Trying to Buy a Lamp

Dear reader, I had wanted to tell you an anecdote about a recent incident in which I considered myself to have been outrageously mistreated, but it occurred to me that you probably would not find the story at all worthy of note. In fact, I fear you would be quite likely to think less of me for complaining in such a melodramatic fashion about something which the prevailing norms of our Society consider quite ordinary and proper. And what authority do I have to insist that it's Society that is in the wrong, and not I?

So I won't tell you. Instead, let me tell you a completely unrelated anecdote about my analogue in an alternate universe not entirely unlike our own. You see, recently, my alternate-universe analogue wanted to buy a table lamp, so he went—or let us say in a manner of speaking that I went—to a store to purchase one.

In the showroom, I found a lamp I liked, flagged down a salesman, and said to him, "I'd like to buy this lamp."

"Have you previously purchased a side table from us before?" he said.

"No," I said, somewhat puzzled by the seemingly irrelevant question.

"Well, you can't buy a lamp unless you already have a table to put it on," said the salesman in a tone of polite condescension.

"Oh, I certainly agree that it simply wouldn't do to get a lamp without having a table to put it on," I said, "but you see, I already have a table."

"So you did buy a table from us."

"No," I said.

"So you don't have a table."

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Interpolating Between Vectorized Green's Theorems

Green's theorem says that (subject to some very reasonable conditions that we need not concern ourselves with here) the counterclockwise line integral of the vector field F = [P Q] around the boundary of a region is equal to the double intregral of \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} over the region itself. It's natural to think of it as a special case of Stokes's theorem in the case of a plane. We can also think of the line integral as the integral of the inner product of the vector field with the unit tangent, leading us to write Green's theorem like this:

 \oint_{\partial D}\vec{\mathbf{F}}\cdot\vec{\mathbf{T}}\, ds=\iint_{D}(\mathrm{curl\,}\vec{\mathbf{F}})\cdot\vec{\mathbf{k}}\, ds

But some texts (I have Mardsen and Tromba's Vector Calculus and Stewart's Calculus: Early Transcendentals in my possession; undoubtedly there are others) point out that we can also think of Green's theorem as a special case of the divergence theorem! Suppose we take the integral of the inner product of the vector field with the outward-facing unit normal (instead of the unit tangent)—it turns out that

\oint_{\partial D}\vec{\mathbf{F}}\cdot\vec{\mathbf{n}}\, ds=\iint_{D}\mathrm{div\,}\vec{\mathbf{F}} ds

—which suggests that there's some deep fundamental sense in which Stokes's theorem and the divergence theorem are really just mere surface manifestations of one and the same underlying idea! (I'm told that it's called the generalized Stokes's theorem, but regrettably I don't know the details yet.)

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Dialogue Concerning Birthdays

"Happy birthday, Synthia!"

"Why," mused Synthia, "do we celebrate birthdays? I fail to see anything special about the calendar day on which a person's age-in-calendar-years passes an integer. Are things supposed to be different now that it's been seven-point-five-seven-four times ten-to-the-eighth seconds since my birth, rather than seven-point-five-seven-three?"

Quiana suppressed a groan. Somehow she had been expecting Synthia to say something "normal," like Thanks! But knowing Synthia, it would have been unusual if she had done so: relative to Quiana's state of knowledge of her friend, this—the general style, though not, of course, the specific questions—was the normal response.
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The Derivative of the Natural Logarithm

Most people learn during their study of the differential and integral calculus that the derivative of the natural logarithm ln x is the reciprocal function 1/x. Indeed, sometimes the natural logarithm is defined as  \int_1^x \frac{1}{t}\,dt. However, on observing the graphs of ln x and 1/x, the inquisitive seeker of knowledge can hardly fail to notice a disturbing anomaly:

y=ln(x) y=1/x

The natural logarithm is only defined for positive numbers; no part of its graph lies in quadrants II or III. But the reciprocal function is defined for all nonzero numbers. So (one cannot help oneself but wonder) how could the latter be the derivative of the former? If the graph of the natural logarithm isn't there to be differentiated in the left half of the plane, how could its derivative be defined in that region?
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