Category Theory II 2: Limits, Higher order functors

March 23, 2017tomcircle Modern Math Leave a comment

1.2 Introduction to Limit

Analogy : Product to Cones (Limit)

2.1 Five categories used to define Limit:

Index category (I)
Category C: Functors (constant , D)
Cones (Lim D)
Functor Category [I, C ]:objects are (constant, D ), morphisms are natural transormations
Set category of Hom-Set Cones [I,C] to Hom-Set C (c , Lim D )

2.2 Naturality

3.1 Examples: Equalizer

CoLimit = duality of Limit (inverted cone = co-Cone)

Functor Continuity = preserve Limit

What exactly is a Limit ?

October 6, 2016tomcircle Education, Modern Math 1 Comment

$\displaystyle\lim_{x\to a}f(x) = L \iff$
$\forall \varepsilon >0, \exists \delta >0$ such that
$\boxed{0<|x-a|<\delta} \implies |f(x)-L|< \varepsilon$

The above scary ‘epsilon-delta’ definition of “Limit” by the French mathematician Cauchy in 19th century is the standard rigorous definition in today’s Analysis textbooks.

It was not taught in my Cambridge GCE A-Level Pure Math in 1970s (still true today), but every French Baccalaureate Math student (Terminale, equivalent to JC 2 or Pre-U 2) knows it by heart. A Cornell University Math Dean recalled how he was told by his high-school teacher to memorise it — even though he did not fully understand — the “epsilon-delta” definition by “chanting”:

“for all epsilon, there is a delta ….”

(French: Quelque soit epsilon, il existe un delta …)

In this video, I am glad someone like Prof N. Wildberger recognised its “flaws” albeit rigorous, by suggesting another more intuitive definition:
◇ Cauchy’s “flaw”: ambiguous
Finding a certain $\delta = f (\varepsilon )$
too counter-intuitive to grasp the idea by most university math students.

◇ Intuitive Alternative:
$\boxed { \displaystyle { \lim_{n \to \infty} P(n) = A} } \iff$
Find any 2 natural numbers m (“Start“), k (“Scale“) such that:
$\boxed { \text {For } m \leq n, \frac {-k}{n} \leq P(n) - A \leq \frac {k}{n} }$

A Simple Analogy in Life:
Let P(n) = Any Person’s lifespan of age n
m = ‘Start’ Age to retire, say 60
k = ‘Scale’ of interval (in years, eg. 1 year)
A = Limit of a person’s lifespan, say 80 (male) or 85 (female)

As we grow older (n increases), from a certain “Start” point (m), our lifespan P(n) approaches the limit A, plus or minus k/n (years).

Limit 极限

March 16, 2016tomcircle Modern Math 1 Comment

Mathematical Rigour:

“Domain of Definition” MUST first be considered prior to tackling :
1) Continuity 连续性
2) Differentiability 可微性
3) Integrability 可积性
4) Limit 极限

Mathematics is linked to Philiosophy! In this life (Domain of Definition ) we have a limit of lifespan (120 years = 2 x 60 years = 2个甲子).

In this same “Domain of Definition” our life is Continuous unless interrupted by unforseen circumstances (accident, diseases, war, …). At certain junctures of life we Differentiate ourselves by having some ‘smooth’ (not abrupt “V- “or “W-” shape) turns of event (eg. graduation from schools and university, National Service in military, marriage, children, jobs, honours/promotions, as well as failures …). It is only in this life you can Integrate these fruits of labor. Beyond this “Domain of Definition” life is meaniningless because we shall return to soil with nothing ….

See : C.I.D .Triangle

https://frankliou.wordpress.com/2013/04/25/微積分極限的一個概念/

Sequence Limit

May 30, 2013tomcircle Modern Math 2 Comments

Definition: $\text{Sequence } (a_n)$
has limit a

$\boxed{\forall \varepsilon >0, \exists N, \forall n \geq N \text { such that } |(a_n) -a| < \varepsilon}$

$\Updownarrow$

$\displaystyle \boxed{ \lim_{n\to\infty} (a_n) = a }$

What if we reverse the order of the definition like this:

∃ N such that ∀ε > 0, ∀n ≥ N,
$|(a_n) -a| < \varepsilon$

This means:

$\boxed {\forall n \geq N, (a_n) = a }$

Example:

$\displaystyle (a_n) = \frac{3n^{2} + 2n +1}{n^{2}-n-3}$

$\displaystyle\text{Prove: } (a_n) \text { convergent? If so, what is the limit ?}$

Proof:
$\displaystyle (a_n) = 3 + \frac{5n +10}{n^{2}-n-3}$

$n \to \infty, (a_n) \to 3$

Let’s prove it.

$\text {Let } \varepsilon >0$
$\text{Choose N such that } \forall n \geq N,$
$\displaystyle |(a_n) -3| = \Bigr|\frac{5n +10}{n^{2}-n-3}\Bigr| < \varepsilon$

$\text{Simplify: } \displaystyle \Bigr|\frac{5n +10}{n^{2}-n-3}\Bigr|$
$\text{Let } n > 10$
$\displaystyle \Bigr|\frac{5n +10}{n^{2}-n-3}\Bigr| < \frac{6n}{\frac{1}{2}n^{2}}= \frac{12}{n} < \varepsilon$

$\text{Choose } N = \max (10, \frac{12}{\varepsilon})$
$\displaystyle\forall n \geq N, |(a_n) -3 | < \frac{12}{n} < \varepsilon$

Therefore,
$\displaystyle \boxed{ \lim_{n\to\infty} (a_n) = 3 }$ [QED]

[Source]: Excellent Introduction in Modern Math:

A Concise Introduction to Pure Mathematics, Third Edition (Chapman Hall/CRC Mathematics Series)

Multi-variable Limit

May 6, 2013tomcircle Modern Math 1 Comment

Analysis is the study of Functions, using the main tool “Limit“.

Limit problems appear in:
1. Continuity
2. Derivative
3. Integral
4. Sequence

Multi-variable Functions have different approach of Limit compared to Single-value functions.
Eg. L’Hôpital Rule is not applicable to Multi-variable Functions.

Case 1: Find the Limit (L) of
$\displaystyle f(x,y)= \frac{xy}{x^{2}+y^{2}} \text{ at point (0,0)}$

Solution:
Consider the point P(x,y) on f(x,y)
$\displaystyle \lim_{x\to 0} f(x,y)=f(0,y)= 0 \text{...(1)}$
$\displaystyle\lim_{y\to 0} f(x,y)=f(x,0)=0 \text{...(2)}$

but when P moves along y=x straight line approaching (0,0),
ie. x->0, y=x->0,
$\displaystyle\lim_{x\to 0} \frac{xy}{x^{2}+y^{2}}= \frac{x^2}{2.x^2} = \frac{1}{2} \text{...(3)}$

From (1),(2),(3) there are 3 limits {0, 0, 1/2}, hence the Limit L does not exist.

Case 2:
Find limit L of
$\displaystyle f(x,y)= \frac{x^{2}+y^{2}}{x^{4}+y^{4}} \text { at point} (\infty, \infty)$

Solution:
Let y= kx
$\displaystyle f(x,y)= \frac{x^{2}+k^{2}y^{2}}{x^{4}+k^{4}y^{4}} = \frac{1+k^{2}}{(1+k^{4})x^2}$
When x -> $\infty$ ,
f(x,y) independent of k => possible limit of 0

Prove: $\displaystyle\lim_{(x,y)\to \infty} f(x,y)= L = 0$

$\displaystyle \left| \frac{x^{2}+y^{2}}{x^{4}+y^{4}} - 0 \right| = \frac{x^{2}+y^{2}}{x^{4}+y^{4}} \leq \frac{x^{2}+y^{2}}{2x^{2}y^{2}} = \frac{1}{2x^{2}} + \frac{1}{2y^{2}}$

Note:
$(x^{2} - y^{2})^2 \geq 0$
$x^{4}-2x^{2}y^{2}+y^{4}\geq 0$
$x^{4}+y^{4} \geq 2x^{2}y^{2}$

$\displaystyle\forall \epsilon > 0 \text{, take } \delta \geq \frac{1}{\sqrt{\epsilon}} \text{ such that}$
$|x|>\delta, |y|>\delta$
$\implies \displaystyle \left| \frac{x^{2}+y^{2}}{x^{4}+y^{4}} - 0 \right| \leq \frac{1}{2x^{2}} + \frac{1}{2y^{2}} < \frac{1}{2{\delta}^{2}}+\frac{1}{2{\delta}^{2}}$

$\implies \displaystyle \left| \frac{x^{2}+y^{2}}{x^{4}+y^{4}} - 0 \right| \leq \frac{1}{2}\left({\frac{1}{\sqrt{\epsilon}}}\right)^{-2}+\frac{1}{2}\left({\frac{1}{\sqrt{\epsilon}}}\right)^{-2} = \frac{\epsilon}{2}+ \frac{\epsilon} {2}$
$\implies \displaystyle \left| \frac{x^{2}+y^{2}}{x^{4}+y^{4}} - 0 \right| \leq \epsilon$
$\implies \displaystyle\lim_{(x,y)\to \infty} f(x,y)= L =0$
[QED]

Source: Prof Zhang ShiZao 章士藻(1940-) “Collected Works of Mathematics Education” 数学教育文集
for Ecole Normale Supérieure in China 高等师范学院

[Video in French ] http://touch.dailymotion.com/video/x89qux

εδ Confusion in Limit & Continuity

April 25, 2013tomcircle Modern Math 2 Comments

1. Basic:
|y|= 0 or > 0 for all y

2. Limit: $\displaystyle\lim_{x\to a}f(x) = L$ ; x≠a
|x-a|≠0 and always >0
hence
$\displaystyle\lim_{x\to a}f(x) = L$
$\iff$
For all ε >0, there exists δ >0 such that
$\boxed{0<|x-a|<\delta}$
$\implies |f(x)-L|< \epsilon$

3. Continuity: f(x) continuous at x=a
Case x=a: |x-a|=0
=> |f(a)-f(a)|= 0 <ε (automatically)
So by default we can remove (x=a) case.

Also from 1) it is understood: |x-a|>0
Hence suffice to write only:
$|x-a|<\delta$

f(x) is continuous at point x = a
$\iff$
For all ε >0, there exists δ >0 such that
$\boxed{|x-a|<\delta}$
$\implies |f(x)-f(a)|< \epsilon$

Limit: ε-δ Analysis

April 6, 2013tomcircle Elementary Math, Modern Math 2 Comments

For x->0, find limit L of

f(x)= (x³+5x)/x

1) guess L:

f(x)= x(x²+5)/x= x²+5

=> L= 5 when x->0

2) epsilon-delta Proof: find δ in function of ε such that:

|f(x)-5| < ε

|(x²+5)-5| <ε

|x|< √ε

Choose δ=√ε

For all ε, there is δ=√ε such that |x-0|< δ =>|f(x)-5|< ε

If ε=0.5, δ=√0.5=0.25

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