Zeno’s paradox

Introduction to knot theory

The Mathematical Association of America has an excellent series of 10-minute lectures on various topics in mathematics that are nevertheless accessible to the general public, including gifted elementary school students. The video below is a gentle introduction to knot theory, including computational issues and 3D printing. From the YouTube description:

Laura Taalman, a professor in the Department of Mathematics and Statistics at James Madison University, discusses using technology to explore mathematics.

Different definitions of e (Part 12): Numerical computation

In this series of posts, we have seen that the number $e$ can be thought about in three different ways.

1. $e$ defines a region of area 1 under the hyperbola $y = 1/x$ .2. We have the limits

$e = \displaystyle \lim_{h \to 0} (1+h)^{1/h} = \displaystyle \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n$ .

These limits form the logical basis for the continuous compound interest formula.

3. We have also shown that $\frac{d}{dx} \left(e^x \right) = e^x$ . From this derivative, the Taylor series expansion for $e^x$ about $x = 0$ can be computed:

$e^x = \displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}$

Therefore, we can let $x = 1$ to find $e$ :

$e = \displaystyle \sum_{n=0}^\infty \frac{1}{n!} = \displaystyle 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \dots$

In yesterday’s post, I showed that using the original definition (in terms of an area under a hyperbola) does not lend itself well to numerically approximating $e$ . Let’s now look at the other two methods.

2. The limit $e = \displaystyle \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n$ gives a somewhat more tractable way of approximating $e$ , at least with a modern calculator. However, you can probably imagine the fun of trying to use this formula without a calculator.

3. The best way to compute $e$ (or, in general, $e^x$ ) is with Taylor series. The fractions $\frac{1}{n!}$ get very small very quickly, leading to rapid convergence. Indeed, with only terms up to $1/6!$ , this approximation beats the above approximation with $n = 1000$ . Adding just two extra terms comes close to matching the accuracy of the above limit when $n = 1,000,000$ .

More about approximating $e^x$ via Taylor series can be found in my previous post.

Different definitions of e (Part 11): Numerical computation

In this series of posts, we have seen that the number $e$ can be thought about in three different ways.

1. $e$ defines a region of area 1 under the hyperbola $y = 1/x$ .2. We have the limits

$e = \displaystyle \lim_{h \to 0} (1+h)^{1/h} = \displaystyle \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n$ .

These limits form the logical basis for the continuous compound interest formula.

3. We have also shown that $\frac{d}{dx} \left(e^x \right) = e^x$ . From this derivative, the Taylor series expansion for $e^x$ about $x = 0$ can be computed:

$e^x = \displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}$

Therefore, we can let $x = 1$ to find $e$ :

$e = \displaystyle \sum_{n=0}^\infty \frac{1}{n!} = \displaystyle 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \dots$

Let’s now consider how the decimal expansion of $e$ might be obtained from these three different methods.

1. Finding $e$ using only the original definition is a genuine pain in the neck. The only way to approximate $e$ is by trapping the value of $e$ using various approximation. For example, consider the picture below, showing the curve $y = 1/x$ and trapezoidal approximations on the intervals $[1,1.8]$ and $[1.8,2.6]$ . (Because I need a good picture, I used Mathematica and not Microsoft Paint.)

Each trapezoid has a (horizontal) height of $h = 0.8$ . Furthermore, the bases of the first trapezoids have length $\displaystyle \frac{1}{1} = 1$ and $\displaystyle \frac{1}{1.8}$ , while the bases of the second trapezoid of length $\displaystyle \frac{1}{1.8}$ and $\displaystyle \frac{1}{2.6}$ . Notice that the trapezoids extend above the hyperbola, so that

$\displaystyle \int_1^{2.6} \frac{dx}{x} < \displaystyle \frac{0.8}{2} \left( 1 + \frac{1}{1.8} \right) + \frac{0.8}{2} \left( \frac{1}{1.8} + \frac{1}{2.6} \right)$

$\displaystyle \int_1^{2.6} \frac{dx}{x} < 0.9983 < 1$

However, the number $e$ is defined to be the place where the area under the curve is exactly equal to $1$ , and so

$\displaystyle \int_1^{2.6} \frac{dx}{x} < \displaystyle \int_1^{e} \frac{dx}{x}$

In other words, we know that the area between $1$ and $2.6$ is strictly less than $1$ , and therefore a number larger than $2.6$ must be needed to obtain an area equal to $1$ .

Great, so $e > 2.6$ . Can we do better? Sadly, with two equal-sized trapezoids, we can’t do much better. If the height of the trapezoids was $h$ and not $0.8$ , then the sum of the areas of the two trapezoids would be

$\displaystyle \frac{h}{2} \left( 1 + \frac{2}{1+h} + \frac{1}{1+2h} \right)$

By either guessing and checking — or with the help of Mathematica — it can be determined that this function of $h$ is equal to 1 at approximately $h \approx 0.8019$ , thus establishing that $e > 1 + 2h \approx 2.6039$ .

We can try to better with additional trapezoids. With four trapezoids, we can establish that $e > 2.6845$ .

With 100 trapezoids, we can show that $e > 2.71822$ .

However, trapezoids can only provide a lower bound on $e$ because the original trapezoids all extend over the hyperbola.

To obtain an upper bound on $e$ , we will use a lower Riemann sum to approximate the area under the curve. For example, notice the following picture of 19 rectangles of width $h = 0.1$ ranging from $x =1$ to $x = 2.9$ .

The rectangles all lie below the hyperbola. The width of each one is $h = 0.1$ , and the heights vary from $\frac{1}{1.1}$ to $\frac{1}{2.9}$ . Therefore,

$\displaystyle \int_1^{2.9} \frac{dx}{x} > \displaystyle 0.1 \left( \frac{1}{1.1}+ \frac{1}{1.2} + \dots + \frac{1}{2.9} \right)$

$\displaystyle \int_1^{2.9} \frac{dx}{x} > 1.0326 > 1$

In other words, we know that the area between $1$ and $2.9$ is strictly greater than $1$ , and therefore a number smaller than $2.9$ must be needed to obtain an area equal to $1$ . So, in a nutshell, we’ve shown that $e < 2.9$ .

Once again, additional rectangles can provide better and better upper bounds on $e$ . However, since rectangles do not approximate the hyperbola as well as trapezoids, we expect the convergence to be much slower. For example, with 100 rectangles of width $h$ , the sum of the areas of the rectangles would be

$h \displaystyle \left( \frac{1}{1+h} + \frac{1}{1+2h} + \dots + \frac{1}{1+100h} \right)$

It then becomes necessary to plug in numbers for $h$ until we find something that’s decently close to $1$ yet greater than $1$ . Or we can have Mathematica do the work for us:

So with 100 rectangles, we can establish that $e < 2.7333$ . With 1000 rectangles, we can establish that $e < 2.71977$ .

Clearly, this is a lot of work for approximating $e$ . With all of the work shown in this post, we have shown that $e = 2.71\dots$ , but we’re not yet sure if the next digit is $8$ or $9$ .

In the next post, we’ll explore the other two ways of thinking about the number $e$ as well as their computational tractability.

Different definitions of e (Part 10): Connecting the two definitions

In this series of posts, I consider how two different definitions of the number $e$ are related to each other. The number $e$ is usually introduced at two different places in the mathematics curriculum:

Algebra II/Precalculus: If $P$ dollars are invested at interest rate $r$ for $t$ years with continuous compound interest, then the amount of money after $t$ years is $A = Pe^{rt}$ .
Calculus: The number $e$ is defined to be the number so that the area under the curve $y = 1/x$ from $x = 1$ to $x = e$ is equal to $1$ , so that

$\displaystyle \int_1^e \frac{dx}{x} = 1$ .

These two definitions appear to be very, very different. One deals with making money. The other deals with the area under a hyperbola. Amazingly, these two definitions are related to each other. In this series of posts, I’ll discuss the connection between the two.

In yesterday’s post, I proved the following theorem, thus completing a long train of argument that began with the second definition of $e$ and ending with a critical step in the derivation of the continuous compound interest formula. Today, I present an alternate proof of the theorem using L’Hopital’s rule.

Theorem. $\displaystyle \lim_{n \to \infty} P \left( 1 + \frac{r}{n} \right)^{nt} = P e^{rt}$ .

Proof #2. Let’s write the left-hand side as

$L = \displaystyle \lim_{n \to \infty} P \left( 1 + \frac{r}{n} \right)^{nt}$ .

Let’s take the natural logarithm of both sides:

$\ln L = \displaystyle \ln \left[ \lim_{n \to \infty} P \left( 1 + \frac{r}{n} \right)^{nt} \right]$

Since $g(x) = \ln x$ is continuous, we can interchange the function and the limit on the right-hand side:

$\ln L = \displaystyle \lim_{n \to \infty} \ln \left[ P \left( 1 + \frac{r}{n} \right)^{nt} \right]$

$\ln L = \displaystyle \lim_{n \to \infty} \left[ \ln P + \ln \left( 1 + \frac{r}{n} \right)^{nt} \right]$

$\ln L = \displaystyle \lim_{n \to \infty} \left[ \ln P + nt \ln \left( 1 + \frac{r}{n} \right)\right]$

$\ln L = \ln P + \displaystyle \lim_{n \to \infty} \frac{t \displaystyle \ln \left(1 + \frac{r}{n} \right)}{1/n}$

$\ln L = \ln P + \displaystyle t \lim_{n \to \infty} \frac{\displaystyle \ln \left(1 + \frac{r}{n} \right)}{1/n}$

The limit on the right-hand side follows the indeterminate form $0/0$ , as so we may apply L’Hopital’s Rule. Taking the derivative of both the numerator and denominator with respect to $n$ , we find

$\ln L = \ln P + \displaystyle t \lim_{n \to \infty} \frac{\displaystyle \frac{1}{1 + \frac{r}{n}} \cdot \frac{-r}{n^2}}{\displaystyle \frac{-1}{n^2}}$

$\ln L = \ln P + \displaystyle t \lim_{n \to \infty}\frac{r}{1 + \frac{r}{n}}$

$\ln L = \ln P + \displaystyle t \frac{r}{1 + 0}$

$\ln L = rt + \ln P$

We now solve for the original limit $L$ :

$L = e^{rt + \ln P}$

$L = e^{rt} e^{\ln P}$

$L = Pe^{rt}$

Different definitions of e (Part 9): Connecting the two definitions

Algebra II/Precalculus: If $P$ dollars are invested at interest rate $r$ for $t$ years with continuous compound interest, then the amount of money after $t$ years is $A = Pe^{rt}$ .
Calculus: The number $e$ is defined to be the number so that the area under the curve $y = 1/x$ from $x = 1$ to $x = e$ is equal to $1$ , so that

$\displaystyle \int_1^e \frac{dx}{x} = 1$ .

We begin with the second definition, which is usually considered the true definition of $e$ . From this definition, I have shown in a previous post that we can derive the differentiation formulas

$\displaystyle \frac{d}{dx} (\ln x) = \frac{1}{x} \qquad$ and $\qquad \displaystyle \frac{d}{dx} \left( e^x \right) = e^x$

beginning with this definition of the number $e$ .

Theorem. $\displaystyle \lim_{n \to \infty} P \left( 1 + \frac{r}{n} \right)^{nt} = Pe^{rt}$ .

Proof #1.In an earlier post in this series, I showed that

$\displaystyle \frac{1}{x} = \displaystyle \lim_{h \to 0} \ln \left(1 + \frac{h}{x} \right)^{1/h}$

Let’s now replace $h$ with $1/n$ . Also, replace $x$ with $1/r$ . Then we obtain

$r = \displaystyle \lim_{n \to \infty} \ln \left(1 + \frac{1/n}{1/r} \right)^{n}$

$r = \displaystyle \lim_{n \to \infty} \ln \left(1 + \frac{r}{n} \right)^{n}$

Multiply both sides by $t$ :

$rt = \displaystyle t \lim_{n \to \infty} \ln \left(1 + \frac{r}{n} \right)^{n}$

$rt = \displaystyle \lim_{n \to \infty} t \ln \left(1 + \frac{r}{n} \right)^{n}$

$rt = \displaystyle \lim_{n \to \infty} \ln \left[ \left(1 + \frac{r}{n} \right)^{n} \right]^t$

$rt = \displaystyle \lim_{n \to \infty} \ln \left(1 + \frac{r}{n} \right)^{nt}$

Since $g(x) = \ln x$ is continuous, we can interchange the function and the limit on the right-hand side:

$rt = \displaystyle \ln \left[ \lim_{n \to \infty} \left(1 + \frac{r}{n} \right)^{nt} \right]$

$e^{rt} = \displaystyle \lim_{n \to \infty} \left(1 + \frac{r}{n} \right)^{nt}$

Finally, we multiply both sides by $P$ :

$P e^{rt} = \displaystyle \lim_{n \to \infty} P \left(1 + \frac{r}{n} \right)^{nt}$

(A second proof of this theorem, using L’Hopital’s Rule, will be presented in tomorrow’s post.)

This firmly established, at last, the connection between the continuous compound interest formula and the area under the hyperbola. I’ve noted that my students feel a certain sense of accomplishment after reaching this point of the exposition.

Different definitions of e (Part 8): Connecting the two definitions

Algebra II/Precalculus: If $P$ dollars are invested at interest rate $r$ for $t$ years with continuous compound interest, then the amount of money after $t$ years is $A = Pe^{rt}$ .
Calculus: The number $e$ is defined to be the number so that the area under the curve $y = 1/x$ from $x = 1$ to $x = e$ is equal to $1$ , so that

$\displaystyle \int_1^e \frac{dx}{x} = 1$ .

Theorem. $\displaystyle \lim_{h \to 0} \left( 1 + h \right)^{1/h} = e$ .

Proof #2. Let’s write the left-hand side as

$L = \displaystyle \lim_{h \to 0} \left( 1 + h \right)^{1/h}$ .

Let’s take the natural logarithm of both sides:

$\ln L = \displaystyle \ln \left[ \lim_{h \to 0} \left( 1 + h \right)^{1/h} \right]$

Since $g(x) = \ln x$ is continuous, we can interchange the function and the limit on the right-hand side:

$\ln L = \displaystyle \lim_{h \to 0} \ln (1+h)^{1/h}$

$\ln L = \displaystyle \lim_{h \to 0} \frac{1}{h} \ln(1+h)$

$\ln L = \displaystyle \lim_{h \to 0} \frac{\ln(1+h)}{h}$

The right-hand side follows the indeterminate form $0/0$ , as so we may apply L’Hopital’s Rule. Taking the derivative of both the numerator and denominator, we find

$\ln L = \displaystyle \lim_{h \to 0} \frac{ \displaystyle ~ \frac{1}{1+h} ~}{1}$

$\ln L = \displaystyle \lim_{h \to 0} \frac{1}{1+h}$

$\ln L = \displaystyle \frac{1}{1+0} = 1$ .

Therefore, the original limit is $L = e^1 = e$ .

Different definitions of e (Part 7): Connecting the two definitions

Algebra II/Precalculus: If $P$ dollars are invested at interest rate $r$ for $t$ years with continuous compound interest, then the amount of money after $t$ years is $A = Pe^{rt}$ .
Calculus: The number $e$ is defined to be the number so that the area under the curve $y = 1/x$ from $x = 1$ to $x = e$ is equal to $1$ , so that

$\displaystyle \int_1^e \frac{dx}{x} = 1$ .

We begin with the second definition, which is usually considered the true definition of $e$ . From this definition, I have shown in a previous post that we can derive the differentiation formulas

$\displaystyle \frac{d}{dx} (\ln x) = \frac{1}{x} \qquad$ and $\qquad \displaystyle \frac{d}{dx} \left( e^x \right) = e^x$

beginning with this definition of the number $e$ .

Theorem. $\displaystyle \lim_{h \to 0} \left( 1 + h \right)^{1/h} = e$ .

Proof #1. Recall the definition of a derivative

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

Let’s apply this to the function $f(x) = \ln x$ :

$\displaystyle \frac{d}{dx} (\ln x) = \displaystyle \lim_{h \to 0} \frac{\ln(x+h) - \ln x}{h}$

$\displaystyle \frac{1}{x} = \displaystyle \lim_{h \to 0} \frac{1}{h} \ln \left( \frac{x+h}{x} \right)$

$\displaystyle \frac{1}{x} = \displaystyle \lim_{h \to 0} \frac{1}{h} \ln \left( 1 + \frac{h}{x} \right)$

$\displaystyle \frac{1}{x} = \displaystyle \lim_{h \to 0} \ln \left(1 + \frac{h}{x} \right)^{1/h}$

(I’ll note parenthetically that I’ll need the above line for a future post in this series.) At this point, let’s substitute $x = 1$ :

$1 = \displaystyle \lim_{h \to 0} \ln (1 + h)^{1/h}$

Let’s now apply the exponential function to both sides:

$e^1 = \exp \left[ \displaystyle \lim_{h \to 0} \ln (1 + h)^{1/h} \right]$

Since $g(x) = e^x$ is continuous, we can interchange the function and the limit on the right-hand side:

$e = \displaystyle \lim_{h \to 0} \exp \left[ \ln (1 + h)^{1/h} \right]$

Finally, since $g(x) = e^x$ and $f(x) = \ln x$ are inverse functions, we can conclude

$e = \displaystyle \lim_{h \to 0} (1 + h)^{1/h}$ .

(A second proof of this theorem, using L’Hopital’s Rule, will be presented in tomorrow’s post.)

The next theorem establishes, at last, the connection between the continuous compound interest formula and the area under the hyperbola. I’ve noted that my students feel a certain sense of accomplishment after reaching this point of the exposition.

Theorem. $\displaystyle \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^{n} = e$ .

Proof. Though a little bit of real analysis is necessary to make this rigorous, we can informally see why this has to be true by letting $n = 1/h$ for $h$ positive. Then the expression $(1+h)^{1/h}$ becomes $\left( 1 + \frac{1}{n} \right)^n$ . Also, as $h \to 0$ , then $n \to \infty$ .

Different definitions of e (Part 6): Continuous compound interest

Algebra II/Precalculus: If $P$ dollars are invested at interest rate $r$ for $t$ years with continuous compound interest, then the amount of money after $t$ years is $A = Pe^{rt}$ .
Calculus: The number $e$ is defined to be the number so that the area under the curve $y = 1/x$ from $x = 1$ to $x = e$ is equal to $1$ , so that

$\displaystyle \int_1^e \frac{dx}{x} = 1$ .

I should say at the outset that the second definition is usually considered the true definition of $e$ . However, compound interest usually appears earlier in the mathematics curriculum than definite integrals, and so an informal definition of $e$ is given at that stage of the curriculum.

In yesterday’s post, I presented an informal derivation of the continuous compound interest formula $A = Pe^{rt}$ from the discrete compound interest formula $A = \displaystyle P \left( 1 + \frac{r}{n} \right)^{nt}$ . In today’s post, I’d like to give the more formal derivation using calculus.

What does it mean for something to compound continuously? In a nutshell, the rate at which the money increases should be proportional to the amount currently present. In other words, $\$10,000$ should earn ten times as much interest as $\$1,000$ . Since $A'(t)$ is the rate at which the money increases and $A(t)$ is the current amount, that means

$A'(t) = r A(t)$

for some constant of proportionality $r$ . This is a differential equation which can be solved using standard techniques. We divide both sides by $A(t)$ and then integrate:

$\displaystyle \frac{A'(t)}{A(t)} = r$

$\displaystyle \int \frac{A'(t)}{A(t)} dt = \int r dt$

$\ln |A(t)| = r t + C$

$|A(t)| = e^{rt+C} = e^rt e^C = C_1 e^{rt}$

$A(t) = \pm C_1 e^{rt}$

$A(t) = C_2 e^{rt}$

(Technically, a better solution would use an integrating factor [see also MathWorld], but I find that the above derivation is much more convincing to students who are a few semesters removed from a formal course in differential equations.) When presenting this in class, I’ll sometimes lazily write $C$ in place of $C_1$ , with the understanding that $e$ to an arbitrary constant is just an arbitrary positive constant. Also, on the last line, plus or minus an arbitrary constant is just an arbitrary constant (which I’ll usually write as $C$ instead of $C_2$ ).

To solve for the missing constant $C_2$ , we use the initial condition $A(0) = P$ :

$A(0) = C_2 e^{r\cdot 0}$

$P = C_2 \cdot 1$

$P = C_2$

Replacing $C_2$ by $P$ , we have arrived at the continuous compound interest formula $A(t) = Pe^{rt}$ .

Different definitions of e (Part 5): Continuous compound interest

Algebra II/Precalculus: If $P$ dollars are invested at interest rate $r$ for $t$ years with continuous compound interest, then the amount of money after $t$ years is $A = Pe^{rt}$ .
Calculus: The number $e$ is defined to be the number so that the area under the curve $y = 1/x$ from $x = 1$ to $x = e$ is equal to $1$ , so that

$\displaystyle \int_1^e \frac{dx}{x} = 1$ .

At this point in the exposition, I have justified the formula $A = \displaystyle P \left(1 + \frac{r}{n} \right)^{nt}$ for computing the value of an investment when interest is compounded $n$ times a year. We have also made the informal definition

$\displaystyle \left(1 + \frac{1}{n} \right) \to e \qquad$ as $\qquad n \to \infty$ .

We are now in position to give an informal derivation of the continuous compound interest formula. Though this derivation is informal, I have found it to be very convincing for my Precalculus students (as well as to my class of future high school teachers).

The basic idea is to rewrite the discrete compound interest formula so that it contains a term like $\displaystyle \frac{1}{\hbox{something}}$ instead of $\displaystyle \frac{r}{\hbox{something}}$ . In this way, we can think like an MIT freshman and reduce to previous work.

To this end, let $n = mr$ . Then the discrete compound interest formula becomes

$A = P\displaystyle \left( 1 + \frac{r}{n} \right)^{nt}$

$A = P\displaystyle \left( 1 + \frac{r}{rm} \right)^{rmt}$

$A = P\displaystyle \left( 1 + \frac{1}{m} \right)^{mrt}$

$A = P\displaystyle \left[ \left( 1 + \frac{1}{m} \right)^m \right]^{rt}$

Inside of the brackets is our familiar friend $\displaystyle \left( 1 + \frac{1}{m} \right)^m$ , except that the name of the variable has changed from $n$ to $m$ . But that’s no big deal: as $n$ tends to infinity, then $m$ does as well since $m = n/r$ and both $n$ and $r$ are positive. Therefore, as interest is compounded more frequently, we have replace the thing inside the brackets with the number $e$ . This leads us to the formula for continuous compound interest:

$A =P e^{rt}$

Again, my experience is that college students have no conceptual understanding of this formula or even a memory of seeing it derived once upon a time. They remember is it as coming out of nowhere, as a number in a formula or as a button on a calculator. It really shouldn’t be this way. The above calculation is perhaps a harder sell to high school students that the other calculations that I’ve posted in this series, but I firmly believe that this explanation is within the grasp of good students at the time that they take Algebra II and Precalculus.
Of course, the above derivation is highly informal. For starters, it rests upon the limit

$\displaystyle \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n = e$ ,

which cannot be formally proven using only the tools of Algebra II and Precalculus. Second, the above computation rests upon the continuity of the function $A(x) = P x^{rt}$ , so that we can simply replace $\displaystyle \left( 1 + \frac{1}{m} \right)^m$ with its limit $e$ . My experience is that students are completely comfortable making this substitution, even though professional mathematicians realize that interchanging limits requires continuity.

So, mathematically speaking, the above argument should not be considered a proper derivation of the continuous compound interest formula. Still, I have found that the above argument to be quite convincing to Algebra II and Precalculus students, appropriate to their current level of mathematical development.