4 ms·

I've always marveled at the exponential function. Here's a 3D visualization of z=x^y that supports the 0^0 is undefined camp: https://s32.postimg.org/v9t4stdl

by etatoby 10y ago

I've always marveled at the exponential function. Here's a 3D visualization of z=x^y that supports the 0^0 is undefined camp: https://s32.postimg.org/v9t4stdl1/x_y.gif https://s32.postimg.org/v9t4stdl1/x_y.gif (Made with Grapher.app and Gimp) You can clearly see that the surface is tangent to the z (vertical) axis, which suggests that any z>=0 satisfies z=0^0, hence 0^0 is undefined. Edit: here's a picture that highlights two important rules: https://s31.postimg.org/4kozbjqbf/x_y_lines.png https://s31.postimg.org/4kozbjqbf/x_y_lines.png The red line is the rule 0^y=0 for all y; the purple line is x^0=1 for all x. It's clear why these two lines would not meet at 0^0. Some will define 0^0=0, others 0^0=1. But the 3D surface shows that it's really a vertical line.

mathgenius 10y ago
Also check out Greg Egan's visualization here: https://johncarlosbaez.wordpress.com/2016/06/25/a-quirky-function/#comment-80933 https://johncarlosbaez.wordpress.com/2016/06/25/a-quirky-fun...
dahart 10y ago
This is a great example. Wikipedia has the same one with isolines demonstrating this point in the 0^0 subsection of the exponentiation article, here: https://en.m.wikipedia.org/wiki/Exponentiation#Continuous_exponents https://en.m.wikipedia.org/wiki/Exponentiation#Continuous_ex... With the caveat: "However, under certain conditions, such as when f and g are both analytic functions and f is positive on the open interval (0, b) for some positive b, the limit approaching from the right is always 1.[26][27][28]" Also, definitely take note of Knuth's argument, because while you might feel strongly that indeterminate is the right answer right now, there are big implications for allowing other choices. (And NB: indeterminate vs undefined) "The debate stopped there, apparently with the conclusion that 0^0 should be undefined. But no, no, ten thousand times no! Anybody who wants the binomial theorem to hold for at least one nonnegative integer n must believe that 0^0 = 1, for we can plug in x = 0 and y=1 to get 1 on the left and 0^0 on the right. The number of mappings from the empty set to the empty set is 0^0. It has to be 1." http://arxiv.org/pdf/math/9205211v1.pdf http://arxiv.org/pdf/math/9205211v1.pdf

by etatoby 10y ago

mathgenius 10y ago

Also check out Greg Egan's visualization here: https://johncarlosbaez.wordpress.com/2016/06/25/a-quirky-function/#comment-80933 https://johncarlosbaez.wordpress.com/2016/06/25/a-quirky-fun...

dahart 10y ago

This is a great example. Wikipedia has the same one with isolines demonstrating this point in the 0^0 subsection of the exponentiation article, here: https://en.m.wikipedia.org/wiki/Exponentiation#Continuous_exponents https://en.m.wikipedia.org/wiki/Exponentiation#Continuous_ex... With the caveat: "However, under certain conditions, such as when f and g are both analytic functions and f is positive on the open interval (0, b) for some positive b, the limit approaching from the right is always 1.[26][27][28]" Also, definitely take note of Knuth's argument, because while you might feel strongly that indeterminate is the right answer right now, there are big implications for allowing other choices. (And NB: indeterminate vs undefined) "The debate stopped there, apparently with the conclusion that 0^0 should be undefined. But no, no, ten thousand times no! Anybody who wants the binomial theorem to hold for at least one nonnegative integer n must believe that 0^0 = 1, for we can plug in x = 0 and y=1 to get 1 on the left and 0^0 on the right. The number of mappings from the empty set to the empty set is 0^0. It has to be 1." http://arxiv.org/pdf/math/9205211v1.pdf http://arxiv.org/pdf/math/9205211v1.pdf