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November 24

Intersection of Hyperplance and Hypercube

Let $A=\{x\in \mathbb {R} ^{n}|a^{T}x=b\}$ (for $a\in \mathbb {R} ^{n},b\in \mathbb {R}$ ) be a hyperplane, and $B=[0,1]^{n}$ be a hypercube.

Does the intersection $A\cap B$ equal the convex hull of two $n-2$ -dimensional hypercubes? David Frid (talk) 12:52, 24 November 2017 (UTC)[reply]

Even in the case of a 3-D cube you can get a hexagon by such a cut and you can't get six outside points as thehull of two lines with four end points in total. Dmcq (talk) 16:33, 24 November 2017 (UTC)[reply]

Thank you David Frid (talk) 07:58, 25 November 2017 (UTC)[reply]

Maximum Distance

Let $A$ be the intersection of some hyperplane and hypercube. Let $p\in \mathbb {R} ^{n}$ . Let $f(x)={\begin{cases}d(x,p)&x\in A\\0&\mathrm {else} \end{cases}}$ ( $d$ denotes the Euclidean distance).

Does $f$ have a local maximum, which is not a global maximum too? David Frid (talk) 13:25, 24 November 2017 (UTC)[reply]

You can have a local maximum which is not a global maximum. For instance just consider a plane intersecting a cube to give a square. The distance to the two far corners from a point near the middle of one side will be local maxima. On the other hand with the point close to one corner there will just be a single maximum to the opposite corner. Dmcq (talk) 16:39, 24 November 2017 (UTC)[reply]

(edit conflict) Let n=2. The hypercube is a square (including the interior), and the hyperplane is a line intersecting the square in the line segment AB, with endpoints A and B on the square’s boundary. Let P be a point in the plane such that its nearest point on AB is the point C = (2/3)A +(1/3)B. Then f has maxima at A and B with the global maximum at B. Loraof (talk) 16:45, 24 November 2017 (UTC)[reply]

Thank you!

What happens if we also assume that the hyperplane is

\{x\in \mathbb {R} ^{n}|a^{T}x=b\}

for some

a\in \mathbb {\mathbb {R} } _{+}^{n},b\in \mathbb {R} _{+}

, and the hypercube is

[0,1]^{n}

, and

p=(0.5,0.5,\dots )

is the hypercube's center.

Under these extra assumptions, can we have a local maximum that is not a global maximum too? (the above examples do not apply for these extra assumptions)David Frid (talk) 07:45, 25 November 2017 (UTC)[reply]

Yes lots. A plane going through p will have the origin and the farthest point of the hypercube from the origin as maxima. Some other plane going through the origin but just a little away from it will have a local maximum to a point near the farthest away point of the hypercube from the origin. Dmcq (talk) 22:04, 25 November 2017 (UTC)[reply]

Can we upper bound the number of local-and-not-global maxima? For example, can we have more than 2n such maxima? David Frid (talk) 08:49, 28 November 2017 (UTC)[reply]

Difficulty Simplying Terms in Lagrange Polynomial

I'm teaching myself Lagrange interpolation, and I can't seem to get the correct interpolating polynomial for the study exercises. I'm sure that I'm setting up the calculation correctly, but even with just 4 data points, the resulting expression is so complicated that I'm probably making mistakes when I simplify it. The data points are (-1,0), (2,1), (3,1), and (5,2). The resulting polynomial terms are:

$P_{3}(x)={\frac {(x+1)(x-3)(x-5)}{9}}-{\frac {(x+1)(x-2)(x-5)}{8}}+{\frac {(x+1)(x-2)(x-3)}{18}}$

By combining the first and third terms, I get:

${\frac {-x^{2}+x+12}{6}}$

Next, I find the LCD for that expression and the second term:

${\frac {-4x^{2}+4x-48}{24}}-{\frac {(3x+3)(x^{2}-7x+10)}{24}}$

Adding and simplifying results in the expression:

$P_{3}(x)={\frac {-x^{3}}{8}}+{\frac {7x^{2}}{12}}+{\frac {-5x}{24}}-{\frac {13}{4}}$

However, the answer (from Wolfram Alpha) is:

$P_{3}(x)={\frac {x^{3}}{24}}-{\frac {x^{2}}{4}}+{\frac {11x}{24}}+{\frac {3}{4}}$

Can anyone see where I'm making a mistake? OldTimeNESter (talk) 20:21, 24 November 2017 (UTC)[reply]

Well, you're missing an

x^{3}

term on the first step, so that'll throw things off at least. But overall, it would probably be easiest to simplify systematically: just multiply out each term and then combine them all together at the end (by factoring out 1/72). --Deacon Vorbis (talk) 20:40, 24 November 2017 (UTC)[reply]

The simplest way is to note that there is a common factor

x+1

in all three terms. Ruslik_Zero 20:57, 24 November 2017 (UTC)[reply]

... so take out a common factor of (x+1)⁄72 at the start, then combine the quadratic terms: 8(x-3)(x-5) - 9(x-2)(x-5) + 4(...... There was nothing wrong with your original method except for the error in multiplying, but you might find it easier to start this way. Dbfirs 21:23, 24 November 2017 (UTC)[reply]

Shoot, Deacon Vorbis, you're right: I don't know how I missed that. I usually do just multiply everything out, too, but I was trying to do it in fewer steps to avoid making mistakes (like the one I did make!). OldTimeNESter (talk) 18:33, 25 November 2017 (UTC)[reply]

I often make mistakes when I try to multiply out cubics (lack of practice), so combining terms with the factor outside makes errors less likely (for me at least). Dbfirs 23:02, 25 November 2017 (UTC)[reply]

As you have 4 data points you must also have 4 terms in the Lagrange Polynomial expansion. You have forgotten one term. (The article Lagrange interpolation stupidly counts the 4 terms from 0 to 3. Count from 1 to 4). Bo Jacoby (talk) 08:19, 1 December 2017 (UTC).[reply]