How do you find the maclaurin series expansion of #f(x) =ln(1 + x) #?

Question

3494 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-13T16:39:01

The series is #=sum_(k=1)^n((-1)^(k+1)x^k)/k#

The Maclaurin series is

#f(x)=sum_(n=0)^(oo)(f^(n)(0))/(n!)x^n#

#=f(0)+(f'(0))/(1!)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+ (f^(iv)(0))/(4!)x^4+......#

#f(x)=1n(1+x)#, #=>#, #f(0)=0#

#f'(x)=1/(1+x)#, #=>#, #f'(0)=1#

#f''(x)=-1/(1+x)^2#, #=>#, #f'(0)=-1#

#f'''(x)=2/(1+x)^3#, #=>#, #f'(0)=2#

#f^(iv)(x)=-6/(1+x)^4#, #=>#, #f'(0)=-6#

Therefore,

#ln(1+x)=0+1*x-x^2/2+x^3/3-x^4/4+...#

#=x-x^2/2+x^3/3-x^4/4+...#

#=sum_(k=1)^n((-1)^(k+1)x^k)/k#