Why is $\frac{\frac{1}{x}}{1+\frac{1}{x^2}}=\frac{x}{1+x^2}$ true? [closed]

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?

• 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? Jan 24, 2021 at 4:46
• I was surprised that the ratio $\frac{x}{1+x^2}$ was the same value for both $x$ and $\frac{1}{x}$. Jan 24, 2021 at 4:53
• 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. Jan 24, 2021 at 4:55
• 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 $x = 0$. Mar 25, 2021 at 23:15

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:

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

• Thank you. That explains a lot. Never even thought of dividing both numerator and denominator by $x$. 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.

• That's kind of how I derived it. It still surprised me. I wanted to dig a little deeper. Jan 24, 2021 at 4:40
• 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! Jan 24, 2021 at 5:22
• Except $x=0$, of course. Jan 24, 2021 at 13:56
• @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. Mar 11, 2021 at 4:29