TITLE: The limit of a recurrence relation (with resistors)
QUESTION [6 upvotes]: Background to problem (not too important):

My proposed solution:
The infinitely long element,

, however complex, can be represented as a single resistor of resistance $R$.
Remembering the initial resistor near $A$, we know $R_{AB}= r+R$. However, as this is an infinitely long element, it is equivalent to a resistor of resistance $R$ attached  to the right of two resistors of resistance $r$ (the resistance $R$ is an intrinsic property of the element, so is unaffected by the fact that the further to the right it is, the lower the current passing through it).
Thus, taking $R$ in series with $r$, then the result in parallel with $r$, then in series with $r$: $$R_{AB}=  r + \left (\frac{1}{\frac{1}{R+r}+\frac{1}{r}}  \right )$$
On the second iteration (moving $R$ further to the right):
$$R_{AB}= r+ \frac{1}{\frac{1}{r}+\frac{1}{r + \left (\frac{1}{\frac{1}{R+r}+\frac{1}{r}}  \right )}}$$
Ad infinitum.
I understand this may not be the fastest solution, but I'd like to know a little more about it nonetheless.
The mathematics
$$u_{1}=r+R$$
$$\large u_{n+1}=r+\frac{1}{\frac{1}{r}+\frac{1}{u_n}}$$
Does $\lim_{n \rightarrow \infty} (u_n)$ exist (important: is the limit a function of $R$?), and, if so, what is it?
First cases
$$u_{2}=\frac{3r^2+2rR}{R+2r}$$
$$u_{3}=\frac{8r^2+5rR}{3R+5r}$$
$$u_{4}=\frac{21r^2+13rR}{8R+13r}$$
$$u_{4}=\frac{55r^2+34rR}{21R+34r}$$
$$\lim _{n \rightarrow \infty} (u_n)\stackrel{?}{=}\varphi r$$
Seems the doing of Fibonnaci. How does one take the limit of this (I assume it requires knowledge of knowledge of the form of $f(n)=F_n$).
Intuitively, why does Fibonnaci appear here? What are the rabbits in this case?
Link to Wolfram's computation.

REPLY [2 votes]: I didn't dare look at this earlier, but here's another (transcripted) solution (I prefer Aryabhata's):
Algebraically this equivalence can be written as
$$R_{AB}=r+\frac{1}{\frac{1}{r}+\frac{1}{R_{AB}}}$$
Thus
$$R_{AB}^2-rR_{AB}-r^2=0$$
This equation has two solutions:
$$R_{AB}=\frac{1}{2}(1 \pm \sqrt{5})r$$
 The solution corresponding to “-“ in the above formula is negative, while resistance must be positive. So, we reject it. Finally we receive
$$R_{AB}=\frac{1}{2}(1 + \sqrt{5})r= \varphi r$$


TITLE: Integral of the substantial derivative in fluid mechanics?
QUESTION [1 upvotes]: I am studying the book Modern Compressible Flow by John Anderson. He writes

For steady, inviscid, adiabatic, compressible flow with no body forces,
$$
\rho \frac{Dh_o}{Dt}=0
$$
where $h_o$ is the total enthalpy of the flow at a point.
When integrated, this yields $h_0=\text{const}$ along a streamline.

The substantial derivative is defined
$$
\frac{D}{Dt}=\frac{\partial }{\partial t} + u\frac{\partial}{\partial x}+v\frac{\partial}{\partial y}+w\frac{\partial}{\partial z}
$$
My question is, how can I show that if
$$
\frac{Df}{Dt}=0
$$
for a function $f(x,y,z,t)$, if the flow is steady, adiabatic, and inviscid, then $f=\text{const}$ along a streamline?

REPLY [1 votes]: The substantial derivative (also known as the material derivative) is the rate of change of a variable $f$ in a fluid parcel or element. The variable $f$ is specified in the Eulerian way, as a function of time $t$ and current location $\mathbf{x}$ of the fluid parcel. Then the total derivative is
$$\mathrm{d}f\left(\mathrm{x}\left(t\right), t\right) = \frac{\partial f\left(\mathbf{x}\left(t\right), t\right)}{\partial t} \mathrm{d}t + \frac{\partial f\left(\mathbf{x}\left(t\right), t\right)}{\partial \mathbf{x}\left(t\right)} \frac{\partial \mathbf{x}\left(t\right)}{\partial t}\mathrm{d}t$$
where $\partial \mathbf{x}\left(t\right)/\partial t = \mathbf{u}\left(t\right)$ is just the flow velocity. Thus the material derivative is defined as
$$\frac{\mathrm{D}f\left(\mathrm{x}\left(t\right), t\right)}{\mathrm{D}t} \equiv \frac{\partial f\left(\mathbf{x}\left(t\right), t\right)}{\partial t} + \mathrm{grad}\left( f\left(\mathrm{x}\left(t\right), t\right)\right) \cdot \mathbf{u}\left(t\right).$$
If the material derivative, that is the total rate of change of the variable $f$, is zero, then the value of $f$ must be constant along its streamline. This is by definition. Or you could write it as
$$\frac{\mathrm{D}f}{\mathrm{D}t} =\frac{\partial f}{\partial t} + \mathrm{grad}\left( f\right) \cdot \mathbf{u}=0$$
thus
$$\mathrm{d}f = \left[\frac{\partial f}{\partial t} + \mathrm{grad}\left( f\right) \cdot \mathbf{u} \right]\mathrm{d}t = 0 \, \mathrm{d}t$$
and
$$f = \int \mathrm{d}f=\int 0 \, \mathrm{d}t = 0 + C$$
where $C$ is just the integration constant, i.e. the value of $f$ at the beginning of the streamline.
By the way: The conditions steady, adiabatic, and inviscid flow are not needed for this result. A variable $f$ does not change if $\frac{\mathrm{D}f}{\mathrm{D}t}=0$, but $\frac{\partial f}{\partial t} \neq 0$ and thus $\frac{\partial f}{\partial t} = - \mathrm{grad}\left( f\right) \cdot \mathbf{u}$ is still possible!


TITLE: On problems which can be proved easier if we use a different induction step.
QUESTION [3 upvotes]: Say we have a property $P$ defined on the  natural nubers. Usually students are taught that to  pove $P(n)$ is true for all $n\in\mathbb N$ you have to do the following:

make a basis

and use either of the following arguments:

if $P(k)$ is true then  $P(k+1)$ is true
if $P(n)$ is true $\forall n\leq k$ then $P(k+1)$ is true.

But you can also use slightly different methods. In particular I would like to look at a situation like this:

You prove that $P(k)\implies P(k+3)$ and combine this with three induction bases.

(The number $3$ can be any other natural number  of course...)
Now I know this is valid. I was just wondering if anyone knows any problems that would be easier (less time consuming) to solve with a method like this than with the ordinary methods.

REPLY [5 votes]: Try to prove that $12|(k^4 - k^2)$ for every natural number $k$. It can be checked to be true for $k \in \{1,2,3,4,5,6\}$. The induction step $12 | (k^4 - k^2) \implies 12 |((k+6)^4 - (k+6)^2)$ is easy (if not tedious) to verify.

REPLY [2 votes]: There are certainly some situations where it is more natural to use the method you mentioned, for example, when trying to prove
$$
n+(n-2)+(n-4)+\dots=n^2+\frac{n}2+\frac{(-1)^n}{2}n,
$$
where the above sum is only over positive integers. The proof is still pretty short using strong induction, it would only be a couple of words longer.


TITLE: Prove that a subclass of any set is also a set
QUESTION [1 upvotes]: I have been asked, while discussing the ZF subset axioms to prove that a subclass of any set is a set. I have put together the following proof and would be grateful if someone could confirm that it is correct.
By definition of classes, we know that classes and subclasses are given by predicates. Using the language of our axiom, whenever we have a set $A$ along with a predicate $\phi$, there exists a set $B$ which contains exactly the elements of $A$ which satisfy $\phi$. Hence if $A$ is a set then so must be the class $B=\{a\in A \ | \ \phi (a)\}$ which contains all elements of A which satisfy $\phi$.

REPLY [1 votes]: This is exactly true because a class is a predicate and the axiom of separation holds by assumption: it tells us that the intersection of a set and a class is a set, in these terms.


TITLE: Why does a flat degeneration induce equality in the K-Theory?
QUESTION [6 upvotes]: Let $Z_1$ and $Z_2$ be two closed subschemes of a smooth, complex, algebraic variety $X$. We will be interested in the Grothendieck ring of coherent sheaves of $X$, i.e. isomorphism classes of such sheaves modulo exact sequences and a product given by the tensor product of $\mathcal{O}_X$-modules.
Denote by $[\mathcal{O}_{Z_1}]$ the class of $f^\ast\mathcal{O}_{Z_1}$ in the Grothendieck Ring of coherent sheaves on $X$, where $f:Z_1\to X$ is the inclusion.
Assume now that $Z_2$ is a flat degeneration of $Z_1$, i.e. there is a morphism $\pi: Y\to \mathbb{A}^1$ of schemes such that the generic fiber of $\pi$ is isomorphic to $Z_1$ and the fiber of $\pi$ over $0$ is isomorphic to $Z_2$.
I am reading a paper by David Speyer with the title
"A matroid invariant via the K-theory of the Grassmannian" and on page 885, right below equation (3) he says that $[\mathcal{O}_{Z_1}]=[\mathcal{O}_{Z_2}]$ holds in this case. I don't see how to prove this implication and I'd be very grateful if someone could explain it to me.

REPLY [2 votes]: As Allen says, we are talking about flat subfamilies of $X$. I.e. $Y$ is a closed subscheme of $X \times \mathbb{A}^1$, flat over $\mathbb{A}^1$, with $Z_0 \times \{ 0 \}$ and $Z_1 \times \{ 1 \}$ the fibers over $0$ and $1$.
Replacing $Y$ by its closure in $X \times \mathbb{P}^1$, we may assume that we instead have $Y$ closed in $X \times \mathbb{P}^1$, flat over $\mathbb{P}^1$. (Since $\mathbb{P}^1$ is smooth and one dimensional, being flat over it just means all associated primes of $Y$ map to the generic point of $\mathbb{P}^1$.)
On $\mathbb{P}^1$, we have short exact sequences
$$0 \to \mathcal{O}(-1) \to \mathcal{O} \to k_0 \to 0$$
and
$$0 \to \mathcal{O}(-1) \to \mathcal{O} \to k_1 \to 0$$
where $k_0$ and $k_1$ are the skyscraper sheaves of a point at $0$ and $1$ respectively. So the point $0$ and the point $1$ have the same class in $K_0(\mathbb{P}^1)$. Since $Y \to X$ is flat, we can pull this back to an equality $[\mathcal{O}_{Z_0}] = [ \mathcal{O}_{Z_1}]$ on $Y$.
Since $Y$ is closed in $X \times \mathbb{P}^1$, the map $\pi: Y \to X$ is proper and we can push forward this equality to $X$. We need to check that $\pi_{\ast} \mathcal{O}_{Z_i \times \{ i \}} = \mathcal{O}_{Z_i}$ and $R^j \pi_{\ast} \mathcal{O}_{Z_i \times \{ i \}} =0$ in order to know that the $K$-theoretic push forward is the obvious thing.
I'm not sure where to cite this argument to, it is something that everyone I talked to about this sort of thing seemed to view as obvious. Allen does the analogue for equivariant $K_0$, which is more interesting, in Prop 3.5 of his recent paper.