Htykuut

I have to prove that the function $;f:(-1,1)to mathbb{R};$ defined by $f(x)=logsqrt{frac{1+x}{1-x}};$ is bijective.

I have already proved that it is injective:
$$f(x)=f(y)$$
$$logsqrt{frac{1+x}{1-x}}=logsqrt{frac{1+y}{1-y}}$$
$$logsqrt{frac{(1+x)(1-y)}{(1-x)(1+y)}}=0$$
$$frac{1+x-y-xy}{1-x+y-xy}=1$$
$$x=y$$

But now, how can I prove that the function is surjective?

We have that $f(x)$ is defined in $(-1,1)$ and

$$f(x)=logsqrt{frac{1+x}{1-x}}implies f'(x)=frac1{1-x^2}>0$$

then $f(x)$ is injective, moreover

$$lim_{xto 1^-} f(x)=infty quad lim_{xto -1^+} f(x)=-infty$$

and since $f(x)$ is continuous by IVT it is surjective.

Let $yinmathbb{R}$. Then the equation
$$
y=logsqrt{frac{1+x}{1-x}}
$$
becomes
$$
frac{1+x}{1-x}=e^{2y}
$$
that solves as
$$
x=frac{e^{2y}-1}{e^{2y}+1}
$$
This can be rewritten as $x=tanh y$, but is not relevant. Note that
$$
-1<frac{e^{2y}-1}{e^{2y}+1}<1
$$
if and only if
$$
-e^{2y}-1<e^{2y}-1<e^{2y}+1
$$
and both inequalities are obviously true for every $y$.

This can be simplified if you know about hyperbolic function: you need to prove that $-1<tanh y<1$, that is, $tanh^2y<1$ or $sinh^2y<cosh^2y$, which is clear because $sinh^2y=cosh^2y-1$.

The proof of injectivity can be shortened by noticing that both the logarithm and the square root are injective, so you just have to prove that
$$
frac{1+x}{1-x}=frac{1+y}{1-y}
$$
implies $x=y$. The given equality becomes
$$
1+x-y-xy=1+y-x-xy
$$
that's exactly $x=y$.

$f:(-1,1) to mathbb R^+, f(x) = frac {1+x}{1-x}$ is continuous, monotonically increasing, and a bijection.

$g: mathbb R^+to mathbb R^+, g(x) = sqrt{x}$ is continuous, monotonically increasing, and a bijection.

$h: mathbb R^+to mathbb R, h(x) = ln x$ is continuous, monotonically increasing, and a bijection.

The composition of bijections gives a bijection.

You might also note that:

$sqrt frac{1+x}{1-x}$

substituting $x = cos theta$ gives

$sqrt frac {1+cos theta}{1-costheta} = cot frac theta2$

We can show that $f$ is surjective using a proof by contradiction and a few basic facts about $+, -, exp, log, sqrt{cdots} dots $ .

$renewcommand{isa}{mathop{:}} renewcommand{opdot}{mathop{.}}$Our hypothesis that $f(x) = logsqrt{frac{1+x}{1-x}}$ where $text{-1} < x < 1$ can be written using quantifiers as:

$$ forall y isa mathbb{R} opdot exists x isa, (text{-1}, 1) opdot log sqrt{frac{1+x}{1-x}} = y tag H $$

Let's take our hypothesis and negate it ... and then show that the negation is inconsistent.

$$ exists y isa mathbb{R} opdot forall x isa, (text{-1}, 1) opdot log sqrt{frac{1+x}{1-x}} = y tag{NG1} $$

We can assume that we have a counterexample, let's name it $c$ . We only get one counterexample though, so after we assume (NG2), (NG1) can't be used anymore.

$$ forall x isa, (text{-1}, 1) opdot log sqrt{frac{1+x}{1-x}} ne c tag{NG2} $$

For the sake of conciseness, let's leave off $x isa, (text{-1}, 1)$ on our intermediate steps.

$$ log sqrt{frac{1+x}{1-x}} ne c tag{1} $$

We can apply an injective function $g$ to both sides of the inequality and get back another true inequality $a ne b iff g(a) ne g(b)$. Let's start with $exp$ .

$$ exp log sqrt{frac{1+x}{1-x}} ne exp{c} tag{2} $$

$exp$ is the inverse of $log$

$$ sqrt{frac{1+x}{1-x}} ne exp{c} tag{3} $$

The function that sends $x$ to its square is not injective on the reals. However, $exp(c)$ is positive and $sqrt{cdots}$ is real and hence non-negative. Therefore in this particular case, squaring will preserve $(ne)$ .

$$ frac{1+x}{1-x} ne exp{2c} tag{4} $$

multiplying by $1-x$ is injective regardless unless $x=1$, but $x$ is defined to be in $(text{-1}, 1)$ .

$$ 1+x ne (1-x)exp{2c} tag{5} $$

$$ 1+x ne (exp{2c}) - xexp{2c} tag{6} $$

Adding a constant $xexp{2c}$ and $-1$ is injective.

$$ x + xexp{2c} ne (exp{2c}) - 1 tag{7} $$

Division by a positive constant $(exp{2c})+ 1$ is injective

$$ x ne frac{(exp{2c})-1}{(exp{2c})+1} tag{8} $$

$$ bot $$

In order to produce a contradiction, we pick $x = frac{(exp{2c})-1}{(exp{2c})+1}$ .

Our choice of $x$ lies in the open interval $(text{-1}, 1)$, but we need to prove it.

In order to show $text{-1} < x < 1$, we'll show that $x le -1$ and $x ge 1$ are absurd.

$$ frac{(exp{2c})-1}{(exp{2c})+1} le text{-1} tag{NG3} $$

multiplication by a positive constant is monotonic, preserve $(le)$.

$$ (exp{2c}) -1 le -1 - exp{2c} tag{9} $$

addition by a constant is monotonic.

$$ (exp{2c}) + exp{2c} le 0 tag{10} $$

simplify and divide by two.

$$ exp{2c} le 0 tag{11} $$

$$ bot $$

Contradiction, because $exp{2c}$ is positive.

Now to the other boundary case.

$$ frac{(exp{2c}) - 1}{(exp{2c})+1} ge 1 tag{NG4} $$

$$ (exp{2c}) - 1 ge (exp{2c}) + 1 tag{12} $$

$$ 0 ge 2 tag{13} $$

$$ bot $$

Contradiction.

Therefore, if there were any $c in mathbb{R}$ that weren't hit by $f$, then $frac{(exp{2c}) - 1}{(exp{2c}) + 1}$, which is in $(text{-1}, 1)$, wouldn't be in the domain of definition of $f$ . However, $f$ is surjective and therefore total.