A supermutation is a number that contains all the possible permutations of a set of digits. For instance, 123121321 contains all the possible permutations of 1, 2, and 3 (123, 132, 213, 231, 312, and 321). This video is an interesting exploration of permutations, beyond what we’re doing in class.

# Category Archives: In Depth

# Synthetic Division with a Quadratic Divisor

In class and in the book, it is said that synthetic division only works if the divisor is a linear function, i.e., something that can be written in the form (x + k). That’s not true, but synthetic division with higher-order polynomials is a little more complicated.

Here’s an example of synthetic division using a linear divisor. Consider \((3x^3 + 2x^2 – x + 7)\div (x – 2)\). Here it is using synthetic division: \[\begin{array}{cccccc}2&|&3&2&-1&7\\&|&\downarrow&6&16&30\\&&——&——&——&——\\&&3&8&15&37\end{array}\]

The quotient is \(3x^2 + 8 + 15 + \frac{37}{x – 2}\).

Synthetic division for higher order divisors relies on the same concept, but needs a new line for each coefficient. For instance, a quadratic divisor uses two lines instead of one; a cubic divisor uses three lines; and so on.

For instance, \((x^2 + 3x – 7)(x^2 – 5x + 2) = x^4 – 2x^3 -20x^2 + 41x – 14\). To get a remainder, we’ll change the last two terms. To the left, we’ll reverse the signs of both of the coefficients of the lower terms of the divisor; as before, we’ll write all of the coefficients of the dividend: \[\begin{array}{cccccccc}-3&7&|&1&-2&-20&40&-10\\ & &|&\downarrow&-3&15&-6& \\ & &|&\downarrow&&7&-35&14\\ & & &——&——&——&——&——\\ & & &1&-5&2&-1&4\end{array}\]

In the second row, we use opposite of the coefficient from the \(x\) term (that is, -3) as a multiplier; in the third row, we use opposite of the constant (that is, 7).

This gives us a result of \(x^2 – 5x + 2 + \frac{-x + 4}{x^2 + 3x – 7}\), which is the same thing we get from long division.

Notice that, regardless, the lead coefficient (\(a\)) of the divisor must be 1. You can use synthetic division with, say, a divisor of \(4x – 5\), but you’d have to use \(x – 5/4\) instead, and then divide each coefficient of the result by 4.

For example, consider \((2x^3 + x^2 – 17x + 14)/(2x + 7)\). This is what synthetic division yields: \[\begin{array}{cccccc}-7/2&|&2&1&-17&14\\&|&\downarrow&-7&21&-14\\&&——&——&——&——\\&&2&-6&4&0\end{array}\]

The actual quotient is \(x^2 – 3x + 2\), with no remainder, which is what we get from dividing each of the resulting values by 2.

# Quadratic roots in the complex hyperplane

This video has a wonderful visualization on what complex roots look like. It is the first in a series of videos on imaginary numbers.

# Other Cultures and Math

I’ve mentioned in class that German uses slightly different symbols than we do. Here’s a screen shot from a German video on linear functions.

Most of this is recognizable. The 1s are different than ours, but there are two bigger differences:

- The original problem is: Given the point \(P(1, 2)\) and the slope \(m = -2\), what is the graph? In this case, the point is given as \(P(1|2)\).
- In the last step of solving for b, instead of writing \(+2\) on both sides, the instructor wrote \(| +2\) to the right of the entire equation.

I find it interesting to look at how other countries and cultures represent mathematics. The notation is indeed a language, and like any language there are variations around the world. The underlying concepts are universal, though.

# Graphing calculators for your smartphone

Here are three options for free graphing calculators on your tablet or smart phone. If none of these are to your liking, just search the store for “graphing calculators”. The information below is based on the Android versions running on my tablet.

- Desmos (available for iOS and Android, as well as in browsers)
- Basic functions are available on the main screen.
- To get the third root, press “functions” -> “misc” and pick the nth root button. Then use 3 for n.

- GeoGebra (available for iOS and Android, as well as in browsers)
- Basic functions are available on the main screen.
- To get the third root, press “f(x)” and pick the nth root button. Then use 3 for the first blank.

- Wabbitemu (available for Android, as well as computer installation)
- This allows you to load a calculator that works just like our class calculators! I would recommend this for tablets, but it’s probably too small for a phone.
- Once you’ve installed and run the app, select “Help me create a ROM using open source software” and click “Next”.
- Select “TI-84 Plus C SE” and click “Next”.
- You can now touch buttons just like the class calculators.

# Math videos

Here are some videos I’ve been watching today, on math teaching and the deeper beauty of mathematics.

# Vi Hart on Logarithms

In this video, Vi Hart talks about the multiplication scale, which gives us logarithms. She speaks quickly and poetically; don’t expect to understand all of it at this point. If you just want to hear the part about logarithms (which applies to the slide rule that I was discussing today), jump to the 5:00 mark.

# The History of Imaginary Numbers

In this video, Barry Mazur discusses an early appearance of imaginary numbers in mathematical history, and how the mathematician responded to it.

I highly recommend all Numberphile videos as interesting explorations on mathematics and mathematical history.

# Simplifying Radicals on the TI-84 CE

Unfortunately, there doesn’t seem to be a pre-existing function on the TI-84 CE to present a simplified radical. You can write a program, which I provide here, but this takes a lot of work to simply enter into the calculator that I don’t advise it. However, I’m presenting it here to give you an idea of how to program the calculator, in case you’re interested.

To create a program of your own, including entering this one, press the **prgm** button, then select **NEW** and **1:Create New**. Give it a name (I called this **SIMPRAD** for “Simplify Radical”).

Then enter the code below, not including the line-initial colons. These colons represent the start of a new line.

:Input "RADICAND? ",D :1→C :"+"→Str3 :If D<0 :Then :-D→D :"i"→Str3 :End :If D>0 and fPart(D)=0 :Then :While fPart(D/4)=0 :C*2→C :D/4→D :End :For(E,3,√(D),2) :While fPart(D/E^2)=0 :C*E→C :D/E^2→D :End :End :"? :For(A,1,1+log(C)) :sub("0123456789",ipart(10*fPart(C*10^(-A)))+1,1)+Ans :End :sub(Ans,1,length(Ans)-1→Str1 :"? :For(A,1,1+log(D)) :sub("0123456789",ipart(10*fPart(D*10^(-A)))+1,1)+Ans :End :sub(Ans,1,length(Ans)-1→Str2 :If Str3="i" :Str1+Str3→Str1 :If D>1 :Str1+"√("+Str2+")"→Str1 :Disp Str1 :Else :If D=0 :Then :Disp "0" :Else :Disp "INVALID" :End :End

I won’t go through all the entry details; some of the characters can be entered from keys on the calculator, while others require going to specific menus. If you really do want to enter this into your calculator, search around or ask me for specific items.

Let’s look at how each section of this code works.

:Input "RADICAND? ",D :1→C :"+"→Str3

The lines above ask the user for the number to be simplified. For instance, if you want to simplify \(\sqrt{412}\), you would enter **412**. When the program is done, **D** will hold the radicand and **C** will hold the coefficient. **Str3** will let us know if the initial radicand is negative.

:If D<0 :Then :-D→D :"i"→Str3 :End

The lines above allow for imaginary roots.

:If D>0 and fPart(D)=0 :Then

We will only process positive integers this way; 0 and non-integers will be handled separately.

:While fPart(D/4)=0 :C*2→C :D/4→D :End

There are two approaches we could use: Have a list of primes that we walk through, or test 2 and then all odd integers greater than 1. For ease of programming, I’ll do the latter. So these lines divide the radicand by \(2^2 = 4\) until doing so results in a non-integer.

:For(E,3,√(D),2) :While fPart(D/E^2)=0 :C*E→C :D/E^2→D :End :End

These lines divide the radicand by \(3^2 = 9\), \(5^2 = 25\), and so on up to the square root of the radicand, moving on to each new odd number when dividing results in a non-integer.

At this point, we have what we need. If we were willing to have ugly output, we could pretty much stop here. Most of the rest of the code is to make the output attractive. Because the TI-84 CE couldn’t easily convert a number into a string (characters on a screen that aren’t numbers), and couldn’t connect a number to a string, we have to do this. A recent OS update changed this, but I’m providing code that works for all the calculators we have in the room.

:"? :For(A,1,1+log(C)) :sub("0123456789",ipart(10*fPart(C*10^(-A)))+1,1)+Ans :End :sub(Ans,1,length(Ans)-1→Str1

The lines above convert the coeefficient from the number **C** into the string **Str1**.

:"? :For(A,1,1+log(D)) :sub("0123456789",ipart(10*fPart(D*10^(-A)))+1,1)+Ans :End :sub(Ans,1,length(Ans)-1→Str2

The lines avove convert the radicand into **Str2**.

:If Str3="i" :Str1+Str3→Str1 :If D>1 :Str1+"√("+Str2+")"→Str1 :Disp Str1

The lines above create a string like **4i√3**. The rest of the code handles 0 (in which case, just display 0) and non-integers (in which case, display “INVALID”).

:Else :If D=0 :Then :Disp "0" :Else :Disp "INVALID" :End :End

I think this gives an interesting overview to programming the TI 84. If you have a personal calculator and want to store this, feel free.

# Deriving the Quadratic Formula

So where does the quadratic formula come from, anyway?

The formula gives us the solutions of a quadratic equation of the form \[ax^2 + bx + c = 0\] It tells us that this equation is true when \[x = \frac{-b\pm\sqrt{b^2 – 4ac}}{2a}\]

But where did such a strange formula come from?

It comes from solving the quadratic equation for \(x\), but that requires some substitution. We can’t directly solve an equation that contains the same variable to different powers. Technically, \(x^2\) is a different variable than \(x\).

Instead, we need to rewrite the quadratic equation into a form that only has \(x\) one time. Consider the expression \((dx + e)^2\): This satisfies that condition. We can solve \((dx + e)^2 – f = 0\) in terms of \(x\): \[(dx + e)^2 – f = 0 \Rightarrow \\ (dx + e)^2 = f \Rightarrow \\ dx + e = \pm \sqrt{f} \Rightarrow \\ dx = -e \pm \sqrt{f} \Rightarrow \\ x = \frac{-e \pm \sqrt{f}}{d} = \frac{-e}{d} \pm \frac{\sqrt{f}}{d} \]

This looks similar to the quadratic formula, but simpler. If we could find a way to rewrite \(d\), \(e\), and \(f\) in terms of \(a\), \(b\), and \(c\), we’d be all set.

And we can do that. Let’s assume that the two forms of the quadratic equation represent an identical function, that is, \[\color{red}{a}x^2 + \color{blue}{b}x + \color{green}{c} = (dx + e)^2 – f\] If we expand the right hand side, we get \[(dx + e)^2 – f = \color{red}{d^2}x^2 + \color{blue}{2de}x + \color{green}{e^2 – f}\] This means: \[\color{red}{a = d^2} \\ \color{blue}{b = 2de} \\ \color{green}{c = e^2 – f}\]

The first line is straightforward: \(d = \sqrt{a}\) (we’ll assume that d is positive; we could still derive the formula without this assumption, but it’s more confusing).

The second line then becomes \(b = 2e\sqrt{a}\), so \(e = \frac{b}{2\sqrt{a}}\).

The third line then becomes \(c = \frac{b^2}{4a} – f\), so \(f = \frac{b^2}{4a} – c = \frac{b^2 – 4ac}{4a} \).

At this point, we can substitute each variable. Start with \(\color{red}{f}\): \[ \frac{-e}{d} \pm \frac{\sqrt{\color{red}{f}}}{d} = \frac{-e}{d} \pm \frac{\sqrt{\color{red}{\frac{b^2 – 4ac}{4a}}}}{d} \\ = \frac{-e}{d} \pm \frac{\sqrt{\color{red}{b^2 – 4ac}}}{\color{red}{2}d\color{red}{\sqrt{a}}} \]

Next, replace \(\color{red}{e}\): \[\frac{-\color{red}{e}}{d} \pm \frac{\sqrt{b^2 – 4ac}}{2d\sqrt{a}} = \frac{-\color{red}{\frac{b}{2\sqrt{a}}}}{d} \pm \frac{\sqrt{b^2 – 4ac}}{2d\sqrt{a}} \\ = \frac{-\color{red}{b}}{\color{red}{2}d\color{red}{\sqrt{a}}} \pm \frac{\sqrt{b^2 – 4ac}}{2d\sqrt{a}} = \frac{-b \pm \sqrt{b^2 – 4ac}}{2d\sqrt{a}}\]

Finally, replace \(\color{red}{d}\): \[\frac{-b \pm \sqrt{b^2 – 4ac}}{2\color{red}{d}\sqrt{a}} = \frac{-b \pm \sqrt{b^2 – 4ac}}{2\color{red}{\sqrt{a}}\sqrt{a}} = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\]

Recall that we started with an equation involving \(d\), \(e\), and \(f\) which represented the values of \(x\) that made the equation true. That is, \[x = \frac{-e \pm \sqrt{-f}}{d} = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\] which is the quadratic formula in its typical form.