I created this forum to discuss the Engineering Q-Vectors (New Math Construct) by Robert Distinti.
For the beginning, make sure to watch the introduction video:
emv049: Introduction to the New Math Construct
http://youtu.be/BiyX1v04oqI
So what do you think about it?
I'm very curious about the improvements of the mathematically description of the ether/particle coupling.
Getting started: Engineering Q-Vectors (New Math Construct)
- SebastianG
- Site Admin
- Posts: 228
- Joined: Fri Sep 14, 2012 2:50 am
- Location: Germany
- SebastianG
- Site Admin
- Posts: 228
- Joined: Fri Sep 14, 2012 2:50 am
- Location: Germany
Re: Getting started: New Math Construct
Here is the playlist for the Distinti University Videos about engineering Q-Vectors:
http://www.youtube.com/playlist?list=PL ... ITF-5ZcOsX
http://www.youtube.com/playlist?list=PL ... ITF-5ZcOsX
Re: Getting started: Engineering Q-Vectors (New Math Constru
Hi Robert,
In one of your more recent videos you mentioned that science doesn't really know what “i”, the imaginary number, is. If that is true, it is the fault of mathematicians for not educating. This subject came up in an advanced linear algebra class in grad school. Here are the facts according to an algebraist( the professor, not me).
The imaginary numbers are not numbers but a set of ordered pairs(vectors) of numbers with a special type of multiplication. C is R2. The “i” is there to remind us how to do that multiplication. “i” is not the square root of -1. If it were, you could prove that 1 = -1.
Like so:
-1 = i*i = (-1)^(1/2) * (-1)^(1/2) = (-1)^(2/2) = 1^(1/2) = 1
Hopefully this is one less thing for you to worry about.
One other unrelated fact that is not widely taught . The cross product only exists in 2, 3, and 7 dimensions.
Looking forward to seeing the Q algebra in action.
Frank
In one of your more recent videos you mentioned that science doesn't really know what “i”, the imaginary number, is. If that is true, it is the fault of mathematicians for not educating. This subject came up in an advanced linear algebra class in grad school. Here are the facts according to an algebraist( the professor, not me).
The imaginary numbers are not numbers but a set of ordered pairs(vectors) of numbers with a special type of multiplication. C is R2. The “i” is there to remind us how to do that multiplication. “i” is not the square root of -1. If it were, you could prove that 1 = -1.
Like so:
-1 = i*i = (-1)^(1/2) * (-1)^(1/2) = (-1)^(2/2) = 1^(1/2) = 1
Hopefully this is one less thing for you to worry about.
One other unrelated fact that is not widely taught . The cross product only exists in 2, 3, and 7 dimensions.
Looking forward to seeing the Q algebra in action.
Frank
Re: Getting started: Engineering Q-Vectors (New Math Constru
Hi Frank,
Thank you for bringing this to my attention.
I would have to disagree with the notion that "i' is not the square root of -1 -- see Note 1 below. Also the derivation presented has a slight fault -- explained in note 2.
Note 1
Electronic circuits become resonant when the roots of the equation become complex; the quadratic formula which is often used
x=(-b +/- sqrt(b^2-4ac))/2a
when 4ac>b^2 the roots become complex and a circuit becomes resonant or bridges begin to vibrate etc.
Also, the derivation of Euler's equation e^(i*theta)=cos(theta)+i*sin(theta) follows the simple infinite series expansion of e using the definition that i=sqrt(-1)
-- there are many other examples.
Note 2: The equation that your professor used tries to say the following are equal (shown as functions for clarity)
square(sqrt(-1))=sqrt(square(-1))
This is problematic because the square function is non-invertible by definition, this causes the order of operation to matter
In other words, the square(2) and square(-2) both result in 4; therefore, because 4 contains no other information, the reciprocal process (sqrt) defaults to 2 by convention (this is cause of the non-invertiblity)
On the other hand, the sqrt function is invertible because sqrt(-2) and sqrt(2) have two different answers (i1.414 and 1.414) which allows the reciprocal process to restore the correct answer.
because one of the processes is not invertible, the order of operation is important.
Ironically, the problem is not with the sqrt(-1), the problem is with the square(-1)
This does not prove that sqrt(-1) is not a number, it just shows that the classical arithmetic definition of exponentiation loses information.
Note 3: I agree, the correct solution is to replace the old 'non-dimensional" arithmetic representations with vectors. This is the solution offered by Q-algebra.
In later videos, you will see that the AB matrix follows the behavior of complex numbers without the need to define sqrt(-1)
Note4:
The lossy and ambiguous nature of antique arithmetic is the cause of many problems. For example the matrix solution to vector multiplication was overlooked was due to the lossy nature of arithmetic multiplication -- this was in the slides of the U_Q1_04 but I forgot to mention it.
Thank you for bringing this to my attention.
I would have to disagree with the notion that "i' is not the square root of -1 -- see Note 1 below. Also the derivation presented has a slight fault -- explained in note 2.
Note 1
Electronic circuits become resonant when the roots of the equation become complex; the quadratic formula which is often used
x=(-b +/- sqrt(b^2-4ac))/2a
when 4ac>b^2 the roots become complex and a circuit becomes resonant or bridges begin to vibrate etc.
Also, the derivation of Euler's equation e^(i*theta)=cos(theta)+i*sin(theta) follows the simple infinite series expansion of e using the definition that i=sqrt(-1)
-- there are many other examples.
Note 2: The equation that your professor used tries to say the following are equal (shown as functions for clarity)
square(sqrt(-1))=sqrt(square(-1))
This is problematic because the square function is non-invertible by definition, this causes the order of operation to matter
In other words, the square(2) and square(-2) both result in 4; therefore, because 4 contains no other information, the reciprocal process (sqrt) defaults to 2 by convention (this is cause of the non-invertiblity)
On the other hand, the sqrt function is invertible because sqrt(-2) and sqrt(2) have two different answers (i1.414 and 1.414) which allows the reciprocal process to restore the correct answer.
because one of the processes is not invertible, the order of operation is important.
Ironically, the problem is not with the sqrt(-1), the problem is with the square(-1)
This does not prove that sqrt(-1) is not a number, it just shows that the classical arithmetic definition of exponentiation loses information.
Note 3: I agree, the correct solution is to replace the old 'non-dimensional" arithmetic representations with vectors. This is the solution offered by Q-algebra.
In later videos, you will see that the AB matrix follows the behavior of complex numbers without the need to define sqrt(-1)
Note4:
The lossy and ambiguous nature of antique arithmetic is the cause of many problems. For example the matrix solution to vector multiplication was overlooked was due to the lossy nature of arithmetic multiplication -- this was in the slides of the U_Q1_04 but I forgot to mention it.
Re: Getting started: Engineering Q-Vectors (New Math Constru
Distinti wrote:Hi Frank,
Thank you for bringing this to my attention.
I would have to disagree with the notion that "i' is not the square root of -1 -- see Note 1 below. Also the derivation presented has a slight fault -- explained in note 2.
Note 1
Electronic circuits become resonant when the roots of the equation become complex; the quadratic formula which is often used
x=(-b +/- sqrt(b^2-4ac))/2a
when 4ac>b^2 the roots become complex and a circuit becomes resonant or bridges begin to vibrate etc.
Also, the derivation of Euler's equation e^(i*theta)=cos(theta)+i*sin(theta) follows the simple infinite series expansion of e using the definition that i=sqrt(-1)
-- there are many other examples.
Note 2: The equation that your professor used tries to say the following are equal (shown as functions for clarity)
square(sqrt(-1))=sqrt(square(-1))
This is problematic because the square function is non-invertible by definition, this causes the order of operation to matter
In other words, the square(2) and square(-2) both result in 4; therefore, because 4 contains no other information, the reciprocal process (sqrt) defaults to 2 by convention (this is cause of the non-invertiblity)
On the other hand, the sqrt function is invertible because sqrt(-2) and sqrt(2) have two different answers (i1.414 and 1.414) which allows the reciprocal process to restore the correct answer.
because one of the processes is not invertible, the order of operation is important; that's why square(sqrt(-1))!=sqrt(square(-1))
Ironically, the problem is not with the sqrt(-1), the problem is with the square(-1)
This exercise shows that the classical arithmetic definition of exponentiation loses information. The solution is in note 3.
Note 3: I agree, the correct solution is to replace the old 'non-dimensional" arithmetic representations with vectors. This is the solution offered by Q-algebra.
In later videos, you will see that the AB matrix follows the behavior of complex numbers without the need to define sqrt(-1)
Note4:
The lossy and ambiguous nature of antique arithmetic is the cause of many problems. For example the matrix solution to vector multiplication was overlooked was due to the lossy nature of arithmetic multiplication -- this was in the slides of the U_Q1_04 but I forgot to mention it.
Who is online
Users browsing this forum: No registered users and 1 guest