## ParthKohli one year ago Continuing what I left off yesterday.

1. ParthKohli

@ganeshie8 urgent...

2. ParthKohli

$x_1 + x_2 + x_3 = 20$where$-3 \le x_1 , x_2 , x_3$

3. ParthKohli

That reduces to the coefficient of $$x^{20}$$ in$(x^{-3} + x^{-2} + \cdots + x^{20})^3$

4. ParthKohli

How do I calculate the above?

5. ParthKohli

6. Compassionate

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

$=\left( x^{-3}\cdot\dfrac{1-x^{30}}{1-x}\right)^3$

8. ParthKohli

$= x^{-9}(1-x^{30})^3(1-x)^{-3 }$

9. ParthKohli

$x^{-9} (1 - x^{90} - 3x^{30} + 3x^{60})(1-x)^{-3}$

10. ParthKohli

$= (x^{-9} - x^{81}-3x^{21} + 3x^{51} )(1-x)^{-3}$

11. ParthKohli

now the coefficient of $$x^r$$ in $$(1 - x)^{-n}$$ is$\binom{n+r-1}{r-1}$

12. ganeshie8

you could also simply use stars and bars

13. ganeshie8

$x_1 + x_2 + x_3 = 20$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

hmm, yeah. is it $$y_1 + 3$$ or $$y_1 - 3$$?

15. ganeshie8

$$y_i=x_i+3$$ so yeah it should be $$y_i-3$$

16. ParthKohli

ok, that's nice.$\binom{n-1}{k-1}$

17. dan815

watcha doing?

18. dan815

u want int solutions?

19. ParthKohli

$\binom{19}{2}$

20. ParthKohli

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

id just use stars and bars

22. dan815

but sure ill bite, how does it come out to be the coeff of x^k again

23. ParthKohli

oh sorry, I missed all the threes.

24. ParthKohli

$y_1 + y_2 + y_3 = 29$

25. ParthKohli

should the number of nonnegative solutions be$\binom{29-1}{3-1}$

26. dan815

oh nvm i get whwy its the coeff of x^20 in that expansion kinda neat trick

27. ParthKohli

yeah, it is.

28. dan815

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

30. dan815

do you know how to solve that problem

31. dan815

allowed -3 what u basicaaly did is just allow more 1s in each place

32. dan815

by introducing -3 you just shifted your number index

33. ParthKohli

ik, ik. there are various tricks to find the coefficient.

34. dan815

ok so watcha looking for now

35. dan815

ya x1+x2+x3=29 is right

36. dan815

for x1,x2,x3 non negative solutions

37. ParthKohli

ok, so nonnegative solutions|dw:1444026999220:dw|

38. dan815

hahahaha

39. ParthKohli

eh, that thing was just meant to illustrate stars-and-bars for x1 + x2 + x3 = 4

40. dan815

ya i know

41. ParthKohli

hey shut up

42. dan815

but yes parth

43. dan815

heres a very easy way to work it out

44. ParthKohli

$\binom{29+3-1}{3-1}$

45. dan815

suppose x1,x2,x3 = 20 but u allow no 0 x1,x2,x3>=1

46. dan815

what does this really mean?

47. dan815

48. dan815

1 +k1+ 1 + k2 +1 +k3 = 20 so your new zero beginning equation is now k1+k2+k3=17

49. dan815

now u see how we went from the 1 case to 0 case

50. dan815

same logic from 0 to -3 case

51. dan815

just number indices here they are just symbols

52. ParthKohli

i know what's happening here jesus

53. dan815

dont sweat

54. dan815

swear

55. ParthKohli

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

ohhh I got where I went wrong lol

57. Astrophysics

the n?

58. Astrophysics

Oh wait..

59. Astrophysics

$\binom{29+3-1}{3-1}$ this should be fixed no?

60. ParthKohli

i need $$(x^{-3} + x^{-2} + \cdots + x^{26})^3$$

61. ParthKohli

yay 465 is right baby

62. ParthKohli
63. Astrophysics

Yeah see it should be $\binom{29+3-1}{29}$ ye?

64. ParthKohli

well both of them are equal so it doesn't even matter

65. Astrophysics

Ah I see, you're too good Mr. Kohli!

66. ParthKohli

how many solutions to $$x+y+z+t=29$$ when $$x\ge 1, y\ge 2, z\ge 3, t\ge0$$

67. dan815

same thing

68. ParthKohli

again we transform in the following manner$x=a+1$$y=b+2$$z=c+3$

69. dan815

yea

70. ParthKohli

how many solutions to $$x_1 + x_2 + x_3 + 4x_4 = 20$$

71. ParthKohli

coefficient of $$x^{20}$$ in$(1 + x + \cdots + x^{20})^3(1 + x^{4}+x^8+\cdots + x^{20})$

72. ParthKohli

looks like you're back

73. ParthKohli

I'm looking at a sample problem where they removed $$1-x^{21}$$ because it doesn't contribute to $$x^{20}$$.

74. ParthKohli

so can you use that trick here? I haven't got the real feel for it.

75. ParthKohli

$\frac{\left(1 - x^{21}\right)^3 }{(1-x)^{3}} \cdot \frac{1- x^{20}}{1-x^{4} }$

76. ParthKohli

now can we remove $$(1-x^{21})^3$$ because it doesn't contribute to $$x^{20}$$?

77. ParthKohli

$(1-x)^{-3} \left(1 + x^4 +x^8 + x^{12}+x^{16}+x^{20}\right)$

78. ParthKohli

now the coefficient of $$x^r$$ in $$(1-x)^{-n}$$ is$\binom{n+r-1}{r-1}$is that right?

79. ParthKohli

or is it the opposite... coefficient of $$x^n$$ in $$(1-x)^{-r}$$ is $$\binom{n+r-1}{r-1}$$

80. ParthKohli

$\frac{1}{(1-x)^s}=\sum_{k=0}^{\infty} \binom{s+k-1}{s-1}x^k$

81. ParthKohli

$\frac{1}{(1-x)^s}=\sum_{k=0}^{\infty} \binom{s+k-1}{k }x^k$

82. ganeshie8

$\dbinom{-n}{r} = (-1)^r\dbinom{n+r-1}{r-1}$

83. ParthKohli

that's nice

84. ParthKohli

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?

85. ganeshie8

$(1-x)^{-s}=\sum_{k=0}^{\infty} (-1)^k\binom{s+k-1}{s-1}x^k$ right ?

86. ParthKohli

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