Thanks

Suppose without loss of generality that x > y. Then we have:

x <= z = x * (1 + (y/x)^n)^(1/n) <= x*(1 + (y/x)^n) -> x*(1 + 0) = x as n -> infinity.

So z -> x as n -> infinity as required.

Thankyou

sfwc
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Also, it'seems that in general, n, z, x1, x2, x3, ..., xi being real numbers, the higher the value of n, the closer the value of z in

z^n = x1^n + x2^n + x3^n + ... + xi^n

is to max( x1, x2, x3, ..., xi ).

Is this a trivial statement, is it wrong, has it been proved, or is it a conjecture?

If a > 1, then in the limit as n->inf, (1+a^n) -> a^n
and z^n -> (ax)^n so z -> ax = max(x,y)

If a = 1 then (1 plus a^n) = 2, irrespective of n
and z^n = 2*x^n and z = x*2^(1/n)
In the limit z->x = y.

If a < 1, then in the limit as n->inf, (1+a^n) -> 1
and z^n -> x^n so z -> x = max(x,y)

Let x >= y, then

2xⁿ >= zⁿ >= xⁿ, or

2^1/nx >= z >= x.

As n goes to infinity, 1/n approaches 0, and, with it, 2^1/n approaches 1.

This is probably what you meant anyway, but it is good to have the assumptions stated explicitly.

By the way, if you are willing to write your equation as

z^n = |x|^n + |y|^n where x and y are real and z is positive real

then you can see an interesting geometric picture of how this limit occurs. Of course, the proofs already given in this thread show that the result is true, and this is NOT a proof, but it is a nice visualization:

Now fix the value of z to be 1 (this merely rescales x and y, and you can undo the rescaling at the end of the process if you wish since all three variables are raised to the same power). For n=2, you have the equation of the unit circle. As n increases, the circle bulges outward and gradually assumes a square shape. The points (x,y) = (0,1), (0,-1), (1,0) and (-1,0) do not move, as the circle is already tangent to the unit'square there. Points on the unit circle where |x| > |y| move to the x=1 or x=-1 sides of the square depending on the sign of x, and those with |y| > |x| similarly go to the y=1 or y=-1 sides of the square. Therefore |y| > |x| implies |y|=1=z, and
|x| > |y| implies |x|=1=z. So z = max(|x|, |y|).

By the way, Graham's proof in response #3 works even if x, y, and z are complex, provided that the conditions on "a" are replaced with conditions on "|a|" and provided we choose the correct n-th root to get z. Then z = x or y, whichever has the larger modulus.