Terrence Howard Math?

D

Dg1

New member
Joined
Apr 7, 2024
Messages
11
Has anyone watched “Terrance Howard: This is The Best Kept SECRET in The ENTIRE WORLD!”
It’s a you tube video some of its about math. I can’t figure out what he’s rambling on about.
 
I gave it a shot, but I think it is a waste of time. If someone keeps saying "I challenge physicists, I challenge cosmologists, ..." and is more concerned about what people think of his ideas than publishing it in scientific journals rather than on youtube then it is 99% a nutjob.
 
fresh_42 said:
....is more concerned about what people think of his ideas than publishing it in scientific journals rather than on youtube then it is 99% a nutjob.
Click to expand...
I agree totally it is a waste of time ............ 100%
 
What's wrong with the following?

[imath]\displaystyle \lim_{x \to \infty} \sqrt [x]2 = 1[/imath]
[imath]2 = 1^\infty[/imath]
[imath]1 \times 1 = 1^\infty[/imath]
Ergo,
[imath]2 = 1 \times 1[/imath]
 
blamocur said:
The second equality.
Click to expand...
Is [imath]1^ \infty[/imath] like [imath]\frac{0}{0}[/imath], indeterminate, instead of 1? We're not talking about any ol' number here, this is [imath]\infty[/imath].

Perhaps 1 is an asymptote (close, but no cigar). 😀
 
Related is, cogito, my error, in thinking a number e.g. 9 is the 1 × 1 matrix [9].

1 is a threshold for multiplication
For a > 1, b > 1, ab > 1, ab > a and ab > b
For a = 1, b = 1, ab = 1
For a < 1, b < 1, ab < 1, ab < a, ab < b
For a = 1, for any b, ab = b
 
Agent Smith said:
Is [imath]1^ \infty[/imath] like [imath]\frac{0}{0}[/imath], indeterminate, instead of 1? We're not talking about any ol' number here, this is [imath]\infty[/imath].

Perhaps 1 is an asymptote (close, but no cigar). 😀
Click to expand...
This is a recurring topic in this forum. My answer: there is no [imath]1^\infty[/imath] or [imath]\frac{0}{0}[/imath]. Those symbols are occasionally used as names of classes of certain non-trivial limits. For example:
[math]\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n = e \;\;\;\;\; \text{and} \;\;\;\; \lim_{n\rightarrow \infty}\left(1+\frac{1}{n^2}\right)^n = 0[/math]Both limts are of the type [imath]1^\infty[/imath], but the values are difference. I.e. [imath]1^\infty[/imath] is not a number.
 
blamocur said:
This is a recurring topic in this forum. My answer: there is no [imath]1^\infty[/imath] or [imath]\frac{0}{0}[/imath]. Those symbols are occasionally used as names of classes of certain non-trivial limits. For example:
[math]\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n = e \;\;\;\;\; \text{and} \;\;\;\; \lim_{n\rightarrow \infty}\left(1+\frac{1}{n^2}\right)^n = 0[/math]Both limts are of the type [imath]1^\infty[/imath], but the values are difference. I.e. [imath]1^\infty[/imath] is not a number.
Click to expand...
It's a relief to me that you mentioned limits.
 
blamocur said:
This is a recurring topic in this forum. My answer: there is no [imath]1^\infty[/imath] or [imath]\frac{0}{0}[/imath]. Those symbols are occasionally used as names of classes of certain non-trivial limits. For example:
[math]\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n = e \;\;\;\;\; \text{and} \;\;\;\; \lim_{n\rightarrow \infty}\left(1+\frac{1}{n^2}\right)^n = 0[/math]Both limts are of the type [imath]1^\infty[/imath], but the values are difference. I.e. [imath]1^\infty[/imath] is not a number.
Click to expand...
Almost. [imath] \lim_{n\rightarrow \infty}\left(1+\frac{1}{n^2}\right)^n = 1.[/imath] And we can certainly define [imath]1^\infty :=\lim_{n \to \infty} 1^n = 1[/imath], but it has to be said what is meant. It actually even has to be said whether we are on the ground of standard analysis or non-standard analysis or in the range of hyperreals.

Any notation has to be defined! Even [imath] 1+1=0 [/imath] is allowed in some areas, computers for example. However, nobody wants to add "if the characteristic is zero" every time we use [imath] 1+1=2, [/imath] so we have the convention that the characteristic is zero if not noted otherwise. This covers 99.% of all cases and is therefore a useful convention. Hence, we have a lot of conventions to make life easier. I once was asked why we write a matrix as [imath] A=(a_{i,j}) [/imath] and not as [imath] A=(a_{j,i}) .[/imath] There is no mathematical reason, but it is convenient to use [imath](a_{j,i})=A^\tau [/imath] for the transposed matrix.

So if we take a convention out of that 99% pool, strip off the context, and put it into another context where it has never been intended to use, we almost automatically get nonsense.

If we play around with symbols then we should assume that we are dealing with formal languages where such symbols are letters. But even then, we have to make sure that we use the same syntax!
 
Perhaps, in terms of multiclause functions

[imath]f(x) = \begin {cases} x^2 \text{ if } x \ne 1 \\ 2 \text{ if } x = 1 \end {cases}[/imath]
 
By the way, cogito, in modular arithmetic, 1 full turn = n full turns i.e. [imath]0 = 2 \mod 1[/imath]

Maybe someone should tell Mr. Howard that [imath]2 \times \frac{1}{2} = 1 = 1 \times 1[/imath]. There's the increase which Mr. Howard claims is missing with 1 × 1
 
You must log in or register to reply here.
Top