Proof without words: The difference of consecutive cubes

Source: https://www.facebook.com/photo.php?fbid=451328078334029&set=a.416585131808324.1073741827.416199381846899&type=1&theater

For a more conventional algebraic proof, notice that

$(n+1)^3 - n^3 = n^3 + 3n^2 + 3n + 1 - n^3 = 3n(n+1) + 1$

The product $n(n+1)$ is always an even number times an odd number: if $n$ is even, then $n+1$ is odd, but if $n$ is odd, then $n(n+1)$ . So $n(n+1)$ is a multiple of 2, and so $3n(n+1)$ is a multiple of 6. Therefore, $3n(n+1)+1$ is one more than a multiple of 6, proving the theorem.

Arithmetic with big numbers (Part 3)

In the previous two posts, we considered the use of base- $10^n$ arithmetic so that a calculator can solve addition and multiplication problems that it ordinarily could not handle. Today, we turn to division. Let’s now consider the decimal representation of $\displaystyle \frac{8}{17}$ .

There’s no obvious repeating pattern. But we know that, since 17 has neither 2 nor 5 as a factor, that there has to be a repeating decimal pattern.

So… what is it?

When I ask this question to my students, I can see their stomachs churning a slow dance of death. They figure that the calculator didn’t give the answer, and so they have to settle for long division by hand.

That’s partially correct.

However, using the ideas presented below, we can perform the long division extracting multiple digits at once. Through clever use of the calculator, we can quickly obtain the full decimal representation even though the calculator can only give ten digits at a time.

Let’s now return to where this series began… the decimal representation of $\displaystyle \frac{1}{7}$ using long division. As shown below, the repeating block has length $6$ , which can be found in a few minutes with enough patience. By the end of this post, we’ll consider a modification of ordinary long division that facilitates the computation of really long repeating blocks.

Because we arrived at a repeated remainder, we know that we have found the repeating block. So we can conclude that $\displaystyle \frac{1}{7} = 0.\overline{142857}$ .

Students are taught long division in elementary school and are so familiar with the procedure that not much thought is given to the logic behind the procedure. The underlying theorem behind long division is typically called the division algorithm. From Wikipedia:

Given two integers $a$ and $b$ , with $b \ne 0$ , there exist unique integers $q$ and $r$ such that $a = bq+r$ and $0 \le r < |b|$, where $|b|$ denotes the absolute value of $b$ .

The number $q$ is typically called the quotient, while the number $r$ is called the remainder.

Repeated application of this theorem is the basis for long division. For the example above:

Step 1.

$10 = 1 \times 7 + 3$ . Dividing by $10$ , $1 = 0.1 \times 7 + 0.3$

Step 2.

$30 = 4 \times 7 + 2$ . Dividing by $100$ , $0.3 = 0.04 \times 7 + 0.02$

Returning to the end of Step 1, we see that

$1 = 0.1 \times 7 + 0.3 = 0.1 \times 7 + 0.04 \times 7 + 0.02 = 0.14 \times 7 + 0.02$

Step 3.

$20 = 2 \times 7 + 6$ . Dividing by $1000$ , $0.02 = 0.002 \times 7 + 0.006$

Returning to the end of Step 2, we see that

$1 = 0.14 \times 7 + 0.02 = 0.14 \times 7 + 0.0002 \times 7 + 0.006 = 0.142 \times 7 + 0.006$

And so on.

By adding an extra zero and using the division algorithm, the digits in the decimal representation are found one at a time. That said, it is possible (with a calculator) to find multiple digits in a single step by adding extra zeroes. For example:

Alternate Step 1.

$1000 = 142 \times 7 + 6$ . Dividing by $1000$ , $1 = 0.142 \times 7 + 0.006$

Alternate Step 2.

$6000 = 587 \times 7 + 1$ . Dividing by $100000$ , $0.006 = 0.000587 \times 7 + 0.000001$

Returning to the end of Alternate Step 1, we see that

$1 = 0.142 \times 7 + 0.006= 0.142 \times 7 + 0.000587\times 7 + 0.000001 = 0.142857 \times 7 + 0.000001$

So, with these two alternate steps, we arrive at a remainder of $1$ and have found the length of the repeating block.

The big catch is that, if $a = 1000$ or $a = 6000$ and $b = 7$ , the appropriate values of $q$ and $r$ have to be found. This can be facilitated with a calculator. The integer part of $1000/7$ and $6000/7$ are the two quotients needed above, and subtraction is used to find the remainders (which must be less than $7$ , of course).

At first blush, it seems silly to use a calculator to find these values of $q$ and $r$ when a calculator could have been used to just find the decimal representation of $1/7$ in the first place. However, the advantage of this method becomes clear when we consider fractions who repeating blocks are longer than 10 digits.

Let’s now return to the question posed at the top of this post: finding the decimal representation of $\displaystyle \frac{8}{17}$ . As noted in Part 6 of this series, the length of the repeating block must be a factor of $\phi(17)$ , where $\phi$ is the Euler toitent function, or the number of integers less than $17$ that are relatively prime with $17$ . Since $17$ is prime, we clearly see that $\phi(17) = 16$ . So we can conclude that the length of the repeating block is a factor of $16$ , or either $1$ , $2$ , $4$ , $8$ , or $16$ .

Here’s the result of the calculator again:

We clearly see from the calculator that the repeating block doesn’t have a length less than or equal to $8$ . By process of elimination, the repeating block must have a length of $16$ digits.

Now we perform the division algorithm to obtain these digits, as before. This can be done in two steps by multiplying by $10^8 = 100,000,000$ .

So, by the same logic used above, we can conclude that

$\displaystyle \frac{8}{17} = 0.\overline{4705882352941176}$

In other words, through clever use of the calculator, the full decimal representation can be quickly found even if the calculator itself returns only ten digits at a time… and had rounded the final $2941176$ of the repeating block up to $3$ .

(Note: While this post continues exploring the unorthodox use of a calculator to handle arithmetic problems, it also appeared in a previous series on the decimal expansions of rational numbers.)

Proof without words: A relation between the sides of a right trapezoid

Source: https://www.facebook.com/photo.php?fbid=494947463972090&set=a.416585131808324.1073741827.416199381846899&type=1&theater

Inverse Functions: Logarithms and Complex Numbers (Part 30)

Ordinarily, there are no great difficulties with logarithms as we’ve seen with the inverse trigonometric functions. That’s because the graph of $y = a^x$ satisfies the horizontal line test for any $0 < a < 1$ or $a > 1$ . For example,

$e^x = 5 \Longrightarrow x = \ln 5$ ,

and we don’t have to worry about “other” solutions.

However, this goes out the window if we consider logarithms with complex numbers. Recall that the trigonometric form of a complex number $z = a+bi$ is

$z = r(\cos \theta + i \sin \theta) = r e^{i \theta}$

where $r = |z| = \sqrt{a^2 + b^2}$ and $\tan \theta = b/a$ , with $\theta$ in the appropriate quadrant. This is analogous to converting from rectangular coordinates to polar coordinates.

Over the past few posts, we developed the following theorem for computing $e^z$ in the case that $z$ is a complex number.

Definition. Let $z = r e^{i \theta}$ be a complex number so that $-\pi < \theta \le \theta$ . Then we define

$\log z = \ln r + i \theta$ .

Of course, this looks like what the definition ought to be if one formally applies the Laws of Logarithms to $r e^{i \theta}$ . However, this complex logarithm doesn’t always work the way you’d think it work. For example,

$\log \left(e^{2 \pi i} \right) = \log (\cos 2\pi + i \sin 2\pi) = \log 1 = \ln 1 = 0 \ne 2\pi i$ .

This is analogous to another situation when an inverse function is defined using a restricted domain, like

$\sqrt{ (-3)^2 } = \sqrt{9} = 3 \ne -3$

$\sin^{-1} (\sin \pi) = \sin^{-1} 0 = 0 \ne \pi$ .

The Laws of Logarithms also may not work when nonpositive numbers are used. For example,

$\log \left[ (-1) \cdot (-1) \right] = \log 1 = 0$ ,

but

$\log(-1) + \log(-1) = \log \left( e^{\pi i} \right) + \log \left( e^{\pi i} \right) = \pi i + \pi i = 2\pi i$ .

green line

This material appeared in my previous series concerning calculators and complex numbers: https://meangreenmath.com/2014/07/09/calculators-and-complex-numbers-part-21/

Proof without words: Pythagoras for a clipped rectangle

Source: https://www.facebook.com/photo.php?fbid=494951053971731&set=a.416585131808324.1073741827.416199381846899&type=1&theater

2014 MacArthur “Genius” grant winner and the twin prime conjecture

From http://www.macfound.org/fellows/927/:

Prime numbers have inspired great intrigue over the last centuries, and one of the most basic unanswered questions has been the spacing between two consecutive prime numbers, or the twin prime conjecture, which states that there are infinitely many pairs of primes that differ by two. Despite many efforts at proving this conjecture, mathematicians were not able to rule out the possibility that the gaps between primes continue to expand, eventually exceeding any particular bound. Zhang’s work shows that there are infinitely many consecutive primes, or pairs of primes, closer than 70 million. In other words, as you go to larger and larger numbers, the primes will not become further and further apart—you will keep finding prime pairs that differ by less than 70 million.

His work has generated significant collaborations across the community to expand on his effort, and within months of his discovery that number was reduced from 70 million to less than 5,000.

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

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