(1) Explain the features of the graph of the relation $|x|+|y|=1.$

If the above equation is true then neither $|x|$ nor $|y|$ can be greater than 1. If say $1+h=|x|>1$, we have: $1+h+|y|=1$ and so $|y|=-h<0$, a contradiction, as $|a|\ge 0$ for any $a$. Therefore, this function is not defined for values of $x$ and $y$ greater than 1.

I now quote without proof the fact that if $|a|=t$ then $a=t$ or $a=-t$. This is useful when applied in this problem.

If $|y|=1-|x|$ then $y=1-|x|$ or $y=-1+|x|$. If $y=1-|x|$ then $y=1+x$ or $y=1-x$. Similarly, if $y=-1+|x|$, then $y=x-1$ or $y=-x-1$.

The graph is the line segments (where $-1<x<1$ and $-1<y<1$) given by: $y=1-x$ in the first quadrant; $y=1+x$ in the second quadrant; $y=-1-x$ in the third quadrant and $y=x-1$ in the fourth quadrant.

Editor's note: By symmetry if $(a,b)$ satisfies this relation then so does

${\displaystyle (b,a),\hspace{0.5em}(-a,b)\hspace{0.5em}(-b,a)\hspace{0.5em}(-a,-b),\hspace{0.5em}(-b,-a),\hspace{0.5em}(a,-b),\hspace{0.5em}\mathrm{and}\hspace{1em}(b,-a)}$

so the graph is the square consisting of the four line segments described.

The line segments are perpendicular and intersect at (0,1), (-1,0) (0,-1) and (1,0). The square shape with vertices at the above co-ordinates, has symmetry in $y=x$, $y=-x$, $y=0$ and $x=0$. I refer to this as a graph, not a function, due to one-to-many relation.

2) As $n/(n+1)$ can be written as $1/(1+1/n)$, by dividing both numerator and denominator by $n$ we consider:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{{(1+\frac{1}{n})}^n}& =1+n\left(\frac{1}{n}\right)+\frac{n(n-1)}{2}{\left(\frac{1}{n}\right)}^2+\dots \\ \multicolumn{1}{c}{}& =2+\dots \ge 2>1.\end{array}}$

Taking the nth root

${\displaystyle 1+\frac{1}{n}\ge 2^{1/n}>1^{1/n}=1}$

Hence

${\displaystyle \frac{1}{(1+1/n)}=\frac{n}{(n+1)}\le \frac{1}{2^{1/n}}<1.}$

(3) Taking the simple case $x^2+y^2=1$ (the circle), and considering the first quadrant, it will be shown that $y$ decreases as $x$ increases. As this is in the first quadrant, the positive root alone is considered: $y=(1-x^2{)}^{1/2}.$ As $x$ increases to a maximum value of 1 on the interval $0\le x\le 1$ so $x^2$ increases monotonically and $f(x)=1-x^2$ decreases monotonically (to confirm, check f'(x)=-2x).

If we have $f(x_1)>f(x_2)$ (with $x_1<x_2$), then taking the positive square root of both sides, we have ${f(x_1)}^{1/2}>{f(x_2)}^{1/2}$. So as $x$ increases, $y$ decreases.

This same method could be extended to all values of $n$. For $g(x)=1-x^n$ then $g(x+h)<g(x)$ for $h>0$ because $x^n$ increases monotonically on the interval $0<x<1$. Clearly, if $x_1<x_2$ then $x_1^n<x_2^n$ so $1-x_1^n>1-x_2^n$ and hence $g(x_1)>g(x_2)$. Taking the positive nth root of both sides, we get: $y_1={g(x_1)}^{1/n}>{g(x_2)}^{1/n}=y_2$. So $y_1>y_2$ for $x_1<x_2$ ; hence as $x$ increases in the first quadrant, $y$ decreases.

Editor's note: Alternatively we can use calculus to prove this. Fixing one value of $n$ in $x^n+y^n=1$:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{y^n}& =1-x^n\\ \multicolumn{1}{c}{{\mathrm{ny}}^{n-1}\frac{\mathrm{dy}}{\mathrm{dx}}}& =-{\mathrm{nx}}^{n-1}\\ \multicolumn{1}{c}{\frac{\mathrm{dy}}{\mathrm{dx}}}& =-\frac{x^{n-1}}{y^{n-1}}<0.\end{array}}$

As $x$ and $y$ are positive in the first quadrant we see that the derivative is negative which shows that $y$ decreases as $x$ increases.

When $x=y$, on the general graph $x^n+y^n=1$ we get $2x^n=1$ so $x=y=1/(2^{1/n})$. From earlier in this question, we have for $n>1$,

${\displaystyle \frac{n}{n+1}<\frac{1}{2^{1/n}}}$

so $x=y>n/(n+1)$. Therefore, as there is only one solution to $2x^n=1$ in the first quadrant, the graph should lie outside the square with coordinates:

${\displaystyle (0,0),\hspace{0.5em}(0,n/(n+1)),\hspace{0.5em}(n/(n+1),n/(n+1)),\hspace{0.5em}(n/(n+1),0)}$

In the first quadrant, as $x,y\ge 0$, and on $x^n+y^n=1$, neither $x$ nor $y$ may be greater than 1, there are no co-ordinates on the curve for which $x,y>1$. If $x>1$ then $x^n=1+g$, for $g>0$ so $y^n=-g$, which is not allowed in the first quadrant. The curve must lie inside (or on at the co-ordinate axes) the square with coordinates (1,0), (0,1), (1,1), (0,0) NB, the curve will touch the square only at vertices (1,0) and (0,1).

(4) As $n\to \infty$, $1/(1+1/n)=n/(1+n)\to 1$.

The square which the curve must lie within is fixed (as above). However, as $n$ increases, the square which the curve must lie outside of gets increasingly large. As the curve is squeezed between two squares, its shape becomes increasingly more squared.

For curves where $n$ is even, the curve becomes squeezed in all four quadrants because, by symmetry (exactly as in part (1) of this question) the graph is symmetric about $y=x$, $y=-x$, $y=0$ and about $x=0$.

The same features occur for the whole family of graphs of $x^n+y^n=1$ for odd $n$.

The graphs of $|x^n|+|y^n|=1$ for odd $n$ would have the symmetrical squarish shape, reflecting the family of graphs in the first quadrant into the remaining quadrants (see part 1).