Wikipedia:Reference desk/Archives/Mathematics/2009 October 29

= October 29 =

Differential operators of order 4 - help me before I die of boredom
Hi there,

I want to show that the 4th order differential operator $$\mathcal{L}=\sum_{r=0}^4\,p_r(x)\frac{d^r}{dx^r}$$, with $$p_r$$ real functions, is self adjoint iff $$p_3=2p_4',\,p_1=p_2'-p_4'''$$ - the only way I've seen of doing this on my course was for the Sturm-Liouville (2nd order) operator, and it involved nothing more than churning through 20 integrations by parts. Not only does this seem like pretty much the ugliest possible approach to the question, it is also long and tedious for order 2 and undoubtedly far more tedious in 4th order operators - can someone please tell me any more concise or cleaner way to solve this problem or at the very least link me to some sort of proof so I can spend my time learning something a little more beneficial rather than sitting up plugging things into "$$\int y_2\mathcal{L}y_1=\int y_1\mathcal{L}y_2$$"?

I don't mean to sound lazy and I certainly don't just want other people to do all my work for me, I'd just rather actually be learning something new instead of what seems to be little more than an exercise in checking how consistent my algebra is and I'm sure someone out there knows how to save me some time! :)

Thankyou very much for the help, Otherlobby17 (talk) 00:56, 29 October 2009 (UTC)


 * I understand your point, but if it's just a matter of characterizing when $$ \mathcal{L}$$ is symmetric (say on $$\mathcal{D}(\R)$$ wrto the $$L^2$$ inner product), after all the plain computation is not that bad.  With a bit of concentration you can just write down immediately what should be the result of: integrating by parts, using the Leibnitz' rule, and equating the integrands of the two sides of $$\scriptstyle\int y_2\mathcal{L}y_1=\int y_1\mathcal{L}y_2$$ :
 * $$p_0=+p_0-p_1^{(1)}+p_2^{(2)}-p_3^{(3)}+p_4^{(4)}$$
 * $$p_1=-p_1+2p_2^{(1)}-3p_3^{(2)}+4p_4^{(3)}$$
 * $$p_2=+p_2-3p_3^{(1)}+6p_4^{(2)}$$
 * $$p_3=-p_3+4p_4^{(1)}$$
 * $$\textstyle p_4=+p_4$$
 * that gives you the condition you wrote. As a consequence, the bilinear form associated with $$\mathcal{L}$$ via the $$L^2$$ inner product, is symmetric iff it is of the form
 * $$\textstyle\int_{\R}u \mathcal{L}v\, dx= \int_{\R }\left( a_0uv+a_1u'v'+a_2uv\right)  dx.$$
 * If you already know this fact for other reasons (which?), you may also go back to the corresponding operator $$\mathcal{L}$$ and find the condition for being symmetric in another way (and quicker). The interesting question is, how all that generalize to any order. The answer should be: the order is even and you can freely choose the coefficients $$p_r$$ with even indices r, while the ones with odd r are determined. (I guess). --pma (talk) 15:13, 29 October 2009 (UTC)

Logarithmic differentiation
I'll make a comment rather than ask a question, but maybe after that I'll ask a question too, if I feel like it two minutes from now.

Today I found myself before a class of about 25 students, and one of them asked how to use logarithmic differentiation to find something like
 * $$ \frac{d}{dx} \left( (2x + 1)^5 (x^4 - 3)^6 \right), $$

and while I'm going through this someone asked a more intelligent question than what I expect of students in this particular course: doesn't this involve taking a logarithm of a negative number? Of course, one can say that
 * $$ \frac{d}{dx} \log |y| = \frac{dy/dx}{y}, $$

and then one is taking the logarithm of the absolute value of a negative number.

But there's still the question of dy/dx at points where y = 0. I expect to think about that when I separate variables in a differential equation and find dy/y, and I integrate that to get log |y|, but in such cases I deal with that by asking separately whether "y = 0 everywhere" is a solution.

I could tell the student that the derivative in a case like this must be a polynomial, and a polynomial is determined by its values at all points except the finitely many where it is 0. That still might leave a question of what happens in more general situations than polynomials.

OK, do I feel like ending with a question? Maybe I'll leave it as an exercise to say what questions should be asked at this point...... Michael Hardy (talk) 01:23, 29 October 2009 (UTC)
 * ..... so now I notice that the book does say "If &fnof;(x) < 0 for some values of x, then.....". I hardly ever pay close attention to what the book says..... Michael Hardy (talk) 01:39, 29 October 2009 (UTC)


 * A first remark that comes to my mind is that negative (or even complex) values are not a problem, for we don't need logarithms to define the logarithmic derivative of u as u'/u; nor we need them to prove properties like (uv)'/(uv) = u'/u + v'/v . The zero set of u is clearly a problem in general (e.g. if u is a constant I do not see what use could be u'/u ), but as in your example, a computation with logarithmic derivatives may lead to a perfectly correct result; for instance, one finds an equality that certainly holds at any point where all the functions involved in the computation do not vanish, but then the validity may be extended by continuity everywhere. Of course this works very well with polynomials and more in general with analytic functions. --pma (talk) 09:00, 29 October 2009 (UTC)
 * That's all very well but you try to explain all that to a class full of freshmen your instructor evals will tank. Pull the one student who asked aside and give them pma's explanation. (You might also see if you can get them become a math major.) For the other students just refer them to the book.--RDBury (talk) 13:39, 29 October 2009 (UTC)
 * To the 24 freshmen I would just say: the "logarithmic derivative" of u at x is defined as u'(x)/u(x), when u(x)≠0 and u'(x) exists. The name is suggested by the identity (log u)'=u'/u when u is also a positive function. As to the problem with the 25th, it is easily solved telling the class that he is going to convince you to explain everything in terms of operator calculus ;-) --pma (talk) 15:37, 29 October 2009 (UTC)

help this formulae identify itself
For a range data if we apply the formulae (X-min(X))/(max(X)-min(X))=?.What is this referred to as and what is the significance of doing such operation on data sets.(X represents any one data value.) 220.225.98.251 (talk) sci-hunter —Preceding unsigned comment added by 220.225.98.251 (talk) 18:53, 29 October 2009 (UTC)
 * it's 0 if X is the min; it's 1 if X is the max; it's 1/2 if X is half-way between the min and the max; it's 1/4 if X is 1/4 on the way from the min and the max. So it just tells you the relative position of X in the range between the min and the max in a scale from 0 (X=min) to 1 (X=max). Just a change of scale for jour data, that puts everything in [0,1]. --78.13.140.32 (talk) 19:10, 29 October 2009 (UTC)
 * Another term for this kind of operation is normalization. -- Meni Rosenfeld (talk) 20:26, 29 October 2009 (UTC)

Algebraic long division
At least, I think that's what I need. I have to divide $$b^3-a^3$$ by $$b-a$$ as part of deriving the variance of the uniform distribution from first principles. Can anyone help?  It Is Me Here   t / c 19:00, 29 October 2009 (UTC)
 * b2+ab+a2 --78.13.140.32 (talk) 19:03, 29 October 2009 (UTC)


 * Thanks, but how did you get there?  It Is Me Here   t / c 19:05, 29 October 2009 (UTC)
 * You're welcome. How I did: because I knew. Check:
 * (b-a)(bb+ab+aa)=
 * b(bb+ab+aa)-a(bb+ab+aa)=
 * bbb+abb+aab-abb-aab-aaa=
 * bbb-aaa.
 * sorry for not writing the exponents; just too lazy to clik on the "Supersctipt" thing. Note: you can do as well (b7-a7)/(a-b) or any other exponent, in the same way (otherwise check polynomial long division and Synthetic division if you want more theory).--78.13.140.32 (talk) 19:22, 29 October 2009 (UTC)
 * Also, with some algebraic manipulation, $$\frac{b^3-a^3}{b-a}$$ matches the form of the sum of a geometric series, so you only need to work in reverse to find the corresponding terms of the series. Telescoping series is also related. -- Meni Rosenfeld (talk) 20:24, 29 October 2009 (UTC)
 * Factorizing small polynomials like that is a standard part of highschool algebra. If you're taking a serious probability course and you're not already reasonably skilled with that sort of thing, you may be in for a difficult time.  Anyway, in general,
 * $$a^n-b^n = (a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots+b^{n-1}).$$
 * If you multiply it all out, you'll see how all the cross-terms in the middle cancel each other. 69.228.171.150 (talk) 21:07, 29 October 2009 (UTC)


 * First of all, we observe that if a=b then both those expressions are zero, this tells us that a-b is a factor of a3-b3, so we expect to get a polynomial as the answer (as opposed to a Laurent polynomial - one with negative exponents). a-b has both terms having degree 1 and a3-b3 has both terms having degree 3. That tells us we're looking for a polynomial with all the terms having degree 2. There are 3 possible such terms - a2, b2 and ab (don't forget that last one!). So, the answer will be of the form ma2+nab+pb2. The easiest way to find m, n and p is just to multiply it out and equate coefficents - (a-b)(ma2+nab+pb2)=a3-b3. --Tango (talk) 22:22, 29 October 2009 (UTC)


 * The real refdesk answer, what the poster actually asked for, is the article polynomial long division. Meni mentioned it in his response, but this was the answer that should have been given right off. --Trovatore (talk) 22:24, 29 October 2009 (UTC)
 * The OP asked how to solve the problem and guessed that polynomial long division might be the best way. I disagree, I think the the method I gave is better in this case. Long division is good if you aren't expecting a perfect polynomial for the answer, but if you are then there are better ways. --Tango (talk) 23:35, 29 October 2009 (UTC)


 * I tend to think that what the poster asked for was exactly the plain result of $$(b^3-a^3)/(b-a)$$, not general theories nor tricks... --pma (talk) 00:05, 30 October 2009 (UTC)
 * This is indeed the first thing the OP was given. The title and his request for clarification indicate he has at least some interest in the general technique. -- Meni Rosenfeld (talk) 06:06, 30 October 2009 (UTC)
 * Actually, the mention of polynomial long division was part of anon's answer. I was going to mention it before I noticed that anon beat me to it. -- Meni Rosenfeld (talk) 06:06, 30 October 2009 (UTC)


 * Yeah, Tango's method was the most helpful - thanks to all of you, though!  It Is Me Here   t / c 16:27, 30 October 2009 (UTC)
 * a nice method for a nice guy! --pma (talk) 18:01, 30 October 2009 (UTC)