Find the Taylor series of 3 sin(x)/cos^2(x) + sin(2x) at x = 0 up to degree 5.

Step-by-step solution

1. 1

   Turn the quotient into a product of known series. Direct repeated differentiation of $3\sin x/\cos^2 x$ is painful, so write it as $3\sin x\cdot\sec^{2}x$ and use two standard Maclaurin expansions: $\sin x=x-\frac{x^{3}}{6}+\frac{x^{5}}{120}+O(x^{7}),\qquad \sec^{2}x=1+x^{2}+\frac{2}{3}x^{4}+O(x^{6}).$

2. 2

   Multiply, keeping only terms up to x^5. Collecting the products whose total degree is at most $5$: $x\cdot 1=x$; $x\cdot x^{2}=x^{3}$; $-\frac{x^{3}}{6}\cdot 1=-\frac{x^{3}}{6}$; $x\cdot\frac23x^{4}=\frac23x^{5}$; $-\frac{x^{3}}{6}\cdot x^{2}=-\frac{x^{5}}{6}$; $\frac{x^{5}}{120}\cdot 1=\frac{x^{5}}{120}$. Hence $\sin x\,\sec^{2}x=x+\frac56x^{3}+\left(\frac23-\frac16+\frac1{120}\right)x^{5}+O(x^{6})=x+\frac56x^{3}+\frac{91}{120}x^{5}+O(x^{6}).$

3. 3

   Scale by 3. Multiplying through gives $\frac{3\sin x}{\cos^{2}x}=3x+\frac{5}{2}x^{3}+\frac{91}{40}x^{5}+O(x^{6}).$

4. 4

   Expand the second term by substitution. Replacing $x$ by $2x$ in the sine series, $\sin(2x)=2x-\frac{(2x)^{3}}{6}+\frac{(2x)^{5}}{120}+O(x^{7})=2x-\frac43x^{3}+\frac{4}{15}x^{5}+O(x^{7}).$

5. 5

   Add the two series coefficient by coefficient. The $x$ terms give $3+2=5$; the $x^{3}$ terms give $\frac52-\frac43=\frac{15-8}{6}=\frac76$; the $x^{5}$ terms give $\frac{91}{40}+\frac{4}{15}=\frac{273+32}{120}=\frac{305}{120}=\frac{61}{24}$. Only odd powers survive, as expected since $f$ is odd.

6. 6

   Check the polynomial numerically. Evaluating $f(x)-\left(5x+\frac76x^{3}+\frac{61}{24}x^{5}\right)$ at $x=0.01,\,0.02,\,0.05$ gives $-7.5\times10^{-11}$, $-2.40\times10^{-9}$ and $-2.34\times10^{-7}$; dividing each by $x^{7}$ gives roughly the same constant, confirming the error is $O(x^{7})$ and no lower-order term was dropped.