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A problem out of Strang's Calculus asks:

Where does $ \tan (h)=1.01 h$? Where does $\tan (h)=h? $

The second question seems pretty straight-forward; $\tan (h) =h$ when $h$ is $0$.

The first question doesn't seem like it has a sensible answer when I use linear approximation, though (I keep getting zero, or something insane like $1.01=1$).

Any guidance is much appreciated.

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    Does Newton's method work for the function $f(h)=\tan h-1.01h$? – 2017-01-05

3 Answers 3

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This is trivial:

$$\tan(0)=a(0)$$

for arbitrary $a$.

If you insist on linear approximation,

$$\tan(h)=\tan(x)+\sec^2(x)(h-x)+\mathcal O(h^2)$$

for any $x$. Just choose $x$ values close to where you think a solution is...

For example, there is one solution near $4.5$, so with $x_0=4.5$,

$$\tan(x_1)\approx\tan(x_0)+\sec^2(x_0)(x_1-x_0)=1.01x_1$$

Solving for $x_1$,

$$x_1=\frac{\tan(x_0)-\sec^2(x_0)x_0}{1.01-\sec^2(x_0)}=4.495704457$$

Repeat the process to find $x_2$,

$$x_2=\frac{\tan(x_1)-\sec^2(x_1)x_1}{1.01-\sec^2(x_1)}=4.495612941$$

and $h=\lim_{n\to\infty}x_n$

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    Trivial? what about the non-zero solutions to $\tan(h) = h $? – 2017-01-05
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    @RobArthan In that scenario, we apply the second half of my post. – 2017-01-05
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    Just to clarify: you're telling me that zero is the correct soluton for both questions? – 2017-01-05
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    @HandsomeGorilla Yes, have you tried plugging it in? – 2017-01-05
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    Zero is a solution to both equations. A glance at the graph of $\tan$ shows that there are infinitely many other solutions. – 2017-01-05
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    Well I derived it from using linear approximation without assuming was zero, but the solution seemed _too_ trivial. My intuition told me that there should have been a non-zero solution to the first question (I don't believe there are non-zero solutions to the second) – 2017-01-05
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    @HandsomeGorilla Ok, I'll add that. – 2017-01-05
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    @HandsomeGorilla There we go, I gave you one more solution. :) – 2017-01-05
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    @HandsomeGorilla: you are wrong about non-zero solutions to $\tan(h) = h$. Take a glance at the graph of the tangent function over the whole real line. You will see that it intersects the line $y = x$ infinitely many times and that gives you infinitely many non-zero solutions to $\tan(h) = h$. – 2017-01-05
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    All of those intersections occur at nonzero multiples of pi. Nonzero arguments of tangent producing zero as the result do not meet the criterion. – 2017-01-05
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    @HandomeGorilla: $\tan(h) = h$ **is** the criterion you quoted from the textbook. You are confusing the $x$-axis with the line $x = y$. Sketch a graph of $\tan$ and $x = y$ and you will see that there are infinitely many values of $h$ that meet that criterion and only one of them has $\tan(h) = 0$. – 2017-01-05
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    @SimpleArt what made you pick 4.5? – 2017-01-06
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    @HandsomeGorilla DougM's method plus my comment below his answer rounded to a decimal. – 2017-01-06
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$\tan x \approx x + \frac 13 x^3\\ x + \frac 13 x^3 = 1.01x\\ x(\frac 13 x^2 - 0.01) = 0\\ x=\pm 0.1 \sqrt 3\approx \pm 0.173\\ \tan(0.1 \sqrt 3)\approx 0.17496\\ 1.01(0.1\sqrt3)\approx 0.17494$

And I haven't investigated outside of the interval $(-\frac \pi2, \frac \pi2)$

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    For $x$ outside that interval, take advantage that $\tan(x)$ is periodic. – 2017-01-05
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    Not sure that helps. Certainly there is one point in every other interval. But I couldn't tell you more than that with the approach I have taken, using a Taylor polynomial centered at 0. That polynomial doesn't converge outside this interval. I need new polynomial for other intervals. – 2017-01-05
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    Or just realize that $\tan(x)=\tan(x+k\pi)$, reducing our problem as follows:$$\frac13x^2+k\pi-1.01=0$$Which provide decent approximations. – 2017-01-05
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If you consider the first positive root, you could have a better approximation using Padé approximants (they approximate functions better than Taylor series).

The simplest, built around $h=0$, would be $$\tan(h)=\frac{h}{1-\frac{h^2}{3}}$$ and then $$\tan(h)=ah\implies h= \sqrt{3\,\frac{a-1}{a}}$$ So, for $a=1.01$, the result would be $h\approx 0.172345$ while, as Doug M answered, using Taylor expansion would lead to $h\approx 0.173205$ for an "exact" solution $h\approx 0.172175$.

Even better would be $$\tan(h)=\frac{h-\frac{h^3}{15}}{1-\frac{2 h^2}{5}}$$ from which $$\tan(h)=ah\implies h=\sqrt{15\frac {a-1}{6a-1}}$$ which would lead to $h\approx 0.172175$ which is the "exact" solution for six significant figures.