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Derivations and the curvature tensor

Let ${M}$ be a Riemannian manifold with Riemannian connection ${\nabla}$. A connection is a thing that knows how to differentiate a vector field ${Y}$ in the direction of a vector field ${X}$; the result is denoted by ${\nabla_X Y}$ and is also a vector field. For consistency of notation, it is convenient to write ${\nabla_X f }$ for the derivative of scalar function ${f}$ in the direction ${X}$, even though this derivative does not need a connection: vector fields are born with the ability to differentiate functions.

The pairs ${(f,Y)}$, with ${f}$ a scalar function and ${Y}$ a vector field, form a funky nonassociative algebra ${\mathcal{A}}$ described in the previous post. And ${\nabla_X}$ is a derivation on this algebra, because

• ${\nabla_X(fY) = f\nabla_X Y + (\nabla_X f)Y }$ by the definition of a connection
• ${\nabla_X\langle Y, Z\rangle = \langle \nabla_X Y, Z\rangle + \langle Y, \nabla_X Z\rangle}$ by the metric property of the Riemannian connection.

Recall that the commutator of two derivations is a derivation. Or just check this again: if ${{}'}$ and ${{}^\dag}$ are derivations, then

${ {(ab)'}^\dag = (a'b+ab')^\dag = {a'}^\dag b+a'b^\dag +a^\dag b'+a{b'}^\dag }$
${ {(ab)^\dag}' = (a^\dag b+ab^\dag)' = {a^\dag }' b+a^\dag b' +a' b^\dag +a{b^\dag}' }$

and the difference ${{(ab)'}^\dag-{(ab)^\dag}'}$ simplifies to what it should be.

Thus, for any pair ${X,Y}$ of vector fields the commutator ${\nabla_X \nabla_Y-\nabla_Y\nabla_X}$ is a derivation on ${\mathcal{A}}$. The torsion-free property of the connection tells us how it works on functions:

$\displaystyle (\nabla_X \nabla_Y-\nabla_Y\nabla_X) f = \nabla_{[X,Y]}f=\nabla_{\nabla_XY}f -\nabla_{\nabla_YX}f$

Subtracting ${\nabla_{[X,Y]}f}$ from the commutator, we get a derivation that kills scalar functions,

$\displaystyle R(X,Y) = \nabla_X \nabla_Y-\nabla_Y\nabla_X - \nabla_{[X,Y]}$

But a derivation that kills scalar functions is linear over functions:

$\displaystyle R(X,Y)(fZ) = R(X,Y)(f)\, Z + f\,R(X,Y)Z = f\,R(X,Y)Z$

In plain terms, ${R(X,Y)}$ processes any given vector field ${Z}$ pointwise, applying some linear operator ${L_p}$ to the vector ${Z_p}$ at every point ${p}$ of the manifold. No derivatives of ${Z}$ are actually taken, either of first or of second order.

Moreover, the derivation property immediately implies that ${R(X,Y)}$ is a skew-symmetric operator: for any vector fields ${Z,W}$

$\displaystyle \langle R(X,Y)Z,W\rangle + \langle R(X,Y)W,Z\rangle = R(X,Y)\langle Z,W\rangle =0$

because ${R(X,Y)}$ kills scalar functions.

The other kind of skew-symmetry was evident from the beginning: ${R(X,Y)=-R(Y,X)}$ by definition.

What is not yet evident is that ${R(X,Y)}$ is also a tensor in ${X}$ and ${Y}$, that is, it does not differentiate the direction fields themselves. To prove this, write ${R(X,Y)=\nabla_{X,Y}^2-\nabla_{Y,X}^2}$ where $\displaystyle \nabla_{X,Y}^2 = \nabla_X \nabla_Y - \nabla_{\nabla_X Y}$ should be thought of as the pointwise second-order derivative in the directions ${X,Y}$ (i.e., the result of plugging two direction vectors into the Hessian matrix). By symmetry, it suffices to show that ${\nabla_{X,Y}^2}$ is a tensor in ${X}$ and ${Y}$. For ${X}$, this is clear from the definition of connection. Concerning ${Y}$, we have

${ \nabla_{X,fY}^2 = \nabla_X (f\nabla_{Y}) - \nabla_{(\nabla_X f) Y+f\nabla_X Y} }$
${= f \nabla_X \nabla_{Y} + (\nabla_X f )\nabla_{Y} - (\nabla_X f) \nabla_{ Y} - f \nabla_{\nabla_X Y} }$
${= f \nabla_{X,Y}^2 }$

That’s it, we have a tensor that takes three vector fields ${X,Y,Z}$ and produces another one, denoted ${R(X,Y)Z}$. Now I wonder if there is a way to use the language of derivations to give a slick proof of the first Bianchi identity, ${R(X,Y)Z+R(Y,Z)X+R(Z,X)Y=0}$

To avoid having two picture-less posts in a row, here is something completely unrelated:

New design of the Command key?

This is the image of the unit circle ${|z|=1}$ under the polynomial ${z^3-\sqrt{3}\,\bar z}$. Which area is larger: red or green? Answer hidden below.

They are equal.

Posted in Uncategorized | | 2 Comments

Derivations

A map ${D}$ is a derivation if it satisfies the Leibniz rule: ${D(ab)=D(a)b+aD(b)}$. To make sense out of this, we need to be able to

• multiply arguments of ${D}$ together
• multiply values of ${D}$ by arguments of ${D}$

For example, if ${D\colon R\to M}$ where ${R}$ is a ring and ${M}$ is a two-sided module over ${R}$, then all of the above makes sense. In practice it often happens that ${M=R}$. In this case, the commutator (Lie bracket) of two derivations ${D_1,D_2}$ is defined as ${[D_1,D_2]=D_1\circ D_2-D_2\circ D_1}$ and turns out to be a derivation as well. If ${R}$ is also an algebra over a field ${K}$, then ${K}$-linearity of ${D}$ can be added to the requirements of being a derivation, but I am not really concerned about that.

What I am concerned about is that two of my favorite instances of the Leibniz rule are not explicitly covered by the ring-to-module derivations. Namely, for smooth functions ${\varphi\colon{\mathbb R}\rightarrow{\mathbb R}}$, ${F\colon{\mathbb R}\rightarrow{\mathbb R}^n}$ and ${G\colon{\mathbb R}\rightarrow{\mathbb R}^n}$ we have

$\displaystyle (\varphi F)' = \varphi' F + \varphi F' \quad \text{and} \quad (F\cdot G)' = F'\cdot G+F\cdot G' \ \ \ \ \ (1)$

Of course, ${{\mathbb R}^n}$ could be any ${{\mathbb R}}$-vector space ${V}$ with an inner product.

It seems that the most economical way to fit (1) into the algebraic concept of derivation is to equip the vector space ${{\mathbb R}\oplus V}$ with the product

$\displaystyle (\alpha,u)(\beta,v)= (\alpha\beta+u\cdot v, \alpha v+\beta u) \ \ \ \ \ (2)$

making it a commutative algebra over ${{\mathbb R}}$. Something tells me to put ${-u\cdot v}$ there, but I resist. Actually, I should have said “commutative nonassociative algebra”:

$\displaystyle \{(\alpha,u)(\beta,v)\}(\gamma,w) = (\alpha\beta+u\cdot v, \alpha v+\beta u) (\gamma,w) \\ \\ = (\alpha\beta\gamma+\gamma u\cdot v+ \alpha v\cdot w+\beta u\cdot w, \alpha\beta w + \gamma \alpha v+ \gamma \beta u +(u\cdot v) w)$

Everything looks nice, except for the last term ${(u\cdot v) w}$, which destroys associativity.

Now we can consider maps ${{\mathbb R}\rightarrow {\mathbb R}\oplus V}$, which are formal pairs of scalar functions and vector-valued functions. The derivative acts component-wise ${(\varphi,F)'=(\varphi',F')}$ and according to (1), it is indeed a derivation:

$\displaystyle \left\{(\varphi,F)(\psi,G)\right\}'= (\varphi,F)'(\psi,G)+(\varphi,F)(\psi,G)' \ \ \ \ \ (3)$

Both parts of (1) are included in (3) as special cases ${(\varphi,0)(0,F)}$ and ${(0,F)(0,G)}$.

If (2) has a name, I do not know it. Clifford algebras do a similar thing and are associative, but they are also larger. If I just want to say that (1) is a particular instance of a derivation on an algebra, (2) looks like the right algebra structure to use (maybe with ${-u\cdot v}$ if you insist). If ${V}$ has no inner product, the identity ${(\varphi F)' = \varphi' F + \varphi F'}$ can still be expressed via (2) using the trivial inner product ${u\cdot v=0}$.

Fourth order obstacle problem

Having solved the obstacle problem for a string, let us turn to a more difficult one, in which an elastic string is replaced with an elastic rod (or plate, if we are looking at a cross-section). Elastic rods resist bending much in the same way that strings don’t. This can be modeled by minimizing the bending energy

$\displaystyle B(u) = \frac12 \int_{-2}^2 u''(x)^2 \,dx$

subject to the boundary conditions ${u(-2)=0=u(2)}$, ${u'(-2)=0=u'(2)}$, and the same obstacle as before: ${u(x)\le -\sqrt{1-x^2}}$. The boundary conditions for ${u'}$ mean that the rod is clamped on both ends.

As before, the obstacle permits one-sided variations ${u+\varphi}$ with ${ \varphi\le 0}$ smooth and compactly supported. The linear term of ${B(u+\varphi)}$ is ${\int u''\varphi''}$, which after double integration by parts becomes ${\int u^{(4)} \varphi}$. Since the minimizer satisfies ${E(u+\varphi)-E(u)\ge 0}$, the conclusion is ${\int u^{(4)} \varphi \ge 0 }$ whenever ${\varphi\le 0}$. Therefore, ${u^{(4)}\le 0}$ everywhere, at least in the sense of distributions. In the parts where the rod does not touch the obstacle, we can do variation of either sign and obtain ${u^{(4)}=0}$; that is, ${u}$ is a cubic polynomial there.

So far everything looks similar to the previous post. But the fourth derivative of the obstacle function ${-\sqrt{1-x^2}}$ is ${3(4x^2+1)/(1-x^2)^{7/2}}$, which is positive. Since the minimizer ${u}$ must satisfy ${u^{(4)}\le 0}$, it cannot assume the shape of the obstacle. The contact can happen only at isolated points.

Therefore, ${u}$ is a cubic spline with knots at the contact points and at ${\pm 2}$. The distributional derivative ${u^{(4)}}$ consists of negative point masses placed at the contact points. Integrating twice, we find that ${u''}$ is a piecewise affine concave function; in particular it is continuous. The minimizer will be ${C^2}$-smooth in ${(-2,2)}$.

How many contact points are there? If only one, then by symmetry it must be at ${x=0}$, and the only three-knot cubic spline that satisfies the boundary conditions and passes through ${(0,-1)}$ with zero derivative is ${(-1/4)(1+|x|)(2-|x|)^2}$. But it does not stay below the obstacle:

Three-knot spline fails the obstacle condition

With a smaller circle, or a longer bar, the one-contact (three-knot) spline would work. For example, on ${[-3,3]}$:

With a longer bar, a three-knot spline solves the problem

But with our parameters we look for two contact points. By symmetry, the middle piece of the spline must be of the form ${q(x)=cx^2+d}$. The other two will be ${p(x)=(ax+b)(2-x)^2}$ and ${p(-x)}$, also by symmetry and to satisfy the boundary conditions at ${\pm 2}$. At the positive knot ${x_0}$ the following must hold:

$\displaystyle p(x_0)=q(x_0)=-\sqrt{1-x_0^2}, \quad p'(x_0)=q'(x_0)=\frac{-x_0}{\sqrt{1-x_0^2}}, \quad p''(x_0)=q''(x_0)$

where the last condition comes from the fact that ${u''}$ is concave and therefore continuous. With five equations and five unknowns, Maple finds solutions in closed form. One of them has ${x_0=0}$, ${a=b=-1/4}$ as above, and is not what we want. The other has ${x_0=(\sqrt{10}-2)/3\approx 0.3874}$ and coefficients such as ${a=-\frac{1}{3645}\sqrt{505875+164940\sqrt{10}}}$. Ugly, but it works:

Minimizer

This time, the bar does stay below the obstacle, touching it only at two points. The amount by which it comes off the obstacle in the middle is very small. Here is the difference ${u(x)-(-\sqrt{1-x^2})}$:

Deviation from the obstacle

And this is the second derivative ${u''}$.

Second derivative is Lipschitz continuous

Again, the minimizer has a higher degree of regularity (Lipschitz continuous second derivative) than a generic element of the function space in which minimization takes place (square-integrable second derivative).

If the rod is made shorter (and the obstacle stays the same), the two-contact nature of the solution becomes more pronounced.

Shorter rod

Assuming the rod stays in one piece, of course.

Second order obstacle problem

Imagine a circular (or cylindrical, in cross-section) object being supported by an elastic string. Like this:

Obstacle problem

To actually compute the equilibrium mass-string configuration, I would have to take some values for the mass of the object and for the resistance of the string. Instead, I simply chose the position of the object: it is the unit circle with center at ${(0,0)}$. It remains to find the equilibrium shape of the string. The shape is described by equation ${y=u(x)}$ where ${u}$ minimizes the appropriate energy functional subject to boundary conditions ${u(-2)=0=u(2)}$ and the obstacle ${u(x)\le -\sqrt{1-x^2}}$. The functional could be the length

$\displaystyle L(u) = \int_{-2}^2 \sqrt{1+u'(x)^2}\,dx$

$\displaystyle E(u) = \frac12 \int_{-2}^2 u'(x)^2 \,dx$

The second one is nicer because it yields linear Euler-Lagrange equation/inequality. Indeed, the obstacle permits one-sided variations ${u+\varphi}$ with ${ \varphi\le 0}$ smooth and compactly supported. The linear term of ${E(u+\varphi)}$ is ${\int u'\varphi'}$, which after integration by parts becomes ${-\int u'' \varphi}$. Since the minimizer satisfies ${E(u+\varphi)-E(u)\ge 0}$, the conclusion is ${\int u'' \varphi \le 0 }$ whenever ${\varphi\le 0}$. Therefore, ${u''\ge 0}$ everywhere (at least in the sense of distributions), which means ${u}$ is a convex function. In the parts where the string is free, we can do variation of either sign and obtain ${u''=0}$; that is, ${u}$ is an affine function there.

The convexity of ${u}$ in the part where it touches the obstacle is consistent with the shape of the obstacle: the string can assume the same shape as the obstacle.

The function ${u}$ can now be determined geometrically: the only way the function can come off the circle, stay convex, and meet the boundary condition is by leaving the circle along the tangents that pass through the endpoint ${(\pm 2,0)}$. This is the function pictured above. Its derivative is continuous: Lipschitz continuous, to be precise.

First derivative is Lipschitz continuous

The second derivative does not exist at the transition points. Still, the minimizer has a higher degree of regularity (Lipschitz continuous derivative) than a generic element of the function space in which minimization takes place (square-integrable derivative).

As a bonus, the minimizer of energy ${E}$ turns out to minimize the length ${L}$ as well.

All in all, this was an easy problem. Next post will be on its fourth-order version.

Biased and unbiased mollification

When we want to smoothen (mollify) a given function ${f:{\mathbb R}\rightarrow{\mathbb R}}$, the standard recipe suggests: take the ${C^{\infty}}$-smooth bump function

$\displaystyle \varphi(t) = \begin{cases} c\, \exp\{1/(1-t^2)\}\quad & |t|<1; \\ 0 \quad & |t|\ge 1 \end{cases}$

where ${c}$ is chosen so that ${\int_{{\mathbb R}} \varphi=1}$ (for the record, ${c\approx 2.2523}$).

Standard bump

Make the bump narrow and tall: ${\varphi_{\delta}(t)=\delta^{-1}\varphi(t/\delta)}$. Then define ${f_\delta = f*\varphi_\delta}$, that is

$\displaystyle f_\delta(x) = \int_{\mathbb R} f(x-t) \varphi_\delta(t)\,dt = \int_{\mathbb R} f(t) \varphi_\delta(x-t)\,dt$

The second form of the integral makes it clear that ${f_\delta}$ is infinitely differentiable. And it is easy to prove that for any continuous ${f}$ the approximation ${f_\delta}$ converges to ${ f}$ uniformly on compact subsets of ${{\mathbb R}}$.

The choice of the particular mollifier given above is quite natural: we want a ${C^\infty}$ function with compact support (to avoid any issues with fast-growing functions ${f}$), so it cannot be analytic. And functions like ${\exp(-1/t)}$ are standard examples of infinitely smooth non-analytic functions. Being nonnegative is obviously a plus. What else to ask for?

Well, one may ask for a good rate of convergence. If ${f}$ is an ugly function like ${f(x)=\sqrt{|x|}}$, then we probably should not expect fast convergence. But is could be something like ${f(x)=|x|^7}$; a function that is already six times differentiable. Will the rate of convergence be commensurate with the head start ${f\in C^6}$ that we are given?

No, it will not. The limiting factor is not the lack of seventh derivative at ${x=0}$; it is the presence of (nonzero) second derivative at ${x\ne 0}$. To study this effect in isolation, consider the function ${f(x)=x^2}$, which has nothing beyond the second derivative. Here it is together with ${f_{0.1}}$: the red and blue graphs are nearly indistinguishable.

Good approximation

But upon closer inspection, ${f_{0.1}}$ misses the target by almost ${2\cdot 10^{-3}}$. And not only around the origin: the difference ${f_{0.1}-f}$ is constant.

But it overshoots the target

With ${\delta=0.01}$ the approximation is better.

Better approximation

But upon closer inspection, ${f_{0.01}}$ misses the target by almost ${2\cdot 10^{-5}}$.

Still overshoots

And so it continues, with the error of order ${\delta^2}$. And here is where it comes from:

$\displaystyle f_\delta(0) = \int_{\mathbb R} t^2\varphi_\delta(t)\,dt = \delta^{-1} \int_{\mathbb R} t^2\varphi(t/\delta)\,dt = \delta^{2} \int_{\mathbb R} s^2\varphi(s)\,ds$

The root of the problem is the nonzero second moment ${\int_{\mathbb R} s^2\varphi(s)\,ds \approx 0.158}$. But of course, this moment cannot be zero if ${\varphi}$ does not change sign. All familiar mollifiers, from Gaussian and Poisson kernels to compactly supported bumps such as ${\varphi}$, have this limitation. Since they do not reproduce quadratic polynomials exactly, they cannot approximate anything with nonzero second derivative better than to the order ${\delta^2}$.

Let’s find a mollifier without such limitations; that is, with zero moments of all orders. One way to do it is to use the Fourier transform. Since ${\int_{\mathbb R} t^n \varphi(t)\,dt }$ is a multiple of ${\widehat{\varphi}^{(n)}(0)}$, it suffices to find a nice function ${\psi}$ such that ${\psi(0)=1}$ and ${\psi^{(n)}(0)=0}$ for ${n\ge 1}$; the mollifier will be the inverse Fourier transform of ${\psi}$.

As an example, I took something similar to the original ${\varphi}$, but with a flat top:

$\displaystyle \psi(\xi) = \begin{cases} 1 \quad & |\xi|\le 0.1; \\ \exp\{1-1/(1-(|\xi|-0.01)^2)\} \quad & 0.1<|\xi|<1.1\\ 0\quad & |\xi|\ge 1.1 \end{cases}$

Fourier transform of unbiased mollifier

The inverse Fourier transform of ${\psi}$ is a mollifier that reproduces all polynomials exactly: ${p*\varphi = p}$ for any polynomial. Here it is:

Unbiased mollifier

Since I did not make ${\psi}$ very carefully (its second derivative is discontinuous at ${\pm 0.01}$), the mollifier ${\varphi}$ has a moderately heavy tail. With a more careful construction it would decay faster than any power of ${t}$. However, it cannot be compactly supported. Indeed, if ${\varphi}$ were compactly supported, then ${\widehat{\varphi}}$ would be real-analytic; that is, represented by its power series centered at ${\xi=0}$. But that power series is

$\displaystyle 1+0+0+0+0+0+0+0+0+0+\dots$

The idea of using negative weights in the process of averaging ${f}$ looks counterintuitive, but it’s a fact of life. Like the appearance of negative coefficients in the 9-point Newton-Cotes quadrature formula… but that’s another story.

Credit: I got the idea of this post from the following remark by fedja on MathOverflow:

The usual spherical cap mollifiers reproduce constants and linear functions faithfully but have a bias on quadratic polynomials. That’s why you cannot go beyond ${C^2}$ and ${\delta^2}$ with them.

Posted in Uncategorized | | 2 Comments

Entropic uncertainty

Having considered the SMBC version of the Fourier transform, it is time to take a look at the traditional one:

$\displaystyle \widehat{f}(\xi)=\int_{{\mathbb R}}f(x)e^{-2\pi i \xi x}\,dx$

(I am not going to worry about the convergence of any integrals in this post.) It is obvious that for any ${\xi\in{\mathbb R}}$

$\displaystyle |\widehat{f}(\xi)|\le \int_{{\mathbb R}}|f(x)|\,dx$

which can be tersely stated as ${\|\widehat{f}\|_\infty\le \|f\|_1}$ using the ${L^p}$ norm notation. A less obvious, but more important, relation is ${\|\widehat{f}\|_2= \|f\|_2}$. Interpolating between ${p=1}$ and ${p=2}$ we obtain the Hausdorff-Young inequality ${\|\widehat{f}\|_q\le \|f\|_p}$ for ${1\le p\le 2}$. Here and in what follows ${q=p/(p-1)}$.

Summarizing the above, the function ${r(p)=\|\widehat{f}\|_q/\|f\|_p}$ does not exceed ${1}$ on the interval ${[1,2]}$ and attains the value ${1}$ at ${p=2}$. This brings back the memories of Calculus I and the fun we had finding the absolute maximum of a function on a closed interval. More specifically, it brings the realization that ${r'(2)\ge 0}$. (I do not worry about differentiability either.)

What does the inequality ${r'(2)\ge 0}$ tell us about ${f}$? When writing it out, it is better to work with ${(\log r)' = r'/r}$, avoiding another memory of Calculus I: the Quotient Rule.

$\displaystyle \log r = \frac{1}{q} \int_{{\mathbb R}} |\widehat{f}(\xi )|^q \,d\xi -\frac{1}{p} \int_{{\mathbb R}} |f(x)|^{p} \,dx$

To differentiate this, we have to recall ${(a^p)'=a^p \log a}$, but nothing more unpleasant happens:

$\displaystyle (\log r)'(2) = - \frac12 \int_{{\mathbb R}} |\widehat{f}(\xi )|^2 \,\log |\widehat{f}(\xi)|\,d\xi - \frac12 \int_{{\mathbb R}} |f(x)|^2 \,\log |f(x)|\,dx$

Here the integral with ${\widehat{f}}$ gets the minus sign from the chain rule: ${(p/(p-1))'=-1}$ at ${p=2}$. In terms of the Shannon entropy ${H(\phi)=-\int |\phi|\log |\phi| }$, the inequality ${(\log r)'(2)\ge 0}$ becomes simply

$\displaystyle H(|f|^2)+H(|\widehat{f}|^2)\ge 0 \ \ \ \ \ (1)$

Inequality (1) was proved by I. Hirschman in 1957, and I followed his proof above. The left side of (1) is known as the entropic uncertainty (or Hirschman uncertainty) of ${f}$. As Hirschman himself conjectured, (1) is not sharp: it can be improved to

$\displaystyle H(|f|^2)+H(|\widehat{f}|^2)\ge 1-\log 2 \ \ \ \ \ (2)$

The reason is that the Hausdorff-Young inequality ${r(p)\le 1}$ is itself not sharp for ${1. It took about twenty years until W. Beckner proved the sharp form of the Hausdorff-Young inequality in his Ph.D. thesis (1975):

$\displaystyle r(p) \le \sqrt{p^{1/p}/q^{1/q}} \ \ \ \ \ (3)$

Here is the plot of the upper bound in (3):

Sharp constant

Since the graph of ${r}$ stays below this curve and touches it at ${(2,1)}$, the derivative ${r'(2)}$ is no less than the slope of the curve at ${p=2}$, which is ${(1-\log 2)/4}$. Recalling that ${H(|f|^2)+H(|\widehat{f}|^2)=4r'(2)}$, we arrive at (2).

The best known form of the uncertainty principle is due to H. Weyl:

$\displaystyle \|\,x f\,\|_2 \cdot \|\,\xi \widehat{f}\,\|_2 \ge \frac{1}{4\pi}\|f\|_2^2 \ \ \ \ \ (4)$

Although (4) can be derived from (2), this route is rather inefficient: Beckner’s theorem is hard, while a direct proof of (4) takes only a few lines: integration by parts ${\int |f(x)|^2\,dx = -\int x \frac{d}{dx}|f(x)|^2\,dx }$, chain rule and the Cauchy-Schwarz inequality.

But we can take another direction and use (1) (not the hard, sharp form (2)) to obtain the following inequality, also due to Hirschman: for every ${\alpha>0}$ there is ${C_\alpha>0}$ such that

$\displaystyle \|\,|x|^\alpha f\,\|_2 \cdot \|\,|\xi|^\alpha \widehat{f}\,\|_2 \ge C_\alpha \|f\|_2^2 \ \ \ \ \ (5)$

It is convenient to normalize ${f}$ so that ${\|f\|_2=1}$. This makes ${\rho =|f|^2}$ a probability distribution (and ${|\widehat f |^2 }$ as well). Our goal is to show that for any probability distribution ${\rho}$

$\displaystyle H(\rho) \le \alpha^{-1} \log \int |x|^\alpha \rho(x)\,dx + B_\alpha \ \ \ \ \ (6)$

where ${B_\alpha}$ depends only on ${\alpha}$. Clearly, (1) and (6) imply (5).

A peculiar feature of (6) is that ${x}$ appears in the integral on the right, but not on the left. This naturally makes one wonder how (6) behaves under scaling ${\rho_\lambda(x)=\lambda \rho (\lambda x)}$. Well, wonder no more—

$\displaystyle H(\rho_\lambda)= H(\rho) - \log \lambda \int \rho = H(\rho)- \log\lambda$

and

$\displaystyle \int |x|^\alpha \rho_\lambda (x)\,dx =\lambda^{-\alpha} \int |x|^\alpha \rho (x)\,dx$

Thus, both sides of (6) change by ${-\log \lambda}$. The inequality passed the scaling test, and now we turn scaling to our advantage by making ${\int |x|^\alpha \rho(x)\,dx =1}$. This reduces (6) to ${H(\rho)\le B_\alpha}$.

Now comes a clever trick (due to Beckner): introduce another probability measure ${d\gamma = A_\alpha \exp(-|x|^\alpha/\alpha)\,dx}$ where ${A_\alpha}$ is a normalizing factor. Let ${\phi(x) = A_\alpha^{-1}\exp( |x|^\alpha/\alpha)\,\rho (x)}$, so that ${\int \phi\,d\gamma=1}$. By Jensen’s inequality,

$\displaystyle \int \phi \log \phi \,d\gamma \ge \int \phi \,d\gamma \cdot \log \int \phi \,d\gamma =0$

On the other hand,

$\displaystyle \int \phi \log \phi \,d\gamma = \int \rho \log \phi \,dx = -\log A_\alpha+\alpha^{-1} -H(\rho)$

and we have desired bound ${H(\rho)\le B_\alpha}$.

Biographical note

Although Wikipedia refers to Isidore Isaac Hirschman, Jr. as a living person, he died in 1990. From the MAA Digital Library:

Halmos photographed analyst Isidore Hirschman (1922-1990) in June of 1960. Hirschman earned his Ph.D. in 1947 from Harvard with the dissertation “Some Representation and Inversion Problems for the Laplace Transform,” written under David Widder. After writing ten papers together, Hirschman and Widder published the book The Convolution Transform in 1955 (Princeton University Press; now available from Dover Publications). Hirschman spent most of his career (1949-1978) at Washington University in St. Louis, Missouri, where he published mainly in harmonic analysis and operator theory.

1. Beckner, William. “Inequalities in Fourier analysis.” Ann. of Math.(2) 102.1 (1975): 159-182.
2. Cowling, Michael G., and John F. Price. “Bandwidth versus time concentration: the Heisenberg-Pauli-Weyl inequality.” SIAM Journal on Mathematical Analysis 15.1 (1984): 151-165.
3. Folland, Gerald B., and Alladi Sitaram. “The uncertainty principle: a mathematical survey.” Journal of Fourier Analysis and Applications 3.3 (1997): 207-238.
4. Hirschman Jr, I. I. “A note on entropy.” American Journal of Mathematics 79.1 (1957): 152-156.

Having exported the Shared/Starred items from Google Reader, I considered several ways of keeping them around, and eventually decided to put them right here (see “Reading Lists” in the main navigation bar). My reasons were:

• This site is unlikely to disappear overnight, since I am paying WordPress to host it.
• Simple, structured HTML is not much harder for computers to parse than JSON, and is easier for humans to use.
• Someone else might be interested in the stuff that I find interesting.

First step was to trim down shared.json and starred.json, and flatten them into a comma-separated list. For this I used Google Refine, which, coincidentally, is another tool that Google no longer develops. (It is being continued by volunteers as an open-source project OpenRefine). The only attributes I kept were

• Title
• Author(s)
• Feed name
• Timestamp
• Summary

Springer feeds gave me more headache than anything else in this process. They bundled the authors, abstract, and other information into a single string, within which they were divided only by presentational markup. I cleaned up some of Springer-published entries, but also deleted many more than I kept. “Sorry, folks – your paper looks interesting, but it’s in Springer.”

Compared to the clean-up, conversion to HTML (in a Google Spreadsheet) took no time at all. I split the list into chunks 2008-09, 2010-11, 2012, and 2013, the latter being updated periodically. Decided not to do anything about LaTeX markup; much of it would not compile anyway.

Last question: how to keep the 2013 list up-to-date, given that Netvibes offers no export feature for marked items? jQuery to the rescue:

Now with a freehand circle!

The script parses the contents of Read Later tab, extracts the items mentioned above, and wraps them into the same HTML format as the existing reading list.

$(document).ready(function() { function exportSaved() { var html='';$('.hentry.readlater').each(function() {
var title = $('.entry-innerTitle', this).html().split('<span')[0]; var hyperref =$('.entry-innerTitle', this).attr('href');
var summary = ($('.entry-content', this).html()+'').replace('\n',' '); var timestamp =$('.entry-innerPublished', this).attr('time');
var feed = $('.entry-innerFeedName', this).html(); var author = ($('.author', this).html()+'').slice(5);
var date = new Date(timestamp*1000);
var month = ('0'+(date.getMonth()+1)).slice(-2);
var day = ('0'+date.getDate()).slice(-2);
html = html+"<div class='list-item'><h4 class='title'><strong><a href='"+hyperref+"'>"+title+"</a></strong></h4><h4 class='author'>"+author+"</h4><h5 class='source'>"+feed+" : "+readableDate+"</h5><div class='summary' style='font-size:80%;line-height:140%;'>"+summary+"</div></div>";
});
console.log(html);
}
$('export').insertAfter($('#switch-view')).click(function() { exportSaved(); });
});

The script accesses the page via an extension in Google Chrome. It labels HTML elements with classes for the convenience of future parsing, not for styling. I had to use inline styles to keep WordPress happy. Dates are formatted according to ISO 8601, making it possible to quickly jump to any month using in-page search for “yyyy-mm”. Which would not work with mm/dd/yyyy, of course.

Unfortunately, some publishers still insert the authors’ names in the summary. Unlike arXiv, which serves reasonably structured feeds, albeit with unnecessary hyperlinks in the author field. While I am unhappy with such things (and the state of the web in general), my setup works for me, and this concludes my two-post detour from mathematics.

xkcd 743: Infrastructures


text:
This was part of Systemic Risk and Stress Testing

## While Stability Lasts: A Stochastic Model of Stablecoins

Andreea Minca, Cornell University
Wednesday, April 6, 2022

Abstract: The Black Thursday' crisis in cryptocurrency markets demonstrated deleveraging risks in over-collateralized stablecoins. We develop a stochastic model that helps explain such crises. In our model, the stablecoin supply is decided by speculators who optimize the profitability of a leveraged position while incorporating the forward-looking cost of collateral liquidations, which involves the endogenous price of the stablecoin. We formally characterize regimes that are interpreted as stable and unstable for the stablecoin. We prove bounds on quadratic variation and the probability of large deviations in the stable domain and we demonstrate distinctly greater price variance in the unstable domain. We identify a deflationary deleveraging spiral by means of a submartingale.   These deleveraging spirals, which resemble short squeezes, lead to faster collateral drawdown (and potential shortfalls) and are accompanied by higher price variance, as experienced on Black Thursday. We also demonstrate perfect' stability results in idealized settings and discuss mechanisms which could bring realistic settings closer to such idealized stable settings.

format: plain


text:
# operator efficiency

This topic is 5394 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

hi all, I have a question about operator overloading and the implications of using auxiliary operators over member functions. For the sake of simplicity and argument I have a class, say CPosition, with x, y and z doubles. to add two positions together I do:
...
CPos a(...), b(...), c(...);
...

CPos &CPos::Add(const CPos& l, const CPos& r) {
x = l.x + r.x;
y = l.y + r.y;
z = l.z + r.z;
return *this;
}

which enables me to call the very (what I think is) efficient:
//assign a + b to C

now if I make an auxiliary operator for this, I do something along the lines of:
const CPos operator+ (const CPos &l, const CPos &r) {
}

//so the user has the option of doing:

a = b + c;  //slow
a = b + 1;  //slow
a = 1 + b;  //slow works because of NON-member operator
a.(b,c);    //fastest

I admit my knowledge of this area is shaky, so please point out any stupid mistakes i have made, but my query is basically along the lines of: can I make an operator that works just as fast as the member function? I know it is slower because it returns by value rather than reference. Can operators return by reference? Thanks.

##### Share on other sites
Hi,

optimization too much.

number of redundant function calls to a minimum.

Here is how i would re-implement your system:
class CPos {public: float x, y, z; // add appropriate constructor code here. const CPos operator +(const CPos& rhs ) {   CPos r;   r.x = x + rhs.x;   r.y = y + rhs.y;   r.z = z + rhs.z;   return r; //de-reference 'this' pointer and return; }};

do you understand whats going on above ?

EDIT: dont write code when you have the flu.

dont be affraid to use the features of C++, when used
appropriatly and sensibly, performace (in comparison with other
OO languages) is not even an issue.

Also remember, that the clearer and cleaner your code is at
showing the compiler its intentions, the better job the compiler
can do at optimizing it (in most cases, far better than a human
ever could)

Cheers,
Danu

PS. me thinks we need a small FAQ section on optimization or something.

[Edited by - silvermace on October 10, 2004 12:44:37 AM]

##### Share on other sites
Hi, thanks for your reply. I just want to get the very basic stuff optimized like crazy :)

I understand that what you've written is going to be as fast as it gets (being an inlined member operator returning an address) - is that right?

I think part of my worry is balancing maintainability with efficiency... as in your case I believe my second example of what the user can do:

a = 1 + b;// will not work (as it will try)a.operator=(1.operator+(b));or something to this effect

Well thanks for clearing this up for me :) Being a graphics programming student we really are drilled into our heads how important efficiency at every step of the way is!

##### Share on other sites
Quote:
 Original post by silvermaceHi,You shouldn't be worried too much about this level of optimization too much.but, to answer your question, you'd probably want to keep thenumber of redundant function calls to a minimum.Here is how i would re-implement your system:*** Source Snippet Removed ***do you understand whats going on above ?

This is a bad idea. You have redefined the fundamental behavior of operator+ to be mutating. Instead of using operator+, you should be using operator +=.
Quote:
 dont be affraid to use the features of C++, when used appropriatly and sensibly, performace (in comparison with otherOO languages) is not even an issue.

Yes, and with that in mind, always make sure that your operators behave as one would expect them to. operator+ should be non-mutating, for instance.

##### Share on other sites
I think silvermace is correct in proposing this function, because the '+' operator is returning the resulting value, very much like '+='.
The difference is that '+' returns the result as an expression which may or may not be used, whereas '+=' stores the result into the assigned variable.

Edit: This is wrong, don't believe it! See next post
const CPos &operator+(const CPos &rvalue) {
x += rvalue.x;
y += rvalue.y;
z += rvalue.z;
return *this;
}

This enables you to execute "b + c", which will be executed like "b.operator+(c)"

In order to execute "a = b + c", you additionally need the assignment operator defined, like so:

const CPos &operator=(const CPos &rvalue) {
x = rvalue.x;
y = rvalue.y;
z = rvalue.z;
return *this;
}

"a = b + c" will then be executed like "a.operator=(b.operator+(c))"

You could also implement the '+=' operator if you'd like to use it. Other than that, I would strongly recommend not to implement any operators which do not behave exactly as they are supposed to behave with primitive types, because nobody other than you (and at a later time maybe not even you) will expect them behave in a "wrong" way.
If you need to have other behaviour, best stick with normal named functions like "CPos::assignSum(const CPos& value1, const CPos& value2)"

This is all just about one extra member-function call, which is really very little in execution time. I don't know how serious performance is to you, but I can hardly imagine that the difference is going to matter.

See, if speed really is that important, it might be better not to use objects at all - and rather do it all with primitive integers in pure C style. The benefit of an object model like you are using is readability - readabilty traded against a tiny little bit of speed. So does it really make sense then to give away readability only to spare a single function call, just to regain a hardly measurable fraction of that speed?

AV

[Edited by - beefsteak on October 9, 2004 8:12:43 PM]

##### Share on other sites
I gave it a second though, and I think Washu is right about the '+' operator, and I was wrong. :-)
I mustn't change the values of the object instance itself, because after "b + c", b must still have the same value inside.

##### Share on other sites
Quote:
 Original post by beefsteakI gave it a second though, and I think Washu is right about the '+' operator, and I was wrong. :-)I mustn't change the values of the object instance itself, because after "b + c", b must still have the same value inside.

I take back all of the nasty things I said about your name just a second ago...well, ok, not all of them.

It is important to make sure that code is always consistent. When you come back to a code base even a few months later, then something stupid like making operator+ mutating would be completely forgotten. So you would make obvious mistakes, and bugs would run rampant.

##### Share on other sites
Quote:
 Original post by StrikenI just want to get the very basic stuff optimized like crazy :)
Bad idea. You should want to get the basic stuff as correct as possible. No matter how much you work on your Add method, it is likely not going to be the bottleneck in your application.

Never optimize without profiling first. At this stage, focus on writing syntactically and semantically correct code. Once that builds and performs correctly, then you want to profile your application to determine what portions spend the most time executing. You'll want to examine those portions to determine if you can make any simple changes to improve efficiency, followed by algorithm changes, finally culminating in code-intensive but probably platform-specific optimizations.

##### Share on other sites
Washu has it right. Non-mutable operators should retain that status. Take the folowing example with operator + acting as silvermace described:

Cpos a(1,1,1);
Cpos b(2,2,2);

Cpos c = a + b;

now, c = (3,3,3) and b = (2,2,2) but guess what: a also = (3,3,3), which is incorrect because a new value was never assigned to a.

##### Share on other sites
Heh, I'm glad I just recognize my error before getting the head chopped off. :-)

So, now I tried to write up a '+' function that may actually work, but don't slap me if I'm wrong again. I just gave it a shot. :-)

The following function needs both an assignment operator ('=') and an empty constructor for CPos defined in order to work properly

const CPos operator+(const CPos &rvalue) const {
CPos result;
result.x = x + rvalue.x;
result.y = y + rvalue.y;
result.z = z + rvalue.z;
return result;
}

AV

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text:
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text:
SKYRIM
Nightingale Dock by reggert
Updated: 28/02/2012 - 03:26PM

28 Endorsements

615 Unique D/Ls

940 Total D/Ls

17,952 Total Views

## Description

Last updated at 15:26, 28 Feb 2012 Uploaded at 15:26, 28 Feb 2012

This is just an extension of the dock connected to Honeyside in Riften. It adds no new functionality, it's just there to look cool. Nightingale/Nocturnal fans should like it.

Install:
Add plug-in NightingaleDock to C:\Program Files (x86)\Steam\steamapps\common\skyrim\Data

Unintall:
Remove plug-in NightingaleDock from C:\Program Files (x86)\Steam\steamapps\common\skyrim\Data

Notes: This was my CK learning project. I learned i have a love-hate relationship with navmesh.
After all is said and done i believe i got the navmesh down. Please inform me of any issues, as of now I see none.

My Other Mods:
Freeroam Housecarls


text:
## Assignment: Financial Literacy Problem Set

### Skills – Create a Calculator

Create a spreadsheet calculator using the Microsoft Excel™ FV function (Future Value)

Excel™ and other spreadsheets use a built-in function called future value to calculate the future value of an investment or savings account given information about a fixed interest rate and periodic payments.

To be able to use future value to answer multiple problems, you’ll need to set up some cells to collect the input data and a cell to use the function to calculate the output. Open a new spreadsheet and being by setting up cells to collect input data. You’ll need the following values (the Excel™ functions codes are in parentheses by the name of each value.

FV(rate, nper, pmt, [pv], [type]), where rate is the annual % rate,/ periods per year, nper is number of years * periods per year, pmt is (-)payment, [pv] is (optional) present value, and [type] is (optional) whether payment is applied to beginning or end of the period (will will omit).

Annual % rate: the interest rate per period, not in decimal form (for example, if the rate is 5%, enter 5)

Periods per year: The number of payments per year, or the number of compoundings per year.

Number of years: The number of years you’ll make payments. (Use months/12 to enter a partial year)

Payment: The amount you save per payment (enter a positive value). The amount you pay on a loan per payment (enter a negative value). The negative sign used in FV to represent payments to savings deposits has been embedded in the function formula, so this instruction is opposite the one you would find in Excel™ documentation.

Present Value: Any lump sum already existing in the account before regular payments begin, such as a loan balance or a lump sum you would use to open a savings account. Follow the same convention given above for payments vs dividends.

Future Value: Somewhere nearby the above values, place a Future Value cell. This is the cell that will compute the future value of your investment or loan based on your input values.

The following instruction assumes the input values are in the following cells.

Annual % Rate: B3

Periods per year: B4

Number of years: B5

Payment: B6

Present Value: B7

Future Value: into the empty cell where you want the future value computation to appear, type

=FV(B3%/B4, B5*B4, -B6, -B7)

### Skills – Use the Calculator

Use the either the spreadsheet calculators your built in the chapter or your new future value calculator to answer each the following problems

1. You borrow $1,200 from your brother to buy a new laptop and agree to pay it back with a one-time 10% interest by making monthly payments for one year. (a) How much is each monthly payment? (b) How much total interest will you pay? 1. Suppose you deposit$750 into a savings account paying 1.8% annual interest and leave it there for 3 years. How much will you have in the account (principal plus interest) at the end of 3 years?
2. A Treasury bond matures in 30 years and pay an annual yield of 4% twice a year for the entire term. At the end of the term, you receive the original amount of the bond back. (You cannot use the FV function for this problem without modification, as it assumes annual compounding. Use the appropriate spreadsheet). How much interest will you have received in total if you allow a $10,000 bond to mature? 3. You deposit$2,750 into a 5 year CD that pays 1.2% interest, compounded daily. How much money will you have when the CD matures?
4. What amount would you need to deposit into a CD paying 1.89% compounded monthly for 3 years to have $5,000 at the end of the term? Use your spreadsheet calculator to answer this question exactly. 5. You deposit$150 per month for 10 years into a savings annuity that pays 4.23%. How much will you have in the account after 10 years?
6. You’ve saved $300,000 for retirement and are looking at an annuity with a rate of 3%. (a) If you want to withdraw$1,750 per month, how much money will still be in the account 20 years later?

(b)  How much money will you have received in total over the 20 years

1. You’d like to purchase a house for $325,000. You’ve found a 30 year mortgage at 3.99% and have$35,000 to put down. Use your loans calculator to answer this question.

(a) How much will you need to borrow?

(b) How much will your monthly payment be?

(c) How much will you have paid the bank by the end of the 30 years?

(d) If you increase your monthly payment by $500, how many years will it take to pay off the loan? 1. You have a credit card with a$9,000 balance on it and would like to pay it off in 3 years. The interest rate is 11.5%. How much will you need to pay each month if you don’t accrue any new charges?
2. You’ve looked at your budget, and you can afford $425 per month for a car payment. You have a trade-in worth$7,000. How expensive a car can you purchase if the loan terms were 5% interest for 4 years?

### Concepts and Explorations

1. A CD-ladder is a way to be able to earn higher interest on longer term CDs while still keeping some of your money more accessible in the short run. To build a CD ladder, take some amount of money you’d like to invest and spread it out over several CDs with different maturity terms. Then, as the CDs mature, take the money and reinvest it in the highest interest CD.

Say you receive a lump sum of $20,000 and would like to invest it but don’t know if you’ll need access to any of it in a few years. So, you decide to build a CD ladder by investing 5 equal parts of the sum into 5 separate CDs, maturing in 1, 2, 3, 4, and 5 years, respectively. Assume all CDs compound daily. (a) If the CDs have the interest rates shown below, how much money would you have in 10 years, assuming that when you get to the 5 year CD, you reinvest it in for another 5 years? 1-year CD rate 0.9%; 2-year CD rate 1.02%; 3-year CD rate 1.4%; 4-year CD rate 1.6 %. 5-year CD rate 1.89% (b) If you had invested the entire$20,000 into a 10 year CD paying 1.89%, how much more money would you have had?

(c) Look online or in your local newspaper to find banks advertising CD rates. Make a CD ladder of your own with an imaginary windfall. It doesn’t have to be $20k. Imagine you received a large commission on a sale or a hiring bonus. Is there a different savings or investment product that you can find that looks interesting? Speculate and compare rates to see what might earn the bigger return. 1. You take out a 30 year mortgage at 5% interest for$150,000 to purchase a house.

(a) How much are your monthly payments?

(b) How much will you still have to pay on the loan after paying on it for 20 years?

(c) Find the first month’s payment in which your payment applies more money to your principal than to the interest. (Hint: use an amortization schedule).

1. You purchased two pairs of jeans, new basketball shoes, and a hoodie at your favorite store last month. It was a good deal. You applied for the store credit card and received 30% off when you were approved. The total charge came to $263.41. That’s$355 for the sale, less 30% plus tax. The card comes with an APR of 23%.

(a) How much will you need to pay each month if you want to pay the card off in 10 months?

(b) If you take 10 months to pay the card off, will the interest you pay overcome the 30% discount on the purchase? How much will you have paid in interest?

(c) If you took 10 months to pay off the card, how much of a discount did you end up with? (Subtract the interest paid from the original discount, then find the percentage)

(d) If you only made the minimum payment of $10 per month, how long would it take to pay off the card? (e) How much in interest would you have paid making only the minimum monthly payment of$10?

(f) Discuss the pros and cons of using a high-interest store card to make purchases. Look up the terms for the cards for stores that you shop and compare. Which of your favorite stores offers the best deal?  Include considerations such as discounts and privileges that come with these cards as well as how it could help or hurt someone trying to build a strong credit history.