Finding inverses of two functions in Lambda Notation

• CGandC
In summary, the first function is injective and its inverse is given by ## 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} ##. The second function is also injective and its inverse is given by ## f^{-1} = \lambda g \in \mathbb{R} \longrightarrow \mathbb{R}. \lambda x \in \mathbb{R} . g(x-1) ##. Both inverses can be
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.

CGandC said:
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##.

Last edited:
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.

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) ##

stevendaryl

What is Lambda Notation?

Lambda Notation is a mathematical notation used to represent functions in a concise and precise manner. It is commonly used in computer science and mathematics to define and manipulate functions.

What is an inverse function?

An inverse function is a function that undoes the action of another function. In other words, if a function f(x) maps an input x to an output y, then the inverse function f^-1(y) maps the output y back to the input x.

How do you find the inverse of two functions in Lambda Notation?

To find the inverse of two functions in Lambda Notation, you need to use the lambda operator (λ) and the dot operator (.) to define the functions. Then, you can use the inverse function notation (f^-1) to represent the inverse function.

What is the process for finding the inverse of two functions in Lambda Notation?

The process for finding the inverse of two functions in Lambda Notation involves swapping the input and output variables of the original function and then solving for the new output variable. This will give you the inverse function in Lambda Notation.

Why is it important to find the inverse of two functions in Lambda Notation?

Finding the inverse of two functions in Lambda Notation allows you to solve for the original input value when given the output value. This can be useful in various mathematical and computational applications, such as solving equations and creating efficient algorithms.

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