(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| ³ 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

(b,a), (-a,b) (-b,a) (-a,-b), (-b,-a), (a,-b), and  (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:

æ ç è 1+ 1 n ö ÷ ø n   = 1 + n æ ç è 1 n ö ÷ ø + n(n-1) 2 æ ç è 1 n ö ÷ ø 2   + ¼ = 2 + ¼ ³ 2 > 1 .

Taking the nth root

1+ 1 n ³ 21/n > 11/n = 1

Hence

1 (1+1/n) = n (n+1) £ 1 21/n < 1.

(3) Taking the simple case x² + y² = 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-x²)^1/2. As x increases to a maximum value of 1 on the interval 0 £ x £ 1 so x² increases monotonically and f(x)=1-x² decreases monotonically (to confirm, check f'(x)=-2x).

If we have f(x₁) > f(x₂) (with x₁ < x₂), then taking the positive square root of both sides, we have f(x₁)^1/2 > f(x₂)^1/2. So as x increases, y decreases.

This same method could be extended to all values of n. For g(x) = 1-xⁿ then g(x+h) < g(x) for h > 0 because xⁿ increases monotonically on the interval 0 < x < 1. Clearly, if x₁ < x₂ then x₁ⁿ < x₂ⁿ so 1-x₁ⁿ > 1-x₂ⁿ and hence g(x₁) > g(x₂). Taking the positive nth root of both sides, we get: y₁ = g(x₁)^1/n > g(x₂)^1/n=y₂. So y₁ > y₂ for x₁ < x₂ ; 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 xⁿ+yⁿ=1:

yn = 1-xn nyn-1 dy dx = -nxn-1 dy dx = - xn-1 yn-1 < 0.

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 xⁿ + yⁿ = 1 we get 2xⁿ=1 so x=y=1/(2^1/n). From earlier in this question, we have for n > 1,

n n+1 < 1 21/n

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

(0,0), (0, n/(n+1)), (n/(n+1), n/(n+1)), (n/(n+1),0)

In the first quadrant, as x,y ³ 0, and on xⁿ + yⁿ = 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 xⁿ=1+g, for g > 0 so yⁿ=-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 ® ¥, 1/(1+1/n) = n/(1+n) ® 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 xⁿ + yⁿ=1 for odd n.

The graphs of |xⁿ| + |yⁿ| = 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).