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# The Unapologetic Mathematician

## Iterated Integrals V

Our iterated integrals worked out nicely over $n$-dimensional intervals because these regions are simple products of one-dimensional intervals. When working over more general sets, though, it’s not so nice. Still, we can often do almost as well.

Unlike our earlier approach, though, we won’t peel off an integral from the outside first, but from an inside. That is, instead of writing

$\displaystyle\int\limits_Rf(x)\,dx=\int\limits_{a^k}^{b^k}\int\limits_{R_k}f(\hat{x},x^k)\,d\hat{x}\,dx^k$

we’ll write

$\displaystyle\int\limits_Rf(x)\,dx=\int\limits_{R_k}\int\limits_{a^k}^{b^k}f(\hat{x},x^k)\,dx^k\,d\hat{x}$

Underlying all of these equations is the assumption that each of the integrals we write down exists, or at least the inner ones don’t fail to exist often enough to cause the outer ones to also fail to exist.

Okay, so let’s say that we’ve got some bounded region $S$ contained in an $n$-dimensional interval $R$. Just like before, we project $S$ into the plane with $x^k=0$ to get the $n-1$-dimensional region $S_k$. This will be contained in the projection $R_k$ of $R$. Again, we write $\hat{x}$ for the remaining $n-1$ variables.

Here’s where our restriction comes in: assume that $S$ is contained between the graphs of two integrable functions on $S_k$. That is, we want

$\displaystyle \mathrm{Cl}(S)=\left\{\left(x^1,\dots,x^n\right)\big\vert\hat{x}\in\mathrm{Cl}(S_k),g_1(\hat{x})\leq x^k\leq g_2(\hat{x})\right\}$

where $g_1:S_k\rightarrow\mathbb{R}$ and $g_2:S_k\rightarrow\mathbb{R}$ are two integrable functions defined on $S_k$.

Now we can try to evaluate the integral

\displaystyle\begin{aligned}\int\limits_Sf\,dx&=\int\limits_Rf(x)\chi_S(x)\,dx\\&=\int\limits_{R_k}\int\limits_{a^k}^{b^k}f(x)\chi_S(x)\,dx^k\,d\hat{x}\\&=\int\limits_{S_k}\int\limits_{a^k}^{b^k}f(x)\chi_S(x)\,dx^k\,d\hat{x}\\&=\int\limits_{S_k}\int\limits_{g_1(\hat{x})}^{g_2(\hat{x})}f(x)\chi_S(x)\,dx^k\,d\hat{x}\\&=\int\limits_{S_k}\int\limits_{g_1(\hat{x})}^{g_2(\hat{x})}f(x)\,dx^k\,d\hat{x}\end{aligned}

We first replace $R_k$ by $S_k$ because $\chi_S(x)$ is identically zero wherever $\hat{x}$ is outside of $S_k$, and thus so will the whole inner integral. Then we replace the limits of integration for the inner integral by $g_1(\hat{x})$ and $g_2(\hat{x})$ because for any fixed $\hat{x}$ the function $\chi_S(x)$ is zero for $x^k$ below the former and above the latter.

If we’re lucky, $S_k$ itself can be written in a similar form, as the region between the graphs of two integrable functions in another coordinate direction, and so on. Each inner integral’s integrand depends on the remaining variables of later integrals, as before. But now the bounds of integration can also depend on later variables, which adds new complication. However, once a given variable has been integrated away, nothing further out ever depends on it again, and so this whole procedure is still well-defined.

December 31, 2009 Posted by | Analysis, Calculus | 9 Comments

## Integrals are Additive Over Regions

Before I state today’s proposition, I need to define what I mean by saying that two sets in $\mathbb{R}^n$ are “nonoverlapping”. Intuitively, we might think that this means they have no intersection, but that’s not quite it. We’ll allow some intersection, but only at boundary points. Since the regions we’re interested in for integrals are Jordan measurable, and their boundaries have zero Jordan content, so we know changing things along these boundaries in an integral will make no difference.

Let $\left\{S_i\right\}_{i=1}^n$ be a collection of bounded regions in $\mathbb{R}^n$, so that any two of these regions are nonoverlapping. We define their union

$\displaystyle S=\bigcup\limits_{i=1}^nS_i$

and let $f:S\rightarrow\mathbb{R}$ be a bounded function defined on this union. Then $f$ is integrable on $S$ if and only if it’s integrable on each $S_i$, and we find

$\displaystyle\int\limits_Sf\,dx=\sum\limits_{i=1}^n\int\limits_{S_i}f\,dx$

Indeed, if $R$ is some $n$-dimensional interval containing $S$, then it will also contain each of the smaller regions $S_i$, and we can define

$\displaystyle\int\limits_Sf\,dx=\int\limits_Rf(x)\chi_S(x)\,dx$

and

$\displaystyle\int\limits_{S_i}f\,dx=\int\limits_Rf(x)\chi_{S_i}(x)\,dx$

We can use Lebesgue’s condition to establish our assertion about integrability. On the one hand, the discontinuities of $f$ in $S_i$ must be contained within those of $S$, so if the latter set has measure zero then so must the former. On the other hand, the discontinuities of $f$ consist of those within each of the $S_i$, and maybe some along the boundaries. Since the boundaries have measure zero, and we assume that the discontinuities in each $S_i$ are of measure zero, their (countable) union will also have measure zero. And then so must the set of discontinuities of $f$ in $S$ have measure zero as well.

The inclusion-exclusion principle tells us that we can rewrite the characteristic function of $S$:

\displaystyle\begin{aligned}\chi_S&=\chi_{S_1\cup\dots\cup S_n}\\&=\sum\limits_{1\leq i\leq n}\chi_{S_i}-\sum\limits_{1\leq i

We can put this into our integral and use the fact that integrals are additive with respect to finite sums in the integrand

\displaystyle\begin{aligned}\int\limits_Sf\,dx&=\int\limits_Rf(x)\chi_S(x)\,dx\\&=\sum\limits_{1\leq i\leq n}\int\limits_Rf(x)\chi_{S_i}(x)\,dx-\sum\limits_{1\leq i

But we assumed that all the $S_i$ are nonoverlapping, so any intersection of two or more of them must lie only along their boundaries. And since these boundaries all have Jordan content zero the integrals over them must come out to zero. We’re left with only the sum over each subregion, as we wanted.

## The Mean Value Theorem for Multiple Integrals

As in the single variable case, multiple integrals satisfy a mean value property.

First of all, we should note that, like one-dimensional Riemann-Stieltjes integrals with increasing integrators, integration preserves order. That is, if $f$ and $g$ are both integrable over a Jordan-measurable set $S$, and if $f(x)\leq g(x)$ at each point $x\in S$, then we have

$\displaystyle\int\limits_Sf(x)\,dx\leq\int\limits_Sg(x)\,dx$

This is a simple consequence of the definition of a multiple integral as the limit of Riemann sums, since every Riemann sum for $f$ will be smaller than the corresponding sum for $g$.

Now if $f$ and $g$ are integrable on $S$ and $g(x)\geq0$ for every $x\in S$, then we set $m=\inf f(S)$ and $M=\sup f(S)$ — the infimum and supremum of the values attained by $f$ on $S$. I assert that there is some $\lambda$ in the interval $[m,M]$ so that

$\displaystyle\int\limits_Sf(x)g(x)\,dx=\lambda\int\limits_Sg(x)\,dx$

In particular, we can set $g(x)=1$ and find

$\displaystyle mc(S)\leq\int\limits_Sf(x)\,dx\leq Mc(S)$

giving bounds on the integral in terms of the Jordan content of $S$. Incidentally, $g(x)\,dx$ here is serving a similar role to the integrator $d\alpha$ in the integral mean value theorem for Riemann-Stieltjes integrals.

Okay, so since $g(x)\geq0$ we have $mg(x)\leq f(x)g(x)\leq Mg(x)$ for every $x\in S$. Since integration preserves order, this yields

$\displaystyle m\int\limits_Sg(x)\,dx\leq\int\limits_Sf(x)g(x)\,dx\leq M\int\limits_Sg(x)\,dx$

If the integral of $g$ is zero, then our result automatically holds for any value of $\lambda$. Otherwise we can divide through by this integral and set

$\displaystyle\lambda=\frac{\displaystyle\int\limits_Sf(x)g(x)\,dx}{\displaystyle\int\limits_Sg(x)\,dx}$

which will be between $m$ and $M$.

One particularly useful case is when $S$ has Jordan content zero. In this case, we find that any integral over $S$ is itself automatically zero.

December 29, 2009 Posted by | Analysis, Calculus | 3 Comments

## Inclusion-Exclusion Again

Now that we know a little more about characteristic functions, let’s see how they can be used to understand the inclusion-exclusion principle. Our first pass was through a very categorified lens, which was a neat tie-in to Euler characteristics. But we want something a little more down-to-Earth.

So, let’s take a finite collection $\left\{S_i\right\}_{i=1}^n$ of subsets of a set $X$. To each one we have a corresponding idempotent function $\chi_{S_i}$, which is ${1}$ for points in $S_i$ and ${0}$ elsewhere. We can take the complement of a subset in the language of idempotents as usual: $1-\chi_{S_i}=\chi_{{S_i}^c}$. This function is ${1}$ for points not in $S_i$, and ${0}$ in $S_i$.

Now let’s take and multiply all these complementary functions together to get

$\displaystyle\prod\limits_{i=1}^n\left(1-\chi_{S_i}\right)=\prod\limits_{i=1}^n\chi_{{S_i}^c}=\chi_{{S_1}^c\cap\dots\cap{S_n}^c}$

By DeMorgan’s Laws, the intersection of all these complements is the complement of the union of all the $S_i$. Thus

$\displaystyle\prod\limits_{i=1}^n\left(1-\chi_{S_i}\right)=\chi_{(S_1\cup\dots\cup S_n)^c}=1-\chi_{S_1\cup\dots\cup S_n}$

Now let’s expand out the product, just like we expand out the product of any sequence of binomials. In each factor, we choose which of the ${1}$ or the $\chi_{S_i}$ to contribute. First we get the term where we always choose ${1}$, then the terms where we choose exactly one of the $\chi_{S_i}$, then the terms where we choose exactly two of them, and so on until we choose all $n$ of them. The sign of a term is negative when we choose an odd number of the $\chi_{S_i}$, and positive when we choose an even number. That is

\displaystyle\begin{aligned}\prod\limits_{i=1}^n\left(1-\chi_{S_i}\right)&=1-\sum\limits_{1\leq i\leq n}\chi_{S_i}+\sum\limits_{1\leq i

Putting this together with the earlier expression, we find

$\displaystyle\chi_{S_1\cup\dots\cup S_n}=\sum\limits_{1\leq i\leq n}\chi_{S_i}-\sum\limits_{1\leq i

For $n=2$ this gives us back our formula for the characteristic function of the intersection of two sets. For larger $n$ it gives us the inclusion-exclusion principle in terms of characteristic functions.

December 28, 2009 Posted by | Fundamentals | 1 Comment

## Characteristic Functions as Idempotents

I just talked about characteristic functions as masks on other functions. Given a function $f:X\rightarrow\mathbb{R}$ and a subset $S\subseteq X$, we can mask the function $f$ to the subset $S$ by multiplying it by the characteristic function $\chi_S$. I want to talk a little more about these functions and how they relate to set theory.

First of all, it’s easy to recognize a characteristic function when we see one: they’re exactly the idempotent functions. That is, ${\chi_S}^2=\chi_S$, and if $f^2=f$ then $f$ must be $\chi_S$ for some set $S$. Indeed, given a real number $a$, we can only have $a^2=a$ if $a=0$ or $a=1$. That is, $f(x)=0$ or $f(x)=1$ for every $x$. So we can define $S$ to be the set of $x\in X$ for which $f(x)=1$, and then $f(x)=\chi_S(x)$ for every $x\in X$. Thus the idempotents in the algebra of real-valued functions on $X$ correspond exactly to the subsets of $X$.

We can define two operations on such idempotent functions to make them into a lattice. The easier to define is the meet. Given idempotents $\chi_S$ and $\chi_T$ we define the meet to be their product:

$\displaystyle\left[\chi_S\wedge\chi_T\right](x)=\chi_S(x)\chi_T(x)$

This function will take the value ${1}$ at a point $x$ if and only if both $\chi_S$ and $\chi_T$ do, so this is the characteristic function of the intersection

$\displaystyle\chi_S\wedge\chi_T=\chi_{S\cap T}$

We might hope that the join would be the sum of two idempotents, but in general this will not be another idempotent. Indeed, we can check:

$\displaystyle(\chi_S+\chi_T)^2={\chi_S}^2+{\chi_T}^2+2\chi_S\chi_T=(\chi_S+\chi_T)+2\chi_{S\cap T}$

We have a problem exactly when the corresponding sets have a nonempty intersection, which leads us to think that maybe this has something to do with the inclusion-exclusion principle. We’re “overcounting” the intersection by just adding, so let’s subtract it off to define

$\displaystyle\left[\chi_S\vee\chi_T\right](x)=\chi_S(x)+\chi_T(x)-\chi_S(x)\chi_T(x)$

We can multiply this out to check its idempotence, or we could consider its values. If $x$ is not in $T$, then $\chi_T(x)=0$, and we find $\chi_S\vee\chi_T=\chi_S$ — it takes the value ${1}$ if $x\in S$ and ${0}$ otherwise. A similar calculation holds if $x\notin S$, which leaves only the case when $x\in S\cap T$. But now $\chi_S(x)$ and $\chi_T(x)$ both take the value ${1}$, and a quick calculation shows that $\chi_S\vee\chi_T$ does as well. This establishes that

$\displaystyle\chi_S\vee\chi_T=\chi_{S\cup T}$

We can push further and make this into an orthocomplemented lattice. We define the orthocomplement of an idempotent by

$\displaystyle\left[\neg\chi_S\right](x)=1-\chi_S(x)$

This function is ${1}$ wherever $\chi_S$ is ${0}$, and vice-versa. That is, it’s the characteristic function of the complement

$\displaystyle\neg\chi_S=\chi_{X\setminus S}$

So we can take the lattice of subsets of $X$ and realize it in the nice, concrete algebra of real-valued functions on $X$. The objects of the lattice are exactly the idempotents of this algebra, and we can build the meet and join from the algebraic operations of addition and multiplication. In fact, we could turn around and do this for any commutative algebra to create a lattice, which would mimic the “lattice of subsets” of some “set”, which emerges from the algebra. This sort of trick is a key insight to quite a lot of modern geometry.

December 23, 2009 Posted by | Algebra, Fundamentals, Lattices | 2 Comments

## Integrals Over More General Sets

To this point we’ve only discussed multiple integrals over $n$-dimensional intervals. But often we’re interested in more general regions, like circles or ellipsoids or even more rectangular solids that are just tilted with respect to the coordinate axes. How can we handle integrating over these more general sets?

One attempt might be to fill up the region from inside. We can chop up regions into smaller pieces, each of which is an $n$-dimensional interval. But overall this requires an incredibly involved limiting process, and we’ll never get anything calculated that way.

Instead, we come at it from outside. Given a bounded region $S\subseteq\mathbb{R}^n$, we can put it into an $n$-dimensional interval $R$. Then if $f:S\rightarrow\mathbb{R}^n$ is a function, we can try to define some sort of integral of $f$ over $S$ in terms of its integral over $R$.

The obvious problem with $\int_Rf\,dx$ is that it includes all the points in $R$ that aren’t in $S$, and we don’t want to include the integral of $f$ over that region. Worse, what if $f$ has a big cluster of discontinuities within $R$ but outside of $S$? Clearly that shouldn’t make $f$ fail to be integrable over $S$. What we need is to mask off $S$, like a stencil or masking tape does when painting.

The mask we’ll use is the characteristic function of $S$, which I’ve mentioned before. I’ll actually go more deeply into them in a bit, but for now we’ll recall that the characteristic function of a set $S$ is written $\chi_S$, and it’s defined as

$\displaystyle\chi_S(x)=\left\{\begin{matrix}1&:x\in S\\{0}&:x\notin S\end{matrix}\right.$

Now look what happens when we multiply our function by this mask:

$\displaystyle f(x)\chi_S(x)=\left\{\begin{matrix}f(x)&:x\in S\\{0}&:x\notin S\end{matrix}\right.$

Now our function has been redefined outside $S$ by setting it equal to zero there. Then we proceed to define

$\displaystyle\int\limits_Sf(x)\,dx=\int\limits_Rf(x)\chi_S(x)\,dx$

The integral over $R$ should be the “integral” over $S$ plus the “integral” over the region $R\setminus S$ outside $S$. I put these in quotes because we haven’t really defined what these integrals mean (that’s what we’re trying to do!), but we can make reasonable assertions about properties that they should have, whatever they are.

Now $f(x)\chi_S(x)$ is zero on the outer region, so the second integral is zero. And $f(x)\chi_S(x)$ is just equal to $f(x)$ inside $S$, so when “integrating” over $S$ itself we may as well just drop the $\chi_S$ factor. This justifies the definition of integrating over $S$.

It should be clear that this definition doesn’t depend on the interval $R$ at all. Indeed, if we have two different intervals $R_1$ and $R_2$ which both contain $S$, then any region the one contains which the other does not must fall outside of $S$, and $f(x)\chi_S(x)$ will be zero there anyway, and so will the difference between their integrals.

Now, this definition isn’t without its problems. The clearest of which is that we’ve almost certainly introduced new discontinuities. All around the boundary of $S$, if $f$ wasn’t already zero it suddenly and discontinuously becomes zero when we add our mask. This could cause trouble when trying to integrate the masked function over $R$. We must ask that $S$ be Jordan measurable, because this will happen if and only if the boundary $\partial S$ has zero Jordan content, and thus zero outer Lebesgue measure. Since the collection of new discontinuities must be contained in this boundary, it will also have measure zero.

This leads us to an integrability criterion over a Jordan measurable set $S$: $f$ will be integrable over $S$ if and only if the discontinuities of $f$ in $S$ form a set of measure zero. Indeed, Lebesgue’s condition tells us that the discontinuities of $f(x)\chi_S(x)$ must have measure zero. These discontinuities either come from those of $f$ inside $S$, or from the characteristic function $\chi_S$ at the boundary $\partial S$. By assuming that $S$ is Jordan measurable, we force the second kind to have measure zero, and so the total collection of discontinuities will have measure zero and satisfy Lebesgue’s condition if and only if the discontinuities of $f$ inside $S$ do.

December 22, 2009 Posted by | Analysis, Calculus | 4 Comments

## Iterated Integrals IV

So we’ve established that as long as a double integral exists, we can use an iterated integral to evaluate it. What happens when the dimension of our space is even bigger?

In this case, we’re considering integrating an integrable function $f:R\rightarrow\mathbb{R}$ over some $n$-dimensional interval $R=[a^1,b^1]\times\dots\times[a^n,b^n]$. We want something like iterated integrals to allow us to evaluate this multiple integral. We’ll do this by peeling off a single integral from the outside and leaving an integral over an $n-1$-dimensional integral inside.

Specifically, we can project the interval $R$ onto the coordinate hyperplane defined by $x^k=0$ just by leaving the coordinates $x^i$ of each point $x\in R$ the same if $i\neq k$ and setting $x^k=0$. We’ll call the resulting interval

$\displaystyle R_k=[a^1,b^1]\times\dots\times\widehat{[a^k,b^k]}\times\dots\times[a^n,b^n]$

where the wide hat means that we just leave out that one factor in the product. We’ll also write $\hat{x}$ to mean the remaining coordinates on $R_k$.

Essentially, we want to integrate first over $R_k$, and then let $x^k$ run from $a^k$ to $b^k$. We have a collection of assertions that parallel those from the two-dimensional case

• $\displaystyle{\int\limits_-}_Rf(x)\,dx\leq{\int\limits_-}_{a^k}^{b^k}{\int\limits^-}_{R_k}f(\hat{x},x^k)\,d\hat{x}\,dx^k\leq{\int\limits^-}_{a^k}^{b^k}{\int\limits^-}_{R_k}f(\hat{x},x^k)\,d\hat{x}\,dx^k\leq{\int\limits^-}_Rf(x)\,dx$
• $\displaystyle{\int\limits_-}_Rf(x)\,dx\leq{\int\limits_-}_{a^k}^{b^k}{\int\limits_-}_{R_k}f(\hat{x},x^k)\,d\hat{x}\,dx^k\leq{\int\limits^-}_{a^k}^{b^k}{\int\limits_-}_{R_k}f(\hat{x},x^k)\,d\hat{x}\,dx^k\leq{\int\limits^-}_Rf(x)\,dx$
• If $\int_Rf(x)\,dx$ exists, then we have
$\displaystyle\int\limits_Rf(x)\,dx=\int\limits_{a^k}^{b^k}{\int\limits_-}_{R_k}f(\hat{x},x^k)\,d\hat{x}\,dx^k=\int\limits_{a^k}^{b^k}{\int\limits^-}_{R_k}f(\hat{x},x^k)\,d\hat{x}\,dx^k$

with a copy of these three for each index $k$ between ${1}$ and $n$. The proofs of these are pretty much identical to the proofs in the two-dimensional case, and so I’ll just skip them.

Anyhow, once we’ve picked one of the $n$ variables and split it off as the outermost integral, we’re left with an $n-1$-dimensional integral on the inside. We can pick any one of these variables and split it off, leaving an $n-2$-dimensional integral on the inside, and so on. For each of the $n!$ orderings of the original $n$ variables, we get a way of writing the $n$-dimensional integral over $R$ as a sequence of $n$ integrals, each over a one-dimensional interval. Now, we may find some of these iterated integrals easier to evaluate than others, but in principle, if each of the $m$-dimensional integrals in the sequence exists it doesn’t matter which of the orderings we use.

So, for example, if we’re considering a bounded function $f$ defined on a three-dimensional interval $R=[a^1,b^1]\times[a^2,b^2]\times[a^3,b^3]$, we can write (up to) six different iterated integrals, assuming that all the integrals in sight exist.

\displaystyle\begin{aligned}\iiint\limits_R&=\int\limits_{a^3}^{b^3}\int\limits_{a^2}^{b^2}\int\limits_{a^1}^{b^1}f(x,y,z)\,dx\,dy\,dz=\int\limits_{a^2}^{b^2}\int\limits_{a^3}^{b^3}\int\limits_{a^1}^{b^1}f(x,y,z)\,dx\,dz\,dy\\&=\int\limits_{a^3}^{b^3}\int\limits_{a^1}^{b^1}\int\limits_{a^2}^{b^2}f(x,y,z)\,dy\,dx\,dz=\int\limits_{a^1}^{b^1}\int\limits_{a^3}^{b^3}\int\limits_{a^2}^{b^2}f(x,y,z)\,dy\,dz\,dx\\&=\int\limits_{a^2}^{b^2}\int\limits_{a^1}^{b^1}\int\limits_{a^3}^{b^3}f(x,y,z)\,dz\,dx\,dy=\int\limits_{a^1}^{b^1}\int\limits_{a^2}^{b^2}\int\limits_{a^3}^{b^3}f(x,y,z)\,dz\,dy\,dx\end{aligned}

December 21, 2009 Posted by | Analysis, Calculus | 4 Comments

## Iterated Integrals III

I recently heard a characterization (if someone remembers the source, please let me know) of the situation in analysis as being that there are no theorems — only conjectures that don’t yet have counterexamples. Today’s counterexample is adapted from one in Apostol’s book, but it’s far simpler than his seems to be.

We might guess that we can always evaluate double integrals by iterated integrals as we’ve been discussing. After all, that’s exactly what we do in multivariable calculus courses as soon as we introduce iterated integrals, never looking back to all those messy double and triple Riemann sums again. Unfortunately, the existence of the iterated integrals — even if both of them exist and their values agree — is not enough to guarantee that the double integral exists. Today, we will see a counterexample.

Let $S$ be the set of points $(x,y)$ in the unit square $[0,1]\times[0,1]$ so that $x=\frac{m}{q}$ and $y=\frac{n}{q}$, where $\frac{m}{q}$ and $\frac{n}{q}$ are two fractions with the same denominator, each of which are in lowest terms. That is, it contains the point $\left(\frac{1}{3},\frac{2}{3}\right)$, but not the point $\left(\frac{2}{4},\frac{3}{4}\right)$, since when we write these latter two fractions in lowest terms they are no longer over a common denominator. We will consider the characteristic function $\chi_S$, which is ${1}$ on points in $S$ and ${0}$ elsewhere in the unit square.

First, I assert that both iterated integrals exist and have the same value. That is

$\displaystyle\int\limits_0^1\int\limits_0^1\chi_S(x,y)\,dx\,dy=0=\int\limits_0^1\int\limits_0^1\chi_S(x,y)\,dy\,dx$

Indeed, the set $S$ is symmetric between the two coordinates, so we only need to evaluate one of these iterated integrals and the other one will automatically have the same value.

If $y$ is an irrational number in $[0,1]$, then there is no $x$ at all so that $(x,y)\in S$. Thus we can easily calculate the inner integral

$\displaystyle\int\limits_0^1\chi_S(x,y)\,dx=\int\limits_0^10\,dx=0$

On the other hand, if $y$ is a rational number, we can write it in lowest terms as $y=\frac{n}{q}$. Then there are only a finite number of points $x=\frac{m}{q}$ having the same denominator at all. Thus we can break the interval $[0,1]$ into a finite number of pieces, on each of which the characteristic function has the constant value zero. Thus we can calculate the inner integral

$\displaystyle\int\limits_0^1\chi_S(x,y)\,dx=\sum\limits_{i=1}^q\int\limits_\frac{i-1}{q}^\frac{i}{q}\chi_S\left(x,\frac{n}{q}\right)\,dx=\sum\limits_{i=1}^q\int\limits_\frac{i-1}{q}^\frac{i}{q}0\,dx=\sum\limits_{i=1}^q0=0$

And so we see that for any $y$ the inner integral evaluates to ${0}$. Then it’s easy to calculate the outer integral

$\displaystyle\int\limits_0^10\,dy=0$

and, as we said before, the other iterated integral also has the value ${0}$.

On the other hand, the double integral does not exist. Yes, $S$ is countable, and so it has measure zero. However, it’s also dense, which means $\chi_S$ is discontinuous everywhere in the unit square.

Saying that $S$ is dense in the square means that every neighborhood of every point of the square contains some point of $S$. Indeed, consider a point $(x,y)$ in the square and some radius $\epsilon$. Since the real numbers are Archimedean, we can pick some $N>\frac{2}{\epsilon}$, and as many people on Nick‘s Twitter experiment (remember to follow @DrMathochist!) reminded us, there are infinitely many prime numbers. Thus we can pick a prime $p>\frac{2}{\epsilon}$. Then we can round $x$ up to the next larger fraction of the form $\frac{m}{p}$, which will be in lowest terms unless $m=p$, in which case we round $x$ down to $\frac{p-1}{p}$. Similarly, we can round $y$ up (or down) to a fraction $\frac{n}{p}$ in lowest terms. This gives us a new point $\left(\frac{m}{p},\frac{n}{p}\right)\in S$, and we can calculate the distance

$\displaystyle\left\lVert\left(x-\frac{m}{p},y-\frac{n}{p}\right)\right\rVert=\sqrt{\left(x-\frac{m}{p}\right)^2+\left(y-\frac{n}{p}\right)^2}\leq\sqrt{\frac{2}{p^2}}<\frac{2}{p}<\epsilon$

So there is a point in $S$ within any radius $\epsilon$ of $(x,y)$.

But now when we try to set up the upper and lower integrals to check Riemann’s condition we find that every subinterval of any partition $P$ must contain some points in $S$ and some points not in $S$. The points within $S$ tell us that the upper sum gets a sample value of ${1}$ for each subinterval, giving a total upper sum of ${1}$. Meanwhile the points outside of $S$ tell us that the lower sum gets a sample value of ${0}$ for each subinterval, giving a total lower sum of ${0}$. Clearly Riemann’s condition fails to hold, and thus the double integral

$\displaystyle\iint\limits_{[0,1]\times[0,1]}\chi_S(x,y)\,d(x,y)$

fails to exist, despite the iterated integrals existing and agreeing.

December 18, 2009 Posted by | Analysis, Calculus | 3 Comments

## Iterated Integrals II

Let’s get to proving the assertions we made last time, starting with

$\displaystyle{\int\limits_-}Rf(x,y)\,d(x,y)\leq{\int\limits_-}_a^b{\int\limits^-}_c^df(x,y)\,dy\,dx\leq{\int\limits^-}_a^b{\int\limits^-}_c^df(x,y)\,dy\,dx\leq{\int\limits^-}Rf(x,y)\,d(x,y)$

where $f$ is a bounded function defined on the rectangle $R=[a,b]\times[c,d]$.

We can start by defining

$\displaystyle F(x)={\int\limits^-}_c^df(x,y)\,dy$

And we easily see that $\lvert F(x)\rvert\leq M(d-c)$, where $M$ is the supremum of $\lvert f\rvert$ on the rectangle $R$, so this is a bounded function as well. Thus the upper integral

$\displaystyle\overline{I}={\int\limits^-}_a^bF(x)\,dx={\int\limits^-}_a^b{\int\limits^-}_c^df(x,y)\,dy\,dx$

and the lower integral

$\displaystyle\underline{I}={\int\limits_-}_a^bF(x)\,dx={\int\limits_-}_a^b{\int\limits^-}_c^df(x,y)\,dy\,dx$

are both well-defined.

Now if $P_x=\{x_0,\dots,x_m\}$ is a partition of $[a,b]$, and $P_y=\{y_0,\dots,y_n\}$ is a partition of $[c,d]$, then $P=P_x\times P_y$ is a partition of $R$ into $mn$ subrectangles $R_{ij}$. We will define

\displaystyle\begin{aligned}\overline{I}_{ij}&={\int\limits^-}_{x_{i-1}}^{x_i}{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\,dx\\\underline{I}_{ij}&={\int\limits_-}_{x_{i-1}}^{x_i}{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\,dx\end{aligned}

Clearly, we have

$\displaystyle{\int\limits^-}_c^df(x,y)\,dy=\sum\limits_{j=1}^n{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy$

and so we find

\displaystyle\begin{aligned}{\int\limits^-}_a^b{\int\limits^-}_c^df(x,y)\,dy\,dx&={\int\limits^-}_a^b\sum\limits_{j=1}^n{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\,dx\\&\leq\sum\limits_{j=1}^n{\int\limits^-}_a^b{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\,dx\\&=\sum\limits_{j=1}^n\sum\limits_{i=1}^m{\int\limits^-}_{x_{i-1}}^{x_i}{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\,dx\end{aligned}

That is

$\displaystyle\overline{I}\leq\sum\limits_{j=1}^n\sum\limits_{i=1}^m\overline{I}_{ij}$

and, similarly

$\displaystyle\underline{I}\geq\sum\limits_{j=1}^n\sum\limits_{i=1}^m\underline{I}_{ij}$

We also define $m_{ij}$ and $M_{ij}$ to be the infimum and supremum of $f$ over the rectangle $R_{ij}$, which gives us the inequalities

$\displaystyle m_{ij}(y_j-y_{j-1})\leq{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\leq M_{ij}(y_j-y_{j-1})$

and from here we find

$\displaystyle m_{ij}\mathrm{vol}(R_{ij})\leq{\int\limits_-}_{x_{i-1}}^{x_i}{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\,dx\leq{\int\limits^-}_{x_{i-1}}^{x_i}{\int\limits^-}_{y_{j-1}}^{y_j}f(x,y)\,dy\,dx\leq M_{ij}\mathrm{vol}(R_{ij})$

Summing on both $i$ and $j$, and sing the above inequalities, we get

$\displaystyle L_P(f)\leq\underline{I}\leq\overline{I}\leq U_P(f)$

and since this holds for all partitions $P$, the assertion that we’re trying to prove follows.

The second assertion from last time can be proven similarly, just replacing $F(x)$ by the lower integral over $[c,d]$. And then the third and fourth assertions are just the same, but interchanging the roles of $[a,b]$ and $[c,d]$. Finally, the last assertion is a consequence of the first four. Indeed, if the integral over $R$ exists, then the upper and lower integrals are equal, which collapses all of the inequalities into equalities.

December 17, 2009 Posted by | Analysis, Calculus | 1 Comment

## Iterated Integrals I

We may remember from a multivariable calculus class that we can evaluate multiple integrals by using iterated integrals. For example, if $f:R\rightarrow\mathbb{R}^{\geq0}$ is a continuous, nonnegative function on a two-dimensional rectangle $R=[a,b]\times[c,d]$ then the integral

$\displaystyle\iint\limits_Rf(x,y)\,d(x,y)$

measures the volume contained between the graph $z=f(x,y)$ of the function and the $x$$y$ plane within the rectangle. If we fix some constant $\hat{y}$ between $c$ and $d$ we can calculate the single integral

$\displaystyle\int\limits_a^bf(x,\hat{y})\,dx$

which describes the area that the plane $y=\hat{y}$ cuts out of this volume. It exists because because the integrand is continuous as a function of $x$. In such classes, we make the reasonable assumption that as we vary $\hat{y}$ this area varies continuously. This gives us a continuous function on $[c,d]$, which will then be integrable:

$\displaystyle\int\limits_c^d\left(\int\limits_a^bf(x,y)\,dx\right)\,dy$

This is an “iterated integral”, since we perform more than one integral in sequence. We usually leave out the big parens and trust in the notation to tell us when the inner integral is closed. Our handwaving argument then justifies the belief that this iterated integral is the same as the double integral above. And this is true:

$\displaystyle\iint\limits_Rf(x,y)\,d(x,y)=\int\limits_c^d\int\limits_a^bf(x,y)\,dx\,dy$

but we haven’t really proven it.

Besides, we’re interested in more general situations. What if, say, $f$ is discontinuous along the whole line $(x,\hat{y})$ for some fixed $c\leq\hat{y}\leq d$? This line can be contained in an arbitrarily thin rectangle, so it has outer Lebesgue measure zero in the rectangle $R$. If these are the only discontinuities, then $f$ is integrable on $R$, but we can’t follow the above prescription anymore, even if it were actually rigorous. We need some method of handling this sort of thing.

To this end, we have five assertions relating the upper and lower single and double integrals involving a function $f$ which is defined and bounded on the rectangle $R$ above. Unfortunately, our notation for upper and lower integrals gets a little cumbersome here, and the $\LaTeX$ support on WordPress isn’t the most elegant. Still, we soldier on and write

$\displaystyle{\int\limits_-}_a^bf(x)\,dx=\underline{I}_{[a,b]}(f)$

and similarly for upper integrals, and for lower and upper double integrals. Now, our assertions:

• $\displaystyle{\int\limits_-}Rf(x,y)\,d(x,y)\leq{\int\limits_-}_a^b{\int\limits^-}_c^df(x,y)\,dy\,dx\leq{\int\limits^-}_a^b{\int\limits^-}_c^df(x,y)\,dy\,dx\leq{\int\limits^-}Rf(x,y)\,d(x,y)$
• $\displaystyle{\int\limits_-}Rf(x,y)\,d(x,y)\leq{\int\limits_-}_a^b{\int\limits_-}_c^df(x,y)\,dy\,dx\leq{\int\limits^-}_a^b{\int\limits_-}_c^df(x,y)\,dy\,dx\leq{\int\limits^-}Rf(x,y)\,d(x,y)$
• $\displaystyle{\int\limits_-}Rf(x,y)\,d(x,y)\leq{\int\limits_-}_c^d{\int\limits^-}_a^bf(x,y)\,dx\,dy\leq{\int\limits^-}_c^d{\int\limits^-}_a^bf(x,y)\,dx\,dy\leq{\int\limits^-}Rf(x,y)\,d(x,y)$
• $\displaystyle{\int\limits_-}Rf(x,y)\,d(x,y)\leq{\int\limits_-}_c^d{\int\limits_-}_a^bf(x,y)\,dx\,dy\leq{\int\limits^-}_c^d{\int\limits_-}_a^bf(x,y)\,dx\,dy\leq{\int\limits^-}Rf(x,y)\,d(x,y)$
• If $\int_Rf(x,y)\,d(x,y)$ exists, then we have
\displaystyle\begin{aligned}\int\limits_Rf(x,y)\,d(x,y)&=\int\limits_a^b{\int\limits_-}_c^df(x,y)\,dy\,dx=\int\limits_a^b{\int\limits^-}_c^df(x,y)\,dy\,dx\\&=\int\limits_c^d{\int\limits_-}_a^bf(x,y)\,dx\,dy=\int\limits_c^d{\int\limits^-}_a^bf(x,y)\,dx\,dy\end{aligned}

Okay, as ugly as all those are, they’re what we’ll prove next time.

December 16, 2009 Posted by | Analysis, Calculus | 10 Comments


text:
0
Research Papers: Heat Exchangers

# Investigation of Vortex Generator Enhanced Double-Fin and Tube Heat Exchanger

[+] Author and Article Information
Shobhana Singh

Department of Energy Technology,
Aalborg University,
Pontoppidanstræde 111,
Aalborg East 9220, Denmark
e-mail: ssi@et.aau.dk

Kim Sørensen

Department of Energy Technology,
Aalborg University,
Pontoppidanstræde 111,
Aalborg East 9220, Denmark

Thomas Condra

Department of Energy Technology,
Aalborg University,
Pontoppidanstræde 111,
Aalborg East 9220, Denmark

1Corresponding author.

Contributed by the Heat Transfer Division of ASME for publication in the JOURNAL OF HEAT TRANSFER. Manuscript received March 23, 2018; final manuscript received November 7, 2018; published online December 13, 2018. Assoc. Editor: Danesh K. Tafti.

J. Heat Transfer 141(2), 021802 (Dec 13, 2018) (13 pages) Paper No: HT-18-1172; doi: 10.1115/1.4042050 History: Received March 23, 2018; Revised November 07, 2018

## Abstract

In the present work, a numerical analysis of conjugate heat transfer and fluid flow in vortex generator (VG) enhanced double-fin and tube heat exchanger is carried out. The enhanced design aims to improve the heat transfer performance of a conventional double-fin and tube heat exchanger for waste heat recovery applications. A three-dimensional (3D) numerical model is developed using ANSYS cfx to simulate fluid flow and conjugate heat transfer process. Numerical simulations with rectangular winglet vortex generators (RWVGs) at five different angles of attack ($−20deg≤α≤20deg$) are performed for the Reynolds number range of $5000≤Re≤11,000$. Salient performance characteristics are analyzed in addition to the temperature distribution and flow fields. Based on the numerical results, it is concluded that the overall performance of the double-fin and tube heat exchanger can be improved by 27–91% by employing RWVGs at $α=−20deg$ for the range of Reynolds number investigated. The study provides useful design information and necessary performance data that can be adopted for the design development of the heat exchanger at a lower manufacturing cost.

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## Figures

Fig. 1

A pictorial view of double-fin and tube heat exchanger

Fig. 2

Illustration of the conventional double-fin and tube heat exchanger unit: (a) conventional plain fin and (b) enhanced fin with RWVG

Fig. 3

Geometric details of the enhanced double-fin and tube heat exchanger design: (a) RWVG punched on the horizontal fin surface and (b) enhanced double-fin and tube heat exchanger with RWVGs

Fig. 4

Simulated computational geometry with set boundary conditions

Fig. 5

Illustration of the boundary layer mesh

Fig. 6

A plot of Nusselt number versus number of mesh elements

Fig. 7

The comparison of model results with the correlation values

Fig. 8

Temperature distribution on the lower fin and tube surface: (a) plain fin with no RWVG, (b) fin with RWVG at α = 0 deg, (c) fin with RWVG at α = 10 deg, (d) fin with RWVG at α = 20 deg, (e) fin with RWVG at α = −10 deg, and (f) fin with RWVG at α = −20 deg

Fig. 9

Velocity streamlines starting from the leading edge of the winglet of first pair inside the fluid domain at Re=5000: (a) fin with RWVG at α = 0 deg, (b) fin with RWVG at α = 10 deg, (c) fin with RWVG at α = 20 deg, (d) fin with RWVG at α = −10 deg, and (e) fin with RWVG at α = −20 deg

Fig. 10

Velocity streamlines at Re = 5000 on the outlet plane (x=0.15 m) of the fluid domain: (a) plain fin with no RWVG, (b) fin with RWVG at α = 0 deg, (c) fin with RWVG at α = 10 deg, (d) fin with RWVG at α = 20 deg, (e) fin with RWVG at α = −10 deg, and (f) fin with RWVG at α = −20 deg

Fig. 11

Variation of average synergy angle with the angle of attack and Reynolds number

Fig. 12

Variation in heat transfer characteristics with the angle of attack and Reynolds number: (a) Nusselt number and (b) Colburn j factor

Fig. 13

Variation in pressure loss characteristics with the angle of attack and Reynolds number: (a) Euler number and (b)friction factor

Fig. 14

Variation of the normalized volume goodness factor with the angle of attack and Reynolds number: (a) Euler number and (b) friction factor

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# Upper bounds on option prices

I am having troubles with deriving the upper bounds on option prices. For example, I am trying to determine the upper bound on a European put option.

If I define the put price as $P$, the terminal stock price as $S_T$, the strike as $X$, the risk free rates as $r$, and the time to expiry as $T$, I am able to create the following portfolio;

Short 1 Put, and invest the proceeds.

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At expiry, if $S_T > X$, I have zero cash flow from the short put position, and have positive cash flow of $P\text{e}^{rT}$. My net position is $P\text{e}^{rT}$, which is a non negative amount.

If however at expiry $S_T < X$, I have a cash flow of $-(X-S_T)$ from the short position and $P\text{e}^{rT}$ from the invested proceeds. This is a net position of $P\text{e}^{rT} - X + S_T$.

From my understanding, for this to satisfy a no arbitrage condition, we need to have $P\text{e}^{rT} - X + S_T < 0$. Rearranging this should show that the upper bound for the put option needs to be $P < X\text{e}^{rT} - S_T\text{e}^{rT}$.

However, when I consult sources (such as Hull), I am told the upper bound is $P < X\text{e}^{rT}$.

Where is the mistake in my work? This will also help me understand the upper bounds for European call options! Thanks!

First, I think you got discounting and compounding wrong. The upper bound for the put price is $P_0 < X e^{-r T}$. This has to hold as long as $S_0 > 0$.

Your approach is generally correct though. Assume that $P_0 = X e^{-r T}$. Then you invest $P_0$ at time $t = 0$ and have $P_0 e^{r T} = X$ at time $t = T$. Now there are three possible states in time $t = T$:

1. The put option expires worthless. You you have no further obligations and are left with $V_T = X$.
2. The option expires in-the-money but $S_T > 0$. Then you have to pay out $P_T < X$ from your short put position, leaving you again with a strictly positive cash-flow $V_T > 0$.
3. We have $S_T = 0$ and thus $P_T = X$. Your net cash-flow is now zero - $V_T = 0$.

As long as $\mathbb{P} \left\{ S_T = 0 \right\} < 1$, this represents a free lottery arbitrage. I.e. you have a zero initial cash-flow $V_0 = 0$, a non-negative terminal cash-flow with probability one $\mathbb{P} \left\{ V_T \geq 0 \right\} = 1$ and a strictly positive terminal cash-flow with a strictly positive probability $\mathbb{P} \left\{ V_T > 0 \right\} > 0$.

But when $\mathbb{P} \left\{ S_T = 0 \right\} = 1$, then any $S_0 \neq 0$ would already represent an arbitrage by itself.

Have you considered implied volatility ? It might be really high for some volatile stocks, so put price might goes to extremely high values due to iv ...

• This question is about model independent no-arbitrage bounds. I.e. bounds that have to be satisfied in any model an no matter what the unobservable market parameters such as implied volatility are. $P_0 < X e^{-r T}$ holds in the Black-Scholes model even if $\sigma \rightarrow \infty$. – LocalVolatility Mar 11 '17 at 19:12


text:
# Create an R package from a single R file with roxyPackage

Documenting code can be a bit of a pain. Yet, the older (and wiser?) I get, the more I realise how important it is. When I was younger I said 'documentation is for people without talent'. Well, I am clearly loosing my talent, as I sometimes struggle to understand what I programmed years ago. Thus, anything that soothes the pain of writing and maintaining documentation must be good and should help me to better understand my 'old me' in the future.

Ideally I want my R code and documentation in as few files as possible. A good start to achieve this is using roxygen2, an R package which has been around for some time. It allows me to tie R code together with the documentation in the same file and helps considerably in maintaining R packages.

The roxyPackage by Meik Michalke goes a step further, building on roxygen2. Meik presented his package at the Cologne R user group meeting a few weeks ago and I was intrigued by it. As I said in my notes to the meeting, with roxyPackage I can create a package from a single file of R code and documentation.

Here is an example of one R file to build a package using roxyPackage. For my toy example I wrote a doughnut plot function in R, something which is clearly missing ;)

I took the basic pie chart function and amended it to plot another white disc in the middle. On top of the function code I wrote the help file documentation using the roxygen2 syntax.

In my next step I use roxyPackage to turn this file into its own standalone Doughnut package. Following the roxyPackage vignette I have to pass a data frame with the details of a typical DESCRIPTION file to roxyPackage for it to turn the R file into a package.

The arguments pck.source.dir, R.libs, repo.root specify the folders in which I store my R file, my R libraries and output package respectively.

Job done. After executing the above commands I end up with my package build and installed on my system. Additionally, roxyPackage created binary packages for Windows and Mac in my very own repository (my own little CRAN). In the future I only have to update the version number in the above data frame, rather than DESCRIPTION file, package.Rd help file and NEW/ChangeLog. Fantastic!

### Closing remarks

This post is dedicated to my colleagues, who I have pestered with end user computing (EUC) standards over the last months. I believe with roxygen2 and roxyPackage it won't be too burdensome anymore to comply with those standards.

Indeed, I think roxygen2 and roxyPackage can help novice R users and those who start programming to think a little bit like professional IT guys (I know, I haven't even mentioned testing or version control yet). Often users just want to get their job done and this is were the above approach can help them to document their work without too much pain.

### Session Info

R version 2.15.2 (2012-10-26)
Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit)

locale:
[1] en_GB.UTF-8/en_GB.UTF-8/en_GB.UTF-8/C/en_GB.UTF-8/en_GB.UTF-8

attached base packages:
[1] tools     stats     graphics  grDevices utils     datasets  methods   base

other attached packages:
[1] roxyPackage_0.03-1 XiMpLe_0.03-16     roxygen2_2.2.2     digest_0.6.1

loaded via a namespace (and not attached):
[1] brew_1.0-6    stringr_0.6.2

siefert said...

Neat. Can you explain which text you had to type yourself and which you just copied from the existing function and documentation. It looks like you added this:

Pin <- t2xy(seq.int(0, 1, length.out = n*nx),

polygon(Pin$x, Pin$y, density = density[i],

angle = angle[i], border = border[i],

col = "white", lty = lty[i])

in the R code. But what were you able to take from the existing documentation of pie() and what did you have to add yourself?

Joe Cheng said...

Very nice! It would be great if the myDescription info could be encoded as roxygen in the original file as well.

Markus Gesmann said...

Indeed, I added the code you mentioned above. Additionally, type 'pie' into the R console and you will find the original code. You will notice that I split the argument radius into inner.radius and outer.radius. Hence, I could recycle most of the original content of the pie help file as well.

m.eik said...

hi joe,

i'm not involved in the roxygen development myself, so i'm afraid i cannot help much here. but what i usually do is to include the data.frame in the package itself, as a non-exported object, and reference it as myPackage:::myDescription in the roxy.package() call. (which of course only works after the package was installed for the first time)

internally, the package uses write.dcf() to write the DESCRIPTION file, so it should be in data.frame format sooner or later. i split the contents of that file into two parts: firstly, version number and date, and secondly, everything else ;-)

this is because the "everything else" part usually doesn't change so often, but version number and release date change all time. therefore those two are supplied as direct arguments to the roxy.package() function call, whereas all the rest can be put into one rather static object in the package.

seifert said...

Thanks, I saw this on the R-code, but what about the roxygen material, lines 1-57 in your example?

Yogalakshmi J said...

Hi

Thanks for this post. It was a very useful post for new users like me. I have created my package. I would also like to know, how to include other packages into my package. For example, inside my package, I would like to call a funciton called "hcluster" which is inside the package "amap". Right now, I am loading it through "library(amap)" code written inside my R script. Is this the right way, if not, kindly let me know of the formal way to include other packages into my package.

Thanks,
Yogalakshmi


text:
ML p(r)ior | On the Structure and Efficient Computation of IsoRank Node Similarities

On the Structure and Efficient Computation of IsoRank Node Similarities

2016-02-01
1602.00668 | q-bio.MN
The alignment of protein-protein interaction (PPI) networks has many applications, such as the detection of conserved biological network motifs, the prediction of protein interactions, and the reconstruction of phylogenetic trees [1, 2, 3]. IsoRank is one of the first global network alignment algorithms [4, 5, 6], where the goal is to match all (or most) of the nodes of two PPI networks. The IsoRank algorithm first computes a pairwise node similarity metric, and then generates a matching between the two node sets based on this metric. The metric is a convex combination of a structural similarity score (with weight $\alpha$) and an extraneous amino-acid sequence similarity score for two proteins (with weight $1 - \alpha$). In this short paper, we make two contributions. First, we show that when IsoRank similarity depends only on network structure ($\alpha = 1$), the similarity of two nodes is only a function of their degrees. In other words, IsoRank similarity is invariant to any network rewiring that does not affect the node degrees. This result suggests a reason for the poor performance of IsoRank in structure-only ($\alpha = 1$) alignment. Second, using ideas from [7, 8], we develop an approximation algorithm that outperforms IsoRank (including recent versions with better scaling, e.g., [9]) by several orders of magnitude in time and memory complexity, despite only a negligible loss in precision.

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text: Braggs Law and Laue equation for diffraction, Concept of Reciprocal Lattice, constructing the Miller Indices from reciprocal lattice, Braggs law in reciprocal lattice. an arbitary phase origin: contained in the unit cell, we can Φ = 2π (h x + k y + l z) radians (Formula 1) Analytical expression of the structure factor. Of course, Now consider a second atom added to the unit cell. The structure factor may be expressed as $\mathbf{F}_{hkl} = F_{hkl}\exp(i\alpha_{hkl}) = \sum_j f_j\exp[2\pi i (hx_j + ky_j + lz_j)]$ $\qquad = \sum_j f_j\cos[2\pi (hx_j + ky_j + lz_j)] + i\sum_{j} f_j\sin[2\pi (hx_j + ky_j + lz_j)]$ \qquad = A… Structure-factor equations like Eq. From this initial model, structure factors are calculated and compared with those experimentally observed. S) and integrating over the volume of diffraction space, dvr, we get an expression for the electron density of the unit cell. , so that its expression is modified with The structure factor. dispersion (diffraction), caused by all the atoms of the cell, in a speaking, each structure factor can be considered as a vector, with its the so-called. A vector of desired length is positioned that it any crystalline model can be decomposed into as many simple lattices as We have calculated above that for the 111 reflection, sinθ / λ = 0.154 Å −1. We know by now how to calculate the individual atomic scattering factors f(j) which we need in the calculation of the structure factors : (1) The structure factors F(hkl) are directly related to the Intensity I (hkl) of the corresponding reflection h,k,l: (2) LP is a combined geometry and polarization factor which depends on the particular experimental setup. For X-ray crystallography they are multiples of the unit of scattering by a single electron, ; for neutron scattering by atomic nuclei the unit of scattering length of, information contact us at info@libretexts.org, status page at https://status.libretexts.org. It plays a central role in the solution and refinement of crystal structures because it represents the quantity related to the intensity of the reflection which depends on the structure giving rise to that reflection … Whatever you may or may not have understood during the theory and explanations of these last two sections, it is vital that you do at least learn how to calculate predicted intensities from a known structure. From the above expressions it can clearly be seen that the structure factors of crystallographic structures composed of several atoms in the unit cell are the sum of individual complex numbers, each one of them having a length equal to the atom's scattering factor f j and phase built out of the atom's position. Unit Cell. The crystal structure can be described as a Bravais lattice with a group of atoms, called the basis, placed at every lattice point; that is, [crystal structure] = [lattice] {\displaystyle \ast } [basis]. This presentation shows how the structure factor equation arises from the positions of atoms in the unit cell. We know by now how to calculate the individual atomic scattering factors f(j) which we need in the calculation of the structure factors : (1) The structure factors F(hkl) are directly related to the Intensity I (hkl) of the corresponding reflection h,k,l: (2) LP is a combined geometry and polarization factor which depends on the particular experimental setup. As much of crystallography is involved with determining atomic structure from the intensity of diffraction spots/peaks, then this equation acts as the basic computational link between the two. The units of the structure-factor amplitude depend on the incident radiation. It is seen that the path difference between waves diffracted from the two planes shown differs by just one wave cycle. A structure factor represents the resultant X-ray scattering power of the whole crystal structure, though, since the whole structure consists of a large number of unit cells all scattering in phase'with each other, the resultant scattering power is … If Bragg's equation is the most well-known equation, then the structure factor equation must be the second most well-known. and represents the total wave resulting from the co-operative reflected on planes of the same color, shifted (. As much of crystallography is involved with determining atomic structure from the intensity of diffraction spots/peaks, then this equation acts as … In general, if F C is the structure factor computed from a model, the centroid of the probability distribution for the true structure will be obtained by multiplying F C by a parameter D (which arises for similar reasons as d j above). It starts with the definition of the structure factor appropriate for X-ray and neutron scattering, and includes the derivation of the appropriate expressions. K is a factor that puts the experimental structure factors, (F rel), measured on a relative scale (which depends on the power of the X-ray source, crystal size, etc.) the structure factor amplitue|F|, which is equal to the SQRT (A2+B2)with resultant phase ( α ) = Atan2(B,A), or tan‐1 B/A. arising from the phase difference due to the offset in position between the two sets of diffracting atoms. This is the determinant of an Hermitian matrix and hence its value is real. that measures the X-ray scattering power of each atom. The structure factor. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. between the waves scattered from the first and the new sets of atoms. Of course, F C could equally be replaced by (say) the structure factor from an isomorphous crystal. Analytically The phases differ by the angle φ1 . to the scheme shown on the left, the module of Crystallographic Directions is a vector connecting the coordinate origin and a specific point of a unit cell. coordinates (that is, assuming, Once the (5.15), p. 98, present the structure factor as a sum of terms each containing the exponential element e2πi(hx+ky+lz). But now there is a phase difference φ1 between the waves scattered from the first and the new sets of atoms. Normally the phase is converted from radians to degrees. anomalous fraction of the atomic scattering factor... , This chapter summarizes the mathematical development of the structure-factor formalism. In this section we will look at the effect of crystallographic symmetry on the structure factors, with a single 2-fold screw axis. So this sub-section will take you step by step … Structure factor for a centrosymmetric crystal; Structure factor for a C-centred unit cell; Reflection conditions; Screw axes and glide planes is equivalent to considering that the waves can also be represented as. where the sum is over all atoms in the unit cell, xj,yj,zj are the positional coordinates of the jth atom, fj is the scattering factor of the jth atom, and $$α_{hkl}$$ is the phase of the diffracted beam. Structure Factors: 2 2 mm me f h π = φ 2 atoms i m m m F fe= ∑ π⋅gd g Atomic Form Factors: The Fourier components of the crystal potential are normalized by the unit-cell volume. Beyond the isolated-atom case, the atom-centred spherical harmonic (multipole) model is treated in detail. In crystallography such vector is defined by three directional indices [u n w] Rules for indexes determination: 1. be: and its phase, referred to show the location of the atoms, that is, the internal structure of the crystals. The value is also sometimes called the discrepancy … [ "article:topic", "Structure factor", "showtoc:no", "license:ccbyncsa" ], https://chem.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fchem.libretexts.org%2FBookshelves%2FInorganic_Chemistry%2FModules_and_Websites_(Inorganic_Chemistry)%2FCrystallography%2FX-rays%2FStructure_Factor, Now consider a second atom added to the unit cell. mathematical expression of the phase is established in terms of the 10-24. a crystal formed by the repetition of the atomic model constituted by For example, you can look up examples of the derivation of systematic absences using these equations in any of the text books I have suggested. shape of the crystalographic model, represents the dispersion of the red atoms, and, for the blue atoms (figure on the left). 1e e eik l ih l ih k() () () F f hkl =⋅+ + +π+ π+ π+  It is usually … The fundamental fact that the crystallographic structure factor phase information is present in EM … They vary with sinθ / λ for reasons that were explained earlier (the "atomic form factor"). The structure factor may be expressed as (1) F h k l = F h k l exp (i α h k l) = ∑ j f j exp The amplitudes of the waves are proportional to the atomic scattering factors, . into an absolute scale, which is to say, the scale of the calculated (theoretical) structure factors (if we could know them from the real structure, Formula 2 above). As in the case of two atoms, the resultant diffracted wave is obtained from the linear superposition of the wave vectors scattered from each different atom. independent space directions: Combining the three Of course, F C could equally be replaced by (say) the structure factor from an isomorphous crystal. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. 2. In general, if F C is the structure factor computed from a model, the centroid of the probability distribution for the true structure will be obtained by multiplying F C by a parameter D (which arises for similar reasons as d j above). For X-ray crystallography they are multiples of the unit of scattering by a single electron ($$2.82 \times 10^{-15}\;m$$); for neutron scattering by atomic nuclei the unit of … The atomic scattering factors for Na + and Cl − ions can be obtained from the International Crystallographic Tables. of both atom types will be. 1. , a. phase shifts, and generalizing to the three dimensions: Finally, taking fractional In structure determination, phases are estimated and an initial description of the positions and anisotropic displacements of the scattering atoms is deduced. left, please, , , Click here to let us know! which modifies the atomic dispersion factor. Each original atom is now accompanied by a companion atom of the new type, offset by a displacement vector r1. phases), We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. B, ch. Structure Factor (F hkl) • The amplitude of the resultant wave is given by a ratio of amplitudes:ratio of amplitudes: amplitude of the wave scattered by all atoms of a UC hkl li d f h d b l F • The intensity of the diffracted wave is proportional amplitudeof the wave scatteredby one electron The intensity of the diffracted wave is proportional to |F Introduction to the Calculation of Structure Factors Introduction to the Calculation of Structure Factors S. C. Wallwork University of Nottingham, England In X-ray crystallography the structure factor F(hkl) of any X-ray reflection (diffracted beam) hkl is the quantity that expresses both the amplitude and the phase of that reflection. Remember that these exponential elements can also be expressed trigonometrically as [cos 2π (hx + ky + lz) + i … vectors These vectors are usually described by three integer components as [ h k l ] (note the square brackets), where h is the component along the a axis, k is the component along the b axis and l is the component along the c axis. (or equivalently in 2-dimensions, a glide plane). \begin{align} \mathbf{F}_{hkl} &= F_{hkl}\exp(i\alpha_{hkl}) = \sum_j f_j\exp[2\pi i (hx_j + ky_j + lz_j) \\[4pt] &= \sum_j f_j\cos[2\pi (hx_j + ky_j + lz_j)] + i\sum_{j} f_j\sin[2\pi (hx_j + ky_j + lz_j)] \\[4pt] &= A_{hkl} + iB_{hkl} \end{align}. Introduction to the Calculation of Structure Factors Introduction to the Calculation of Structure Factors S. C. Wallwork University of Nottingham, England In X-ray crystallography the structure factor F(hkl) of any X-ray reflection (diffracted beam) hkl is the quantity that … this vector sum will The theoretical background and practical procedures are explained with hundreds of figures. For such a system, only a set of specific values for {\displaystyle \mathbf {q} } can give scattering, and the scattering amplitude for all other values is zero. Adopted a LibreTexts for your class? The resultant vector represents the two-atom structure factor with amplitude. in the direction of the. 32 Point groups + 14 Bravais lattice →230 space group. This equation is used to show how systematic absences occur, where classes of reflections are required to have zero intensity by symmetry. A structure factor is calculated by summing up scattering factors of each atom with multiplying the phases at each atomic position in a unit cell as described previously in an equation. . The intensity of a diffracted beam is directly related to the amplitude of the structure factor, but the phase must normally be deduced by indirect means. If you do not see a menu on the Iterative refinement procedures attempt to minimize the difference between calculation and experiment, until a satisfactory fit has been obtained. two terms. , The outer point of the vector intercepts the cell surface at the reciprocals: 1/ h, 1/ k, and 1/ l. amplitude and phase (referred to an arbitrary origin of Module 31-36: Scattering Power of an atom, Structure Factor Equation, Derivation and Interpretation of Structure Factor Equation. OR. It is a facile approach to calculate the structure factor; however, the calculation becomes complicated when the unit cell includes the large number of atoms. The map and structure factors now show a 3-atom structure with a 2-fold screw axis parallel to the b axis. structure factor (not quite so easy). oscillation frequency of the electrons of a given atom, there occurs below, a structure factor, , is the resultant of all waves scattered which represent the real and imaginary components, respectively, of the The central problem in crystallography arises because the experimental data yield only the modulus of the structure factor, ∣ F hkl ∣ and not the phase. The structure factor [math]\mathbf{F}_{hkl} is a mathematical function describing the amplitude and phase of a wave diffracted from crystal lattice planes characterised by Miller indices $h, k, l$. the incident X-ray radiation has a frequency close to the natural immediatly calculate the structure factors, and their numerical values, corresponding to the so-called absolute a. applied to the three Introduction to the Calculation of Structure Factors S. C. Wallwork University of Nottingham, England In X-ray crystallography the structure factor F(hkl) of any X-ray reflection (diffracted beam) hkl is the quantity that expresses both the amplitude and the phase of that reflection. From the definition of the structure factor, equation it can be seen that U(-h) = U(h) *, where the star indicates 'complex conjugate', so that atoms (look at the two lattices drawn on the left, in red and blue). The structure factor F h k l is a mathematical function describing the amplitude and phase of a wave diffracted from crystal lattice planes characterized by Miller indices h,k,l. The structure factor may be expressed as (31.5.1) F h k l = F h k l exp (i α h k l) = ∑ j f j exp Argand diagram; Linking F with the structure; The structure factor F in practice . The red X-ray beams, which are being The units of the structure-factor amplitude depend on the incident radiation. The structure factor F h k l is a mathematical function describing the amplitude and phase of a wave diffracted from crystal lattice planes characterized by Miller indices h,k,l. We can measure the amplitude of the true structure factor, F, but not its phase, so we know that it lies somewhere on a circle. Each original atom is now accompanied by a companion atom of the new type, offset by a displacement vector, from these new scatterers (since they occupy planes parallel to those originally drawn). is the resultant of all waves scattered in the direction of the. Note that there is a net phase φ arising from the phase difference due to the offset in position between the two sets of diffracting atoms. International Tables for Crystallography (2006). given direction of space. that applies to both centrosymmetric and non-centrosymmetric structure factors. Have questions or comments? Units. Vol. Friedel’s law; Typical applications of the structure factor equation . In crystallography, the R-factor (sometimes called residual factor or reliability factor or the R-value or R Work) is a measure of the agreement between the crystallographic model and the experimental X-ray diffraction data. IV.H Solving the Phase Problem. are the fundamental quantities on which the function of electron If the lattice is infinite and completely regular, the system is a perfect crystal. The amplitudes of the waves are proportional to the atomic scattering factors f0 and f1. The fcc structure can be generated from a sc lattice with a four-atom basis. The units of the structure-factor amplitude depend on the incident radiation. easily calculated using the following "rule of three" proportions, Legal. For X-ray crystallography they are multiples of the unit of scattering by a single electron ($$2.82 \times 10^{-15}\;m$$); for neutron scattering by atomic nuclei the unit of scattering length of $$10^{-14}\; m$$ is commonly used. Calculating the Intensity of Diffraction Using the Structure Factor Equation This is a very important sub-section. This is known as the “crystallographic phase problem”. This is the first textbook describing crystal structure determination (especially inorganic) from high-resolution transmission electron microscopy (HRTEM) and electron diffraction (ED). P. Coppens. In a crystal, the constitutive particles are arranged periodically, with translational symmetry forming a lattice. of the electron density function. . The resultant vector represents the two-atom structure factor with amplitude Fhkl. An interesting and useful consequence of the structure factor equations is that the phases found in centro-symmetric crystals are only on the real axis, thus the phase α is either 0 or π. In X-ray crystallography the structure factor F(hkl) of any X-ray reflection (diffracted beam) hkl is the quantity that expresses both the amplitude and the phase of that reflection. same manner will behave the blue beams on the blue planes, Said in other words, It is a facile approach to calculate the structure factor; however, the calculation becomes complicated when the unit cell includes the large number of atoms. In fact, this offset reflected on the red planes of indices, , fulfill the Bragg's Law, and in the The incident X-ray beam will also diffract from these new scatterers (since they occupy planes parallel to those originally drawn). • Ittellsuswhichreflections(ie peaksIt tells us which reflections (i.e., peaks , hkl)to) to expect in a diffraction pattern. The determinant can be of any order. The map and structure factors now show a 3-atom structure with a 2-fold screw axis parallel to the b axis. Formula of crystallography: Local (point)symmetry + translational symmetry →spatial symmetry. , The structure factor $$\mathbf{F}_{hkl}$$is a mathematical function describing the amplitude and phase of a wave diffracted from crystal lattice planes characterized by Miller indices h,k,l. crystallography. Suppose Jeffrey R. Deschamps, Judith L. Flippen-Anderson, in Encyclopedia of Physical Science and Technology (Third Edition), 2002. Consider Bragg's law for an array of atom scatterers in a primitive lattice with just one atom at each lattice point. r. Remember that a dot product can be interpreted as the projection of one vector on We can measure the intensities and determine the unit cell from the locations of the reflections, but it is very difficult to actually measure the relative phase of the reflections. An incident X-ray wave of wavelength λ diffracts strongly through an angle $$2θ$$ where the perpendicular distance between two lattice planes $$d_{hkl}$$ satisfies the relation. The structure factor calculated from a model, F C, can be used to supply a phase, giving the structure factor that combines the observed amplitude with the calculated phase, |F|exp(iα C). Scattering by a crystal: Laue and Bragg equations . But now there is a phase difference. A method of using linear prediction analysis to define a first structure factor component for a first reflection from x-ray crystallography data, the x-ray crystallography data comprising a set of cognizable reflections, the method comprising: expressing the first structure factor component as a first linear equation in which the first structure factor component is equal to a sum of a first … If Bragg's equation is the most well-known equation, then the structure factor equation must be the second most well-known. In other words, as we shall see Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. 1.2, pp. (or equivalently in 2-dimensions, a glide plane). If if Bragg's Law is fulfilled, the phase shift between waves can be the pair of atoms (red and blue) shown in the left figure. scale, since they are calculated with the dispersion factors (, Scattering and diffraction. In this section we will look at the effect of crystallographic symmetry on the structure factors, with a single 2-fold screw axis. density, depends. The phase is required in order to evaluate the electron … and states that the structure factor of a reflection hkl can be calculated as a function of structure factors whose Laue indices sum to the desired values of hkl. The total resultant dispersion According Structure Factor (Fhkl) 2( ) 1 ij i N ihu kv lw hkl i i Ffe • Describes how atomic arrangement (uvw) influences the intensity of the scattered beam. In other words, it is a measure of how well the refined structure predicts the observed data. Structure factors provide both intensity and phases for the diffracted beams. This set of values forms a lattice, called the reciprocal lattice, which is the Fourier transform of the real-space crystal lattice. These are very important magnitudes, since the maxima 2. me F U hv = Φ= π. g. Structure Function: gg A structure factor is calculated by summing up scattering factors of each atom with multiplying the phases at each atomic position in a unit cell as described previously in an equation. • In 3D space the unit cells are replicated by three noncoplanar translation vectors. A displacement vector r1 calculation and experiment, until a satisfactory fit has been obtained a measure of how the! Structure-Factor formalism are explained with hundreds of figures new scatterers ( since occupy... That is, the system is a measure of how well the refined structure the! Then the structure factor equation must be the second most well-known derivation and Interpretation of structure factor equation be! New scatterers ( since they occupy planes parallel to the b axis by …. 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Resistors and Ohm's Law Video Lessons

Concept

# Problem: When the switch S in the figure is closed, the reading of the voltmeter V willa. increase.b. decrease.c stay the same.

###### FREE Expert Solution

From Ohm's law:

$\overline{){\mathbf{V}}{\mathbf{=}}{\mathbf{i}}{\mathbf{R}}}$, where V is voltage, i is current, and R is resistance.

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###### Problem Details

When the switch S in the figure is closed, the reading of the voltmeter V will

a. increase.

b. decrease.

c stay the same.