Lemma: Derivative of Increasing and Decreasing Functions

Let f be a continuous function on (a,b) where the at no point f'(x)=0.

If for all x on [a,b], f(x) is increasing, then f'(x) is positive.

If for all x on [a,b], f(x) is decreasing, then f'(x) is negative.

Proof:

(1) From the definition of derivatives (see Definition 1, here):

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

So, the sign of f'(x) is the sign of [f(x + Δx) - f(x)]/Δx and we can assume that is is nonzero.

(2) Case I: Δx is positive

If f(x) is strictly increasing, then f(x + Δx) - f(x) is positive and f'(x) is positive.

If f(x) is striclty decreasing, then f(x + Δx) - f(x) is negative and f'(x) is negative.

(3) Case II: Δx is negative

If f(x) is strictly increasing, then f(x + Δx) - f(x) is negative and f'(x) is positive

If f(x) is strictly decreasing, then f(x + Δx) - f(x) is positive and f'(x) is negative.

QED

## Saturday, February 07, 2009

## Tuesday, February 03, 2009

### An Inequality Lemma for the Cauchy Bound of Real Roots

The following result is used my proof of Cauchy's Bound for real roots.

Lemma 1: abs(c)

Proof:

(1) Assume that c is nonnegative

(2) Then, c

(3) Then, abs(c

(4) Since abs(c) = c, it follows that: c

(5) Assume that c is negative

(6) We can assume that n is odd

[Otherwise, c

(8) abs(c

QED

Lemma 2:

Let:

a

Proof:

(1) Using the Triangle Inequality (see Lemma 4, here), we know that:

abs(-a

so that:

abs(a

(2) Using a basic property of inequalities (see Lemma 1, here):

abs(a

and likewise:

abs(-a

...

(3) So we have:

abs(a

(4) Using Lemma 1 above, we have:

abs(a

QED

Lemma 1: abs(c)

^{n}= abs(c^{n})Proof:

(1) Assume that c is nonnegative

(2) Then, c

^{n}is nonnegative(3) Then, abs(c

^{n}) = c^{n}(4) Since abs(c) = c, it follows that: c

^{n}= abs(c)^{n}(5) Assume that c is negative

(6) We can assume that n is odd

[Otherwise, c

^{n}= (-c)^{n}= abs(c)^{n}= abs(c^{n}) ](8) abs(c

^{n}) = -c^{n}= (-1)^{n}*c^{n}= (-c)^{n}= abs(c)^{n}QED

Lemma 2:

Let:

a

_{n}c^{n}= -a_{n-1}c^{n-1}+ .... + -a_{0}Then:_{}abs(a_{n})*abs(c)^{n}≤ abs(a_{n-1})*abs(c)^{n-1}+ ... + abs(a_{0})Proof:

(1) Using the Triangle Inequality (see Lemma 4, here), we know that:

abs(-a

_{n-1}c^{n-1}+ .... + -a_{0}) ≤ abs(-a_{n-1}c^{n-1}) + ... + abs(-a_{0})so that:

abs(a

_{n}c^{n}) ≤ abs(-a_{n-1}c^{n-1}) + ... + abs(-a_{0})(2) Using a basic property of inequalities (see Lemma 1, here):

abs(a

_{n})*abs(c^{n}) = abs(a_{n}c^{n})and likewise:

abs(-a

_{n-1})*abs(c^{n-1}) = abs(-a_{n-1}c^{n-1})...

(3) So we have:

abs(a

_{n})*abs(c^{n}) ≤ abs(-a_{n-1})*abs(c^{n-1}) + ... + abs(-a_{0})(4) Using Lemma 1 above, we have:

abs(a

_{n})*abs(c)^{n}≤ abs(a_{n-1})*abs(c)^{n-1}+ ... + abs(a_{0})QED

### Triangle Inequality

Definition 1: Absolute Value

abs(a) = a if a is nonnegative or abs(a)=-a if a is negative.

So for example:

abs(5) = 5

abs(0) = 0

abs(-1) = 1

Now, let's look at some basic properties

Lemma 1: abs(ab) = abs(a)*abs(b)

Proof:

Case I: both a,b positive

abs(ab) = ab = abs(a)*abs(b)

Case II: both a,b negative

abs(ab) = ab = (-a)*(-b) = abs(a)*abs(b)

Case III: one negative, one positive

Assume a is positive, b is negative (since a,b are symmetrical, we can switch them as necessary)

abs(ab) = -ab = a*(-b) = abs(a)*abs(b)

QED

Lemma 2: -abs(a) ≤ a ≤ abs(a)

Proof:

Case I: a is nonnegative

-a ≤ a ≤ a

so

-abs(a) ≤ a ≤ abs(a)

Case II: a is negative

a ≤ a ≤ -a

so

-abs(a) ≤ a ≤ abs(a)

QED

Lemma 3: abs(a) ≤ b if and only if -b ≤ a and a ≤ b.

Proof:

(1) Assume abs(a) ≤ b

Case I: a is nonnegative

abs(a) = a

Since abs(a) ≤ b, it follows that a ≤ b and b is nonnegative

Since b is nonnegative and a is nonnegative, then it -b ≤ a.

Case II: a is negative

Since abs(a) ≤ b, it follows that -a ≤ b which is the same as -b ≤ a and therefore b must be nonnegative.

Since b is nonnegative, it follows that a ≤ b.

(2) Assume that -b ≤ a and a ≤ b.

Case I: a is nonnegative

Since a ≤ b, it follows that b is nonnegative

So abs(a) ≤ b.

Case II: a is negative

Since -b ≤ a, it follows that b ≥ -a.

Since a is negative, -a is positive, and we have:

abs(a) ≤ b.

QED

Lemma 4: Triangle Inequality

For all real numbers a,b

abs(a + b) ≤ abs(a) + abs(b)

Proof:

(1) For all real numbers a,b (from Lemma 1 above)

-abs(a) ≤ a ≤ abs(a)

-abs(b) ≤ b ≤ abs(b)

(2) Adding these two conditions together gives us:

-[abs(a) + abs(b)] ≤ a + b ≤ abs(a) + abs(b)

(3) Let c = a+b and d =abs(a) + abs(b)

(4) Using Lemma 3, we know that:

abs(c) ≤ d if and only if -d ≤ c and c ≤ d.

(5) But using step #2, we know that:

-d = -[abs(a) + abs(b)] ≤ c = a + b

and

c = a + b ≤ d = abs(a) + abs(b)

(6) So, using step #4 we get:

abs(c) ≤ d

which is equivalent to:

abs(a + b) ≤ abs(a) + abs(b)

QED

abs(a) = a if a is nonnegative or abs(a)=-a if a is negative.

So for example:

abs(5) = 5

abs(0) = 0

abs(-1) = 1

Now, let's look at some basic properties

Lemma 1: abs(ab) = abs(a)*abs(b)

Proof:

Case I: both a,b positive

abs(ab) = ab = abs(a)*abs(b)

Case II: both a,b negative

abs(ab) = ab = (-a)*(-b) = abs(a)*abs(b)

Case III: one negative, one positive

Assume a is positive, b is negative (since a,b are symmetrical, we can switch them as necessary)

abs(ab) = -ab = a*(-b) = abs(a)*abs(b)

QED

Lemma 2: -abs(a) ≤ a ≤ abs(a)

Proof:

Case I: a is nonnegative

-a ≤ a ≤ a

so

-abs(a) ≤ a ≤ abs(a)

Case II: a is negative

a ≤ a ≤ -a

so

-abs(a) ≤ a ≤ abs(a)

QED

Lemma 3: abs(a) ≤ b if and only if -b ≤ a and a ≤ b.

Proof:

(1) Assume abs(a) ≤ b

Case I: a is nonnegative

abs(a) = a

Since abs(a) ≤ b, it follows that a ≤ b and b is nonnegative

Since b is nonnegative and a is nonnegative, then it -b ≤ a.

Case II: a is negative

Since abs(a) ≤ b, it follows that -a ≤ b which is the same as -b ≤ a and therefore b must be nonnegative.

Since b is nonnegative, it follows that a ≤ b.

(2) Assume that -b ≤ a and a ≤ b.

Case I: a is nonnegative

Since a ≤ b, it follows that b is nonnegative

So abs(a) ≤ b.

Case II: a is negative

Since -b ≤ a, it follows that b ≥ -a.

Since a is negative, -a is positive, and we have:

abs(a) ≤ b.

QED

Lemma 4: Triangle Inequality

For all real numbers a,b

abs(a + b) ≤ abs(a) + abs(b)

Proof:

(1) For all real numbers a,b (from Lemma 1 above)

-abs(a) ≤ a ≤ abs(a)

-abs(b) ≤ b ≤ abs(b)

(2) Adding these two conditions together gives us:

-[abs(a) + abs(b)] ≤ a + b ≤ abs(a) + abs(b)

(3) Let c = a+b and d =abs(a) + abs(b)

(4) Using Lemma 3, we know that:

abs(c) ≤ d if and only if -d ≤ c and c ≤ d.

(5) But using step #2, we know that:

-d = -[abs(a) + abs(b)] ≤ c = a + b

and

c = a + b ≤ d = abs(a) + abs(b)

(6) So, using step #4 we get:

abs(c) ≤ d

which is equivalent to:

abs(a + b) ≤ abs(a) + abs(b)

QED

## Sunday, February 01, 2009

### Polynomials are continuous

For a definition of polynomials, see Definition 1, here. For a definition of continuous functions, see Definition 1, here.

Lemma 1: f(x)=x is continuous

Proof:

(1) Let ε be any arbitrary value.

(2) Let δ = ε

(3) For any point c, it is clear that if x lies in (c - δ, c + δ), then f(x)=x lies in (f(c) - ε, f(c) + ε )

QED

Corollary 1.1 : Polynomials are continuous

Proof:

(1) The function f(x)=x is continuous. [See Lemma 1 above]

(2) Since the product of continuous functions is continuous [See Lemma 3, here], then f(x)=x

(3) Since f(x)=C is continuous [See Lemma 1, here], it follows that any function of the form cx

(4) Since the addition of continuous functions is continuous [See Lemma 2, here], it follows that any polynomial function is continuous since it consists of the form:

f(x) = c

where each c

QED

Lemma 2: The Derivative of a polynomial is itself a polynomial

Proof

(1) The derivative of each term of a polynomial is itself a term of a polynomial [See the Lemma 2, here]

(2) So, it follows that the derivative itself is also a polynomial. [See Definition 1, here]

QED

Corollary 2.1: The derivative of a polynomial is a continuous function.

Proof:

This follows directly from Lemma 2 above and Corollary 1.1 above.

QED

Lemma 1: f(x)=x is continuous

Proof:

(1) Let ε be any arbitrary value.

(2) Let δ = ε

(3) For any point c, it is clear that if x lies in (c - δ, c + δ), then f(x)=x lies in (f(c) - ε, f(c) + ε )

QED

Corollary 1.1 : Polynomials are continuous

Proof:

(1) The function f(x)=x is continuous. [See Lemma 1 above]

(2) Since the product of continuous functions is continuous [See Lemma 3, here], then f(x)=x

^{n}where n is a positive integer is also continuous.(3) Since f(x)=C is continuous [See Lemma 1, here], it follows that any function of the form cx

^{n}is also continuous.(4) Since the addition of continuous functions is continuous [See Lemma 2, here], it follows that any polynomial function is continuous since it consists of the form:

f(x) = c

_{0}+ c_{1}x + c_{2}x^{2}+ ... + c_{n}x^{n}where each c

_{i}is a constant.QED

Lemma 2: The Derivative of a polynomial is itself a polynomial

Proof

(1) The derivative of each term of a polynomial is itself a term of a polynomial [See the Lemma 2, here]

(2) So, it follows that the derivative itself is also a polynomial. [See Definition 1, here]

QED

Corollary 2.1: The derivative of a polynomial is a continuous function.

Proof:

This follows directly from Lemma 2 above and Corollary 1.1 above.

QED

### Interval of a Function with Simple Roots

Lemma: Interval of a Function with Simple Roots

Let f be a function with simple roots such that f(c)=0

Then there exists an interval (a,b) such that:

c is in (a,b)

for all x in (a,b), f'(x) is all positive or all negative

Proof:

(1) Since f has only simple roots and f(c)=0, then it follows that f'(c) ≠ 0. [See Corollary 1.1, here]

(2) Since f is a polynomial, we know that f'(x) is continuous. [See Corollary 2.1, here]

(3) Since f'(c) is nonzero, let ε be nonzero and less than abs{ f'(c) }.

(4) Since f'(x) is continuous at c, there exists a number δ such that if x is in the (c - δ, c + δ), then f'(x) is in (f'(c)-ε, f'(c)+ε). [By the definition of a continuous function]

(5) Since ε is less than abs{f'(c) }, it follows that for x in (c -δ, c + δ), f'(x) is either entirely positive or entirely negative.

QED

Let f be a function with simple roots such that f(c)=0

Then there exists an interval (a,b) such that:

c is in (a,b)

for all x in (a,b), f'(x) is all positive or all negative

Proof:

(1) Since f has only simple roots and f(c)=0, then it follows that f'(c) ≠ 0. [See Corollary 1.1, here]

(2) Since f is a polynomial, we know that f'(x) is continuous. [See Corollary 2.1, here]

(3) Since f'(c) is nonzero, let ε be nonzero and less than abs{ f'(c) }.

(4) Since f'(x) is continuous at c, there exists a number δ such that if x is in the (c - δ, c + δ), then f'(x) is in (f'(c)-ε, f'(c)+ε). [By the definition of a continuous function]

(5) Since ε is less than abs{f'(c) }, it follows that for x in (c -δ, c + δ), f'(x) is either entirely positive or entirely negative.

QED

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