Gamer.Site Web Search

Search results

  1. Results From The WOW.Com Content Network
  2. Easy way to remember Taylor Series for log (1+x)?

    math.stackexchange.com/questions/1262181

    To remember log(1 − x) l o g (1 − x) just do log(1 − x) = − ∫ dx 1 − x = − ∫(1 + x + x2 + ⋯)dx = − (x + x2 2 + x3 3 + ⋯) = − ∞ ∑ n = 1xn n. l o g (1 − x) = = = = − ∫ d x 1 − x − ∫ (1 + x + x 2 + ⋯) d x − (x + x 2 2 + x 3 3 + ⋯) − ∑ n = 1 ∞ x n n. – Kian. Commented May 2, 2015 at 16:15. Is ...

  3. Which is correct? log in, log on, log into, log onto [duplicate]

    english.stackexchange.com/questions/290874

    For my money, log on to a system or log in to a system are interchangeable, and depend on the metaphor you are using (see comment on your post). I suppose there is a small bit of connotation that "log on" implies use, and "log in" implies access or a specific user.

  4. verbs - "log in to" or "log into" or "login to" - English...

    english.stackexchange.com/questions/5302

    Nov 7, 2018 at 22:13. 5. "To log in" and "to log into" are Reflexive Separable Phrasal Verbs which often have the reflection omitted. They mean the same thing but have slightly different grammatical construction. "To log in" requires a prepositional phrase to describe what a person is logging into.

  5. Taylor Series for $\\log(x)$ - Mathematics Stack Exchange

    math.stackexchange.com/questions/585154

    the Taylor series for ln (x) is relatively simple : 1/x , -1/x^2, 1/x^3, -1/x^4, and so on iirc. log (x) = ln (x)/ln (10) via the change-of-base rule, thus the Taylor series for log (x) is just the Taylor series for ln (x) divided by ln (10). – correcthorsebatterystaple. Mar 18 at 14:35.

  6. The difference between log and ln - Mathematics Stack Exchange

    math.stackexchange.com/.../the-difference-between-log-and-ln

    1. $\begingroup$. According to the international standard ISO 31-11 "ln" stands for base-e natural logarithm; "lg" is for base-10 common logarithm; and "lb" is for the base-2 binary one. "log" is a generic notation for a logarithm of an arbitrary base that needs to be specified.

  7. How do I square a logarithm? - Mathematics Stack Exchange

    math.stackexchange.com/questions/1271168

    You could, however, do a change of base with the logs and put them in base 10. We have the formula logbx = logax logab where a can be any base you want. Most common base is 10. So we have, (log23)2 = (log103 log102)2 = (log3 log2)2 = log23 log22. Share.

  8. Logging in or on? - English Language & Usage Stack Exchange

    english.stackexchange.com/questions/2002

    log in, log out, log on, log off are verbs. for developers, this will probably be helpful in deciding which term to use: Just an observation, but the more casual the site, the more likely it’s going to use Sign in/Sign out. The paradigm of a visitor making an announcement of presence to a casual gathering of people.

  9. What is the difference between log in, sign in; register, sign...

    english.stackexchange.com/questions/211729/what-is-the...

    log in, sign in, login, log on, logon, authenticate. All of these words are more-or-less synonyms. "Login"/"logon" may or may not be acceptable (check your style guide). If they are, they may be nouns instead of verbs (referring to the action of logging in or the state of being logged in).

  10. logarithms - Is there an approximation to the natural log...

    math.stackexchange.com/questions/977586

    If the number whose logarithm you require is real and positive, Borchardt's algorithm, which is similar in structure to Gauss' AGM algorithm, if started with initial values 1 and x, will converge to (x-1)/Ln(x) for large values of x, although many steps will be required, each involves a square root, and convergence is not quadratic.

  11. What is the best way to calculate log without a calculator?

    math.stackexchange.com/questions/820094

    $\begingroup$ If you know the log of a few prime numbers, you can find the log of a number that is close to the desired one. So if you have log(x) and you want log(x+d), just add 0.4343*d/(x+d/2) to log(x) and you will be close enough for gubbermint work. Example: compute log(10).