Show that $f(x) = \arctan\left(\frac{x}{x+1}\right) + \arctan\left(\frac{x+1}{x}\right) = \frac{\pi}{2}$
Show that $$f(x) = \arctan\left(\frac{x}{x+1}\right) + \arctan\left(\frac{x+1}{x}\right) = \frac{\pi}{2} \quad \forall x \in (-\infty, -1)\large\cup (0, \infty)$$
I observed this feature graphically, but am looking for a way to prove it. My attempt is as follows:
We can show that $$\arctan(A) + \arctan(B) = \arctan\left(\frac{A+B}{1 - AB}\right)$$ It follows that
$$\displaylines{f(x) = \arctan\left(\frac{x}{x+1}\right) + \arctan\left(\frac{x+1}{x}\right) = \arctan\left(\frac{\frac{x}{x+1} + \frac{x+1}{x}}{1 - \left(\frac{x}{x+1}\right)\left(\frac{x+1}{x}\right)}\right) \cr = \arctan\left(\frac{\frac{x}{x+1} + \frac{x+1}{x}}{1 - 1}\right) = \arctan\left(\frac{\frac{x}{x+1} + \frac{x+1}{x}}{0}\right) = \arctan(\infty) = \boxed{\frac{\pi}{2}}}$$
My attempt has a few questionable steps. The first is the deduction that division by 0 yields $\infty$. Secondly, I have been unable to consider the appropriate domain. Although my steps exclude for $-1$ and $0$, I don’t think they accurately consider values in between. I would appreciate if someone can make this "intuitive" proof more rigorous/concrete. I am also open to alternate solutions, but I would also appreciate comments on my attempt.
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If— | (no comment) | Aug 3, 2022 at 08:55 |
$\infty$ is not a number so $\arctan(\infty)$ isn't meaningful, nor does it have a reasonably standard interpretation. You can already tell something is wrong, because there's nothing in your argument that keeps it from applying for all values of $x$. One way to attempt to make an argument along the lines you sketch would be to perturb the problem slightly, then take the limit as the perturbation approaches $0$.
For example, $$\begin{align} \arctan\left(\frac{x+\varepsilon}{x+1}\right) + \arctan\left(\frac{x+1}{x}\right) & = \arctan\left(\frac{\frac{x+\varepsilon}{x+1} + \frac{x+1}{x}}{1 - \left(\frac{x+\varepsilon}{x+1}\right)\left(\frac{x+1}{x}\right)}\right) \\ & = \arctan\left(\frac{\frac{x+\varepsilon}{x+1} + \frac{x+1}{x}}{\frac{-\varepsilon}{x}}\right) \\ & = \arctan\left(-\frac{x^2}{\varepsilon(x+1)} - \frac{x}{x+1} - \frac{x+1}{\varepsilon}\right) \\ & = \arctan\left(-\frac{x^2+(x+1)^2}{\varepsilon(x+1)} - \frac{x}{x+1}\right) \\ \end{align}$$
Since the limit of $z^{-1}$ approaches $\infty$ when $z$ approaches $0$ from the right and $-\infty$ when approached from the left, the limit of the above expression is dependent on the sign of $\frac{x^2 + (x+1)^2}{x+1}$ and the direction we have $\varepsilon$ approach $0$. With this we know that the answer is at least $\pm\frac{\pi}{2}$.
Short of making the choices that give the right answer, it's not obvious how we should decide how $\varepsilon$ approaches $0$. This is the point where the fact that $$\arctan(A)+\arctan(B)=\arctan\left(\frac{A+B}{1-AB}\right)$$ is false starts being relevant. You can see this simply by setting $A=1$ and $B=2$. Because $\arctan$ is an odd, increasing function, the left hand side is necessarily positive, but the right hand side is $\arctan(-3)$ which is necessarily negative. The formula is only valid when $0 < 1 - AB$. For the above formulas to be valid, then, $\frac{-\varepsilon}{x}$ must be positive. For the given ranges, this means $\varepsilon$ must have the opposite sign as both $x$ and $x+1$ giving the desired positive $\frac{\pi}{2}$. We also see that for $x\in(-1,0)$, the constraint requires $\varepsilon$ to be positive but $x+1$ will positive as well which will lead to $-\frac{\pi}{2}$ which is indeed what you see in a plot.
As an addendum, here's a geometric/trigonometric argument. First note that the domain constraints ensure that $x$ and $x+1$ always have the same sign, so the arguments to $\arctan$ are always positive. For $a$ and $b$ both positive, $\theta = \arctan(\frac{b}{a})$ is in the interval $[0,\frac{\pi}{2})$. We can view $a$ and $b$ as, respectively, the $x$ and $y$ coordinates of a point. $\theta$ is then the angle between the $x$-axis and the ray from the origin through the point $(a, b)$. If we then consider $\varphi = \arctan(\frac{a}{b})$, we can use the same logic to view it as the angle between the $x$-axis and the ray through $(b,a)$. But $(b,a)$ is the reflection of $(a,b)$ through the line $y=x$, so the angle between the $y$-axis (note, $y$-axis, not $x$-axis) and the ray through $(b,a)$ is $\theta$, therefore $\varphi = \frac{\pi}{2}-\theta$.
You can use vector algebra, complex exponentials, or trigonometric identities if you want a more symbolic rendition of this.
You're right that division by zero is a problem if you want to use that identity. Rather than salvage that, I would recommend this identity instead:
$$ \arctan a + \arctan{a^{-1}} = \frac\pi2\qquad\text{for positive $a$} $$
Proving this identity is easy: draw a $1 \times a$ rectangle with a diagonal and consider the two triangles thus formed.
Then show that $\frac{x}{x+1}$ is positive for all $x$ in the domain of $f$ and you're done.
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