The Algebraic and Geometric Meaning of Derivative

Note: This is the first part of the Derivative Concept Series. The second part is Derivative in Real Life Context and the third part is Derivative and the Maximum Area Problem.

***

If we want to get the slope of a line, we need two points. Suppose the points have coordinates $(x_1,y_1)$ and $(x_2,y_2)$, we have learned that the slope is described by the formula $\displaystyle\frac{y_2-y_1}{x_2-x_1}$.

In Figure 1, we have line $l$ tangent to the function $f$ at point $P$ where the coordinates of $P$ are $(x,f(x))$.  The problem that gave birth to calculus is getting the slope of this tangent line. There is, however, a problem.  We need two points to compute for the slope but we have only one point.

Note that the word tangent in this problem is different from the definition of tangent on a circle because it is clear that line $l$ will intersect the graph in more than one point.

Figure 1 - Line l tangent to the function f at point P.

Using the concept of limits we can remedy this problem. First, we create point $Q$ with x coordinate $h$ units to the right of the x-coordinate of $P$. We then draw line $PQ$, a secant line to the function $f$.

Figure 2 - A secant line is drawn through P.

In effect, the coordinates of $Q$ would be $(x+h, f(x+h))$ and it is clear that the slope of the secant line $PQ$ is described by the formula

$\displaystyle\frac{f(x+h)-f(x)}{x+h - x} = \frac{f(x+h)-f(x)}{h}$

If we want to approximate the slope of the tangent line, it is reasonable that we move $Q$ towards $P$ with $P$ fixed. Click here to explore the diagram above using GeoGebra.

From the GeoGebra exploration above, if we move $Q$ towards $P$, we observe the following:

1.)    The value of $h$ approaches $0$.

2.)    The inclination of the secant line approaches the inclination of the tangent line.

3.)    The slope of the secant line approaches the slope of the tangent line.

4.)    If point $Q$ coincides with point $P$, then the slope of the secant line and is equal to the slope of the tangent line.

If we let $m$ be the slope of the secant line and$f'(x)$ be the slope of the tangent line, focusing on observations 1 and 4, we can say the following equivalent statements:

• The limit of the slope of the secant line $m$ as $Q$ approaches $P$ is equal to $f'(x)$.
• The limit of the slope of the secant line $m$ as $h$ approaches $0$ is equal to $f'(x)$.
• The limit of $\displaystyle \frac{f(x+h)-f(x)}{h}$ as $h$ approaches $0$ is equal to $f'(x)$.

Using the limit notation, we can say that

$f'(x) = \lim_{h \to 0} \displaystyle\frac{f(x+h)-f(x)}{h}$

From the above discussion, we can see that the derivative of a function at  a particular point is the slope of the line tangent to that function at that particular point.

In the next post, we will discuss the meaning of derivative in real life situations.

Proof Tutorial 2: Proving Square Root of 2 is Irrational by Contradiction

One of the most difficult proof strategies in mathematics is proof by contradiction. If P, for example, is a statement or a conjecture, one strategy to prove that P is true is to assume that P is not true  and find a contradiction so that the statement not P does not hold. If not P does not hold, it follows that P is true.

One well-known proof that uses proof by contradiction is proof of the irrationality of $\sqrt{2}$.  If we consider P to be the statement “$\sqrt{2}$ is irrational”, then not P is the opposite statement or “$\sqrt{2}$ is rational”.  To use proof by contradiction, we assume that $\sqrt{2}$ is rational, and find a contradiction somewhere. If this happens, then we would have shown that $\sqrt{2}$ is indeed irrational.

Before proceeding, recall that a rational number is a fraction with non-zero denominator.  We know that all fractions can be expressed in lowest term.  A fraction in $\displaystyle\frac{a}{b}$ is said to be in lowest term if $a$ and $b$ have no common divisors except $1$.

On the other hand, irrational numbers cannot be expressed as fractions. They are decimal numbers that do not end and do not repeat. For example, $0.10100100010000...$ is an irrational number (the three dots means and so on which means that the number does not end). The most popular irrational number is $\pi$.

Now, we prove our conjecture.

Conjecture: The $\sqrt{2}$ is irrational.

Proof:

Suppose $\sqrt{2}$ is rational, then it can be expressed in fraction form $\displaystyle\frac{a}{b}$ . Let us assume that our fraction is in lowest term, i.e., their only common divisor is $1$. Then,

$\sqrt{2} = \displaystyle\frac{a}{b}$

Squaring both sides, we have

$2= \displaystyle\frac{a^2}{b^2}$

Multiplying both sides by $b^2$ yields

$2b^2= a^2$*

Since $a^2 = 2b^2$, we can conclude that $a^2$ is even because whatever the value of $b^2$ has to be multiplied by $2$. If $a^2$ is even, then $a$ is also even. Since $a$ is even, no matter what the value of $a$ is, we can always find an integer that if we divide $a$ by $2$, it is equal to that integer. If we let that integer be $k$, then $\displaystyle\frac{a}{2} = k$ which means that $a = 2k$.

Substituting the value of $2k$ to $a$ in *, we have $2b^2= (2k)^2$ which means that $2b^2=4k^2$.  Dividing both sides by $2$, we have $b^2 = 2k^2$. That means that the value $b^2$ is even, since whatever the value of $k$ you have to multiply it by $2$.  Again, if $b^2$ is even, then $b$ is even.

This implies that both $a$ and $b$ are even, which means that both the numerator and the denominator of our fraction are divisible by $2$. This contradicts our assumption that $\displaystyle\frac{a}{b}$ has no common divisor except $1$. Since we found a contradiction, our assumption is, therefore, false. Hence, the theorem is true.

Notice that I have highlighted the word suppose and assume in the proof. This is one unique feature of proof by contradiction. You can always assume, most of the time, the opposite of the conjecture as long as the following statements are logically valid.

Why is 0! equals 1?

Many books will tell you that $0!$ equals $1$ is a definition. There are actually a few reasons why this is so – the two of which are shown below.

Explanation 1:

Based on mypost, we can conclude that $n$ taken $n$ at a time is equal to $1$. This means, that there is only one way that you can group $n$ objects from $n$ objects. For example, we can only form one group consisting of $4$ letters from AB, C and D using all the 4 letters.

From above, we know that the $\displaystyle{n \choose n} = 1$. But, $\displaystyle{n \choose n} = \displaystyle\frac{n!}{(n-n)!n!} = \displaystyle\frac{n!}{0!n!} = 1$. To satisfy the equation, $0!$ must be equal to $1$.

Explanation 2:

We can also use the fact that $n! = n(n-1)!$. Dividing both sides by $n$, we have $\displaystyle\frac{n!}{n} = (n-1)!$. If we let $n = 1$, we have $\displaystyle\frac{1!}{1} = (1 - 1)! \Rightarrow 1 = 0!$ which is what we want to show.

Explanation 3

We can also use the following pattern. We know that $n! = n(n-1)(n-2)(n-3) \cdots 3(2)(1)$ which means that

$n! = n(n-1)!$.

Dividing both sides of the equation by $n$, we have

$(n - 1)! = \frac{n!}{n}$

Using this fact, we can check the following pattern.

$4! = \displaystyle \frac{5!}{5} = \frac{(5)(4)(3)(2)(1)}{5} = 24$

$3! = \displaystyle \frac{4!}{4} = \frac{(4)(3)(2)(1)}{4} = 6$

$2! = \displaystyle \frac{3!}{3} = \frac{(3)(2)(1)}{(3)} = 2$

$1! = \displaystyle \frac{2!}{2} = \frac{(2)(1)}{(2)}$

Now, we go to $0!$

$0! = \displaystyle \frac{1!}{1} = 1$

As we can see from the 3 examples, $0! = 1$.

1 26 27 28 29 30