1. Is the graph $y^2=x^2$ the same as $y=x$? I think not since $(2{)}^2=(-2{)}^2$ but I agree it should but I am still unsure.

   Tell me what is right and wrong, please.

2. Studying the nature of $y^{2n}+x^{2n}=r^{2n}$ where $n$ is a positive integer and $y$, $x$, and $r$ are real. If $n$ is 1 then the graph is a circle as $n$ increases the graph appears to become more like a square but never reaches perfection.

   Tell me, is $y^{\infty }+x^{\infty }=1$ the equation for a square, and why or why not?

Dear James,

Q1) Consider what any graph actually represents. For any given x the value (or possibly values) of y that will be plotted against it will be the values that satisfy the equation of the graph. If the equation has more than one solution for a given x than all solutions y should be plotted.

So now consider the equation you gave y² = x² . Put x equal to a number so we can see what is going on; say x=1. The values of y you should plot then, are those that satisfy y² = 1. i.e. y=+1, -1. You see that you will get two distinct lines on your graph.

Now to find the whole line we consider the same equation for general x. Put x=t, where t is any number. Then y must satisfy y² =t² . Hence y= +t, -t. So you have the parametric equation of the graph:
x=t, y=t
x=t, y=-t
So the graph is two straight lines, through the orgin with gradients 1 and -1. Compare this with the graph of y=x.

What I have done above is very general. If it is just this problem you wish to solve, then you must simply consider the multivalued nature of the square root. y² =x² is equivalent to y=±x.

Q2) What you need here is a limiting argument. Writing down $y^{\infty }$ etc. doesn't really mean anything since infinity is not a number. Noone could tell you what your equation involving infinity meant unless you could define infinity well enough to know how it behaves, and it turns out that that is not the best way to go about things anyway.

You may have seen things like: what is 1/n as n tends to $\infty$? As opposed to: what is $1/\infty$? This is called taking a limit, and if you haven't encountered it before then this could be rather confusing for you. If it is then please write back for further clarification.
The correct way to write you last question is: What is the shape of the graph of y²ⁿ +x²ⁿ =r²ⁿ as n tends to infinity?

The answer is that it is a square.

You show this by considering the important parts of the graph one at a time, and using limiting arguments in each case. The important points to consider are the vertical edges of the square, i.e. where x is roughly equal to -r and r, and the bit in the middle. I leave it to an exercise for you to show that there is no y solution beyond -r and r.

Okay, first the vertical edges. For x=r and x=-r, y²ⁿ =0 so y=0. Now how about just to the right of x=-r, say x=-r+e, where e is positive and can be made as small as you like. The equation becomes y²ⁿ +(-r+e)²ⁿ =r²ⁿ . Binomially expand (-r+e)²ⁿ to get r²ⁿ -r^(2n-1) *e*(2n)+...

Note that you can ignore the other terms in the binomial expansion since they all contain factors of e² or e³ etc. and you are taking e to be small, so they are insignificant compared to the first two terms.

Substituting this expansion into y²ⁿ +(-r+e)²ⁿ =r²ⁿ , we get:
y²ⁿ =(2n)*e*r^2n-1
and so
y=+-(2n)^1/2n *e^1/2n *r^(2n-1)/2n


Now consider what happens to each of these terms as n tends to infinity.
(2n)^1/2n tends to 1
e^1/2n tends to 1
r^(2n-1)/2n tends to r^2n/2n = r.

So y tends to ±r, as it should if the graph was a square. Note that the smaller you take e, the longer the e^1/2n term will take to converge to 1 for increasing n. See if you can work out what this means graphically.

I leave the other case, (x=r) to you. The demonstration is along exactly the same lines.

This explanation contains many concepts that may be new to you, such as binomial expansions and limiting arguments, so if you don't understand them don't worry. They are core concepts for more advanced maths, but as such need teaching properly in a way that can not be done on this website.

I hope this helps, if you do have any questions please write back, and we'll do our best to answer them.

Best wishes,
Richard.