# 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]  Attachments:
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Posted 3 months ago
 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 Attachments:
Posted 3 months ago
Posted 3 months ago
 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 wonÂ´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
Posted 3 months ago
 It follows a jepg of the text sent before in docx Attachments:
Posted 3 months ago
 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.