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binarymimic

prove if possible (cos x)^n <= cos (nx) if 0 <= x <= pi/2 and 0 < n < 1

  • one year ago
  • one year ago

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  1. binarymimic
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    \[\cos^{n} x \le \cos(nx)\]

    • one year ago
  2. IsTim
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    Not sure if I got this right, but have you used trial numbers?

    • one year ago
  3. binarymimic
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    to test the inequality ?

    • one year ago
  4. IsTim
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    I have no clue.

    • one year ago
  5. IsTim
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    Grade 12 Advanced Functions?

    • one year ago
  6. binarymimic
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    f'(x) - g'(x) > 0 implies f'(x) > g'(x) right? i think i found a method to figure it out

    • one year ago
  7. binarymimic
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    f(x) = cos (nx) g(x) = cos^n(x) f(x) - g(x) = cos (nx) - cos^n(x) if x = 0 then f(x) - g(x) = 0 - 0 =0 this is our starting point, the point at which both functions are equal maybe if we show f'(x) > g'(x) on the interval then that is sufficient to prove the original inequality

    • one year ago
  8. IsTim
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    I have no clue.

    • one year ago
  9. experimentX
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    showing (cos(nx))' .>= (cos^n(x))' is sufficient after showing they begin at some point.

    • one year ago
  10. binarymimic
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    Ok so f(x) = cos (nx) f'(x) = -n sin(nx) g(x) = cos^n(x) g'(x) = -n cos^(n-1)(x) * sin x f'(x) > g'(x) implies f'(x) - g'(x) = 0 so . . . if we can show -n sin(nx) - (-n cos^(n-1)(x) * sin x) = \[-n \sin(nx) + n \cos^{n-1}(x) * \sin x > 0\] is true on the interval.. hm

    • one year ago
  11. binarymimic
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    ok .. this becomes \[n \left( -\sin(nx) + \cos^{p - 1}(x) * \sin x \right)\]

    • one year ago
  12. binarymimic
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    \[n \left( -\sin (nx) + \frac{ \sin x }{ \cos^{1 - p} (x) } \right)\] reversing the order of the exponent's difference and putting cos in the denominator

    • one year ago
  13. binarymimic
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    ok since sin is increasing on the interval (0, pi/2), if we focus just on the two sin functions... -sin(nx) + sin x we know that n is fractional, so this portion of the expression will always be positive also we know cos is positive on the interval (0, pi/2) so we can conclude with certainty that -sin(nx) + sin x / cos^(1 - p)(x) > 0 if we multiply this expression by positive n, we will show f'(x) - g'(x) > 0 which is what we set out to prove. seems ok to me

    • one year ago
  14. ZeHanz
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    It looks quite ingenious to me! I think p should be n, btw... You have done a good job (yourself, that is...)!

    • one year ago
  15. binarymimic
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    youre right, the question uses p for the exponent but i'm used to n to denote an exponent . or k

    • one year ago
  16. binarymimic
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    wow i should seriously review what i type before i post it. lets just say p = n

    • one year ago
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