text:
# Zero space between bars of the same interval in pgfplots ybar interval plots

I need a bar plot (containing multiple \addplot) with following requirements:

1. ticks between x labels (and, in this respect, my question differs from Adjusting width of ybar interval separator to width of histogram bars),
2. non zero space between bars of adjacent x coordinates,
3. zero space between bars corresponding to the same x coordinate.

I could manage the requirements:

1. by using a ybar interval plot,
2. by using a value <1 for the ybar interval option.

as shown by the following MWE:

\documentclass{article}
\usepackage{pgfplots}
\usepackage{filecontents}
%
\pgfplotsset{compat=1.11}
%
\begin{filecontents}{data.txt}
A B C D
0 13 9 19
1 0 1 5.5
2 0 4 4
3 1 3 14.5
4 3 8 6
5 1 8 6.5
6 2 5 5.5
7 0 7 14
8 8 14 6
9 0 5 12.5
10 0 14 17.5
\end{filecontents}
%
\begin{document}
\begin{tikzpicture}
\begin{axis}[%
ybar interval=0.5,%
width=\textwidth%
]
\end{axis}
\end{tikzpicture}
\end{document}

But I don't know how to remove the horizontal space between the bars of a same "interval". How could I achieve this?

Here I present a solution without ybar interval. For details on how it works, please have a look at the comments in the code.

% used PGFPlots v1.14
\begin{filecontents}{data.txt}
A B C D
0 13 9 19
1 0 1 5.5
2 0 4 4
3 1 3 14.5
4 3 8 6
5 1 8 6.5
6 2 5 5.5
7 0 7 14
8 8 14 6
9 0 5 12.5
10 0 14 17.5
\end{filecontents}
\documentclass[border=5pt]{standalone}
\usepackage{pgfplots}
\begin{document}
\begin{tikzpicture}
\begin{axis}[
width=\textwidth,
% set space between adjacent bars to zero
ybar=0pt,
% adjust bar width so the bars are not overlapping with the bars
% of another x value
% (this depends on the chosen plot width', xmin' and xmax' values
bar width=7pt,
% to show each x value
xtick distance=1,
% set xmin' and xmax' values manually so the bars aren't clipped
xmin=-0.5,
xmax=10.5,
% set the tick length of the xticks' to zero ...
xtick style={
% (I this key has to be prefixed by /pgfplots, because
%  normally here are just expected tikz keys)
/pgfplots/major tick length=0pt,
},
% but show them *between* the x values together with the grid
extra x ticks={-0.5,0.5,...,10.5},
extra x tick labels=\empty,
extra x tick style={
grid=major,
% reset the tick length to the default value
% (which otherwise would be the same as for the normal ticks
%  which is set to zero in this case --> see above)
xtick style={
/pgfplots/major tick length=4pt,
},
},
]
\addplot table [x=A,y=B] {data.txt};
\addplot table [x=A,y=C] {data.txt};
\addplot table [x=A,y=D] {data.txt};
\end{axis}
\end{tikzpicture}
\end{document}

• Nice! But I was hoping a solution which doesn't involve manual intervention. – Denis Bitouzé Dec 11 '16 at 17:20
• When you mean with "manual intervention" that bar width is automatically calculated depending on ... (this will be a long list) then I guess that there (currently, i.e. with PGFPlots v1.14) is no such feature. – Stefan Pinnow Dec 11 '16 at 17:24


text:
# Magnetic field density

(Redirected from Magnetic flux density)

Magnetic field density, otherwise known as magnetic flux density, is essentially what the layman knows as a magnetic field—akin to a gravitational or electric field. The SI unit of magnetic flux density is the tesla. 1 tesla = 1 weber per square metre.

It can be more easily explained if one works backwards from the equation: [itex]B=\frac {F} {I L} \,[itex]

where

B is the magnitude of flux density in teslas
F is the force in newtons experienced by a wire carrying
I amperes of current
L metres in length
Demonstration of the left hand rule

So, one can see for a magnetic flux density to equal 1 tesla, a force of 1 newton must act on a wire of length 1 metre carrying 1 ampere of current.

1 newton is a lot of force, and is not easily accomplished. To put it in perspective: the most powerful superconducting electromagnets in the world have flux densities of 'only' 20 T.

This is true obviously for both electromagnets and natural magnets, but a magnetic field can only act on moving charge—hence the current, I, in the equation.

Indeed, the equation can be adjusted to incorporate moving single charges, ie protons, electrons, and so on via

[itex]F = BQv \,[itex]

where

Q is 1 coulomb of charge
v is the velocity of that charge in metre per second

Fleming's left hand rule can be used to determine the direction of motion/current/polarity from any two of those, as seen in the example.

Other units of magnetic flux density are

• 1 gauss = 10-4 teslas = 100 microteslas (T)
• 1 gamma = 10-9 teslas = 1 nanotesla (nT)

## SI units

• Art and Cultures
• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)
• Space and Astronomy

format: plain


text:
# Tutte matrix

Main Article
Discussion
Related Articles  [?]
Bibliography  [?]
Citable Version  [?]

This editable Main Article is under development and subject to a disclaimer.

In graph theory, the Tutte matrix ${\displaystyle A}$ of a graph G = (V,E) is a matrix used to determine the existence of a perfect matching: that is, a set of edges which is incident with each vertex exactly once.

If the set of vertices V has 2n elements then the Tutte matrix is a 2n × 2n matrix A with entries

${\displaystyle A_{ij}={\begin{cases}x_{ij}\;\;{\mbox{if}}\;(i,j)\in E{\mbox{ and }}ij\\0\;\;\;\;{\mbox{otherwise}}\end{cases}}}$

where the xij are indeterminates. The determinant of this skew-symmetric matrix is then a polynomial (in the variables xij, i<j ): this coincides with the square of the pfaffian of the matrix A and is non-zero (as a polynomial) if and only if a perfect matching exists. (It should be noted that this is not the Tutte polynomial of G.)

The Tutte matrix is a generalisation of the Edmonds matrix for a balanced bipartite graph.


text:
# Is the first principles?

• Apr 21st 2010, 03:38 AM
piglet
Is the first principles?
Im trying to prove function $
f:\mathbb{C}\longrightarrow\mathbb{C}$
defined by $f(z) = z^{2}+z$ is differentiable everywhere in $C$ using derivative as a limit

Is this asking me to use first principles to show this?

• Apr 21st 2010, 03:42 AM
undefined
Quote:

Originally Posted by piglet
Im trying to prove function $
f:\mathbb{C}\longrightarrow\mathbb{C}$
defined by $f(z) = z^{2}+z$ is differentiable everywhere in $C$ using derivative as a limit

Is this asking me to use first principles to show this?

It sounds like you're supposed to write the definition of derivative in terms of a limit, then plug in f(z), then show that the limit exists for all z in C.
• Apr 21st 2010, 04:41 AM
HallsofIvy
Yes, show that $\lim_{h\to 0}\frac{(z+ h)^2+ (z+h)- (z^2+ z)}{h}$ exists.


text:
# How do you convert 250% into decimals and fractions?

250%=2.5=25/10=250/100...
Percent is based on "out of a hundred". In an area such as probability, we often use probabilities in decimals, where 1=100% chance of occurring. So when you have a multiple of 100%, just think about it in terms of $1$.
So 250% must be 2.5 as a decimal, but there is likely an infinite number of ways to describe it as a fraction- so I only gave a few.


text:
# Natural Growth and Logistic Models

1. May 4, 2005

### Beez

Hi, I have a question regarding natural growth and logistic models. According to my textbook, any exponential equations, P(t) = e^kt can be expressed logistically as P(t) = K/(1 + Ae^-kt) where A = (K-P0)/P0. When I applied this rule to P(t) = e ^0.0015t, and P(t) = 6000 / (1 + 5e^-0.0015t) it worked fine. The results were so close. But when I tried to do the same thing for P(t) = 400e^1.0986t and P(t) = 10000/(1 + 24e^-1.0986t), the results were so different from each other; 1199.98 and 1111.098 for P(1). Did I do something wrong here or they don't work if k>1?

I would appreciate it if I can get any help with this problem. Thank you in advance.

Naoko

2. May 4, 2005

### arildno

No, you have misuderstood the issue here:

The logistic model and the exponential model are nearly identical AT SMALL TIMES!!!
That is, it is only when time has gone a bit that the logistic growth restriction kicks in, and flattens the population level to K.

3. May 4, 2005

### Beez

Problem remains

I understood there is nothing to do with the value of k. Thank you.
But then how come the second logistic equation does not work for only t=1?

N0=400
N1=1200 (after a year the population tripled)
The lake can hold up to 10,000 fish
P(1)= 400e^(1.09861*1)=1199.99... =N1

P(1) = 10,000/(1+24e^(-1.09861*1))=1111.10... where 24 = (10000-400)/400

Am I doing something wrong here?

Thanks.

4. May 5, 2005

### arildno

Do you understand that the natural growth model is a DIFFERENT model than the logistic model; that is, there are some effects included in the logistic model that the natural growth model does not take into account?

5. May 5, 2005

### Beez

Yes I do

Yes I do. The logistic model takes the maximum population, in this case 10,000, and A, in this case (K-P0)/Po, in to account. But what I understood was while t is very small, they really don't have much effect and the exponential model and logistic model are almost identical. I used t=1 for the problem and the results were already so different. I used t=50 for the other problems that I mentioned earlier, but they still showed identical answers. That is why I thought the value of k or A might be the factors which determine if a size of the population satisfies both the exponential and logistic models or the only exponential model. But you told that it was nothing to do with them. Thus my question remained.

I hope I described my questions more clearly this time.

So, according to your explanation, the size of the population in question does not satisfy the logistic model. Am I right?

Thanks.

6. May 6, 2005

### arildno

First, wherever have I stated that some of the parameters don't matter???
I've never said anything to that effect!

Let's take it in full:
Let's call the exponential work model $$p_{E}(t)=P_{0}e^{kt}$$
And the logistic model $$p_{L}(t)=\frac{K}{1+\frac{K-P_{0}}{P_{0}}e^{-kt}}$$
Rewrite the logistic model as $$P_{L}(t)=\frac{P_{0}e^{kt}}{1+\frac{P_{0}(e^{kt}-1)}{K}}=\frac{P_{E}(t)}{1+\frac{P_{0}(e^{kt}-1)}{K}}$$
Thus, if we have: $$\frac{P_{0}(e^{kt}-1)}{K}<<1$$, then it follows $$P_{L}(t)\approx{P}_{E}(t)$$
Let us make this into a condition on "t":
We rewrite, and find:
$$t<<\frac{1}{k}ln(1+\frac{K}{P_{0}})$$
This condition determines what a "small time" is, in this particular context.

If you plug into your values of $$k,P_{0},K$$ here, you'll see why the approximation holds in one case but not in the other..
(that is, the "small times"-regions are different)

Note that a concept like "small time" is not really meaningful if you haven't specified a REFERENCE time.
That is, "small times" means :the time is small compared with the reference time for the problem.

I think it was this usage of mine of "small times" which confused you.
I hope that I have made it more explicit now.

Last edited: May 6, 2005
7. May 6, 2005

### Beez

Thank you

Thank you so much for providing me the detailed explanation.
Yes, I was taking "small amount" literally.
Now my sky is as blue as it could be.
Thank you for your help again.


text:
Abstract

We study the linear syzygies of a homogeneous ideal $I \subseteq S = {\rm Sym}_{{\bf k}}(V)$, focussing on the graded betti numbers $$b_{i,i+1} = {\rm dim}_{{\bf k}} {\rm Tor}_i (S/I,{\bf k})_{i+1}.$$ For a variety $X$ and divisor $D$ with $V = H^0(D)$, what conditions on $D$ ensure that $b_{i,i+1} \ne 0$? Eisenbud has shown that a decomposition $D \sim A +B$ such that $A$ and $B$ have at least two sections gives rise to determinantal equations (and corresponding syzygies) in $I_X$; and conjectured that if $I_2$ is generated by quadrics of rank $\le 4$, then the last nonvanishing $b_{i,i+1}$ is a consequence of such equations. We describe obstructions to this conjecture and prove a variant. The obstructions arise from toric specializations of the Rees algebra of Koszul cycles, and we give an explicit construction of toric varieties with minimal linear syzygies of arbitrarily high rank. This gives one answer to a question posed by Eisenbud and Koh.


text:
# Group scheme over a DVR whose special fibre is the image of points under reduction mod p

Let $R$ be a complete discrete valuation ring with maximal ideal $\mathfrak{p}$ and algebraically closed residue field $k$. Denote the field of fractions of $R$ by $F$. Let $G$ be an affine flat group scheme of finite type over $R$, and assume that the generic fibre $G_{F}$ is smooth. Since $G$ is not necessarily smooth over $R$ the reduction mod $\mathfrak{p}$ map on the points $\rho:G(R)\longrightarrow G(k)$ is not necessarily surjective. Let $H_{k}$ denote the image of $\rho$ provided with its canonical affine reduced structure.

Does there exist a group scheme $G'$ of finite type over $R$ such that $G'_{F}\cong G_{F}$ (i.e., the generic fibres coincide) and $G'_{k}\cong H_{k}$ (i.e., the special fibre of $G'$ is $H_{k}$)?

An affirmative answer would give some information about the image $H_{k}$, for example, we would have $\dim H_{k}\geq\dim G_{F}$ by Chevalley's upper semicontinuity theorem. Therefore any example where $\dim H_{k}<\dim G_{F}$ would provide a negative answer.

A way to see that $H_{k}$ is closed in $G(k)$ is by using a theorem of Greenberg (in "Rational points in Henselian discrete valuation rings") which implies that there exist two integers $r\geq s\geq1$ such that the image of any $x\in G(R/\mathfrak{p}^{r})$ in $G(R/\mathfrak{p}^{s})$ has a lift to $G(R)$. This implies that $\mathrm{Im}(G(R)\rightarrow G(R/\mathfrak{p}^{s}))=\mathrm{Im}(G(R/\mathfrak{p}^{r})\rightarrow G(R/\mathfrak{p}^{s}))$. Using the Greenberg functor, we can consider each $G(R/\mathfrak{p}^{i})$ as (the $k$-points of) a linear algebraic group over $k$ (equipped with its reduced structure in case $G$ is not smooth over $R$), and by another theorem of Greenberg ($\S\,5$, Prop. 2 and Coroll. 5 in "Schemata over local rings I") each map $G(R/\mathfrak{p}^{i})\rightarrow G(R/\mathfrak{p}^{j})$ with $i\geq j\geq1$ induces an algebraic homomorphism. Thus $X:=\mathrm{Im}(G(R)\rightarrow G(R/\mathfrak{p}^{s}))$ is a closed subgroup of $G(R/\mathfrak{p}^{s})$ and hence $H_{k}=\mathrm{Im}(X\rightarrow G(k))$ is a closed subgroup of $G(k)$.

-

Note that the question does not make sense until we know that the image of $\rho$ is closed, and such closedness is not obvious since not even construcibility is apparent at the outset (e.g., Chevalley's theorem on constructibility is not relevant here, as $\rho$ is in no sense an algebraic morphism). Also, it isn't said if $G'$ is supposed to be separated, perhaps even affine, but presumably you want $G'$ to be (quasi-)affine?

The only way I know to see the closedness of the image of $\rho$ is by a procedure that answers the entire question. Namely, consider the canonically associated group smoothening $f:\widehat{G} \rightarrow G$ in the sense of section 7.1 of the book "Neron Models". Since $G$ is a flat affine $R$-group of finite type, and the group smoothening is an $R$-homomorphism built from a "dilatation" process that entails forming a certain relatively affine open chart in a blow-up, it is immediate from the construction that the smooth group $\widehat{G}$ is affine, and by design $\widehat{G}(R) = G(R)$ and $f_F$ is an isomorphism. Consequently, since $\widehat{G}(R) \rightarrow \widehat{G}(k)$ is surjective by smoothness of $\widehat{G}$, it follows that $\rho(G(R))$ is the image on $k$-points of the special fiber $k$-homomorphism $f_k:\widehat{G}_k \rightarrow G_k$. Now we can apply our experience with morphisms of algebraic groups over fields to know that $\rho(G(R))$ is closed, since (i) it is just the set of $k$-points of the image of $f_k$ since $k$ is algebraically closed and (ii) the image of $f_k$ is closed (as for any homomorphism between finite type group schemes over a field).

This argument gives more, namely that $\dim H_k \le \dim \widehat{G}_k = \dim \widehat{G}_F = \dim G_F$, whereas you noted that if the $G'$ you seek is to exist then $\dim H_k \ge \dim G_F$, so we would be forced to have $\dim H_k = \dim G_F$, but $\dim G_F = \dim G_k$ by $R$-flatness of $G$, so in such cases $H_k$ would have to contain the identity component $G_k^0$ and then $f_k$ has finite kernel and it has image containing $G_k^0$. So we see that in such cases $H_k$ is a union of some of the connected components of $G_k$, and hence $G'$ can be taken to be the open subscheme of $G$ given by removing the components of $G_k$ not in the image of $f_k$.

The upshot is that the answer is affirmative if and only if $\rho(G(R))$ contains $G_k^0(k)$, and that this is also equivalent to the condition that the special fiber of the group smoothening morphism has finite kernel. (It also makes me wonder if perhaps you would prefer simply to work with $\widehat{G}$ in general and forget about $G'$.)

--

We can say more, by relating the group smoothening to the normalization of $G$. Note that the normalization map $\widetilde{G} \rightarrow G$ is finite since $R$ is excellent, so $\widetilde{G}$ is finite type over $R$ and visibly $R$-flat with the same generic fiber as $G$ over $F$.

Suppose we're in the case with an affirmative answer, so $f_k$ is quasi-finite, and hence $f$ is quasi-finite (as $f_F$ is an isomorphism). Since $f$ is an isomorphism between (possibly disconnected) generic fibers and $\widehat{G}$ is smooth (hence normal), $f$ uniquely factors through the normalization, say via $h:\widehat{G} \rightarrow \widetilde{G}$. But $f$ is quasi-finite, so $h$ is also quasi-finite. Yet $h_F$ is an isomorphism, so by Zariski's Main Theorem and normality considerations we see that $h$ is an open immersion. The quasi-finite $f_k$ must be finite (as for any quasi-finite homomorphism between finite type group schemes over a field), so $h_k$ is finite too. Yet $h_k$ is an open immersion (since $h$ is), so $h_k$ identifies $\widehat{G}_k$ with a union of connected components of $\widetilde{G}_k$ (with their natural open subscheme structure). Since $\widetilde{G}$ is $R$-flat (of finite type), so smooth points in its special fiber are precisely where the Zariski-open $R$-smooth locus of $\widetilde{G}$ meets $\widetilde{G}_k$, and all smooth $k$-points of $\widetilde{G}_k$ lift to $\widetilde{G}(R) = G(R) = \widehat{G}(R)$ (first equality by finiteness of normalization), we conclude that $\widehat{G}$ is exactly the smooth locus in $\widetilde{G}$ in these cases.

But in fact we need much less about the normalization to deduce the same conclusion: suppose that the $R$-flat $\widetilde{G}$ merely has non-empty smooth locus in its special fiber, or in other words that the Zariski-open $R$-smooth locus has non-empty special fiber. That provides a dense open locus of smooth points in some irreducible component $X$ of the special fiber of $\widetilde{G}$, and those all lift to $\widetilde{G}(R)=G(R)$ yet clearly $X$ maps finitely onto an irreducible component $Y$ of $G_k$ (as $G_k$ and $\widetilde{G}_k$ are equidimensional of the same dimension, due to $R$-flatness and equidimensionality of their common generic fiber $G_F$), so a dense open locus in $Y(k)$ lifts to $G(R)$, forcing $\rho(G(R))$ to contain a dense open locus in some irreducible component (= connected component) of $G_k$. Hence, that would force $\dim H_k = \dim G_k$.

To summarize, the following conditions are equivalent: $G'$ exists, $H_k$ contains $G_k^0(k)$, $f_k$ has finite kernel, $\widetilde{G}_k$ has non-empty smooth locus, $\widehat{G}$ is an open subscheme of $\widetilde{G}$. In such cases, $\widehat{G}_k \rightarrow G_k$ is a finite homomorphism with image $H_k$, and $\widehat{G} \rightarrow G$ has open image that we can take as $G'$.

-
Greenberg Theorem implies indeed that the image of $\rho$ is constructible. – ACL Jun 19 '13 at 7:41
@ACL: What is a reference for this particular "Greenberg Theorem"? I wondered about trying to use the Greenberg functor to prove the constructibility, but when passing from artinian quotients of $R$ to the inverse limit $R$ itself I got stuck on issues of non-emptiness of a countable descending intersection when $k$ is countable. – user29283 Jun 19 '13 at 8:52
The (rather, A) theorem of Greenberg says that there exists an integer $n$ such that any point of $G(k)$ which lifts to a point modulo $\pi^n$ ($pi$ being an uniformizer) lifts to $G(R)$. Therefore, calling $G_n$ the image of $G$ by the Greenberg functor, the desired group $H$ is the image of $G_n$ by the projection $G_n\to G_1$. – ACL Jun 19 '13 at 13:56
I have added an argument which uses at least two theorems of Greenberg. Perhaps this is similar to what ACL had in mind? – A Stasinski Jun 19 '13 at 13:56
@A.Stasinski: The equality $\widehat{G}(R)=G(R)$ is the map induced on $R$-points by the $R$-morphism $f:\widehat{G}\rightarrow G$ (and $f_F$ is the identification of $F$-fibers, which identifies $F$-points too). So the commutativity just expresses the fact that the "functor of points" of a scheme is functorial in the scheme. – user29283 Jun 19 '13 at 14:33


text:
• What are the objectives of annealing processes, such as normalizing, soft annealing, coarse grain annealing, recrystallisation annealing, diffusion annealing, solution annealing and stress-relief annealing?
• In which temperature ranges are the annealing processes carried out?
• Why is an improvement in machinability also achieved with soft annealing?
• How can the generally poorer properties of a coarse-grained microstructure be subsequently removed?
• What is the significance of recrystallisation annealing for transformation-free steels?
• What is hot forming or cold forming?
• Why is diffusion annealing relatively cost-intensive?
• Why does the workpiece have to be cooled slowly after stress relief annealing?

## Normalizing

As already explained in the chapter deformability of metals, fine roundish grains generally lead to better toughness and strength values compared to large grains. A uniform microstructure is desirable, which has similarly small grains over the entire microstructure. This is the only way to ensure that the material meets the strength requirements at every point to the same extent.

However, achieving a homogeneous microstructure already during solidification requires high demands, since the solidification conditions will not be identical across the entire melt. For example, cast steel may cool faster at the contact points to the mould wall than inside the melt. While finer grains will develop in the edge areas due to the stronger undercooling, larger grains will form in the middle of the melt.

A heterogeneous grain microstructure can also occur during forging, as the grain boundaries shift due to diffusion processes and the grains can unite. This results in a new grain formation with the consequence of a heterogeneous structure. The same effect of heterogeneity can also be seen in welded workpieces in the area of the joint.

For this reason it is necessary to standardize (homogenize) a non-uniform microstructure by means of a heat treatment. This gives the steel its “normal” properties, which are always reproducible. For this reason, the targeted homogenization of a steel microstructure is also called normalizing.

The aim of normalizing is to achieve a uniform homogeneous microstructure with reproducible properties!

During normalizing, the steel is heated to just above the GSK-line so that the pearlite is completely converted to austenite. The austenitized steel is then slowly cooled in air. Since the grains form anew during the $$\gamma$$-$$\alpha$$-transformation, grain refinement occurs and makes the microstructure homogeneous.

The temperature during normalizing should not exceed approx. 30 °C above the GSK-line, otherwise there is a risk of coarse grain formation. The reason for this is that, from an energy point of view, large roundish grains are more favourable than many small ones. Therefore, the microstructure always strives to form a single large grain. This requires, among other things, diffusion processes which are favoured by higher temperatures. Therefore, it is important to keep the temperature as low as possible during normalizing in order to avoid the formation of coarse grains. For this reason, hypereutectoid steels are not completely heated up to the austenite region (above the SE-line).

Normalizing is preferably used for hypoeutectoid steels whose microstructure has been negatively influenced by manufacturing processes such as forging, rolling, casting, welding, etc. During rolling, normalizing can already be carried out during the rolling process (normalizing rolling). A normalized microstructure is generally characterized by very good toughness and strength values due to its homogeneous, fine structure.

Normalizing is often carried out on hypoeutectoid steels after forging, rolling, casting or welding!

## Soft annealing

Not every material has to be designed to withstand high mechanical forces. With a curved sheet metal with milled out slots, for example, it is not necessary for the material to be able to absorb high forces. Rather, the focus in the selection of materials is on good formability and machinability of the steel. This plays an important role, especially in automated production with large batch sizes, in order to make production economical.

For this reason, it may be necessary to adapt the microstructure of a steel in such a way that it can be better formed and/or machined. Particularly with regard to formability, it is therefore necessary to produce a correspondingly soft microstructure. This can be achieved by so-called soft annealing.

The aim of soft annealing is to improve formability and machinability!

During soft annealing, hypoeutectoid steels are heated to just below the PS-line so that the cementite does not just yet decompose. The lamellar cementite now has enough time to transform through diffusion processes into the thermodynamically more favourable, roundish form. The strip cementite in pearlite transforms into spherical cementite (spheroidal cementite). After the cementite has disintegrated into the round shape, the steel is slowly cooled. In contrast to hypoeutectoid steels, hypereutectoid steels are heated during soft annealing just above or oscillating around the PSK-line.

A particularly homogeneous microstructure with finely divided spheroidal cementite can be achieved by hardening the steel before soft annealing. The spherical cementite is then formed from the relatively homogeneous martensite microstructure.

After soft annealing, the steel shows much better formability due to the spherical cementite shape. The cause lies in the facilitated dislocation movement. While the strip-shaped cementite lamellas partially extend completely from one end of the grain to the other, the cementite spheres are only occasionally present in the grain. The dislocation movement is thus less strongly hindered by the spheroidal cementite than with the completely penetrating strip cementite. The deformability increases accordingly, while the hardness decreases.

This facilitates subsequent rolling, bending, deep drawing, etc. due to reduced forming forces. In addition, the spherical cementite achieves better machinability, as the cementite spheres offer less resistance to the tool cutting edge compared to the lamellar cementite form. This increases the durability of the tool accordingly.

Spheroidal cementite improves the machinability of the microstructure compared to lamellar cementite!

The micrograph below shows a soft-annealed steel C45 with the cementite lamellae disintegrating into small roundish spheres.

Hypoeutectoid steels with a carbon content below approx. 0.3 % carbon are generally not soft annealed, as these are relatively soft anyway. Although these steels already have good formability, their machinability is unfavourable due to the tendency to form built-up cutting edges at the tool edges. In order to give these low-carbon steels good machinability, the coarse-grain annealing described below can be used as heat treatment.

## Coarse-grain annealing

In general a coarse-grained steel microstructure is undesirable due to the relatively low toughness and strength values. The only advantage of a coarser grain is the resulting better machinability, which is due to the increased brittleness of the coarse grain (note that brittleness and toughness always behave in reverse). Especially for low-carbon steels with a carbon content of less than 0.3 %, coarse grain annealing is an alternative to soft annealing in order to improve machinability.

The aim of coarse grain annealing is to improve machinability!

During coarse grain annealing, the steel is annealed in the range between 950 °C and 1100 °C. At these high temperatures, diffusion processes can take place to a sufficient extent so that the atoms can reattach to the grain boundaries and thus cause them to grow. The driving force is the reduction in surface energy, which is accompanied by a larger grain instead of many small ones. Since the diffusion processes take time, depending on the thickness of the workpiece, the annealing process takes several hours.

Due to the generally unfavourable mechanical strength properties, coarse grain annealing is limited to low-carbon steels and is only rarely used (e.g. in high-temperature applications). After the coarse grain microstructure has been machined, it can be removed by normalization to regain better strength properties.

The generally poorer strength values of a coarse-grained structure can be removed by normalizing!

## Recrystallization annealing

The microstructure of rolled, bent or deep-drawn workpieces is strongly deformed by the high forming forces. This also changes the material properties. In the case of rolled sheets, this can lead to a strong anisotropy through the elongated crystals, also known as rolling texture. In addition, strain hardening takes place in the forming area, which increases the strength and reduces the deformability accordingly. If the workpiece is to be further formed in this condition, the risk of cracking increases. Multi-stage forming processes are therefore not possible without further ado. However, many workpieces and semi-finished products have to be formed several times in the course of their production in order to reach their final state. For example, a steel block that is several centimetres thick cannot be rolled in a single pass down to a few millimetres.

The aim must therefore be to restore the deformed crystals (grains) of a reshaped microstructure to their original shape before every multi-stage forming process. This can be achieved by recrystallisation annealing.

The aim of recrystallisation annealing is to restore a deformed microstructure to improve its deformability!

During recrystallisation annealing, the steel is annealed below the PSK-line in the range between 550 °C and 700 °C. Therefore, no lattice transformation takes place, as is the case with normalizing or partially also with soft annealing, although a recrystallisation effect also occurs with these two processes. During recrystallisation annealing, the grain boundaries can migrate through diffusion processes and the grains thus form anew. The deformed grains regain their original shape and the material regains its deformability.

In addition to the annealing time and temperature, the size of the recrystallized grains depends in particular on how strongly the individual grains were deformed before. A high degree of deformation with very fine, elongated crystals allows the microstructure to recrystallize rather fine-grained. A lower degree of deformation leads accordingly to a coarser grain after recrystallisation. Especially for a slightly deformed microstructure, however, there is also the danger of coarse grain formation. This risk can occur particularly for low-carbon steels with carbon concentrations below 0.2 %, so that normalizing may be more suitable for the formation of new crystals.

Recrystallization annealing is the only method for transformation-free steels (where $$\gamma$$-$$\alpha$$-transformation is completely suppressed by alloying elements) to achieve grain refinement.

In order to always maintain the ductility of the material in multi-stage forming processes, the microstructure must be recrystallized between each forming step. This process is then also called intermediate annealing.

The effect of recrystallisation can also be used during the forming process itself by forming above the recrystallisation temperature. This is known as hot forming. However, if the material is formed below the recrystallisation temperature (e.g. at room temperature) it is called cold forming. Hot forming places much higher demands on the machines involved, so that economic efficiency must always be checked.

In hot forming, the workpiece is formed above the recrystallisation temperature, in cold forming, however, below the recrystallisation temperature!

## Diffusion annealing

When steels with high alloy concentrations solidify, the alloying elements may not be distributed homogeneously in the microstructure or in the individual crystals. Such concentration differences within the individual crystals are also called microsegregations.

With different alloy concentrations, there are also different properties within a grain which may weaken the microstructure. That is why such microsegregations must always be prevented. However, the formation of such concentration differences cannot always be prevented from the outset due to the finite cooling rates. For this reason, differences in concentration within a microstructure must be eliminated by subsequent heat treatment. This can be done by diffusion annealing.

The aim of diffusion annealing is to compensate for concentration differences!

During diffusion annealing, the steel is annealed to relatively high temperatures in the range between 1050 °C and 1300 °C. This ensures that the diffusion processes can take place to a sufficient extent so that the atoms can cover the relatively long diffusion paths. However, this usually requires several hours of annealing time.

Disadvantage of diffusion annealing is the formation of coarse grains due to the high temperatures. Although this could be reduced by reducing the temperature, longer annealing times would then be necessary and diffusion annealing would no longer be economical. If coarse grain formation cannot be prevented during diffusion annealing, the coarse grain structure must be subsequently removed again. This can be achieved by normalizing.

Since microsegregations occur during the solidification of a steel, they are usually removed directly in the steel mill on the ingot. Especially as the process is very energy-intensive and places high demands due to the high temperatures and long annealing times.

## Solution annealing

During welding or hot forming, microstructural changes occur in the steel due to the influence of heat and uncontrolled cooling. This is particularly the case with austenitic steels, where the high temperatures in the range of 500 °C to 800 °C can cause carbide precipitation at the grain boundaries. This leads to intergranular corrosion (intercrystalline corrosion) due to the different electrochemical properties. To avoid this, the formed precipitations must be dissolved again. This can be achieved by solution annealing in the range of approximately 1000 °C to 1100 °C. In the case of deformed steel microstructures, the effect of recrystallisation also occurs.

The aim of solution annealing is to dissolve formed precipitates!

Solution annealing is also used as an intermediate step in the value chain to temporarily improve machinability. The precipitates responsible for the poor machinability are dissolved by annealing. If the workpiece is then rapidly cooled, a supersaturated (metastable) solid solution microstructure without precipitates is formed. In this condition, a better machinability of the material is temporarily achieved. The precipitates are then formed by subsequent cold ageing or warm ageing and the material regains its original properties. This process is used, for example, in the precipitation hardening of some special aluminium alloys.

## Stress relief annealing

If a steel is heated or cooled unevenly, internal stresses can arise, also known as residual stresses. Such residual stresses are often induced during welding, for example, because the workpiece is heated not evenly but only locally at a certain point and then cooled down. However, residual stresses can also occur in the workpiece during milling or turning, as high temperatures can occur in the machining area of the workpiece. During hardening, the quench distortion is also based on the residual stresses caused by the uneven cooling.

Such residual stresses can, among other things, reduce the strength of the workpiece. Furthermore, residual stresses during a subsequent machining process can also lead to distortion of the workpiece if the residual stresses are suddenly released as a result. Therefore, in some cases it is necessary to eliminate residual stresses in the workpiece. This is achieved by stress relief annealing.

The aim of stress-relief annealing is to release residual stresses!

In stress relief annealing, the workpiece is annealed below the PSK-line in the range between 550 °C and 650 °C. The effect of the stress relief is based on the fact that the strength of the heated material decreases at a higher temperature. If the yield strength (hot yield strength) falls below the value of the residual stresses, these are reduced by plastic deformation. The dislocations begin to move accordingly.

The residual stresses can therefore only ever be reduced to a maximum of the corresponding hot yield point, never completely. After annealing, the workpiece must then be cooled slowly to avoid a renewed occurrence of stress. In most cases, the workpiece remains stationary in the switched-off annealing furnace.


text:
# Start a new discussion

## Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

## Site Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

1. copied reference from HoTT wiki

Anonymous

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeJun 9th 2022

I have taken the liberty of (hyperlinking the abstract and) re-formatting as follows:

on the join of maps in homotopy type theory.

Abstract. In homotopy type theory we can define the join of maps as a binary operation on maps with a common co-domain. This operation is commutative, associative, and the unique map from the empty type into the common codomain is a neutral element. Moreover, we show that the idempotents of the join of maps are precisely the embeddings, and we prove the ’join connectivity theorem’, which states that the connectivity of the join of maps equals the join of the connectivities of the individual maps. We define the image of a map $f:A \to X$ in $U$ via the join construction, as the colimit of the finite join powers of $f$. The join powers therefore provide approximations of the image inclusion, and the join connectivity theorem implies that the approximating maps into the image increase in connectivity. A modified version of the join construction can be used to show that for any map $f:A\to X$ in which $X$ is only assumed to be locally small, the image is a small type. We use the modified join construction to give an alternative construction of set-quotients, the Rezk completion of a precategory, and we define the $n$-truncation for any $n: \mathbb{N}$. Thus we see that each of these are definable operations on a univalent universe for Martin-Löf type theory with a natural numbers object, that is moreover closed under homotopy coequalizers.


text:
5 of 5 replaced http://codegolf.stackexchange.com/ with https://codegolf.stackexchange.com/

Inspired by Golf me an ASCII Alphabet, of which this challenge is (almost) a direct inverse.

Take a string of ASCII-art text and output the content of the text as regular ASCII text.

### Input:

String of ASCII-art text.

Input will only contain instances of ASCII character #, spaces and 4 or 5 newlines (a trailing newline is optional). All lines have the same length. (That is, the last ASCII-art letter is padded with trailing spaces.) You can use some other printable ASCII character instead of # in the input if you wish.

The input will contain ASCII-art letters A-Z and ASCII-art spaces (a 5x5 block of whitespace). No punctuation. There is only one line of ASCII-art text (5 actual lines). There will be no trailing or leading ASCII-art spaces, nor will there be adjacent ASCII-art spaces.

Letter size is 5x5 characters. There is a 1x5 space between each letter. Space between words is a 5x5 block of whitespace (+ 1x5 space on each side, because it is just another letter). There will be no 1x5 space at the end or at the beginning, only between ASCII-art letters.

### Output:

String containing the text as ASCII characters A-Z + spaces. The output can be in lowercase also, if that is somehow easier for your solution. Mixed case is also allowed.

### The ASCII-art letters:

 ###  ####   ###  ####  ##### #####  ###  #   # ##### ##### #   # #     #   #
#   # #   # #   # #   # #     #     #     #   #   #     #   #  #  #     ## ##
##### ####  #     #   # ####  ####  #  ## #####   #     #   ###   #     # # #
#   # #   # #   # #   # #     #     #   # #   #   #   # #   #  #  #     #   #
#   # ####   ###  ####  ##### #      ###  #   # ##### ###   #   # ##### #   #

#   #  ###  ####   ###  ####   ###  ##### #   # #   # #   # #   # #   # #####
##  # #   # #   # #   # #   # #       #   #   # #   # #   #  # #   # #     #
# # # #   # ####  #   # ####   ###    #   #   #  # #  # # #   #     #     #
#  ## #   # #     #  #  #   #     #   #   #   #  # #  ## ##  # #    #    #
#   #  ###  #      ## # #   #  ###    #    ###    #   #   # #   #   #   #####

The space:

     |
| A 5x5 square of spaces.
| (Padded with |s to make it appear in this post.)
|
|

### Examples:

Input:

#   # ##### #     #      ###        #   #  ###  ####  #     ####
#   # #     #     #     #   #       #   # #   # #   # #     #   #
##### ####  #     #     #   #       # # # #   # ####  #     #   #
#   # #     #     #     #   #       ## ## #   # #   # #     #   #
#   # ##### ##### #####  ###        #   #  ###  #   # ##### ####

Output: HELLO WORLD

Input:

 ###   ###   ###  ##### #####
#   # #     #   #   #     #
#####  ###  #       #     #
#   #     # #   #   #     #
#   #  ###   ###  ##### #####

Output: ASCII

Input:

####  ####   ###   ###
#   # #   # #   # #
####  ####  #     #  ##
#     #     #   # #   #
#     #      ###   ###

Output: PPCG

This is , so the shortest answer in bytes wins.


text:
The characteristic impedance of free space is

Free Practice With Testbook Mock Tests

Options:

1. 397 Ω

2. 367 Ω

3. 387 Ω

4. 377 Ω

Correct Answer: Option 4 (Solution Below)

This question was previously asked in

UPPCL AE EC 2019 Official Paper (Held On 5 November 2019)

EKT EE Official Paper 1: 01/2014

AFCAT EKT EE Official Paper: Set 01/2014

AFCAT EKT EE Official Paper (Held on Jan 2014)

Solution:

Concept:

The characteristic impedance of a wave travelling in a lossy medium is given by

$$Z = \sqrt {\frac{{j\omega μ }}{{\sigma + j\omega ϵ}}}$$

Put σ = 0 (free space and lossless medium)

$$Z = \sqrt {\frac{μ_0 }{ϵ_0}}$$

Where,

μ0 = Permeability of free space = 4π x 10-7 H / m

ϵ0 = Permittivity of free space = 8.85 x 10-12 F/m

Calculation:

$$Z = \sqrt {\frac{4\pi \times10^{-7} }{8.85 \times10^{-12}}}$$

∴ Z = 377 Ω


text:
Chapter 13.II, Problem 29RE

### Contemporary Mathematics for Busin...

8th Edition
Robert Brechner + 1 other
ISBN: 9781305585447

Chapter
Section

### Contemporary Mathematics for Busin...

8th Edition
Robert Brechner + 1 other
ISBN: 9781305585447
Textbook Problem

# Calculate the finance charge and the monthly payment for the following loans by using the APR table, Table 13-1. Amount Number of Table Finance Monthly Financed Payments APR Factor Charge Payment 29. $1,800 12 11.25% To determine To calculate: The table factor, finance charge and monthly payment if annual percentage rate (APR) is 11.25%, installment loan is for 12 months and amount financed is$1,800.

Explanation

Given Information:

The annual percentage rate (APR) is 11.25%, installment loan is for 12 months and amount financed is $1,800. Formula used: Follow the steps to compute the finance charge and monthly payment through APR tables. Step 1: Through APR table, scan the table factor at the intersection of the APR column and the Number-of-payments row. This will represent the finance charge per$100 financed.

Step 2: Compute the total charge by multiplying the amount financed with the table factor, then dividing the amount by 100.

Finance charge=Amount financed×Table factor100

Step 3: Compute the monthly payment by adding the amount financed with the finance charge and then dividing it by number of months of the loan,

Monthly payment=Amount financed+Finance chargeNumber of months of the loan

Calculation:

Consider that annual percentage rate (APR) is 11.25%, installment loan is for 12 months and amount financed is $1,800. The table factor at the intersection of 11.25% APR column and 12 payments row is$6

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started


text:
Definition 9.14.7. Let $E/F$ be an algebraic field extension. Let $E_{sep}$ be the subextension found in Lemma 9.14.6.

1. The integer $[E_{sep} : F]$ is called the separable degree of the extension. Notation $[E : F]_ s$.

2. The integer $[E : E_{sep}]$ is called the inseparable degree, or the degree of inseparability of the extension. Notation $[E : F]_ i$.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 030L. Beware of the difference between the letter 'O' and the digit '0'.


text:
# Square Meter To Acre Calculator

Square Meter (m2):
Acre :

The Square Meter to Acre Calculator is a free online tool that shows the conversion from square meters to acre for the given input. BYJU’S Square Meter to Acre Calculator tool makes calculations faster and it displays the result in a fraction of seconds.

## How to Use the Square meter To Acre Calculator?

The procedure to use the square meter to acre calculator is as follows:
Step 1: Enter the number of square meters in the input field
Step 2: Now click the button “Solve” to get the time period
Step 3: Finally, the number of acres will be displayed in the output field

### What is Meant by the Square Meter to Acre Conversion?

Square meter is the unit of measurement of a square whose side is given as one meter. In this case, the area of the square is one square meter. In simple words, it is the size of a square whose side is one meter. Acres are used to measure the area of the land. An acre is a unit of area equal to 4046.8564224 square meters (sq.m). Also, 1 sq.m = 0.00024710538146717 acre. The calculator given here, provides the conversion values rounded to three decimal points.


text:
Please use this identifier to cite or link to this item: http://hdl.handle.net/1813/6935
 Title: Provably Good Approximation Algorithms for Optimal Kinodynamic Planning for Cartesian Robots and Open Chain Manipulators Authors: Donald, Bruce RandallXavier, Patrick G. Keywords: computer sciencetechnical report Issue Date: Feb-1990 Publisher: Cornell University Citation: http://techreports.library.cornell.edu:8081/Dienst/UI/1.0/Display/cul.cs/TR90-1095 Abstract: We consider the following problem: given a robot system, find a minimal-time trajectory from a start state to a goal state, while avoiding obstacles by a speed-dependent safety-margin and respecting dynamics bounds. In [CDRX] we developed a theoretical, provably good approximation algorithm for the minimum-time trajectory problem for a robot system with decoupled dynamics bounds (e.g. a point robot in). This algorithm differs from previous work in three ways. It is possible (1) to bound the goodness of the approximation by an error term $\epsilon$; (2) to polynomially bound the computational complexity of our algorithm; and (3) to express the complexity as a polynomial function of the error term. Hence, given the geometric obstacles, dynamics bounds, and the error term $\epsilon$, the algorithm returns a solution that is $\epsilon$-close to optimal and requires only a polynomial (in ($\frac{1}{\epsilon}$)) amount of time. We extend the results of [CDRX] in two ways. First, we reanalyze the [CDRX] algorithm for robots with decoupled dynamics bounds. We halve the exponent in the polynomial bounds and prove a better approximation accuracy. These new results indicate that an implementation of the theoretical algorithm could be reasonable. We report on a preliminary implementation of the extended algorithm and experiments. Second, we extend [CDRX] to $d$-link, revolute-joint 3D robots will full rigid body dynamics. Specifically, we first prove a generalized trajectory- tracking lemma for robots with coupled dynamics bounds. Then, using this result we describe polynomial-time approximation algorithms for Cartesian robots obeying $L_{2}$ dynamics bounds and open kinematic chain manipulators with revolute and prismatic joints; the latter class includes most industrial manipulators. We obtain a general $O(n^{2}$(log$n$) ($\frac{1}{\epsilon})^{6d-1})$ algorithm, where $n$ is the geometric complexity. URI: http://hdl.handle.net/1813/6935 Appears in Collections: Computer Science Technical Reports

Files in This Item:

File Description SizeFormat


text:
# Proper custom PDF bookmarks with changed page numeration

I have following code in which I use custom heading to:

1. Create TOC entry
2. Create custom bookmark

And I want first PDF bookmark to point to TOC page.

Since \addcontentsline{toc}{chapter}{#1} automatically creates PDF bookmarks if either hyperref or bookmark packages are loaded, I used \hypersetup{bookmarksdepth=-1} to supress those PDF bookmarks that are created with it. Thanks to it, I can create all bookmarks that I need with bookmark package.

\documentclass[final,11pt]{book}
\usepackage{blindtext}
\usepackage{hyperref}
\hypersetup{bookmarksdepth=-1}
\usepackage{bookmark}
\chapter*{#1}
\bookmark[page=\thepage,level=-1]{My own bookmark for #1}
}
\begin{document}
\bookmark[page=\thepage,level=-1]{My own TOC bookmark}
\tableofcontents
\blindtext[10]
\blindtext[10]
\blindtext[10]
\blindtext[10]
\blindtext[10]
\end{document}

Code above produces document in which both TOC and PDF bookmarks point to the right places.

However, when I try to change page numbering, things start to go south.

When I use:

\documentclass[final,11pt]{book}
\usepackage{blindtext}
\usepackage{hyperref}
\hypersetup{bookmarksdepth=-1}
\usepackage{bookmark}
\chapter*{#1}
\bookmark[page=\thepage,level=-1]{My own bookmark for #1}
}
\begin{document}
\frontmatter
% \bookmark[page=\thepage,level=-1]{My own TOC bookmark} % This is commented out, because adding it will cause compilation error.
\tableofcontents
\mainmatter
\blindtext[10]
\blindtext[10]
\blindtext[10]
\blindtext[10]
\blindtext[10]
\end{document}

TOC works just fine. But PDF bookmarks generated by bookmarks package got confused about page numbers of respective headings and they point to wrong place as a result: "Two" bookmark points to page 3 instead of 5, "Three" points to page 7 instead of 9 and so on.

And using bookmark for TOC causes compilation error (so I had to comment it out).

I was able to deal with this particular issue of compilation error by using \setcounter{page}{1} instead of \frontmatter/\mainmatter:

\documentclass[final,11pt]{book}
\usepackage{blindtext}
\usepackage{hyperref}
\hypersetup{bookmarksdepth=-1}
\usepackage{bookmark}
\chapter*{#1}
\bookmark[page=\thepage,level=-1]{My own bookmark for #1}
}
\begin{document}
\bookmark[page=\thepage,level=-1]{My own TOC bookmark}
\tableofcontents
\setcounter{page}{1}
\blindtext[10]
\blindtext[10]
\blindtext[10]
\blindtext[10]
\blindtext[10]
\end{document}

but it makes bookmark for TOC to have same page number (1) as bookmark for "One" and all other PDF bookmarks pointing to wrong page.

How can I create proper custom bookmarks pointing to right page (as in first code sample), while at the same time changing numeration for everything before "One" to roman numerals (with TOC being on page "i") and "One" being on page 1.

(Of course TOC should also point to right pages, but it doesn't seem to be an issue.)

• The page counter has the value 1 twice, since \frontmatter etc. changes the pagenumbering – user31729 Oct 21 '15 at 15:54
• I suggest to use dest=... rather with some uniquely defined hyperref anchor instead of the page=... way – user31729 Oct 21 '15 at 16:06

• \pdfbookmark can be used for the table of contents.

• \texorpdfstring can be used to get different texts into the table of contents and the bookmarks. Then pdfbookmarksdepth=-1 is not needed.

Example:

\documentclass[final,11pt]{book}
\usepackage{blindtext}
\usepackage{hyperref}
\usepackage{bookmark}
\chapter*{#1}
}
\begin{document}
\frontmatter
\pdfbookmark[0]{My own TOC bookmark}{toc}
\tableofcontents
\mainmatter
\blindtext[10]
\blindtext[10]
\blindtext[10]
\blindtext[10]
\blindtext[10]
\end{document}

• Thank you: this solves my problem. Though, I am curious why bookmark package gets confused by \frontmatter/\mainmatter or any other change in page numeration. You are probably the best person to answer this since you wrote both hyperref and bookmarks packages :) (Ps. With your solution, bookmarks package does not need to be loaded at all, even though you included it in the example). – Rafal Oct 22 '15 at 10:44
• @Rafal I would always load package bookmark, then the bookmarks get faster updated. – Heiko Oberdiek Oct 22 '15 at 17:40
• @Rafal Option page expects the absolute page number. \thepage will only work, if the page number equals the absolute page number and its expansion is a plain number. – Heiko Oberdiek Oct 22 '15 at 17:42

Here's a proposition using dest= instead of page= - label usage

\documentclass[final,11pt]{book}
\usepackage{blindtext}
\usepackage{hyperref}
\usepackage[depth=-1]{bookmark}

\newcounter{fakecntr}

\refstepcounter{fakecntr}
\phantomsection
\chapter*{\hypertarget{chap::\number\value{fakecntr}}{#1}}
\bookmark[dest=chap::\number\value{fakecntr},level=-1]{My own bookmark for #1}
}
\begin{document}
\frontmatter
\bookmark[dest=toc::bookmark,level=-1]{My own TOC bookmark}
\hypertarget{toc::bookmark}{\tableofcontents}
\mainmatter
\cleardoublepage


text:
Modulus of rigidity of a liquid

(a) Non zero constant

(b) Infinite

(c) Zero

(d) Can not be predicted

Difficulty Level:

A cube of aluminium of sides 0.1 m is subjected to a shearing force of 100 N. The top face of the cube is displaced through 0.02 cm with respect to the bottom face. The shearing strain would be

(a) 0.02                                   (b) 0.1

(c) 0.005                                 (d) 0.002

High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

The upper end of a wire of radius 4 mm and length 100 cm is clamped and its other end is twisted through an angle of 30°. Then angle of shear is

(a) $12°$                                      (b) $0.12°$

(c) $1.2°$                                     (d) $0.012°$

Concept Questions :-

Stress-strain
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

A rod of length l and radius r is joined to a rod of length l/2 and radius r/2 of same material. The free end of small rod is fixed to a rigid base and the free end of larger rod is given a twist of $\theta$, the twist angle at the joint will be

(a) $\theta /4$                          (b) $\theta /2$

(c) $5\theta /6$                         (d) $8\theta /9$

Concept Questions :-

Stress-strain
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

Shearing stress causes a change in-

(c)   Shape                                (d)   Volume

Concept Questions :-

Stress-strain

Difficulty Level:

To break a wire, a force of ${10}^{6}N/{m}^{2}$ is required. If the density of the material is , then the length of the wire which will break by its own weight will be -

(a) 34 m                             (b) 30 m

(c) 300 m                            (d) 3 m

Concept Questions :-

Stress-strain
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

One end of a uniform wire of length L and of weight $W$ is attached rigidly to a point in the roof and a weight ${W}_{1}$ is suspended from its lower end. If S is the area of cross-section of the wire, the stress in the wire at a height 3L/4 from its lower end is

1. $\frac{{W}_{1}}{S}$                                  2. $\frac{{W}_{1}+\left(W/4\right)}{S}$

3. $\frac{{W}_{1}+\left(3W/4\right)}{S}$                    4. $\frac{{W}_{1}+W}{S}$

Concept Questions :-

Stress-strain
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

The strain-stress curves of three wires of different materials are shown in the figure. P, Q and R are the elastic limits of the wires. The figure shows that

(a) Elasticity of wire P is maximum

(b) Elasticity of wire Q is maximum

(c) Tensile strength of R is maximum

(d) None of the above is true

Concept Questions :-

Stress-strain curve
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

The diagram shows a force-extension graph for a rubber band. Consider the following statements

I. It will be easier to compress this rubber than expand it

II. Rubber does not return to its original length after it is stretched

III. The rubber band will get heated if it is stretched and released

Which of these can be deduced from the graph?

(a)   III only                               (b)   II and III

(c)   I and III                             (d)   I only

Concept Questions :-

Stress-strain curve
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

The adjacent graph shows the extension $\left(∆l\right)$ of a wire of length 1m suspended from the top of a roof at one end with a load W connected to the other end. If the cross sectional area of the wire is ${10}^{-6}{m}^{2}$ calculate the young’s modulus of the material of the wire

(a) $2×{10}^{11}N/{m}^{2}$

(b) $2×{10}^{-11}N/{m}^{2}$

(c) $3×{10}^{-12}N/{m}^{2}$

(d) $2×{10}^{-13}N/{m}^{2}$

Concept Questions :-

Stress-strain curve


text:
# Search by Topic

#### Resources tagged with Addition & subtraction similar to Rolling Along the Trail:

Filter by: Content type:
Stage:
Challenge level:

### Hot Pursuit

##### Stage: 3 Challenge Level:

The sum of the first 'n' natural numbers is a 3 digit number in which all the digits are the same. How many numbers have been summed?

### Largest Number

##### Stage: 3 Challenge Level:

What is the largest number you can make using the three digits 2, 3 and 4 in any way you like, using any operations you like? You can only use each digit once.

### The Patent Solution

##### Stage: 3 Challenge Level:

A combination mechanism for a safe comprises thirty-two tumblers numbered from one to thirty-two in such a way that the numbers in each wheel total 132... Could you open the safe?

### Chameleons

##### Stage: 3 Challenge Level:

Whenever two chameleons of different colours meet they change colour to the third colour. Describe the shortest sequence of meetings in which all the chameleons change to green if you start with 12. . . .

### As Easy as 1,2,3

##### Stage: 3 Challenge Level:

When I type a sequence of letters my calculator gives the product of all the numbers in the corresponding memories. What numbers should I store so that when I type 'ONE' it returns 1, and when I type. . . .

### Being Collaborative - Primary Number

##### Stage: 1 and 2 Challenge Level:

Number problems at primary level to work on with others.

### Cows and Sheep

##### Stage: 2 Challenge Level:

Use your logical reasoning to work out how many cows and how many sheep there are in each field.

### Digit Sum

##### Stage: 3 Challenge Level:

What is the sum of all the digits in all the integers from one to one million?

### Book Codes

##### Stage: 2 Challenge Level:

Look on the back of any modern book and you will find an ISBN code. Take this code and calculate this sum in the way shown. Can you see what the answers always have in common?

### Being Determined - Primary Number

##### Stage: 1 and 2 Challenge Level:

Number problems at primary level that may require determination.

### Number Tracks

##### Stage: 2 Challenge Level:

Ben’s class were cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see?

### Arrange the Digits

##### Stage: 3 Challenge Level:

Can you arrange the digits 1,2,3,4,5,6,7,8,9 into three 3-digit numbers such that their total is close to 1500?

### Abundant Numbers

##### Stage: 2 Challenge Level:

48 is called an abundant number because it is less than the sum of its factors (without itself). Can you find some more abundant numbers?

### Zios and Zepts

##### Stage: 2 Challenge Level:

On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?

### A Mixed-up Clock

##### Stage: 2 Challenge Level:

There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?

### GOT IT Now

##### Stage: 2 and 3 Challenge Level:

For this challenge, you'll need to play Got It! Can you explain the strategy for winning this game with any target?

### Scoring with Dice

##### Stage: 2 Challenge Level:

I throw three dice and get 5, 3 and 2. Add the scores on the three dice. What do you get? Now multiply the scores. What do you notice?

### Neighbours

##### Stage: 2 Challenge Level:

In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square?

### Like Powers

##### Stage: 3 Challenge Level:

Investigate $1^n + 19^n + 20^n + 51^n + 57^n + 80^n + 82^n$ and $2^n + 12^n + 31^n + 40^n + 69^n + 71^n + 85^n$ for different values of n.

### Got it for Two

##### Stage: 2 Challenge Level:

Got It game for an adult and child. How can you play so that you know you will always win?

### Throw a 100

##### Stage: 2 Challenge Level:

Can you score 100 by throwing rings on this board? Is there more than way to do it?

##### Stage: 2 Challenge Level:

Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon?

### Sums and Differences 2

##### Stage: 2 Challenge Level:

Find the sum and difference between a pair of two-digit numbers. Now find the sum and difference between the sum and difference! What happens?

### Super Shapes

##### Stage: 2 Short Challenge Level:

The value of the circle changes in each of the following problems. Can you discover its value in each problem?

### Twenty Divided Into Six

##### Stage: 2 Challenge Level:

Katie had a pack of 20 cards numbered from 1 to 20. She arranged the cards into 6 unequal piles where each pile added to the same total. What was the total and how could this be done?

### The Dice Train

##### Stage: 2 Challenge Level:

This dice train has been made using specific rules. How many different trains can you make?

### Bean Bags for Bernard's Bag

##### Stage: 2 Challenge Level:

How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?

### Machines

##### Stage: 2 Challenge Level:

What is happening at each box in these machines?

### On Target

##### Stage: 2 Challenge Level:

You have 5 darts and your target score is 44. How many different ways could you score 44?

### One Wasn't Square

##### Stage: 2 Challenge Level:

Mrs Morgan, the class's teacher, pinned numbers onto the backs of three children. Use the information to find out what the three numbers were.

### Napier's Bones

##### Stage: 2 Challenge Level:

The Scot, John Napier, invented these strips about 400 years ago to help calculate multiplication and division. Can you work out how to use Napier's bones to find the answer to these multiplications?

### Sums and Differences 1

##### Stage: 2 Challenge Level:

This challenge focuses on finding the sum and difference of pairs of two-digit numbers.

### The Pied Piper of Hamelin

##### Stage: 2 Challenge Level:

This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!

### Clock Face

##### Stage: 2 Challenge Level:

Where can you draw a line on a clock face so that the numbers on both sides have the same total?

### Hubble, Bubble

##### Stage: 2 Challenge Level:

Winifred Wytsh bought a box each of jelly babies, milk jelly bears, yellow jelly bees and jelly belly beans. In how many different ways could she make a jolly jelly feast with 32 legs?

### The Puzzling Sweet Shop

##### Stage: 2 Challenge Level:

There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money?

### Make 100

##### Stage: 2 Challenge Level:

Find at least one way to put in some operation signs (+ - x ÷) to make these digits come to 100.

### Penta Post

##### Stage: 2 Challenge Level:

Here are the prices for 1st and 2nd class mail within the UK. You have an unlimited number of each of these stamps. Which stamps would you need to post a parcel weighing 825g?

### Two Primes Make One Square

##### Stage: 2 Challenge Level:

Can you make square numbers by adding two prime numbers together?

### The Clockmaker's Birthday Cake

##### Stage: 2 Challenge Level:

The clockmaker's wife cut up his birthday cake to look like a clock face. Can you work out who received each piece?

### Seven Square Numbers

##### Stage: 2 Challenge Level:

Add the sum of the squares of four numbers between 10 and 20 to the sum of the squares of three numbers less than 6 to make the square of another, larger, number.

### Dividing a Cake

##### Stage: 2 Challenge Level:

Annie cut this numbered cake into 3 pieces with 3 cuts so that the numbers on each piece added to the same total. Where were the cuts and what fraction of the whole cake was each piece?

##### Stage: 2 Challenge Level:

Can you put plus signs in so this is true? 1 2 3 4 5 6 7 8 9 = 99 How many ways can you do it?

### The Deca Tree

##### Stage: 2 Challenge Level:

Find out what a Deca Tree is and then work out how many leaves there will be after the woodcutter has cut off a trunk, a branch, a twig and a leaf.

### Sealed Solution

##### Stage: 2 Challenge Level:

Ten cards are put into five envelopes so that there are two cards in each envelope. The sum of the numbers inside it is written on each envelope. What numbers could be inside the envelopes?

### Wild Jack

##### Stage: 2 Challenge Level:

A game for 2 or more players with a pack of cards. Practise your skills of addition, subtraction, multiplication and division to hit the target score.

### Jugs of Wine

##### Stage: 3 Challenge Level:

You have four jugs of 9, 7, 4 and 2 litres capacity. The 9 litre jug is full of wine, the others are empty. Can you divide the wine into three equal quantities?

### Magic Triangle

##### Stage: 2 Challenge Level:

Place the digits 1 to 9 into the circles so that each side of the triangle adds to the same total.

### Six Ten Total

##### Stage: 2 Challenge Level:

This challenge combines addition, multiplication, perseverance and even proof.


text:
# Equilibrium: Electrostatic force and Gravitational force

1. May 10, 2006

### Swagger

Two identical balls of mass 38 g are suspended from threads of length 1.5 m and carry equal charges of 16 nC as shown in the figure. Assume that θ is so small that its tangent can be replaced by its sine and find the value of x.

(picture attached)

I know that the sum of all the forces must equal zero.

Fy: Tcosθ-mg=0
Fx: Tsinθ-[(k*q^2)/(x^2)]=0

Are these correct? I'm also confused on where to go from here.

#### Attached Files:

The figure does not show so I can't tell for sure but this looks correct at the condition that $\theta$ is defined between a string and the vertical. And I assume that x is the total (horizontal) distance between the two charges. The next step is to solve for x. The best thing is to isolate the tension from one equation and to replace this in th esecond equation. You will end up with one equation without any tension (but the angle theta will still be there...it wil appear in the form of $tan(\theta)$). Now use geometry. You may replace than by sine, and from the drawing you can get an expression for the sine of the angle (in terms of the length of the string and of x)


text:
# The distributive property

Posted by: Gary Ernest Davis on: January 10, 2010

This post is about the following fact: $a\times(b+c) = a\times b + a\times c$ for whole numbers $a, b, c$.

Sometimes this is called the distributive property of multiplication over addition, and sometimes it is called the distributive law. Some people are uncomfortable with use of the word “law”, as if it were passed by fiat and we darned well better follow it or else.  That’s not what the word “law” means in mathematics, but to avoid those sort of associations, let’s just call it the distributive property.

In the set of natural numbers the distributive property can be proven true from the standard Peano axioms.

However, school mathematics does not deal with natural numbers. That comes later, in college, when students learn about foundational issues in mathematics, such as what are sets, axioms for sets, axioms of infinity, and so on.  In school we deal with the counting numbers $1, 2, 3, 4, ...$.

What we know about the counting numbers has to come from acts of counting, not by inductive proof. So the distributive property in school mathematics needs to be placed in the context of counting, since we are dealing, in school , with counting numbers, a much less formal notion than natural numbers.

A simple version of the distributive property is $a\times (b+1) = a\times b + a$ for all counting numbers $a \textrm{ and }b$.

This simple version of the distributive property has a very nice, simple interpretation in terms of units of the number “a”. To illustrate this through a concrete example, let’s look at $13\times 11 = 13\times 10 + 13$:

Here, as adults, we can see the column of 13 small squares as a unit that is repeated 11 times. The very simple form of the distributive property says that we can split off 1 of those units, and count 10 lots of the unit 13, and then add in the extra unit of 13. Sort of obvious when you see it this way.

However, many children, even in grades 3-6, are not entirely stable or comfortable counting in units. Often young children, 6 years old, will recite count by 2’s: $2, 4, 6, 8, 10, 12, 14, 16$, … If they are able to do this they will happily show you that they can do so.

However, they may not yet be able to count by units of 2. For example, if a child can recite by 2’s, tell them you were counting by 2’s to 24 and ask them how many times you counted. Some children, not all, will answer 12, usually tapping two fingers as they count, or perhaps nodding their head as they recite 2’s. These children are counting their acts of reciting by 2’s, and are able to count by units of 2. The ability to do this develops gradually, and by grades 3-6 most, but not all, children will be confident counting by units. If they are not, the distributive property is going to cause them some difficulty.

Additionally, young children, in grades 1- 3, often cannot see rectangular arrays the way adults see them: as repeated rows, or repeated columns, where the rows, or columns, are themselves built from repeated squares.

Lyne Outhred & Michael Mitchelmore

Seminal studies by Lynne Outhred and her doctoral adviser Michael Mitchelmore show convincingly the developmental steps involved in young children’s ability to recognize and reproduce rectangular arrays as repeated rows of repeated squares. Here is an example from a paper of theirs [Student’s_structuring_of_rectangular_arrays] which illustrates some of the development steps children go though in being able to communicate in a drawing what constitutes an array:

So this simple form of the distributive property, so clearly illustrated for an adult by a rectangular array, may not be so clear to a child who (a) is not yet confidently counting by units or (b) is still developing the skill of seeing a rectangular array as composed of rows and columns.

As adults we take these two developmental skills for granted, because it’s hard to imagine a time when we did not have them. But clearly, at age 1 month we did not, and somewhere before age 21 years we probably did. These skills did not just pop into our heads fully formed, so we went through a period of developmental in gaining these skills. So, do all children.  It is advantageous for us, as parents and teachers, to figure out where children are before we begin “explaining” the distributive property to them.

The more general distributive property  $a\times (b+c)=a\times b + a\times c$ also has a simple interpretation in terms of rectangular arrays – simple for an adult that is.  Here is an array decomposition  illustrating $11\times 17 = 11\times 10+ 11 \times 7$:

Repeating the distributive property allows us to see that for all counting numbers $a, b, c, d$ we have $(a+b)\times(c+d)=(a+b)\times c+(a+b)\times d$

$= a\times c+ a\times d + b\times c +b\times d$.

Again this can illustrated nicely by a decomposition of a rectangular array. For example, here’s an array decomposition illustrating $13\times 17 = 10\times 10+ 10 \times 7 + 3\times 10 + 3\times 7$:

But we have to bear in mind, As I wrote above, that just because these are  simple and compelling explanations of the distributive property for us, as adults, it does not follow that a child will have the necessary skills to see what we see so clearly.

The distributive property is very important for algebra. One reason is because of its applications, for example in multiplying binomial expressions $(ax+b)(cx+d)$. Another reason is that a grasp of the distributive property is a good indication of a student’s number sense and that, in turn, is an excellent indicator of how readily a student can grasp algebraic ideas.

### 1 Response to "The distributive property"

I made a video that shows the “freshman dream” aka the biggest mistake kids (and adults!) make when using the distributive property. I’d like to make another video showing that distribution is just a fast way to ass a bunch of (a+b)’s. I’ll get to that one at some point!

http://zerosumruler.wordpress.com/


text:
# Calculus question involving an infinite series

The series from n = 1 infinity of 1/(n*(3^n)) must

A) converge to a value greater than 1/4
B) converge to a value greater than 1/9
C) Converge to a value less than 1/8
D) converge to a value less than 1/2
E) diverge.

I know the series definitely does not diverge because the series (1/3)^n is a geometric series which converges and the series is question is smaller than that geometric series so it much converge by the direct comparision test.

But I am not sure how to see what it would converge to. Can someone help?

The series from n = 1 infinity of 1/(n*(3^n)) must

A) converge to a value greater than 1/4
B) converge to a value greater than 1/9
C) Converge to a value less than 1/8
D) converge to a value less than 1/2
E) diverge.

I know the series definitely does not diverge because the series (1/3)^n is a geometric series which converges and the series is question is smaller than that geometric series so it much converge by the direct comparision test.

But I am not sure how to see what it would converge to. Can someone help?

Given the multiple choices, just look at the first term in your infinite series. What can you conclude?

lanedance
Homework Helper
so you've ruled out E)

couple of things to consider...
- what does 1/3^n converge to? then what can you say about each term of your series relative to that series

- what do the first few terms add upto, can you rule out any other answers?

lanedance
Homework Helper
welcome to PF by the way