How do you find the taylor series for #lnx# in powers of #x-1#?

2 Answers
Jan 13, 2017

#lnx = sum_(n=1)^oo (-1)^(n-1) (x-1)^n/n#

Explanation:

The general formula for the Taylor series of a function #f(x)# around #x=1# is:

#f(x) = sum_(n=0)^oo (f^((n))(1))/(n!)(x-1)^n#

we can immediately note that:

#f^((0))(1) = lnx |_(x=1) = 0#

so the constant term is null. For the following terms, we have to calculate the derivatives of #f(x) = lnx# for all orders:

#d/(dx)lnx = 1/x = x^(-1)#

#(d^2)/(dx^2) lnx = -x^(-2)#

#(d^3)/(dx^3) lnx = 2x^(-3)#

and we can easily see that in general:

#(d^n)/(dx^n) lnx = (-1)^(n-1)(n-1)!x^(-n)#

and for #x=1#

#f^((n))(1) =[(d^n)/(dx^n) lnx]_(x=1) = (-1)^(n-1)(n-1)!#

The Taylor series is then:

#lnx = sum_(n=1)^oo (-1^(n-1)(n-1)!)/(n!)(x-1)^n = sum_(n=1)^oo (-1)^(n-1) (x-1)^n/n#

As a bonus, we can note that this expansion can be used to calculate the sum of the alternating harmonic series:

#1-1/2+1/3-1/4+... = sum_(n=1)^oo (-1)^(n-1)/n#

In fact if we use the Taylor series above for #x=2# we have:

#ln2 = sum_(n=1)^oo (-1)^(n-1) (2-1)^n/n = sum_(n=1)^oo (-1)^(n-1)/n#

Jan 13, 2017

#ln x=(x-1)-(x-1)^2/2-(x-1)^3/3-,,,.-1<x-1<=1# that gives #-2<x<=2#

Explanation:

Using #X = x-1#, the The Maclaurin series

# ln(1+X)=X-X^2/2+X^3/3-..., -1<X<1 #

Is the Taylor series for

# ln x = ln(1+(x-1))=(x-1)-(x-1)^2/2-(x-1)^3/3-,,,.-1<x-1<=1 # that gives #-2<x<=2#