Re: JSH: Logic and paradox

Jesse F. Hughes wrote:

Vladimir Dergachev <volodya@xxxxxxxxxxxxxx> writes:

Joshua Cranmer wrote:

JSH wrote:
Like consider 1 = 1, a simple tautological statement which is called
an identity in mathematics, and notice, the equal sign means you have
the same thing on the left of the equals as on the right.

x = O(x^2).

This sentence is not "x is equal to the order of growth of x^2" but "x
grows at most as fast as x^2". An equals sign does not necessarily
indicate equality.


Well, actually in this case it does.

O(x^2) is a common notation that means a function (any function !) that
grows at most a Cx^2. As f(x)=x is such a function this is equality.

This is a strange interpretation. It is as if you claim that the
equation y = 7 is "true", since y stands for any number and 7 is a
number.


No, what I say is that there is a function that satisfies this property.

I.e. sin(x)=x+O(x^2) should be interpreted as an equation

sin(x)=x+f(x) , where |f(x)|<=Cx^2 for some C

together with the statement that solution does exist. Each time you use
O(x^n) you use a different variable: f1, f2, etc.

To hagman: you made a typo x is not O(x^2). A correct way is to say:

x^5=O(x^2)
x^5=O(x^3)
(subtract)
0=O(x^2)-O(x^3)
0=O(x^2)

What this means is that the equation above is satisfied up to first order.
Here is the same but substituting unknown functions explicitly:

x^5=f1(x) |f1(x)|<=Cx^2
x^5=f2(x) |f2(x)|<=Cx^3
(subtract)
0=f1(x)-f2(x) |f1(x)|<=Cx^2, |f2(x)|<=Cx^3
0=f3(x) |f3(x)|<=Cx^2

Another way to think of the above calculation is to concider algebra of
pencils in a particular point (say a).

A pencil in point a of order n for a function f(x) is the set of all
functions g(x) such that |f(x)-g(x)|<=C(x-a)^n for some constant C.

Let us denote such pencil as f(x)+O((x-a)^n) (from now on, I will assume
a=0 for easy typing). Then one can easily prove the following properties:

(f(x)+O(x^n))+(g(x)+O(x^m))=( (f(x)+g(x)) + O(x^(min(n, m)) )

(f(x)+O(x^n))*(g(x)+O(x^m))=( f(x)*g(x)+ O(x^(min(n, m))

Also, this pencil is a module of the algebra of functions, with the
following properties:

(f(x)+O(x^n))*g(x)=f(x)*g(x)+O(x^n)
(f(x)+O(x^n))*x^m=f(x)*x^m+O(x^(n+m))

Additionally we have a family of projections from algebra of functions into
pencils: f(x) ---> f(x)+O(x^n)

Thus we can do: sin(x)=x-x^3/3!+TaylorRemainder implies
(sin(x)+O(x^4))=(x+O(x^4))-(x^3/3! + O(x^4))+ (TaylorRemainder+O(x^4))
and
(sin(x)+O(x^4))=(x+O(x^4))-(x^3/3! + O(x^4))+ (0+O(x^4))
as we know that |TaylorRemainder|<=C(x^4) from theorem about Taylor series.

It also makes sense to simply use O(x^n) for the pencil (0+O(x^n)) derived
from function f(x)=0.

Note that it does not make sense to add pencils defined in different points:
O(x^n)+O((x-1)^m) = nonsense.

best

Vladimir Dergachev

.

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