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

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.

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

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

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

$u'(0) = 0$ ,

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).

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.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 6c: Solving New Differential Equation with Variation of Parameters

Under general relativity, the motion of a planet around the Sun —in polar coordinates $(r,\theta)$ , with the Sun at the origin — satisfies 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 variation of parameters. First, we solve the associated homogeneous differential equation

$u''(\theta) + u(\theta) = 0$ .

The characteristic equation of this differential equation is $r^2 + 1 = 0$ , which clearly has the two imaginary roots $r = \pm i$ . Therefore, two linearly independent solutions of the associated homogeneous equation are $u_1(\theta) = \cos \theta$ and $u_2(\theta) = \sin \theta$ .

(As an aside, this is one answer to the common question, “What are complex numbers good for?” The answer is naturally above the heads of Algebra II students when they first encounter the mysterious number $i$ , but complex numbers provide a way of solving the differential equations that model multiple problems in statics and dynamics.)

According to the method of variation of parameters, the general solution of the original nonhomogeneous differential equation

$u''(\theta) + u(\theta) = g(\theta)$

$u(\theta) = f_1(\theta) u_1(\theta) + f_2(\theta) u_2(\theta)$ ,

where

$f_1(\theta) = -\displaystyle \int \frac{u_2(\theta) g(\theta)}{W(\theta)} d\theta$ ,

$f_2(\theta) = \displaystyle \int \frac{u_1(\theta) g(\theta)}{W(\theta)} d\theta$ ,

and $W(\theta)$ is the Wronskian of $u_1(\theta)$ and $u_2(\theta)$ , defined by the determinant

$W(\theta) = \displaystyle \begin{vmatrix} u_1(\theta) & u_2(\theta) \\ u_1'(\theta) & u_2'(\theta) \end{vmatrix} = u_1(\theta) u_2'(\theta) - u_1'(\theta) u_2(\theta)$ .

Well, that’s a mouthful.

The only good news is that $W$ is easy to compute. Since $u_1(\theta) = \cos \theta$ and $u_2(\theta) = \sin \theta$ , we have

$W(\theta) = (\cos \theta)(\cos \theta) - (\sin \theta)(-\sin \theta) = \cos^2 \theta + \sin^2 \theta = 1$

from the usual Pythagorean trigonometric identity. Therefore, the denominators in the integrals for $f_1(\theta)$ and $f_2(\theta)$ essentially disappear.

Unfortunately, computing $f_1(\theta)$ and $f_2(\theta)$ , using

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

is a beast, requiring the creative use of multiple trigonometric identities. We begin with $f_1(\theta)$ , using the substitution $t = \cos \theta$ :

$f_1(\theta) = -\displaystyle \int \left[ \frac{1}{\alpha} + \delta \left( \frac{1 + \epsilon \cos \theta}{\alpha} \right)^2 \right] \sin \theta \, d\theta$

$= \displaystyle \int \left[ \frac{1}{\alpha} + \delta \left( \frac{1 + \epsilon t}{\alpha} \right)^2 \right] \, dt$

$= \displaystyle \frac{t}{\alpha} + \frac{\alpha\delta}{3\epsilon} \left( \frac{1 + \epsilon t}{\alpha} \right)^3 + a$

$= \displaystyle \frac{\cos \theta}{\alpha} + \frac{\alpha\delta}{3\epsilon} \left( \frac{1 + \epsilon \cos \theta}{\alpha} \right)^3 + a$ ,

where we use $+a$ for the constant of integration instead of the usual $+C$ . Second,

$f_2(\theta) = \displaystyle \int \left[ \frac{1}{\alpha} + \delta \left( \frac{1 + \epsilon \cos \theta}{\alpha} \right)^2 \right] \cos\theta \, d\theta$ .

Unfortunately, this is not easily simplified with a substitution, so we have to expand the integrand:

$f_2(\theta) = \displaystyle \int \left[ \frac{\cos \theta}{\alpha} + \frac{\delta \cos \theta}{\alpha^2} + \frac{2 \delta \epsilon \cos^2 \theta}{\alpha^2} + \frac{\delta \epsilon^2 \cos^3 \theta}{\alpha^2} \right] \, d\theta$

$= \displaystyle \int \left[ \frac{\cos \theta}{\alpha} + \frac{\delta \cos \theta}{\alpha^2} + \frac{\delta \epsilon (1 + \cos 2 \theta)}{\alpha^2} + \frac{\delta \epsilon^2 \cos \theta \cos^2 \theta}{\alpha^2} \right] \, d\theta$

$= \displaystyle \int \left[ \frac{\cos \theta}{\alpha} + \frac{\delta \cos \theta}{\alpha^2} + \frac{\delta \epsilon}{\alpha^2} + \frac{\delta \epsilon \cos 2 \theta}{\alpha^2}+ \frac{\delta \epsilon^2 \cos \theta (1- \sin^2 \theta)}{\alpha^2} \right] \, d\theta$

$= \displaystyle \int \left[ \frac{\cos \theta}{\alpha} + \frac{\delta (1 + \epsilon^2) \cos \theta}{\alpha^2} + \frac{\delta \epsilon}{\alpha^2} + \frac{\delta \epsilon \cos 2 \theta}{\alpha^2} - \frac{\delta \epsilon^2 \cos \theta \sin^2 \theta}{\alpha^2} \right] \, d\theta$

$= \displaystyle \frac{\sin\theta}{\alpha} + \frac{\delta (1 + \epsilon^2) \sin\theta}{\alpha^2} + \frac{\delta \epsilon \theta}{\alpha^2} + \frac{\delta \epsilon \sin 2 \theta}{2\alpha^2} - \frac{\delta \epsilon^2 \sin^3 \theta}{3\alpha^2} + b$ ,

using $+b$ for the constant of integration. Therefore, by variation of parameters, the general solution of the nonhomogeneous differential equation is

$u(\theta) = f_1(\theta) u_1(\theta) + f_2(\theta) u_2(\theta)$

$= \displaystyle \left( \frac{\cos \theta}{\alpha} + \frac{\delta}{3\alpha^2\epsilon} \left( 1 + \epsilon \cos \theta \right)^3 + a \right) \cos \theta$

$+ \displaystyle \left(\frac{\sin\theta}{\alpha} + \frac{\delta (1 + \epsilon^2) \sin\theta}{\alpha^2} + \frac{\delta \epsilon \theta}{\alpha^2} + \frac{\delta \epsilon \sin 2 \theta}{2\alpha^2} - \frac{\delta \epsilon^2 \sin^3 \theta}{3\alpha^2} + b \right) \sin \theta$

$= a\cos \theta + b \sin \theta + \displaystyle \frac{\cos^2 \theta + \sin^2 \theta}{\alpha} + \frac{\delta \cos \theta}{3\alpha^2\epsilon} + \frac{\delta \cos^2 \theta}{\alpha^2} + \frac{\delta \epsilon \cos^3 \theta}{\alpha^2}+ \frac{\delta\epsilon^2 \cos^4 \theta}{3\alpha^2}$

$+ \displaystyle \frac{\delta \sin^2 \theta}{\alpha^2} + \frac{\delta \epsilon^2 \sin^2\theta}{\alpha^2} + \frac{\delta \epsilon \theta \sin \theta}{\alpha^2} + \frac{\delta \epsilon \sin \theta\sin 2 \theta}{2\alpha^2} - \frac{\delta \epsilon^2 \sin^4 \theta}{3\alpha^2}$

$= a\cos \theta + \displaystyle \frac{\delta \cos \theta}{3\alpha^2\epsilon} + b \sin \theta + \displaystyle \frac{1}{\alpha} + \frac{\delta (\cos^2 \theta+\sin^2 \theta)}{\alpha^2} + \frac{\delta \epsilon \cos^3 \theta}{\alpha^2}+ \frac{\delta\epsilon^2 \cos^4 \theta}{3\alpha^2}$

$\displaystyle + \frac{\delta \epsilon^2 \sin^2\theta}{\alpha^2} + \frac{\delta \epsilon \theta \sin \theta}{\alpha^2} + \frac{\delta \epsilon \sin \theta\sin 2 \theta}{2\alpha^2} - \frac{\delta \epsilon^2 \sin^4 \theta}{3\alpha^2}$

$= a\cos \theta+ \displaystyle \frac{\delta \cos \theta}{3\alpha^2\epsilon} + b \sin \theta + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon \theta \sin \theta}{\alpha^2} + \frac{\delta \epsilon \cos^3 \theta}{\alpha^2} + \frac{\delta \epsilon \sin \theta\sin 2 \theta}{2\alpha^2}$

$\displaystyle + \frac{\delta \epsilon^2 \sin^2\theta}{\alpha^2} + \frac{\delta \epsilon^2 (\cos^4 \theta -\sin^4 \theta)}{3\alpha^2}$

$= a\cos \theta + \displaystyle \frac{\delta \cos \theta}{3\alpha^2\epsilon} + b \sin \theta + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon \theta \sin \theta}{\alpha^2} + \frac{\delta \epsilon \cos^3 \theta}{\alpha^2} + \frac{\delta \epsilon \sin \theta \cdot 2 \sin \theta \cos \theta}{2\alpha^2}$

$\displaystyle + \frac{\delta \epsilon^2 (1-\cos 2\theta)}{2\alpha^2} + \frac{\delta \epsilon^2 (\cos^2 \theta +\sin^2\theta)(\cos^2 \theta - \sin^2 \theta)}{3\alpha^2}$

$\displaystyle + \frac{\delta \epsilon^2}{2\alpha^2} -\frac{\delta \epsilon^2 \cos 2\theta}{2\alpha^2} + \frac{\delta \epsilon^2 \cos 2 \theta}{3\alpha^2}$

$= a\cos \theta + \displaystyle \frac{\delta \cos \theta}{3\alpha^2\epsilon}+ b \sin \theta + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{\delta \epsilon \theta \sin \theta}{\alpha^2} -\frac{\delta \epsilon^2 \cos 2\theta}{6\alpha^2}$

$\displaystyle + \frac{\delta \epsilon \cos^3 \theta}{\alpha^2} + \frac{\delta \epsilon \sin^2 \theta\cos \theta}{\alpha^2}$

$= a\cos \theta + \displaystyle \frac{\delta \cos \theta}{3\alpha^2\epsilon}+ b \sin \theta + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{\delta \epsilon \theta \sin \theta}{\alpha^2} -\frac{\delta \epsilon^2 \cos 2\theta}{6\alpha^2}$

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

$= a\cos \theta + \displaystyle \frac{\delta \cos \theta}{3\alpha^2\epsilon}+ b \sin \theta + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{\delta \epsilon \theta \sin \theta}{\alpha^2} -\frac{\delta \epsilon^2 \cos 2\theta}{6\alpha^2} + \frac{\delta \epsilon \cos \theta}{\alpha^2}$

$= \displaystyle \left(a + \displaystyle \frac{\delta}{3\alpha^2\epsilon} + \frac{\delta \epsilon}{\alpha^2} \right) \cos \theta + b \sin \theta + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{\delta \epsilon \theta \sin \theta}{\alpha^2} -\frac{\delta \epsilon^2 \cos 2\theta}{6\alpha^2}$

$= \displaystyle A \cos \theta + b \sin \theta + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{\delta \epsilon \theta \sin \theta}{\alpha^2} -\frac{\delta \epsilon^2 \cos 2\theta}{6\alpha^2}$ ,

where $A$ is another arbitrary constant.

Next, we use the initial conditions to find the constants $A$ and $b$ . From the initial condition $u(0) = \displaystyle \frac{1}{P} = \frac{1+\epsilon}{\alpha}$ , we obtain

$u(0) = \displaystyle A \cos 0 + b \sin 0 + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{\delta \epsilon \cdot 0 \cdot \sin 0}{\alpha^2} -\frac{\delta \epsilon^2 \cos 0}{6\alpha^2}$

$\displaystyle \frac{1+\epsilon}{\alpha} = A + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} -\frac{\delta \epsilon^2}{6\alpha^2}$

$\displaystyle \frac{\epsilon}{\alpha} = A + \displaystyle \frac{3\delta +\delta \epsilon^2}{3\alpha^2}$ ,

so that

$A = \displaystyle \frac{\epsilon}{\alpha} - \frac{\delta(3 + \epsilon^2)}{3\alpha^2}$ .

Next, we compute $u'(\theta)$ and use the initial condition $u'(0) = 0$ :

$u'(\theta) = \displaystyle -A \sin \theta + b \cos \theta + \frac{\delta \epsilon}{\alpha^2} (\sin \theta + \theta \cos \theta) + \frac{\delta \epsilon^2 \sin 2\theta}{3\alpha^2}$

$u'(0) = \displaystyle -A \sin 0 + b \cos 0 + \frac{\delta \epsilon}{\alpha^2} (\sin 0 + 0 \cos 0) + \frac{\delta \epsilon^2 \sin 0}{3\alpha^2}$

$0 = b$ .

Substituting these values for $A$ and $b$ , we finally arrive at the solution

$u(\theta) = \displaystyle \left(\frac{\epsilon}{\alpha} - \frac{\delta(3 + \epsilon^2)}{3\alpha^2} \right) \cos \theta + \displaystyle \frac{1}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{\delta \epsilon \theta \sin \theta}{\alpha^2} -\frac{\delta \epsilon^2 \cos 2\theta}{6\alpha^2}$

$= \displaystyle \frac{1 + \epsilon \cos \theta}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{ \delta\epsilon}{\alpha^2} \theta \sin \theta - \frac{ \delta \epsilon^2}{6\alpha^2} \cos 2\theta - \frac{\delta(3+\epsilon^2)}{3\alpha^2} \cos \theta$ .

Confirming Einstein’s Theory of General Relativity With Calculus, Part 6b: Checking Solution of New Differential Equation with Calculus

In the last post, we showed that if the motion of a planet around the Sun is expressed in polar coordinates $(r,\theta)$ , with the Sun at the origin, then under general relativity the motion of the planet follows the initial-value problem

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

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

$u'(0) = 0$ ,

I won’t sugar-coat it; the solution is a big mess:

$u(\theta) = \displaystyle \frac{1 + \epsilon \cos \theta}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{\epsilon \delta}{\alpha^2} \theta \sin \theta - \frac{\epsilon^2 \delta}{6\alpha^2} \cos 2\theta - \frac{\delta(3+\epsilon^2)}{3\alpha^2} \cos \theta$ .

That said, it is an elementary, if complicated, exercise in calculus to confirm that this satisfies all three equations above. We’ll start with the second one:

$u(0) = \displaystyle \frac{1 + \epsilon \cos 0}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{\epsilon \delta}{\alpha^2} \cdot 0 \sin 0 - \frac{\epsilon^2 \delta}{6\alpha^2} \cos 0 - \frac{\delta(3+\epsilon^2)}{3\alpha^2} \cos 0$

$= \displaystyle \frac{1 + \epsilon}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} - \frac{\epsilon^2 \delta}{6\alpha^2} - \frac{\delta(3+\epsilon^2)}{3\alpha^2}$

$= \displaystyle \frac{ 6\alpha(1+\epsilon) + 6\delta + 3 \delta \epsilon^2 - \delta \epsilon^2 - 2\delta (3+\epsilon^2)}{6\alpha^2}$

$= \displaystyle \frac{ 6\alpha(1+\epsilon) + 6\delta + 3 \delta \epsilon^2 - \delta \epsilon^2 - 6\delta - 2\delta \epsilon^2}{6\alpha^2}$

$= \displaystyle \frac{ 6\alpha(1+\epsilon)}{6\alpha^2}$

$= \displaystyle \frac{ 1+\epsilon}{\alpha}$

$= \displaystyle \frac{1}{P}$ ,

where in the last step we used the equation $P = \displaystyle \frac{\alpha}{1 + \epsilon}$ that was obtained earlier in this series.

Next, to check the initial condition $u'(0) = 0$ , we differentiate:

$u'(\theta) = \displaystyle -\frac{\epsilon \sin \theta}{\alpha} + \frac{\epsilon \delta}{\alpha^2} (\sin \theta + \theta \cos \theta) + \frac{\epsilon^2 \delta}{3\alpha^2} \sin 2\theta + \frac{\delta(3+\epsilon^2)}{3\alpha^2} \sin\theta$

$u'(0) = \displaystyle -\frac{\epsilon \sin 0}{\alpha} + \frac{\epsilon \delta}{\alpha^2} (\sin 0 + 0 \cdot \cos 0) + \frac{\epsilon^2 \delta}{3\alpha^2} \sin 0 + \frac{\delta(3+\epsilon^2)}{3\alpha^2} \sin 0 = 0$ .

Finally, to check the differential equation itself, we compute the second derivative:

$u''(\theta) = \displaystyle -\frac{\epsilon \cos \theta}{\alpha} + \frac{\epsilon \delta}{\alpha^2} (2 \cos \theta - \theta \sin \theta) + \frac{2\epsilon^2 \delta}{3\alpha^2} \cos 2\theta + \frac{\delta(3+\epsilon^2)}{3\alpha^2} \cos\theta$ .

Adding $u''(\theta)$ and $u(\theta)$ , we find

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

$\displaystyle - \frac{\epsilon^2 \delta}{6\alpha^2} \cos 2\theta + \frac{2\epsilon^2 \delta}{3\alpha^2} \cos 2\theta - \frac{\delta(3+\epsilon^2)}{3\alpha^2} \cos \theta + \frac{\delta(3+\epsilon^2)}{3\alpha^2} \cos\theta$ ,

which simplifies considerably:

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

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

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

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

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

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

where we used the power-reduction trigonometric identity

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

on the second-to-last step.

While we have verified the proposed solution of the initial-value problem, and the steps for doing so lie completely within the grasp of a good calculus student, I’ll be the first to say that this solution is somewhat unsatisfying: the solution appeared seemingly out of thin air, and we just checked to see if this mysterious solution actually works. In the next few posts, I’ll discuss how this solution can be derived using standard techniques from first-semester differential equations.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 5d: Deriving Orbits under Newtonian Mechanics Using Variation of Parameters

We previously showed that if the motion of a planet around the Sun is expressed in polar coordinates $(r,theta)$ , with the Sun at the origin, then under Newtonian mechanics (i.e., without general relativity) the motion of the planet follows the differential equation

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha}$ ,

where $u = 1/r$ and $\alpha$ is a certain constant. We will also impose the initial condition that the planet is at perihelion (i.e., is closest to the sun), at a distance of $P$ , when $\theta = 0$ . This means that $u$ obtains its maximum value of $1/P$ when $\theta = 0$ . This leads to the two initial conditions

$u(0) = \displaystyle \frac{1}{P} \qquad \hbox{and} \qquad u'(0) = 0$ ;

the second equation arises since $u$ has a local extremum at $\theta = 0$ .

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 variation of parameters. First, we solve the associated homogeneous differential equation

$u''(\theta) + u(\theta) = 0$ .

According to the method of variation of parameters, the general solution of the original nonhomogeneous differential equation

$u''(\theta) + u(\theta) = g(\theta)$

$u(\theta) = f_1(\theta) u_1(\theta) + f_2(\theta) u_2(\theta)$ ,

where

$f_1(\theta) = -\displaystyle \int \frac{u_2(\theta) g(\theta)}{W(\theta)} d\theta$ ,

$f_2(\theta) = \displaystyle \int \frac{u_1(\theta) g(\theta)}{W(\theta)} d\theta$ ,

and $W(\theta)$ is the Wronskian of $u_1(\theta)$ and $u_2(\theta)$ , defined by the determinant

$W(\theta) = \displaystyle \begin{vmatrix} u_1(\theta) & u_2(\theta) \\ u_1'(\theta) & u_2'(\theta) \end{vmatrix} = u_1(\theta) u_2'(\theta) - u_1'(\theta) u_2(\theta)$ .

Well, that’s a mouthful.

Fortunately, for the example at hand, these computations are pretty easy. First, since $u_1(\theta) = \cos \theta$ and $u_2(\theta) = \sin \theta$ , we have

$W(\theta) = (\cos \theta)(\cos \theta) - (\sin \theta)(-\sin \theta) = \cos^2 \theta + \sin^2 \theta = 1$

from the usual Pythagorean trigonometric identity. Therefore, the denominators in the integrals for $f_1(\theta)$ and $f_2(\theta)$ essentially disappear.

Since $g(\theta) = \displaystyle \frac{1}{\alpha}$ , the integrals for $f_1(\theta)$ and $f_2(\theta)$ are straightforward to compute:

$f_1(\theta) = -\displaystyle \int u_2(\theta) \frac{1}{\alpha} d\theta = -\displaystyle \frac{1}{\alpha} \int \sin \theta \, d\theta = \displaystyle \frac{1}{\alpha}\cos \theta + a$ ,

where we use $+a$ for the constant of integration instead of the usual $+C$ . Second,

$f_2(\theta) = \displaystyle \int u_1(\theta) \frac{1}{\alpha} d\theta = \displaystyle \frac{1}{\alpha} \int \cos \theta \, d\theta = \displaystyle \frac{1}{\alpha}\sin \theta + b$ ,

using $+b$ for the constant of integration. Therefore, by variation of parameters, the general solution of the nonhomogeneous differential equation is

$u(\theta) = f_1(\theta) u_1(\theta) + f_2(\theta) u_2(\theta)$

$= \left( \displaystyle \frac{1}{\alpha}\cos \theta + a \right) \cos \theta + \left( \displaystyle \frac{1}{\alpha}\sin\theta + b \right) \sin \theta$

$= a \cos \theta + b\sin \theta + \displaystyle \frac{\cos^2 \theta + \sin^2 \theta}{\alpha}$

$= a \cos \theta + b \sin \theta + \displaystyle \frac{1}{\alpha}$ .

Unsurprisingly, this matches the answer in the previous post that was found by the method of undetermined coefficients.

For the sake of completeness, I repeat the argument used in the previous two posts to determine $a$ and $b$ . This is require using the initial conditions $u(0) = \displaystyle \frac{1}{P}$ and $u'(0) = 0$ . From the first initial condition,

$u(0) = a \cos 0 + b \sin 0 + \displaystyle \frac{1}{\alpha}$

$\displaystyle \frac{1}{P} = a + \frac{1}{\alpha}$

$\displaystyle \frac{1}{P} - \frac{1}{\alpha} = a$

$\displaystyle \frac{\alpha - P}{\alpha P} = a$

From the second initial condition,

$u'(\theta) = -a \sin \theta + b \cos \theta$

$u'(0) = -a \sin 0 + b \cos 0$

$0 = b$ .

From these two constants, we obtain

$u(\theta) = \displaystyle \frac{\alpha - P}{\alpha P} \cos \theta + 0 \sin \theta + \displaystyle \frac{1}{\alpha}$

$= \displaystyle \frac{1}{\alpha} \left( 1 + \frac{\alpha-P}{P} \cos \theta \right)$

$= \displaystyle \frac{1 + \epsilon \cos \theta}{\alpha}$ ,

where $\epsilon = \displaystyle \frac{\alpha - P}{P}$ .

Finally, since $r = 1/u$ , we see that the planet’s orbit satisfies

$r = \displaystyle \frac{\alpha}{1 + \epsilon \cos \theta}$ ,

so that, as shown earlier in this series, the orbit is an ellipse with eccentricity $\epsilon$ .

Confirming Einstein’s Theory of General Relativity With Calculus, Part 5c: Deriving Orbits under Newtonian Mechanics with the Method of Undetermined Coefficients

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha}$ ,

$u(0) = \displaystyle \frac{1}{P} \qquad \hbox{and} \qquad u'(0) = 0$ ;

the second equation arises since $u$ has a local extremum at $\theta = 0$ .

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 undetermined coefficients. First, we solve the associated homogeneous differential equation

$u''(\theta) + u(\theta) = 0$ .

The characteristic equation of this differential equation is $r^2 + 1 = 0$ , which clearly has the two imaginary roots $r = \pm i$ . Therefore, the solution of the associated homogeneous equation is $u_h(\theta) = a \cos \theta + b \sin \theta$ .

Next, we find a particular solution to the original differential equation. Since the right-hand side is a constant and $r=0$ is not a solution of the characteristic equation, this leads to trying something of the form $U(\theta) = A$ as a solution, where $A$ is a soon-to-be-determined constant. (Guessing this form of $U(\theta)$ is a standard technique from differential equations; later in this series, I’ll give some justification for this guess.)

Clearly, $U'(\theta) = 0$ and $U''(\theta) = 0$ . Substituting, we find

$U''(\theta) + U(\theta) = \displaystyle \frac{1}{\alpha}$

$0 + A = \displaystyle \frac{1}{\alpha}$

$A = \displaystyle \frac{1}{\alpha}$

Therefore, $U(\theta) = \displaystyle \frac{1}{\alpha}$ is a particular solution of the nonhomogeneous differential equation.

Next, the general solution of the nonhomogeneous differential equation is found by adding the general solution to the associated homogeneous differential equation and the particular solution:

$u(\theta) = u_h(\theta) + U(\theta) = a \cos \theta + b\sin \theta + \displaystyle \frac{1}{\alpha}$ .

Finally, to determine $a$ and $b$ , we use the initial conditions $u(0) = \displaystyle \frac{1}{P}$ and $u'(0) = 0$ . From the first initial condition,

$u(0) = a \cos 0 + b \sin 0 + \displaystyle \frac{1}{\alpha}$

$\displaystyle \frac{1}{P} = a + \frac{1}{\alpha}$

$\displaystyle \frac{1}{P} - \frac{1}{\alpha} = a$

$\displaystyle \frac{\alpha - P}{\alpha P} = a$

From the second initial condition,

$u'(\theta) = -a \sin \theta + b \cos \theta$

$u'(0) = -a \sin 0 + b \cos 0$

$0 = b$ .

From these two constants, we obtain

$u(\theta) = \displaystyle \frac{\alpha - P}{\alpha P} \cos \theta + 0 \sin \theta + \displaystyle \frac{1}{\alpha}$

$= \displaystyle \frac{1}{\alpha} \left( 1 + \frac{\alpha-P}{P} \cos \theta \right)$

$= \displaystyle \frac{1 + \epsilon \cos \theta}{\alpha}$ ,

where $\epsilon = \displaystyle \frac{\alpha - P}{P}$ .

Finally, since $r = 1/u$ , we see that the planet’s orbit satisfies

$r = \displaystyle \frac{\alpha}{1 + \epsilon \cos \theta}$ ,

so that, as shown earlier in this series, the orbit is an ellipse with eccentricity $\epsilon$ .

The reader will notice that this solution is pretty much a carbon-copy of the previous post. The difference is that calculus students wouldn’t necessarily be able to independently generate the solutions of the associated homogeneous differential equation and a particular solution of the nonhomogeneous differential equation, and so the line of questioning is designed to steer students toward the answer.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 5b: Deriving Orbits under Newtonian Mechanics with Calculus

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha}$ ,

$u(0) = \displaystyle \frac{1}{P} \qquad \hbox{and} \qquad u'(0) = 0$ ;

the second equation arises since $u$ has a local extremum at $\theta = 0$ .

In the previous post, we confirmed that

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

solved this initial-value problem. However, the solution was unsatisfying because it gave no indication of where this guess might have come from. In this post, I suggest a series of questions that good calculus students could be asked that would hopefully lead them quite naturally to this solution.

Step 1. Let’s make the differential equation simpler, for now, by replacing the right-hand side with 0:

$u''(\theta) + u(\theta) = 0$ ,

or

$u''(\theta) = -u(\theta)$ .

Can you think of a function or two that, when you differentiate twice, you get the original function back, except with a minus sign in front?

Answer to Step 1. With a little thought, hopefully students can come up with the standard answers of $u(\theta) = \cos \theta$ and $u(\theta) = \sin \theta$ .

Step 2. Using these two answers, can you think of a third function that works?

Answer to Step 2. This is usually the step that students struggle with the most, as they usually try to think of something completely different that works. This won’t work, but that’s OK… we all learn from our failures. If they can’t figure out, I’ll give a big hint: “Try multiplying one of these two answers by something.” In time, they’ll see that answers like $u(\theta) = 2\cos \theta$ and $u(\theta) = 3\sin \theta$ work. Once that conceptual barrier is broken, they’ll usually produce the solutions $u(\theta) = a \cos \theta$ and $u(\theta) = b \sin \theta$ .

Step 3. Using these two answers, can you think of anything else that works?

Answer to Step 3. Again, students might struggle as they imagine something else that works. If this goes on for too long, I’ll give a big hint: “Try combining them.” Eventually, we hopefully get to the point that they’ll see that the linear combination $u(\theta) = a \cos \theta + b \sin \theta$ also solves the associated homogeneous differential equation.

Step 4. Let’s now switch back to the original differential equation $u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha}$ . Let’s start simple: $u''(\theta) + u(\theta) = 5$ . Can you think of an easy function that’s a solution?

Answer to Step 4. This might take some experimentation, and students will probably try unnecessarily complicated guesses first. If this goes on for too long, I’ll give a big hint: “Try a constant.” Eventually, they hopefully determine that if $u(\theta) = 5$ is a constant function, then clearly $u'(\theta) = 0$ and $u''(\theta) = 0$ , so that $u''(\theta) + u(\theta) = 5$ .

Step 5. Let’s return to $u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha}$ . Any guesses on an answer to this one?

Answer to Step 5. Hopefully, students quickly realize that the constant function $u(\theta) = \displaystyle \frac{1}{\alpha}$ works.

Step 6. Let’s review. We’ve shown that anything of the form $u(\theta) = a\cos \theta + b \sin \theta$ is a solution of $u''(\theta) + u(\theta) = 0$ . We’ve also shown that $u(\theta) = \displaystyle\frac{1}{\alpha}$ is a solution of $u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha}$ . Can you think use these two answers to find something else that works?

Answer to Step 6. Hopefully, with the experience learned from Step 3, students will guess that $u(\theta) = a\cos \theta + b\sin \theta + \displaystyle \frac{1}{\alpha}$ will work.

Step 7. OK, that solves the differential equation. Any thoughts on how to find the values of $a$ and $b$ so that $u(0) = \displaystyle \frac{1}{P}$ and $u'(0) = 0$ ?

Answer to Step 7. Hopefully, students will see that we should just plug into $u(\theta)$ :

$u(0) = a \cos 0 + b \sin 0 + \displaystyle \frac{1}{\alpha}$

$\displaystyle \frac{1}{P} = a + \frac{1}{\alpha}$

$\displaystyle \frac{1}{P} - \frac{1}{\alpha} = a$

$\displaystyle \frac{\alpha - P}{\alpha P} = a$

To find $b$ , we first find $u'(\theta)$ and then substitute $\theta = 0$ :

$u'(\theta) = -a \sin \theta + b \cos \theta$

$u'(0) = -a \sin 0 + b \cos 0$

$0 = b$ .

From these two constants, we obtain

$u(\theta) = \displaystyle \frac{\alpha - P}{\alpha P} \cos \theta + 0 \sin \theta + \displaystyle \frac{1}{\alpha}$

$= \displaystyle \frac{1}{\alpha} \left( 1 + \frac{\alpha-P}{P} \cos \theta \right)$

$= \displaystyle \frac{1 + \epsilon \cos \theta}{\alpha}$ ,

where $\epsilon = \displaystyle \frac{\alpha - P}{P}$ .

Finally, since $r = 1/u$ , we see that the planet’s orbit satisfies

$r = \displaystyle \frac{\alpha}{1 + \epsilon \cos \theta}$ ,

so that, as shown earlier in this series, the orbit is an ellipse with eccentricity $\epsilon$ .

Confirming Einstein’s Theory of General Relativity With Calculus, Part 5a: Confirming Orbits under Newtonian Mechanics with Calculus

We previously showed that if the motion of a planet around the Sun is expressed in polar coordinates $(r,\theta)$ , with the Sun at the origin, then under Newtonian mechanics (i.e., without general relativity) the motion of the planet follows the differential equation

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha}$ ,

$u(0) = \displaystyle \frac{1}{P} \qquad \hbox{and} \qquad u'(0) = 0$ ;

the second equation arises since $u$ has a local extremum at $\theta = 0$ .

In the next few posts, we’ll discuss the solution of this initial-value problem. Today’s post would be appropriate for calculus students, which is confirming that

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

solves this initial-value problem, where $\epsilon = \displaystyle \frac{\alpha-P}{P}$ . Since $r$ is the reciprocal of $u$ , we infer that

$r = \displaystyle \frac{\alpha}{1 + \epsilon \cos \theta}$ .

As we’ve already seen in this series, this means that the orbit of the planet is a conic section — either a circle, ellipse, parabola, or hyperbola. Since the orbit of a planet is stable and $\epsilon = 0$ is extremely unlikely, this means that the planet orbits the Sun in an ellipse, with the Sun at one focus of the ellipse.

So, for a calculus student to verify that planets move in ellipses, one must check that

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

is a solution of the initial-value problem

$u''(\theta) + u(\theta) = \displaystyle \frac{1}{\alpha}$ ,

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

$u'(0) = 0$ .

The second line is easy to check:

$u(0) = \displaystyle \frac{1 + \epsilon \cos 0}{\alpha}$

$= \displaystyle \frac{1 + \epsilon}{\alpha}$

$= \displaystyle \frac{1 + \displaystyle \frac{\alpha-P}{P}}{\alpha}$

$= \displaystyle \frac{1}{\alpha} \frac{P + \alpha - P}{P}$

$= \displaystyle \frac{1}{\alpha} \frac{\alpha}{P}$

$= \displaystyle \frac{1}{P}$ .

The third line is also easy to check:

$u'(\theta) = \displaystyle \frac{-\epsilon \sin \theta}{\alpha}$

$u'(0) = \displaystyle \frac{-\epsilon \sin 0}{\alpha} = 0$ .

To check the first line, we first find $u''(\theta)$ :

$u''(\theta) = \displaystyle \frac{-\epsilon \cos \theta}{\alpha}$ ,

so that

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

thus confirming that $u(\theta) = \displaystyle \frac{1 + \epsilon \cos \theta}{\alpha}$ solves the initial-value problem.

While the above calculations are well within the grasp of a good Calculus I student, I’ll be the first to admit that this solution is less than satisfying. We just mysteriously proposed a solution, seemingly out of thin air, and confirmed that it worked. In the next post, I’ll proposed a way that calculus students can be led to guess this solution. Then, we talk about finding the solution of this nonhomogeneous initial-value problem using standard techniques from differential equations.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 1c: Outline of Argument

This is going to be a very long series, so I’d like to provide a tree-top view of how the argument will unfold.

We begin by using three principles from Newtonian physics — the Law of Conservation of Angular Momentum, Newton’s Second Law, and Newton’s Law of Gravitation — to show that the orbit of a planet, under Newtonian physics, satisfies the initial-value problem

$u''(\theta) + u(\theta) = \displaystyle \frac{GMm^2}{\ell^2}$ ,

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

$u'(0) = 0$ .

In these equations:

The orbit of the planet is in polar coordinates $(r,\theta)$ , where the Sun is placed at the origin.
The planet’s perihelion — closest distance from the Sun — is a distance of $P$ at angle $\theta = 0$ .
The function $u(\theta)$ is equal to $\displaystyle \frac{1}{r(\theta)}$ .
$G$ is the gravitational constant of the universe.
$M$ is the mass of the Sun.
$m$ is the mass of the planet.
$\ell$ is the angular momentum of the planet.

The solution of this differential equation is

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

so that

$r(\theta) = \displaystyle \frac{\alpha}{1 + \epsilon \cos \theta}$ .

In polar coordinates, this is the graph of an ellipse. Substituting $\theta = 0$ , we see that

$P = \displaystyle \frac{1 + \epsilon}{\alpha}$ .

In the solution for $u(\theta)$ , we have $\alpha = \displaystyle \frac{\ell^2}{GMm^2}$ and $\epsilon = \displaystyle \frac{\alpha - P}{P}$ . The number $\epsilon$ is the eccentricity of the ellipse, while $\alpha = \displaystyle \frac{P}{1+\epsilon}$ is proportional to the size of the ellipse.

Under general relativity, the governing initial-value problem changes to

$u''(\theta) + u(\theta) = \displaystyle \frac{GMm^2}{\ell^2} + \frac{3GM}{c^2} [u(\theta)]^2$ ,

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

$u'(0) = 0$ ,

where $c$ is the speed of light. We will see that the solution of this new differential equation can be well approximated by

$u(\theta) = \displaystyle \frac{1 + \epsilon}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{\delta\epsilon}{\alpha^2} \theta \sin \theta - \frac{\delta \epsilon^2}{6\alpha^2} \cos 2\theta - \frac{\delta(3+\epsilon^2)}{3\alpha^2} \cos \theta$

$\approx \displaystyle \frac{1}{\alpha} \left[1 + \epsilon \cos \left(\theta - \frac{\delta \theta}{\alpha} \right) \right]$ .

This last equation describes a spiral that precesses by approximately

$\displaystyle \frac{2\pi \delta}{\alpha} \quad$ radians per orbit

$\displaystyle \frac{6\pi G M}{a c^2 (1-\epsilon^2)} \quad$ radians per orbit,

where $a$ is the length of the semimajor axis of the orbit.

This matches the amount of precession in Mercury’s orbit that is not explained by Newtonian physics, thus confirming Einstein’s theory of general relativity.

To the extent possible, I will take the perspective of a good student who has taken Precalculus and Calculus I. However, I will have to break this perspective a couple of times when I discuss principles from physics and derive the solutions of the above differential equations.

Here we go…

What is your go-to dance move?

This is frivolous but fun, and it made me smile. Well done.

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Engaging students: Deriving the double angle formulas for sine, cosine, and tangent

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 Morgan Mayfield. His topic, from Precalculus: deriving the double angle formulas for sine, cosine, and tangent.

How could you as a teacher create an activity or project that involves your topic? I want to provide some variety for opportunities to make this an engaging opportunity for Precalculus students and some Calculus students. Here are my three thoughts: IDEA 1: For precalculus students in a regular or advanced class, have them derive this formula in groups. After students are familiar with the Pythagorean identities and with angle sum identities, group students and ask them to derive a formula for double angles Sin(2θ), Cos(2θ), Tan(2θ). Let them struggle a bit, and if needed give them some hints such as useful formulas and ways to represent multiplication so that it looks like other operations. From here, encourage students to simplify when they can and challenge students to find the other formulas of Cos(2θ). Ask students to speculate instances when each formula for Cos(2θ) would be advantageous. This gives students confidence in their own abilities and show how math is interconnected and not just a bunch of trivial formulas. Lastly, to challenge students, have them come up with an alternative way to prove Tan(2θ), notably Sin(2θ)/Cos(2θ). This would make an appropriate activity for students while having them continue practicing proving trigonometric identities. IDEA 2: This next idea should be implemented for an advanced Precal class, and only when there is some time to spare. Euler was an intelligent man and left us with the Euler’s Formula: $e^{ix}=\cos x + i \sin x$ . Have Precalculus students suspend their questions about where it comes from and what it is used for. This is not something they would use in their class. Reassure them that for what they will do, all they need to understand is imaginary numbers, multiplying imaginary numbers, and laws of exponents. Have them plug in x = A + B and simplify the right-hand side of the equation so that we get: $\cos(A+B)+i\sin(A+B)= a + bi$ where $a$ and $b$ are two real numbers. The goal here is to get $\cos(A+B)+i\sin(A+B)= \cos \theta \cos \theta - \sin \theta \sin \theta + (\sin \theta \cos \theta + \cos \theta \sin \theta)i$ . All the steps to get to this point is Algebra, nothing out of their grasp. Now, the next part is to really get their brains going about what meaning we can make of this. If they are struggling, have them think about the implications of two imaginary numbers being equal; the coefficient of the real parts and imaginary parts must be equal to each other. Lastly, ask them if these equations seem familiar, where are they from, and what are they called…the angle sum formulas. From here, this can lead into what if x=2A? Students will either brute force the formula again, and others will realize x = A + A and plug it in to the equation they just derived and simplify. This idea is a 2-in-1 steal for the angle sum formulas and double angle formulas. It’s biggest downside is this is for Sin(2θ) and Cos(2θ). IDEA 3: Take IDEA 2, and put it in a Calculus 2 class. Everything that the precalculus class remains, but now have the paired students prove the Euler’s Formula using Taylor Series. Guide them through using the Taylor Series to figure out a Taylor Series representation of $e^x$ , $sin x$ , and $cos x$ . Then ask students to find an expanded Taylor Series of to 12 terms with ellipses, no need to evaluate each term, just the precise term. Give hints such as $i^2= -1$ and to consider $i^3=i^2 \cdot i = -i$ and other similar cases. Lastly, ask students to separate the extended series in a way that mimics $a + bi$ using ellipses to shows the series goes to infinity. What they should find is something like this:

Look familiar? Well it is the addition of two Taylor Series that represent Sin(x) and Cos(x). This is the last connection students need to make. Give hints to look through their notes to see why the “a” and “b” in the imaginary number look so familiar. This, is just one way to prove Euler’s Formula, then you can continue with IDEA 2 until your students prove the angle sum formulas and double angle formulas.

How does this topic extend what your students should have learned in previous courses? Students in Texas will typically be exposed to the Pythagorean Theorem in 8^th grade. At this stage, students use $a^2+b^2=c^2$ to find a missing side length. Students may also be exposed to Pythagorean triples at this stage. Then at the Geometry level or in a Trigonometry section, students will be exposed to the Pythagorean Identity. The Identity is $\sin^2 \theta + \cos^2 \theta = 1$ . I think that this is not fair for students to just learn this identity without connecting it to the Pythagorean Theorem. I think it would be a nice challenge student to solve for this identity by using a right triangle with hypotenuse c so that Sin (θ) = b/c and cos (θ) = a/c, one could then show either $c^2 \sin^2 \theta + c^2 \cos^2 \theta = c^2$ and thus $c^2(\sin^2 \theta + \cos^2 \theta) = c^2$ or one could show $(a/c)^2 + (b/c)^2 = (c/c)^2 = 1$ (using the Pythagorean theorem). From here, students learn about the angle addition and subtraction formulas in Precalculus. This is all that they need to derive the double angle formulas.

$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$

$\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$

$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

$\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$

$\tan(\alpha + \beta) = \displaystyle \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$

$\tan(\alpha - \beta) = \displaystyle \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$

This would be a good challenge exercise for students to do in pairs. Sin(2θ) = Sin(θ + θ), Cos(2 θ) = Cos(θ + θ), Tan(2θ) = Tan(θ + θ). Now we can apply the angle sum formula where both angles are equal: Sin(2θ) = sin(θ)cos(θ) + cos(θ)sin(θ) = 2sin(θ)cos(θ) Cos(2θ) = cos(θ)cos(θ) – sin(θ)sin(θ) = (We use a Pythagorean Identity here) Tan(2θ) = $\displaystyle \frac{\tan \theta + \tan \theta}{1 - \tan^2 \theta} = \frac{2 \tan \theta}{1-\tan^2 \theta}$ Bonus challenge, use Sin(2θ) and Cos(2θ) to get Tan(2θ). Well, if $\tan \theta = \displaystyle \frac{\sin \theta}{\cos \theta}$ , then

$\tan 2\theta = \displaystyle \frac{\sin 2\theta}{\cos 2\theta}$

$= \displaystyle \frac{2 \sin \theta \cos \theta}{\cos^2 \theta - \sin^2 \theta}$

$= \displaystyle \frac{ \frac{2 \sin \theta \cos \theta}{\cos^2 \theta} }{ \frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta} }$

$= \displaystyle \frac{2 \tan \theta}{1 - \tan^2 \theta}$

The derivations are straight forward, and I believe that many students get off the hook by not being exposed to deriving many trigonometric identities and taking them as facts. This is in the grasp of an average 10^th to 12^th grader.

What are the contributions of various cultures to this topic? I have included four links that talk about the history of Trigonometry. It seemed that ancient societies would need to know about the Pythagorean Identities and the angles sum formulas to know the double angle formulas. Here is our problem, it’s hard to know who “did it first?” and when “did they know it?”. Mathematical proofs and history were not kept as neatly written record but as oral traditions, entertainment, hobbies, and professions. The truth is that from my reading, many cultures understood the double angle formula to some extent independently of each other, even if there was no formal proof or record of it. Looking back at my answer to B2, it seems that the double angle formula is almost like a corollary to knowing the angle sum formulas, and thus to understand one could imply knowledge of the other. Perhaps, it was just not deemed important to put the double angle formula into a category of its own. Many of the people who figured out these identities were doing it because they were astronomers, navigators, or carpenters (construction). Triangles and circles are very important to these professions. Knowledge of the angle sum formula was known in Ancient China, Ancient India, Egypt, Greece (originally in the form of broken chords theorem by Archimedes), and the wider “Medieval Islamic World”. Do note that that Egypt, Greece, and the Medieval Islamic World were heavily intertwined as being on the east side of the Mediterranean and being important centers of knowledge (i.e. Library of Alexandria.) Here is the thing, their knowledge was not always demonstrated in the same way as we know it today. Some cultures did have functions similar to the modern trigonometric functions today, and an Indian mathematician, Mādhava of Sangamagrāma, figured out the Taylor Series approximations of those functions in the 1400’s. Greece and China for example relayed heavily on displaying knowledge of trigonometry in ideas of the length of lines (rods) as manifestations of variables and numbers. Ancient peoples didn’t have calculators, and they may have defined trigonometric functions in a way that would be correct such as the “law of sines” or a “Taylor series”, but still relied on physical “sine tables” to find a numerical representation of sine to n numbers after the decimal point. How we think of Geometry and Trigonometry today may have come from Descartes’ invention of the Cartesian plane as a convenient way to bridge Algebra and Geometry. References: https://www.mathpages.com/home/kmath205/kmath205.htm https://en.wikipedia.org/wiki/History_of_trigonometry https://www.ima.umn.edu/press-room/mumford-and-pythagoras-theorem