Timeline for Homogeneity of a function within a region
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 19, 2019 at 12:58 | comment | added | Hugh | We need more clarity in what you are doing. What you call your first function seems to be a list of trajectories plotted within a circular region. This is not a function but a list of trajectories. A function would have the form f[r,theta] = ... | |
| Feb 19, 2019 at 9:08 | comment | added | Alper91 | Because it will pass the circle discretely. The spiral rotates around the z axis by 3 degrees at a time. The function is in polar r=e^((θ+θr)*0.5) θ: -Pi to Pi and θr: 0 to 2Pi. Intersection circle is: {(Exp[Pi*0.5] + 1)/2, 0} with radius (Exp[Pi*0.5] - 1)/2 in the first function. | |
| Feb 19, 2019 at 8:40 | comment | added | Hugh | Best would be to get an expression for the function within the region. Then take its gradient. Can you give the expression for the function? It is not clear to me why you are representing your function as a set of lines. | |
| Feb 19, 2019 at 7:18 | comment | added | Alper91 | @Hugh Do you think the areas between the lines may help me with my purpose? I edited the question. | |
| Feb 19, 2019 at 7:16 | history | edited | Alper91 | CC BY-SA 4.0 |
better explanation
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| Feb 18, 2019 at 12:00 | comment | added | Alper91 | @Hugh Yes, actually change in the gradient is a good point to prove homogenity. I will edit the question. | |
| Feb 18, 2019 at 11:58 | comment | added | Hugh | We need a definition of what you mean by homogeneous. Perhaps the gradient is constant or the gradient is approximately constant compared to...? | |
| Feb 18, 2019 at 11:19 | history | asked | Alper91 | CC BY-SA 4.0 |