19.2 Exp,sin,cos[2D7]

E406

[1M3]Prerequisites:2,1, 2, 3.It is customary to define

\[ e^ z =∑_{k=0}^∞ \frac 1{k!} z^ k \]

for \(z∈ℂ\). We want to reflect on this definition.

• First, for each \(z∈ℂ\), we can actually define

  \[ f(z) =∑_{k=0}^∞ \frac 1{k!} z^ k \]

  (Note that the radius of convergence is infinite — as it easily occurs using the root criterion 217).

• We note that \(f(0)=1\); we define \(e=f(1)\) which is Euler’s number ¹

• Show that \(f(z+w)=f(z)f(w)\) for \(z,w∈ℂ\).

• It is easy to verify that \(f(x)\) is monotonic increasing for \(x∈(0,∞)\); by the previous relation, \(f(x)\) is monotonic increasing for \(x∈ℝ\).

• Then show that, for \(n,m{\gt}0\) integer, \(f(n/m) = e^{n/m}\) (for the definition of \(e^{n/m}\) see 2).

• Deduce that, for every \(x∈ℝ\), \(f(x) = e^{x}\) (for the definition of \(e^{x}\) see 3)

Hidden solution: [UNACCESSIBLE UUID ’1M4’]

E406

[1M5]Prerequisites:3.

Given \(z∈ℂ\), show that

\begin{equation} \lim _{N→∞}\left(1+\frac z N\right)^ N=e^ z\label{eq:lim_ 1+l/n^ n} \end{equation}

407

and that the limit is uniform on compacts sets. Hidden solution: [UNACCESSIBLE UUID ’1M6’]

E406

[1M7]If \(z=x+iy\) with \(x,y∈ℝ\), then we can express the complex exponential as a product \(e^{z}=e^ xe^{iy}\). Use power series developments to show Euler’s identity \(e^{iy}=\cos y + i \sin y~ .\)

Hidden solution: [UNACCESSIBLE UUID ’1M8’]

E406

[1M9]Conversely, note then that  \(\cos y =\frac{e^{iy}+e^{-iy}} 2~ ~ ,~ ~ \sin y =\frac{e^{iy}-e^{-iy}}{i2}.\)

E406

[1MB]Use the above formula to verify the identities

\[ \sin (x+y) = \cos x \sin y + \cos y \sin x \]

\[ \cos (x+y) = \cos x \cos y - \sin y \sin x \]

Hidden solution: [UNACCESSIBLE UUID ’1MC’]

E406

[1MD]We define the functions hyperbolic cosine ²

\[ \cosh y =\frac{e^{y}+e^{-y}} 2 \]

and hyperbolic sine

\[ \sinh y =\frac{e^{y}-e^{-y}}{2}. \]

• Verify that

  \[ (\cosh x) ^ 2 - (\sinh x) ^ 2 =1 \]

  (which justifies the name of ”hyperbolic”).

• Prove the validity of these power series expansion

  \[ \cosh (x) = 1+\frac 1 2 x^ 2+ \frac 1{4!} x^ 4+\frac 1{6!} x^ 6+ \ldots \]
  \[ \sinh (x) = x+\frac 1{3!} x^ 3+ \frac 1{5!} x^ 5+\frac 1{7!} x^ 7+ \ldots \]

• Check that

  \[ \cosh '=\sinh ~ ~ ,~ ~ \sinh '=\cosh \]

• Check the formulas

  \[ \sinh (x+y) = \cosh x \sinh y + \cosh y \sinh x ~ ~ \]
  \[ \cosh (x+y) = \cosh x \cosh y + \sinh y \sinh x ~ . \]