1=01^\infty=0

For any number xx, we can express the difference between xx and xx raised to some power nn as:

y=xxny=x-x^n

To find which number xx most exceeds xnx^n, we can find the maximum point of the above function.

First the derivative:

y=1nx(n1)y\prime=1-nx^{(n-1)}

And then the maximum (assuming that the power, nn, is greater than 1):

1nx(n1)=01-nx^{(n-1)}=0

nx(n1)=1nx^{(n-1)}=1

x(n1)=1nx^{(n-1)}=\frac{1}{n}

x=(1n)1/(n1)x=(\frac{1}{n})^{1/(n-1)}

We can relabel this final equation as a function in xx and yy which will except a power and yield the number which most exceeds itself raised to that power:

y=(1x)1/(x1)y=(\frac{1}{x})^{1/(x-1)}

This function can be shown to converge to 11:

limx (1x)1/(x1)=limx (1x)0=limx 1=1\lim_{x \to \infty}~(\frac{1}{x})^{1/(x-1)} = \lim_{x \to \infty}~(\frac{1}{x})^{0}=\lim_{x \to \infty}~1=\boxed{1}

This is equivalent to saying that the number xx which most exceeds xx^{\infty} is 11 (11 most exceeds 11^{\infty}). Remember that (because it gets weirder!).

Now the function describing the difference between these "greatest" numbers and their respective powers is:

y=(1x)1/(x1)[(1x)1/(x1)]xy=(\frac{1}{x})^{1/(x-1)}-[(\frac{1}{x})^{1/(x-1)}]^x

Simplifying:

y=(1x)1/(x1)[(1x)x/(x1)]y=(\frac{1}{x})^{1/(x-1)}-[(\frac{1}{x})^{x/(x-1)}]

y=(1x)1/(x1)[1(1x)(x1)/(x1)]y=(\frac{1}{x})^{1/(x-1)}[1-(\frac{1}{x})^{(x-1)/(x-1)}]

y=(1x)1/(x1)[1(1x)1]y=(\frac{1}{x})^{1/(x-1)}[1-(\frac{1}{x})^{1}]

y=(1x)1/(x1)[1(1x)]y=(\frac{1}{x})^{1/(x-1)}[1-(\frac{1}{x})]

y=(11x)(1x)1/(x1)y=(1-\frac{1}{x})(\frac{1}{x})^{1/(x-1)}

This can also be shown to converge to 11:

limx (11x)(1x)1/(x1)=limx (10)(1x)0=limx 1=1\lim_{x \to \infty}~(1-\frac{1}{x})(\frac{1}{x})^{1/(x-1)} = \lim_{x \to \infty}~(1-0)(\frac{1}{x})^{0}=\lim_{x \to \infty}~1=\boxed{1}

This is equivalent to saying that the difference between the number xx which most exceeds xx^{\infty} (11) and xx^{\infty} (11^{\infty}) is 11.

This is understandably counter-intuitive, but taken in this context, it seems provable!

To finish, here is the grand consequence of this discussion:

1+1=11^{\infty}+1=1

Which means:

1=0\boxed{1^{\infty}=0}

¨\ddot\smile

Note by David Stiff
2 weeks, 4 days ago

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