1. **Problem statement:** Find the first few coefficients $C_n$ of the Taylor series for $f(x) = \ln(\sec(x))$ centered at $a=0$, where the series is given by $\sum C_n x^n$. 2. **Recall the function and series:** We want to expand $f(x) = \ln(\sec(x))$ around $x=0$. Note that $\sec(x) = \frac{1}{\cos(x)}$. 3. **Rewrite the function:** $$ f(x) = \ln(\sec(x)) = \ln\left(\frac{1}{\cos(x)}\right) = -\ln(\cos(x)) $$ 4. **Use the Taylor series for $\cos(x)$:** $$ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots $$ 5. **Expand $\ln(\cos(x))$ using the series for $\ln(1+u)$:** Let $u = \cos(x) - 1 = -\frac{x^2}{2} + \frac{x^4}{24} - \cdots$ The Taylor series for $\ln(1+u)$ is: $$ \ln(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \frac{u^4}{4} + \cdots $$ 6. **Substitute $u$ and simplify terms up to $x^6$:** - First term: $u = -\frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720}$ - Second term: $-\frac{u^2}{2} = -\frac{1}{2} \left(-\frac{x^2}{2} + \frac{x^4}{24}\right)^2 + \cdots$ Calculate $u^2$ up to $x^6$: $$ \left(-\frac{x^2}{2}\right)^2 = \frac{x^4}{4}, \quad 2 \times \left(-\frac{x^2}{2}\right) \times \frac{x^4}{24} = -\frac{x^6}{24} $$ So, $$ u^2 = \frac{x^4}{4} - \frac{x^6}{12} + \cdots $$ Then, $$ -\frac{u^2}{2} = -\frac{1}{2} \left(\frac{x^4}{4} - \frac{x^6}{12}\right) = -\frac{x^4}{8} + \frac{x^6}{24} $$ - Third term: $\frac{u^3}{3}$ is of order $x^6$ and higher. Approximate $u^3$ using $u \approx -\frac{x^2}{2}$: $$ u^3 \approx \left(-\frac{x^2}{2}\right)^3 = -\frac{x^6}{8} $$ So, $$ \frac{u^3}{3} \approx \frac{-x^6/8}{3} = -\frac{x^6}{24} $$ 7. **Sum the terms:** $$ \ln(\cos(x)) \approx u - \frac{u^2}{2} + \frac{u^3}{3} = \left(-\frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720}\right) + \left(-\frac{x^4}{8} + \frac{x^6}{24}\right) + \left(-\frac{x^6}{24}\right) $$ Combine like terms: - $x^2$ term: $-\frac{x^2}{2}$ - $x^4$ term: $\frac{x^4}{24} - \frac{x^4}{8} = -\frac{x^4}{12}$ - $x^6$ term: $-\frac{x^6}{720} + \frac{x^6}{24} - \frac{x^6}{24} = -\frac{x^6}{720}$ 8. **Recall $f(x) = -\ln(\cos(x))$:** $$ f(x) = -\ln(\cos(x)) \approx \frac{x^2}{2} + \frac{x^4}{12} + \frac{x^6}{720} $$ 9. **Conclusion:** The first few coefficients $C_n$ in the Taylor series $\sum C_n x^n$ are: - $C_0 = 0$ - $C_1 = 0$ - $C_2 = \frac{1}{2}$ - $C_3 = 0$ - $C_4 = \frac{1}{12}$ - $C_5 = 0$ - $C_6 = \frac{1}{720}$ These coefficients correspond to the even powers only, as expected from the even function. **Final answer:** $$ f(x) = \ln(\sec(x)) = \frac{x^2}{2} + \frac{x^4}{12} + \frac{x^6}{720} + \cdots $$