B Question about Euler's formula

[jaydnul]

Euler gave us the below equations:

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?

 

[PeroK]

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?

 

[Dale]

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.

 

[jbriggs444]

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, $\cos 2\pi = 1$.

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.