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

Dec 11, 2016

$f \left(x\right) = \left(x - 1\right) - \frac{1}{2} {\left(x - 1\right)}^{2} + \frac{1}{3} {\left(x - 1\right)}^{3} - \frac{1}{4} {\left(x - 1\right)}^{4} + \ldots$

Explanation:

The Taylor series of $f \left(x\right)$ 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 +...

$f \left(x\right) = \left(x - 1\right) - \frac{1}{2} {\left(x - 1\right)}^{2} + \frac{1}{3} {\left(x - 1\right)}^{3} - \frac{1}{4} {\left(x - 1\right)}^{4} + \ldots$

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
$\text{ } = \frac{{\left(- 1\right)}^{n - 1}}{n} {\left(x - 1\right)}^{n}$ for $n \ge 1$