All posts tagged square root

Engaging students: Square roots

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 Jessica Martinez. Her topic, from Algebra: square roots.

How has this topic appeared in the news?

There is a (sort of) holiday for square root days; sort of because square root days only come 9 time every century and this year we celebrated 4/4/16. Since it’s not as frequent as Pi Day, it’s a lesser known “holiday”, but even then, it still pops up in the news. I found this online article for a UK news site that described other square root-related fun facts in history. It also included a post from Good Morning America with the hashtag #squarerootday, which gave me this idea: I would like to encourage my students to participate in all of the fun square root-related activities that celebrate this day (if there was one that school year). The founder of square root day has suggestions that include but are not limited to: square dancing, drinking root beer out of square glasses, or even taking a drive on route 66. In the days leading up to this fantastic math-related day, I would consider giving my kids an extra credit point for posting a picture of themselves doing something square root related on the class twitter with the tag #squarerootday (or a post on some other class social media). If there wasn’t a square root day during that academic year, I still think it would be fun to tell my students about this holiday.

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

My students should have already learned about perfect squares and their multiplication tables up to 12 or 13, at least. For a simple refresher, I could have my students color/highlight perfect squares on multiplication tables. Then taking the square root of something is the inverse of creating perfect squares, unless what’s under the square root sign isn’t a perfect square. Then what’s under the radical is something that they need to divide into its prime factors so that they can simplify. My students should have also at least learned about prime numbers, if not prime factoring. A way to solve square roots would be pairing up the prime factors under the square root so that you can “take it out” from under the radical; for my students, I could have them think of the square root sign as a jail cell, and the only way that the numbers could “get out” of the cell is if they had a “prime partner” to escape with (i.e. a pair of 2s, a pair of 3s etc.).

green line

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

So one of the oldest records of square roots in history would be The Old Babylonian tablet YBC 7289, which dates back anywhere from 2000-1600 BC. It depicts a square with two diagonals drawn and on the diagonals are numbers; when they are calculated, you get a very close approximation of the square root of 2 for the diagonal. Their value for the square root of two was about 1.41421297; I could have my students quickly calculate the square root of two (about 1.41421356) and mention to my students that this is pretty impressive for a civilization without modern day technology. The fact that they used clay tablets for math calculations shows how little they had to work with. Yet Babylon was also one of the most famous ancient cities in Mesopotamia; it’s mentioned multiple times in the bible and they were pretty advanced in mathematics for their area, despite the lack of resources we have today. They used a sexagesimal number system, which is base 60; they could solve algebra problems and work with what we now call Pythagorean triples; they could also solve equations with cubes.

References

A Visual Approach to Simplifying Radicals (A Get Out of Jail Free Card). (2012, January 15). Retrieved September 09, 2016, from https://reflectionsinthewhy.wordpress.com/2012/01/15/a-visual-approach-to-simplifying-radicals-a-get-out-of-jail-free-card/

Babylon and the Square Root of 2. (2016). Retrieved September 09, 2016, from https://johncarlosbaez.wordpress.com/2011/12/02/babylon-and-the-square-root-of-2/

Buncombe, A. (16, April 4). Square Root Day: There are only nine days this century like this. Retrieved September 09, 2016, from http://www.independent.co.uk/news/world/americas/square-root-day-there-are-only-nine-days-this-century-like-this-a6967991.html

Fowler, D., & Robson, E. (n.d.). Square Root Approximations in Old Babylonian Mathematics: YBC 7289 in Context. Historical Mathematica, 366-378. Retrieved September 9, 2016, from https://math.berkeley.edu/~lpachter/128a/Babylonian_sqrt2.pdf.

Mark, J. J. (2011, April 28). Babylon. Retrieved September 09, 2016, from http://www.ancient.eu/babylon/

Lessons from teaching gifted elementary students (Part 8g)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprisingly depth of mathematical knowledge.

Here’s a question I once received, in the students’ original handwriting. They wanted me to add adjacent numbers on the bottom row to produce the number on the next row, building upward until I reached the apex of the triangle. Then, after I reached the top number, they wanted me to take the square root of that number. (Originally, they wanted me to first multiply by 80 before taking the square root, but evidently they decided to take it easy on me.)

And, just to see if I could do it, they wanted me to do all of this without using a calculator. But they were nice and allowed me to use pencil and paper.

So far, I’ve used Pascal’s triangle to obtain

$y = \displaystyle \sum_{k=0}^{11} (k+1)^2 {11 \choose k}$

$= \displaystyle \sum_{k=2}^{11} k(k-1) {11 \choose k} + \sum_{k=1}^{11} 3k {11 \choose k} + \sum_{k=0}^{11} {11 \choose k}$ .

$= \displaystyle \sum_{k=2}^{11} k(k-1) \left( \frac{11!}{k!(11-k)!} \right) + \sum_{k=1}^{11} 3k \left( \frac{11!}{k!(11-k)!} \right) + \sum_{k=0}^{11} \left( \frac{11!}{k!(11-k)!} \right)$ .

$= \displaystyle \sum_{k=2}^{11} \frac{11!}{(k-2)!(11-k)!} + 3 \sum_{k=1}^{11} \frac{11!}{(k-1)!(11-k)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$ .

$= \displaystyle \sum_{i=0}^{9} \frac{11!}{i!(9-i)!} + 3 \sum_{j=0}^{10} \frac{11!}{j!(10-j)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$ .

$= \displaystyle \sum_{i=0}^{9} 11 \times 10 \frac{9!}{i!(9-i)!} + 3 \sum_{j=0}^{10} 11 \frac{10!}{j!(10-j)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$

$= \displaystyle 110 \sum_{i=0}^{9} {9 \choose i} + 33 \sum_{j=0}^{10} {10 \choose j} + \sum_{k=0}^{11} {11 \choose k}$

$= 110 \times 2^9 + 33 \times 2^{10} + 2^{11}$

$= 92,160$ .

I’m almost done… except my students wanted me to find the square root of this number without using a calculator.

There are a couple ways to do this; the method I chose was directly extracting the square root by hand… a skill that was taught to children in previous generations but has fallen out of pedagogical disfavor with the advent of handheld calculators. I lost my original work, but it would have looked something like this (see the above website for details on why this works):

And so I gave my students their answer: $x \approx 303.578\dots$

1 Comment

by John Quintanilla on September 14, 2016 • Permalink

Posted in Algebra II, Precalculus

Tagged binomial series, Pascal's triangle, square root

Posted by John Quintanilla on September 14, 2016

https://meangreenmath.com/2016/09/14/lessons-from-teaching-gifted-elementary-students-part-8g/

Engaging students: Square roots

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 Tiffany Jones. Her topic, from Algebra: square roots.

B.1 How can this topic be used in your students’ future courses in mathematics or science?

One area of mathematics I wish I had more practice with in grade school is numerical reasoning. I feel that, as a student, I was allowed to use my calculator too much and am struggling to remove my calculator crutch. I hope to encourage my students to sharpen their numerical reasoning skills and to not rely on their calculator. Does this number make sense? Is it too high, too low? Is a negative result valid given the scenario of the problem? The following video introduces a method to estimate the square root of non-perfect squares to the nearest tenth by hand:

“Estimating Square Roots To the Nearest Tenth by Hand” by Fort Bend Tutoring

It gives the students another tool for their toolbox of numerical reasoning, practice using formulas, reviews long division by hand, and strongly encourages students to remember the perfect squares.

I think that introducing this idea as an engage could intrigue student to wonder why the formula works and to wonder what else they are able to do quickly by hand.

Fort Bend Tutoring’s YouTube channel offers videos on a wide verity of high school mathematics topics and courses. The videos cover several examples. They are engaging, not dry and there is also a “theme song” to the videos. I feel that these videos can sever as a great addition to lessons as extra help to the students.

D.1 What interesting things can you say about the people who contributed to the discovery and/or the development of this topic?

The following story was first told to me in a calculus one course. While the telling of the story was to serve as amusement and did not directly relate to the topic of the day, it stuck with me. It comes to mind frequently when working with the Pythagorean theorem and with irrational squares. And when given this assignment, I saw square roots as an option, this story again came to mind. I think having an interesting story cross my mind makes a problem overall more fun. I would want to give that to my students. The article “The Dangerous Ratio” by Brain Clegg does a wonderful job of telling the story, its implications, and gives a mock dialogue so reads can work through the logic. At the end of the article, there is a link to an activity about the proof that the square root of 2 is irrational.

E.1

How can technology be used to effectively engage students with this topic?

I really like the idea of a flipped classroom and hope to be able to practice it in my classroom. While a completely flipped classroom will take some time to implement, videos such as Math Antics’ “Exponents & Square Roots” will be a great place to start.

This particular video address a previously learned topic, namely exponents and relates it to the new topic. It provides definitions and visuals to remember how the terms relate to each other and how to read the symbols. It goes through several examples of varying level and shows the viewer how to use technology such as calculators to solve hard problems. In addition, the video addresses some common misconceptions such as mistaking the root sign and the division sign. Moreover, it ties everything together with a quick review at the end.

One of my favorite aspects so of flipped classrooms, is that the student can review the video over and over. Math Antics does an excellent job of talking the math out to the viewer. The animations are amusing yet helpful. While a lot of information is covered, the video is not dry.

Resources:

“Estimating Square Roots To the Nearest Tenth by Hand” by Fort Bend Tutoring – https://www.youtube.com/watch?v=bUh7Hj-3dkw

“The Dangerous Ratio” by Brian Clegg – http://nrich.maths.org/2671

“Exponents & Square Roots” by Math Antics –https://www.youtube.com/watch?v=C_iKTTI1E34

How I Impressed My Wife: Part 5h

Earlier in this series, I gave three different methods of showing that

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.$

Using the fact that $Q$ is independent of $a$ , I’ll now give a fourth method.

Since $Q$ is independent of $a$ , I can substitute any convenient value of $a$ that I want without changing the value of $Q$ . As shown in previous posts, substituting $a =0$ yields the following simplification:

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\pi}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 b^2 - 2) u^2 + 1}$

The four roots of the denominator satisfy

$u^2 = \displaystyle 1 - 2b^2 \pm 2|b| \sqrt{b^2 - 1}$

So far, I’ve handled the cases $|b| = 1$ and $|b| > 1$ . In today’s post, I’ll start considering the case $|b| < 1$ .

Factoring the denominator is a bit more complicated if $|b| < 1$ . Using the quadratic equation, we obtain

$u^2 = \displaystyle 1 - 2b^2 \pm 2|b| i \sqrt{1-b^2}$

However, unlike the cases $|b| \ge 1$ , the right-hand side is now a complex number. So, To solve for $u$ , I’ll use DeMoivre’s Theorem and some surprisingly convenient trig identities. Notice that

$(1-2b^2)^2 + (2|b| \sqrt{1-b^2})^2 = 1 - 4b^2 + b^4 + 4b^2 (1 - b^2) = 1 - 4b^2 + b^4 + 4b^2 - b^4 = 1$ .

Therefore, the four complex roots of the denominator satisfy $|u^2| = 1$ , or $|u| = 1$ . This means that all four roots can be written in trigonometric form so that

$u^2 = \cos 2\phi + i \sin 2\phi$ ,

where $2\phi$ is some angle. (I chose the angle to be $2\phi$ instead of $\phi$ for reasons that will become clear shortly.)

I’ll begin with solving

$u^2 = \displaystyle 1 - 2b^2 + 2|b| i \sqrt{1-b^2}$ .

Matching the real and imaginary parts, we see that

$\cos 2\phi = 1-2b^2$ ,

$\sin 2\phi = 2|b| \sqrt{1-b^2}$

This completely matches the form of the double-angle trig identities

$\cos 2\phi = 1 - 2\sin^2 \phi$ ,

$\sin 2 \phi = 2 \sin\phi \cos \phi$ ,

and so the problem reduces to solving

$u^2 = \cos 2\phi + i \sin 2\phi$ ,

where $\sin \phi = |b|$ and $\cos \phi = \sqrt{1-b^2}$. By De Moivre’s Theorem, I can conclude that the two solutions of this equation are

$u = \pm(\cos \phi + i \sin \phi)$ ,

$u = \pm( \sqrt{1-b^2} + i |b|)$ .

I could re-run this argument to solve $u^2 = \displaystyle 1 - 2b^2 - 2|b| i \sqrt{1-b^2}$ and get the other two complex roots. However, by the Conjugate Root Theorem, I know that the four complex roots of the denominator $u^4 + (4 b^2 - 2) u^2 + 1$ must come in conjugate pairs. Therefore, the four complex roots are

$u = \pm \sqrt{1-b^2} \pm i |b|$ .

Therefore, I can factor the denominator as follows:

$u^4 + (4 b^2 - 2) u^2 + 1 = (u - [\sqrt{1-b^2} + i|b|])(u - [\sqrt{1-b^2} - i|b|])$

$\qquad \times (u - [-\sqrt{1-b^2} + i|b|])(u - [-\sqrt{1-b^2} + i|b|])$

$= (u - \sqrt{1-b^2} - i|b|)(u - \sqrt{1-b^2} + i|b])$

$\qquad \times (u +\sqrt{1-b^2} + i|b|)(u +\sqrt{1-b^2} + i|b|)$

$= ([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)$

To double-check my work, I can directly multiply this product:

$([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)$

$= (u^2 - 2u \sqrt{1-b^2} + 1 - b^2 + b^2) (u^2 + 2u \sqrt{1-b^2} + 1 - b^2 + b^2)$

$= ([u^2 +1] - 2u\sqrt{1-b^2})([u^2+1] + 2u\sqrt{1-b^2})$

$= [u^2+1]^2 - [2u\sqrt{1-b^2}]^2$

$= u^4 + 2u^2 + 1 - 4u^2 (1-b^2)$

$= u^4 + u^2 (2 - 4[1-b^2]) + 1$

$= u^4 + u^2 (4b^2 - 2) + 1$ .

So, at last, I can rewrite the integral $Q$ as

$Q = \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{ ([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)}$

I’ll continue with this fourth evaluation of the integral, continuing the case $|b| < 1$ , in tomorrow’s post.

1 Comment

by John Quintanilla on August 20, 2015 • Permalink

Posted in Calculus, Precalculus

Tagged complex numbers, cosine, De Moivre's Theorem, double angle trig identities, integral, polar coordinates, sine, square root, trigonometry

Posted by John Quintanilla on August 20, 2015

https://meangreenmath.com/2015/08/20/how-i-impressed-my-wife-part-5h/

How I Impressed My Wife: Part 4f

Previously in this series, I have used two different techniques to show that

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.$

Originally, my wife had asked me to compute this integral by hand because Mathematica 4 and Mathematica 8 gave different answers. At the time, I eventually obtained the solution by multiplying the top and bottom of the integrand by $\sec^2 x$ and then employing the substitution $u = \tan x$ (after using trig identities to adjust the limits of integration).

But this wasn’t the only method I tried. Indeed, I tried two or three different methods before deciding they were too messy and trying something different. So, for the rest of this series, I’d like to explore different ways that the above integral can be computed.

Previously in this series, I have used two different techniques to show that

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.$

Here’s my progress so far:

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$= \displaystyle \int_0^{2\pi} \frac{2 \, dx}{1+\cos 2x + 2 a \sin 2x + (a^2 + b^2)(1-\cos 2x)}$

$= 2 \displaystyle \int_0^{2\pi} \frac{d\theta}{(1+a^2+b^2) + 2 a \sin \theta + (1 - a^2 - b^2) \cos \theta}$

$= 2 \displaystyle \int_{0}^{2\pi} \frac{d\theta}{S + R \cos (\theta - \alpha)}$

$= 2 \displaystyle \int_{0}^{2\pi} \frac{d\phi}{S + R \cos \phi}$

$= \displaystyle -\frac{4i}{R} \oint_C \frac{dz}{z^2 + 2\frac{S}{R}z + 1}$ ,

where this last integral is taken over the complex plane on the unit circle, a closed contour oriented counterclockwise. In these formulas, $R = \sqrt{(2a)^2 + (1-a^2-b^2)^2}$ and $S = 1 + a^2 + b^2$ . (Also, $\alpha$ is a certain angle that is now irrelevant at this point in the calculation).

This contour integral looks complicated; however, it’s an amazing fact that integrals over closed contours can be easily evaluated by only looking at the poles of the integrand. In recent posts, I established that there was only one pole inside the contour, and the residue at this pole was equal to $\displaystyle \frac{R}{ 2 \sqrt{S^2-R^2} }$ .

This residue can be used to evaluate the contour integral. Ordinarily, integrals are computed by subtracting the values of the antiderivative at the endpoints. However, there is an alternate way of computing a contour integral using residues. It turns out that the value of the contour integral is $2\pi i$ times the sum of the residues within the contour; see Wikipedia and Mathworld for more information.

Therefore,

$Q = \displaystyle -\frac{4i}{R} \oint_C \frac{dz}{(z - r_1)(z- r_2)}$

$= \displaystyle -\frac{4i}{R} \cdot 2\pi i \cdot \frac{R}{ 2 \sqrt{S^2-R^2} }$

$= \displaystyle \frac{4\pi}{\sqrt{S^2-R^2}}$

Next, I use some algebra to simplify the denominator:

$S^2 - R^2 = (1+a^2+b^2)^2 - (1-a^2-b^2)^2 - (2a)^2$

$S^2 - R^2 = [(1 + a^2 + b^2) + (1-a^2-b^2)][(1 + a^2 + b^2) - (1 - a^2 -b^2)] - 4a^2$

$S^2 - R^2 = 2[2 a^2 + 2b^2] - 4a^2$

$S^2 - R^2 = 4b^2$

Therefore,

$Q = \displaystyle \frac{4\pi}{\sqrt{4b^2}} = \displaystyle \frac{4\pi}{2|b|} = \frac{2\pi}{|b|}$

Once again, this matches the solution found with the previous methods… and I was careful to avoid a common algebraic mistake.

In tomorrow’s post, I’ll discuss an alternative way of computing the residue.

1 Comment

by John Quintanilla on August 9, 2015 • Permalink

Posted in Calculus, Precalculus

Tagged complex numbers, geometric series, integral, residue, square root

Posted by John Quintanilla on August 9, 2015

https://meangreenmath.com/2015/08/09/how-i-impressed-my-wife-part-4f/

Inverse Functions: Index

I’m doing something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my series on the different definitions on inverse functions that appear in Precalculus and Calculus.

Square Roots, nth Roots, and Rational Exponents

Part 1: Simplifying $\sqrt{x^2}$

Part 2: The difference between $\sqrt{t}$ and solving $x^2 = t$

Part 3: Definition of an inverse function and the horizontal line test

Part 4: Why extraneous solutions may occur when solving algebra problems involving a square root

Part 5: Defining $\sqrt{x}$

Part 6: Consequences of the definition of $\sqrt{x}$ : simplifying $\sqrt{x^2}$

Part 7: Defining $\sqrt[n]{x}$ if $n$ is odd or even

Part 8: Rational exponents if the denominator of the exponent is odd or even

Arcsine

Part 9: There are infinitely many solutions to $\sin x = 0.8$

Part 10: Defining arcsine with domain $[-\pi/2,\pi/2]$

Part 11: Pedagogical thoughts on teaching arcsine.

Part 12: Solving SSA triangles: impossible case

Part 13: Solving SSA triangles: one way of getting a unique solution

Part 14: Solving SSA triangles: another way of getting a unique solution

Part 15: Solving SSA triangles: continuation of Part 14

Part 16: Solving SSA triangles: ambiguous case of two solutions

Part 17: Summary of rules for solving SSA triangles

Arccosine

Part 18: Definition for arccosine with domain $[0,\pi]$

Part 19: The Law of Cosines and solving SSS triangles

Part 20: Identifying impossible triangles with the Law of Cosines

Part 21: The Law of Cosines provides an unambiguous angle, unlike the Law of Sines

Part 22: Finding the angle between two vectors

Part 23: A proof for why the formula in Part 22 works

Arctangent

Part 18: Definition for arctangent with domain $(-\pi/2,\pi/2)$

Part 24: Finding the angle between two lines

Part 25: A proof for why the formula in Part 24 works.

Arcsecant

Part 26: Defining arcsecant using $[0,\pi/2) \cup (\pi/2,\pi]$

Part 27: Issues that arise in calculus using the domain $[0,\pi/2) \cup (\pi/2,\pi]$

Part 28: More issues that arise in calculus using the domain $[0,\pi/2) \cup (\pi/2,\pi]$

Part 29: Defining arcsecant using $[0,\pi/2) \cup [pi,3\pi/2)$

Logarithm

Part 30: Logarithms and complex numbers

4 Comments

by John Quintanilla on February 21, 2015 • Permalink

Posted in Calculus, Precalculus

Tagged index to a series of posts, inverse function, square root, trigonometry

Posted by John Quintanilla on February 21, 2015

https://meangreenmath.com/2015/02/21/inverse-functions-index/

Square roots and Logarithms Without a Calculator: Index

I’m using the Twelve Days of Christmas (and perhaps a few extra days besides) to do something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my series on computing square roots and logarithms without a calculator.

Part 1: Method #1: Trial and error.

Part 2: Method #2: An algorithm comparable to long division.

Part 3: Method #3: Introduction to logarithmic tables. At the time of this writing, this is the most viewed page on my blog.

Part 4: Finding antilogarithms with a table.

Part 5: Pedagogical and historical thoughts on log tables.

Part 6: Computation of square roots using a log table.

Part 7: Method #4: Slide rules

Part 8: Method #5: By hand, using a couple of known logarithms base 10, the change of base formula, and the Taylor approximation $ln(1+x) \approx x$ .

Part 9: An in-class activity for getting students comfortable with logarithms when seen for the first time.

Part 10: Method #6: Mentally… anecdotes from Nobel Prize-winning physicist Richard P. Feynman and me.

Part 11: Method #7: Newton’s Method.

3 Comments

by John Quintanilla on December 29, 2014 • Permalink

Posted in Algebra II, Precalculus, Technology

Tagged Feynman, index to a series of posts, logarithm, square root

Posted by John Quintanilla on December 29, 2014

https://meangreenmath.com/2014/12/29/square-roots-and-logarithms-without-a-calculator-index/

A Curious Square Root: Index

I’m using the Twelve Days of Christmas (and perhaps a few extra days besides) to do something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my series on expressions containing nested square roots that nevertheless can be simplified.

Part 1: Simplifying $\sqrt{5 - \sqrt{6} + \sqrt{22+8\sqrt{6}}}$ .

Part 2: DIfferent ways of calculating $\sin 15^\circ$ .

Inverse Functions: Rational Exponents (Part 8)

In this series of posts, we have seen that the definition of $\sqrt[n]{x}$ and saw that the definition was a little different depending if $n$ is even or odd:

If $n$ is even, then $y = \sqrt[n]{x}$ means that $x = y^n$ and $y \ge 0$ . In particular, this is impossible (for real $y$ ) if $x < 0$ .
If $n$ is odd, then $y = \sqrt[n]{x}$ means that $x = y^n$ . There is no need to give a caveat on the possible values of $y$ .

Let’s now consider the definition of $x^{m/n}$ , where $m$ and $n$ are positive integers greater than 1. Ideally, we’d like to simply defined

$x^{m/n} = \left[ x^{1/n} \right]^m$

This definition reduces to previous work (like a good MIT freshman), using prior definition for raising to powers that are either integers or reciprocals of integers. Indeed, if $x \ge 0$ , there is absolutely no ambiguity about this definition.

Unfortunately, if $x < 0$ , then a little more care is required. There are four possible cases.

Case 1. $m$ and $n$ are odd. In this case, there is no ambiguity if $x < 0$ is negative. For example,

$(-32)^{3/5} = \left[ (-32)^{1/5} \right]^3 = [-2]^3 = -8$

Case 2: $m$ is even but $n$ is odd. Again, there is no ambiguity if $x< 0$ is negative. For example,

$(-32)^{4/5} = \left[ (-32)^{4/5} \right]^3 = [-2]^4 = 16$

Case 3: $m$ is odd but $n$ is even. In this case, $x^{m/n}$ is undefined if $x < 0$ . For example, we would like $(-16)^{3/2}$ to be equal to $\left[ (-16)^{1/2} \right]^3$ , but ${-16}^{1/2} = \sqrt{-16}$ is undefined (using real numbers).

Case 4. $m$ and $latex $n$ are both even. This is perhaps the most interesting case. For example, how should we evaluate $(-8)^{4/6}$ ?. There are two legitimate choices… which lead to different answers!

Option #1: If we just apply the proposed definition of $x^{m/n}$ , we find that

$(-8)^{2/6} = \left[ (-8)^2 \right]^{1/6} = [64]^{1/6} = 2$

Option #2: We could first reduce $2/6$ to lowest terms:

$(-8)^{2/6} = (-8)^{1/3} = -2$

So… which is it?!?!?!?! The rule that mathematicians have chosen is that simplifying the exponent takes precedence over the above definition. In other words, the definition $x^{m/n} = \left[ x^{1/n} \right]^m$ should only be applied in $m/n$ has been reduced to lowest terms in order to remove the above ambiguity.

For the sake of completeness, I note that the above discussion restricts our attention to real numbers. If complex numbers are permitted, then things become a lot more interesting. If we repeat a few of the above calculations using complex numbers, we get answers that are different!

The explanation for this surprising result is not brief, but I discussed it in a previous series of posts:

https://meangreenmath.com/2014/06/19/calculators-and-complex-numbers-part-1/

https://meangreenmath.com/2014/06/20/calculators-and-complex-numbers-part-2/

https://meangreenmath.com/2014/06/21/calculators-and-complex-numbers-part-3/

https://meangreenmath.com/2014/06/22/calculators-and-complex-numbers-part-4/

https://meangreenmath.com/2014/06/23/calculators-and-complex-numbers-part-5/

https://meangreenmath.com/2014/06/24/calculators-and-complex-numbers-part-6/

https://meangreenmath.com/2014/06/25/calculators-and-complex-numbers-part-7/

https://meangreenmath.com/2014/06/26/calculators-and-complex-numbers-part-8/

https://meangreenmath.com/2014/06/27/calculators-and-complex-numbers-part-9/

https://meangreenmath.com/2014/06/28/calculators-and-complex-numbers-part-10/

https://meangreenmath.com/2014/06/29/calculators-and-complex-numbers-part-11/

https://meangreenmath.com/2014/06/30/calculators-and-complex-numbers-part-12/

https://meangreenmath.com/2014/07/01/calculators-and-complex-numbers-part-13/

https://meangreenmath.com/2014/07/02/calculators-and-complex-numbers-part-14/

https://meangreenmath.com/2014/07/03/calculators-and-complex-numbers-part-15-2/

https://meangreenmath.com/2014/07/04/calculators-and-complex-numbers-part-16/

https://meangreenmath.com/2014/07/05/calculators-and-complex-numbers-part-17/

https://meangreenmath.com/2014/07/06/calculators-and-complex-numbers-part-18/

https://meangreenmath.com/2014/07/07/calculators-and-complex-numbers-part-19/

https://meangreenmath.com/2014/07/08/calculators-and-complex-numbers-part-20/

https://meangreenmath.com/2014/07/09/calculators-and-complex-numbers-part-21/

https://meangreenmath.com/2014/07/10/calculators-and-complex-numbers-part-22/

https://meangreenmath.com/2014/07/11/calculators-and-complex-numbers-part-23/

https://meangreenmath.com/2014/07/12/calculators-and-complex-numbers-part-24/

1 Comment

by John Quintanilla on October 8, 2014 • Permalink

Posted in Precalculus

Tagged complex numbers, inverse function, square root

Posted by John Quintanilla on October 8, 2014

https://meangreenmath.com/2014/10/08/inverse-functions-rational-exponents-part-8/

Inverse Functions: nth Roots (Part 7)

In the previous posts of this series, I carefully considered the definition of $f(x) = \sqrt{x} = x^{1/2}$ . Let’s now repeat this logic to consider the definition of $f(x) = \sqrt[n]{x} = x^{1/n}$ , where $n \ge 3$ is an integer. We begin with $n$ even.

A typical case is $n =4$ ; the graph of $y = x^4$ is shown below.

The full graph of $y= x^n$ fails the horizontal line test if $n$ is even. Therefore, we need to apply the same logic that we used earlier to define $y = \sqrt[n]{x}$ . In particular, we essentially erase the left half of the graph. By restricting the domain to $[0,\infty)$ , we create a new function that does satisfy the horizontal line test, so that the graph of $y = \sqrt[n]{x}$ is found by reflecting through the line $y = x$ .

Written in sentence form,

If $n$ is even, then $y = \sqrt[n]{x}$ means that $x = y^n$ and $y \ge 0$ . In particular, this is impossible for real $y$ if $x < 0$ .

green line
We now turn to the case of $n$ odd. Unlike before, the full graph of $y= x^n$ (in thick blue) satisfies the horizontal line test. Therefore, there is no need to restrict the domain to define the inverse function. (shown in thin purple).

In other words,

If $n$ is odd, then $y = \sqrt[n]{x}$ means that $x = y^n$ . There is no need to give a caveat on the possible values of $y$ .

In particular, $\sqrt{-8}$ and $\sqrt[4]{-8}$ are both undefined since there is no (real) number $x$ so that $x^2 = -8$ or $x^4 = -8$ . However, $\sqrt[3]{-8}$ is defined and is equal to $-2$ since $(-2)^3 = -8$ .