Re: cos(x) >= 1 - x^2/2! + x^4/4! - x^6/6!
From: Oleg (anonymous_at_mathforum.org)
Date: 08/23/04
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Date: Mon, 23 Aug 2004 00:50:02 +0000 (UTC)
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.
On 22 Aug 2004, Dae-jung Yoo wrote:
>f_0(x) = 1
>f_1(x) = x^2/2!
>f_2(x) = x^4/4!
>f_3(x) = x^6/6!
>f_n(x) = x^(2n) / (2n)!
>We know that cos(x) = f_0(x) - f_1(x) + f_2(x) - f_3(x) + ... +
>(-1)^n*f_n(x) + ... for all x.
>I wonder if it is true that:
> cos(x) >= f_0(x) - f_1(x) + .... + (-1)^(2n+1)*f_(2n+1)(x) for all x.
>This inequality is true for very small x, because the sequence ( f_0(x),
>f_1(x), f_2(x), ... ) is a decreasing sequence when -1<x<1.
>This inequality is true for very large x, because then
>f_0(x) - f_1(x) + .... + (-1)^(2n+1)*f_(2n+1)(x)
>is a negative number far away from 0 while cos(x) is bounded between -1 and
>1.
>Therefore we can guess that this inequality may be also true for other
>values of x.
>This inequality comes from a book called "Problem-Solving Through Problems".
>The author of the book assumes the inequality to be true, for the only
>reason that the sequence f_0(x), -f_1(x), f_2(x), -f_3(x), ... is an
>alternating sequence. ( That the tayler series for cos(x) is an alternating
>series.) But the absolut values of the sequence is not steadily decreasing.
>it is not true that f_n(x) >= f_(n+1)(x) for all x and for all nonnegative
>integer n.
>I failed to prove or disprove the inequality. Can somebody help me on this?
>
>Thanks.
>
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