# How do you find the first 3 non-zero terms in the Taylor series of the function cosh(x) about the point x=ln8 and use this series to estimate cosh(2)?

Oct 9, 2017

${T}_{3} f \left(2\right) \approx 3.7667$

#### Explanation:

Taylor Series Socratic Question

https://socratic.org/questions/how-do-you-find-the-first-3-non-zero-terms-in-the-taylor-series-of-the-function-

The formula for a third degree Taylor Series polynomial is:

T_3 f(x) = f(a) + f'(a)(x-a) +f''(a)frac{(x-a)^2}{2!}+f'''(a)frac{(x-a)^3}{3!}

So to find this, we need to determine the first, second, and third derivatives of $f \left(x\right)$.
$f \left(x\right) = \cosh x = \frac{1}{2} {e}^{x} + \frac{1}{2} {e}^{- x}$
$f \left(\ln 8\right) = \frac{1}{2} {e}^{\ln 8} + \frac{1}{2} {e}^{\ln \left(\frac{1}{8}\right)} = 4 + \frac{1}{16} = \frac{65}{16}$

$f ' \left(x\right) = \frac{1}{2} {e}^{x} - \frac{1}{2} {e}^{- x}$
$f ' \left(\ln 8\right) = \frac{1}{2} {e}^{\ln 8} - \frac{1}{2} {e}^{\ln \left(\frac{1}{8}\right)} = 4 - \frac{1}{16} = \frac{63}{16}$

$f ' ' \left(x\right) = \frac{1}{2} {e}^{x} + \frac{1}{2} {e}^{- x}$
$f ' ' \left(\ln 8\right) = \frac{65}{16}$

$f ' ' ' \left(x\right) = \frac{1}{2} {e}^{x} - \frac{1}{2} {e}^{- x}$
$f ' ' ' \left(\ln 8\right) = \frac{63}{16}$

T_3 f(x) = f(ln8) + f'(ln8)(x-ln8) +f''(ln8)frac{(x-ln8)^2}{2!}+f'''(ln8)frac{(x-ln8)^3}{3!}

T_3 f(x) = frac{65}{16} + frac{63}{16}(x-ln8) +frac{65}{16}frac{(x-ln8)^2}{2!}+ frac{63}{16} * frac{(x-ln8)^3}{3!}

Finally, use this Taylor Series approximation to estimate $\cosh \left(2\right)$:
${T}_{3} f \left(2\right) = \frac{65}{16} + \frac{63}{16} \left(2 - \ln 8\right) + \frac{65 \cdot {\left(2 - \ln 8\right)}^{2}}{16 \cdot 2} + \frac{63 {\left(2 - \ln 8\right)}^{3}}{16 \cdot 6}$

This would be tedious to simplify, so I'll plug this in to my calculator to get:

${T}_{3} f \left(2\right) \approx 3.7667$

Compare this to the actual value: $\cosh \left(2\right) \approx 3.7622$. This shows that our approximation is accurate to the nearest hundreths place.