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November 5[edit]

why so many 2nd order pdes in physics?[edit]

Now jerk is 3rd deriv., which reminds me...people debate/discuss about whether Nature has a preference for linearity--great. But how about more debate/exploration on Nature's "preference", if the preference exists, for second order differential equations? One way to explore it would be to check if there is something mathematically significant that happens or doesn't happen with linear third order partial differential equations. Does all hell break loose in the solutions or what? This is like asking why we have 3 space dimensions instead of 38 or 190 spatial dimensions.Rich 00:02, 22 April 2007 (UTC) I moved this question from jerk talk page.Rich 04:15, 5 November 2007 (UTC)[reply]

It turns out linear ODEs of any order are solvable, so at least in the ODE case we have no mathematical preference other than simplicity of examples. In the nonlinear case, we have a much easier time with certain classes of nonlinear 2nd-order ODEs, and again we have explicit solvability for 1st-order nonlinear one-dimensional ODEs given certain general conditions. For PDEs, things get more complicated and we get a lot of nonanalytic cases that are heavily dependent on boundary conditions at second order. Higher orders become much worse, but luckily in many cases there are tricks that we can use to solve them. So there is really no magic number in the case of DiffEq solving, but the lower order the better.

So the answer actually comes from Noether's Theorem, in which we get conservation laws from symmetries in the universe. It is from these conservation laws and symmetries that we define our dimensions: energy, location, time, etc., and we see in mechanics and electrodynamics that these are often manifest in terms of second-order differential equations. Equivalently, this has to do with the r-square nature of these fundamental forces. It doesn't matter how many spatial or temporal dimensions we have, as this only increases the number of dimensions in a PDE, not the order. Higher-order differential equations may have simpler conditions for chaos (I don't know), but the Lorenz attractor equations are first-order, and that is a situation in nature where "all hell breaks loose". SamuelRiv 04:45, 5 November 2007 (UTC)[reply]

Newton's 2'nd law is usually expressed as second order differential equations of space coordinates with respect to time, but it can also be expressed as Hamilton's first order differential equation of space and momentum coordinates with respect to time. So the order of the differential equation is not a property of nature. Bo Jacoby 11:11, 5 November 2007 (UTC).[reply]

You may see a difference between ODE's for fundamental concepts, electro-magnetism etc. an those for less fundamental concepts. For example the Korteweg–de Vries equation describes a model of waves on shallow water surfaces. Which can be expressed as a third order ODE. A google search for "third order differential equation" has some other interesting examples. --Salix alba (talk) 14:34, 5 November 2007 (UTC)[reply]

Name of y = x³[edit]

I know that y = x² is a parabola, but is there any specific name for y = x³? Also, when the exponents of x are even powers greater than 2 (such as y = x⁴, y = x⁶, etc.), are those also known as parabolas? I know what all these functions look like; I just what to know if they have specific names. —Preceding unsigned comment added by 71.107.147.127 (talk) 05:11, 5 November 2007 (UTC)[reply]

y=x^3 (and in fact ax^3 + bx^2 + cx + d) is generally called a "cubic", and x^4 and x^5 are sometimes called "quartics" and "quintics", respectively. After that, there isn't a special name, although all fall under the class of polynomials, and in particular x^n is a polynomial of degree n. Confusing Manifestation 06:01, 5 November 2007 (UTC)[reply]

For all quadratic functions the graph is a parabola, and all parabolas are similar. That is not true of all cubic curves or even the graphs of all cubic functions. --Lambiam 07:53, 5 November 2007 (UTC)[reply]

It may also be refered to as the Cubing Function since it represents the cube of x, but it is more likely going to be refered to by it's polynomial name of Cubic Function. The graph of

y=x^{3}

has 180 degree rotational symmetry about (0,0) and has three roots x=0,0,0 or a triple root at x=0. Whereas,

y=x^{2}

has reflective symmetry about the vertical line x=0. It should be noted that the name Parabola actually comes from the Study of Conic Sections not from polynomials. Which were studied extensively by the Greeks. I doubt there is a intriguing focus-directrix type Locus definition for the curve of a

y=x^{3}

as there is for

y=x^{2}

. Calculus allows for a more detailed comparison between the two curves via The First and Second Derivatives. A math-wiki 08:37, 5 November 2007 (UTC)[reply]

In earlier times curves were individually cherished and given special names. The conic sections — circle, ellipse, parabola, hyperbola — were studied (and named) in ancient Greece, but the idea of describing a curve by an algebraic equation was more than a millennium away. With algebra, we opened the door to so many functions that naming became less manageable. We do have intermediate possibilities like

y^{2}=x^{3},\,\!

called a semicubical parabola, but that's an unusual case. It wouldn't surprise me if some antique names can be found for some of the higher degree curves, but I am aware of nothing special in common use. --KSmrq^T 14:44, 5 November 2007 (UTC)[reply]

y=x^2n is a parabola. so, if n=1, or course, but if n=2, then y=x^4 is a parabola. because its y=x^2^2. n=3, y=x^6, or y=u^2 where u=x^3, is a parabola. Cubic Plane Curve is an order 3 curve.

http://staff.jccc.net/swilson/planecurves/cubics.htm

Artoftransformation 13:11, 9 November 2007 (UTC)[reply]

gcd[edit]

Plz help me. Given that (a,4)=2, (b,4)=2 show that (a+b,4)=4.--58.68.97.34 09:05, 5 November 2007 (UTC)[reply]

Hint: If gcd(a,4)=2, then a is congrunet to 2 mod 4, so a=4n+2 for some integer n ... Gandalf61 09:35, 5 November 2007 (UTC)[reply]

Congruence is in the next chapter of the book. So can you derive a =4n + 2 without using congruence relation. That will be sufficient.--58.68.97.34 09:41, 5 November 2007 (UTC)[reply]

Yes. Recall that the GCD is a common divisor, so 2 is a divisor of a. It is the greatest common divisor, so 4>2 is not a common divisor, but as 4|4 you have that 4 is not a divisor of a. What does all this tell you? -- Meni Rosenfeld (talk) 13:03, 5 November 2007 (UTC)[reply]

I was the person above. Thanks for the help. I assume we now proceed by a = 4n + r, 0<r<4 (division algorithm) and note that as 2|r so r=2.--Shahab 14:53, 5 November 2007 (UTC)[reply]

2|r, by itself, doesn't tell you that r=2 (r could be 0). You need to also use the fact that 4 doesn't divide a. -- Meni Rosenfeld (talk) 17:21, 5 November 2007 (UTC)[reply]

Yet another LaTeX question[edit]

Is there an easy way of marking a function in "function" text? Say, $\cos {\frac {\theta }{\sigma }}$ compared with $cos{\frac {\theta }{\sigma }}$ ? This is because I am using a function that isn't part of the standard functions so renders as if it was a variable. I tried using \textrm but that didn't help the spacing, so I was just wondering if there is equivalent markup for a function. x42bn6 Talk Mess 14:24, 5 November 2007 (UTC)[reply]

One recommended way in general is \operatorname. Perhaps there are issues on wikipedia. A discussion took place over a year ago at wiki project math. Here is a sample:

\exp(x)\cdot \operatorname {cosh} (y)

. JackSchmidt 14:37, 5 November 2007 (UTC)[reply]

If TeX did not have a special case for cosine, here are some things we could try, compared to the "real thing".

`cos \theta`	$cos\theta \,\!$
`\textrm{cos} \theta`	${\textrm {cos}}\theta \,\!$
`\mathrm{cos} \theta`	$\mathrm {cos} \theta \,\!$
`\operatorname{cos} \theta`	$\operatorname {cos} \theta \,\!$
`\mathrm{cos}\, \theta`	$\mathrm {cos} \,\theta \,\!$
`\cos \theta`	$\cos \theta \,\!$

Obviously \cos has a little extra magic built in to give the correct spacing when not followed by a parenthesis. --KSmrq^T 14:54, 5 November 2007 (UTC)[reply]

Either \operatorname{} or \mbox{}——you know there is a TeX formula help page, don't you? ~Kaimbridge~15:10, 5 November 2007 (UTC)[reply]

I'm not bothered with the Wikipedia "version" of LaTeX. Just wondering if it is available for normal LaTeX. x42bn6 Talk Mess 15:53, 5 November 2007 (UTC)[reply]

Sure, just \usepackage{amsmath}. For example \documentclass{article}\usepackage{amsmath}\begin{document}$\operatorname{cosh}(x)$\end{document} JackSchmidt 16:07, 5 November 2007 (UTC)[reply]

Far better than using \mathoperator directly, I would recommend \DeclareMathOperator{\foo}{foo}. Donald Hosek 17:53, 5 November 2007 (UTC)[reply]

\DeclareMathOperator is the appropriate solution. (It handles spacing issues as well.) Tesseran 00:19, 6 November 2007 (UTC)[reply]

Such solutions cannot be demonstrated within Wikipedia. When genuine (La)TeX is available, it is clearly better to use one declaration rather than fixes for each instance. --KSmrq^T 04:11, 6 November 2007 (UTC)[reply]

Math[edit]

what is multiplying exponents ? —Preceding unsigned comment added by 4.236.129.227 (talk) 23:17, 5 November 2007 (UTC)[reply]

Power of a power, e.g. (x^m)^n = x^(mn)?…86.132.235.192 00:09, 6 November 2007 (UTC)[reply]

2+4 —Preceding unsigned comment added by 4.236.129.227 (talk) 00:15, 6 November 2007 (UTC)[reply]

6? 2+4 redirects to Treaty on the Final Settlement with Respect to Germany (or the Two Plus Four Agreement). PrimeHunter 01:44, 6 November 2007 (UTC)[reply]

70+9 —Preceding unsigned comment added by 4.236.129.227 (talk) 00:16, 6 November 2007 (UTC)[reply]

79? (I know we shouldn't do people's homework but would somebody operating a computer and writing English be needing homework help for that?). PrimeHunter 01:44, 6 November 2007 (UTC)[reply]

i have been having trouble with products in school can you help me —Preceding unsigned comment added by 216.174.52.150 (talk) 01:50, 6 November 2007 (UTC)[reply]

We can't figure out what it is that you want from us. Give us more information. risk —Preceding comment was added at 02:34, 6 November 2007 (UTC)[reply]

I'm going to assume by product you mean product of powers. The most important thing to remember is that $(x^{m})^{n}=x^{mn}$ . This can be shown to be a reasonable relationship, I will give an example as to help clairify. say $x=2,m=5,n=3$ . Then the above equality says that $(2^{5})^{3}=2^{5\cdot 3}$ . So starting with the left using the rule $a^{b}\cdot a^{c}=a^{b+c}$ I will show that $(2^{5})^{3}=2^{5\cdot 3}$ .

$(2^{5})^{3}=2^{5}\cdot 2^{5}\cdot 2^{5}=2^{5+5+5}=2^{15}=2^{5\cdot 3}$

You can pick any other integers and show that the above process still works, although what I did is not a proof. Using those methods on the general case, it can be shown to be true for all integers and even in more general cases such as rational numbers. A math-wiki 09:14, 6 November 2007 (UTC)[reply]