1. **Problem:** Show that $$\frac{x - a}{a} + \frac{1}{2} \left( \frac{a - x}{a} \right)^2 + \frac{1}{3} \left( \frac{a - x}{a} \right)^3 = \log a - \log x$$. 2. **Formula and Explanation:** This expression resembles the expansion of the logarithm function using Taylor series around $a$. Recall the Taylor series for $\log x$ about $a$: $$\log x = \log a + \frac{1}{a}(x - a) - \frac{1}{2a^2}(x - a)^2 + \frac{1}{3a^3}(x - a)^3 - \cdots$$ Rearranging terms and dividing by $a$ as in the problem, we can write: $$\log a - \log x = -\frac{x - a}{a} - \frac{1}{2} \left( \frac{x - a}{a} \right)^2 - \frac{1}{3} \left( \frac{x - a}{a} \right)^3 + \cdots$$ 3. **Intermediate Work:** Note that the problem uses $\frac{a - x}{a}$ instead of $\frac{x - a}{a}$. Since $\frac{a - x}{a} = -\frac{x - a}{a}$, substitute this into the series: $$\frac{x - a}{a} + \frac{1}{2} \left( \frac{a - x}{a} \right)^2 + \frac{1}{3} \left( \frac{a - x}{a} \right)^3 = \frac{x - a}{a} + \frac{1}{2} \left(-\frac{x - a}{a}\right)^2 + \frac{1}{3} \left(-\frac{x - a}{a}\right)^3$$ Simplify powers of $-1$: $$= \frac{x - a}{a} + \frac{1}{2} \left( \frac{x - a}{a} \right)^2 - \frac{1}{3} \left( \frac{x - a}{a} \right)^3$$ 4. **Compare with the series for $\log x - \log a$:** The Taylor expansion of $\log x$ about $a$ is: $$\log x = \log a + \frac{1}{a}(x - a) - \frac{1}{2a^2}(x - a)^2 + \frac{1}{3a^3}(x - a)^3 - \cdots$$ Dividing both sides by $1$ and rearranging: $$\log x - \log a = \frac{x - a}{a} - \frac{1}{2} \left( \frac{x - a}{a} \right)^2 + \frac{1}{3} \left( \frac{x - a}{a} \right)^3 - \cdots$$ 5. **Final Step:** The expression in the problem is the negative of this series, so: $$\frac{x - a}{a} + \frac{1}{2} \left( \frac{a - x}{a} \right)^2 + \frac{1}{3} \left( \frac{a - x}{a} \right)^3 = \log a - \log x$$ **Answer:** The given expression equals $\log a - \log x$ as required. --- 5. **Problem:** If $\alpha, \beta, \gamma$ are roots of $$x^3 - 6x^2 + 11x - 6 = 0,$$ find $\sum \alpha^2$. 6. **Formula and Explanation:** For a cubic equation $x^3 + px^2 + qx + r = 0$ with roots $\alpha, \beta, \gamma$: - Sum of roots: $\alpha + \beta + \gamma = -p$ - Sum of products of roots two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = q$ - Product of roots: $\alpha\beta\gamma = -r$ We want $\sum \alpha^2 = \alpha^2 + \beta^2 + \gamma^2$. Recall: $$\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)$$ 7. **Intermediate Work:** Given equation is: $$x^3 - 6x^2 + 11x - 6 = 0$$ So, $p = -6$, $q = 11$, $r = -6$. Calculate: $$\alpha + \beta + \gamma = 6$$ $$\alpha\beta + \beta\gamma + \gamma\alpha = 11$$ 8. **Calculate $\sum \alpha^2$:** $$\sum \alpha^2 = 6^2 - 2 \times 11 = 36 - 22 = 14$$ **Answer:** $\sum \alpha^2 = 14$. --- 9. **Problem:** Find the $n^{th}$ derivative of $e^x$. 10. **Formula and Explanation:** The function $e^x$ is unique because its derivative is itself. 11. **Intermediate Work:** - First derivative: $\frac{d}{dx} e^x = e^x$ - Second derivative: $\frac{d^2}{dx^2} e^x = e^x$ By induction, the $n^{th}$ derivative is: $$\frac{d^n}{dx^n} e^x = e^x$$ **Answer:** The $n^{th}$ derivative of $e^x$ is $e^x$. --- **Summary:** - Problem 1 proved the logarithmic expansion. - Problem 5 found $\sum \alpha^2 = 14$ for the cubic roots. - Problem 9 found the $n^{th}$ derivative of $e^x$ is $e^x$.