0
$\begingroup$

I found a class paper that mentioned that if we have $1-1$ function $f$ that takes a vector, $(X_1, X_2, \ldots, X_n)$ and returns:

$$ f(X_1, \ldots, X_n) = \left(\sum_{i=1}^{n}X_i, X_2, \ldots, X_n\right) $$

then the inverse is defined to be:

$$ f^{-1}(y, y_2, \ldots, y_n) = \left(y-\sum_{i=2}^{n}y_i, y_2, \ldots, y_n\right) $$

where $y$ is defined as: $y = f(X_1, \ldots, X_k) = \sum_{i=1}^{n}X_i$.

I am wondering how I can find the inverse and where exactly the subtraction of the first entry of the inverse comes in. Would anyone have any ideas? Thanks.

$\endgroup$
  • 1
    $\begingroup$ "where $y$ is defined as..." No, $y$ is not defined as what follows in your post, actually $y$ is simply part of the argument for the function $f^{-1}$ hence $y$ does not need to be defined. $\endgroup$ – Did Sep 6 '16 at 7:09
1
$\begingroup$

We set :

$$f(X_1, \ldots, X_n) = \left(\sum_{i=1}^{n}X_i, X_2, \ldots, X_n\right)=(y, y_2, \ldots, y_n).$$

Then we have the following linear system :

$$\left\{\begin{matrix} \sum_\limits{i=1}^{n}X_i = y_1\\ X_2 = y_2 \\ \vdots\\ X_n = y_n \end{matrix}\right. \Leftrightarrow\left\{\begin{matrix} X_1+\sum_\limits{i=2}^{n} y_i = y_1\\ X_2 = y_2 \\ \vdots \\ X_n = y_n \end{matrix}\right.$$

Since $\forall i, n\geq i \geq 2, X_i=y_i $. Then the system is equivalent to : $$\left\{\begin{matrix} X_1 = y_1-\sum_\limits{i=2}^{n} y_i\\ X_2 = y_2 \\ \vdots \\ X_n = y_n \end{matrix}\right.$$

Then $$f^{-1}(y_1,\dots,y_n)=(X_1,\dots,X_n)=\left(y_1-\sum_\limits{i=2}^{n} y_i,y_2,\dots,y_n\right).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.