Find the Taylor series expansion formula of #f(x)=\lnx# at #a=3#?

Question

#f'=1/x=x^-1#
#f''=-1/x^2=(-1)x^-2#
#f'''=2/x^3=(-1)(-2)x^-3#
#f^4=(-1)(-2)(-3)x^-4...#

must find #C_n# before finding #f(x)# expansion

Isabella Ackerman · Accepted Answer

#f(x)=ln3+\sum_(n=1)^\infty(-1)^(n+1)1/(n3^n)(x-3)^n# #f'(3)=(3)^-1# #f''(3)=(-1)3^-2# #f'''(3)=(-1)(-2)(3)^-3# #f^4(3)=(-1)(-2)(-3)(3)^-4# In general, for #n>=1# we have #f^n(3)=(-1)^n(n-1)!3^(-n)=((-1)^n(n-1)!)/3^n# #C_n=f^(n)(3)/(n!)=(1/(n!))*((-1)^(n-1)(n-1)!)/3^n=(-1)^(n-1)1/(3^n(n))# We also have #\color(red)(f(3) = ln 3)# and so #\color(red)(C_0 = ln(3))# #\rArrf(x)= sum_{n=0}^oo C_n(x-3)^n# #qquad qquad = \color(red)(ln(3))+\sum_(n=1)^\infty(-1)^(n+1)1/(n3^n)(x-3)^n#

Aurora Alvarado · Answer

undefined To find the Taylor series expansion formula of f(x) = ln(x) at a = 3, we first need to find the derivatives of ln(x) and evaluate them at x = 3.

The derivatives of ln(x) are:
f'(x) = 1/x
f''(x) = -1/x^2
f'''(x) = 2/x^3
f''''(x) = -6/x^4

Now, let's evaluate these derivatives at x = 3:
f'(3) = 1/3
f''(3) = -1/9
f'''(3) = 2/27
f''''(3) = -6/81 = -2/27

Now, let's write the Taylor series expansion formula using these derivatives:
f(x) ≈ f(3) + f'(3)(x - 3) + f''(3)(x - 3)^2/2! + f'''(3)(x - 3)^3/3! + f''''(3)(x - 3)^4/4!

Substitute the values we found:
f(x) ≈ ln(3) + (1/3)(x - 3) - (1/9)(x - 3)^2/2 + (2/27)(x - 3)^3/6 - (2/27)(x - 3)^4/24

This is the Taylor series expansion formula of f(x) = ln(x) at a = 3.

Find the Taylor series expansion formula of #f(x)=\lnx# at #a=3#?

#f'=1/x=x^-1# #f''=-1/x^2=(-1)x^-2# #f'''=2/x^3=(-1)(-2)x^-3# #f^4=(-1)(-2)(-3)x^-4...# must find #C_n# before finding #f(x)# expansion

#f'=1/x=x^-1#
#f''=-1/x^2=(-1)x^-2#
#f'''=2/x^3=(-1)(-2)x^-3#
#f^4=(-1)(-2)(-3)x^-4...#

must find #C_n# before finding #f(x)# expansion