14. 테일러 급수Taylor Series

Solution. The $n$th derivative of $f(x)=1/x$ is $f^{(n)}(x) = (-1)^n n!\, x^{-(n+1)}$, so $f^{(n)}(2) = (-1)^n n! \cdot 2^{-(n+1)}$. Hence the Taylor series at $x=2$ is $$\sum_{n=0}^\infty \frac{f^{(n)}(2)}{n!}(x-2)^n = \sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}}(x-2)^n.$$

This is a geometric series with first term $1/2$ and common ratio $r=-\dfrac{x-2}{2}$, so it converges when $|r|\lt1$, i.e., $|x-2|\lt2$, with sum $$\frac{1/2}{1-r} = \frac{1/2}{1+\dfrac{x-2}{2}} = \frac{1}{2+(x-2)} = \frac{1}{x} = f(x).$$ Hence the series converges to $f(x)$ exactly when $|x-2|\lt2$, i.e., $0\lt x\lt4$.

Solution.

(1) For $f(x)=e^x$, $f^{(n)}(x)=e^x$ for every $n$, so $f^{(n)}(0)=1$. Hence the Taylor series is $$\sum_{n=0}^\infty \frac{x^n}{n!},$$ and the $n$th-degree Taylor polynomial is $$P_n(x) = \sum_{k=0}^n \frac{x^k}{k!} = 1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}.$$

(2) For $f(x)=\cos x$, the values $f^{(n)}(0)$ for $n=0,1,2,3,4,\ldots$ cycle through $1,0,-1,0,1,\ldots$, so $f^{(n)}(0)=(-1)^k$ when $n=2k$ and $f^{(n)}(0)=0$ when $n$ is odd. Hence the Taylor series is $$\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!},$$ and since only the even-degree terms up to $\lfloor n/2\rfloor$ survive, the $n$th-degree Taylor polynomial is $$P_n(x) = \sum_{k=0}^{\lfloor n/2\rfloor} \frac{(-1)^k x^{2k}}{(2k)!}.$$

Solution.

(1) For $f(x)=x^4+x^2+1$, $$f(x)=x^4+x^2+1,\quad f'(x)=4x^3+2x,\quad f''(x)=12x^2+2,\quad f'''(x)=24x,\quad f^{(4)}(x)=24,$$ and $f^{(n)}(x)=0$ for $n\geq5$. At $x=-2$, $$f(-2)=21,\ f'(-2)=-36,\ f''(-2)=50,\ f'''(-2)=-48,\ f^{(4)}(-2)=24,$$ so $$\frac{f(-2)}{0!}=21,\ \frac{f'(-2)}{1!}=-36,\ \frac{f''(-2)}{2!}=25,\ \frac{f'''(-2)}{3!}=-8,\ \frac{f^{(4)}(-2)}{4!}=1.$$ Since $f$ is a degree-$4$ polynomial, all derivatives of order $5$ and higher vanish, so the Taylor series is the (finite) polynomial $$f(x) = 21-36(x+2)+25(x+2)^2-8(x+2)^3+(x+2)^4.$$

(2) Every derivative of $f(x)=e^x$ is $f^{(n)}(x)=e^x$, so $f^{(n)}(2)=e^2$. Hence the Taylor series at $x=2$ is $$\sum_{n=0}^\infty \frac{e^2}{n!}(x-2)^n = e^2\sum_{n=0}^\infty \frac{(x-2)^n}{n!}.$$

Proof. The $n$th derivative of $f(x)=e^x$ is $f^{(n)}(x)=e^x$. By Taylor's Theorem, for each $n$ there exists some $c$ between $0$ and $x$ such that $$f(x) = \sum_{k=0}^n \frac{f^{(k)}(0)}{k!}x^k + R_n(x),\qquad R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}=\frac{e^c}{(n+1)!}x^{n+1}.$$

Since $c$ lies between $0$ and $x$, $e^c\leq e^{|x|}$, so $$|R_n(x)| \leq e^{|x|}\frac{|x|^{n+1}}{(n+1)!}.$$

Fix $x$ and let $a_n=|x|^n/n!$. Then $\dfrac{a_{n+1}}{a_n}=\dfrac{|x|}{n+1}\to0$, so by the Ratio Test $\sum a_n$ converges, hence $a_n\to0$. Thus $|R_n(x)|\to0$ as $n\to\infty$, for every real number $x$.

Therefore $$\sum_{k=0}^n \frac{f^{(k)}(0)}{k!}x^k = f(x)-R_n(x) \xrightarrow{n\to\infty} f(x)-0=f(x),$$ so the Taylor series of $e^x$ at $x=0$ converges to $f(x)$ for every real number $x$.

Proof.

(1) For $f(x)=\sin x$, $f^{(n)}(x)$ is always one of $\pm\sin x,\pm\cos x$, so $|f^{(n)}(x)|\leq1$ for all $n$ and $x$. By Taylor's Theorem, for each $n$ there is some $c$ between $0$ and $x$ with remainder $R_n(x)=\dfrac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$, so $$|R_n(x)| \leq \frac{|x|^{n+1}}{(n+1)!}.$$ By the same argument as in Example 02 (the Ratio Test), $\dfrac{|x|^{n+1}}{(n+1)!}\to0$ for fixed $x$, so $R_n(x)\to0$. Hence the Taylor series of $\sin x$ at $x=0$ converges to $f(x)$ for every real number $x$.

(2) For $f(x)=\cos x$, likewise $f^{(n)}(x)$ is always one of $\pm\sin x,\pm\cos x$, so $|f^{(n)}(x)|\leq1$, and the same argument gives $R_n(x)\to0$. Hence the Taylor series of $\cos x$ at $x=0$ converges to $f(x)$ for every real number $x$.

Solution. Since $\sin x = x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\cdots$, the polynomial $x-\dfrac{x^3}{3!}$ is also the 4th-degree Taylor polynomial of $\sin x$ (its $x^4$ coefficient is $0$). Hence the remainder term is $$R_4(x) = \frac{f^{(5)}(c)}{5!}x^5,$$ and since $f^{(5)}(x)=\cos x$, $|f^{(5)}(c)|\leq1$. That is, $$\left|\sin x - \left(x-\frac{x^3}{3!}\right)\right| = |R_4(x)| \leq \frac{|x|^5}{5!}.$$

For the error to be less than $3\times10^{-4}$ on $|x|\lt R$, it suffices to have $\dfrac{R^5}{5!}\leq3\times10^{-4}$. Taking $R=\dfrac12$, $$\frac{(1/2)^5}{5!} = \frac{1/32}{120} = \frac{1}{3840} \approx 2.6\times10^{-4} \lt 3\times10^{-4},$$ so $R=\dfrac12$ satisfies the requirement.

Solution. Since $\cos x = 1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\cdots$, the polynomial $1-\dfrac{x^2}{2}$ is also the 3rd-degree Taylor polynomial of $\cos x$ (its $x^3$ coefficient is $0$). The remainder term is $$R_3(x) = \frac{f^{(4)}(c)}{4!}x^4,$$ and since $f^{(4)}(x)=\cos x$, $|f^{(4)}(c)|\leq1$. Hence $$\left|\cos x - \left(1-\frac{x^2}{2}\right)\right| = |R_3(x)| \leq \frac{|x|^4}{4!}.$$

Since $|x|\lt\dfrac12$, the bound on the error is $$\frac{(1/2)^4}{4!} = \frac{1/16}{24} = \frac{1}{384} \approx 0.0026.$$

Proof.

(1) Suppose the power series $\displaystyle f(x)=\sum_{n=0}^\infty a_n x^n$ converges for $|x|\lt R$. A power series may be differentiated term by term within its interval of convergence, so for each $k\geq0$, $$f^{(k)}(x) = \sum_{n=k}^\infty n(n-1)\cdots(n-k+1)\,a_n x^{n-k}, \qquad |x|\lt R.$$ Setting $x=0$, only the $n=k$ term on the right survives, giving $$f^{(k)}(0) = k!\,a_k.$$ Hence $a_k = \dfrac{f^{(k)}(0)}{k!}$, which means the Taylor series $\displaystyle\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$ agrees term by term with the original series $\displaystyle\sum_{n=0}^\infty a_n x^n$. That is, the Taylor series of $f(x)$ is $f(x)$ itself.

(2) Since $\cos x = \displaystyle\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}$, $$x\cos x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n)!} = x - \frac{x^3}{2!}+\frac{x^5}{4!}-\cdots,$$ so $$2x+x\cos x = 2x + x + \sum_{n=1}^\infty \frac{(-1)^n x^{2n+1}}{(2n)!} = 3x + \sum_{n=1}^\infty \frac{(-1)^n x^{2n+1}}{(2n)!}.$$ Hence $$f(x) = \frac13(2x+x\cos x) = x + \sum_{n=1}^\infty \frac{(-1)^n x^{2n+1}}{3(2n)!} = x-\frac{x^3}{6}+\frac{x^5}{72}-\frac{x^7}{2160}+\cdots.$$

(3) Since $e^{ix}=\cos x+i\sin x$, $e^x\cos x = \operatorname{Re}\!\left(e^x e^{ix}\right) = \operatorname{Re}\!\left(e^{(1+i)x}\right)$. Writing $1+i=\sqrt2\,e^{i\pi/4}$, we get $(1+i)^n = 2^{n/2}\left(\cos\dfrac{n\pi}{4}+i\sin\dfrac{n\pi}{4}\right)$, and taking the real part of $$e^{(1+i)x} = \sum_{n=0}^\infty \frac{(1+i)^n}{n!}x^n$$ gives $$e^x\cos x = \sum_{n=0}^\infty \frac{2^{n/2}\cos(n\pi/4)}{n!}x^n = \sum_{n=0}^\infty \frac{(\sqrt2)^n\cos(n\pi/4)}{n!}x^n.$$ Computing the coefficients for $n=0,1,2,3,4$ gives $1,\ 1,\ 0,\ -\dfrac13,\ -\dfrac16$, so $$e^x\cos x = 1+x-\frac{x^3}{3}-\frac{x^4}{6}+\cdots.$$

Solution.

(1) $e^{-x^2+\ln5} = e^{\ln5}\cdot e^{-x^2} = 5e^{-x^2}$, and $e^{-x^2} = \displaystyle\sum_{n=0}^\infty \frac{(-x^2)^n}{n!} = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{n!}$, so $$e^{-x^2+\ln5} = 5\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{n!}.$$

(2) $\cos^2x = \dfrac{1+\cos2x}{2}$, and $\cos2x = \displaystyle\sum_{n=0}^\infty \frac{(-1)^n(2x)^{2n}}{(2n)!} = \sum_{n=0}^\infty \frac{(-1)^n2^{2n}x^{2n}}{(2n)!}$, so $$\cos^2x = \frac12+\frac12\sum_{n=0}^\infty \frac{(-1)^n2^{2n}x^{2n}}{(2n)!} = \frac12+\frac12+\frac12\sum_{n=1}^\infty \frac{(-1)^n2^{2n}x^{2n}}{(2n)!} = 1+\sum_{n=1}^\infty \frac{(-1)^n2^{2n-1}x^{2n}}{(2n)!}.$$

(3) $\ln(1+x) = \displaystyle\sum_{n=1}^\infty \frac{(-1)^{n-1}x^n}{n}$ and $\ln(1-x) = -\displaystyle\sum_{n=1}^\infty \frac{x^n}{n}$, so $$\ln\!\left(\frac{1+x}{1-x}\right) = \ln(1+x)-\ln(1-x) = \sum_{n=1}^\infty \frac{\left[(-1)^{n-1}+1\right]x^n}{n}.$$ This vanishes for even $n$ and equals $2x^n/n$ for odd $n$, so substituting $n=2k+1$, $$\ln\!\left(\frac{1+x}{1-x}\right) = \sum_{k=0}^\infty \frac{2x^{2k+1}}{2k+1}.$$

(4) Substituting $u=x^2$ into $\tan^{-1}u = \displaystyle\sum_{n=0}^\infty \frac{(-1)^nu^{2n+1}}{2n+1}$ gives $\tan^{-1}x^2 = \displaystyle\sum_{n=0}^\infty \frac{(-1)^nx^{4n+2}}{2n+1}$, so $$x\tan^{-1}x^2 = \sum_{n=0}^\infty \frac{(-1)^nx^{4n+3}}{2n+1}.$$

Solution.

(1) Substituting $u=x^2$ into $\sin u = \displaystyle\sum_{n=0}^\infty \frac{(-1)^nu^{2n+1}}{(2n+1)!}$ gives $$\sin(x^2) = \sum_{n=0}^\infty \frac{(-1)^nx^{4n+2}}{(2n+1)!},$$ so integrating term by term, $$\int\sin(x^2)\,dx = \sum_{n=0}^\infty \frac{(-1)^nx^{4n+3}}{(4n+3)(2n+1)!}+C.$$

(2) Every term of the series above vanishes at $x=0$, so $$\int_0^1\sin(x^2)\,dx = \sum_{n=0}^\infty \frac{(-1)^n}{(4n+3)(2n+1)!}.$$ Computing the first few terms, $$\frac{1}{3\cdot1!}=\frac13,\quad \frac{1}{7\cdot3!}=\frac1{42},\quad \frac{1}{11\cdot5!}=\frac1{1320},\quad \frac{1}{15\cdot7!}=\frac1{75600}.$$ This is an alternating series whose terms decrease monotonically to $0$, so the error of a partial sum is bounded by the magnitude of the next term. Summing the first three terms, $$\int_0^1\sin(x^2)\,dx \approx \frac13-\frac1{42}+\frac1{1320} \approx 0.3103,$$ with error at most $\dfrac1{75600}\approx1.3\times10^{-5}\lt0.001$, which meets the required precision.

Proof.

(1) By the definition of the generalized binomial coefficient, $$\binom{-1}{k} = \frac{(-1)(-2)(-3)\cdots(-1-k+1)}{k!} = \frac{(-1)(-2)\cdots(-k)}{k!} = \frac{(-1)^kk!}{k!} = (-1)^k.$$

On the other hand, for $|x|\lt1$, the geometric series gives $$\frac{1}{1+x} = \sum_{n=0}^\infty(-x)^n = \sum_{n=0}^\infty(-1)^nx^n = \sum_{n=0}^\infty\binom{-1}{n}x^n,$$ so $(1+x)^{-1} = \displaystyle\sum_{n=0}^\infty(-1)^nx^n$.

(2) Let $f(x)=(1+x)^{1/2}$. For $n\geq1$, $$f^{(n)}(x) = \frac12\left(\frac12-1\right)\left(\frac12-2\right)\cdots\left(\frac12-n+1\right)(1+x)^{1/2-n},$$ so $$f^{(n)}(0) = \frac12\left(-\frac12\right)\left(-\frac32\right)\cdots\left(\frac12-n+1\right).$$ Hence the Taylor coefficients are $a_n=f^{(n)}(0)/n!=\dbinom{1/2}{n}$, with $a_0=1,\ a_1=\dfrac12,\ a_2=-\dfrac18,\ a_3=\dfrac1{16},\ldots$, so $$(1+x)^{1/2} = \sum_{n=0}^\infty\binom{1/2}{n}x^n = 1+\frac12x-\frac18x^2+\frac1{16}x^3-\cdots.$$

Solution.

(1) $f(x)=\dfrac{x}{\sqrt[3]{1+x}} = x(1+x)^{-1/3}$. By the binomial series, $$(1+x)^{-1/3} = \sum_{n=0}^\infty\binom{-1/3}{n}x^n,$$ and $$\binom{-1/3}{0}=1,\quad \binom{-1/3}{1}=-\frac13,\quad \binom{-1/3}{2}=\frac{(-1/3)(-4/3)}{2!}=\frac29,$$ so $$(1+x)^{-1/3} = 1-\frac13x+\frac29x^2-\cdots.$$ Multiplying by $x$, $$f(x) = x-\frac13x^2+\frac29x^3-\cdots,$$ so the 3rd-degree Taylor polynomial of $f(x)$ is $$P_3(x) = x-\frac13x^2+\frac29x^3.$$

(2) For $f(x)=\sin^{-1}x$, $f'(x)=(1-x^2)^{-1/2}$. Substituting $u=-x^2$ into the binomial series, $$(1-x^2)^{-1/2} = \sum_{n=0}^\infty\binom{-1/2}{n}(-x^2)^n = \sum_{n=0}^\infty\binom{-1/2}{n}(-1)^nx^{2n},$$ and $$\binom{-1/2}{n} = \frac{(-1/2)(-3/2)\cdots(-1/2-n+1)}{n!} = \frac{(-1)^n(2n-1)!!}{2^nn!},$$ so $$\binom{-1/2}{n}(-1)^n = \frac{(2n-1)!!}{2^nn!} = \frac{(2n)!}{4^n(n!)^2}$$ (the last equality uses $(2n-1)!!=\dfrac{(2n)!}{2^nn!}$). Hence $$(1-x^2)^{-1/2} = \sum_{n=0}^\infty \frac{(2n)!}{4^n(n!)^2}x^{2n},$$ and since $\sin^{-1}0=0$, integrating term by term gives $$\sin^{-1}x = \sum_{n=0}^\infty \frac{(2n)!}{4^n(n!)^2(2n+1)}x^{2n+1}.$$ Computing the coefficients for $n=0,1,2,3$ gives $$1,\quad \frac16,\quad \frac3{40},\quad \frac{5}{112},$$ so $$\sin^{-1}x = x+\frac{x^3}6+\frac{3x^5}{40}+\frac{5x^7}{112}+\cdots.$$ Since $\sin^{-1}x$ is an odd function, all its even-degree coefficients vanish, so the 8th-degree Taylor polynomial is $$P_8(x) = x+\frac{x^3}6+\frac{3x^5}{40}+\frac{5x^7}{112}.$$

Proof. Substituting $\alpha=-1/2$ into the binomial series, $$(1-u)^{-1/2} = \sum_{n=0}^\infty\binom{-1/2}{n}(-u)^n,$$ and, as computed in Exercise 04-1(2), $$\binom{-1/2}{n}(-1)^n = \frac{(2n-1)!!}{(2n)!!},$$ so $$(1-u)^{-1/2} = \sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!!}u^n, \qquad |u|\lt1.$$

Setting $u = k^2\sin^2\theta$, since $|k|\lt1$ we have $|u|\leq k^2\lt1$, so $$\frac{1}{\sqrt{1-k^2\sin^2\theta}} = \sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!!}k^{2n}\sin^{2n}\theta.$$ Let $f_n(\theta) = \dfrac{(2n-1)!!}{(2n)!!}k^{2n}\sin^{2n}\theta$. Then $|f_n(\theta)|\leq M_n := \dfrac{(2n-1)!!}{(2n)!!}k^{2n}$ for $0\leq\theta\leq\pi/2$, and $\displaystyle\sum_{n=0}^\infty M_n = (1-k^2)^{-1/2}\lt\infty$. Hence, by the given fact (term-by-term integration), $$K = \int_0^{\pi/2}\sum_{n=0}^\infty f_n(\theta)\,d\theta = \sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!!}k^{2n}\int_0^{\pi/2}\sin^{2n}\theta\,d\theta.$$

By Wallis' formula, $\displaystyle\int_0^{\pi/2}\sin^{2n}\theta\,d\theta = \dfrac{(2n-1)!!}{(2n)!!}\cdot\dfrac\pi2$, so $$K = \sum_{n=0}^\infty \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 k^{2n}\cdot\frac\pi2 = \frac\pi2\left[1+\left(\frac12\right)^2k^2+\left(\frac{1\cdot3}{2\cdot4}\right)^2k^4+\left(\frac{1\cdot3\cdot5}{2\cdot4\cdot6}\right)^2k^6+\cdots\right].$$

Solution. To find the Taylor series of $\ln x$ at $x=1$, substitute $t=x-1$: $$\ln(1+t) = t-\frac{t^2}2+\frac{t^3}3-\cdots,$$ so $$\ln x = (x-1)-\frac{(x-1)^2}2+\frac{(x-1)^3}3-\cdots.$$ Hence $$\frac{\ln x}{x-1} = 1-\frac{x-1}2+\frac{(x-1)^2}3-\cdots,$$ and as $x\to1$, $x-1\to0$, so $$\lim_{x\to1}\frac{\ln x}{x-1} = 1.$$

Solution.

(1) Since $\sin x = x-\dfrac{x^3}6+O(x^5)$ and $\cos x = 1-\dfrac{x^2}2+O(x^4)$, $$\frac1{\cos x} = 1+\frac{x^2}2+O(x^4)$$ (substituting $u=\frac{x^2}2+O(x^4)$ into the geometric series $\frac1{1-u}=1+u+O(u^2)$). Hence $$\tan x = \sin x\cdot\frac1{\cos x} = \left(x-\frac{x^3}6+O(x^5)\right)\left(1+\frac{x^2}2+O(x^4)\right) = x+\frac{x^3}2-\frac{x^3}6+O(x^5) = x+\frac{x^3}3+O(x^5).$$ Therefore $$\sin x-\tan x = \left(x-\frac{x^3}6\right)-\left(x+\frac{x^3}3\right)+O(x^5) = -\frac{x^3}2+O(x^5),$$ so $$\lim_{x\to0}\frac{\sin x-\tan x}{x^3} = -\frac12.$$

(2) $$\frac1{\sin x}-\frac1x = \frac{x-\sin x}{x\sin x}.$$ Since $x-\sin x = \dfrac{x^3}6+O(x^5)$ and $x\sin x = x^2+O(x^4)$, $$\frac{x-\sin x}{x\sin x} = \frac{x^3/6+O(x^5)}{x^2+O(x^4)} = \frac x6+O(x^3),$$ so $$\lim_{x\to0}\left(\frac1{\sin x}-\frac1x\right) = 0.$$

(3) Since $e^x = 1+x+\dfrac{x^2}2+O(x^3)$, $$e^x-(1+x) = \frac{x^2}2+O(x^3),$$ so $$\lim_{x\to0}\frac{e^x-(1+x)}{x^2} = \frac12.$$

(4) Since $\ln(1+x^2) = x^2+O(x^4)$ and $1-\cos x = \dfrac{x^2}2+O(x^4)$, $$\frac{\ln(1+x^2)}{1-\cos x} = \frac{x^2+O(x^4)}{x^2/2+O(x^4)} = \frac{1+O(x^2)}{1/2+O(x^2)} \to \frac1{1/2}=2,$$ so $$\lim_{x\to0}\frac{\ln(1+x^2)}{1-\cos x} = 2.$$

Solution. Since $\sin u = u-\dfrac{u^3}6+O(u^5)$, $$\sin(ax) = ax-\frac{a^3x^3}6+O(x^5).$$ Hence $$\sin(ax)-\sin x-x = \left(ax-\frac{a^3x^3}6\right)-\left(x-\frac{x^3}6\right)-x+O(x^5) = (a-2)x+\frac{1-a^3}6x^3+O(x^5),$$ so $$\frac{\sin(ax)-\sin x-x}{x^3} = \frac{a-2}{x^2}+\frac{1-a^3}6+O(x^2).$$

For this to converge as $x\to0$, the $\dfrac1{x^2}$ term must not blow up, so we need $a-2=0$, i.e., $a=2$. (This is necessary, and it is also sufficient: when $a=2$ the $\dfrac1{x^2}$ term vanishes, and the remaining term converges to $\dfrac{1-a^3}6$.)

So $a=2$, and the resulting limit is $$\frac{1-2^3}6 = \frac{1-8}6 = -\frac76.$$

Solution.

(1) By Euler's formula, $$e^{-i\pi} = \cos(-\pi)+i\sin(-\pi) = -1+0 = -1.$$

(2) $$e^{i\pi/4} = \cos\frac\pi4+i\sin\frac\pi4 = \frac{\sqrt2}2+i\frac{\sqrt2}2 = \frac{\sqrt2}2(1+i).$$

(3) $$e^{1-i\pi/2} = e\cdot e^{-i\pi/2} = e\left(\cos\left(-\frac\pi2\right)+i\sin\left(-\frac\pi2\right)\right) = e(0-i) = -ie.$$

Proof. By Euler's formula, $e^{i\theta}=\cos\theta+i\sin\theta$, and since $\cos$ is even and $\sin$ is odd, $$e^{-i\theta} = \cos(-\theta)+i\sin(-\theta) = \cos\theta-i\sin\theta.$$

(1) $$\frac{e^{i\theta}+e^{-i\theta}}2 = \frac{(\cos\theta+i\sin\theta)+(\cos\theta-i\sin\theta)}2 = \frac{2\cos\theta}2 = \cos\theta.$$

(2) $$\frac{e^{i\theta}-e^{-i\theta}}{2i} = \frac{(\cos\theta+i\sin\theta)-(\cos\theta-i\sin\theta)}{2i} = \frac{2i\sin\theta}{2i} = \sin\theta.$$

(3) For a complex number $z$, $\cosh z := \dfrac{e^z+e^{-z}}2$ by definition, so by part (1), $$\cosh i\theta = \frac{e^{i\theta}+e^{-i\theta}}2 = \cos\theta.$$

(4) For a complex number $z$, $\sinh z := \dfrac{e^z-e^{-z}}2$ by definition, so $$\sinh i\theta = \frac{e^{i\theta}-e^{-i\theta}}2 = i\cdot\frac{e^{i\theta}-e^{-i\theta}}{2i} = i\sin\theta,$$ where the last equality follows from part (2).

Proof. For $x,y\gt0$ and complex numbers $z,w$, define $x^z := e^{z\ln x}$ (where $\ln x$ is the real natural logarithm).

(1) $$x^zx^w = e^{z\ln x}e^{w\ln x} = e^{z\ln x+w\ln x} = e^{(z+w)\ln x} = x^{z+w}$$ (the third equality uses the property $e^ae^b=e^{a+b}$ of the complex exponential function).

(2) Since $e^{-w\ln x}e^{w\ln x}=e^0=1$, we have $e^{-w\ln x}=1/e^{w\ln x}=1/x^w$. Hence $$\frac{x^z}{x^w} = x^z\cdot\frac1{x^w} = e^{z\ln x}\cdot e^{-w\ln x} = e^{(z-w)\ln x} = x^{z-w}.$$

(3) Since $c$ is real, $x^c=e^{c\ln x}$ is a positive real number, and by the property of the real logarithm, $\ln(x^c)=c\ln x$. Hence $$(x^c)^w = e^{w\ln(x^c)} = e^{w\cdot c\ln x} = e^{cw\ln x} = x^{cw}.$$

(4) By the property of the real logarithm, $\ln(xy)=\ln x+\ln y$, so $$(xy)^z = e^{z\ln(xy)} = e^{z(\ln x+\ln y)} = e^{z\ln x+z\ln y} = e^{z\ln x}e^{z\ln y} = x^zy^z.$$