Prequisite knowledge Flashcards
a function f : I → R is smooth if
all its derivatives f′(t), f′′(t), f′′′(t), . . . exist ∀ t ∈ I).
E.g. polynomials, trigonometric functions (sin, cos etc.), exponentials, logarithms, hyperbolic trig functions (sinh, cosh etc.) are all smooth.
We say that the map γ is smooth if every one of its component functions γ_i : I → R,
i = 1, 2, . . . , n is smooth
γ(t) = (γ_1(t), γ_2(t), . . . , γ_n(t)) ∈ R^ n
not smooth example
f : R → R such that f(t) = t^(4/3) is not smooth. Check:
f′(t) = (4/3)t^(1/3) ⇒ f′′(t) = (4/9)t^(− 2/3)
so f′′(0) does not exist.
A parametrized curve in R^n
regular point
singular point
A parametrized curve in R
^n is a smooth map γ : I → R^n.
A time t ∈ I is a regular point of γ if γ′(t) ≠ 0.
If γ′(t) = 0, then t ∈ I is a singular
point of γ.
If every t ∈ I is regular then γ is said to be a regularly parametrized
ie if and only if there does not exist a time t ∈ I such that γ′ (t) = 0 = (0, 0, . . . , 0).
*A PC is a smooth map γ : I → R^n, but it does not necessarily represent a “smooth” curve! A RPC does,
parametrisation of a 3D sphere
spherical coordinates
S^2=
{(x,y,z} ∈R³: x²+y²+z²=1}
x(u,v)= cosv cosu
y(u,v)= cosv sinu
z(u,v)= sinv
v in (-pi/2, pi/2)
u in (0,2pi)
Ω = (0,2pi) X (-pi/2, pi/2)
Ω must be open interior points
image doesnt cover whole surface
subspace of whole space
also is preimage
f-1({1}) of
f:R^2 to R^2
f(x,y,z)x^2+y^2+z^2
x(u,v)= cosv cosu
y(u,v)= cosv sinu
z(u,v)= sinv
v in (-pi/2, pi/2)
u in (0,2pi)
Ω = (0,2pi) X (-pi/2, pi/2)
Ω must be open interior points
drawbacks
This image however doesn’t cover the whole surface as boundary points are missing
for example
u=0
u=2pi
any points satisfying
v=-pi/2
v=-pi/2
missing the poles and missing an arc joining them
Thus it represents a subset of the whole sphere (with a cut)
We can’t use an injective continous map to show the whole sphere
If we are taking this surface and deforming it then we need the whole object/surface
curvature of sphere =1
whenever we deform we will always have curvature, but only a section of the sphere we can deform to e.g. flat 0 curvature
closed curves
are deformations of circles
image set
The image set of a curve γ is the range of the mapping, that is,
γ(I) = {γ(t) ∈ R^n : t ∈ I} ⊂ R^n
E.g consider parabola x_2 = x_1 ^2
There are infinitely many parametrised curves whose image set is this parabola.
γ : R → R^2 , γ(t) = (t, t^2)
δ : R → R^2, δ(t) = (t^3, t^6).
γ is a regularly parametrized curve: γ′(t) = (1, 2t) ≠ (0, 0) ∀t as 1 ≠0
But δ is not: δ′(t) = (3t^2, 6t^5) = (0, 0) when t = 0
γ : R → R^2 , γ(t) = (t^2, t^3) is a smooth map,
is a parametrised curve
but the image set is not smooth (cusp) γ is not a RPC:
γ′(t) = (2t, 3t^2) = (0, 0) when t = 0
Note also that the nasty point in γ(I) oc-curs precisely where γ′(t) = 0, that is, at the singular point of γ.
preimage defn
For any subset B ⊂ Y the subset
u⁻¹(B) := {x ∈ X : u(x) ∈ B} ⊂ X
is called the pre-image of B under u.
The pre-image of a one-point set B = {b} we call the pre-image of a point b, and denote it by u⁻¹(b).
The use doesn’t assume invertibility; defined for any map u : X → Y and any subset B ⊂ Y .
Example I.24.
Let f : R → R be a function given by the formula f(t) = t². Then the pre-image of the interval [1, 2] is
the set
f⁻¹([1, 2])
= [−√2, −1] ∪ [1, √2].
Self-Check Question I.13. Let u : X → Y be a map, and let A ⊂ X be a subset. Is the relation
**u⁻¹ (u(A)) = A **
true? What is the correct statement that relates A and u⁻¹(u(A))?
WHEN INJECTIVE TRUE
u^{-1}(u(A)) = A not always true.
A ⊆ u^{-1}(u(A))
A is a subset of the preimage of its image under u. regardless of the properties of u.
u^{-1}(u(A)) = A if and only if u is injective (one-to-one).
If u is injective, then each point in X has a unique image in Y, and thus the preimage of u(A) contains precisely the elements of A.
if u is not injective, other elements outside of A may also map to u(A).
the whole sphere isn’t a preimage of a..
S^2=
{(x,y,z} ∈R³: x²+y²+z²=1} subset of R^3
considering
f: R³→ R
f(x,y,z) = x²+y²+z² in R
f⁻¹({1})
f⁻¹({1}) =S^2
e.g the set of all real roots
of the polynomial pₙ(t) = a_nt
n + · · · + a_1t + a_0 is the precisely the pre-image
p⁻¹ₙ(0)
where pₙ is map pₙ: R → R
The kernel of a linear map
L : V → W is precisely the pre-image
L⁻¹(0). In particular, a plane H ⊂ R³ can be described as the pre-image L⁻¹(0) of the orthogonal projection
L : R³ → R³ ,
x → (x · v)v, where v is a unit length vector orthogonal to H.
This way of describing geometric objects as pre-images plays an important role in the sequel chapters
