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e.g. The function $e^x$ reflected through $y=x$ is $\ln x$. Is this always true OR just in some cases?

Later
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RHS
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5 Answers5

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Yes, but it is the graph of the function that is reflected, not the function itself.

The graph of a function $f$ is the set of all pairs $(x,y)$ with $y=f(x)$. If $f$ has an inverse function $g$, then $y=f(x)$ is equivalent to $x=g(y)$, so when $(x,y)$ belongs to the graph of $f$, then $(y,x)$ belongs to the graph of $g$ and vice versa.

Interchanging $x$ and $y$ in the pair $(x,y)$, that is replacing it by $(y,x)$, can be described as reflection throught the diagonal. And that explains the phenomenon in general.

Harald Hanche-Olsen
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    I know that it is not uncommon to say that the graph is the function. But for the sake of this discussion, I thought it better to leave that to one side. – Harald Hanche-Olsen May 10 '13 at 12:57
  • Could you explain more about the interchange of the x and y mathematically? – RHS May 10 '13 at 13:05
  • I don't see any point in explaining further unless you give me a sense of why my explanation is insufficient. – Harald Hanche-Olsen May 10 '13 at 13:18
  • Nvm, got it. $y' \equiv g(x')$, g is the inverse of f. W/ the original y values of f we could get back the x's w/ g. So, $x'=y$ & $y'=x$. – RHS May 10 '13 at 13:25
  • I don't know what that would mean. Possibly you could use the term about replacing $f(x)$ by $f(-x)$ (reflection through the $y$ axis) or by $-f(-x)$ (reflection through the origin). – Harald Hanche-Olsen May 10 '13 at 15:22
  • Right. The trace. I was asking about "What the reflect of func. would mean?" but accidentally removed it. – RHS May 10 '13 at 15:27
  • There is still one logical gap, although now we know why there is a interchange in term of f & g. Still the interchange in term of reflection is left unexplained. – RHS May 10 '13 at 15:32
  • Let $(x_2,y_2)$ be the image of $(x_1, y_1)$ (above $y=x$) w/ dist. d from $(x_3, x_3)$. We have $x_1 = x_3 - d/\sqrt{2} = y_2$. Similarly for $x_2 = y_1$. – RHS May 10 '13 at 16:08
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    The midpoint between $(x,y)$ and $(y,x)$ is $(\frac12(x+y),\frac12(x+y))$ on the diagonal. And the vector $(y,x)-(x,y)=(y-x,x-y)$ is orthogonal to the vector $(1,1)$ pointing along the diagonal. Thus $(y,x)$ is the reflection of $(x,y)$ through the diagonal. – Harald Hanche-Olsen May 10 '13 at 16:12
  • Let $(x_2,y_2)$ be the image of $(x_1, y_1)$ (above $y=x$) w/ dist. d from $(x_3, x_3)$. We have $y_2 = x_3 - d/\sqrt{2} = x_1$. Similarly for $x_2 = y_1$. So both reflection of f alone $y=x$ & plotting the inverse func. of it involved interchange of x & y. – RHS May 10 '13 at 16:16
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Suppose that $D(a,b)$ is a point on the graph of a one-to-one function defined by $y=f(x)$. Then $b=f(a)$ This means that $a=f^{-1}(b)$, so $D_{1}(b,a)$ is a point on the graph of the inverse function $f^{-1}$. Now, two points are said to be symmetric with respect to any line if the line is perpendicular to the segment that links both points in its midpoint.

Therefore, the demonstration reduces to prove that $y=x$ is perpendicular to the segment $DD_{1}$ in its midpoint $M$. The midpoint of $DD_{1}$ is $M$($\frac{a+b}{2}$,$\frac{a+b}{2}$), which also belongs to $y=x$. Then the line $y=x$ intersects the segment $DD_1$ at its midpoint $M$. Now, two nonvertical lines are perpendicular if and only if the product of their slopes is $-1$.The slope $m$ of $y=x$ is $m=1$. The slope $m_1$ of the line $DD_1$ is $m_1=\frac{b-a}{a-b}=\frac{-(a-b)}{a-b}=-1$. It follows that the line $y=x$ is perpendicular to the segment $DD_1$in its midpoint $M$.

Q.E.D.

user152370
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Reflection with respect to the diagonal $y=x$ means substitution of $x$ with $y$. So, start from $x$, apply $f$ and get $y$, then substitute $y$ with $x$, obtaining $x$. Conversely, start with $y$, apply substitution and get $x$, then apply $f$ to obtain $y$. As you can see, substitution is right and left inverse of $f$, hence is its inverse.

Federica Maggioni
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Consider the reflection matrix:

$$\begin {bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}$$

This matrix reflects a point over the y = x axis. Now consider a point on a function, (x,f(x)). With respect to the origin, it can be expressed as a vector:

$$\begin {bmatrix} x \\ f(x) \\ \end{bmatrix}$$

Now apply the transformation matrix to the vector:

$$\begin {bmatrix} f(x) \\ x \\ \end{bmatrix}=\begin {bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}\begin {bmatrix} x \\ f(x) \\ \end{bmatrix}$$

Now, in order to normalize this point (i.e. make x-coordinate x and the y coordinate a function of x), we need to apply the inverse operation to both coordinates:

$$\begin {bmatrix} f(x) \\ x \\ \end{bmatrix}=\begin {bmatrix} x \\ f^{-1}(x) \\ \end{bmatrix}$$$$

rurouniwallace
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Here is a summary of my comments on Harald Hanche-Olsen's ans.

To proof g is the inverse of f iff they reflect each other along $y=x$.

First, we need $y′\equiv g(x′)$, g is the inverse of f. W/ the original y values of f we could get back the x's w/ g. So, $x′=y$ & $y′=x$.

Second, we need to let $(x_2,y_2)$ be the image of $(x_1,y_1)$ (above y=x) w/ dist. d from $(x_3,x_3)$. We have $y2=x3−d/\sqrt{2}=x1$. Similarly for $x_2=y_1$.

So $(x_0,y_0)$ of f, corresponds $(y_0,x_0)$ of g, which is also where it's image locate.

Same for other points of f.

So proofed.

RHS
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