Previous Newton's Basin Discussion

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Back to Newton's Basin page and current Newton's Basin page

Contents

Checklist

  1. Are the words that you define bold? YES
  2. Do you have references and other interesting links? YES
  3. Did you cite your pictures, or say if you created them? YES
  4. Have you considered all of the comments on the discussion page? YES
  5. Have you looked over everyone else’s pages, and linked to the relevant ones? YES




Chris 7/9

Great page. I bet both the concept and the video will be enticing to my students.

Newton's Method:

  • Are you concerned only with the x-component of the starting point? (It seems so, since you use x as the prime mover in your formula.) Is there a way to clarify that you are concerned either only or mostly with the x-value?
Yes. We are concerned with the x-value for estimating the root, but need the y-value for finding the tangent line. I'll try to make this more clear.

Newton's Basin:

  • I love this sentence; it's so clear: Each complex number is applied to the equation and iterated continually with the output of the previous iteration becoming the input of the next iteration.
  • ⁋4: "Each root has a set of complex numbers x(o) that converge to the root." When I see x(o), I think of one specific number , not a set of numbers.
I see what you mean. I'll re-word that.
  • ⁋4 The shading description needs work. Is it the darker the shading, the more iterations needed to reach the root location?
Yes, darker does mean more iterations to converge. I'll make this more clear

I like having the equation for the image. I could plug it in on Grapher and see the connection between the "real" 2D graph and its complex "map." Is it possible for you to show the equation used to generate the image in the self-similarity section?

Sure, I'll put that up. Ryang1 (7/9)



Anna 7/6

In this math,

f'(z_n) = \frac{\mathrm{\Delta p}}{\mathrm{\Delta z}} = \frac{f(z_n)}{z_n - z_{n+1}}
z_{n+1} = x_n - \frac{f(z_n)}{f'(z_n)}

I'm pretty sure you mean z_{n+1} = z_n - \frac{f(z_n)}{f'(z_n)}

given that it's a recursion relationship (ie  z_{n+1} depends on  z_n

You're right, thanks.

Also, can you move this picture http://mathforum.org/mathimages/index.php/Image:Roots.gif over? I can't read the equation that supposed to be the caption (I think the thumbnail may just be too small).

Sure, I didn't catch that. Ryang1 (7/9)




Maria said ...

Test.

--Mkelly1 15:08, 10 June 2009 (EDT)

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