## Mathochism: The same B

One woman’s attempt to revisit the math that plagued her in school. But can determination make up for 25 years of math neglect?

*UPDATED WITH TEST PROBLEMS THAT ARE PROBABLY QUITE EASY, YET BEFUDDLED ME EARLIER IN THE WEEK*
I got my test back, and, as it turns out, I suck at guessing.

I got all three of the questions that were puzzling me wrong, plus a bonus one I really thought I’d gotten right. I clearly need to go over bounds and Cartesian signs again!

So — (sigh) — another B.

The Youthful Professor refused to go over the test (he may have caught the predatory look in my eye) outside of office hours. So I suppose I will have to stalk him next week. I still can’t figure out what I did wrong.

But we have moved on. And moving on means revisiting the dreaded logarithm. Ick, logarithms. I was delighted to find, however, that they seem less puzzling this time around. I like the way this instructor broke them down.

He started by revisiting inverse functions, where you take a polynomial, swap the xs for ys, then solve again for y. Such equations, when graphed, tend to be mirror images of each other.

Then, he introduced the concept of y = b to the x, graphed that (kind of a swoopy line), and set out the rules for b. B cannot equal zero, or less than zero. B cannot equal 1. B, however, can equal something between zero and 1.

Then, he introduced y = b to the x’s inverse, which, surprise!, is the the logarithm. So, log sub b of y equals x is the inverse of b to the x equals y. And here’s the beauty of it: the Bs are the same! And they follow the same rules as above, so logs cannot equal 1, zero or less than zero (no negs here, folks), but they can land somewhere between 1 and zero.

This is the kind of tidy explanation I love, and the kind of tidy explanation I’ve missed getting since the days of the Dapper Professor. Yes, math can be complicated, and sometimes messy. But it is moments like these that keep me going. And make me aspire to more than a B.

And now, at the request of helpful commenter Dave W.:

1) Use Descartes’ rule of signs to find how many positive or negative zeros are in this polynomial:
x^8 – x^5 + x^4 – x^3 + x^2 – x + 8

Answer: No negatives, 0, 2, 4 or 6 positives.

Now, I count only 5, because I thought the constant at the end didn’t count. But maybe they do?

2) There is a guaranteed zero in p(x) = x^3 + 2x^2 – x + 1. In which interval can you find it?

Looking at this problem, it thought it was a bounds thing, as in the second theorem of bounds, which was all YP required for this exam. The biggest coefficient is 2. Put that over the leading coefficient, which is 1, and you get 2. Add 1, and you get 3. Shouldn’t the interval then be [-3.3}? Where does the -2 come in?

All text copyrighted by A.K. Whitney, and cannot be used without permission.

• Can you reproduce the questions and your answers here? Either typographically, or just a jpeg or something? If you could, some of us might be able to comment usefully about them (says the former math tutor). A polynomial like x cubed minus 3 x squared plus 4 can be represented typographically as x^3-3x^2+4.

• akwhitney

I suppose I could. I’d like to puzzle over it a bit longer, though. One problem involves the second rule of bounds, which I thought was the highest number over the leading coefficient plus one. But apparently not?
I’m on deadlines for journo work today, though, so I’m taking a break from pre-calc for a few hours. I appreciate the offer, and will post the questions I just don’t get soon!

• I’m not sure about what the “second rule of bounds” is in your source material – I’m pretty sure it’s just a local name for something in your text or course notes, not a universal name like “Decartes’ Rule of Signs”. Can you describe what you think it does? The calculation you describe doesn’t sound familiar to me, so I’m guessing you’ve probably remembered it slightly wrong.

• This (http://oakroadsystems.com/math/polysol.htm) looks like a pretty good web page on solving polynomial equations, with a bunch of useful links.

• akwhitney

Thanks, I’ll take a look at it! And I will get back to this and update with a few questions as soon as I’m off deadline.

• If you don’t mind my asking, was #2 a multiple choice question? If so, then it’s a nice little application of the Intermediate Value Theorem. I just taught on that a few weeks ago, so feel free to pick my brain once you’re off deadline.

(It’s also entirely possible that there is another solution to this problem, as with so many other problems in mathematics!)

• akwhitney

It was multiple choice!
The thing about the Intermediate Value Theorem is that the book (and the professor) just sort of glossed over it, and gave problems and examples that gave you a polynomial, some values and then expected you to find an interval. It wasn’t the other way around. How do you apply it the other way?

• 1) Yes, the constant counts when you are counting sign changes. One suggestion that I have found helpful over the years: if you are faced with remembering a general rule and don’t remember some detail (like whether or not the constant counts here), see if you can construct a really simple example where that detail applies and see which interpretation you need to make it work.

In this case, consider the polynomial x-1, which has a positive root at 1. It has one sign change if you count the constant, none if you don’t. So for Decartes’ Rule to apply, you need to count the constant here to get the one sign change that would allow for a single positive root, which should tell you that you need to count the constant in general. If you like, you can go on to look at slightly more complicated examples, like (x+2)(x-1) = x^2+x-2. That has roots at -2 and +1 (by construction from the factors), so it has one positive root, and again, one sign change only if you count the constant.

2) What the Intermediate Value Theorem says is that if you have a continuous function f(x) (and all polynomials are continuous) such that for one value a, f(a) 0, then there must be some value c between a and b such that f(c) = 0. Here, if you check the endpoints of the interval [-3, -2], you find that p(-3) = -5 and p(-2) = +3. So the conditions of the theorem apply, and you can conclude that there must be a root between -3 and -2. If this was multiple choice, then you should have found for all the other intervals that p(x) was either positive at both endpoints or negative at both endpoints.

“Continuous function” is one of those terms that gets defined more precisely when you take calculus, but intuitively, it basically means that you can trace its plot over any given interval without lifting your pen from the paper – there are no gaps, jumps, or holes anywhere the function is defined. So here, you know that the function is below the x-axis at x=-3, and above the x-axis at x=-2, and (because all polynomials are continuous), you should be able to trace the graph of the function from one point to the other without lifting your pen from the paper. That means that there has to be some point (maybe more than one, but at least one) in between the two points where your pen actually crosses the x-axis – and that point will be a root of the polynomial, the point c that I mentioned above. The theorem doesn’t tell you exactly where that point is, it just says it has to happen somewhere in that interval. That’s the intuition behind the Intermediate Value Theorem.

• It looks like the formulas in my statement of the Intermediate Value Theorem got misparsed as html. Let’s try that again:

What the Intermediate Value Theorem says is that if you have a continuous function f(x) (and all polynomials are continuous) such that for one value a, f(a) &lt 0, and for another value b, f(b) &gt 0, then there must be some value c between a and b such that f(c) = 0.

• Ok, let’s say it in words:

What the Intermediate Value Theorem says is that if you have a continuous function f(x) (and all polynomials are continuous) such that for one value a, f(a) is less than 0, and for another value b, f(b) is greater than 0, then there must be some value c between a and b such that f(c) = 0.

• Another thing: have you covered synthetic division in your class? If not, or if you need some review, there are some links about synthetic division from that web page I cited earlier.

You can use synthetic division not only to verify roots of a polynomial, but also to evaluate the polynomial at any value fairly quickly (when you use r as the divisor, the remainder at the end is f(r)). If you understand how it works, it’s possible to use the properties of synthetic division to explain why Descartes’ Rule works.

Also, I can use the properties of synthetic division to show that 1 plus (the absolute value of the largest coefficient divided by the leading coefficient) satisfies the upper bound rule they give on that web page, and that the negative of that value satisfies the lower bound rule. So yeah, that would mean that on problem 2, any real roots of p(x) would have to fall in the interval [-3, 3], which is I think what your second theorem of bounds is saying. Note that this doesn’t guarantee that there are any roots at all in this interval, it just says that *if* there are real roots, they have to fall in there. To guarantee a root, you would have to evaluate it at the endpoints and use the intermediate value theorem, as I explained above. In this case, it would show that there is a root in [-3, 3] as you expected, but given that it was multiple choice, the fact that you can guarantee a root in [-3, -2] provides a tighter bound and the best answer to the problem.

• akwhitney

Thanks, Dave W. I do know how to do synthetic division, but must admit I prefer long division (it’s a matrix thing). I’m realizing what happened here was that I had an incomplete understanding of both Descartes’ rule of signs and the Intermediate Value Theorem. Like I said in a previous entry, this current book is not very helpful. And the professor is good, but I should have asked him more questions to make sure I had it all clear in my head. I wasn’t shy on that when it came to end behavior or inverse functions, and I aced those!

• Any chance you could buy/borrow/check out a copy of your previous book? If its explanations are better than your current text, it might make a good supplement to what you have.