4 ms·

(x2-2.5x+1.5)/(x-1.5)

23 points by alexk 17y ago

alexk 17y ago
Taken from here: 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 http://translate.google.ru/translate?js=y&prev=_t&hl...
- jpwagner 17y ago
  good old Tungsten-Alpha :)
  - qeorge 17y ago
    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 http://tungstenalpha.com
- zitterbewegung 17y ago
  So what is the actual solution?
  - teilo 17y ago
    The solution is exactly what Alpha computed: x > 1. The numerator factors to (x-1.5)(x-1), which reduces the problem to x - 1 > 0. Solve for x: x > 1. Not sure what the fuss here is.
    - RiderOfGiraffes 17y ago
      The fuss is that the inequality is undefined when x=1.5 and so there is no definition of what happens in that case. Performing "obvious" completions of incompletely defined functions is assuming you know the context, and sometimes you don't. This is a sloppy engineering solution that is not guaranteed to be right in all contexts. Without the context, the correct solution is to say that the question is ill-defined/undefined when x=1.5, and all values of x such that x>1 and x.ne.1.0 satisfy the given inequality.
- alexk 17y ago
  Sorry for the bias, the original page is in Russian, so I've used Google translate link to make it possible for others to read.
jeffcoat 17y ago
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 17y ago
  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 17y ago
    It's not just the graph. The solution given in the third box -- "x > 1" -- doesn't mention the gap either.
mquander 17y ago
If you consider this a bug, perhaps you should email Wolfram Research instead of making this post?
- cythrawll 17y ago
  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 http://community.wolframalpha.com
deleted 17y ago
[deleted]
bbq 17y ago
I found this interesting: http://www06.wolframalpha.com/input/?i=50%25+of+60%25+of+%2420 http://www06.wolframalpha.com/input/?i=50%25+of+60%25+of+%24... Whereas it gets this completely fine: http://www06.wolframalpha.com/input/?i=60%25+of+%2420 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%20squared%20US%20dollars http://www.wolframalpha.com/input/?i=60000%20percent%20squar...
- boredguy8 17y ago
  http://www06.wolframalpha.com/input/?i=50%25+of+%2860%25+of+%2420%29 http://www06.wolframalpha.com/input/?i=50%25+of+%2860%25+of+...
mnemonicsloth 17y ago
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/L%27H%C3%B4pital%27s_rule http://en.wikipedia.org/wiki/Removable_singularity http://en.wikipedia.org/wiki/Removable_singularity http://xkcd.com/123/ http://xkcd.com/123/
- lmkg 17y ago
  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.
- dfranke 17y ago
  A removable singularity is still a singularity. If I'm asking for f(x) and it's giving me lim(i->x,f(i)), it should tell me so.
chancegarcia 17y ago
they have a place for people to submit bugs like this. community.wolframalpha.com
frevd 17y ago
wow, nice tool. rofl: http://www06.wolframalpha.com/input/?i=42234098029384023409823409823094823411 http://www06.wolframalpha.com/input/?i=422340980293840234098... 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...
simon_ 17y ago
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 17y ago
  The vertical axis is f(x), and the shaded region represents f(x) > 1.
ambition 17y ago
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 17y ago
  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 17y ago
    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 17y ago
      (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 17y ago
        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.
      - teilo 17y ago
        Now, interestingly, Wolfram Alpha does not agree with you: http://www.wolframalpha.com/input/?i=xy/y+where+y%3D0+and+x+%3D+0.5 http://www.wolframalpha.com/input/?i=xy/y+where+y%3D0+and+x+...
jws 17y ago
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.
RiderOfGiraffes 17y ago
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.

23 points by alexk 17y ago

alexk 17y ago

Taken from here: 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 http://translate.google.ru/translate?js=y&prev=_t&hl...

jpwagner 17y ago
good old Tungsten-Alpha :)
- qeorge 17y ago
  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 http://tungstenalpha.com
zitterbewegung 17y ago
So what is the actual solution?
- teilo 17y ago
  The solution is exactly what Alpha computed: x > 1. The numerator factors to (x-1.5)(x-1), which reduces the problem to x - 1 > 0. Solve for x: x > 1. Not sure what the fuss here is.
  - RiderOfGiraffes 17y ago
    The fuss is that the inequality is undefined when x=1.5 and so there is no definition of what happens in that case. Performing "obvious" completions of incompletely defined functions is assuming you know the context, and sometimes you don't. This is a sloppy engineering solution that is not guaranteed to be right in all contexts. Without the context, the correct solution is to say that the question is ill-defined/undefined when x=1.5, and all values of x such that x>1 and x.ne.1.0 satisfy the given inequality.
alexk 17y ago
Sorry for the bias, the original page is in Russian, so I've used Google translate link to make it possible for others to read.

jeffcoat 17y ago

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 17y ago
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 17y ago
  It's not just the graph. The solution given in the third box -- "x > 1" -- doesn't mention the gap either.

mquander 17y ago

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

cythrawll 17y ago
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 http://community.wolframalpha.com

deleted 17y ago

[deleted]

bbq 17y ago

I found this interesting: http://www06.wolframalpha.com/input/?i=50%25+of+60%25+of+%2420 http://www06.wolframalpha.com/input/?i=50%25+of+60%25+of+%24... Whereas it gets this completely fine: http://www06.wolframalpha.com/input/?i=60%25+of+%2420 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%20squared%20US%20dollars http://www.wolframalpha.com/input/?i=60000%20percent%20squar...

boredguy8 17y ago
http://www06.wolframalpha.com/input/?i=50%25+of+%2860%25+of+%2420%29 http://www06.wolframalpha.com/input/?i=50%25+of+%2860%25+of+...

mnemonicsloth 17y ago

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/L%27H%C3%B4pital%27s_rule http://en.wikipedia.org/wiki/Removable_singularity http://en.wikipedia.org/wiki/Removable_singularity http://xkcd.com/123/ http://xkcd.com/123/

lmkg 17y ago
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.
dfranke 17y ago
A removable singularity is still a singularity. If I'm asking for f(x) and it's giving me lim(i->x,f(i)), it should tell me so.

chancegarcia 17y ago

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

frevd 17y ago

wow, nice tool. rofl: http://www06.wolframalpha.com/input/?i=42234098029384023409823409823094823411 http://www06.wolframalpha.com/input/?i=422340980293840234098... 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...

simon_ 17y ago

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 17y ago
The vertical axis is f(x), and the shaded region represents f(x) > 1.

ambition 17y ago

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 17y ago
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 17y ago
  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 17y ago
    (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 17y ago
      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.
    - teilo 17y ago
      Now, interestingly, Wolfram Alpha does not agree with you: http://www.wolframalpha.com/input/?i=xy/y+where+y%3D0+and+x+%3D+0.5 http://www.wolframalpha.com/input/?i=xy/y+where+y%3D0+and+x+...

jws 17y ago

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.

RiderOfGiraffes 17y ago

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.