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imqwerty

  • one year ago

emergency!!!

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  1. imqwerty
    • one year ago
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  2. ganeshie8
    • one year ago
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    which q

  3. ParthKohli
    • one year ago
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    Last one x = y = z = 4

  4. imqwerty
    • one year ago
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    any question that u like...these r the question which i wsn't able to solve

  5. Empty
    • one year ago
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    First one probably some fermat's little theorem nonsense, the second one looks like part of the Riemann zeta function so you could probably just multiply two geometric series or something together for it, and 3rd one looks like it's probably a telescoping series, just my first guesses

  6. imqwerty
    • one year ago
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    @ParthKohli how did u get x=y=z=4

  7. ParthKohli
    • one year ago
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    It's a symmetric situation. Apply all inequalities you please, but you'll finally get to this.

  8. Empty
    • one year ago
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    Simplest way to think of it (not a proof) is that \[n^2 > (n-a)(n+a)\] when a is not equal to zero, so that's the max when they're all the same.

  9. ganeshie8
    • one year ago
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    first q doesnt look easy

  10. Empty
    • one year ago
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    I think we can split that 3rd one up using partial fractions, then it becomes a telescoping series that way.

  11. Empty
    • one year ago
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    Good questions by the way!

  12. imqwerty
    • one year ago
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    thanks @Empty @ParthKohli i can't get to the result x=y=z. can u tell how to start ..

  13. ParthKohli
    • one year ago
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    I've only given a qualitative argument here so far - the whole situation you have here is symmetric in nature. x can be interchanged with y, y with z or z with x, and the situation won't change. So if the triplet \((a, b , c)\) gives you a maximum value then so should \((b, a, c)\) or \((b, c, a)\) or \((c, a, b)\), which means that \(x = y = z \) when the given expression is maximum. Let me think of an inequality that can be applied here. You'll see that we'll find the maximum when the equality of two expressions is considered, which is almost always when all variables are equal.

  14. Empty
    • one year ago
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    Ok so for the second one I'm getting: $$\frac{1}{1-9}\frac{1}{1-4}\frac{5}{6}$$ Does that match your answer?

  15. imqwerty
    • one year ago
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    wait m doing the 1st one

  16. ganeshie8
    • one year ago
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    for first one you may want to try factoring it http://www.wolframalpha.com/input/?i=factor+5%5E5%5E%28n%2B1%29%2B5%5E5%5En%2B1

  17. Empty
    • one year ago
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    Yeah I think it is more clear how to solve the first one if you do that, here to make factoring easier substitute: $$5^{5^n}=a$$ then you end up with the expression: $$a^5+a+1$$ which somehow wolfram is able to factor out into: $$(a^2+a+1)(a^3-a^2+1)$$

  18. imqwerty
    • one year ago
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    @ParthKohli if we are given a question like - x+y=4 and x belongs to positive integers. and we are supposed to find out maximum of xy + x/y +y/x then by symmetric approach x=y=2 but in this question x=3 and y=1 gives the maximum so the symmetric approach is violated.

  19. imqwerty
    • one year ago
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    @Empty we can't write 5^5^(n+1) as a^5 if a=5^5^n

  20. ParthKohli
    • one year ago
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    OK, I didn't say that x = y gives you the maximum. I said it gives you the maximum if it exists. Here, x = y is most probably giving you the minimum.

  21. imqwerty
    • one year ago
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    yes so how do we know that in which case x=y gives the max

  22. Empty
    • one year ago
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    $$5^{5^{n+1}}=5^{5*5^n} = (5^{5^n})^5$$ @imqwerty

  23. ParthKohli
    • one year ago
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    You have to check if x = y is giving you the maximum/minimum by entering another pair of values. If that pair gives you a higher value, then x = y gives you the minimum.

  24. ganeshie8
    • one year ago
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    i feel #1 is still a stupid q, you're not gona learn anything frm it

  25. imqwerty
    • one year ago
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    wait sry yes we can put a^5...

  26. imqwerty
    • one year ago
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    yes it is @ganeshie8 :D

  27. ParthKohli
    • one year ago
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    Y'know, qwerty, it's just like that with single variable functions. You can differentiate a function and equate it to zero, but you still have to check if that gives you the maxima or minima by other methods.

  28. Empty
    • one year ago
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    Ok so I also found a good way of solving the second one: We only want odd total powers, so the way we can do that is only combine even powers with odd and odd with evens. So in order to do that, we multiply the geometric series' together: $$\left(\sum_{i=0}^{\infty} \frac{1}{2^{2i}} \right) \left(\sum_{j=0}^{\infty} \frac{1}{3^{2j+1}} \right) +\left(\sum_{i=0}^{\infty} \frac{1}{3^{2i}} \right) \left(\sum_{j=0}^{\infty} \frac{1}{2^{2j+1}} \right) $$ factor out a 1/3 and 1/2 so you get only geometric series in terms of 1/4 and 1/9, and you're done!

  29. ParthKohli
    • one year ago
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    Isn't that the standard way to do it though? How did you originally do it?

  30. Empty
    • one year ago
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    I factored it

  31. imqwerty
    • one year ago
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    the 1st one is solved thanks guys :)

  32. ganeshie8
    • one year ago
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    \[\begin{align} x^5 + x + 1 & = (x^5-x^2)+(x^2+x+1)\\ \\ & = x^2(x^3 - 1) + (x^2 + x + 1) \\ \\ & = x^2(x-1)\color{blue}{\bf (x^2 + x + 1)} + \color{blue}{\bf (x^2 + x + 1)}\end{align}\] http://math.stackexchange.com/questions/477295/factor-x5-x-1

  33. ParthKohli
    • one year ago
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    To be fair, that solution was most likely obtained by working backwards. :P

  34. ganeshie8
    • one year ago
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    that is okay for a proof

  35. ParthKohli
    • one year ago
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    Yes of course, but you can't really think of that on your own.

  36. ganeshie8
    • one year ago
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    that all goes as "scratch work" :P

  37. Empty
    • one year ago
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    ahhh thanks @ganeshie8 that's a clever trick I think I can even remember that as adding fancy form of zero and then abusing the geometric series yet again lol

  38. ParthKohli
    • one year ago
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    I'm ashamed that I have no other ideas other than telescoping for #3.

  39. Empty
    • one year ago
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    It wants to telescope, why force it to be anything it's probably not? :P

  40. Empty
    • one year ago
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    When you separate it out, the coefficients are both 1, it's truly a beautiful result

  41. ganeshie8
    • one year ago
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    I'm either dumb or too much drunk to notice it but why do you think telescoping is a bad idea ?

  42. Empty
    • one year ago
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    Yeah you're not drunk enough ganeshie, cause this baby wants to see the moon

  43. ParthKohli
    • one year ago
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    I didn't say telescoping is a bad idea. I just said that this series doesn't seem to be telescoping nicely and I have no other ideas.

  44. imqwerty
    • one year ago
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    m really confused nd annoyed nd kinda drunk ..cheers @ganeshie8 @Empty ..can u please tell me what u did with the 2nd question

  45. imqwerty
    • one year ago
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    my mind is jst nt wrkin normal today :(

  46. ParthKohli
    • one year ago
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    OK, it's actually telescoping really nicely.

  47. Empty
    • one year ago
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    Here I made it smaller, part of the total sum, just write out all these terms (it's not tooo many, and it's good for you) \[\left(\sum_{i=0}^{3} \frac{1}{2^{2i}} \right) \left(\sum_{j=0}^{3} \frac{1}{3^{2j+1}} \right) +\left(\sum_{i=0}^{3} \frac{1}{3^{2i}} \right) \left(\sum_{j=0}^{3} \frac{1}{2^{2j+1}} \right) \]

  48. imqwerty
    • one year ago
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    no i mean hw did u split that

  49. Empty
    • one year ago
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    No, I _created_ this

  50. Empty
    • one year ago
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    so by multiplying it out you can kinda see how I came up with this too... I'll draw a picture of how I think about it

  51. ParthKohli
    • one year ago
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    \[\frac{6^k}{(3^k - 2^k)(3^{k+1 } - 2^{k+1})}\]\[= \frac{3^{k+1}}{2^{k+1} - 3^{k+1}} - \frac{3^k}{3^k - 2^k}\]

  52. ParthKohli
    • one year ago
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    There's a typo.

  53. Empty
    • one year ago
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    |dw:1438453754914:dw| So this is the first half of the terms! @imqwerty

  54. ParthKohli
    • one year ago
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    Supposed to be\[= \frac{3^{k+1}}{2^{k+1} - 3^{k+1}} - \frac{3^k}{2^k - 3^k}\]

  55. Empty
    • one year ago
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    Specifically I knew to write that because the only way for i+j=odd is when one is even and the other is odd since odd+odd=even and even+even=even. So combine all the 2s with even powers with all the 3s with odd powers, and add it to all the 2s with odd powers with all the 3s with even powers.

  56. Empty
    • one year ago
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    so that's where the 2i and 2i+1 is coming from, keeping the powers even and odd: \[\left(\sum_{i=0}^{\infty} \frac{1}{2^{2i}} \right) \left(\sum_{j=0}^{\infty} \frac{1}{3^{2j+1}} \right) +\left(\sum_{i=0}^{\infty} \frac{1}{3^{2i}} \right) \left(\sum_{j=0}^{\infty} \frac{1}{2^{2j+1}} \right) \]

  57. imqwerty
    • one year ago
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    thank u so much @Empty :)

  58. ParthKohli
    • one year ago
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    OK, so we're sorta done with all of the questions.

  59. ganeshie8
    • one year ago
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    whats the ansur for q2

  60. Empty
    • one year ago
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    Are you asking me or asking imqwerty?

  61. imqwerty
    • one year ago
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    m calculating the answer...

  62. ParthKohli
    • one year ago
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    @Empty Did you check out the solution to the problem I gave you yesterday?

  63. Empty
    • one year ago
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    Oh no I didn't, also I ended up getting 2 to that, I'll go check real fast

  64. ParthKohli
    • one year ago
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    2 is correct.

  65. imqwerty
    • one year ago
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    :) a big THANKS TO ALL OF YOU @Empty @ParthKohli @ganeshie8 :)

  66. ganeshie8
    • one year ago
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    #2 im getting 5/144

  67. ParthKohli
    • one year ago
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    Ganeshie and Qwerty haven't done that problem so I'll ask them to do it too.\[\sum_{n=1}^{\infty}\frac{F_n}{2^n}\]where \(F_n\) is the \(n\)-th Fibonacci number. \(F_1 = F_2 = 1 \).

  68. imqwerty
    • one year ago
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    my answer the #2 is 121/144

  69. ganeshie8
    • one year ago
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    I am familiar with that problem \[\sum_{n=1}^{\infty}F_nx^n= \frac{x}{1-x-x^2}\] plugin \(x=1/2\)

  70. Empty
    • one year ago
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    Here's the solution: \[\left(\sum_{i=0}^{\infty} \frac{1}{2^{2i}} \right) \left(\sum_{j=0}^{\infty} \frac{1}{3^{2j+1}} \right) +\left(\sum_{i=0}^{\infty} \frac{1}{3^{2i}} \right) \left(\sum_{j=0}^{\infty} \frac{1}{2^{2j+1}} \right) \] \[\left(\sum_{i=0}^{\infty} \frac{1}{2^{2i}} \right) \frac{1}{3}\left(\sum_{j=0}^{\infty} \frac{1}{3^{2j}} \right) +\left(\sum_{i=0}^{\infty} \frac{1}{3^{2i}} \right) \frac{1}{2} \left(\sum_{j=0}^{\infty} \frac{1}{2^{2j}} \right) \] \[\left( \frac{1}{3} + \frac{1}{2} \right) \left(\sum_{i=0}^{\infty} \frac{1}{3^{2i}} \right) \left(\sum_{j=0}^{\infty} \frac{1}{2^{2j}} \right) \] \[\left( \frac{1}{3} + \frac{1}{2} \right) \left( \frac{1}{1-3^2} \right) \left(\frac{1}{1-2^2} \right) \] \[\frac{5}{144}\]

  71. imqwerty
    • one year ago
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    damn my mind is nt ok today ....instead of starting with i=0 i started with i=1. :( feeling bad at the silly mistakes m doin :(

  72. Empty
    • one year ago
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    Haha nah it's fine that's why I wrote it all out for you. And hey, if you want you can always come back tomorrow and look, no reason to feel bad XD

  73. Empty
    • one year ago
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    I'm pretty sure "feeling bad" is not supposed to be a step in solving problems haha, even though sometimes it does tend to be when those problems are proofs in real analysis for me lolol

  74. ganeshie8
    • one year ago
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    sometimes i hate problems from indian competitive exams/tests mostly because they are meaning less, how do they expect a 10/12th grader to know about generating functions and stuff ?

  75. ganeshie8
    • one year ago
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    they just want the exams to be tough, they dont bother to look at the syllabus of 12th grader

  76. imqwerty
    • one year ago
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    the reason y m feeling bad and annoyed - today the results of the last test held at of coaching institute came- the 1st ranker got 360/360 nd i ws like O_o nd i ws 12th with score 329. i could've done way better if i wuld have rechecked my paper after completing it instead of looking at other ppl faces :(

  77. imqwerty
    • one year ago
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    at my home m nicknamed - 'silly mistake'

  78. ganeshie8
    • one year ago
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    trust me, you and PK are way better than me when i was in my 12th

  79. Empty
    • one year ago
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    Yeah I don't understand the obsession with rote memorizing all these useless one-time-use tricks for problems mankind invented calculators and wolfram alpha for. Time to change the curriculum to a modern age I think.

  80. imqwerty
    • one year ago
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    @Empty so true^ .

  81. Empty
    • one year ago
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    I always see these questions like "find the number of roots of this polynomial" or "show this ridiculous trig identity is true without any geometric picture that it came from" ...why?

  82. Empty
    • one year ago
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    It's like teachers want people to believe math is useless AND ugly.

  83. ParthKohli
    • one year ago
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    Whut, qwerty? The top-ranker got 360?! Surely the questions must have been really simple. The topper at my coaching (me) got 307/360 which was an exception to the rule that the previous years' toppers got no more than 250-260.

  84. ParthKohli
    • one year ago
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    humblebrag

  85. Empty
    • one year ago
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    lol duuuude gotta go for that \(2 \pi\) score man!

  86. imqwerty
    • one year ago
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    the paper wasn't enough easy to get 360/360. the guys are too crazy...they are like 24 x 7 studying.

  87. Empty
    • one year ago
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    Their parents must be awful people

  88. Empty
    • one year ago
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    Ok ok maybe I'm being pessimistic here haha

  89. imqwerty
    • one year ago
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    The guy who got 1st rank is a vry good frnd of mine. He says his parents think that he can do whatever he wants but he must remain healthy eating proper food.

  90. imqwerty
    • one year ago
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    his parents don't interfere his personal life

  91. Empty
    • one year ago
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    Woah he must be a genius :O

  92. imqwerty
    • one year ago
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    this guy got 1st rank in 2/4 test held till date

  93. ParthKohli
    • one year ago
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    Kaunsa institute hai ye?

  94. ganeshie8
    • one year ago
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    see you cant study more than 14 hours each day, you need at least 6 hours of sleep and 4 hours for maintenance

  95. imqwerty
    • one year ago
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    allen

  96. ParthKohli
    • one year ago
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    Kota wala?

  97. imqwerty
    • one year ago
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    han

  98. ParthKohli
    • one year ago
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    Tumhe internet mil jaata hai?

  99. imqwerty
    • one year ago
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    mai to kota ka hi hu

  100. ParthKohli
    • one year ago
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    Oh, badiya yaar.

  101. imqwerty
    • one year ago
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    :) badiya bhi hai aur thodi prblems bhi hai....roz ghar par koi na koi guest aa jate hai...roz tv etc etc ka shore distrb karta hai...nd many more prblems

  102. ParthKohli
    • one year ago
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    Guest kyun aa jaate hain?

  103. imqwerty
    • one year ago
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    pata nai... btw mere relatives paas me hi rahte hai to wo dadi dada se milne aa jate hai.

  104. imqwerty
    • one year ago
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    btw @ganeshie8 did u got the answer to the question - dIzI/dz z=complex number

  105. ganeshie8
    • one year ago
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    common sense : common lannguage guys, when others are there in the thread

  106. imqwerty
    • one year ago
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    ok :) sry

  107. ganeshie8
    • one year ago
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    thats okay, just saying..

  108. ParthKohli
    • one year ago
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    あなたは何について話していますか?理解できません。

  109. ganeshie8
    • one year ago
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    dIzI/dz derivative doesnt exist

  110. ParthKohli
    • one year ago
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    What kind of derivative is this?

  111. ganeshie8
    • one year ago
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    z is a complex number f(z) = |z| is a real valued function that takes complex number as input

  112. ParthKohli
    • one year ago
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    Yeah, I can see that. But it's a complex number. How do you compute derivatives of complex-valued functions? Is it that instead of taking a small interval, you take a small circular area?

  113. ganeshie8
    • one year ago
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    Exactly, the "interval on real number line" becomes a "ball in argand plane"

  114. ParthKohli
    • one year ago
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    @imqwerty knows all this?! o.O

  115. imqwerty
    • one year ago
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    not too much..

  116. ParthKohli
    • one year ago
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    Well, there goes my self-esteem.

  117. ganeshie8
    • one year ago
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    if you're in 12th grade and if you say you know all this, then you're just lying

  118. ParthKohli
    • one year ago
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    He's in 11th grade!

  119. ParthKohli
    • one year ago
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    So am I.

  120. EclipsedStar
    • one year ago
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    Me too...lol

  121. imqwerty
    • one year ago
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    i asked my teacher to explain this but he refused :( he said - its of no use

  122. ParthKohli
    • one year ago
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    That's stupid. "It's not there in the JEE examination" should not be restated as "it's of no use".

  123. imqwerty
    • one year ago
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    ^true that

  124. ParthKohli
    • one year ago
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    I think complex derivatives are a hell lot more useful that the four questions you gave us.

  125. imqwerty
    • one year ago
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    yes!! the teaching methodology nd the mindset of ppl must change.

  126. ganeshie8
    • one year ago
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    hmm your teacher knows more than you, he might be thinking that teaching you about complex derivatives is useless because you ppl are not ready yet

  127. ParthKohli
    • one year ago
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    What if he straight-up doesn't know what they are? That's a possibility too. I don't know much about Allen teachers but I don't expect mine to give a satisfactory explanation.

  128. imqwerty
    • one year ago
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    es this can be the reason @ganeshie8 nd we nt taught complex numbers till nw ..but still he said that its far away frm JEE syllabus. but i'll ask him when the chapter gets over. I asked Anna sir [HOD math allen] nd m sure he knws it

  129. ParthKohli
    • one year ago
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    Haha, of course it's beyond the reaches of JEE syllabi. It's not just the complex numbers we're taught in 11th and 12th.

  130. ParthKohli
    • one year ago
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    Wow, I just searched him up and he looks pretty young, like a passout.

  131. imqwerty
    • one year ago
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    m enrolled in vibrant academy too i'll go and ask Vikas gupta sir (HOD math vibrant).i think he will gimme the explanation.

  132. ParthKohli
    • one year ago
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    I think it's better to not ask for it...

  133. ganeshie8
    • one year ago
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    If you get a satisficatory explanation, please do share wid us

  134. imqwerty
    • one year ago
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    haahaha once i got kicked out (frnkly) fr asking such doubts.. nd yes i'll definately share it.

  135. imqwerty
    • one year ago
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    pk have u joined any institute?

  136. ParthKohli
    • one year ago
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    Yeah, Vidyamandir.

  137. imqwerty
    • one year ago
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    oh nice :)

  138. anonymous
    • one year ago
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    consider \(x^2+x+1\) divides into both \(x^5+x+1,x^{25}+x^5+1\). this is sufficient to prove \(5^{5^{n+1}}+5^{5^n}+1\) is not prime for positive integer \(n\)

  139. anonymous
    • one year ago
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    for the next problem, you just break it into an iterated sum, though if you don't set a lower bound on \(i,j\) individually you will run into problems

  140. imqwerty
    • one year ago
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    thanks @oldrin.bataku heres the new question that m solvin -

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  141. ParthKohli
    • one year ago
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    Eh, this problem isn't that hard.

  142. ParthKohli
    • one year ago
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    It may be calculative, but it's not hard.

  143. ganeshie8
    • one year ago
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    looks that fraction evaluates to \(\left(\dfrac{n+1}{n}\right)^2\)

  144. imqwerty
    • one year ago
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    Yes it has lil bit calculation prt

  145. Empty
    • one year ago
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    You can do the gauss fliperoo method there I think.

  146. anonymous
    • one year ago
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    $$\sum_{k=1}^n (2k)^2=4\sum_{k=1}^n k^2=4\cdot\frac{n(n+1)(2n+1)}6\\\sum_{k=0}^n(2k+1)^2=\sum_{k=0}^n (4k^2+4k+1)=4\sum_{k=1}^n k^2+4\sum_{k=1}^n k+\sum_{k=0}^n 1$$ so the second sum gives us: $$4\cdot\frac{n(n+1)(2n+1)}6+4\cdot\frac{n(n+1)}2+(n+1)$$

  147. anonymous
    • one year ago
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    oops, the second should be the sum up to \(n-1\), actually

  148. Empty
    • one year ago
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    @imqwerty You might like Mathematical Methods in the Physical Sciences by Mary Boas. I think you might like it, it introduces complex numbers and complex analysis in separate chapters, check it out: https://faculty.psau.edu.sa/filedownload/doc-4-pdf-0a187866618ca3049030ec5014860ae8-original.pdf It's very accessible and if you have any questions feel free to ask, that book covers so much more though than just that.

  149. imqwerty
    • one year ago
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    ok i am getting 151 as the answer

  150. imqwerty
    • one year ago
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    thank u so much @Empty ^^

  151. imqwerty
    • one year ago
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    wait no m getting 150 as the answer

  152. anonymous
    • one year ago
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    so $$\frac23\cdot n(n-1)(2n-1)+2 n(n-1)+n$$ which reduces the rational expression as $$\frac{2n+2}{2n-1}>1.01\\2n+2>1.01(2n-1)\\2+1.01>2(1.01-1)n\\3.01>0.02n\\n<50\cdot 3.01=150.5$$ so \(n\le 150\)

  153. imqwerty
    • one year ago
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    correct @oldrin.bataku :)

  154. anonymous
    • one year ago
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    for the last one \(xyz+xy+yz+zx\) links us to $$(x+1)(y+1)(z+1)=(x+1)(yz+y+z+1)\\\quad =xyz+xy+zx+yz+x+y+z+1$$ so we ahve that $$(x+1)(y+1)(z+1)-(x+y+z)-1=xyz+xy+zx+yz$$ now, we know \(x+y+z=12\) so: $$(x+1)(y+1)(z+1)-13=xyz+xy+zx+yz$$ now, let's use the AM-GM on \(x+1,y+1,z+1\): $$\frac{(x+1)+(y+1)+(z+1)}3\ge \sqrt[3]{(x+1)(y+1)(z+1)}$$ so it follows: $$(x+1)(y+1)(z+1)\le \left[1+\frac{x+y+z}3\right]^3$$ which gives us: $$(x+1)(y+1)(z+1)\le \left[1+4\right]^3=125$$ so we have that $$(x+1)(y+1)(z+1)-13\le125-13=112$$

  155. imqwerty
    • one year ago
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    thank u so much!! @oldrin.bataku now all the questions r solved ^~^

  156. imqwerty
    • one year ago
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    @Empty in the second question in the last step u took the common ratios of the 2 infinite GPs wrong

  157. anonymous
    • one year ago
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    complex differentiation is not beyond what a high schooler can know; I used the residue theorem in calculus II which was in 11th grade

  158. anonymous
    • one year ago
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    anyways, \(|z|=z\bar z\) is not differentiable on \(\mathbb{C}\) because it depends on \(\bar z\)

  159. anonymous
    • one year ago
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    $$\sum_{k=0}^N\left(\frac{3^{k+1}}{2^{k+1} - 3^{k+1}} - \frac{3^k}{2^k - 3^k}\right)=\frac{3^{N+1}}{2^{N+1}-3^{N+1}}-\frac{3^0}{2^0-3^0}$$ the first term behaves like: $$\frac{1}{(2/3)^{N+1}-1}\to \frac1{-1}=-1$$as \(N\to\infty\)

  160. anonymous
    • one year ago
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    also it should be \(k=1\) giving: $$-\frac{3^1}{2^1-3^1}=-\frac{3}{2-3}=-\frac{3}{-1}=3$$ so \(-1+3=2\)

  161. anonymous
    • one year ago
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    @imqwerty

  162. imqwerty
    • one year ago
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    thanks @oldrin.bataku ^~^ :)

  163. imqwerty
    • one year ago
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    @dan815

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