ago
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    \[\binom{19}{2}\]

  20. ParthKohli
    • one year ago
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    I really wasn't trying to solve the problem ... I just want to know the approach to calculate the coefficient of \(x^k\) in huge expressions like the above one.

  21. dan815
    • one year ago
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    id just use stars and bars

  22. dan815
    • one year ago
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    but sure ill bite, how does it come out to be the coeff of x^k again

  23. ParthKohli
    • one year ago
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    oh sorry, I missed all the threes.

  24. ParthKohli
    • one year ago
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    \[y_1 + y_2 + y_3 = 29\]

  25. ParthKohli
    • one year ago
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    should the number of nonnegative solutions be\[\binom{29-1}{3-1}\]

  26. dan815
    • one year ago
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    oh nvm i get whwy its the coeff of x^20 in that expansion kinda neat trick

  27. ParthKohli
    • one year ago
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    yeah, it is.

  28. dan815
    • one year ago
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    you would still use stars and bars to solve for the coeff i guess now, since theres no real nice way of expanding that

  29. dan815
    • one year ago
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    the way to think about this problem is the same as x1+x2+x3=k x1,x2,x3 >= 0

  30. dan815
    • one year ago
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    do you know how to solve that problem

  31. dan815
    • one year ago
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    allowed -3 what u basicaaly did is just allow more 1s in each place

  32. dan815
    • one year ago
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    by introducing -3 you just shifted your number index

  33. ParthKohli
    • one year ago
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    ik, ik. there are various tricks to find the coefficient.

  34. dan815
    • one year ago
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    ok so watcha looking for now

  35. dan815
    • one year ago
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    ya x1+x2+x3=29 is right

  36. dan815
    • one year ago
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    for x1,x2,x3 non negative solutions

  37. ParthKohli
    • one year ago
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    ok, so nonnegative solutions|dw:1444026999220:dw|

  38. dan815
    • one year ago
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    hahahaha

  39. ParthKohli
    • one year ago
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    eh, that thing was just meant to illustrate stars-and-bars for x1 + x2 + x3 = 4

  40. dan815
    • one year ago
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    ya i know

  41. ParthKohli
    • one year ago
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    hey shut up

  42. dan815
    • one year ago
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    but yes parth

  43. dan815
    • one year ago
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    heres a very easy way to work it out

  44. ParthKohli
    • one year ago
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    \[\binom{29+3-1}{3-1}\]

  45. dan815
    • one year ago
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    suppose x1,x2,x3 = 20 but u allow no 0 x1,x2,x3>=1

  46. dan815
    • one year ago
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    what does this really mean?

  47. dan815
    • one year ago
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    that means we must start with 1 in x1 x2 x3

  48. dan815
    • one year ago
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    1 +k1+ 1 + k2 +1 +k3 = 20 so your new zero beginning equation is now k1+k2+k3=17

  49. dan815
    • one year ago
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    now u see how we went from the 1 case to 0 case

  50. dan815
    • one year ago
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    same logic from 0 to -3 case

  51. dan815
    • one year ago
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    just number indices here they are just symbols

  52. ParthKohli
    • one year ago
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    i know what's happening here jesus

  53. dan815
    • one year ago
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    dont sweat

  54. dan815
    • one year ago
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    swear

  55. ParthKohli
    • one year ago
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    ok so by stars and bars, it turns out to be 465 however http://www.wolframalpha.com/input/?i=%28x%5E%28-3%29+%2B+x%5E%28-2%29+%2B+...+%2B+x%5E%2820%29%29%5E3 says x^20's coefficient is 402

  56. ParthKohli
    • one year ago
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    ohhh I got where I went wrong lol

  57. Astrophysics
    • one year ago
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    the n?

  58. Astrophysics
    • one year ago
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    Oh wait..

  59. Astrophysics
    • one year ago
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    \[\binom{29+3-1}{3-1} \] this should be fixed no?

  60. ParthKohli
    • one year ago
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    i need \((x^{-3} + x^{-2} + \cdots + x^{26})^3 \)

  61. ParthKohli
    • one year ago
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    yay 465 is right baby

  62. Astrophysics
    • one year ago
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    Yeah see it should be \[\binom{29+3-1}{29}\] ye?

  63. ParthKohli
    • one year ago
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    well both of them are equal so it doesn't even matter

  64. Astrophysics
    • one year ago
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    Ah I see, you're too good Mr. Kohli!

  65. ParthKohli
    • one year ago
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    how many solutions to \(x+y+z+t=29\) when \(x\ge 1, y\ge 2, z\ge 3, t\ge0 \)

  66. dan815
    • one year ago
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    same thing

  67. ParthKohli
    • one year ago
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    again we transform in the following manner\[x=a+1\]\[y=b+2\]\[z=c+3\]

  68. dan815
    • one year ago
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    yea

  69. ParthKohli
    • one year ago
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    how many solutions to \(x_1 + x_2 + x_3 + 4x_4 = 20\)

  70. ParthKohli
    • one year ago
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    coefficient of \(x^{20}\) in\[(1 + x + \cdots + x^{20})^3(1 + x^{4}+x^8+\cdots + x^{20}) \]

  71. ParthKohli
    • one year ago
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    looks like you're back

  72. ParthKohli
    • one year ago
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    I'm looking at a sample problem where they removed \(1-x^{21}\) because it doesn't contribute to \(x^{20}\).

  73. ParthKohli
    • one year ago
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    so can you use that trick here? I haven't got the real feel for it.

  74. ParthKohli
    • one year ago
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    \[\frac{\left(1 - x^{21}\right)^3 }{(1-x)^{3}} \cdot \frac{1- x^{20}}{1-x^{4} }\]

  75. ParthKohli
    • one year ago
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    now can we remove \((1-x^{21})^3\) because it doesn't contribute to \(x^{20}\)?

  76. ParthKohli
    • one year ago
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    \[(1-x)^{-3} \left(1 + x^4 +x^8 + x^{12}+x^{16}+x^{20}\right)\]

  77. ParthKohli
    • one year ago
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    now the coefficient of \(x^r\) in \((1-x)^{-n}\) is\[\binom{n+r-1}{r-1}\]is that right?

  78. ParthKohli
    • one year ago
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    or is it the opposite... coefficient of \(x^n\) in \((1-x)^{-r}\) is \(\binom{n+r-1}{r-1}\)

  79. ParthKohli
    • one year ago
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    \[\frac{1}{(1-x)^s}=\sum_{k=0}^{\infty} \binom{s+k-1}{s-1}x^k \]

  80. ParthKohli
    • one year ago
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    \[ \frac{1}{(1-x)^s}=\sum_{k=0}^{\infty} \binom{s+k-1}{k }x^k \]

  81. ganeshie8
    • one year ago
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    \[\dbinom{-n}{r} = (-1)^r\dbinom{n+r-1}{r-1}\]

  82. ParthKohli
    • one year ago
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    that's nice

  83. ParthKohli
    • one year ago
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    i'm appearing for an all-india test tomorrow and i'm really nervous :( so many expectations! how do I get the best out of myself?

  84. ganeshie8
    • one year ago
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    \[(1-x)^{-s}=\sum_{k=0}^{\infty} (-1)^k\binom{s+k-1}{s-1}x^k \] right ?

  85. ParthKohli
    • one year ago
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    hmm, but that \((-1)^k\) shouldn't be there because there's another \((-1)^k\) attached to \(x^k\) so they cancel out in a way

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