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November 19

Question about polynomial rings

I'm kind of suck on a question where I have to prove that a common divisor of the highest degree of two arbitrary elements f(x) and g(x) in the polynomial ring, lets call it h(x). I have to prove that the greatest common divisor, d(x) (which is defined as a monic polynomial) is an associate of h(x). So far I'm trying to argue that since h(x) is a common divisor of the highest degree. What property should I use to show that they have the same degree? —Preceding unsigned comment added by 142.244.143.26 (talk) 02:44, 19 November 2010 (UTC)[reply]

To show that d and h are associates is equivalent to showing that each divides the other (or to showing that one divides the other, and they are the same degree). How exactly you do that depends on what definitions you are using (since apparently you mean "common divisor of the highest degree" and "greatest common divisor" to mean different things). If you tell us what definitions you are using, and where you are getting stuck on the proof, we may be able to help you better. (I'll be away from the internet for a while so it'll have to be someone else, though.) Also please be more careful with your grammar, it can be hard to understand you. Good luck. Eric. 82.139.80.197 (talk) 23:43, 19 November 2010 (UTC)[reply]

poles problem

I saw one of those puzzles online today, I'll try to describe it. There are two poles of equal height with pegs at the same position on each of them. A rope of length 10 m is hung on these pegs such that the lowest point on the rope is 5 m below the pegs in the y-direction. How far apart are the poles? Intuitively I thought it would be 5 m because 10-5=5. I'd like to know how to do this with the arc length integrals and stuff to check my answer. How would I? Is my answer right? Thanks. 24.92.78.167 (talk) 04:09, 19 November 2010 (UTC)[reply]

You don't need any arc lengths. Just draw a picture. 69.111.192.233 (talk) 05:32, 19 November 2010 (UTC)[reply]

The rope should take the shape of a catenary if left hanging under uniform gravity, if I've understood your setup. Take the rope as y(x), the center of the system at x=0, and the poles at x=+r or -r. Say that y(0)=0 and y(r)=y(-r)=5. Translating the catenary vertically so its lowest point is at x=0, we have y(x) = a cosh(x/a) + b and y(0) = a + b = 0 so a = -b. Also, y(r) = 5 = a (cosh(r/a) - 1). The length of y from -r to r is given by the usual arc length integral,

\int _{-r}^{r}{\sqrt {1+f'(x)^{2}}}\,{dx}

. Since

f'(x)=\sinh(x/a)

, the integrand is just

|\cosh(x/a)|=\cosh(x/a)

for our purposes, so the arc length is

2a\sinh(r/a)=10

. Combining this with the above and asking Wolfram Alpha to solve the system of two equations in two unknowns gives no solutions. Either I've made a mistake or these constants don't work out. Slightly different constants--replacing 10 with 15, for instance, gives a = 3.125 and r ~= 5.02949, so that the poles must have been 2r ~= 10.05898 m apart. 67.158.43.41 (talk) 06:08, 19 November 2010 (UTC)[reply]

No solutions with the given constants makes sense to me. The rope has to go down 5m from the left peg to the lowest point and up 5m from there to the right peg, only considering vertical distance. Assuming it wastes no distance moving left or right, it's at least 10m long. At precisely 10m of length, this forces r=0, so any solution would satisfy y(0)=0=5=y(r), a contradiction. 67.158.43.41 (talk) 06:18, 19 November 2010 (UTC)[reply]

There's not exactly a contradiction, r=0 is the obvious and necessary answer. The pegs are touching each other. 69.111.192.233 (talk) 09:03, 19 November 2010 (UTC)[reply]

I disagree. r=0 would force the "left" half and "right" half of the rope to occupy the same space, even in the infinitely thin idealization the question seems to be asking about. In that case its length would be 5. By my "contradiction" above, I mostly meant the catenary equation I used would produce a contradiction if it were solvable with the given constants. Sorry I wasn't clearer. 67.158.43.41 (talk) 09:10, 19 November 2010 (UTC)[reply]

If you consider it as a set, yes, but not if you consider it as a curve. -- Meni Rosenfeld (talk) 10:02, 21 November 2010 (UTC)[reply]

differentiate x^x

Hello. I am trying to differentiate y = x^x. I get literally within ±1 of the correct answer, which WolframAlpha says is x^x(ln x + 1). I used the derivative formula, and when I get to $(x+\Delta x)^{x+\Delta x}$ I use binomial expansion. This reduces it to $\lim _{\Delta x\to 0}{\frac {x^{x}(x^{\Delta x}-1)}{\Delta x}}$ and by L'Hopital's rule I find $x^{x}(\ln {x})$ Where will the +1 come from? 24.92.78.167 (talk) 23:15, 19 November 2010 (UTC)[reply]

Rewrite

x^{x}

as what it is, namely

e^{x\ln x}

. Then take the derivative. — Carl (CBM · talk) 23:24, 19 November 2010 (UTC)[reply]

Doesn't use of l'Hopital's rule to evaluate that limit operate under the assumption that you already know the derivative of

x^{x}

? --Kinu ^t/_c 23:42, 19 November 2010 (UTC)[reply]

No, because x is constant in that limit. The derivative is taken wrt Δx. -- Meni Rosenfeld (talk) 15:59, 20 November 2010 (UTC)[reply]

facepalm*... ah, but of course. Thank you. --Kinu ^t/_c 23:06, 20 November 2010 (UTC)[reply]

The error was in dropping the second term in the binomial expansion, so the limit should be

\lim _{\Delta x\to 0}{\frac {x^{x}(x^{\Delta x}-1)}{\Delta x}}+(x+\Delta x)x^{x+\Delta x-1}.

The limit of the second term is easy to work out and you can use l'Hopital's rule on the first term since it's the derivative of a different expression. Not the easiest way to do it but it does work.--RDBury (talk) 07:04, 20 November 2010 (UTC)[reply]

The binomial expansion was incorrectly done here. The following is correct:

(x+\Delta x)^{x+\Delta x}=(x+\Delta x)^{x}\cdot (x+\Delta x)^{\Delta x}\,

but that's not the binomial expansion—you would still have to expand after that.

However, you get the answer more easily by logarithmic differentiation, thus:

{\begin{aligned}y&=x^{x}\\{\text{therefore }}\log _{e}y&=\log _{e}(x^{x})=x\log _{e}x.\\{\text{Hence }}{\frac {d}{dx}}\log _{e}y&={\frac {d}{dx}}(x\log _{e}x).\\{\frac {y\,'}{y}}&=x{\frac {d}{dx}}\log _{e}x+(\log _{e}x){\frac {d}{dx}}x\end{aligned}}

and so on..... Michael Hardy (talk) 03:10, 21 November 2010 (UTC)[reply]