Two Equations, Two Unknowns and the Beta Function

I spotted this little problem on Twitter (see Figure 1):

Figure 1

The proximity of \(+31\) to \(+32\) should alert us to the possibility that either \(x=+2\) or \(y=+2\) is a solution. After that, the corresponding \(x\) and \(y\) values fall into place. However, let's suppose that we didn't spot this and I have to confess I didn't. In that case, we follow the conventional method. 

At first sight, it looks like you will end up with a quintic equation after substituting \(y=1-x\) into \(x^5+y^5=31\) and there is no general formula for solving quintic equations. However, the \(x^5\) terms cancels out nicely leaving us with a quartic equation. Let's see how:$$\begin{align} x^5+(1-x)^5&=31\\x^5+(1-5x+10x^2-10x^3+5x^4-x^5)&=31\\5x^4-10x^3+10x^2-5x-30&=0\\5x^3(x-2)+5(2x^2-x-6)&=0\\5x^3(x-2)+5(2x+3)(x-2)&=0\\(x-2)(5x^3+10x+15)&=0\\5(x-2)(x^3+2x+3)&=0\\5(x-2)(x+1)(x^2-x+3)&=0 \end{align}$$Thus \(x=2\) and \(y=-1\) or \(x=-1\) and \(y=2\). Using GeoGebra, this can be confirmed visually as shown in Figure 2:

Figure 2

Differentiation reveals a little more about the graph of \(x^5+y^5=31\):$$\begin{align} y'&=\frac{-x^4}{y^4}\\&=-\frac{x^4}{\sqrt[5]{31-x^5}} \end{align} $$It can be seen that \(y'\) is undefined when \(x=\sqrt[5]{31} \approx 1.9873 \) and thus the graph is vertical at this point. If we zoom in a little, this becomes clearer (see Figure 3). There is a point of horizontal inflection when \(x=0\).

Figure 3

This was a useful little exercise as I got to revisit the binomial expansion for \( (a+b)^5\) and the factor theorem in order to factorise \(x^3+2x+3\). Using 1 rather than 31, I was also prompted to consider the area under the curve for \( x^5+y^5 =1\) between 0 and 1. See Figure 4.

Figure 4

In this graph, the vertical tangent occurs at \(x=1\) and this got me thinking about the area under the curve between 0 and 1. In other words, what is the value of:$$ \int_0^1 \sqrt[5]{1-x^5} \, dx$$Using SageMathCell, the integral evaluates to:$$ 0.2 \cdot \beta \! \left (0.2, 1.2 \right )$$Now I vaguely remember coming across a so-called \( \beta \) function that is in some way, I think, related to the \(\Gamma \) function. Further investigation revealed that they are indeed related. In fact:$$ 0.2 \cdot \beta(0.2, 1.2)= 0.2 \cdot \frac{ \Gamma(0.2) \cdot \Gamma(1.2)}{ \Gamma(0.2 +1.2)} \approx 0.950150138988437 $$This result looks about right but what is going on with this \( \beta \) function? Well, the beta function is actually an integral between 0 and 1 involving two different values, let's say \(a\) and \(b\), so that we have:$$\beta(a,b)=\int_0^1 t^{a-1} \cdot (1-t)^{b-1} dt$$If we can get our integral into that form then we can integrate it. Firstly, we make a substitution. Let \(t=x^5\) and so \(dt/dx=5x^4\) and \(dx=0.2 x^{-4}\). We can write \(x^{-4}=t^{-0.8}\) and the integral now transforms into:$$ \begin{array} \displaystyle \displaystyle \int_0^1 \sqrt[5]{1-x^5} \, dx &=\text{0}.2 \displaystyle \int_0^1 t^{-0.8} (1-t)^{0.2} \, dt \\&=0.2 \displaystyle \int_0^1 t^{0.2-1} (1-t)^{1.2-1} \, dt\\&=0.2 \, \beta(0.2,1.2) \end{array} $$So this was quite an interesting post that led me in some unexpected directions. I also was interested in the area between the curves \(x^2+y^2=1\) and \(x^5+y^5=1\) in the range from 0 to 1. See Figure 5 where eq2 is \(x^2+y^2=1\) and the curve \(x^5+y^5=1\) is a tangent to the circle at \(x=0\) and \(x=1\).

Figure 5

Clearly, the area between the two curves from \(x=0 \text{ to }1\) is given by \( \beta(0.2,1.2) - \dfrac{\pi}{4} \). What I've learned from all this is that the \(\beta\) function is a very useful tool to have in one's integration armoury.