text:
# Locking by archiving¶

Pre-condition: there exists a contract that needs to be locked and unlocked. In this section, Coin is used as the original contract to demonstrate locking and unlocking.

template Coin
with
owner: Party
issuer: Party
amount: Decimal
delegates : [Party]
where
signatory issuer, owner
observer delegates

    choice Transfer : ContractId TransferProposal
with newOwner: Party
controller owner
do
create TransferProposal
with coin=this; newOwner

    --a coin can only be archived by the issuer under the condition that the issuer is the owner of the coin. This ensures the issuer cannot archive coins at will.
choice Archives
: ()
controller issuer
do assert (issuer == owner)

Archiving is a straightforward choice for locking because once a contract is archived, all choices on the contract become unavailable. Archiving can be done either through consuming choice or archiving contract.

## Consuming choice¶

The steps below show how to use a consuming choice in the original contract to achieve locking:

• Add a consuming choice, Lock, to the Coin template that creates a LockedCoin.
• The controller party on the Lock may vary depending on business context. In this example, owner is a good choice.
• The parameters to this choice are also subject to business use case. Normally, it should have at least locking terms (eg. lock expiry time) and a party authorized to unlock.
    choice Lock : ContractId LockedCoin
with maturity: Time; locker: Party
controller owner
do create LockedCoin with coin=this; maturity; locker

• Create a LockedCoin to represent Coin in the locked state. LockedCoin has the following characteristics, all in order to be able to recreate the original Coin:

• The signatories are the same as the original contract.
• It has all data of Coin, either through having a Coin as a field, or by replicating all data of Coin.
• It has an Unlock choice to lift the lock.
template LockedCoin
with
coin: Coin
maturity: Time
locker: Party
where
signatory coin.issuer, coin.owner
observer locker

    choice Unlock
: ContractId Coin
controller locker
do create coin

## Archiving contract¶

In the event that changing the original contract is not desirable and assuming the original contract already has an Archive choice, you can introduce another contract, CoinCommitment, to archive Coin and create LockedCoin.

• Examine the controller party and archiving logic in the Archives choice on the Coin contract. A coin can only be archived by the issuer under the condition that the issuer is the owner of the coin. This ensures the issuer cannot archive any coin at will.
    --a coin can only be archived by the issuer under the condition that the issuer is the owner of the coin. This ensures the issuer cannot archive coins at will.
choice Archives
: ()
controller issuer
do assert (issuer == owner)

• Since we need to call the Archives choice from CoinCommitment, its signatory has to be Issuer.
template CoinCommitment
with
owner: Party
issuer: Party
amount: Decimal
where
signatory issuer
observer owner

• The controller party and parameters on the Lock choice are the same as described in locking by consuming choice. The additional logic required is to transfer the asset to the issuer, and then explicitly call the Archive choice on the Coin contract.
• Once a Coin is archived, the Lock choice creates a LockedCoin that represents Coin in locked state.
    nonconsuming choice LockCoin
: ContractId LockedCoin
with coinCid: ContractId Coin
maturity: Time
locker: Party
controller owner
do
inputCoin <- fetch coinCid
assert (inputCoin.owner == owner && inputCoin.issuer == issuer && inputCoin.amount == amount )
--the original coin firstly transferred to issuer and then archivaed
prop <- exercise coinCid Transfer with newOwner = issuer
do
id <- exercise prop AcceptTransfer
exercise id Archives
--create a lockedCoin to represent the coin in locked state
create LockedCoin with
coin=inputCoin with owner; issuer; amount
maturity; locker


text:
Finding $\prod_{n=1}^{999}\sin\left(\frac{n \pi}{1999}\right)$ [closed]

I would appreciate if somebody could help me with the following problem. How can we find the product

$$\prod_{n=1}^{999}\sin\left(\frac{n \pi}{1999}\right)$$

closed as off-topic by user21820, Claude Leibovici, The Phenotype, TheSimpliFire, Parcly TaxelFeb 18 '18 at 11:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Claude Leibovici, The Phenotype, TheSimpliFire, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.

• No idea how to do this exactly. If a decimal approximation is an acceptable answer you could use Sage. – Jim Mar 8 '13 at 6:24
• @wildcow: whats source ? – M.H Mar 8 '13 at 6:36
• source : high school problem – Young Mar 8 '13 at 11:18
• @Young do you attend the NUS high school? This looks like a SMO problem. – anegligibleperson Jun 7 '13 at 9:53
• Using complex numbers and eulers formula seem apparent – Torsten Hĕrculĕ Cärlemän Jul 15 '13 at 11:25

$$\prod_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}\right)=\frac{n}{2^{n-1}}$$ Here, let $$x=\prod_{k=1}^{999}\sin\left(\frac{k\pi}{1999}\right) \tag{1}$$

Since $\sin t=\sin (\pi-t)$, therefore,

$$x=\prod_{k=1}^{999}\sin\left(\frac{(1999-k)\pi}{1999}\right)=\prod_{k=1000}^{1998}\sin\left(\frac{k\pi}{1999}\right) \tag{2}$$

Multiplying equation $(1)$ by equation $(2)$ gives,

$$x^2=\prod_{k=1}^{1998}\sin\left(\frac{k\pi}{1999}\right)=\frac{1999}{2^{1998}}$$ $$\implies x=\frac{\sqrt{1999}}{2^{999}}$$

We took $x>0$ because all angles are in $1^{st}$ and $2^{nd}$ quadrant.

• Thank you so much for your concern. But how can I prove the first indentity: $$\prod_{k=1}^{n}\sin\left(\frac{k\pi}{n}\right)=\frac{n}{2^{n-1}}$$ – Young Mar 8 '13 at 6:51
• That's pretty neat! – Inceptio Mar 8 '13 at 6:51
• Can you please explain how u got $\prod_{k=1}^{n-1}\sin\big(\frac{k\pi}{n}\big)=\frac{n}{2^{n-1}}$ – Phani Raj Mar 8 '13 at 6:53
• @PhaniRaj: I updated my answer. Have a look. – Aang Mar 8 '13 at 6:58
• @Avatar Nice answer! – Start wearing purple Jul 15 '13 at 11:17

I have an even more precise bound: the product is less than

$$(999)! \left(\frac{\pi}{1999}\right)^{999}$$

Use Stirling approximation to get that this in turn is less than

$$\sqrt{999 e} \left ( \frac{999 \pi}{1999 e}\right )^{999} \approx 4 \times 10^{-237}$$

The first digits of your number are:

0.00

To get this estimate notice that all terms of the product are less than $1$ and that $|\sin(x)| \le |x|$.

To be more precise your number is between $0$ and $10^{-80}$.

• Hows this helpful? – Inceptio Mar 8 '13 at 6:34
• @Inceptio: I know more digits of this number than of $\pi$... – Emanuele Paolini Mar 8 '13 at 6:38

format: plain


text:
Unique - Maple Help

MmaTranslator[Mma]

 Unique
 generate a new symbol

 Calling Sequence Unique(arguments)

Parameters

 arguments - Maple translation of the Mathematica command arguments

Description

 • The Unique command generates a new symbol with its name in \$symbol form.

Examples

 > $\mathrm{with}\left(\mathrm{MmaTranslator}\left[\mathrm{Mma}\right]\right):$

Use the command with the Maple translation.

 > $\mathrm{Unique}\left(\right)$
 ${\mathrm{dollar1}}$ (1)

Alternatively, you can use the FromMma command with the evaluate option specified.

 > $\mathrm{with}\left(\mathrm{MmaTranslator}\right):$
 > $\mathrm{FromMma}\left(\mathrm{Unique\left[\right]},\mathrm{evaluate}\right)$
 ${\mathrm{dollar2}}$ (2)


text:
## Error message

Notice: Undefined offset: 0 in include() (line 35 of /home/it/www/www-icts/sites/all/themes/riley/templates/views/views-view-fields--related-file-field-collection-view.tpl.php).
Seminar
Speaker
Shradha Mishra (Indian Institute of Technology (BHU), Varanasi)
Date & Time
Thu, 05 November 2020, 15:00 to 16:30
Venue
Online seminar
Resources
Abstract

We study the dynamics of a particle moving in a two-dimensional Lorentz lattice-gas. The underlying lattice-gas is occupied by two kinds of rotators, "right-rotator (R)" and "left-rotator (L)" and some of the sites are empty {\it {viz.}} vacancy" V". The density of R and L are the same and density of V is one of the key parameters of our model. The rotators deterministically rotate the direction of a particle's velocity to the right or left and vacancies leave it unchanged. We characterise the dynamics of particle motion for different densities of vacancies. The probability of the particle being in a closed or open orbit is a function of the density of vacancies and time. As the density of vacancies increases in the system, the tendency to form a closed orbit reduces. Since the system is deterministic, the particle forms a closed orbit asymptotically. We calculate the probability of the particle to get trapped in a closed orbit and extract the Fisher's exponent in the diffusive regime. We also calculate the fractal dimension and observe a consistent change in the values as we increase the number of vacancies in the system. The system follows the hyper-scaling relation between the Fisher's exponent and fractal dimension for a finite vacancy density.


text:
# Board $7\times 7$ problem

An aid in this problem:

On a board of $$7 \times 7$$ each square is painted red or blue so that every square on the board has at least two neighboring squares blue. determine the minimum number of blue boxes on the board. Two cells are neighbors if they have a common side.

My progress is trying to have painted boxes that most have only two neighboring blue boxes, but my problem is:

How do I know I've got the minimum?

Will there be other settings with fewer blue boxes? Thank you.

• Having almost all of the squares have two neighbours painted blue is not a sufficient condition for the minimum. For instance, for a $3 \times 3$ board, one of those solutions would be $$\begin{matrix} B & B & B \\ B & - & B \\ B & B & B \end{matrix}$$ while a solution with fewer blue squares would be $$\begin{matrix} B & B & - \\ B & B & B \\ - & B & B \end{matrix}$$ So I think knowing when you have a minimum is not so easy. – TMM Jul 24 '13 at 23:05
• Yes TMM , thats the problem – giancarlo Jul 24 '13 at 23:10
• I'm curious what your solution is. I can get to 31 blue boxes; can you go any lower? – TMM Jul 24 '13 at 23:15
• Yes my solution is 31 , but my problem is how to justify that is the solution. – giancarlo Jul 24 '13 at 23:18
• Hmm. The best lower bound I got was 29. – Michael Biro Jul 25 '13 at 1:00

I got a bit carried away, but the minimum seems to be $31$ indeed. Unfortunately my method does not seem to generalize very well, but maybe that's too ambitious.

We have a total of $49$ boxes, each of which must have $2$ edges painted blue, adding up to a total of $98$ edges to be painted blue. As the corner boxes have only $2$ edges available for painting, these two edges must be painted blue. This yields $8$ boxes painted blue, each having $3$ edges painted bue. This makes for a total of $24$ edges painted blue, leaving $74$ edges yet to be painted blue. The image on the left shows these $8$ boxes:

$\hspace{60pt}$$\hspace{100pt} Note that there are 24 boxes yet to be painted that have 'excess' edges, i.e. edges that do not need to be painted blue. They are shown in green in the second picture. These are the 4 corners, having 2 excess edges each, the 12 remaining side boxes, having 1 excess edge each, and 8 boxes adjacent to the 4 interior boxes that already have 2 blue edges, having 1 excess edge each. This yields a total of 28 excess edges, but we will encounter a few more excess edges later on. The square is divided into 4 quadrants by its diagonals. We will show that in each quadrant, at least 4 excess edges must be painted blue. To this end we will paint the top quadrant, attempting to paint fewer than 4 excess edges blue. We distinguish 5 cases, painting towards a contradiction. Both blue boxes in the top quadrant are adjacent to 2 boxes with an excess edge, and at least 1 of these 2 boxes must be painted blue, hence at least 2 more boxes in the top row must be painted blue. If we paint 4 more boxes in the top row blue, then we paint 4 excess edges blue, and we are done. So we restrict to the cases where we paint 2 or 3 more boxes in the top row blue. After painting the top row, every box in the top row must have at least one blue edge. Painting only two boxes in the top row, there is only one way to do so; see the picture on the left. \hspace{60pt}$$\hspace{100pt}$

Then in order for our newly painted boxes to have $2$ blue edges, we must paint the boxes immediately below them blue, see the picture on the right. Then we have painted $4$ excess edges blue, and we are done.

Finally we paint precisely $3$ boxes in the top row blue. Up to symmetry, there are $4$ ways to do so, keeping in mind that each box must have at least $1$ blue edge. They are shown in the image below.

$\hspace{25pt}$

Whenever a box in the top row has only $1$ blue edge, the box directly below it must be painted blue. These necessarily blue boxes are shown in the image below.

$\hspace{25pt}$

In the left two cases, we see that $5$ excess edges are painted blue, and we are done. It remains to show that in the right two cases, at least one more excess edge must be painted blue.

In the middle right case, we are actually done already; in the left image below, the box highlighted in green has $3$ blue edges, so it has $1$ excess edge painted blue. With a total of $4$ excess edges painted blue, this case is closed.

$\hspace{60pt}$

For the rightmost and final case, we continue painting edges blue without painting $3$ edges of the same box blue, towards a contradiction. In the middle image above, the only way for the highlighted box to have $2$ blue edges without any other boxes having $3$ blue edges, we must paint the box directly below it blue. For this newly painted box to have $2$ blue edges, at least one of the boxes to its left, right, or top must be painted blue. These are highlighted in orange in the right image above. But in each case one or both of the boxes highlighted in green will have $3$ blue edges, hence an excess edge painted blue.

We conclude that in each quadrant, there are at least $4$ excess edges painted blue. With $4$ quadrants, this yields a total of $16$ excess edges to be painted blue. Hence in total we must paint at least $74+16=90$ edges blue, after having painted the $8$ initial boxes blue. But as each box has only $4$ edges, this requires at least $\tfrac{90}{4}>22$ more boxes to be painted blue, i.e. at least $23$ more boxes. Together with the inital $8$ boxes, this yields a minimum of $8+23=31$ boxes to be painted blue. This is in fact the minimum:

$\hspace{25pt}$

I wonder how many minimal solutions there are...

• Via integer linear programming, I confirmed that the minimum is 31, and there are 263 solutions that attain that minimum. – RobPratt Apr 7 '20 at 15:51
• @RobPratt That's interesting! Are those solutions up to rotation and reflection? Or is there some highly symmetrical solution to yield an odd total number of solutions? I'd be interested to see the latter. – Servaes Apr 7 '20 at 15:54
• Sorry, I meant 264, and this is treating rotations and reflections as distinct solutions. – RobPratt Apr 7 '20 at 16:01
• There are 48 optimal solutions up to symmetry. – RobPratt Apr 7 '20 at 16:30
• I added a comment to that question just now. – RobPratt Apr 8 '20 at 15:29


text:
# In an isosceles triangle, if the measure of a base angle is 74°, what is the measure of the vertex angle?

Jan 29, 2016

The vertex angle would be ${32}^{o}$

#### Explanation:

In an isosceles triangle the base angles would be congruent and therefore both have measures of ${74}^{o}$

The three angles of any triangle must add up to ${180}^{o}$

Therefore

${74}^{o} + {74}^{o} + v = {180}^{o}$
${148}^{o} + v = {180}^{o}$
$v = {32}^{o}$


text:
# Is Strong law of Large numbers equal to pointwise convergence?

1. function pointwise convergence

$$\{f_n\}$$ converges to $$f$$ pointwise on $$E$$ if the following equation holds, $$f(x) = \lim_{n\to\infty}f_n(x) ,$$ for every $$x\in E$$.

2. Strong Law of Large Numbers

A sequence of random variables $$\{X_n\}$$ converges almost surely to a random variable $$X$$ if for almost every $$\omega \in \Omega$$, $$\lim_{n\to\infty}{X_n(\omega)}=X(\omega).$$

3. The relation of above

I found out that these two are amazingly similar. When the almost is removed, they become exactly the same.

The first definition is from Principles of Mathematical Analysis Definition 7.1. The second definition is from What is the difference between the weak and strong law of large numbers?.

We say that $$f_n \to f$$ $$\mu$$-almost everywhere if $$\mu\{\omega: f_n(\omega) \not\to f(\omega)\} = 0$$.
It is obvious that pointwise convergence implies convergence almost surely, since if $$f_n \to f$$ pointwise, then $$\mu \{\omega: f_n(\omega) \not\to f(\omega)\} = \mu(\emptyset) = 0$$.
For a counterexample of the other direction, we work with the lebesgue measure $$\mu = \lambda$$. Consider your favorite pointwise convergence limit $$f_n \to f$$ on the reals and change the value of $$f$$ in your favorite point and denote this adaption of $$f$$ with $$g$$. Then $$f_n \to g$$ almost surely (since singeltons have Lebesgue measure zero) but not pointwise because we destroyed the convergence in the point where we changed $$f$$.
• So $f_n \to f$ almost everywhere means there can be a measure zero exceptional set, but $f_n \to f$ pointwise means there is no single exceptional point. Jan 20 '20 at 15:35


text:
# Name of this identity? $\int e^{\alpha x}\cos(\beta x) \space dx = \frac{e^{\alpha x} (\alpha \cos(\beta x)+\beta \sin(\beta x))}{\alpha^2+\beta^2}$

Again:

$$\int e^{\alpha x}\cos(\beta x) \space dx = \frac{e^{\alpha x} (\alpha \cos(\beta x)+\beta \sin(\beta x))}{\alpha^2+\beta^2}$$

Also the one for $\sin$:

$$\int e^{\alpha x}\sin(\beta x) \space dx = \frac{e^{\alpha x} (\alpha \sin(\beta x)-\beta \cos(\beta x))}{\alpha^2+\beta^2}$$

Both help avoid integration by parts. I find these two proofs very useful and I'd like to know what their names are.

-
You can use \sin and \cos in LaTeX to make your problem more readable. –  Tyler Feb 1 '11 at 2:04
@tyler: ah thanks! I should've thought of that. It's my first time actually writing something complicated. –  JustcallmeDrago Feb 1 '11 at 2:07
@Tyler Bailey: Just wondering, how did you edit that post? Don't you need 2000 reputation? –  PEV Feb 1 '11 at 2:12
What you have given here is an identity, and can ask for it's name. You haven't actually given any proof... –  Aryabhata Feb 1 '11 at 2:26
I seriously doubt those equalities have a name: they are just indefinite integrals, and there are way too many of those to name them... –  Mariano Suárez-Alvarez Feb 1 '11 at 3:52

They are the real and imaginary parts of the following identity: $$\int{e^{(\alpha+i\beta)x}dx} = \frac{e^{(\alpha+i\beta)x}}{\alpha+i\beta} = \frac{(\alpha - i\beta)e^{(\alpha+i\beta)x}}{\alpha^2 + \beta^2}.$$

-
Does it have a name? –  JustcallmeDrago Feb 1 '11 at 3:04
and also: Wait, what? Could you give some explanation on how that relates to my identity? I don't understand how trig becomes something containing $i$ –  JustcallmeDrago Feb 1 '11 at 3:07
@JustcallmeDrago Take a look at Euler's formula and you'll see the relationship. –  Adrián Barquero Feb 1 '11 at 3:15
@mjqxxxx Now that is a very clever way to find them! –  Pedro Tamaroff Apr 29 '12 at 1:01

I answered this in another question, but I think it might apply better here:

In general, if you want to find $$\int e^{ax}\cdot \sin{bx}\cdot dx$$ you can argue as follows:

Note that for any $\alpha$ or $\beta$, you have

\eqalign{ & \frac{d}{{dx}}\left( {{e^{\alpha x}}\sin \beta x} \right) = \alpha {e^{\alpha x}}\sin \beta x + \beta {e^{\alpha x}}\cos \beta x \cr & \frac{d}{{dx}}\left( {{e^{\alpha x}}\cos \beta x} \right) = \alpha {e^{\alpha x}}\cos \beta x - \beta {e^{\alpha x}}\sin \beta x \cr}

so that any integral of the form

$$\int e^{\alpha x}\cdot \sin{\beta x}\cdot dx$$

is a linear combination of the former functions. Let's then find $c_1$ and $c_2$ such that

$$\frac{d}{{dx}}\left( {{c_1}{e^{\alpha x}}\sin \beta x + {c_2}{e^{\alpha x}}\cos \beta x} \right) = {e^{\alpha x}}\sin \beta x$$

$${c_1}\alpha {e^{\alpha x}}\sin \beta x + {c_1}\beta {e^{\alpha x}}\cos \beta x + {c_2}\alpha {e^{\alpha x}}\cos \beta x - {c_2}\beta {e^{\alpha x}}\sin \beta x = {e^{\alpha x}}\sin \beta x$$

This means we need

\eqalign{ & {c_1}\alpha - {c_2}\beta = 1 \cr & {c_1}\beta + {c_2}\alpha = 0 \cr}

This will yield with little work

\eqalign{ & {c_1} = \frac{\alpha }{{{\alpha ^2} + {\beta ^2}}} \cr & {c_2} = - \frac{\beta }{{{\alpha ^2} + {\beta ^2}}} \cr}

which means that, in general:

$$\int {{e^{\alpha x}}} \cdot\sin \beta x\cdot dx = {e^{\alpha x}}\frac{{\alpha \sin \beta x - \beta \cos \beta x}}{{{\alpha ^2} + {\beta ^2}}} + C$$

Analogously, you will get that

$$\int {{e^{\alpha x}}} \cdot\cos \beta x\cdot dx = {e^{\alpha x}}\frac{{\alpha \cos \beta x + \beta \sin \beta x}}{{{\alpha ^2} + {\beta ^2}}} + C$$

This has no formal name, it is just a formula for finding a primitive.

-

HINT

Generally, if we denote $A=\int e^{a x} \cdot \cos (b x) \ dx$ and $B=\int e^{a x} \cdot \sin (b x) \ dx$ and integrate by parts A and B, then we obtain an easy-to-solve system in $A$ and $B$. But this is just another possible approach. However, mjqxxxx's approach is by far a better one.

Chris.

-


text:
Unit Circle Table

The unit circle table lists the coordinates of the points on the unit circle that correspond to common angles. The unit circle demonstrates the output of the trigonometric functions sine and cosine as discussed on this page. The table below shows angles measured using both degrees and radians and can be visualized by this chart.

Note: This website uses the constant (tau) instead of (pi) as the default circle constant. The substitution can be used to translate between the two constants.

Common Angles

Fractions of 8

This table shows the angles and their corresponding points formed from dividing the unit circle into equal parts.

Fractions of 12

This table shows the angles and their corresponding points formed from dividing the unit circle into equal parts.

Explanation

A point on the unit circle that corresponds to the angle (theta) is given by the sine and cosine of the angle as discussed on this page. This concept is visualized in the illustration and equations below.

Lookup Table

Historically, the trigonometric ratios represented by the trigonometric functions cosine and sine would have been looked up in a table. For, example the lookup table below shows the trigonometric ratios for angles in the first quadrant of the circle.

0° 0.000 TAU 0.000 1.000
5° 0.014 TAU 0.087 0.996
10° 0.028 TAU 0.174 0.985
15° 0.042 TAU 0.259 0.966
20° 0.056 TAU 0.342 0.940
25° 0.069 TAU 0.423 0.906
30° 0.083 TAU 0.500 0.866
35° 0.097 TAU 0.574 0.819
40° 0.111 TAU 0.643 0.766
45° 0.125 TAU 0.707 0.707
50° 0.139 TAU 0.766 0.643
55° 0.153 TAU 0.819 0.574
60° 0.167 TAU 0.866 0.500
65° 0.181 TAU 0.906 0.423
70° 0.194 TAU 0.940 0.342
75° 0.208 TAU 0.966 0.259
80° 0.222 TAU 0.985 0.174
85° 0.236 TAU 0.996 0.087
90° 0.250 TAU 1.000 0.000


text:
# QGrain

An open-source and easy-to-use software for the comprehensive analysis of grain-size distributions

0%

The interface of SSU is the most complex between these modules. But don’t be worry, these parameters and panels are designed to make you have the ability to handle the most difficult samples, and it’s no need to do tedious adjusting work for all samples.

Click the Load Dataset button to load the grain-size distributions. Click here to see the tutorial on loading the dataset. After loading, you can see the number of samples ($N_{samples}$) has been displayed. Also, the name of the first sample will be displayed. You can adjust the Sample Index to switch the current sample to be fitted.

You can select the SSU Resolver, one option is the neural SSU, another is the classic SSU. The neural SSU is similar to NNEMMA. Usually, the classic SSU has better performance, but the fitting process of neural SSU is normal, this makes that the neural SSU is more suitable for the analysis of the natures of the distance function. You can click the Configure Fitting Algorithm button to open the dialog to adjust the algorithm settings. The dialogs for the neural and classic SSU algorithms are different. Click the Distribution Type combo box to select an appropriate elementary distribution, and adjust the $N_{components}$ to an appropriate value.

Click the Single Test button to try fitting with default settings. You can see the fitting result in the chart below, and the result will be added to the right Result table. You can select the result in the table and right-click to pop up the menu. In this menu, you can display the variations of distance function, the distribution chart, and the animation of the fitting process.

If it did not find the proper components (end-members), you can adjust the parameters and retry it. Increase the precision and enable the global optimization (GO) process will help it to find the components. If your sample has more than 3 components and the proportions of some components are very low, it’s recommended to click the Manual Test to fitting the current sample manually. At the manual fitting panel, you can give some suggestions to each component. No need to be very accurate, for most conditions, you just need to move the mean values to the approximate locations. And click the Try button to fit the sample with your suggestions. If you are satisfied with this result, you can click the Confirm button to save this result and close this panel. If the result is not good, set the parameters more accurately and consider that whether the $N_{components}$ is enough.

If it could find the proper components, you can click the Test Next button to fit the next sample. Or you can click the Continuous Test button to fit all samples in order.

If your dataset is very complex, for example, the $N_{components}$ is not constant. You should use the reference mechanism to reduce the workload. You can fit some typical samples manually, and select these correct results, and right-click to mark them as references. And then, you can click the Continuous Test button to fit all samples automatically. The algorithm will calculate the similarities between the current sample and references, and use the most similar result as a reference to fit the current sample. Note: if there is only one marked result in the Reference table, all samples will use this result as the reference.

If all samples have been fitted, you can right-click in the Result table to select the actions to detect the outliers. If there are some outliers, you can discard or retry them.

Finally, you can select the Save Excel action to save the results into an Excel file. It’s recommended to also save the results into a binary dump file because the dump file could record the fitting history and could be checked later.


text:
# How do you evaluate  11 - 4 * (-7) ?

Jun 7, 2015

$11 - 4 \cdot - 7 = 39$

Multiplication comes before subtraction.

$11 - 4 \cdot - 7$

Multiply $4 \cdot - 7$ to get $- 28$.

Subtract $- 28$ from $11$.

$11 - \left(- 28\right)$ =

$11 + 28 = 39$


text:
# Typs of alloys

• What are alloys and why are they needed?
• What are binary alloy systems?
• What characterizes a solid solution alloy?
• What types of solid solutions can be distinguished?
• What is a complete solid solution series?
• What characterizes a mixture of pure crystals?

## Introduction

In many technical fields, high demands are placed on the materials used, for example in aeronautical engineering. In some combustion chambers the temperature can exceed 2000 °C. The materials must therefore not only withstand high mechanical loads but also thermal stresses.

Simple metals often do not meet these requirements. For this reason, several metals are usually melted together in order to obtain completely new properties after solidification. Such mixtures of two or more metals are also known as alloys. On a chemical level, alloys are therefore characterized by their metallic bonds.

Mixtures of substances with metallic character are called alloys!

In order to specifically influence the desired material properties, profound knowledge of alloying is required. The basics of alloying will therefore be explained in this article. Due to their complexity, only alloys which are consist of two components are considered (also referred to as binary systems). The total of the possible mixture concentrations of a two-component alloy is also referred to as alloying system.

Binary systems are alloy systems consisting of two components!

In general, alloys are obtained by melting, mixturing and subsequent solidification. For this purpose, a certain amount of an alloying element B (solute) is added to a host material A (solvent) in the liquid state. In the liquid state, the atoms of the substances involved are only weakly bound to each other. In general, the substances can therefore be mixed relatively well.

During solidification, this solubility can then either be retained completely (complete solid solution) or be completely lost (mixture of pure crystals). Partial solubilities of the substances can also occur during solidification (mixture of solid solutions).

Depending on the solubility of the two components A and B in the solid state, alloys can thus be divided into three different types.

## Solid solution

If the two components A and B of an alloy remain completely soluble in one another in the solid state, the atoms of the alloying component B are incorporated in the host crystal (host matrix) of the host component A. The atoms A and B then form a common lattice structure. Such a crystal structure of mixed atoms in a common lattice is called solid solution.

A complete solubility of the alloy components in the solid state is referred to as a solid solution!

Figuratively, the components of a solid solution behave like a mixture of water and alcohol, in which the alcohol particles can also be completely dissolved in the water.

In a solid solution, the alloying atoms B can accumulate in the host lattice of the host substance A in two different ways during crystallization. Accordingly, a distinction can be made between substitutional solid solutions and interstitial solid solutions. These are explained in more detail in the corresponding chapters.

### Substitutional solid solution

If the atoms of the alloying element B occupy regular positions in the host lattice of the element A during crystallization, this is referred to as a substitutional solid solution. A comparison between the host lattice before melting down and the common crystal lattice after solidification shows that individual A atoms were simply substituted by B atoms.

In general, the two alloy components have different atom radii and chemical properties. Therefore, in reality, there is a lattice distortion within the crystalls. The lattice distortion increases with the number of substituted atoms and finally leads to the fact that the host atoms cannot be replaced to an unlimited extent by alloying atoms. The solubility of the alloying element in the host material is therefore generally limited (partial solubility of the components in the solid state).

Only under the conditions that the alloying component B, compared to the host component A, has

• the same lattice structure,
• similar atomic radii (differ less than 15 %) and
• similar chemical properties,

the alloying atoms B can occupy regular lattice sites of the host crystal A “unnoticed” over the entire mixing range. Any alloy concentration can ultimately be produced without a so-called miscibility gap by means of such a substitutional solid solution.

A solubility over the entire mixing range is also referred to as a complete solid solution series, i.e. a perfect solubility of the components in the solid state. In principle, the copper-nickel alloy system forms such a complete solid solution series.

An alloy system which shows a complete solubility of the components in the solid state over the entire concentration range is also referred to as a “complete solid solution series”!

The conditions mentioned above to form a solid solution are also called Hume-Rothery rules.

### Interstitial solid solution

If the atoms B of the alloying element are relatively small compared to the atoms A of the host material (maximum diameter ratio 0.4), there is another possibility of atomic arrangement in the lattice. Due to their small size, the alloying atoms B can then also be placed in interstitial lattices sites of the host crystal structure.

Such a mixture of atoms at intermediate lattice sites is also known as an interstitial solid solution. Due to their dominant role, the regular lattice sites are reserved exclusively for the atoms of the host lattice.

Since only the intermediate lattice sites are available in an interstitial solid solution, a complete solubility of the components only occurs within a strongly limited concentration range (usually only a few percent). If alloying is carried out beyond this solubility limit, the “excess” of alloying atoms B will precipitate and form its own crystal in the microstructure. This crystal can in turn partly contain atoms of the host component A. Therefore, only a partial solubility of the components in the solid state is obtained for interstitial solid solutions.

## Mixture of pure crystals

If the conditions for the formation of a solid solution are not met, the alloying atoms B may not be able to occupy regular lattice sites or intermediate lattice sites as well. This is the case when the alloying component B, in comparison to the host component A, has

• another lattice structure or
• has very different chemical properties.

The atoms are then practically displaced from the other lattice structure during solidification and are forced to form their own (“pure”) crystals. Each type of atom then forms its own crystal structure, so that no alloying atoms can be found in the host lattice and no host atoms ca be found in the alloying lattice (complete insolubility of the components in the solid state). The microstructure consists of a mixture of completely different crystallites (grains).

If the components of an alloy are not soluble, each component will form its own crystal structure!

Figuratively, the components of such a alloy behave like a mixture of water and oil, whose components cannot be mixed either – neither the water particles in oil nor the oil particles in water.

Note that just because the atoms of a mixed crystal alloy cannot be mixed in a common lattice structure, this does not mean that the alloy is less stable than a solid solution alloy! Very high interatomic forces also act between the pure crystals of such an alloy to ensure cohesion.

A mixture of pure crystallites will not be found in reality, since the components can always be mixed to a certain degree, even if the solubility is very low. But in the bismuth-cadmium alloy system, for example, the solubility is so low that a complete insolubility of the components can be assumed for simplicity.

## Mixture of solid solutions

Perfect solubility (solid solution) or complete insolubility of the components (crystal mixture) are only special cases. In general, the components are neither completely miscible nor immiscible.

In reality, an alloying component B can always be dissolved to a certain degree in the host component A and vice versa. In general, therefore, a limited solubility of the components in the solid state is always obtained.

Figuratively, partial solubility can be compared with a water-sugar mixture in which the solubility of sugar in water is also limited. The water can only dissolve the sugar in it to a certain extent, the undissolved sugar will eventually settle.

In the lattice structure of the host component A (solvent), B atoms (solute) will also be found to a certain extent. Depending on their chemical properties, the B atoms can either occupy regular lattice sites in the host lattice A or be stored at intermediate lattice sites. It is then either a substitutional solid solution or an interstitial solid solution.

Such a solid solution, which primarily consists of the host element A (in particular of the lattice structure of component A) and contains only small amounts of alloying element B, is also referred to as $$\alpha$$-solid solution (alpha solid solution).

Conversely, at very high concentrations of B atoms, the crystallites consist mainly of the lattice structure of component B (solvent), while small amounts of atoms A (solute) will be deposited therein. In such a case one speaks of a $$\beta$$-solid solution (beta solid solution).


text:
# Intraclass Correlation

-

Intraclass correlation(ICC) refers to the correlation within the groups or classes. It tells us how the members of a group are correlated with respect to a particular characteristic. The correlation between the heights of members of a family is an example of intraclass correlation.

We compare two different variables when calculating the usual correlation coefficient, whereas we use the same characteristic for the two variables when calculating intraclass correlation. For example, when calculating the intraclass correlation between the heights of family members, we take both of the variables X and Y to be equal to the height.

### Intraclass Correlation Coefficient Formula:

Suppose that our data has ‘n’ families/groups each of size ‘k’. Then the intraclass correlation coefficient can be calculated using the formula,

\text{ICC } = \frac{k\sigma_m^2 - \sigma^2}{(k-1)\sigma^2}

where \sigma^2 denotes the variance of all data values and \sigma_m^2 denotes the variance of the means of the families.

### Interpretation of Intraclass Coefficient:

The value of the ICC always lies within the following range:

\frac{-1}{k-1} \leq \text{ICC}\leq 1

So we see the values of the intraclass coeffiecient are skewed to the right. Values closer to +1 indicate higher positive intraclass correlation. If the value of ICC is close to zero then it implies that the data points within a group are independent of each other.

### What is a good intraclass correlation coefficient?

Values of ICC lying between 0.75 and 1 are considered to be good. It indicates that the data has high reliability. If the value of ICC is less than 0.5 then we conclude that our data has less reliability. Thus values less than 0.5 are considered to be bad.

### Calculating ICC using statistical software:

1. We can calculate ICC using R software by using the command icc(ratings, model, type, unit).
2. We calculate the ICC in SPSS by clicking on the following commands: Analyze > Scale > Reliability Analysis > Intraclass Correlation Coefficient.

Hey 👋

I have always been passionate about statistics and mathematics education.

I created this website to explain mathematical and statistical concepts in the simplest possible manner.

If you've found value from reading my content, feel free to support me in even the smallest way you can.

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text:
# How do you find the radius of a circle given its area of a circle is 616cm^2?

Mar 21, 2017

$r = 14 c m$

#### Explanation:

Area of a circle$= \pi {r}^{2}$

$\therefore \pi {r}^{2} = 616 c {m}^{2}$ given

$\therefore {r}^{2} = \frac{616}{\pi}$

$\therefore {r}^{2} = \frac{616}{3.141592654}$

$\therefore {r}^{2} = 196.0788899$

$\therefore r = \sqrt{196.0788899}$

$\therefore r = 14.00281721 c m$

$\therefore r = 14 c m$ to the nearest cm


text:
Extracting information from simultaneous boolean AND, XOR, and NOT

I'm looking to extract some information from a series of equations with AND, XOR and NOT. I've already covered all of the easy parts using various boolean identities, so I'm looking to now determine if there are any non-obvious sources of information.

Right now, I've got this simultaneous equation which can yield extra information where c1 and c2 are known, and a and b are unknown:

a ^ b = c1
a && b = c2

a b c1 c2
0 0 0  0
1 0 1  0
0 1 1  0
1 1 0  1

When both c1 and c2 are false, then a and b must also be false. However, I've been unable to determine any other scenarios, simultaneous or otherwise, which can yield additional information.

What I would like to know is

1. Are there any other ways of extracting information?
2. Can I prove that I have all of it?

Edit: What I'm looking for, exactly, are other simultaneous equations or identities of this, or a similar, form, which may (depending on the values of constants) allow me to prove additional values for unknowns. I have already exhausted the avenues through single-operator equations.

-

1. If $c_2$ is true, then both $a$ and $b$ are true.
2. Otherwise, if $c_1$ is false, then both $a$ and $b$ are false.
3. Otherwise exactly one of $a$ and $b$ is true, but you don't know which.


text:
# How do you simplify -3/4 + 1/3 div-2/5 using PEMDAS?

Mar 9, 2016

No Parentheses, no Exponents, so Division is first.

#### Explanation:

We take the $\frac{1}{3} \div - \frac{2}{5}$ part first.
Dividing by a fraction is multiplying by its inverse:
$\frac{1}{3} \times - \frac{5}{2} = - \frac{5}{6}$

Now the original changes into:
$- \frac{3}{4} - \frac{5}{6}$

For a common denominator:
$- \left(\frac{3}{4} \times \frac{3}{3}\right) - \left(\frac{5}{6} \times \frac{2}{2}\right) = - \frac{9}{12} - \frac{10}{12}$

And this is:
$- \frac{19}{12} = - 1 \frac{7}{12}$


text:
Math Help - Volumes by Cross-Sections

1. Volumes by Cross-Sections

A solid has at its base the region bounded by the parabola x-y^2 = -8 and the left branch of the hyperbola x^2-y^2-4=0. The vertical slices perpendicular to the x-axis are squares. Find the volume of the solid. This is the full problem.

I was able to find out the points of intersection, which were x=-3 and x=4. I am not sure what the cross-sectional area is.

2. Originally Posted by dalbir4444
A solid has at its base the region bounded by the parabola x-y^2 = -8and the left branch of the hyperbola x^2-y^2-4=0. The vertical slices perpendicular to the x-axis are squares. Find the volume of the solid.
Could you please post the problem as it is written and show your attempt at the solution?

3. hi dalbir4444,

the left side of the hyperbola is $x=-\sqrt{y^2+4}$

You've identified the x co-ordinate of the points of intersection.
x=4 is unnecessary, as the parabola intersects the rightside part there.

To calculate the area of the bounded region, you only need the area of the part above the x-axis and multiply by 2.

If you also calculate where the parabola and hyperbola cut the x-axis to the left of the y-axis, then you can integrate under the parabola from x=-8 to -3
and integrate under the hyperbola from x=-3 to x=-2.

That will be one way to calculate the area between the curves.

I can't visualise the object itself however, as your question mentions a square.


text:
## Manipulation of expressions: Iterate over the deeper values in a nested association

Is it possible to iterate over the values of the deepest layer of keys in a nested association? For example:

``````assoc = <| "a"->0, "b" -> <| "c"->"one", "d" -> 2, "e" -> 3 |>, "f" -> 4 |>
``````

Would it be possible to iterate over the values of the deepest level in this example? `{0, "one", 2,3,4}`? Or is it possible to iterate over the deeper level of the keys so that you can test the type for the value of each key?

## Google sheets – Iterate through the range, copy the value of the cell to the single cell reference

I have a sheet where I would like to track inventory transfers and update the inventory master sheet accordingly. I'm a little new in this. I have a range that consists of a column of new values (calculated with formulas) and a column with cell references where I would like to copy those values (on another sheet).

Currently, my code only works for the first row, but it does not repeat itself in the range and I can not understand why. I thought that using an "index" variable would correctly reassign the values for sourceCell and sourceRef, but maybe it's redundant?

Any help would be appreciated.

``````copypaste () {function
var sheet1 = ss.getSheetByName ("FMGXfers");
var sheet2 = ss.getSheetByName ("FMGInventory");
var numRows = sheet1.getRange (& # 39; R1 & # 39;). getDisplayValue ();

index var = 3
var sourceCell = & # 39; T & # 39; + index;
var sourceRef = & # 39; U & # 39; + index;
var targetCell = sheet1.getRange (sourceRef) .getDisplayValue ();
var sourceValue = sheet1.getRange (sourceCell) .getValues ();

for (var i = 1; i <numRows; i ++) {
sheet2.getRange (targetCell) .setValues (sourceValue);
index var = index ++;
var sourceCell = & # 39; T & # 39; + index;
var sourceRef = & # 39; U & # 39; + index;
}
}
``````

## complexity theory – NL – iterate all the edges of a graph in the registry space

There is absolutely no problem to iterate over all the edges in a graph in the log space (even in the deterministic log space!). The details depend on how the graphic is encoded. For example, if the graph is coded as an adjacency matrix, then it can simply go over all the pairs of vertices.

However, this does not give an NL algorithm to click. The problem is that you need to iterate $$k$$-tuples of vertices, not constant. $$k$$. This is something you can not do in the registration space.

NP is sometimes described in terms of witnesses. However, the "official" definitions are using non-deterministic Turing machines, which can have more than one correct movement at any given time. The class of languages accepted by non-deterministic Turing machines do not coincide with the class of languages to which a polynomial-sized token is assigned and can be verified in polynomial time. Unfortunately, there is no such alternative description for non-deterministic record space Turing machines.

## react – How to iterate a URL in Javascript

In the case that it is a text file, it must be taken into account that, within a require (url), someone has some idea?

## python – Iterate over a matrix and save in a dictatorship

Since this matrix:

``````array ([[0, 0, 0, ..., 0, 0, 0],
[0, 0, 0, ..., 0, 0, 0],
[0, 0, 0, ..., 0, 0, 0],
...
[0, 0, 0, ..., 0, 0, 0],
[0, 0, 0, ..., 0, 0, 0],
[0, 0, 0, ..., 0, 0, 0]], dtype = uint8)
``````

I'm wanting to do it to count values on each line and then save them by line.
The following code is doing the general sum. I would like to have the account by line.

``````def count (image):
array = np.array (image)
training[[ array == 0 ]]= 1
training[[ array == 255 ]]= 0
for row in matrix:
unique, counts = np.unique (array[row,] , return_counts = True)
d = dict (zip (unique, counts))
return new
``````

The result:

``````{0: 234710, 1: 515}
``````

## Tensioning objects are only iterable when avid execution is enabled. To iterate over this tensor use tf.map_fn

I created a custom loss function and tried to execute it, which caused me an error.

mse = Custom_loss (y_real, y_pred, df_for_loss, price)

def Custom_loss (y_true, y_pred, df_for_loss, price, sample_weight = None, multioutput = & # 39; uniform_average & # 39;):

``````df_for_loss['y_pred']= y_pred
df_for_loss['agg_closing_stock'] = np.where (df_for_loss.agg_closing_stock> 0, (price / 2) * df_for_loss.agg_closing_stock, df_for_loss.agg_closing_stock)

df_for_loss['agg_closing_stock'] = np.where (df_for_loss.agg_closing_stock == 0, price * df_for_loss.y_pred, df_for_loss.agg_closing_stock)
df_for_loss.agg_closing_stock = df_for_loss.agg_closing_stock.astype (& # 39; float32 & # 39 ;, raise_on_error = False)
y_pred = y_pred.astype (& # 39; float32 & # 39 ;, raise_on_error = False)
data = tf.convert_to_tensor (df_for_loss.agg_closing_stock)
y_pred = K.tf.math.multiply (y_pred, data)

output_errors = np.average ((y_true - y_pred) ** 2, axis = 0,
weights = sample_weight)
if it is an instance (multioutput, string_types):
If multioutput == & # 39; raw_values & # 39;
return output_errors
elif multioutput == & # 39; uniform_average & # 39 ;:
# pass None as weights to np.average: uniform average
multioutput = None

returns np.average (output_errors, weights = multioutput)
``````

## python 3.x – Iterate through groups of rows with different indexed values

The data looks like this:

Data = {& # 39; group_id: [‘1′,’1′,’1′,’1′,’2′,’2′,’2’],
& # 39; Gravity & # 39; [4,2,7,4,8,9,3,5]}

I need:

1) Take the first row of the Severity Code of each group
2) Obtain the absolute value of all the rows (difference) of the identified severity code of each group (from # 1). Example: Group severity code 1 (4) … first row diff = 0; second row diff = 2; third row diff = 3; The same for group 2.
3) In each group, find the closest neighbor of each source, up to the severity of the first row.

I have identified the first row and indexed the severity code. When iterating, the code only uses the last indexed severity code to calculate the difference.

``````df = pd.DataFrame (Data)
first_row = b.groupby (['group_id']).First()
for the row in first1.itertuples (index = True, name = & # 39; Pandas & # 39;):
value = getattr (row, & # 39; Severity & # 39;)
df['dif'] = (df['Severity'] - value) .abs ()
``````

I hope the output is in a Dataframe with a column & # 39; dif & # 39; added I can extract nearest neighbors in each group for each source where True. Repeat the process: extract the rows where True and pass the False to find additional rows with a new severity of the first row. Repeat again until there are no rows, or all rows are False.

## c # – How to iterate efficiently in the data list and search data in another datatable?

I have a list that contains 25000 data and I am iterating in all the data and in each iteration, I am looking for data in another data table. All my routine is taking a long time to finish.

``````PrivateComid UpdateCommentFirst (string strCommentPath, string TickerName)
{
counter of int = 0;
bool QcViewAllFileExist = false;
bool QcCommentFileExist = false;
bool AllowUpdate = false;
string savepath = Convert.ToString (ConfigurationManager.AppSettings["OutputPath"]) .Trim () + TickerName + "\" + TickerName + "_QC-ViewwAll.xml";
DataSet QCCommentstmp = new DataSet ();
DataSet QCViewAlltmp = new DataSet ();

if (File.Exists (strCommentPath))
{
QcCommentFileExist = true;
}

if (File.Exists (savepath))
{
QcViewAllFileExist = true;
}

if (QcCommentFileExist && QcViewAllFileExist)
{
{
{
AllowUpdate = true;
}
}

if (AllowUpdate)
{

.Select (row => new clsCommentPopup
{
// BrokerFor, Formula, LineItem, Section, PeriodCollection
bolFollowUP = (row.Field("FollowUP")) == null? false: Convert.ToBoolean ((row.Field("Follow"))),
bolThisPeriod = (row.Field("ThisPeriod")) == null? false: Convert.ToBoolean ((row.Field("This period"))),
Formula = (row.("Formula")) == null? string.Empty: (row.Field("Formula")),
ModelValue = (row.Field("ModelValue")) == null? string.Empty: (row.Field("ModelValue")),
ExternalComment = (row.Field("ExternalComment")) == null? string.Empty: (row.Field("ExternalComment")),
InternalComment = (row.Field("InternalComment")) == null? string.Empty: (row.Field("InternalComment")),
strEndPeriod = (row.Field("EndPeriod")) == null? string.Empty: (row.Field("EndPeriod")),
strStartPeriod = (row.Field("StartPeriod")) == null? string.Empty: (row.Field("StartPeriod")),
PeriodType = (row.Field("PeriodType")) == null? string.Empty: (row.Field("PeriodType")),
SectionFor = (row.Field("Section")) == null? string.Empty: (row.Field("Section")),
LiFor = (row.Field("LineItem")) == null? string.Empty: (row.Field("Elemento en línea")),
QcPeriodFor = (row.Field("QcPeriod")) == null? string.Empty: (row.Field("QcPeriod")),
BrokerFor = (row.Field("BrokerFor")) == null? string.Empty: (row.Field("BrokerFor")),
PeriodCollection = (row.Field("PeriodCollection")) == null? string.Empty: (row.Field("PeriodCollection")),
boolIgnoreValue = (row.Field("IgnoreValue")) == null? false: Convert.ToBoolean ((row.Field("IgnoreValue"))),
IgnoreData = (! QCCommentstmp.Tables[0].Columns.Contains ("IgnoreData")? string.Empty: (row.Field("IgnoreData") == null? string.Empty: row.Field("IgnoreData")))
}).To list();

if (QCCommentlist! = null)
{
foreach (comment var in QCCommentlist)
{
string section = comment.SectionFor;
string li = comment.LiFor;
string broker = comment.BrokerFor;
string period = comment.PeriodCollection;
string strQCPeriodValue = "";

if (comment.boolIgnoreValue && period.Trim ()! = "")
{
var QcViewColumnName = QCViewAlltmp.Tables[0]. Columns.Cast() .AsParalelo ()
.Where (x => x.ColumnName.Contains (period))
.Select (x => new {x.ColumnName}). FirstOrDefault ();

if (QcViewColumnName! = null)
{
period = QcViewColumnName.ColumnName;

if (period.Trim ()! = "")
{
var datarow = QCViewAlltmp.Tables[0].AsEnumerable (). AsParallel ()
.Where (row => row .field("GroupKey"). Split (& # 39; ~ & # 39;)[0].ToUpper () == section.ToUpper ()
&& row.Field("GroupKey"). Split (& # 39; ~ & # 39;)[1].ToUpper () == li.ToUpper ()
&& row.Field("Section") .ToUpper () == broker.ToUpper ());

if (datarow! = null && datarow.Count ()> 0)
{
strQCPeriodValue = (datarow.FirstOrDefault ()[period] ! = null? datarow.FirstOrDefault ()[period].ToString (): string.Empty);
if (strQCPeriodValue.Trim ()! = string.Empty)
{
comment.IgnoreData = strQCPeriodValue;
counter ++;
}
}
}
}
}
}
}

SerializeQcComment (QCCommentlist);
toolTip1.Hide (this);
}
}
}
``````

Yes ignored field is there in the data set table QCCommentstmp `QCCommentstmp.Tables[0].Columns.Contains ("IgnoreData")` then deserialize the data from the QCCommentstmp table to list them `List QCCommentlist`

iterate in the QCCommentlist data using for looping and find data in QCViewAlltmp.Tables[0] data table for each iteration

when QCCommentlist has 25000 data then I'm iterating through all the 25000 data and finding data in another data table. If data is found, I am updating the data in the list. This process is getting very slow and the code is taking a long time to complete the entire iteration and data search in the data table.

please check my code and tell me how to restructure my code, as a result, there will be an improvement in the speed of code execution.

If my approach is incorrect, guide me with the correct approach and also tell me the relevant code that I can use in my previous code. As a result, my routine will take a minimum time to complete if I repeat more than 25,000 data. Searching for suggestions and better code to achieve the same task.

## javascript: Iterate the HTML table to highlight the differences where the cells have multiple comparison elements

I want to highlight the differences between the first row of a table and all the other rows in the column.

I have discovered how to achieve this when each cell in the table only has 1 element / comparison. But I would like to extend this to multiple comparisons per cell separated by "," s.

Here is the code for the only element per cell. https://jsfiddle.net/t19Lqbkn/
using the following code:

``````    var table = document.getElementById ("mytab1");
for (var i = 1, row; row = table.rows[i]; i ++) {
var matc = table.rows[1]
for (var j = 0, col; col = row.cells[j]; j ++) {
if (col.innerHTML! == matc.cells[j].innerHTML) {col.innerHTML = ""+ col.innerHTML +"";}

}
}
``````

And here is a table with several elements per cell.
https://jsfiddle.net/6c7s9mky/

As you can see in the second link. The first column of the second row, only the element "Eva" should be red, and in the last row of the first column there should be no red text.

## java – Retrieve sum data from the database or iterate at run time

I have an object (Shopping cart) that has a list of CartItems, which contains the related productstrong text and the amount purchased.

``````Public class ShoppingCart extends BaseEntity {

Private list cartItems = new ArrayList <> ();

@DateTimeFormat (pattern = "dd / MM / yyyy hh: MM: ss")
private LocalDateTime dateTime;

@Enumerated
private PaymentMethod paymentMethod = PaymentMethod.CASH;

public class CartItem extends BaseEntity {

@ManyToOne
Private product of the product;

@ManyToOne (fetch = FetchType.LAZY)
@JoinColumn
Private ShoppingCart ShoppingCart;

Public class product extends BaseEntity {

@Not empty
name of the private chain;

Private double price;

int private amount;
``````

I have a view that summarizes all the shopping carts, and I do not know if I should save that amount within the ShoppingCart entity, or iterate to find the sum.
Here is the code, right now I am iterating.

``````@GetMapping ("/")
Public string getProduct (@ModelAttribute ("reportDto") ReportDto reportDto, Model model) {

double total = 0, cash = 0, credit = 0, debit = 0;
int quantity
List results
if (reportDto.getBeginDate () == null || reportDto.getEndDate () == null) {
results = shoppingCartService.findAll ();
} else {
results = shoppingCartService.findByDateTimeBetween (reportDto.getBeginDate (), reportDto.getEndDate ());
}
total = results. current ()
.mapToDouble (shoppingCart -> shoppingCart.getCartItems (). stream ()
.mapToDouble (cartItem -> cartItem.getProduct (). getPrice () * cartItem.getQuantity ()). sum ())
.sum();
quantity = results.stream (). mapToInt (
shoppingCart -> shoppingCart.getCartItems (). stream (). mapToInt (cartItem -> cartItem.getQuantity ()). sum ())
.sum();

cash = results.stream (). filter (shoppingCart -> shoppingCart.getPaymentMethod (). isCash ())
.mapToDouble (shoppingCart -> shoppingCart.getCartItems (). stream ()
.mapToDouble (cartItem -> cartItem.getProduct (). getPrice () * cartItem.getQuantity ()). sum ())
.sum();

credit = results.stream (). filter (shoppingCart -> shoppingCart.getPaymentMethod (). isCredit ())
.mapToDouble (shoppingCart -> shoppingCart.getCartItems (). stream ()
.mapToDouble (cartItem -> cartItem.getProduct (). getPrice () * cartItem.getQuantity ()). sum ())
.sum();

debit = results.stream (). filter (shoppingCart -> shoppingCart.getPaymentMethod (). isDebit ())
.mapToDouble (shoppingCart -> shoppingCart.getCartItems (). stream ()
.mapToDouble (cartItem -> cartItem.getProduct (). getPrice () * cartItem.getQuantity ()). sum ())
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text:
# Tag Info

There are a few issues with the derivation: (not necessarily a mistake, but) please note that if you mean the Coulomb Hamiltonian, the RHS of (1) should involve the reciprocals of $|r_i|$ etc., not $r_i^2$ etc. The Coulomb interaction enters the Hamiltonian in the form of potential energy, rather than in the form of force. Please ignore this comment if you ...


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# WEEK 8 – MATH 11 – GRAPHING TRINOMIALS

In this week we learned how to graph trinomials.

if the graph is Y=X^2
notice how this graph has its axis at (0,0) and how it is going up 1,3,5

if the graph is Y=X^2 +2
The + 2 makes the parabola just go up to spaces so the axis is now (0,2) and still has the same pattern 1,3,5

if the graph is y= (x-2)^2
the -2 means to go to the right 2 so its the opposite of the symbol to which side you go so if it was +2 you would have gone 2 to the left

if the graph is y=2x^2
the 2 in front of the x makes it stretched out upwards because before there is an imaginary 1 in front of the x and now its 2 so it goes from 1,3,5 to 2,6,10 which makes the graph look skinner

# Week 6 – Math 11 – Solving Trinomials

In this week we learned how to balance trinomials, to do this we use this formula,

$X=$ $\frac{-B \sqrt{(b)^{2}-4(a)(c)}}{2(a)}$

so lets say we have this as our equation,
$3X^{2} + 8X + 24=0$

first we must identify A, B and C,
$3$ is A
$8$ is B
$24$ is C

So using these values we must replace them in the spots of the formula with the matching letters,
$X=$ $\frac{-8 \sqrt{(8)^{2}-4(3)(24)}}{2(3)}$

Now we will evaluate the equation,
$X=$ $\frac{-8 \sqrt{64-288}}{6}$
then,
$X=$ $\frac{-8 \sqrt{-224}}{6}$

after you get to this point we must find a common factor to divide everything with,
$X=$ $\frac{-8 \sqrt{-224}}{6}$ $/$ $\frac{2}{2}$

Giving us,
$X=$ $\frac{-4 \sqrt{-56}}{3}$

simplified again,
$X=$ $\frac{-4 -2\sqrt{14}}{3}$

# WEEK 5 – MATH 11 – Simplifying Binomials and Trinomials

This week we learned stuff along the lines of  CDPEU and other stuff to help us simplify binomials and trinomials.

for example,

lets say I have something that looks like this,
$(X)^{2}-16=0$

this binomial would look somehing like this once it is simplyfied
$(X+n)(X-n)=0$

Now, what time what makes 16,
$4$
So it would look like this
$(X+4)(X-4)=0$

$X * X=(X)^{(2)}$
$4 * -4=-16$
$X * 4=4X$
$X * -4=-4X$

once you combine all of these you get,
$(X)^{(2)}-16=0$

# Chemistry 11 – Mole lab – Weigh boat

These are the picture I took of my weigh boat,
I was given 0.150 mol of MgO.

# Week 4 – Math 11 – Simplifying Radicals

To find the simplest form of $(/5+2)_{2}$ we follow these steps…

First, $(/5+2)$ $(/5+2)$ make two of the expressions because it was squared.

then we will start by multiplying each number by the other
$/5 * 2 = 2/5$
$/5 * /5 = /25$

$2 * /5 = 2/5$
$2 * 2 = 4$

once done all this combine like factors

$/25 + 4/5 + 4$

then we will get rid of any hole numbers

$4/5 + 5 + 4$

and we will end up with…

$4/5 + 9$

PS.  I don’t know the code for the square root symbol so I just used (/) as a replacement.

# Week 3 – Math 11 – Radicals

in this week we of math we learned about radicals and to sum up what a radical is it basically turns a negative into a positive.

so |-5| becomes 5

and something like |-8+5| would be |-3| which makes it 3

but if its -|-5| it becomes -5.

# Week 2 – Math 11 – Geometric series

To find a term in a geometric sequence you must use the formula

$t_{n}$ = a $(r)^{(n-1)}$

to give an example here is my pattern  -3, 9, -27, 81…
to find  $t_{10}$ follow these steps….

First, $(r)^{(n-1)}$ becomes $(r)^{(10-1)}$ wich is $(r)^{9}$ leaving you with…

$t_{10}$ = a $(r)^{9}$

Then you want to add in (a), replace (a) with -3 because it’s your first term. making your equation look like…

$t_{10}$ = -3 $(r)^{9}$

once you have used all the given numbers you have. find the common ratio (r). if you follow the pattern of the sequence you will notice it is being multiplied by (-3) each time. so replace (r) with (-3)

$t_{10}$ = -3 $(-3)^{9}$

$(-3)^{9}$ is -19683

-3 x -19683 is 59,049

therefore $t_{10}$ = 59,049

$Part 2$
to find the Geometric series you need to use this formula…

$S_{n}$ = $\frac{a (r^{n}-1)}{r-1}$

using the pattern above we will find the series of  $S_{10}$. first lets add in the values we were given straight of the bat. replace (a) with -3, and replace (n) with 10…

$S_{10}$ = $\frac{-3 (r^{10}-1)}{r-1}$

if you remember, above we found out that the common ratio was -3, so replace (r) with -3…

$S_{10}$ = $\frac{-3 (-3^{10}-1)}{-3-1}$

first, we will do$(-3)^{10}$ wich is,  59,049. then subtract 1 from this giving you, 59,048. now add this in…

$S_{10}$ = $\frac{-3 (59,048)}{-3-1}$

after that multiply 59,048 by -3 giving you, -177,144. now add this to your equation…

$S_{10}$ = $\frac{-177,144}{-3-1}$

now we must solve the denominator, -3-1 = -4.  your equation should look like this…

$S_{10}$ = $\frac{-177,144}{-4}$

now divide these…

$S_{10}$ = 44,286

# Mutation Story

I’m the FG Syndrome. I infected my male carrier  by the X-linked recessive pattern. I change his X chromosome and ive caused many mutations in the MED12. I’ve not only effected that but I’ve also changed his intelligence and behavior. I effect intelligence disability’s just like my siblings, witch can vary from minor to severe. in my case I’m near the minor side. my host is now very hyper, and cant really focus, he’s very social but sometimes struggles with verbal communication and would rather just play. his muscles are week, he has large thumbs, large fists, and toes, he spends a long time in the bathroom and half the time is in pain due to constipation, his face is a little warped. with a abnormal foreheads, small ears and some wide set eyes. we first met when he was 6 but his parents didn’t notice till he was about 7 on a annual doctors appointment. after this he was put in a special class. he couldn’t focus at school and his relation ship at home changed. I was caused by a simple recessive gene in  the X-chromosome.

to research the fg syndrome I looked what the sg syndrome was/ is

to finish this project I asked how to format it because I was very confused. luckily I bookmarked these websites so could see them later.

I checked four websites but ended up only using two

I compared it to others to see witch had the same or similar information

I could have published this  mush earlier

https://ghr.nlm.nih.gov/condition/fg-syndrome#genes

# Engineering Brightness, First look

Our project was 3D printed over the weekend un supervised. over the weekend it printed and failed. with out any supports and a very odd shape it toppled over. we will be reprinting it with supports soon.

We will keep you posted on new updates

# First Peoples Principals

First peoples principles  are what we are doing in engineering brightness and the SSEP project.

The first persons principal is, learning is embedded in memory, history and story. because there correct memory is all threw ought history, because over the ages we have learned these things by what they did back then. and what they wrote down or passed down threw story’s to the next generations till they were able to spread this knowledge to others. the next one that I thought was a big one to is, learning involves practice and time. this is a very big one because it dose take time, when I am studying for a test I practice and study what I’m learning and repeat that over and over again taking hours of my time but drilling this information in to my brain. so that when comes test time ill be ready. the final one in my three would have to  be, learning involves generational roles and responsibility’s. this is the one I chose because just as I said in the first principal these story’s and knowledge were passed down for generations until there was a way to give this knowledge to the world. passing down this information could be as simple as telling your son a story your parents told you or singing your daughter to sleep with your family’s lullaby. were would we be if no one were able to or even bothered to pass down this information, would we be a hole generation behind in technology?

I use the first peoples principals a little bit, not all the time but a little bit. if everyone used these principals we might have a sturdier foundation for the future generations to build there life off of, like less drugs and under aged drinking to help our brains stay as full of knowledge as possible. what we are doing with engineering brightness is not only getting us more practice but helping those in need of light to benefiting the bot of us. one example of how I am using the first peoples principals to work, would be in our engineering brightness project. we are using our recourses and knowledge to help those in need.


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# What does the geodesic effect amount to for the axis of the Earth in its orbit...

###### Question:

What does the geodesic effect amount to for the axis of the Earth in its orbit round the Sun? Take Mo = 2 x 1030 kg,r = 1.5 x 1011cm, (G = 6.67 x 10-11 N mềkg-1, and c-3 x 108m s-1)

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# Math Help - Find Equation of Tangent Line

1. ## Find Equation of Tangent Line

Find the equation of a tangent line at the given value (in form Ax+Bx+C=0)

$y^3+xy-y=8x^4$ ; x=1

2. first you have to find the slope of the function by finding the derivative which with this function you have to do implictly then plugging in 1and whatever why value we get by plugging 1 into the original function

so
$3y^{2}\frac{dy}{dx}+(y+x\frac{dy}{dx})-\frac{dy}{dx}=32x^3$

now group our dy/dx terms to the left and non dy\dx terms to the right

$3y^2\frac{dy}{dx}+x\frac{dy}{dx}-\frac{dy}{dx}=32x^3-y$

factor out dy/dx

$\frac{dy}{dx}(3y^2+x-1)=32x^3-y$

$\frac{dy}{dx}=\frac{32x^3-y}{3y^2+x-1}$

now plug in your x and y values to get the slope and then use point slope form to get the linear equation for the tangent line

3. I plug in the x=1 and y=o but the slope i get isn't what the answer should be. The answer in the back of the book says it should be $y=\frac{5}{2}x-\frac{1}{2}$ The math you did though is what i got too. Can anyone find out how to get this answer?

Thanks!

4. Still stuck on this, its probably an addition error but anyone who can help?

5. Originally Posted by Jim Marnell
I plug in the x=1 and y=o but the slope i get isn't what the answer should be. The answer in the back of the book says it should be $y=\frac{5}{2}x-\frac{1}{2}$ The math you did though is what i got too. Can anyone find out how to get this answer?

Thanks!
Why are you taking y= 0? At x= 1, your equation. $y^3+xy-y=8x^4$ becomes $y^3+y- y= y^3= 2$. y= 0 does not satisfy that.

6. Originally Posted by HallsofIvy
Why are you taking y= 0? At x= 1, your equation. $y^3+xy-y=8x^4$ becomes $y^3+y- y= y^3= 2$. y= 0 does not satisfy that.
oo ok, so i had 2 plug x=1 into the original equation to get y. But if i plug 1 into 8x^4 wouldnt i get 8 so it would be y^3=8 which gives us y=2


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##### A light year, the distance light travels in 1 year, is approximately 5.9×1012 miles. A nearby galaxy

label Algebra
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5

A light year, the distance light travels in

1 year, is approximately 5.9×1012 miles. A nearby galaxy is approximately 8.5×108 light years from our galaxy. Find the distance in miles between our galaxy and the nearby galaxy.
May 17th, 2015

I am sorry but I cannot read the distance between the galaxies.  However, I will solve the questions as best as I can without.  If you would like me to do it with the correct number, please provide the distance to the nearby galaxy in light years.

Let's assume that the distance to is 200 light years (I have no idea if this is even close, but you can replace the correct number in the formula)

So we start with that number and multiply by the number of miles in a lightyear

200 light years * 5.9x10^12 miles / light year = 1.18x10^15 miles.

May 18th, 2015

...
May 17th, 2015
...
May 17th, 2015
Oct 21st, 2017
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# process of glass extraction

## Extraction by De Dietrich Process Systems

Borosilicate glass 3.3 is an ideal material for extraction equipment as the process can be optimized while visually observing the process. Batch mode - Mixers/Reactors Using a mixer in batch mode offers a huge flexibility to optimize the mass transfer by varying multiple aspects of the process, including ratio of the liquid phases, the type of ...

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## Synthesis of high purity silica from low cost water glass

Mar 29, 2017 Figure 1 shows experimental procedure for synthesis of silica via sol–gel process and purification through Soxhlet extraction. We first prepared water glass (Sigma Aldrich) with SiO 2 of 26.5 wt% and diluted HNO 3 (Sigma Aldrich) with DI water (Daejung Chemicals and Metals, Korea) to 0.3 N. Then, the water glass was added into the diluted HNO 3 solution.

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## Start of the espresso extraction process from bottomless

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## Silica Sand Manufacturing Process Silica Sand Extraction

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## Raw Materials Used in Glass Manufacturing Process

It is a process of slowly cooling hot glass objects after it was formed. It is the process to relieve residual internal stresses introduced during manufacture. In glass manufacturing process, quality is checked before packaging process. The final glass product is rejected if the …

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## Essential Oils from Steam Distillation

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## DMT extraction using lye sodium hydroxide and naphtha

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## Extraction of Pd by liquid tin from molten borosilicate

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## The middle of the espresso extraction process from

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## Bead Beating an overview ScienceDirect Topics

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## 11.13 Glass Fiber Manufacturing

11.13 Glass Fiber Manufacturing 11.13.1 General1-4 Glass fiber manufacturing is the high-temperature conversion of various raw materials (predominantly borosilicates) into a homogeneous melt, followed by the fabrication of this melt into glass fibers. The 2 basic types of glass fiber products, textile and wool, are manufactured by similar ...

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## DNA Extraction SERC

Jan 15, 2021 This bead beater is used in the breaking apart or lysing of cells in the early steps of extraction in order to make the DNA accessible. Glass beads are added to an eppendorph tube containing a sample of interest and the bead beater vigorously vibrates the solution causing the glass beads to physically break apart the cells.

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Using Maxos glass in a standard process view will severely compromise the safety factor as the PTFE gasket and nut do NOT reduce the unsupported diameter sufficiently to meet our Safety Factor values. Our full line of sight glass process views can be found here. Extraction Disclaimer:

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## Thermal Solutions for Separation De Dietrich Process

We use QVF borosilicate glass 3.3, De Dietrich glass-lined steel, SiC, PTFE, PVDF, and corrosion-resistant metals such as Tantalum, Zirconium and Titanium. The droplet density is determined by the process so that for the a given column diameter different types of liquid distributors have to be realized.

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## Extraction Columns Pfaudler

Extraction Columns are used for liquid-liquid extraction, which is an important separation technology in the chemical and pharmaceutical industry. The fields of application are the separation of azeotropic mixtures, of thermosensitive products, of mixtures with very high or very low boiling temperatures, where there are low amounts of target ...

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## Fibre extraction SlideShare

Apr 06, 2018 Fibre products like jute bags are no hidden names today.Fibre extraction process has brought ease in modern world. Prominent fibre extraction includes those of jute hemp, cotton & sugarcane.The presentation describes the various unit operations involved in processing ,recycling process & Some facts about it .

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## The Basics RNA Isolation Thermo Fisher Scientific US

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## De Dietrich Process Systems Chemical Process Solutions

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## Efficient and low energy mechanochemical extraction of

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## DNA Purification DNA Extraction Methods Promega

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## EP2678279A1 Process for melting and refining soda lime

A process for making soda-lime glass includes calcinating calcium carbonate in solid phase and at elevated temperature to form calcium oxide and release gases such as carbon dioxide. Sodium silicate glass is formed separately in liquid phase while releasing gaseous reaction products. The calcium oxide and the sodium silicate glass intermediate products are mixed in liquid phase to form a soda ...

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## The making of glass fiber CompositesWorld

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## Preparation of Sodium Silicate Solutions and Silica

Jun 20, 2019 The second process is a direct alkaline leaching (DAL) process, which consists of a single-stage elemental extraction from ash using NaOH, i.e. without the preceding acid leaching step used in SAAL. The two processes generated Na2SiO3 solutions with identical pH (11.8), similar silicon (10.2–10.3 g/L), iron (ca. 200 mg/L) and potassium (ca ...

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## Coffee Powder Extraction Process Methods Techniques

Mar 25, 2019 The extraction process of coffee: Raw Coffee Beans. Harvest maturity: Coffee berries or fruit matures from May to October, depending on the climatic zone & the cultivars. Coffee berries are mature when they are red and soft. The coffee berries are complete 8 to 9 months after the plant flowers. Harvesting methods for coffee extraction process:

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Internal angle

Geometry Level 3

Let $$n(x)$$ be the probability that a polygon with $$x$$ sides has at least one internal angle equal to $$180^{\circ}$$. For which of the following shapes is $$n(x)$$ a positive real number?

×


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# showing a sequence of random variables is a martingale

Let $(X_n)$ be a sequence of IID random variables with $P(X_i = 1) = P(X_i = -1) = 1/2$. Let $(\mathcal{F}_n)$ be the natural filtration. Define $(S_n)$ by $S_0 = 0$ and $S_n = S_{n-1} + X_N \sqrt{1+S_{n-1}^2}$ if $n \geq 1$.

I am trying to show that $S_n$ is a martingale:

$E(S_n | \mathcal{F}_{n-1}) = E(S_{n-1} | \mathcal{F}_{n-1}) + E(X_n\sqrt{1+S_{n-1}^2} | \mathcal{F}_{n-1})$ now, I don't know what to do, as we don't know anything about $S_n$.

Start from your equation, we could have $E(S_n |F_{n-1})$ = $E(S_{n-1} |F_{n-1})$ + $1/2 \sqrt{1 + S_{n-1}^2} - 1/2 \sqrt{1 + S_{n-1}^2}$.
This is because of the probability distribution of $X_i$.
Then we have $E(S_n |F_{n-1})$ = $E(S_{n-1} |F_{n-1}) = S_{n-1}$ which is a martingale.


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occasional meanderings in physics' brave new world

Name:
Location: New Zealand

Marni D. Sheppeard

## Wednesday, January 23, 2008

### Associativity

It is easy enough to invent associative algebraic structures that quickly lead to associated non-associative structures. Unfortunately, ordinary numbers are almost always used as models in the sense that addition is assumed to be commutative. Let us consider a system with two binary operations, by convention called addition and multiplication, neither of which is commutative. Let us assume that scalars are associative and distributive, so that

$x(y + z) = xy + xz \neq xz + xy = x(z + y)$

What happens with $2 \times 2$ matrices over these scalars? Matrix multiplication is well defined by the usual rule, but one must be careful about ordering scalars. In a triple product of matrices $ABC$, associativity is lost, because the first element of the product $A(BC)$ is given by

$A_{11} B_{11} C_{11} + A_{11} B_{12} C_{21} + A_{12} B_{21} C_{11} + A_{12} B_{22} C_{21}$

which is distinct from

$A_{11} B_{11} C_{11} + A_{12} B_{21} C_{11} + A_{11} B_{12} C_{21} + A_{12} B_{22} C_{21}$

in $(AB)C$ by non-associativity of addition. Commutativity of addition would restore associativity for all matrices.

phil said...

You said: "It is easy enough to invent associative algebraic structures that quickly lead to associated non-associative structures", however you have not actually constructed one and I am not convinced that it is "easy" to do so. I.e. can you construct an algebra whose matrix maultiplication is non-associative but whose addition and multiplication is assocative and distributive and prove it?

The reason I think this is difficult is that
you can show associativity for matrix multiplication with only minimal additional rules. E.g. if you have a cancellation law for addition

$x + y = x + z => y = z$ and
$y + x = z + x => y = z$

then use

$(a + b)(c + d) = (a + b)c + (a + b)d = ac + bc + ad + bd = a(c + d) + b(c + d) = ac + ad + bc + bd$

therefore
$ac + bc + ad + bd = ac + ad + bc + bd$
now using the cancellation law on each side
$bc + ad = ad + bc$

so addition is then always commutative for composite numbers which is enough to show that your two expressions for the element of the product are equal and matrix multiplication is associative.

(I used the right distributive law even though you only stated the left distributive law, but you had used the right distributive law in your matrix multiplication so I think you meant to include it)

January 23, 2008 11:18 PM
Kea said...

Thanks, Phil! But cancellation laws only apply to monic (or epic) arrows in a category and do not in general hold, so I might want to ditch them.

January 24, 2008 9:21 AM
kneemo said...

It is easy enough to invent associative algebraic structures that quickly lead to associated non-associative structures

I think Kea had the Jordan algebra of (complex) n x n Hermitian matrices in mind. The Jordan product in this case is given by:
A o B = 1/2(AB+BA), for arbitrary Hermitian matrices A, B and "AB" and "BA" denoting ordinary matrix multiplication. Playing with this Jordan product you'll see it is nonassociative, while the full algebra of n x n complex matrices and the complex numbers are associative and distributive.

January 25, 2008 4:07 PM


text:
# Approximately how much x-ray to gamma radiation penetrates the earth's current atmosphere?

I am a science fiction writer with an undergrad level of understanding in this particular area. I am trying to figure out how the changes in the atmosphere will affect the higher end of the radiation spectrum. I know that as of now the amount of x-ray to gamma radiation that the atmosphere allows through is minute, but as the damage to the atmosphere continues that amount is surely going to change. I am particularly interested in specific numbers related to damage to the human body. I know the amount of radiation that is harmful to fatal to humans but I am looking at a future situation.(I know there are equations out there to calculate electron volts to rad or sievert but the math is a bit beyond me.{I also know that electron volts is a measurement of energy where sievert or rad is a measurement of energy per mass}). The current level of radiation on the higher end of the spectrum that penetrates the atmosphere would give me a good starting point, but the more information the better.

Basically, would it be possible for the atmosphere to decay to the amount that the sun's gamma rays would harm humans, or would the environmental damage decimate life before that could even occur?

(Most of this is completely theoretical, but if you could give specific current numbers that would be very helpful)

P.S. If I am completely off base with my understanding of this topic please be kind in your corrections; I am totally open to constructive criticism and I am ultimately here for knowledge.

• Look at the projections of what the atmosphere will be composed of in say 100 years, if we continue with fossil fuels. Then that will tell you how much CO and CO2, etc there will be. Then see how the radiation interacts with the individual atmospheric constituents. But this is really a question for Earth ScienceSE or similar, in my opinion. Gamma radiation from the sun nasa.gov/feature/goddard/2017/… – user154420 Jun 19 '17 at 13:13
• It's not clear to me whether you specifically intend to ask for gamma rays and X-rays (that is, photonic radiation) or whether you are generally interested in cosmic radiation that might pose health risks for future humans. The answers might come out vastly different, because as far as I know, gamma and X-ray radiation are not really considered as major health threats as far as radiation from space is considered. – carsten Jun 19 '17 at 13:16
• Most of the gamma rays hitting us right now are actually secondary products of massive particles (such as electrons and protons) creating showers in the atmosphere. Hence, losing our atmosphere might actually reduce gamma and X-ray radiation, but increase radiation from other sources. – carsten Jun 19 '17 at 13:16
• What "damage to the atmosphere", and how would it relate to absorption in the x-ray and gamma regimes? Similarly, if you're thinking of the Sun as the source for this radiation, you should make that very explicit (since, as explained in the answers, the fraction of gamma radiation on the surface that comes directly from the Sun is somewhere between small and negligible). – Emilio Pisanty Jun 19 '17 at 17:27

You can get a pretty good idea of the answer by looking at the radiation exposure that people get when flying on transcontinental planes: they might be at an altitude of 10,000 to 12,000 m, and thus they are experiencing significantly less shielding than on earth (at 10 km, the pressure is about 26% of the pressure at ground level).

It is estimated that flying from New York to LA exposes you to about 40 uSv (see this beautiful chart) . Assuming the flight takes 6 hours, that would mean that an entire year at that altitude gives you a dose of 58 mSv - approximately the dose permitted for "US radiation workers" (that is, considered to be low risk of causing problems).

The problem is much worse near the poles, because charged cosmic ray particles (mostly from the Sun) are very effectively shielded by the Earth's magnetic field - except that they can "thread their way to Earth" along the magnetic field lines. This is of course the origin of the Northern lights.

By the time the atmosphere is so depleted that the atmospheric attenuation is a problem, people will choke from lack of oxygen before dying of radiation exposure: but if the magnetic field should fail, we're all going to be in trouble because of the much higher radiation levels that would penetrate (charged particles - which is not what your question was about).

To calibrate you - the solar wind at 1 AU is about $3\times 10^8$ particles/cm$^2$/s - mostly protons. The energy is roughly 1.5 erg/cm$^2$/sec (same source, table 2) or 1.5 mJ/m$^2$/s. The effect of this is nicely summarized in this article that shows that 6 months in the ISS gives you about 75 mSv of radiation dose - or 150 mSv per year. That's starting to be a little high... and that is in the protective capsule of the ISS which will stop quite a few of the incident rays.

In short - if you want a good story, make the earth's magnetic field fail...

• The altitudinal variation of exposure may be what the OP was looking for, but has essentially nothing to due with solar winds. The cosmic rays that generate the muons (which make tertiary particles) that make it into the troposphere are much higher in energy than the vast bulk of the solar wind (I mean, the document you linked discusses the speed of the solar protons in kilometers per second—it's not even properly relativistic). – dmckee Jun 19 '17 at 17:03

The cosmic radiation that penetrates to the troposphere is secondary to high energy particle interactions with the upper layers of the atmosphere. That means that some highly energetic particle hits the upper atmosphere and generates a spray of particles. Most of those are very short lived and decay away in a few meters at most, but the muons (with their low mass, long lifetime, and relatively low interaction cross-section) are able to travel for kilometers or even tens of kilometers before decaying or undergoing another energetic interaction, so a non-trivial fraction of these make it to the ground or spawn tertiary radiation that does.

However, the particles that initiate these events must be orders of magnitude more energetic that either the direct ionizing photons from the sun or the solar wind.1 They originate in energetic astrophysical events elsewhere in the galaxy or even from extra-galactic sources.

The threat to the habitability of the planet from solar activity doesn't come from direct or secondary exposure of the biosphere to ionization, it comes from the ability of solar sources to remove atoms and molecules in the upper-most regions of the atmosphere from Earth orbit—from dribbling the atmosphere away into inter-planetary space. Earth escape velocity only corresponds to about $9 \,\mathrm{eV}$ of kinetic energy for a nitrogen atom which is well within the range that can be supplied by solar activity. Human life on the surface will require atmospheric support before direct ionization from solar sources is a threat.

1 The mean energy of secondary muons making it to sea level is around $1\,\mathrm{GeV}/c^2$ meaning that they have a Lorentz factor of 7–8: they are highly relativistic. Solar wind protons are not relativistic (meaning their kinetic energy is much less than $1\,\mathrm{GeV}/c^2$, and the energy of ionizing solar photons are typically measured in $\mathrm{keV}/c^2$).