Hi, could anybody please help with this inequality? I've tried it several times, but my answers are always absurd.

Find a constant A for which abs(log(cos(x))+(x^2)/2) =< Ax^4 for -pi/3 <= x <= pi/3

Any help at all would be greatly appreciated.

Printable View

- October 1st 2012, 09:11 PMHalcyonNeed help with an inequality
Hi, could anybody please help with this inequality? I've tried it several times, but my answers are always absurd.

Find a constant A for which abs(log(cos(x))+(x^2)/2) =< Ax^4 for -pi/3 <= x <= pi/3

Any help at all would be greatly appreciated. - October 1st 2012, 11:02 PMchiroRe: Need help with an inequality
Hey Halcyon.

One suggestion that comes to mind is to use the triangle inequality which says that |a+b| <= |a| + |b| and if this is <= A*x^4 then you're done.

Now for cos(x) when you have pi/3, log(cos(x)) = log(1/2) and -pi/3 log(cos(x)) = log(1/2) and |log(cos(x))| will take on every value between 0 and ln(2). So |log(cos(x))| <= |log(1/2)| = |ln(2)| and |x^2/2|. So now find an A such that it fits this identity.

Also another hint: log(x) < x^2 for x > 1 and recall that |log(1/x)| = log(x) if x > 0. To show that log(x) < x^2 show that the initial value satisfies this (i.e. log(1) = 0 < 1) and that the derivative is always less for log(x) than for x^2 (which is easy since 1/x < 2x implies 1 < 2x^2 which is true when x>=1).