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.