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June 20[edit]

Approximation[edit]

Hi. Just did a show that involving an approximation and I'm not convinced that what I've done is rigorous. Could someone please check it for me?

"Let ${\frac {1}{y}}={\frac {2+3x^{2}}{3(1+x^{2})^{1.5}}}+{\frac {1}{3}}$ . Show that, for large positive x, $y\approx 3-{\frac {9}{x}}$ ."

My method goes like this. for large positive x $1+x^{2}\approx x^{2}$ and so $(1+x^{2})^{1.5}\approx x^{3}$ . Also for large positive x, $2+3x^{2}\approx 3x^{2}$ so ${\frac {2+3x^{2}}{3(1+x^{2})^{1.5}}}\approx {\frac {3x^{2}}{3x^{3}}}={\frac {1}{x}}$ so ${\frac {1}{y}}\approx {\frac {1}{x}}+{\frac {1}{3}}$ . Rearranging this gives $y\approx {\frac {3x}{3+x}}={\frac {3x+9}{3+x}}-{\frac {9}{3+x}}=3-{\frac {9}{3+x}}$ . Finally, for large positive x, $3+x\approx x$ so $y\approx 3-{\frac {9}{x}}.$

Is this a solid argument? Thanks 92.2.16.39 (talk) 10:09, 20 June 2009 (UTC)[reply]

Yes, I think that's spot on.. Rkr1991 (talk) 14:15, 20 June 2009 (UTC)[reply]

Yes, that looks fine to me. The statement "for large positive x,

y\approx 3-{\frac {9}{x}}

" isn't a rigorous statement, so there is no way your "proof" could be rigorous. What you've got is good enough, in my opinion. If the question had been to show that

y=3-{\frac {9}{x}}+O({\frac {1}{x^{2}}})

or something (or whatever the appropriate notation is, I can never remember the differences between big-O, little-O, etc. without looking them up), then you could get a proper rigorous proof (probably along the same lines as the argument you gave). --Tango (talk) 14:26, 20 June 2009 (UTC)[reply]

Hm, either two curves are asymptotic or they're not, meseems. —Tamfang (talk) 04:54, 25 June 2009 (UTC)[reply]

You may prefer to simplify first and approximate afterwards.

y={\frac {3(1+x^{2})^{1.5}}{2+3x^{2}+{(1+x^{2})^{1.5}}}}=3-9x^{-1}\cdot {\frac {2x^{-2}+3}{6x^{-3}+9x^{-1}+3{(1+x^{-2})^{1.5}}}}\approx 3-9x^{-1}

.

Bo Jacoby (talk) 14:37, 20 June 2009 (UTC).[reply]

Car dynamics[edit]

Please tell me about the algorithm of finding International Roughness Index(of a road surface)from a longitudinal profile data. —Preceding unsigned comment added by 113.199.158.140 (talk) 12:36, 20 June 2009 (UTC)[reply]

Googleing 'International Roughness Index' leads to the fortran program at http://www.umtri.umich.edu/content/IRIMain.f . Bo Jacoby (talk) 13:51, 20 June 2009 (UTC).[reply]

I wanted to have the algorithm and not the program. —Preceding unsigned comment added by 113.199.170.182 (talk) 02:47, 21 June 2009 (UTC)[reply]

What's the difference? A programming language is just a language for writing algorithms. --Tango (talk) 03:09, 21 June 2009 (UTC)[reply]

The program contains the following comment: 'For more information about the IRI and how this program works, see Sayers, M. W., "On the Calcluation of International Roughness Index from Longitudinal Road Profile." Transportation Research Record 1501, (1995) p. 1-12. See http://spyder.umtri.umich.edu/erd_soft/erd_file.html on the web for a description of ERD files'. So now it is time for the OP to do some homework. Bo Jacoby (talk) 10:20, 21 June 2009 (UTC).[reply]

Necessary and sufficient conditions[edit]

I'm a little hazy on N&S conditions so wanted to check my thoughts with people who can tell me if I'm right or wrong. Let $f(x)=x^{2}-2bx+c$ . I have to find N&S conditions on b and c a) for f(x) to have two distinct real roots and b) for f(x) to have two distinct positive real roots.

For a) I get that $c-b^{2}<0$ just from using the fact that if the turning point is below the x axis, you've got two distinct roots.

For b) I again get $c-b^{2}<0$ , for the same reason as before. I then get c>0, which ensures that the roots are of the same sign and b>0, which says the minimum point occurs in the first quadrant (at least I think that's its name; I mean bottom right). So the y-intercept is positive, both the roots are of the same sign and the turning point is in the first quadrant, so between them, they say that the roots are distinct and positive.

Are they necessary and sufficient or not? Thanks 92.4.255.16 (talk) 17:52, 20 June 2009 (UTC)[reply]

"A is necessary for B" means "B implies A". "A is sufficient for B" means "A implies B". Therefore to prove that something is necessary and sufficient you need to prove it in both directions. Sometimes you can do both directions at once, but if you're not sure it is best to do them separately. You need to come up with a sequence of implications from A to B and then a sequence from B to A. While considering the graphs is a good idea in order to get an intuitive answer, your rigorous proof should probably be purely algebraic. --Tango (talk) 18:01, 20 June 2009 (UTC)[reply]

The quadratic formula gives you the roots of f as

x={\frac {2b\pm {\sqrt {4b^{2}-4c}}}{2}}=b\pm {\sqrt {b^{2}-c}}

. The answer to both questions follow naturally from there. For question a you are correct; the expression in the radical must be positive for the radical to be real. For question b, it is sufficient and necessary for the smaller root to be real and positive. I'll let you work the algebra to see where that gets you. --COVIZAPIBETEFOKY (talk) 18:24, 20 June 2009 (UTC)[reply]

~~Your solution to (b) is almost right. The x-coordinate of the vertex is ${\frac {-b}{2a}}$ (note the minus sign). So to be in the fourth quadrant, you need $b<0$~~ . -- Meni Rosenfeld (talk) 19:37, 20 June 2009 (UTC)[reply]

Where has the 'a' come from? Are you possibly confusing the f(x) I gave above with the general form of a quadratic equation? 92.7.54.6 (talk) 19:41, 20 June 2009 (UTC)[reply]

Yes, COVIZAPIBETEFOKY was correct, and Meni is correct to say that you meant the fourth quadrant, and your conclusions were also correct, but see Tango's suggestion for a rigorous proof. Dbfirs 00:12, 21 June 2009 (UTC)[reply]

Of course, silly me. I was thinking about the general case, and didn't notice the differences in your case, most importantly that your b is already negated. So indeed it should be

b>0

. -- Meni Rosenfeld (talk) 08:50, 21 June 2009 (UTC)[reply]

Maximizing the volume in a certain class of sets[edit]

Has anybody already seen an optimization problem like this one:

Find the maximum value of the 3-dimensional volume of the subset of the unit cube:

{\hat {A}}:=\{(x,y,z)\in [0,1]\times [0,1]\times [0,1]\,:(x,y)\in A,(y,z)\notin A\}

among all measurable subsets $\textstyle A$ of the unit square $[0,1]\times [0,1]$ .

Incidentally, I know the answer (it is hidden here in case somebody prefers not to see it), but I'd like to settle this and other similar but more difficult problems into a general theory/method, if there is any available. --pma (talk) 21:17, 20 June 2009 (UTC)[reply]

Cute problem. I don't recall having seen it before. Maybe I'll try to work out where it came from. I wonder if thinking of it in the language of probability could shed any light. Michael Hardy (talk) 17:48, 21 June 2009 (UTC)[reply]

In the most general sense, this seems like calculus of variations. But that answer is probably useless. 67.122.209.126 (talk) 08:53, 22 June 2009 (UTC)[reply]