Talk:Logarithm

Notation

• ln = logarithm to the base e
• lg = logarithm to the base 2
• log = ambiguous (engineers: base 10; mathematicians & scientists: base e; computer scientists: base 2)

ln is universally recognised, so it should stay. Whenever log is used, the base should be specified.

The above seems to be a good summary of what was here before.

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And I agree with the above, but I think that lg should not be used (even if it is the correct notation) because it is too rare. I myself have read some calculus books that say 10 should be the "default" base, and some that say e should be the "default" base.

Brianjd 06:11, 2004 Jun 19 (UTC)

For the old discussion of notation, see Talk:Logarithm/Notation. - dcljr 23:32, 8 Nov 2004 (UTC)

Two things: (1) For me, `lg' is just as ambiguous as `log'. Log base 2 can be referred to unambiguously (I hope) as `ld', but even that isn't very common in areas where base 2 is dominant (computer science, information theory). (2) Specifying an explicit base for `log' is a good idea in many cases, but there are situation when this is not necessary. For example, in computer science O(log n) is unambiguous, since the choice of base does not matter. MarkSweep 08:38, 2004 Jun 20 (UTC)

Most UK A-level textbooks I've seen like to use lg for log₁₀ and ln for log_e (the rest use log for log₁₀, and I've never seen lg used for log₂). Style is up to the article writer, but I propose that, if only one base is used in the article, the first occurrence of log includes the base. Any further occurrences don't need to include the base, since the default base is implied. But, really, we don't mean sin(x°) when we say sin(x), because measuring angles in radians is much more useful (namely, you can use the Taylor series, and sin'(x) = cos(x)). Similarly, log'(x) = 1/x. If we define log = log₁₀, then we're stuck with log'(x) = 1/(x ln 10), which is ugly at best. Of course, log(2) is another special number (namely, it's important when calculating logs of big numbers). --Elektron 15:30, 2004 Jun 30 (UTC)

Definition

I thought:

• x^(r/s) = (the rth root of x)^s = the rth of (x^s).

• The logarithmic function is any function which satisfies four conditions:
  1. Domain is all of R.
  2. It is nonconstant.
  3. It is differentiable.
  4. For any real numbers x, y, L(xy) = L(x) + L(y).

• The natural logarithm of x, denoted ln x, is the definite integral of 1/x with the lower limit 1 and the upper limit x.

• The exponential functions are then inverses of the logarithmic functions. This is the only way to define exponentials for irrational exponents.

I have not seen anything like this on Wikipedia. Only that e = lim (n->infinity) (1+1/n)^n and logarithms are inverses of exponents, which is absolute rubbish according to what I have read elsewhere.

Brianjd 08:25, 2004 Jun 18 (UTC)

That anything is the only way to define exponentials of irrationals is surprising, to say the least. Seldom, if ever, is there only one way to do something in mathematics. The power-series

<math>\exp(z)=\sum_{n=0}^\infty{z^n \over n!}<math>

defines exponentials, at least to base-e, for all complex numbers, including irrationals. If you want other bases, you may want to bring in already-defined natural logarithms, and then perhaps you want to proceed as you indicate above, but any assertion that that is the only way is dubious.

Contrary to your assertion, the domain of the natural logarithm function is not all reals, but only positive reals. There is a "multiple-valued" (if that expression can be forgiven) logarithm function defined for all complex numbers except 0. But how did you propose to define the logarithm of 0, if the domain is all reals? Michael Hardy 21:39, 18 Jun 2004 (UTC)

What I have read seems to imply that defining exponentials that way would be seen to be, by most people, as absurd as defining pi by a power series.

Yes, the domain of the natural logarithm is all positive reals. My sources were correct - I posted incorrectly.

Anyway I have not read Wikipedia pages very thoroughly but I remember seeing basically "logarithms are inverses of exponentials; exponentials are inverses of logarithms". Is there any solid definition?

Michael Hardy posted the same response to my talk page. Can we keep the discussion where it belongs?

Also, don't post to someones talk page with the section heading "Definition". When things are taken out of context like that they become gibberish.

Brianjd 05:52, 2004 Jun 19 (UTC)

Exponentials at least to the base e seem to be defined properly...but the proofs seem very complicated.

I have seen a calculus book with much easier proofs of the product rule, quotient rule, and the various proofs related to logarithms/exponentials (logarithms require some knowledge of integration though; as the book used my definition above).

Brianjd 05:55, 2004 Jun 19 (UTC)

I see that my definition is indeed buried in the middle of the natural logarithm page. I think that all common definitions (and mine qualifies as a pretty common definition) should be listed in the intro, or in the first section, with their equivalence proved. Brianjd 06:01, 2004 Jun 19 (UTC)

Don't define the logarithm. As with anything in math, there are too many equivalent definitions to call any of them the definition. Rather, list the properties of the log function, with a side note that they are equivalent. Also, no need to prove this in an article.--Sean Kelly 20:05, 24 Dec 2004 (UTC)

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e is just a number that satisfies a certain property. From first principles (with shamelessly incorrect use of dx),

<math>\frac {d}{dx} a^{x} = \frac {a^{x+dx} - a^{x}}{dx} = \frac {a^{x} a^{dx} - a^{x}}{dx} = \frac {a^{dx} - 1}{dx} a^{x}<math>

So we define e as the number such that the first derivative of e^x is e^x. That is,

<math>\frac {e^{dx} - 1}{dx} e^{x} = e^{x}<math>

<math>e^{dx} - 1 = dx\,\!<math>

<math>e^{dx} = 1 + dx\,\!<math>

<math>e = \left(1 + dx\right)^\frac{1}{dx}<math>

If we define dx = 1/n,

<math>e = \left(1 + \frac {1}{n}\right)^{n}<math>

Alternatively, if we let f(x) = e^x, it's trivial to prove by induction that

<math>f^{(n)}\left(x\right) = e^{x}<math>

Which is where you get the taylor expansion for e^x

The basic definition is, really, just d/dx e^x = e^x. I believe I've just proved that the standard definitions of e are equivalent. --Elektron 16:17, 2004 Jun 30 (UTC)

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Um, could someone put the basic formula of a logarithm in big letters somewhere? Like (but using variables) log(base 2)8 = 3 and 2^3=8 are equal. I know that example is used in the article, but it would be great if a more general one was included that used the big math font. --pie4all88

You mean,

<math>\log_b x = n \qquad \mbox{means} \qquad b^n = x<math>

? Also, it shouldn't be called the "basic formula" but rather one way of algebraically defining it.--Sean Kelly 20:05, 24 Dec 2004 (UTC)

Formula moved from article

I moved the formula

<math>

\sum_{j=1}^{k-1}\left( \frac{-2^{-1 + j}}{\left( -j + k \right) \,\left( -1 + k \right) !} \right) +

 \frac{2^{-1 + k}\,\ln (2)}{\left( -1 + k \right) !}=\sum_{n = 0}^{\infty }\frac{{\left( -1 \right) }^n\,n!}{\left( n + k \right) !}

<math> here, because it is not clear to me what it is doing at that place in the article. What connection does it have to the algorithm? Furthermore, I'd like to see a reference (I'm too lazy to check it myself). -- Jitse Niesen 14:55, 17 May 2005 (UTC)