Would this require a Taylor Series Proof

[Bachelier]

Show that

abs[ sin (x) - 6x/(6+x^2) ] <= x^5/24, for all x in [0,2]

I tried to use the sine function taylor expansion but I get stuck

 

[chiro]

Hey Bachelier.

What are the taylor series for both sin(x) and 6x/(6 + x^2)?

 

[Bachelier]

for sine it is:


sin x = x - x^3/3! + x^5/5! - ... + (-1)^(n-1) x^(2n-1)/(2n-1)! + (-1)^n x^(2n+1)/(2n+1)! cos(y)

for a y in [0,2]

and f(x)= (6x /(x^2 + 6)) = x - x^3/6 + x^5/6^2 -x^7/6^3 + x^9/6^4 - x^11/ 6^5 + ... + (-1)^(n-1) x^(2n-1)/6^(n-1) + (-1)^n x^(2n+1)/6^n * f^n(0)

f^n(0) may equal 0 or -,+1 depending on n

 

[Bachelier]

Ok adding both terms we get:

abs [sin x - (6x /(x^2 + 6))] = |-7x^2/360 + 67/15120 x^7 -...|

≈ 7x^2/360 which is clearly ≤ x^2/24

is this good enough or do I have to bound my function?

 

[lurflurf]

A monotone alternating series' error is less than the first term omitted.
so yes
abs[ sin (x) - 6x/(6+x^2) ] <=7 x^5/360<x^5/24