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# Newton polynomial divided differences

When trying to interpolate data points $$(t_i,y_i)$$, one can use Newton's polynomials which are spanned by the basis of $$P_{n-1}$$ given by polynomials of the form

$$t-t_1, (t-t_1)(t-t_2), ..., \prod_{j=1}^{n-1} (t-t_j)$$

So, in the newton basis, an interpolant looks as

$$\phi_{n-1}(t) = \alpha_1 + \alpha_2(t-t_1) + ... + \alpha_n (t-t_1)(t-t_2)...(t-t_{n-1})$$

now, one can do $$\phi_{n-1}(t_i) = y_i$$ to find $$\alpha_i$$ which requires to solve a lower triangular matrix.

Now, if we introduce the notion of divided differences,

$$f[t_k] = y_k, f[t_1,t_2,...,t_k] = \frac{ f[t_2,t_3,...,t_k] - f[t_1,t_2,...,t_{k-1}] }{t_k-t_1}$$

observe that

$$\phi_{n-1}(t_1) = \alpha_1 = y_1 = f[t_1]$$

Also,

$$\phi_{n-1}(t_2) = \alpha_1 + \alpha_2 (t_2 - t_1) = y_2$$

since $$\alpha_1 = y_1$$, we observe that $$\alpha_2 = \dfrac{ y_2 - y_1 }{t_2 - t_1} = \dfrac{ f[t_2]-f[t_1]}{t_2-t_1} = f[t_1,t_2]$$

So, we can claim that in general, $$\alpha_k = f[t_1,...,t_k]$$. Im trying to prove this by induction but I dont see a plan on how to use the induction hypothesis to check for $$\alpha_{k+1}$$. We can assume that result holds for $$k$$ so that we can write the interpolant $$\phi_k(t)$$ as

$$\phi_k(t) = f[t_1] + f[t_1,t_2](t-t_1) + ..+ f[t_1,...,t_k] (t-t_1)...(t-t_{k-1}) + \alpha_{k+1} (t-t_1)...(t-t_{k})$$

but, then how can we take from there?

## 2 Answers

The subject is better caught using a matricial reperesentation, and it is preferrable to index from $$0$$, as it provides a neater handling of the sums and products.

Given an assigned sequence $$\left\{ {t_{\,0} ,t_{\,1} , \cdots ,t_{\,h} , \cdots } \right\}\quad \left| {\;t_{\,k} \ne t_{\,j} } \right.$$, let's indicate by $$\pi _{\,n} (t)$$ the $$n$$-th degree polynomial $$\pi _{\,n} (x,{\bf t}) = \prod\limits_{0\, \le \,j\, \le \,n-1} {\left( {x - t_{\,j} } \right)}$$ so that, with the convention of the null product, we have $$\pi _{\,0} (x,{\bf t}) = 1\quad \pi _{\,1} (x,{\bf t}) = \left( {x - t_{\,0} } \right)\quad \cdots$$

We want to construct a polynomial of degree $$h$$ such that at the $$h+1$$ points $$(t_0, \cdots,t_h)$$ it takes the values $$(f_0, \cdots,f_h)$$, that is $$p_{\,h} \left( {x,{\bf t},{\bf f}} \right)\;:\;\;p_{\,h} \left( {t_{\,k} ,{\bf t},{\bf f}} \right) = f_{\,k} \quad \left| {\,0 \le k \le h} \right.$$ While we could express it in terms of Lagrange polynomials, we want instead
to express it as a linear combination of the $$\pi _{\,k} (x,{\bf t})$$ polynomials \eqalign{ & p_{\,h} \left( {x,{\bf t},{\bf f}} \right) = \left( {\matrix{ {\pi _{\,0} (x,{\bf t})} & {\pi _{\,1} (x,{\bf t})} & \cdots & {\pi _{\,h} (x,{\bf t})} \cr } } \right)\left( {\matrix{ {\beta _{\,0} } \cr {\beta _{\,1} } \cr \vdots \cr {\beta _{\,h} } \cr } } \right) = \cr & = \overline {{\bf \pi }_{\,h} } (x,{\bf t})\;{\bf \beta }_{\,h} \cr} note that the vector $${\bf \pi }$$ includes polynomials of varying degree.

Since the above holds for whichever value of $$x$$, it shall hold in particular for each $$x=t_k$$, and we get the LT matrix equation you already know \bbox[lightyellow] { \eqalign{ & \left( {\matrix{ {f_{\,0} } \cr {f_{\,1} } \cr \vdots \cr {f_{\,h} } \cr } } \right) = \left( {\matrix{ 1 & 0 & \cdots & 0 \cr 1 & {\left( {t_1 - t_0 } \right)} & \cdots & 0 \cr \vdots & \vdots & \ddots & \vdots \cr 1 & {\prod\limits_{0\, \le \,j\, \le \,0} {\left( {t_h - t_{\,j} } \right)} } & \cdots & {\prod\limits_{0\, \le \,j\, \le \,h - 1} {\left( {t_h - t_{\,j} } \right)} } \cr } } \right)\left( {\matrix{ {\beta _{\,0} } \cr {\beta _{\,1} } \cr \vdots \cr {\beta _{\,h} } \cr } } \right) = \cr & = {\bf f} = {\bf P}({\bf t})\;{\bf \beta } \cr} }\tag{1}

Consider now that the Divided Difference of a sequence $$y_k$$ wrt $$x_k$$ $$\left\{ \matrix{ \left[ {y_{\,q} } \right] = y_{\,q} \hfill \cr \left[ {y_{\,q} ,\, \ldots ,\,y_{\,q + n} } \right] = {{\left[ {y_{\,q + 1} ,\, \ldots ,\,y_{\,q + n} } \right] - \left[ {y_{\,q} ,\, \ldots ,\,y_{\,q + n - 1} } \right]} \over {x_{\,q + n} - x_{\,q} }} \hfill \cr} \right.$$ can be expressed as a linear combination of the $$y_k$$ values as follows $$\bbox[lightyellow] { \left[ {y_{\,q} ,\, \ldots ,\,y_{\,q + n} } \right] = \sum\limits_{0\, \le \,m\, \le \,n} {{{y_{m + q} } \over {\prod\limits_{0\, \le \,k\, \ne \,m\, \le \,n} {\left( {x_{\,m + q} - x_{\,k + q} } \right)} }}} }\tag{2}$$ because in fact, taking $$q=0$$ wlog, it is \eqalign{ & \left[ {y_{\,1} ,\, \ldots ,\,y_{\,n + 1} } \right] - \left[ {y_{\,0} ,\, \ldots ,\,y_{\,n} } \right] = \cr & = \sum\limits_{0\, \le \,m\, \le \,n} {{{y_{m + 1} } \over {\prod\limits_{0\, \le \,k\, \ne \,m\, \le \,n} {\left( {x_{\,m + 1} - x_{\,k + 1} } \right)} }}} - \sum\limits_{0\, \le \,m\, \le \,n} {{{y_m } \over {\prod\limits_{0\, \le \,k\, \ne \,m\, \le \,n} {\left( {x_{\,m} - x_{\,k} } \right)} }}} = \cr & = \sum\limits_{1\, \le \,m + 1\, \le \,n + 1} {{{y_{m + 1} } \over {\prod\limits_{1\, \le \,k + 1\, \ne \,m + 1\, \le \,n + 1} {\left( {x_{\,m + 1} - x_{\,k + 1} } \right)} }}} - \sum\limits_{0\, \le \,m\, \le \,n} {{{y_m } \over {\prod\limits_{0\, \le \,k\, \ne \,m\, \le \,n} {\left( {x_{\,m} - x_{\,k} } \right)} }}} = \cr & = \sum\limits_{1\, \le \,m\, \le \,n + 1} {{{y_m } \over {\prod\limits_{1\, \le \,k\, \ne \,m\, \le \,n + 1} {\left( {x_{\,m} - x_{\,k} } \right)} }}} - \sum\limits_{0\, \le \,m\, \le \,n} {{{y_m } \over {\prod\limits_{0\, \le \,k\, \ne \,m\, \le \,n} {\left( {x_{\,m} - x_{\,k} } \right)} }}} = \cr & = \sum\limits_{1\, \le \,m\, \le \,n} {{{y_m } \over {\prod\limits_{1\, \le \,k\, \ne \,m\, \le \,n + 1} {\left( {x_{\,m} - x_{\,k} } \right)} }}} + {{y_{n + 1} } \over {\prod\limits_{1\, \le \,k\, \le \,n} {\left( {x_{\,n + 1} - x_{\,k} } \right)} }} - {{y_0 } \over {\prod\limits_{1\, \le \,k\, \le \,n} {\left( {x_{\,0} - x_{\,k} } \right)} }} - \sum\limits_{1\, \le \,m\, \le \,n} {{{y_m } \over {\prod\limits_{0\, \le \,k\, \ne \,m\, \le \,n} {\left( {x_{\,m} - x_{\,k} } \right)} }}} = \cr & = - {{y_0 } \over {\prod\limits_{1\, \le \,k\, \le \,n} {\left( {x_{\,0} - x_{\,k} } \right)} }} + \sum\limits_{1\, \le \,m\, \le \,n} {y_m \left( {{1 \over {\prod\limits_{1\, \le \,k\, \ne \,m\, \le \,n + 1} {\left( {x_{\,m} - x_{\,k} } \right)} }} - {1 \over {\prod\limits_{0\, \le \,k\, \ne \,m\, \le \,n} {\left( {x_{\,m} - x_{\,k} } \right)} }}} \right)} + {{y_{n + 1} } \over {\prod\limits_{1\, \le \,k\, \le \,n} {\left( {x_{\,n + 1} - x_{\,k} } \right)} }} = \cr & = - {{\left( {x_{\,0} - x_{\,n + 1} } \right)y_0 } \over {\prod\limits_{1\, \le \,k\, \le \,n + 1} {\left( {x_{\,0} - x_{\,k} } \right)} }} + \sum\limits_{1\, \le \,m\, \le \,n} {y_m \left( {{{\left( {x_{\,m} - x_{\,0} } \right)} \over {\prod\limits_{0\, \le \,k\, \ne \,m\, \le \,n + 1} {\left( {x_{\,m} - x_{\,k} } \right)} }} - {{\left( {x_{\,m} - x_{\,n + 1} } \right)} \over {\prod\limits_{0\, \le \,k\, \ne \,m\, \le \,n + 1} {\left( {x_{\,m} - x_{\,k} } \right)} }}} \right)} + {{\left( {x_{\,n + 1} - x_{\,0} } \right)y_{n + 1} } \over {\prod\limits_{0\, \le \,k\, \le \,n} {\left( {x_{\,n + 1} - x_{\,k} } \right)} }} = \cr & = \left( {x_{\,n + 1} - x_{\,0} } \right)\sum\limits_{0\, \le \,m\, \le \,n + 1} {{{y_m } \over {\prod\limits_{0\, \le \,k\, \ne \,m\, \le \,n + 1} {\left( {x_{\,m} - x_{\,k} } \right)} }}} = \cr & = \left( {x_{\,n + 1} - x_{\,0} } \right)\left[ {y_{\,0} ,\, \ldots ,\,y_{\,n + 1} } \right] \cr}

Therefore, in matrix notation we will have \bbox[lightyellow] { \eqalign{ & \left( {\matrix{ {\left[ {f_{\,0} } \right]} \cr {\left[ {f_{\,0} ,f_{\,1} } \right]} \cr \vdots \cr {\left[ {f_{\,0} , \ldots ,f_{\,h} } \right]} \cr } } \right) = \left( {\matrix{ 1 & 0 & \cdots & 0 \cr {{1 \over {\left( {t_{\,0} - t_{\,1} } \right)}}} & {{1 \over {\left( {t_{\,1} - t_{\,0} } \right)}}} & \cdots & 0 \cr \vdots & \vdots & \ddots & \vdots \cr {{1 \over {\prod\limits_{0\, \le \,k\, \ne \,0\, \le \,h} {\left( {t_{\,0} - t_{\,k} } \right)} }}} & {{1 \over {\prod\limits_{0\, \le \,k\, \ne \,1\, \le \,h} {\left( {t_{\,1} - t_{\,k} } \right)} }}} & \cdots & {{1 \over {\prod\limits_{0\, \le \,k\, \ne \,h\, \le \,h} {\left( {t_{\,h} - t_{\,k} } \right)} }}} \cr } } \right) \left( {\matrix{ {f_{\,0} } \cr {f_{\,1} } \cr \vdots \cr {f_{\,h} } \cr } } \right) = \cr & = {\bf D}_{\,{\bf div}} ({\bf t})\;{\bf f} \cr} }\tag{3}

The crucial point is that $$\bbox[lightyellow] { {\bf D}_{\,{\bf div}} ({\bf t}) = {\bf P}({\bf t})^{\, - \,{\bf 1}} }\tag{4}$$ which in connection with (1) demonstrates the thesis.

In fact \eqalign{ & \left[ {y_{\,0} ,\, \ldots ,\,y_{\,n} } \right]\quad \left| {\;y = f(x) = \prod\limits_{0\, \le \,j\, \le \,g - 1} {\left( {x - x_{\,j} } \right)} } \right.\quad = \cr & = \sum\limits_{0\, \le \,m\, \le \,n} {{{f\left( {x_m } \right)} \over {\prod\limits_{0\, \le \,k\, \ne \,m\, \le \,n} {\left( {x_{\,m} - x_{\,k} } \right)} }}} = \cr & = \left[ {g \le n} \right]\sum\limits_{g\, \le \,m\, \le \,n} {{{f\left( {x_m } \right)} \over {\prod\limits_{0\, \le \,k\, \ne \,m\, \le \,n} {\left( {x_{\,m} - x_{\,k} } \right)} }}} = \cr & = \left[ {g \le n} \right]\sum\limits_{g\, \le \,m\, \le \,n} {{{\prod\limits_{0\, \le \,j\, \le \,g - 1} {\left( {x_{\,m} - x_{\,j} } \right)} } \over {\prod\limits_{0\, \le \,k\,\, \le \,g - 1} {\left( {x_{\,m} - x_{\,k} } \right)} \prod\limits_{g\, \le \,k\, \ne \,m\, \le \,n} {\left( {x_{\,m} - x_{\,k} } \right)} }}} = \cr & = \left[ {g \le n} \right]\sum\limits_{g\, \le \,m\, \le \,n} {{1 \over {\prod\limits_{g\, \le \,k\, \ne \,m\, \le \,n} {\left( {x_{\,m} - x_{\,k} } \right)} }}} = \cr & = \left[ {g \le n} \right]\sum\limits_{0\, \le \,m\, \le \,n - g} {{1 \over {\prod\limits_{0\, \le \,k\, \ne \,m\, \le \,n - g} {\left( {x_{\,m + g} - x_{\,k + g} } \right)} }}} = \cr & = \left[ {g \le n} \right]\left[ {1_{\,g} ,\, \ldots ,\,1_{\,g + \left( {n - g} \right)} } \right] = \left[ {g = n} \right] \cr} where the second step follows from being $$f(x_m)=0$$ for $$m and where $$[g \le n],\,[g = n]$$ denotes the Iverson bracket.

I would like to show how the missing induction step can be carried out. We have $$n+1$$ distinct nodes $$x_0, x_1, \dotsc, x_n$$. Let $$p$$ denote the polynomial of degree at most $$n-1$$ which interpolates $$f$$ at the first $$n$$ nodes, i.e., $$x_0, x_1, \dotsc x_{n-1}$$. Then by our induction hypothesis $$p(x) = f[x_0,x_1,\dotsc,x_{n-1}] x^{n-1} + r(x)$$ where $$r$$ is a polynomial of degree at most $$n-2$$. Let $$q$$ denote the polynomial of degree at most $$n-1$$ which interpolates $$f$$ at the last $$n$$ nodes, i.e., $$x_1, x_2, \dotsc x_n$$. Then by our induction hypothesis $$q(x) = f[x_1,x_2,\dotsc,x_n] x^{n-1} + s(x)$$ where $$s$$ is a polynomial of degree at most $$n-2$$. Now, consider the polynomial $$t$$ given by $$t(x) = p(x)\frac{x_n-x}{x_n-x_0} + q(x)\frac{x-x_0}{x_n-x_0}$$ I claim that $$t$$ is unique polynomial which interpolates $$f$$ at all $$n+1$$ nodes. We have $$t(x_0) = p(x_0) \cdot 1 + q(x_0) \cdot 0 = p(x_0) = f(x_0),$$ because $$p$$ interpolates $$f$$ at $$x_0$$. We have $$t(x_n) = p(x_n) \cdot 0 + q(x_n) \cdot 1 = q(x_n) = f(x_n),$$ because $$q$$ interpolates $$f$$ at $$x_n$$. For $$1 we have $$t(x_i) = q(x_i)\left(\frac{x_n-x_i}{x_n-x_0}\right) + q(x_i)\left(\frac{x_i-x_0}{x_n-x_0}\right) = f(x_i) \frac{(x_n-x_i) + (x_i-x_0)}{x_n-x_0} = f(x_i),$$ because $$p$$ and $$q$$ both interpolate $$f$$ at $$x_i$$. This shows that $$t$$ is the polynomial of degree at most $$n$$ which interpolates $$f$$ at the $$n+1$$ nodes. We have $$t(x) = c_n t^n + u(x)$$ where $$c_n$$ is a constant and $$u$$ is a polynomial of degree at most $$n-1$$. The formula for $$c_n$$ follows readily from the expressions for $$p$$, $$q$$ and $$t$$. We have \begin{align} t(x) &= \frac{\Big(f[x_1,x_2,\dotsc,x_n]x^{n-1} +r(x)\Big)(x-x_0) - \Big(f[x_0,x_1,\dotsc,x_{n-1}]x^{n-1}+s(x)\Big)(x-x_n)}{x_n-x_0} \\ &= \left(\frac{f[x_1,x_2,\dotsc,x_n]-f[x_0,x_1,\dotsc,x_{n-1}]}{{x_n-x_0}} \right) x^n + \text{lower order terms.} \end{align} We see that the choice of $$f[x_0,x_1,\dots,x_n] = \frac{f[x_1,x_2,\dotsc,x_n]-f[x_0,x_1,\dotsc,x_{n-1}]}{{x_n-x_0}}$$ is exactly what is needed to complete our induction step.

The construction of $$t$$ in terms of $$p$$ and $$q$$ is the foundation for Neville's algorithm for evaluating the interpolating polynomial.

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# how to give a group structure

suppose you have two sets $G_1$ and $G_2$ with same cardinality

in $G_2$ you have the group structure and there is a bijective map from $G_1$ to $G_2$ this is just a set map. can we define a binary operation on $G_1$ with the help of binary operation on $G_2$ so that $G_1$ also become a group?

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This is a standard "transport of structure" argument. Conceptually, $G_1$ and $G_2$ are the "same" for all (non-set-theoretic) intents and purposes because they are in bijection. –  Zhen Lin May 20 '12 at 10:31
Yes. If $f:G_1\to G_2$ is the given bijection, then letting $\circ_1$ be defined by $x\circ_1 y=f^{-1}(f(x)\circ_2 f(y))$ gives you a group structure on $G_1$. –  Michael Greinecker May 20 '12 at 10:32
but how I will be sure that $x._1y$ is in $G_1$?I mean closoure property holds? –  La Belle Noiseuse May 20 '12 at 10:38
I don't see your problem. $f(x)$ and $f(y)$ are in $G_2$, so $f(x)circ_2 f(y)$ is in $G_2$, since $G_2$ has a group structure. And since $f$ is bijective, $f^{-1}(f(x)circ_2 f(y))$ is a unique element in $G_1$. –  Michael Greinecker May 20 '12 at 10:41
thank you dear sir :) –  La Belle Noiseuse May 20 '12 at 10:42

HINT: There’s really only one reasonable thing to try, and it works. Let $f:G_1\to G_2$ be your bijection, and let $\cdot$ be the group operation in $G_2$. You want to use $f$ to define a group operation $\odot$ on $G_1$. Suppose that $x,y\in G_1$; what would $x\odot y$ have to be in order for $f$ to be a group isomorphism?


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Lesson: Electricity

# Question 1

A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R', then the ratio $\frac{R}{R\text{'}}$ is:

(a) $\frac{1}{25}$
(b)
$\frac{1}{5}$
(c) 5
(d) 25

d

# Question 2

Which of the following terms does not represent electrical power in a circuit?

(a) ${I}^{2}R$
(b)
$I{R}^{2}$
(c)
$VI$
(d)
$\frac{{V}^{2}}{R}$

a

# Question 3

An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be:

(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W

d

# Question 4

Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be:

(a) 1:2

(b) 2:1

(c) 1:4

(d) 4:1

d

# Question 5

How is a voltmeter connected in the circuit to measure the potential difference between two points?

## Solution:

A voltmeter should be connected in parallel to measure the potential difference between two points.

# Question 6

A copper wire has diameter 0.5 mm and resistivity of . What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

## Solution:

Area of cross-section of the wire, $A=\pi {\left(\frac{D}{2}\right)}^{2}$.

Resistance,
.
The resistance is given by,

$R=\rho \frac{l}{A}$. Therefore the length of the wire,

$l=\frac{RA}{\rho }$

$=\frac{10×3.14×\frac{{\left(0.0005\right)}^{2}}{{2}^{2}}}{1.6×{10}^{-8}}$

$=\frac{10×3.14×25}{1.6×4}$

If the diameter of the wire is doubled,

the new diameter

Therefore, the new resistance

$=\rho \frac{l}{A}$

$=\frac{1.6×{10}^{-8}×122.72×4}{\pi ×{\left(\frac{1}{2}×{10}^{-3}\right)}^{2}}$

$=\frac{1.6×{10}^{-8}×122.72×4}{3.14×{10}^{-5}}$

$=250.2×{10}^{-2}$

$=25\text{\hspace{0.17em}}\Omega$

# Question 7

The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below:

I(amperes)       0.5,      1.0,      2.0,      3.0,      4.0

V (volts)          1.6,      3.4,      6.7,      10.2,    13.2

Plot a graph between V and I and calculate the resistance of that resistor.

## Solution:

The coordinates for the graph is given below.

 V (volts) 1.6 3.4 6.7 10.2 13.2 I (amperes) 0.5 1 2 3 4

The corresponding graph is given below.

Resistance of the resistor:

The resistance of the resistor is equal to the slope of the line formed by the V-I graph.

So, the slope

.

Therefore, the resistance is 3.4 Ω.

# Question 8

When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

## Solution:

Resistance of a resistor,
Here the potential difference, V = 12 V and the current in the circuit,

Therefore,

$\begin{array}{l}R=\frac{1.2}{2.5×{10}^{-3}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4.8×{10}^{3\text{\hspace{0.17em}}}\text{\hspace{0.17em}}\Omega \end{array}$

Therefore, the resistance of the resistor is $4.8×{10}^{3}\Omega$.

# Question 9

A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistors?

## Solution:

Since all the resistors are connected in series, the current flow through all the component is the same. The equivalent resistance R

As per Ohm’s law:

Potential difference,

As the current flow through each resistance will be the same in a series circuit, the current that will flow through the 12 Ω resistor is 0.671 A.

# Question 10

How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220V line?

## Solution:

As per Ohm’s law V = IR

Here, the supply voltage,
V and the current,  A
Let n be the number of resistors connected in parallel.

The equivalent resistance of the combination = R, is given as:

$\begin{array}{l}\frac{1}{R}=n×\frac{1}{176}\\ ⇒R=\frac{176}{n}\end{array}$

From Ohm’s law,

$\begin{array}{l}\frac{V}{I}=\frac{176}{n}\\ n=\frac{176×I}{V}\\ =\frac{176×5}{220}\\ =4\end{array}$

Therefore, four resistors of 176 Ω are required to draw 5A of current on a 220 V line.

# Question 11

Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.

## Solution:

a)   Connecting all the resistors in series:

The equivalent resistance

b)   Connecting all the resistors in parallel:

The equivalent resistance will be

As both the results are not desired, a combination of series and parallel

connections would be required.

c)   Connecting two resistors in parallel:

When two 6 Ω resistors are connected in parallel. Their equivalent resistance is:

$\begin{array}{l}=\frac{1}{\frac{1}{6}+\frac{1}{6}}\\ =\frac{6×6}{6+6}\\ =\text{\hspace{0.17em}}3\text{\hspace{0.17em}}\Omega \end{array}$

Connecting the third resistor in series with 3 Ω, we get the equivalent resistance

d)      Connecting two resistors in series:

When two 6 Ω resistors are in series, the equivalent resistance will be their sum,

Connecting the third resistor in parallel with 12 Ω, we get the equivalent resistance as:

$\begin{array}{l}=\frac{1}{\frac{1}{12}+\frac{1}{6}}\\ =\frac{12×6}{12+6}\\ =\text{\hspace{0.17em}}4\text{\hspace{0.17em}}\Omega \end{array}$

The combination of resistors, as done in the previous two cases, give the desired resistance.

# Question 12

Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

## Solution:

Given:
Supply voltage,
V
Maximum allowable current,
A
Rating of an electric bulb,
$P=10$ watt.

Let R1 be the resistance of each lamp.
As the resistance,
$R=\frac{{V}^{2}}{P}$

$⇒{R}_{1}=\frac{{\left(220\right)}^{2}}{10}=4840\text{\hspace{0.17em}}\Omega$

According to Ohm’s law,

$V=IR$

Let R be the total resistance of the circuit for n number of electric bulbs

$⇒R=\frac{V}{I}=\frac{220}{5}=44\text{\hspace{0.17em}}\Omega$

Since the resistance of each electric bulb, ${R}_{1}=4840\text{\hspace{0.17em}}\Omega$

Therefore, the number of electric bulbs connected in parallel are 110.

# Question 13

A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistances, which may be used separately, in series, or in parallel. What are the currents in the three cases?

## Solution:

Given:

The supply voltage,
The resistance of one coil,

Case 1: When the coils are used separately

Let I1 be the current flowing through the coil.

According to Ohm's law,
$V={I}_{1}{R}_{1}$

Thus,9.16 A current will flow through the coil when used separately.

Case 2: When the coils are connected in series

Let the total resistance of the circuit be R2and the total current flowing be I2.

The total resistance,

According to Ohm's law,

Thus, when the coils are connected in series, 4.58A current will flow through the circuit.

Case 3: When the coils are connected in parallel

Let the total resistance of the circuit be R3 and the current through it be I3.

The total resistance, R3 is given as

$\frac{1}{\frac{1}{24}+\frac{1}{24}}=\frac{24}{2}=12\text{\hspace{0.17em}}\Omega$

According to Ohm's law,

$V={I}_{3}{R}_{3}$

Thus, 18.33 A current will flow through the circuit.

# Question 14

Compare the power used in the 2 Ω resistor in each of the following circuits:

(i)   a 6 V battery in series with 1 Ω and 2 Ω resistor, and

(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

## Solution:

Case (i)

Given:

·         Potential difference,  V

·         1 Ω and 2 Ω resistors are connected in series.

The equivalent resistance of the circuit,

Let I be the current through the circuit.
According to Ohm’s law,

So, the current flowing through the 2 Ω resistor is 2 A.

The power used

Case (ii)

Given:

·       Potential difference,  V

·       12 Ω and 2 Ω resistors are connected in parallel.

Since the resistors are connected in parallel, the voltage across 2 Ω resistor is 4 V.

The power consumed by the resistor

# Question 15

Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

## Solution:

Since both the bulbs are connected in parallel, the potential difference across each of them is same and is equal to 220 V.

The current drawn by the bulb of rating 60 W $=\frac{\text{power}}{\text{voltage}}=\frac{60}{220}\text{A}$.

The current drawn by the bulb of rating 100 W

Therefore, the total current drawn

# Question 16

Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

## Solution:

If P is the power of the appliance and t is the time for which it is in use,

then, the energy consumed by the electrical appliance H= Pt.
Therefore, the energy consumed by the TV set of power 250 W in 1 h

The energy consumed by the toaster of power 1200 W in 10 minutes

Thus, the energy consumed by a 250 W TV set in 1 h is more than the energy consumed by a toaster of power 1200 W in 10 minutes.

# Question 17

An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

## Solution:

If I is the current flowing through the circuit and R is the resistance of the circuit, the rate of heat produced by a device is the expression for power of the device which is equal to ${I}^{2}R$.

Given:

The resistance of the electric heater,
The current drawn,

Therefore, the power,

Thus, the heat produced by the heater is at the rate of 1800 J/s.

# Question 18

Explain the following:

(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and

electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electricity

transmission?

## Solution:

(a) Tungsten is an alloy. It has a very high melting point and a very high resistivity. Thus, it offers very high resistance when current flows through it. Due to the high resistance, the tungsten becomes very hot and glows.

(b) The heating element of the heater is made up of alloy. An alloy offers very high resistance when current flows through it. Due to the high resistance, the alloy becomes very hot and glows. The heating elements are not made up of pure metal because metals are good conductor of electricity. Thus, they offer very little resistance in flow of electric current. Very little heat is produced when current passes through it.

(c) Series arrangements are not used in the domestic circuits for the following reasons:

i)  In a series connection, the overall resistance of the circuit becomes high due to which the current supply from the power source is very low.

ii) If one appliance in series connection stops working, the other appliances too stop working. Similarly, if we have to use only appliance, the other appliances will also start working.

iii) All the appliances connected in series connection do not get the same voltage as supplied by power supply.

(d) When the area of cross section increases, the resistance decreases, and vice versa. Thus, the resistance (R) of a wire is inversely proportional to its area of cross-section (A).

(e) Copper and aluminium are good conductors of electricity. Thus, during transmission, the loss of current would be least. That is the reason why copper and aluminium are usually employed for electricity transmission.


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# Radioactive Dating with Potassium Argon

1. Dec 9, 2007

[SOLVED] Radioactive Dating with Potassium Argon

1. The problem statement, all variables and given/known data
The technique known as potassium-argon dating is used to date old lava flows. The potassium isotope $$^{40}{\rm K}$$ has a 1.28 billion year half-life and is naturally present at very low levels. $$^{40}{\rm K}$$ decays by beta emission into $$^{40}{\rm Ar}$$. Argon is a gas, and there is no argon in flowing lava because the gas escapes. Once the lava solidifies, any argon produced in the decay of $$^{40}{\rm K}$$ is trapped inside and cannot escape. A geologist brings you a piece of solidified lava in which you find the $$^{40}{\rm Ar}/^{40}{\rm K}$$ ratio to be 0.350.

t = ? [billions of years]

2. Relevant equations
Any of these I suppose:
N = N_0 e^(-t/T)
T = time constant = 1/r
r = decay rate = [per seconds]
(t/2) = half-life = 1.28 billion years
Beta-plus decay: X becomes Y (A same, Z-1) + e^+1 + energy

3. The attempt at a solution
N = given ratio of Ar/K = .350
N_0 = 1

ln(1/2) = -(t/2) / T
T = -t/2 / ln(.5) = 1.846... years
r = 1 / T = 5.41 * 10^-10 [yr^-1]

.350 = 1e^(-rt)
t = ln .350 / -r = 1,938,653,661
t = 1.94 billion years

different attempt using
N = N_0 * (.5)^t/(t/2)
t/2 = given halflife, N = ratio = .350, N_0 = 1
t = 1.94 billion years

I'm guessing I shouldn't be putting in the ratio of Ar to K in for N. But my book only goes into details about Carbon-dating, so I'm not sure where to go from here.

Cheers.

2. Dec 9, 2007

### Kurdt

Staff Emeritus
Bit of a tricky question. What you can do however is say that:

$$N_{Ar}/N = 0.35$$ where N is the amount of potassium after decay, and,

$$N_0 = N_{Ar} + N$$


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# Lars Eighner's Homepage

## LarsWiki

### Math Formulas

#### Laws of Exponents

Generally these can be relied upon only for real numbers and positive bases (x and y). Exponents (n and m) can be irrational, but what some of these formulas mean in that case is not clear. {$0^0$} is indeterminate, but is defined as 1 in some restricted contexts. These are actually theorems and many of them can be derived from a relatively small subset. Some obvious extensions have been omitted.

{$$x^0 = 1$$}

{$$x^{-n} = {1 \over x^n}$$}

{$$x^n = \prod^n x$$}

{$$x^n x^m = x^{n+m}$$}

{$$(xy)^n = x^n y^n$$}

{$$(x^n)^m = x^{nm}$$}

{$$\left( x \over y \right)^n = {x^n \over y^n}$$}

{$$\sqrt[n] {x} = x^{ \frac {1} {n}}$$}

#### Binomial Coefficients

{$${n \choose k} = \frac{n!}{k!\,(n-k)!} = \prod_{\ell=1}^k \frac{n-\ell+1}{\ell}$$}

#### Binomial Theorem

{$$(x+y)^n = \sum_{k=0}^n {n \choose k}x^{n-k}y^k = \sum_{k=0}^n {n \choose k}x^{k}y^{n-k}$$}

#### Factors of Cubic Expressions

##### Difference of Two Cubes

{$$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$}

Sources:

Recommended:

Category: Math

This is a student's notebook. I am not responsible if you copy it for homework, and it turns out to be wrong.

Figures are often enhanced by hand editing; the same results may not be achieved with source sites and source apps.

### March 16, 2018

• HomePage
• WikiSandbox

Lars

Contact by Snail!

Lars Eighner
APT 1191
8800 N IH 35
AUSTIN TX 78753
USA

Help


text:
# Orbital mechanics 1

4 articles
1 challenge
1 video
2 skills
1 talkthrough
How do planets move? An introduction to orbital mechanics and the work of Johannes Kepler.

ARTICLE

ARTICLE

### Problem solving with area and circumference of circles

PRACTICE PROBLEMS
Practice interesting area and circumference problems.

ARTICLE

CHALLENGE

ARTICLE

### Introduction to ellipses

VIDEO 14:14 minutes
Sal introduces ellipses and shows how their standard equation relates to their center and radii (ellipses have two radii: major and minor).

### Find the center and radii of an ellipse from its equation

PRACTICE PROBLEMS
Find the center and major and minor radius of an ellipse given its equation.

### INTERACT: Elliptical orbits

TALKTHROUGH
Click and drag either foci to change orbit. You can also mouseover the terms below for definitions.


text:
This is a testing site only. See the live Public Lab site here »

gsoc-2017 gsoc-2017

Author Comment Last activity Moderation
rakib_uddin "Oh, Thank you so much. I was looking for something like this. I like Open Source Software and I have a huge interest in it. Hope one day I will bec..." | Read more » about 3 years ago
icarito "I appreciate to have followed your progress thru this season and look forward to continue collaborating more and more! Regards, Sebastian " | Read more » about 5 years ago
warren "@stella @ryzokuken @Ashan @ccpandhare @mridulnagpal @david-days @icarito @stevie @liz @Ujitha @ananyo2012 -- thank you so much! I posted a wrap-up ..." | Read more » about 5 years ago
warren "@stella @ryzokuken @Ashan @ccpandhare @mridulnagpal @david-days @icarito @stevie @liz @Ujitha @ananyo2012 -- thank you!!!!! " | Read more » about 5 years ago
warren "Sorry for late reply -- yes, feel free to add the Display segment here. It was a cool last-minute addition and it's great that you're doing it! " | Read more » about 5 years ago
warren "Hi, @stella - i agree with Ujitha, perhaps you can add a bit more detail, for example linking to each of your PR urls, and talking about some of th..." | Read more » about 5 years ago
warren "Hi, @ryzokuken - this looks great. I still feel you could link more clearly to https://github.com/publiclab/plotsbot/#submitting-behaviors -- i lov..." | Read more » about 5 years ago
ccpandhare "Great! Ping me when you want to version bump. Or if you'd like I can see if i can give you permissions for this? If you think that's appropriate :..." | Read more » about 5 years ago
warren "Maybe we should move some of the extra links to the bottom of the document, and lead with the introduction, to make it more readable? " | Read more » about 5 years ago
warren "Great! Ping me when you want to version bump. Or if you'd like I can see if i can give you permissions for this? " | Read more » about 5 years ago
ananyo2012 "@mridulnagpal Yes it looks good now. Can you try putting the code snippet in ` (3 backticks ) blocks. That would make it look better. " | Read more » about 5 years ago


text:
# statsmodels.duration.hazard_regression.PHReg.hessian¶

method

PHReg.hessian(params)[source]

Returns the Hessian matrix of the log partial likelihood function evaluated at params.


text:
<meta http-equiv="refresh" content="1; url=/nojavascript/">

# 1.15: Understand the Problem-Solving Plan

Difficulty Level: At Grade Created by: CK-12
0%
Progress
Practice Problem-Solving Models
Progress
0%

Credit: Lori L. Stalteri
Source: https://www.flickr.com/photos/llstalteri/7530318478/in/photolist-ctqRkd-7Tb1dH-2moFcA-2mjjhe-ctqSc5-UmpLS-NEyPL-bVLPGQ-a92RsP-fwzNuT-fvar4E-8yWhwo-fyrNcH-fvatRY-8WzmeP-cFFFUN-PgNnq-cPQjXh-NEyzE-yEXY3-bMn1Rv-ctqQZm-ftUmKJ-bysjW5-fvatH5-nYZHdu-ftDYWD-9ZUaau-ftUn4d-gsVeaZ-ftE2bK-ftUmAU-4QLfAe-fwzNG8-fw6TSj-fuV8mv-icx2gh-ccXwdj-nD61n9-79h4Zn-cfavtE-ccXt6U-fqxJD1-fwQ8Ew-fvRBo6-fdA4QS-fyrMZX-fuV9yH-cCS3Af-gsU2rb

Julien is designing flower gardens with his many nieces and nephews. He has to construct rectangular gardens where all the kids can plant their flower seeds. Each of the gardens is 6 feet by 12 feet. Each garden will have room for four kids to plant. How can Julien figure out how much space each of his nieces and nephews will have in the garden beds?

In this concept, you will learn to understand and use a problem-solving plan.

### Guidance

When making a plan to solve a problem, you may choose one of several strategies, including:

• Act it Out
• Make a Model
• Look for a Pattern
• Guess and Check
• Make a Table
• Work Backwards
• Write an Equation
• Write a Proportion

When you read a problem, it is helpful to underline or highlight any important information. Important information can include words that identify an operation. You can also identify key words such as distance, time, speed, area or perimeter. All of these parts help you to identify a problem and what the problem is asking you to solve.

Let’s look at a situation.

Mollie is planning to meet a group of friends at the movies on Saturday night at 6:00 p.m. Mollie is incharge of driving a few friends to the movies. Mollie lives 15 minutes from Sara’s house. It takes 10 minutes to get from Sara’s house to Madison’s house. It is another 20 minutes to the movie theatre from Madison’s house. It takes Mollie one hour to get dressed and ready for the evening. At what time should Mollie begin to get ready for the evening?

First, what is the problem asking you to find out?

The problem is asking you to determine the time at which Mollie should begin getting ready for the evening if she is to beat the movies at 6:00 p.m. You must take into consideration the amount of travel time to each of her friends houses and the amount of time it takes for Mollie to get dressed.

This problem involves time. You need to backup the time that it takes for Molly to get to the movie theater.

Next, make a plan.

Mollie needs to beat the movies at 6:00 p.m. Work backward to determine the time Mollie should begin getting ready. If you work backwards, you will be able to figure out the time that she needs to get there.

Then, solve the problem.

Time of Movie – Time to Dress – Time to Movies – Time to Madison’s – Time to Sara’s

6:00 – 1:00 – 0:20 – 0:10 – 0:15

5:00 – 0:20 – 0:10 – 0:15

4:40 – 0:10 – 0:15

4:30 – 0.15

4:15

Therefore, with all that she has todo, Mollie should begin getting ready at 4:15 p.m.

Finally, check the results.

If Mollie begins getting dressed at 4:15 p.m. and takes 1 hour, she will be ready to leave at 5:15 p.m.

Since it takes Mollie 15 minutes to get to Sara’s house, she’ll arrive at 5:30 p.m.

10 minutes later at 5:40 Mollie will arrive at Madison’s house.

Since it takes 20 minutes to drive to the movies, Mollie will arrive at the movies promptly at 6:00 p.m.

Therefore, Mollie needs to begin getting ready at 4:15 p.m.

### Guided Practice

The bowling alley charges $12 for one person to play,$10 for the second person to play, $8 for the third to person to play, and so on. What is the total cost for a family of five to bowl? How much money does the family save bowling together rather than separately? First, consider what the problem is asking. The problem is asking you to analyze the pattern to determine the cost for a family of five to bowl. After determining the cost of a family of five to bowl, determine the difference between the amount the family is spending and the amount it would cost for five individuals to play. Next, make a plan. To better observe the pattern, organize the information on a table. You know that every time you add a family member, the cost goes down by$2.

 Group size Cost to play 1st person $12 2nd person$10 3rd person $8 Then, solve the problem. Continue the pattern for the fourth and fifth family members. Add the cost to play for each family member to determine the total cost.  Group size Cost to play 1st person$12 2nd person $10 3rd person$8 4th person $6 5th person$4 Total $40 Then, solve for the amount you save by playing as a family. To determine the amount saved playing as a family, subtract the total cost for a family of five from the total cost for five individuals. To play as an individual it costs$12. Therefore the cost for five individuals is 12×5\begin{align*}\12 \times 5\end{align*} or60.

To play as a family, it costs $40. Cost to Play IndividuallyCost to Play as a Family$60$40==Amount Saved$20

The answer is $20. Therefore the cost for a family of five to bowl is$40. A family of five saves 20 playing together rather than individually. ### Examples Use this situation to answer the following example questions. For every day that Jesse harvested vegetables in the garden, he collected 4 pounds of vegetables. #### Example 1 If Jesse continued this for 45 days, how many total pounds of vegetables will he have collected? First, what is the problem asking you to find out? The problem is asking you to determine the number of pounds Jesse collects in 45 days. Next, make a plan. Jesse can collect four pounds of vegetables a day. Set up an equation to solve for the number of pounds he can collect in 45 days. # pounds=4 poundsday×45 days Then, solve the problem. # pounds# pounds==4 poundsday×45 days180 The answer is 180. Therefore, Jesse can collect 180 pounds of vegetables. #### Example 2 If Jesse collects 20 pounds a week, how many days does he work a week? First, what is the problem asking you to find out? The problem is asking you to determine the number of days Jesse works a week. Next, make a plan. Use the equation from the previous example to solve for the number of days Jesse works in a week. # 20 pounds=4 poundsday×x days Then, solve the problem. # 20 poundsx daysx days===4 poundsday×x days20 pounds×1 day4 pounds5 days The answer is 15. Therefore, Jesse works 5 days a week. #### Example 3 If Jesse collected vegetables for 90 days, how many pounds would he gather at this same rate? First, what is the problem asking you to find out? The problem is asking you to determine the number of pounds Jesse collects in 90 days. Next, make a plan. Jesse can collect four pounds of vegetables a day. Set up an equation to solve for the number of pounds he can collect in 90 days. # pounds=4 poundsday×90 days Then, solve the problem. # pounds# pounds==4 poundsday×90 days360 The answer is 360. Therefore, Jesse can collect 360 pounds of vegetables. ### Follow Up Credit: Swallowtail Garden Seeds Source: https://www.flickr.com/photos/swallowtailgardenseeds/15664082779/in/photolist-pSbw5K-mLf1Hg-nYZHdu-8qYvNs-otYyvW-4VaCuG-8qLJ8L-q9YddN-8SRhk7-8qLBq5-82gnR5-eW18JV-71vWeT-oHF3jm-cS6ZMo-fvTGsh-ehaJtM-tkEA7-seB6Up-ti6BGN-ouPwLY-p2no8B-rCHrrF-3zUYDV-rX35ph-5Chj7w-tcj77a-fDJhW2-rXzpdS-rjt9jR-MyTp-sBZc4Q-9voxve-2Qvzr-nCba7-oWSmCs-bTZaD6-tFCtbu-7GcAgi-9CmmYU-6EdgPq-p1AUNJ-51mw5J-tnym4-4LHqw9-5v9yPo-4enNHv-gCQWBi-6FZd1H-fdoYZM License: CC BY-NC 3.0 Remember Julien and the kids’ garden space? First, find the total area of one of the rectangular gardens. The formula for area is A=l×w\begin{align*}A = l \times w\end{align*}. You know that the length of each garden is 12 feet and the width of each garden is 6 feet. Substitute these values into the formula and solve for the area of the garden. AAA===l×w12×672 So the area of each garden is 72 ft2\begin{align*}72 \ ft^2\end{align*}. Next, each garden needs to be divided into four equal spaces so that each child can have the same space. 724=18\begin{align*}\frac{72}{4} = 18\end{align*} The answer is 18. Therefore, each child will have 18 square feet of garden to plant their flowers. ### Video Review ### Explore More Read each problem and then answer the questions following each problem. At the end of a phone call home, Brad had0.85. The initial cost of the phone call was $0.75 plus$0.12 per minute. If Brad spoke on the phone for 20 minutes, how much money did he have before making the phone call home?

1. Should you use working backwards or writing a proportion for this problem?

2. Why wouldn’t you use a proportion for this problem?

3. You will need to multiply one part of this problem, which part?

4. What equation could you write to solve this problem?

5. How much money did Brad have before making the phone call?

Suppose you have $75 in your savings account. You plan to save an additional$25 per week. After how many weeks will you have saved $500? 6. Which strategy makes the most sense look for a pattern or working backwards? 7. What is the unknown quantity that you are trying to figure out? 8. What equation can you use to solve this problem? 9. How many weeks will it take to save$500.00?

An online music store charges $1.90 to download 2 songs. Determine the cost of downloading 13 songs. 10. Which strategy would you use look for a pattern or use a proportion? 11. Why would you use that method? 12. What is the cost for the 13 songs? 13. What would be the cost for double the songs? 14. If the cost had been$2.25 per song, how much would 2 songs cost?

15. What would be the cost for 4 songs?

16. If six friends each downloaded four songs, how much would the total cost be?

### Vocabulary Language: English

Area

Area

Area is the space within the perimeter of a two-dimensional figure.
Model

Model

A model is a mathematical expression or function used to describe a physical item or situation.
Perimeter

Perimeter

Perimeter is the distance around a two-dimensional figure.
Problem Solving

Problem Solving

Problem solving is using key words and operations to solve mathematical dilemmas written in verbal language.
Proportion

Proportion

A proportion is an equation that shows two equivalent ratios.
Volume

Volume

Volume is the amount of space inside the bounds of a three-dimensional object.

## Date Created:

Dec 19, 2012

Aug 10, 2015
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text:
# Problem of the Week

## Updated at Jul 11, 2022 2:15 PM

This week's problem comes from the algebra category.

How can we factor $$4{z}^{2}-8z-12$$?

Let's begin!

$4{z}^{2}-8z-12$

 1 Find the Greatest Common Factor (GCF).1 What is the largest number that divides evenly into $$4{z}^{2}$$, $$-8z$$, and $$-12$$?It is $$4$$.2 What is the highest degree of $$z$$ that divides evenly into $$4{z}^{2}$$, $$-8z$$, and $$-12$$?It is 1, since $$z$$ is not in every term.3 Multiplying the results above,The GCF is $$4$$.To get access to all 'How?' and 'Why?' steps, join Cymath Plus!GCF = $$4$$2 Factor out the GCF. (Write the GCF first. Then, in parentheses, divide each term by the GCF.)$4(\frac{4{z}^{2}}{4}+\frac{-8z}{4}-\frac{12}{4})$3 Simplify each term in parentheses.$4({z}^{2}-2z-3)$4 Factor $${z}^{2}-2z-3$$.1 Ask: Which two numbers add up to $$-2$$ and multiply to $$-3$$?$$-3$$ and $$1$$2 Rewrite the expression using the above.$(z-3)(z+1)$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$4(z-3)(z+1)$Done4*(z-3)*(z+1)


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# Skills

The Discworld skill model is large and complex. It is broken up into eight branches.

## Skill System

### Trees and branches

Discworld uses a skill system, which is divided into eight skilltrees:

A tree can branch into subskills and sub-subskills.

What you actually use is the bonus, not the level. But to increase your bonus you need to increase your level or change your stats, which can be done temporarily with stat effects.

To refer to a skill you have to provide the path to it, for example fighting.defence.dodging.

• You can abbreviate skills by only listing a few letters: skills fi.de.do.
• Note that if two skills could be referred by such reduced version the mud will use one, but this might not be the one you want. For example, ma.me.me.ch is channeling, not charming, so you'd have to use skills ma.me.me.char instead.

Languages are special, you start with one and need to learn others from specific teachers before you can improve them. They do not use stats.

### Calculation of levels

The level is of the skilltree is calculated by taking the average of the levels of the main branches and round this down to the nearest integer. On its turn, the level of a (sub) branch is calculated the same way, by taking the average of the levels of the sub (sub) branches and round this down to the nearest integer.

### Calculation of guild level

A player's guild level is not the skilltree level of his specialisation (shown by 'skills primaries'). The guild level is calculated by taking the average of the level of the outermost branches and round this down to the nearest integer. For a wizard, the outermost branches are e.g. magic.methods.mental.animating and magic.methods.mental.channeling, not magic.methods.mental nor magic.methods.

## Improving your skills

There are 4 methods of improving your level, and hence your bonus, in a skill.

To gain levels from the taskmaster you don't need to do anything, just using skills has a chance to advance your skills!

• The taskmaster is the system that performs skill checks on Discworld. It decides if you pass or fail a skill check, the degree to which you do so, and whether you are to be awarded a skill increase as part of your attempts.
• To TM is to gain a level from the taskmaster.
• It is often said that you have the most chance to get a TM when you have 50% chance to pass a skillcheck.

Cost: Skill increases awarded by taskmasker are "free" advancement, they have no XP or financial cost.

### Advance: Instruction by Guild

Every guild has a teacher who will advance you to some level in all skills; however, the level to which they are willing to teach you in various skills will depend upon your choice of guild and, in some cases, your specialisation within the guild.

• This is done with the advance command (in the presence of a guild teacher).
• Use the cost command (in the presence of a guild teacher) to see how much one or several skill advance will cost.
• For example: cost primaries, cost magic, cost ma.points, etc.
• Guilds will advance primaries up to level 300; all other skills are taught to a lower level.
• For example, Priests may advance all faith skills up to level 300 within the guild, but may only advance magic skills up to level 5.
• Examples: advance magic, advance ma.sp, advance ma.sp.offensive
• To advance a skill tree, find the advancement room/person and type advance [skill]. So to advance your skill tree for magic type "advance magic" without the quotation marks.
• To advance a branch of a skill, simply type advance [skill.branch]

Cost: Advancing within your guild is the second cheapest in terms of XP cost after taskmaster skill increases, and costs AM\$ 12.50 or equivalent per 100k XP spent on skills.

### Teachers

All players are able to teach their skill to other players, subject to their bonus in the relevant skills being high enough.

• Learning from another player is considerably more expensive than advancing from a guild, but always cheaper (sometimes not by much) than learning from yourself (see below).

Cost: Learning from a teacher has a high XP cost, but generally has no financial cost. The teacher receives a portion of the XP you spend in return for teaching you.

### Self Teaching

Self teaching is the application of your XP to teaching yourself skills.

Cost: This is the most expensive method, but it has the benefit of always being available, and can also be done no matter what level you currently have in the skill.

## Skills tree table

See also the TM Project page for a more compact list of skills.

Adventuring Con Dex Int Str Wis
movement 2 2 0 1 0
swimming 2 2 0 1 0
riding 2 2 0 1 0
camel 2 2 0 1 0
horse 2 2 0 1 0
climbing 2 2 0 1 0
rock 2 2 0 1 0
rope 2 2 0 1 0
tree 2 2 0 1 0
sailing 2 2 0 1 0
following 2 2 0 1 0
pursuit 2 2 0 1 0
evade 2 2 0 1 0
acrobatics 1 2 0 2 0
tumbling 1 2 0 2 0
vaulting 1 2 0 2 0
balancing 1 2 0 2 0
evaluating 0 0 4 0 1
weapons 0 0 4 0 1
armour 0 0 4 0 1
perception 0 0 2 0 3
direction 0 2 2 0 1
health 4 0 0 1 0
points 1 1 1 1 1
Covert Con Dex Int Str Wis
stealth 0 3 1 1 0
inside 0 3 1 1 0
outside 0 3 1 1 0
underwater 0 3 1 1 0
hiding 0 2 2 1 0
person 0 2 2 1 0
object 0 2 2 1 0
lockpick 0 4 1 0 0
doors 0 4 1 0 0
safes 0 4 1 0 0
traps 0 4 1 0 0
manipulation 0 2 1 2 0
palming 0 2 1 2 0
passing 0 2 1 2 0
sleight-of-hand 0 2 1 2 0
stealing 0 2 1 2 0
casing 0 1 2 0 2
person 0 1 2 0 2
place 0 1 2 0 2
items 0 1 4 0 0
poisons 0 1 4 0 0
traps 0 1 4 0 0
weapons 0 1 4 0 0
points 1 2 2 0 0
Crafts Con Dex Int Str Wis
smithing 0 2 2 1 0
gold 0 2 2 1 0
silver 0 2 2 1 0
black 0 2 2 1 0
tools 0 2 2 1 0
weapons 0 2 2 1 0
armour 0 2 2 1 0
gem 0 2 2 1 0
cutting 0 2 2 1 0
polishing 0 2 2 1 0
setting 0 2 2 1 0
locks 0 2 2 1 0
mining 0 1 2 2 0
gem 0 1 2 2 0
ore 0 1 2 2 0
panning 0 1 2 2 0
mineral 0 1 2 2 0
hunting 0 2 3 0 0
tracking 0 2 3 0 0
fishing 0 2 3 0 0
trapping 0 2 3 0 0
foraging 0 2 3 0 0
carpentry 0 2 2 1 0
furniture 0 2 2 1 0
coopering 0 2 2 1 0
turning 0 2 2 1 0
whittling 0 2 2 1 0
pottery 0 3 2 0 0
forming 0 3 2 0 0
throwing 0 3 2 0 0
shaping 0 3 2 0 0
glazing 0 3 2 0 0
staining 0 3 2 0 0
firing 0 3 2 0 0
materials 0 2 2 1 0
tanning 0 2 2 1 0
leatherwork 0 2 2 1 0
weaving 0 2 2 1 0
spinning 0 2 2 1 0
needlework 0 2 2 1 0
dyeing 0 2 2 1 0
husbandry 0 0 3 0 2
plant 0 0 3 0 2
edible 0 0 3 0 2
herbal 0 0 3 0 2
milling 0 0 3 0 2
tree 0 0 3 0 2
animal 0 0 3 0 2
grooming 0 0 3 0 2
breeding 0 0 3 0 2
slaughtering 0 0 3 0 2
culinary 0 2 3 0 0
cooking 0 2 3 0 0
baking 0 2 3 0 0
butchering 0 2 3 0 0
preserving 0 2 3 0 0
brewing 0 2 3 0 0
distilling 0 2 3 0 0
medicine 0 1 2 0 2
firstaid 0 1 2 0 2
diagnosis 0 1 2 0 2
treatment 0 1 2 0 2
disease 0 1 2 0 2
injury 0 1 2 0 2
poison 0 1 2 0 2
arts 0 1 4 0 0
design 0 1 4 0 0
calligraphy 0 1 4 0 0
drawing 0 1 4 0 0
painting 0 1 4 0 0
printing 0 1 4 0 0
sculpture 0 1 4 0 0
theatre 0 1 4 0 0
music 0 1 4 0 0
instruments 0 1 4 0 0
percussion 0 1 4 0 0
wind 0 1 4 0 0
stringed 0 1 4 0 0
keyboard 0 1 4 0 0
vocal 0 1 4 0 0
special 0 1 4 0 0
theory 0 1 4 0 0
performance 0 1 4 0 0
points 0 2 2 0 1
Faith Con Dex Int Str Wis
rituals 0 0 1 1 3
offensive 0 0 1 2 2
area 0 0 1 2 2
target 0 0 1 2 2
defensive 0 2 1 0 2
area 0 2 1 0 2
self 0 2 1 0 2
target 0 2 1 0 2
curing 2 0 1 0 2
self 2 0 1 0 2
target 2 0 1 0 2
misc 0 0 2 0 3
area 0 0 2 0 3
self 0 0 2 0 3
target 0 0 2 0 3
special 0 0 1 1 3
items 0 1 2 0 2
rod 0 1 2 0 2
scroll 0 1 2 0 2
points 1 0 2 0 2
Fighting Con Dex Int Str Wis Comment
melee 1 2 0 2 0 is used to determine a player's ability when using melee weapons.
dagger 0 4 0 1 0 is the skill used to determine a player's ability in melee combat with a dagger.
sword 0 3 0 2 0 is the skill used to determine a player's ability in melee combat with a sword.
heavy-sword 1 1 0 3 0 is the skill used to determine a player's ability in melee combat when using a heavy sword.
axe 1 1 0 3 0 is the skill used to determine a player's ability in melee combat with an axe.
mace 2 1 0 2 0 is the skill used to determine a player's ability in melee combat with maces and hammers.
flail 1 2 0 2 0 is the skill used to determine a player's ability in melee combat with a flail.
polearm 2 0 0 3 0 is the skill used to determine a player's ability in melee combat with a polearm.
misc 1 2 0 2 0 is the skill used to determine a player's ability in melee combat with a miscellaneous weapons - that is, melee weapons that do not fit into any other other fighting.melee skill branches.
range 0 3 0 2 0 is used to determine a player's ability when using ranged weapons.
thrown 0 3 0 2 0 determines one's ability to hit a target with a thrown object.
fired 0 4 0 1 0 determines one's ability to hit a target with projectile launched from any weapon.
unarmed 0 2 0 2 1 is used to determine a player's ability at performing unconventional unarmed manoeuvers in combat, beyond simply punching and kicking.
striking 0 2 0 1 2 is the skill used to determine the magnitude of unarmed special attacks.
grappling 0 2 0 2 1 is the skill used to determine a player's ability to wrestle with opponents.
defence 0 2 0 2 1 is used to determine a player's ability to defend themselves against physical attacks - whether they come from melee or ranged distance.
parrying 0 3 0 1 1 is the skill used primarily to determine one's chance of successfully parrying an incoming physical attack with a weapon.
blocking 0 2 0 2 1 is the skill used primarily to determine one's chance of successfully blocking an incoming physical attack with a shield.
dodging 0 4 0 0 1 is the skill used primarily to determine one's chance of successfully dodging an incoming physical attack.
special 1 1 2 1 0 is used to determine various miscellaneous aspects relating to combat,
weapon 0 1 3 1 0 is the skill used to determine whether or not a special attack can be prepared with a weapon
unarmed 0 2 3 0 0 is the skill used to determine whether or not a special attack can be prepared while unarmed.
tactics 0 0 3 0 2 is a skill which has a strong influence on how rapidly one is able to act in combat.
mounted 2 2 0 0 1 is a skill related to preparing special attacks when mounted on a horse (and perhaps other mounts in future).
points 2 1 0 2 0 determines the number of fighting guild points a player has.
Magic Con Dex Int Str Wis
spells 0 1 2 0 2
offensive 0 0 2 2 1
defensive 2 0 2 0 1
misc 0 2 2 0 1
special 0 0 2 0 3
methods 0 1 3 0 1
elemental 3 0 2 0 0
air 3 0 2 0 0
earth 3 0 2 0 0
fire 3 0 2 0 0
water 3 0 2 0 0
mental 0 0 5 0 0
animating 0 0 5 0 0
channeling 0 0 5 0 0
charming 0 0 5 0 0
convoking 0 0 5 0 0
cursing 0 0 5 0 0
physical 0 3 2 0 0
binding 0 3 2 0 0
brewing 0 3 2 0 0
chanting 0 3 2 0 0
dancing 0 3 2 0 0
enchanting 0 3 2 0 0
evoking 0 3 2 0 0
healing 0 3 2 0 0
scrying 0 3 2 0 0
spiritual 0 0 2 0 3
abjuring 0 0 2 0 3
banishing 0 0 2 0 3
conjuring 0 0 2 0 3
divining 0 0 2 0 3
summoning 0 0 2 0 3
items 0 1 2 0 2
held 0 1 2 0 2
wand 0 1 2 0 2
rod 0 1 2 0 2
staff 0 1 2 0 2
broom 0 1 2 0 2
worn 0 1 2 0 2
amulet 0 1 2 0 2
ring 0 1 2 0 2
scroll 0 1 2 0 2
talisman 0 1 2 0 2
points 0 0 2 1 2
People Con Dex Int Str Wis
trading 0 0 4 0 1
buying 0 0 4 0 1
selling 0 0 4 0 1
valueing 0 0 4 0 1
gems 0 0 4 0 1
jewellery 0 0 4 0 1
weapons 0 0 4 0 1
armour 0 0 4 0 1
teaching 5/7 8/7 10/7 5/7 7/7
covert 1 2 2 0 0
crafts 0 2 2 0 1
faith 1 0 2 0 2
fighting 1 2 0 2 0
magic 0 0 2 1 2
people 1 1 1 1 1
adventuring 1 1 1 1 1
culture 0 0 3 0 2
ankh-morporkian 0 0 3 0 2
klatchian 0 0 3 0 2
agatean 0 0 3 0 2
lancrastian 0 0 3 0 2
genuan 0 0 3 0 2
points 1 1 1 1 1
Languages
 Agatean Brindisian Djelian Dwarfish Ephebian Klatchian Morporkian Uberwaldean


text:
Volume 9 (2013) Article 19 pp. 617-651
Complete Convergence of Message Passing Algorithms for Some Satisfiability Problems
by
Revised: May 4, 2013
Published: June 15, 2013


text:
# Integral on sequence

1. Sep 10, 2011

### burak100

1. The problem statement, all variables and given/known data
show that holds,

$\sum\limits_{n=1}^{\infty}\int\limits_{0}^{\frac{\pi}{2}}\frac{(2n-1) sin(2n-1) x}{n^2(n+1)}dx = \sum\limits_{n=2}^{\infty}\frac{1}{n^2}$

2. Relevant equations

3. The attempt at a solution

Actually, I have no idea how should I start.

2. Sep 10, 2011

### micromass

Staff Emeritus
This reduces to

$$\sum_{n=1}^{+\infty}{\frac{2n-1}{n^2(n+1)}\int_0^{\pi/2}{\sin((2n-1)x)dx}}$$

Now, calculate the integral.

3. Sep 10, 2011

### burak100

thank you very much for your help, I calculate the integral;

$= \sum\limits_{n=1}^{\infty}\frac{(2n-1)}{n^2(n+1)}\Big( \frac{-1}{(2n-1)} \cos((2n-1)x) \vert_{0}^{\frac{\Pi}{2}} \Big)$

$= \sum\limits_{n=1}^{\infty}\frac{-1}{n^2(n+1)} \Big( cos((2n-1)\frac{\Pi}{2}) -1 \Big)$

$= \sum\limits_{n=1}^{\infty} \frac{1- cos((2n-1)\frac{\Pi}{2}) }{n^2(n+1)}$

for $n=1$, this summation becomes $\frac{1}{2}$

so,

$\frac{1}{2} +\sum\limits_{n=2}^{\infty} \frac{1- cos((2n-1)\frac{\Pi}{2}) }{n^2(n+1)}$

it becomes like that. but here how should I continue to show that this equation equal to

$\sum\limits_{n=2}^{\infty} \frac{1}{n^2}$

???????

4. Sep 10, 2011

### micromass

Staff Emeritus
OK, but $\cos((2n-1)\pi/2)$ can be calculated, right??

I have no idea why you drop the summation here...

5. Sep 10, 2011

### burak100

Ok if I rewrite it, finally it becomes
$\frac{}{}$
$$=\sum\limits_{n=1}^{\infty}\frac{1- \cos((2n-1)\frac{\Pi}{2})}{n^2(n+1)}$$
$$=\sum\limits_{n=1}^{\infty}\frac{1- \cos(n \Pi - \frac{\Pi}{2})}{n^2(n+1)}$$
$$=\sum\limits_{n=1}^{\infty}\frac{1- \Big(\cos(n\Pi)\cos(\frac{\Pi}{2}) + \sin(n\Pi)\sin(\frac{\Pi}{2})\Big)}{n^2(n+1)}$$
$$=\sum\limits_{n=1}^{\infty}\frac{1}{n^2(n+1)}$$

actually I can't go on ..

how it is equal to
$$=\sum\limits_{n=2}^{\infty}\frac{1}{n^2}$$
????

6. Sep 10, 2011

### micromass

Staff Emeritus
OK, now split

$$\frac{1}{n^2(n+1)}$$

into partial fractions...

7. Sep 10, 2011

### burak100

ok now I think we found, can you check,

$=\sum\limits_{n=1}^{\infty}\frac{1}{n^2(n+1)}$
$=\sum\limits_{n=1}^{\infty}\frac{1}{n^2} - \sum\limits_{n=1}^{\infty}\frac{1}{n(n+1)}$
$=\sum\limits_{n=1}^{\infty}\frac{1}{n^2} - \Big( \sum\limits_{n=1}^{\infty}\frac{1}{n}- \sum\limits_{n=1}^{\infty}\frac{1}{(n+1)} \Big)$
$=\sum\limits_{n=1}^{\infty}\frac{1}{n^2} - 1$
$=\sum\limits_{n=2}^{\infty}\frac{1}{n^2}$

thank you very much....

8. Sep 10, 2011

### micromass

Staff Emeritus
Nonono, you can't write that. $\sum_{n=1}^{+\infty}{\frac{1}{n}}$ is infinity. So what's standing there is infinity - infinity. This doesn't make any sense.

You'll need another way to calculate

$$\sum_{n=1}^{+\infty}{\frac{1}{n}-\frac{1}{n+1}}$$

Write out the first 10 terms of the sum and see if you notice anything...

9. Sep 10, 2011

### burak100

$=\sum\limits_{n=1}^{\infty}\frac{1}{n(n+1)}$
$=\sum\limits_{n=1}^{\infty}\Big( \frac{1}{n} - \frac{1}{n+1} \Big)$
$= \Big( \frac{1}{1} - \frac{1}{2} \Big)$
$~~+ \Big( \frac{1}{2} - \frac{1}{3} \Big)$
$~~+ \Big( \frac{1}{3} - \frac{1}{4} \Big)$
$~~+ \Big( \frac{1}{4} - \frac{1}{5} \Big)$
........
and so on,
then it is equal to $1$
..

10. Sep 10, 2011

### micromass

Staff Emeritus
Yes, that is the correct reasoning!!

11. Sep 10, 2011

### burak100

Thank you very much..


text:
## Generalized separation algebras

Abbreviation: GSepAlg

### Definition

A generalized separation algebra is a cancellative partial monoid such that

$\cdot$ is conjugative: $\exists w, \ x\cdot w=y \iff \exists w, \ w\cdot x=y$.

##### Morphisms

Let $\mathbf{A}$ and $\mathbf{B}$ be cancellative partial monoids. A morphism from $\mathbf{A}$ to $\mathbf{B}$ is a function $h:A\rightarrow B$ that is a homomorphism: $h(e)=e$ and if $x\cdot y\ne *$ then $h(x \cdot y)=h(x) \cdot h(y)$.

Example 1:

### Properties

Classtype first-order

### Finite members

$\begin{array}{lr} f(1)= &1\\ f(2)= &2\\ f(3)= &3\\ f(4)= &8\\ f(5)= &14\\ f(6)= &48\\ f(7)= &172\\ f(8)= &\\ f(9)= &\\ f(10)= &\\ \end{array}$


text:
# Il See Periodic Table See Hin If all of the energy from burning 254.0 g of...

###### Question:

Il See Periodic Table See Hin If all of the energy from burning 254.0 g of propane (AHcomb.cana = -2220 kJ/mol) is used to heat water, how many liters of water can be heated from 20.0°C to 100.0°C? (Assume that the density of water is 1.00 g/mL) L of water

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text:
• ### Determination of N* amplitudes from associated strangeness production in p+p collisions(1703.01978)

Sept. 6, 2018 nucl-ex
We present the first determination of the energy-dependent production amplitudes of N$^{*}$ resonances with masses between 1650 MeV/c$^{2}$ and 1900 MeV/c$^{2}$ for an excess energy between $0$ and $600$ MeV. A combined Partial Wave Analysis of seven exclusively reconstructed data samples for the reaction p+p $\rightarrow pK\Lambda$ measured by the COSY-TOF, DISTO, FOPI and HADES collaborations in fixed target experiments at kinetic energies between 2.14 and 3.5 GeV is used to determine the amplitude of the resonant and non-resonant contributions.
• ### Theory of the two-loop self-energy correction to the $\boldsymbol g$ factor in non-perturbative Coulomb fields(1804.05733)

April 16, 2018 quant-ph, physics.atom-ph
Two-loop self-energy corrections to the bound-electron $g$ factor are investigated theoretically to all orders in the nuclear binding strength parameter $Z\alpha$. The separation of divergences is performed by dimensional regularization, and the contributing diagrams are regrouped into specific categories to yield finite results. We evaluate numerically the loop-after-loop terms, and the remaining diagrams by treating the Coulomb interaction in the electron propagators up to first order. The results show that such two-loop terms are mandatory to take into account for projected near-future stringent tests of quantum electrodynamics and for the determination of fundamental constants through the $g$ factor.
• ### Access to improve the muon mass and magnetic moment anomaly via the bound-muon $g$ factor(1801.02501)

Jan. 5, 2018 hep-ph, physics.atom-ph
A theoretical description of the $g$ factor of a muon bound in a nuclear potential is presented. One-loop self-energy and multi-loop vacuum polarization corrections are calculated, taking into account the interaction with the binding potential exactly. Nuclear effects on the bound-muon $g$ factor are also evaluated. We put forward the measurement of the bound-muon $g$ factor via the continuous Stern-Gerlach effect as an independent means to determine the free muons magnetic moment anomaly and mass. The scheme presented enables to increase the accuracy of the mass by more than an order of magnitude.
• ### Extraction of the electron mass from $g$ factor measurements on light hydrogenlike ions(1703.10649)

March 30, 2017 quant-ph, physics.atom-ph
The determination of the electron mass from Penning-trap measurements with $^{12}$C$^{5+}$ ions and from theoretical results for the bound-electron $g$ factor is described in detail. Some recently calculated contributions slightly shift the extracted mass value. Prospects of a further improvement of the electron mass are discussed both from the experimental and from the theoretical point of view. Measurements with $^4$He$^+$ ions will enable a consistency check of the electron mass value, and in future an improvement of the $^4$He nuclear mass and a determination of the fine-structure constant.
• Substantial experimental and theoretical efforts worldwide are devoted to explore the phase diagram of strongly interacting matter. At LHC and top RHIC energies, QCD matter is studied at very high temperatures and nearly vanishing net-baryon densities. There is evidence that a Quark-Gluon-Plasma (QGP) was created at experiments at RHIC and LHC. The transition from the QGP back to the hadron gas is found to be a smooth cross over. For larger net-baryon densities and lower temperatures, it is expected that the QCD phase diagram exhibits a rich structure, such as a first-order phase transition between hadronic and partonic matter which terminates in a critical point, or exotic phases like quarkyonic matter. The discovery of these landmarks would be a breakthrough in our understanding of the strong interaction and is therefore in the focus of various high-energy heavy-ion research programs. The Compressed Baryonic Matter (CBM) experiment at FAIR will play a unique role in the exploration of the QCD phase diagram in the region of high net-baryon densities, because it is designed to run at unprecedented interaction rates. High-rate operation is the key prerequisite for high-precision measurements of multi-differential observables and of rare diagnostic probes which are sensitive to the dense phase of the nuclear fireball. The goal of the CBM experiment at SIS100 (sqrt(s_NN) = 2.7 - 4.9 GeV) is to discover fundamental properties of QCD matter: the phase structure at large baryon-chemical potentials (mu_B > 500 MeV), effects of chiral symmetry, and the equation-of-state at high density as it is expected to occur in the core of neutron stars. In this article, we review the motivation for and the physics programme of CBM, including activities before the start of data taking in 2022, in the context of the worldwide efforts to explore high-density QCD matter.
• ### Muonic vacuum polarization correction to the bound-electron $g$-factor(1610.01340)

Oct. 5, 2016 physics.atom-ph
The muonic vacuum polarization contribution to the $g$-factor of the electron bound in a nuclear potential is investigated theoretically. The electric as well as the magnetic loop contributions are evaluated. We found these muonic effects to be observable in planned trapped-ion experiments with light and medium-heavy highly charged ions. The enhancement due to the strong Coulomb field boosts these contributions much above the corresponding terms in the free-electron $g$-factor. Due to their magnitude, muonic vacuum polarization terms are also significant in planned determinations of the fine-structure constant from the bound-electron $g$-factor.
• ### Centrality dependence of subthreshold $\phi$ meson production in Ni+Ni collisions at 1.9A GeV(1602.04378)

June 9, 2016 nucl-ex
We analysed the $\phi$ meson production in central Ni+Ni collisions at the beam kinetic energy of 1.93A GeV with the FOPI spectrometer and found the production probability per event of $[8.6 ~\pm~ 1.6 ~(\text{stat}) \pm 1.5 ~(\text{syst})] \times 10^{-4}$. This new data point allows for the first time to inspect the centrality dependence of the subthreshold $\phi$ meson production in heavy-ion collisions. The rise of $\phi$ meson multiplicity per event with mean number of participants can be parameterized by the power function with exponent $\alpha = 1.8 \pm 0.6$. The ratio of $\phi$ to $\text{K}^-$ production yields seems not to depend within the experimental uncertainties on the collision centrality, and the average of measured values was found to be $0.36 \pm 0.05$.
• ### Strange meson production in Al+Al collisions at 1.9A GeV(1512.06988)

May 17, 2016 nucl-ex
The production of K$^+$, K$^-$ and $\varphi$(1020) mesons is studied in Al+Al collisions at a beam energy of 1.9A GeV which is close or below the production threshold in NN reactions. Inverse slopes, anisotropy parameters, and total emission yields of K$^{\pm}$ mesons are obtained. A comparison of the ratio of kinetic energy distributions of K$^-$ and K$^+$ mesons to the HSD transport model calculations suggests that the inclusion of the in-medium modifications of kaon properties is necessary to reproduce the ratio. The inverse slope and total yield of $\phi$ mesons are deduced. The contribution to K$^-$ production from $\phi$ meson decays is found to be [17 $\pm$ 3 (stat) $^{+2}_{-7}$ (syst)] %. The results are in line with previous K$^{\pm}$ and $\phi$ data obtained for different colliding systems at similar incident beam energies.
• ### Influence of $\phi$ mesons on negative kaons in Ni+Ni collisions at 1.91A GeV beam energy(1412.4493)

Dec. 15, 2014 nucl-ex
$\phi$ and K$^-$ mesons from Ni+Ni collisions at the beam energy of 1.91A GeV have been measured by the FOPI spectrometer, with a trigger selecting central and semi-central events amounting to 51% of the total cross section. The phase space distributions, and the total yield of K$^-$, as well as the kinetic energy distribution and the total yield of $\phi$ mesons are presented. The $\phi$\K$^-$ ratio is found to be $0.44 \pm 0.07(\text{stat}) ^{+0.18}_{-0.12} (\text{syst})$, meaning that about 22% of K$^-$ mesons originate from the decays of $\phi$ mesons, occurring mostly in vacuum. The inverse slopes of direct kaons are up to about 15 MeV larger than the ones extracted within the one-source model, signalling that a considerable share of gap between the slopes of K$^+$ and K$^-$ could be explained by the contribution of $\phi$ mesons to negative kaons.
• ### Azimuthal Emission Patterns of $K^{+}$ and of $K^{-}$ Mesons in Ni + Ni Collisions near the Strangeness Production Threshold(1403.1504)

July 4, 2014 nucl-ex
Azimuthal emission patterns of $K^\pm$ mesons have been measured in Ni + Ni collisions with the FOPI spectrometer at a beam kinetic energy of 1.91 A GeV. The transverse momentum $p_{T}$ integrated directed and elliptic flow of $K^{+}$ and $K^{-}$ mesons as well as the centrality dependence of $p_{T}$ - differential directed flow of $K^{+}$ mesons are compared to the predictions of HSD and IQMD transport models. The data exhibits different propagation patterns of $K^{+}$ and $K^{-}$ mesons in the compressed and heated nuclear medium and favor the existence of a kaon-nucleon in-medium potential, repulsive for $K^{+}$ mesons and attractive for $K^{-}$ mesons.
• ### Systematics of azimuthal asymmetries in heavy ion collisions in the 1 A GeV regime(1112.3180)

Dec. 14, 2011 nucl-ex
Using the large acceptance apparatus FOPI, we study central and semi-central collisions in the reactions (energies in A GeV are given in parentheses): 40Ca+40Ca (0.4, 0.6, 0.8, 1.0, 1.5, 1.93), 58Ni+58Ni (0.15, 0.25, 0.4), 96Ru+96Ru (0.4, 1.0, 1.5), 96Zr+96Zr (0.4, 1.0, 1.5), 129Xe+CsI (0.15, 0.25, 0.4), 197Au+197Au (0.09, 0.12, 0.15, 0.25, 0.4, 0.6, 0.8, 1.0, 1.2, 1.5). The observables include directed and elliptic flow. The data are compared to earlier data where possible and to transport model simulations. A stiff nuclear equation of state is found to be incompatible with the data. Evidence for extra-repulsion of neutrons in compressed asymmetric matter is found.
• ### Systematics of central heavy ion collisions in the 1A GeV regime(1005.3418)

Sept. 13, 2010 nucl-ex
Using the large acceptance apparatus FOPI, we study central collisions in the reactions (energies in A GeV are given in parentheses): 40Ca+40Ca (0.4, 0.6, 0.8, 1.0, 1.5, 1.93), 58Ni+58Ni (0.15, 0.25, 0.4), 96Ru+96Ru (0.4, 1.0, 1.5), 96Zr+96Zr (0.4, 1.0, 1.5), 129Xe+CsI (0.15, 0.25, 0.4), 197Au+197Au (0.09, 0.12, 0.15, 0.25, 0.4, 0.6, 0.8, 1.0, 1.2, 1.5). The observables include cluster multiplicities, longitudinal and transverse rapidity distributions and stopping, and radial flow. The data are compared to earlier data where possible and to transport model simulations.
• ### Measurement of the in-medium K0 inclusive cross section in pi- -induced reactions at 1.15 GeV/c(0807.3361)

March 20, 2009 nucl-ex
The K0 meson production by pi- mesons of 1.15 GeV/c momentum on C, Al, Cu, Sn and Pb nuclear targets was measured with the FOPI spectrometer at the SIS accelerator of GSI. Inclusive production cross-sections and the momentum distributions of K0 mesons are compared to scaled elementary production cross-sections and to predictions of theoretical models describing the in-medium production of kaons. The data represent a new reference for those models, which are widely used for interpretation of the strangeness-production in heavy-ion collisions. The presented results demonstrate the sensitivity of the kaon production to the reaction amplitudes inside nuclei and point to the existence of a repulsive KN-potential of 20+-5 MeV at normal nuclear matter density.


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# Why does \left| insert unecessary vertical space below its argument?

When using \left| and \right| to typeset absolute value, I get unnecessary space as follows

\documentclass[11pt, a4paper]{IEEEtran}

\usepackage{amsmath}
\usepackage{amssymb, amsfonts}

\begin{document}

$$SDSCR_{i} = \frac{\left| V_{R, \thinspace i} \right|^{2}}{ \left| S_{eq, \thinspace i}^{\adjustbox{raise=0.1\baselineskip}{*}} \right| \left| Z_{RR, \thinspace ii} \right|}$$

\end{document}

• That is because delimiters are symmetric w.r.t. the math axis, and there is only a discrete set of sizes – Bernard Jan 18 at 19:46
• Why do you raise the * superscript even higher? That the main cause... – Werner Jan 18 at 19:49
• @Werner For aesthetics because the variable would look crowded. – Al-Motasem Aldaoudeyeh Jan 18 at 19:52
• If SDSCR is an identifier/variable, it should not be typeset in math italic -- it looks awful that way – Christian Hupfer Jan 18 at 19:57
• @ChristianHupfer. Most people here seem to not like italic math, but in electrical engineering, italic math for variables and even subscripts somtimes is the norm. – Al-Motasem Aldaoudeyeh Jan 18 at 20:03

delimiters are centred on the math axis, so by making the * extra high you are forcing the delimiters to have extra depth, but it is usually better to use fixed size delimiter. Also never use the math italic font for multi-letter identifiers;

\documentclass[11pt, a4paper]{IEEEtran}

\usepackage{amsmath}
\usepackage{amssymb, amsfonts}

\begin{document}

$$\mathit{SDSCR}_{i} = \frac{\bigl| V_{\mathit{f}R, i} \bigr|^{2}}{ \bigl| S_{\mathit{eq}, i}^{*} \bigr| \bigl| Z_{\mathit{RR}, ii} \bigr|}$$

\end{document}


text:
# TCP Proxy¶

TCP Proxy configuration overview.

## config.filter.network.tcp_proxy.v2.TcpProxy¶

[config.filter.network.tcp_proxy.v2.TcpProxy proto]

{
"stat_prefix": "...",
"cluster": "...",
"weighted_clusters": "{...}",
"idle_timeout": "{...}",
"access_log": [],
"max_connect_attempts": "{...}"
}

stat_prefix

(string, REQUIRED) The prefix to use when emitting statistics.

cluster

(string) The upstream cluster to connect to.

Precisely one of cluster, weighted_clusters must be set.

weighted_clusters

(config.filter.network.tcp_proxy.v2.TcpProxy.WeightedCluster) Multiple upstream clusters can be specified for a given route. The request is routed to one of the upstream clusters based on weights assigned to each cluster.

Precisely one of cluster, weighted_clusters must be set.

(core.Metadata) Optional endpoint metadata match criteria. Only endpoints in the upstream cluster with metadata matching that set in metadata_match will be considered. The filter name should be specified as envoy.lb.

idle_timeout

(Duration) The idle timeout for connections managed by the TCP proxy filter. The idle timeout is defined as the period in which there are no bytes sent or received on either the upstream or downstream connection. If not set, the default idle timeout is 1 hour. If set to 0s, the timeout will be disabled.

Warning

Disabling this timeout has a highly likelihood of yielding connection leaks due to lost TCP FIN packets, etc.

access_log

(config.filter.accesslog.v2.AccessLog) Configuration for access logs emitted by the this tcp_proxy.

max_connect_attempts

(UInt32Value) The maximum number of unsuccessful connection attempts that will be made before giving up. If the parameter is not specified, 1 connection attempt will be made.

## config.filter.network.tcp_proxy.v2.TcpProxy.WeightedCluster¶

[config.filter.network.tcp_proxy.v2.TcpProxy.WeightedCluster proto]

Allows for specification of multiple upstream clusters along with weights that indicate the percentage of traffic to be forwarded to each cluster. The router selects an upstream cluster based on these weights.

{
"clusters": []
}

clusters

(config.filter.network.tcp_proxy.v2.TcpProxy.WeightedCluster.ClusterWeight, REQUIRED) Specifies one or more upstream clusters associated with the route.

## config.filter.network.tcp_proxy.v2.TcpProxy.WeightedCluster.ClusterWeight¶

[config.filter.network.tcp_proxy.v2.TcpProxy.WeightedCluster.ClusterWeight proto]

{
"name": "...",
"weight": "..."
}

name

(string, REQUIRED) Name of the upstream cluster.

weight

(uint32) When a request matches the route, the choice of an upstream cluster is determined by its weight. The sum of weights across all entries in the clusters array determines the total weight.