How do you find the first two nonzero terms in Maclaurin's Formula and use it to approximate #f(1/3)# given #f(x)=sqrt(x^2+1)#?

Question

Ezekiel Alexander · Accepted Answer

Please see the explanation below

The growth is #f(x)=sqrt(1+x^2)=1+x^2/2+.....# Consequently, #f(1/3)=sqrt(1+1/9)=sqrt10/3=1.054# #f(1/3)=1+(1/3)^2/2=1+1/18=19/18=1.056#

Paisley Johnson · Answer

undefined To find the first two nonzero terms in Maclaurin's Formula, first, compute the derivatives of $ f(x) = \sqrt{x^2 + 1} $ up to the second derivative. Then, evaluate these derivatives at $ x = 0 $.

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

Now, substitute $ x = 0 $ into these expressions to find the values of the derivatives at $ x = 0 $:

1. $ f(0) = \sqrt{0^2 + 1} = 1 $
2. $ f'(0) = \frac{0}{\sqrt{0^2 + 1}} = 0 $
3. $ f''(0) = -\frac{0^2}{{(0^2 + 1)}^{\frac{3}{2}}} + \frac{1}{\sqrt{0^2 + 1}} = 1 $

Thus, the first two nonzero terms in Maclaurin's Formula are $ f(0) = 1 $ and $ f''(0) = 1 $.

Now, we use these terms to approximate $ f\left(\frac{1}{3}\right) $:

$$ f(x) \approx f(0) + f'(0)x + \frac{f''(0)x^2}{2!} $$

Substitute the values:

$$ f\left(\frac{1}{3}\right) \approx 1 + 0\left(\frac{1}{3}\right) + \frac{1\left(\frac{1}{3}\right)^2}{2!} $$

$$ f\left(\frac{1}{3}\right) \approx 1 + \frac{1}{18} = \frac{19}{18} $$