Wikipedia:Reference desk/Archives/Mathematics/2014 September 13

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September 13

3rd and 4th degree Bézier curve

Hi,
the computer's fonts use Bézier curve to represent glyphs.
The creator of the font divide the complex curve to sub-curve, which he then describe with Bézier curve technique.
My question is whether there is a technique that can take 4-th degree collection of Bézier curve, and create 3-rd degree collection of Bézier curve that looks the same.Exx8 (talk) 01:10, 13 September 2014 (UTC)

You can't in general turn a 4th degree Bézier into exact 3rd degree ones. So your 'looks the same' needs to be some sort of approximation. They do lend themselves to mathematical reasoning - for instance you can always subdivide a Bézier curve exactly into smaller ones of the same degree and your rule may lend itself to checking easily if the sub curve can be approximated closely enough. A computing solution might be to simply generate lots of points close to each other and then approximate them with cubic Bézier curves with smoothing just like one would for freehand drawing. There could easily be a computer package already written to do this. Dmcq (talk) 06:27, 13 September 2014 (UTC)

TrueType fonts just use quadratic Bézier curves and OpenType fonts uses cubic Bézier curves. You should be able to get a good enough approximation to your curves for use with fonts. I would suspect most font designers would do this by eye adjusting control points to get the look they want rather than use an deterministic algorithm. A good discussion of Bezier curves is A Primer on Bézier Curves sections 29 and 30 near the end might help.--Salix alba (talk): 08:18, 13 September 2014 (UTC)

My naïve suggestion: Cut up your original curve so that there is a control point wherever the tangent angle is a multiple of π/4, and wherever the curvature reaches an extreme or crosses zero; these are the points most important to the appearance of the glyph. On each segment of the arc, make the derivatives of the cubic match (in angle and magnitude) those of the original curve at the control points. (I have a sporadic project involving even more arcane curves, and this is how I plan to convert them to cubic fonts.) —Tamfang (talk) 02:05, 14 September 2014 (UTC)

Verifying that the cts dual space is subspace of the algebraic dual space

Hi,

It's mentioned on the page for dual spaces that if a vector space is topological, the continuous dual space is a linear subspace of the algebraic dual space.

I tried to verify this directly from the subspace axioms:

Let $V$ be a topological v.s over $F$. Assume $f\in Hom(V,F)$ is cts, and let $k\in F$. Then

$(kf)(x)=k(f(x))=(\phi \circ f)(x)$

(where $\phi$ is multiplication by $k$), by the definition of scalar multiplication on $Hom(V,F)$

Since scalar multiplication on a topological field is cts, as is the composition of cts functions, it follows that $kf$ is continuous.

But I can't seem to show closure under addition. Have I overlooked something, or is there a better approach?

Neuroxic (talk) 06:28, 13 September 2014 (UTC)

Perhaps I'm missing something, this isn't really my area, but by definition of a topological field, addition is continuous, so if f and g are V->k, then f + g is continuous. Clearly, f + g is a linear functional, so the continuous dual is closed under addition. That the continuous dual is a subset of the algebraic dual is immediate, so it is a linear subspace. Again, I apologize if there is something that I am glossing over, I haven't touched any of this stuff in a year, or two.Phoenixia1177 (talk) 06:36, 13 September 2014 (UTC)

"by definition of a topological field, addition is continuous, so if f and g are V->k, then f + g is continuous" this is the bit I'm having trouble with. Would you be able to expand on this more precisely?

Neuroxic (talk) 10:15, 13 September 2014 (UTC)

The definition of a topological field k includes that the basic field operations are continuous. For any topological space X and continuous f,g:X -> k, f + g is continuous because it is composition of functions; explicitly: +(f, g). Continuous linear functionals are linear maps V -> k that are continuous. Thus, pointwise addition will yield, another, continuous map. It is obvious that the sum of any two linear functionals is a linear functional, thus, the sum of two continuous linear functionals is another one. Hence, closure under addition. --Essentially, functions add using pointwise addition, since addition is continuous, addition of functions outputs continuous functions.Phoenixia1177 (talk) 10:28, 13 September 2014 (UTC)

"For any topological space X and continuous f,g:X -> k, f + g is continuous because it is composition of functions; explicitly: +(f, g)."

This part doesn't make sense to me, and was one of the reasons I was stuck. I tried to use the idea of composing cts functions to preserve continuity (that worked to show closure under scalar multiplication) but wasn't able to do it. I thought I couldn't write $(f+g)(x)=({\tilde {f}}\circ g)(x)$ for some ${\tilde {f}}$ because ${\tilde {f}}$ depends on $g(x)$, not on $x$. How do you write $f(x)+g(x)$ as a composition of two cts functions?

Neuroxic (talk) 12:24, 13 September 2014 (UTC)

The function +:k x k -> k is continuous, so f + g is +(f, g), it is a continuous function of two variables composed with f and g. Since all the operations are continuous, the end result is. As a more basic example, consider continuous f,g from a topological space into the reals, their sum is continuous for the same reason - it's the continuity of addition that gives this result. The same principle applies here, just X replaced with the dual space and the reals with a topofield.Phoenixia1177 (talk) 12:32, 13 September 2014 (UTC)

Epic fail. I forgot that a function $(f,g)$ is cts if $f,g$ are cts. I know the proof now. Thanks.

Neuroxic (talk) 12:37, 13 September 2014 (UTC)