Euler’s Handy Formula

Use Euler’s formula in this question without proof.

Theorem 1 (Euler’s Formula). For any $t \in \mathbb R$ ,

$e^{it} = \cos(t) + i \sin(t).$

Remark 1. Setting $t = \pi$ yields Euler’s famous identity $e^{i \pi} + 1 = 0$ .

Problem 1. Let $T$ be a linear transformation. Prove that

$\begin{aligned} T(\cos(t)) &= \frac 12 (T(e^{it}) + T(e^{-it})), \\ T(\sin(t)) &= \frac 1{2i} (T(e^{it}) - T(e^{-it})). \end{aligned}$

(Click for Solution)

Solution. By Euler’s formula,

$\begin{aligned} e^{it} &= \cos(t) + i \sin(t), \\ e^{-it} &= \cos(t) - i \sin(t). \end{aligned}$

Adding (resp. subtracting) then dividing by $2$ (resp. $2i$ ) yields

$\displaystyle \cos(t) = \frac{e^{it} + e^{-it}}{2},\quad \sin(t) = \frac{e^{it} + e^{-it}}{2i}.$

The result follows since $T$ is linear over $\mathbb C$ .

Let $p$ be any polynomial and $\alpha \in \mathbb C$ be a constant. Define the linear transformation $\displaystyle \frac 1{p(\mathcal D)}$ by

$\displaystyle \frac 1{p(\mathcal D)}(e^{\alpha t} \cdot V(t)) := e^{\alpha t} \cdot \frac 1{p(\mathcal D + \alpha)} (V(t)).$

Furthermore, define $\displaystyle \frac 1{p(\mathcal D)}(1) := \frac 1{p(0)}$ if $p(0) \neq 0$ and $\displaystyle \frac 1{\mathcal D}(1) := t$ .

Problem 2. For any polynomial $F$ such that $F(\alpha) \neq 0$ , evaluate $\displaystyle \frac 1{F(\mathcal D)}(e^{\alpha x})$ .

(Click for Solution)

Solution. By the given result using $V(x) = 1$ ,

$\displaystyle \frac 1{p(\mathcal D)}(e^{\alpha t} \cdot 1) = e^{\alpha t} \cdot \frac 1{p(\mathcal D + \alpha)} (1).$

Defining the polynomial $q$ by $q(x) := p(x+\alpha)$ , we have $q(0) = p(\alpha) \neq 0$ , so that the right-hand side simplifies to

$\displaystyle \frac 1{p(\mathcal D)}(e^{\alpha t}) = e^{\alpha t} \frac 1{q(\mathcal D)} (1) = e^{\alpha t} \cdot \frac 1{q(0)} = e^{\alpha t} \cdot\frac 1{p(\alpha)} = \frac 1{p(\alpha)} \cdot e^{\alpha t}.$

Problem 3. For any polynomial $F$ such that $F(-k^2) \neq 0$ , evaluate $\displaystyle \frac 1{F(\mathcal D^2)}(\cos kt)$ and $\displaystyle \frac 1{F(\mathcal D^2)}(\sin kt)$ .

(Click for Solution)

Solution. By Problem 1,

$\begin{aligned} \frac 1{F(\mathcal D^2)}(\cos kt) &= \frac 12 \cdot \left( \frac 1{F(\mathcal D^2)} (e^{(ik)t}) + \frac 1{F(\mathcal D^2)} (e^{(-ik)t}) \right) \\ &= \frac 12 \cdot \left( \frac 1{G(\mathcal D)} (e^{(ik)t}) + \frac 1{G(\mathcal D)} (e^{(-ik)t}) \right) \end{aligned}$

where we defined the polynomial $G$ by $G(x) := F(x^2)$ . We observe that

$G(\pm ik) = F((\pm ik)^2) = F((ik)^2) = F(i^2 \cdot k^2) = F(-k^2) \neq 0.$

By Problem 2,

$\begin{aligned} \frac 1{F(\mathcal D^2)}(\cos kt) &= \frac 12 \cdot \left( \frac 1{G(\mathcal D)} (e^{(ik)t}) + \frac 1{G(\mathcal D)} (e^{(-ik)t}) \right) \\ &= \frac 12 \cdot \left( \frac 1{G(ik)} (e^{(ik)t}) + \frac 1{G(-ik)} (e^{(-ik)t}) \right) \\ &= \frac 12 \cdot \left( \frac 1{F(-k^2)} (e^{(ik)t}) + \frac 1{F(-k^2)} (e^{(-ik)t}) \right) \\ &= \frac 1{F(-k^2)} \left(\frac {e^{i(kt)} + e^{-i(kt)}}{2} \right) = \frac 1{F(-k^2)} \cos(kt). \end{aligned}$

In an analogous manner, $\displaystyle \frac 1{F(D^2)}(\sin kt) = \frac 1{F(-k^2)} \sin(kt)$ .

Problem 4. Evaluate $\displaystyle \frac 1{\mathcal D^2 + k^2}(\cos kt)$ and $\displaystyle \frac 1{\mathcal D^2 + k^2}(\sin kt)$ .

(Click for Solution)

Solution. We will take advantage of the factorisation $z^2 + k^2 = (z-ik)(z+ik)$ . We observe that

$\begin{aligned} \frac{1}{\mathcal D^2 + k^2}(e^{ikt}) &= \frac{1}{(\mathcal D - ik)(\mathcal D + ik)}(e^{ikt}) \\ &= \frac{1}{\mathcal D - ik}\left( \frac{1}{\mathcal D + ik} (e^{ikt}) \right) \\ &= \frac{1}{\mathcal D - ik} \left( \frac{1}{ik + ik} (e^{ikt}) \right) \\ &= \frac 1{2ik} \cdot \frac{1}{\mathcal D - ik} (e^{ikt}) \\ &= \frac 1{2ik} \cdot e^{ikt} \cdot \frac{1}{(\mathcal D + ik) - ik} (1) \\ &= \frac 1{2ik} \cdot e^{ikt} \cdot \frac{1}{\mathcal D}(1) = \frac 1{2ik} te^{ikt}. \end{aligned}$

Similarly, $\displaystyle \frac{1}{\mathcal D^2 + k^2}(e^{-ikt}) = -\frac 1{2ik} te^{-ikt}$ . By Problem 1,

$\begin{aligned} \frac{1}{\mathcal D^2 + k^2}(\cos kt) &= \frac 12 \cdot \left(\frac{1}{\mathcal D^2 + k^2}(e^{ikt}) + \frac{1}{\mathcal D^2 + k^2}(e^{-ikt}) \right) \\ &= \frac 12 \cdot \left( \frac 1{2ik} te^{ikt} -\frac 1{2ik} te^{-ikt} \right) \\ &= \frac {t}{2k} \cdot \left( \frac {e^{ikt} - e^{-ikt}}{2i} \right) = \frac {t \sin kt}{2k}.\end{aligned}$

Analogously, $\displaystyle \frac{1}{\mathcal D^2 + k^2}(\sin kt) = -\frac {t \cos kt}{2k}$ .

Problem 5. Suppose $\mathcal L$ is a linear transformation such that for any constant $\alpha \in \mathbb C$ ,

$\displaystyle \mathcal L \{e^{\alpha t} \} = \frac 1{s-\alpha }.$

For any constant $a \in \mathbb R$ , evaluate the real-valued functions $\mathcal L \{\cos(at)\}$ and $\mathcal L \{\sin(at)\}$ .

(Click for Solution)

Solution. Replacing $\alpha$ with $i a$ ,

$\begin{aligned}\mathcal L \{e^{i a t} \} = \frac 1{s-ia} &= \frac {s+ia}{(s-ia)(s+ia) } \\ &= \frac{s+ia}{s^2 + a^2} = \frac{s}{s^2 + a^2} + i \cdot \frac{a}{s^2 + a^2}. \end{aligned}$

By Euler’s formula and the linearity of $\mathcal L$ ,

$\mathcal L \{ e^{i a t} \} = \mathcal L\{ \cos(at) \} + i \cdot \mathcal L\{ \sin(at)\}$

Hence,

$\displaystyle \mathcal L\{ \cos(at) \} + i \cdot \mathcal L\{ \sin(at)\} = \frac{s}{s^2 + a^2} + i \cdot \frac{a}{s^2 + a^2}.$

Comparing real and imaginary parts,

$\displaystyle \mathcal L \{ \cos(at) \} = \frac{s}{s^2 + a^2},\quad \mathcal L\{ \sin(at) \} = \frac{a}{s^2 + a^2}.$

Remark 2. One can explore the feasibility of these definitions for trigonometric functions defined using complex arguments:

$\displaystyle \cos(z) = \frac{e^{iz} + e^{-iz}}{2},\quad \sin(z) = \frac{e^{iz} - e^{-iz}}{2i}.$

I haven’t done so, so perhaps that’s an open question.

—Joel Kindiak, 19 May 25, 2113H

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The Inverse-D Recipe – KindiakMath

May 19, 2025 at 10:22 pm

[…] a somewhat more elegant proof, of some of these identities, see this post. Finally, to see that this inverse- recipe really works, let’s revisit our tedious […]

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Euler’s Handy Formula

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