While trying to adapt Euler's series for $\tan^{-1}{x}$ to be used for $\cot^{-1}{x}$ so I could more efficiently compute $\pi$ with Machin-like formulae, I discovered the following identity: $$\frac{\frac{1}{x}}{1+\frac{1}{x^2}}=\frac{x}{1+x^2}$$ I was surprised to discover that it's true because it seems counterintuitive. Why should the expression $\frac{x}{1+x^2}$ be the exact same value for both $x$ and its reciprocal, regardless of the value of $x$? It's not exactly the first example that comes to mind when you consider when you see an identity like $$f\left(\frac{1}{x}\right)=f(x)$$ What's the principle behind it?

  • 1
    $\begingroup$ How is it counterintuitive? Would it be more intuitive like this: $\frac{1/x}{1+1/x^2}=\frac{x}{x^2+1}$ (that is, not swapping the order of terms in the denominator)? Also, in another comment you admitted how you derived it, so you know the principle behind it; what other "principle" do you need? $\endgroup$
    – David K
    Jan 24, 2021 at 4:46
  • $\begingroup$ I was surprised that the ratio $\frac{x}{1+x^2}$ was the same value for both $x$ and $\frac{1}{x}$. $\endgroup$ Jan 24, 2021 at 4:53
  • $\begingroup$ Sometimes mathematics surprises us. That's a good thing. It would be so much less interesting if every result and its reasons were obvious before you even started to think about them. $\endgroup$
    – David K
    Jan 24, 2021 at 4:55
  • $\begingroup$ Divide both numerator and denominator by $x^2$ gets you the former from the latter...? I mean, not wanting to sound condescending but this is basic algebra, right? And of course it is not true when $x = 0$. $\endgroup$
    – xuq01
    Mar 25, 2021 at 23:15

2 Answers 2


Consider the following function of $x$:

$$ f(x) = \frac{1}{x + \frac 1x}. \tag1$$

But since $\frac{1}{1/x} = x,$ we find that

$$ f\left(\frac1x\right) = \frac{1}{\frac 1x + x} = f(x). $$

In other words, we have a formula that is symmetric in $x$ and $\frac1x,$ so of course we can interchange $x$ and $\frac1x$ and get the same result.

If you graph $y = f(x)$ on log-log graph paper (or even semi-log graph paper using the logarithmic axis for $x$) you will get a kind of bell-shaped symmetric curve. In general, any function $\phi$ whose graph on such paper is left-to-right symmetric will satisfy $\phi\left(\frac1x\right) = \phi(x).$

Your two formulas, $\frac{1/x}{1+1/x^2}$ and $\frac{x}{x^2+1},$ are just the right-hand side of Equation $(1)$ multiplied top and bottom by either $\frac1x$ or $x.$ That is, you have two alternative formulas for $f(x).$ And as we know that $f\left(\frac1x\right) = f(x),$ we know you can substitute $\frac1x$ for $x$ (or $x$ for $\frac1x$) in any formula of $f(x)$ and get back another formula of $f(x).$

For some more fun, consider

$$ \DeclareMathOperator{\sech}{sech} g(x) = \frac12 \sech x = \frac{1}{e^x + e^{-x}}. $$

This function is a kind of bell-shaped curve that is symmetric left-to-right across the $y$ axis. It also happens to be related to the function $f$ defined in Equation $(1)$ by the relationship

$$ f(x) = g(\ln x). $$

Also note that the property $\phi\left(\frac1x\right) = \phi(x)$ is a property of any formula that uses $x$ and $\frac1x$ in a completely symmetric fashion. For example:

$$ \phi_1(x) = x + 1 + \frac1x, \qquad \phi_1\left(\frac1x\right) = \phi_1(x). $$ $$ \phi_2(x) = x^3 + x + \frac1x + \frac1{x^3}, \qquad \phi_2\left(\frac1x\right) = \phi_2(x). $$ $$ \phi_3(x) = \sqrt x + \sqrt{\frac1x}, \qquad \phi_3\left(\frac1x\right) = \phi_3(x). $$

On the other hand, we can write a formula that uses $x$ and $\frac1x$ in an antisymmetric way; for example:

$$ \phi_4(x) = \frac{1}{x - \frac1x}, \qquad \phi_4\left(\frac1x\right) = -\phi_4(x). $$

  • $\begingroup$ Thank you. That explains a lot. Never even thought of dividing both numerator and denominator by $x$. $\endgroup$ Jan 24, 2021 at 5:55

Just multiply the numerator and denominator by $x^2$.

$$\frac{\frac{1}{x}}{1+\frac{1}{x^2}} = \frac{\frac{1}{x}}{1+\frac{1}{x^2}} \cdot \underbrace{\frac{x^2}{x^2}}_{=1} = \frac{\frac{1}{x} \cdot x^2}{\left(1+\frac{1}{x^2}\right) \cdot x^2} = \frac{x}{x^2+1}$$

Nothing particularly special going on here.

  • $\begingroup$ That's kind of how I derived it. It still surprised me. I wanted to dig a little deeper. $\endgroup$ Jan 24, 2021 at 4:40
  • $\begingroup$ I was surprised that the expression $\frac{x}{1+x^2}$ should be the same value for both $x$ and $\frac{1}{x}$, no matter what the value of $x$ is! $\endgroup$ Jan 24, 2021 at 5:22
  • 1
    $\begingroup$ Except $x=0$, of course. $\endgroup$ Jan 24, 2021 at 13:56
  • $\begingroup$ @BrianJ.Fink it seems that just the nature of that function. the same value for it's reciprocal. when you look at the graph here the graph at the right of x=1 is just the expanded version of the graph at the left x=1. $\endgroup$
    – acegs
    Mar 11, 2021 at 4:29

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