

Using Wolfram Alpha to prove 1=0 - DiabloD3
http://www.xamuel.com/using-wolfram-alpha-to-prove-10/

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Jach
Cache:
[http://webcache.googleusercontent.com/search?q=cache%3Ahttp%...](http://webcache.googleusercontent.com/search?q=cache%3Ahttp%3A%2F%2Fwww.xamuel.com%2Fusing-
wolfram-alpha-to-prove-10%2F)

Unfortunately the author seems uninformed. He claims that 0^0 = 1, not
indeterminate which is what Wolfram Alpha claims (and as anyone knows, if you
use indeterminates you can prove any number equals any other number, which the
author does in uncached screenshots). The problem though is that 0^0 changes
its meaning depending on the context. If you have an function f(x,y) = x^y,
it's not continuous at (0,0) because the limit of x -> 0 is 0 and the limit y
-> 0 is 1, making the value undefined! (Which is still different from 1, 0, or
indeterminate.) I used to know of a better blog post on the issue, but here's
<http://mathforum.org/dr.math/faq/faq.0.to.0.power.html>

Oh wait, I found it: <http://www.askamathematician.com/?p=4524> "Zero raised
to the zero power is one. Why? Because mathematicians said so."

Sort of in the same character of dispute is the fact that 0 and 1 aren't
probabilities: <http://lesswrong.com/lw/mp/0_and_1_are_not_probabilities/>

I'm also unsure what the author is entering since the screenshot isn't cached.
If I try
[http://www.wolframalpha.com/input/?i=summation+x^n%2Fn!+from...](http://www.wolframalpha.com/input/?i=summation+x^n%2Fn!+from+n%3D0+to+infinity+where+x+%3D+0)
wolfram gives the correct e^0 = 1.

Edit: Ah, he's using
[http://www.wolframalpha.com/input/?i=summation+0^n%2Fn!+from...](http://www.wolframalpha.com/input/?i=summation+0^n%2Fn!+from+n%3D0+to+infinity)
(substituting 0 in explicitly instead of instructing Wolfram to substitute on
x) which is quite an interesting quirk to give 0.

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vog
Although slightly off-topic, it is interesting to see that some kinds of
mistakes never die. As a schoolboy, I once thought that 0^0 can only take the
values 0 and 1. (which is quite similar to the author's mistake to think that
0^0 is always 1)

I thought so because for x → 0,

0^x → 0

x^0 → 1

x^x → 1

and I couldn't find any other expressions of the type "0^0" that have a limit
different from 0 and 1. I even wrote a "proof" for it, which of course was
flawed, but it took others (and me) quite some time to find the exact mistake
in my "proof".

0^0 is completely indefinite in the sense that you can find "0^0" expressions
for any limit _a_ you want it to be. Here's an example for that kind of
expressions:

x ^ (ln a / ln x) → a

Although I wasn't able to come up with those as a schoolboy, this is an easy
exercise for any math student.

------
vog
Unfortunately, the site is currently down. Here's a Google Cache version of
it:

[http://webcache.googleusercontent.com/search?q=cache:nY9J5bU...](http://webcache.googleusercontent.com/search?q=cache:nY9J5bUtcV8J:www.xamuel.com/using-
wolfram-alpha-to-prove-10/+http://www.xamuel.com/using-wolfram-alpha-to-
prove-10/&cd=1&hl=en&ct=clnk&client=iceweasel-a&source=www.google.com)

~~~
kurokikaze
Still, there is no images. You may want to look at comment here from the
author of this post:

[http://www.mathteacherctk.com/blog/2011/04/00-mnchausens-
syn...](http://www.mathteacherctk.com/blog/2011/04/00-mnchausens-
syndrome/#comment-3115)

------
mrfu
_Riddle me this: what value does the power series Σn=0∞xn/n! have, in terms of
x? Of course, the answer is ex_

From my (fuzzy) math memories, I thought that:

exp(x)= 1 + (x^1/1!) + (x^2/2!) + ... + (x^n/n!) + o(x^n) _near_ x=0

