# Double Helix and Euler's Formula

1. May 6, 2015

### Jimster41

I was just trying for the ninetieth time to try and understand the sort of physicality Euler's formula and the role $i$ plays in the way so many equations work - it's always been such an obstacle to reading so many of them. I mean what the heck is that $i$ doing there. Progress, maybe, but... no luck

so it's a totally well-known thing, or a completely coincidental irrelevance that the double helix looks like the plot of euler's formula I'd never seen Euler plotted this way, and it did sort give me a visual of the way ${ e }^{ ix }$ works I had not had before.

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

And I gotta ask this question before I'm going to be able to keep studying.

2. May 6, 2015

### Staff: Mentor

The plot of Euler's formula that you refer to is really a single helix, not a double helix.

3. May 6, 2015

### Jimster41

I was looking at the two orthogonal axes, the lighter gray trends of the Re and Imaginary projections? That's a double helix isn't it?

I guess I was just looking for an answer as to why it looks like that those, if there was any connection, therefore to Euler's formula. Like, I was expecting someone to chime in with something over my head but confirming, "yes yes, the mathematical behavior described by ${ e }^{ ix }$ describes accurately the the something something process of base pair chirality something something of the tertiary structure, due to optimization of the Gibbs free energy of the PI bonds something something." Or for someone to say... "what are you talking about?"

So now I have one data point.

The overall quest I'm currently on is to get a feel for what ${ e }^{ ix }$ means as mechanism, which is just so ubiquitous. I can feel what an integral means, what a derivative, a square root, etc, I am even starting to get a feel for what the ${ U }^{ a }{ \nabla }_{ a }{ U }^{ b }$ means. But the darn ${ e }^{ ix }$ is just so perplexing. I mean I know euler's equation. I took signals and systems and did the Fourier transform using it, but I was just following orders. I want to understand more intuitively what it's operational function is. My only cartoon at present is that it "induces periodic aspect".

I thought maybe this was an opportunity to nail that down in a way that I could keep as a picture in my head.

Last edited: May 6, 2015
4. May 6, 2015

### Staff: Mentor

No, those are two sinusoidal waves that are perpendicular to each other. The helix is the red curve. As $x$ increases, $e^{ix}$ goes around in a circle, forming a helix. And it's just one helix, not two.

Not really, no.

This would be my response.

5. May 6, 2015

### Jimster41

I was imagining that the Helix might represent the minimum of some energy minimization function circling, or centered inside of those other two.

Anyway, now I know.

Carry on sir.

6. May 6, 2015

### billy_joule

EDIT: Too late. I received no "Other have posted in this thread, would you like to review them before posting?" type message, does PF not have this feature?

The gray trends are 2d projections of the helix. They are just sin waves (out of phase by pi/2). A helix is three dimensional, DNA is the classic example of a double helix, this is another striking example:

7. May 6, 2015

### Jimster41

Yeah, that's beautiful. Singapore. Gotta show that to the wife. She's got an Asia uber-vacation plan going. I know Singapore is on the short list.

8. May 6, 2015

### Jimster41

I did know that part, it's the "just" in the sentence that that gives me trouble.

9. May 6, 2015

### Jimster41

If you follow the helix, and let it rotate the axes, then aren't they a double helix. I guess that's what my eye was doing, filling in those little bars going around in a chase after the red trend. Seemed to make a DNA thing.

10. May 6, 2015

### Staff: Mentor

A double helix would have two red lines. This is a single helix.

11. May 6, 2015

### Jimster41

Starting at bottom left and going up to top right. I think if you allow the red single helix to rotate the orthogonal coordinates you, you get a double helix from the grey lines. But maybe I'm imagining it.

12. May 6, 2015

### Staff: Mentor

Then you don't have the plot of $e^{i \phi}$ any more. I don't get the point.

13. May 6, 2015

### Jimster41

Dude I just thought that that graph was pretty clever, the way it overlaid a track (the red helix). I started with the first crossbar and started filling them in following it with my eye. At the end It sure seemed like double helix. I wasn't thinking very carefully about the steps, or if was defeating the purpose. I thought that what the little rungs were for.

Last edited: May 6, 2015
14. May 6, 2015

### Jimster41

If you start with bottom left-most line connecting the two light grey sine waves. Go to the middle of it then go to the middle of the next one, down the middle of n imagined identical connectors, then do the same to the middle of the next one, all the way through to the last one, it traces out a helix with a radius of 1/2 the red helix, running through a double helix. No?​

15. May 6, 2015