# Engaging students: Computing the determinant of a matrix

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 again comes from my former student Brendan Gunnoe. His topic: computing the determinant of a matrix.

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

When students learn about the determinant of a matrix, they only learn about computing it and don’t learn about the applications of the determinant or what they signify. One interesting use of the determinant is finding the eigenvectors of a matrix. A visual understanding of what an eigenvector is can be done by showing what a matrix does to the any generic vector, and what it does to the eigenvectors. For a generic vector that is different from an eigenvector, the matrix knocks the vector off the span of the original vector. What makes an eigenvector special is the fact that the matrix transformation keeps the eigenvector on its span but rescales said eigenvector by its eigenvalue. For example, take the matrix

$\left[ \begin{array}{cc} 5 & 3 \\ 3 & 5 \end{array} \right]$.

This matrix’s eigenvectors are $\left[ \begin{array}{c} 1 \\ 1 \end{array} \right]$ and $\left[ \begin{array}{c} 1 \\ -1 \end{array} \right]$ with eigenvalues 8 and 2 respectively. That is,

$\left[ \begin{array}{cc} 5 & 3 \\ 3 & 5 \end{array} \right] \left[ \begin{array}{c} 1 \\ 1 \end{array} \right] = \left[ \begin{array}{c} 8 \\ 8 \end{array} \right] = 8 \left[ \begin{array}{c} 1 \\ 1 \end{array} \right]$

and

$\left[ \begin{array}{cc} 5 & 3 \\ 3 & 5 \end{array} \right] \left[ \begin{array}{c} 1 \\ -1 \end{array} \right] = \left[ \begin{array}{c} 2 \\ -2 \end{array} \right] = 2 \left[ \begin{array}{c} 1 \\ -1 \end{array} \right]$.

Eigenvectors have many useful applications in future math and science classes including electronics, linear algebra, differential equations and mechanical engineering.

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.

The YouTube channel 3Blue1Brown has a fantastic video on determinates and linear transformations. Grant, the channel owner, uses animations to visualize what a matrix transformation does to the plane . He starts by showing what a transformation does to a single square then shows why the change of change of that one area shows what happens to the area of any region. He also gives multiple explanations for what a negative determinate means in terms of orientation of the axes. Then he explains what happens when the determinate is 0. All of this is already extremely useful for understanding what a 2×2 matrix does, but Grant continues and extends all the same things for 3×3 transformations. Lastly, Grant gives a few explanations on why the formula for the determinate is what it is and poses a small puzzle for the viewer to contemplate. This video explains what and why we use determinates and how they can be useful all while showing pleasing visual examples and other explanations.

What interesting word problems using this topic can your students do now?

Using determinates to see if a set of vectors is a basis.

The determinant can tell you when a matrix squishes space into a lower dimensional space like a line or a plane. Thus, taking the determinate of a matrix composed of a set of vectors tells you if those vectors are a basis for the matrix’s dimension.

Part 1. A 3D printer’s axes are set up in such a way that the print head can only travel in the direction $\left[ \begin{array}{c} 1 \\ 1 \end{array} \right]$ and $\left[ \begin{array}{c} -1 \\ 1 \end{array} \right]$. Assume that the place where the print head is right now is the origin $\left[ \begin{array}{c} 0 \\ 0 \end{array} \right]$. Can you move the print head to the location $\left[ \begin{array}{c} x \\ y \end{array} \right]$ and $\left[ \begin{array}{c} 1 \\ -1 \end{array} \right]$ by only moving in the directions of $\left[ \begin{array}{c} 1 \\ 1 \end{array} \right]$ and $\left[ \begin{array}{c} -1 \\ 1 \end{array} \right]$?

Hint: Try to solve $\left[ \begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array} \right] \left[ \begin{array}{c} a \\ b \end{array} \right] = \left[ \begin{array}{c} x \\ y \end{array} \right]$ . Does this always have a solution $\left[ \begin{array}{c} a \\ b \end{array} \right]$?

Part 2. Next time you turn on your 3D printer, one of the motor’s broke and now the print head can only move in the direction of $\left[ \begin{array}{c} 1 \\ 0 \end{array} \right]$. Assume that the place where the print head is right now is the origin $\left[ \begin{array}{c} 0 \\ 0 \end{array} \right]$. Can you move the print head to the location  by only moving in the direction of $\left[ \begin{array}{c} 1 \\ 0 \end{array} \right]$?

Hint: Try to solve $\left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right] \left[ \begin{array}{c} a \\ b \end{array} \right] = \left[ \begin{array}{c} x \\ y \end{array} \right]$ . Does this always have a solution $\left[ \begin{array}{c} a \\ b \end{array} \right]$?

Part 3. You buy a new 3D printer that it can move in all three directions: up/down, left/right, forward/backwards. When you test out the movement of the print head, you see that each axis moves in the directions of $\left[ \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right]$, $\left[ \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right]$, and $\left[ \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right]$. Can you use your new 3D printer to go to any location $\left[ \begin{array}{c} x \\ y \\ z \end{array} \right]$, inside the printing space?

Hint: Think about solving $\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \left[ \begin{array}{c} a \\ b \\ c \end{array} \right] = \left[ \begin{array}{c} x \\ y \\ z \end{array} \right]$ . Does this always have a solution $\left[ \begin{array}{c} a \\ b \\ c \end{array} \right]$? How do you know?

Part 4. Your little sibling messed around with your new 3D printer and now it moves in the directions $\left[ \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right]$, $\left[ \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right]$, and $\left[ \begin{array}{c} 2 \\ 1 \\ 1 \end{array} \right]$. Is your 3D printer able to get to some point $\left[ \begin{array}{c} x \\ y \\ z \end{array} \right]$ inside the printing space as is, or do you need to fix your printer?

Hint: Think about solving $\left[ \begin{array}{ccc} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{array} \right] \left[ \begin{array}{c} a \\ b \\ c \end{array} \right] = \left[ \begin{array}{c} x \\ y \\ z \end{array} \right]$. Does this always have a solution $\left[ \begin{array}{c} a \\ b \\ c \end{array} \right]$? How do you know?

# Thoughts on Numerical Integration (Part 11): Simpson’s Rule and exploration of error analysis

Numerical integration is a standard topic in first-semester calculus. From time to time, I have received questions from students on various aspects of this topic, including:

• Why is numerical integration necessary in the first place?
• Where do these formulas come from (especially Simpson’s Rule)?
• How can I do all of these formulas quickly?
• Is there a reason why the Midpoint Rule is better than the Trapezoid Rule?
• Is there a reason why both the Midpoint Rule and the Trapezoid Rule converge quadratically?
• Is there a reason why Simpson’s Rule converges like the fourth power of the number of subintervals?

In this series, I hope to answer these questions. While these are standard questions in a introductory college course in numerical analysis, and full and rigorous proofs can be found on Wikipedia and Mathworld, I will approach these questions from the point of view of a bright student who is currently enrolled in calculus and hasn’t yet taken real analysis or numerical analysis.

In the previous post in this series, I discussed three different ways of numerically approximating the definite integral $\displaystyle \int_a^b f(x) \, dx$, the area under a curve $f(x)$ between $x=a$ and $x=b$.

In this series, we’ll choose equal-sized subintervals of the interval $[a,b]$. If $h = (b-a)/n$ is the width of each subinterval so that $x_k = x_0 + kh$, then the integral may be approximated as

$\int_a^b f(x) \, dx \approx h \left[f(x_0) + f(x_1) + \dots + f(x_{n-1}) \right] \equiv L_n$

using left endpoints,

$\int_a^b f(x) \, dx \approx h \left[f(x_1) + f(x_2) + \dots + f(x_n) \right] \equiv R_n$

using right endpoints, and

$\int_a^b f(x) \, dx \approx h \left[f(c_1) + f(c_2) + \dots + f(c_n) \right] \equiv M_n$

using the midpoints of the subintervals. We have also derived the Trapezoid Rule

$\int_a^b f(x) \, dx \approx \displaystyle \frac{h}{2} [f(x_0) + 2f(x_1) + \dots + 2f(x_{n-1}) + f(x_n)] \equiv T_n$

and Simpson’s Rule (if $n$ is even)

$\int_a^b f(x) \, dx \approx \displaystyle \frac{h}{3} \left[y_0 + 4 y_1 + 2 y_2 + 4 y_3 + \dots + 2y_{n-2} + 4 y_{n-1} + y_{n} \right] \equiv S_n$.

In the previous post in this series, we saw that both the left-endpoint and right-endpoint rules have a linear rate of convergence: if twice as many subintervals are taken, then the error appears to go down by a factor of 2. If ten times as many subintervals are used, then the error should go down by a factor of 10. However, both the Midpoint Rule and the Trapezoid Rule have a quadratic rate of convergence: if twice as many subintervals are taken, then the error appears to go down by a factor of 4. If ten times as many subintervals are used, then the error should go down by a factor of 100. Moreover, it appears that the error from the Midpoint Rule is about half that of the Trapezoid Rule if the same number of subintervals are used.

Let’s now explore the results of Simpson’s Rule applied to $\displaystyle \int_1^2 x^9 \, dx = 102.3$ using different numbers of subintervals. The results are summarized in the table below.

The first immediate observation is that these approximations are far better than even the Midpoint and Trapezoid Rules! Indeed, we see that $S_{20} \approx 102.301$, using only 20 subintervals, is a better approximation than (from previous posts) either $M_{100} = 102.290$ or $T_{100} \approx 102.319$ using 100 subintervals! The lesson to learn again: Simpson’s Rule is a bit more difficult to compute than the Trapezoid Rule or the Midpoint Rule because of the different weights. Nevertheless, choosing a good algorithm is often far better than simply doing lots of computations.

There’s a second observation: the rate of convergence appears to be much, much faster. Indeed, the data points appear to fit a quartic polynomial very well. Said another way, if twice as many subintervals are taken, then the error appears to go down by a factor of 16. We can actually see this in the figure, looking at the lines with 10, 20, 40, and 80 subintervals.

Error with 10 subintervals = $|102.3174904 - 102.3| = 0.0174904$.

Error with 20 subintervals = $|102.3011002 - 102.3| = 0.0011002$.

Error with 40 subintervals = $|102.3000689 - 102.3| = 0.0000689$.

Error with 80 subintervals = $|102.3000043 - 102.3| = 0.0000043$.

In all cases, the error on the next line is about the error on the previous line divided by 16.

This illustrates quartic convergence, as opposed to the mere linear convergence of the left- and right-endpoint rules or the quadratic convergence of the Midpoint and Trapezoid Rules.

# Engaging students: Finding the equation of a circle

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 Noah Mena. His topic, from Precalculus: finding the equation of a circle.

The equation of a circle relies on knowing the definition of a circle, knowing the radius and deciding where the circle is centered at. All of these come into play when a student has to find the equation of a circle. It takes basic understanding of the cartesian grid and understanding the coordinate system.  The equation of a circle also builds on students being able to manipulate the equation to get it into standard form and identifying the equation of a circle when it is expanded out. The shape of a circle should also be known, which means with the equation of a circle, students should be able to construct the perfect circle according to the given specifications in the equation.

Learning to write the equation of a circle can be difficult. For one of my teaches last semester my mentor teacher suggested the use of a desmos paired with a worksheet to allow the students to explore what changes the standard equation of a circle. The worksheet had the students enter certain coordinates into the graphing calculator and write down what they thought was the equation of a circle. The next part of the assignment was student driven by having them share their conclusions on what the equation for a circle would be when it is centered at the origin vs. centered at (h,k). The worksheet shows that the students drove their own learning and came to their own conclusions which enhanced engagement through the lesson.

This topic can come up again in trigonometry, upper level calculus and in math modeling. In my TNTX math modeling course, we took a closer look at the derivation of this equation and the subtleties of the standard form. This topic may also be used in physics calculations or in general, science labs.  For a physics word problem, it may ask you to calculate the net force and acceleration of a moving object around a circle. In this instance, it would suffice to just know the definition and general shape of a circle to complete these calculations. The definition of a circle is also needed to calculate centripetal force.

# Thoughts on Numerical Integration (Part 10): Trapezoid Rule and exploration of error analysis

Numerical integration is a standard topic in first-semester calculus. From time to time, I have received questions from students on various aspects of this topic, including:

• Why is numerical integration necessary in the first place?
• Where do these formulas come from (especially Simpson’s Rule)?
• How can I do all of these formulas quickly?
• Is there a reason why the Midpoint Rule is better than the Trapezoid Rule?
• Is there a reason why both the Midpoint Rule and the Trapezoid Rule converge quadratically?
• Is there a reason why Simpson’s Rule converges like the fourth power of the number of subintervals?

In this series, I hope to answer these questions. While these are standard questions in a introductory college course in numerical analysis, and full and rigorous proofs can be found on Wikipedia and Mathworld, I will approach these questions from the point of view of a bright student who is currently enrolled in calculus and hasn’t yet taken real analysis or numerical analysis.

In the previous post in this series, I discussed three different ways of numerically approximating the definite integral $\displaystyle \int_a^b f(x) \, dx$, the area under a curve $f(x)$ between $x=a$ and $x=b$.

In this series, we’ll choose equal-sized subintervals of the interval $[a,b]$. If $h = (b-a)/n$ is the width of each subinterval so that $x_k = x_0 + kh$, then the integral may be approximated as

$\int_a^b f(x) \, dx \approx h \left[f(x_0) + f(x_1) + \dots + f(x_{n-1}) \right] \equiv L_n$

using left endpoints,

$\int_a^b f(x) \, dx \approx h \left[f(x_1) + f(x_2) + \dots + f(x_n) \right] \equiv R_n$

using right endpoints, and

$\int_a^b f(x) \, dx \approx h \left[f(c_1) + f(c_2) + \dots + f(c_n) \right] \equiv M_n$

using the midpoints of the subintervals. We have also derived the Trapezoid Rule

$\int_a^b f(x) \, dx \approx \displaystyle \frac{h}{2} [f(x_0) + 2f(x_1) + \dots + 2f(x_{n-1}) + f(x_n)] \equiv T_n$

and Simpson’s Rule (if $n$ is even)

$\int_a^b f(x) \, dx \approx \displaystyle \frac{h}{3} \left[y_0 + 4 y_1 + 2 y_2 + 4 y_3 + \dots + 2y_{n-2} + 4 y_{n-1} + y_{n} \right] \equiv S_n$.

In the previous post in this series, we saw that both the left-endpoint and right-endpoint rules have a linear rate of convergence: if twice as many subintervals are taken, then the error appears to go down by a factor of 2. If ten times as many subintervals are used, then the error should go down by a factor of 10. However, the Midpoint Rule has a quadratic rate of convergence: if twice as many subintervals are taken, then the error appears to go down by a factor of 4. If ten times as many subintervals are used, then the error should go down by a factor of 100.

Let’s now explore the results of the Trapezoid Rule applied to $\displaystyle \int_1^2 x^9 \, dx = 102.3$ using different numbers of subintervals. The results are summarized in the table below.

Once again, the data points fit a quadratic polynomial well, indicating quadratic convergence.

More subtly, it appears that the Trapezoid Rule isn’t quite as good as the Midpoint Rule. Here are the results from the Midpoint Rule (which also appeared in the previous post in this series):

For $n = 100$ subintervals, the error of the Trapezoid Rule is $|102.3191-102.3| = 0.0191$, which the error from the Midpoint Rule is $|102.2904-102.3| = 0.0096$. In other words, while both of these methods are superior to the left- and right-endpoint rules, it appears that the error from the Midpoint Rule is about half of the error from the Trapezoid Rule. The Midpoint Rule appears to be better.

To me, this is far from an obvious conclusion. Geometrically, it’s far from clear that the rectangles from the Midpoint Rule…

… provide a better approximation than using trapezoids …

… yet it appears that’s exactly what happened. This can be rigorously proven, as we’ll explore later in this series.

# Engaging students: Solving exponential equations

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 Austin Stone. His topic, from Precalculus: solving exponential equations.

What interesting (i.e., uncontrived) word problems using this topic can your students do now?

Exponential equations can be used in lots of different kinds of word problems. One that is pretty common but is very useful for students involves interest rate. “Megan has $20,000 to invest for 5 years and she found an interest rate of 5%. How much money will she have at the end of 5 years if the interest rate compounds monthly?” I would give them the formula A=P(1+r/n)rt. It is pretty easy to convince students that this is a real-world problem and would get the students engaged about exponential equations. You can also reword the problem to ask for how much Megan started with, what the rate is, or how much time the money was in there. That way students get used to solving equations when the variable is in the exponent and when it is not. This also can lead into or us prior knowledge of natural log to solve for the variable in the exponent. How could you as a teacher create an activity or project that involves your topic? Using the basis of the problem I mentioned above, a teacher could create a Project Based Instruction lesson using this idea. The teacher can set up a scenario where, over the course of a week or two, the students would have to decide which bank to make an investment in by calculating how much money they would profit at each bank. The students would have to research different banks and their interest rate. The teacher could also give each group different scenarios where some groups have more money to invest. Students would have to figure out how long they would like to invest. The teacher would give Do It Yourselves and Workshops that deal with solving exponential equations and also getting used to natural log. They would then make a presentation explaining what bank they have chosen and why. They would also have to explain the math that they would have used. How has this topic appeared in the news? To say that exponential equations have been in the news lately would be an understatement. It has virtually been the news this year. COVID-19 is a virus and viruses spread exponentially. This would get students engaged immediately because the topic would be relatable to their own lives. Doctors and scientists try to figure out different ways to “flatten the curve”, which essentially means to make the spread of the virus not exponential anymore. We have all heard people on the news telling the public how to stop the virus from spreading and how not make people around you at risk of contracting it (contributing the exponential spread). We all have most likely seen a doctor or scientist show a graph of the virus’s spread and their predictions on how it will look in the upcoming weeks. This would give students a chance to see that what they are learning can be applied to very crucial things going on in the world around them. References Exponential Functions # Thoughts on Numerical Integration (Part 9): Midpoint rule and exploration of error analysis Numerical integration is a standard topic in first-semester calculus. From time to time, I have received questions from students on various aspects of this topic, including: • Why is numerical integration necessary in the first place? • Where do these formulas come from (especially Simpson’s Rule)? • How can I do all of these formulas quickly? • Is there a reason why the Midpoint Rule is better than the Trapezoid Rule? • Is there a reason why both the Midpoint Rule and the Trapezoid Rule converge quadratically? • Is there a reason why Simpson’s Rule converges like the fourth power of the number of subintervals? In this series, I hope to answer these questions. While these are standard questions in a introductory college course in numerical analysis, and full and rigorous proofs can be found on Wikipedia and Mathworld, I will approach these questions from the point of view of a bright student who is currently enrolled in calculus and hasn’t yet taken real analysis or numerical analysis. In the previous post in this series, I discussed three different ways of numerically approximating the definite integral $\displaystyle \int_a^b f(x) \, dx$, the area under a curve $f(x)$ between $x=a$ and $x=b$. In this series, we’ll choose equal-sized subintervals of the interval $[a,b]$. If $h = (b-a)/n$ is the width of each subinterval so that $x_k = x_0 + kh$, then the integral may be approximated as $\int_a^b f(x) \, dx \approx h \left[f(x_0) + f(x_1) + \dots + f(x_{n-1}) \right] \equiv L_n$ using left endpoints, $\int_a^b f(x) \, dx \approx h \left[f(x_1) + f(x_2) + \dots + f(x_n) \right] \equiv R_n$ using right endpoints, and $\int_a^b f(x) \, dx \approx h \left[f(c_1) + f(c_2) + \dots + f(c_n) \right] \equiv M_n$ using the midpoints of the subintervals. We have also derived the Trapezoid Rule $\int_a^b f(x) \, dx \approx \displaystyle \frac{h}{2} [f(x_0) + 2f(x_1) + \dots + 2f(x_{n-1}) + f(x_n)] \equiv T_n$ and Simpson’s Rule (if $n$ is even) $\int_a^b f(x) \, dx \approx \displaystyle \frac{h}{3} \left[y_0 + 4 y_1 + 2 y_2 + 4 y_3 + \dots + 2y_{n-2} + 4 y_{n-1} + y_{n} \right] \equiv S_n$. In the previous post in this series, we saw that both the left-endpoint and right-endpoint rules have a linear rate of convergence: if twice as many subintervals are taken, then the error appears to go down by a factor of 2. If ten times as many subintervals are used, then the error should go down by a factor of 10. Let’s now explore the results of the midpoint rule applied to $\displaystyle \int_1^2 x^9 \, dx = 102.3$ using different numbers of subintervals. The results are summarized in the table below. The first immediate observation is that these approximations are far better than the left- and right-endpoint rule approximations! Indeed, we see that $M_{10} \approx 101.3$, using only ten subintervals, is a far better approximation than (from the previous post) either $L_{100} = 99.8$ or $R_{100} \approx 104.9$ using 100 subintervals! The lesson to learn: choosing a good algorithm is often far better than simply doing lots of computations. There’s a second observation: the rate of convergence appears to be much, much faster. Indeed, the data points appear to fit a parabola very well instead of a straight line. Said another way, if twice as many subintervals are taken, then the error appears to go down by a factor of 4. If ten times as many subintervals are used, then the error should go down by a factor of 100. This illustrates quadratic convergence, as opposed to the mere linear convergence of the left- and right-endpoint rules. # Engaging students: Compound interest 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 Mason Maynard. His topic, from Precalculus: compound interest. What interesting (i.e., uncontrived) word problems using this topic can your students do now? (You may find resources such as http://www.spacemath.nasa.gov to be very helpful in this regard; feel free to suggest others.) • A deposit of$3000 earns 2% interest compounded semiannually. How much money is in the bank after for 4 years?
• A deposit of $2150 earns 6% interest compounded quarterly. How much money is in the bank after for 6 years? • A deposit of$495 earns 3% interest compounded annually. How much money is in the bank after for 3 years?

These word problems are some of the basic compound interest problems that your students learn how to do where you just plug in the correct values for their corresponding variables.

• If you invested $1,000 in an account paying an annual percentage rate (quoted rate) compounded daily (based on a bank year of 360 days) and you wanted to have$2,500 in your account at the end of your investment time, what interest rate would you need if the investment time were 1 year, 10 years, 20 years, 100 years?
• If you invested $500 in an account paying an annual percentage rate compounded quarterly , and you wanted to have$2,500 in your account at the end of your investment time, what interest rate would you need if the investment time were 1 year, 10 years, 20 years, 100 years?

These are the types of problems that get more difficult for the students. You want them to use compound interest to solve but then they must incorporate logs into their solutions because they are looking for time instead of interest.

How does this topic extend what your students should have learned in previous courses?

With compound interest, students first learn about the simple interest formula. The only main difference is that you start to include exponents with compound interests. Then when you introduce your students to compound interest, you start to get into some more complicated problems. After they learn about compound interest and its basic problems, then you transition into logs with your students. This is used in compound interest and instead of just looking for the interest that will be accumulated after a specific amount of time, you then shift the variable around that you are looking for. The most coming type of problem that refers to this is they give you all of the information except for the amount time it takes to get a certain amount of interest. The last thing that leads up to compound interest in Calculus is when you transition from calculating the amount of interested over specific time intervals and a specific amount of times you compound it to calculating it with compounding it continuously over a specific time interval.

How have different cultures throughout time used this topic in their society?

Interest is something you have to pay on a load. Depending on what side you are and how thinks go, you are either getting some more money back that what you invested or you are paying off a massive debt. Some think that the idea behind charging loans on interest came from the early days of neighbors loaning there cattle to one another. What is really unique about this is that the words in the Egyptian, ancient Greek and Sumerian languages is connected to cattle and their offspring. This leads some to believe that interest came about due to the natural increase of the herd that occurred when you loaned out your cattle.

The first evidence that comes of a compound interest problem dates back to 2000-1700 B.C. in Babylon. A clay tablet was found and the unique thing is that the interest rate use to solve it was not written. Some researchers assume that the rate was 20% due to that mainly all the other compound interest problems dating back closer to this used it. What is really crazy is that 20% worked to solve the problem. The only thing that was wrong was that the time was corresponding to the Babylon calendar of 360 days instead of our 365 days.

In 50 B.C. Cicero writes to a friend in Rome. The letter tells that he would not normally recognize more than 12 percent interest on a loan, even though a decree was passed which required money lenders to charge no more than 12 percent. Cicero would then write a few days later that they will pay back the loan in 6 years will 12 percent interest and more money will be added each year.

Resources:

Compound Interest History:

https://www.cambridge.org/core/services/aop-cambridge-core/content/view/799CB1D40CDD46F3010767BFC60F24DB/S1357321719000254a.pdf/emergence_of_compound_interest.pdf

Word Problems:

https://www.basic-mathematics.com/compound-interest-word-problems.html

http://www.sosmath.com/algebra/logs/log5/log51/log51.html

# Thoughts on Numerical Integration (Part 8): Left and right endpoint rules and exploration of error analysis

Numerical integration is a standard topic in first-semester calculus. From time to time, I have received questions from students on various aspects of this topic, including:

• Why is numerical integration necessary in the first place?
• Where do these formulas come from (especially Simpson’s Rule)?
• How can I do all of these formulas quickly?
• Is there a reason why the Midpoint Rule is better than the Trapezoid Rule?
• Is there a reason why both the Midpoint Rule and the Trapezoid Rule converge quadratically?
• Is there a reason why Simpson’s Rule converges like the fourth power of the number of subintervals?

In this series, I hope to answer these questions. While these are standard questions in a introductory college course in numerical analysis, and full and rigorous proofs can be found on Wikipedia and Mathworld, I will approach these questions from the point of view of a bright student who is currently enrolled in calculus and hasn’t yet taken real analysis or numerical analysis.

In the previous post in this series, I discussed three different ways of numerically approximating the definite integral $\displaystyle \int_a^b f(x) \, dx$, the area under a curve $f(x)$ between $x=a$ and $x=b$.

In this series, we’ll choose equal-sized subintervals of the interval $[a,b]$. If $h = (b-a)/n$ is the width of each subinterval so that $x_k = x_0 + kh$, then the integral may be approximated as

$\int_a^b f(x) \, dx \approx h \left[f(x_0) + f(x_1) + \dots + f(x_{n-1}) \right] \equiv L_n$

using left endpoints,

$\int_a^b f(x) \, dx \approx h \left[f(x_1) + f(x_2) + \dots + f(x_n) \right] \equiv R_n$

using right endpoints, and

$\int_a^b f(x) \, dx \approx h \left[f(c_1) + f(c_2) + \dots + f(c_n) \right] \equiv M_n$

using the midpoints of the subintervals. We have also derived the Trapezoid Rule

$\int_a^b f(x) \, dx \approx \displaystyle \frac{h}{2} [f(x_0) + 2f(x_1) + \dots + 2f(x_{n-1}) + f(x_n)] \equiv T_n$

and Simpson’s Rule (if $n$ is even)

$\int_a^b f(x) \, dx \approx \displaystyle \frac{h}{3} \left[y_0 + 4 y_1 + 2 y_2 + 4 y_3 + \dots + 2y_{n-2} + 4 y_{n-1} + y_{n} \right] \equiv S_n$.

All of the above approximations to $\displaystyle \int_a^b f(x) \, dx$ are precisely that — approximations. That begs the obvious question: how can we get better approximations. One obvious answer is taking more subintervals. The figures below show the left-endpoint approximations using $n = 4$ and $n = 40$ subintervals. Geometrically, it’s clear that the orange rectangles in the second picture do a better job of approximating the area under the curve. Unfortunately, simply taking more subintervals has its limitations. Using a spreadsheet as in the previous post in this series, one can implement 100 or even 1000 subintervals without much difficult. However, as demonstrated in the video below, implementing any of these methods with 10,000 subintervals is pretty time-consuming. (Tl/dw: It can take literally a couple of minutes.)

Instead of relying on sheer computational firepower, let’s instead investigate how good these numerical methods actually are. To begin, let’s explore the left-endpoint rule applied to $\displaystyle \int_1^2 x^9 \, dx = 102.3$ using different numbers of subintervals. The results are summarized in the table below. As $n$ increases, the left endpoint approximations $L_n$ are indeed getting closer and closer to the actual value of $\displaystyle \int_1^2 x^9 \, dx = 102.3$. Interestingly, when the width $h$ is plotted with these approximations, the data points fall almost exactly on a straight line. The same phenomenon occurs when using right endpoints: So, it appears that the errors in both the left- and right-endpoint rules are a linear function of the size of the subintervals. Said another way, if twice as many subintervals are taken, then the error appears to go down by a factor of 2. If ten times as many subintervals are used, then the error should go down by a factor of 10. As we’ll see in the next few posts, the errors for the Midpoint Rule, the Trapezoid Rule, and especially Simpson’s Rule are much better than the errors from these two methods.

# Engaging students: Infinite geometric series

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 Brendan Gunnoe. His topic, from Precalculus: infinite geometric series.

Curriculum:

Students can use the formula for an infinite geometric series to discover the formula for a finite geometric series. The teacher would start by posing the question “Can we use the infinite geometric series to come up with a formula for the finite version?” and writing out a series like so

$\displaystyle \sum_{i=0}^\infty ar^i = ar^0 + ar^1 + ar^2 + \dots + ar^{n-1} + ar^n + ar^{n+1} + \dots$

Next, the instructor could ask questions like “If we’re looking for the sum up to the nth term, where do we need to chop off the terms to get what we want?,” “Does the ending part look familiar?”, and “How can we rewrite the chopped off part so that it looks like what we already know?”. The teacher guides the students into manipulating the formula to get this result

$\displaystyle \sum_{i=0}^\infty ar^i = ar^0 + ar^1 + ar^2 + \dots + ar^{n-1} + ar^n + ar^{n+1} + \dots$

$\displaystyle \sum_{i=0}^\infty ar^i = ar^0 + ar^1 + ar^2 + \dots + ar^{n-1} + ar^n + \sum_{j=n+1}^\infty ar^j$

$ar^0 + ar^1 + ar^2 + \dots + ar^{n-1} + ar^n = \displaystyle \sum_{i=0}^\infty ar^i - \sum_{j=n+1}^\infty ar^j$

The teacher notes that the last sum can be simplified to make it easier to see by doing a substitution of $k = j -n-1$. Adjusting the bounds and substituting in the new index, we get

$ar^0 + ar^1 + ar^2 + \dots + ar^{n-1} + ar^n = \displaystyle \sum_{i=0}^\infty ar^i - \sum_{k=0}^\infty ar^{n+1+k}$

$= \displaystyle \sum_{i=0}^\infty ar^i - \sum_{k=0}^\infty ar^{n+1}r^k$

$= \displaystyle \sum_{i=0}^\infty ar^i - r^{n+1} \sum_{k=0}^\infty ar^k$

Note that the two sums are identical, besides the index name, so we can factor and get

$ar^0 + ar^1 + ar^2 + \dots + ar^{n-1} + ar^n = a(1-r^{n+1}) \displaystyle \sum_{i=0}^\infty r^i$

Lastly, we utilize our formula for an infinite geometric series and get

$ar^0 + ar^1 + ar^2 + \dots + ar^{n-1} + ar^n = a(1-r^{n+1}) \displaystyle \frac{1}{1-r}$

$ar^0 + ar^1 + ar^2 + \dots + ar^{n-1} + ar^n = \displaystyle a\frac{1-r^{n+1}}{1-r}$

$ar^0 + ar^1 + ar^2 + \dots + ar^{n-1} + ar^n = \displaystyle a\frac{r^{n+1}-1}{r-1}$

Although the infinite series requires $|r|<1$ , the finite version works for all real $r$. Although the formal proof that this is the correct formula might be beyond the scope of the intended class, it can easily be done with induction.

Technology: