text:
# I'm stuck trying to prove ∀x : ℕ, 3 | (x + 5x) with Coq

Specifically, I think what's got me is showing that ∀x y z : ℕ, (z|x and z|y) → z|(x + y), or that ∀x y z : ℕ, (x mod y) = 0 → z mod y = (z + x) mod y, depending on how you want to look at it. I know those aren't exactly the same, but I believe either works for this purpose.

This proof is super easy to do on paper with weak induction, especially if you consider the problem as showing a property of pairs in an inductively defined set. One, because the algebra rules (factoring, principally) are given, and two, because I can succinctly write the set of all z|x as x = zk for some natural k.

I'm just having quite a bit of trouble formulating this in Coq terms. Here's what I've got. It feels like distinctly the wrong approach, but it's the best I can do with my very limited knowledge of Coq (and type based proofs in general, to be honest).

Theorem principal : forall x, eqb ((x + (5 * x)) mod 3) 0 = true.
Proof.
induction x.
- reflexivity.
- rewrite special_factor_neutral. rewrite IHx. reflexivity.
Qed.

This obviously expects a special_factor_neutral lemma do do the heavy lifting, which I've been unable to define acceptably. Here it is.

Lemma special_factor_neutral : forall x,
(((S x + 5 * S x)) mod 3) = (((x + 5 * x)) mod 3).

I suspect that with mod is not the ideal way to express divisibility in this problem — and the lemma clearly could be a good deal more general — but it's what I've got so far.

If you could suggest an alternate approach, or a way to make my current approach work, it would be greatly appreciated. To be clear, this is just for fun, so the stakes aren't very high, but I'd like to learn this stuff as well as possible.

Thank you!

Edit: I'm using Coq 8.12.2 in CoqIDE.

• Why not just show (x+5x)=6x?
– Couchy
May 9 at 19:14
• Couchy, thanks for the comment. That does seem like a decent aproach, but then I still have to prove that forall x, 3|(6 * x), which is the bit that I can't seem to figure out. May 9 at 20:19
• what about proving that (6 * x) = 3 * (2 * x)?
– Couchy
May 9 at 20:29
• hint: it's easier to first prove 6 * x = (3 * 2) * x. You don't need induction.
– Couchy
May 9 at 20:34
• Ah, I got it. Thank you for the tips. I guess that was rather obvious. May 9 at 22:48

When proving this, you will likely want to use several lemmas from the standard library, especially those involving mod.

The first step, of course, is to figure out why what you are proving is actually true. Here, we can do a quick algebraic manipulation:

$$(x + 5x) \equiv 6x \equiv 0 \mod 3$$, as $$6 \equiv 0 \mod 3$$

In Coq, we should be able to replay this argument, proving the rules we intuitively used if we get stuck.

The first thing we want to do is change $$x + 5x$$ into $$6x$$. In this case, Coq can directly see that these terms are equal since they can be partially computed, and when you do this you get syntactically equivalent terms. This is called "evaluating to normal form" and happens whenever Coq tries to prove that two things are equal. This kind of equality is called computational equality. The tactic reflexivity (among other things) proves that two things are computationally equal (or fails if they are not).

So, you might extract this into a seperate lemma, and then just prove it using reflexivity. Try to read up on Coq's reduction system if you are unsure why these two terms are computationally equal.

To continue, next we used the fact that $$x*y \equiv x' * y' \mod z \Leftarrow (x \equiv x' \mod z) \wedge (y \equiv y' \mod z)$$.

Unfortunately, that theorem is not that trivial, since it involves proving correctness of the concrete implementation of Coq's division algorithm, which is not particularly pretty. We are lucky, tho, since this is part of the standard library. Just Require Import Coq.Init.Nat Coq.Arith.PeanoNat., and notice that Nat.mul_mod_idemp_l : forall a b n : nat, n <> 0 -> (a mod n * b) mod n = (a * b) mod n is exactly the version of the general equivalence above we need.

The standard library also contains a few more fundamental theorems, which you can use to derive the above. You might want to look at the implementation of div and divmod and in particular use the Nat.divmod_spec lemma, if you feel like this approach is cheating.

The way you find these lemmas is by using Search "mod", or even stronger by searching for ((?a * ?b) mod ?c).

In total, the proof becomes

Lemma total x : (x + (5 * x)) mod 3 = 0 mod 3.
Proof.
assert ((x + 5*x) = 6*x) as -> by reflexivity.
rewrite <- Nat.mul_mod_idemp_l.
- reflexivity.
- discriminate.
Qed.

All reflexivitiy instances are computational equalities. You might want to investigate why each of these actually hold by computation, i.e. reduce them by hand.

discriminate is used to prove that 3 <> 0. This holds in particular since we can find computationally tell these two apart, since both are constructed using different constructors (when considering the inductive definition of nat).

If you prefer an idiomatic short solution:

Require Import Arith. (* Basic definitions. *)
Require Import Ring. (* The ring tactic. *)

Include Nat. (* Part of PeanoNat, to get access to x|y notation. *)

Theorem mobostar : forall x, (3 | x + 5 * x).
Proof.
intro x.
exists (2 * x).
ring.
Qed.
$$$$

Assuming you didn't want to recreate everything from scratch, I highly recommend checking out the ssreflect library, it can really trim down your proofs greatly. It comes bundled with the Coq installer (If you want to search for a proof, you can use the Search command)

For instance, here is the proof in SSReflect:

From mathcomp Require Import all_ssreflect.

Lemma three_div: forall x, 3 %| (x + 5*x).
move => x.
rewrite -{1}(mul1n x) -mulnDl addnE /=.
(have ->: 6 = 3*2 by done); rewrite -mulnA.
exact/dvdn_mulr/dvdnn.
Qed.

First line transforms it into 3 %| 6x, second one writes it into 3 %| 3 * (2*x) then the last one just concludes.

I've figured it out. Not pretty, but it does work.

Lemma plus_0 : forall x, x + 0 = x.
Proof.
induction x.
- reflexivity.
- simpl. rewrite IHx. reflexivity.
Qed.

Lemma plus_comm : forall a b, a + b = b + a.
Proof.
induction a.
- intro b. rewrite plus_0. reflexivity.
- intro b. rewrite <- plus_n_Sm. rewrite <- IHa. reflexivity.
Qed.

Lemma plus_assoc : forall a b c, a + b + c = a + (b + c).
Proof.
induction a.
- reflexivity.
- intros b c. simpl. rewrite IHa. reflexivity.
Qed.

Lemma mul_dist : forall k x y, k * (x + y) = k * x + k * y.
Proof.
induction k.
- intros x y. reflexivity.
- intros x y. simpl. rewrite IHk.
replace (k*x+k*y) with (k*y+k*x).
rewrite <- plus_assoc.
replace (x+y+k*y) with (x + (y+k*y)).
replace (y + k * y) with (S k * y).
rewrite plus_comm.
rewrite <- plus_assoc.
replace (k*x+x) with (x+k*x).
reflexivity.
+ rewrite plus_comm. reflexivity.
+ reflexivity.
+ rewrite plus_assoc. rewrite plus_comm. reflexivity.
+ rewrite plus_comm. reflexivity.
Qed.

Lemma times_0 : forall a, a * 0 = 0.
Proof.
induction a.
- reflexivity.
- simpl. rewrite IHa. reflexivity.
Qed.

Lemma mul_comm : forall a b, a * b = b * a.
Proof.
induction a.
- intro b. rewrite times_0. reflexivity.
- intro b. rewrite <- mult_n_Sm. rewrite <- IHa.
rewrite plus_comm. reflexivity.
Qed.

Lemma mul_assoc : forall a b c, a * b * c = a * (b * c).
Proof.
induction a.
- reflexivity.
- intros b c. simpl. rewrite <- IHa.
rewrite mul_comm. rewrite mul_dist.
replace (c * (a * b)) with (a * b * c). rewrite mul_comm. reflexivity.
+ rewrite mul_comm. reflexivity.
Qed.

Lemma principal : forall x, 6 * x = 3 * (2 * x).
Proof.
intro x. replace 6 with (3 * 2). rewrite mul_assoc. reflexivity.
- reflexivity.
Qed.
`


text:
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# Skeleton (topology)

Article Id: WHEBN0002977921
Reproduction Date:

 Title: Skeleton (topology) Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:

### Skeleton (topology)

In mathematics, particularly in algebraic topology, the n-skeleton of a topological space X presented as a simplicial complex (resp. CW complex) refers to the subspace Xn that is the union of the simplices of X (resp. cells of X) of dimensions mn. In other words, given an inductive definition of a complex, the n-skeleton is obtained by stopping at the n-th step.

These subspaces increase with n. The 0-skeleton is a discrete space, and the 1-skeleton a topological graph. The skeletons of a space are used in obstruction theory, to construct spectral sequences by means of filtrations, and generally to make inductive arguments. They are particularly important when X has infinite dimension, in the sense that the Xn do not become constant as n → ∞.

## In geometry

In geometry, a k-skeleton of n-polytope P (functionally represented as skelk(P)) consists of all i-polytope elements of dimension up to k.[1]

For example:

skel0(cube) = 8 vertices
skel1(cube) = 8 vertices, 12 edges
skel2(cube) = 8 vertices, 12 edges, 6 square faces

## For simplicial sets

The above definition of the skeleton of a simplicial complex is a particular case of the notion of skeleton of a simplicial set. Briefly speaking, a simplicial set $K_*$ can be described by a collection of sets $K_i, \ i \geq 0$, together with face and degeneracy maps between them satisfying a number of equations. The idea of the n-skeleton $sk_n\left(K_*\right)$ is to first discard the sets $K_i$ with $i > n$ and then to complete the collection of the $K_i$ with $i \leq n$ to the "smallest possible" simplicial set so that the resulting simplicial set contains no non-degenerate simplices in degrees $i > n$.

More precisely, the restriction functor

$i_*: \Delta^\left\{op\right\} Sets \rightarrow \Delta^\left\{op\right\}_\left\{\leq n\right\} Sets$

has a left adjoint, denoted $i^*$.[2] (The notations $i^*, i_*$ are comparable with the one of image functors for sheaves.) The n-skeleton of some simplicial set $K_*$ is defined as

### Coskeleton

Moreover, $i_*$ has a right adjoint $i^!$. The n-coskeleton is defined as

For example, the 0-skeleton of K is the constant simplicial set defined by $K_0$. The 0-coskeleton is given by the Cech nerve

$\dots \rightarrow K_0 \times K_0 \rightarrow K_0.$

(The boundary and degeneracy morphisms are given by various projections and diagonal embeddings, respectively.)

The above constructions work for more general categories (instead of sets) as well, provided that the category has fiber products. The coskeleton is needed to define the concept of hypercovering in homotopical algebra and algebraic geometry.[3]

format: plain


text:
A train covers a distance of 90 km at a uniform speed.

Question:

A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hour more, it would have taken 30 minutes less for a journey. Find the original speed of the train.

Solution:

Let the original speed of train be $x \mathrm{~km} / \mathrm{hr}$. Then,

Increased speed of the train $=(x+15) \mathrm{km} / \mathrm{hr}$

Time taken by the train under usual speed to cover $90 \mathrm{~km}=\frac{90}{x} \mathrm{hr}$

Time taken by the train under increased speed to cover $90 \mathrm{~km}=\frac{90}{(x+15)} \mathrm{hr}$

Therefore,

$\frac{90}{x}-\frac{90}{(x+15)}=\frac{30}{60}$

$\frac{\{90(x+15)-90 x\}}{x(x+15)}=\frac{1}{2}$

$\frac{90 x+1350-90 x}{x^{2}+15 x}=\frac{1}{2}$

$2700=x^{2}+15 x$

$x^{2}+15 x-2700=0$

$x^{2}+15 x-2700=0$

$x^{2}-45 x+60 x-2700=0$

$x(x-45)+60(x-45)=0$

$(x-45)(x+60)=0$

So, either

$(x-45)=0$

$x=45$

Or

$(x+60)=0$

$x=-60$

But, the speed of the train can never be negative.

Hence, the original speed of train is $x=45 \mathrm{~km} / \mathrm{hr}$


text:
How to combine two different embeddings in the best way possible?

I have two models which are giving two books embedding

Ml_model_a => book1_embedding [ 1, 200 ]

Ml_model_b => book2_embedding [ 1, 200 ]

I am building a third model which will take these two different embeddings to tell me which book to choose.

Now my final layer is the classification between 0,1 ( which book to choose ). How to learn these embeddings best way possible to classify better?

What I have tried yet :

If I am averaging those embeddings and then sending to one model, then embeddings are losing a lot of information, so I am using concatenation method.

But it's not classifying well, Is there any other model, a technique I could use to enhance the capability of learning to book embedding and predict which book to take?

I would say that the best thing to do here is to concatenate both embeddings and use the concatenated vector as an input for your binary classification model without using the norm -> you loose way too much information, i.e.:

final_layer = concatenate [ book_1_attention, book_2_attention]

I reckon that concatenating without norm will increase the dimension (but that's potentially a problem for which you have solutions - such as regularisation)

You can have a look into the literature about learning to rank, for example this work, which ensures reflexivity, is antisymmetric and transitive.

If you apply the l2 norm to the vectors you loose a lot of informations, i think the best way to act is to evaluate these options:

• concatenate embeddings with/out a linear layer on top of that
• sum the embeddings with/out a linear layer on top of that


text:
# What step have I missed designing this differential amplifier?

First time posting on EE.SE and I hope this question isn't too broad. This will be a long post. I should also point out: I'm asking the process and not the solution the problem. I've did create a circuit that works, but I also found a circuit online that is simpler and probably performs a lot better in real life. The core question here is: what path/thought process could I have taken to arrive at the more concise solution? I understand my design well, and it works, at least in LT Spice, but it uses more components to get there. I understand the math behind both solutions and why they both work, so I don't need that to be explained either. It's only the way/path to achieve the latter solution that eludes me.

The task: Design a circuit that will amplify the difference between two input voltages by two.

My procedure: I cracked open LT Spice, and immediately started of with an op-amp. First I wanted to find a working topology (I can work out the exact values later). I hooked up both input voltages to the two inputs and ran a feedback resistor to give me a gain of two.

Pardon the cluster of a circuit (just a prototype). The difference between Vin1 and Vin2 is about 1 V, but Vout is 3 V. That's obviously wrong, and I quickly realized why. Sure, the difference is amplified by 2, but it is subtracted from Vin2, which gives not 2 V, but 3 V (Vin2 = 5 V, so 5 V - 2 V = 3 V at Vout).

The first attempt made me realize a couple of things. My desired result does appear in this solution, but it is 1) inverted and 2) with respect to Vin2, not ground like I expected. Having analyzed it, it's now obvious, but it wasn't before (beginner brain, oh well). Since the non-inverting input seems to be the reference here, I figured I just have to set it to ground and then feed the difference between Vin1 and Vin2 into one of the inputs.

Combining the voltages is easy, but one of them needs to be inverted so that they cancel out via superposition. The only thing that would be left would be the difference, which can be fed into the amplifier stage. I decided to feed Vin1 through an inverter. The other voltage gets superimposed directly, and the difference is fed into an amplifier with gain 2, giving the circuit like so:

The simulation verifies that the circuit works as expected. The task requirements are satisfied.

After having gone through the design process by myself, I looked up a differential op-amp amplifier online and found this guy (from electronics-tutorials.ws):

What the heck! How did I miss that? This hasn't even crossed my mind! After failing with feeding Vin2 into the non-inverting input, I reworked my design based on my observations. I observed that Vout will be a voltage offset with respect to the non-inverting input, and following that, I arrived at a working solution. But this is also a working solution! What could have I possibly observed about my initial attempt that would have led me to discover the simpler solution presented on electronics-tutorials? I feel like there is a mindset here that is completely lost upon me, and I'm trying to find it. I'm hoping you guys who would have come up with a design like this right away can help by pointing me towards that missing link. Ya'll are much smarter than I. :)

EDIT: I took a couple of days to ponder this problem, and I think I figured out how to get there. As per Reinderien's suggestion, I've taken the initial circuit I had and figured out the transfer function:

$$V_{out}(t) = \frac{R_2}{R_1}(V_{in2}(t)-V_{in1}(t)) + V_{in2}(t)$$

Which is basically what I already expected from the simulation I ran in LT Spice. I get the difference between the two inputs, multiplied by the gain, and offset by the input at the non-inverting op-amp pin.

I started to think about how I could get rid of the offset Vin2. I needed to somehow get both Vin2 terms to cancel out to get just the difference. The method dawned on me when I remembered that this is kind of what you get if you calculate a Thevenin's equivalent circuit. For example, in a simple voltage divider circuit, we end up with the source voltage multiplied by a factor determined by the resistors. The current here isn't of particular importance since the ideal op-amp doesn't draw any current itself, so it does not load the equivalent network (yes, I understand that in the real world, the input impedance would matter considering the voltage source would then be loaded, but for the sake of argument, and considering it is a basic electronics class, assume the planets align just right). So, I got on with the algebra as follows (to simplify latex, Vin1(t) -> V1 and Vin2(t) -> V2):

$$\begin{eqnarray*} \frac{R_2}{R_1}(V_2-V_1)&=&\frac{R_2}{R_1}(\alpha V_2 - V_1) + \alpha V_2\\ &=&\frac{R_2}{R_1}\alpha V_2 + \alpha V_2 - \frac{R_2}{R_1}V_1\\ &=&V_2 (\alpha \frac{R_2}{R_1}+\alpha ) - \frac{R_2}{R_1}V_1\\ &=&\frac{R_2}{R_1}\left[ V_2(\alpha +\frac{\alpha R_1}{R_2}) - V_1\right] \end{eqnarray*}$$

The equation on the left and right are the same, except for the (a*R2/R1 + a) factor. On the left, the matching coefficient is 1. There is only one possibility. Whatever a is, that whole thing needs to work out to 1, so

$$\begin{eqnarray*} 1&=&\alpha + \frac{\alpha R_1}{R_2}\\ R_2&=&R_2\alpha + R_1\alpha\\ R_2&=&(R_2 + R_1)\alpha\\ \frac{R_2}{R_2 + R_1}&=&\alpha \end{eqnarray*}$$

I think that's fairly satisfactory. When we attenuate Vin2 by the complementary ratio (for lack of a better word) of the gain, the sums work out just right.

• Add another constraint to your toolset: the opamp has enormous gain, and forces the Vin- to be the same as Vin+(within a few microVolts). This fact allows the two sides of the opamp inputs (the 4 resistors) to interact. Nov 18 '18 at 5:14

You say you understand the math, so I won't (re)explain it to you. You say you want the process.

The process is -

1. Start with a basic understanding of the theory. One single op-amp can produce an output that's a linear combination of the two inputs. Any resistive divider prior to the op-amp will attenuate its input; you need this behaviour in this case.
2. Build a circuit with symbols instead of values, and calculate the expected output as a function of its inputs.
3. Insert "reasonable values" for some unknown resistors until the solution is not underdetermined. In this case, "reasonable" may be 1k or 10k, for instance, but this also requires knowledge of the application, including the input impedance of the op-amp and output impedance of the prior stage.
4. Solve for the remaining values.
5. Simulate to confirm your circuit.

Quite honestly, after doing work in this field professionally for any length of time, most people memorize the common circuits and skip the first few steps automatically. Whereas professionals understand the theory, they usually don't need to reiterate it once they understand and remember the common solutions.

• Thank you for your answer! Thinking of the output as a linear combination is interesting. I have never seen it described that way. I tried to recombine the inputs on paper, but looking at the equation, I thought maybe I could have just added another feedback resistor going to the non-inverting input. After carrying out the math, I can see it doesn't really work out. In your first step, would you be able elaborate a little bit more on why I need to attenuate the input? Nov 17 '18 at 19:44
• Try the circuit without the attenuation. See what the transfer function ends up being. Nov 18 '18 at 4:07
• Hello @Reinderien, sorry for replying so late, school and work get in the way. Anyway, going over the transfer function, I ended up with Vo = (R2/R1)*(Vin2-Vin1) + Vin2, but that just gets me where I started in the first example. I can see that I have the difference amplified by R2/R1 and offset by Vin2, which is what I ended up incorporating into design 2. It worked, but I would never have guessed to feed Vin2 through a voltage divider to get the same effect, barring any other factors like in/out impedance. Nov 20 '18 at 2:38
• Alrighty. I took a little bit to ponder this whole thing and eventually found an answer based on your information Too long to summarize in the comments, so I added it as an edit to the original question. Thank you! Nov 21 '18 at 17:28

Note that your solution is not “wrong” it is simply not the most efficient or the one with the most desirable input characteristics in some applications. As engineers we are often dealing with under-constrained problems and we have to make decisions that impose their own constraints.

If the problem had called for the “highest possible input impedance” (something that the solution you think is ‘better’ does not satisfy) it is very likely you would have come up with a very different solution. One with non-inverting amplifiers at the input.

As a matter of fact, your solution seems to be the exact opposite of what that could look like. Take a look at a common instrumentation amplifier configuration . From Analog devices.

Which is not what instrumentation amplifiers traditionally look like] . From Wikipedia.

All different solutions to the same problem, each one of them with different design choices and constraints.

• Thank you Edgar for your reply. I understand where you're coming from, but I'm not really sure if it answers my question per se. The problem given to me only asked for a differential amplifier that amplifies the difference by two. It's a contrived classroom example. Anyway, I'm not so much concerned to figure out how to get the best design. I'm only trying to figure out how I could have arrived at the design that I found online without resorting to randomly adding components until it worked. Sorry if I was being unclear. Thanks again! Nov 20 '18 at 2:44
• @KamilJarosz The main difference between an expert and a novice is that the expert has failed many more times. What you are missing is the intuition that comes from experience. Nowadays I seldom write equations when coming up with a circuit, I can achieve most of the design just by doing "schematic algebra in my head." Moving sources around, doing Thevenin/Norton equivalents, estimating time constants, doing Miller transformations, and a bit of basic algebra is all that is needed. I only need the equations when it comes time to verify my intuitions. That can only be achieved with experience. Nov 21 '18 at 17:41


text:
# Lieb-Thirring inequalities

Inequalities concerning the negative

eigenvalues of the

Schrödinger operator

(cf. also

\begin{equation*} H = - \Delta + V ( x ) \end{equation*}

on

$L ^ { 2 } ( \mathbf{R} ^ { n } )$,

$n \geq 1$.

With

$e _ { 1 } \leq e _ { 2 } \leq \ldots < 0$

denoting the negative eigenvalue(s) of

$H$

(if any), the Lieb–Thirring

inequalities state that for suitable

$\gamma \geq 0$

and constants

,

$$\tag{a1} \sum _ { j \geq 1 } | e _ { j } | ^ { \gamma } \leq L _ { \gamma , n } \int _ { \mathbf{R} ^ { n } } V _ { - } ( x ) ^ { \gamma + n / 2 } d x$$

with

$V _ { - } ( x ) : = \operatorname { max } \{ - V ( x ) , 0 \}$.

When

$\gamma = 0$,

the left-hand side is just the number of negative eigenvalues. Such an

inequality

(a1)

can hold if and only if

$$\tag{a2} \left\{ \begin{array} { l l } { \gamma \geq \frac { 1 } { 2 } } & { \text { for } n= 1, } \\ { \gamma > 0 } & { \text { for }n = 2, } \\ { \gamma \geq 0 } & { \text { for } n\geq 3. } \end{array} \right.$$

The cases

$\gamma > 1 / 2$,

$n = 1$,

$\gamma > 0$,

$n \geq 2$,

were established by

E.H. Lieb

and

W.E. Thirring

in connection with their proof of

stability of matter.

The case

$\gamma = 1 / 2$,

$n = 1$,

was established by

T. Weidl

The case

$\gamma = 0$,

$n \geq 3$,

was established independently by

M. Cwikel

Lieb

and

G.V. Rosenbljum

by different methods and is known as the

CLR bound;

the smallest known

value (as of

1998)

for

$L _ { 0 , n }$

is in

[a6],

[a7].

Closely associated with the inequality

(a1)

is the

semi-classical approximation

for

$\sum | e | ^ { \gamma }$,

which serves as a heuristic motivation for

(a1).

It is

(cf.

[a14]):

\begin{equation*} \sum _ { j \geq 1 } | e | ^ { \gamma } \approx \end{equation*}

\begin{equation*} \approx ( 2 \pi ) ^ { - n } \int _ { {\bf R} ^ { n } \times {\bf R} ^ { n } } [ p ^ { 2 } + V ( x ) ] _ { - } ^ { \gamma } d p d x = \end{equation*}

\begin{equation*} = L _ { \gamma , n } ^ { c } \int _ { {\bf R} ^ { n } } V _ { - } ( x ) ^ { \gamma + n / 2 } d x, \end{equation*}

with

\begin{equation*} L _ { \gamma , n } ^ { c } = 2 ^ { - n } \pi ^ { - n / 2 } \frac { \Gamma ( \gamma + 1 ) } { \Gamma ( \gamma + 1 + n / 2 ) }. \end{equation*}

Indeed,

$L _ { \gamma , n } ^ { c } < \infty$

for all

$\gamma \geq 0$,

whereas

(a1)

holds only for the range given in

(a2).

It is easy to prove (by considering

$V ( x ) = \lambda W ( x )$

with

$W$

smooth and

$\lambda \rightarrow \infty$)

that

\begin{equation*} L _ { \gamma , n } \geq L _ { \gamma , n } ^ { c }. \end{equation*}

An interesting, and mostly open

(1998)

problem is to determine the sharp

value

of the constant

,

especially to find those cases in which

$L _ { \gamma , n } = L _ { \gamma , n } ^ { c }$.

M. Aizenman

and

Lieb

proved that the ratio

is a monotonically non-increasing function of

$\gamma$.

Thus, if

$R _ { \Gamma , n } = 1$

for some

$\Gamma$,

then

$L _ { \gamma , n } = L _ { \gamma , n } ^ { c }$

for all

$\gamma \geq \Gamma$.

The equality

$L _ { \frac { 3 } { 2 } ,\, n } = L _ { \frac { 3 } { 2 } ,\, n } ^ { c }$

was proved for

$n = 1$

in

and for

$n > 1$

in

by

A. Laptev

and

Weidl.

[a1].)

The following sharp constants are known:

$L _ { \gamma , n } = L _ { \gamma , n } ^ { c }$,

all

$\gamma \geq 3 / 2$,

[a3],

[a2];

$L _ { 1 / 2,1 } = 1 / 2$,

There is strong support for the

conjecture

that

$$\tag{a3} L _ { \gamma , 1 } = \frac { 1 } { \sqrt { \pi } ( \gamma - \frac { 1 } { 2 } ) } \frac { \Gamma ( \gamma + 1 ) } { \Gamma ( \gamma + 1 / 2 ) } \left( \frac { \gamma - \frac { 1 } { 2 } } { \gamma + \frac { 1 } { 2 } } \right) ^ { \gamma + 1 / 2 }$$

for

$1 / 2 < \gamma < 3 / 2$.

Instead of considering all the negative eigenvalues as in

(a1),

one can consider just

$e_1$.

Then for

$\gamma$

as in

(a2),

\begin{equation*} | e _ { 1 } | ^ { \gamma } \leq L _ { \gamma , n } ^ { 1 } \int _ { \mathbf{R} ^ { n } } V _ { - } ( x ) ^ { \gamma + n / 2 } d x. \end{equation*}

Clearly,

$L _ { \gamma , n } ^ { 1 } \leq L _ { \gamma ,n }$,

but equality can hold, as in the cases

$\gamma = 1 / 2$

and

$3 / 2$

for

$n = 1$.

Indeed, the conjecture in

(a3)

amounts to

$L _ { \gamma , 1 } ^ { 1 } = L _ { \gamma , 1 }$

for

$1 / 2 < \gamma < 3 / 2$.

The sharp value

(a3)

of

$L _ { \gamma , n} ^ { 1 }$

is obtained by solving a differential equation

It has been conjectured that for

$n \geq 3$,

$L _ { 0 ,\, n } = L _ { 0 ,\, n } ^ { 1 }$.

In any case,

B. Helffer

and

D. Robert

showed that for all

$n$

and all

$\gamma < 1$,

$L _ { \gamma , n } > L _ { \gamma , n } ^ { c }$.

The sharp constant

$L _ { 0 , n } ^ { 1 }$,

$n \geq 3$,

is related to the sharp constant

$S _ { n }$

in the

Sobolev inequality

$$\tag{a4} \| \nabla f \| _ {{ L } ^ 2 ( \mathbf{R} ^ { n } ) } \geq S _ { n } \| f \| _ { L ^{ 2 n / ( n - 2 ) } ( \mathbf{R} ^ { n } ) }$$

by

$L _ { 0 , n } ^ { 1 } = ( S _ { n } ) ^ { - n }$.

By a

"duality argument"

the case

$\gamma = 1$

in

(a1)

can be converted into the following bound for the

$\Delta$.

This bound is

referred to as a

Lieb–Thirring kinetic energy inequality

and its most important application is to the

stability of matter

[a8],

Let

$f _ { 1 } , f _ { 2 } , \ldots$

be

any

orthonormal sequence (finite or infinite, cf. also

in

$L ^ { 2 } ( \mathbf{R} ^ { n } )$

such that

$\nabla f _ { j } \in L ^ { 2 } ( \mathbf{R} ^ { n } )$

for all

$j \geq 1$.

Associated with this sequence is a

"density"

$$\tag{a5} \rho ( x ) = \sum _ { j \geq 1 } | f _ { j } ( x ) | ^ { 2 }.$$

Then, with

$K _ { n } : = n ( 2 / L _ { 1 , n } ) ^ { 2 / n } ( n + 2 ) ^ { - 1 - 2 / n }$,

$$\tag{a6} \sum _ { j \geq 1 } \int _ { \mathbf{R} ^ { n } } | \nabla f _ { j } ( x ) | ^ { 2 } d x \geq K _ { n } \int _ { \mathbf{R} ^ { n } } \rho ( x ) ^ { 1 + 2 / n } d x.$$

This can be extended to

anti-symmetric functions

in

$L ^ { 2 } ( \mathbf{R} ^ { n N } )$.

If

$\Phi = \Phi ( x _ { 1 } , \dots , x _ { N } )$

is such a function, one defines, for

$x \in \mathbf{R} ^ { n }$,

\begin{equation*} \rho ( x ) = N \int _ { \mathbf{R} ^ { n ( N - 1 ) } } | \Phi ( x , x _ { 2 } , \ldots , x _ { N } ) | ^ { 2 } d x _ { 2 } \ldots d x _ { N }. \end{equation*}

Then, if

$\int _ { \mathbf{R} ^ { n N } } | \Phi | ^ { 2 } = 1$,

$$\tag{a7} \int _ { R ^ { n N } } | \nabla \Phi | ^ { 2 } \geq K _ { n } \int _ { {\bf R} ^ { n } } \rho ( x ) ^ { 1 + 2 / n } d x.$$

Note that the choice

$\Phi = ( N ! ) ^ { - 1 / 2 } \operatorname { det } f _ { j } ( x _ { k } ) | _ { j , k = 1 } ^ { N }$

with

$f_j$

orthonormal reduces the general case

(a7)

to

(a6).

If the conjecture

$L _ { 1,3 } = L _ { 1,3 } ^ { c }$

is correct, then the bound in

(a7)

equals the

Thomas–Fermi kinetic energy Ansatz

(cf.

and hence it is a challenge to prove this conjecture. In the meantime,

see

[a7],

for the best available constants to date

(1998).

Of course,

$\int ( \nabla f ) ^ { 2 } = \int f ( - \Delta f )$.

Inequalities of the type

(a7)

can be found for other powers of

$- \Delta$

than the first power. The first example of this kind, due to

I. Daubechies

and one of the most important physically, is to replace

$- \Delta$

by

$\sqrt { - \Delta }$

in

$H$.

Then an inequality similar to

(a1)

holds with

$\gamma + n / 2$

replaced by

$\gamma + n$

(and with a different

$L _ { \gamma , n _ { 1 }}$,

of course). Likewise there is an analogue of

(a7)

with

$1 + 2 / n$

replaced by

$1 + 1 / n$.

All proofs of

(a1)

(except

and

actually

proceed by finding an upper bound to

$N _ { E } ( V )$,

the number of eigenvalues of

$H = - \Delta + V ( x )$

that are below

$- E$.

Then, for

$\gamma > 0$,

\begin{equation*} \sum | e | ^ { \gamma } = \gamma \int _ { 0 } ^ { \infty } N _ { E } ( V ) E ^ { \gamma - 1 } d E. \end{equation*}

Assuming

$V = - V _ { - }$

(since

$V _ { + }$

only raises the eigenvalues),

$N _ { E } ( V )$

is most accessible via the positive semi-definite

Birman–Schwinger kernel

(cf.

[a4])

\begin{equation*} K _ { E } ( V ) = \sqrt { V _ { - } } ( - \Delta + E ) ^ { - 1 } \sqrt { V _ { - } }. \end{equation*}

$e < 0$

is an eigenvalue of

$H$

if and only if

$1$

is an eigenvalue of

$K _ { |e| } ( V )$.

Furthermore,

$K _ { E } ( V )$

is

operator

that is monotone decreasing in

$E$,

and hence

$N _ { E } ( V )$

equals the number of eigenvalues of

$K _ { E } ( V )$

that are greater than

$1$.

An important generalization of

(a1)

is to replace

$- \Delta$

in

$H$

by

$| i \nabla + A ( x ) | ^ { 2 }$,

where

$A ( x )$

is some arbitrary vector field in

${\bf R} ^ { n }$

(called a

magnetic vector potential).

Then

(a1)

still holds, but it is not known if the sharp value of

changes. What is known is that all

presently

(1998)

known values of

are unchanged. It is also known that

$( - \Delta + E ) ^ { - 1 }$,

as a kernel in

$\mathbf{R} ^ { n } \times \mathbf{R} ^ { n }$,

is pointwise greater than the absolute value of the kernel

$( | i \nabla + A | ^ { 2 } + E ) ^ { - 1 }$.

There is another family of inequalities for orthonormal functions,

which is closely related to

(a1)

and to the CLR

bound

[a9].

As before, let

$f _ { 1 } , \dots , f _ { N }$

be

$N$

orthonormal functions in

$L ^ { 2 } ( \mathbf{R} ^ { n } )$

and set

\begin{equation*} u _ { j } = ( - \Delta + m ^ { 2 } ) ^ { - 1 / 2 } f _ { j }, \end{equation*}

\begin{equation*} \rho ( x ) = \sum _ { j = 1 } ^ { N } | u _ { j } ( x ) | ^ { 2 }. \end{equation*}

$u _ { j }$

is a

($m = 0$)

or a

($m > 0$)

of

$f_j$.

If

$n = 1$

and

$m > 0$,

then

$\rho \in C ^ { 0,1 / 2 } ( \mathbf{R} ^ { n } )$

and

$\| \rho \| _ { L^\infty ( {\bf R} )} \leq L / m$.

If

$n = 2$

and

$m > 0$,

then for all

$1 \leq p < \infty$,

$\| \rho \| _ { L ^ { p } ( R ^ { 2 } ) } \leq B _ { p } m ^ { - 2 / p } N ^ { 1 / p }$.

If

$n \geq 3$,

$p = n / ( n - 2 )$

and

$m \geq 0$

(including

$m = 0$),

then

$\| \rho \| _ { L ^ { p } ( \mathbf{R} ^ { n } ) } \leq A _ { n } N ^ { 1 / p }$.

Here,

$L$,

$B _ { p }$,

$A _ { n }$

are universal constants. Without the orthogonality,

$N ^ { 1 / p }$

would have to be replaced by

$N$.

Further generalizations are

possible

[a9].

#### References

 [a1] R. Benguria, M. Loss, "A simple proof of a theorem of Laptev and Weidl" Preprint (1999) [a2] A. Laptev, T. Weidl, "Sharp Lieb–Thirring inequalities in high dimensions" Acta Math. (in press 1999) [a3] M.A. Aizenman, E.H. Lieb, "On semiclassical bounds for eigenvalues of Schrödinger operators" Phys. Lett. , 66A (1978) pp. 427–429 [a4] B. Simon, "Functional integration and quantum physics" , Pure Appl. Math. , 86 , Acad. Press (1979) [a5] Ph. Blanchard, J. Stubbe, "Bound states for Schrödinger Hamiltonians: phase space methods and applications" Rev. Math. Phys. , 8 (1996) pp. 503–547 [a6] E.H. Lieb, "The numbers of bound states of one-body Schrödinger operators and the Weyl problem" , Geometry of the Laplace Operator (Honolulu, 1979) , Proc. Symp. Pure Math. , 36 , Amer. Math. Soc. (1980) pp. 241–251 [a7] E.H. Lieb, "On characteristic exponents in turbulence" Comm. Math. Phys. , 92 (1984) pp. 473–480 [a8] E.H. Lieb, "Kinetic energy bounds and their applications to the stability of matter" H. Holden (ed.) A. Jensen (ed.) , Schrödinger Operators (Proc. Nordic Summer School, 1988) , Lecture Notes Physics , 345 , Springer (1989) pp. 371–382 [a9] E.H. Lieb, "An $L ^ { p }$ bound for the Riesz and Bessel potentials of orthonormal functions" J. Funct. Anal. , 51 (1983) pp. 159–165 [a10] G.V. Rosenbljum, "Distribution of the discrete spectrum of singular differential operators" Dokl. Akad. Nauk SSSR , 202 (1972) pp. 1012–1015 ((The details are given in: Izv. Vyss. Uchebn. Zaved. Mat. 164 (1976), 75-86 (English transl.: Soviet Math. (Izv. VUZ) 20 (1976), 63-71))) [a11] D. Hundertmark, E.H. Lieb, L.E. Thomas, "A sharp bound for an eigenvalue moment of the one-dimensional Schrödinger operator" Adv. Theor. Math. Phys. , 2 (1998) pp. 719–731 [a12] B. Helffer, D. Robert, "Riesz means of bound states and semi-classical limit connected with a Lieb–Thirring conjecture, II" Ann. Inst. H. Poincaré Phys. Th. , 53 (1990) pp. 139–147 [a13] I. Daubechies, "An uncertainty principle for fermions with generalized kinetic energy" Comm. Math. Phys. , 90 (1983) pp. 511–520 [a14] E.H. Lieb, W. Thirring, "Inequalities for the moments of the eigenvalues of the Schrödinger Hamiltonian and their relation to Sobolev inequalities" E. Lieb (ed.) B. Simon (ed.) A. Wightman (ed.) , Studies in Mathematical Physics: Essays in Honor of Valentine Bargmann , Princeton Univ. Press (1976) pp. 269–303 ((See also: W. Thirring (ed.), The stability of matter: from the atoms to stars, Selecta of E.H. Lieb, Springer, 1977)) [a15] M. Cwikel, "Weak type estimates for singular values and the number of bound states of Schrödinger operators" Ann. Math. , 106 (1977) pp. 93–100 [a16] T. Weidl, "On the Lieb–Thirring constants $L_{ \gamma , 1}$ for $\gamma \geq 1 / 2$" Comm. Math. Phys. , 178 1 (1996) pp. 135–146

Elliott H. Lieb


text:
# Tag Info

## Hot answers tagged bjt

4

Adapting Your Schematic Here's how I might adapt your schematic: I knew (because I looked) that the 2N2222 has $\beta\approx 200$. So I selected an appropriate base resistor, given the collector resistor I first selected. Other than that, there's not much to the adaptation except that I'm using two voltage sources instead of one so that I can sweep out ...

3

You are correct, emitter current is base current + collector current. In fact, emitter current is I(emitter) = I(base) + I(collector) And you can calculate Beta from measured values... Beta = I(collector) / I(base) Then calculate exactly how much current is in each part of the transistor.

3

The current in the emitter region is indeed larger than in the collector, because the emitter current has to be equal to the collector current plus the base current. That's just KCL. Just like $β$ is defined as $\frac{I_c}{I_b}$, you can also define $α=\frac{I_c}{I_e}$, which is always less than one--i.e., $I_c<I_e$. (α is also often seen defined ...

3

Prelude In the section titled Question 1 in this answer of mine, I qualitatively discuss the rule of $\frac{\textbf{1}}{\textbf{10}}$ths. That answer was in response to a question on this very same page of the very same book (page 439 of "Practical Electronics for Inventors".) We recently engaged a short discussion (found in the comments at the ...

3

It may be just slightly easier to see it drawn this way: simulate this circuit – Schematic created using CircuitLab (I've left things to use $V_{\!_\text{CC}}$ rather than $9\:\text{V}$ to keep it more abstract for now, despite the specific resistor and capacitor values still present. I've kept all your part name-designations, though.) Static ...

3

To simulate BJT output curves like this: source I would use a schematic like: simulate this circuit – Schematic created using CircuitLab IB: constant current or stepped (parametric sweep). For example 1 uA or 1 uA to 10 uA in steps of 1 uA VCE: DC sweep 0 V to 12 V Note how I use the DC simulation (.DC) and not the transient (.tran) simulation! I don'...

2

Capacitors aren't even connected to negative rail, how will they ever charge? They are. The negative terminal of each capacitor is connected to ground through the base-emitter junction of the transistors. The positive terminal is connected to +9 V via the LED and its current limiting resistor. Due to imbalances in the component tolerances and the ...

1

This circuit is called an astable multivibrator. The very short description is, once the circuit gets going, each capacitor charges through the transistor collector and discharges through the opposite transistor base. The two transistors take turns doing this as one triggers the other into its opposite state. A more detailed breakdown of this circuit is here:...

1

That circuit is a transistor astable multivibrator. Start with Q2 having just been switched on with its base at 0.7V and its collector at 0V. Q1’s base will be at -8.3V and will be slowly rising as C2 charges through R3. Simultaneously the collector of Q1 will be rising as C1 charges via R1. The collector of Q1 will reach 9V before its base reaches 0.7V ...

1

You have a transistor with about 430uA flowing into the base. The transistor will act more-or-less like a constant current sink until the voltage across the transistor drops below 200mV or so. The current is hFE dependent, but somewhere around 100mA with a large possible variation. The resistor R1 fights that current sink, but only affects it by 5% or so ...

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November 30, 2022, 01:14:27 PM
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### Topic: How do we make our own metal paint  (Read 3087 times)

0 Members and 1 Guest are viewing this topic.

#### abdulloh

• Regular Member
• Posts: 14
• Mole Snacks: +0/-0
##### How do we make our own metal paint
« on: May 15, 2018, 09:52:19 AM »
How could we make our own metal (metal-surface) paint, at least as close as to such in effect, and it's feasible.

#### P

• Full Member
• Posts: 641
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• I am what I am
##### Re: How do we make our own metal paint
« Reply #1 on: May 16, 2018, 05:36:25 AM »
Did I answer this on the other forum or was that someone else asking the same question   -  I will ask the same thing:

What metal?  Steel? I'll presume steel and also presume that you want some kind of rust preventing/converting primer?

Is it for commercial formulation or your home use?

There are many phosphates and things used as oxygen scavengers which get the O2 before the metal does - so rusting is prevented in the substrate.

Tonight I’m going to party like it’s on sale for $19.99! - Apu Nahasapeemapetilon #### abdulloh • Regular Member • Posts: 14 • Mole Snacks: +0/-0 ##### Re: How do we make our own metal paint « Reply #2 on: May 17, 2018, 05:20:17 AM » It's absolutely for my own use and need. The metal is iron with 0 - 2 % carbon proportion.. ..and Yes it's intended to prevent corrodible metal from being corroded. Thank you #### P • Full Member • Posts: 641 • Mole Snacks: +64/-15 • Gender: • I am what I am ##### Re: How do we make our own metal paint « Reply #3 on: May 17, 2018, 07:23:29 AM » OK - Then like I suggested on the other site - I would just buy a tin of primer for steel from the shop. I, and others there, asked what the application was - is it indoor? Warm place? Outdoor, Exposed area, sea side? Read the tins - if you need a strong rust prevention then get a high zinc, zinc phosphate or aluminium tripolyphosphate loaded primer - all the info you need will be on the tin. Where is the substrate? External/internal, sheltered/exposed? Tonight I’m going to party like it’s on sale for$19.99!

- Apu Nahasapeemapetilon


text:
# For the function g(x) = 32/(x+3), what is (g^-1 cdot g)(5)?

Jul 1, 2017

$\left({g}^{-} 1 \cdot g\right) \left(5\right) = \frac{68}{5}$

#### Explanation:

Let's start by finding an equation for the inverse.

$y = \frac{32}{x + 3}$

Switch the x and the y.

$x = \frac{32}{y + 3}$

$x \left(y + 3\right) = 32$

$x y + 3 x = 32$

$x y = 32 - 3 x$

$y = \frac{32 - 3 x}{x}$

Now let's complete the required operation.

$\left({g}^{-} 1 \cdot g\right) \left(5\right) = \frac{32 - 3 x}{x} \cdot \frac{32}{x + 3}$

$\left({g}^{-} 1 \cdot g\right) \left(5\right) = \frac{32 - 15}{5} \cdot \frac{32}{5 + 3}$

$\left({g}^{-} 1 \cdot g\right) \left(5\right) = \frac{17}{5} \cdot 4$

$\left({g}^{-} 1 \cdot g\right) \left(5\right) = \frac{68}{5}$

Hopefully this helps!

Jul 1, 2017

$\left({g}^{- 1} \cdot g\right) \left(5\right) = 5$

#### Explanation:

It should be obvious that irrespective of the definition of the function $g$ then the composition:

$\left({g}^{- 1} \cdot g\right) \left(x\right) \equiv x$

That is, in words, if we apply any function to a variable $x$, then apply the inverse function the we end up with what we started with, i.e. $x$

If you are not convinced, then we can easily demonstrate this by determining an expression for the inverse function ${g}^{- 1} \left(x\right)$:

$g \left(x\right) = \frac{32}{x + 3} \implies g = \frac{32}{x + 3}$
$\therefore x + 3 = \frac{32}{g}$
$\therefore x = \frac{32}{g} - 3$

So for the function $g \left(x\right) = \frac{32}{x + 3}$, the inverse function ${g}^{- 1} \left(x\right)$ is defined by:

${g}^{- 1} \left(x\right) = \frac{32}{x} - 3$

Now let us find $g \left(5\right)$

$g \left(5\right) = \frac{32}{5 + 3} = \frac{32}{8}$

Now let us find the composition $\left({g}^{- 1} \cdot g\right) \left(5\right)$

$\left({g}^{- 1} \cdot g\right) \left(5\right) = {g}^{- 1} \left(g \left(5\right)\right)$
$\text{ } = {g}^{- 1} \left(\frac{32}{8}\right)$
$\text{ } = \frac{32}{\frac{32}{8}} - 3$
$\text{ } = 8 - 3$
$\text{ } = 5$


text:
zbMATH — the first resource for mathematics

Bounds on the probability of radically different opinions. (English) Zbl 07204036
Summary: We establish bounds on the probability that two different agents, who share an initial opinion expressed as a probability distribution on an abstract probability space, given two different sources of information, may come to radically different opinions regarding the conditional probability of the same event.
MSC:
 6e+16 Inequalities; stochastic orderings
Full Text:
References:
 [1] Viktor Beneš and Josef Štěpán, Extremal solutions in the marginal problem, Advances in probability distributions with given marginals (Rome, 1990), Math. Appl., vol. 67, Kluwer Acad. Publ., Dordrecht, 1991, pp. 189-206. · Zbl 0738.60001 [2] Patrick Billingsley, Probability and measure, third ed., Wiley Series in Probability and Mathematical Statistics, John Wiley & Sons, Inc., New York, 1995, A Wiley-Interscience Publication. · Zbl 0822.60002 [3] Krzysztof Burdzy, The search for certainty, World Scientific Publishing Co. Pte. Ltd., Hackensack, NJ, 2009, On the clash of science and philosophy of probability. · Zbl 1330.60003 [4] Krzysztof Burdzy, Resonance—from probability to epistemology and back, Imperial College Press, London, 2016. · Zbl 1338.60002 [5] Krzysztof Burdzy and Soumik Pal, Contradictory predictions, (2019), preprint, arXiv:1912.00126. [6] R. Casarin, G. Mantoan, and F. Ravazzolo, Bayesian calibration of generalized pools of predictive distributions, Econometrics 4 (2016), no. 1. [7] A. P. Dawid, M. H. DeGroot, and J. Mortera, Coherent combination of experts’ opinions, Test 4 (1995), no. 2, 263-313. · Zbl 0844.62002 [8] A. P. Dawid and J. Mortera, A note on prediction markets, Available at arxiv[math.ST] arXiv:1702.02502, 2018. [9] M. H. DeGroot and J. Mortera, Optimal linear opinion pools, Management Science 37 (1991), no. 5, 546-558. · Zbl 0728.62008 [10] Morris H. DeGroot, A Bayesian view of assessing uncertainty and comparing expert opinion, J. Statist. Plann. Inference 20 (1988), no. 3, 295-306. · Zbl 0666.62003 [11] Lester E. Dubins, David Gilat, and Isaac Meilijson, On the expected diameter of an $$L_2$$-bounded martingale, Ann. Probab. 37 (2009), no. 1, 393-402. · Zbl 1159.60020 [12] Lester E. Dubins and Jim Pitman, A divergent, two-parameter, bounded martingale, Proc. Amer. Math. Soc. 78 (1980), no. 3, 414-416. · Zbl 0433.60045 [13] Lester E. Dubins and Jim Pitman, A maximal inequality for skew fields, Z. Wahrsch. Verw. Gebiete 52 (1980), no. 3, 219-227. · Zbl 0427.60053 [14] Kenny Easwaran, Luke Fenton-Glynn, Christopher Hitchcock, and Joel D. Velasco, Updating on the credences of others: Disagreement, agreement, and synergy, Philosophers’ Imprint 16 (2016), 1-39. [15] Simon French, Aggregating expert judgement, Rev. R. Acad. Cienc. Exactas Fís. Nat. Ser. A Mat. RACSAM 105 (2011), no. 1, 181-206. · Zbl 1217.62004 [16] Christian Genest, A characterization theorem for externally Bayesian groups, Ann. Statist. 12 (1984), no. 3, 1100-1105. · Zbl 0541.62003 [17] Tilmann Gneiting and Roopesh Ranjan, Combining predictive distributions, Electron. J. Stat. 7 (2013), 1747-1782. · Zbl 1294.62220 [18] Sam Gutmann, J. H. B. Kemperman, J. A. Reeds, and L. A. Shepp, Existence of probability measures with given marginals, Ann. Probab. 19 (1991), no. 4, 1781-1797. · Zbl 0739.60001 [19] Pierre Henry-Labordère, Jan Obłój, Peter Spoida, and Nizar Touzi, The maximum maximum of a martingale with given $$n$$ marginals, Ann. Appl. Probab. 26 (2016), no. 1, 1-44. · Zbl 1337.60078 [20] G. Kapetanios, J. Mitchell, S. Price, and N. Fawcett, Generalised density forecast combinations, J. Econometrics 188 (2015), no. 1, 150-165. · Zbl 1337.62292 [21] Younes Kchia and Philip Protter, Progressive filtration expansions via a process, with applications to insider trading, Int. J. Theor. Appl. Finance 18 (2015), no. 4, 1550027, 48. · Zbl 1344.60037 [22] Davar Khoshnevisan, Multiparameter processes, Springer Monographs in Mathematics, Springer-Verlag, New York, 2002, An introduction to random fields. · Zbl 1005.60005 [23] Fabian Krüger and Ingmar Nolte, Disagreement versus uncertainty: Evidence from distribution forecasts, Journal of Banking & Finance 72 (2016), S172-S186, IFABS 2014: Bank business models, regulation, and the role of financial market participants in the global financial crisis. [24] Steffen L. Lauritzen, Rasch models with exchangeable rows and columns, Bayesian statistics, 7 (Tenerife, 2002), Oxford Univ. Press, New York, 2003, pp. 215-232. [25] Jan Lorenz, Heiko Rauhut, Frank Schweitzer, and Dirk Helbing, How social influence can undermine the wisdom of crowd effect, Proceedings of the National Academy of Sciences 108 (2011), no. 22, 9020-9025. [26] Annette Möller and Jörgen Groß, Probabilistic temperature forecasting based on an ensemble autoregressive modification, Quarterly Journal of the Royal Meteorological Society 142 (2016), no. 696, 1385-1394. [27] Enrique Moral-Benito, Model averaging in economics: An overview, Journal of Economic Surveys 29 (2015), no. 1, 46-75. [28] Adam Osȩkowski, Estimates for the diameter of a martingale, Stochastics 87 (2015), no. 2, 235-256. · Zbl 1326.60059 [29] Adam Osȩkowski, Method of moments and sharp inequalities for martingales, Inequalities and extremal problems in probability and statistics, Academic Press, London, 2017, pp. 1-27. · Zbl 0931.41017 [30] Jim Pitman, Bounds on the probability of radically different opinions, Unpublished, 2014. [31] Roopesh Ranjan and Tilmann Gneiting, Combining probability forecasts, J. R. Stat. Soc. Ser. B Stat. Methodol. 72 (2010), no. 1, 71-91. · Zbl 1411.62270 [32] Ville A. Satopää, Robin Pemantle, and Lyle H. Ungar, Modeling probability forecasts via information diversity, J. Amer. Statist. Assoc. 111 (2016), no. 516, 1623-1633. · Zbl 1348.62017 [33] V. Strassen, The existence of probability measures with given marginals, Ann. Math. Statist. 36 (1965), 423-439. · Zbl 0135.18701 [34] James W. Taylor and Jooyoung Jeon, Probabilistic forecasting of wave height for offshore wind turbine maintenance, European J. Oper. Res. 267 (2018), no. 3, 877-890. · Zbl 1403.90275 [35] Cédric Villani, Optimal transport, Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], vol. 338, Springer-Verlag, Berlin, 2009, Old and new. · Zbl 1156.53003 [36] Theodore Zhu, On coherent distributions, (forthcoming), 2020.
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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# Different undistorting results (first rotate, then undistort OR first undistort, then rotate)

Hello,

i don't know why, but i get different results if i first rotate the image and undistort it, or if i first undistort and then rotate it.
There is a very small difference at both results... (In the middle of the image it seems to be same, but at the upper and lower borders of the images the difference is higher)

My minimal code is shown below:

Mat camera_matrix, distCoeffs;

char* out_file = "C:\\Users\\Bob\\Documents\\Visual Studio 2013\\Projects\\Samples\\camera.txt";

fs["K"] >> camera_matrix;
fs["D"] >> distCoeffs;

Mat image1 = imread("C:\\Users\\Bob\\Documents\\Visual Studio 2013\\Projects\\Samples\\1.jpg");
Mat image2 = imread("C:\\Users\\Bob\\Documents\\Visual Studio 2013\\Projects\\Samples\\2.jpg");

Mat image1_undistorted, image2_undistorted;
Mat image1_rot, image2_rot, image1_flip, image2_flip;

//First workflow (first undistort image, then turn)
///////////////////////////////////////////////////////////////////////

undistort(image1, image1_undistorted, camera_matrix, distCoeffs);
undistort(image2, image2_undistorted, camera_matrix, distCoeffs);

transpose(image1_undistorted, image1_rot);
flip(image1_rot, image1_undistorted, 0);
transpose(image2_undistorted, image2_rot);
flip(image2_rot, image2_undistorted, 0);

imwrite("image1_undistorted_FIRSTWORKFLOW.jpg", image1_undistorted);
imwrite("image2_undistorted_FIRSTWORKFLOW.jpg", image2_undistorted);

//Second workflow (first turn image, then undistort)
///////////////////////////////////////////////////////////////////////

//Adjusting the camera matrix and distortion coefficients
double fx = camera_matrix.at<double>(0, 0);
double fy = camera_matrix.at<double>(1, 1);
double cx = camera_matrix.at<double>(0, 2);
double cy = camera_matrix.at<double>(1, 2);

camera_matrix.at<double>(0, 0) = fy;
camera_matrix.at<double>(1, 1) = fx;
camera_matrix_new.at<double>(0, 2) = cy;
camera_matrix_new.at<double>(1, 2) = image1.size().width - cx;

double p1 = distCoeffs.at<double>(0, 2);
double p2 = distCoeffs.at<double>(0, 3);

distCoeffs.at<double>(0, 2) = p2;
distCoeffs.at<double>(0, 3) = p1;

transpose(image1, image1_rot);
flip(image1_rot, image1_flip, 0);
transpose(image2, image2_rot);
flip(image2_rot, image2_flip, 0);

undistort(image1_flip, image1_undistorted, camera_matrix, distCoeffs, Mat());
undistort(image2_flip, image2_undistorted, camera_matrix, distCoeffs, Mat());

imwrite("image1_undistorted_SECONDWORKFLOW.jpg", image1_undistorted);
imwrite("image2_undistorted_SECONDWORKFLOW.jpg", image2_undistorted);

The results are shown below:

The first image (both workflow results)

The second image (both workflow results)

---------------------------------------------------------------------------------------------------------------------------------------

And if i calculate the differences of the images resulting from the different workflows, then i get these:

image1_undistorted_FIRSTWORKFLOW - image1_undistorted_SECONDWORKFLOW =

image2_undistorted_FIRSTWORKFLOW - image2_undistorted_SECONDWORKFLOW =

You can see the in the middle of the "difference images" it seems to be there is no difference...
But on the up and down side of the "difference image" there is a bigger difference...

Can someone tell me what im doing wrong? Or why i get different results in both workflows?

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I think this is because image distortion is expressed with respect to the center of an image as the position of undistorted pixel is dependent on the parameter r, which is the distance to the image center:

If the image dimensions are odd numbers, then the position of the image center is always in the same point regardless of the orientation. If the numbers are even (like in your case), then the position of the center depends on the image orientation.

In case of your image the dimensions are 526x390. Image center is probably at [263, 195], but these coordinates correspond to two different points when the image is flipped or not, which affects the r parameter for particular pixels.

You see the differences in the corners, because distortion impact is proportional to the distance to the image center. The further from the distance, the more distorted the image is.

If you want to make sure, take some image whose dimensions are both odd numbers, and try your code on it. If there is no difference between the results regardless of the order of operations, you've got the answer.

more

Official site

GitHub

Wiki

Documentation


text:
# nLab jet bundle

### Context

#### Differential geometry

differential geometry

synthetic differential geometry

# Contents

## Idea

A jet can be thought of as the infinitesimal germ of a section of some bundle or of a map between spaces. Jets are a coordinate free version of Taylor-polynomials and Taylor series.

## Definition

### Concrete

For $p : P \to X$ a surjective submersion of smooth manifolds and $k \in \mathbb{N}$, the bundle $J^k P \to X$ of $k$- order jets of sections of $p$ is the bundle whose fiber over a point $x \in X$ is the space of equivalence classes of germs of sections of $p$, where two germs are considered equivalent if their first $k$ partial derivatives at $x$ coincide.

In the case when $p$ is a trivial bundle $p:X\times Y \to X$ its sections are canonically in bijection with maps from $X$ to $Y$ and two sections have the same partial derivatives iff the partial derivatives of the corresponding maps from $X$ to $Y$ agree. So in this case the jet space $J^k P$ is called the space of jets of maps from $X$ to $Y$ and commonly denoted with $J^k(X,Y)$.

### General abstract

We discuss a general abstract definition of jet bundles.

Let

For $X \in \mathbf{H}$, write $\Im(X)$ for the corresponding de Rham space object.

Notice that we have the canonical morphism, the $X$-component of the unit of the $\Im$-monad

$i \colon X \to \Im(X)$

(“inclusion of constant paths into all infinitesimal paths”).

The corresponding base change geometric morphism is

$(i^\ast \dashv i_\ast) \;\colon\; \mathbf{H}_{/X} \stackrel{\overset{i^*}{\leftarrow}}{\underset{Jet := i_*}{\to}} \mathbf{H}_{/\Im(X)}$
###### Definition

$i^\ast i_\ast \;\colon\; \mathbf{H}_{/X} \longrightarrow \mathbf{H}_{/X}$
###### Remark

Since base change gives even an adjoint triple $(i_! \dashv i^\ast \dashv i_\ast)$, there is a left adjoint $T_{inf} X \times_X (-)$ to the jet comonad of def. 1,

$T_{inf}X \times_X (-) \;\dashv\; Jet$

where $T_{inf} X$ is the infinitesimal disk bundle of $X$, see at differential cohesion – infinitesimal disk bundle – relation to jet bundles

###### Remark

In the context of differential geometry the fact that the jet bundle construction is a comonad was explicitly observed in (Marvan 86, see also Marvan 93, section 1.1, Marvan 89). It is almost implicit in (Krasil’shchik-Verbovetsky 98, p. 13, p. 17, Krasilshchik 99, p. 25).

In the context of synthetic differential geometry the fact that the jet bundle construction is right adjoint to the infinitesimal disk bundle construction is (Kock 80, prop. 2.2).

In the context of algebraic geometry and of D-schemes as in (BeilinsonDrinfeld, 2.3.2, reviewed in Paugam, section 2.3), the base change comonad formulation inf def. 1 was noticed in (Lurie, prop. 0.9).

In as in (BeilinsonDrinfeld, 2.3.2, reviewed inPaugam, section 2.3) jet bundles are expressed dually in terms of algebras in D-modules. We now indicate how the translation works.

###### Remark

In terms of differential homotopy type theory this means that forming “jet types” of dependent types over $X$ is the dependent product operation along the unit of the infinitesimal shape modality

$jet(E) \coloneqq \underset{X \to \Im X}{\prod} E \,.$
###### Definition

A quasicoherent (∞,1)-sheaf on $X$ is a morphism of (∞,2)-sheaves

$X \to Mod \,.$

We write

$QC(X) := Hom(X, Mod)$

A D-module on $X$ is a morphism of (∞,2)-sheaves

$\Im (X) \to Mod \,.$

We write

$DQC(X) := Hom(\Im (X), Mod)$

for the stable (∞,1)-category of D-modules.

The Jet algebra functor is the left adjoint to the forgetful functor from commutative algebras over $\mathcal{D}(X)$ to those over the structure sheaf $\mathcal{O}(X)$

$(Jet \dashv F) : Alg_{\mathcal{D}(X)} \stackrel{\overset{Jet}{\leftarrow}}{\underset{F}{\to}} Alg_{\mathcal{O}(X)} \,.$

## Application

Typical Lagrangians in quantum field theory are defined on jet bundles. Their variational calculus is governed by Euler-Lagrange equations.

Examples of sequences of local structures

geometrypointfirst order infinitesimal$\subset$formal = arbitrary order infinitesimal$\subset$local = stalkwise$\subset$finite
$\leftarrow$ differentiationintegration $\to$
smooth functionsderivativeTaylor seriesgermsmooth function
curve (path)tangent vectorjetgerm of curvecurve
smooth spaceinfinitesimal neighbourhoodformal neighbourhoodgerm of a spaceopen neighbourhood
function algebrasquare-0 ring extensionnilpotent ring extension/formal completionring extension
arithmetic geometry$\mathbb{F}_p$ finite field$\mathbb{Z}_p$ p-adic integers$\mathbb{Z}_{(p)}$ localization at (p)$\mathbb{Z}$ integers
Lie theoryLie algebraformal grouplocal Lie groupLie group
symplectic geometryPoisson manifoldformal deformation quantizationlocal strict deformation quantizationstrict deformation quantization

## References

Textbook accounts and lecture notes include

• David Saunders, The geometry of jet bundles, London Mathematical Society Lecture Note Series 142, Cambridge Univ. Press 1989.

• Joseph Krasil'shchik in collaboration with Barbara Prinari, Lectures on Linear Differential Operators over Commutative Algebras, 1998 (pdf)

• Shihoko Ishii, Jet schemes, arc spaces and the Nash problem, arXiv:math.AG/0704.3327

• G. Sardanashvily, Fibre bundles, jet manifolds and Lagrangian theory, Lectures for theoreticians, arXiv:0908.1886

• Peter Olver, Lectures on Lie groups and differential equation, chapter 3, Jets and differential invariants, 2012 (pdf)

Early accounts include

• Hubert Goldschmidt, Integrability criteria for systems of nonlinear partial differential equations, J. Differential Geom. Volume 1, Number 3-4 (1967), 269-307 (Euclid)

Discussion of the Frechet manifold structure on jet bundles includes

• Floris Takens, A global version of the inverse problem of the calculus of variations, J. Differential Geom. Volume 14, Number 4 (1979), 543-562. (Euclid)

• M. Bauderon, Differential geometry and Lagrangian formalism in the calculus of variations, in Differential Geometry, Calculus of Variations, and their Applications, Lecture Notes in Pure and Applied Mathematics, 100, Marcel Dekker, Inc., N.Y., 1985, pp. 67-82.

• Andrew Lewis, The bundle of infinite jets 2006 (pdf)

Discussion in tems of synthetic differential geometry is in

• Anders Kock, Formal manifolds and synthetic theory of jet bundles, Cahiers de Topologie et Géométrie Différentielle Catégoriques (1980) Volume: 21, Issue: 3 (Numdam)

• Anders Kock, section 2.7 of Synthetic geometry of manfiolds, Cambridge Tracts in Mathematics 180 (2010). (pdf)

The comonad structure on the jet operation in the context of differential geometry is made explicit in

• Michal Marvan, A note on the category of partial differential equations, in Differential geometry and its applications, Proceedings of the Conference August 24-30, 1986, Brno (pdf)

(notice that prop. 1.3 there is wrong, the correct version is in the thesis of the author)

with further developments in

• Michal Marvan On the horizontal cohomology with general coefficients, 1989 (web announcement, web archive)

Abstract: In the present paper the horizontal cohomology theory is interpreted as a special case of the Van Osdol bicohomology theory applied to what we call a “jet comonad”. It follows that differential equations have well-defined cohomology groups with coefficients in linear differential equations.

• Michal Marvan, section 1.1 of On Zero-Curvature Representations of Partial Differential Equations, (1993) (web)

In the context of algebraic geometry, the abstract characterization of jet bundles as the direct images of base change along the de Rham space projection is noticed on p. 6 of

The explicit description in terms of formal duals of commutative monoids in D-modules is in

An exposition of this is in section 2.3 of

A discussion of jet bundles with an eye towards discussion of the variational bicomplex on them is in chapter 1, section A of

• Ian Anderson, The variational bicomplex (pdf)

The de Rham cohomology of jet bundles is discussed in

• G.Giachetta, L.Mangiarotti, G.Sardanashvily, Cohomology of the variational bicomplex on the infinite order jet space, Journal of Mathematical Physics 42, 4272-4282 (2001) (arXiv:math/0006074)

Discussion of jet-restriction of the Haefliger groupoid is in

• Arne Lorenz, Jet Groupoids, Natural Bundles and the Vessiot Equivalence Method, Thesis (pdf)

Discussion of jet bundles in supergeometry includes

• Arthemy V. Kiselev, Andrey O. Krutov, appendix of On the (non)removability of spectral parameters in $\mathbb{Z}_2$-graded zero-curvature representations and its applications (arXiv:1301.7143)

Revised on January 10, 2016 06:59:57 by David Corfield (84.92.190.125)


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# Is there a pH range for "neutral" detergents?

There exist many home made pH scales like these

But do any of you know if there is an "official" scale that defines if a neutral detergent is 7 sharp or within a range _______?

• Strictly speaking, anything other than pH 7 is not neutral, because it has an imbalance in the ratio of $\ce{H+}$ to $\ce{OH-}$. Nov 6 '14 at 8:12
• I know that chemically this is the definition :-) Nov 6 '14 at 8:16
• Well, I would argue that chemically$=$officially Nov 6 '14 at 8:20
• I'm more thinking like public affairs or expert panel that is generally accepted as giving guidelines. Like some EU group og FDA declaration. What I could find was GSA gsa.gov/portal/content/113006 but not exactly hitting the spot Nov 6 '14 at 11:44
• I have dissected the definition of 'neutral' in this question which might help. Dec 11 '14 at 2:51

## 2 Answers

Neutral detergents are not neccesarily at pH 7, and I know of no legal restriction (at least in the EU) on labelling detergents as neutral based on pH. For example, this patent claims protection for a set of neutral detergent formulae with pH of "from about 5.5 to about 8.5".

An example of a commercial product is Proctor and Gamble's "Tide", which has a pH of 8.1-8.5 according to their safety data. This is advertised as "100%" neutral !

• Thank you for your answer. Listing commerical products is not advertisement in my oppion, its just showing that there are no clear standard and companies are allowed to interpret neutral as they like. Dec 11 '14 at 6:51

I want to add some imformation to J.LS answer. According to U.S. General Services Administration, neutral detergents are produced with strong alkalis and weak acids. They are the most common soapless detergent and are usually alkylbenzenesulfonates. So, neutral detergent is not that neutral after all.


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# zbMATH — the first resource for mathematics

On Bach-flat gradient shrinking Ricci solitons. (English) Zbl 1277.53036
A complete Riemannian manifold $$(M^n,g_{ij})$$ is called a gradient Ricci soliton if there exists a smooth function $$f$$ on $$M^n$$ such that the Ricci tensor $$R_{ij}$$ of the metric $$g$$ satisfies the equation $R_{ij}+\nabla_i\nabla_j f=\rho g_{ij}$ for some constant $$\rho$$. For $$\rho=0$$ the Ricci soliton is steady, for $$\rho>0$$ it is shrinking, and for $$\rho<0$$ expanding. In this paper, the authors investigate an interesting class of complete gradient shrinking Ricci solitons: those with vanishing Bach tensor. On any manifold $$(M^n,g_{ij})$$, $$n\geqslant 4$$, the Bach tensor is defined by $B_{ij}=\frac{1}{n-3}\nabla^{k}\nabla^{l} W_{ikjl}+\frac{1}{n-2}R_{kl}W_{i_j}{}^{k l},$ $$W_{ikjl}$$ being the Weyl tensor. It is easy to see that if $$(M^n,g_{ij})$$ is either locally conformally flat or Einstein, then $$(M^n,g_{ij})$$ is Bach-flat: $$B_{ij}=0$$. The authors show that Bach-flat 4-dimensional gradient shrinking Ricci solitons are either Einstein or locally conformally flat. More generally, for $$n\geqslant 5$$, they prove that a Bach-flat gradient shrinking Ricci soliton is either Einstein, or a finite quotient of the Gaussian shrinking soliton $$\mathbb R^n$$ or the product $$N^{n-1}\times\mathbb R$$, where $$N^{n-1}$$ is an Einstein manifold of positive scalar curvature.

##### MSC:
 53C21 Methods of global Riemannian geometry, including PDE methods; curvature restrictions 53C20 Global Riemannian geometry, including pinching 53C25 Special Riemannian manifolds (Einstein, Sasakian, etc.)
Full Text:


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# TANCET 2008 Solution : Work Time

This aptitude practice question is a Work Time problem solving question. The question appeared in TANCET 2008 and is an easy question. Concept tested: Framing work time equations when two people work together in parallel to complete a task.

## Question

A father can do a certain job in x hours. His son takes twice as long to do the job. Working together, they can do the job in 6 hours. How many hours does the father take to do the job?

1. 9
2. 18
3. 12
4. 20
5. 16

Correct Answer      Choice (1). The father takes 9 hours to complete the task.

The father completes the job in x hours.
Son takes twice as long to finish the job. So, the son will take 2x hours to complete the same job.
In an hour, the father will complete $$frac{1}{x}$ of the total job. In an hour, the son will complete $\frac{1}{2x}$ of the total job. So, if the father and son work together, in an hour they will complete $\frac{1}{x}$ + $\frac{1}{2x}$ of the job. i.e., in an hour they will complete $\frac{3}{2x}$ of the job. The question states that they complete the entire task in 6 hours if they work together. i.e., they complete $\frac{1}{6}^{th}$ of the task in an hour. Equating the two information, we get $\frac{3}{2x}$ = $\frac{1}{6}$ Cross multiplying and solving for x we get 2x = 18 or x = 9. 'x' is the time taken by the father to complete the task. The father takes 9 hours to complete the job. The correct answer is Choice$1).

## Online TANCET MBA CourseTry it Free!

Register in 2 easy steps and start learning in 5 minutes!

## TANCET MBA Coaching in Chennai

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text:
# What if we had commercial fusion power?

• Featured

#### phyzguy

Science Advisor
There have been several recent threads on the feasibility of fusion power. There is reason to be hopeful. This study from MIT claims, due to the breakthroughs in high-temperature superconductors, that an economically feasible Tokamak can be built in the foreseeable future.

For the purpose of this thread, let's not debate that. Let's assume that the technical problems were solved and we had a working design for a fusion reactor with similar economics to existing fission reactors. Then my question is, is there any reason to believe that there would be greater public acceptance of this technology than there is of fission technology?

It's true that the radioactive waste produced by a fusion reactor will be less and less long-lived than that produced by a fission reactor. However, the radioactive inventory and waste stream from a fusion reactor will still be enormous. Public opposition to fission power is not based on quantitative arguments, but is more based on an "all radiation is bad," mindset. I worry that even if we had a working fusion reactor, the public response would be, "Wait a minute, I thought you told us this was clean technology? Now we find out that it still generates large amounts of radioactive waste."

This article highlights my concern. In many ways fusion reactors will have the same problems that lead to the public opposition to fission power. Comments?

Last edited by a moderator:
Related Nuclear Engineering News on Phys.org

#### gleem

Science Advisor
Well there you go. Same issues as fission. Still unsolved problems and who know what we will find when we actually start running them. So I see a significant cry of "not in my back yard". Is the devil we know better than the one that we don't?

Well all that said I think we should have a demonstration unit running for a significant period of time to fully assess the problems that must be addressed before going to the general public.

#### russ_watters

Mentor
There have been several recent threads on the feasibility of fusion power. There is reason to be hopeful. This study from MIT claims, due to the breakthroughs in high-temperature superconductors, that an economically feasible Tokamak can be built in the foreseeable future. For the purpose of this thread, let's not debate that.
Fair enough, but I'm still curious about it -- I'm not seeing the economic feasibility discussion in that link. Could you be more specific about where it is?
....is there any reason to believe that there would be greater public acceptance of this technology than there is of fission technology?
I think fusion is and will be viewed as the panacea solar is. I don't think acceptance is something to be concerned about.
Public opposition to fission power is not based on quantitative arguments, but is more based on an "all radiation is bad," mindset. I worry that even if we had a working fusion reactor, the public response would be, "Wait a minute, I thought you told us this was clean technology? Now we find out that it still generates large amounts of radioactive waste."
If 60 years and the prospect of imminent doom are not enough to turn the public's opinion on fission, I don't expect fusion's can be turned easily either.
It's true that the radioactive waste produced by a fusion reactor will be less and less long-lived than that produced by a fission reactor. However, the radioactive inventory and waste stream from a fusion reactor will still be enormous.

This article highlights my concern. In many ways fusion reactors will have the same problems that lead to the public opposition to fission power. Comments?
Blasphemy! Of course, fusion is the power of the sun and like solar is perfectly safe, clean, plentiful, scaleable and above all, completely free!

#### Tiran

One item missing from the list is the relative possibility of a meltdown. Given that fusion isn't based on a radioactive pile, I would imagine this is a rather unlikely possibility.

The unlikelihood of a Chernobyl or Fukashima alone may greatly affect public opinion in the positive.

Personally, I think that orbital (or regular) solar collectors would be a better way of powering the energy grid, but fusion would be an excellent replacement for powerplants on submarines and ships - or spacecraft.

#### anorlunda

Mentor
Practical fusion power plants are decades away (No, really. Real soon now.) But everything associated with electric power is undergoing dramatic change.

Thirty years from now significant changes are likely. Central power plants may be illegal. Power utilities as a business model may disappear. The public's thirst for energy may be quenched. The grid as we know it may not exist. blah blah.

All that makes it completely pointless to speculate on future public opinion on a particular power plant design.

By analogy, it would be like a public poll at the time of the Wright Brothers about airport noise 80 years into the future.

#### phyzguy

Science Advisor
Practical fusion power plants are decades away (No, really. Real soon now.) But everything associated with electric power is undergoing dramatic change.

Thirty years from now significant changes are likely. Central power plants may be illegal. Power utilities as a business model may disappear. The public's thirst for energy may be quenched. The grid as we know it may not exist. blah blah.

All that makes it completely pointless to speculate on future public opinion on a particular power plant design.

By analogy, it would be like a public poll at the time of the Wright Brothers about airport noise 80 years into the future.
Well, in that case, maybe there is no point in continuing this thread. No objection from me if you close it.

#### gmax137

Well, in that case, maybe there is no point in continuing this thread.
Hey, not so fast! On a technical note, can someone compare the decay heat from a fission reactor to the corresponding residual heat from a fusion reactor? I just don't know anything about fusion, is there any appreciable decay heat? In a fission plant, safety is all about removing decay heat once the reactor is shutdown.
Is that even an issue with a fusion machine?

#### Klystron

Gold Member
[snip]
Thirty years from now significant changes are likely. Central power plants may be illegal.
[...snip...]
By analogy, it would be like a public poll at the time of the Wright Brothers about airport noise 80 years into the future.
@anorlunda Please elaborate on the possible illegality of central power plants unless you mean this an example of specious argument?

Your 'Wright bro's' analogy has merit as for instance newspaper and popular science publications from that era (circa 1905+) often discuss the drawbacks of automobiles frightening livestock; not envisioning that automobile traffic would rapidly replace horse-drawn vehicles even on farms. The concepts of horse and horsepower achieved near-mystical standing among the public; a mystique that quickly accrued to automobiles. Modern industries such as solar try to accrue similar mystique to sell their products.

Consider also that the typical pre-20th landowner likely owned horses, understood equine basics (e.g.; feeding hay vs. oats), and measured useful energy in horse units. A 21st homeowner can purchase solar technology, install panels on their house, even feed excess electricity back to the public power grid. Similar to recycling, adopting solar panels provides intrinsic benefits and status to the homeowner analogous to pre-20th horse ownership.

If these analogies have merit, public acceptance of fusion power sources might improve as the technology scales to more compact units such as in public transport and personal vehicles. Meanwhile the 'fission dirty; fusion clean', 'fission dangerous; fusion safe' rubrics may have to do within the end user community.

#### PeterDonis

Mentor
One item missing from the list is the relative possibility of a meltdown. Given that fusion isn't based on a radioactive pile, I would imagine this is a rather unlikely possibility.
It's an impossibility. There is no significant stored energy inside the reactor once it stops.

#### Tiran

It's an impossibility. There is no significant stored energy inside the reactor once it stops.
That doesn't mean that fusion reactor can't have some non-fission accident that would release considerable energy and atomize radioactive structure sending contaminates into the air.

#### phyzguy

Science Advisor
That doesn't mean that fusion reactor can't have some non-fission accident that would release considerable energy and atomize radioactive structure sending contaminates into the air.
I think the most worrisome accident from a fusion reactor would be a leak that released a large amount of tritium. A fusion reactor would likely have on the order of 10 kg of tritium in inventory. If I calculate correctly, this is ~ 100X the radiation released at Fukushima, so an accident that released even 1% of the tritium inventory would be an equal amount to Fukushima. Also, since tritium is easily taken up by the body, it can have significant biological effects.

The metal structure of a fusion reactor becomes radioactive through neutron activation, but it's hard to imagine an accident that would disperse the solid structure of the reactor into the environment.

#### PeterDonis

Mentor
I think the most worrisome accident from a fusion reactor would be a leak that released a large amount of tritium.
This is the type of incident that secondary containment structures prevent.

#### phyzguy

Science Advisor
This is the type of incident that secondary containment structures prevent.
If the secondary containment were always successful, we wouldn't have had a radiation release at Fukushima.

#### PeterDonis

Mentor
If the secondary containment were always successful, we wouldn't have had a radiation release at Fukushima.
The reason the secondary containment failed at Fukushima was loss of decay heat removal, due to poor siting of the backup switchgear (and also to the secondary containment not being designed to handle that amount of stress--since the design assumed decay heat removal would be present). A fusion reactor has no decay heat, so no decay heat removal.

#### Klystron

Gold Member
There have been several recent threads on the feasibility of fusion power. There is reason to be hopeful. This study from MIT claims, due to the breakthroughs in high-temperature superconductors, that an economically feasible Tokamak can be built in the foreseeable future.

[snip] Comments?
The picture above the super-conductor article says much about this thread. The blue-shadow human figure gives a rough size factor of current technology. The cut-away view of the torus reminds me of the dee's in early cyclotron diagrams. Here's 2 2019!

#### Tiran

Without using the word "meltdown", can you describe what risk of a fusion reactor you are referring to in the quote below, that you think is missing from the list?
To clarify, I meant that a fusion reactor will be perceived to not have a likelihood of having the equivalent of Chernobyl or Fukushima type radiation containment failure, and that is why that item is rightfully missing from the list. And since it is missing from the list, a fusion reactor might garner support that a fission reactor lacks.

But I said that I "imagine it is a rather unlikely possibility" because the inside of a fusion reactor is the temperature of the inside of a star, and I have no idea what sort of possible fusion reactor failure modes could lead to a containment failure serious enough to unleash the energy necessary to destroy the reactor structure. It just seems likely that the answer is not 'zero'. Whether that failure is a full on nuclear detonation or merely enough instantaneous heat to slag a building, both would produce results of great public concern because radioactive material would be released.

It is a similar conundrum to battery powered cars. They obviously are more efficient and should pollute less than fuel burning cars, but that thinking contains the assumption that battery production and replacement pollution is negligible. Again, I don't have that information, but it is also unlikely to be zero. The battery car bandwagon will grow, but if some unexpected level of environmental impact is brought to light, public opinion could quickly turn.

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#### Lord Jestocost

Gold Member
2018 Award
1. A fusion reactor is much less likely to have an event like a "meltdown". You said it is "impossible", but phyzguy agreed with me in post 12.
On the hompage of the ITER organization (https://www.iter.org/sci/Fusion) one reads:

"No risk of meltdown: A Fukushima-type nuclear accident is not possible in a tokamak fusion device. It is difficult enough to reach and maintain the precise conditions necessary for fusion—if any disturbance occurs, the plasma cools within seconds and the reaction stops. The quantity of fuel present in the vessel at any one time is enough for a few seconds only and there is no risk of a chain reaction."

#### PeterDonis

Mentor
the inside of a fusion reactor is the temperature of the inside of a star
Yes, but temperature is not the same as heat content. The inside of a fusion reactor is a plasma with a very low density, and even at fusion temperature, there simply isn't that much energy there--many orders of magnitude less energy than is stored inside a fission reactor.

Whether that failure is a full on nuclear detonation or merely enough instantaneous heat to slag a building
Consider this: a power plant producing a typical amount of electrical power (1000 MW or so) is producing roughly a ton of TNT worth of heat every second. This is true regardless of the type of fuel it is burning; it's just simple numbers--a factor of 3 to 4 loss in the conversion from thermal to electrical power means 3000 to 4000 MW thermal, which is about 4 billion Joules every second, and a ton of TNT is $4.184 \times 10^9$ Joules.

For any power plant that injects fuel and burns it right away, which includes oil, natural gas, and fusion, the only potential heat source inside it at any given time is the fuel needed to produce a few seconds or so worth of heat when burned (the quote that @Lord Jestocost gave from the ITER page indicates a few seconds' worth). So at most there is a few tons of TNT's worth of heat in the fuel inside the reactor, if it is burned.

A nuclear bomb has the explosive energy of tens to hundreds of thousands to millions of tons of TNT, all released in a tiny fraction of a second. So clearly a fusion reactor has many orders of magnitude too little fuel inside it to produce a nuclear detonation of this kind.

A bomb containing roughly a ton of TNT (or the equivalent) can collapse an ordinary building if exploded. But it doesn't do that by melting the building; it does it by converting the energy in the TNT into a high pressure shock wave in the air that causes the building's structure to fail. Combustion of fuel inside a fusion reactor doesn't do that: there is no air, and the plasma inside the reactor has much too low a density to form a shock wave. Also, the energy produced by fusion is mostly carried by neutrons, which don't interact with the plasma anyway, they just hit the first wall, which is designed for that. So nothing like a conventional explosion that could collapse a building can happen inside a fusion reactor even while it is running.

And, of course, if the reactor is shut down, fuel stops flowing and there is no more heat produced.

Here's another way to look at it: I said above that a power plant producing a typical amount of electrical power is producing roughly a ton of TNT worth of heat every second, regardless of the type of fuel it is burning. That means that conventional oil or natural gas power plants are doing that too. (They produce the heat at a lower temperature than a fusion reactor--a few thousand degrees vs. roughly ten million--but the density inside them is correspondingly higher, so the heat content is the same.) So if those plants don't have a failure mode where the entire building gets collapsed because something goes wrong with the reaction, a fusion reactor won't either.

The reason fission reactors have this unique sort of failure mode, then, is evidently that there is a lot more stuff inside the reactor than just a few seconds' worth of fuel at any given time. There is a few years' worth (in a typical pressurized water reactor) of fuel inside, plus all of the fission products. So there is a huge source of heat that is still there even after the reactor is shut down. (Newer designs like pebble bed reactors are an attempt to address this problem.)

#### anorlunda

Mentor
The reason fission reactors have this unique sort of failure mode, then, is evidently that there is a lot more stuff inside the reactor than just a few seconds' worth of fuel at any given time. There is a few years' worth (in a typical pressurized water reactor) of fuel inside, plus all of the fission products. So there is a huge source of heat that is still there even after the reactor is shut down. (Newer designs like pebble bed reactors are an attempt to address this problem.)
Very good explanations Peter. But that last paragraph is unclear.

The primary safety problem with LWRs is not the quantity of fuel in the core but rather the quantity of fission products. Fission products produce decay heat, thus requiring continuous cooling. The large mass of unburned fuel is not a heat source after shutdown.

If the containment is breached and everything inside were to leak out (not realistic), it is the fission products, not the unburned fuel that creates hazards. U235 and U238 are not a radiation hazard, even if leaked.

Pebble bed reactors have the advantage of storing most unburned fuel and most fission products external to the reactor vessel. They also operate at high temperatures, so that even if not cooled following shutdown, they won't melt.

However, like all designs, pebble beds have their own safety problems.
1. Rainer Moormann (2008). "A safety re-evaluation of the AVR pebble bed reactor operation and its consequences for future HTR concepts". Forschungszentrum Jülich, Zentralbibliothek, Verlag. Berichte des Forschungszentrums Jülich JUEL-4275.
2. ^ Rainer Moormann (1 April 2009). "PBR safety revisited". Nuclear Engineering International. Archived from the original on 30 May 2012.

#### PeterDonis

Mentor
The primary safety problem with LWRs is not the quantity of fuel in the core but rather the quantity of fission products. Fission products produce decay heat, thus requiring continuous cooling. The large mass of unburned fuel is not a heat source after shutdown.
Yes, this is correct. But there is still a connection with the quantity of fuel in the core: the quantity of fission products is only an issue because the whole fuel assembly stays in the reactor for a few years--i.e., because the reactor is only fueled once every few years. If only a few seconds' worth of fuel was in the reactor at any given time, the fission products would be removed along with the burnt fuel every few seconds. (As you note, designs like the pebble bed reactor are basically trying to achieve something like this.) In such a design, decay heat after shutdown would not be an issue.

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