What is the Maclaurin series for ln(5-x)ln(5x)?

2 Answers
Jul 16, 2017

ln(5-x) = 4 - sum_{n=1}^{\infty}((-4)^n/n(\sum_{r=1}^{\infty}{::}_nC_r(-4)^{-r}))x^n)ln(5x)=4n=1((4)nn(r=1nCr(4)r))xn).

Explanation:

ln(5-x)=ln(1-(x-4))ln(5x)=ln(1(x4))

Then, ln(1-x) = -(sum_{n=1}^{\infty}x^n/n)ln(1x)=(n=1xnn),
ln(1-(x-4)) = -(sum_{n=1}^{\infty}(x-4)^n/n)ln(1(x4))=(n=1(x4)nn),
ln(5-x) = -sum_{n=1}^{\infty}sum_{r=1}^{n}(({::}_nC_r (-4)^{n-r})/n x^r )ln(5x)=n=1nr=1(nCr(4)nrnxr),
ln(5-x) = -(-4 + sum_{n=1}^{\infty}((-4)^n/n(\sum_{r=1}^{\infty}{::}_nC_r(-4)^{-r}))x^n)ln(5x)=(4+n=1((4)nn(r=1nCr(4)r))xn),
ln(5-x) = 4 - sum_{n=1}^{\infty}((-4)^n/n(\sum_{r=1}^{\infty}{::}_nC_r(-4)^{-r}))x^n)ln(5x)=4n=1((4)nn(r=1nCr(4)r))xn).

Expansion of ln(1-x)ln(1x) holds for abs(x)<1|x|<1. Then our expansion for ln(5-x)ln(5x) holds for abs(x-4)<1|x4|<1 or 3<x<53<x<5.

Jul 16, 2017

ln(5-x) = ln(5) - x/5-x^2/50-x^3/375-x^4/2500-x^5/15625- ...

And in sigma notation, this is

ln(5-x) = ln5 -sum_(n=1)^oo (x^n)/(n5^n)

Explanation:

The power series for ln(1-x) is often quoted in examination formula books etc, and is as follows:

ln(1-x) = -x-x^2/2-x^3/3-x^4/4-x^5/5 + ...
" " = -sum_(n=1)^oo x^n/n

So for the series we require, we have:

ln(5-x) = ln(5(1-x/5))
" " = ln(5) + ln (1-x/5)

So then if we substitute x for x/5 in the initial expansion we get:

ln(5-x) = ln(5) + {-(x/5)-(x/5)^2/2-(x/5)^3/3-(x/5)^4/4-(x/5)^5/5} + ...

" " = ln(5) - {x/5+(x^2/25)/2+(x^3/125)/3+(x4/625)/4+(x^5/3125)/5} + ...

" " = ln(5) - x/5-x^2/50-x^3/375-x^4/2500-x^5/15625- ...

And in sigma notation, this is

ln(5-x) = ln5 -sum_(n=1)^oo (x^n)/(n5^n)