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# Calculus: Other Calculus Problems – #27077

Question: #28) Estimate the error using the indicated partial sum

${{S}_{80}},\,\,\,\sum\limits_{k=1}^{\infty }{\frac{2}{{{k}^{2}}+1}}$

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## Number of non-isomorphic groups of order 10 = ?

| 53 views
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Ans will be 4, with lagrange theorem
0

srestha  mam how ...

How langrage theorem applied here ??

+1
Every group of prime order has no proper ans nontrivial subgroup.

So, here order 1,3,5 forms proper trivial subgroup and also nonisomorphic

Now, order 10 also trivial subgroup.

So, total 4 subgroup, which are trivial too

So, 4 groups
0

I think answer should be 1.

The number of Abelian groups of order  P^k (P is prime) is the number of partitions of  k.
When we prime factorize 10 = 2^1 * 5^1.

Now we find partitions of the powers which are 1 and 1.

We multiply them 1*1=1

Correct me if i am wrong.

i referred to this question:-

https://gateoverflow.in/1219/gate2007-21

+1 vote
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NTNUJAVA Virtual Physics LaboratoryEnjoy the fun of physics with simulations! Backup site http://enjoy.phy.ntnu.edu.tw/ntnujava/
July 23, 2018, 10:22:18 pm

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 Author Topic: Hydrosystem with two tanks connected one after another  (Read 31684 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
pavwell
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 « Embed this message on: March 17, 2010, 03:10:03 pm » posted from:Zilina,Zilina,Slovakia

Hello ! Could anyone help me with creating EJS aplication of Hydrosystem with two tanks connected one after another ?
Thanks a lot !!
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Fu-Kwun Hwang
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 « Embed this message Reply #1 on: March 17, 2010, 04:17:57 pm » posted from:Taipei,T'ai-pei,Taiwan

Please describe the whole system in detail!
If I can not understand the system and the physics model involved, there is no way I can create a simulation.
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pavwell
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 « Embed this message Reply #2 on: March 17, 2010, 04:52:27 pm » posted from:Zilina,Zilina,Slovakia

I will send you email where i subcribe my problem.
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Fu-Kwun Hwang
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 « Embed this message Reply #3 on: March 17, 2010, 09:35:42 pm » posted from:Taipei,T'ai-pei,Taiwan

$s_1 \frac{dh_1}{dt}=q_1 -\alpha_1 \sqrt{h_1-h_2}$

$s_2 \frac{dh_2}{dt}=\alpha_1 \sqrt{h_1-h_2}-\alpha_2\sqrt{h_2}$

You did provide me information about:
$\alpha_1=0.0000208, s1=0.002025m^2, 0.0000163, s2=0.002025m^2$

However, you did not provide range for $q_1, h_1,h_2$ and maximum height for both tank.

Embed a running copy of this simulation

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Full screen applet or Problem viewing java?Add http://www.phy.ntnu.edu.tw/ to exception site list
• Please feel free to post your ideas about how to use the simulation for better teaching and learning.
• Post questions to be asked to help students to think, to explore.
• Upload worksheets as attached files to share with more users.
Let's work together. We can help more users understand physics conceptually and enjoy the fun of learning physics!
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pavwell
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 « Embed this message Reply #4 on: March 17, 2010, 11:07:38 pm » posted from:Lutila,Banska Bystrica,Slovakia

Thanks a lot for these solution.
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pavwell
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 « Embed this message Reply #5 on: March 18, 2010, 03:04:51 pm » posted from:Zilina,Zilina,Slovakia

Please, if it possible, could you modify this project to change open loop system to close loop system with simple PID controller ?
And one more task is possible modify system to turn off automaticly inflow when liquid levels of tank1 and tank 2 get maximum range ?
Maximum range both of tanks set to 0,14
 « Last Edit: March 18, 2010, 03:22:35 pm by pavwell » Logged
Fu-Kwun Hwang
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 « Embed this message Reply #6 on: March 18, 2010, 05:18:30 pm » posted from:Taipei,T'ai-pei,Taiwan

Please explain in detail what it mean to be a close loop with simple PID controller?
It will be easier for me to design the simulation if you can provide me detail information as possible.
You can add picture as attached file.

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pavwell
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 « Embed this message Reply #7 on: March 18, 2010, 05:47:01 pm » posted from:Zilina,Zilina,Slovakia

Please, could you modify system as I write in next two tasks ??

1.is possible modify system to turn off automaticly inflow when liquid levels of tank1 and tank 2 get maximum range ?
2.Please set maximum range both of tanks set to 0,14 and please, reconfigure range of visualization to better see line of h1 . it is too high
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Fu-Kwun Hwang
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 « Embed this message Reply #8 on: March 18, 2010, 05:57:08 pm » posted from:Taipei,T'ai-pei,Taiwan

I can modify it according to your request.
However, you never give me information about the height , and q1 for the system.
The default value are the best I can guest.
It will  be better if you give me detail information about your system.
Height and range of possible q1 values.
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pavwell
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 « Embed this message Reply #9 on: March 18, 2010, 06:24:15 pm » posted from:Zilina,Zilina,Slovakia

O.k. I try to discribe Controlled close loop system . I set in input value box some value of level h2 and by heigh of level h2 I can control value of inflow. h2 is in feedback to inflow. Is this clearle to you ?
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Fu-Kwun Hwang
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 « Embed this message Reply #10 on: March 18, 2010, 08:36:12 pm » posted from:Taipei,T\'ai-pei,Taiwan

I do not understand what you just described???

Quote
2.Please set maximum range both of tanks set to 0,14 and please, reconfigure range of visualization to better see line of h1 . it is too high
You said that h1 is too high. Would you please provide a suitable value for h1?
And please provide me a suitable value for maximum height!
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pavwell
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 « Embed this message Reply #11 on: March 18, 2010, 08:40:02 pm » posted from:Zilina,Zilina,Slovakia

Meaning too high i think not value of h1 but value of plot window where is shown h1. This plot window i would like to see larger

And please, set the maximum value of h1 on 0,14 and maximum of h2 also on 0,14.

Just another question for you..Do you have any experience with cooperating EJS and MATLAB ??
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Fu-Kwun Hwang
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 « Embed this message Reply #12 on: March 18, 2010, 11:29:52 pm » posted from:Taipei,T\'ai-pei,Taiwan

Quote

Meaning too high i think not value of h1 but value of plot window where is shown h1. This plot window i would like to see larger Smiley
I did not changed it because I do not know how much larger you need.
I can not guest what you want in your mind.  Please describe it in detail.
However, I did change the way it was displayed (scale).

Quote
And please, set the maximum value of h1 on 0,14 and maximum of h2 also on 0,14.
I just modified the code so that q1 will be turn off when h1>0.140
And q1 will be tuen on when h1
Quote
Just another question for you..Do you have any experience with cooperating EJS and MATLAB ??
I do not have MATLAB so I have never use that feature.
However, it is only available for EJS version 3.47 or earlier.
I have heard that some users from swissland did use MATLAB with EJS for their research work.
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pavwell
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 « Embed this message Reply #13 on: March 19, 2010, 12:00:43 am » posted from:Lutila,Banska Bystrica,Slovakia

Thanks for your very good work Dont have by the way any contact on swissland studnets ??
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Fu-Kwun Hwang
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 « Embed this message Reply #14 on: March 19, 2010, 01:19:48 pm » posted from:Taipei,T\'ai-pei,Taiwan

Sorry! I do not know.
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pavwell
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 « Embed this message Reply #15 on: March 19, 2010, 01:24:57 pm » posted from:Zilina,Zilina,Slovakia

Hello
Please, I have big appeal on you. If it possible, could you write this hydro system in EJS version 3.4 especialy for me But I need one version of system at EJS able to communicate with MATLAB. And as you wrote to me, EJS 4.2 have not this option.
THANKS A LOT !!!
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Fu-Kwun Hwang
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 « Embed this message Reply #16 on: March 19, 2010, 01:37:09 pm » posted from:Taipei,T\'ai-pei,Taiwan

Because you have to write code to connect to MATLAB with EJS.
So you need to learn to use EJS.

Right click and select open ejs model to open EJS and learn how it was created.
And you can do similar thing with EJS 3.4 or earlier EJS version (which support MATHLAB).

I do not have any experience with EJS connect to MATHLAB, so you should learn EJS in order to do it.
And you can use the version I created as an example to learn it.
I will try to help when you learn to use EJS. But I would not be able to help when EJS need to communicate with MATHLAB. So you better learn to create it with EJS 3.4 by yourself. I think this is the best for you to do it by yourself.
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pavwell
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 « Embed this message Reply #17 on: March 19, 2010, 01:43:34 pm » posted from:Zilina,Zilina,Slovakia

Of course I tried do this yesterday, but for i have problem with creating both of tanks and inflow at EJS old version. I have EJS 4.2 as new version and EJS 3.4 as old version. And i tried to do this yesterday at EJS 3.4. But in this version is not options like in 4.2 version for building polygon and other model parts Therefore i appeal you for help. Ploting is o.k. This part of project I built in old version without problems. But with creating tanks i have problem
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Fu-Kwun Hwang
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 « Embed this message Reply #18 on: March 19, 2010, 02:07:51 pm » posted from:Taipei,T\'ai-pei,Taiwan

I do not have EJS 3.4 but I do have EJS3.45

The tank created in the above simulations was using polygon element.
The height h1/h2 was change by changing y coordinate for 4 points belongs to polygon.
You should be able to do it with EJS 3.45 and above.
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pavwell
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 « Embed this message Reply #19 on: March 19, 2010, 02:16:57 pm » posted from:Zilina,Zilina,Slovakia

...EJS 3.45 is also o.k could you make in this version thank you so much !!!
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Fu-Kwun Hwang
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 « Embed this message Reply #20 on: March 19, 2010, 02:45:21 pm » posted from:Taipei,T\'ai-pei,Taiwan

I think you should learn how to use it if you really want to connect to MathLAB.

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pavwell
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 « Embed this message Reply #21 on: March 19, 2010, 02:47:28 pm » posted from:Zilina,Zilina,Slovakia

O.K. I will upload my xml file couple of hours later because now I am at work and on this project i work at home.
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pavwell
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 « Embed this message Reply #22 on: March 20, 2010, 02:36:54 am » posted from:Lutila,Banska Bystrica,Slovakia

Hello...I found simple example on PI controller of level tank...I attach you a xml file..If it possible just like that use it for my model , please add similiar PI controller to my model..Thanks !
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Fu-Kwun Hwang
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 « Embed this message Reply #23 on: March 20, 2010, 09:22:03 am » posted from:Taipei,T\'ai-pei,Taiwan

You did a nice job with the EJS view. I am impressed.
I move two tab pages in evolution to constrain page.

Normally, we only put relations which will changed with time in evolution page.
like the rate equation. And the solver will use suitable to calculate the best value for next time step.
I also change the solver from midpoint method to Runge-Kutta 4th order method.
You should use this one for most of the simulation. Midpoint method is only good for constant acceleration or similar relations. It is not good for other cases (numerical error will accumulate).

I also change code so that the "play" button will toggle between "play" and "pause".

You sould be albe to create a two level system from this one.

Add one more equation in evolution page and rearrange GUI elements to model two level tanks!
Good job! Keep going!
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pavwell
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 « Embed this message Reply #24 on: March 20, 2010, 01:34:14 pm » posted from:Lutila,Banska Bystrica,Slovakia

No no This is not my job...I found this model at directory _examples in EJS install directory...I am not good enough in EJS I am very beginer...So could you change your two tanks model by this one tank model, please Thanks a lot !!
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Fu-Kwun Hwang
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 « Embed this message Reply #25 on: March 20, 2010, 03:10:45 pm » posted from:Taipei,T\'ai-pei,Taiwan

I can do it easily. However, I am not going to do it. Because it should be your job. (It is not my job).

If you are a beginer, then you should try to do it by yourself. And I will try to help.
I will feel sorry if I did the homework for you.

Adding a view element is not difficult. Just select an element(click it) from right hand side element list,
then click it inside the drawingPanel. Then assign proper value to it's property (coordinate...etc)
Please check out EJS tutorial (How to create simulation with EJS) (there are step by step examples).
or check out SWF movie show step by step instructions to create java simulation with EJS.

I will try to help if you try to lean to use it. Otherwise, there is no way you can learn to use it to communicate with MATHLAB. If you learn to do it by yourself, then you can do it for other problems, too! That is the way student learn and gain ability. I should not do all the work for you.

Do you feel if fair, if I did all the work for your classmate and he get better grade because someone did all the homework for him? (Actually, for long time effect, the classmate will be loss in the job market, because he did not learn the require ability).

I promise to help if you try to learn it. However, I should not do all the work for you. otherwise, you would learn nothing.

Try to modified from the above examples and you will find it is not really so difficult as you thought.
If you do find problem, write it down and attached the xml file and I will try to help.
This is the best I can do! O.K!
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pavwell
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 « Embed this message Reply #26 on: March 20, 2010, 03:15:45 pm » posted from:Lutila,Banska Bystrica,Slovakia

Thanks a lot !!!! I will try Have a nice day
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pavwell
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 « Embed this message Reply #27 on: March 21, 2010, 03:18:57 am » posted from:Lutila,Banska Bystrica,Slovakia

I attach my xml file which I created at EJS 3.47. Please check it because it is not exactly as in your model created at EJS 4.2. I dont know how set to fill tank with blue color and another options. Thanks.
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Fu-Kwun Hwang
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 « Embed this message Reply #28 on: March 21, 2010, 09:27:05 am » posted from:Taipei,T\'ai-pei,Taiwan

Those two polygons did not show up because you did not set "number of points"
It was different from current, which you need to set it.
The current version use all the number of elements from x,y array.

I also remove marker property, unless you really want those markers.
Please check out the version I have moified.
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 « Embed this message Reply #29 on: April 16, 2010, 04:09:17 am »

Hi,
I am a french student and I think I am doing the same project as you. But I am blocked I don't how to link the view and the equations... Could you help me please ?
Here is my work...
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Never underestimate others. Never overestimate oneself. ...Wisdom


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# dual space is a vector space

I was wondering if some one could please shed some light on why or how a dual space itself becomes a vector space over the Field. The "Finite Dimensional Vector spaces" book by Paul Halmos states."To every vector space V we make correspond the dual space V' consisting of all the linear functionals on V; " The book goes on to present the defining property of a linear functional and the definition of the linear operations for linear functionals.

Also for the sake of completion a linear functional is defined by the text as "A linear functional on a vector space v is a scalar-valued function y defined for every vector x, with the property that (identically in the vectors $x_{1}$ and $x_{2}$ and the scalar $\alpha _{1}$ and $\alpha _{2}$)

$y\left( \alpha _{1}x_{1}+\alpha _{2}x_{2}\right) =\alpha _{1}y\left( x_{1}\right) +\alpha _{2}y\left( x_{2}\right)$

Based on these definitions is n't V' composed of scalar valued functions with the above property. I fail to see any vectors present in V'. Yet the book later assumes that we must know that and starts defining a dual space V'' of a dual space V' of a vector space V.

Any help would be much appreciated.

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A vector is an element of a vector space. Thus, if $V'$ has a vector space structure all its elements are vectors. –  azarel Feb 20 '12 at 18:53
Halmos surely gives the axioms for a vector space at some point. Check that the "linear operations for linear functionals" make $V'$ into a vector space. –  Dylan Moreland Feb 20 '12 at 19:04

Let's go back further:

Let $\mathbf{V}$ and $\mathbf{W}$ be any two vector spaces over the same field $\mathbf{F}$. Let $\mathcal{L}(\mathbf{V},\mathbf{W})$ be the set of linear transformations $T\colon \mathbf{V}\to\mathbf{W}$.

We will make $\mathcal{L}(\mathbf{V},\mathbf{W})$ into a vector space over $\mathbf{F}$. In order to do this, we need to define an "addition of linear transformations" and a "scalar multiplication of elements of $\mathbf{F}$ by linear transformations" (that is, our "vectors" will be linear transformations from $\mathbf{V}$ to $\mathbf{W}$; remember that a vector space is just a set with a "vector addition" and a "scalar multiplication" that satisfy certain properties, and we call the elements of the set "vectors"; they don't have to be "tuples" in the usual sense).

So, given two linear transformations $T,U\colon \mathbf{V}\to\mathbf{W}$, we need to define a new linear transformation that is called the "sum of $T$ and $U$". I'm going to write this as $T\oplus U$, to distinguish the "sum of linear transformations" from the sum of vectors. Since we want $T\oplus U$ to be a linear transformation (which is a special kind of function) from $\mathbf{V}$ to $\mathbf{W}$, in order to specify it we need to say what the value of $T\oplus U$ is at every $\mathbf{v}\in \mathbf{V}$. My definition is: $$(T\oplus U)(\mathbf{v}) = T(\mathbf{v}) + U(\mathbf{v}),$$ where the sum on the right is taking place in $\mathbf{W}$. This makes sense, because $T$ and $U$ are already functions from $\mathbf{V}$ to $\mathbf{W}$, so $T(\mathbf{v})$ and $U(\mathbf{v})$ are vectors in $\mathbf{W}$, which we can add.

Is $T\oplus U$ a linear transformation from $\mathbf{V}$ to $\mathbf{W}$? First, it is a function from $\mathbf{V}$ to $\mathbf{W}$. Now, to check that it is a linear transformation, we need to check that for all $\mathbf{v}_1,\mathbf{v}_2\in\mathbf{V}$ and all $\alpha\in \mathbf{F}$, we have $$(T\oplus U)(\mathbf{v}_1+\mathbf{v}_2) = (T\oplus U)(\mathbf{v}_1)+(T\oplus U)(\mathbf{v}_2)\quad\text{and}\quad (T\oplus U)(\alpha\mathbf{v}_1) = \alpha((T\oplus U)(\mathbf{v}_1)).$$ Indeed, since $T$ and $U$ are themselves linear transformations, we have: \begin{align*} (T\oplus U)(\mathbf{v}_1+\mathbf{v}_2) &= T(\mathbf{v}_1+\mathbf{v}_2) + U(\mathbf{v}_1+\mathbf{v}_2) &\text{(by definition of }T\oplus U\text{)}\\ &= T(\mathbf{v}_1)+T(\mathbf{v}_2) + U(\mathbf{v}_1)+U(\mathbf{v}_2) &\text{(by linearity of }T\text{ and }U\text{)}\\ &= T(\mathbf{v}_1)+U(\mathbf{v}_1) + T(\mathbf{v}_2)+U(\mathbf{v}_2)\\ &= (T\oplus U)(\mathbf{v}_1) + (T\oplus U)(\mathbf{v}_2) &\text{(by definition of }T\oplus U\text{)}\\ (T\oplus U)(\alpha\mathbf{v}_1) &= T(\alpha\mathbf{v}_1) + U(\alpha\mathbf{v}_1) &\text{(by definition of }T\oplus U\text{)}\\ &= \alpha T(\mathbf{v}_1) + \alpha U(\mathbf{v}_1) &\text{(by linearity of }T\text{ and }U\text{)}\\ &= \alpha(T(\mathbf{v}_1) + U(\mathbf{v}_1))\\ &= \alpha((T\oplus U)(\mathbf{v}_1)) &\text{(by definition of }T\oplus U\text{)} \end{align*} so $T\oplus U$ is indeed an element of $\mathcal{L}(\mathbf{V},\mathbf{W})$.

I'll let you verify that $(S\oplus T)\oplus U = S\oplus (T\oplus U)$ for all $S,T,U\in\mathcal{L}(\mathbf{V},\mathbf{W})$ (since this is an equality of functions, you need to check that they have the same value at every $\mathbf{v}\in \mathbf{V}$). That $T\oplus U=U\oplus T$ for all $T,U\in\mathcal{L}(\mathbf{V},\mathbf{W})$; that if $\mathbf{0}$ is the linear transformation that sends every $\mathbf{v}\in\mathbf{V}$ to $\mathbf{0}\in\mathbf{W}$, then $T\oplus\mathbf{0}=T$ for all $T$; and that given $T\in\mathcal{L}(\mathbf{V},\mathbf{W})$, and we define $-T$ to be the function $(-T)(\mathbf{v}) = -(T(\mathbf{v}))$, then $T\oplus (-T) = \mathbf{0}$.

Now we define a scalar multiplication, which I will denote by $\odot$ (again, to avoid confusion with the scalar multiplication from $\mathbf{V}$ and $\mathbf{W}$. Given $T\colon \mathbf{V}\to\mathbf{W}$ and $\alpha\in\mathbf{F}$, define $(\alpha\odot T)$ to be the function $$(\alpha\odot T)(\mathbf{v}) = \alpha T(\mathbf{v}).$$ I will let you verify that this definition works, in that $\alpha\odot T$ is a linear transformation when $T$ is a linear transformation; and that it satisfies the necessary properties:

• $\alpha\odot(\beta\odot T) = (\alpha\beta)\odot T$;
• $1\odot T = T$;
• $(\alpha + \beta)\odot T = (\alpha\odot T)\oplus (\beta\odot T)$;
• $\alpha\odot(T\oplus U) = (\alpha\odot T)\oplus (\alpha\odot U)$.

So $(\mathcal{L}(\mathbf{V},\mathbf{W}),\oplus,\odot)$ is a vector space over $\mathbf{F}$ whenever $\mathbf{V}$ and $\mathbf{W}$ are vector spaces over $\mathbf{F}$.

So now, dual spaces: Note that $\mathbf{F}$ is always a vector space over itself, by defining vector addition to be the same as the addition of $\mathbf{F}$, and scalar multiplication to be the same as multiplication in $\mathbf{F}$.

So if $\mathbf{V}$ is any vector space over $\mathbf{F}$, then we can consider $\mathcal{L}(\mathbf{V},\mathbf{F})$: this makes sense, because both $\mathbf{V}$ and $\mathbf{F}$ are vector spaces over $\mathbf{F}$; and this is itself a vector space over $\mathbf{F}$ with vector addition $\oplus$ and scalar multiplication $\odot$ as defined above.

This vector space, $\mathcal{L}(\mathbf{V},\mathbf{F})$, is called the dual space of $\mathbf{V}$. We write $\mathbf{V}^*$ instead of $\mathcal{L}(\mathbf{V},\mathbf{F})$, and the elements of $\mathbf{V}^*$ are called "functionals".

By abuse of notation, we usually write $+$ instead of $\oplus$ (just like we use the same symbol for the addition of $\mathbf{V}$ and the addition of $\mathbf{W}$), and $\cdot$ or just juxtaposition instead of $\odot$.

The equation you have, $$y(\alpha_1 x_1 + \alpha_2x_2) = \alpha_1y(x_1) + \alpha_2y(x_2)$$ is just telling you that the function $y$ is a linear transformation from $\mathbf{V}$ to $\mathbf{F}$.

It is traditional to use boldface lower case letters like $\mathbf{f}$, $\mathbf{g}$, $\mathbf{h}$ to represent functionals. This to remind us that even though they are vectors in the vector space $\mathbf{V}$, they are "really" functions (when they are at home).

In fact, you could go back even further. If $\mathbf{W}$ is a vector space over $\mathbf{F}$, and $X$ is any set, then we can look at $$\mathcal{F}(X,\mathbf{W}) = \{f\colon X\to\mathbf{W}\mid f\text{ is a function}\}.$$ Then $\mathcal{F}(X,\mathbf{W})$ is a vector space, with addition $(f\oplus g)(x) = f(x)+g(x)$ and scalar multiplication $(\alpha\odot f)(x) = \alpha f(x)$. The case of $\mathcal{L}(\mathbf{V},\mathbf{W})$ corresponds to looking at a subspace of $\mathcal{F}(\mathbf{V},\mathbf{W})$ consisting of linear transformations.

This is a standard construction in abstract algebra. Whenever $A$ is an algebra (in the sense of General Algebra; a group, semigroup, ring, vector space, lattice, etc), and $X$ is a set, the collection of all function $f\colon X\to A$ becomes an algebra of the same type under "pointwise operations". In fact, this is nothing more than a "direct power" (a direct product in which every factor is the same) indexed by $X$.

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You should really write a book on linear algebra; I'm confident it would become an instant best-seller –  ItsNotObvious Feb 20 '12 at 19:28
Wow, Thanks so much for that insight. I think i understand finally what vector spaces are and how generic they are and hence why so many other fields rely on vector spaces. –  Hardy Feb 20 '12 at 19:40
A really great answer. Thanks - this really helped me also. –  user60088 Jan 29 '13 at 11:59


text:
# Another question

1. Apr 25, 2006

### beanryu

another question....

Solve the seperable differential equation for u
du/dt = e^(2u-16t)
Use the following initial condition: u(0) = 0

HINT: To determine the constant of integration after you integrated both sides, DO NOT take natural logs, but rather just set u = 0 and t = 0 with u and t both still in the exponents. After determining the constant, then you need to take logs on both sides to solve for u.

here is what I did
du/dt = e^(2u)/e^(16t)
du/e^(2u) = dt/e^(16t)
ln(e^(2u))/2 = ln(e^(16t))/2+C
if u(0)=0... everything will be zero, including C

I know I am missing the hint
can someone point it out to me?! please...

one more last question
(2 pts) A tank contains 2300 liters of pure water. Beginning at time 0, solution containing 0.02 kg of sugar per liter enters the tank at a rate of 7 L/min. The solution is mixed and the mixed solution drains out of the tank at the same rate. Let y(t) be the amount of sugar (in kg) in the tank.
(a) Give the differential equation for y'(t) (in terms of the variable y)
all I know is that y=0.02*t
all i dont know how to y'(t) interms of y.... enlighten me....

Last edited: Apr 25, 2006
2. Apr 25, 2006

### AKG

This is not right. It's true that:

$$\int \frac{du}{u} = ln(u)$$

but it's not true that:

$$\int \frac{du}{e^u} = ln(e^u)$$

or anything like it.

3. Apr 26, 2006

### beanryu

gentlemens

I got the 1st question
but I need help on the second one
the tank one


text:
Symmetry groups and Caley tables

1. Aug 30, 2007

benedwards2020

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

I have a shape about the origin. It has rotational symmetry but not reflectional symmetry (its an odd star shape!).

I have to write down in standard notation the elements of the symmetry group and I have to construct a caley table under composition of symmetries.

I think I'm getting mixed up with some of the terminology here. By standard notation does it mean

S(Q) = {e, Rpi/2, Rpi,.....) etc?

And the caley table under composition symmetries, does it mean simply the table constructed with each element across the top and down the side and then calculated in the normal way?

Thanks

2. Aug 30, 2007

matt grime

The definition of 'Standard notation' is almost certainly purely local, and definitely not standard. Sorry. You need to use whatever notation has been set in your book or lecture course.

The definition of Cayley table is precisely what you wrote.

3. Aug 30, 2007

benedwards2020

Many thanx.... Sometimes these maths books can be a bit vague


text:
# Total nuclear spin of deuteron

1. Feb 24, 2012

### Demon117

Hello all, I am having trouble understanding how this works. In Krane there arises a discussion on total angular momentum I of the deuteron. While it is true it has three components, namely the individual spins of the neutron and proton, but also the orbital angular momentum l of the nucleons as they move about their common center of mass. This total angular momentum can be denoted by

$I=s_{p} + s_{n} + l$

He continues on to talk about the different ways to couple these contributions and states there are only four possibilities. I can see the first two possibilities for total angular momentum I=1, but the other two make no sense. These are the possibilities:

(1) $s_{n}$ and $s_{p}$ are parallel with $l=0$
(2) $s_{n}$ and $s_{p}$ are antiparallel with $l=1$
(3) $s_{n}$ and $s_{p}$ are parallel with $l=1$
(4) $s_{n}$ and $s_{p}$ are parallel with $l=2$

One can see why (1) and (2) hold by inspection but (3) and (4) make my brain hurt. Perhaps I am just not seeing the correct orientation. Any suggestions?

Last edited: Feb 24, 2012
2. Feb 25, 2012

### Bill_K

matumich26, Do you know how to couple two angular momenta? You don't just add them. When you couple J1 to J2 the combined system can have any value of J in the range from their sum J1 + J2 to their difference |J1 - J2|. You must include all of these possibilities. For example when you coupled the two spins together, 1/2 ⊗ 1/2 = 1 ⊕ 0. That's the way we write it, and it means the coupled system can have either S = 1 (parallel) or S = 0 (antiparallel).

Ok, you wanted to look at the cases with S = 1. When you further couple S with L,

1 ⊗ 0 = 1
1 ⊗ 1 = 2 ⊕ 1 ⊕ 0
1 ⊗ 2 = 3 ⊕ 2 ⊕ 1

You can see in all three of these cases, I = 1 is one of the possibilities.


text:
Creating a Cell from an Existing GDSII file¶

This sample illustrates how to import an existing GDSII file into a PCell, and assign ports and a circuit model to it.

There are cases when you want to integrate an existing old design into a new Ipkiss design, but where only the GDSII output is available. Ipkiss can wrap this GDSII file into an Ipkiss cell, and then you can manually add the ports, netlist and model so you can use it into a larger circuit.

Note that all the layers used in the GDSII must first be defined in the Technology file. This may not always be the case. (see example Creating a Cell from an Existing GDSII file with a new Technology).

Importing from GDSII

Illustrates

1. Write a Cell to a GDSII file
2. Import a GDSII file into an IPKISS component
3. Assign a CircuitModel to the imported component

Export a grating coupler to GDSII¶

We first write a fiber coupler to gdsii so we can import it again later. Note that all the layers used in the gdsii must be defined in the technology file.

In this example we will use technologies.silicon_photonics, which is an example TECH shipped with IPKISS.

from technologies import silicon_photonics
from ipkiss3 import all as i3
import numpy as np
import pylab as plt

Then, we instantiate a grating coupler, and write it to my_grating.gds:

from picazzo3.fibcoup.curved.cell import FiberCouplerCurvedGrating # Here we use a fiber_coupler from Picazzo
my_grating = FiberCouplerCurvedGrating(name="unique_grating_name_used_in_GDSII")  # We give a unique name to the cell that will be used in the gdsii file.
my_grating_layout = my_grating.Layout()
my_grating_layout_ports = my_grating_layout.ports
my_grating_layout.visualize()
my_grating_layout.write_gdsii("my_grating.gds")

We now define a grating coupler simulation model. It is a simple gaussian model with a fixed reflection parameter (which is useful to model light that reflects back and forth in the circuit because of the grating coupler).

class GratingModel(i3.CompactModel):

parameters = [
'bandwidth_3dB',
'peak_transmission',
'center_wavelength',
'reflection'
]

terms = [
i3.OpticalTerm(name='in'),
i3.OpticalTerm(name='vertical_in'),
]

def calculate_smatrix(parameters, env, S):
sigma = parameters.bandwidth_3dB / 2.35482 # fwhm => sigma
power_S = parameters.peak_transmission * np.exp(-(env.wavelength - parameters.center_wavelength)**2.0 / (2.0 * (sigma**2.0)))
S['vertical_in', 'in'] = S['in', 'vertical_in'] = np.sqrt(power_S)
S['in', 'in'] = parameters.reflection

Importing the GDSII file¶

We use i3.GDSCell to import the GDSII file. In addition, we also

• Assign it ports
• Assign it a simulation model based on GratingModel that we defined in the code above.
class ImportedGrating(i3.GDSCell):

def _default_filename(self):
return 'my_grating.gds' # path to the gdsii file that contains the cell to be imported

def _default_cell_name(self):
return 'unique_grating_name_used_in_GDSII' # name of the cell to be imported inside the gdsii file.

class Layout(i3.GDSCell.Layout):

def _generate_ports(self, ports):
ports += i3.OpticalPort(name='in', position=(20.0, 0.0), angle=0.0) # We have to manually set the ports as this info is not in the gdsii file yet
ports += i3.VerticalOpticalPort(name="vertical_in",
position=(0.0,0.0),
inclination=90.0,
angle=0.0)  # For the fiber a vertical port is used.

return ports

# Here we use our compact model. The netlist is automatically inferred from the layout.
class CircuitModel(i3.CircuitModelView):

center_wavelength = i3.PositiveNumberProperty(default=1.55, doc="center wavelength [um]")
bandwidth_3dB = i3.PositiveNumberProperty(default=0.060, doc="3dB bandwidth [um]")
peak_transmission = i3.NonNegativeNumberProperty(default=1.0, doc="peak transmission (0 to 1)")
reflection = i3.ComplexFractionProperty(default=0.0, doc="Complex reflection back into the waveguide")

def _generate_model(self):
return GratingModel(center_wavelength=self.center_wavelength,
bandwidth_3dB=self.bandwidth_3dB,
peak_transmission=self.peak_transmission,
reflection=self.reflection)

Instantiate the imported cell with its Layout view

im_cell = ImportedGrating()
im_layout = im_cell.Layout()
im_layout.visualize()

Instantiate the CircuitModel view and do a simulation

wavelengths = np.arange(1.5, 1.6, 0.001)
im_caphemodel = im_cell.CircuitModel(bandwidth_3dB=0.1, peak_transmission=0.6)
S = im_caphemodel.get_smatrix(wavelengths=wavelengths)
plt.figure()
plt.plot(wavelengths, np.abs(S['in', 'vertical_in'])**2)
plt.show()