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Math Help - ooh a tricky question

  1. #1
    Member grgrsanjay's Avatar
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    ooh a tricky question

    i don't know weather its grouped correctly(sorry)
    for 0 < \theta < \frac{\pi}{2} if
     x = \displaystyle\Sigma_{n=0}^{\infty}(\cos ^{2n} \theta)
    y = \displaystyle\Sigma_{n=0}^{\infty}(\sin^{2n} \theta)
    z = \displaystyle\Sigma_{n=0}^{\infty}(\sin^{2n} \theta \cos^{2n} \theta)
    then which of the following are true(multiple choice)

    (A) \frac{1}{x} + \frac{1}{y} = 1
    (B)  x + y +xy =0
    (C) xyz = xy + z
    (D) xyz = x + y + z
    Last edited by grgrsanjay; September 4th 2010 at 04:11 AM.
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  2. #2
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    You may try using mathematical induction.
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    THE ACTUAL TIME GIVEN FOR PROBLEM IS 3 MINUTES using mathematical induction it takes 10 minutes at least

    i think we should expand it and take the G.P sum of n terms and do something might work

    this is a tricky sum for me
    Last edited by grgrsanjay; September 4th 2010 at 03:41 AM.
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    Quote Originally Posted by grgrsanjay View Post
    i don't know weather its grouped correctly(sorry)
    for 0 < \theta < \frac{\pi}{2} if
     x = \displaystyle\Sigma_{n=0}^{\infty}(\cos ^2n \theta)
    y = \displaystyle\Sigma_{n=0}^{\infty}(\sin^2n \theta)
    z = \displaystyle\Sigma_{n=0}^{\infty}(\sin^2n \theta \cos^2n \theta)
    then which of the following are true(multiple choice)

    (A) \frac{1}{x} + \frac{1}{y} = 1
    (B)  x + y +xy =0
    (C) xyz = xy + z
    (D) xyz = x + y + z
    There is something wrong with this question, because x+y = \displaystyle\Sigma_{n=0}^{\infty}(\cos ^2n \theta + \sin^2n \theta) = \Sigma_{n=0}^{\infty}1, which is infinite. So at least one of the series for x and y fails to converge. If for example \theta = \pi/4 then the three series for x, y, z all diverge, so none of the conditions (A), (B), (C), (D) is true.
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    Member grgrsanjay's Avatar
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    oh sorry it was just a typing error i corrected the question actually it sin to the power of 2n
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    Quote Originally Posted by grgrsanjay View Post
    oh sorry it was just a typing error i corrected the question actually it sin to the power of 2n
    In that case, these are geometric series, and the sums are given by x = \dfrac1{1-\cos^2\theta} = \dfrac1{\sin^2\theta} and y = \dfrac1{1-\sin^2\theta} = \dfrac1{\cos^2\theta}, from which it should be clear that (A) is correct.
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    Member grgrsanjay's Avatar
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    yea i got (A) correct so any other options are correct??? but when substitute a value (C) and (D) also seem correct to me
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    Quote Originally Posted by grgrsanjay View Post
    yea i got (A) correct so any other options are correct??? but when substitute a value (C) and (D) also seem correct to me
    Is this a multiple choice question or isn't it?
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    Quote Originally Posted by grgrsanjay View Post
    which of the following are true(multiple choice)says it is multiple

    (A) \frac{1}{x} + \frac{1}{y} = 1
    (B)  x + y +xy =0
    (C) xyz = xy + z
    (D) xyz = x + y + z
    me too getting c and d as answer with a(when substituting)
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    MHF Contributor undefined's Avatar
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    Quote Originally Posted by ggn View Post
    says it is multiple
    Yes thank you I can read, my point is that in a valid multiple choice question it's not possible to have more than one correct option, therefore if A is correct then B,C,D are automatically incorrect. If the OP wanted to call into question the validity of the problem, the OP could have specified that.
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    Quote Originally Posted by grgrsanjay View Post
    yea i got (A) correct so any other options are correct??? but when substitute a value (C) and (D) also seem correct to me
    That's right! I only looked at (A), because when I saw that was correct I assumed that it would be the only correct choice. But (C) and (D) are also correct.
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    yea but how to prove (C) and (D) are correct?? tell me please
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    Quote Originally Posted by grgrsanjay View Post
    yea but how to prove (C) and (D) are correct?? tell me please
    The formulas for x, y and z are x = \dfrac1{\sin^2\theta},\quad y = \dfrac1{\cos^2\theta},\quad z = \dfrac1{1-\sin^2\theta\cos^2\theta}. Substitute those into the equations (C) and (D), and check that both sides agree.
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  14. #14
    Member grgrsanjay's Avatar
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    what have you used to get that formula??
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    Quote Originally Posted by grgrsanjay View Post
    what have you used to get that formula??
    Sum of a infinite geometric series, \displaystyle\sum_{n=0}^\infty x^n = \frac1{1-x}.
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