4 ms·
It's discontinuous if the value is undefined at the origin. That happens here because the denominator of (x + t) yields a division by zero. A nonzero slope wo
by T-hawk 12y ago
It's discontinuous if the value is undefined at the origin. That happens here because the denominator of (x + t) yields a division by zero. A nonzero slope would give a kink in the curve, but not a discontinuity.
One approach to avoiding the whole question of abs(x) would be to use cosine instead of sine, since cosine is symmetric across x = 0.
- JadeNB 12y ago> One approach to avoiding the whole question of abs(x) would be to use cosine instead of sine, since cosine is symmetric across x = 0. I don't think that this would work, since the division by an odd function would just make it odd again (that is, `sin(x)/x` is even precisely because `sin` is odd). Note, however, that blaming `abs` for discontinuity (or non-differentiability) is a bit of a red herring; any even function on the real line, no matter how smooth, can be written as a composition with `abs`.
- fenomas 12y agoHurmph. I don't see how it makes much difference whether he uses sine or cosine, since he's also adding t to the operand and then animating over it, causing the waveform to move along the x-axis. He can use one and start from t=0 or use the other and start from t=pi/2, as you like. My point was simply that the "spike" the article goes on about (which I earlier called a discontinuity, oops) shows up because the author is reflecting his plot around the Y axis. The article makes it sound like the spike was some physically significant feature of the function being graphed.
- JadeNB 12y agoI agree that it doesn't make any difference when you allow phase shifts (I was intentionally focussing on the `t = 0` case), so I'm puzzled by your suggestion that the author should have used cosine instead. My point was more that there's nothing inherently 'spiky' about even functions; as your own example of cosine shows, they can be just as smooth (or spiky) as any other function.
- fenomas 12y agoI think you're conflating people, if that was in reply to my immediately previous post.
- JadeNB 12y agoYes, you're right—I thought that you had made the post that was actually made by T-hawk (https://news.ycombinator.com/item?id=9022743 https://news.ycombinator.com/item?id=9022743 )—but I think that it doesn't change my point.