Intermediate Value Theorem

Intermediate Value

Suppose

f : [a, b] \to ℝ

is continuous and

z \in ℝ

such that

f (a) < z < f (b)

then there exists a point

c \in (a, b)

such that

f (c) = z

Path

A path in

S \subseteq ℝ^{n}

from

a

b

(elements of

S

) is the image of a continuous function

γ : [0, 1] \to S

such that

γ (0) = a

and

γ (1) = b

Intermediate Value for Paths

Suppose that

S \subseteq ℝ^{n}

and

f

is a continuous real-valued function on

S

. If there is a path from

a

b

S

and

z \in ℝ

such that

f (a) < z < f (b)

then there is a point

c

on the path such that

f (c) = z

The circle into R is not One to One

Suppose that

f

is a continuous function from a circle into

ℝ

then

f

cannot be one to one.

Select 3 distinct points on the circle, $x, y, z \in C \subseteq ℝ^{2}$ , if any of $f (x), f (y), f (z)$ were equal than we would be done, so assume they are distinct, and without loss of generality that $f (x) < f (y) < f (y)$ .

Consider the paths defined by the line segments tracing the circle from $x$ to $y$ then $y$ to $z$ and $z$ back to $x$ , call these $P_{1}, P_{2}, P_{3} \subseteq C$ respectively.

Since $f$ is continuous we can see that $f (P_{1}) = [f (x), f (y)]$ , $f (P_{2}) = [f (y), f (z)]$ and that $f (P_{3}) = [f (x), f (z)]$

Todo formalize the above, but pretty much get a contradiction because the intersection of their images is non-empty

We can make the Absolute value of a Polynomial as Small as we Want

Let

ϵ \in ℝ^{+}

then there exists a

δ \in ℝ^{+}

such that

| x | < δ ⟹ | p (x) | < ϵ

Suppose that

p (x) = \sum_{i = 0}^{n} a_{i} x^{i}

we note that

| p (x) | \leq \sum_{i = 0}^{n} | a_{i} | | x^{i} |

Therefore by selecting

δ = \min (1, \frac{ϵ}{(n + 1) \cdot \max {| a_{i} | : i \in [1, \dots, n]}})

choosing the min of one in the above so that we have the property that for any

i \in ℕ_{0}

we we have

{| x |}^{i} \leq | x |

showing us that

\begin{matrix} | p (x) | & \leq \sum_{i = 0}^{n} | a_{i} | | x^{i} | \\ \leq \sum_{i = 0}^{n} | a_{i} | | x | \\ \leq \sum_{i = 0}^{n} \frac{| a_{i} |}{\max {| a_{i} | : i \in [1, \dots, n]}} (\frac{ϵ}{n + 1}) \\ \leq \sum_{i = 0}^{n} 1 (\frac{ϵ}{n + 1}) \\ \leq ϵ \end{matrix}

We can make the Absolute value of a Polynomial as small as we Want Inverse Edition

Let

ϵ \in ℝ^{+}

then there exists a

δ \in ℝ^{+}

such that

| x | > δ ⟹ | p (x^{- 1}) | < ϵ

Let

ϵ \in ℝ^{+}

by the above we obtain some

δ^{'} \in ℝ^{+}

such that if

| x | < δ^{'}

then

| p (x) | < ϵ

, take

δ = \frac{1}{δ^{'}}

and assume that

| x | > \frac{1}{δ^{'}}

therefore

δ^{'} > \frac{1}{| x |} = | \frac{1}{x} |

therefore we have that

| p (x^{- 1}) | < ϵ

as needed.

A Polynomial of Odd Degree has at Least One Root

Any monic polynomial

p (x) \in ℝ [x]

such that

\deg (p) = 2 k + 1

for some

k \in ℕ_{0}

has at least one real root.

We approach with the intermediate value theorem in mind, we note that

\frac{p (x)}{x^{2 k + 1}} = 1 + \sum_{n = 0}^{2 k} a_{n} x^{n - (2 k + 1)}

and that

n - (2 k + 1) \leq - 1

therefore let

q (x) = \sum_{n = 0}^{2 k} a_{n} x^{2 k + 1 - n}

so that we have

\frac{p (x)}{x^{2 k + 1}} = 1 + q (x^{- 1})

Now note that for any

ϵ \in (0, 1)

there is some

δ \in ℝ^{+}

such that if

| x | > δ

we have

| q (x^{- 1}) | < ϵ

, so that

q (x^{- 1}) \geq - ϵ

to allow us to say:

\frac{p (x)}{x^{2 k + 1}} = 1 + q (x^{- 1}) > 1 - ϵ

so that we conclude

p (x) > x^{2 k + 1} - ϵ \cdot x^{2 k + 1} = (1 - ϵ) x^{2 k + 1}

noticing that

1 - ϵ \in ℝ^{+}

thus by taking

a = δ + 1

(so that

| x | > δ

we see that

p (a) > (1 - ϵ) a^{2 k + 1} > 0

. We also take a moment to note that for symmetrical reasons

p (- a) < 0

therefore since

p (- a) < 0 < p (0)

there exists a point

c \in (- a, a)

such that

p (c) = 0

, as needed.