Wikipedia:Reference desk/Archives/Mathematics/2011 November 13

= November 13 =

logarithms revisited
hey again. the general logarithmic function can be given by $$f(x)=a+b\log{(x)}$$, where log is the natural log and a and b are constant parameters. I know that for a general exponential function $$f(x)=ab^{x}$$ a can be interpreted as the inital condition (at x=0, x usually time but sometimes other stuff) and b the rate of increase. Is there a similar interpretation for the parameters of the general log function? Thanks.24.92.85.35 (talk) 02:59, 13 November 2011 (UTC)
 * The inverse of the function $$f(x)=ab^{x}$$ has the form you describe as the general logarithm. In particular $$f^{-1}(y) = (\log y - \log a)/\log b$$, so $$f^{-1}(y) = c + d\log y$$ where $$c = -\log a/\log b$$ and $$d = 1/\log b$$.  If you want to think of f(x) as the growth of some quantity over time, then f-1(x) is the amount of time it will take for that quantity to reach a particular level. Rckrone (talk) 03:12, 13 November 2011 (UTC)

no I'm sorry I think I have misexplained. I am interested only in the interpretation of the parameters a and b, not in the functions themselves (I am OK there). thanks for your response though. 24.92.85.35 (talk) 03:15, 13 November 2011 (UTC)
 * Well, a is its value at x = 1, and b is sort of a scaling factor.—PaulTanenbaum (talk) 03:40, 13 November 2011 (UTC)
 * Alternatively, $$b=\frac{1}{\log B}$$ where B is the base of the logarithm. -- Meni Rosenfeld (talk) 06:07, 13 November 2011 (UTC)

Integrability of fg
This is part of a proof that f and g integrable implies fg integrable on an interval [a,b]. Take f, g to be bounded and non-negative on [a,b]. Let P = {t0, ..., tn} be a partition of [a,b], and denote by U(f,P) and L(f,P), respectively, the upper and lower sums of f in P on [a,b]. Also take Mi', Mi, and Mi as the respective maximum values of f, g, and fg on each of the intervals [ti&minus;1, ti], with the same notation but using a lowercase m'' representing the minimum values. I've shown that
 * $$M_i \le M_i ' M_i  \quad \mathrm{and} \quad m_i \ge m_i ' m_i  \quad \mathrm{implies} \quad U(fg,P) - L(fg,P) \le \sum_{i = 1}^n (M_i ' M_i  - m_i ' m_i )(t_i - t_{i-1}).$$

The next step, according to my textbook, is to establish that if f and g are bounded above by M, then we have
 * $$U(fg,P) - L(fg,P) \le M \left [ \sum_{i = 1}^n (M_i ' - m_i ')(t_i - t_{i-1}) + \sum_{i = 1}^n (M_i  - m_i )(t_i - t_{i-1}) \right ].$$

I'm not sure how to move from the second last inequality to this one, though, or how otherwise to establish the fact. I recognise the sums in the brackets as U(f,P) &minus; L(f,P) and U(g,P) &minus; L(g,P) but little else comes to mind. Some suggestions would be great. Thanks. — Anonymous Dissident  Talk 07:36, 13 November 2011 (UTC)
 * First, a good technique to use to find the right strategy is to solve a simple special case and then extrapolate that solution to the general case. In this problem try the case where the interval is [0,1] and it's partition is a trivial partition into the interval [0,1]. Then the inequality you must prove is M′M″-m′m″≤M(M′-m′)+M(M″-m″). But this follows from M″≤M and m′≤M; add and subtract M″m′ on the left hand side. With the special case established, it should be simple matter to add the indices back in, then multiply by the ti - ti-1 and sum everything. Note, by "integrable" it's clear from the problem that you mean Darboux integrable, but be aware there several types of integrability.--RDBury (talk) 21:14, 13 November 2011 (UTC)


 * Also, since $$fg=\big((f+g)^2-f^2-g^2\big)/2$$, it is sufficient to show that the square of a (Riemann) integrable function $$f$$ is integrable (once it is known that linear combinations of integrable functions are integrable). As to the squaring operation, starting from
 * $$f^2(x)-f^2(y)=(f(x)+f(y))(f(x)-f(y))\le 2\|f\|_\infty|f(x)-f(y)|$$
 * you can easily prove the inequality:
 * $$U(f^2,P)-L(f^2,P)\le 2\|f\|_\infty (U(f,P)-L(f,P))$$
 * whence it is apparent that $$f^2$$ is Riemann integrable whenever $$f$$ is integrable. The same argument works more generally to show that if $$f$$ is integrable and $$\phi$$ is locally Lipschitz continuous, then $$\phi \circ f $$ is integrable. In fact, it is also true if $$\phi$$ is just continuous, though the proof is somehow less easy. --pm a 22:09, 13 November 2011 (UTC)

Modular division
Is there an efficient way to compute (a/b) mod m, if it is known that a mod b == 0? 184.98.169.135 (talk) 18:23, 13 November 2011 (UTC)
 * By "efficient" I mean, being able to take the modulus before doing the division. 184.98.169.135 (talk) 18:29, 13 November 2011 (UTC)


 * Because a mod b=0, you know a=cb for some constant integer c. So, a/b = cb/b = c. All you know about c is that it is smaller than a. It could be a/2 or a/3 or a/4 or a/5... So, that doesn't provide a lot of help calculating c mod m. -- k a i n a w &trade; 20:43, 13 November 2011 (UTC)


 * Let x, y, z be the reductions of a, b, c modulo m. If a = bc, then x = yz up to factors of m. When b is relatively prime to m, the second equation can be solved uniquely for z; that is, you can perform the modulus before performing the division. If b is not relatively prime to m, you cannot perform the modulus first as you will get multiple answers and can't distinguish which is correct. Eric. 151.48.27.93 (talk) 10:41, 14 November 2011 (UTC)