# Tag Archives: polynomial

## Infinite Ways to an Infinite Geometric Sum

One of my students, K, and I were reviewing Taylor Series last Friday when she asked for a reminder why an infinite geometric series summed to $\displaystyle \frac{g}{1-r}$ for first term g and common ratio r when $\left| r \right| < 1$.  I was glad she was dissatisfied with blind use of a formula and dove into a familiar (to me) derivation.  In the end, she shook me free from my routine just as she made sure she didn’t fall into her own.

STANDARD INFINITE GEOMETRIC SUM DERIVATION

My standard explanation starts with a generic infinite geometric series.

$S = g+g\cdot r+g\cdot r^2+g\cdot r^3+...$  (1)

We can reason this series converges iff $\left| r \right| <1$ (see Footnote 1 for an explanation).  Assume this is true for (1).  Notice the terms on the right keep multiplying by r.

The annoying part of summing any infinite series is the ellipsis (…).  Any finite number of terms always has a finite sum, but that simply written, but vague ellipsis is logically difficult.  In the geometric series case, we might be able to handle the ellipsis by aligning terms in a similar series.  You can accomplish this by continuing the pattern on the right:  multiplying both sides by r

$r\cdot S = r\cdot \left( g+g\cdot r+g\cdot r^2+... \right)$

$r\cdot S = g\cdot r+g\cdot r^2+g\cdot r^3+...$  (2)

This seems to make make the right side of (2) identical to the right side of (1) except for the leading g term of (1), but the ellipsis requires some careful treatment. Footnote 2 explains how the ellipses of (1) and (2) are identical.  After that is established, subtracting (2) from (1), factoring, and rearranging some terms leads to the infinite geometric sum formula.

$(1)-(2) = S-S\cdot r = S\cdot (1-r)=g$

$\displaystyle S=\frac{g}{1-r}$

STUDENT PREFERENCES

I despise giving any formula to any of my classes without at least exploring its genesis.  I also allow my students to use any legitimate mathematics to solve problems so long as reasoning is justified.

In my experiences, about half of my students opt for a formulaic approach to infinite geometric sums while an equal number prefer the quick “multiply-by-r-and-subtract” approach used to derive the summation formula.  For many, apparently, the dynamic manipulation is more meaningful than a static rule.  It’s very cool to watch student preferences at play.

K’s VARIATION

K understood the proof, and then asked a question I hadn’t thought to ask.  Why did we have to multiply by r?  Could multiplication by $r^2$ also determine the summation formula?

I had three nearly simultaneous thoughts followed quickly by a fourth.  First, why hadn’t I ever thought to ask that?  Second, geometric series for $\left| r \right|<1$ are absolutely convergent, so K’s suggestion should work.  Third, while the formula would initially look different, absolute convergence guaranteed that whatever the “$r^2$ formula” looked like, it had to be algebraically equivalent to the standard form.  While I considered those conscious questions, my math subconscious quickly saw the easy resolution to K’s question and the equivalence from Thought #3.

Multiplying (1) by $r^2$ gives

$r^2 \cdot S = g\cdot r^2 + g\cdot r^3 + ...$ (3)

and the ellipses of (1) and (3) partner perfectly (Footnote 2), so K subtracted, factored, and simplified to get the inevitable result.

$(1)-(3) = S-S\cdot r^2 = g+g\cdot r$

$S\cdot \left( 1-r^2 \right) = g\cdot (1+r)$

$\displaystyle S=\frac{g\cdot (1+r)}{1-r^2} = \frac{g\cdot (1+r)}{(1+r)(1-r)} = \frac{g}{1-r}$

That was cool, but this success meant that there were surely many more options.

EXTENDING

Why stop at multiplying by r or $r^2$?  Why not multiply both sides of (1) by a generic $r^N$ for any natural number N?   That would give

$r^N \cdot S = g\cdot r^N + g\cdot r^{N+1} + ...$ (4)

where the ellipses of (1) and (4) are again identical by the method of Footnote 2.  Subtracting (4) from (1) gives

$(1)-(4) = S-S\cdot r^N = g+g\cdot r + g\cdot r^2+...+ g\cdot r^{N-1}$

$S\cdot \left( 1-r^N \right) = g\cdot \left( 1+r+r^2+...+r^{N-1} \right)$  (5)

There are two ways to proceed from (5).  You could recognize the right side as a finite geometric sum with first term 1 and ratio r.  Substituting that formula and dividing by $\left( 1-r^N \right)$ would give the general result.

Alternatively, I could see students exploring $\left( 1-r^N \right)$, and discovering by hand or by CAS that $(1-r)$ is always a factor.  I got the following TI-Nspire CAS result in about 10-15 seconds, clearly suggesting that

$1-r^N = (1-r)\left( 1+r+r^2+...+r^{N-1} \right)$.  (6)

Math induction or a careful polynomial expansion of (6) would prove the pattern suggested by the CAS.  From there, dividing both sides of (5) by $\left( 1-r^N \right)$ gives the generic result.

$\displaystyle S = \frac{g\cdot \left( 1+r+r^2+...+r^{N-1} \right)}{\left( 1-r^N \right)}$

$\displaystyle S = \frac{g\cdot \left( 1+r+r^2+...+r^{N-1} \right) }{(1-r) \cdot \left( 1+r+r^2+...+r^{N-1} \right)} = \frac{g}{1-r}$

In the end, K helped me see there wasn’t just my stock approach to an infinite geometric sum, but really an infinite number of parallel ways.  Nice.

FOOTNOTES

1) RESTRICTING r:  Obviously an infinite geometric series diverges for $\left| r \right| >1$ because that would make $g\cdot r^n \rightarrow \infty$ as $n\rightarrow \infty$, and adding an infinitely large term (positive or negative) to any sum ruins any chance of finding a sum.

For $r=1$, the sum converges iff $g=0$ (a rather boring series). If $g \ne 0$ , you get a sum of an infinite number of some nonzero quantity, and that is always infinite, no matter how small or large the nonzero quantity.

The last case, $r=-1$, is more subtle.  For $g \ne 0$, this terms of this series alternate between positive and negative g, making the partial sums of the series add to either g or 0, depending on whether you have summed an even or an odd number of terms.  Since the partial sums alternate, the overall sum is divergent.  Remember that series sums and limits are functions; without a single numeric output at a particular point, the function value at that point is considered to be non-existent.

2) NOT ALL INFINITIES ARE THE SAME:  There are two ways to show two groups are the same size.  The obvious way is to count the elements in each group and find out there is the same number of elements in each, but this works only if you have a finite group size.  Alternatively, you could a) match every element in group 1 with a unique element from group 2, and b) match every element in group 2 with a unique element from group 1.  It is important to do both steps here to show that there are no left-over, unpaired elements in either group.

So do the ellipses in (1) and (2) represent the same sets?  As the ellipses represent sets with an infinite number of elements, the first comparison technique is irrelevant.  For the second approach using pairing, we need to compare individual elements.  For every element in the ellipsis of (1), obviously there is an “partner” in (2) as the multiplication of (1) by r visually shifts all of the terms of the series right one position, creating the necessary matches.

Students often are troubled by the second matching as it appears the ellipsis in (2) contains an “extra term” from the right shift.  But, for every specific term you identify in (2), its identical twin exists in (1).  In the weirdness of infinity, that “extra term” appears to have been absorbed without changing the “size” of the infinity.

Since there is a 1:1 mapping of all elements in the ellipses of (1) and (2), you can conclude they are identical, and their difference is zero.

## Probability, Polynomials, and Sicherman Dice

Three years ago, I encountered a question on the TI-Nspire Google group asking if there was a way to use CAS to solve probability problems.  The ideas I pitched in my initial response and follow-up a year later (after first using it with students in a statistics class) have been thoroughly re-confirmed in my first year teaching AP Statistics.  I’ll quickly re-share them below before extending the concept with ideas I picked up a couple weeks ago from Steve Phelps’ session on Probability, Polynomials, and CAS at the 64th annual OCTM conference earlier this month in Cleveland, OH.

BINOMIALS:  FROM POLYNOMIALS TO SAMPLE SPACES

Once you understand them, binomial probability distributions aren’t that difficult, but the initial conjoining of combinatorics and probability makes this a perennially difficult topic for many students.  The standard formula for the probability of determining the chances of K successes in N attempts of a binomial situation where p is the probability of a single success in a single attempt is no less daunting:

$\displaystyle \left( \begin{matrix} N \\ K \end{matrix} \right) p^K (1-p)^{N-K} = \frac{N!}{K! (N-K)!} p^K (1-p)^{N-K}$

But that is almost exactly the same result one gets by raising binomials to whole number powers, so why not use a CAS to expand a polynomial and at least compute the $\displaystyle \left( \begin{matrix} N \\ K \end{matrix} \right)$ portion of the probability?  One added advantage of using a CAS is that you could use full event names instead of abbreviations, making it even easier to identify the meaning of each event.

The TI-Nspire output above shows the entire sample space resulting from flipping a coin 6 times.  Each term is an event.  Within each term, the exponent of each variable notes the number of times that variable occurs and the coefficient is the number of times that combination occurs.  The overall exponent in the expand command is the number of trials.  For example, the middle term– $20\cdot heads^3 \cdot tails^3$ –says that there are 20 ways you could get 3 heads and 3 tails when tossing a coin 6 times. The last term is just $tails^6$, and its implied coefficient is 1, meaning there is just one way to flip 6 tails in 6 tosses.

The expand command makes more sense than memorized algorithms and provides context to students until they gain a deeper understanding of what’s actually going on.

FROM POLYNOMIALS TO PROBABILITY

Still using the expand command, if each variable is preceded by its probability, the CAS result combines the entire sample space AND the corresponding probability distribution function.  For example, when rolling a fair die four times, the distribution for 1s vs. not 1s (2, 3, 4, 5, or 6) is given by

The highlighted term says there is a 38.58% chance that there will be exactly one 1 and any three other numbers (2, 3, 4, 5, or 6) in four rolls of a fair 6-sided die.  The probabilities of the other four events in the sample space are also shown.  Within the TI-Nspire (CAS or non-CAS), one could use a command to give all of these probabilities simultaneously (below), but then one has to remember whether the non-contextualized probabilities are for increasing or decreasing values of which binomial outcome.

Particularly early on in their explorations of binomial probabilities, students I’ve taught have shown a very clear preference for the polynomial approach, even when allowed to choose any approach that makes sense to them.

TAKING POLYNOMIALS FROM ONE DIE TO MANY

Given these earlier thoughts, I was naturally drawn to Steve Phelps “Probability, Polynomials, and CAS” session at the November 2014 OCTM annual meeting in Cleveland, OH.  Among the ideas he shared was using polynomials to create the distribution function for the sum of two fair 6-sided dice.  My immediate thought was to apply my earlier ideas.  As noted in my initial post, the expansion approach above is not limited to binomial situations.  My first reflexive CAS command in Steve’s session before he share anything was this.

By writing the outcomes in words, the CAS interprets them as variables.  I got the entire sample space, but didn’t learn gain anything beyond a long polynomial.  The first output– $five^2$ –with its implied coefficient says there is 1 way to get 2 fives.  The second term– $2\cdot five \cdot four$ –says there are 2 ways to get 1 five and 1 four.  Nice that the technology gives me all the terms so quickly, but it doesn’t help me get a distribution function of the sum.  I got the distributions of the specific outcomes, but the way I defined the variables didn’t permit sum of their actual numerical values.  Time to listen to the speaker.

He suggested using a common variable, X, for all faces with the value of each face expressed as an exponent.  That is, a standard 6-sided die would be represented by $X^1+X^2+ X^3+X^4+X^5+X^6$ where the six different exponents represent the numbers on the six faces of a typical 6-sided die.  Rolling two such dice simultaneously is handled as I did earlier with the binomial cases.

NOTE:  Exponents are handled in TWO different ways here.  1) Within a single polynomial, an exponent is an event value, and 2) Outside a polynomial, an exponent indicates the number of times that polynomial is applied within the specific event.  Coefficients have the same meaning as before.

Because the variables are now the same, when specific terms are multiplied, their exponents (face values) will be added–exactly what I wanted to happen.  That means the sum of the faces when you roll two dice is determined by the following.

Notice that the output is a single polynomial.  Therefore, the exponents are the values of individual cases.  For a couple examples, there are 3 ways to get a sum of 10 $\left( 3 \cdot x^{10} \right)$, 2 ways to get a sum of 3 $\left( 2 \cdot x^3 \right)$, etc.  The most commonly occurring outcome is the term with the largest coefficient.  For rolling two standard fair 6-sided dice, a sum of 7 is the most common outcome, occurring 6 times $\left( 6 \cdot x^7 \right)$.  That certainly simplifies the typical 6×6 tables used to compute the sums and probabilities resulting from rolling two dice.

While not the point of Steve’s talk, I immediately saw that technology had just opened the door to problems that had been computationally inaccessible in the past.  For example, what is the most common sum when rolling 5 dice and what is the probability of that sum?  On my CAS, I entered this.

In the middle of the expanded polynomial are two terms with the largest coefficients, $780 \cdot x^{18}$ and $780 \cdot x^{19}$, meaning a sums of 17 and 18 are the most common, equally likely outcomes when rolling 5 dice.  As there are $6^5=7776$ possible outcomes when rolling a die 5 times, the probability of each of these is $\frac{780}{7776} \approx 0.1003$, or about 10.03% chance each for a sum of 17 or 18.  This can be verified by inserting the probabilities as coefficients before each term before CAS expanding.

With thought, this shouldn’t be surprising as the expected mean value of rolling a 6-sided die many times is 3.5, and $5 \cdot 3.5 = 17.5$, so the integers on either side of 17.5 (17 & 18) should be the most common.  Technology confirms intuition.

ROLLING DIFFERENT DICE SIMULTANEOUSLY

What is the distribution of sums when rolling a 4-sided and a 6-sided die together?  No problem.  Just multiply two different polynomials, one representative of each die.

The output shows that sums of 5, 6, and 7 would be the most common, each occurring four times with probability $\frac{1}{6}$ and together accounting for half of all outcomes of rolling these two dice together.

A BEAUTIFUL EXTENSION–SICHERMAN DICE

My most unexpected gain from Steve’s talk happened when he asked if we could get the same distribution of sums as “normal” 6-sided dice, but from two different 6-sided dice.  The only restriction he gave was that all of the faces of the new dice had to have positive values.  This can be approached by realizing that the distribution of sums of the two normal dice can be found by multiplying two representative polynomials to get

$x^{12}+2x^{11}+3x^{10}+4x^9+5x^8+6x^7+5x^6+4x^5+3x^4+2x^3+x^2$.

Restating the question in the terms of this post, are there two other polynomials that could be multiplied to give the same product?  That is, does this polynomial factor into other polynomials that could multiply to the same product?  A CAS factor command gives

Any rearrangement of these eight (four distinct) sub-polynomials would create the same distribution as the sum of two dice, but what would the the separate sub-products mean in terms of the dice?  As a first example, what if the first two expressions were used for one die (line 1 below) and the two squared trinomials comprised a second die (line 2)?

Line 1 actually describes a 4-sided die with one face of 4, two faces with 3s, and one face of 2.  Line 2 describes a 9-sided die (whatever that is) with one face of 8, two faces of 6, three faces of 4, two faces of 2, and one face with a 0 ( $1=1 \cdot x^0$).  This means rolling a 4-sided and a 9-sided die as described would give exactly the same sum distribution.  Cool, but not what I wanted.  Now what?

Factorization gave four distinct sub-polynomials, each with multitude 2.  One die could contain 0, 1, or 2 of each of these with the remaining factors on the other die.  That means there are $3^4=81$ different possible dice combinations.  I could continue with a trail-and-error approach, but I wanted to be more efficient and elegant.

What follows is the result of thinking about the problem for a while.  Like most math solutions to interesting problems, ultimate solutions are typically much cleaner and more elegant than the thoughts that went into them.  Problem solving is a messy–but very rewarding–business.

SOLUTION

Here are my insights over time:

1) I realized that the $x^2$ term would raise the power (face values) of the desired dice, but would not change the coefficients (number of faces).  Because Steve asked for dice with all positive face values.  That meant each desired die had to have at least one x to prevent non-positive face values.

2) My first attempt didn’t create 6-sided dice.  The sums of the coefficients of the sub-polynomials determined the number of sides.  That sum could also be found by substituting $x=1$ into the sub-polynomial.  I want 6-sided dice, so the final coefficients must add to 6.  The coefficients of the factored polynomials of any die individually must add to 2, 3, or 6 and have a product of 6.  The coefficients of $(x+1)$ add to 2, $\left( x^2+x+1 \right)$ add to 3, and $\left( x^2-x+1 \right)$ add to 1.  The only way to get a polynomial coefficient sum of 6 (and thereby create 6-sided dice) is for each die to have one $(x+1)$ factor and one $\left( x^2+x+1 \right)$ factor.

3) That leaves the two $\left( x^2-x+1 \right)$ factors.  They could split between the two dice or both could be on one die, leaving none on the other.  We’ve already determined that each die already had to have one each of the x, $(x+1)$, and $\left( x^2+x+1 \right)$ factors.  To also split the $\left( x^2-x+1 \right)$ factors would result in the original dice:  Two normal 6-sided dice.  If I want different dice, I have to load both of these factors on one die.

That means there is ONLY ONE POSSIBLE alternative for two 6-sided dice that have the same sum distribution as two normal 6-sided dice.

One die would have single faces of 8, 6, 5, 4, 3, and 1.  The other die would have one 4, two 3s, two 2s, and one 1.  And this is exactly the result of the famous(?) Sicherman Dice.

If a 0 face value was allowed, shift one factor of x from one polynomial to the other.  This can be done two ways.

The first possibility has dice with faces {9, 7, 6, 5, 4, 2} and {3, 2, 2, 1, 1, 0}, and the second has faces {7, 5, 4, 3, 2, 0} and {5, 4, 4, 3, 3, 2}, giving the only other two non-negative solutions to the Sicherman Dice.

Both of these are nothing more than adding one to all faces of one die and subtracting one from from all faces of the other.  While not necessary to use polynomials to compute these, they are equivalent to multiplying the polynomial of one die by x and the other by $\frac{1}{x}$ as many times as desired. That means there are an infinite number of 6-sided dice with the same sum distribution as normal 6-sided dice if you allow the sides to have negative faces.  One of these is

corresponding to a pair of Sicherman Dice with faces {6, 4, 3, 2, 1, -1} and {1,5,5,4,4,3}.

CONCLUSION:

There are other very interesting properties of Sicherman Dice, but this is already a very long post.  In the end, there are tremendous connections between probability and polynomials that are accessible to students at the secondary level and beyond.  And CAS keeps the focus on student learning and away from the manipulations that aren’t even the point in these explorations.

Enjoy.

## Number Bases and Polynomials

About a month ago, I was working with our 5th grade math teacher to develop some extension activities for some students in an unleveled class.  The class was exploring place value, and I suggested that some might be ready to explore what happens when you allow the number base to be something other than 10.  A few students had some fun learning to use their basic four algorithms in other number bases, but I made an even deeper connection.

When writing something like 512 in expanded form ($5\cdot 10^2+1\cdot 10^1+2\cdot 10^0$), I realized that if the 10 was an x, I’d have a polynomial.  I’d recognized this before, but this time I wondered what would happen if I applied basic math algorithms to polynomials if I wrote them in a condensed numerical form, not their standard expanded form.  That is, could I do basic algebra on $5x^2+x+2$ if I thought of it as $512_x$–a base-x “number”?  (To avoid other confusion later, I read this as “five one two base-x“.)

Following are some examples I played with to convince myself how my new notation would work.  I’m not convinced that this will ever lead to anything, but following my “what ifs” all the way to infinite series was a blast.  Read on!

If I wanted to add $(3x+5)$$(2x^2+4x+1)$, I could think of it as $35_x+241_x$ and add the numbers “normally” to get $276_x$ or $2x^2+7x+6$.  Notice that each power of x identifies a “place value” for its characteristic coefficient.

If I wanted to add $3x-7$ to itself, I had to adapt my notation a touch.  The “units digit” is a negative number, but since the number base, x, is unknown (or variable), I ended up saying $3x-7=3(-7)_x$.  The parentheses are used to contain multiple characters into a single place value.  Then, $(3x-7)+(3x-7)$ becomes $3(-7)_x+3(-7)_x=6(-14)_x$ or $6x-14$.  Notice the expanding parentheses containing the base-x units digit.

The last example also showed me that simple multiplication would work.  Adding $3x-7$ to itself is equivalent to multiplying $2\cdot (3x-7)$.  In base-x, that is $2\cdot 3(-7)_x$.  That’s easy!  Arguably, this might be even easier that doubling a number when the number base is known.  Without interactions between the coefficients of different place values, just double each digit to get $6(-14)_x=6x-14$, as before.

What about $(x^2+7)+(8x-9)$?  That’s equivalent to $107_x+8(-9)_x$.  While simple, I’ll solve this one by stacking.

and this is $x^2+8x-2$.  As with base-10 numbers, the use of 0 is needed to hold place values exactly as I needed a 0 to hold the $x^1$ place for $x^2+7$. Again, this could easily be accomplished without the number base conversion, but how much more can we push these boundaries?

Level 3–Multiplication & Powers:

Compute $(8x-3)^2$.  Stacking again and using a modification of the multiply-and-carry algorithm I learned in grade school, I got

and this is equivalent to $64x^2-48x+9$.

All other forms of polynomial multiplication work just fine, too.

From one perspective, all of this shifting to a variable number base could be seen as completely unnecessary.  We already have acceptably working algorithms for addition, subtraction, and multiplication.  But then, I really like how this approach completes the connection between numerical and polynomial arithmetic.  The rules of math don’t change just because you introduce variables.  For some, I’m convinced this might make a big difference in understanding.

I also like how easily this extends polynomial by polynomial multiplication far beyond the bland monomial and binomial products that proliferate in virtually all modern textbooks.  Also banished here is any need at all for banal FOIL techniques.

Level 4–Division:

What about $x^2+x-6$ divided by $x+3$? In base-x, that’s $11(-6)_x \div 13_x$. Remembering that there is no place value carrying possible, I had to be a little careful when setting up my computation. Focusing only on the lead digits, 1 “goes into” 1 one time.  Multiplying the partial quotient by the divisor, writing the result below and subtracting gives

Then, 1 “goes into” -2 negative two times.  Multiplying and subtracting gives a remainder of 0.

thereby confirming that $x+3$ is a factor of $x^2+x-6$, and the other factor is the quotient, $x-2$.

Perhaps this could be used as an alternative to other polynomial division algorithms.  It is somewhat similar to the synthetic division technique, without its  significant limitations:  It is not limited to linear divisors with lead coefficients of one.

For $(4x^3-5x^2+7) \div (2x^2-1)$, think $4(-5)07_x \div 20(-1)_x$.  Stacking and dividing gives

So $\displaystyle \frac{4x^3-5x^2+7}{2x^2-1}=2x-2.5+\frac{2x+4.5}{2x^2-1}$.

CONCLUSION

From all I’ve been able to tell, converting polynomials to their base-x number equivalents enables you to perform all of the same arithmetic computations.  For division in particular, it seems this method might even be a bit easier.

In my next post, I push the exploration of these base-x numbers into infinite series.

## Extending graph control

This article takes my idea from yesterday’s post about using $g(x)=\sqrt \frac{\left | x \right |}{x}$ to control the appearance of a graph and extends it in two ways.

• Part I below uses Desmos to graph $y=(x+2)^3x^2(x-1)$ from the left and right simultaneously
• Part II was inspired by my Twitter colleague John Burk who asked if this control could be extended in a different direction.

Part I: Simultaneous Control

When graphing polynomials like $y=(x+2)^3x^2(x-1)$, I encourage my students to use both its local behavior (cubic root at $x=-2$, quadratic root at $x=0$, and linear root at $x=1$) and its end behavior (6th degree polynomial with a positive lead coefficient means $y\rightarrow +\infty$ as $x\rightarrow\pm\infty$). To start graphing, I suggest students plot points on the x-intercepts and then sketch arrows to indicate the end behavior.  In the past, this was something we did on paper, but couldn’t get technology to replicate it live–until this idea.

In class last week, I used a minor extension of yesterday’s idea to control a graph’s appearance from the left and right simultaneously.  Yesterday’s post suggested  multiplying  by $\sqrt \frac{\left | a-x \right |}{a-x}$ to show the graph of a function from the left for $x.  Creating a second graph multiplied by $\sqrt \frac{\left | x-b \right |}{x-b}$ gives a graph of your function from the right for $b.  The following images show the polynomial’s graph developing in a few stages.  You can access the Desmos file here.

First graph the end behavior (pull the a and b sliders in a bit to see just the ends of the graph) and plot points at the x-intercepts.

From here, you could graph left-to-right or right-to-left.  I’ll come in from the right to show the new right side controller. The root at $x=1$ is linear, so decreasing the b slider to just below 1 shows this.

Continuing from the right, the next root is a bounce at $x=0$, as shown by decreasing the b slider below 0.  Notice that this forces a relative minimum for some $0.

Just because it’s possible, I’ll now show the cubic intercept at $x=2$ by increasing the a slider above 2.

All that remains is to connect the two sides of the graph, creating one more relative minimum in $-2.

The same level of presentation control can be had for any function’s graph.

Part II: Vertical Control

I hadn’t thought to extend this any further until my colleague asked if a graph could be controlled up and down instead of left and right.  My guess is that the idea hadn’t occurred to me because I typically think about controlling a function through its domain.  Even so, a couple minor adjustments accomplished it.  Click here to see a vertical control of the graph of $y=x^3$ from above and below.

Enjoy.

## Quadratics, Statistics, Symmetry, and Tranformations

A problem I assigned my precalculus class this past Thursday ended up with multiple solutions by the time we finished.  Huzzah for student creativity!

The question:

Find equations for all polynomial functions, $y=f(x)$, of degree $\le 2$ for which $f(0)=f(1)=2$ and $f(3)=0$.

After they had worked on this (along with several variations on the theme), four very different ways of thinking about this problem emerged.  All were valid and even led to a lesson I hadn’t planned–proving that, even though they looked different algebraically, all were equivalent.  I present their approaches (and a few extras) in the order they were offered in our post-solving debriefing.

The commonality among the approaches was their recognition that 3 non-collinear points uniquely define a vertical parabola, so they didn’t need to worry about polynomials of degree 0 or 1.  (They haven’t yet heard about rotated curves that led to my earlier post on rotated quadratics.)

Solution 1–Regression:  Because only 3 points were given, a quadratic regression would derive a perfectly fitting quadratic equation.  Using their TI-Nspire CASs, they started by entering the 3 ordered pairs in a Lists&Spreadsheets window.  Most then went to a Calculator window to compute a quadratic regression.  Below, I show the same approach using a Data&Statistics window instead so I could see simultaneously the curve fit and the given points.

The decimals were easy enough to interpret, so even though they were presented in decimal form, these students reported $y=-\frac{1}{3}x^2+\frac{1}{3}x+2$.

For a couple seconds after this was presented, I honestly felt a little cheated.  I was hoping they would tap the geometric or algebraic properties of quadratics to get their equations.  But I then I remembered that I clearly hadn’t make that part of my instructions.  After my initial knee-jerk reaction, I realized this group of students had actually done exactly what I explicitly have been encouraging them to do: think freely and take advantage of every tool they have to find solutions.  Nothing in the problem statement suggested technology or regressions, so while I had intended a more geometric approach, I realized I actually owed these students some kudos for a very creative, insightful, and technology-based solution.  This and Solution 2 were the most frequently chosen approaches.

Solution 2–Systems:  Equations of quadratic functions are typically presented in standard, factored, or vertex form.  Since neither two zeros nor the vertex were explicitly given, the largest portion of the students used the standard form, $y=a\cdot x^2+b\cdot x+c$ to create a 3×3 system of equations.  Some solved this by hand, but most invoked a CAS solution.  Notice the elegance of the solve command they used, working from the generic polynomial equation that kept them from having to write all three equations, keeping their focus on the form of the equation they sought.

This created the same result as Solution 1, $y=-\frac{1}{3}x^2+\frac{1}{3}x+2$.

CAS Aside: No students offered these next two solutions, but I believe when using a CAS, it is important for users to remember that the machine typically does not care what output form you want.  The standard form is the only “algebraically simple” approach when setting up a solution by hand, but the availability of technology makes solving for any form equally accessible.

The next screen shows that the vertex and factored forms are just as easily derived as the standard form my students found in Solution 2.

I was surprised when the last line’s output wasn’t in vertex form, $y=-\frac{1}{3}\cdot \left ( x-\frac{1}{2} \right )^2+\frac{25}{12}$, but the coefficients in its expanded form clearly show the equivalence between this form and the standard forms derived in Solutions 1 and 2–a valuable connection.

Solution 3–Symmetry:  Two students said they noticed that $f(0)=f(1)=2$ guaranteed the vertex of the parabola occurred at $x=\frac{1}{2}$.  Because $f(3)=0$ defined one real root of the unknown quadratic, the parabola’s symmetry guaranteed another at $x=-2$, giving potential equation $y=a\cdot (x-3)(x+2)$.  They substituted the given (0,2) to solve for a, giving final equation $y=-\frac{1}{3}\cdot (x-3)(x+2)$ as confirmed by the CAS approach above.

Solution 4–Transformations:  One of the big lessons I repeat in every class I teach is this:

If you don’t like how a question is posed.  Change it!

Notice that two of the given points have the same y-coordinate.  If that y-coordinate had been 0 (instead of its given value, 2), a factored form would be simple.  Well, why not force them to be x-intercepts by translating all of the given points down 2 units?

The transformed data show x-intercepts at 0 and 1 with another ordered pair at $(3,-2)$.  From here, the factored form is easy:  $y=a\cdot (x-0)(x-1)$.  Substituting $(3,-2)$ gives $a=-\frac{1}{3}$ and the final equation is $y=-\frac{1}{3}\cdot (x-0)(x-1)$ .

Of course, this is an equation for the transformed points.  Sliding the result back up two units, $y=-\frac{1}{3}\cdot (x-0)(x-1)+2$, gives an equation for the given points.  Aside from its lead coefficient, this last equation looked very different from the other forms, but some quick expansion proved its equivalence.

Conclusion:  It would have been nice if someone had used the symmetry noted in Solution 3 to attempt a vertex-form answer via systems.  Given the vertex at $x=\frac{1}{2}$ with an unknown y-coordinate, a potential equation is $y=a\cdot \left ( x-\frac{1}{2} \right )^2+k$.  Substituting $(3,0)$ and either $(0,2)\text{ or }(1,2)$ creates a 2×2 system of linear equations, $\left\{\begin{matrix} 0=a\cdot \left ( 3-\frac{1}{2} \right )^2+k \\ 2=a\cdot \left ( 0-\frac{1}{2} \right )^2+k \end{matrix}\right.$.  From there, a by-hand or CAS solution would have been equally acceptable to me.

That the few alternative approaches I offered above weren’t used didn’t matter in the end.  My students were creative, followed their own instincts to find solutions that aligned with their thinking, and clearly appreciated the alternative ways their classmates used to find answers.  Creativity and individual expression reigned, while everyone broadened their understanding that there’s not just one way to do math.

It was a good day.

## Cubics and CAS

Here’s a question I posed to one of my precalculus classes for homework at the end of last week along with three solutions we developed.

Suppose the graph of a cubic function has an inflection point at (1,3) and passes through (0,-4).

1. Name one other point that MUST be on the curve, and
2. give TWO different cubic equations that would pass through the three points.

SOLUTION ALERT!  Don’t read any further if you want to solve this problem for yourself.

The first question relies on the fact that every cubic function has 180 degree rotational symmetry about its inflection point.  This is equivalent to saying that the graph of a cubic function is its own image when the function’s graph is reflected through its inflection point.

That means the third point is the image of (0,-4) when point-reflected through the inflection point (1,3), which is the point (2,10) as shown graphically below.

From here, my students came up with 2 different solutions to the second question and upon prodding, we created a third.

SOLUTION 1:  Virtually every student assumed $y=a\cdot x^3$ was the parent function of a cubic with unknown leading coefficient whose “center” (inflection point) had been slid to (1,3).  Plugging in the given (0,-4) to $(y-3)=a\cdot (x-1)^3$ gives $a=7$.  Here’s their graph.

SOLUTION 2:  Many students had difficulty coming up with another equation.  A few could sketch in other cubic graphs (curiously, all had positive lead coefficients) that contained the 3 points above, but didn’t know how to find equations.  That’s when Sara pointed out that if the generic expanded form of a cubic was $a\cdot x^3+b\cdot x^2 +c\cdot x+d$ , then any 4 ordered pairs with unique x-coordinates should define a unique cubic.  That is, if we picked any 4th point with x not 0, 1, or 2, then we should get an equation.  That this would create a 4×4 system of equations didn’t bother her at all.  She knew in theory how to solve such a thing, but she was thinking on a much higher plane.  Her CAS technology expeditiously did the grunt work, allowing her brain to keep moving.

A doubtful classmate called out, “OK.  Make it go through (100,100).”  Following is a CAS screen roughly duplicating Sara’s approach and a graph confirming the fit.  The equation was onerous, but with a quick copy-paste, Sara had moved from  idea to computation to ugly equation and graph in just a couple minutes.  The doubter was convinced and the “wow”s from throughout the room conveyed the respect for the power of a properly wielded CAS.

In particular, notice how the TI-Nspire CAS syntax in lines 1 and 3 keep the user’s focus on the type of equation being solved and eliminates the need to actually enter 4 separate equations.  It doesn’t always work, but it’s a particularly lovely piece of scaffolding when it does.

SOLUTION 3:  One of my goals in Algebra II and Precalculus courses is to teach my students that they don’t need to always accept problems as stated.  Sometimes they can change initial conditions to create a much cleaner work environment so long as they transform their “clean” solution back to the state of the initial conditions.

In this case, I asked what would happen if they translated the inflection point using $T_{-1,-3}$ to the origin, making the other given point (-1,-7).  Several immediately called the 3rd point to be (1,7) which “untranslating” — $T_{1,3}(1,7)=(2,10)$ — confirmed to be the earlier finding.

For cases where the cubic had another real root at $x=r$, then symmetry immediately made $x=-r$ another root, and a factored form of the equation becomes $y=a\cdot (x)(x-r)(x+r)$ for some value of a.  Plugging in (-1,-7) gives a in terms of r.

The last line slid the initially translated equation using $T_{1,3}$ to re-position the previous line according to the initial conditions.  While unasked for, notice how the CAS performed some polynomial division on the right-side expression.

I created a GeoGebra document with a slider for the root using the equation from the last line of the CAS image above.  The image below shows one possible position of the retranslated root.  If you want to play with a live version of this, you will need a free copy of GeoGebra to run it, but the file is here.

What’s nice here is how the problem became one of simple factors once the inflection point was translated to the origin.  Notice also that the CAS version of the equation concludes with $+7x-4$, the line containing the original three points.  This is nice for two reasons.  The numerator of the rational term is $-7x(x-2)(x-1)$ which zeros out the fraction at x=0, 1, or 2, putting the cubic exactly on the line $y=7x-4$ at those points.

The only r-values are in the denominator, so as $r\rightarrow\infty$, the rational term also becomes zero.  Graphically, you can see this happen as the cubic “unrolls” onto $y=7x-4$ as you drag $|x|\rightarrow\infty$.  Essentially, this shows both graphically and algebraically that $y=7x-4$ is the limiting degenerate curve the cubic function approaches as two of its transformed real roots grow without bound.

## Recognizing Patterns

I’ve often told my students that the best problem solvers are those who recognize patterns from past problems in new situations.  So, the best way to become a better problem solver is to solve lots of problems, study the ways others have solved the problems you’ve already cracked (or at least attempted), and to keep pushing your boundaries because you never know what parts of what you learn may end up providing unexpected future insights.

I lay no claims to being a great problem solver, but I absolutely benefited from problem solving exposure when I encountered @jamestanton‘s latest “Playing with Numbers” puzzler from his May 2012 Cool Math Newsletter.  (Click here to access all of Jim’s newsletters).  (BTW, Jim’s Web page is chock full of amazing videos and insights both on the problem-solving front and for those interested in curriculum discussions.)  Here’s what Jim posed:

Write the numbers 1 though 10 on the board. Pick any two numbers, erase them, and replace them with the single number given by their sum plus their product. (So, if you choose to erase the numbers a and b, replace them with a + b + ab .) You now have nine numbers on the board.

Do this again: Pick any two numbers, erase them, and replace them with their sum plus product. You now have eight numbers on the board.

Do this seven more times until you have a single number on the board.

Why do all who play this game end up with the same single number at the end?  What is that final number?

SOLUTION ALERT:  Don’t read any further if you want to solve this yourself.

I started small and general.  If the first two numbers (a and b) produce $a+b+ab$, then adding c to the original list gives $[(a+b+ab)+c]+ [(a+b+ab)\cdot c]$ which can be rewritten as $a+b+c+ab+ac+bc+abc$.  That’s when the intuition struck.  Notice that the rewritten form is the sum of every individual number in the list, AND every possible pair of those numbers, AND concludes with the product of the three numbers. I’ve seen that pattern before!

Given an nth degree polynomial with roots $r_1, r_2, \ldots r_n$, it’s factored form is $(x-r_1)\cdot (x-r_2)\cdot\ldots\cdot (x-r_n)$ which can be expanded and rewritten as

$x^n-(r_1+r_2+\ldots+r_n)\cdot x^{n-1}+$
$+(\text{every pair-wise product of } r_1 \ldots r_n)\cdot x^{n-2}-$
$-(\text{every three-way product of } r_1 \ldots r_n)\cdot x^{n-3}+\ldots$
$\ldots \pm (r_1\cdot r_2\cdot \ldots\cdot r_n)$

Where the sign of the final term is positive for even n and negative for odd n.  I saw this in many algebra textbooks when I started teaching over 20 years ago, but haven’t encountered it lately.  Then again, I haven’t been looking for it.

Here’s the point … other than the alternating signs, the coefficients of the expanded polynomial are exactly identical to the sums I was getting from Jim’s problem.  That’s when I rewrote my original problem on my CAS to $(x+a)\cdot (x+b)\cdot (x+c)$ and expanded it (see line 1 below).  The coefficients of $x^2$ are the individual numbers, the coefficients of $x$ are all the pair-wise combinations, and the constant term drifting off the end of the line is $a\cdot b\cdot c$.  And I eliminated the $\pm$ sign challenge by individually adding the numbers instead of subtracting them as I had in the polynomial root example that inspired my insight.  Let $x=1$ to clean up and make the pattern more obvious.

So, creating a polynomial with the given numbers and substituting $x=1$ will create a number one more than the sum I wanted (note the extra +1 resulting from the coefficient of $x^n$).  Using pi notation, substituting $x=1$, and subtracting 1 gives the answer.  In case you didn’t know, pi notation works exactly the same as sigma notation, but you multiply the terms instead of adding them.

The solution–39,916,799–is surprisingly large given the initial problem statement, and while my CAS use confirmed my intuition and effortlessly crunched the numbers, its tendency to multiply numbers whenever encountered has actually hidden something pretty. From the top line of the last image, substituting $x=1$ would have created the product $2\cdot 3\cdot 4\cdot\ldots \cdot 11$ which the penultimate line computed to be 39916800.  But before the product, that number was $11!$, making $11!-1$ a far more revealing version of the solution.

MORAL:  Even after you have an answer, take some time to review what has happened to give yourself a chance to learn even more.

EXTENSION 1:  Instead of 1 to 10 as the initial numbers, what if the list went from 1 to any positive integer n?  Prove that the final number on the board is $(n+1)!-1$.

EXTENSION 2:  While a positive integer sequence starting at 1 (or 0) produces a nice factorial in the answer, this approach can be used with any number list.  For example, follow Jim’s rules with 37, 5, -2, and 7.9.  Use the polynomial approach below for a quick solution of -2030.2.  Confirm using the original rules if you need.

EXTENSION 3:  By now it should be obvious that any list of numbers can be used in this problem.  Prove that every list of numbers which includes -1 has the same solution.

EXTENSION 4:  Before explaining some lovely extensions of problems like this to generalized commutativity and associativity, Jim’s May 12 Cool Math Newsletter asks what would happen if instead of $a+b+ab$, the rule for combining a and b was $\displaystyle\frac{a\cdot b}{a+b}$.  You can show that with three terms, this would become $\displaystyle\frac{abc}{ab+ac+bc}$, and four terms would give $\displaystyle\frac{abcd}{abc+abd+acd+bcd}$. In other words, this rule would morph n original numbers into a fraction whose numerator is the product of the numbers and whose denominator is the sum of all possible products of any (n-1)-sized groups of those numbers.  For the original integers 1 to 10, I know the numerator and denominator terms are the last two coefficients in a polynomial expansion.

The final fraction simplifies, but I think $\displaystyle\frac{3628800}{10628640}$ is slightly more informative.

Happy thoughts, problems, solutions, and connections to you all.