structure of complex plane a function of equal and opposite exponentials?

[LHanawalt]

This question is part of an amateur (very) cosmology project.

I have noticed that the real number line is the real part of i*ln''(x), the second derivative of the imaginary logarithm. See: http://www.wolframalpha.com/input/?i=i+ln''(x)

It seems to me that the real number line must be the arithmetic(?) mean between e^(ix) and i*ln''(x). Can you verify that I am correct about this?

Is it accurate to say that i*ln(x) is the inverse of e^(ix)?

Is i*ln’’(x) the inverse of –e^(ix), the second derivative of the complex exponential function?

I have other questions but I'll start there.

Larry

 

[Deleted member 4993]

This question is part of an amateur (very) cosmology project.

I have noticed that the real number line is the real part of i*ln''(x), the second derivative of the imaginary logarithm. See: http://www.wolframalpha.com/input/?i=i+ln''(x)

It seems to me that the real number line must be the arithmetic(?) mean between e^(ix) and i*ln''(x). Can you verify that I am correct about this?

Is it accurate to say that i*ln(x) is the inverse of e^(ix)? ............. No

Inverse of e^x is ln(x) → Inverse of e^ix is ln(ix) → ln(i) + ln(x) → i * π/2 + ln(x)

Is i*ln’’(x) the inverse of –e^(ix), the second derivative of the complex exponential function?

I have other questions but I'll start there.

Larry

.

 

[pka]

This question is part of an amateur (very) cosmology project.I have noticed that the real number line is the real part of i*ln''(x), the second derivative of the imaginary logarithm. It seems to me that the real number line must be the arithmetic(?) mean between e^(ix) and i*ln''(x). Can you verify that I am correct about this? Is it accurate to say that i*ln(x) is the inverse of e^(ix)?Is i*ln’’(x) the inverse of –e^(ix), the second derivative of the complex exponential function?

Albeit a half a century ago, I do have a double major in mathematics & philosophy.It is still difficult to see how this is related to a cosmology project.I think you are simply confused about the definitions and/or notations.If you are actually thinking of a complex function $\displaystyle f(z)=i\ln(x)$, then that function is in $\displaystyle u(x,y)+iv(x,y)$ form. Where $\displaystyle u(x,y)=0~\&~v(x,y)=\ln(x)$.But that means that has no derivative because the Cauchy-Riemann equations are not satisfied

Now if you actually mean to use the complex logarithm then $\displaystyle f(z) = i\log (z)$ and the derivatives exists.
$\displaystyle f''(z) = \dfrac{{ - i}}{{{z^2}}}$.

The same statements can be applied to your use of $\displaystyle e^{ix}$.

Do you see your confusion?

 

[LHanawalt]

PKA, yes, thank you very much. Here's the cosmology. I once had an experience of light as constituted by equal and opposite exponentials. I am hoping to articulate accurately a mathematical analogy. The second derivative of the complex exponential function and the second derivative of the complex logarithmic function i*ln(z) seem to work.

It seems to me that -i/z^2 (as the location of the observer, the model-maker, the measurer) might represent the inverse square relationship that we attribute to gravity--that "gravity" might be a function of perception, not force. (The Weber-Fechner Law and other evidence support the idea that all perception might be logarithmic.)

It's crazy, I know--but it would eliminate the need for dark energy and dark matter. And a complex exponential "space" viewed from a complex logarithmic perspective ("time") would give us a flat, steady-state rather than an exponentially expanding universe. (It would also be an anti-entropic universe. All a very hard sell. But since the current macro-model doesn't work, there may be some tolerance for crazy.)

The flat, steady-state aspect is indicated mathematically by the fact that the real number line is the real part of -i/z^2...the real number line representing a mean between exponential and logarithmic curves. http://www.wolframalpha.com/input/?i=i+ln''(z)

At this point I just want to see if the model works mathematically.

Thanks very much for responding.