The latest Problem of the Week has now been posted. This one (and next week's as well) involve calculus. Fun!
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My recent travels, to Parsippany, NJ via Baltimore, MD, which involved three talks in two days, followed by multiple games of chess, bookended by two long drives, came to a dramatic conlcusion yesterday when I had to drive home in the snow. Not fun! There was so much snow on the road that you…
Better late than never! A small technical SNAFU yesterday interfered with the well-oiled machine that is Problem of the Week. But now we're back on track! The fourth problem has now been posted. The official solution to last week's problem will be up by tomorrow.
On Thursday I'll be heading up to Baltimore to give a talk at Johns Hopkins University. I'll be discussing an old favorite: The Monty Hall Problem! Actually, it's been about two years since I've given a talk on that particular subject, so it will be nice to have an excuse to revisit it.
From…
The second Problem of the Week has now been posted. It's a harder version of last week's problem, but perhaps it is good for a bit of amusement. By its nature, it might be a bit hard to describe the solution in a comment, but have a look anyway.
What goes wrong here is that the number of terms on the right-hand side is not constant, so a correct differentiation requires you to apply the chain rule. The derivative of (1 + 1 + ... + 1) [x times] with respect to x is 1, not 0. So the derivative of (x + x + ... + x) [x times] is actually x + x, or 2x, in agreement with the left-hand side: one x from the fact that dx/dx = 1 which multiplies (1 + 1 + ... + 1) [x times], and the other from the fact that d/dx{(1 + 1 + ... + 1) [x times] = 1, which multiplies x.
You can see why this must be so because multiplication-as-repeated-addition is only a sensible definition for positive integers. Actual multiplication is the generalization of this process to arbitrary real numbers.
One of my hosts in Bemidji, MN this weekend was a mathematician named Eric Lund. When I met him I remarked that I enjoyed the comments that he leaves at my blog. He replied that he had no idea I had a blog.
That's when I thought to check your e-mail address. I see you're at the University of New Hampshire, not Bemidji State University!
Bemidji has a killer live webcam:
http://bemidji.org/pages/ChamberWebcams
If you don't remember the chain rule, there are other ways of going about this, but they all come back to the fact that, as Eric Lund pointed out, the number of terms on the right side varies. Then it's a question of what the = sign is actually doing here; part of the problem is that we're used to seeing it as f(x) = 10x (for instance) when defining a function where x is free to vary, and something like 4x = x + 15, where there is a solution for x. In the latter case, we have two functions, one on the left and one on the right, and the solution is the point of intersection (where x = 5); differentiating both sides with respect to x would give us the point at which the tangent slope of each is the same, but doing so yields 4 = 1 which is contradictory, so there is no such point.
Back to the problem: for any positive integer value a, we end up with:
x^2 = x + x + x... a times, or x^2 = a*x. These are two functions with a point of intersection at x = a. Differentiating them gives 2x = a, which states that the tangent slopes are equal at x = a/2.
I just saw the solution to POTW 7. I had been assuming that it was simply a discrete function with an integer domain. I thought those could be differentiated in some circumstances.