

(x2-2.5x+1.5)/(x-1.5) - alexk
http://www06.wolframalpha.com/input/?i=%28x2-2.5x%2B1.5%29%2F%28x-1.5%29%3E0

======
mnemonicsloth
Not a bug. Alpha is right. There is no singularity.

let a=1.5 for brevity and let f(x) = (x-a)(x-1)/(x-a).

Relabel f(x)= N(x)/D(x) (N for numerator, D for denominator)

N(x) and D(x) are analytic and so tend to zero as x--> a, so f(x) tends to the
indeterminate form 0/0.

In that case, according to L'hopital's rule, f(a) = N'(a) / D'(a).

N'(x) = (x-a) + (x-1) by the product rule. D'(x) = 1.

Plugging in x=a:

f(a) = (a-1) / 1 = .5

Which is what Alpha implies.

See also: <http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule>

<http://en.wikipedia.org/wiki/Removable_singularity>

<http://xkcd.com/123/>

~~~
lmkg
Well, kinda. But not really, sometimes.

L'Hopital's rule says that the _limit_ of the function exists, not the
function itself. A removable singularity is still a singularity, just one
that's more well-behaved than a pole (or gods forbid, an essential
singularity). While the function can be analytically extended to be defined at
the singularity, that's an extension of the function, rather than the function
itself...

... sorta. Since extensions of analytic function are unique, we usually treat
the extension and the function as the same thing. Whether this is revealing a
deeper truth about the function, or sloppy notation, is up to the individual
practitioner, and the context. Certainly, the definition of the function that
has the removable singularity is sufficient to define the function that has
the singularity removed. For analysis they're basically the same and we
usually extend as far as we can. But for topology, a single point separates
open and closed sets, which are worlds apart, and the extended function is a
very different beast from the original one. This just goes to show how limits
identify the points that are most important to a (structured) set.

However, for the purposes of Wolfram|alpha, I don't fault them for making that
mistake. Removable singularities can be safely extended over for the purposes
of computational mathematics.

------
mquander
If you consider this a bug, perhaps you should email Wolfram Research instead
of making this post?

~~~
cythrawll
There is an official place for people to submit bugs for Wolfram|Alpha, to
talk them over, and to vote bugs up and down.

<http://community.wolframalpha.com>

------
RiderOfGiraffes
Mathematicians will tell you that the inequality is undefined for x=1.5, and
is otherwise satisfied for x>1.0.

Yes, the inequality simplifies, but when x=1.5 you are multiplying both sides
by 0 which is not a permitted operation.

In this case nothing goes wrong if you blindly simplify and use the simpler
inequality. Mathematicians go ballistic when you do that because there are
other occasions where similar operations create nonsense. The usual proof that
1=2 is an example.

Trying to show why this sort of thing is wrong is like trying to show that a
fallacious proof of a true theorem is wrong. Counter-examples are hard to
produce because the theorem is true.

And that's why it's a general rule. If you are dividing by something you have
to preclude the possibility that the denominator is 0.

------
ambition
The numerator is equivalent to (x-1.5)*(x-1). Cancel out the (x-1.5) from
numerator and denominator, leaving x - 1 > 0.

I'm not sure what the point of the post is, or why the discussions here are
using so many words I don't understand.

~~~
ajscherer
Type these two expressions into a calculator or spreadsheet:

((1.5 - 1.5) * (1.5 - 1))/(1.5 - 1.5)

(1.5 - 1)

Did you get the same answer? If not then there must either a bug in your
calculator/spreadsheet or in wolfram alpha.

~~~
teilo
Wrong. You are invoking division by zero to prove what, exactly?

0.5 * 0 / 0 = 0.5

Excel or your calculator will freak out, but it's still true.

~~~
dagw
(0.5 * 0)/0 does not in any way equal 0.5 (assuming real numbers and all
normal definition of operators). Sure the limit of (0.5 * x)/x =0.5 as x->0,
but that is not the same thing.

~~~
teilo
This is then my ignorance showing, in that I made a classic blunder of
accepting the premise of division by zero in a test case.

What I should have said is that if you intentionally craft the circumstances
of an equation such that you are dividing by zero, it is your premise that is
nonsense, NOT the equation itself.

------
bbq
I found this interesting:
[http://www06.wolframalpha.com/input/?i=50%25+of+60%25+of+%24...](http://www06.wolframalpha.com/input/?i=50%25+of+60%25+of+%2420)

Whereas it gets this completely fine:
<http://www06.wolframalpha.com/input/?i=60%25+of+%2420>

It seems to forget what % means..

edit:

And then
[http://www.wolframalpha.com/input/?i=60000%20percent%20squar...](http://www.wolframalpha.com/input/?i=60000%20percent%20squared%20US%20dollars)

~~~
boredguy8
[http://www06.wolframalpha.com/input/?i=50%25+of+%2860%25+of+...](http://www06.wolframalpha.com/input/?i=50%25+of+%2860%25+of+%2420%29)

------
simon_
Maybe I'm missing something obvious, but I don't understand what that graph is
trying to tell me. If the horizontal axis is x... what is the vertical axis?
What does the shaded wedge represent? It certainly doesn't represent "x>1" in
any way I'm familiar with. (A vertical line at x=1, with the space to the
right shaded, I could see...)

~~~
ericwaller
The vertical axis is f(x), and the shaded region represents f(x) > 1.

------
alexk
Taken from here:

[http://translate.google.ru/translate?js=y&prev=_t&hl...](http://translate.google.ru/translate?js=y&prev=_t&hl=ru&ie=UTF-8&layout=1&eotf=1&u=http%3A%2F%2Fhabrahabr.ru%2Fblogs%2Fnigma%2F78531%2F&sl=ru&tl=en)

~~~
jpwagner
good old Tungsten-Alpha :)

~~~
qeorge
This cracked me up, so I decided to see if the domain was available. Not only
is it registered, the guy is trying to sell it:

<http://tungstenalpha.com>

------
jeffcoat
What are you pointing out here? That it doesn't mention the gap at x=1.5?

Okay: what did we learn from that?

~~~
weaksauce
Even that is a mediocre bug at best. Since wa deals with the infinitesimals as
standard practice it could follow that they are showing the gap at 1.5 but it
is so small as not to display correctly or that the gap is smaller than the
step size of the graph.

~~~
jcl
It's not just the graph. The solution given in the third box -- "x > 1" --
doesn't mention the gap either.

------
frevd
wow, nice tool.

rofl:
[http://www06.wolframalpha.com/input/?i=422340980293840234098...](http://www06.wolframalpha.com/input/?i=42234098029384023409823409823094823411)

full name: 42 undecillion, 234 decillion, 98 nonillion, 29 octillion, 384
septillion, 23 sextillion, 409 quintillion, 823 quadrillion, 409 trillion, 823
billion, 94 million, 823 thousand and 411...

------
jws
Darn it you made me click an Alpha link. Now I'm bound by a 2777 word license
if I need to use this equation. Thanks a lot.

------
chancegarcia
they have a place for people to submit bugs like this.
community.wolframalpha.com

