POTW 7 Posted

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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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.

By Eric Lund (not verified) on 26 Oct 2015 #permalink

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!

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.

By Another Matt (not verified) on 28 Oct 2015 #permalink

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.

By Another Matt (not verified) on 02 Nov 2015 #permalink