Science:Math Exam Resources/Courses/MATH100 B/December 2024/Question 09 (a)/Solution 1

The second order Taylor polynomial for $\cos (x)$ is given by $$\cos (x)\approx 1-\frac{x^2}{2}.$$ Similarly, the second order Taylor polynomial for $e^y$ is given by $$e^y\approx 1+y+\frac{y^2}{2},$$ and setting $y=3x$ gives $$e^{5x}\approx 1+3x+\frac{9x^2}{2}.$$ Multiplying these polynomials together and truncating at the second order gives $$e^{3x}\cos (x)\approx (1+3x+\frac{9x^2}{2})(1-\frac{x^2}{2})\approx 1+3x+(\frac{9}{2}-\frac{1}{2})x^2=1+3x+4x^2.$$

Alternatively, we could obtain the second Taylor polynomial directly from the definition. Letting $f(x)=e^{3x}\cos (x)$ we compute $$f^{\prime }(x)=e^{3x}(3\cos (x)-\sin (x))\text{and}{f}^{″}(x)=3e^{5x}(3\cos (x)-\sin (x))-e^{3x}(3\sin (x)+\cos (x)).$$ Evaluating at $x=0$ gives $$f(0)=1,f^{\prime }(0)=3,\text{and}{f}^{″}(0)=28.$$ Plugging these values into the formula $T_2(x)=f(0)+f^{\prime }(0)+\frac{f^{″}(0)}{2}$ gives $$e^{3x}\cos (x)\approx 1+3x+4x^2.$$