69 lines
2.1 KiB
Markdown
69 lines
2.1 KiB
Markdown
|
# Inverse functions
|
||
|
|
|
||
|
|
## Injectivity
|
||
|
|
|
||
|
|
A function $f$ is called injective if for all $x_1,x_2 \in \mathrm{Dom}(f), \space x_1 \neq x_2$ implies that $f(x_1) \neq f(x_2).$ Meaning that for every $y \in \mathrm{Rang}(f)$ there is precisely one $x \in \mathrm{Dom}(f)$ such that $y = f(x)$. Meaning, every $x$ has an unique $y$.
|
||
|
|
|
||
|
|
## Inverse function
|
||
|
|
|
||
|
|
If $f$ is injective, then it has an inverse function $f^{-1}$. The value of $f^{-1}(x)$ is the unique number $y$ in the domain of $f$ for which $f(y) = x$. Thus,
|
||
|
|
|
||
|
|
$$
|
||
|
|
y = f^{-1}(x) \iff x = f(y)
|
||
|
|
$$
|
||
|
|
|
||
|
|
Suppose $f$ is a continuous function, $f$ is injective if $f$ is strictly increasing or decreasing. That is, $f' \leq 0 \vee f' \geq 0$.
|
||
|
|
|
||
|
|
### Derivative of inverse function
|
||
|
|
|
||
|
|
When $f$ is differentiable and injective $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$.
|
||
|
|
|
||
|
|
**Proof:**
|
||
|
|
|
||
|
|
$$f(y) = x \implies f'(y) \frac{dy}{dx} = 1$$
|
||
|
|
|
||
|
|
$$\frac{dy}{dx} = \frac{1}{f'(y)} = \frac{1}{f'(f^{-1}(x))}$$
|
||
|
|
|
||
|
|
Without knowing the inverse function a value of the inverse derivative may be determined.
|
||
|
|
|
||
|
|
## The arcsine function
|
||
|
|
|
||
|
|
Always $\arcsin$ not $\sin^{-1}$ that is wrong since $\sin$ is not injective.
|
||
|
|
|
||
|
|
For $x \in [-\frac{\pi}{2},\frac{\pi}{2}] \space \arcsin(\sin x) = x$
|
||
|
|
|
||
|
|
For $x \in [-1,1] \space \sin(\arcsin x) = x$
|
||
|
|
|
||
|
|
The arccosine function is similar.
|
||
|
|
|
||
|
|
## Example question
|
||
|
|
|
||
|
|
Prove that $\forall x \geq 0$: $\arctan(x + 1) - \arctan(x) < \frac{1}{1 + x^2}$.
|
||
|
|
|
||
|
|
For $x = 0$: $\frac{\pi}{4} < 1$.
|
||
|
|
|
||
|
|
For $x > 0$: Consider the function $f(t) = \arctan(t)$ on the interval $[x, x+1]$. Apply the [Mean-value theorem](../differentation.md/#mean-value-theorem) of $f$ at the interval $[x,x+1]$,
|
||
|
|
|
||
|
|
$$\frac{f(x+1) - f(x)}{(x+1) - 1} = f'(c).$$
|
||
|
|
|
||
|
|
Let $\arctan(c) = y$ then, $c = \tan y$,
|
||
|
|
|
||
|
|
$$
|
||
|
|
\begin{array}{ll}
|
||
|
|
\frac{dy}{dc} (c = \tan y) &\implies 1 = \sec^2 (y) \frac{dy}{dc} = (\tan^2 y + 1) \frac{dy}{dc} \\
|
||
|
|
&\implies 1 = (c^2 + 1) \frac{dy}{dc} \\
|
||
|
|
&\implies \frac{dy}{dc} = \frac{1}{c^2 + 1}.
|
||
|
|
\end{array}
|
||
|
|
$$
|
||
|
|
|
||
|
|
Obtaining,
|
||
|
|
|
||
|
|
$$\arctan(x+1) - \arctan(x) = f'(c) = \frac{1}{c^2 + 1}.$$
|
||
|
|
|
||
|
|
For some $c \in (x,x+1)$, since $c > x$
|
||
|
|
|
||
|
|
$$\frac{1}{1 + c^2} < \frac{1}{1 + x^2},$$
|
||
|
|
|
||
|
|
thereby
|
||
|
|
|
||
|
|
$$\arctan(x+1) - \arctan(x) < \frac{1}{1 + x^2}.$$
|