18.2 Exp,sin,cos[2D7]

E405

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

$$e^z={\sum }_{k=0}^{\infty }\frac{1}{k!}{z}^k$$

for $z\in ℂ$. We want to reflect on this definition.

• First, for each $z\in ℂ$, we can actually define

  $$f(z)={\sum }_{k=0}^{\infty }\frac{1}{k!}{z}^k$$

  (Note that the radius of convergence is infinite — as it easily occurs using the root criterion 216).

• 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 $z,w\in ℂ$.

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

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

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

Hidden solution: [UNACCESSIBLE UUID ’1M4’]

E405

[1M5]Prerequisites:3.

Given $z\in ℂ$, show that

$$\underset{N\to \infty }{lim}{(1+\frac{z}{N})}^N=e^z$$

406

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

E405

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

Hidden solution: [UNACCESSIBLE UUID ’1M8’]

E405

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

E405

[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’]

E405

[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

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

• Check that

  $${\cosh }^{\prime }=\sinh ,{\sinh }^{\prime }=\cosh$$

• Check the formulas

  $$\sinh (x+y)=\cosh x\sinh y+\cosh y\sinh x$$
  $$\cosh (x+y)=\cosh x\cosh y+\sinh y\sinh x.$$