@@bertik2326 I think there might be an issue with the series, particularly because . To verify this, you could try solving directly, or refer to my video where I’ve explained how to derive in detail. This approach might help identify the source of the problem in the series expansion.
1st you wrote "Your theory" ... it's not his. 2nd there's a proof for 1+2+3+ ... = -1/12 ... and plenty phd mathematicians and physicists are okay with it.
Thank you for your comment, but I disagree. AI cannot invent mathematical concepts-it’s a tool, not a creator. Virtual Numbers are my original work, inspired by extensive study of complex numbers. AI was only used for calculations, similar to a calculator. All concepts have been tested, and I’ll share more videos on their real-life applications. While the editing isn’t perfect, true math lovers will understand my ideas. By the way, feel free to ask AI to create such a concept and share it with me if it really can. Stay tuned for more!
The dual numbers, which are the real numbers along with a new unit ε such that ε^2 = 0, already exists. Most of the results you find in this video are valid in the theory of the dual numbers (especially the result e^ε = 1 + ε), However, your main result that ln(-1) = ε, is still false as ln(-1) is still undefined. The biggest problem with defining ln(-1) is that -1 * -1 = 1, which means that ln(-1) + ln(-1) = ln(1) j + j = 0 j = 0 This could be solved by saying that the natural logarithm is not a function (as the complex numbers do) or by saying that ln does not follow the property ln(ab) = ln(a) + ln(b), but both options are not really that appealing or useful
I think there’s some misunderstanding here. My proposal is to use a virtual number system instead of the traditional complex number system. I believe virtual numbers make things simpler in many ways compared to complex numbers. Regarding the argument that we "cannot assume log(-1) = j"-why not? If we accept the square root of -1 as "i" in the complex number system, even though it’s undefined in the real number system, then why can't we introduce log(-1) = j in the virtual number system?
@@log_menus_1 Err, did you even read his comment? He just showed that ln(-1) = j leads to the logical conclusion that j = 0, if one uses the usual property ln(ab) = ln(a) + ln(b) of logarithms.
Virtual numbers are not the same as dual numbers, as they are not based on real numbers. I clearly explained in the video that k * j = 0 when k is even. In your example, ln(-1) + ln(-1) = 2j = 0, which follows the property of virtual numbers. You might want to rewatch the video to understand how this fits within the virtual number system. It’s a different concept from dual numbers, and the properties I mentioned apply specifically to virtual numbers.
I clearly explained in the video that k * j = 0 when k is even. In your example, ln(-1) + ln(-1) = 2j = 0, which is a property of virtual numbers, not a contradiction. This follows the rules specific to virtual numbers, which differ from the standard properties of logarithms in complex numbers. You might want to rewatch the video to see how this fits within the virtual number system.
@@log_menus_1 " In your example, ln(-1) + ln(-1) = 2j = 0, which is a property of virtual numbers, not a contradiction." And why can't we conclude from this that j = 0? Simply divide both sides of the equation 2j = 0 by 2, or equivalently, multiply by the inverse of 2, i. e. by 1/2. Are the virtual numbers not a field, or why can't we multiply the equation 2j = 0 with 1/2?
Yes, ln(-1) has a value in the complex number framework, but the foundation of complex numbers is the square root of -1. In contrast, Virtual Numbers are built on ln(-1) as their foundation. In the complex number framework: sqrt(-1) = i ln(-1) = iπ In the Virtual Number framework: ln(-1) = j sqrt(-1) = e^(j/2) Just as ln(-1) is expressed using the complex number system, sqrt(-1) is expressed using the Virtual Number framework. Both systems tackle similar concepts but have entirely different foundations.
@@log_menus_1 If there's a formula for converting between i and j, then virtual numbers aren't really a new system, just a simple transformation applied to an existing system. That is to say, any theorem is provable with virtual numbers if and only if it is provable with complex numbers. That doesn't necessarily mean it's useless, mind you; it's just not fundamentally distinct from what already exists.
What I mean is that if complex numbers didn’t exist, virtual numbers could serve as a complete alternative. With virtual numbers, we can perform all the tasks typically done using complex numbers. For example, any square root of a negative number can be represented and handled within the virtual number system effectively.
@@matthewjohnson6092 I think there might be an issue with the series, particularly because . To verify this, you could try solving directly, or refer to my video where I’ve explained how to derive in detail. This approach might help identify the source of the problem in the series expansion.
Thanks! I’m glad you appreciate it. Math is full of possibilities, and sometimes thinking outside the box leads to the most interesting ideas. Stay tuned for more!
@@Ten2378 But that would be assuming exponentiation is associative, when it is not. In general, (a^b)^c is not equal to a^(b^c). (For example use a = b = 2 and c = 3.) However, the argument you presented assumes (a^b)^c = a^(b^c) in assuming that (e^ln(-1))^2 = e^(ln(-1)^2)
@@MuffinsAPlenty whene k tried to solve j² or or higher power , there was no formula for LnA* LnB , one of my video I create formula for this product and then put A=B=-1 so I got a result 0 so If j² = 0 then all other powers will be zero
@@log_menus_1 I see. So using ln(A)+ln(B)=ln(AB), you developed the formula: ln(A)*ln(B) = (1/4)*[log(AB)^2 - log(A/B)^2]. So, essentially, you're wishing to preserve this equation, even when A and B are negative numbers. The complex number approach does _not_ preserve this equation when A and B are not positive real numbers, specifically because ln(A)+ln(B) = ln(AB) is not preserved for complex numbers which are not positive real numbers. I will think about this a little bit more and see if I can find any potential issues with your approach. But I do thank you for your explanation! It gives me something to think about!
Confusing. Should have another letter chosen for virtual number! In most electrical engineering text the letter "j = sqrt (-1) " is already used instead of "i".
this really demands a proof that j^2 = 0. j^2 = j*j = j*log-1 = log((-1)^j) log(j^2) = 2*log j = -2 you can't have the laws of logarithms if you head down that path
Not exactly! Dual numbers are a different concept. They're written as "a + bε", where ε² = 0, but ε ≠ 0, and are used in automatic differentiation and other areas of math. Virtual numbers, on the other hand, are based on a different approach-where I define them using logarithms , not the same way dual numbers work. So, while they may seem similar at first, the underlying principles are quite different!
If we are trying to extend the natural logarithm to the negatives with log(-1)=j then we should have -1=exp(log(-1))=exp(j) But you define exp(j)=1+j Combining both we get j=-2 Also ming showing any work for any of those results? Most probably have similar issues, but I don't have time now as my bus is arriving soon
The expression e^j = -1 is different from e^j = 1 + j. The second equation comes from the expansion of e^x when the domain is extended to include undefined values like j. However, the first equation, e^j = -1, is simply a mathematical representation of -1 using the exponential function. When we extend the domain of a function, its behavior can change. For example: In the factorial function, 0! = 1. But when the factorial is extended to non-integer values using the Gamma function, we get (1/2)! ≈ 0.886. This shows that the properties of a function at specific points may differ from its behavior in the extended domain. For instance, 0! must be less than (1/2)!, just as 2! is less than 3!. Another issue arises when manipulating Euler's identity incorrectly. Euler's identity, e^(πi) = -1, implies: πi = ln(-1). However, multiplying by constants can cause contradictions. For example: 2πi = 2ln(-1). Using logarithmic properties, 2ln(-1) = ln((-1)^2) = ln(1) = 0. This creates a contradiction: π = 0.
If complex numbers didn't exist, virtual numbers could provide an alternative. The key difference is in how they're defined and the unique properties they introduce, not just relying on Euler's identity. It’s a new conceptual approach, offering different ways to think about mathematical operations. So while ln(-1) = iπ is valid in the complex system, virtual numbers offer an entirely new perspective that could replace complex numbers in certain contexts.
Glad you found it interesting! 😊 Virtual Numbers are definitely a unique concept, and I believe they could open up new ways to understand math and physics.
1. e^(i*pi) + 1 = 0 already contains everything you're trying to do here and a lot more. 2. e^x = x + 1 is only the first order approximation of the exponential (stress on approximation) and, also, it's valid only for |x|
great job , nice new concept , you have to do something with the 2kj=0 and (2k+1)j=j when k is an integer ,it's such a strange result , keep the hard work and good luck finding an application for this
I don't know enough about mathematics yet to be a critic of the creative process. I don't know if this is right or wrong but you got a new subscriber! 😊
Genuinely one of the most unique troll channels I've seen. Good use of AI to piss people off, plus the use of math that's clearly wrong but almost looks like it could work. Then, claiming it to be something with potential over complex numbers, kinda like the tau vs pi people? It's beautiful.
Thanks for your comment! The equations I presented are mathematically correct, though some functions may seem unconventional with virtual numbers-just like how 1 + 2 + 3 + ... = -1/12 seems odd but is valid via analytical continuation. As for AI, it’s just a tool I used to simplify equations, like a calculator. The concepts and research are entirely my own. I understand virtual numbers are new and challenge traditional views, but I encourage keeping an open mind. By the way, my next video on singularity numbers will blow your mind-so you still have time to unsubscribe! 😉
Are you a professor of mathematics or D. or scientist , because your ideas are amazing . I saw a video about x² = 0 x≠0 I didn't think about ln(-1) since it's not defined I want to add something : if j²=0 so j^2n=0 but j^(2n+1)=j and lnj=ln√j²=2ln√j=2lne^√j/e = 2(√j-1) = 2√j-2 . finally : we know e^iπ=-1 , so iπ=ln(-1) so iπ=j we can connect virtual and complex numbers . and ij=-π ! ( very real ) . but we have a problem in our link , since j=iπ j²=-π²≠0 so you should set some laws to connect j & i , if j = zi , it will not have a sence . because -z²≠0 ( if z=0 =⟩ j=0 , doesn't make any sense ) .
Yes j = πi and if u take square, j² = -π² By definition j² = 0 So 0 = -π² this is a paradox but it's true , let me prove it .. Consider Euler identity e^πi = -1 and take ln πi = ln(-1) Now multiply by 2 2πi = 2ln(-1) 2πi = 0 π=0 π² = 0 By definition of virtual numbers cj = 0 for c us even If virtual numbers are not consistent then complex numbers will be inconsistent as well ... Complex numbers produced using rotation by 90⁰ and while virtual numbers created using rotation by 180⁰
@log_menus_1 so you should set the limits like : √x is defined if x≥0 ( i≠√-1 it's i²=-1 ) and some applications , complex numbers are useful , for exemple un shrodinger equation we have ih/2 ..
U can remove iota i from the Schrodinger equation and convert it into a virtual number as √-1 = e ^j/2 then that equation will give a new meaning, it will be interesting 😃
The Schrödinger equation describes the behavior of waves. As we know, the concept of iota (or i, the imaginary unit) in complex numbers originates from a rotation by 90°. Furthermore, all waves can be represented as polynomial equations and expressed in 2D geometry. In the Schrödinger equation, i essentially represents the geometric nature of wave functions. But what happens if we replace complex numbers with virtual numbers? Virtual numbers are based on a 180° rotation, and their virtual unit (j) is defined as e^(j/2), which is intriguing due to the involvement of e with the virtual unit’s power. While I’m not a physicist, I can confidently say that the role of i in the Schrödinger equation is primarily to represent the geometry of wave functions.
If ln(-1) = j, then e^j = -1, but you also stated, that e^j = 1 + j, that would mean, that -1 = 1 + j => j = -2. Your theory has some holes in it
@@bertik2326 I think there might be an issue with the series, particularly because . To verify this, you could try solving directly, or refer to my video where I’ve explained how to derive in detail. This approach might help identify the source of the problem in the series expansion.
@@log_menus_1Or your "number system" is just flawed and has no actual rigor
1st you wrote "Your theory" ... it's not his. 2nd there's a proof for 1+2+3+ ... = -1/12 ... and plenty phd mathematicians and physicists are okay with it.
You asked an AI to invent a new type of math and it did exactly as poorly as I would have expected it to
Thank you for your comment, but I disagree. AI cannot invent mathematical concepts-it’s a tool, not a creator. Virtual Numbers are my original work, inspired by extensive study of complex numbers. AI was only used for calculations, similar to a calculator.
All concepts have been tested, and I’ll share more videos on their real-life applications. While the editing isn’t perfect, true math lovers will understand my ideas. By the way, feel free to ask AI to create such a concept and share it with me if it really can. Stay tuned for more!
@@log_menus_1😂
@@log_menus_1 just discard comments like that , it was a respectful one but doesn't worth being replied
What if he tried the half-order of the natural logarithm of negative the squareroot of pi?
The dual numbers, which are the real numbers along with a new unit ε such that ε^2 = 0, already exists. Most of the results you find in this video are valid in the theory of the dual numbers (especially the result e^ε = 1 + ε), However, your main result that ln(-1) = ε, is still false as ln(-1) is still undefined.
The biggest problem with defining ln(-1) is that -1 * -1 = 1, which means that
ln(-1) + ln(-1) = ln(1)
j + j = 0
j = 0
This could be solved by saying that the natural logarithm is not a function (as the complex numbers do) or by saying that ln does not follow the property ln(ab) = ln(a) + ln(b), but both options are not really that appealing or useful
I think there’s some misunderstanding here. My proposal is to use a virtual number system instead of the traditional complex number system. I believe virtual numbers make things simpler in many ways compared to complex numbers.
Regarding the argument that we "cannot assume log(-1) = j"-why not? If we accept the square root of -1 as "i" in the complex number system, even though it’s undefined in the real number system, then why can't we introduce log(-1) = j in the virtual number system?
@@log_menus_1 Err, did you even read his comment? He just showed that ln(-1) = j leads to the logical conclusion that j = 0, if one uses the usual property ln(ab) = ln(a) + ln(b) of logarithms.
Virtual numbers are not the same as dual numbers, as they are not based on real numbers. I clearly explained in the video that k * j = 0 when k is even. In your example, ln(-1) + ln(-1) = 2j = 0, which follows the property of virtual numbers. You might want to rewatch the video to understand how this fits within the virtual number system. It’s a different concept from dual numbers, and the properties I mentioned apply specifically to virtual numbers.
I clearly explained in the video that k * j = 0 when k is even. In your example, ln(-1) + ln(-1) = 2j = 0, which is a property of virtual numbers, not a contradiction. This follows the rules specific to virtual numbers, which differ from the standard properties of logarithms in complex numbers. You might want to rewatch the video to see how this fits within the virtual number system.
@@log_menus_1 " In your example, ln(-1) + ln(-1) = 2j = 0, which is a property of virtual numbers, not a contradiction."
And why can't we conclude from this that j = 0? Simply divide both sides of the equation 2j = 0 by 2, or equivalently, multiply by the inverse of 2, i. e. by 1/2.
Are the virtual numbers not a field, or why can't we multiply the equation 2j = 0 with 1/2?
Greetings Sir. Could you show me how you obtein the expression √j = e^√j / e ? Thank you for you kind response.
In my future videos
ln-1 is i*pi. this number already exists.
Yes, ln(-1) has a value in the complex number framework, but the foundation of complex numbers is the square root of -1. In contrast, Virtual Numbers are built on ln(-1) as their foundation.
In the complex number framework:
sqrt(-1) = i
ln(-1) = iπ
In the Virtual Number framework:
ln(-1) = j
sqrt(-1) = e^(j/2)
Just as ln(-1) is expressed using the complex number system, sqrt(-1) is expressed using the Virtual Number framework. Both systems tackle similar concepts but have entirely different foundations.
@@log_menus_1 If there's a formula for converting between i and j, then virtual numbers aren't really a new system, just a simple transformation applied to an existing system. That is to say, any theorem is provable with virtual numbers if and only if it is provable with complex numbers. That doesn't necessarily mean it's useless, mind you; it's just not fundamentally distinct from what already exists.
@@log_menus_1 if j=i*pi and j*j=0 then -pi^2=0?🙄. Show your work.
What I mean is that if complex numbers didn’t exist, virtual numbers could serve as a complete alternative. With virtual numbers, we can perform all the tasks typically done using complex numbers. For example, any square root of a negative number can be represented and handled within the virtual number system effectively.
@@matthewjohnson6092 I think there might be an issue with the series, particularly because . To verify this, you could try solving directly, or refer to my video where I’ve explained how to derive in detail. This approach might help identify the source of the problem in the series expansion.
I really like this idea, love when people do something weird with math like this
Thanks! I’m glad you appreciate it. Math is full of possibilities, and sometimes thinking outside the box leads to the most interesting ideas. Stay tuned for more!
Why is j^2 = 0? How does that follow from j being defined as ln(-1)?
@@Ten2378 But that would be assuming exponentiation is associative, when it is not. In general, (a^b)^c is not equal to a^(b^c). (For example use a = b = 2 and c = 3.)
However, the argument you presented assumes (a^b)^c = a^(b^c) in assuming that (e^ln(-1))^2 = e^(ln(-1)^2)
bro trying to make dual numbers
@@Ten2378 e^ln(-1)*e^ln(-1) = [e^ln(-1)]^2 = e^2ln(-1), you got it wrong
@@MuffinsAPlenty whene k tried to solve j² or or higher power , there was no formula for LnA* LnB , one of my video I create formula for this product and then put A=B=-1 so I got a result 0 so
If j² = 0 then all other powers will be zero
@@log_menus_1 I see. So using ln(A)+ln(B)=ln(AB), you developed the formula:
ln(A)*ln(B) = (1/4)*[log(AB)^2 - log(A/B)^2].
So, essentially, you're wishing to preserve this equation, even when A and B are negative numbers. The complex number approach does _not_ preserve this equation when A and B are not positive real numbers, specifically because ln(A)+ln(B) = ln(AB) is not preserved for complex numbers which are not positive real numbers.
I will think about this a little bit more and see if I can find any potential issues with your approach. But I do thank you for your explanation! It gives me something to think about!
Confusing. Should have another letter chosen for virtual number! In most electrical engineering text the letter "j = sqrt (-1) " is already used instead of "i".
Lett v?
this really demands a proof that j^2 = 0.
j^2 = j*j = j*log-1 = log((-1)^j)
log(j^2) = 2*log j = -2
you can't have the laws of logarithms if you head down that path
Well...
0 = j^2 = j*j = j*ln-1 = ln((-1)^j) = ln(1 + j*ln(-1)) = ln(1 + j*j) = ln(1+ 0) = ln(1) = 0
...does follow.
Also, ln(j^2) would be ln(0), which was not defined anyway, so... You're kind of misusing the notation like that.
I think the dual numbers already does what you're trying to achieve here.
Not exactly! Dual numbers are a different concept. They're written as "a + bε", where ε² = 0, but ε ≠ 0, and are used in automatic differentiation and other areas of math.
Virtual numbers, on the other hand, are based on a different approach-where I define them using logarithms , not the same way dual numbers work. So, while they may seem similar at first, the underlying principles are quite different!
If we are trying to extend the natural logarithm to the negatives with log(-1)=j then we should have
-1=exp(log(-1))=exp(j)
But you define
exp(j)=1+j
Combining both we get
j=-2
Also ming showing any work for any of those results? Most probably have similar issues, but I don't have time now as my bus is arriving soon
The expression e^j = -1 is different from e^j = 1 + j. The second equation comes from the expansion of e^x when the domain is extended to include undefined values like j. However, the first equation, e^j = -1, is simply a mathematical representation of -1 using the exponential function.
When we extend the domain of a function, its behavior can change. For example:
In the factorial function, 0! = 1.
But when the factorial is extended to non-integer values using the Gamma function, we get (1/2)! ≈ 0.886.
This shows that the properties of a function at specific points may differ from its behavior in the extended domain. For instance, 0! must be less than (1/2)!, just as 2! is less than 3!.
Another issue arises when manipulating Euler's identity incorrectly. Euler's identity, e^(πi) = -1, implies:
πi = ln(-1).
However, multiplying by constants can cause contradictions. For example:
2πi = 2ln(-1).
Using logarithmic properties, 2ln(-1) = ln((-1)^2) = ln(1) = 0. This creates a contradiction:
π = 0.
if j=ln(-1), e^j=-1. and since e^iπ, j=iπ which is basically an imaginary number.
j = πi can be written in virtual system as i or negative square root of 1 = e^j/2
So j =π( e^j/2)
you need to decide if you want either j = ln(-1) or j² = 0, because ln(-1)² is NOT equal to 0
to find j^2 = 0 u need to find lgA *lgB
ln(-1) = iπ
An already complex number. ln(-1) is not new or revolutionary. It's simply comes from eulers identity.
let say complex numbers dont exist
@@log_menus_1 okay and?
If complex numbers didn't exist, virtual numbers could provide an alternative. The key difference is in how they're defined and the unique properties they introduce, not just relying on Euler's identity. It’s a new conceptual approach, offering different ways to think about mathematical operations. So while ln(-1) = iπ is valid in the complex system, virtual numbers offer an entirely new perspective that could replace complex numbers in certain contexts.
@@log_menus_1 How do you equate ln(-1) = i\pi without Euler's identity?
use binomial theorem to expand such numbers
You should next explore quaternion formulations using the virtual numbers 💯
What in the world! is this! :D Mind blowing. Thank you Virtual Numbers ln -1. Looks very futuristic.
Are you an extraterrestrial or an advanced AI?
yes .... HUMAN
Glad you found it interesting! 😊 Virtual Numbers are definitely a unique concept, and I believe they could open up new ways to understand math and physics.
the virtual number j is an interesting idea but it just can't equal ln(-1)
Like √ -1 ≠ i
also explain more practical and graphical approach
you can try your self , Ej =0 , Oj=j , E is even O is odd and j^n=0 for n>1
1. e^(i*pi) + 1 = 0 already contains everything you're trying to do here and a lot more.
2. e^x = x + 1 is only the first order approximation of the exponential (stress on approximation) and, also, it's valid only for |x|
U can check by putting x = ln(-1) in e^x expansion
ln of -1 is i*π cuz e^iπ= -1
Then, how to represent ln(0), please?
I'll explain this in more detail in my next video, but for now, let me tell you that
ln(0) is related to the zeta function.
Great
Can I share these properties of Virtual Numbers in my TikTok account ?
Yes, sure
and send me your id or link
@log_menus_1 OK, I scheduled the post for this Saturday, hopefully there would be nothing that can happen to change my mind
Ln(-1) =-π^2 and e^jshould be zero
Why j²=0 ?
watch second video of e^j video
im pretty sure these are isomorphic to complex numbers
just use phi:C->V with phi(a+bi)=a+(b/pi)*j
Terryology expansion
Bold ideas need bold names-I'll take it! 😄
Yet another example of AI fundamental rule : an AI is as dumb as the dumbest of its programmers
why j²=0? ln(-1)²=0?
Watch my other video title e^j = 1+j
great job , nice new concept , you have to do something with the 2kj=0 and (2k+1)j=j when k is an integer ,it's such a strange result , keep the hard work and good luck finding an application for this
Even *lnx=0 and ODD*lnx=lnx , for x=-1 thats why when ln(-1) = i*Pi for
2ln(-1) = 2I*Pi = 0
pi=0
j = i*π
Multiply by 2
I don't know enough about mathematics yet to be a critic of the creative process. I don't know if this is right or wrong but you got a new subscriber! 😊
Thanks
Plot it
Next videos
Bro is going too far
This has to be fake I swear to god
j=ln(-1)
j=iπ
j²=-π² 😂not 0
It's correct that
Ln(-1) = πi
Multiply both sides by 2 ..
2Ln(-1) = 2πi
0 = 2πi
Also j² = 0.if you watch my video on e^j = 1+j u will understand
Genuinely one of the most unique troll channels I've seen. Good use of AI to piss people off, plus the use of math that's clearly wrong but almost looks like it could work. Then, claiming it to be something with potential over complex numbers, kinda like the tau vs pi people? It's beautiful.
Thanks for your comment! The equations I presented are mathematically correct, though some functions may seem unconventional with virtual numbers-just like how 1 + 2 + 3 + ... = -1/12 seems odd but is valid via analytical continuation.
As for AI, it’s just a tool I used to simplify equations, like a calculator. The concepts and research are entirely my own. I understand virtual numbers are new and challenge traditional views, but I encourage keeping an open mind.
By the way, my next video on singularity numbers will blow your mind-so you still have time to unsubscribe! 😉
Umm, how is Tau vs Pi a troll?
More AI trash
AI is a tool not a creator
such a silly thing hhhh
Are you a professor of mathematics or D. or scientist , because your ideas are amazing .
I saw a video about
x² = 0
x≠0
I didn't think about ln(-1) since it's not defined
I want to add something :
if j²=0 so j^2n=0
but j^(2n+1)=j
and lnj=ln√j²=2ln√j=2lne^√j/e
= 2(√j-1) = 2√j-2 .
finally : we know e^iπ=-1 , so iπ=ln(-1)
so iπ=j
we can connect virtual and complex numbers .
and ij=-π ! ( very real ) .
but we have a problem in our link , since j=iπ
j²=-π²≠0 so you should set some laws to connect j & i , if j = zi , it will not have a sence .
because -z²≠0 ( if z=0 =⟩ j=0 , doesn't make any sense ) .
Yes j = πi and if u take square,
j² = -π²
By definition j² = 0
So 0 = -π² this is a paradox but it's true , let me prove it ..
Consider Euler identity
e^πi = -1
and take ln
πi = ln(-1)
Now multiply by 2
2πi = 2ln(-1)
2πi = 0
π=0
π² = 0
By definition of virtual numbers cj = 0 for c us even
If virtual numbers are not consistent then complex numbers will be inconsistent as well ...
Complex numbers produced using rotation by 90⁰ and while virtual numbers created using rotation by 180⁰
@log_menus_1 so you should set the limits like :
√x is defined if x≥0
( i≠√-1 it's i²=-1 )
and some applications , complex numbers are useful , for exemple un shrodinger equation we have ih/2 ..
U can remove iota i from the Schrodinger equation and convert it into a virtual number as √-1 = e ^j/2 then that equation will give a new meaning, it will be interesting 😃
@@log_menus_1 what is this new meaning ?
The Schrödinger equation describes the behavior of waves. As we know, the concept of iota (or i, the imaginary unit) in complex numbers originates from a rotation by 90°. Furthermore, all waves can be represented as polynomial equations and expressed in 2D geometry.
In the Schrödinger equation, i essentially represents the geometric nature of wave functions. But what happens if we replace complex numbers with virtual numbers? Virtual numbers are based on a 180° rotation, and their virtual unit (j) is defined as e^(j/2), which is intriguing due to the involvement of e with the virtual unit’s power.
While I’m not a physicist, I can confidently say that the role of i in the Schrödinger equation is primarily to represent the geometry of wave functions.
ln(-1) is equal to i(2n+1)π where n∈ℤ