MathFilter! Re Lemniscate of Bernoulli
August 2, 2022 4:30 PM Subscribe
The Cartesian equation of the Lemniscate of Bernoulli given at MacTutor differs from the one at Wolfram MathWorld (eq. 3 has a 2 that's not in MacTutor), but both sites give the same polar equation. How can that be?
To expand a bit, the condition specified in the preamble at Wolfram (that the product of the distances from two fixed points (-a,0) and (a,0) is a constant a^2) is not consistent with the statement in Eq. 4 of Wolfram. This is probably where the extra 2 (or sqrt(2) in the linear size of the lemniscate) is coming from.
posted by heatherlogan at 5:14 PM on August 2, 2022
posted by heatherlogan at 5:14 PM on August 2, 2022
Also, the polar expression makes no sense as it stands at either site for values of theta other than zero. The cos(2 theta) should probably be a cos^2(theta) or something.
In summary, don't eat candy you find on the ground (i.e., don't use formulas you haven't checked yourself).
posted by heatherlogan at 5:37 PM on August 2, 2022
In summary, don't eat candy you find on the ground (i.e., don't use formulas you haven't checked yourself).
posted by heatherlogan at 5:37 PM on August 2, 2022
I wouldn't normally characterize MathWorld as stuff found on the ground, but they do in fact seem to have conflated the two parameters (half-width and focal distance) which Wikipedia correctly distinguishes. Someone should drop them a note about that!
The polar equations are fine, though. What is your objection, heatherlogan?
posted by aws17576 at 5:51 PM on August 2, 2022 [3 favorites]
The polar equations are fine, though. What is your objection, heatherlogan?
posted by aws17576 at 5:51 PM on August 2, 2022 [3 favorites]
Best answer: Just to clarify a little further, we only need one parameter to characterize a lemniscate of Bernoulli; either focal distance or half-width will do, since each one can be recovered from the other. MathWorld makes the mistake of using the same symbol to refer to both, hence the inconsistency between the equations there.
A useful analogy is that a circle (of known center) can be identified by just its radius, or by just its diameter, since each can be determined from the other (r = D/2, D = 2r). We could even use the circumference as our parameter of choice. But whichever one we choose, we should stick to it or, if we want to switch for some reason, we should obviously be clear about that and use a different symbol. This is where the MathWorld article fails: in equations (1)–(3), a stands for the focal distance, but from (4) onward (at least as far down as I checked), they switch to using a for the half-width.
That's a serious editing mistake, but the folks at MathWorld generally know their subject and I suspect they'd like to get this fixed; I'll send a correction myself.
posted by aws17576 at 6:14 PM on August 2, 2022 [4 favorites]
A useful analogy is that a circle (of known center) can be identified by just its radius, or by just its diameter, since each can be determined from the other (r = D/2, D = 2r). We could even use the circumference as our parameter of choice. But whichever one we choose, we should stick to it or, if we want to switch for some reason, we should obviously be clear about that and use a different symbol. This is where the MathWorld article fails: in equations (1)–(3), a stands for the focal distance, but from (4) onward (at least as far down as I checked), they switch to using a for the half-width.
That's a serious editing mistake, but the folks at MathWorld generally know their subject and I suspect they'd like to get this fixed; I'll send a correction myself.
posted by aws17576 at 6:14 PM on August 2, 2022 [4 favorites]
aws17576 - I got hung up on what happens when the cosine is negative. But I think you are right -- there just aren't lobes there. My bad!
posted by heatherlogan at 6:27 PM on August 2, 2022
posted by heatherlogan at 6:27 PM on August 2, 2022
Response by poster: Many thanks to you both! I'll be posting another lemniscate question soon(ish) so please keep an eye out for it.
posted by mpark at 8:18 PM on August 2, 2022
posted by mpark at 8:18 PM on August 2, 2022
Haha, looking forward to it! While posting my answer, I was wondering "What is mpark doing that inspired this question?"
posted by aws17576 at 11:00 PM on August 2, 2022 [1 favorite]
posted by aws17576 at 11:00 PM on August 2, 2022 [1 favorite]
Just to illustrate the inconsistency, here are Eqs. (3) and (6) from MathWorld plotted in Desmos.
posted by Johnny Assay at 5:05 AM on August 3, 2022 [2 favorites]
posted by Johnny Assay at 5:05 AM on August 3, 2022 [2 favorites]
« Older film about siblings who find compromising photo of... | Fashion resources for your 13 going on 30 year old Newer »
This thread is closed to new comments.
To demonstrate this, consider the point as far as possible to the right on the x axis. In the polar expression (assuming that theta starts from zero at the x axis), this is at r = a (or in Cartesian coordinates, x = a and y = 0). The Cartesian equation at MacTutor is consistent with this, but the one at Wolfram is not.
posted by heatherlogan at 5:11 PM on August 2, 2022 [3 favorites]