User talk:GFauxPas/Archive1

Change to MathWorld citation template
I noticed (based on One-to-One and Strictly Between) that some pages on MathWorld are credited to different authors from Eric Weisstein, and so require that author to be included in the citation.

I have fixed the template (which is now "MathWorld" not "Mathworld", that's just me tidying up) so as to be able to include the author (which, if not given, defaults to the "Weisstein, Eric W." format as per normal).

What you need to do is add "author=author-name" and "authorpage=author-pagename" where "author-name" is the displayname of the author and "author-pagename" is the name of the html file on MathWorld (not including the full path, not including the extension).

An example:

which gives:
 *  


 * {{MathWorld|One-to-One|One-to-One|author=Barile, Margherita|authorpage=Barile

If the page is given as written by "Weisstein, Eric W." then you should not add the "author" and "authorpage" tags.

I have included this info in the usage section of the Template:MathWorld page itself, but I'm bringing it to your attention because I know you've been active in using it.

Chx. --prime mover 02:55, 31 December 2011 (CST)

Source Review
Just a quick heads up to draw your attention to Definition:Random Variable, which has an open SourceReview call listed on your name. &mdash; Lord_Farin (talk) 11:42, 5 April 2014 (UTC)


 * I have that book packed away; I'll have to dig it up. --GFauxPas (talk) 11:20, 8 April 2014 (UTC)

Mathjax
Is there a way to take the code of a given page and copy paste it into some external program or into some website, and end up with a pdf?

--GFauxPas (talk) 14:43, 8 April 2014 (UTC)

zeta 2
If you've been watching my sandbox, you might have seen me work on this:


 * $\displaystyle \int_{\to 0}^{\to 1} \ln x \ln \left({1-x}\right) \, \mathrm dx = 2 - \zeta\left({2}\right)$

Should I put this up as a theorem, or is it too specific an exercise to be worth sharing?

--GFauxPas (talk) 11:42, 9 April 2014 (UTC)


 * I think it's worth sharing, specially after all the hard work you put into it. The name needs to be considered carefully. Might want to put it into an "/Examples" subpage of either Logarithms or Integral Calculus (or both) ... once I've reached that part of my Spiegel handbook where there are hundreds of examples of integrations I will be in there to flesh it out, but it's something I haven't got to yet. --prime mover (talk) 04:26, 10 April 2014 (UTC)


 * Or maybe under Basel Problem... --GFauxPas (talk) 11:03, 10 April 2014 (UTC)


 * No, rather not under Basel problem -- the latter is a specific problem with a specific solution (i.e. the one Euler came up with, which explains its name), this just happens to have a result which includes the same zeta instance. --prime mover (talk) 16:43, 10 April 2014 (UTC)

Real and Imaginary Parts
Are these theorems?


 * $\displaystyle \int \operatorname{Im}\,\left({f\left({s}\right)}\right)\, \mathrm ds = \operatorname{Im}\,\left({\int

f\left({s}\right)\,\mathrm ds}\right)$


 * $\displaystyle \int \operatorname{Re}\,\left({f\left({s}\right)}\right)\, \mathrm ds = \operatorname{Re}\,\left({\int f\left({s}\right)\,\mathrm ds}\right)$

for real $s$? For complex $s$?

--GFauxPas (talk) 12:14, 5 May 2014 (UTC)


 * If $f$ is real then it's true trivially. If $f$ is complex then I believe (although I would have to go back to my source works to prove it) that it's not.


 * The question for whether it is true for real or complex $s$ (as opposed to $f$) I'm not sure how much sense it makes, but consider the fact that while $s$ may be real then it may not be the case that so is $f(s)$.


 * I don't know without thinking about it for more than 5 seconds but, as I say, I don't believe so. --prime mover (talk) 12:52, 5 May 2014 (UTC)

Gamma function Proof
Regarding the proof for:


 * $\displaystyle \mathcal L \left \{ {t^p} \right \} = \frac {\Gamma \left({p+1}\right)} { s^{p+1} }$

for $\operatorname{Re}\left({p}\right) > -1$, some sites are doing this. Is it justified?


 * $\displaystyle \int_0^{\to +\infty} t^p e^{-st} \, \mathrm dt = \int_{0}^{?} t^p e^{-u} \frac {\mathrm dt}{\mathrm du}\, \mathrm du$

where $u = st$, and then doing various stuff with the limits of integration. But it seems to me like a mistake; if $u$ is complex, then how can the limits of integration be complex? If $u$ is real, how can we replace a complex variable with a real variable?

I asked someone about this and he suggested that maybe they're making the integral into a contour integral, but that doesn't help me make it look like $\Gamma$... or does it? --GFauxPas (talk) 14:19, 26 May 2014 (UTC)


 * It's justified. You can prove by using the contour of two rays going to infinity, connected by an ever larger arc, that the "direction" of the infinity does not influence the value of the integral. This is because the integral over the arc goes to zero, because of the exponential. 't May be beyond the current scope of PW, though. &mdash; Lord_Farin (talk) 17:46, 26 May 2014 (UTC)


 * Once I get to it, I intend this *not* to be beyond the current scope. It's all in my complex analysis source notes -- but these are all buried way down in the substrata of the intellectual sedimentary rock that my office resembles ... and anyway, I'm mucking about with primitives at the moment. --prime mover (talk) 19:35, 26 May 2014 (UTC)


 * Ah, so, it's not $u = st$ so much as $u = rt$ where $s = re^{i\theta}$? And with $\displaystyle \int_C t^p e^{-u}\, \mathrm du$, $C$ is a curve connecting any two rays connected by an arc, as the length of the rays go to $+\infty$? and $\arg s$ doesn't affect anything? --GFauxPas (talk) 11:09, 27 May 2014 (UTC)


 * Yes, I think that's about it. &mdash; Lord_Farin (talk) 20:48, 27 May 2014 (UTC)

Modeling question
So I'm taking a class on modeling. What's the explanation or justification of this process?:

Given a sequence $\langle a \rangle : \N \to \R$, we create a function $f:\R \to \R$ such that $f \restriction_{\N} = \langle a \rangle$.

The objective is to find $\sup (\langle a \rangle)$ or $\inf(\langle  a \rangle)$.

To do so, we find the local extrema of $f$ (say at $f(c)$), and the assumption is that $\langle a \rangle$ attains its supremum/infimum at either $a_i$ or $a_{i+i}$, where $i \le c \le i + 1$.

Why and when is this method kosher? --GFauxPas (talk) 22:57, 11 September 2014 (UTC)


 * Very quickly, thinking as I type ...


 * If you've identified *all* the local extrema of $f$ (assuming that there are a finite number), then it can intuitively seen that one of the points either side of the *most* extreme one is going to be "as close as you can get" to it. This of course requires that the function does not have great big peaks at the $f \left({n + \dfrac 1 2}\right)$ points. Consider $f = \sin \left({\pi x}\right)$ as an awkward case which does not fit this intuitive notion but which *does* fulfil the conditions of the statement. --prime mover (talk) 05:29, 12 September 2014 (UTC)

HW help again
Help please with a problem from my modelling and optimization class. We're doing 2 variable optimization using Lagrange Multipliers. We're also discussing shadow prices.

The first part of this problem is to maximize profit using the price and advertising budget assumptions and data.

The data and assumptions are things like:

The cost of manufacture is 600 dollars per unit, and the wholesale price is 950 dollars

Other data are that units sold per month is 10 000, maximum advertising budget allowed by management is 100 000 per month.

Here are the data I don't know how to turn into workable function or equations, and could use help please:

In a test market, lowering the price by 100 increased sales by 50%

The advertising agency claims that increasing the advertising budget by 10 000 a month would result in a sales increase of 200 units a month.

How do I use these last data?


 * A manufacturer of PCs currently sells 10,000 units per month of a basic model. The cost of manufacture is 700/unit, and the wholesale price is 950. The cost of manufacture is 700/unit, and the wholesale price is 950. During the last quarter the manufacturer lowered the price 100 dollars in a few test markets, and the result was a 50% increase in sales. The company has been advertising its product nationwide at a cost of 50000 per month. The advertising agency claims that increasing the advertising budget by 10000 a month would result in a sales increase of 200 units a month. Managemeny has agreed to consider an increase in the advertising budget to no more than 100000 a month.

I have to find the price and advertising budget that maximizes profit.

If I can get the equations I know how to use Lagrange multpliers but the way it's presented is a mess. --GFauxPas (talk) 16:40, 13 October 2014 (UTC)


 * Assuming the pricing kicks in after the advertisement effect and all dependencies are linear (note the nontrivial assumptions!), we would have the following (with $10000a$ the additional money spent on advertising and $100d$ the lowered cost):


 * Units sold: $u = (10000+200a)(1+0.5d)$
 * Price: $p = 950-100d$
 * Cost: $700u + 50000 + 10000a$
 * Revenue: $pu$


 * The relevant constraints indicate that $0 \le a \le 5$, and a sanity check is that $d < 250$. As you can see, I approached this by simply defining auxiliary quantities that were easier to work with. I hope you understand, and that this is sufficient for you to complete the exercise. &mdash; Lord_Farin (talk) 17:41, 13 October 2014 (UTC)


 * Very helpful, and I get the price cost and revenue. It's the unit equation that I'm baffled by. $(10000 + 200a)$ makes sense to me. But I'm baffled by how $u$ varies with $p$. --GFauxPas (talk) 18:05, 13 October 2014 (UTC)


 * Also, the if the constraint is a region and not a curve, how do I set up lagrange multipliers? --GFauxPas (talk) 18:17, 13 October 2014 (UTC)


 * In a region, you can use differentiation. Since the exercise is about LM, it might be that the lower price is also part of the marketing budget; in that case, the cost would include an additional term $100ud$, which would correlate the two variables $a$ and $p$. Then, you could use LM on the curve "marketing budget = 100k" to solve for the optimal distribution. But this is all a lot of guesswork, so you might be off better by consulting your professor. &mdash; Lord_Farin (talk) 08:19, 14 October 2014 (UTC)


 * Additionally, the $1+0.5d$ (yes, I made a typo before, it was $d$ instead of $p$) represents a $50\%$ increase for a $\$100$ discount. &mdash; Lord_Farin (talk) 08:20, 14 October 2014 (UTC)


 * Thanks for the suggestions. I'll ask my professor, but I have a feeling the answer will be I can do whatever I think makes sense, as long as I state what assumptions I'm making. --GFauxPas (talk) 10:36, 14 October 2014 (UTC)

Thanks a lot for the help LF. My prof said I should make any guesses and assumptions I need to to make the problem make sense. I just have to explain what it is I'm doing.

Putting the explanation here, in case any reader is curious:

I should set up what's called a feasible region in which the solution lies. Solving $\nabla(\text{revenue - cost}) = \mathbf 0$ will get me potential solutions; I'm allowed to use graphing technology to determine if these are relative extrema rather than using Hessians etc.

Here's the rub:

The potential solutions will hopefully lie within the feasible region. If the region is described by a curve such that, uh, "that part" of the region is closed, then that curve needs to be checked with Lagrange multipliers. Any interesting points that that method produces need to be checked.

Here, using the model we're using, my feasible region will be defined by $a = 0$, $a = 5$, $d < \frac {950} {100}$ (because the computers can't be given away for free).

So what are the lines $a = 0$ and $a = 5$ called in describing $\{\left({a,d}\right) \in \R^2: 0 \le a \le 5, d < \frac {950} {100} \}$ on the $ad$-plane? Is there a name for the solid-line defining part of a region rather than a dotted-line defining part of a region? And is there a name for the "sharp corners" part of a region? E.g. the points $P = (0,\frac{950}{100})$ and $P = (5,\frac{950}{100})$ in this problem. He said to check those, too. --GFauxPas (talk) 22:41, 14 October 2014 (UTC)


 * Progress: https://www.dropbox.com/s/t0xycxt1oqp2b1d/p52prob6draft1.pdf?dl=0 I also realize there's no need to check the corner points for this one, I'll have to take that part out. --GFauxPas (talk) 15:37, 15 October 2014 (UTC)