Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

binarymimic

  • 3 years ago

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

  • This Question is Closed
  1. binarymimic
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  2. IsTim
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Not sure if I got this right, but have you used trial numbers?

  3. binarymimic
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    to test the inequality ?

  4. IsTim
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I have no clue.

  5. IsTim
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Grade 12 Advanced Functions?

  6. binarymimic
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    f'(x) - g'(x) > 0 implies f'(x) > g'(x) right? i think i found a method to figure it out

  7. binarymimic
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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

  8. IsTim
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I have no clue.

  9. experimentX
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    showing (cos(nx))' .>= (cos^n(x))' is sufficient after showing they begin at some point.

  10. binarymimic
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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

  11. binarymimic
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  12. binarymimic
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  13. binarymimic
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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

  14. ZeHanz
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  15. binarymimic
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    youre right, the question uses p for the exponent but i'm used to n to denote an exponent . or k

  16. binarymimic
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    wow i should seriously review what i type before i post it. lets just say p = n

  17. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy