Implicitly differentiating the vanishing real part of the hyperbolic tangent of one plus the square of the Hardy Z function

  • #1
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Summary:

Let $$Y(t)=tanh(ln(1+Z(t)^2))$$ where Z(t) is the Hardy Z function; How to calculate the implicit derivative of the curve defined by $$Re(Y(t(u)+is(u)))=0$$?
Let $$Y(t)=tanh(ln(1+Z(t)^2))$$ where Z is the Hardy Z function; I'm trying to calculate the pedal coordinates of the curve defined by $$L = \{ (t (u), s (u)) : {Re} (Y (t (u) + i s (u)))_{} = 0 \}$$ and $$H = \{ (t (u), s (u)) : {Im} (Y (t (u) + i s (u)))_{} = 0 \}$$ , and for that I need to calculate the derivative of $$t(u)$$ and $$s(u)$$ for which the parametric form is not available. When I tried to apply the chain rule, I ended up with
$$\frac{dy}{dx} = - \frac{\frac{d}{dx} Y(x + iy)}{\frac{d}{dy}(Y(x + iy))} = i$$
which I don't think is correct. I have written an article about it with more information at The Hyperbolic Tangent of the Logarithm of One Plus The Square of The Hardy Z Function

Here is a composite image generated by the real and imaginary parts of Y from 7004.5-0.5i to 7005.5+0.5i (showing a region of the neighborhood surrounding the first Lehmer pair) generated with some modifications to arblib's complex_plot program :

YrealImagCompositeFrom7004.5-0.5I..7005.5+0.5I.png


Here is the imaginary part vanishing alone over the same range, with aspect ratio maintained
YimagFrom7004.5-0.5I..7005.5+0.5I.png

Here is the same range where the real part vanishes
YrealFrom7004.5-0.5I..7005.5+0.5I.png







Does anyone have any ideas on how to differentiate these curves ? Specifically, where the real part of Y vanishes?

Does there exist an expression for the tangent line of $$L = \{ (t (u), s (u)) : {Re} (Y (t (u) + i s (u))) = 0 \}$$ ?

There is a maple procedure to calculate the path at Maple script to calculate path of curve where the real part of Y vanishes around the 1st root


Actually, I just realized I can use finite differences.. it doesnt appear that they are pedal curves of one another.. or if they are, the pedal point is not the root
 
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Answers and Replies

  • #2
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No replies? you people are too boring and conventional.
 
  • #3
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No replies? you people are too boring and conventional.
Well, this has probably several reasons:
  1. It is too long.
  2. You struck everybody with the pictures so people barely read the rest.
  3. What are your questions? Probably somewhere, but who wants to search for them.
  4. The Hardy function is a very specific tool in number theory, a branch which is not very popular.
  5. Especially not among physicists.
  6. You have several media breaks: changing sites only to be able to understand is a bad method.
  7. You used far too many double $ instead of inline formulas.
  8. Are you sure you didn't simply want to promote your paper? Where is it published?
  9. Our rules require a published reference.
  10. It is too long.
The main reason, however, is likely the lack of interest in problems around the Hardy Z function.
 
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  • #4
berkeman
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Unfortunately, @qbar went nutzoid at the helpful reply by @fresh_42 and that profane rant (now deleted) resulted in qbar's ban. Thread is now closed.
 
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