What is the Maclaurin series for ln(5-x)ln(5−x)?
2 Answers
Explanation:
Then,
Expansion of
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) = -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
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)