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- #1

- Feb 14, 2012

- 3,601

- Thread starter anemone
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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,601

- Apr 22, 2018

- 251

By Taylor’s theorem, $\sin x=x-\dfrac{x^2}2\sin\xi$ for some $0<\xi<x$ (using the Lagrange form of the remainder). Thus:

$\begin{array}{rcl}\sin x &=& x-\dfrac{x^2}2\sin{\xi} \\\\ {} &>&x-\dfrac{x^2}2\xi \\\\ {} &>& x-\dfrac{x^3}2.\end{array}$

Hence $\sin\left(\dfrac1{1993}\right)>\dfrac1{1993}-\dfrac{\left(\frac1{1993}\right)^3}2 = \dfrac{7944097}{15832587314}>\dfrac1{1994}$, as required. So the smallest natural number is $\boxed{n=60}$.

- Oct 18, 2017

- 245

For $0<x<1$, we have $\sin(x)<x$. This shows that the inequality is satisfied for $n=60$. Taking $n=59$, we find:

$$

\sin\left(\frac{1}{1993}\right)\approx 0.00050176> \frac{1}{1994}\approx 0.00050150

$$

As $\sin\left(\dfrac{1}{n+1934}\right)$ is a decreasing function of $n$ for $n>0$, this shows that the smallest $n$ that satifies the inequality is $60$.