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Limit of a derivative to the imaginary

Posted 3 months ago
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In the attempt to calculate the limit of a derivative to imaginary x I found a limit of the radius of a derivative function of 2pir, no matter how big the radius the limit of the function will allwys tend to 0.81i(imaginary ), further investigating it i found to be usefull in making a wormhole from two hyperboloids separated from each other just be equaling the equation to negative Pi, it opens a hole in between the hyperboles and comunicate it one to the other. I further wanted to manipulate the value of the right part of the equation to open up a possibility of seeing that the expansion of the hole leads to a finite /infinite no border limit universe, but it is just a sketch. Hope you see the novelty aproach to resolve a derivative of a complex number in a new manner.

c=300000000
n=RandomInteger[100,{200}]
r=RandomReal[5.29*10^-11,{200}]
f=(((Pi+1)*r)*Sqrt[(-2*Pi*r)/((Pi+1)*r)])/((Sqrt[(2*Pi*r)^2+2*Pi*r]))
g=2*Pi*(r+f)
gg=2*Pi*r
h=((c-c*Power[c, (c)^-1])/f)^2
Plot[f^-1,{f,-Pi,Pi}]
t=Table[Log[f,g],Pi]
Plot[t,{t,-Pi,Pi}]
ParametricPlot[g^-1*gg,{g,-Pi,Pi}]
ContourPlot[f*g,{f,0,4Pi},{g,0,4Pi}]
ContourPlot3D[f*h-gg^2==Pi,{f,-2Pi,2Pi},{h,-2Pi,2Pi},{gg,-2Pi,2Pi}] 
Play[f,{f,0,90}]

c=300000000
n=RandomInteger[100,{200}]
r=RandomReal[5.29*10^-11,{200}]
f=(((Pi+1)*r)*Sqrt[(-2*Pi*r)/((Pi+1)*r)])/((Sqrt[(2*Pi*r)^2+2*Pi*r]))
g=2*Pi*(r+f)
gg=2*Pi*r
h=((c-c*Power[c, (c)^-1])/f)^2
Plot[f^-1,{f,-Pi,Pi}]
t=Table[Log[f,g],Pi]
Plot[t,{t,-Pi,Pi}]
ParametricPlot[g^-1*gg,{g,-Pi,Pi}]
ContourPlot[f*g,{f,0,4Pi},{g,0,4Pi}]
ContourPlot3D[f*h-gg^2==-Pi,{f,-2Pi,2Pi},{h,-2Pi,2Pi},{gg,-2Pi,2Pi}] 
Play[f,{f,0,90}]
![a way to avoid singulaities][1]![making th value of the equation equals negative pi\]\[2\![enter image description here][2]
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8 Replies

It follows an attached file in doc x for the explanation of the following computation in the lines of the program that explain the purpose of this article

i can't analyze your work at the moment i'm in a hurry. is y y or is y lamda near that 2Pi? i'm unsure what (^+x) is (handwriting wise). you didn't write "Lim" anywhere. you didn't specify the initial problem or if the solution would move toward a different form or a value. it's a little difficult to read. also we work with .nb not .docx (i cannot read docx) (or just posted code) here on Communities (see posting rules). also i can't confirm that your plotting method did not effect the outcome. I can't tell if your comlex or real plane analysis is right because i cannot even follow the arrows on the page (the objectives the steps achieve are not said, the arrows and bars are not used so that one unfamiliar with your problem can easily follow step by step)

but i wanted to add a few tips (having no idea if you'd like to see them):

my advice is if you used L'Hopital's rule recheck it (or don't use it, use the correct methods): L'Hopital's has many cases it yields invalid results for. it is only "a trick" for reducing time to solve in simple cases.

as for imaginary i, if you have f(x) and i is required to solve it, answer may or may not include i, i'm sure you know. but if you replace x with (x*w) and trace through the equation in parallel you'll see it's a different method of using i.

as far as integrating differential equations and having i, you'll see i can dissapear while resolving constants of integration

Hi John, if I may call you for your first name, since I use computation i usually finish the calculation on the lines of the program, I revised the method of deriving x to the limit of i ( imaginary ), i know it is a little messy ,y is just a substitue for x when x + y still remains as zero, so i use it to substitute the value of what would be zero and change it to "i", then i follow by substituting values that are related either to y or to x in the definition of limit, so that i get a total value of 0/0 in such a way that it would be represenative of the minimum division possible , but then i exchange y to an atribute of the function y=2Pi x ,and keep on changing one value for the other since x+y=o it would not change the relation o zero to zero, it gets me a negative y whcih would be the same as considering the square root of negative 1 to be negative 1 so it becomes negative y that still does not hurt the relations in the equations. I turned the limit of x to zero to x to -y so it becomes the limit of x to the imaginary . After that it is just algebra. At the bottom left i then rewrite the limit ox x to i. After that i reintroduce the valure of r , the raidus, to the equation and after computating it gives a a value that no matter how much i use greater values of r the computated value remains 0.81...i. So it is a computational proof that there is an imaginary limit to the equation. Just use the lines of the program and change the value of r ( greater number than what i used, i used the radius of hydrogen electron) and you will see that the expression of f will remains at low 0.81...i , and won´t grow to any higher number. What i am about to do is to right the general formula for that derivative so that it can be apllied to other functions. Tanhk you for your interest and your doubts. I have very few time to work on math since i am just a hobbiest, i work as a physcian on ocupational health and i am preparing my self for exams. Sorry for the doc x document i will try to make another file extension available

It follows a jepg of the text sent before in docx

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The jpeg of the docx text is out of order.the order is first page bottom ,second page top, last page the one in the Midler.

I used l'hopital without knowing It was l'hopital.

In honor to truth i must say that by making a little mistake on the computation there was a limit stablished other wise the value of the equation is simply i( imaginary). By creating a differential proportion to 2pi^2+pi,elevating to 2 it is stablished a limit no matter how great the number of r.

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