1. **Problem statement:** Find the Taylor series expansions for the given functions at the specified points. --- ### Part 2: Taylor series at 0 (Maclaurin series) **i.** Find the Taylor series of $e^{3x}$ at 0. - The Taylor series of $e^u$ at 0 is $$e^u = \sum_{n=0}^\infty \frac{u^n}{n!}$$ - Here, $u=3x$, so $$e^{3x} = \sum_{n=0}^\infty \frac{(3x)^n}{n!} = \sum_{n=0}^\infty \frac{3^n x^n}{n!}$$ **Answer:** $$e^{3x} = 1 + 3x + \frac{9x^2}{2} + \frac{27x^3}{6} + \cdots$$ --- **ii.** Find the Taylor series of $x \sin(2x^3)$ at 0. - Recall the Taylor series for $\sin z$ at 0: $$\sin z = \sum_{k=0}^\infty (-1)^k \frac{z^{2k+1}}{(2k+1)!}$$ - Substitute $z = 2x^3$: $$\sin(2x^3) = \sum_{k=0}^\infty (-1)^k \frac{(2x^3)^{2k+1}}{(2k+1)!} = \sum_{k=0}^\infty (-1)^k \frac{2^{2k+1} x^{6k+3}}{(2k+1)!}$$ - Multiply by $x$: $$x \sin(2x^3) = \sum_{k=0}^\infty (-1)^k \frac{2^{2k+1} x^{6k+4}}{(2k+1)!}$$ **Answer:** $$x \sin(2x^3) = 2x^4 - \frac{8x^{10}}{3!} + \frac{32x^{16}}{5!} - \cdots$$ --- **iii.** Find the Taylor series of $\ln(1 - x^2)$ at 0. - Recall the Taylor series for $\ln(1 - t)$ at 0: $$\ln(1 - t) = - \sum_{n=1}^\infty \frac{t^n}{n}$$ for $|t|<1$ - Substitute $t = x^2$: $$\ln(1 - x^2) = - \sum_{n=1}^\infty \frac{x^{2n}}{n} = -x^2 - \frac{x^4}{2} - \frac{x^6}{3} - \cdots$$ **Answer:** $$\ln(1 - x^2) = -x^2 - \frac{x^4}{2} - \frac{x^6}{3} - \cdots$$ --- **iv.** Find the Taylor series of $\sqrt{1 - \frac{x^2}{3}}$ at 0. - Use the binomial series for $(1 + u)^\alpha$ with $\alpha = \frac{1}{2}$ and $u = -\frac{x^2}{3}$: $$ (1 + u)^{1/2} = \sum_{n=0}^\infty \binom{1/2}{n} u^n $$ - The binomial coefficients for half-integers: $$\binom{1/2}{n} = \frac{(1/2)(1/2 - 1) \cdots (1/2 - n + 1)}{n!}$$ - First few terms: $$\sqrt{1 - \frac{x^2}{3}} = 1 + \frac{1}{2} \left(-\frac{x^2}{3}\right) + \frac{\frac{1}{2}(-\frac{1}{2})}{2!} \left(-\frac{x^2}{3}\right)^2 + \cdots$$ Simplify stepwise: $$= 1 - \frac{x^2}{6} - \frac{1}{8} \frac{x^4}{9} + \cdots = 1 - \frac{x^2}{6} - \frac{x^4}{72} + \cdots$$ **Answer:** $$\sqrt{1 - \frac{x^2}{3}} = 1 - \frac{x^2}{6} - \frac{x^4}{72} + \cdots$$ --- ### Part 3: Taylor series at indicated points **i.** Find the Taylor series of $1 + x^2 + x^4$ at $a = -2$. - Since this is a polynomial, the Taylor series at $a$ is the polynomial itself expressed in powers of $(x - a)$. - Compute derivatives: $$f(x) = 1 + x^2 + x^4$$ $$f(-2) = 1 + 4 + 16 = 21$$ $$f'(x) = 2x + 4x^3$$ $$f'(-2) = 2(-2) + 4(-2)^3 = -4 - 32 = -36$$ $$f''(x) = 2 + 12x^2$$ $$f''(-2) = 2 + 12(4) = 2 + 48 = 50$$ $$f'''(x) = 24x$$ $$f'''(-2) = 24(-2) = -48$$ $$f^{(4)}(x) = 24$$ $$f^{(4)}(-2) = 24$$ - Taylor series formula: $$f(x) = \sum_{n=0}^4 \frac{f^{(n)}(-2)}{n!} (x + 2)^n$$ - Substitute values: $$= 21 - 36(x+2) + \frac{50}{2} (x+2)^2 - \frac{48}{6} (x+2)^3 + \frac{24}{24} (x+2)^4$$ Simplify coefficients: $$= 21 - 36(x+2) + 25 (x+2)^2 - 8 (x+2)^3 + (x+2)^4$$ **Answer:** $$f(x) = 21 - 36(x+2) + 25 (x+2)^2 - 8 (x+2)^3 + (x+2)^4$$ --- **ii.** Find the Taylor series of $\frac{1}{x^2}$ at $a=1$. - Let $f(x) = x^{-2}$. - Compute derivatives: $$f(x) = x^{-2}$$ $$f(1) = 1$$ $$f'(x) = -2 x^{-3}$$ $$f'(1) = -2$$ $$f''(x) = 6 x^{-4}$$ $$f''(1) = 6$$ $$f'''(x) = -24 x^{-5}$$ $$f'''(1) = -24$$ - Taylor series formula: $$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(1)}{n!} (x - 1)^n$$ - First four terms: $$= 1 - 2(x-1) + \frac{6}{2} (x-1)^2 - \frac{24}{6} (x-1)^3 + \cdots$$ Simplify coefficients: $$= 1 - 2(x-1) + 3 (x-1)^2 - 4 (x-1)^3 + \cdots$$ **Answer:** $$\frac{1}{x^2} = 1 - 2(x-1) + 3 (x-1)^2 - 4 (x-1)^3 + \cdots$$ --- **iii.** Find the Taylor series of $2^x$ at $a=1$. - Recall $2^x = e^{x \ln 2}$. - Derivatives: $$f(x) = 2^x$$ $$f^{(n)}(x) = (\ln 2)^n 2^x$$ - Evaluate at $x=1$: $$f^{(n)}(1) = (\ln 2)^n 2$$ - Taylor series: $$2^x = \sum_{n=0}^\infty \frac{f^{(n)}(1)}{n!} (x-1)^n = \sum_{n=0}^\infty \frac{2 (\ln 2)^n}{n!} (x-1)^n$$ **Answer:** $$2^x = 2 + 2 \ln 2 (x-1) + \frac{2 (\ln 2)^2}{2} (x-1)^2 + \cdots$$ --- **Summary:** - Part 2: Maclaurin series for four functions. - Part 3: Taylor series at given points for three functions.