How do you find the Taylor Polynomial of f(x) = ln(x) with a=1, n=4?

1 Answer
Dec 11, 2016

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

Explanation:

The Taylor series of #f(x)# about #x=a# is given by
#f(x) = f(a) + (f'(a))/(1!)(x-a)+(f''(a))/(2!)(x-a)^2 + (f'''(a))/(3!)(x-a)^3+...#

So with a=1 we have:

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

We can tabulate the successive derivatives and values of #f^((n+1))(1))# as follows;

# {: (n, f^((n))(x), f^((n))(1)), ( 0, ln(x), 0 ), ( 1, 1/x, 1 ), ( 2, (-1)/x^2, -1 ), ( 3, (2)/x^3, 2! ), ( 4, (-2*3)/x^4, 3! ), ( vdots, vdots, vdots ), ( n, ((-1)^(n-1)2*3*..*(n-1))/x^n, (-1)^(n-1)(n-1)!) :} #

And so:
# f(x) = 0 + (1)/(1!)(x-1)-(1)/(2!)(x-1)^2 + (2!)/(3!)(x-1)^3-(3!)/(4!)(x-1)^4 +...#

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

General Term
Although we are not asked we can see the general term would be:
# "nth term" = ((-1)^(n-1)(n-1)!)/n! (x-1)^n #
# " " = ((-1)^(n-1)(n-1)!)/((n-1!)n) (x-1)^n #
# " " = ((-1)^(n-1))/(n) (x-1)^n # for #n ge 1#