What Happens When You Differentiate Euler's Formula?

[ADDA]

https://en.wikipedia.org/wiki/Euler's_formula

(1) e^ix = cos(x) + isin(x)
(2) e^ixidx = (-sin(x) + icos(x))dx
(3) e^ix = (-sin(x) + icos(x)) / i
(4) e^ix = cos(x) + isin(x)
Just lost in circles.

Just for fun.. post a solution for x.

 

[Mark44]

https://en.wikipedia.org/wiki/Euler's_formula

(1) e^ix = cos(x) + isin(x)
(2) e^ixidx = (-sin(x) + icos(x))dx
(3) e^ix = (-sin(x) + icos(x)) / i
(4) e^ix = cos(x) + isin(x)
Just lost in circles.

Technically, you didn't differentiate equation 1 -- you took the differential of each side.
If you differentate equation 1 with respect to x, you get
(2) e^ixi = -sin(x) + icos(x)

If you divide both sides by i, and do the usual trick of multiplying $\frac{-\sin(x)}{i}$ by $\frac {-i}{-i}$, you get what you would expect.

Just for fun.. post a solution for x.

Euler's formula is an identity -- it's true for all x, so it's not possible to solve for x.

Added: We know that $\frac d {dx} e^x = e^x$, and you have found that $\frac d {dx} e^{ix} = e^{ix}$ as well. Is that so surprising?

 

[mathman]

Is there a question?

 

[ADDA]

Take item (3) and square both sides:

e^2ix = -1 * (-sin(x) + icos(x))²

expanding the right side:

sin²(x) - i2sin(x)cos(x) - cos²(x)

apply the double angle identity

- isin(2x) - (cos²(x) - sin²(x))

apply the double angle identity

-cos(2x) - isin(2x)

Did I miss the sign above?

​

 

[Mark44]

Take item (3) and square both sides:

e^2ix = -1 * (-sin(x) + icos(x))²

expanding the right side:

sin²(x) - i2sin(x)cos(x) - cos²(x)

No.
You have two sign errors in the line above.

You should have gotten $-\sin^2(x) + 2i \sin(x)\cos(x) + \cos^2(x)$, which can be simplified to $\cos^2(x) - \sin^2(x) + i \cdot 2\sin(x)\cos(x) = \cos(2x) + i \sin(2x)$.
From Euler's Formula directly, $e^{2ix} = \cos(2x) + i\sin(2x)$

apply the double angle identity

- isin(2x) - (cos²(x) - sin²(x))

apply the double angle identity

-cos(2x) - isin(2x)

Did I miss the sign above?

​

 

[ADDA]

Thanks Mark44, I've been experimenting with trying to change the radix of a fast Fourier transform. My solution for x is this:

Since squaring both sides could equal:

e^ix2 = e^ix2

equating the exponents of e would yield

ix² = i * x * 2;

-x² = 2 * i * x;

so x = -2 * i.

Now if I wanted to do a radix three fast Fourier transform, would I cube both sides of item three above?

e^ix3 = ((-sin(x) + icos(x)) / i)³

 

[Mark44]

Thanks Mark44, I've been experimenting with trying to change the radix of a fast Fourier transform. My solution for x is this:

Since squaring both sides could equal:

e^ix2 = e^ix2

No. Squaring $e^{ix}$ does not result in $e^{ix^2}$.
Squaring $e^{ix}$ results in $\left(e^{ix}\right)^2$ which can be simplified to $e^{2ix}$

equating the exponents of e would yield

ix² = i * x * 2;

No, since you have a mistake in your previous work. Taking out the factor of i, you're saying that $x^2 = 2x$, which is true only for x = 2 or x = 0.

-x² = 2 * i * x;

so x = -2 * i.

Again, no, for the reasons stated above.

Now if I wanted to do a radix three fast Fourier transform, would I cube both sides of item three above?

e^ix3 = ((-sin(x) + icos(x)) / i)³

Why don't you back up and see if you can get the work right for squaring? You've had several mistakes throughout this thread.

 

[ADDA]

"To the extent that a fool knows his foolishness, He may be deemed wise. A fool who considers himself wise is indeed a fool."

Thanks again Mark44, modens ponens.

One other question, say I perform a radix 9 transform, in the transform data, would a peak magnitude index represent 1.0 / 9.0 of a usual radix two index which would be 1.0 / 2.0 the transform data?

My discrete formula is this:

index = (frequency * frame_size) / sample_rate

so I'm thinking I would need

index = (frequency * frame_size) / (radix * sample_rate)

I'm able to somewhat reproduce the original audio, and have a video example. Is there a way to upload a video to your site without an external link?

 

[Mark44]

One other question, say I perform a radix 9 transform, in the transform data, would a peak magnitude index represent 1.0 / 9.0 of a usual radix two index which would be 1.0 / 2.0 the transform data?

My discrete formula is this:

index = (frequency * frame_size) / sample_rate

so I'm thinking I would need

index = (frequency * frame_size) / (radix * sample_rate)

I'm not seeing what this has to do with your original question about Euler's formula. The word "radix" has different meanings, with one of them being the symbol √ for square roots, but more commonly referred to as "radical sign."

Are you asking about taking square roots, cube roots, ninth roots, etc.?

 

[ADDA]

Technically, again I am wrong. I looked back in the book. A decimation in time fast Fourier transform starts with many radix-2 transforms then many radix-4 transforms, then many radix-8, continually until frame size / 2 is reached. So I'm attempting a base 3 or base 9 or base 5 transform, I've just termed it radix. So, I've modified the process to allow many radix-3 transforms to many radix-9 transforms to many radix-27 transforms...

 

[Mark44]

Technically, again I am wrong. I looked back in the book. A decimation in time fast Fourier transform starts with many radix-2 transforms then many radix-4 transforms, then many radix-8, continually until frame size / 2 is reached. So I'm attempting a base 3 or base 9 or base 5 transform, I've just termed it radix. So, I've modified the process to allow many radix-3 transforms to many radix-9 transforms to many radix-27 transforms...

You haven't explained what you mean by a radix-2/radix-4-radix-8 transform, nor what you mean by a base-3, base-9, or base-5 transform, so I have no idea what you are saying here.

Also what do you mean by "decimation in time fast Fourier transform"?
The literal meaning of decimation is the removal of every 10th element.

 

[fresh_42]

Technically, again I am wrong. I looked back in the book. A decimation in time fast Fourier transform starts with many radix-2 transforms then many radix-4 transforms, then many radix-8, continually until frame size / 2 is reached. So I'm attempting a base 3 or base 9 or base 5 transform, I've just termed it radix. So, I've modified the process to allow many radix-3 transforms to many radix-9 transforms to many radix-27 transforms...

Are you talking somehow about a discrete FFT?