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Find all solutions of \(e^{2 x}=x+6\), correct to 4 decimal places; use the Newton Method.
in Mathematics by Platinum (131,394 points) | 292 views

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Let \(f(x)=e^{2 x}-x-6\). We want to find where \(f(x)=0\). Note that \(f^{\prime}(x)=2 e^{2 x}-1\), so the Newton Method iteration is
\[
x_{n+1}=x_{n}-\frac{e^{2 x_{n}}-x_{n}-6}{2 e^{2 x_{n}}-1}=\frac{\left(2 x_{n}-1\right) e^{2 x_{n}}+6}{2 e^{2 x_{n}}-1} .
\]
We need to choose an initial estimate \(x_{0}\). This can be done in various ways. We can (if we are rich) use a graphing calculator or a graphing program to graph \(y=f(x)\) and eyeball where the graph crosses the \(x\)-axis. Or else, if (like the writer) we are poor, we can play around with a cheap calculator, a slide rule, an abacus, or scrap paper and a dull pencil.

It is easy to verify that \(f(1)\) is about \(0.389\), and that \(f(0.95)\) is about \(-0.2641\), so by the Intermediate Value Theorem there is a root between \(0.95\) and 1 . And since \(f(0.95)\) is closer to 0 than is \(f(1)\), maybe the root is closer to \(0.95\) than to 1 . Let's make the initial estimate \(x_{0}=0.97\).
The calculator then gives \(x_{1}=0.970870836\), and then \(x_{2}=0.97087002\). Since these two agree to 5 decimal places, we can perhaps conclude with some (but not complete) assurance that the root, to 4 decimal places, is \(0.9709\). If we want greater assurance, we can compute \(f(0.97085)\) and \(f(0.97095)\) and hope for a sign change, which shows that there is a root between \(0.97085\) and \(0.97095\). There is indeed such a sign change: \(f(0.97085)\) is about \(-2.6 \times 10^{-4}\) while \(f(0.97095)\) is about \(10^{-3}\).

But the problem asked for all the solutions. Are there any others?

Draw the graphs of \(y=e^{2 x}\) and \(y=x+6\). The solutions of our equation are the \(x\)-coordinates of all places where the two curves meet. Even a rough picture makes it clear that the curves meet at some negative \(x\). Since \(e^{2 x}\) decays quite rapidly as \(x\) decreases through negative values, it seems reasonable that there will be a single negative root, barely larger than \(-6\). Certainly it cannot be smaller, since to the left of \(-6, x+6\) is negative but \(e^{2 x}\) is not. And it seems plausible that the positive root we found is the only one.

We first estimate the negative root. It is reasonable to start with \(x_{0}=-6 .\) Then \(x_{1}=-5.999993856 .\) We can guess that the root is indeed \(-6\) to 4 decimal places. For certainty, we should check that \(f(x)\) has different signs at \(-6\) and \(-5.9999\). It does.
Let's check that there are no more roots. Note that \(f^{\prime}(x)=2 e^{2 x}-1\). Thus \(f^{\prime}(x)>1\) when \(x>0\), and in particular \(f\) is increasing from \(0 \mathrm{on}\), indeed it starts increasing at \(x=-(1 / 2) \ln (2)\). Note also that \(f(0)<0\), and that, for example, \(f(1)>0\). So there is at least one root \(r\) between 0 and 1 . But there can only be one root there. For \(f(x)\) is increasing in the first quadrant, so can cross the \(x\)-axis only once.
A similar argument shows that there is a single negative root. For since \(f(x)\) is negative in the interval \((-\infty,(1 / 2) \ln 2)\), the function \(f\) is decreasing in this interval, so can cross the \(x\)-axis at most once in this interval. We saw already that it crosses the \(x\)-axis near \(x=-6\).
Note. There are many other ways of solving the problem. For example our equation is equivalent to \(2 x=\ln (x+6)\), and we could apply the Newton Method to \(2 x-\ln (x+6)\). Or we can use basically the same approach as above, but let \(y=2 x\). We end up solving \(e^{y}=y / 2+6\). If we are doing the calculations by hand, this saves some arithmetic.
by Platinum (131,394 points)

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