How do you find the nth Taylor polynomials for f(x) = ln x centered about a=1?

Question

Evelyn Agard · Accepted Answer

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

We have:

#f(x) = lnx#

#f'(x) = 1/x = x^-1#

#f''(x) = -x^-2#

and in general:

#f^((n)) = (-1)^(n-1)(n-1)!x^-n#

So that:

#f(1) = 0#

#f^((n)) (0) = (-1)^(n-1)(n-1)!#

The Taylor series is then:

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

Luke Alvarez · Answer

undefined To find the nth Taylor polynomial for $ f(x) = \ln x $ centered about $ a = 1 $, you would first need to find the nth derivative of $ f(x) $ and evaluate each derivative at $ x = 1 $. The nth Taylor polynomial is then given by the formula:

$$ P_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n $$

For $ f(x) = \ln x $, the nth derivative is:

$$ f^{(n)}(x) = (-1)^{n-1} (n-1)! \cdot \frac{1}{x^n} $$

Evaluating at $ x = 1 $, you get:

$$ f^{(n)}(1) = (-1)^{n-1} (n-1)! $$

Thus, the nth Taylor polynomial for $ f(x) = \ln x $ centered about $ a = 1 $ is:

$$ P_n(x) = \ln 1 + \frac{1}{1}(x-1) - \frac{1}{2 \cdot 1}(x-1)^2 + \frac{1}{3 \cdot 1}(x-1)^3 - \cdots + (-1)^{n-1} \frac{(x-1)^n}{n} $$

Simplified to:

$$ P_n(x) = (x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 - \cdots + (-1)^{n-1} \frac{(x-1)^n}{n} $$

Paisley Johnson · Answer

undefined To find the nth Taylor polynomial for $ f(x) = \ln x $ centered about $ a = 1 $, you can use the Taylor series expansion formula:

$$ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots $$

For $ f(x) = \ln x $, the derivatives are:

$$ f(x) = \ln x $$
$$ f'(x) = \frac{1}{x} $$
$$ f''(x) = -\frac{1}{x^2} $$
$$ f'''(x) = \frac{2}{x^3} $$
$$ f^{(4)}(x) = -\frac{6}{x^4} $$
$$ \cdots $$

Evaluate these derivatives at $ x = a = 1 $ to find the coefficients for the Taylor polynomial. Since $ f(1) = \ln 1 = 0 $, the Taylor polynomial simplifies:

$$ P_n(x) = 0 + \frac{1}{1}(x-1) - \frac{1}{2}(x-1)^2 + \frac{2}{3}(x-1)^3 - \frac{6}{4!}(x-1)^4 + \cdots $$

You can continue this pattern to find the nth Taylor polynomial for $ f(x) = \ln x $ centered about $ a = 1 $.