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June 4

In cubic interpolation, does the cubic curve pass through the outer points?

Image from 2016

Hello,

In diagrams I've seen e.g. http://researchgate.net/figure/Bi-cubic-spline-interpolation_fig1_280062913 , the cubic curves pass through all 4 points from which the cubic curve is derived, and that's what I've drawn in this diagram. I've come to realise that if that's the case, it's not possible to guarantee that the adjacent cubic curves have continuous derivative. This image also shows the curve passing through only the middle two points: http://paulinternet.nl/?page=bicubic . Is this interpretation correct?

Thanks,
cmɢʟee⎆τaʟκ 22:13, 4 June 2020 (UTC)

There are lots of different ways to do it. See for example Bézier curve. 2602:24A:DE47:BB20:50DE:F402:42A6:A17D (talk) 22:17, 4 June 2020 (UTC)

See also Cubic Hermite spline. -- ToE 00:43, 5 June 2020 (UTC)

There's an old technique which I call 'condition counting' which, while not 100% rigorous, is useful from a heuristic point of view. For a cubic polynomial there are four coefficients so you get to set four condition, and in four point interpolation you set the four conditions to be that the curve passes through the four points. For two adjacent curve you can set eight conditions, but an additional condition, namely that the derivatives match an the common point, would make 9, which is more than you're allowed. For Bezier and Hermite splines the four condition you set are the values and derivatives at the two endpoints. (For cubic curves, Bezier and Hermite splines are pretty much the same thing, just written in different ways.) So if you just have to pick the same derivatives at the endpoints to be the same and you're back down to the 8 conditions you're allowed. If you just specify that the derivatives are equal but not what they are, then you're down to 7 conditions and you need an additional condition to get a unique solution; the Hermite article gives some different choices for what this condition might be. --RDBury (talk) 03:46, 5 June 2020 (UTC)

Many thanks, RDBury, ToE and 2602:24A:DE47:BB20:50DE:F402:42A6:A17D. From Cubic_Hermite_spline#Catmull–Rom_spline, I see that a centripetal Catmull–Rom spline goes through all four points, if I understood it correctly. Good to know the diagram needn't be amended. Cheers, cmɢʟee⎆τaʟκ 11:42, 6 June 2020 (UTC)

To fit a piecewise cubic to a one-dimensional sequence, set the derivative at each sample to what it would be for the quadratic that fits it and the two adjacent points. Then you have for each interval four constraints: the two given points and their two derivatives. These define a cubic. (For the first and last intervals, one derivative is missing, so these intervals can be quadratic.) I imagine that an analogous approach works for a bicubic, but I have not tried it. —Tamfang (talk) 00:32, 7 June 2020 (UTC)

Thanks, Tamfang. cmɢʟee⎆τaʟκ 17:42, 7 June 2020 (UTC)

(continuing) The logic for this is that the curve for interval $t_{1}..t_{2}$ is defined by the values at $t_{0},t_{1},t_{2},t_{3}$, and the curve for interval $t_{2}..t_{3}$ is defined by the values at $t_{1},t_{2},t_{3},t_{4}$, so the properties of the curve at $t_{2}$ must be determined only by the intersection of these two sets of inputs. —Tamfang (talk) 00:47, 11 June 2020 (UTC)

June 6

Why does the generalized continuum hypothesis implies the following equality for cardinal exponentiation?

The article about the continuum hypothesis states that the the generalized continuum hypothesis implies the following:

$\aleph _{\alpha }^{\aleph _{\beta }}=\aleph _{\alpha }$ when β+1 < α and $\aleph _{\beta }<\operatorname {cf} (\aleph _{\alpha })$, where cf is the cofinality operation.

How does one prove that?

Thanks!

Dan Gluck (talk) 12:59, 6 June 2020 (UTC)

See Talk:Continuum hypothesis#Proofs of certain consequences of GCH where I justify the claims in the article. Here I extract the relevant portion:

$\aleph _{\alpha }=\aleph _{\alpha }^{1}\leq \aleph _{\alpha }^{\aleph _{\beta }}\,.$

Since we are in the case that $\aleph _{\beta }<\operatorname {cf} (\aleph _{\alpha })$ where cf is the cofinality operation, then any function from $\aleph _{\beta }\,$ to $\aleph _{\alpha }\,$ must be bounded above by some $\gamma <\aleph _{\alpha }\,.$

And γ has a cardinality $\vert \gamma \vert =\aleph _{\delta }\,$ where δ < α. :The cardinality of the set of functions so bounded by γ is

$\aleph _{\delta }^{\aleph _{\beta }}\leq 2^{(\aleph _{\delta }\cdot \aleph _{\beta })}=\aleph _{\max(\delta ,\beta )+1}\leq \aleph _{\alpha }\,.$

Adding these together for the $\aleph _{\alpha }\,$ possible values of γ gives

$\aleph _{\alpha }^{\aleph _{\beta }}\leq \aleph _{\alpha }\cdot \aleph _{\alpha }=\aleph _{\alpha }\,$

which means

$\aleph _{\alpha }^{\aleph _{\beta }}=\aleph _{\alpha }\,.$

OK? JRSpriggs (talk) 04:42, 7 June 2020 (UTC)

June 11

The deduction theorem in provability logic

Does the deduction theorem holds in GL logic? Specifically, when Necessitation is defined as it is here, in that it can only be applied with no assumptions before the $\vdash$ (so from $\{\phi \}\vdash \phi$ one may not deduce $\{\phi \}\vdash \square \phi$). The article indicates the answer might be yes, but I have been told by someone whom I am inclined to believe that the answer is no. For some context, arithmetic soundness and completeness mean that GL logic is in some way the same as talking about provability in PA, which (I believe) does have the deduction theorem. — crh23 (Talk) 08:52, 11 June 2020 (UTC)

(I hope the following is not foot-in-mouth, as I am straying from the areas I'm comfortable with.) For the deduction theorem to be applicable, you need a logic that allows assumptions to the left of the turnstyle. Starting with the assumption $p$, unless we have a perverse logic, it should then be the case that

$p\vdash p.$

From the necessitation rule of GL logic given in the Provability logic article, we obtain

$p\vdash \Box p.$

So, if the deduction theorem holds, we obtain now

$\vdash p\rightarrow \Box p.$

Is this a theorem of GL logic? If not, the deduction theorem does not hold. I see no rule for condition introduction other than the distribution axiom and Löb's axiom, and either one requires a box in the antecedent of the conditional, so if I take the definition of GL logic in the Provability logic article for gospel, this cannot be a theorem.  --Lambiam 11:13, 11 June 2020 (UTC)

I think the normal consensus is that $\vdash p\rightarrow \Box p$ is not a theorem in GL logic. However, the necessitation rule does not allow the deduction you state, since it may only be used when there is no assumption to the left of the $\vdash$: the rule is specifically written

• Necessitation: From $\vdash$ p conclude $\vdash$ □p

rather than

• Necessitation: From p conclude □p

(which is how the rule for modus ponens is denoted) --- see section 2 of .— crh23 (Talk) 13:39, 11 June 2020 (UTC)