The reason why we reach Zeros at each natural logarithm and why some Zeta functions have zeros at 0.5. mainly because of the natural logarithm for the imaginary unit number [i] = pi/2.
This is far from a proof, and log(z) is a multi valued function, so ln(i)=π/2i+2πki. I don't know you may have a more precise written expression, but to call it a proof is rather bold. Interesting stuff though
Ln(i) = pi/2 *i + 2pi * i * k , please note that the term that is representing actual value for the imaginary unit value is the first part(pi/2* i) the second part in the term is actually gave you none imaginary values if added because it will be in full fold and based on the value of K you get the position of the fold. But to get the value for (i) the least fold you can get this value is by ln(i)= pi*i/2 only the second part if you had extra folds. To proof my point try (e) to the power of (pi*i/2+ 2*pi*0.25*i) , if k = 0.25 you will not going to get imaginary number at all. The seconnd part for folding only ,the part that control the imaginary value is (pi/2)
@@shaimaasoltan5334 Thanks for your explanation. I dont really understand what you mean, but for k=0.3 you get soemthing with real and imaginary. Dunno, I almost agree with you in this, but I think I'm really understanding it
Thank you so much for your comment and veiwingmy video; I also took some time to grasp the folding concept; one thing to check here on your equalition is, if the mathematic difference equal zero then they are mathimatilcally equal but they are not; Ln(i)-pi/2-2*pi*i*k not equal zero. It is not giving natural log value of (i) it only describe the concept of domain folding. At k=0.3 it means you are doing partial folding but not the value of the imaginary unit number it self . The imaginary unit number (i) value in the complex plane is at each value pi/2 or -pi/2, this the complex plane definition for (i). Because (i) is a unit in orthognal folding
☺️final part to add to this comment hope it will clear the confusion. There is a difference between the imaginary numbers and the imaginary UNIT NUMBER ; I think in your equality talking about the imaginary numbers (folding) and not the UNIT NUMBER (I) Ln(imaginary number) not equal ln(UNIT NUMBER) Your equation is correct for the imaginary numbers but not for the UNIT NUMBER ln(i) equal pi/2 i Same as int 1 not equal to 1+0i
0:55 what happened here e^{ipi} = -1 then it becomes 1??!! Also you have to be careful with the complex log, it’s not as simple as doing exp(ln(z)) with complex numbers as it’s a multivalued function in C
Thank you @ramy4697 for your comment you are correct for -1 the text should be the same as the equation -1, Will add the correction to the Video Thank you. But for the log part, it is natural log exponent for base e, so by difinition they are a function and its inverse so the complex number will not going to change only flipping the singes for the complex number(z) because of the exponent rotation effect. Please check the equation in 13 decimal place and 15 decimal places and let me know if you see any case that it is not matching the equation I used Geogebra , if you have another tool to check please do and let me know
@@shaimaasoltan5334 the complex exponential function is not injective, it doesn’t necessarily have an inverse function, you have to restrict the domain at least, check the Wikipedia page for more information
@@ramy4697 Thank you for your comment Ramy, my work combines both functions (e & ln) together at the same time and not doing separate analysis for each function properties alone. If you apply both ( e and ln) both together not each one alone hope this clears any confuse. The buty of this way is that its domain is any natural number positive or negative (shown in the video fot T value changes). I will be more than happy to discuss with you any counter case for any equation.(The Domain have been tested for all real number values for T ) I only included the Natural values in the Video record.
@@shaimaasoltan5334 the way you use exp(ln(z)) = z has to be justified as z is a complex number, this relation is not true for all complex numbers, because exp is not injective for all numbers in C, multiple inputs can give the same value, which means it has no inverse in this domain, unless you restrict it, for example if z= exp(i(pi/4)) then z = exp(i(pi/4+2pi) since 2pi is a full rotation, which shows that the complex exponential is not injective so it doesn’t have an inverse for the domain C
1 is not equal to e to the i* pi. This is an error in your work or a typo.
Awesome work man!!! I would check out the Zeta Orbital Eq. Next!
Thank you! by the way my name is Shaimaa Soltan and I am Canadian woman leaves currently in Mississauga
please check my github as well
This is far from a proof, and log(z) is a multi valued function, so ln(i)=π/2i+2πki. I don't know you may have a more precise written expression, but to call it a proof is rather bold. Interesting stuff though
Thank you, appreciate your opinion please review other posted videos for the proof as well
Ln(i) = pi/2 *i + 2pi * i * k , please note that the term that is representing actual value for the imaginary unit value is the first part(pi/2* i) the second part in the term is actually gave you none imaginary values if added because it will be in full fold and based on the value of K you get the position of the fold. But to get the value for (i) the least fold you can get this value is by ln(i)= pi*i/2 only the second part if you had extra folds. To proof my point try (e) to the power of (pi*i/2+ 2*pi*0.25*i) , if k = 0.25 you will not going to get imaginary number at all. The seconnd part for folding only ,the part that control the imaginary value is (pi/2)
@@shaimaasoltan5334 Thanks for your explanation. I dont really understand what you mean, but for k=0.3 you get soemthing with real and imaginary. Dunno, I almost agree with you in this, but I think I'm really understanding it
Thank you so much for your comment and veiwingmy video; I also took some time to grasp the folding concept; one thing to check here on your equalition is, if the mathematic difference equal zero then they are mathimatilcally equal but they are not; Ln(i)-pi/2-2*pi*i*k not equal zero. It is not giving natural log value of (i) it only describe the concept of domain folding. At k=0.3 it means you are doing partial folding but not the value of the imaginary unit number it self . The imaginary unit number (i) value in the complex plane is at each value pi/2 or -pi/2, this the complex plane definition for (i). Because (i) is a unit in orthognal folding
☺️final part to add to this comment hope it will clear the confusion. There is a difference between the imaginary numbers and the imaginary UNIT NUMBER ; I think in your equality talking about the imaginary numbers (folding) and not the UNIT NUMBER (I)
Ln(imaginary number) not equal ln(UNIT NUMBER)
Your equation is correct for the imaginary numbers but not for the UNIT NUMBER ln(i) equal pi/2 i
Same as int 1 not equal to 1+0i
0:55 what happened here e^{ipi} = -1 then it becomes 1??!! Also you have to be careful with the complex log, it’s not as simple as doing exp(ln(z)) with complex numbers as it’s a multivalued function in C
Thank you @ramy4697 for your comment you are correct for -1 the text should be the same as the equation -1, Will add the correction to the Video Thank you.
But for the log part, it is natural log exponent for base e, so by difinition they are a function and its inverse so the complex number will not going to change only flipping the singes for the complex number(z) because of the exponent rotation effect. Please check the equation in 13 decimal place and 15 decimal places and let me know if you see any case that it is not matching the equation I used Geogebra , if you have another tool to check please do and let me know
@@shaimaasoltan5334 check out the complex logarithm on Wikipedia, functions and their inverses don’t work the same in C
@@shaimaasoltan5334 the complex exponential function is not injective, it doesn’t necessarily have an inverse function, you have to restrict the domain at least, check the Wikipedia page for more information
@@ramy4697 Thank you for your comment Ramy, my work combines both functions (e & ln) together at the same time and not doing separate analysis for each function properties alone. If you apply both ( e and ln) both together not each one alone hope this clears any confuse. The buty of this way is that its domain is any natural number positive or negative (shown in the video fot T value changes). I will be more than happy to discuss with you any counter case for any equation.(The Domain have been tested for all real number values for T ) I only included the Natural values in the Video record.
@@shaimaasoltan5334 the way you use exp(ln(z)) = z has to be justified as z is a complex number, this relation is not true for all complex numbers, because exp is not injective for all numbers in C, multiple inputs can give the same value, which means it has no inverse in this domain, unless you restrict it, for example if z= exp(i(pi/4)) then z = exp(i(pi/4+2pi) since 2pi is a full rotation, which shows that the complex exponential is not injective so it doesn’t have an inverse for the domain C