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May 20

finding the zeroes from fractions

Hi, can anyone help me with my homework and help me find the zero(s) of x/4 + 1/x^2? I saw the instructions that say that I should prove that I've tried, but I'm totally stuck...I don't know how to start. I mean, I tried putting them all under the lowest common denominator of 4x^2, and tried a couple of other things to try and make them fit but it doesn't seem right...if someone can kindly help me get started that would be nice. 174.5.131.59 (talk) 02:04, 20 May 2011 (UTC)[reply]

Multiply

{\frac {x}{4}}+{\frac {1}{x^{2}}}=0

through by x^2 to get

{\frac {x^{3}}{4}}+1=0

, which sould be easier to solve. ―JuPitEer (talk) 08:03, 20 May 2011 (UTC)[reply]

Thank you! I had myself confused that I was only allowed to multiply both terms by something that equaled to 1 (like x^2/x^2). How come this rule doesn't apply to this case? 174.5.131.59 (talk) 09:53, 20 May 2011 (UTC)[reply]

I believe the risk is that you "lose" solutions - there might be more solutions to the original equation than there were to the new one. Here, you might consider putting the fractions over a single denominator:

{\frac {x^{3}}{4x^{2}}}+{\frac {4}{4x^{2}}}=0\implies {\frac {x^{3}+4}{4x^{2}}}=0

and going from there; in this instance, it is clearer that both sides can be multiplied by

4x^{2}

without losing solutions. Grandiose (me, talk, contribs) 10:43, 20 May 2011 (UTC)[reply]

forgive me for being confused, but I finally managed to figure out the zeros from Jupiter's method. They are 0 and [-(cuberoot of 4)], yikes that was troublesome. How do I find them using your method, Grandiose? 174.5.131.59 (talk) 10:56, 20 May 2011 (UTC)[reply]

Unfortunately, 0 isn't a solution (assuming you mean "finding the solutions to this equals 0" when you say "zeroes". If you follow Jupiter's:

{\frac {x^{3}}{4}}+1=0\implies x^{3}=-4

which has three solutions; I'm going to guess that [-(cuberoot of 4)] is the only one you're expected to give (the other two are complex numbers). If you took mine,(

{\frac {x^{3}+4}{4x^{2}}}=0

) and apply the rule that a fraction can be zero if and only if its numerator is zero, then we get

x^{3}+4=0\implies x^{3}=-4

as before. You can times both sides of an equation by anything, so long as you do the same to both sides (subject to a couple of small things which don't apply here); or you can apply something equal to one to either side, or both sides. Grandiose (me, talk, contribs) 11:25, 20 May 2011 (UTC)[reply]

The denominator has to be nonzero as well, for the faction to be zero. --COVIZAPIBETEFOKY (talk) 14:35, 20 May 2011 (UTC)[reply]

The reason you can multiply by things other than 1/1 is that you aren't trying to keep the fraction the same. You are trying the keep the equation true. That means the rule is that you have to multiply both sides of the equation by the same thing. Since zero times anything is just zero, that means you can multiply the other side by anything you like and it will still equal zero.

{\frac {x}{4}}+{\frac {1}{x^{2}}}

and

{\frac {x^{3}}{4}}+1

aren't equal to each other but, if one of them equals zero for a given x, then they both will (although you do have to be careful that the solutions make sense for both - if one involves dividing by zero, then it doesn't work). --Tango (talk) 20:00, 21 May 2011 (UTC)[reply]

http://www.wolframalpha.com/input/?i=x%2F4%2B1%2Fx^2%3D0 does the trick automatically. It's amazing. Bo Jacoby (talk) 12:28, 21 May 2011 (UTC).[reply]

It's amazing, certainly. It's also completely useless for the OP's purposes. This is a homework question, which exists to help the OP learn how to manipulate algebraic expressions. The point of the homework isn't the find out the answer. --Tango (talk) 20:00, 21 May 2011 (UTC)[reply]

I don't think it is useless. Experimenting with expressions of varying complexity in wolframalpha give a nice feeling for the subject. Bo Jacoby (talk) 22:52, 21 May 2011 (UTC).[reply]

Critical points

The critical points of $f(x,y)={\frac {xy}{2+x^{4}+y^{4}}}$ . These occur where the partial derivatives are both equal to zero, so I have derived the partial derivatives: $f_{x}(x,y)={\frac {(2+x^{4}+y^{4})y-4x^{4}y}{(2+x^{4}+y^{4})^{2}}}$ and $f_{y}(x,y)={\frac {(2+x^{4}+y^{4})x-4xy^{4}}{(2+x^{4}+y^{4})^{2}}}$ . (0,0) satisfies this condition by inspection. Where are the other critical points and how to you find them (and is there an analytical approach)? Widener (talk) 11:38, 20 May 2011 (UTC)[reply]

Critical point (mathematics) is relevant but doesn't answer the specific question. -- SGBailey (talk) 11:52, 20 May 2011 (UTC)[reply]

For the partial derivatives to be 0, the numerators need to be 0. so we have (2+x⁴+y⁴)y-4x⁴y=0 and (2_x⁴+y⁴)x-4xy⁴=0. These simplify to 2+y⁴=3x⁴ and 2+x⁴=3y⁴ which has to be easier to solve. -- SGBailey (talk) 11:59, 20 May 2011 (UTC)[reply]

Indeed, this easily reduces to

x^{4}=y^{4}=1

. -- Meni Rosenfeld (talk) 12:18, 20 May 2011 (UTC)[reply]

Thanks. Widener (talk) 02:05, 21 May 2011 (UTC)[reply]

I get the numerators of ∂ƒ/∂x and ∂ƒ/∂y to be y·(3y⁴ – x⁴ – 2) and x·(x⁴ – 3y⁴ + 2), respectively. There are some details to take care of. First of all x = y = 0 is a critical point. If y = 0 then ∂ƒ/∂x = 0 and ∂ƒ/∂y = x·(x⁴ + 2). We already have x = y = 0, so ignore x = 0, but what about x⁴ + 2 = 0; does that have any real solutions? If x = 0 then ∂ƒ/∂y = 0 and ∂ƒ/∂x = y·(3y⁴ – 2). We already have x = y = 0, so ignore y = 0, but what about 3y⁴ – 2 = 0; does that have any real solutions? (The answer is no in both cases, but you should show why not.) Finally, consider 3y⁴ – x⁴ – 2 = x⁴ – 3y⁴ + 2 = 0. Solve these as tw simultaneous equations in x⁴ and y⁴ to give x⁴ = 1 and y⁴ = 1. This gives you (x,y) = (1,1), (–1,1), (1,–1) and (–1,–1). The critical points are then (x,y) = (0,0), (1,1), (–1,1), (1,–1) and (–1,–1). — Fly by Night (talk) 14:08, 22 May 2011 (UTC)[reply]

Monster group

How to prove that the Monster group cannot act doubly transitive on any set? --84.61.186.230 (talk) 20:15, 20 May 2011 (UTC)[reply]