Slightly Incorrect Ugly Mathematical Christmas T-Shirts: Part 2

This was another T-shirt that I found in my search for the perfect ugly mathematical Christmas sweater: https://www.amazon.com/Pascals-Triangle-Math-Christmas-shirt/dp/B07KJS5SM2/I love the artistry of this shirt; the “ornaments” at the corners of the hexagons and the presents under the tree are nice touches.

There’s only one small problem:

\displaystyle {8 \choose 3} = \displaystyle {8 \choose 5} = \displaystyle \frac{8!}{3! \times 5!} = 56.

Oops.

Engaging students: Using Pascal’s triangle

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 Rachel Delflache. Her topic, from Precalculus: using Pascal’s triangle.

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How does this topic expand what your students would have learned in previous courses?

In previous courses students have learned how to expand binomials, however after (x+y)^3 the process of expanding the binomial by hand can become tedious. Pascal’s triangle allows for a simpler way to expand binomials. When counting the rows, the top row is row 0, and is equal to one. This correlates to (x+y)^0 =1. Similarly, row 2 is 1 2 1, correlating to (x+y)^2 = 1x^2 + 2xy + 1y^2. The pattern can be used to find any binomial expansion, as long as the correct row is found. The powers in each term also follow a pattern, for example look at (x+y)^4:

1x^4y^0 + 4x^3y^1 + 6x^2y^2 + 4x^1y^3 + 1x^0y^4

In this expansion it can be seen that in the first term of the expansion the first monomial is raised to the original power, and in each term the power of the first monomial decreases by one. Conversely, the second monomial is raised to the power of 0 in the first term of the expansion, and increases by a power of 1 for each subsequent term in the expansion until it is equal to the original power of the binomial.

 

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Sierpinski’s Triangle is triangle that was characterized by Wacław Sieriński in 1915. Sierpinski’s triangle is a fractal of an equilateral triangle which is subdivided recursively. A fractal is a design that is geometrically constructed so that it is similar to itself at different angles. In this particular construction, the original shape is an equilateral triangle which is subdivided into four smaller triangles. Then the middle triangle is whited out. Each black triangle is then subdivided again, and the patter continues as illustrated below.

Sierpinski’s triangle can be created using Pascal’s triangle by shading in the odd numbers and leaving the even numbers white. The following video shows this creation in practice.

 

 

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What are the contributions of various cultures to this topic?

The pattern of Pascal’s triangle can be seen as far back as the 11th century. In the 11th century Pascal’s triangle was studied in both Persia and China by Oman Khayyam and Jia Xian, respectively. While Xian did not study Pascal’s triangle exactly, he did study a triangular representation of coefficients. Xian’s triangle was further studied in 13th century China by Yang Hui, who made it more widely known, which is why Pascal’s triangle is commonly called the Yanghui triangle in China. Pascal’s triangle was later studies in the 17th century by Blaise Pascal, for whom it was named for. While Pascal did not discover the number patter, he did discover many new uses for the pattern which were published in his book Traité du Triangle Arithméthique. It is due to the discovery of these uses that the triangle was named for Pascal.

Reference:
https://en.wikipedia.org/wiki/Pascal%27s_triangle
http://mathforum.org/workshops/usi/pascal/images/fill.comb.gif
https://www.britannica.com/biography/Blaise-Pascal#toc445406main
https://en.wikipedia.org/wiki/Sierpinski_triangle

Happy Fibonacci Day!

Today is 11/23, and 1, 1, 2, 3 are the first four terms of the Fibonacci sequence.

Courtesy Kris McCoy’s Pointless Math Fact of the Day.

Decimal Approximations of Logarithms (Part 5)

While some common (i.e., base-10) logarithms work out evenly, like \log_{10} 10,000, most do not. Here is the typical output when a scientific calculator computes a logarithm:

In today’s post, I’ll summarize the past few posts to describe how talented Algebra II students, who have just been introduced to logarithms, can develop proficiency with the Laws of Logarithms while also understanding that the above answer is not just a meaningless jumble of digits. The only tools students will need are

To estimate \log_{10} 5.1264, Algebra II students can try to find a power of 5.1264 that is close to a power of 10. In principle, this can be done by just multiplying by 5.1264 until an answer decently close to 5.1264 arises. For the teacher who’s guiding students through this exploration, it might be helpful to know the answer ahead of time.

One way to do this is to use Wolfram Alpha to find the convergents of \log_{10} 5.1264. If you click this link, you’ll see that I entered

Convergents[Log[10,5.1264],15]

A little explanation is in order:

  • Convergents, predictably, is the Mathematica command for finding the convergents of a given number.
  • Log[10,5.1264] is the base-10 logarithm of 5.1264. By contrast, Log[5.1264] is the natural logarithm of 5.1264. Mathematica employs the convention of that \log should be used for natural logarithms instead of \ln, as base-10 logarithms are next to useless for mathematical researchers. That said, I freely concede that this convention is confusing to students who grew up thinking that \log should be used for base-10 logarithms and \ln for natural logarithms. (See also my standard joke about using natural logarithms.) Naturally, the 5.1264 can be changed for other logarithms.
  • The 15 means that I want Wolfram Alpha to give me the first 15 convergents of \log_{10} 5.1264. In most cases, that’s enough terms to provide a convergent whose denominator is at least six digits long. In the rare instance when this doesn’t happen, a number larger than 15 can be entered.

From Wolfram Alpha, I see that \displaystyle \frac{22}{31} is the last convergent with a numerator less than 100. For the purposes of this exploration, I interpret these fractions as follows:

  • The best suitable power of 5.1264 for an easy approximation on a scientific calculation will be (5.1264)^{31}. In this context, “best” means something that’s close to a power of 10 but less than 10^{100}. Students entering (5.1264)^{31} into a calculator will find

(5.1264)^{31} \approx 1.009687994 \times 10^{22}

(5.1264)^{31} \approx 10^{22}

In other words, the denominator of the convergent \displaystyle \frac{22}{31} gives the exponent for 5.1264, while the numerator gives the exponent for the approximated power of 10. Continuing with the Laws of Logarithms,

\log_{10} (5.1264)^{31} \approx \log_{10} 10^{22}

31 \log_{10} 5.1264 \approx 22

\log_{10} 5.1264 \approx \displaystyle \frac{22}{31}

\log_{10} 5.1264 \approx 0.709677\dots

A quick check with a calculator shows that this approximation is accurate to three decimal places. This alone should convince many students that the above apparently random jumble of digits is not so random after all.

While the above discussion should be enough for many students, some students may want to know how to find the rest of the decimal places with this technique. To answer this question, we again turn to the convergents of \log_{10} 5.1264 from Wolfram Alpha. From this list, we see that \displaystyle \frac{89,337}{125,860} is the first convergent with a denominator at least six digits long. The student therefore has two options:

Option #1. Ask the student to use Wolfram Alpha to raise 5.1264 to the denominator of this convergent. Surprisingly to the student, but not surprisingly to the teacher who knows about this convergent, the answer is very close to a power of 10: 10^{89,337}. The student can then use the Laws of Logarithms as before:

\log_{10} (5.1264)^{125,860} \approx \log_{10} 10^{89,337}

125,860 \log_{10} 5.1264 \approx 89,337

\log_{10} 5.1264 \approx \displaystyle \frac{89,337}{125,860}

\log_{10} 5.1264 \approx 0.70981249006\dots,

which matches the output of the calculator.

Option #2. Ask the student to “trick” a hand-held calculator into finding (5.1264)^{125,860}. This option requires the use of the convergent with the largest numerator less than 100, which was \displaystyle \frac{22}{31}.

  • Option #2A: Use the Microsoft Excel spreadsheet that I’ve written to perform the calculations that follow.
  • Option #2B: The student divides the smaller denominators into the larger denominator and finds the quotient and remainder. It turns out that 125,860 = 31 \times 4060 + 0. (This is a rare case where there happens to be no remainder.) Next, the student uses a hand-held calculate to compute

\displaystyle \left( \frac{(5.1264)^{31}}{10^{22}} \right)^{4060} \times (5.1264)^0

In this example, the \times (5.1264)^0 is of course superfluous, but I include it here to show where the remainder should be placed. Entering this in a calculator yields a result that is close to 10^{17}. (The teacher should be aware that some of the last few digits may differ from the more precise result given by Wolfram Alpha due to round-off error, but this discrepancy won’t matter for the purposes of the student’s explorations.) In other words,

\displaystyle \left( \frac{(5.1264)^{31}}{10^{22}} \right)^{4060} \times (5.1264)^0 \approx 10^{17},

which may be rearranged as

(5.1264)^{125,860} \approx 10^{89,337}

after using the Laws of Exponents. From this point, the derivation follows the steps in Option #1.

Decimal Approximations of Logarithms (Part 4)

While some common (i.e., base-10) logarithms work out evenly, like \log_{10} 10,000, most do not. Here is the typical output when a scientific calculator computes a logarithm:

To a student first learning logarithms, the answer is just an apparently random jumble of digits; indeed, it can proven that the answer is irrational. With a little prompting, a teacher can get his/her students wondering about how people 50 years ago could have figured this out without a calculator. This leads to a natural pedagogical question:

Can good Algebra II students, using only the tools at their disposal, understand how decimal expansions of base-10 logarithms could have been found before computers were invented?

Here’s a trial-and-error technique — an exploration activity — that is within the grasp of Algebra II students. It’s simple to understand; it’s just a lot of work. The only tools that are needed are

  • The Laws of Logarithms
  • A hand-held scientific calculator
  • Access to the Wolfram Alpha website (optional)
  • A lot of patience multiplying x by itself repeatedly in a quest to find integer powers of x that are close to powers of 10.

In the previous post in this series, we found that

3^{153} \approx 10^{73}

and

3^{323,641} \approx 10^{154,416}.

Using the Laws of Logarithms on the latter provides an approximation of \log_{10} 3 that is accurate to an astounding ten decimal places:

\log_{10} 3^{323,641} \approx \log_{10} 10^{154,416}

323,641 \log_{10} 3 \approx 154,416

\log_{10} 3 \approx \displaystyle \frac{154,416}{323,641} \approx 0.477121254723598\dots.

Compare with:

\log_{10} 3 \approx 0.47712125471966\dots

Since hand-held calculators will generate identical outputs for these two expressions (up to the display capabilities of the calculator), this may lead to the misconception that the irrational number \log_{10} 3 is actually equal to the rational number \displaystyle \frac{154,416}{323,641}, so I’ll emphasize again that these two numbers are not equal but are instead really, really close to each other.

We now turn to a question that was deferred in the previous post.

Student: How did you know to raise 3 to the 323,641st power?

Teacher: I just multiplied 3 by itself a few hundred thousand times.

Student: C’mon, really. How did you know?

While I don’t doubt that some of our ancestors used this technique to find logarithms — at least before the discovery of calculus — today’s students are not going to be that patient. Instead, to find suitable powers quickly, we will use ideas from the mathematical theory of continued fractions: see Wikipedia, Mathworld, or this excellent self-contained book for more details.

To approximate \log_{10} x, the technique outlined in this series suggests finding integers m and n so that

x^n \approx 10^m,

or, equivalently,

\log_{10} x^n \approx \log_{10} 10^m

n \log_{10} x \approx m

\log_{10} x \approx \displaystyle \frac{m}{n}.

In other words, we’re looking for rational numbers that are reasonable close to \log_{10} x. Terrific candidates for such rational numbers are the convergents to the continued fraction expansion of \log_{10} x. I’ll defer to the references above for how these convergents can be computed, so let me cut to the chase. One way these can be quickly obtained is the free website Wolfram Alpha. For example, the first few convergents of \log_{10} 3 are

\displaystyle \frac{1}{2}, \frac{10}{21}, \frac{21}{44}, \frac{52}{109}, and \frac{73}{153}.

A larger convergent is \frac{154,416}{323,641}, our familiar friend from the previous post in this series.

As more terms are taken, these convergents get closer and closer to \log_{10} 3. In fact:

  • Each convergent is the best possible rational approximation to \log_{10} 3 using a denominator that’s less than the denominator of the next convergent. For example, the second convergent \displaystyle \frac{10}{21} is the closest rational number to \log_{10} 3 that has a denominator less than 44, the denominator of the third convergent.
  • The convergents alternate between slightly greater than \log_{10} 3 and slightly less than \log_{10} 3.
  • Each convergent \displaystyle \frac{m}{n} is guaranteed to be within \displaystyle \frac{1}{n^2} of \log_{10} 3. (In fact, if \displaystyle \frac{m}{n} and \displaystyle \frac{p}{q} are consecutive convergents, then \displaystyle \frac{m}{n} is guaranteed to be within \displaystyle \frac{1}{nq} of \log_{10} 3.)
  • As a practical upshot of the previous point: if the denominator of the convergent \displaystyle \frac{m}{n} is at least six digits long (that is, greater than 10^5), then \displaystyle \frac{m}{n} must be within \displaystyle \frac{1}{(10^5)^2} = 10^{-10} of \log_{10} 3… and it’ll probably be significantly closer than that.

So convergents provide a way for teachers to maintain the illusion that they found a power like 3^{323,641} by laborious calculation, when in fact they were quickly found through modern computing.

 

 

 

 

Decimal Approximations of Logarithms (Part 3)

While some common (i.e., base-10) logarithms work out evenly, like \log_{10} 10,000, most do not. Here is the typical output when a scientific calculator computes a logarithm:

To a student first learning logarithms, the answer is just an apparently random jumble of digits; indeed, it can proven that the answer is irrational. With a little prompting, a teacher can get his/her students wondering about how people 50 years ago could have figured this out without a calculator. This leads to a natural pedagogical question:

Can good Algebra II students, using only the tools at their disposal, understand how decimal expansions of base-10 logarithms could have been found before computers were invented?

Here’s a trial-and-error technique — an exploration activity — that is within the grasp of Algebra II students. It’s simple to understand; it’s just a lot of work.

To approximate \log_{10} x, look for integer powers of x that are close to powers of 10.

In the previous post in this series, we essentially used trial and error to find such powers of 3. We found

3^{153} \approx 9.989689095 \times 10^{72} \approx 10^{73},

from which we can conclude

\log_{10} 3^{153} \approx \log_{10} 10^{73}

153 \log_{10} 3 \approx 73

\log_{10} 3 \approx \displaystyle \frac{73}{153} \approx 0.477124.

This approximation is accurate to five decimal places.

By now, I’d imagine that our student would be convinced that logarithms aren’t just a random jumble of digits… there’s a process (albeit a complicated process) for obtaining these decimal expansions. Of course, this process isn’t the best process, but it works and it only uses techniques at the level of an Algebra II student who’s learning about logarithms for the first time.

If nothing else, hopefully this lesson will give students a little more appreciation for their ancestors who had to perform these kinds of calculations without the benefit of modern computing.

We also saw in the previous post that larger powers can result in better and better approximation. Finding suitable powers gets harder and harder as the exponent gets larger. However, when a better approximation is found, the improvement can be dramatic. Indeed, the decimal expansion of a logarithm can be obtained up to the accuracy of a hand-held calculator with a little patience. For example, let’s compute

3^{323,641}

Predictably, the complaint will arise: “How did you know to try 323,641?” The flippant and awe-inspiring answer is, “I just kept multiplying by 3.”

I’ll give the real answer that question later in this series.

Postponing the answer to that question for now, there are a couple ways for students to compute this using readily available technology. Perhaps the most user-friendly is the free resource Wolfram Alpha:

3^{323,641} \approx 9.999970671 \times 10^{154,415} \approx 10^{154,416}.

That said, students can also perform this computation by creatively using their handheld calculators. Most calculators will return an overflow error if a direct computation of 3^{323,641} is attempted; the number is simply too big. A way around this is by using the above approximation 3^{153} \approx 10^{73}, so that 3^{153}/10^{73} \approx 1. Therefore, we can take large powers of 3^{153}/10^{73} without worrying about an overflow error.

In particular, let’s divide 323,641 by $153$. A little work shows that

\displaystyle \frac{323,641}{153} = \displaystyle 2115 \frac{46}{153},

or

323,641 = 153 \times 2115  + 46.

This suggests that we try to compute

\displaystyle \left( \frac{3^{153}}{10^{73}} \right)^{2115} \times 3^{46},

and a hand-held calculator can be used to show that this expression is approximately equal to 10^{21}. Some of the last few digits will be incorrect because of unavoidable round-off errors, but the approximation of 10^{21} — all that’s needed for the present exercise — will still be evident.

By the Laws of Exponents, we see that

\displaystyle \left( \frac{3^{153}}{10^{73}} \right)^{2115} \times 3^{46} \approx 10^{21}

\displaystyle \frac{3^{153 \times 2115 + 46}}{10^{73 \times 2115}} \approx 10^{21}

\displaystyle \frac{3^{323,641}}{10^{154,395}} \approx 10^{21}

3^{323,641} \approx 10^{154,395} \times 10^{21}

3^{323,641} \approx 10^{154,395+21}

3^{323,641} \approx 10^{154,416}.

Whichever technique is used, we can now use the Laws of Logarithms to approximate \log_{10} 3:

\log_{10} 3^{323,641} \approx \log_{10} 10^{154,416}

323,641 \log_{10} 3 \approx 154,416

\log_{10} 3 \approx \displaystyle \frac{154,416}{323,641} \approx 0.477121254723598\dots.

This approximation matches the decimal expansion of \log_{10} 3  to an astounding ten decimal places:

\log_{10} 3 \approx 0.47712125471966\dots

Since hand-held calculators will generate identical outputs for these two expressions (up to the display capabilities of the calculator), this may lead to the misconception that the irrational number \log_{10} 3 is actually equal to the rational number \displaystyle \frac{154,416}{323,641}, so I’ll emphasize again that these two numbers are not equal but are instead really, really close to each other.

Summarizing, Algebra II students can find the decimal expansion of \log_{10} x can be found up to the accuracy of a hand-held scientific calculator. The only tools that are needed are

  • The Laws of Logarithms
  • A hand-held scientific calculator
  • Access to the Wolfram Alpha website (optional)
  • A lot of patience multiply x by itself repeatedly in a quest to find integer powers of x that are close to powers of 10.

While I don’t have a specific reference, I’d be stunned if none of our ancestors tried something along these lines in the years between the discovery of logarithms (1614) and calculus (1666 or 1684).

 

Decimal Approximations of Logarithms (Part 2)

While some common (i.e., base-10) logarithms work out evenly, like \log_{10} 10,000, most do not. Here is the typical output when a scientific calculator computes a logarithm:

To a student first learning logarithms, the answer is just an apparently random jumble of digits; indeed, it can proven that the answer is irrational. With a little prompting, a teacher can get his/her students wondering about how people 50 years ago could have figured this out without a calculator. This leads to a natural pedagogical question:

Can good Algebra II students, using only the tools at their disposal, understand how decimal expansions of base-10 logarithms could have been found before computers were invented?

Here’s a trial-and-error technique — an exploration activity — that is within the grasp of Algebra II students. It’s simple to understand; it’s just a lot of work. While I don’t have a specific reference, I’d be stunned if none of our ancestors tried something along these lines in the years between the discovery of logarithms (1614) and calculus (1666 or 1684).

To approximate \log_{10} x, look for integer powers of x that are close to powers of 10.

I’ll illustrate this idea with \log_{10} 3.

3^1 = 3

3^2 = 9

Not bad… already, we’ve come across a power of 3 that’s decently close to a power of 10. We see that

3^2 = 9 < 10^1

and therefore

\log_{10} 3^2 < 1

2 \log_{10} 3< 1

\log_{10} 3< \displaystyle \frac{1}{2} = 0.5

Let’s keep going. We just keep multiplying by 3 until we find something close to a power of 10. In principle, these calculations could be done by hand, but Algebra II students can speed things up a bit by using their scientific calculators.

3^3 = 27

3^4 = 81

3^5 = 243

3^6 = 729

3^7 = 2,187

3^8 = 6,561

3^9 = 19,683

3^{10} = 59,049

3^{11} = 177,147

3^{12} = 531,441

3^{13} = 1,594,323

3^{14} = 4,782,969

3^{15} = 14,348,907

3^{16} = 43,046,721

3^{17} = 129,140,163

3^{18} = 387,420,489

3^{19} = 1,162,261,467

3^{20} = 3,486,784,401

3^{21} = 10,460,353,203

This looks pretty good too. (Students using a standard ten-digit scientific calculator, of course, won’t be able to see all 11 digits.) We see that

3^{21} > 10^{10}

and therefore

\log_{10} 3^{21} > \log_{10} 10^{10}

21 \log_{10} 3 > 10

\log_{10} 3 > \displaystyle \frac{10}{21} = 0.476190\dots

Summarizing our work so far, we have

0.476190\dots < \log_{10} 3 < 0.5.

We also note that this latest approximation actually gives the first two digits in the decimal expansion of \log_{10} 3.

To get a better approximation of \log_{10} 3, we keep going. I wouldn’t blame Algebra II students a bit if they use their scientific calculators for these computations — but, ideally, they should realize that these calculations could be done by hand by someone very persistent.

3^{22} = 31,381,059,609

3^{23} = 94,143,178,827

3^{24} = 282,429,536,481

3^{25} = 847,288,609,443

3^{26} = 2,541,865,828,329

3^{27} = 7,625,597,484,987

3^{28} = 22,876,792,454,961

3^{29} = 68,630,377,364,883

3^{30} = 205,891,132,094,649

3^{31} = 617,673,396,283,947

3^{32} = 1,853,020,188,851,841

3^{33} = 5,559,060,566,555,523

3^{34} = 16,677,181,699,666,569

3^{35} = 50,031,545,098,999,707

3^{36} = 150,094,635,296,999,121

3^{37} = 450,283,905,890,997,363

3^{38} = 1,350,851,717,672,992,089

3^{39} = 4,052,555,153,018,976,267

3^{40} = 12,157,665,459,056,928,801

3^{41} = 36,472,996,377,170,786,403

3^{42} = 109,418,989,131,512,359,209

3^{43} = 328,256,967,394,537,077,627

3^{44} = 984,770,902,183,611,232,881

Using this last line, we obtain

3^{44} < 10^{21}

and therefore

\log_{10} 3^{44} < \log_{10} 10^{21}

44 \log_{10} 3 < 21

\log_{10} 3 < \displaystyle \frac{21}{44} = 0.477273\dots

Summarizing our work so far, we have

0.476190\dots < \log_{10} 3 < 0.477273\dots.

A quick check with a calculator shows that \log_{10} 3 = 0.477121\dots. In other words,

  • This technique actually works!
  • This last approximation of 0.477273\dots actually produced the first three decimal places of the correct answer!

With a little more work, the approximations

3^{109} \approx 1.014417574 \times 10^{52} > 10^{52}

3^{153} \approx 9.989689095 \times 10^{72} < 10^{73}

can be found, yielding the tighter inequalities

\displaystyle \frac{52}{109} < \log_{10} 3 < \displaystyle \frac{73}{153},

or

0.477064\dots < \log_{10} 3 < 0.477124.

Now we’re really getting close… the last approximation is accurate to five decimal places.

Decimal Approximations of Logarithms (Part 1)

My latest article on mathematics education, titled “Developing Intuition for Logarithms,” was published this month in the “My Favorite Lesson” section of the September 2018 issue of the journal Mathematics Teacher. This is a lesson that I taught for years to my Precalculus students, and I teach it currently to math majors who are aspiring high school teachers. Per copyright law, I can’t reproduce the article here, though the gist of the article appeared in an earlier blog post from five years ago.

Rather than repeat the article here, I thought I would write about some extra thoughts on developing intuition for logarithms that, due to space limitations, I was not able to include in the published article.

While some common (i.e., base-10) logarithms work out evenly, like \log_{10} 10,000, most do not. Here is the typical output when a scientific calculator computes a logarithm:

To a student first learning logarithms, the answer is just an apparently random jumble of digits; indeed, it can proven that the answer is irrational. With a little prompting, a teacher can get his/her students wondering about how people 50 years ago could have figured this out without a calculator. This leads to a natural pedagogical question:

Can good Algebra II students, using only the tools at their disposal, understand how decimal expansions of base-10 logarithms could have been found before computers were invented?

Students who know calculus, of course, can do these computations since

\log_{10} x = \displaystyle \frac{\ln x}{\ln 10},

and the Taylor series

\ln (1+t) = t - \displaystyle \frac{t^2}{2} + \frac{t^3}{3} - \frac{t^4}{4} + \dots,

a standard topic in second-semester calculus, can be used to calculate \ln x for values of x close to 1. However, a calculation using a power series is probably inaccessible to bright Algebra II students, no matter how precocious they are. (Besides, in real life, calculators don’t actually use Taylor series to perform these calculations; see the article CORDIC: How Hand Calculators Calculate, which appeared in College Mathematics Journal, for more details.)

In this series, I’ll discuss a technique that Algebra II students can use to find the decimal expansions of base-10 logarithms to surprisingly high precision using only tools that they’ve learned in Algebra II. This technique won’t be very efficient, but it should be completely accessible to students who are learning about base-10 logarithms for the first time. All that will be required are the Laws of Logarithms and a standard scientific calculator. A little bit of patience can yield the first few decimal places. And either a lot of patience, a teacher who knows how to use Wolfram Alpha appropriately, or a spreadsheet that I wrote can be used to obtain the decimal approximations of logarithms up to the digits displayed on a scientific calculator.

I’ll start this discussion in my next post.

Engaging students: Equations of two variables

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 Trent Pope. His topic, from Algebra: equations of two variables.

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A1. 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.)

I found a website that has many word problems where students can solve for two variables. An example of one of these problems is “If a student were to buy a certain number of $5 scarfs and $2 hats for a total amount of $100, how many scarfs and hats did they buy?”. This example would give students a real world application of how we use two variable equations. It would show students that there are multi variable problems when we go to the store to shop for things, like food or clothing. An instance for food would be when a concession stand sells small and large drinks at a sporting event and want to know how many drinks they have sold at the end of the night. After using a two variable linear equation and knowing the price of the cups, total amount earned, and total cups sold, students would be able to solve for the number of small cups as well as large cups sold.
https://sites.google.com/site/harlandclub/Home/math/algebra/word2var

 

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B2. How does this topic extend what your students should have learned in previous courses?

This topic extends on the students’ ability to graph and solve a linear equation, which should have been taught in their previous classes. The only difference is that the variable, y, that you solved for in Pre-Algebra is now on the same side as the other variable. For instance, the equation y =(-1/4) x + 4 is the same as x + 4y = 16. We see that we solve for the same variables, but they are both on the same side. This is because you are solving the same linear equation. A linear equation can be written in multiple forms, as long as the forms have matching solutions. This is something that students could prove to you by graphing and solving the equations. They would solve the equations to see that they have the same variables. This makes students more aware that they need to be able to compute for other variables besides x if the question asks for it.

 

 

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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.

The most effective way to engage a student about this topic is by using a graphing calculator. This is to help students make the visual connection with the topic and check to see if they have graphed the equations the correct way. Students learn more effectively through visual demonstration. Because students are the ones to solve for the equation and plug it into the calculator to check their work, they are going to be able to make that connection, and we will be able to verify that they understand the material. As teachers, we need to incorporate more technology into the ways of learning because we are surrounded by it daily. Using graphing calculators would be a great way to show and check the work of a two variable equation. This gives students a chance to see what mistakes they have made and what lose ends need to be tied up.

References

Solving Word Problems using a system with 2 variables. n.d. <https://sites.google.com/site/harlandclub/Home/math/algebra/word2var&gt;.

My Favorite One-Liners: Part 113

I tried a new wisecrack when teaching my students about Euler’s formula. It worked gloriously.

Source: https://www.facebook.com/MathematicalMemesLogarithmicallyScaled/photos/a.1605246506167805.1073741827.1605219649503824/2062654510427000/?type=3&theater