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December 24

Why does this get so ugly so fast? (factoring cubic equations)

I've tried to factor $x^{3}-x-1$ and wolfram gives me this HUGE crazy equation:

http://www.wolframalpha.com/input/?i=factor+x^3+-+x+-+1

Why does it happen like that?

$x^{2}-x-1$ is relatively simple in contrast... but the cubic becomes a nightmare!!!

Can someone explain this?

(also, wolfram refuses to answer if I put it to the fourth or fifth degree instead. What's up with that?)--99.179.21.131 (talk) 05:04, 24 December 2010 (UTC)[reply]

Compare the complexity of Cubic function#General formula of roots to the simplicity of Quadratic function#Roots. I don't know how Wolfram Alpha works but Quartic function#Solving a quartic equation gives further complications. For a quintic equation, the Abel–Ruffini theorem says there is no general algebraic solution. PrimeHunter (talk) 05:26, 24 December 2010 (UTC)[reply]

Note that if you drop the word "factor" and give it a quartic polynomial, it will give you its roots, and you can click on "exact form" to get the symbolic expressions. -- Meni Rosenfeld (talk) 05:40, 24 December 2010 (UTC)[reply]

Why can't I take the indefinite integral of the reciprocate of the Log Integral function?

The Log Integral is $li(x)=\int {\frac {1}{ln(x)}}dx$

When I try to put $g(x)=\int {\frac {1}{li(x)}}dx$ into wolfram, it says it cannot be found. Who has studied this function and where can I find more information about it? Thanks.--99.179.21.131 (talk) 19:32, 24 December 2010 (UTC)[reply]

See Nonelementary integral. It is unlikely that anyone has studied that integral, Li is already a name for a special integral rather than being an elementary function. Dmcq (talk) 23:32, 24 December 2010 (UTC)[reply]

Condition of a circle within another circle

Hello, I am solving a problem where I have to find the condition that one of the circles $x^{2}+y^{2}+2ax+c=0$ and $x^{2}+y^{2}+2bx+c=0$ lies within the other. The four options given are (a) $ab>0,\,c<0$ (b) $ab<0,\,c<0$ (c) $ab>0,\,c>0$ and (d) $ab<0,\,c>0$ . I start by finding the distance between the centres of circles which are at $(-a,0)$ and $(-b,0)$ . So the distance is $|a-b|$ . This distance should be less than the absolute difference in the radii of the circle. So, the equation would become something like $|a-b|<|{\sqrt {a^{2}-c}}-{\sqrt {b^{2}-c}}|$ , but from here to reach any of the options given in the questions is causing me problems. Could somebody help me or give me different approach? Thanks - DSachan (talk) 22:31, 24 December 2010 (UTC)[reply]

Since you only have 4 possible answers, you can just pick some numbers and try out each scenario on graph paper (or a graphing program/calculator). This might help you understand the equation of a circle a bit better, too. StuRat (talk) 23:02, 24 December 2010 (UTC)[reply]

Thanks, I drew them in Mathematica, and the answer I got was option (c). But I am still getting mightily confused in removing the modulus and making tons of cases. There ought to be a more elegant way to solve it. I somehow feel it. - DSachan (talk) 00:52, 25 December 2010 (UTC)[reply]

Square both sides of the inequality, both sides are positive so this is valid. When you simplify you get

ab-c>{\sqrt {a^{2}-c}}{\sqrt {b^{2}-c}}

. This is equivalent to

ab-c>0

and

(ab-c)^{2}>(a^{2}-c)(b^{2}-c)

. The second inequality becomes

(a-b)^{2}c>0

or simply c>0. Given that, and the fact that you know a²>c and b²c, it's easy to see (waving hands vigorously) that the first condition is equivalent to ab>0.--RDBury (talk) 11:13, 25 December 2010 (UTC)[reply]