Intervals for bisection method

Question

I have this function below: $$f(x) = \tan(x)\frac{(e^{2x} - 1)}{(e^{2x} + 1)} + 1$$ and I want to find the intervals to use the bisection method. The first interval I think is $f(0) = 1 >0$ but i can't find the $f()<0$. Does anybody have an idea about that?

Answer 1

since the denominator of the function has several roots, implies the function has several vertical asymptotes (indeed we should show that the domain of the function is an infinite set, otherwise it's possible for a function to have infinitely many undefined points, but the domain of the function is in a way which restricts the functions so that the function would not lie between any two vertical asymptotes, in this case despite having infinitely many undefined points (infinitely many roots of the denominator) the functions does not have any vertical asymptotes), like the function $\frac{\sqrt[]{1-x}+\sqrt[]{1+x}}{\cos\left(x\right)}$.

we are allowed to use Bisection method if the given function $f$ is continuous or at least it's continuous between the interval $\left[a,b\right]$( suppose $a$,$b$ are in the domain of the function $f$) and then the second conditions of Bolzano's Theorem holds, notice that in these cases we might can't be able to find all the possible answers.

here we have to find the vertical asymptotes, so first compute the roots of the denominator containing two portions:

$e^{2x}+1=0$ which has no real roots and $cos(x)=0$ which its real roots are $\frac{\left(2k+1\right)\pi}{2}$ for some $k∈ℤ$,for simplicity we consider the two roots $\frac{\pi}{2}$ and $\frac{3\pi}{2}$, now we should find two numbers between the interval $\left(\frac{\pi}{2},\frac{3\pi}{2}\right)$, the number $\pi$ easily can be found,note that $sgn(f(\pi))>0$, now you should find another number which its sign is opposite to $sgn(f(\pi))$, With a little effort $\frac{3\pi}{5}$ with $sgn\left(f\left(\frac{3\pi}{5}\right)\right)<0$ can be found.

now bisect the interval $\left[\frac{3\pi}{5},\pi\right]$ and compute the middle point of the given interval:

$$\frac{\frac{3\pi}{5}\ +\pi}{2}=\frac{4\pi}{5}$$

since $sgn\left(f\left(\frac{4\pi}{5}\right)\right)>0$ therefore take the opposite sign, means compute the middle point of the interval $\left[\frac{3\pi}{5},\frac{4\pi}{5}\right]$:

$$\frac{\frac{3\pi}{5}+\frac{4\pi}{5}}{2}=\frac{7\pi}{10}$$

since $sgn\left(f\left(\frac{7\pi}{10}\right)\right)<0$ therefore take the opposite sign, means compute the middle point of the interval $\left[\frac{7\pi}{10},\frac{4\pi}{5}\right]$:

$$\frac{\frac{7\pi}{10}+\frac{8\pi}{10}}{2}=\frac{3\pi}{4}$$

here the Approximate root is $\frac{3\pi}{4}=2.35619$. you can continue the method and it's clear that the more the steps of the method continue the closer the answer will be to the real root.

Added:

as mentioned above the function is not continuous over any arbitrary interval, so over what intervals we are allowed to use Bisection method? hence we should determine the points of discontinuity in order to identify the intervals which the function is not continuous over them,the points of discontinuity are the points whose first derivative does not exist,therefore take the derivative:

$\frac{dy}{dx}=\frac{\left(\left(1+\sec^2\left(x\right)\right)\left(e^{2x}-1\right)+\left(2e^{2x}\right)\left(\tan\left(x\right)\right)\right)\left(e^{2x}+1\right)-\left(2e^{2x}\right)\left(\tan\left(x\right)\left(e^{2x}-1\right)\right)}{\left(e^{2x}+1\right)^2}$,

since the derivative is undefined on the points which are the roots of its denominator, hence we just need simplify the derivative in order to get the denominator:

the denominator has two portions: first $(e^{2x}+1)^2$ whic has no real root and the second part is $(cos^2(x))$ which its real roots are all in the form $\frac{\left(2k+1\right)\pi}{2}$ for some $k∈ℤ$,since the domain of the function is $ℝ-\left\{\frac{\left(2k+1\right)\pi}{2},\left(k∈ℤ\right)\right\}$,implies the derivative has infinitely many undefined points, and since the only points of discontinuity which the function is not continuous are the points which the function has vertical asymptotes,therefore the function has infinitely many vertical asymptotes and as it's shown we know that between any two consecutive vertical asymptotes the function is continuous and the first condition of Bolzano's Theorem holds (note that for using Bisection method the second condition also should holds).

in your example it's clear that $0$ lies in the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, based on above explanations the functions is continuos over the given interval, so the first condition of Bolzano's Theorem holds, but how about the second one? not finding two points $a$,$b$ in the interval such that $f(a)f(b)<0$ cannot guarantee either such points exist or they don't.

since the interval is infinite, hence it's not possible to find all points $a$,$b$s, therefore we have to draw an approximate graph of the function between the interval:

$$\large\lim_{x \to \frac{\pi}{2}^-} \tan\left(x\right)\ \frac{\left(e^{2x}-1\right)}{\left(e^{2x}+1\right)}+1 = \lim_{x \to \frac{\pi}{2}^-}\frac{\sin\left(x\right)}{\cos\left(x\right)}·\lim_{x \to \frac{\pi}{2}^-}\frac{\left(e^{2x}-1\right)}{\left(e^{2x}+1\right)}+1$$

$$=\large\lim_{x \to \frac{\pi}{2}^-}\frac{\sin\left(\frac{\pi}{2}^-\right)}{\cos\left(\frac{\pi}{2}^-\right)}·\lim_{x \to \frac{\pi}{2}^-}\frac{\left(e^{^{\pi^-}}-1\right)}{\left(e^{^{\pi^-}}+1\right)}+1$$

by the assumption that $\bbox[5px,border:2px solid #C0A000]{\frac{\left(e^{\pi^-}-1\right)}{\left(e^{\pi^-}+1\right)}+1=z}$ we have:

$$=\large\underbrace{\lim_{x\rightarrow\frac{\pi}{2}^-}\frac{\sin\left(1\right)}{\cos\left(0^+\right)}}_\textrm{+∞}·\underbrace{\lim_{x\rightarrow\frac{\pi}{2}^-}\frac{\left(e^{\pi^-}-1\right)}{\left(e^{\pi^-}+1\right)}+1}_\textrm{sgn(z)>0}$$

$$=\large\lim_{x \to \frac{\pi}{2}^-} +∞·z = +∞$$

the same for $$\large\lim_{x \to \frac{\pi}{2}^+} \tan\left(x\right)\ \frac{\left(e^{2x}-1\right)}{\left(e^{2x}+1\right)}+1 =+∞$$

Note:in general we also should find horizontal asymptote(s), but since the function has a trigonometric section, hence there is no need to do that (indeed this refers to the periodicity property of trigonometric functions).

then compute the roots of first derivative:

$\large\frac{dy}{dx}=\frac{\left(\left(1+\sec^2\left(x\right)\right)\left(e^{2x}-1\right)+\left(2e^{2x}\right)\left(\tan\left(x\right)\right)\right)\left(e^{2x}+1\right)-\left(2e^{2x}\right)\left(\tan\left(x\right)\left(e^{2x}-1\right)\right)}{\left(e^{2x}+1\right)^2}=\frac{\left(e^{4x}-1\right)\left(\sec^2\left(x\right)+1\right)}{\left(e^{2x}+1\right)^2}$

it's clear that the only root is $0$, and $f(0)=1$.

this concludes that the function is always positive between $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ and never would meet the $x$-axis, in other words the function does not have any real root in this interval.

the only root of the derivative is $0$ and $0∈\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, also the both limits in the interval is $+∞$, but it does give any information about the existence of a real root, indeed it depends on $f(0)$ which is $1$, hence the function never would meet the $x$-axis, in other words it does not have any real root in this interval.

and about the rest of the intervals,it should be said that all of them are in the form $\left[\frac{\left(2k-1\right)\pi}{2},\frac{\left(2k+1\right)\pi}{2}\right]$($k∈ℤ$), and either one of $$\lim_{x \to \left(\large\frac{\left(2k-1\right)\pi}{2}\right)^{^{\large\pm}}} f(x)$$ $$or$$ $$\lim_{x \to \left(\large\frac{\left(2k+1\right)\pi}{2}\right)^{^{\large\pm}}} f(x)$$ is $+∞$ and the other is $-∞$,since the fucntion between any two consecutive vertical asymptotes is continuous, then we have a sign change, implies the second condition of Bolzano's Theorem holds,in other words between any interval expect $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ there exist at least one real root,but since the function between any interval except $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ is strictly monotone (the sign of the derivative between this kind of intervals does not change), therefore between any this kind of intervals there exist just one real root.

since the intervals are all in the form $\frac{\left(2k+1\right)\pi}{2}$ for some $k∈ℤ$, so the intervals except $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ are labeled with the infinite set $ℤ-\left\{-1,0\right\}$, hence there exist infinitely many vertical asymptotes,concludes there exist infinitely many roots.