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.