Real Analysis/Applications of Derivatives

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Real Analysis Applications of Derivatives

Riemann integration→

Derivatives are also used in theorems. Even though this chapter is titled "Applications of Derivatives", the following theorems will only serve as much application as any other mathematical theorem does in relation to the whole of mathematics. The following theorems we will present are focused on illustrating features of functions which are useful in an identification sort-of-sense. Since graphical analysis is constructed using a different set of analyses, the theorems presented here will instead be applicable to only functions. However, all of what this chapter will discuss on has a graphical component, which this chapter may make reference to in order to more easily bridge a connection. In Real Analysis, graphical interpretations will generally not suffice as proof.

Higher Order Derivatives

To begin our construction of new theorems relating to functions, we must first explicitly state a feature of differentiation which we will use from time to time later on in this chapter.

Suppose $f:\mathbb {R} \to \mathbb {R}$ be differentiable

Let $f'(x)$ be differentiable for all $x\in \mathbb {R}$ . Then, the derivative of $f'(x)$ is called the second derivative of $f$ and is written as $f''(a)$ .

What we have stated is that there may exist a second derivative, which is a derivative of a derivative. We will further define second derivative at a to refer to a derivation of a derivative only at the value a.

Similarly, we can define the n^th-derivative of $f$ , written as $f^{(n)}(x)$

Foundation Theorems

This short section will first introduce some intriguing properties that differentiation has to offer. Concepts here will help shed insight into the latter theorems. In the previous chapter, you have been introduced to the concept that being able to take the derivative of a function implies continuity at that point. This is important to remember as each theorem will use concepts that continuity offers to justify their proof.

In this section, we will provide more groundwork.

Minimum and Maximum Points

We will now introduce two new concepts about functions, both of which you may be familiar with. We will then justify its existence by creating a theorem to go along with it.

Definitions

We will define a maximum point for a function $f$ on an interval A as such:

$\exists \alpha \in A:f(\alpha )\geq f(x)\quad \forall x\in A$

We will define a minimum point for a function $f$ on an interval A as such:

$\exists \zeta \in A:f(\zeta )\leq f(x)\quad \forall x\in A$

Both definitions complement each other, as they refer to opposing inequalities.

Existence of Minimum and Maximum Points Theorem

This proof will justify the minimum and maximum point definition by relating it to differentiation, namely through this statement

Theorem Given a function $f$ defined on $(a,b)$ which is both differentiable at $x$ and it is a maximum or minimum point, its derivative at $x$ is equal to 0

The proof is straightforward: it invokes the definition of differentiation to assert the result of the theorem.

Proof that a Maximum Point has a derivative of 0 for any Given Defined Function

Application of definition. Note that the variable $h$ is any number such that the addition of it with α is still within the interval $(a,b)$ .	$f(\alpha )\geq f(\alpha +h)$
Algebraic Manipulations	$f(\alpha +h)-f(\alpha )\leq 0$
Divide by h, which means two cases because it may be negative (and that means an inequality change!)	$h>0$	$h<0$
	${\frac {f(\alpha +h)-f(\alpha )}{h}}\leq 0$	${\frac {f(\alpha +h)-f(\alpha )}{h}}\geq 0$
One sided limits are applied, which is algebraically valid if applied to both sides (which it is; limit of 0 is 0)	$\lim _{h\to 0^{+}}{\frac {f(\alpha +h)-f(\alpha )}{h}}\leq 0$	$\lim _{h\to 0^{-}}{\frac {f(\alpha +h)-f(\alpha )}{h}}\geq 0$
Merge the one-sided limits together to form a full limit, which demands an equal limit to be valid. Only one value is equal: 0.	$\lim _{h\to 0}{\frac {f(\alpha +h)-f(\alpha )}{h}}=0$
	$f'(\alpha )=0$
$\blacksquare$

To prove the case for the minimum point, you simply reverse the initial inequality. Note that the proof will still yield the same result, namely that the derivative of the minimum point is also 0.

Convexity and Concavity

An example of a convex function at an interval $[x_{1},x_{2}]$

An example of a concave function at an interval $[x,y]$

Likewise, we will now create two more complementary definitions related to functions. These are the convexity and concavity definitions, a method of describing functions based on how the function relates to some given reference point, in this case a line. Convexity and concavity mirror the visual, physical description of convex and concave, although they are described in reference to how the line protrudes out from the function instead of the other way around, so that graphically the definitions appear reversed.

Definitions

We will define convexity of a function $f$ over an interval $A$ as such:

Given $A=[a,b]$ , the secant line joining $(a,f(a))$ and $(b,f(b))>f$ for all $x\in (a,b)$

We will define concavity of a function $f$ over an interval $A$ as such:

Given $A=[a,b]$ , the secant line joining $(a,f(a))$ and $(b,f(b))<f$ for all $x\in (a,b)$

If you take the definition of convexity as is, the mathematical description will appear as such

${\frac {f(b)-f(a)}{b-a}}(x-a)+f(a)>f(x)$

Which can be expressed differently so that although it appears more foreign, it is more applicable to theorems.

${\begin{aligned}{\dfrac {f(b)-f(a)}{b-a}}(x-a)+f(a)&>f(x)\\{\dfrac {f(b)-f(a)}{b-a}}(x-a)&>f(x)-f(a)\\{\dfrac {f(b)-f(a)}{b-a}}&>{\dfrac {f(x)-f(a)}{x-a}}\end{aligned}}$

Likewise, the definition of concavity will simply have the inequality reversed.

This definition is the one we will use in future convexity/concavity theorems in this chapter.

Corollary

The first corollary that comes from this definition is a simple, yet often not explained facet of properties coming from negating a function

Corollary Given a convex function $f$ over the interval $A$ , $-f$ is concave over the interval $A$

Proof that Convexity is Negated Concavity

Application of convexity definition.	${\dfrac {f(b)-f(a)}{b-a}}<{\dfrac {f(x)-f(a)}{x-a}}$
Supposing its negation and after some algebraic manipulations	${\begin{aligned}{\dfrac {f(b)-f(a)}{b-a}}&<{\dfrac {f(x)-f(a)}{x-a}}\\{\dfrac {-f(b)-(-f(a))}{b-a}}&<{\dfrac {-f(x)-(-f(a))}{x-a}}\\-{\dfrac {f(b)-f(a)}{b-a}}&<-{\dfrac {f(x)-f(a)}{x-a}}\end{aligned}}$
The definition of concavity is reached.	${\dfrac {f(b)-f(a)}{b-a}}>{\dfrac {f(x)-f(a)}{x-a}}$
$\blacksquare$

Tangent of a Function

Conceptually, finding the derivative means finding the slope of the tangent line to the function. Thus the derivative can be thought of as a linear, or first-order, approximation of the function. This can be graphically represented by creating a tangential line using the derivative of $f$ at $a$ through this formula

$y=\lim _{h\to 0}{\frac {f(a+h)-f(a)}{h}}(x-a)-f(a)$

Which is clearly the equation of a line translated to the point $(a,f(a))$ through function manipulations.

"Differentiation Produces Tangents" Theorem

This unnamed theorem is the theorem that proves our definition earlier is not wrong. Using it, our understanding of what differentiation graphically represents; a tangent line, should become rigorously clear. It is so ingrained that just in the section above, we defined what a tangent line is in reference to differentiation. There, we will prove the validity of such a statement.

How do we prove this notion? Well, we simply prove the physical concept of what being tangential is: if we have a perfectly round surface, we can imagine balancing a ruler on it and that the ruler is only balancing on one point. This physical description implies that there is a gap everywhere else, and this concept of a "gap" will be expressed mathematically by the inequality between the value of the tangent line to the curved function.

Theorem Given a function $f$ and any tangent line through some point $(a,f(a))$ , this tangent line will be exclusively greater than or less than the function except for $(a,f(a))$ , which will be equal

To begin our proof, we must first simplify the problem and assign a direction to our proof. Overall, we will [ADD OVERALL OBJECTIVE OF PROOF]

Now, we will assign cases to allow us to use our known properties properties.

Case 1: Convex

The first case we will tackle is when the function is convex over some interval $A$ .

Proof of the Existence of Tangent Lines for Convex Functions

Suppose we animate the definition of a derivative. This means that given the definition, we analyze all values of h as it approaches 0.	${\dfrac {f(a+h)-f(a)}{h}}$
By definition, this is not a tangent line, but a secant line.	${\dfrac {f(a+h_{1})-f(a)}{h_{1}}}<{\dfrac {f(b)-f(a)}{b-a}}$
	${\dfrac {f(a+h_{2})-f(a)}{h_{2}}}<{\dfrac {f(b)-f(a)}{b-a}}$
and because	${\dfrac {f(a+h_{1})-f(a)}{h_{1}}}<{\dfrac {f(b)-f(a)}{b-a}}$
	${\dfrac {f(a+h_{2})-f(a)}{h_{2}}}<{\dfrac {f(b)-f(a)}{b-a}}$

Theorems about Tangents

These three theorems all relate to how a tangent relates to intersections of the function and the graph.

Rolle's Theorem

Rolle's Theorem is the introductory theorem for derivatives. This proof will open up access to many other proofs in this chapter, especially the next big differentiation theorem found in the next section, Mean Value Theorem. Rolle's Theorem is described below

Theorem Given a function $f:[a,b]\to \mathbb {R}$ which is both continuous on $[a,b]$ and differentiable on $(a,b)$ and $f(a)=f(b)$ , there exists $c\in (a,b)$ such that $f'(c)=0$

Proof

We will present two proofs for Rolle's Theorem. The first utilizes the Existence of Minimum and Maximum Points Theorem, while the second utilizes a contradiction.

Proof 1

We will first remove an obvious case. If the function $f$ is a constant function, then $f'(x)=0\quad \forall x\in [a,b]$ . $\blacksquare$

If the function $f$ is not a constant function, let $\alpha \in [a,b]$ such that $f(c)=\sup f([a,b])$. In other words, we know that there exists a supremum that exists in the continuous function as a consequence of the Minimum-maximum theorem. This supremum mimics the maximum point by definition.

Since it mimics the definition of a maximum, we can apply the Existence of Minimum and Maximum Points Theorem to prove the statement $f'(\alpha )=0$ .

The minimum works the same way, except that we will find an infimum instead of a supremum and that the infimum definition will match up with the minimum point. The Existence of Minimum and Maximum Points Theorem still holds.

$\blacksquare$

Proof 2

We will first remove an obvious case. If the function $f$ is a constant function, then $f'(x)=0\quad \forall x\in [a,b]$ . $\blacksquare$

If the function $f$ is not a constant function, let $\alpha \in [a,b]$ such that $f(c)=\sup f([a,b])$ . In other words, we know that there exists a supremum that exists in the continuous function as a consequence of the Minimum-maximum theorem. This supremum mimics the maximum point by definition.

Without loss of generality, we can state that $f(\alpha )>f(a)=f(b)$ . In other words, because we have, though an application of a theorem, found a supremum that is greater than every value in the function and we ruled out the case that the function is constant, we can safely assert that this supremum is greater than at least the endpoints, by definition.

Assume that $f'(c)>0$ is valid. Thus, $\lim _{x\to c}{\frac {f(x)-f(c)}{x-c}}>0$ and hence, there exists $x_{1}\in [a,b]$ such that $f(x_{1})>f(c)$ contradicting the fact that $f(c)$ is a maximum.

Similarly, we can show that the assumption $f'(c)<0$ leads to a contradiction. Thus, $f'(c)=0$ .

$\blacksquare$

Mean Value Theorem

This theorem is a stronger version of Rolle's Theorem. This theorem's statement is boxed below

Theorem Given a function $f:[a,b]\to \mathbb {R}$ which is both continuous on $[a,b]$ and differentiable on $(a,b)$ , there exists $c\in (a,b)$ such that $f'(c)={\frac {f(b)-f(a)}{b-a}}$

It is named the mean value theorem because graphically, this theorem states that if you have a function and its secant line, there will always be a tangent line somewhere on the function that is parallel to the secant line.

Note that it is a generalized version of Rolle's Theorem applied to secant lines connecting $f(a)$ and $f(b)$ instead of a horizontal secant line. You can see this in the theorem's equation by making $f(a)=f(b)$ . Because it is so similar, the proof is surprisingly similar in structure as well. However, it will require more algebra due to the nature of working with secant line segments instead of horizontal ones that can be easily zeroed.

Proof

The proof reduces the problem into one which can be solved using Rolle's Theorem by, in a sense, normalizing the graph based on the line i.e. the function in question is analyzed in the perspective of the line instead of the Cartesian grid.

Proof of the Mean Value Theorem

We create a new function based on the function $f$ and a linear function that uses the same slope as the secant line between the endpoints $(a,f(a))$ and $(b,f(b))$ . Note that this linear function is not the secant line. This new function, based on continuity and differentiation theorems, maintains all continuity and differentiation properties given to the function $f$ .	$h=f-{\frac {f(b)-f(a)}{b-a}}(x-a)$
We will now check to see if Rolle's Theorem can be applied to this new function.	$x=a$	$x=b$
	$h(a)=f(a)$	${\begin{aligned}h(b)&=f(b)-f(b)+f(a)\\&=f(a)\end{aligned}}$
The conditions of Rolle's Theorem is true - we can assert its conclusion, namely ${\textstyle \exists c:h'(c)=0}$ and work from there.	${\begin{aligned}h'(x)&=f'(x)-\left[{\dfrac {f(b)-f(a)}{b-a}}(x-a)\right]'\\&=f'(x)-{\dfrac {f(b)-f(a)}{b-a}}\\0&=f'(c)-{\dfrac {f(b)-f(a)}{b-a}}\\\end{aligned}}$
	$f'(c)={\dfrac {f(b)-f(a)}{b-a}}$
$\blacksquare$

Cauchy Mean Value Theorem

This theorem is an even stronger version of the Mean Value Theorem, extending the secant line used in the Mean Value Theorem into other types of functions.

Theorem If $f(x),g(x)$ are continuous on $[a,b]$ and differentiable on $(a,b)$ (without both having infinite derivatives at the same point) then there exists $c\in [a,b]$ such that ${\frac {f'(c)}{g'(c)}}={\frac {f(b)-f(a)}{g(b)-g(a)}}$

Proof

Define the function $h:[a,b]\to \mathbb {R}$ as

$h(x)=f(x){\big (}g(b)-g(a){\big )}-g(x){\big (}f(b)-f(a){\big )}$

Obviously, this function satisfies $h(a)=h(b)$ , and by Rolle's theorem, there is a $c\in [a,b]$ such that $h'(c)=0$ .

Theorems about Change

Before we prove any more things about derivatives, we should define some facets about functions that we will use in the following proofs.

Definition of a "function increasing on an interval $A$" Given a function $f$ and two numbers $a,b\in A$ such that $a<b\ \Rightarrow \ f(a)<f(b)$

Definition of a "function decreasing on an interval $A$" Given a function $f$ and two numbers $a,b\in A$ such that $a<b\ \Rightarrow \ f(a)>f(b)$

If there is no interval mentioned, the interval is defaulted to the real numbers.

Armed with both of these definitions, we will now explore the simplistic concept that we learned derivatives represent: change. The following proofs will solidify these concepts on mathematically rigorous grounds and provide new tools to verify the properties of functions and lay out some tools that can be reformed to prove new concepts.

"Increasing Functions have Positive Change" Theorem

This theorem, which has no official name, is the proof that verifies the simplistic definition of derivatives as "measuring change". It states

Theorem Given a function $f$ such that $f(x)>0$ for some interval $A$ , $f$ must also be increasing on the interval $A$

Proof

This proof relies on asserting the conditions and deriving the conclusion using a theorem, namely the Mean Value Theorem.

We start by choosing any two points $a,b\in A$ such that $a<b$ and assuming that $f'(x)>0$ for the entire interval.

Because the function $f$ is differentiable over $A$ , it is also continuous over $A$ . Therefore, all the conditions of the Mean Value Theorem are satisfied and we can use it.

$\exists x:f'(x)={\frac {f(b)-f(a)}{b-a}}>0$

After some algebraic manipulations,

${\begin{aligned}f'(x)={\dfrac {f(b)-f(a)}{b-a}}&>0\\f(b)-f(a)&>0\\f(b)&>f(a)\end{aligned}}$

As a reminder, $b-a>0$ .

Let this proof iterate for all possible $a,b$ combinations in the interval A. All $a<b$ implies $f(a)<f(b)$ , given the conditions.

$\blacksquare$

Corollary

A similar proof works for a decreasing function. In fact,

Theorem Given a function $f$ such that $f'(x)<0$ for some interval $A$ , $f$ must also be decreasing on the interval $A$

the proof is a mirror image of the previous proof.

$\blacksquare$

"Second Derivative Indicates Local Min/Max" Theorem

This theorem, which has no official name, is the proof that allows you to know whether any critical points you found are local minimums or maximums. It states

Theorem Given a function $f$ such that $f'(x)=0$ and $f''(x)>0$ for some $x$ . $f(x)$ is a local minimum

Proof

The proof involves using the conditions given to invoke the previous theorem's proof in order to infer the result.

We first invoke the definition of the second derivative $f''(x)=\lim _{h\to 0}{\frac {f'(x+h)-f'(x)}{h}}>0$

From there, we note that $f'(x)=0$ , which means that the equation now becomes $f''(x)=\lim _{h\to 0}{\dfrac {f'(x+h)}{h}}>0$

Given that $f''(x)>0$ , that means that the inequality needs to ensure its positive nature, which would mean that

$h\to 0^{+}$	$h\to 0^{-}$
$f(x+h)>0$	$f(x+h)<0$
Because in order for $f''(x)>0,f'(x+h)>0$ , we can use the previous proof's result to state that because $f'(x+h)>0$ , $f$ must be increasing.	Because in order for $f''(x)>0,f'(x+h)<0$ , we can use the previous proof's result to state that because $f'(x+h)<0$ , $f$ must be decreasing.

Since the continuous function goes from decreasing to increasing as $h$ increases, it must have stopped at the defined local minimum.

$\blacksquare$

Corollary

A similar proof works when deducing whether local maximums can be done in the same manner. In fact,

Theorem Given a function ƒ such that $f'(x)=0$ and $f''(x)<0$ for some x. ƒ(x) is a local maximum

the proof is a mirror image of the previous proof.

$\blacksquare$

Taylor's Theorem

Let $f:[a,x]\to \mathbb {R}$

Let $f^{(n-1)}$ be differentiable on $[a,x]$

Then, there exists $c\in (a,x)$ such that

$f(x)=f(a)+{\frac {(x-a)f'(a)}{1!}}+{\frac {(x-a)^{2}}{2!}}f''(a)+\cdots +{\frac {(x-a)^{n-1}}{(n-1)!}}f^{(n-1)}(a)+{\frac {(x-a)^{n}}{n!}}f^{(n)}(c)$

Proof

For the proof, we use the technique known as "Telescopic Sum"

Consider the function $\phi :[a,x]\to \mathbb {R}$ , given by,

$\phi (y)=f(y)+{\tfrac {(x-y)f'(y)}{1!}}+{\tfrac {(x-y)^{2}f''(y)}{2!}}+\ldots +{\tfrac {A(x-y)^{n}}{n!}}$, where the constant $A\in \mathbb {R}$ is chosen so as to satisfy $\phi (a)=\phi (x)$

By Rolle's theorem, we have that there exists $c\in (a,x)$ such that $\phi '(c)=0$

Expanding, we have (be careful while applying the product rule!)

$0=\phi '(c)$

$=f'(c)+\left[{\tfrac {(x-c)f''(c)}{1!}}-f'(c)\right]+\left[{\tfrac {(x-c)^{2}f'''(c)}{2!}}-{\tfrac {(x-c)f''(c)}{1!}}\right]+\ldots$

$+\left[{\tfrac {(x-c)^{n-1}f^{(n)}(c)}{(n-1)!}}-{\tfrac {(x-c)^{n-2}f^{(n-1)}}{(n-2)!}}\right]-{\tfrac {A(x-c)^{n-1}}{(n-1)!}}$

Which can be rearranged to give the telescopic sum:

$\left[f'(c)-f'(c)\right]+\left[{\tfrac {(x-c)f''(c)}{1!}}-{\tfrac {(x-c)f''(c)}{1!}}\right]+\ldots +\left[{\tfrac {(x-c)^{n-1}f^{(n)}(c)}{(n-1)!}}-{\tfrac {A(x-c)^{n-1}}{(n-1)!}}\right]=0$

That is, ${\frac {(x-c)^{n-1}f^{(n)}(c)}{(n-1)!}}={\frac {A(x-c)^{n-1}}{(n-1)!}}$, or $A=f^{(n)}(c)$

Now, we can easily see that $\phi (x)=f(x)$ and that $\phi (a)=f(a)+{\tfrac {(x-a)f'(a)}{1!}}+{\tfrac {(x-a)^{2}f''(a)}{2!}}+\ldots +{\tfrac {(x-a)^{n-1}f^{(n-1)}(a)}{(n-1)!}}+{\tfrac {(x-a)^{n}f^{n}(c)}{n!}}$

but by choice, $\phi (x)=\phi (a)$ and hence we have:

$f(x)=f(a)+{\tfrac {(x-a)f'(a)}{1!}}+{\tfrac {(x-a)^{2}f''(a)}{2!}}+\ldots +{\tfrac {(x-a)^{n-1}f^{(n-1)}(a)}{(n-1)!}}+{\tfrac {(x-a)^{n}f^{n}(c)}{n!}}$

QED