The standard notation for a function is the form f(x,y,z) where x,y,z are the arguments. In most of the examples of this blog, I am only using one argument so this type of function is represented as f(x) where x is the argument.

An equation such as e=mc

^{2}is a function. In this case, c is a constant since it is the speed of light so e=mc

^{2}is a function with only one argument m. So, we could represent Einstein's equation as: f(m) = mc

^{2}

A function is said to be continuous if as a function moves through input moves through a continuum of values, the output also moves through a continuum of values. The important idea is that there are no gaps in a continuous function. If a function has gaps, then it is not continuous.

Definition 1: Continuous at a Point

A function is continuous at a point c if and only if:

for any arbitrary value ε, there exists another a positive value δ such that:

if x is a point that lies between c - δ and c + δ, then f(x) lies between f(c) - ε and f(c) + ε

The important idea here is that if a function is continuous at c, then we can choose an arbitary interval ε such that we know that f(x) lies between f(c)-ε and f(c) + ε

Definition 2: Continuous Function

A function is said to continuous if it is continuous at all points in its domain.

Definition 3: Closed Interval [α, β]

A closed interval can be thought of as a set of elements that make up a subset of the domain for a function. For an interval [α, β], this includes all values x such that α ≤ x ≤ β.

Example: a closed interval on the real numbers

For example, we could set up a closed interval [-5,5] this would then include all numbers that are greater or equal to -5 and less than or equal to 5 including -5, -4.5, 0, and 4.

To show the importance of continous functions, let's consider the Weierstrass Intermediate Value Theorem.

Theorem: Weierstrass Intermediate Value Theorem

If f(x) is continuous at all parts of a closed interval [ α, β ] and if f(α) is less than 0 and if f(β) is greater than 0, then there exists a point γ ∈ [α, β] such that f(γ) = 0.

Proof:

(1) Since f(α) is less than 0, we know that there exists a value μ such that for all values x on [α, μ] f(x) is less than 0 and μ ≥ α.

(2) We can assume that if f(x) is less than 0, then it lies on the interval [ α, μ ] since:

(a) If there was a value ζ such that ζ is greater than μ and f(ζ) is less than 0, then it follows that for x greater than μ and less than ζ , f(x) ≥ 0. (Otherwise, we could assume that ζ ∈ [ α, μ ] )

(b) For the interval specified in (a) if f(x)=0, then we are done with the proof so to finish the proof, we can assume that this is not the case.

(c) If in the interval specified in (a) f(x) is greater than 0, then we can set β to a value greater than μ and less than ζ where f(x) is greater than 0.

(d) In this case, we have a smaller interval than the original but we can make the assumption that there is no value ζ greater than μ where f(ζ) is less than 0.

(e) If we prove that f(γ)=0 exists for this smaller interval, we are likewise proving that it exists in the bigger interval.

(3) Let S be the bounded set on [α, μ] (See the note here for the definition of a bounded set)

(4) Since we are dealing with real numbers, we know that there exists a least upper bound γ for S [See here for proof]

(5) Assume f(γ) = v which is greater than 0.

(6) Since we are dealing with a continuous function, by the definition 1 above, we know that there exists a value δ such that at point γ:

if x is between γ - δ and γ + δ, then f(x) lies between f(γ) - v/2 and f(γ) + v/2. [Where I chose ε = v/2 as the arbitrary value.]

(7) From (#6), we know that if x lies between γ - δ and γ + δ, then:

f(x) lies between v - v/2 and v + v/2 [Since f(γ) = v from #5]

(8) From (#7) we have thta:

f(x) lies between v/2 and 3v/2.

(9) Now μ is greater than γ - δ since:

(a) Assume μ ≤ γ - δ

(b) Then, γ - δ is an upper bound for [α,μ] since for all x ∈ [α,μ] x ≤ γ - δ

(c) And γ - δ is less than γ since δ is a positive value.

(d) But γ is the least upper bound from #4

(e) So we have a contradiction and we reject (a).

(10) And μ is less than γ + δ since μ ≤ γ since γ is the least upper bound.

(11) Since μ is in between γ - δ (#9) and γ + δ (#10), we can use (#8) to conclude that:

f(μ) is greater than v/2 which means by step #4 that f(μ) is greater than 0.

(12) But this is a contradiction since f(μ) is less than 0 by our original assumption in #1 so we can reject our assumption in #4 and conclude that f(γ) ≤ 0.

(13) Let's assume that f(γ)=v is less than 0.

(14) Applying #7, we can conclude that if x lies between γ - δ and γ + δ, then:

f(x) is less than 0 since f(x) is less than 3v/2 which ≤ 3(0)/2 ≤ 0 [From #13]

(15) We know that there exists a value ν such that ν is greater than γ and ν is less than γ + δ since:

(a) Let ν = (2*γ + δ)/2. [We can make this assumption be the definition of multiplication, addition, and division on real numbers, see here]

(b) We see ν is less than γ + δ since γ + δ = (2*γ + 2*δ)/2 and 2*δ is greater than δ since δ is a positive value.

(c) We see that ν is greater than γ since γ = 2*γ/2 and 2*γ is less than 2*γ + δ.

(16) Since ν lies between γ - δ and γ + δ, we can use step #13 to conclude that f(ν) is less than 0.

(17) But this also contradicts our assumption in #2 since it presupposes that there exists a value ν which is greater than μ where f(ν) is less than 0.

(18) So we can reject our assumption in #13.

(19) This then gives us that f(γ) =0 since it cannot be greater than 0 (from #12) and it cannot be less than 0 (from #18)

QED

References

- Continuous Function, Wikipedia
- Function, Wikipedia
- Weierstrass Intermediate Value Theorem, MathWorld

## No comments :

Post a Comment