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

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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, ... , t_n} 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 M_i', M_i'', and M_i as the respective maximum values of f, g, and fg on each of the intervals [t_i−1, t_i], with the same notation but using a lowercase m representing the minimum values. I've shown that

$M_{i}\leq M_{i}'M_{i}''\quad \mathrm {and} \quad m_{i}\geq m_{i}'m_{i}''\quad \mathrm {implies} \quad U(fg,P)-L(fg,P)\leq \sum _{i=1}^{n}(M_{i}'M_{i}''-m_{i}'m_{i}'')(t_{i}-t_{i-1}).$

The next step, according to my textbook, is to establish that if f and g are bounded above by M, then we have

$U(fg,P)-L(fg,P)\leq M\left[\sum _{i=1}^{n}(M_{i}'-m_{i}')(t_{i}-t_{i-1})+\sum _{i=1}^{n}(M_{i}''-m_{i}'')(t_{i}-t_{i-1})\right].$

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) − L(f,P) and U(g,P) − 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 t_i - t_i-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

$f^{2}(x)-f^{2}(y)=(f(x)+f(y))(f(x)-f(y))\leq 2\|f\|_{\infty }|f(x)-f(y)|$

you can easily prove the inequality:

$U(f^{2},P)-L(f^{2},P)\leq 2\|f\|_{\infty }(U(f,P)-L(f,P))$

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. --pma 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. -- kainaw™ 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)