Engaging students: Powers and exponents

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 Ashlyn Farley. Her topic, from Pre-Algebra: powers and exponents.

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One class activity that will engage students while reviewing and/or teaching Exponent/Power concepts is “Marshmallow and Toothpicks.” This activity can be used for teaching the basic of exponents, as well as exponent laws. The idea is that the toothpicks are different colors, and the different colors represent different bases, thus the same color means it’s the same base. The marshmallows represent the exponent, i.e. the number of times the student needs to multiply the base. By following a worksheet of questions, the students should be able to solve exponent problems physically, visually, and abstractly. This activity, I believe, is best done with partners or groups so that the students can discuss how they think the exponents/exponent laws work. After the activity, the students are also able to eat their marshmallows, which encourages the students to participate and complete their work.

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Exponents are used in functions, equations, and expressions throughout math, thus having a deep understanding of exponents and their laws is very important. By fully mastering exponents and exponent laws, the students will be able to more easily grasp more difficult material that uses these concepts. Some specific ideas that use exponents and/or exponent laws in future math courses are: multiplying polynomials, finding the volume and surface area of prisms and cylinders, as well as computing the composition of two functions. Exponents are also used in many other situations than just math, such as in science or even in careers. Some careers that consistently use exponents and/or exponent laws are: Bankers, Computer Programmers, Mechanics, Plumbers, and many more.

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How can technology (YouTube, Khan Academy [khanacademy.org], Vi Hart, Geometers Sketchpad, graphing calculators, etc.) be used to effectively engage students with this topic?

An easy way to introduce students who have never seen exponents or exponential growth before is to use a graphing calculator. By plugging in an exponential function into the calculator and viewing the graph and zooming out, students can easily see how quickly numbers start to get The website Legends of Learning focuses on creating educational games for students in kindergarten through 9th grade. One game that goes over exponents, as well as the exponent laws, is Expodyssey. This game has the students solve problems to “fix” a spaceship to get back to Earth. The problems are built upon each other, so it starts by having the student answer what an exponent is, then what multiplying two exponents same base is, and keeps building from there. Each concept has multiple problems to be solved before moving on so that the students can show their mastery of the content. I believe that this game also helps improve cognitive skills by having the students do various activities simultaneously, such as calculating, reading, maneuvering elements and/or filling answers as required.

References:
Blog: Number Dyslexia
Link: https://numberdyslexia.com/top-7-games-for-understanding-math-exponents/

Thoughts on Numerical Integration (Part 23): The normalcdf function on TI calculators

I end this series about numerical integration by returning to the most common (if hidden) application of numerical integration in the secondary mathematics curriculum: finding the area under the normal curve. This is a critically important tool for problems in both probability and statistics; however, the antiderivative of \displaystyle \frac{1}{\sqrt{2\pi}} e^{-x^2/2} cannot be expressed using finitely many elementary functions. Therefore, we must resort to numerical methods instead.

In days of old, of course, students relied on tables in the back of the textbook to find areas under the bell curve, and I suppose that such tables are still being printed. For students with access to modern scientific calculators, of course, there’s no need for tables because this is a built-in function on many calculators. For the line of TI calculators, the command is normalcdf.

Unfortunately, it’s a sad (but not well-known) fact of life that the TI-83 and TI-84 calculators are not terribly accurate at computing these areas. For example:

TI-84: \displaystyle \int_0^1 \frac{e^{-x^2/2}}{\sqrt{2\pi}} \, dx \approx 0.3413447\underline{399}

Correct answer, with Mathematica: 0.3413447\underline{467}\dots

TI-84: \displaystyle \int_1^2 \frac{e^{-x^2/2}}{\sqrt{2\pi}} \, dx \approx 0.1359051\underline{975}

Correct answer, with Mathematica: 0.1359051\underline{219}\dots

TI-84: \displaystyle \int_2^3 \frac{e^{-x^2/2}}{\sqrt{2\pi}} \, dx \approx 0.021400\underline{0948}

Correct answer, with Mathematica: 0.021400\underline{2339}\dots

TI-84: \displaystyle \int_3^4 \frac{e^{-x^2/2}}{\sqrt{2\pi}} \, dx \approx 0.0013182\underline{812}

Correct answer, with Mathematica: 0.0013182\underline{267}\dots

TI-84: \displaystyle \int_4^5 \frac{e^{-x^2/2}}{\sqrt{2\pi}} \, dx \approx 0.0000313\underline{9892959}

Correct answer, with Mathematica: 0.0000313\underline{84590261}\dots

TI-84: \displaystyle \int_5^6 \frac{e^{-x^2/2}}{\sqrt{2\pi}} \, dx \approx 2.8\underline{61148776} \times 10^{-7}

Correct answer, with Mathematica: 2.8\underline{56649842}\dots \times 10^{-7}

I don’t presume to know the proprietary algorithm used to implement normalcdf on TI-83 and TI-84 calculators. My honest if brutal assessment is that it’s probably not worth knowing: in the best case (when the endpoints are close to 0), the calculator provides an answer that is accurate to only 7 significant digits while presenting the illusion of a higher degree of accuracy. I can say that Simpson’s Rule with only n = 26 subintervals provides a better approximation to \displaystyle \int_0^1 \frac{e^{-x^2/2}}{\sqrt{2\pi}} \, dx than the normalcdf function.

For what it’s worth, I also looked at the accuracy of the NORMSDIST function in Microsoft Excel. This is much better, almost always producing answers that are accurate to 11 or 12 significant digits, which is all that can be realistically expected in floating-point double-precision arithmetic (in which numbers are usually stored accurate to 13 significant digits prior to any computations).

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.

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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.

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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.

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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

Thoughts on Numerical Integration (Part 22): Comparison to theorems about magnitudes of errors

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 this series, we have shown the following approximations of errors when using various numerical approximations for \int_a^b x^k \, dx. We obtained these approximations using only techniques within the reach of a talented high school student who has mastered Precalculus — especially the Binomial Theorem — and elementary techniques of integration.

As we now present, the formulas that we derived are (of course) easily connected to known theorems for the convergence of these techniques. These proofs, however, require some fairly advanced techniques from calculus. So, while the formulas derived in this series of posts only apply to f(x) = x^k (and, by an easy extension, any polynomial), the formulas that we do obtain easily foreshadow the actual formulas found on Wikipedia or Mathworld or calculus textbooks, thus (hopefully) taking some of the mystery out of these formulas.

Left and right endpoints: Our formula was

E \approx \displaystyle \frac{k}{2} x_*^{k-1} (b-a)h,

where x_* is some number between a and b. By comparison, the actual formula for the error is

E = \displaystyle \frac{f'(x_*) (b-a)^2}{2n} = \frac{f'(x_*)}{2} (b-a)h.

This reduces to the formula that we derived since f'(x) = kx^{k-1}.
 

Midpoint Rule: Our formula was

E \approx \displaystyle \frac{k(k-1)}{24} x_*^{k-1} (b-a)h,

where x_* is some number between a and b. By comparison, the actual formula for the error is

E = \displaystyle \frac{f''(x_*) (b-a)^3}{24n^2} = \frac{f''(x_*)}{24} (b-a)h^2.

This reduces to the formula that we derived since f''(x) = k(k-1)x^{k-2}.

Trapezoid Rule: Our formula was

E \approx \displaystyle \frac{k(k-1)}{12} x_*^{k-1} (b-a)h,

where x_* is some number between a and b. By comparison, the actual formula for the error is

E = \displaystyle \frac{f''(x_*) (b-a)^3}{12n^2} = \frac{f''(x_*)}{12} (b-a)h^2.

This reduces to the formula that we derived since f''(x) = k(k-1)x^{k-2}.

This reduces to the formula that we derived since f''(x) = k(k-1)x^{k-2}.

Simpson’s Rule: Our formula was

E \approx \displaystyle \frac{k(k-1)(k-2)(k-3)}{180} x_*^{k-4} (b-a)h^4,

where x_* is some number between a and b. By comparison, the actual formula for the error is

E = \displaystyle \frac{f^{(4)}(x_*)}{180} (b-a)h^4.

This reduces to the formula that we derived since f^{(4)}(x) = k(k-1)(k-2)(k-3)x^{k-4}.

Engaging students: Solving two-step algebra problems

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 Chi Lin. Her topic, from Pre-Algebra: solving two-step algebra problems.

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

There is an interesting activity that I found online. It is called mini task cards. However, I want to rename this activity as “Find your partners” as an engage activity in this topic. I am going to create some two-step equations on the cards and give those cards randomly to the students at the beginning of the class. Each student has one mini card. The students will have 5 minutes to solve the equations and they will find the partners who have the same answers as them (there is 2-3 person in each group). The person who has the same answer with them will be the partner that they are working together with in the class. I will set up the answer as their group name (for example, if the answer is 1, then it means the group name is “Group One”). Here is an example that how the card will look like.


Reference:

12 Activities that Make Practicing Two-Step Equations Pop

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

Solving two-step equations is the foundation of solving multi-step equations. Solving two-step equations looks easy but it can become very hard. This topic can be applied in lots of areas such as high-level math classes, computer science, chemistry, physics, engineer, and so on. Most definitely, the students will see lots of problems about solving multi-step equations in different high-level mathematics courses in college, such as pre-calculus, calculus 1-3, differential equations, and so on. Also, the students will use the knowledge when they write the code in computer science class. For example, when they write down the code of two-step or multi-step algebra problems, they need to know which step goes first. If they do the step wrong, then the computer program will compute the wrong result. Moreover, the students will use solving two-step equations in chemistry class. For example, the students will apply this knowledge, when they write down the chemical equations and try to balance the equations.

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How does this topic extend what your students should have learned in previous courses?
First, students should know what linear equations are and how to write down the linear equations. Second, students should know how to solve one-step algebra problems, such as x+8=16 or x/8=16. Students should have learned that when they solve for the one-step equations (addition and subtract), whatever they do to one side of the equation, they need to make sure they add the same thing to the other side. For example, when they solve the equation x+8=16, they can subtract 8 for both sides, which is x+8-8=16-8. Therefore, x=8. Also, student should know that when they solve for the one-step equations (multiplication and division), they need to multiply both side by the reciprocal of the coefficient of the variable. For example, when they solve the equation x/8=16, they need to multiply the reciprocal of 1/8 for both sides, which is x/8*8=16*8. Therefore, x=128. Thus, when they learn to solve two-step equations, they need to combine these rules.

References:
https://www.khanacademy.org/math/algebra-home/alg-basic-eq-ineq/alg-one-step-mult-div-equations/a/one-step-equation-review

Solving Two-Step Equations

 

Thoughts on Numerical Integration (Part 21): Simpson’s rule and global rate of convergence

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 this previous post in this series, we showed that the Simpson’s Rule approximation of \displaystyle \int_{x_i}^{x_i+2h} x^k \, dx has an error of 

-\displaystyle \frac{k(k-1)(k-2)(k-3)}{90} x_i^{k-4} h^5 + O(h^6).

In this post, we consider the global error when integrating on the interval [a,b] instead of a subinterval [x_i,x_i+2h]. The total error when approximating \displaystyle \int_a^b x^k \, dx = \int_{x_0}^{x_n} x^k \, dx will be the sum of the errors for the integrals over [x_0,x_2], [x_2,x_4], through [x_{n-2},x_n]. Therefore, the total error will be

$latex E \approx \displaystyle \frac{k(k-1)(k-2)(k-3)}{90} \left(x_0^{k-4} + x_2^{k-4} + \dots + x_{n-2}^{k-4} \right) h^5.

So that this formula doesn’t appear completely mystical, this actually matches the numerical observations that we made earlier. The figure below shows the left-endpoint approximations to \displaystyle \int_1^2 x^9 \, dx for different numbers of subintervals. If we take n = 100 and h = 0.01, then the error should be approximately equal to

\displaystyle \frac{9 \times 8 \times 7 \times 6}{90} \left(1^5 + 1.02^5 + 1.04^5 + \dots + 1.98^5 \right) (0.01)^5 \approx 0.0000017,

which, as expected, is close to the observed error of 102.3000018 - 102.3 \approx 0.0000018.
Let y_i = x_i^{k-4}, so that the error becomes

E \approx \displaystyle \frac{k(k-1)(k-2)(k-3)}{90} \left(y_0 + y_2 + \dots + y_{n-2} \right) h^5 = \displaystyle \frac{k(k-1)(k-2)(k-3)}{90} \overline{y} \frac{n}{2} h^5,

where \overline{y} = (y_0 + y_2 + \dots + y_{n-2})/(n/2) is the average of the y_i. (We notice that there are only n/2 terms in this sum since we’re adding only the even terms.) Clearly, this average is somewhere between the smallest and the largest of the y_i. Since y = x^{k-4} is a continuous function, that means that there must be some value of x_* between x_0 and x_{k-2} — and therefore between a and b — so that x_*^{k-4} = \overline{y} by the Intermediate Value Theorem. We conclude that the error can be written as

E \approx \displaystyle \frac{k(k-1)(k-2)(k-3)}{180} x_*^{k-4} nh^5,

Finally, since h is the length of one subinterval, we see that nh = b-a is the total length of the interval [a,b]. Therefore,

E \approx \displaystyle \frac{k(k-1)(k-2)(k-3)}{180} x_*^{k-4} (b-a)h^4 \equiv ch^4,

where the constant c is determined by a, b, and k. In other words, for the special case f(x) = x^k, we have established that the error from Simpson’s Rule is approximately quartic in h — without resorting to the generalized mean-value theorem and confirming the numerical observations we made earlier.

Engaging students: Solving one-step algebra problems

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 Emma White. Her topic, from Algebra: solving one-step algebra problems.

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

Solving one-step algebra problems strings into many future scenarios the student may (and will probably) encounter. One-step algebra problems infer that there must be two-step algebra problems and three-step algebra problems and so forth. As mathematicians, we know this to be true. While mathematics in my focus of study, I want to show the importance of learning this concept as it will aid in other classes. Stoichiometry is a concept taught in chemistry that has to do with the “relationship between reactants and products in a reaction” (Washington University in St. Louis, 2005). Chemical reactions require a balance. Essentially, once-step algebra expressions require just the same where both sides of the equations must be equal for the expression to be true. An example of a stoichiometry equation one may see in chemistry would be:

_KMnO{}_4 + _HCl → _MnCl{}_2 + _KCl + _Cl{}_2 + _H{}_2O

In the blanks, a variable can be placed, such that:

aKMnO{}_4 + bHCl → cMnCl{}_2 + dKCl + eCl{}_2 + fH{}_2O

Next, we would apply the Conservation of Mass. This concept deals with the number of atoms that must be on each side for the equation to be balanced. Writing the elements and their balanced equations with the variables, it follows:

K: a = d
Mn: a = c
O: 4a = f
H: b = 2f
Cl: b = 2c + d + 2e

As we can see, there is going to be more expressions and substitutions that must take place. That is something you can solve on your own if you wish. Overall, we see the importance of learning one-step algebra problems because this will be the foundation for solving more complex questions, even more so outside of the math classroom.

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How has this topic appeared in high culture (art, classical music, theatre, etc.)?

Theatre is more than the actors on the stage. While the performance and show are the part most people acknowledge and enjoy, the technical part behind the performance is what allows the show to happen. Algebraic problems are often used in technical theatre, especially when it comes to building a set. A prime example is building a single foundation (usually used in One Act plays where the whole play takes place in one scene). Focusing on a rectangular foundation, if we know the amount of space the actors, set, and featuring décor need, we can use this in an algebraic expression. Furthermore, if we also know dimensions of one of the sides (length or width), a variable can be used for the unknown side (since the area of a rectangle is length times the width). If we want to take this a step further, multiple one-step algebraic expressions can be used when making the foundation. If we know the length and width of the foundation and the length and width of the sheet floorboards to be used, we can write various expressions to determine how many sheet floorboards need to be used lengthwise and widthwise (example shown below).

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How can technology (YouTube, Khan Academy [khanacademy.org], Vi Hart, Geometers Sketchpad, graphing calculators, etc.) be used to effectively engage students with this topic?      

The use of technology is on the rise and the involvement of newer generations is greatly rising as well. Because of this, utilizing online resources is an effective way to capture the attention of the students and make math more engaging. Using algebra tiles is a perfect way to resemble this topic, even more so when it can be done online. Therefore, the teacher does not need to buy any materials and the students (especially high schoolers) don’t have to carry paper resources around or even home where, we all know, they will end up in the trash. Online algebra tiles provide a way to visually see the one-step algebra problem and work accordingly. Even so, these tiles can be an introduction and foundation on what is to come (these tiles are also a great source for solving two-step equations, distribution, polynomials, the perfect square, and so forth). Another insight for using online algebra tiles is in some schools where technology such as tablets/computers are provided, the students can share their screens to a projector (or whatever resources the classroom may have) and describe their thinking process to the class. This builds on the idea of students learning, processing, and being able to teach their peers what they learned as well.

References

End Result for x +4=8: https://technology.cpm.org/general/tiles/?tiledata=b5____g+afx__boy__aaapTtPhF%2B__qvtTauq7tSaurDtSaur5tSausBtSauwisCauwis4auwitAauwit2auwir6auwiuyauwiu0auwirEawq7ukawrDukawr5ukawsBukawwOrEawwOr6awwOsCawwOs4hFProblem%3A%20Solve%20for%20x.%20%20x%20%2B%204%20%3D%208__qPpBhFThere%20is%20one%20x%20left%20on%20the%20left%20sideand%20four%201s%20left%20over%20on%20the%20right.Therefore%2C%20x%20%3D%204.__v-wcgawWwOtBgawWwOt1gawWwOuzgawWwOu0

End Result for 4x=16: https://technology.cpm.org/general/tiles/?tiledata=b5____g+afx__boy__aaatCsnaatCtgaatCviaatCulauvIslauwaslauwGslauw8slauw6tjauwEtjauv8tjauvGtjauvFupauv7upauwDupauw5upauvDvlauv5vlauwBvlauw3vlhF%20%20%20%20%20Look%20at%20one%20x%20on%20the%20left%20side%20and%20seewhat%20it%20is%20paired%20with%20on%20the%20right%20side.We%20see%20that%20one%20x%20is%20paired%20with%20four%201s.Therefore%2C%20x%20%3D%204.__vuwngaqAr8vkgawZxzvnhFProblem%3A%20Solve%20for%20x.%204x%3D16__pEpz
https://chemistrytutor.me/balancing-chemical-equations-algebra/
http://www.chemistry.wustl.edu/~coursedev/Online%20tutorials/Stoichiometry.htm
https://technology.cpm.org/general/tiles/

Thoughts on Numerical Integration (Part 20): Simpson’s rule and local rate of convergence

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 this post, we will perform an error analysis for Simpson’s Rule

\int_a^b f(x) \, dx \approx \frac{h}{3} \left[f(x_0) + 4(x_1) + 2f(x_2) + \dots + 2f(x_{n-2}) + 4f(x_{n-1}) +f(x_n) \right] \equiv T_n

where n is the number of subintervals (which has to be even) and h = (b-a)/n is the width of each subinterval, so that x_k = x_0 + kh.
As noted above, a true exploration of error analysis requires the generalized mean-value theorem, which perhaps a bit much for a talented high school student learning about this technique for the first time. That said, the ideas behind the proof are accessible to high school students, using only ideas from the secondary curriculum (especially the Binomial Theorem), if we restrict our attention to the special case f(x) = x^k, where k \ge 5 is a positive integer.

For this special case, the true area under the curve f(x) = x^k on the subinterval [x_i, x_i +h] will be

\displaystyle \int_{x_i}^{x_i+h} x^k \, dx = \frac{1}{k+1} \left[ (x_i+h)^{k+1} - x_i^{k+1} \right]

= \displaystyle \frac{1}{k+1} \left[x_i^{k+1} + {k+1 \choose 1} x_i^k h + {k+1 \choose 2} x_i^{k-1} h^2 + {k+1 \choose 3} x_i^{k-2} h^3 + {k+1 \choose 4} x_i^{k-3} h^4+ {k+1 \choose 5} x_i^{k-4} h^5+ O(h^6) - x_i^{k+1} \right]

= \displaystyle \frac{1}{k+1} \bigg[ (k+1) x_i^k h + \frac{(k+1)k}{2} x_i^{k-1} h^2 + \frac{(k+1)k(k-1)}{6} x_i^{k-2} h^3+ \frac{(k+1)k(k-1)(k-2)}{24} x_i^{k-3} h^4

+ \displaystyle \frac{(k+1)k(k-1)(k-2)(k-3)}{120} x_i^{k-4} h^5 \bigg] + O(h^6)

= x_i^k h + \displaystyle \frac{k}{2} x_i^{k-1} h^2 + \frac{k(k-1)}{6} x_i^{k-2} h^3 + \frac{k(k-1)(k-2)}{24} x_i^{k-3} h^4 + \frac{k(k-1)(k-2)(k-3)}{120} x_i^{k-4} h^5 + O(h^6)

In the above, the shorthand O(h^6) can be formally defined, but here we’ll just take it to mean “terms that have a factor of h^6 or higher that we’re too lazy to write out.” Since h is supposed to be a small number, these terms will small in magnitude and thus can be safely ignored.
Earlier in this series, we derived the very convenient relationship S_{2n} = \displaystyle \frac{2}{3} M_n + \frac{1}{3} T_n relating the approximations from Simpson’s Rule, the Midpoint Rule, and the Trapezoid Rule. We now exploit this relationship to approximate \displaystyle \int_{x_i}^{x_i+h} x^k \, dx. Earlier in this series, we found the Midpoint Rule approximation on this subinterval to be

M = x_i^k h + \displaystyle \frac{k}{2} x_i^{k-1} h^2  + \frac{k(k-1)}{8} x_i^{k-2} h^3 + \frac{k(k-1)(k-2}{48} x_i^{k-3} h^4

\displaystyle + \frac{k(k-1)(k-2)(k-3)}{384} x_i^{k-4} h^5 + O(h^6)

while we found the Trapezoid Rule approximation to be

 T = x_i^k h + \displaystyle \frac{k}{2} x_i^{k-1} h^2  + \frac{k(k-1)}{4} x_i^{k-2} h^3 + \frac{k(k-1)(k-2)}{12} x_i^{k-3} h^4

\displaystyle + \frac{k(k-1)(k-2)(k-3)}{48} x_i^{k-4} h^5 + O(h^6).

Therefore, if there are 2n subintervals, the Simpson’s Rule approximation of \displaystyle \int_{x_i}^{x_i+h} x^k \, dx — that is, the area under the parabola that passes through (x_i, x_i^k), (x_i + h/2, (x_i +h/2)^k), and (x_i + h, (x_i +h)^k) — will be S = \frac{2}{3}M + \frac{1}{3}T. Since

\displaystyle \frac{2}{3} \frac{1}{8} + \frac{1}{3} \frac{1}{4} = \frac{1}{6},

\displaystyle \frac{2}{3} \frac{1}{48} + \frac{1}{3} \frac{1}{12} = \frac{1}{24},

and

\displaystyle \frac{2}{3} \frac{1}{384} + \frac{1}{3} \frac{1}{48} = \frac{5}{576},

we see that

 S = x_i^k h + \displaystyle \frac{k}{2} x_i^{k-1} h^2  + \frac{k(k-1)}{6} x_i^{k-2} h^3 + \frac{k(k-1)(k-2)}{24} x_i^{k-3} h^4

\displaystyle + \frac{5k(k-1)(k-2)(k-3)}{576} x_i^{k-4} h^5 + O(h^6).

We notice that something wonderful just happened: the first four terms of S perfectly match the first four terms of the exact value of the integral! Subtracting from the actual integral, the error in this approximation will be equal to

\displaystyle \frac{k(k-1)(k-2)(k-3)}{120} x_i^{k-4} h^5 - \frac{5k(k-1)(k-2)(k-3)}{576} x_i^{k-4} h^5 + O(h^6)

= -\displaystyle \frac{k(k-1)(k-2)(k-3)}{2880} x_i^{k-4} h^5 + O(h^6)

Before moving on, there’s one minor bookkeeping issue to deal with. We note that this is the error for S_{2n}, where 2n subintervals are used. However, the value of h in this equal arose from T_n and M_n, where only n subintervals are used. So let’s write the error with 2n subintervals as

-\displaystyle \frac{k(k-1)(k-2)(k-3)}{90} x_i^{k-4} \left( \frac{h}{2} \right)^5 + O(h^6),

where h/2 is the width of all of the 2n subintervals. By analogy, we see that the error for n subintervals will be

-\displaystyle \frac{k(k-1)(k-2)(k-3)}{90} x_i^{k-4} h^5 + O(h^6).

But even after adjusting for this constant, we see that this local error behaves like O(h^5), a vast improvement over both the Midpoint Rule and the Trapezoid Rule. This illustrates a general principle of numerical analysis: given two algorithms that are O(h^3), an improved algorithm can typically be made by taking some linear combination of the two algorithms. Usually, the improvement will be to O(h^4); however, in this example, we magically obtained an improvement to O(h^5).

Engaging students: Finding points on the coordinate plane

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 Pre-Algebra: finding points on the coordinate plane.

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C2: How has this topic appeared in high culture (art, classical music, theatre, etc.)?

One popular art/sport high school students may take part in is marching band. I did four years of marching band in high school and I loved it. One has to wonder: “how does each performer know where they should be?” I’ve included a link from bandtek.com that describes the coordinate system marching bands use. It isn’t quite the same as the coordinate plane in a math class. When starting marching band, you learn how to take appropriately sized “8 to 5” steps, which simply means 8 equally spaced steps for every 5 yards on a football field. Each member will receive little cards that have “sets” on them. A set is a specific point on the field where the performer must be at a specific time of the show. Usually, performers will take straight paths from set to set in a specific amount of 8-5 steps. Looking at a bird eye’s view of the football field, one can see a rough coordinate plane. Like a coordinate plane has 4 quadrants, a football field has a rough 4 quadrant system where a performer is assigned to stand a specified amount of 8-5 steps from a specified yard line either on side 1 or 2 for their horizontal position and a specified amount of 8-5 steps from the front/back hash for vertical position facing the home sideline. Side 1 refers to the left side of the field from the home side perspective, Side 2 refers to the right side of the field from the home side perspective, and the front/back hash refers to the line of dashes that cut through the middle of the field horizontally from the home side perspective.

An example bandtek.com uses is, “4 outside the side 1 45, 3 in front of the front hash” which would mean the following position:

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D1: What interesting things can you say about the people who contributed to the discovery and/or the development of this topic?

René Descartes was a 17th century (1600’s) French mathematician and philosopher. Many people study his work in modern day math and philosophy classes. Some may know him as the man who wrote “cogito, ergo sum” or “I think, therefore I am”. Well, there is a legend about his discovery of the Coordinate Plane. Descartes was often sick as a kid, way before modern medication and technology. He would often have to stay in bed at his boarding school until noon because of his illnesses. This gave him quite a bit of downtime to be observant of his environment. Laying on his bed, he could see a fly crawl around on his ceiling. He thought of ways to describe the location of the fly as it scuttled about the ceiling. Imagine telling a friend where the location of the fly was, “A little to the left of the right wall and a little down from the top wall”. This just isn’t precise enough, nor an easy way to communicate information. However, Descartes realized he could quantify the precise location of the fly from using the distance from a pair of perpendicular walls. Descartes then translated this idea onto a graph where the perpendicular “walls” continued infinitely in both directions and became “axes”. “Flies” then became “points” or “coordinate pairs”. Thus, the coordinate plane was born, and so was a way to describe points in space. Just a little bit of imagination, self-questioning, and observation lead to a fundamental change in Mathematics, a way to tie Algebra and Geometry together.

 

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E1: 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.

 

I believe that https://www.chess.com/vision could be an effective website to engage students on finding points on the coordinate plane in a class that is being introduced to the idea for the first time. Many students won’t know how a chessboard is setup or even know how to play chess. The cool things are that they don’t need to know the fundamentals of chess and that the chessboard is essentially Quadrant I of a coordinate plane (where a1 is in the bottom left corner). The above website tests the player to locate as many squares (points) on a chessboard (coordinate plane) as they can in 30 seconds, given random chess coordinates. There is a way to toggle settings to also test yourself on moves and squares. In a classroom, I would only toggle the setting to list random “black and white squares” where the board is set with a1 at the bottom left corner. Students could start the day with this website as a precursor to formalizing the idea of finding points on a coordinate plane. This website is engaging (with an exclamation point)! The game can be made into a fun little competition amongst students. The time limit and game-y feeling to it encourages active participation. The game takes minimal explanation from the teacher for students to get the hang of it (no chess skills required). The fact that chessboards have one axis in letters and the other axis in numbers aids students in reading the coordinate plane x-axis first, then y-axis like the chess coordinates. I would only have the students run the game for a few rounds, making the activity in total 7 minutes or less.

 

 

References:

http://bandtek.com/how-to-read-a-drill-chart/

https://www.chess.com/vision

https://wild.maths.org/ren%C3%A9-descartes-and-fly-ceiling

https://maths2art.com.sg/2018/01/16/have-you-ever-followed-a-fly