Sep 17, 2011 · log(1+x)=x - x 2 /2 + x 3 /3 - x 4 /4 + ... (Alternate signs) The easiest way to see it is by using an integral representation. log(1+x) = ∫dx/(1+x) Since 1/(1+x) = 1 - x + x 2 - x 3 + ..., …

Oct 21, 2014 · Oh yes I meant if you take (log x)' then that's (1/x). Then when x = 1, the derivative is 1. Using this fact, you can use the definition of the derivative as [log(1+x) - log(1)]/x as x …

Jan 20, 2013 · Taking the derivative of the MacLaurin series gives you $1 -x +x^2 - x^3 + x^4 + \ldots$ Since this is a geometric series with ratio $-x$, it equals $\frac{1}{1 + x}$ when x is in …

Jan 19, 2019 · I suggest you start with the right side and use the rule ##-\log(x) = \log\frac{1}{x}##. You will then have a square root in the denominator, and there is a known trick to remove …

Sep 17, 2011 · Given that \\log{(1+x)}=\\int_0^x\\frac{dt}{1+t}, how would one prove that \\lim_{x \\to 0}\\frac{\\log{(1+x)}}{x}=1?

Mar 17, 2010 · The asymptotic behavior of log(x) is log(x). Log is a special thing, just as is exp(x). The exponential function grows faster than power function x a, no matter how large a is. This …

Jul 13, 2009 · The other way out could be to integrate by parts once and use a series expansion for [itex]\tan^{-1}x[/itex] to express the second integral, [tex]\int \frac{\tan^{-1}x}{1+x}dx[/tex] as …

Jul 21, 2017 · Issues to consider are the radius of convergence and the speed of convergence of the series. An expansion of log(x+1) around x=0 will give values for log(y), y=x+1 from y=0 to …

Aug 7, 2008 · Yes, if y= log 1 (x) then x= 1 y = 1 for all x. The reason I put "functions" in quotes above is that this is not actually a function: strictly speaking, log 1 (x) is not a function. But it is …

Mar 31, 2013 · How to evaluate \int^1_0 log(1-x) dx I am confused as log is not defined at 0. Please help. Insights Blog