How do you find the Maclaurin series of #f(x)=ln(1+x^2)#
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Question

Evelyn Agard · Accepted Answer

Plug in #x^2# for every x in the Maclaurin series for #f(x)=ln(1+x)# So, you have the Maclaurin series for #f(x)=ln(1+x)# defined as: #f(x)=ln(1+x) = x-x^2/2+x^3/3-... =sum_(k=1)^n(((-1)^(k+1)x^k) / k)# The question is essentially asking you to plug in #x^2# for every x here, so: #(x^2)^1-(x^2)^2/2+(x^2)^3/3-(x^2)^4/4+...# which gives... #x^2 - x^4/2+x^6/3-x^8/4+... = sum_(k=1)^n(((-1)^(k+1)(x^2)^k) / k)#

Charles Arocha · Answer

undefined To find the Maclaurin series of $ f(x) = \ln(1 + x^2) $, we'll first find the derivatives of $ f(x) $ and then evaluate them at $ x = 0 $ to obtain the coefficients of the Maclaurin series. The Maclaurin series expansion is given by:

$$ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots $$

First, let's find the derivatives of $ f(x) = \ln(1 + x^2) $:

$$ f'(x) = \frac{2x}{1 + x^2} $$
$$ f''(x) = \frac{2(1 + x^2) - 2x(2x)}{(1 + x^2)^2} = \frac{2 - 2x^2}{(1 + x^2)^2} $$
$$ f'''(x) = \frac{(-4x)(1 + x^2)^2 - 2(2 - 2x^2)(2x)(1 + x^2)}{(1 + x^2)^4} = \frac{-4x - 8x^3 + 8x^3}{(1 + x^2)^3} = \frac{-4x}{(1 + x^2)^3} $$

Now, let's evaluate these derivatives at $ x = 0 $ to find the coefficients of the Maclaurin series:

$$ f(0) = \ln(1) = 0 $$
$$ f'(0) = \frac{2 \cdot 0}{1 + 0^2} = 0 $$
$$ f''(0) = \frac{2 - 2 \cdot 0^2}{(1 + 0^2)^2} = 2 $$
$$ f'''(0) = \frac{-4 \cdot 0}{(1 + 0^2)^3} = 0 $$

Therefore, the Maclaurin series of $ f(x) = \ln(1 + x^2) $ is:

$$ f(x) = 0 + 0x + \frac{2}{2!}x^2 + 0x^3 + \cdots $$

$$ f(x) = x^2 $$

How do you find the Maclaurin series of #f(x)=ln(1+x^2)# ?