How Do Trigonometric and Exponential Functions Connect?

[Trying2Learn]

Hi all:

I really do not know what to ask here, so please be patient as I get a little too "spiritual" (for want of a better word). (This could be a stupid question...)

I get this: e^iθ=cosθ+isinθ
And it is beautiful.

I am struck by the fact that the trig functions manifest harmonic (repetitive) behavior; and the exponential (without the imaginary part) manifest explosive behavior (growth or sinks).

It sort of seems that this equation "connects" (which, I suppose, may be the core of my question: "connects") harmonic behavior (violin strings, vibrations, etc.), with behaviors that grow (explosions) or diminish.

Could anyone elaborate on this?

 

[FactChecker]

You are hinting at something that many people consider to be the most beautiful mathematics there is. At the heart of Laplace transformations and Fourier transformations is that a function, whether it represents an electronic signal, a velocity, opsition, acceleration, etc. can be separated into different frequencies by seeing how it matches it with different exponential functions. (see https://en.wikipedia.org/wiki/Laplace_transform and https://en.wikipedia.org/wiki/Fourier_transform ). Once that is done, the subject of dynamic behavior and stability can be done by analyzing the response of systems and components to different frequencies.

 

[JTC]

You are hinting at something that many people consider to be the most beautiful mathematics there is. At the heart of Laplace transformations and Fourier transformations is that a function, whether it represents an electronic signal, a velocity, opsition, acceleration, etc. can be separated into different frequencies by seeing how it matches it with different exponential functions. (see https://en.wikipedia.org/wiki/Laplace_transform and https://en.wikipedia.org/wiki/Fourier_transform ). Once that is done, the subject of dynamic behavior and stability can be done by analyzing the response of systems and components to different frequencies.

I understand what you are saying and recognize the practical applications of this.

But I am looking for something that might shed some light on the philosophy (or beauty) that this equation connects "harmonic repetitions" with "explosive singularities."

Is it possible that in the physical world, there is something that connects singularities and repetitions?

I think...

 

[FactChecker]

True repititions (cyclic behavior) does not change its amplitude from one cycle to the next. It has a gain of 1. On the other hand, simple exponential gain does not depend on any cyclic behavior. The complex exponential function separates those two aspects cleanly into the gain: $e^{real(z)}$ and the cyclic behavior with gain 1: $e^{i*Im(z)}$. So yes, there is a clear connection between the complex exponential function and the behavior (cyclic, exponential growth, etc.) of a system. This is a deep subject that is studied in great detail in Fourier analysis, stability and control, Laplace transforms, etc.

 

[Trying2Learn]

True repititions (cyclic behavior) does not change its amplitude from one cycle to the next. It has a gain of 1. On the other hand, simple exponential gain does not depend on any cyclic behavior. The complex exponential function separates those two aspects cleanly into the gain: $e^{real(z)}$ and the cyclic behavior with gain 1: $e^{Im(z)}$. So yes, there is a clear connection between the complex exponential function and the behavior (cyclic, exponential growth, etc.) of a system. This is a deep subject that is studied in great detail in Fourier analysis, stability and control, Laplace transforms, etc.

That was what I was looking for. Thanks!

 

[fresh_42]

Hi all:

I really do not know what to ask here, so please be patient as I get a little too "spiritual" (for want of a better word). (This could be a stupid question...)

I get this: e^iθ=cosθ+isinθ
And it is beautiful.

I am struck by the fact that the trig functions manifest harmonic (repetitive) behavior; and the exponential (without the imaginary part) manifest explosive behavior (growth or sinks).

It sort of seems that this equation "connects" (which, I suppose, may be the core of my question: "connects") harmonic behavior (violin strings, vibrations, etc.), with behaviors that grow (explosions) or diminish.

Could anyone elaborate on this?

This is a question one could possibly write an entire book about. I'll try to summarize one chapter of this book.

What we have on the right hand side is a point on the unit circle in $\mathbb{R}^2$, where the axis are labeled by $1 \cdot \mathbb{R}$ and $i \cdot \mathbb{R}$. This is a geometric interpretation, expressed in polar coordinates. Now a circle is the solution of a differential equation: $y'=-\dfrac{x}{y}\,.$ That is the point where the secret is hidden. Let me use a standard way to solve this equation.

We start with setting $y=e^{f(x)}$ and get $y'=f\,'e^f = -xe^{-f}$ which yields $0=2f\,'e^{2f}+2x=(e^{2f})' +2x$ and with $g := e^{2f}$ we have $g'=-2x\,.$ If we integrate this, we get $e^{2f}=g=-x^2+c$ or $y^2=-x^2+c$ which is the equation for our circle. So the question comes down to: Why does an ansatz with the exponential function solve a differential equation?

If we look at the definition of the exponential function, namely the one by its functional formula $e^{a+b}=e^a \cdot e^b$, then we see, that it translates linear (additive) pairs $(a,b)$ into curved (multiplicative) pairs $(e^a,e^b)$. It is what differentiation does in the opposite direction: $(f\cdot g)'=f\,'g+ fg'$. So differentiation and exponentiation are in a way the opposite of each other.

So the answer to your question is: Because of the fundamental and defining functional formula of the exponential function, it is suited to solve differential equations in general, and especially that of a circle. Thus the points on the circle $(\cos \theta , \sin \theta)$ as solutions to a differential equation correspond to the exponentiation $e^{i \theta}$ as seen in the complex plane.

This is one possibility to look at it.

 

[Trying2Learn]

This is a question one could possibly write an entire book about. I'll try to summarize one chapter of this book.

What we have on the right hand side is a point on the unit circle in $\mathbb{R}^2$, where the axis are labeled by $1 \cdot \mathbb{R}$ and $i \cdot \mathbb{R}$. This is a geometric interpretation, expressed in polar coordinates. Now a circle is the solution of a differential equation: $y'=-\dfrac{x}{y}\,.$ That is the point where the secret is hidden. Let me use a standard way to solve this equation.

We start with setting $y=e^{f(x)}$ and get $y'=f\,'e^f = -xe^{-f}$ which yields $0=2f\,'e^{2f}+2x=(e^{2f})' +2x$ and with $g := e^{2f}$ we have $g'=-2x\,.$ If we integrate this, we get $e^{2f}=g=-x^2+c$ or $y^2=-x^2+c$ which is the equation for our circle. So the question comes down to: Why does an ansatz with the exponential function solve a differential equation?

If we look at the definition of the exponential function, namely the one by its functional formula $e^{a+b}=e^a \cdot e^b$, then we see, that it translates linear (additive) pairs $(a,b)$ into curved (multiplicative) pairs $(e^a,e^b)$. It is what differentiation does in the opposite direction: $(f\cdot g)'=f\,'g+ fg'$. So differentiation and exponentiation are in a way the opposite of each other.

So the answer to your question is: Because of the fundamental and defining functional formula of the exponential function, it is suited to solve differential equations in general, and especially that of a circle. Thus the points on the circle $(\cos \theta , \sin \theta)$ as solutions to a differential equation correspond to the exponentiation $e^{i \theta}$ as seen in the complex plane.

This is one possibility to look at it.

Wow... thank you! Perfect.

 

[haushofer]

Without using Taylor series, I motivate Euler's identity to students the following way: imagine you want to define

<br /> f(x) = e^{ix} , \ \ \ \ \ \ \ i^2 \equiv -1 \,.<br />

What should it be? Well, if you want the same rules for differentiation as for real exponentials, you expect that

<br /> f&#039;(x) = ie^{ix} \sim e^{ix}<br />

What's the most general function you can write down which is proportional to itself after differentiation? It's something like (check! This is the subtle part!)

<br /> f(x) = e^{ix} \equiv A \sin{(x)} + B \cos{(x)}.<br />

with A and B constants. Filling in $x=0$ implies $B=1$. Also,

<br /> f&#039;(x) = A \cos{(x)} - B \sin{(x)} \equiv i \Bigl( A \sin{(x)} + B \cos{(x)}. \Bigr)<br />

This implies $A=i$.

 

[forcefield]

<br /> f(x) = e^{ix} \equiv A \sin{(x)} + B \cos{(x)}.<br />

with A and B constants. Filling in $x=0$ implies $B=1$.

Only if one wants
<br /> e^{i0} \equiv 1<br />

 

[haushofer]

Only if one wants
<br /> e^{i0} \equiv 1<br />

Well, yes, but 0*i=0 (the 'real' zero) in the complex plane, and e^0=1, so I took that for granted ;)