find the equation approximate solution , such the root of
$$x^2\sin{\dfrac{1}{x}}=2x-501$$ to an accuracy of $ 0.001$
I think this problem use this $$\sin{x}\approx x-\dfrac{1}{6}x^3$$ then $$x^2(\dfrac{1}{x}-\dfrac{1}{6x^3})\approx 2x-501$$ $$\Longrightarrow x-\dfrac{1}{6x}\approx 2x-501$$ $$x+\dfrac{1}{6x}\approx 501$$
so I guess $$x\approx 501$$
because when $$x=501,\dfrac{1}{6x}\approx \dfrac{1}{3000}\approx 0.0003$$
My methods is true? Have you other methods? Thank you
it is said idea use $$\sin{y}=y-\dfrac{\sin{(\theta y)}}{2}y^2,0<\theta<1\cdots\cdots(1)$$ But this is Taylor formula? and How prove $(1)$ it？
Let $x = 1/t$ and multiply by $t$. The equation becomes $$ \dfrac{\sin(t)}{t} = 2 - 501 t$$ Assuming you're looking for real solutions, $-1 < \dfrac{\sin(t)}{t} < 1$.
So $-1 < 2 - 501 t < 1$, i.e. $\dfrac{1}{501} < t < \dfrac{1}{167}$. With $t$ this close to $0$, $\sin(t)/t$ is close to $1$, so in fact $t \approx \frac{1}{501}$. In fact we have $1 > \sin(t)/t > 1 - t^2/6$, so $$\dfrac{1}{501} < t < 1503 - \sqrt{2259003}$$ and thus $$501 > x > \dfrac{1}{1503 - \sqrt{2259003}} = 500.9996673\ldots$$ In particular, your answer $501$ is within the required accuracy.