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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 $\sin(\xi)$

Warz
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  • 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 $| \sin x | = 1$)? – Sammy Black Mar 11 '14 at 21:19
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    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. –  Mar 11 '14 at 21:22
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    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 ;-) – Bman72 Mar 11 '14 at 21:24
  • @Hurkyl Could you please explain your alternate method for solving the problem? – Warz Mar 11 '14 at 21:37
  • There is equality in $\left |\dfrac{x^5}{5!} \right| \leq \dfrac{2^5}{5!}$ only when ... and check those cases separately. – GEdgar Mar 11 '14 at 21:57
  • @GEdar What do you mean? $|\frac{x^5}{5!}| \leq \frac{2^5}{5!} $$\forall x \in [-2,2]$ – Warz Mar 11 '14 at 22:13
  • @SammyBlack On second thoughts, shouldn't $\sin(\xi)$ actually be $\cos(\xi)$ which then solves all our problems? – Warz Mar 12 '14 at 08:58

2 Answers2

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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 $x=2$.

mookid
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2

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 $0<|x|<6$.

Lutz Lehmann
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