Finding inverses of two functions in Lambda Notation

[CGandC]

Homework Statement:

Find if the following functions are injective and if so find their inverses.

Relevant Equations:

Familiarity with Lambda notation ( from lambda calculus )

I found the following functions ( In lambda notation ) to be injective, and now I am trying to find the inverse functions for them ( the inverse for the Image of $f$ ) but I am stuck and I need help:
1. $f = \lambda n \in \mathbb{N}. (-1)^n + n^2$
2. $f = \lambda g \in \mathbb{R} \rightarrow\mathbb{R}. \lambda x \in \mathbb{R} . g(x+1)$

In 1. This function is indeed injective. My first approach to find the inverse was handwavingly done as follows $(-1)^{2g(n)+1} + (2g(n)+1)^2 =n \longrightarrow g(n) = \frac{\sqrt{n+1} -1}{2}$ ( $n \in \mathbb{N}$ ) , I've done this in order for $(-1)^{2g(n)+1}$ to be equal to $-1$, but clearly $2g(n) + 1 = \sqrt{n+1}$ and it won't always be the case that $\sqrt{n+1}$ is a natural number so clearly, $g$ is not the correct inverse function for $f$.

In 2. I found this function to be injective ( maybe I am wrong? ) but then I was completely stumbled and don't know how to seek the inverse.

 

[stevendaryl]

Homework Statement:: Find if the following functions are injective and if so find their inverses.
Relevant Equations:: Familiarity with Lambda notation ( from lambda calculus )

I found the following functions ( In lambda notation ) to be injective, and now I am trying to find the inverse functions for them ( the inverse for the Image of $f$ ) but I am stuck and I need help:
1. $f = \lambda n \in \mathbb{N}. (-1)^n + n^2$
2. $f = \lambda g \in \mathbb{R} \rightarrow\mathbb{R}. \lambda x \in \mathbb{R} . g(x+1)$

In 1. This function is indeed injective. My first approach to find the inverse was handwavingly done as follows $(-1)^{2g(n)+1} + (2g(n)+1)^2 =n \longrightarrow g(n) = \frac{\sqrt{n+1} -1}{2}$ ( $n \in \mathbb{N}$ ) , I've done this in order for $(-1)^{2g(n)+1}$ to be equal to $-1$, but clearly $2g(n) + 1 = \sqrt{n+1}$ and it won't always be the case that $\sqrt{n+1}$ is a natural number so clearly, $g$ is not the correct inverse function for $f$.

Let's look at the pairs that we know:

$f(0) = 1$ so $g(1) = 0 = \sqrt{1-1}$
$f(1) = 0$ so $g(0) = 1 = \sqrt{0+1}$
$f(2) = 5$ so $g(5) = 2 = \sqrt{5-1}$
$f(3) = 8$ so $g(8) = 3 = \sqrt{8+1}$
$f(4) = 17$ so $g(17) = 4 = \sqrt{17-1}$

So the pattern is not always that $g(n) = \sqrt{n+1}$, but a case split:
If $n$ is even, then $g(n) = \sqrt{n+1}$.
If $n$ is odd, then $g(n) = \sqrt{n-1}$.

The fact that $g(n)$ isn't always equal to an integer is irrelevant. To be an inverse of $f$, all you need is that $g(f(n)) = n$. $f$ and $g$ do not have to have the same range and domain. In general: $f$ maps some set $A$ onto some set $B$. It's inverse would be a function $g$ that maps $B$ onto $A$. What it does on objects that are not in $B$ is irrelevant (or if you want, you can let $g(x)$ be undefined if $x$ is not in $B$). In the case we are talking about, $B$ is the set $0, 1, 5, 8, 17 ...$

In 2. I found this function to be injective ( maybe I am wrong? ) but then I was completely stumbled and don't know how to seek the inverse.

The fact that there are two lambdas is maybe a red herring. Just put it into words:

You have a function $f$ which when given another function $g$ returns a third function $h$ such that $h(x) = g(x+1)$.

The inverse must be a function that given an $h$ returns the corresponding $g$.

 

[CGandC]

Okay I found the inverse for the first function:
Note that $f = \lambda n \in \mathbb{N}. (-1)^n + n^2$ can be written as $f = \lambda n \in \mathbb{N}. \begin{cases} n^2 - 1, & \text{for } n \in \mathbb{N_{odd}} \\ n^2 + 1, & \text{for } n \in \mathbb{N_{even}} \\ \end{cases}$

And the Inverse is:
$g = \lambda n \in Im(f). \begin{cases} \sqrt{n-1}, & \text{for } n \in \mathbb{N_{odd}} \\ \sqrt{n+1} , & \text{for } n \in \mathbb{N_{even}} \\ \end{cases}$
You were right, I should've worked by looking at the image of f from the beginning instead of working with its range.
Right now I'm still trying to think about the second function's inverse.

 

[CGandC]

Ok, I managed to find the second function's inverse, that is:
$f^{-1} = \lambda g \in \mathbb{R} \longrightarrow \mathbb{R}. \lambda x \in \mathbb{R} . g(x-1)$
and so:
$f^{-1} \circ f = f\circ f^{-1} = \lambda g \in \mathbb{R} \longrightarrow \mathbb{R}. \lambda x \in \mathbb{R} . g(x)$