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The task is as follows: Find the biggest negative root of the following equation using the secant method: $$\tan{x}-x+\frac{1}{2}=0$$ with $\epsilon =0.02$ being the max error.

So what I don't understand in this task is the meaning of the biggest negative root. When I graph the function, it has infinite negative roots due to $tan$ function. What is the meaning of biggest negative root?

A6EE
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    I think it's the closest root to $0$, from the left. – Botond Jun 25 '18 at 11:43
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    The root $x_0$ such that $x_0\leq 0$ and $x_k\leq x_0$ for any negative root $x_k$. In other words the negative root closest to zero. – Robert Z Jun 25 '18 at 11:44
  • Oh, thanks. It just came to mind that $-1$ is bigger than $-2$. – A6EE Jun 25 '18 at 11:47
  • If you want to remove the infinite branches, assuming that $\cos(x)\neq=0$, cosider that you look for the solution of $\sin(x)-\left(x-\frac12\right)\cos(x)=0$ – Claude Leibovici Jun 26 '18 at 06:33

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$\tan x-x$ has derivative $\tan^2x$ and is thus monotonically increasing on every interval $I_k=(k\pi-\frac\pi2,k\pi+\frac\pi2)$ with a triple root at $x=0$ and a horizontal inflection point at $x=k\pi$ for $k\ne 0$. Thus there is exactly one solution to $\tan x-x=c$ in each of these intervals, and for $k=0$ the solution for $c=-\frac12$ in $I_0$ is negative, so that this is already the largest negative root.

If you want to solve the problem of finding the root in $I_k$ in a boring way without surprises, which often is the most robust way, you use your method to solve $$ x-k\pi=\arctan(x-\frac12). $$

By using the first Taylor terms one finds first guesses for the root in $I_0$ from $$ x+\frac13x^3-x+\frac12=0\iff x=-\sqrt[\Large3]{\frac32}=-1.14... $$

Lutz Lehmann
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