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Find all solutions of $5 x+\ln x=10000$, correct to 4 decimal places; use the Newton Method.
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Let $f(x)=5 x+\ln x-10000$. We need to approximate the root(s) of the equation $f(x)=0$. The function $f$ is only defined for positive $x$. Note that the function is steadily increasing, since $f^{\prime}(x)=5+1 / x>0$ for all positive $x$. It follows that the function can be 0 for at most one value of $x$. It is easy to verify that $f(1)<0$ and $f(2000)>0$, and therefore the equation has a root in the interval $(1,2000)$.

The Newton Method iteration is easy to set up. We get
$x_{n+1}=x_{n}-\frac{5 x_{n}+\ln x_{n}-10000}{5+1 / x_{n}} .$
We could simplify the right hand side somewhat. This is probably not worthwhile.

Now we need to choose $x_{0}$. The idea is that even when $x$ is large, $\ln x$ is by comparison quite small. So as a first approximation we can forget about the $\ln x$ term, and decide that $f(x)$ is approximately $5 x-10000$. Thus the root of our original equation must be near $x=2000$.

Shall we choose $x_{0}=2000$ ? It is sensible to do so. But we can do better. Note that $\ln (2000)$ is about 7.6. So we can take $5 x_{0} \approx$ $10000-7.6$. Let $x_{0}=1998.48$.

A quick computation gives $x_{1}=1998.479972$. This agrees with $x_{0}$ to 4 decimal places, so the answer, correct to 4 decimal places, should be 1998.4800. If we feel like it, we can show by the usual "sign change" procedure that this answer is indeed correct to 4 places.

Note. If we start with $x_{0}=2000$, it turns out that $x_{1}=1998.479972$, so perhaps the extra thinking that went into starting with $1998.48$ was unnecessary. But it illustrates the fact that in some cases we can get an extremely accurate estimate of a root without bringing out heavy machinery.
by Platinum (93,243 points)

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