Re: cos(x) >= 1 - x^2/2! + x^4/4! - x^6/6!
From: Timothy Murphy (tim_at_birdsnest.maths.tcd.ie)
Date: 08/23/04
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Date: Mon, 23 Aug 2004 11:28:45 +0100
Oleg wrote:
> I think you can clear up this thing considering the function
> h(x)=cos(x)-(1 - x^2/2! + x^4/4! - x^6/6!) and looking for its
> minimums by means of derivatives.
I expect someone has pointed this out -
I haven't been following the thread -
but I think the OP said he knew the result was true for small x -
actually it follows from the form of Taylor's Theorem with remainder
that it is true for -pi/2 < x < pi/2 .
If one assumes that, then it is sufficient to show h(x) has no zero.
If it has a zero then so does h'(x) by the Mean Value Theorem,
and then it follows that so does h''(x) since h'(0) = 0.
But that is the same result with n-2 in place of n.
-- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
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