text:
Artifact Content

Artifact 591a13879c8b022f010956b6d57027e160ac1a7b:

Manifest of check-in [591a13879c] - Provide a build script for the dbus extension, making it available as an optional package that can be included when building a tclkit. by schelte on 2013-08-18 10:20:37.
```C Provide\sa\sbuild\sscript\sfor\sthe\sdbus\sextension,\smaking\sit\savailable\sas\san\noptional\spackage\sthat\scan\sbe\sincluded\swhen\sbuilding\sa\stclkit.
D 2013-08-18T10:20:37.093
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P 98b83d1ed8c4f6173d5c295bfe93b0e1be795f67
R 99ee26c2b8806c50ed1e2016da8048b7
U schelte
Z 588e3f44f1ab2b0fc286d39b936208ae

```


text:
# 【bzoj1689】[Usaco2005 Mar]Checking an Alibi 不在场的证明

## 1681: [Usaco2005 Mar]Checking an Alibi 不在场的证明

Time Limit: 5 Sec   Memory Limit: 64 MB
Submit: 338   Solved: 220
[ Submit][ Status][ Discuss]

## Description

A crime has been comitted: a load of grain has been taken from the barn by one of FJ's cows. FJ is trying to determine which of his C (1 <= C <= 100) cows is the culprit. Fortunately, a passing satellite took an image of his farm M (1 <= M <= 70000) seconds before the crime took place, giving the location of all of the cows. He wants to know which cows had time to get to the barn to steal the grain. Farmer John's farm comprises F (1 <= F <= 500) fields numbered 1..F and connected by P (1 <= P <= 1,000) bidirectional paths whose traversal time is in the range 1..70000 seconds (cows walk very slowly). Field 1 contains the barn. It takes no time to travel within a field (switch paths). Given the layout of Farmer John's farm and the location of each cow when the satellite flew over, determine set of cows who could be guilty. NOTE: Do not declare a variable named exactly 'time'. This will reference the system call and never give you the results you really want.

## Input

* Line 1: Four space-separated integers: F, P, C, and M
* Lines 2..P+1: Three space-separated integers describing a path: F1, F2, and T. The path connects F
1 and F2 and requires T seconds to traverse.
* Lines P+2..P+C+1: One integer per line, the location of a cow. The first line gives the field numb
er of cow 1, the second of cow 2, etc.

## Output

* Line 1: A single integer N, the number of cows that could be guilty of the crime.
* Lines 2..N+1: A single cow number on each line that is one of the cows that could be guilty of the
crime. The list must be in ascending order.

7 6 5 8
1 4 2
1 2 1
2 3 6
3 5 5
5 4 6
1 7 9
1
4
5
3
7

4
1
2
3
4

## Source

[ Submit][ Status][ Discuss]

#include<cstring>
#include<vector>
#include<cstdio>
using namespace std;

struct data
{
int to,w;
};

const int maxn = 2000;
const int maxn_q = 2e5;
vector<data> edge[maxn];
int q[maxn_q],exist[maxn],dis[maxn],ans[maxn],n,m,k,c,tot;

void spfa()
{
int h = 0,t = 1;
for (int i = 1; i <= n; i++) dis[i] = 70001;
dis[1] = 0;
q[t] = 1;
exist[1] = 1;
while (h < t)
{
int u = q[++h];
exist[u] = 0;
for (int i = 0; i < edge[u].size(); i++)
{
data& e = edge[u][i];
if (dis[u] + e.w < dis[e.to])
{
dis[e.to] = dis[u] + e.w;
if (!exist[e.to])
{
exist[e.to] = 1;
q[++t] = e.to;
}
}
}
}
}

int main()
{
scanf("%d%d%d%d",&n,&m,&c,&k);
for (int i = 1; i <= m; i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
edge[u].push_back((data){v,w});
edge[v].push_back((data){u,w});
}
spfa();
for (int i = 1; i <= c; i++)
{
int cow;
scanf("%d",&cow);
if (dis[cow] <= k)
ans[++tot] = i;
}
printf("%d\n",tot);
for (int i = 1; i <= tot; i++)
printf("%d\n",ans[i]);
return 0;
}

09-14 32
07-19 126
09-07 69
08-23 2768
06-15 1万+

format: plain


text:
# Difficulty with the Laplacian

1. Jun 23, 2015

### erb12c

1. The problem statement, all variables and given/known data
Given: |r|=√(x^2+y^2+z^2) r=xi+yj+zk

(i)Find the partial derivative with respect to x of |r|.
(ii) Find the Laplacian of |r|.

2. Relevant equations

3. The attempt at a solution
For (i) I got x/|r|
but then for (ii) I got 2/r which I don't think is correct

2. Jun 23, 2015

### Zondrina

If $| \vec r(x, y, z) | = \sqrt{x^2 + y^2 + z^2}$, then:

$$| \vec r(x, y, z) |_x = \frac{\partial}{\partial x} (x^2 + y^2 + z^2)^{\frac{1}{2}} = \frac{1}{2} (x^2 + y^2 + z^2)^{- \frac{1}{2}} \cdot \frac{\partial}{\partial x} (x^2 + y^2 + z^2)$$

What is the definition of the Laplacian?

3. Jun 23, 2015

### RUber

Part i) is the warm up for part ii).
What did you do to get x/|r|?

If $\frac{\partial}{\partial x } |r| = \frac{x}{|r|}$, then what is $\frac{\partial}{\partial x } \frac{x}{|r|}$?

I think 2/|r| is right.

4. Jun 23, 2015

### Zondrina

If you clean up the computation in the second post:

$$\frac{1}{2} (x^2 + y^2 + z^2)^{- \frac{1}{2}} \cdot \frac{\partial}{\partial x} (x^2 + y^2 + z^2) = \frac{x}{\sqrt{x^2 + y^2 + z^2}} = \frac{x}{| \vec r |}$$

It is, but it would be nice if the OP showed some of the work.


text:
# How is 2.4 million written out in standard notation?

2.4 X ${10}^{5}$ ... check below

#### Explanation:

A million is equals to ${10}^{6}$
You can write 2.4 by = 24$\div$ 10
Thus

Multiplying them, comes 2 400 000

U can now divide it by 1 000 000 so that the result is

2.4 X ${10}^{5}$


text:
# Simple diff.eq

d/dx(y*x^2)=x^2/(1+x^2)

I guess I can't write this as:
dyx^2=x^2dx/(1+x^2)
dy=dx/(1+x^2)
beacuse I don't get the right answer...

So what do I do?

Tide
Homework Helper
You can't just divide out the $x^2$ in the differential. I suggest direct integration of the DE with respect to x.

well, it's the d/dx(y*x^2) that confused me.
we have only learned to solv the types:
dy/dx=f(x)g(y)

what is direct integration?

Thanks

Tide
Homework Helper
Direct integration means this:
$$\int_{x_0}^{x} \frac {d}{dx'}\left(x'^2 y\right) dx' = \int_{x_0}^{x} \frac {x'^2}{1+x'^2} dx'$$

Then I don't think we have learned about direct integration...

dextercioby
Homework Helper
Well, Tide said that, simply

$$\int du = u +\mathcal{C}$$

You must understand what i've written, else why attempt to solve diff. eqns. ?

Daniel.

okey, I understand what you just wrote :P

but I still don't know how to solve my problem...

HallsofIvy
Homework Helper
sony said:
d/dx(y*x^2)=x^2/(1+x^2)
I guess I can't write this as:
dyx^2=x^2dx/(1+x^2)
dy=dx/(1+x^2)
beacuse I don't get the right answer...
So what do I do?

Tide's point was just that you need to divide both sides by x2 to get simply $dy= \frac{dx}{1+x^2}$.

"Direct integration" just meant regular integration!

$$y= \int dy= \int \frac{dx}{1+ x^2}[/itex] Last edited by a moderator: Oh, but I didn't think I could do that when y*x^2 was encapsulated by parantheses. When I solve what you posted I get y=tan^-1(x) but the answer is supposed to be: (pi/4)(1/x^2)-(1/x)-tan^-1(x)/x^2 (the pi/4 comes from the start value...) HallsofIvy Science Advisor Homework Helper sony said: Oh, but I didn't think I could do that when y*x^2 was encapsulated by parantheses. When I solve what you posted I get y=tan^-1(x) but the answer is supposed to be: (pi/4)(1/x^2)-(1/x)-tan^-1(x)/x^2 (the pi/4 comes from the start value...) Oops! Sorry about that. I misread the original equation! It's not that the yx2 was "encapsulated by parentheses" but that here, specifically, the parentheses mean that the entire function yx2 is being differentiated. I read it as $x^2\frac{dy}{dx}$ rather than $\frac{dx^2y}{dx}$. Given that, Tide meant that [tex]x^2y= \int d(x^2y)= \int \frac{x^2dx}{x^2+ 1}$$.

That right hand side is a little more complicated. First divide it out:
$$\frac{x^2}{x^2+1}= 1- \frac{1}{x^2+1}[/itex] and it becomes easy to integrate. Last edited by a moderator: Oh, thanks! Now I see it :) Bah, sorry. I don't see what the integral of "dx^2y" evaluates to :( HallsofIvy Science Advisor Homework Helper sony said: Bah, sorry. I don't see what the integral of "dx^2y" evaluates to :( Fundamental Theorem of Calculus! [tex] \int d(Anything)= Anything+ C$$

$$\int d(x^2y)= x^2y + C$$

Oh! sorry! I see it! bah

hehe, the yx^2 messed up my head :P


text:
# Generating function of bessel function

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Prove the generating function

$$\displaystyle e^{\frac{x}{2}\left(z-z^{-1}\right)}=\sum_{n=-\infty}^{\infty}J_n(x)z^n$$

#### chisigma

##### Well-known member
$\displaystyle e^{\frac{x}{2}\ (z - z^{-1})} = \sum_{m=0}^{\infty} \frac{(\frac{x}{2})^{m}}{m!}\ z^{m} \ \sum_{k=0}^{\infty} (-1)^{k} \ \frac{(\frac{x}{2})^{k}}{k!}\ z^{k} =$

$\displaystyle = \sum_{n = - \infty}^{+ \infty} \{ \sum_{m - k = n} \frac{(-1)^{k}\ (\frac{x}{2})^{m + k}}{m!\ k!}\ \}\ z^{n} = \sum_{n = - \infty}^{+ \infty} \sum_{k=0}^{\infty}^{n} \{ \frac{(-1)^{k}}{(n+k)!\ k!}\ (\frac{x}{2})^ {2 k + n} \} z^{n} = \sum_{n = - \infty}^{+ \infty} J_{n} (x)\ z^{n}$

Kind regards

$\chi$ $\sigma$

#### mathbalarka

##### Well-known member
MHB Math Helper
$$2(n+1)\jmath_{n+1}(x) = x \jmath_{n+2}(x) + x \jmath_n(x)$$

Multiplying by $z^n$ and summing from $-\infty$ to $\infty$ both sides gives

\begin{aligned} \sum_{n = -\infty}^{\infty} 2n \jmath_{n}(x) z^{n-1} &= \sum_{n = -\infty}^{\infty} x \jmath_n(x) z^{n-2} + \sum_{n = -\infty}^{\infty} x \jmath_{n}(x) z^n \\ &= \sum_{n = -\infty}^{\infty} x \left (1 + \frac{1}{z^2} \right ) \jmath_n(x) z^n \end{aligned}

Hence, we have the differential equation :

$$K'(z) = \frac{x}{2} \left (1 + \frac{1}{z^2} \right ) K(z)$$

where $K(z) = \sum_{n = -\infty}^{\infty} \jmath_n(x) z^n$. This results $K(z) = \bar{C} \exp \left (\frac{x}{2} \left ( z - \frac{1}{z} \right) \right )$ after resolving the ODE. Substituting $z = 1$ easily gives $\bar{C} = 1$. Thus, we have

$$\sum_{n = -\infty}^{\infty} \jmath_n(x) z^n = \exp \left (\frac{x}{2} \left ( z - \frac{1}{z} \right) \right ) \;\;\; \blacksquare$$

Balarka
.

Last edited:


text:
# Curl of a function

## Main Question or Discussion Point

if the grad of the curl of a function is always zero does this mean the magnitude of the curl is constant? or am i way off here?

cristo
Staff Emeritus

Last edited:
uart
Firstly note that it is NOT true that the grad of the curl of a function is always zero.

Actually nolan I think you are mixing up two seperate properties of "curl" there to come to this misconception. The properties that you are thinking of are probably these two :

1. The div of the curl of a function is always zero. (div is not the same thing as grad ok).

2. If a given function can be written as the grad of another function then that given function has a curl of zero.

Note that neither of thse two things (taken seperately or together) imply what you have written.

uart
Ok looking at your question again I can see that it's probably the second property that I wrote that is the one you have gotten mixed up. It's the curl of a grad that is zero (a zero vector) not the way you put it, grad of a curl is not neccessarily zero

mathwonk
Homework Helper
that the curl of the grad is zero is essentially trivial. it follows from the trivial fact that the integral of GRAD is zero around a closed curve. this is because that integral is evaluated by subtracting the values of some function at the two endpoints, which are equal.

that in turn is true by the FTC, which holds by the trivial fact that the derivative of the area function is the height functon, which holds trivially because the area of a rectangle is the base times the height.

on the other hand, sticking all these trivialities together, maybe we have a pathologically amazing theorem!

more interesting is the investigation of forms with zero curl which are not gradients, like dtheta.

uart


text:
# Interpolated trajectories¶

This section is focused on the generation of interpolated trajectories by propagators in Nyx.

Interpolated trajectories are computed in parallel leading to a negligible impact in computational performance. All of the event finding requires an interpolated trajectory.

Trajectories store a interpolation splines created from a Hermite interpolation (written from scratch and validated against NumPy). Specifically, the algorithm works by curve fitting the provided states in position, velocity, and fuel mass (when interpolating a spacecraft trajectory instead of just an orbital trajectory). This curve fit will generate a Hermite series which is then converted into a generic polynomial. Note that none of these operations require memory allocations (and rely on Rust's const generic feature).

Technical note

Starting a propagator with the for_duration_with_traj() will return the final state of the propagation and the associated trajectory. This only works for states which also implement the InterpState trait.

The segments in a trajectory are organized in a binary tree map allowing for blazing fast access to any segment of the trajectory.

Limitations

1. To support generating the trajectories in parallell and storing them in a binary tree, it was necessary to choose an indexing method independent of the duration of the interpolation spline. This has been chosen to be the centiseconds (10 milliseconds) since the start of the trajectory, the start being the very first state used to seed the trajectory. This means that if two splines last less than 10 ms, one will overwrite the other in the binary tree, meaning that part of the trajectory will not be queryable.
2. Changing the frame of a trajectory follows the Shannon Nyquist sampling theorem. However, this will introduce some imprecision nevertheless. Multiple conversions (e.g. from EME2000 to Moon J2000 back to EME2000) will accumulate errors quickly.
3. Trajectories can currently only be generated with forward propagation. (Fixed in version 1.1.)

## Ephemeris¶

1. Propagate a LEO orbit with STM for 31 days in two body dynamics and request the propagator to generate the trajectory on the fly.
2. Compute the average SMA of that orbit by sampling the trajectory every 1 day (this ensures that the compiler does not remove the trajectory querying section from the test).
3. Check that the first and last state of the trajectory are strictly equal to the input propagator state and output propagator state.
4. Check that querying a trajectory one nanosecond after its end state returns an error.
5. Request a new propagation from the initial state but without generating the trajectory and read every intermediate state from that propagation. For each state, compute the radius and velocity differences between each of those states and the interpolated states at those exact times.
6. Ensure that the interpolation error in position is less than 1 micrometer and the error in velocity is less than 1 micrometer per second.
7. Convert the whole trajectory to its equivalent in the Moon J2000 frame (called Luna in Nyx). Convert that new trajectory back into the Earth Mean Equator J2000 frame (EME2000).
8. Sample that "double converted" trajectory every five minutes and compare each of those states with the initial interpolated trajectory.
9. Ensure that the interpolation error in position is less than 1 micrometer and the error in velocity is less than 1 micrometer per second.

All of the above is computed in 0.9 seconds on a standard desktop computer.

Note

If propagating with the STM, then the trajectory will also include the STM for that specific step forward.

Validation

$RUST_LOG=info RUST_BACKTRACE=1 /usr/bin/time cargo test traj_ephem --release -- --nocapture INFO nyx_space::propagators::propagator > Propagating for 30 days 23 h 59 min 60 s 0 ms -0.021928026217210607 ns until 2021-02-01T12:00:00 UTC Average SMA: 7712.186 km Should be: 7712.186 [tests/propagation/trajectory.rs:46] sum_sma / cnt - start_state.sma() = 0.000000245786395680625 Ephem: Trajectory from 2021-01-01T12:00:00 UTC to 2021-02-01T12:00:00 UTC (30 days 23 h 59 min 60 s 0 ms -0.021928026217210607 ns, or 2678400.000 s) [4613 splines] INFO nyx_space::propagators::propagator > Propagating for 30 days 23 h 59 min 60 s 0 ms -0.021928026217210607 ns until 2021-02-01T12:00:00 UTC [traj_ephem] Maximum interpolation error: pos: 2.25e-8 m vel: 1.98e-11 m/s INFO nyx_space::md::trajectory::traj > Converted trajectory from Earth J2000 to Moon J2000 in 189 ms ephem_luna Trajectory from 2021-01-01T12:00:00 UTC to 2021-02-01T12:00:00 UTC (30 days 23 h 59 min 60 s 0 ms -0.021928026217210607 ns, or 2678400.000 s) [11203 splines] INFO nyx_space::md::trajectory::traj > Converted trajectory from Moon J2000 to Earth J2000 in 448 ms (...) [traj_ephem] Maximum interpolation error after double conversion: pos: 1.27e-1 m vel: 4.41e-7 m/s test propagation::trajectory::traj_ephem ... ok test result: ok. 1 passed; 0 failed; 1 ignored; 0 measured; 116 filtered out; finished in 0.94s ## Spacecraft trajectory¶ Validation$ RUST_LOG=info RUST_BACKTRACE=1 /usr/bin/time cargo test traj_spacecraft --release -- --nocapture
(...)

[traj_spacecraft] Maximum interpolation error: pos: 1.75e-7 m           vel: 1.35e-6 m/s                fuel: 1.14e-10 g                full state: 3.06e-8 (no unit)
INFO  nyx_space::md::trajectory::traj         > Converted trajectory from Earth J2000 to Moon J2000 in 1 ms
INFO  nyx_space::md::trajectory::traj         > Converted trajectory from Moon J2000 to Earth J2000 in 3 ms
[traj_ephem] Maximum interpolation error after double conversion: pos: 5.76e-2 m                vel: 1.52e-7 m/s
test traj_spacecraft ... ok

test result: ok. 1 passed; 0 failed; 0 ignored; 0 measured; 2 filtered out; finished in 0.35s

Version 1.1 will allow generating navigation as well.

Last update: 2022-12-01


text:
This seems familiar

There is some number x for which the following is known.

$$x^2=x+1$$
$$x^3=2x+1$$

If $$x^8=ax+b$$ for some constant integers $$a$$ and $$b$$, what is $$a + b$$?

×


text:
# How many solutions possible for the equation ${2x}+{3y}+{7z}={r}$

I want to solve this problem with generating function :

How many solutions(non-negative) possible for the equation $${2x}+{3y}+{7z}={r}$$$(r \ge0)$ such that :

1)$x,y,z \ge0$

2)$0 \le z\le 2 \le y \le 8 \le x$

-
If $r$ is given (and $x,y,z$ are unknowns that comprise the solution), you should explain that in the question. The answer "how many solutions" then depends on $r$, right? –  hardmath Dec 27 '12 at 17:22
Also, you explicitly state that the variables are supposed to take integer values, otherwise the question becomes quite boring... –  Ben Millwood Dec 27 '12 at 17:29
I edited question. –  Martin Dec 27 '12 at 17:41

$\textbf{Hint: }$Note that the number of soloutions is given by the coefficient of $z^r$ in the expansion of the following expression : $$\left((z^2)^8+(z^2)^9+(z^2)^{10}+\cdots \right)\left((z^3)^2+(z^3)^3+\cdots +(z^3)^8\right)\left((z^7)^0+(z^7)^1+(z^7)^2\right)$$ and the first expression can be written as $z^{16}(1-z^2)^{-1}.$
Thanks. In part 1, we have :$$\left((z^2)^0+(z^2)^1+(z^2)^{2}+\cdots \right)\left((z^3)^0+(z^3)^1+\cdots \right)\left((z^7)^0+(z^7)^1+(z^7)^2+\cdots\right)$$ , How to get $z^{16}(1-z^2)^{-1}.$,,,, and in part 2 , How to get closed form? –  Martin Dec 27 '12 at 17:57
Let $\xi = x^2$, then by the restriction placed on $x$, we have the generating function $$(\xi^8 + \xi^9 + \xi^{10} + \dots) = \xi^8(1+ \xi^2 + \xi^3 + \dots) = x^{16}/(1-x^2).$$ To handle the $y$ term, let $\eta = x^3$, then \begin{align*} (\eta^ + \eta^3 + \dots + \eta^8) &= \eta^2(1 + \eta + \eta^2 + \dots + \eta^6)\\ &= \eta^2 \frac{1-\eta^7}{1-\eta}\\ &= x^6 \frac{1-x^{21}}{1-x^3} \end{align*} Lastly, to handle the $z$ term, let $\zeta = x^3$, then $$(\zeta^0 + \zeta^1 + \zeta^2) = (1-x^21)/(1-x^7)$$


text:
## k

$\frac{d[R]}{dt}=-k[R]; \ln [R]=-kt + \ln [R]_{0}; t_{\frac{1}{2}}=\frac{0.693}{k}$

Tiffany Cao 1D
Posts: 94
Joined: Fri Sep 29, 2017 7:03 am
Been upvoted: 1 time

### k

When finding the rate constant, should we convert all time units to seconds?

Jessica Yang 1J
Posts: 55
Joined: Fri Sep 29, 2017 7:03 am

### Re: k

I think it depends on the units in the rest of the problem. If it is stated that the rate is mol/L- min, then the k will probably by 1/min. However, if nothing is given in terms on units in the rest of the problem, I think it would be best to use time units as seconds.

Kayla Tchorz-Dis 1F
Posts: 30
Joined: Thu Jul 13, 2017 3:00 am

### Re: k

The units of k are completely dependent on what order you are dealing with. Zero order is M/s, first order is in s^-1, and second order is L/mol●s

CameronJohari1J
Posts: 66
Joined: Fri Sep 29, 2017 7:06 am
Been upvoted: 1 time

### Re: k

In general, it is useful to keep k units for time in terms of seconds. You can always convert to other units of time if the problem specifically asks for it, aka minutes or hours.

Fatima_Iqbal_2E
Posts: 31
Joined: Fri Sep 29, 2017 7:06 am
Been upvoted: 1 time

### Re: k

I think it would be a good idea to convert time to seconds if the problem doesn't ask otherwise. If needed, after finding k (with seconds as your units of time), you can always use conversion factors to change the seconds into another unit of time.

Arshpreet Sandhu 1B
Posts: 32
Joined: Thu Jul 27, 2017 3:00 am

### Re: k

I would keep it in seconds unless another unit is specified in the problem

AnuPanneerselvam1H
Posts: 52
Joined: Fri Sep 29, 2017 7:07 am

### Re: k

Most problems keep k in terms of seconds however it is ok to convert to other units if needed for the problem.

Wenjie Dong 2E
Posts: 53
Joined: Fri Jun 23, 2017 11:40 am

### Re: k

There is no need to do so unless other calculation is needed.

Merzia Subhan 1L
Posts: 33
Joined: Thu Jul 13, 2017 3:00 am

### Re: k

I'm sure that if they use a different unit in the problem like hours or minutes, you could leave the k in that unit instead of in seconds

Alyssa Pelak 1J
Posts: 72
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 1 time

### Re: k

If they want a specific unit they will likely state it in the problem. If not, you can use the easiest unit.

Posts: 53
Joined: Fri Sep 29, 2017 7:03 am

### Re: k

Reaction A: k = 2.3 M-1s-1
Reaction B: k = 1.8 Ms-1
Reaction C: k = 0.75 s-1

How do we know which one is a zero order reaction just given the units?

Gurkriti Ahluwalia 1K
Posts: 37
Joined: Fri Sep 29, 2017 7:07 am

### Re: k

a zero order reaction should have the same units as the rate, mol/L/s or mol x L-1 x s-1

Alex Nechaev 1I
Posts: 50
Joined: Fri Sep 29, 2017 7:04 am

### Re: k

Adam Enomoto 1L wrote:Reaction A: k = 2.3 M-1s-1
Reaction B: k = 1.8 Ms-1
Reaction C: k = 0.75 s-1

How do we know which one is a zero order reaction just given the units?

I would just try to memorize the units for at least the first few orders.

Sophia Bozone 2G
Posts: 51
Joined: Fri Sep 29, 2017 7:07 am

### Re: k

I don't think the math of the problem is messed up, simply on a logistics level, when a different unit is used. its all about keeping track of your units. Also it will be easier to keep the units the same as the units of the answer you are trying to find

Phillip Winters 2F
Posts: 50
Joined: Fri Sep 29, 2017 7:05 am

### Re: k

It isn't always necessary to change the units of k to s^-1, but just make sure that units always cancel

Rishi Khettry 1L
Posts: 29
Joined: Fri Sep 29, 2017 7:05 am

### Re: k

Usually the problem will specify a unit of time and we would want to use that same unit of time for K. Otherwise, using units of time as seconds is usually a safe bet.

Posts: 71
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 1 time

### Re: k

It's dangerous to make blanket statements of ALL PROBLEMS TO SECONDS, but usually the problem will specify with seconds, minutes, or heck, maybe even hours with LARGE half lives. So just read the question and dimensional analysis when necessary.


text:
# Jeff Johns is a staff accountant and has been assigned to the audit of Worldwide Enterprises, Inc. S

Jeff Johns is a staff accountant and has been assigned to the audit of Worldwide Enterprises, Inc. Subsequent to the completion of fieldwork, Jeff was assigned to draft the audit report. The content of one of the paragraphs he has drafted reads as follows:
As explained in Note 2 to the financial statements, Worldwide Enterprises has charged goodwill and certain other intangible assets acquired in two separate acquisitions directly to shareholders' equity. Under generally accepted accounting principles, these intangibles should have been recorded as assets and amortized to income over future periods. Had these intangibles been capitalized, total assets would have increased by $400,000 as of December 31, 2011 and net income and earnings per share would be increased by$380,000 and \$2.25, respectively (assuming a 20-year amortization period).

a. Based on the contents of the paragraph above, which condition requiring a departure from a standard unqualified/unmodified opinion exists in the engagement?
b. Assuming that the engagement partner agrees with the paragraph Jeff has prepared above, where in the auditor's report should the paragraph be placed?
c. How would the materiality of the condition above affect the final choice of opinion?

## Plagiarism Checker

Submit your documents and get free Plagiarism report

Free Plagiarism Checker


text:
#### Archived

This topic is now archived and is closed to further replies.

# [PERL]Text Wrapping etc...

This topic is 5452 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Hey All! I made a code generator that works like this: 1) Read from a header file and produce an intermediate file (named File A) that has tags(inserted by me) that facilitates sorting out the information that is read from the header by me for: Function declaration Actual code for functions with arguments/params etc everything. 2) Read this intermediate file i.e File A and process the information for making another intermediate file (named File B) that I use for another purpose later on. File B uses File A and the result is an intermediate code file that has the following: -Class Declaration -Function Definitions i.e Actual code for functions with arguments/params etc everything (whatever is dictated by the requirements is output to this intermediate file B) along with some information that I need later, 3) Read File B and use a config file named Input.config and use the information from the config file to alter the contents of the File B to produce the final processed file named File C(quite appropriate, isn¡¦t it? ) The input config file has stuff that I need to replace in File B for example Class Name etc. What is the problem: -Formatting is handled as desired by me in File B to get the results I¡¦m looking for and HERE IS THE PROBLEM: 1)What I need in File C is formatted code that wraps after 78 characters and the length of lines SHOULD NOT exceed 78 characters 2)Indented code inside if/for loops etc etc etc etc For Example: File B
class Class_Name:public BaseClass
{
blah blah blah
public:
blah blah blah
}

I CAN in File B control the formatting to some degree. ƒº
What I want is this in File C:
File B

class ClassMyClassNameNoMatterHowLongItIsName:public BaseClass
{
blah blah blah
public:
blah blah blah
}

MyClassNameNoMatterHowLongItIs is read from the config file and right now I use pattern matching and whenever I encounter Class_Name, I replace it with the information read from the input config file , Here in lies the problem, if I just use pattern matching and just *replace* the stuff , then , I have no control on the wrapping etc and in all likelihood, I lose most of the formatting information that I preserved in File B. My questions are: 1) How and what do I do in order to preserve some sort of formatting in File C (my FINAL processed output file) after all the tags etc have been replaced? A possible solution could be tokenizing an entire line that is read from File B (while producing File C) and then priting out each token while keeping in mind that only 78 characters are to be printed out BUT even this solution does not preserve the tab information etc that can no longer be preserved while using this. 2) Should I use a C++ program instead of a PERL script to generate the code in File C instead? Any suggestions are welcome, I must add here: 1) I¡¦m a C++ Programmer ) I don¡¦t know PERL and all that I have managed to accomplish here has been rather painstakingly after debugging with print statements, so please reserve any comments on my illiteracy , I do know that I DON¡¦T know PERL. 3) I don¡¦t have nay other experience in PERL or scripting and this really is my first attempt and it works so some degree Thanks in advance for your time and my apologies for the length of the post. [edited by - hehenoobhehe on October 21, 2003 12:12:09 AM]

##### Share on other sites
If you dont mind another stage you could pipe it through an indentation program e.g Indent.

I suspect I am completely misunderstanding what you are trying to do here though

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text:
Categories:

Data Pipeline DDL

# DROP STREAM¶

Removes a stream from the current/specified schema.

## Syntax¶

DROP STREAM [ IF EXISTS ] <name>

## Parameters¶

name

Specifies the identifier for the stream to drop. If the identifier contains spaces, special characters, or mixed-case characters, the entire string must be enclosed in double quotes. Identifiers enclosed in double quotes are also case-sensitive (e.g. "My Object").

If the stream identifier is not fully-qualified (in the form of db_name.schema_name.stream_name or schema_name.stream_name), the command looks for the stream in the current schema for the session.

## Examples¶

Drop a stream:

SHOW STREAMS LIKE 't2%';

DROP STREAM t2;

SHOW STREAMS LIKE 't2%';

Drop the stream again, but don’t raise an error if the stream does not exist:

DROP STREAM IF EXISTS t2;


text:
# How Tensorflow be used for the gender column, like feature column helping in prediction?

TensorflowServer Side ProgrammingProgramming

The ‘gender’ column will be used in the prediction as well, hence it is important to understand more about it. This can be done by visualizing the ‘age’ column as a horizontal bar plot.

We will use the Keras Sequential API, which is helpful in building a sequential model that is used to work with a plain stack of layers, where every layer has exactly one input tensor and one output tensor.

A neural network that contains at least one layer is known as a convolutional layer. We can use the Convolutional Neural Network to build learning model.

We are using the Google Colaboratory to run the below code. Google Colab or Colaboratory helps run Python code over the browser and requires zero configuration and free access to GPUs (Graphical Processing Units). Colaboratory has been built on top of Jupyter Notebook.

An Estimator is TensorFlow's high-level representation of a complete model. It is designed for easy scaling and asynchronous training.  We will train a logistic regression model using the tf.estimator API. The model is used as a baseline for other algorithms. We use the titanic dataset with the goal of predicting passenger survival, given characteristics such as gender, age, class, etc.

## Example

print("More data about gender")
dftrain.sex.value_counts().plot(kind='barh')
plt.show()

## Explanation

• The gender column will be useful in the prediction process further.
• Hence it is visualized on the console.
Published on 12-Feb-2021 12:34:40


text:
# A gas occupies a volume of 0.2 L at 25 kPa. What volume will the gas occupy at 2.5 kPa?

Mar 15, 2017

$\text{2 L}$

#### Explanation:

Right from the start, assuming, of course, that the temperature and the number of moles of gas remain unchanged, you can say that the volume of the gas will increase as pressure changes from $\text{25 kPa}$ to $\text{2.5 kPa}$.

That is the case because when the temperature and the number of moles of gas remain constant, the pressure of a gas varies indirectly with its volume, as described by Boyle's Law.

In other words, when pressure decreases by factor $k$, volume increases by the same factor $k$.

$P \cdot V = k$

This implies that you can write

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{P}_{1} {V}_{1} = {P}_{2} {V}_{2}}}}$

Here

• ${P}_{1}$ and ${V}_{1}$ represent the pressure and volume of the gas at an initial state
• ${P}_{2}$ and ${V}_{2}$ represent the pressure and volume of the gas at a final state

Rearrange to solve for ${V}_{2}$

${P}_{1} {V}_{1} = {P}_{2} {V}_{2} \implies {V}_{2} = {P}_{1} / {P}_{2} \cdot {V}_{1}$

Plug in your values to find

V_2 = (25 color(red)(cancel(color(black)("kPa"))))/(2.5color(red)(cancel(color(black)("kPa")))) * "0.2 L" = color(darkgreen)(ul(color(black)("2 L")))

The answer is rounded to one significant figure, the number of significant figures you have for the initial volume of the gas.

As predicted, the volume of the gas increased as a result of the decrease in pressure. Moreover, it increased by the same factor!


text:
# Synopsis: Alice and Bob Go Nonlinear

Time-reversing the signals generated by a nonlinear element could form the basis of a new secure communications protocol in the midst of chaotic waves.

Nonlinear devices added to linear systems often yield new, useful phenomena, an example being frequency-doubling crystals that combine two laser photons to create a photon of twice the energy. In a paper in Physical Review Letters, Matthew Frazier and colleagues at the University of Maryland, College Park, report experiments in which they put a nonlinear frequency-multiplying device into a chaotic bath of electromagnetic waves and find signal propagation effects that might launch a new kind of secure communication.

The equations that describe electromagnetic waves are linear and time invariant, which means that signals propagating forward in time can be recorded, played backwards (i.e., time reversed), and sent back along the incoming path, returning exactly to their source. Frazier et al. built a metal box with ports to couple microwave radiation in and out. Two of the ports are equipped with conventional linear antennas, but the third is an antenna incorporating a nonlinear element (in this case a diode). A scattering device in the box creates a chaotic electromagnetic environment to mask the signals and ensure complex signal paths.

A signal sent into one linear port of the chamber will bounce around and eventually hit the nonlinear antenna, which responds by producing signals at new frequencies. These new frequencies are then recorded at the other linear port, and then played backwards into the box, whereupon they reverse their propagation and return precisely focused onto the nonlinear element, regardless of how complex the path is. Among the applications envisaged by the authors is a communications network in which messages broadcast by Alice in a wide area are picked up by Bob with a nonlinear receiver at a secret location (not even known to Alice). Only that location will receive the reversed playback; Eve the eavesdropper will only detect garbled signals from the chaotic wave environment. – David Voss

More Features »

### Announcements

More Announcements »

Fluid Dynamics

Complex Systems

## Related Articles

Atomic and Molecular Physics

### Synopsis: Correcting Hardware Bias in Molecular Spectrometers

More accurate measurements of ${\text{CO}}_{2}$ in the atmosphere can be obtained by accounting for discrepancies in spectroscopy hardware. Read More »

Fluid Dynamics

### Synopsis: Dandelion Fluff Perfected for Flight

Calculations show that the number of white filaments springing from the top of a dandelion seed is optimized for steady flight, allowing the seeds to cruise long distances through the air. Read More »

Energy Research

### Focus: Folded Solar Panel Opens Without Power Source

A spacecraft solar cell design uses a temperature-sensitive polymer to expand the panel’s surface area by 10 times in 40 seconds. Read More »


text:
# Tag Info

Math Formulation + Pyomo model We need to define 3 sets i: 1 to 8 staffs j: 1 to 3 shifts t: 1 to 18 days $X_{i,j,t} \in {0/1}$ Binary There must be 3 people on each shift $\forall t,j$ then $\sum_i X_{i,j,t} =3$ Every individual (i) must work 1 shift per day\ $\forall t,i$ then $\sum_j X_{i,j,t} \geq 1$ Every individual must not work more than 2 ...


text:
board-games-0.1.0.6: Three games for inclusion in a web server

Game.Mastermind

Synopsis

# Documentation

data Eval Source #

Constructors

 Eval Int Int

Instances

 Source # Methods(==) :: Eval -> Eval -> Bool #(/=) :: Eval -> Eval -> Bool # Source # MethodsshowsPrec :: Int -> Eval -> ShowS #show :: Eval -> String #showList :: [Eval] -> ShowS #

evaluate :: Ord a => [a] -> [a] -> Eval Source #

Given the code and a guess, compute the evaluation.

histogram :: Ord a => [a] -> Map a Int Source #

selectFromHistogram :: Ord a => Map a Int -> [(a, Map a Int)] Source #

remainingSimple :: Ord a => Set a -> [a] -> Int -> [[Set a]] Source #

A variant of the game: It is only possible to specify number of symbols at right places.

The results of remaining and remainingSimple cannot be compared.

possibleRightPlaces :: Int -> Int -> [[Bool]] Source #

Combinatorical "choose k from n".

remaining :: (C set, Ord a) => Set a -> [a] -> Eval -> set a Source #

Given a code and an according evaluation, compute the set of possible codes.

The Game.Mastermind game consists of collecting pairs of codes and their evaluations. The searched code is in the intersection of all corresponding code sets.

partitionSizes :: Ord a => Set a -> [a] -> [(Eval, Integer)] Source #

interaction :: (T Char -> StateT state Maybe [Char]) -> state -> Set Char -> Int -> IO () Source #

randomSelect :: (RandomGen g, Monad m) => [a] -> StateT g m a Source #

minimax :: Ord b => [(a, [b])] -> a Source #

minimum of maximums using alpha-beta-pruning

randomizedAttempt :: (C set, RandomGen g, Ord a) => Int -> Set a -> set a -> StateT g Maybe [a] Source #

mixedRandomizedAttempt :: (C set, RandomGen g, Ord a) => Int -> Set a -> set a -> StateT g Maybe [a] Source #

In the beginning we simply choose a random code from the set of possible codes. In the end, when the set becomes small, then we compare different alternatives.

main :: IO () Source #


text:
# No Data Packets in Monitor Mode Capture

I've seen this question a few times before but none of the posted solutions solved my issue. I can see Probes, Beacons, Acknowledgements, Request-to-sends, Clear-to-sends, and null data frames but not any non-null data.

I am running Wireshark on Kali Linux attempting to sniff wireless traffic. I have the following set up:

Capture Device:

Kali Linux LiveBoot USB

OS: Kali 2020.3 - SMP Debian 5.5.17-1 kali1 (2020-04-21) x86_64

NIC: Panda PAU09-RaLink RT5572 - Driver: rt2800usb as wlan0

Wireshark 3.2.3

AP - Raspberry Pi 4 running Raspbian

I configured HostAPD to use my Raspberry Pi as an access point. It is set to run 802.11g with no security. The set up on my sniffing system has been:

ifconfig wlan0 down
iwconfig wlan0 mode Monitor
ifconfig wlan0 up

Start wireshark, check the monitor mode checkbox, restart wireshark, and then begin capture. On the client Pi I am connected to the AP and running a script that periodically curls the Apache server on the AP.

Debugging done so far:

• Use Ubuntu VM instead of Kali Liveboot
• Create wlan0mon interface with airmon-ng and listen on that instead (using sudo airmon-ng start to create the virtual interface wlan0mon and used that as my listening interface in Wireshark)
• Connect to AP after putting device into Monitor mode (which I think just switched the interface to Managed as I was then only able to see my traffic)
• Tried different target devices (Iphone, Windows 10 Desktop)
• Adjusted channel settings through Wireshark's wireless toolbar to match the channel my AP is broadcasting on (7)
• Tried different AP (my home router rather than one I configured myself) and done basic web browsing

Any help or ideas would be greatly appreciated, I am not sure why this isn't working.

edit retag close merge delete

I also have this problem but in using MacOS Catalina. I was trying on capturing WiFi packets on MacOS Catalina but unfortunately there is No showing captured packets. I had follow the recommended guide in installing Wireshark in macOS Catalina and install alone with the “Install ChmodBPF” from the link (https://www.wireshark.org/docs/wsug_h...) but it is not working. I compare it with my Windows OS machine with the same configuration and it work on capturing live WiFi packets.

What would be I have to do to make it work? Any help is really appreciated with this matter.

Thanks,

( 2020-08-28 18:57:19 +0000 )edit

Search on Catalina and several similar questions/answers are available. No good answers.
This one is probably the most recent.

( 2020-08-28 19:34:57 +0000 )edit

In macOS Mojave and later (the problem predates Catalina), at least some Wi-Fi adapters on newer machines cannot capture if they're associated to a Wi-Fi network, so you have to disconnect from the network if you want to capture in monitor mode. I don't know whether this is a hardware problem or a driver problem (driver design decision?).

The Linux problem discussed here my not be related.

( 2020-08-28 21:10:08 +0000 )edit

Yes, you are right that you need to disconnect from the network to be able to sniff the packets. There is Network Diagnostic app in MacOS catalina that gives you .pcap file and can open with wireshark. But comparing to Windows machine, live capture in wireshark gives i think better infos. But until now still looking for solution regarding this matter. Thanks for the response guy.

( 2020-08-29 18:11:28 +0000 )edit

I have this exact problem. Environment is Kali, just installed 2 days ago and updated. I have mesh routers and a Second wifi router, I have searched, and, well, I stop the network connection, or boot with no connection. I use airmon-ng start wlan0 to start monitor mode. I am sudoed to root to run wireshark. I start capturing on wlan0mon, the monitor mode interface created by airmon-ng.

All I see are beacons and sometimes broadcasts. I have tried connecting to my non-mesh router, have double checked the channel, I see the beacons from that router, I put another device on the same wifi, start pinging, and all I see are the beacons and some spanning tree stuff.

I don't see the arp stuff, I don't see the pings. I tried actually connecting to this interface and I get through without an issue, 67% signal strength so it ...(more)

( 2020-11-01 07:26:48 +0000 )edit

Sort by » oldest newest most voted

A sample trace file of what you are seeing would be a big help. For example, it can be used to quickly rule out many basic things that may or may not be an issue - but no way to know for sure without looking.

Otherwise, can only provide general guidance as there are many things that could be affecting this - for example, looking for Data frames when QoS Data frames are in use; sniffing on the wrong channel; too far away; trying to capture with a USB adapter inside a VM; using the adapter at the same time in both managed and monitor mode; trying to capture high data-rate frames with a less-capable adapter; the list goes on.

more

Have you tried turning monitor mode on using the shiny new mac80211 way? See the Linux section of the "monitor mode capture" page in the Wireshark Wiki; search for the part that starts with

The easiest way to turn manually turn monitor mode on or off for an interface is with the airmon-ng script in aircrack-ng; your distribution may already have a package for aircrack-ng.

and then describes command sequences to use with airmon-ng and to use if you don't have airmon-ng.

more

I have already tried this. I used sudo airmon-ng start wlan0 which then created wlan0mon that I then used as my listening interface in Wireshark. I should have been more clear but that was what I meant in my debugging step #2

( 2020-08-28 13:38:24 +0000 )edit


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# Linear and generalised linear models

## Presentation on theme: "Linear and generalised linear models"— Presentation transcript:

Linear and generalised linear models
Purpose of linear models Solution for linear models Some statistics related with the linear model Additive and non-linear models Generlised linear model

Reason for linear models
Purpose of regression is to reveal statistical relations between input and output variables. Statistics cannot reveal functional relationship. It is purpose of other scientific studies. Statistics can help validation various functional relationship. Let us assume that we suspect that functional relationship is where  is a vector of unknown parameters, x=(x1,x2,,,xp) a vector of controllable parameters, and y is output,  is error associated with the experiment. Then we can set for various values of x experiments and get output (or response) for them. If number of experiments is n then we will have n output values. Denote them as a vector y=(y1,y2,,,yn). Purpose of statistics is to evaluate parameter vector using input and output values. If function f is a linear function of the parameters and errors are additive then we are dealing with linear model. For this model we can write Note that linear model is linearly dependent on parameters but not on input variables. For example is a linear model. But is not a linear model.

Assumptions Basic assumptions for analysis of linear model are:
the model is linear in parameters the error structure is additive Random errors have 0 mean and equal variance and errors are uncorrelated. These assumptions are sufficient to deal with linear models. Uncorrelated with equal variance assumption can be removed. Then the treatments becomes a little bit more complicated. Note that for general solution normality assumption is not used. This assumption is necessary to design test statistics. These assumption can be written in a vector form: where y, 0, I,  are vectors and X is a matrix. This matrix is called design matrix, input matrix etc. I is nxn identity matrix.

Solution Solution with given model and assumptions is:
If we use form of the model and write least squares equation (since we want to find solution with minimum least-squares error: and get first and second derivatives and solve the equation then we can see that this solution is correct. This solution is unbiased. If we use formula for the solution and expression of y then we can write: So solution is unbiased. Variance of estimation is: Here we used form of the solution and assumption 3)

Variance To calculate variance we need to be able to calculate 2. Since it is variance of the error term we can find it using form of the solution. For the estimated error (denoted by e) we can write: Using : Immediately gives Since matrix M is idempotent i.e. M2=M=MT we can find for estimation of variance following formula: Wher n is the number of the observations and p is the number of the fitted parameters.

Test of hypothesis Sometimes question arises if some of the parameters are significant. To test this type of hypothesis it is necessary to understand elements of likelihood ratio test. Let us assume that we want to test the following null-hypothesis vs alternative hypothesis: where 1 is the subvector of the parameter vector. It is equivalent to saying that we want to test of one or several parameters are 0 or not. Likelihood ratio test for this case works like that. Let us assume we have the likelihood function for the parameters: where parameters are partitioned into to subvectors: Then maximum likelihood estimators are found for two cases. 1st case is when whole parameter vector is assumed to be variable. 2nd case is when subvector 1 is fixed to a value defined by the null hypothesis. Then values of the likelihood function for this two cases is found and their ratio is calculated. Assume that L0 is value of the likelihood under null hypothesis (subvector is fixed to the given value) and L1 is under the alternative hypothesis. Then ratio of these values is found and statistics related to this ratio is found and is used for testing. Ratio is: If this ratio is sufficiently small then null-hypothesis is rejected. It is not always possible to find distribution of this ratio.

Singular case This forms of the solution is true if matrices X and XTX are non-singular. I.e. rank of matrix X is equal to the number of parameters. If it is not true then either singular value decomposition or eignevalue filtering techniques are used. Fortunately most good properties of the linear model remains. Singular value decomposition: Any nxp matrix can be decomposed in a form: Where U is nxn and V is pxp orthogonal matrices. I.e.multiplication of transpose of the matrix with itself gives unit matrix. D is nxp diagonal matrix of the singular values. If X is singular then number of non-zero diagonal elements of D is less than p. Then for XTX we can write: DTD is pxp diagonal matrix. If the matrix is non-singular then we can write: Since DTD is diagonal its inverse is the diagonal matrix with diagonals inversed. Main trick used in singular value decomposition techniques for equation solution is that when diagonals are 0 or close to 0 then instead of their inversion 0 is used. I.e. if E is the inverse of the DTD then pseudo inverse is calculated:

Likelihood ratio test for linear model
Let us assume that we have found maximum likelihood values for the variances under null and alternative hypothesis and they are: furthermore let us assume that n is the number of the observations, p is the number of all parameters and r is the number of the parameters we want test. Then it turns out that relevant likelihood ratio test statistic for this case is related with F distribution. Relevant random variable is: This random variable has F distribution with (r,n-p) degrees of freedom. It is true if the distribution of the errors is normal. As we know in this case maximum likelihood and least-squares coincide. Note: Distribution becomes F distribution if null-hypothesis is true. If it is not true then distribution becomes non-central F distribution Note: if there are two random variables distributed by 2 distribution with n and m degrees of freedom respectively then their ration has F distribution with (n,m) degrees of freedom.

Let us consider several non-linear models briefly. Additive model. If model is described as then model is called an additive model. Where si can be some set of functions. Usually they are smooth functions. These type of models are used for smoothening. 2) If model is a non-linear function of the parameters and the input variables then it is called non-linear model. In general form it can be written: Form of the function depends on the subject studied. This type of models do not have closed form and elegant solutions. Non-linear least-squares may not have unique solution or may have many local minimum. This type of models are usually solved iteratively. I.e. initial values of the parameters are found and then using some optimisation techniques they are iteratively updated. Statistical properties of non-linear models is not straightforward to derive. Although bootstrap technique can be used to derive some sort of approximations they are not exact in general

Generalised linear model
If the distribution of errors is one of the distributions from the exponential family and some function of the expected value of the observations is linear function of the parameters then generalised linear models are used: Function g is called the link function. Here is list of the popular distribution and corresponding link functions: binomial - logit = ln(p/(1-p)) normal identity Gamma - inverse poisson - log All good statistical packages have implementation of many generalised linear models. To use them finding initial values might be necessary. Additive model can also be generalised. I.e. the function of the expected value of the observation can have the form:

Exercise: linear model
Consider hypothesis testing. We have n observation. Parameter vector has dimension p. We have partitioning of the parameter vector like (dimension of 1 is r): Corresponding partitioning of the design (input) matrix is Assume that all observations are distributed with equal variance normally and they are uncorrelated. Find maximum likelihood estimators for parameters and variance under null and alternative hypothesis: Hint: -loglikelihood function under null hypothesis is (since 1=0): and under the alternative hypothesis: Find minimum of these functions. They will be maximum likelihood estimators.


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Browse Questions

# Which of the following relation is correct?

$\begin{array}{1 1}(a)\;\Delta G^0=-nRTl_nQ;Q\neq K_{eq}\\(b)\;\Delta G^0=-nFE^0_{cell}\\(c)\;\Delta G^0=\Delta H-T\Delta S\\(d)\;None\end{array}$

$\Delta G^0=-nRTl_nQ$ and $Q=K_{eq}$
$\Delta G^0=\Delta H-T\Delta S$
Whereas $\Delta G^0=-nFE^0_{cell}$
Hence (b) is the correct answer.


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## thermodynamics formulas for neet

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For your preparation, you need to refer best Physics Books for NEET Exam. Check Best Android Mobile Apps to Prepare for NEET. Serial No. For quasi-static and reversible processes, the first law of thermodynamics is: d U = δ Q − δ W {\displaystyle dU=\delta Q-\delta W} where δ Q is the heat supplied to the system and δ W is the work done by the system. by Anuj William. Find the energy consumed by the freezer if 5 kg of water at 0oC is converted into ice at 0oC by the freezer. No Result . Practice Now. The empirical formula of an organic compound containing carbon and hydrogen is CH 2.The mass of one litre of this organic gas is exactly equal to that of one litre of N 2.Therefore, the molecular formula of the organic gas is [EAMCET 1985] (1) C 2 H 4 (2) C 3 H 6 (3) C 6 H 12 (4) C 4 H 8 Remember, revision and analysis of mocks are important. Wrong Answer Wrong Explanation … Learn and practice from Thermodynamics quiz, study notes and study tips to help you in NEET Chemistry preparation. 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Simultaneous equations

# Systems of 3 variable Equations

Systems of equations that have three variables are systems of planes. Like systems of linear equations, the solution of a system of planes can be no solution, one solution or infinite solutions.

To solve a system of, write the given system in matrix form. Write all the information contained in the system in the single augmented matrix and choose one of these method to solve the system:

• Gaussian Elimination: Take an augmented matrix for the system and carry it by means of elementary row operations to an equivalent augmented matrix from which the solutions of the system are easily obtained. In particular, we bring the augmented matrix to Row-Echelon Form. At that point, the solutions of the system are easily obtained.

• Matrix inverse method: Write the given system as a single matrix equation AX=B . Solve the equation obtained for the matrix variable X by left-multiplying both sides of the above matrix equation (AX=B) by the inverse of the matrix A. Then, and you have the solution.

• Cramer's Rule: To use this method, you have only to follow the steps:

1. Write the coefficient matrix of the system (call this matrix A); if it is square, you may continue, otherwise Cramer's rule is not applicable here.
2. Compute the determinant of the coefficient matrix, |A|; if |A| is not zero you may continue, otherwise Cramer's rule is not applicable here.
3. Suppose the first variable of the system is x. Then write the matrix Ax as follows: substitute the column of numbers to the right of the equal signs instead of the first (from the left) column of A. Now compute the determinant of Ax, that is |Ax|.
4. The value of x in the solution is now |Ax| / |A|.
5. Repeat steps 3, and 4 with the remaining variables. In each case substitute the column of numbers instead of the column of A that corresponds to the variable you are using. If the variables are x, y, and z, then the values will now be:

 x = |Ax| / |A| y = |Ay| / |A| z = |Az| / |A|

And the solution is

 (x, y, z) = ( |Ax| / |A| , |Ay| / |A| , |Az| / |A| )

Solve:

 $\left{-5x-2y-3z=-21\\-3x-5y+z=-24\\-4x-4y+4z=-24$ x = y = z =