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June 14[edit]

Proof that infinity equals -1[edit]

x=1+2+4+8+16+...

2x=2+4+8+16+...

2x-x=(2+4+8+16+...)-(1+2+4+8+16+...)

x=-1

But x is clearly infinity, so infinity equals -1. Where's the error with this? --76.77.139.243 (talk) 13:26, 14 June 2010 (UTC)[reply]

Interesting, but I believe the error is the step after

2x-x=(2+4+8+16+...)-(1+2+4+8+16+...)

Because infinity - infinity = infinity, the next step should actually be

x = infinity, which is consistent with your premise -- Rick Van Tassel ^{user|talk|contribs} 13:30, 14 June 2010 (UTC)[reply]

We have an article 1 + 2 + 4 + 8 + · · ·. Algebraist 13:35, 14 June 2010 (UTC)[reply]

You can't rearrange the order of summation in an infinite sum. Readro (talk) 13:41, 14 June 2010 (UTC)[reply]

One error is that you can't subtract series in that way unless they converge to a finite number.

Infinity minus infinity is an indeterminate form. That means that if ƒ(x) and g(x) can both be made as large as desired by making x close enough to some specified point (which may be infinite) then ƒ(x) − g(x) may approach anything, depending on which functions ƒ and g are. Michael Hardy (talk) 13:45, 14 June 2010 (UTC)[reply]

The error is to assume that "x is clearly infinity". Bo Jacoby (talk) 14:18, 14 June 2010 (UTC).[reply]

Indeed! minus 1 is the most reasonable answer. Infinity does not really exist, so if you have a divergent expression, it must be the result of some illegal formal manipulations applied to an expression that does correspond to a finite number. You can then guess what the correct expresson is, but any formal method for doing that, must itself contain at least one illegal manipulation, as you have to undo the assumed illegal manipulation that led to the divergent expression in the first place.

It turns out that errors produced by formal manipulations, like interchanging Taylor expansions with integrations when that is not allowed, using series expansion outside the domain of convergence etc. etc., tend to be universal in the sense that they can usually be corrected by some generic resummation technique. Count Iblis (talk) 14:29, 14 June 2010 (UTC)[reply]

There is only one error here. The expression is not a sum. It is shorthand for a limit. Treating it as a sum and attributing some of the standard properties of sums (like distributivity, commutativity and/or associativity) creates the trouble. DVdm (talk) 14:33, 14 June 2010 (UTC)[reply]

Indeed. Those properties only hold for convergent series, so by using them you are assuming x isn't infinity. Once you've assumed that, it is hardly surprising that you get an answer that is consistent with it. --Tango (talk) 14:45, 14 June 2010 (UTC)[reply]

Even for a series which sums to a limit okay you can do some wierd things by rearranging the terms if it isn't absolutely convergent. See the Riemann series theorem. Dmcq (talk) 15:17, 14 June 2010 (UTC)[reply]

The rules for resumming divergent series are discussed in this article. Count Iblis (talk) 15:21, 14 June 2010 (UTC)[reply]

True, but in this case all the terms are positive, so it isn't an issue. --Tango (talk) 15:25, 14 June 2010 (UTC)[reply]

See formal power series. You do not need to assume convergence. The algebraic rules apply nicely. However you are only allowed to insert or remove a finite number of parentheses, so 1−1+1−1+1−1+1−1+... = 1/2 is not equal to (1−1)+(1−1)+(1−1)+(1−1)+... = 0. Also you are only allowed to change the order of a finite number of terms, so 1−1/2+1/3−1/4+1/5−1/6+...=ln(2) is not equal to 1+1/3−1/2+1/5+1/7−1/4+..., even if these series are convergent and contain the same terms.

By the way in p-adic numbers where p is 2 what you have is correct. ...1111111.0 does equal -1. Dmcq (talk) 17:28, 14 June 2010 (UTC)[reply]

Program for creating mathematical graphs[edit]

Cross-posted to the computing desk.

Hi everyone. I'm looking for a program to draw graphs easily (I do mean graphs not charts - essentially circles connected by lines). I'd like a program which has the following:

Graphical User Interface

Ability to drag/drop vertices around.

Preferably ability to resize the vertices, alter colours, alter thickness/colour of the edges, etc.

Does anyone know any software which can do this? -mattbuck (Talk) 17:53, 14 June 2010 (UTC)[reply]

You might try Inkscape. It has it's faults but it's free and you can upload the .SVG files to Wikipedia.

Perhaps one of the packages in the List of mind mapping software or List of concept mapping software… looking through the former now, how about Graphviz? Not used it myself though. Qwfp (talk) 09:07, 15 June 2010 (UTC)[reply]

Try yEd. Barsamin (talk) 14:36, 18 June 2010 (UTC)[reply]

You can try Cytoscape too, especially useful for overlaying data on the graphs.

How do you write integration limits of the solution in TeX?[edit]

How do you write limits of integration after finding what the integral is. For example, let's say if the question was to integrate x^2 from 0 to 1, the (partial) solution to this is x^2/2 evaluated from 0 to 1, how would you write this in TeX? 142.244.151.247 (talk) 21:07, 14 June 2010 (UTC)[reply]

You use the square bracket \left[ and \right]. Make sure you use them in pairs or it won't compile. If you type \left[ \frac{1}{2}x^2 \right]_0^1 you get

\left[{\frac {1}{2}}x^{2}\right]_{0}^{1}

Putting \left before bracket makes LaTeX adjust the size of the bracket to fit the text. You can use \left( \right), \left{ \right}, etc... But make sure you put them in pairs. If you don't want a bracket then use \left. or \right. to omit the left or right bracket. For example \left. \frac{1}{2}x^2 \right]_0^1 gives

\left.{\frac {1}{2}}x^{2}\right]_{0}^{1}

You can even get more exotic brackets, say pointy brackets, by typing \left\langle \right\rangle:

\left\langle {\frac {1}{2}}x^{2}\right\rangle

Without the \left and \right these expressions compile as:

[{\frac {1}{2}}x^{2}]_{0}^{1}\ {\mbox{ and }}\ \langle {\frac {1}{2}}x^{2}\rangle

•• Fly by Night (talk) 21:16, 14 June 2010 (UTC)[reply]

The one I've seen the most is similar to the second one but with a vertical line instead of a bracket. Apparently there is no universal standard though.--RDBury (talk) 02:31, 15 June 2010 (UTC)[reply]

As in \left. \frac{1}{3}x^3 \right|_0^1 ?

\left.{\frac {1}{3}}x^{3}\right|_{0}^{1}

-- 58.147.52.243 (talk) 06:53, 15 June 2010 (UTC)[reply]

That's right. However, if the general integral form contains more than one term (sometimes also when there's a single term, that contains more than two factors, one of which is constant), then the vertical bar alone is not clear enough. The same is even more important if the integral appears as a term in a sum, say something like

f(x)+\int _{2}^{3}y(x)\,dt=7x^{2}+{\tfrac {3}{4}}\ln x+2\cos(x^{2}){\Big \vert }_{2}^{3}

—it's hard to say which terms should be taken at 2 and 3 and then subtracted, and which should not. Putting brackets makes things clear:

f(x)+\int _{2}^{3}y(x)\,dt=7x^{2}+{\Big [}{\tfrac {3}{4}}\ln x+2\cos(x^{2}){\Big ]}_{2}^{3}

CiaPan (talk) 07:46, 15 June 2010 (UTC)[reply]

The standard is square brackets on either side:

\int _{0}^{1}x^{2}\ {\text{d}}x=\left[{\frac {1}{3}}x^{3}\right]_{0}^{1}={\frac {1}{3}}.

I only mentioned the one-sided brackets so the person setting the question could use the functionality in other settings. The single line on the right is usually used as a short hand for evaluation. So ⅓x³ evaluated at x = 0 can be written as

\left.{\frac {1}{3}}x^{3}\right|_{x=0}

•• Fly by Night (talk) 23:04, 15 June 2010 (UTC)[reply]