1 апр. 2013 г. · Mechanically replacing t by x2, we get ln(1+x2)=x2−x42+x63−x84+x105−⋯.

23 июл. 2023 г. · Let F(x)=ln(1−x). Then, ln(1−x2)=F(x2). We know that F(x)=−∫x011−tdt+F(0), and since F(0)=0, we have F(x2)=−∫x2011−tdt. The idea here is ...

9 февр. 2014 г. · Hint. You started well using ln(1+x)=∞∑n=0(−1)n+1xnn. Now change x to −x to get ln(1−x)=−∞∑n=0xnn. Now replace x by t2 and finish.

23 февр. 2018 г. · by using the Maclaurin series expansion for ln(1+x)=∑∞n=1(−1)n+1xnn and replacing x with x2 and multiplying the series with x3.

10 апр. 2013 г. · The MacLaurin series for ln(1+x) is obtained from the series for 11+x by integration. Use this and appropriate substitutions to obtain the ...

2 нояб. 2016 г. · The problem is to expand this ln(1+x+x2) around 0 in power series. I'm confused. What should I use? ln(1+x) and do something with it? power ...

9 нояб. 2021 г. · I am trying to find the power series for ln2(1−x) without directly using the Taylor's series to expand it.

29 мар. 2017 г. · I mean that to find fc(x)=ln(x2+1), you only have to calculate fa(x2), where fa(x) is the result of (a).

16 мар. 2016 г. · Let f(x)=[ln(1+x)]2. Use the series for the logarithm to compute that f(x)=[ln(1+x)]2=∞∑n=2(−1)n(n−1∑k=1 1 (n−k)k

31 мая 2011 г. · But 1/√1+x2 blows up at x=±i, so there is no power series that is valid for |x|>1.