## Saturday, March 11, 2006

### Rolle's Theorem

Rolle's Theorem is one of the basic principles of elementary calculus. It is used to proof the Taylor Series which is used to prove Euler's Formula.

In today's blog, I present a proof based on the maxima and minima property of a continous function.

If you need a review of derivatives, start here.

If you need a review of continuous functions or closed intervals, start here.

If you need a review of the maxima and minima properties of continuous functions over a closed interval, start here.

Theorem: Rolle's Theorem

If a function f(x) is continuous over a closed interval [a,b], differentiable over that interval, with f(a)=0 and f(b)=0, then there exists a point c [a,b] such that f'(c) =0.

Proof:

(1) Because the function f is continuous over the closed interval [a,b], then there exists a maximum value (see Lemma 3 here) and a minimum value (see the Theorem here)

(2) If f has any positive values, then let c = its maximum value over the closed interval [a,b]

(3) Now by assumption in (2), c is not an endpoint since f(a) and f(b) =0 and we are assuming that f(c) is positive. So, we know that c ∈ [a,b]

(4) Since by (2), c is a local maximum (see clarification here for definition), we know by an earlier theorem that f'(c) = 0 (see the Theorem here).

(5) Now, if f has any negative values, then let c = its minimum value over the closed interval [a,b]

(6) Now by assumption (5), c is not an endpoint since f(a) and f(b) = 0 and we are assuming that f(c) is negative. So, we know that c ∈ [a,b]

(7) This will be a local minimum (see clarification here for definition) and we know by an earlier theorem that f'(c) = 0 (see the Corollary here).

(8) If there are no positive or negative values for the function f in [a,b], then we can take any point ∈ [a,b] and know that f(c)=0. We further know that f'(c)=0 since in this case f(x)=C (see Lemma 1 here for the Derivative of f(x)=C).

QED

## Friday, March 10, 2006

### Combining Continuous Functions

Continuous functions can be combined to form new, more complicated continued functions. In today's blog, I will review some basic lemmas on how functions built from continued functions are themselves continued functions. Finally, I will show how these lemmas can be used to establish the minimum property of continuous functions over a closed interval.

If you are not familiar with continuous functions, then should review here.

Lemma 1: Constant Law

f(x) = C where C is a constant is a continuous function

(1) Let δ = 1

(2) Let x,c be any two values in the domain of f(x).

(3) We know that f(x) - f(c) = C - C = 0

(4) So, it is also true, if x-c is between and , for any nonzero value ε, that f(x) - f(c) is between and [Since 0 is between and ]

(5) So, by the definition of continuous functions (see here), f(x)=C is a continuous function.

QED

if f(x) and g(x) are continuous functions, then f(x) + g(x) is a continuous function.

(1) Let ε be any nonzero value.

(2) Let c be any point on the domain of f(x) + g(x).

(3) Since f(x) is continuous, we know that there exists δ1 such that:
if x - c is between 1 and 1, then f(x) - f(c) is between -ε/2 and +ε/2

(4) Since g(x) is continuous, we know that there exists δ2 such that:
if x - c is between 2 and 2, then g(x) - g(c) is between -ε/2 and +ε/2

(5) Let δ = min(δ12)

(6) Now, if x - c is between and , then:

(a) f(x) - f(c) is between -ε/2 and +ε/2

(b) g(x) - g(c) is between - ε/2 and + ε/2

(c) f(x) + g(x) - f(c) - g(c) is between (-ε/2 + -ε/2) and (+ε/2 + +ε/2)

(7) So, f(x) + g(x) is a continous function.

QED

Lemma 3: Multiplication Law
if f(x) and g(x) are continuous functions, then f(x)*g(x) is a continuous function.

(1) Let ε be any nonzero value.

(2) Let c be any point on the domain of f(x)*g(x).

(3) Since f(x) is continuous, we know that there exists δ1 such that:
if x - c is between 1 and 1, then f(x) - f(c) is between and

So there exists a constant A = max(absolute[ε + f(c)], absolute[-ε + f(c)]) such that:
if x - c is between 1 and 1, then f(x) is between -A and +A.

(4) Let B = g(c).

(5) We also know that there exists δ2 such that:
if x - c is between 2 and 2, then f(x) - f(c) is between -ε/(2B) and ε/(2B).

(6) And there exists δ3 such that:
if x - c is between 3 and 3, then g(x) - g(c) is between -ε/(2A) and ε/(2A)

(7) Let δ = min(δ123)

(8) Now, if x - c is between and , then:

(a) f(x) - f(c) is between -ε/(2B) and +ε/(2B)

(b) g(x) - g(c) is between - ε/(2A) and + ε/(2A)

(c) g(c)[f(x) - f(c)] is between (B)[-ε/(2B)] and (B)[+ε/(2B)] which is between -ε/2 and ε/2.

(d) f(x)[g(x) - g(c)] is between (A)[-ε/(2A)] and (A)[+ε/(2A)] which is between -ε/2 and ε/2.

(e) If we add (c) + (d), we get:
f(x)g(c) - f(c)g(c) + f(x)g(x) - f(x)g(c) = f(x)g(x) - f(c)g(c)

(f) So, f(x)g(x) - f(c)g(c) is between (-ε/2 + -ε/2) and (+ε/2 + +ε/2)

(7) So, f(x)g(x) is a continous function.

QED

Lemma 4: Differentiability implies continuity

If f is differentiable at a, then f is continuous at a

Proof:
(1) Since f is differentiable at a, there exists a value c such that f'(a) = c [See here for definition of derivative]

(2) Let ε be any nonzero value

(3) Let δ = Δx

(4) By the definition of the derivative (see here), we know that:
if x - a is between and δ, then f(x) = c so that f(x) - f(c) = 0 which is between ε and

(5) Therefore, by the definition of continuity at a point (see here), f(x) is continuous at point a.

QED

Theorem: Minimum Value Property of Continuous Functions

If a function f is continuous on a closed interval [a,b], then there exists a number c in [a,b] such that f(x) ≥ f(c) for all x in [a,b]

(1) We can construct a continuous function g(x) such that g(x) = (-1)f(x).

We know that g(x) is a continuous function since it is the product of a constant function h(x)=-1 and f(x). [See Lemma 1and Lemma 3 above]

(2) Now, g(x) has a maximum point on the interval [a,b] (from the Maximum Value Property of Continuous Functions, see here)

(3) But since g(x) = -f(x), this corresponds to a minimum point for f(x) since:

(a) Let c be the maximum for g(x)

(b) g(x)g(c) for all x in [a,b]

(c) But then:
-g(x) ≥ -g(c) for all x in [a,b]

(d) Which means that:
f(x) ≥ f(c) for all x in [a,b]

QED

References

## Thursday, March 09, 2006

### Maxima and Minima of Continuous Functions

For a given continuous function on a closed interval, if there is a point in this closed interval where the derivative of f(x) = 0, then this point is either the low point (minima) or high point (maxima) for the continuous function in this interval.

This very basic idea of calculus requires a few lemmas before we are able to prove it.

If you are not familiar with the concept of a function, continuous function, or a closed interval, start here.

A function is said to be bounded if there is a value L such that for all x ∈ [a,b], f(x) ≤ L.

Lemma 1: Nested Interval Property for the real numbers

Suppose that I1, I2, ..., In is a sequence of nested, closed intervals where:
(a) Each Ii+1 is contained with Ii
(b) Each Ii interval is of the form [ai,bi]
(c) The lim (i → inf) (bi - ai) = 0 (See here for review of lim notation and the concept of limit)
Then:
There exists 1 and only 1 point c such that {c} = I1 ∩ I2 ∩ ... ∩ In

Proof:

(1) We know that there can be at most 1 number since lim(i → inf) (bi - ai) = 0 since:

If there were more than 1 number, the lim(i → inf) (bi - ai) would be greater than 0.

The only way that it can be 0 is if bi = c, ai = c, and c-c=0.

(2) We know that there is at least 1 number that is common to all intervals since:

(a) ai has a limit an that fits somewhere in the interval. For all intervals, it is clear that ai ≤ an ≤ bi

(b) Likewise, bi has a limit bn such that ai ≤ bn ≤ bi

(c) So we see that an,bn are both elements of all intervals.

(d) We further note that an = bn since the lim(i → inf)(bi - ai)= 0.

(e) So, if we let c = an = bn, then we are done.

QED

Lemma 2: If a function f is continuous on a closed interval [a,b], then f is bounded there.

(1) Assume that a function f is not bounded on [a,b]

(2) We can bisect the interval [a,b] into two halves which I will label I1 and I2.

(3) We can now pick an interval I which is unbounded. If both intervals are unbounded, then we can pick either one.

(4) We can repeat this process and create a sequence of nested, closed intervals which we can call Ii where each interval selects a subset which is unbounded.

(5) From Lemma 1, we know that there exists a point c which is common to all the intervals in #4.

(6) Because f is continuous, we know that there is a number ε such that f is bounded on the interval c - ε and c + ε [See here for definition of Continuous Functions]

(7) But one of the unbounded values in In must lie within (c - ε, c + ε)

(8) And this is a contradiction since from (#6), it must be bounded.

(9) Therefore, we reject our assumption.

QED

Lemma 3: Maximum value property of continuous functions

If a function f is continuous on the closed interval [a,b], then there exists a number c in [a,b] such that f(x) ≤ f(c) for all x in [a,b]

(1) Let I be the the closed interval [a,b]

(2) From Lemma 2, we know that I is bounded.

(3) Let λ be its least upper bound.

(4) We can divide I in half.

(5) At least one of these halves will have a least upper bound = λ (although it is possible that both have this least upper bound). Let I1 be the division of I which contains λ as the least upper bound.

(6) We can keep dividing up I1 in the same way until we have In which has λ as its upper bound and bn - an = 0.

(7) From Lemma 1, we know that there exists a point c which is common to all these intervals.

(8) It follows from (6) that f(c)=λ since:

(a) There exists a positive value δ such that if x-c is in between and , then f(x)-f(c) is between and ε. [From the definition of a continuous function, see here]

(b) From (a), we have that f(c) - ε is less than f(x) which is less than f(c) + ε

(c) Since ε can be arbitrarily small, we can have f(c) ≤ f(x) ≤ f(c) which means that f(x)=f(c) at some point.

(d) In this case, f(c) cannot be more than λ since λ is an upper bound. [See here for the definition of an upper bound]

(e) Likewise, f(c) cannot be less than λ since λ is the least upper bound. [See here for the definition of a least upper bound]

(f) Therefore, f(c) = λ

QED

Definition 1: Right Hand Limit: lim (x → a+) f(x)

lim (x → a+) f(x) = L if and only if:

if x is between a and a + δ, then f(x) - f(a) is between and ε

Definition 2: Left Hand Limit: lim(x → a-) f(x)

lim(x → a-) f(x) = L if and only if:

if x is between a - δ and a, then f(x) - f(a) is between and ε

Lemma 4: One-sided and two-sided limits

The limit lim (x → a) for f(x) exists and is equal to the number L if and only if the one-sided liimits lim (x → a+) f(x) and lim (x → a-) f(x) both exist and both are equal to the number L.

Proof:

(1) Assume lim(x → a) f(x) = L

(2) Then if x - a is between δ and - δ, then f(x) - f(a) is between and [ By the definition of continuous functions, see here]

(3) Now x - a is between δ and implies that:
x is between a - δ and a + δ.

(4) Since a is greater than a - δ (since δ is a positive value), this implies that:
if x is between a and a + δ, then x is necessarily between a - δ and a + δ.

(5) (#4) combined with (#2) gives us:
lim (x → a+) f(x) = L. [See definition 1 above]

(6) Since a is less than a + δ, this implies that:
if x is between a - δ and a, then x is between a - δ and a + δ.

(7) (#6) combined with (#2) gives us:
lim (x → a-) f(x)= L. [See definition 2 above]

(8) So, that proves the first half of the above theorem.

(9) Assume that lim (x → a+) f(x) = L and lim(x → a-) f(x) = L

(10) From definition 1, we have if x is between a and a + δ, then f(x) - f(a) is between and ε

(11) From definition 2, we have if x is between a-δ and a, then f(x) - f(a) is between and ε

(12) Combining (10) and (11) gives us:
if x is between a-δ and a+δ then f(x) - f(a) is between and ε [For either condition (10) applies or condition (11) applies]

(13) Then, applying the definition of limit (see here), we get:
lim(x → a) f(x) = L

QED

Clarification: Local Maxima and Minima

A maximum or minimum is local if it is true for a given interval. A maximum or minimum is absolute if it is true across a domain. A maximum is value that is greater than all other points in for the range in question. A minimum is a value that is smaller or equal to all other points in the range in question.

Theorem: Local Maxima

If a function f(x) is differentiable at c and is defined as an open interval containing c and if f(c) is a local maximum value of f(x), then f'(c)=0.

Proof:

(1) Assume that f(c) is a local maximum value for f(x) on the open interval (a,b).

(2) Since c is differentiable, it means that right-hand and left-hand limits both exist and are equal to f'(c). [See Lemma 4 above]

lim (Δx → 0+) [f(c + Δx) - f(c)]/Δx = f'(c)

lim (Δx → 0-) [f(c + Δx) - f(c)]/Δx = f'(c)

(3) If Δx is greater than 0, then:

[f(c + Δx) - f(c)]/Δ x ≤ 0

This is true since f(c) ≥ f(c +Δx) for all small positive values of Δx since f(c) is a local maximum by assumption.

(4) If Δx is less than 0, then:

[f(c + Δx) - f(c)]/Δ x ≥ 0

This is true since in this case, we have a negative value f(c + Δx) - f(c) over another negative value Δx.

(5) So, we have from (#2):
f'(c) = lim(Δx → 0+) [f(c + Δx) - f(c)]/Δ x ≤ 0

(6) But we also have from (#2):
f'(c) = lim(Δx → 0-) [f(c + Δx) - f(c)]/Δx ≥ 0

(7) Combining (#5) and (#6), we can conclude that f'(c) = 0 [since f'(c) ≥ 0 and f'(c) ≤ 0 ]

QED

Corollary: Local Minima

If a function f(x) is differentiable at c and is defined as an open interval containing c and if f(c) is a local minimum value of f(x), then f'(c)=0.

Proof:

(1) Assume that f(c) is a local minimum value for f(x) on the open interval (a,b).

(2) Since c is differentiable, it means that right-hand and left-hand limits both exist and are equal to f'(c). [See Lemma 4 above]

lim (Δx → 0+) [f(c + Δx) - f(c)]/Δx = f'(c)

lim (Δx → 0-) [f(c + Δx) - f(c)]/Δx = f'(c)

(3) If Δx is greater than 0, then:

[f(c + Δx) - f(c)]/Δ x ≥ 0

This is true since f(c) ≤ f(c +Δx) for all small positive values of Δx since f(c) is a local minimum by assumption.

(4) If Δx is less than 0, then:

[f(c + Δx) - f(c)]/Δ x ≤ 0

This is true since in this case, we have a positive value f(c + Δx) - f(c) over a negative value Δ x.

(5) So, we have from (#2):
f'(c) = lim(Δx → 0+) [f(c + Δx) - f(c)]/Δ x ≥ 0

(6) But we also have from (#2):
f'(c) = lim(Δx → 0-) [f(c + Δx) - f(c)]/Δx ≤ 0

(7) Combining (#5) and (#6), we can conclude that f'(c) = 0 [since f'(c) ≥ 0 and f'(c) ≤ 0 ]

QED

References

## Wednesday, March 08, 2006

### Mathematical Limits

One of the most important ideas in calculus is the concept of the mathematical limit. Limits relate to continuous functions and the basic idea is that if as a value the argument x approaches the value a, the difference between the limit L and f(x) can be arbitrarily small.

Definition 1: Mathematical Limit:

A function f(x) has a limit of L at point a if given any number ε, there exists a positive number δ such that:
if x-a lies between and , then f(x) - L lies between and ε

This definition is very similar to the definition of a continuous function (see here) and it is not surprising that the two concepts are very closely related.

In today's blog, I will need two definitions in order to prove the Squeeze Law relating to mathematical limits.

Definition 2: Open Interval :

x is an element of an open interval (α, β) if x is greater than α and x is less than β

Definition 3: Deleted Neighborhood

A deleted neighborhood is a set of points that result from deleting a single point in an open interval.

Lemma 1: Constant Law for Limits

if f(x) = C, then lim (x → a) f(x) = C

Proof:

(1) Let δ = 1

(2) if x - a lies between and , we know that f(x) = C.

(3) So, we know that f(x) - C = C - C = 0 which is less than any positive value ε

QED

Lemma 2: Product Law

if lim (x → a) f(x) = L and lim(x → a)g(x) = M, then
lim(x → a)[f(x)*g(x)] = L * M

Proof:

(1) Let ε be any nonzero value. We will prove that f(x)g(x) - LM lies between and

(2) Since the limit of f(x) = L, we know that there exists δ1 such that:
if x - a is between 1 and 1, then f(x) - L is between and

Since by definition, if x - a is between 1 and 1, then f(x) is between -L and +L.

(3) We also know that there exists δ2 such that:
if x - a is between 2 and 2, then f(x) - L is between -ε/(2M) and ε/(2M).

The definition for limits is that for any given positive value (ε), we can find a positive value (δ) to get the result (see above if review is needed).

(4) And there exists δ3 such that:
if x - a is between 3 and 3, then g(x) - M is between -ε/(2L) and ε/(2L)

(5) Let δ = min(δ123)

(6) Now, if x - a is between and , then:

(a) f(x) - L is between -ε/(2M) and +ε/(2M)

(b) g(x) - M is between - ε/(2L) and + ε/(2L)

(c) M[f(x) - L] is between (M)[-ε/(2M)] and (M)[+ε/(2M)] which is between -ε/2 and ε/2.

(d) f(x)[g(x) - M] is between (L)[-ε/(2L)] and (L)[+ε/(2L)] which is between -ε/2 and ε/2.

(e) If we add (c) + (d), we get:
f(x)M - LM + f(x)g(x) - f(x)M = f(x)g(x) - LM

(f) So, f(x)g(x) - LM is between (-ε/2 + -ε/2) and (+ε/2 + +ε/2) which means that it is between and

(7) So LM is the limit for f(x)g(x).

QED

Lemma 3: Squeeze Law

Suppose f(x), g(x), h(x) are functions such that
(a) f(x) ≤ g(x) ≤ h(x) for a deleted neighborhood (α, β) where point a is removed.
(b) lim (x→ a) f(x) = L = lim(x→a)h(x).

Then:
lim (x→ a) g(x) = L

Proof:

(1) Let ε be an arbitary number.

(2) Using the definition of limits, we know that there exists δ1 and δ2 such that:

if x-a lies between 1 and 1, then f(x)-L lies between and

if x-a lies between 2 and 2, then h(x)-L lies between and

(3) Let δ = min(δ12)

(4) We know that δ is greater than 0. [By the definition of mathematical limit]

(5) If x-a in between and , we know that f(x) and h(x) are both points of the open interval (L-ε, L+ε) [Again, from the definition of mathematical limit]

(6) So L-ε is less than f(x) ≤ g(x) ≤ h(x) which is less than L + ε

(7) Combining (#5) and (#6), this gives us that for any given ε, there exists a δ such that:
if x-a is in between and , then g(x)-L is between and

(8) From (#7), L is also a limit for g(x) as x approaches a.

QED

Lemma 4: Substitution Law

If lim (x → a) g(x) = L and lim (x → L) f(x) = f(L), then lim (x → a) f(g(x)) = f(L)

Proof:

(1) Let ε be any positive real value.

(2) Because lim (y → L) f(y) = f(L), we also know that there exists a value δ1 such that:

if (y - L) is between 1 and 1, then f(y) - f(L) is between and

(3) Because lim (x → a) g(x) = L, we know that there exists a value δ2 such that:

if (x - a) is between - δ2 and 2, then g(x) - L is between 1 and 1

(4) But this means if y = g(x), then:

if (x -a ) is between 2 and 2, then y - L is between 1 and 1 and f(g(x)) - f(L) is between and

(5) This then proves that:

lim (x → a) f(g(x)) = f(L)

QED

Lemma 5: lim (x → a) (1/x) = 1/a if a ≠ 0

Proof:

(1) Let ε be any positive real number.

(2) Assume that a is greater than 0.

(3) abs(1/x - 1/a) = abs([a - x]/ax) = abs([x -a]/ax) = (1/a)abs(x-a)/abs(x)

(4) Let us assume that abs(a-x) is less than a/2.

We can do this since abs(x-a) approaches 0 as x moves toward a.

(5) Then x -a is between -a/2 and +a/2 which means that x is between a/2 and 3a/2.

(6) This gives us that abs(x) is greater than a/2 and 1/abs(x) is less than 2/a.

(7) So that abs(1/x - 1/a) = abs(x-a)*(1/a)*abs(1/x) which is less than abs(x-a)*(1/a)(2/a) = 2/a2 * abs(x-a)

(8) Let δ be the minimum of a/2 and a2ε/2

(9) Then if x - a is between and , then:

abs(1/x - 1/a) is less than (2/a2)(a2ε/2) = ε

This then proves that lim (x → a) (1/x) = 1/a for when a is greater than 0.

(10) Assume that a is less than 0

(11) Then

abs(1/x - 1/a) = abs(x-a)/(-a)*1/abs(x)

(12) If we assume that abs(x-a) is less than -a/2, then:

x - a is between -a/2 and +a/2, then x is between -3a/2 and -a/2.

(13) So abs(x) is greater than -a/2.

(14) So 1/abs(x) is less than -2/a.

(15) In this case, then:

abs(1/x - 1/a) = abs(x-a)/(-a)*1/abs(x) which is less than 1/(-a)*(-2/a)*abs(x-a) = 2/a2*abs(x-a)

(16) Let δ be the minimum of -a/2 and a2ε/2

(17) Then if x - a is between and , then:

abs(1/x - 1/a) is less than (2/a2)(a2ε/2) = ε

This then proves that lim (x → a) (1/x) = 1/a for when a is less than 0.

QED

Lemma 6: Reciprocal Law

if lim(x → a) g(x) = L and L ≠ 0, then lim (x → a) 1/g(x) = 1/L

Proof:

(1) Let f(x) = 1/x

(2) lim (x → a) f(x) = lim (x → a) (1/x)

(3) Using Lemma 5 above, we have:

lim (x → a) f(x) = 1/L = f(L)

(4) Applying the Substitution Law (Lemma 4 above) gives us:

lim (x → a) 1/g(x) = lim (x → a) f(g(x)) = f(L) = 1/L

QED

Lemma 7: Quotient Law

if lim (x → a) f(x) = L and lim (x → a) g(x) = M ≠ 0, then:

lim (x → a) f(x)/g(x) = L/M

Proof:

(1) Using the Product Law above, we have:

lim (x → a) f(x)/g(x) = lim (x → a)f(x) * lim(x → a)1/g(x)

(2) Using the Reciprocal Law above:

lim (x → a) 1/g(x) = 1/M

(3) Combining step #1 and step #2 gives us:

lim (x → a) f(x)/g(x) = L*(1/M) = L/M

QED

Lemma 8: abs(a + b - c - d)) ≤ abs(a - c) + abs(b - d)

Proof:

(1) If (a-c),(b-d) are the same sign, then abs(a + b - c -d) = abs(a -c) + abs(b - d)

(2) If (a-c),(b-d) are not the same sign, then abs(a - c + b - d) is less than abs(a -c) + abs(b-d).

QED

if lim (x → a) f(x) = L and lim(x → a)g(x) = M, then
lim(x → a)[f(x)+g(x)] = L + M

Proof:

(1) Let ε be any nonzero value.

(2) Since the limit of f(x) = L, we know that there exists δ1 such that:
if x - a is between 1 and 1, then f(x) - L is between -ε/2 and +ε/2

Since by definition, if x - a is between 1 and 1, then f(x) is between -L and +L.

(3) Since the limit of g(x) = L, we know that there exists δ2 such that:
if x - a is between 2 and 2, then g(x) - M is between -ε/2 and +ε/2

Since by definition, if x - a is between 2 and 2, then g(x) is between -M and +M.

(4) Let δ = min(δ12)

(5) Now, if x - a is between and , then:

(a) f(x) - L is between -ε/2 and +ε/2

(b) g(x) - M is between - ε/2 and + ε/2

(c) By Lemma 8 above, abs([f(x) + g(x)] - (L + M)) ≤ abs(f(x) - L) + abs(g(x) - M)

(d) abs(f(x) - L) + abs(g(x) - M) ≤ η/2 + η/2 = η

(7) So L+M is the limit for f(x)+g(x).

QED

References

## Tuesday, March 07, 2006

### Continous Functions

In mathematics, a function is a mapping from a set of input to an output. Each input is called an argument and the set of all arguments that can be put into this function is called the domain. The set of all outputs is called the range.

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=mc2 is a function. In this case, c is a constant since it is the speed of light so e=mc2 is a function with only one argument m. So, we could represent Einstein's equation as: f(m) = mc2

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

## Monday, March 06, 2006

### Dedekind Cut

The Dedekind Cut is mathematical construction created by Richard Dedekind to provide a definition for the real numbers.

The Dedekind Cut itself is defined in terms of the rational numbers.

Definition 1 - Dedekind Cut

A Dedekind cut α is defined as the subset of the rational integers Q (ratios of integers) which is less than α.

NOTE: Q is used to present the set of all rational numbers; R is used to represent the set of all real numbers, and Z is used to represent the set of all integers.

Example: π

We could create a Dedekind Cut around π. In this case, we could think of 3, 22/7, -4, etc. as elements of the Dedekind Cut. On the other hand, 32/10, 4, etc. would not be elements of the cut.

Definition 2 - Set of Real Numbers R

The set of real numbers is the set of Dedekind cuts α that have the following properties:

(a) α is not empty

(b) α contains no greatest element

For any element xα, there exists y ∈ α such that x is less than y.

(c) If x,y are rational integers where y is less than x, then x ∈ α → y ∈ α.

With these definitions, we have enough the construct the properties of the real numbers.

0 is defined as the set { x ∈ Q such that x is less than 0 }

Definition 4: Multiplicative Identity: 1

1 is defined as the set { x ∈ Q such that x is less than 1 }

α + β is defined as the set { x + y such that x ∈ α, y ∈ β }

Definition 6: Subtraction

α - β is defined as the set { x - y such that x ∈ α, y ∈ β }

Definition 7: Multiplication

α * β is defined as the set { x * y such that x ∈ α, y ∈ β }

Definition 8: Division

α / β is defined as the set { x/y such that β ≠ 0, x ∈ α, y ∈ β }

Definition 9: Irrational Numbers

A real number α is said to be irrational if α ∩ Q does not have a least element.

Now, I will present a proof for a fundamental property of reals using the Dedekind Cut.

Theorem 1: Every real number that is bounded above has a least upper bound.

(1) Let Α be a set of real numbers that is bounded by γ such that α ∈ Α → α ≤ γ

NOTE: Bounded above just means that there is an element γ that is greater or equal to all the elements that make up A.

(2) The union of all the sets that make up Α are themselves a real number because:

(a) Since each real number α is a Dedekind cut, we know that each α is the set of x ∈ Q such that x is less than some real number.

(b) Since each α is not empty, the union of all elements that make up α is not empty.

(c) We know that the union does not have a greatest element since if it did, this element would likewise be the greatest element for whichever real number that it is an element for but this is impossible since by definition, none of the real numbers have a greatest element.

(d) If x ∈ the Union and y is less than x, then y is necessarily an element of the Union, since y would necessarily be an element of the real number that x is an element of.

(3) The union of all sets that make up A is an upper bound for A since every element of A ⊆ Union of all sets of A.

(4) Now, γ is an arbitrary upper bound, so all we need to prove is that A ⊆ γ since this shows that A is necessarily less than or equal to any given upper bound.

(5) But this is easy to prove since x ∈ A → x ∈ γ since we defined each element of A as less than γ.

QED

Theorem 2: For any positive real number ε, there exists a natural number n such that:

0 is less than 1/n is less than ε

Proof:

(1) We can find a natural number n such that:

(n-1) ≤ (1/ε) ≤ n.

(2) Taking the reciprocal for each gives us:

1/(n-1) ≥ ε ≥ 1/n.

QED

Corollary 2.1: Between any two distinct real numbers, there exists a rational number.

Proof:

(1) Let x,y be real numbers such that y is greater than x.

(2) By Theorem 2 above, there exists an integer n such that: 1/n is less than y - x.

(3) Let m = floor(n*x) where floor(n*x) returns the highest integer that is less than n*x.

(4) From this, we know that:

(m/n) ≤ x and x is less than (m+1)/n.

(5) Further, we have:

(m + 1)/n = m/n + 1/n ≤ x + 1/n is less than x + y - x = y.

(6) So that, we have:

x is less than (m+1)/n is less than y.

QED

References