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Prove $ \left |\sin(x) - x + \dfrac{x^3}{3!} \right | < \dfrac{4}{15}$ $\forall x \in [-2,2]$

By Maclaurin's formula and Lagrange's remainder we have $\sin(x) = x - \dfrac{x^3}{3!} + \dfrac{\sin(\xi)}{5!}x^5$ for some $0<\xi<2$

subbing this in we get $\left|\dfrac{\sin(\xi)}{5!}x^5 \right| \leq \left |\dfrac{x^5}{5!} \right| \leq \dfrac{2^5}{5!} = \dfrac{4}{15}$, but the question has $<$ rather than $\leq$ - where have I done wrong?

edit: thinking the $\cos(\xi)$ should be there rather than $\sin(\xi)$

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  • $\begingroup$ Perhaps the estimate $| \sin(\xi) | \le 1$ is too weak. Can you argue that $\xi \ne \pm \frac{\pi}{2}$ (the only values in $[-2, 2]$ where $| \sin x | = 1$)? $\endgroup$ Mar 11, 2014 at 21:19
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    $\begingroup$ You haven't done anything wrong: you just have to do more. You'll have to analyze something more carefully; e.g. as @Sammy mentions. as an aside, another way to do the problem is to recognize that the remainder of the Taylor series is an alternating series whose terms are strictly decreasing in magnitude. $\endgroup$
    – user14972
    Mar 11, 2014 at 21:22
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    $\begingroup$ keep in mind that $sin(x) = x-{x^3 \over 3!} + {x^5 \over 120}$ - ${x^7 \over 7!}$ since $\frac{x^5}{5!}= 4/15$ for x = 2 if you subtract something than it would be lower ;-) $\endgroup$
    – Bman72
    Mar 11, 2014 at 21:24
  • $\begingroup$ @Hurkyl Could you please explain your alternate method for solving the problem? $\endgroup$
    – Warz
    Mar 11, 2014 at 21:37
  • $\begingroup$ There is equality in $\left |\dfrac{x^5}{5!} \right| \leq \dfrac{2^5}{5!}$ only when ... and check those cases separately. $\endgroup$
    – GEdgar
    Mar 11, 2014 at 21:57

2 Answers 2

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Use the exact form of the Taylor formula: $$ \sin x - x + \frac{x^3}3 = \int_0^x \frac{(x-t)^4}{4!}\cos(t) dt \\ \left| \sin x - x + \frac{x^3}3 \right| = \left| \int_0^x \frac{(x-t)^4}{4!}\cos(t) dt \right| \le\int_0^x \left| \frac{(x-t)^4}{4!}\cos(t) \right|dt \\ \le\int_0^x \left| \frac{(x-t)^4}{4!} \right|dt = \int_0^x \frac{t^4}{4!} dt = \frac{2^5}{5!} $$ Now if there is equality anywhere, every inequality becomes an equality, but considering the first implies that $x=0$, and the last implies that $x=2$.

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  • $\begingroup$ Thanks for this, but I've never seen this exact form before, so I guess we have yet to cover it. $\endgroup$
    – Warz
    Mar 11, 2014 at 23:07
  • $\begingroup$ just do a few integrations by parts :) $\endgroup$
    – mookid
    Mar 11, 2014 at 23:10
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From the Leibniz rule it is known that, if the sequence of $\frac{x^4}{5!},\frac{x^6}{7!},\frac{x^8}{9!},...$ is decreasing, which is the case for $x^2<6\cdot7=42$, $|x|\le6$ to get a round number, then $$ 0\le\frac{x^4}{5!}-\frac{x^6}{7!} \le \frac{\sin x}{x}-1+\frac{x^2}{3!} \le\frac{x^4}{5!}-\frac{x^6}{7!}+\frac{x^8}{9!} =\frac{x^4}{5!}-\frac{x^6}{7!}\left(1-\frac{x^2}{72}\right). $$ Since under the assumed restrictions $1-\frac{x^2}{72}\ge\frac12$, we get $$ \left|\sin x-x+\frac{x^3}{3!}\right|\le\frac{|x|^5}{5!}\left(1-\frac{x^2}{84}\right)<\frac{|x|^5}{5!} $$ for $0<|x|<6$.

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