Here's the question you clicked on:
binarymimic
prove if possible (cos x)^n <= cos (nx) if 0 <= x <= pi/2 and 0 < n < 1
\[\cos^{n} x \le \cos(nx)\]
Not sure if I got this right, but have you used trial numbers?
to test the inequality ?
Grade 12 Advanced Functions?
f'(x) - g'(x) > 0 implies f'(x) > g'(x) right? i think i found a method to figure it out
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
showing (cos(nx))' .>= (cos^n(x))' is sufficient after showing they begin at some point.
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
ok .. this becomes \[n \left( -\sin(nx) + \cos^{p - 1}(x) * \sin x \right)\]
\[n \left( -\sin (nx) + \frac{ \sin x }{ \cos^{1 - p} (x) } \right)\] reversing the order of the exponent's difference and putting cos in the denominator
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
It looks quite ingenious to me! I think p should be n, btw... You have done a good job (yourself, that is...)!
youre right, the question uses p for the exponent but i'm used to n to denote an exponent . or k
wow i should seriously review what i type before i post it. lets just say p = n