Confirming Einstein’s Theory of General Relativity With Calculus, Part 9: Pedagogical Thoughts

At long last, we have reached the end of this series of posts.

The derivation is elementary; I’m confident that I could have understood this derivation had I seen it when I was in high school. That said, the word “elementary” in mathematics can be a bit loaded — this means that it is based on simple ideas that are perhaps used in a profound and surprising way. Perhaps my favorite quote along these lines was this understated gem from the book Three Pearls of Number Theory after the conclusion of a very complicated proof in Chapter 1:

You see how complicated an entirely elementary construction can sometimes be. And yet this is not an extreme case; in the next chapter you will encounter just as elementary a construction which is considerably more complicated.

Here are the elementary ideas from calculus, precalculus, and high school physics that were used in this series:

Physics
- Conservation of angular momentum
- Newton’s Second Law
- Newton’s Law of Gravitation
Precalculus
- Completing the square
- Quadratic formula
- Factoring polynomials
- Complex roots of polynomials
- Bounds on $\cos \theta$ and $\sin \theta$
- Period of $\cos \theta$ and $\sin \theta$
- Zeroes of $\cos \theta$ and $\sin \theta$
- Trigonometric identities (Pythagorean, sum and difference, double-angle)
- Conic sections
- Graphing in polar coordinates
- Two-dimensional vectors
- Dot products of two-dimensional vectors (especially perpendicular vectors)
- Euler’s equation
Calculus
- The Chain Rule
- Derivatives of $\cos \theta$ and $\sin \theta$
- Linearizations of $\cos x$ , $\sin x$ , and $1/(1-x)$ near $x \approx 0$ (or, more generally, their Taylor series approximations)
- Derivative of $e^x$
- Solving initial-value problems
- Integration by $u-$ substitution

While these ideas from calculus are elementary, they were certainly used in clever and unusual ways throughout the derivation.

I should add that although the derivation was elementary, certain parts of the derivation could be made easier by appealing to standard concepts from differential equations.

One more thought. While this series of post was inspired by a calculation that appeared in an undergraduate physics textbook, I had thought that this series might be worthy of publication in a mathematical journal as an historical example of an important problem that can be solved by elementary tools. Unfortunately for me, Hieu D. Nguyen’s terrific article Rearing Its Ugly Head: The Cosmological Constant and Newton’s Greatest Blunder in The American Mathematical Monthly is already in the record.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 6i: Rationale for Method of Undetermined Coefficients VI

In this series, I’m discussing how ideas from calculus and precalculus (with a touch of differential equations) can predict the precession in Mercury’s orbit and thus confirm Einstein’s theory of general relativity. The origins of this series came from a class project that I assigned to my Differential Equations students maybe 20 years ago.

In the last few posts, I’ve used a standard technique from differential equations: to solve the $n$ th order homogeneous differential equation with constant coefficients

$a_n y^{(n)} + \dots + a_3 y''' + a_2 y'' + a_1 y' + a_0 y = 0$ ,

we first solve the characteristic equation

$a_n r^n + \dots + a_3 r^3 + a_2 r^2 + a_1 r + a_0 = 0$

using techniques from Precalculus. The form of the roots $r$ determines the solutions of the differential equation.

While this is a standard technique from differential equations, the perspective I’m taking in this series is scaffolding the techniques used to predict the precession in a planet’s orbit using only techniques from Calculus and Precalculus. So let me discuss why the above technique works, assuming that the characteristic equation does not have repeated roots. (The repeated roots case is a little more complicated but is not needed for the present series of posts.)

We begin by guessing that the above differential equation has a solution of the form $y = e^{rt}$ . Differentiating, we find $y' = re^{rt}$ , $y'' = r^2 e^{rt}$ , etc. Therefore, the differential equation becomes

$a_n r^n e^{rt} + \dots + a_3 r^3 e^{rt} + a_2 r^2 e^{rt} + a_1 r e^{rt} + a_0 e^{rt} = 0$

$e^{rt} \left(a_n r^n + \dots + a_3 r^3 + a_2 r^2 + a_1 r + a_0 \right) = 0$

$a_n r^n + \dots + a_3 r^3 + a_2 r^2 + a_1 r + a_0 = 0$

The last step does not “lose” any possible solutions for $r$ since $e^{rt}$ can never be equal to $0$ . Therefore, solving the differential equation reduces to finding the roots of this polynomial, which can be done using standard techniques from Precalculus.

For example, one of the differential equations that we’ve encountered is $y''+y=0$ . The characteristic equation is $r^2+1=0$ , which has roots $r=\pm i$ . Therefore, two solutions to the differential equation are $e^{it}$ and $e^{-it}$ , so that the general solution is

$y = c_1 e^{it} + c_2 e^{-it}$ .

To write this in a more conventional way, we use Euler’s formula $e^{ix} = \cos x + i \sin x$ , so that

$y = c_1 (\cos t + i \sin t) + c_2 (\cos (-t) + i \sin (-t))$

$= c_1 \cos t + i c_1 \sin t + c_2 \cos t - i c_2 \sin t$

$= (c_1 + c_2) \cos t + (ic_1 - ic_2) \sin t$

$= C_1 \cos t + C_2 \sin t$ .

Likewise, in the previous post, we encountered the fourth-order differential equation $y^{(4)}+5y''+4y = 0$ . To find the roots of the characteristic equation, we factor:

$r^4 + 5r^2 + 4r = 0$

$(r^2+1)(r^2+4) = 0$

$r^2 +1 = 0 \qquad \hbox{or} \qquad \hbox{or} r^2 + 4 = 0$

$r = \pm i \qquad \hbox{or} \qquad r = \pm 2i$ .

Therefore, four solutions of this differential equation are $e^{it}$ , $e^{-it}$ , $e^{2it}$ , and $e^{-2it}$ , so that the general solution is

$y = c_1 e^{it} + c_2 e^{-it} + c_3 e^{2it} + c_4 e{-2it}$ .

Using Euler’s formula as before, this can be rewritten as

$y = C_1 \cos t + C_2 \sin t + C_3 \cos 2t + C_4 \sin 2t$ .

Confirming Einstein’s Theory of General Relativity With Calculus, Part 6h: Rationale for Method of Undetermined Coefficients V

We have shown that the motion of a planet around the Sun, expressed in polar coordinates $(r,\theta)$ with the Sun at the origin, under general relativity follows the initial-value problem

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{2\delta \epsilon \cos \theta}{\alpha^2} + \frac{\delta \epsilon^2 \cos 2\theta}{2\alpha^2}$ ,

$u(0) = \displaystyle \frac{1}{P}$ ,

$u'(0) = 0$ ,

where $u = \displaystyle \frac{1}{r}$ , $\displaystyle \frac{1}{\alpha} = \frac{GMm^2}{\ell^2}$ , $\delta = \displaystyle \frac{3GM}{c^2}$ , $G$ is the gravitational constant of the universe, $m$ is the mass of the planet, $M$ is the mass of the Sun, $\ell$ is the constant angular momentum of the planet, $c$ is the speed of light, and $P$ is the smallest distance of the planet from the Sun during its orbit (i.e., at perihelion).

In the two previous posts, we derived the method of undetermined coefficients for the simplified differential equations

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2}$ .

and

$u''(\theta) + u(\theta) = \displaystyle \frac{2\delta \epsilon \cos \theta}{\alpha^2}$ .

In this post, we consider the simplified differential equation if the right-hand side has only the fifth term,

$u''(\theta) + u(\theta) = \displaystyle \frac{\delta \epsilon^2 \cos 2\theta}{2\alpha^2}$ .

Let $v(\theta) = \displaystyle \frac{\delta \epsilon^2 }{2\alpha^2} \cos 2\theta$ . Then $v$ satisfies the new differential equation $v'' + 4v = 0$ . Also, $v = u'' + u$ . Substituting, we find

$(u''+u)'' + 4(u''+u) = 0$

$u^{(4)} + u'' + 4u'' + 4u = 0$

$u^{(4)} + 5u'' + 4u = 0$

The characteristic equation of this new differential equation is

$r^4 + 5r^2 + 4 = 0$

$(r^2 + 1)(r^2 + 4) = 0$

$r^2 + 1 = 0 \qquad \hbox{or} \qquad r^2 + 4 = 0$

$r = \pm i \qquad \hbox{or} \qquad r = \pm 2i$

Therefore, the general solution of the new differential equation is

$u(\theta) = c_1 \cos \theta + c_2 \sin \theta + c_3 \cos 2\theta + c_4 \sin 2\theta$ .

The constants $c_3$ and $c_4$ can be found by substituting back into the original differential equation:

$u''(\theta) + u(\theta) = \displaystyle \frac{\delta \epsilon^2 \cos 2\theta}{2\alpha^2}$

$-c_1 \cos \theta - c_2 \sin \theta - 4c_3 \cos 2\theta - 4c_4 \sin 2\theta + c_1 \cos \theta + c_2 \sin \theta + c_3 \cos 2\theta + c_4 \sin 2\theta = \displaystyle \frac{\delta \epsilon^2 \cos 2\theta}{2\alpha^2}$

$- 3c_3 \cos 2\theta - 3c_4 \sin 2\theta = \displaystyle \frac{\delta \epsilon^2 \cos 2\theta}{2\alpha^2}$

Matching coefficients, we see that $c_3 = \displaystyle -\frac{\delta \epsilon^2}{6\alpha^2}$ and $c_4 = 0$ . Therefore, the solution of the simplified differential equation is

$u(\theta) = c_1 \theta + c_2 \theta \displaystyle -\frac{\delta \epsilon^2}{6\alpha^2} \cos 2\theta$ .

In particular, setting $c_1 = 0$ and $c_2 = 0$ , we see that

$u(\theta) = \displaystyle -\frac{\delta \epsilon^2}{6\alpha^2} \cos 2\theta$

is a particular solution to the simplified differential equation.

In the next post, we put together the solutions of these three simplified differential equations to solve the original differential equation,

Confirming Einstein’s Theory of General Relativity With Calculus, Part 6g: Rationale for Method of Undetermined Coefficients IV

We have shown that the motion of a planet around the Sun, expressed in polar coordinates $(r,\theta)$ with the Sun at the origin, under general relativity follows the initial-value problem

$u(0) = \displaystyle \frac{1}{P}$ ,

$u'(0) = 0$ ,

In this post, we will use the guesses

$u(\theta) = f(\theta) \cos \theta \qquad \hbox{or} u(\theta) = f(\theta) \sin \theta$

that arose from the technique/trick of reduction of order, where $f(\theta)$ is some unknown function, to find the general solution of the differential equation

$u^{(4)} + 2u'' + u = 0$ .

To do this, we will need to use the Product Rule for higher-order derivatives that was derived in the previous post:

$(fg)'' = f'' g + 2 f' g' + f g''$

and

$(fg)^{(4)} = f^{(4)} g + 4 f''' g' + 6 f'' g'' + 4f' g''' + f g^{(4)}$ .

In these formulas, Pascal’s triangle makes a somewhat surprising appearance; indeed, this pattern can be proven with mathematical induction.

We begin with $u(\theta) = f(\theta) \cos \theta$ . If $g(\theta) = \cos \theta$ , then

$g'(\theta) = - \sin \theta$ ,

$g''(\theta) = -\cos \theta$ ,

$g'''(\theta) = \sin \theta$ ,

$g^{(4)}(\theta) = \cos \theta$ .

Substituting into the fourth-order differential equation, we find the differential equation becomes

$(f \cos \theta)^{(4)} + 2 (f \cos \theta)'' + f \cos \theta = 0$

$f^{(4)} \cos \theta - 4 f''' \sin \theta - 6 f'' \cos \theta + 4 f' \sin \theta + f \cos \theta + 2 f'' \cos \theta - 4 f' \sin \theta - 2 f \cos \theta + f \cos \theta = 0$

$f^{(4)} \cos \theta - 4 f''' \sin \theta - 6 f'' \cos \theta + 2 f'' \cos \theta = 0$

$f^{(4)} \cos \theta - 4 f''' \sin \theta - 4 f'' \cos \theta = 0$

The important observation is that the terms containing $f$ and $f'$ cancelled each other. This new differential equation doesn’t look like much of an improvement over the original fourth-order differential equation, but we can make a key observation: if $f'' = 0$ , then differentiating twice more trivially yields $f''' = 0$ and $f^{(4)} = 0$ . Said another way: if $f'' = 0$ , then $u(\theta) = f(\theta) \cos \theta$ will be a solution of the original differential equation.

Integrating twice, we can find $f$ :

$f''(\theta) = 0$

$f'(\theta) = c_1$

$f(\theta) = c_1 \theta + c_2$ .

Therefore, a solution of the original differential equation will be

$u(\theta) = c_1 \theta \cos \theta + c_2 \cos \theta$ .

We now repeat the logic for $u(\theta) = f(\theta) \sin \theta$ :

$(f \sin \theta)^{(4)} + 2 (f \sin \theta)'' + f \sin \theta = 0$

$f^{(4)} \sin \theta + 4 f''' \cos \theta - 6 f'' \sin \theta - 4 f' \cos\theta + f \sin \theta + 2 f'' \sin \theta + 4 f' \cos \theta - 2 f \sin \theta + f \sin \theta = 0$

$f^{(4)} \sin\theta + 4 f''' \cos \theta - 6 f'' \sin \theta + 2 f'' \sin \theta = 0$

$f^{(4)} \sin\theta - 4 f''' \cos\theta - 4 f'' \sin\theta = 0$ .

Once again, a solution of this new differential equation will be $f(\theta) = c_3 \theta + c_4$ , so that $f'' = f''' = f^{(4)} = 0$ . Therefore, another solution of the original differential equation will be

$u(\theta) = c_3 \theta \sin \theta + c_4 \sin \theta$ .

Adding these provides the general solution of the differential equation:

$u(\theta) = c_1 \theta \cos \theta + c_2 \cos \theta + c_3 \theta \sin \theta + c_4 \sin \theta$ .

Except for the order of the constants, this matches the solution that was presented earlier by using techniques taught in a proper course in differential equations.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 6f: Rationale for Method of Undetermined Coefficients III

We have shown that the motion of a planet around the Sun, expressed in polar coordinates $(r,\theta)$ with the Sun at the origin, under general relativity follows the initial-value problem

$u(0) = \displaystyle \frac{1}{P}$ ,

$u'(0) = 0$ ,

In the previous post, I used a standard technique from differential equations to find the general solution of

$u^{(4)} + 2u'' + u = 0$ .

to be

$u(theta) = c_1 \cos \theta + c_2 \sin \theta + c_3 \theta \cos \theta + c_4 \theta \sin \theta$ .

However, as much as possible in this series, I want to take the perspective of a talented calculus student who has not yet taken differential equations — so that the conclusion above is far from obvious. How could this be reasonable coaxed out of such a student?

To begin, we observe that the characteristic equation is

$r^4 + 2r^2 + 1 = 0$ ,

$(r^2 + 1)^2 = 0$ .

Clearly this has the same roots as the simpler equation $r^2 + 1 = 0$ , which corresponds to the second-order differential equation $u'' + u = 0$ . We’ve already seen that $u_1(\theta) = \cos \theta$ and $u_2(\theta) = \sin \theta$ are solutions of this differential equation; perhaps they might also be solutions of the more complicated differential equation also? The answer, of course, is yes:

$u_1^{(4)} + 2 u_1'' + u_1 = \cos \theta - 2 \cos \theta + \cos \theta = 0$

and

$u_2^{(4)} + 2u_2'' + u_2 = \sin \theta - 2 \sin \theta + \sin \theta = 0$ .

The far trickier part is finding the two additional solutions. To find these, we use a standard trick/technique called reduction of order. In this technique, we guess that any additional solutions much have the form of either

$u(\theta) = f(\theta) \cos \theta \qquad \hbox{or} \qquad u(\theta) = f(\theta) \sin \theta$ ,

where $f(\theta)$ is some unknown function that we’re multiplying by the solutions we already have. We then substitute this into the differential equation $u^{(4)} + 2u'' + u = 0$ to form a new differential equation for the unknown $f$ , which we can (hopefully) solve.

Doing this will require multiple applications of the Product Rule for differentiation. We already know that

$(fg)' = f' g + f g'$ .

We now differentiate again, using the Product Rule, to find $(fg)''$ :

$(fg)'' = ( [fg]')' = (f'g)' + (fg')'$

$= f''g + f' g' + f' g' + f g''$

$= f'' g + 2 f' g' + f g''$ .

We now differential twice more to find $(fg)^{(4)}$ :

$(fg)''' = ( [fg]'')' = (f''g)' + 2(f'g')' + (fg'')'$

$= f'''g + f'' g' + 2f'' g' + 2f' g'' + f' g'' + f g'''$

$= f''' g + 3 f'' g' + 3 f' g'' + f g'''$ .

A good student may be able to guess the pattern for the next derivative:

$(fg)^{(4)} = ( [fg]''')' = (f'''g)' + 3(f''g')' +3(f'g'')' + (fg''')'$

$= f^{(4)}g + f''' g' + 3f''' g' + 3f'' g'' + 3f'' g'' + 3f'g''' + f' g''' + f g^{(4)}$

$= f^{(4)} g + 4 f''' g' + 6 f'' g'' + 4f' g''' + f g^{(4)}$ .

In this way, Pascal’s triangle makes a somewhat surprising appearance; indeed, this pattern can be proven with mathematical induction.

In the next post, we’ll apply this to the solution of the fourth-order differential equation.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 6e: Rationale for Method of Undetermined Coefficients II

We have shown that the motion of a planet around the Sun, expressed in polar coordinates $(r,\theta)$ with the Sun at the origin, under general relativity follows the initial-value problem

$u(0) = \displaystyle \frac{1}{P}$ ,

$u'(0) = 0$ ,

In the previous post, we derived the method of undetermined coefficients for the simplified differential equation

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2}$ .

In this post, we consider the simplified differential equation if the right-hand side has only the fourth term,

$u''(\theta) + u(\theta) = \displaystyle \frac{2\delta \epsilon }{\alpha^2}\cos \theta$ .

Let $v(\theta) = \displaystyle \frac{2\delta \epsilon }{\alpha^2}\cos \theta$ . Then $v$ satisfies the new differential equation $v'' + v = 0$ . Since $u'' + u = v$ , we may substitute:

$(u''+u)'' + (u'' + u) = 0$

$u^{(4)} + u'' + u'' + u = 0$

$u^{(4)} + 2u'' + u = 0$ .

The characteristic equation of this homogeneous differential equation is $r^4 + 2r^2 + 1 = 0$ , or $(r^2+1)^2 = 0$ . Therefore, $r = i$ and $r = -i$ are both double roots of this quartic equation. Therefore, the general solution for $u$ is

$u(\theta) = c_1 \cos \theta + c_2 \sin \theta + c_3 \theta \cos \theta + c_4 \theta \sin \theta$ .

Substituting into the original differential equation will allow for the computation of $c_3$ and $c_4$ :

$u''(\theta) + u(\theta) = -c_1 \cos \theta - c_2 \sin \theta - 2c_3 \sin \theta - c_3 \theta \cos \theta + 2c_4 \cos \theta - c_4 \theta \sin \theta$

$+ c_1 \cos \theta + c_2 \sin \theta + c_3 \theta \cos \theta + c_4 \theta \sin \theta$

$\displaystyle \frac{2\delta \epsilon }{\alpha^2}\cos \theta = - 2c_3 \sin \theta+ 2c_4 \cos \theta$

Matching coefficients, we see that $c_3 = 0$ and $c_4 = \displaystyle \frac{\delta \epsilon }{\alpha^2}$ . Therefore,

$u(\theta) = c_1 \cos \theta + c_2 \sin \theta + \displaystyle \frac{\delta \epsilon }{\alpha^2} \theta \sin \theta$

is the general solution of the simplified differential equation. Setting $c_1 = c_2 = 0$ , we find that

$u(\theta) = \displaystyle \frac{\delta \epsilon }{\alpha^2} \theta \sin \theta$

is one particular solution of this simplified differential equation. Not surprisingly, this matches the result is the method of undetermined coefficients had been blindly followed.

As we’ll see in a future post, the presence of this $\theta \sin \theta$ term is what predicts the precession of a planet’s orbit under general relativity.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 6d: Rationale for Method of Undetermined Coefficeints I

We have shown that the motion of a planet around the Sun, expressed in polar coordinates $(r,\theta)$ with the Sun at the origin, under general relativity follows the initial-value problem

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha} + \delta \left( \frac{1 + \epsilon \cos \theta}{\alpha} \right)^2$ ,

$u(0) = \displaystyle \frac{1}{P}$ ,

$u'(0) = 0$ ,

We now take the perspective of a student who is taking a first-semester course in differential equations. There are two standard techniques for solving a second-order non-homogeneous differential equations with constant coefficients. One of these is the method of constant coefficients. To use this technique, we first expand the right-hand side of the differential equation and then apply a power-reduction trigonometric identity:

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{2\delta \epsilon \cos \theta}{\alpha^2} + \frac{\delta \epsilon^2 \cos^2 \theta}{\alpha^2}$

$= \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{2\delta \epsilon \cos \theta}{\alpha^2} + \frac{\delta \epsilon^2}{\alpha^2} \frac{1 + \cos 2\theta}{2}$

$= \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{2\delta \epsilon \cos \theta}{\alpha^2} + \frac{\delta \epsilon^2 \cos 2\theta}{2\alpha^2}$

This is now in the form for using the method of undetermined coefficients. However, in this series, I’d like to take some time to explain why this technique actually works. To begin, we look at a simplified differential equation using only the first three terms on the right-hand side:

$u''(\theta) + u(\theta) = \displaystyle\frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2}$ .

Let $v(\theta) =\displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2}$ . Since $v$ is a constant, this function satisfies the simple differential equation $v' = 0$ . Since $u''+u=v$ , we can substitute:

$(u'' + u)' = 0$

$u''' + u' = 0$

(We could have more easily said, “Take the derivative of both sides,” but we’ll be using a more complicated form of this technique in future posts.) The characteristic equation of this differential equation is $r^3 + r = 0$ . Factoring, we obtain $r(r^2 + 1) = 0$ , so that the three roots are $r = 0$ and $r = \pm i$ . Therefore, the general solution of this differential equation is

$u(\theta) = c_1 \cos \theta + c_2 \sin \theta + c_3$ .

Notice that this matches the outcome of blindly using the method of undetermined coefficients without conceptually understanding why this technique works.

The constants $c_1$ and $c_2$ are determined by the initial conditions. To find $c_3$ , we observe

$u''(\theta) +u(\theta) = -c_1 \cos \theta - c_2 \sin \theta +c_1 \cos \theta + c_2 \sin \theta + c_3$

$\displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} = c_3$ .

Therefore, the general solution of this simplified differential equation is

$u(\theta) = c_1 \cos \theta + c_2 \sin \theta + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2}$ .

Furthermore, setting $c_1 = c_2 = 0$ , we see that

$u(\theta) = \displaystyle\frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2}$

is a particular solution to the differential equation

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2}$ .

In the next couple of posts, we find the particular solutions associated with the other terms on the right-hand side.

Engaging students: Finding prime factorizations

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Bri Del Pozzo. Her topic, from Pre-Algebra: finding prime factorizations.

How could you as a teacher create an activity or project that involves your topic?

An activity that I would create for my students involving Prime Factorization is based on an example that I saw on Pinterest. I would prepare an activity where students would be given a picture of a tree and assigned a two-digit number. I would then have students decorate their tree and at the base of the tree, they would write their assigned number. Then, as the roots expand down, students would be able to write the factors of their number as a factor tree until they are left with only prime factors (based on the image from https://www.hmhco.com/blog/teaching-prime-factorization-of-36). In the example from Pinterest, the teacher focused on finding the greatest common divisors between two numbers and used the factors trees as guidance. For my activity, I would assign some students the same number and emphasize that some numbers (such as 24, 36, 72, etc.) can be factored in multiple ways, so the roots of the trees could look different depending on how the student decides to factor their number.

How can this topic be used in your students’ future courses in mathematics or science?

There are a few ways that Prime Factorization can be used in my students’ future math courses. Prime Factorization is incredibly useful when learning how to simplify fractions. By practicing Prime Factorization, students become more familiar with the factors of large numbers, which becomes helpful when simplifying fractions. In the instance that a fraction is not in its simplest form, students will have an easier time recognizing such and will feel more confident in simplifying the fraction. Additionally, Prime Factorization prepares students for finding Greatest Common Divisors. Knowing how to find Greatest Common Divisors can be useful when solving real-world problems as well as in simplifying fractions. At a higher level of math, Prime Factorization allows students to practice the skills needed to prepare themselves for factoring things more complicated than numbers. For example, the idea of factoring can be applied to factoring a common factor out of an expression, factoring quadratic equations, and factoring polynomials with complex numbers.

How can technology (YouTube, Khan Academy [khanacademy.org], Vi Hart, Geometers Sketchpad, graphing calculators, etc.) be used to effectively engage students with this topic? Note: It’s not enough to say “such-and-such is a great website”; you need to explain in some detail why it’s a great website.

Khanacademy.org would be a fantastic website to engage students in this topic because of the inclusion of multiple representations. This website allows students to work through multiple practice problems where they can find the Prime Factorization of a number. When the student gets the question correct, they can move on to the next question, or they have the option to view a brief explanation on how to arrive at the correct answer. If students get a problem incorrect, they can retry the problem or get help on the question. The “get help” feature also provides students with a brief explanation, with options in video form and picture/written form, of how to solve the problem. Another important feature of this website is the ability for students to write out their thoughts as they work through the problem. Khan Academy allows students the option to use an online “whiteboard” feature that appears directly below the problem. This “whiteboard” feature allows students to write out their work and also offers a walkthrough of how to draw a factor tree.

Resources:
https://www.khanacademy.org/math/pre-algebra/pre-algebra-factors-multiples/pre-algebra-prime-factorization-prealg/e/prime_factorization
https://www.hmhco.com/blog/teaching-prime-factorization-of-36

Engaging students: Finding prime factorizations

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Brendan Gunnoe. His topic, from Pre-Algebra: finding prime factorizations.

How can prime factorization be used in curriculum?

The teacher starts the class by asking students how they would find the least common multiple and greatest common divisor for two numbers. For the LCM, the most basic answer is listing the multiples of both denominators until they share a common multiple. For GCD, the most basic answer is listing out the factors of both numerator and denominator and finding the largest one in common.

Both processes can be made faster when using prime factorization, especially for larger numbers. First, do the process of prime factorization for both numbers. Then, for each prime, take the highest power on the lists and multiply everything together.

For example, take 12 and 45.

$12 = 2^2 \times 3^1$

$45 =3^2 \times 5^1$

$\lcm(12,45) = 2^2 \times 3^2 \times 5^1 = 180$

The process for finding the GCF is similar. Start off by doing the prime factorization for both numbers. Then, for each shared prime factor, take the smallest power and multiply everything together.

For example, take 12 and 30.

$12 = 2^2 \times 3^1$

$30 =2^1 \times 3^1 \times 5^1$

$\gcd(12,45) = 2^1 \times 3^1 = 6$

This process generalizes very easily for any amount of input numbers.

GCF and LCM are incredibly important when working with fractions and are used when reducing and adding fractions. Because fractions have loads of misconceptions associated with them, giving students another way to understand fractions can be very beneficial.

Technology

https://youtube.com/watch?v=2JM2oImb9Qg%3Ffeature%3Doembed

Have you ever wondered why we use 60 seconds in a minute and 60 minutes in an hour? Or why there is 24 hours in a day? What about why there is 360 degrees in a circle? One explanation is because these numbers can be divided evenly by loads of smaller numbers that we use often. In other words, these numbers have lots of factors in them. These kinds of numbers are called highly composite numbers.

A great video showcasing highly composite numbers is Numberphile’s video “5040 and other Anti-Prime Numbers,” hosted by Dr. James Grimes. This video is extremely dense with informative as Dr. Grimes explains what a highly composite number is, shows properties of these numbers, explains why they have these properties, and gives examples of how highly composite numbers are used both in math and in real life. Dr. Grimes also gives a few historical uses of highly composite numbers, which answer some of the questions listed above.

Prime factorization is the foundation of highly composite numbers. Highly composite numbers can be an interesting and exciting application of prime factorization.

Application

Semiprime numbers were also used in the making of the Arecibo message. Because the message is composed of 1679 bits, there is only four ways of decomposing the message into a rectangle. All possible decompositions of 1679 into a rectangle are 1×1679, 73×23, 23×73 and 1679×1. If decoded correctly, then the message forms a picture which contains loads of information about the solar system and life on Earth.

For a way to make semiprime numbers into an engaging activity for students, the teacher could have students create their own mini version of the Arecibo message and show them off in class. Students can be made into groups and each group get assigned a certain semiprime. Then, each group gets to decide what information goes in their mini message and draw their message onto a sheet of poster paper with a grid on it. Finally, they present their message to the class, representing the students sending their message off into space for extraterrestrial life to decode.

References:

https://topdrawer.aamt.edu.au/Fractions/Misunderstandings

https://www.youtube.com/watch?v=2JM2oImb9Qg

https://en.wikipedia.org/wiki/Semiprime

https://en.wikipedia.org/wiki/Arecibo_message

Factorization of 2021

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