B Question about Euler's formula

AI Thread Summary
Euler's formula connects complex exponentials with trigonometric functions, but confusion arises when evaluating specific values like cos(2π). The discussion clarifies that using complex exponentiation, e^(2iπ) simplifies to 1, aligning with the known value of cos(2π). It emphasizes the importance of understanding complex numbers and their properties rather than relying solely on calculators. The equation's significance extends beyond mere numerical evaluation, as it plays a crucial role in fields like Fourier transforms and communications systems. Ultimately, the formula illustrates deeper mathematical principles rather than just providing direct numerical outputs.
jaydnul
Messages
558
Reaction score
15
Euler gave us the below equations:

1694459421013.png


But this doesn't actually give me a number value for where the y value is when you plug in a number for x. For example, if i plug in 2pi for x, i know cosx should be 1. But that equation gives me (e^2i*pi +e^-2i*pi)/2. This doesn't give me 1. So what really is the point of the equation if you have to use the taylor series representation of sin and cos to interpret the results in the first place?
 
Physics news on Phys.org
jaydnul said:
Euler gave us the below equations:

View attachment 331823

But this doesn't actually give me a number value for where the y value is when you plug in a number for x. For example, if i plug in 2pi for x, i know cosx should be 1. But that equation gives me (e^2i*pi +e^-2i*pi)/2. This doesn't give me 1. So what really is the point of the equation if you have to use the taylor series representation of sin and cos to interpret the results in the first place?
Perhaps there's more to mathematics than plugging numbers into an equation?
 
  • Like
Likes vanhees71, Vanadium 50 and Dale
jaydnul said:
what really is the point of the equation
The point is that it is true.

It turns out to be useful in the Fourier transform, and important for communications systems.
 
jaydnul said:
Euler gave us the below equations:

View attachment 331823

For example, if i plug in 2pi for x, i know cosx should be 1. But that equation gives me (e^2i*pi +e^-2i*pi)/2. This doesn't give me 1.
Perhaps it is time to learn the rules for complex exponentiation. Windows Calculator does not know about complex numbers. So it cannot do the job for you.

Let us first work on ##\cos x = Re(e^{ix}) = \frac{e^{ix} + e^{-ix}}2## for ##x=2\pi##.

In particular, let us work on evaluating ##e^{2i\pi}##.

That exponent is a complex number. Its real part is zero. Its imaginary part is ##2\pi##.

When you raise a [real] number to a complex power, you use the real part of the exponent and the imaginary part of the exponent differently.

1. You raise the real part of the root (##e## in this case) to the power of the real part of the exponent. In this case, that gives you ##e^0 = 1##. That is ##(1 + 0i)##

2. You place that result on the complex plane and rotate it through the angle given by the imaginary part. In this case, the rotation angle is ##2\pi##. So there is effectively no rotation.

You conclude that ##e^{2i\pi} = 1 + 0i##.

You can repeat the process and conclude that ##e^{-2i\pi} = 1 + 0i##.

Now substitute back into the formula and evaluate ##\frac{e^{ix} + e^{-ix}}2 = \frac{1 + 1}{2} = 1##

By no coincidence, ##\cos 2\pi = 1##.

jaydnul said:
So what really is the point of the equation if you have to use the taylor series representation of sin and cos to interpret the results in the first place?
Instead of using De Moivre's formula as I did above, you could use the Taylor series. Did you actually try that? I would expect some extremely helpful cancellation.
 
Last edited:
  • Like
  • Love
Likes vanhees71, topsquark, nasu and 2 others
Consider an extremely long and perfectly calibrated scale. A car with a mass of 1000 kg is placed on it, and the scale registers this weight accurately. Now, suppose the car begins to move, reaching very high speeds. Neglecting air resistance and rolling friction, if the car attains, for example, a velocity of 500 km/h, will the scale still indicate a weight corresponding to 1000 kg, or will the measured value decrease as a result of the motion? In a second scenario, imagine a person with a...
Dear all, in an encounter of an infamous claim by Gerlich and Tscheuschner that the Greenhouse effect is inconsistent with the 2nd law of thermodynamics I came to a simple thought experiment which I wanted to share with you to check my understanding and brush up my knowledge. The thought experiment I tried to calculate through is as follows. I have a sphere (1) with radius ##r##, acting like a black body at a temperature of exactly ##T_1 = 500 K##. With Stefan-Boltzmann you can calculate...
Thread 'Griffith, Electrodynamics, 4th Edition, Example 4.8. (First part)'
I am reading the Griffith, Electrodynamics book, 4th edition, Example 4.8 and stuck at some statements. It's little bit confused. > Example 4.8. Suppose the entire region below the plane ##z=0## in Fig. 4.28 is filled with uniform linear dielectric material of susceptibility ##\chi_e##. Calculate the force on a point charge ##q## situated a distance ##d## above the origin. Solution : The surface bound charge on the ##xy## plane is of opposite sign to ##q##, so the force will be...
Back
Top