Science:Math Exam Resources/Courses/MATH100/December 2015/Question 11 (a)/Solution 1

By Hint(1) and (2), we need to apply product rule to find the coefficients for the series for $h$ given by

$T_{2_{h}}(x)=h(0)+h'(0)x+{\dfrac {h''(0)}{2!}}x^{2}$.

We have $h(x)=e^{x}\cdot g(x)$, so

${\begin{aligned}h'(x)&=(e^{x})'\cdot g(x)+e^{x}\cdot g'(x)\\&=e^{x}\cdot g(x)+e^{x}\cdot g'(x)\end{aligned}}$

Similarly applying product rule to $h'$ we have:

${\begin{aligned}h''(x)&=\left[(e^{x})'\cdot g(x)+e^{x}\cdot g'(x)\right)]+\left[(e^{x})'\cdot g'(x)+e^{x}\cdot g''(x)\right]\\&=e^{x}\cdot g(x)+e^{x}\cdot g'(x)+e^{x}\cdot g'(x)+e^{x}\cdot g''(x)\end{aligned}}$

Therefore; $h(0)=g(0)$

$h'(0)=g(0)+g'(0)$ and

$h''(0)=g(0)+2g'(0)+g''(0)$.

As you see, we need $g(0),g'(0),g''(0)$. We compare the given Maclaurin series for $g$ with the actual formula as the following:

$T_{2_{g}}(x)=g(0)+g'(0)x+{\dfrac {g''(0)}{2!}}x^{2}=5-{\frac {x}{3}}+2x^{2}$

to see that

$g(0)=5,g'(0)=-{\frac {1}{3}}$ and ${\frac {g''(0)}{2!}}=2\Rightarrow g''(0)=4$

which gives

$h(0)=5$

$h'(0)=5-{\frac {1}{3}}={\frac {14}{3}}$ and

$h''(0)=5-{\frac {2}{3}}+4={\frac {25}{3}}$.

Substituting these into the very first formula gives:

${\color {blue}T_{2_{h}}(x)=5+{\frac {14}{3}}x+{\frac {25}{6}}x^{2}}$.