Triangular Roots

While everyone has heard of a "square root", I for one had not heard of a "triangular root" defined for a number \(x\) as:$$ \sqrt [\strut \triangle] {x} = \frac{-1 \pm \sqrt{1+8x}}{2}$$Suppose we want to find the triangular root of 6. This gives:$$  \begin{align} \sqrt [\strut \triangle] {6} &= \frac{-1 \pm \sqrt{1+8 \times 6}}{2}\\  \\ &=  \frac{-1 \pm \sqrt{1+48}}{2} \\  \\ &= \frac{-1 \pm \sqrt{49}}{2} \\  \\ &= \frac{-1 \pm 7}{2}\\ \\ &= 3 \text{ } \text{  or} -4 \end{align}$$The formula arises from the definition of a triangular number \(x\) and the solution of the resulting quadratic equation:$$ \begin{align}  \frac{n \, (n+1)}{2} &= x \\ \\ n^2 + n -2x &=0 \\ \\  n &= \frac{-1 \pm \sqrt{1+8x}}{2} \end{align} $$This method is no different to what we do when finding the square root of a number where we have:$$ \begin{align} n^2 &= x \\ n &= \pm \sqrt{x}  \end{align}$$By way of comparison it can be noted that triangular roots are only real if \(x \geq -1/8 \) whereas square roots are only real if \(x \geq 0 \). I can thank Dr. Barker's YouTube video for prompting this post. 

Figure 1 shows the situation for \(6 = \dfrac{n \, (n+1)}{2}\) where \(n=-4 \text{ or } 3\).

Figure 1:  -4 and 3 are the values of \( \sqrt [\strut \triangle] {6}\)

Complex Chicanery

I came across an interesting video on YouTube, uploaded today on a channel named Higher Mathematics that posed the following problem:$$ \text{Solve for } x:\\1^x=5$$Clearly the problem has no solution amongst the real numbers but there turns out to be an infinity of solutions once we introduce complex numbers. The solution (with \(k=1,2,3 \dots \) ) then unfolds:$$\begin{align} e^{i \theta} &= \cos{\theta}+i \sin{\theta}\\e^{i 2k \pi} &= \cos{2k \pi}+i \sin{2k \pi}  \\&=1 \\ \therefore (e^{i 2k \pi)})^x &= 5\\ e^{i 2k \pi x} &= 5 \\\ln {e^{i 2k \pi x} }&= \ln{5} \\i 2k \pi x &= \ln{5} \\x &= \frac{ \ln{5}}{i 2k \pi} \\\ &=\frac{-i \ln{5}}{2k \pi}\end{align} $$The complex numbers for values of x given by \(k=1, 2,3\) are shown in Figure 1. They all map to the point \(5+0i\).

Figure 1

Thus we have a function \( f(x) \) with its domain D being the countably infinite set of points:$$ \left( \Big( 0,\dfrac{-i \ln{5}} {2k \pi} \Big) \text{ where k=1, 2, 3, ...} \right) $$ that maps all these points to \( (5, 0i) \) so that:$$ f(x) \rightarrow (5, 0i) \text{ for all x in D}$$

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Let's look now at a slightly different problem:$$2^x=x$$Right from the start, it can be said once again that there are no real solutions to this equality. This is clear once we graph the line \(y=2^x\) and \(y=x\) where we see that there are no points of intersection. See Figure 2.

Figure 2

This problem was posed on the same YouTube channel mentioned earlier. Here is a link to the video. So how do we find if any complex numbers satisfy this equality? Firstly, we take the natural logarithms of both sides and proceed from there:$$ \begin{align} 2^x &= x \\ \ln{2^x} &= \ln{x} \\ x \ln{2} &= \ln{x} \\ \frac{\ln{x}}{x} &= \ln{2} \\ \frac{\ln{x}}{e^{\, \ln{x}} }&= \ln{2} \\ \ln{x} \, {e^{\, -\ln{x}} } &= \ln{2} \\ -\ln{x} \, {e^{\, -\ln{x}} } &= -\ln{2} \\ W(-\ln{x} \, {e^{\, -\ln{x}} } ) &= W(-\ln{2}) \\ -\ln{x} &= W(-\ln{2} ) \\ \ln{x} &= -W(-\ln{2}) \\ x &= e^{-W(-\ln{2})} \\ x &= \frac{1}{e^{W(-\ln{2})}} \\ x &\approx -0.37927 - 0.72087 i\end{align}$$where W is the Lambert W function that I've written about in previous posts. See The Omega Constant and the Lambert W Function (June 24th 2020) and More on the Lambert W Function (February 16th 2021). To evaluate \(x\) using Wolfram Alpha the command ProductLog[-ln(2)] needs to be used.