Wikipedia:Reference desk/Archives/Mathematics/2012 September 17

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September 17

L'Hopital's rule

How do I apply L'Hopital's rule to a limit function with two or more variables? Plasmic Physics (talk) 10:45, 17 September 2012 (UTC)

With two or more independent variables, I think you need to define the path along which you approach the limit point (and so reduce the problem to a one-variable problem) before applying L'Hopital's rule. This is because the limit can depend on the path taken. To take a simple example (which doesn't really need L'Hopital's rule, although you can use it), consider the limit of the function

$f(x,y)={\frac {x+2y}{2x+y}}$

at (0,0). If we approach (0,0) along the line x = 0 then the limit is 2; along the line y = 0 then the limit is 1/2; and along the line y = x then the limit is 1. Gandalf61 (talk) 11:11, 17 September 2012 (UTC)

What if I use a mixed double partial derivatives of both the numerator and denominator? If I do that for the limit of f(x,y) as (x,y)->(0,0), then this approach would give 2/2, which simplifies to 1. Is this just a coincidence? Plasmic Physics (talk) 13:17, 17 September 2012 (UTC)

Not sure what you mean by "mixed double partial derivatives". If you mean

${\frac {\partial }{\partial x}}+{\frac {\partial }{\partial y}}$

then you are implictly choosing the path y = x (or at least, a path that is tangent to y = x at (0,0)). Gandalf61 (talk) 13:51, 17 September 2012 (UTC)

No, I mean the product not the sum, as in ${\frac {\partial f}{\partial x}}{\frac {\partial f}{\partial y}}$ over ${\frac {\partial g}{\partial x}}{\frac {\partial g}{\partial y}}$. Plasmic Physics (talk) 21:16, 17 September 2012 (UTC)

Roughly, expand f and g in their multivariate Taylor series at (0,0). If the Taylor series have a common factor, then you can cancel this common factor and evaluate the limit. Of course, this works much less often than in one variable, since two functions can be zero at a point yet have no common factor (e.g., f(x,y)=x and g(x,y)=y are both zero at the origin, with no common factor). Sławomir Biały (talk) 00:03, 18 September 2012 (UTC)

To show that a limit exists in the multivariate context, though, you need to show that the limit is the same along all paths. I'm not sure l'Hopital's rule is going to be much help (it's really just a short-cut anyway - if you can find a limit using the rule then you can find it using elementary techniques almost as quickly). --Tango (talk) 23:12, 17 September 2012 (UTC)

Definite integral from -1 to 1 of 1/x

Since 1/x is discontinuous and undefined at x = 0, my understanding is that the definite integral $\int _{-1}^{1}{\frac {1}{x}}\,dx$ therefore has no limit and is undefined (and claiming that

$\int _{-1}^{1}{\frac {1}{x}}\,dx=\int _{-1}^{0}{\frac {1}{x}}\,dx+\int _{0}^{1}{\frac {1}{x}}\,dx=-\infty +\infty =0$

is sketchy, dubious, and unrigorous). However, I have heard that it is possible to rigorously evaluate $\int _{-1}^{1}{\frac {1}{x}}\,dx$ so that the result is not undefined. Can someone explain how this is done? —SeekingAnswers (reply) 15:34, 17 September 2012 (UTC)

See Cauchy principal value. Your integral is discussed in the Examples section. --Wrongfilter (talk) 16:18, 17 September 2012 (UTC)

See also Principal value for the complex number case and list of integrals#Integrals with a singularity which has a note about both cases. Dmcq (talk) 16:42, 17 September 2012 (UTC)

simple Differential equation

hi,

I would like to know the solution for:

f(x)*f^2(x)=1 15:50, 17 September 2012 (UTC) — Preceding unsigned comment added by Exx8 (talk • contribs)

What have you tried so far on solving the problem? —SeekingAnswers (reply) 16:28, 17 September 2012 (UTC)

Is this supposed to be $f(x)f''(x)=1$? Looie496 (talk) 21:40, 17 September 2012 (UTC)

no, it is $f^{2}(x)f''(x)=1$ Exx8 (talk) —Preceding undated comment added 23:16, 17 September 2012 (UTC)

Really the second derivative? This would be easy to deal with if it was the first derivative, but with the second derivative, it is essentially the equation for the motion of a charged particle in a repulsive inverse square field, which doesn't have a solution that can be written in closed form (as far as I know). Looie496 (talk) 19:23, 18 September 2012 (UTC)

I agree that it seems implausible that this is correct equation. It does have a solution though: $-{({9 \over 2})}^{1/3}x^{2/3}$. Dragons flight (talk) 19:41, 18 September 2012 (UTC)

Correct you are!! Do you have some software or you did it alone? Thank you! Exx8 (talk) 21:26, 19 September 2012 (UTC)

Skewed distribution with given median, mean and SD

For what skewed generalizations of the normal distribution can parameters be estimated given the mean, median and standard deviation of a sample but no other information about the sample? NeonMerlin 16:37, 17 September 2012 (UTC)

Are you asking for named types of distributions? Otherwise the answer is banal: you could do it for almost any family of distributions that has three parameters. Looie496 (talk) 18:36, 17 September 2012 (UTC)

These are not skewed, but the elliptical distributions are a generalization of the normal distribution in which one parameter is the median (which equals the mean if the mean exists) and the other parameter is the standard deviation. Duoduoduo (talk) 18:53, 17 September 2012 (UTC)

If you type three parameter distribution into the Wikipedia search box, a number of promising hits come up. Duoduoduo (talk) 19:03, 17 September 2012 (UTC)

When does additivity imply homogeneity?

There are usually two requirements for R-linearity of a function f : U → V, where U and V are modules over a ring R (here x, y ∈ U, α ∈ R):

• f(x + y) = f(x) + f(y) (additivity)
• f(αx) = αf(x) (homogeneity).

The two requirements are clearly closely related (partially redundant conditions). When does the first imply the second? For example, if R = ℤ, it clearly does. Wherever R is not commutative, it clearly doesn't. Does the implication extend to when R = ℚ? To other commutative rings? The reverse implication clearly holds whenever the module U can be generated by a single element (i.e. when it is one-dimensional over R). — Quondum 21:15, 17 September 2012 (UTC)

Yes, it holds for ℚ. As you noted, $f(nx)=nf(x)$ for $n$ an integer. So $f(n(1/n)x)=nf((1/n)x)$, and thus $(1/n)f(x)=f(x/n)$. So $f((p/q)x)=pf((1/q)x)=(p/q)f(x)$.

It doesn't hold for $\mathbb {R}$, however. Let $B$ be a basis for $\mathbb {R}$ over $\mathbb {Q}$, and let $f$ be projection onto the first basis element.--121.73.35.181 (talk) 22:38, 17 September 2012 (UTC)

Resolved

Thanks. Nice counterexample, which by analogy severely restricts the rings for which the implication holds. — Quondum 03:44, 18 September 2012 (UTC)