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Find, correct to 5 decimal places, the $x$-coordinate of the point on the curve $y=\ln x$ which is closest to the origin. Use the Newton Method.
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Let $(x, \tan x)$ be a general point on the curve, and let $S(x)$ be the square of the distance from $(x, \tan x)$ to the origin. Then
$S(x)=x^{2}+\ln ^{2} x .$
We want to minimize the distance. This is equivalent to minimizing the square of the distance. Now the minimization process takes the usual route. Note that $S(x)$ is only defined when $x>0$. We have
$S^{\prime}(x)=2 x+2 \frac{\ln x}{x}=\frac{2}{x}\left(x^{2}+\ln x\right) .$

Since $\ln x$ is increasing, and $x^{2}$ is increasing for $x>0$, it follows that $S^{\prime}(x)$ is always increasing. It is clear that $S^{\prime}(x)<0$ for a while, for example at $x=1 / 2$, and that $S^{\prime}(1)>0$. It follows that $S^{\prime}(x)$ must be 0 at some place $r$ between $1 / 2$ and 1 , and that $S^{\prime}(x)<0$ if $x<r$ and $S^{\prime}(x)>0$ for $x>r$. We conclude that $S(x)$ is decreasing up to $x=r$ and then increasing. Thus the minimum value of $S(x)$, and hence of the distance, is reached at $x=r$.
Our problem thus comes down to solving the equation $S^{\prime}(x)=0$. We can use the Newton Method directly on $S^{\prime}(x)$, but calculations are more pleasant if we observe that $S^{\prime}(x)=0$ is equivalent to $x^{2}+\ln x=0$. Let $f(x)=x^{2}+\ln x$. Then $f^{\prime}(x)=2 x+1 / x$ and we get the recurrence
$x_{n+1}=x_{n}-\frac{x_{n}^{2}+\ln x_{n}}{2 x_{n}+1 / x_{n}} .$
We need to find a suitable starting point $x_{0}$. Experimentation with a calculator suggests that we take $x_{0}=0.65$. Then $x_{2}=0.6529181$, and $x_{2}=0.65291864$. Since $x_{1}$ agrees with $x_{2}$ to 5 decimal places, we can perhaps decide that, to 5 places, the minimum distance occurs at $x=0.65292$. If we have doubt, we can try to see whether $f(x)$ has different signs at $0.652915$ and $0.652925$. It does.
by Platinum (130,878 points)

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