How do we show 1−cos⁡x≥x23 for |x|≤1? My first idea was to write1−cos⁡x=12|eix−1|2which is true...

Algotssleeddynf

Algotssleeddynf

Answered

2021-12-28

How do we show 1cosxx23 for |x|1? My first idea was to write
1cosx=12|eix1|2
which is true for all xR, but I don't have a suitable lower bound for the right-hand side at hand.

Answer & Explanation

Ethan Sanders

Ethan Sanders

Expert

2021-12-29Added 35 answers

For |x|1 the Taylor series
cos(x)=1x22!+x44!x66!+
is an alternating series with terms that decrease in absolute value. It follows that for these x
cos(x)1x22!+x44!1x22!+x24!=11124x2
and therefore
1cos(x)1124x2
That is an even better estimate since 1124>13
The same approach can be used to show that 1cos(x)x23 holds on the larger interval [−2,2]:
cos(x)1x22!+x44!1x22!+4x24!=113x2
because the x2n(2n)! terms decrease in absolute value for n1
Cheryl King

Cheryl King

Expert

2021-12-30Added 36 answers

Since sinx is concave for x[0,π2], if x[0,1] then sinxxsin11cosxx212sin1. The x2 coefficient approximates 0.42. The generalization x[1,1] follows from 1cosx,cx2 being even.
Vasquez

Vasquez

Expert

2022-01-09Added 457 answers

f(x)=1cosxx2 is an even function, so we only look at x[0,1]. It's easy to prove f(x) is continuous and differentiable at x=0, and
f(x)=xsinx+2cosx2x3
Now
xsinx+2cosx22tanx2sinx4sin2x2
=2tanx22sinx2cosx24sin2x2=0,x[0,1]
Hence f(x)f(1)=1cos1.46>13,  1cosx(1cos1)x2>13x2

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