Why is the following function NOT invertible?

$$y[n] = \frac 1 {x[n-1]^2}$$

A function is invertible if two distinct inputs give two distinct outputs. Is there a quick way to check for invertibility?

  • \$\begingroup\$ Draw a graph. And if you can draw a horizontal line over two or more points in it, it is not invertible. And for the function in your example y[n]=1/(x[n-1])^2 = 1/(-x[n-1])^2. So it has the same value for any pair of negative/positive value. \$\endgroup\$ – Eugene Sh. Oct 21 '15 at 19:40
  • \$\begingroup\$ any other ways ? I won't be able to draw on the exam \$\endgroup\$ – user65652 Oct 21 '15 at 19:43
  • \$\begingroup\$ In the case of the example the function is clearly even. Even functions are non-invertible. \$\endgroup\$ – Eugene Sh. Oct 21 '15 at 19:45
  • \$\begingroup\$ how can you tell it is even by just looking at it? \$\endgroup\$ – user65652 Oct 21 '15 at 19:50
  • 1
    \$\begingroup\$ When you see x to the power of even number it will always lead to a same result for both x and -x. \$\endgroup\$ – Eugene Sh. Oct 21 '15 at 19:52

The principle here is that you can't get information from nothing. If a function throws away information, the inverse function would need to magically reproduce it. In this case, your function is throwing away the sign of the input value. Let's look at two examples. In the first, x[n] = 1 for all values of n:

$$x[n-1] = 1$$ $$y[n] = \frac 1 {x[n-1]^2} = \frac 1 {(1)^2} = 1$$

In the second example, x[n] = -1:

$$x[n-1] = -1$$ $$y[n] = \frac 1 {x[n-1]^2} = \frac 1 {(-1)^2} = 1$$

In both cases, y[n] = 1.

Now, imagine that you're told y[n] = 1, and you want to find x[n]. How can you tell which of the two example sequences to choose? You can't. The difference between them is the sign, but squaring the input gets rid of the sign! Thus, the function is not invertible.


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