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# Yang Mills
Standard
To define the Yang-Mills Lagrangian, we need to define the ‘Trace’ of an End(E) valued form. Recall that the Trace of a matrix is the sum of its diagonal enTries. The Trace is independent of the choice of basis – an invariant notion that is independent of the choice of basis. A definition of the Trace that mkes this clear is as follows. Consider ${End(V) \simeq V \otimes V^*}$ – an isomorphism that does not depend on any choice of basis – so the pairing between V and ${V^*}$ defines a linear map $Tr: End(V) \rightarrow \mathbb{R}$ $v \otimes f \mapsto f(v)$ To see that this v is really a Trace, pick ${e_i}$ of V and let ${\epsilon^j}$ be a dual basis of ${V^*}$. Writing ${T \;\in\; End(V)}$ as $T = T^i_j e_i \otimes \epsilon^j$ We have $Tr(T) = T^i_je_i(\epsilon^j) = T^i_j \delta_i^j = T^i_i$ which is of course the sum of the diagonal enTries. \newline This implies that if we have a section T of ${End(E),}$ we can define a funciton Tr(T) on the base manifold M whose value at ${p \in M}$ is the Trace of the endomorphism T(p) of the fiber ${E_p}$: $Tr(T)(p) = Tr(T(p))$ If ${T \in \Gamma(End(E))}$ and ${\omega \in \Omega^p(M)}$ we define
$\displaystyle Tr(T \otimes \omega) = Tr(T)\omega$
Now we can write down the Yang-Mills Lagrangian: If D is a connection on E, this is the n-form given by
$\displaystyle \mathcal{L}_{YM} = \frac{1}{2} Tr(F \wedge \textasteriskcentered F) \ \ \ \ \ (1)$
where F is the curvature of D. Note that by the defintion of the hodge star operator (also in this collection of notes), we can write this in local co-ordinates as
$\displaystyle \mathcal{L}_{YM} = \frac{1}{4} Tr (F_{\mu \nu}F^{\mu \nu})vol \ \ \ \ \ (2)$
If we integrate ${\mathcal{L}_{YM}}$ over M we get the Yang-Mills action
$\displaystyle S_YM = \frac{1}{2} \int_M Tr (F \wedge \textasteriskcentered F) \ \ \ \ \ (3)$
This needs some elaboration. So let us explain these formulas better. We choose the physics convention ${F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu -ig[A_\mu,A_\nu]}$ where the generators of the Lie algebra are Hermitian.
$\displaystyle \mathcal{L}_{YM} = \mathcal{I} = - \int Tr(F \wedge \textasteriskcentered F) \ \ \ \ \ (4)$
by another convention (there is a lot of ambiguity of signs in this subject). The first thing to note is that F has vector and Lie algebra indices. The Trace is over the Lie algebra, not over the vector indices. The vector indices are just those of the field sTrength in QED. In Yang-Mills the curvature form is Lie Algebra valued. \newline In this case ${F_{\mu \nu} = F^{a}_{\mu \nu}T^a}$ where the summation convention is used,and where ${T^a}$ are the generators of ${\mathfrak{su}(n)}$. To be explicit, F has not only tensor components but maTrix components $(F_{\mu \nu})_{ij} = F^{a}_{\mu \nu}T^a T^a_{ij}$ The inner product of F with itself ${ = \int F \wedge \textasteriskcentered F}$ where \textasteriskcentered is the Hodge \textasteriskcentered – operator. Thus we are calculating ${\mathcal{I} = -Tr}$. It is a standard exercise to find the exterior product of two r-forms. We find $F \wedge \textasteriskcentered F = \frac{1}{2!}F_{\mu \nu}F^{\mu \nu} dx^1 \wedge \cdots \wedge dx^4$ Note that the differential forms don’t ‘know’ the Lie algebra. The algebra hasn’t come into the calcuation yet. ${Tr(F_{\mu \nu}F^{\mu \nu})=Tr(F^{a}_{\mu \nu}T^aF^{\mu \nu b}T^b}$ ${= Tr(T^aT^b)F^{a}_{\mu \nu}F^{\mu \nu b}}$ ${= \frac{1}{2}\delta^{ab}F^{a}_{\mu \nu}F^{\mu \nu b}}$ ${=\frac{1}{2}F^{a}_{\mu \nu}F^{\mu \nu a}}$ where e have used the standard normalization convention for the ${T^a}$, ${Tr T^a T^b = \frac{1}{2}\delta^{ab}}$. (This comes from the fact we want ${\mathfrak{su}(2)}$ to live in ${\mathfrak{su}(n)}$, and the generators of ${\mathfrak{su}(2)}$ are taken to be ${T^a = \frac{\sigma^a}{2}}$ where ${\sigma^a}$ are the Pauli matrices.) Thus, we find $\mathcal{I} = - Tr (F \wedge \textasteriskcentered F)$ which can be written as $\frac{1}{4} \int d^4 x F^{a}_{\mu \nu}F^{a \mu \nu}$
# Tony Judt on Society
Standard
I posted the following on my Facebook account last night
‘Thinking “economistically,” as we have done now for thirty years, is not intrinsic to humans. There was a time when we ordered our lives differently.’
– Tony Judt
A friend of mine challenged me to provide some analysis, so here I provide some.
Tony Judt was writing a book aimed at young people who en masse seem to have lost a political activism that previous generations have. There are serious problems caused by rampant neo-liberalism and the domination of policy making by economic concerns. Someone quipped to me that the religion of our age is ‘pop culture, economics, business and money making’
http://www.nybooks.com/articles/archives/2010/apr/29/ill-fares-the-land/?pagination=false
Tony was thinking of the fact that policy considerations are largely dominated by economic concerns. Not to mention the economic dogma of the Chicago School – .Empirically the ‘efficient market hypothesis’ is false. We can have an argument about how false it is, some other time. Also I think Judt makes a very clear point (see any of his articles on this on the NYRB on this set of topics) that economic considerations dominating political discussion, aren’t a natural occurrence they are a matter of taste. What about ‘is it right’. I don’t think notions of ‘fairness’ or ‘morality’ should be neglected in political discourse. And this is very important for some defence of the social democratic model.
My friend Sam posted this “In the modern age of policy, economic analysis supersedes other decision-making criteria that most of us use every day such as ethics, morality, and principles such as robustness and precaution. As a result we have entered in to a political paradigm which totally relies on models to justify a government’s supposedly utilitarian agenda. The choice and subsequent blame becomes not that of the elected decision maker but put squarely on the models and the limits of human ability in building such a model. Blame becomes diffuse, as does responsibility. This makes for bad policy making.” on his blog a few years ago. It seemed profound then and it seems profound now.
# On Fubini-Study Metrics
Standard
1. A Little Complex Analysis
We want to introduce the notion of a ‘Fubini-Study’ metric which is important in Complex Manifold Theory and Differential Geometry (and the associated theories such as Mathematical Physics). But first we need to introduce a little Complex Analysis. The source is of course Griffiths and Harris. Let M be a complex manifold, ${p \in M}$ any point, and ${z=(z_{1},\cdots,z_{n})}$ a holomorophic co-ordinate system around p. There are three different notions of a tangent space to M at p,which we now describe:
• ${T_{\mathbb{R},p}(M)}$ is the usual real tangent space to M at p,when we consider M a real manifold of dimension 2n. ${T_{\mathbb{R},p}(M)}$ can be realized as the space of ${\mathbb{R}-}$linear derivations on the ring of real-valued ${C^{\infty}}$-functions in a neighbourhood of p; if we write ${z_i = x_i + iy_i}$, ${T_{\mathbb{R},p}(M) = \mathbb{R}(\frac{\partial}{\partial x_{i}}, \frac{\partial}{\partial y_i}}$.
• ${T_{\mathbb{C},p}(M) = T_{\mathbb{R},p}(M)\otimes_{\mathbb{R}} \mathbb{C}}$ is called the complexified tangent space to M at p. It can be realized as the space of ${\mathbb{C} -}$ linear derivations in the ring of complex valued ${C^{\infty}}$-functions on M around p. We can write ${T_{\mathbb{C},p}(M) = \mathbb{C}{\frac{\partial}{\partial x_{i}},\frac{\partial}{\partial y_i}}}$
=${\mathbb{C}{\frac{\partial}{\partial z_{i}},\frac{\partial}{\partial \bar{z}_i}}}$
• ${T'_p(M)= \mathbb{C}{\frac{\partial}{\partial z_{i}}}\subset T_{\mathbb{C}, p}(M)}$ is called the holomorphic tangent space to M at p. It can be realized as the subspace of ${T_{\mathbb{C},p}(M)}$ consisting of derivations that vanish on antiholomorphic functions (i.e. F such that T is holomorphic), and so is independent of the holomorphic co-ordinate system chosen. The subspace ${T''_p(M)= \mathbb{C}{\frac{\partial}{\partial \bar{z}_{i}}}}$ is called the antiholomorphic tangent space to M at p; clearly ${T_{\mathbb{C},p}(M) = T'_p(M) \oplus T''_p(M)}$
Now we consider some Calculus on Complex Manifolds. Let M be a complex manifold of dimension n. A hermitian metric on M is given by a positive definite hermitian inner product ${(,)_z: T'_z(M) \otimes T'_z(M) \rightarrow \mathbb{C}}$ on the holomorphic tangent space at z for each ${z \in M}$,
depending smootly on z – that is, such that for local co-ordinates z on M the function
${h_ij(z) = (\frac{\partial}{\partial z_i},\frac{\partial}{\partial z_j})_z}$ are ${C^{\infty}}$
Writing ${(,)_z}$ in terms of the basis ${{dz_i \otimes d\bar{z}_j}}$ for ${(T'_z(M) \otimes \bar{T'_z(M)}^{\textasteriskcentered} = T^{\textasteriskcentered\textquoteright}_z(M) \otimes T^{* \textquotedblright}_{z}(M)}$, the hermitian metric is given by ${ds^{2} = \sum_{i,j} h_{ij}(z) dz_i \otimes d \bar{z}_j}$ So let us describe the Fubini-Study Metric Let ${z_0,\cdots,z_n}$ be co-ordinates on ${\mathbb{C}^{n+1}}$ and denote by ${\pi:\mathbb{C}^{n+1} -{0} \rightarrow \mathbb{P}^n}$ the standard projection map. Let ${U \subset \mathbb{P}^{n}}$ be an open set and ${Z: U \rightarrow \mathbb{C}^{n-1} - {0}}$ a lifting of U, i.e. a holomorphic map with ${\pi \circ z = id}$; consider the differential form
${\omega = \dfrac{i}{2\pi}\partial \bar{\partial}log\|z\|^{2}}$ If ${Z':U \rightarrow \mathbb{C}^{n-1} - {0}}$ is another lifting, then ${Z' = f.Z}$ with f a nonzero holomorphic function, so that
${\dfrac{i}{2\pi}\partial \bar{\partial}log\|z\|^{2} = \frac{i}{2 \pi}\partial \bar{\partial} (log\|z\|^{2} + log f + log \tilde{f})}$
${= \omega + \dfrac{i}{2\pi}(\partial \bar{\partial}log f - \bar{\partial} \partial log \tilde{f})}$ = ${\omega}$ Therefore ${\omega}$is independent of the lifting chosen; since liftings always exist locally, ${\omega}$ is a globally defined differential form in ${\mathbb{P}^{n}}$. (By the sheaf properties of differential forms) Clearly ${\omega}$ is of type (1,1). To see that ${\omega}$ is positive, first note that the unitary group ${U(n+1)}$ acts transitively on ${\mathbb{P}^{n}}$ and leaves the form ${\omega}$ positive everywhere if it is positive at one point. Now let ${{w_i = z_i/z_0}}$ be co-oridnates on the open set ${U_{0} = (z_0 \neq 0)}$in ${\mathbb{P}^{n}}$ and use the lifting ${Z = (1,w_1,\cdots,w_n)}$ on ${U_0}$ ; we have (after some substitutions
${\omega = \dfrac{i}{2 \pi} [\frac{\sum dw_i \wedge d\bar{w}_i}{1 + \sum w_i \bar{w}_i} - \frac{(\sum \bar{w}_i dw_i \wedge \sum w_i d\bar{w}_i)}{(1 + \sum w_i \bar{w}_i)^{2}}]}$ At the point ${[1,0,\cdots,0]}$, \\ ${\omega = \frac{i}{2\pi} \sum dw_i \wedge d \bar{w}_i > 0}$ Thus ${\omega}$ defines a particular hermitian metric on the projective complex space called the Fubini-Study metric. That was the aim of the article!
Cosma Shalizi, has an excellent talk on Academic talks.
I merely quote my favourite part:
1. The point of the talk is not to please you, by reminding yourself of what a badass you are, but to tell your audience something useful and interesting. (Note to graduate students: It is important that you internalize that you are, in fact, a badass, but it is also important that then you move on. Needing to have your ego stroked by random academics listening to talks is a sign that you have not yet reached this stage.) Unless something matters to your actual message, it really doesn’t belong in the main body of the talk.
2. You can stick an arbitrary amount of detail in the “I’m glad you asked that” slides, which go after the one which says “Thank you for your attention! Any questions?”.
3. You also can and should put all these details in your paper, and the people who really care, to whom it really matters, will go read your paper. Once again, think of an academic talk as an extended oral abstract.
Internalise that you are in fact a bad ass. I wish more Professors gave advice like that.
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2015 ENERGY CONFERENCE SUMMARY & PRESENTATION RELEASE DATES ENERGY CONFERENCE
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#31
10-19-2010, 10:44 AM
Mario Senior Member Join Date: Apr 2007 Posts: 279
Ok, I want to calculate it anyway .
We have:
C1 (now in series) which is 0.0235F with a charge of 0.0235F @ 20V = 0.47J
C2 which is 0.094F with a charge of 0.094F @ 10V = 0.94J
To find the final voltage:
Ctot.= 0.0235F + 0.094F = 0.1175F
Qtot.= 0.47Q + 0.94Q = 1.41Q
Vtot. = 1.41Q / 0.1175F = 12V
So you were right, even if we were not talking about the same exact setup.
This confirms the loss I am seeing by measurement.
regards,
Mario
__________________
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#32
10-19-2010, 12:54 PM
Armagdn03 Silver Member Join Date: Oct 2007 Posts: 902
Quote:
Originally Posted by Mario Ok, I want to calculate it anyway . We have: C1 (now in series) which is 0.0235F with a charge of 0.0235F @ 20V = 0.47J C2 which is 0.094F with a charge of 0.094F @ 10V = 0.94J To find the final voltage: Ctot.= 0.0235F + 0.094F = 0.1175F Qtot.= 0.47Q + 0.94Q = 1.41Q Vtot. = 1.41Q / 0.1175F = 12V So you were right, even if we were not talking about the same exact setup. This confirms the loss I am seeing by measurement. regards, Mario
You are not seeing a loss in measurement, unless you are not completing the cycle.
For example at this point:
C1 = .094F @ 12V
C2 = .0235F @ 12V
If we switch C2 into parallel again, we have .094F @ 6V.
C1 = .094F @ 12V
C2 = .094F @ 6V.
Which gives us a charge of
C1= 0.094c * 12v = 1.128q
C2 = 0.094c * 6v = 0.564q
Qtot = 1.128 + 0.564 = 1.844
And we have a capacitance total of .094 + .094 = 0.188
Now knowing we that Vtot = Qtot/Ctot 1.844 / 0.188 = 9.85V across C1 and C2, and they started at 10v each.
which is 98.5% of the original energy, and I would be willing to bet that the 1.5% missing is due to rounding error.
So where is this loss?
__________________
#33
10-19-2010, 03:45 PM
Mario Senior Member Join Date: Apr 2007 Posts: 279
Hi Andrew,
there's a mistake in your calculation:
Qtot = 1.128 + 0.564 = is 1.692 not 1.844
Capacitance total of .094 + .094 = 0.188
So Vtot = Qtot/Ctot 1.1.692 / 0.188 = 9V across C1 and C2.
So we started with C1+C2 (0.094F + 0.094F)= 0.188F @ 10V = 9.4 Joules
After we switched C1 to series and back to parallel we end up with C1 + C2 = 0.188F @ 9V = 7.614 Joules
This means in one complete cycle we've lost 19% of the initial energy. Btw 9V is what I actually measured when connecting back to parallel.
If we summarize the individual steps:
At start C1 is in parallel, total energy of C1+C2 0.188F @ 10V is 9.4 Joules.
1) C1 gets switched to series (20V) and discharges into C2 (10V) resulting in 12V across C1-C2. Energy left is 0.5*((0.0235F + 0.094F)*12V^2) = 8.46 Joules. In one passage we have lost 10% of the initial energy.
2) C1 gets switched back to parallel (6V) and receives discharge coming back from C2 (12V) resulting in 9V across C1-C2. Energy left is 0.5*((0.094F + 0.094F)*9V^2) = 7.614 Joules. In this step too we have lost 10% of energy with regards to the 8.46 Joules of the previous step. But in one complete cycle (back and forth) we have lost 19% of energy since we are left with 7.614 Joules with regards to 9.4 Joules we had at start.
Does this sound right?
regards,
Mario
__________________
#34
10-19-2010, 04:22 PM
Jetijs Gold Member Join Date: Aug 2007 Posts: 2,134
Hi guys
I am following this thread with interest. I think that charging caps and putting them in series and parallel is not the same as physically changing a capacitance of a single cap. There mist ne a difference. To test the concept out I would use a variable capacitor that is made so that it can rotate around full 360 degrees and put a small high speed motor on the shaft. Then I would put this small cap in parallel with a standard cap through a low impedance load (transformer primary). I guess to see the difference with a capacity that small, I would need to use several hundereds of volts.
Thanks,
Jetijs
__________________
It's better to wear off by working than to rust by doing nothing.
#35
10-19-2010, 05:33 PM
Mario Senior Member Join Date: Apr 2007 Posts: 279
Hi Jetijs,
this is actually the next thing I wanted to ask Andrew, maybe moving plates is different than switching caps, then again the calculations seem to work for the switched caps too and reflect what I measured in the real world. So, who knows...
regards,
Mario
__________________
#36
10-19-2010, 05:44 PM
Armagdn03 Silver Member Join Date: Oct 2007 Posts: 902
Quote:
Originally Posted by Mario Hi Andrew, there's a mistake in your calculation: Qtot = 1.128 + 0.564 = is 1.692 not 1.844 Capacitance total of .094 + .094 = 0.188 So Vtot = Qtot/Ctot 1.1.692 / 0.188 = 9V across C1 and C2. So we started with C1+C2 (0.094F + 0.094F)= 0.188F @ 10V = 9.4 Joules After we switched C1 to series and back to parallel we end up with C1 + C2 = 0.188F @ 9V = 7.614 Joules This means in one complete cycle we've lost 19% of the initial energy. Btw 9V is what I actually measured when connecting back to parallel. If we summarize the individual steps: At start C1 is in parallel, total energy of C1+C2 0.188F @ 10V is 9.4 Joules. 1) C1 gets switched to series (20V) and discharges into C2 (10V) resulting in 12V across C1-C2. Energy left is 0.5*((0.0235F + 0.094F)*12V^2) = 8.46 Joules. In one passage we have lost 10% of the initial energy. 2) C1 gets switched back to parallel (6V) and receives discharge coming back from C2 (12V) resulting in 9V across C1-C2. Energy left is 0.5*((0.094F + 0.094F)*9V^2) = 7.614 Joules. In this step too we have lost 10% of energy with regards to the 8.46 Joules of the previous step. But in one complete cycle (back and forth) we have lost 19% of energy since we are left with 7.614 Joules with regards to 9.4 Joules we had at start. Does this sound right? regards, Mario
Thank you for catching my mistake! This exchange has been very fruitful, thanks again Mario!
While the system that you (mario) are describing is different, i couldn't put my finger on why, but thanks to your checking my numbers and forcing me to do some calculations I believe I have nailed it.
First I setup a spreadsheet, and checked various values.
The 3 columns represent the various switching of the capacitors. Column 1 is our starting conditions capacitors equal.
Column 2 is after capacity is reduced by factor of 4, and voltage is doubled, as you can see the joules stay equal as they should. The 3rd column represents the switching back to the original position.
At the bottom of each column you can see I have the total joules for each step, and as you can see each step throws away 10% of the total energy, going from 1000 to 900 to 810.
Good catch Mario.
Here is the large difference between Mario's example, and what I propose:
When capacitors are switched from parallel to series, and then allowed to discharge by an amount, A certain amount of charge is nullified, never to return.
If you imagine a capacitor which has (generating random number)...+10 charge on one plate and -10 charge on the other.
This capacitor has +10 and -10 charges to nullify into each other to produce work.
If you place two in series (C1,2) (see image above) The outer plates have +10 and -10, and the inner plates have +10 and -10. As the outer plates discharge into our other capacitor (previously labled C3,4) the charge is conserved. However the inner plates nullify directly into each other, nullifying charge.
The outer plates conserve charge shuttling into the other cap, while the inner plates nullify their charge into each other to an extent.
Thus each time you go into series mode, and discharge a portion, you "loose" some of your total charge...Thus what you have shown is NOT as charge CONSERVING capacitive spring.
In my proposition, Charge is 100% conserved omitting leakage.
__________________
#37
10-19-2010, 05:46 PM
Armagdn03 Silver Member Join Date: Oct 2007 Posts: 902
Quote:
Originally Posted by Mario Hi Jetijs, this is actually the next thing I wanted to ask Andrew, maybe moving plates is different than switching caps, then again the calculations seem to work for the switched caps too and reflect what I measured in the real world. So, who knows... regards, Mario
Quote:
Hi guys I am following this thread with interest. I think that charging caps and putting them in series and parallel is not the same as physically changing a capacitance of a single cap. There mist ne a difference. To test the concept out I would use a variable capacitor that is made so that it can rotate around full 360 degrees and put a small high speed motor on the shaft. Then I would put this small cap in parallel with a standard cap through a low impedance load (transformer primary). I guess to see the difference with a capacity that small, I would need to use several hundereds of volts. Thanks, Jetijs
Hopefully my previous response outlines the difference between proposed systems.
Cheers!
__________________
#38
10-19-2010, 06:01 PM
Armagdn03 Silver Member Join Date: Oct 2007 Posts: 902
Quote:
Originally Posted by DrStiffler @Armagdn03 People have tried 10k RPM motors driving disks etc., and I have never heard of one going farther than the initial work and observation. Theory look nice, yet what is the missing part?
Quote:
To test the concept out I would use a variable capacitor that is made so that it can rotate around full 360 degrees and put a small high speed motor on the shaft. Then I would put this small cap in parallel with a standard cap through a low impedance load (transformer primary). I guess to see the difference with a capacity that small, I would need to use several hundereds of volts.
@DrStiffler
@Jetis
This is a photo of Chris Carson device. According to Dr. Lindemann...
Quote:
The machine has two sections where stator plates and rotor plates rotate in-between each other, varying the capacitance. These two sections are 180* out of phase with each other, so when the rotor plates are connected to a source of 5,000 volts, and spun at 10,000 rpm, the two sections of the stator produce an AC signal between them. Drawing power from the machine produces no drag on the input motor.
http://www.free-energy.ws/images/ccvcg2.jpg
If there were a second capacitor, with a load in series, this would be a great rotational candidate for a Charge Conserving Capacitive Spring. Beautiful model, however I do not plan to take the mechanical route, as the limitations are......big.
__________________
Last edited by Armagdn03; 10-19-2010 at 06:06 PM.
#39
10-19-2010, 09:26 PM
Mario Senior Member Join Date: Apr 2007 Posts: 279
Hi again guys,
Andrew, just trying to be logic: If the theory and calculations we made work for the cap switching setup (in line with my actual measurements), does this mean that theory doesn't apply to your system? Can't work for both if you say your system doesn't have the losses my setup has....
About your explanation about the inner plates being the cause of the loss, not sure about that. I just did a few more tests and calculations following the last ones I did. I wanted to replace the C1 in series (0,0235F) of the previous setup with a similar sized cap also charged to 20V, to avoid the inner plates and just have two plates. I didn't have a 23'500uF cap so I used a 47'000uF cap, C2 still being a 94'000uF cap. I made 3 tests with C1 at 3 different voltages discharging into the same C2 @ 10V as the previous setup.
1. C1 charged to 20V
2. C1 charged to 25V
3. C1 charged to 15V
I'm not going to rewrite all the calculations but they follow the ones of the previous setup.
results:
C1 @ 20V: Start energy 14.1 Joules. Charge at end of discharge across the system: 12.527 Joules. Loss: 11.15%
C1 @ 25V: Start energy 19.388 Joules. Final charge: 15.862 Joules. Loss: 18.183%
C1 @ 15V: Start energy 9.988 Joules. Final charge 9.596 Joules. Loss: 3.92%.
The calculations are in line with the measurements I made. This means the losses are not due to the inner plates since C1 in this case is only one capacitor.
These tests also confirm what I said about the losses getting bigger the greater the voltage difference between C1 and C2.
regards,
Mario
__________________
#40
10-20-2010, 01:02 AM
Armagdn03 Silver Member Join Date: Oct 2007 Posts: 902
Quote:
Originally Posted by Mario Hi again guys, Andrew, just trying to be logic: If the theory and calculations we made work for the cap switching setup (in line with my actual measurements), does this mean that theory doesn't apply to your system? Can't work for both if you say your system doesn't have the losses my setup has.... About your explanation about the inner plates being the cause of the loss, not sure about that. regards, Mario
Mario,
My setup has losses, like all do, leakage etc. When I spoke of a difference between the two I was referring to the loss of charge due to the direct action of the circuit. The capacitor switching device nullifies part of its charge as the series connected capacitors are discharged. This can be shown looking at this sheet again.
http://i210.photobucket.com/albums/b...tyswitcher.jpg
As you can see the system starts out in equilibrium with a charge of 200 coulombs. Durring the next cycle we have 150 coulombs, and we end at 180 coulombs. As you can see the charge does not stay constant through the cycle.
Also note that begining = 200q and ending state = 180q, so where did the 20q go?
When one set of caps is switched to series configuration, it goes from having 100 coulombs on it to 50 across it with a doubling of voltage bringing it 20v. When discharged into the second cap (of parallel configuration) the voltage drops from 20v, to 12v.
The difference in potential is obviously 8v,
If we only look at the series configured cap, we have 2.5Farads, and an 8v loss...Q=VC so, 2.5c*8v = 20 coulombs. Thus, the 8v drop caused a 20 coulomb loss in this capacitor.
The other capacitor is at 10f and 12v giving it a 20 coulomb gain.
however when we continue to work through,
Now we switch it back to parallel, giving it 6v @ 10farads = 60 coulombs
The parallel cap is siting at 12v @ 10 farads, = 120 coulombs
Combined = 60c + 120C = 180C total when all is said and done. The 20 coulombs lost from switching to series follows through to the end, for a loss of 20 coulombs for the cycle. This means that when the next cycle starts, there is less than there was previously to work with, constituting loss.
Think good and hard about why switching to series makes 50% of the charge appear to go missing. About how any apparent loss of coulombs from this state, is actually Doubled, since what ever is lost on the outer plates is also nullified on the inner plates as they discharge directly into, nullifying one another. (example, parallel cap has 100q, switched to series it has 50q, discharge 20 of that and you have 30q. Switch back to parallel and you will have 60q only. Though you lost 20 in series, you really lost 40 total.
In my version however, the charge stays 100% the same through the ENTIRE process, since there is no series switching, no charge can ever be nullified.
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#41
10-20-2010, 09:48 AM
Mario Senior Member Join Date: Apr 2007 Posts: 279
Andrew, ok. But did you read the whole last post I made? I replaced C1 comprised of two caps by C1 made of only one cap. No inner plates, and when I discharge this cap to C2 I still have the losses I had in the previous setup. Please read my last post and tests again.
I also repeat the question: If the theory and calculations work for my setup (and they do since measurement confirmed it), does it mean they don't apply to yours since you don't seem to have losses except minimal ones? How can the same theory support both our systems?
regards,
Mario
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#42
10-20-2010, 12:56 PM
Armagdn03 Silver Member Join Date: Oct 2007 Posts: 902
Quote:
Originally Posted by Mario Andrew, ok. But did you read the whole last post I made? I replaced C1 comprised of two caps by C1 made of only one cap. No inner plates, and when I discharge this cap to C2 I still have the losses I had in the previous setup. Please read my last post and tests again. regards, Mario
@ Mario,
There will always be loss whenever a cap is discharged into another cap, this is not my argument.
My arguement is that action and reaction are separated by time with my setup, so that they may both be used.
step 1. Rise the energy state of the system. Work is accomplished. Work input = work output.
Step 2) let the system free, and let it return to its own equilibrium. again work is accomplished.
note: The energy of my system is exactly the same when the process is completed as when it started. If you start with X volts and Y joules, you end with X volts and Y joules.
When the capacitor switching system completes one cycle, it is at a lower energy state than it began, that is a large clue as to how these are very different systems.
Quote:
I also repeat the question: If the theory and calculations work for my setup (and they do since measurement confirmed it), does it mean they don't apply to yours since you don't seem to have losses except minimal ones? How can the same theory support both our systems?
You are correct. What applies to yours, does not apply to mine and vice versa. I wanted to work on what you proposed anyways because it is similar, but I could not pin point why charge went missing for a bit. Once I did you can see that I posted that they were not the same, and the same maths, analysis can not be used. As for theory, one theory should describe everything, but different things will have different explanations!
Quote:
Armagdn03: Thus each time you go into series mode, and discharge a portion, you "loose" some of your total charge...Thus what you have shown is NOT as charge CONSERVING capacitive spring.
Your method allows a nullification of charge, mine does not, therefore they are fundamentally different.
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#43
10-20-2010, 02:36 PM
Jbignes5 Gold Member Join Date: Mar 2009 Location: NY, USA Posts: 1,014
Why not?
Quote:
Originally Posted by Armagdn03 @ Mario, There will always be loss whenever a cap is discharged into another cap, this is not my argument. My arguement is that action and reaction are separated by time with my setup, so that they may both be used. step 1. Rise the energy state of the system. Work is accomplished. Work input = work output. Step 2) let the system free, and let it return to its own equilibrium. again work is accomplished. note: The energy of my system is exactly the same when the process is completed as when it started. If you start with X volts and Y joules, you end with X volts and Y joules. When the capacitor switching system completes one cycle, it is at a lower energy state than it began, that is a large clue as to how these are very different systems. You are correct. What applies to yours, does not apply to mine and vice versa. I wanted to work on what you proposed anyways because it is similar, but I could not pin point why charge went missing for a bit. Once I did you can see that I posted that they were not the same, and the same maths, analysis can not be used. As for theory, one theory should describe everything, but different things will have different explanations! Your method allows a nullification of charge, mine does not, therefore they are fundamentally different.
Why not do the twin variable cap setup and loose nothing except for the losses in the small traces that are linking the caps. Have two caps as tanks and use the variable caps to slosh the bucket so to say and generate from the side to side motion you create with the variable caps. It would be like a 10 foot bucket with a small radius. Only variate a small portion on the top. If the caps were of a higher quality I bet you would loose little from the process. You could even design a way to make the variable cap better as well by vacuum or oil filled methods. With the oil filled the best in dielectric strength but slower to turn.
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#44
10-20-2010, 05:14 PM
Armagdn03 Silver Member Join Date: Oct 2007 Posts: 902
Quote:
Originally Posted by Jbignes5 Why not do the twin variable cap setup and loose nothing except for the losses in the small traces that are linking the caps. Have two caps as tanks and use the variable caps to slosh the bucket so to say and generate from the side to side motion you create with the variable caps. It would be like a 10 foot bucket with a small radius. Only variate a small portion on the top. If the caps were of a higher quality I bet you would loose little from the process. You could even design a way to make the variable cap better as well by vacuum or oil filled methods. With the oil filled the best in dielectric strength but slower to turn.
you are spot on.
Two variable capacitors would be ideal for increase in power output, however for sake of simplicity, and ease of explanation, I am describing the system with only one variable capacitor.
I have designed my own special variable capacitance for this. It requires no moving parts, however the voltage across my system is 10KV - 12KV before the decrease in capacitance takes place, meaning that even with good insulation, leakage will be something to watch.
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#45
10-20-2010, 05:28 PM
Jbignes5 Gold Member Join Date: Mar 2009 Location: NY, USA Posts: 1,014
Quote:
Originally Posted by Armagdn03 you are spot on. Two variable capacitors would be ideal for increase in power output, however for sake of simplicity, and ease of explanation, I am describing the system with only one variable capacitor. I have designed my own special variable capacitance for this. It requires no moving parts, however the voltage across my system is 10KV - 12KV before the decrease in capacitance takes place, meaning that even with good insulation, leakage will be something to watch.
Ok there is a big difference in a single variable cap and twin ones. One would be harmonic(twin) and the other non harmonic(single). The harmonic one would be like putting a teeter tawter in the system and the other would be like jacking the one side up. The harmonic one would be the less leaky since it would have an increase with the associated decrease on the other side. This would make it more powerful as well and you wont have to have extreme high voltages like you are talking then.
On the subject of leakage you could always shield any wires with static shielding(not connected to anything) This would reflect the leaks back to the source of the leak and contain it better especially if you stay with higher voltages. Cheap static shielding would be a heavy aluminum foil tape or just foil glued onto the wires. If you do them in bands and overlap the bands it should make an effective static shielding for your high voltage app..
Also drop by our thread in the "don't kill dipole" I am working on the same kind of system to get an ac output with a high voltage pair of leyden jars type setup with twin variable caps on a motorized rotor. Weather or not this will work it would be nice to have your setup to compare to as well.
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Last edited by Jbignes5; 10-20-2010 at 05:36 PM.
#46
10-20-2010, 05:51 PM
Armagdn03 Silver Member Join Date: Oct 2007 Posts: 902
Quote:
Originally Posted by Jbignes5 Ok there is a big difference in a single variable cap and twin ones. One would be harmonic(twin) and the other non harmonic(single). The harmonic one would be like putting a teeter tawter in the system and the other would be like jacking the one side up. The harmonic one would be the less leaky since it would have an increase with the associated decrease on the other side. This would make it more powerful as well and you wont have to have extreme high voltages like you are talking then. On the subject of leakage you could always shield any wires with static shielding(not connected to anything) This would reflect the leaks back to the source of the leak and contain it better especially if you stay with higher voltages. Cheap static shielding would be a heavy aluminum foil tape or just foil glued onto the wires. If you do them in bands and overlap the bands it should make an effective static shielding for your high voltage app.. Also drop by our thread in the "don't kill dipole" I am working on the same kind of system to get an ac output with a high voltage pair of leyden jars type setup with twin variable caps on a motorized rotor. Weather or not this will work it would be nice to have your setup to compare to as well.
I agree with the difference and utility of using two variable capacitors. What I was initially trying to get across is the barriers of understanding in "action reaction separation" and the concept that the time constants dictated by the load impedance and capacitance at a point in time (involves differential and integral calculus to calculate) and how this affects your maximum attainable frequency.
For example, with a resistive load, you may have a time constant of 1 second. This means in 5 seconds you would have moved 99 percent of the available movable charge.
However within the first second 63% of the charge is moved. During the remaining 4 seconds only 36% of charge moves. Therefore, your time constants might dictate an operational frequency of X, but in reality you will accomplish more work per unit time if you reduce your frequency 80% or 1/5th of what it was.
Frequency also must be taken into consideration with respect to how fast you can switch the capacitance. if you can only do it a few thousand times per second, and you are in the PicoFarad range, you are not going anywhere fast.
I will perhaps show a video soon showing how to find the ionization time of fluorescent tubes, as this is my frequency limiting factor.
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#47
10-20-2010, 06:22 PM
Jbignes5 Gold Member Join Date: Mar 2009 Location: NY, USA Posts: 1,014
Quote:
Originally Posted by Armagdn03 I agree with the difference and utility of using two variable capacitors. What I was initially trying to get across is the barriers of understanding in "action reaction separation" and the concept that the time constants dictated by the load impedance and capacitance at a point in time (involves differential and integral calculus to calculate) and how this affects your maximum attainable frequency. For example, with a resistive load, you may have a time constant of 1 second. This means in 5 seconds you would have moved 99 percent of the available movable charge. However within the first second 63% of the charge is moved. During the remaining 4 seconds only 36% of charge moves. Therefore, your time constants might dictate an operational frequency of X, but in reality you will accomplish more work per unit time if you reduce your frequency 80% or 1/5th of what it was. Frequency also must be taken into consideration with respect to how fast you can switch the capacitance. if you can only do it a few thousand times per second, and you are in the PicoFarad range, you are not going anywhere fast. I will perhaps show a video soon showing how to find the ionization time of fluorescent tubes, as this is my frequency limiting factor.
What about ultra caps? They can store and discharge even faster and really since Wimshursts machines are just caps that are variable you get the same effect. The trick would be to increase the dielectric value and lessen the gap to increase the capacity ratings. Vacuum does a good job and so will oils that are reduced by boiling to expunge contaminants. Although boiling of the oil will raise the viscosity I'm sure disk design would help in getting the blades to move through the oil very effectively.
You do have a point about charge response and that needs to be looked at more closely. As to reducing the frequency of the switching of the variable cap you make sense but that only will increase the resistive reactance of the cap. Meaning that the longer the charge stays in the cap the more self discharge that would happen increasing the loss.
Anyways my system should employ twin variable caps that will only have to variate 120-160v back and forth. This might allow a lower operating base voltage to apply the swing to but I have not built a test rig yet. I want to stay in the high range for voltage but we will see what static like voltages when applied to a variable capacitor will do.
Using a vacuum does present a problem of maintaining that vacuum but that might be workable. Using oil is a simpler method but might be harder to turn. I'll have to see for myself i guess...
Tesla did make an open ended tube on one of his devices and could maintain a vacuum with that method I'll have to research that more to see if it is possible to use that method.
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#48
10-20-2010, 08:16 PM
Armagdn03 Silver Member Join Date: Oct 2007 Posts: 902
Quote:
Originally Posted by Jbignes5 What about ultra caps? They can store and discharge even faster and really since Wimshursts machines are just caps that are variable you get the same effect. The trick would be to increase the dielectric value and lessen the gap to increase the capacity ratings. Vacuum does a good job and so will oils that are reduced by boiling to expunge contaminants. Although boiling of the oil will raise the viscosity I'm sure disk design would help in getting the blades to move through the oil very effectively. Agreed, however I would need a very large and fast Wimshursts for what I want. but as a method of attack not a bad idea for people to peruse....one disk can control 2 capacitors very easily......."Testatika" ??? You do have a point about charge response and that needs to be looked at more closely. As to reducing the frequency of the switching of the variable cap you make sense but that only will increase the resistive reactance of the cap. Meaning that the longer the charge stays in the cap the more self discharge that would happen increasing the loss. I think you misunderstood me, I was pointing to increasing frequency, by using only 1 time constant vs 5, thus decreasing the reactance, besides this it also moves more coulombs per second. My wording was poor now that I look at it "my bad". Anyways my system should employ twin variable caps that will only have to variate 120-160v back and forth. This might allow a lower operating base voltage to apply the swing to but I have not built a test rig yet. I want to stay in the high range for voltage but we will see what static like voltages when applied to a variable capacitor will do. Look forward to your results! Using a vacuum does present a problem of maintaining that vacuum but that might be workable. Using oil is a simpler method but might be harder to turn. I'll have to see for myself i guess... these are a few of the reasons i do not want a mechanical method, besides my lack of fabrication ability, it is just overall more difficult. Tesla did make an open ended tube on one of his devices and could maintain a vacuum with that method I'll have to research that more to see if it is possible to use that method. This was a clever way to draw a vacuum with HV, but i cannot say if this is any more effective than a vacuum pump. I opened up a CRT analogue o-scope the other day, the the cathode ray tube still had a perfect seal and negative internal pressure and it was over 35 years old, so perhaps with quality manufacturing this is not a problem, but like I said before, i don't have the tools!!!
My response is in red. This is obvious, but it is making me type some sort of a 10 character plus message besides the quote to submit this post, lol.
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Last edited by Armagdn03; 10-20-2010 at 08:26 PM.
#49
12-02-2010, 07:36 PM
Armagdn03 Silver Member Join Date: Oct 2007 Posts: 902
Hello all,
Progress has been made in this area although a bit slow. I have a good friend who is helping with an incredible driving circuit, for this very device.
However, what I want to post, and what other people can use, for many parametric purposes, is this modification I have designed based on the Hiddink Patent mentioned earlier in the thread, and alluded to as the preferred method.
Quote:
you have two caps, 4 sets of plates, inner and outer. If you remove the inner plates, you have much much less capacitance, because the remaining capacitor is dictated by the two outer plates. But how to physically remove the inner plates???? With a modified version of: Capacity Changer Why not add and remove a virtual plate. This of course throws out all of the "force on plates" mumbo jumbo, but that was still important to realize the mechanism in its simplest embodiment.
So, the problem with the patent as it stands, is that there needs to be a complicated switching scheme to remove completely one of the sources of power. The patent actually dictates two separate power sources!
Well among the many people I have talked to who have replicated this, (Dr. Stiffler has videos of his replication I believe), the switching has always been the problem with bringing this patent to realization.
Well, there is a solution!
YouTube - Solid State Capacity switching Tube 2
If a person could build a circuit which operates around one of the large capacitance jumps (correlated to the phase changes of the particles in the tube) then you have an easy capacity switching device, to run all sorts of parametric circuits.
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Last edited by Armagdn03; 04-19-2011 at 07:14 PM.
#50
04-19-2011, 07:26 PM
Armagdn03 Silver Member Join Date: Oct 2007 Posts: 902
testing connection,
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#51
05-29-2011, 05:52 AM
Web000x Silver Member Join Date: Apr 2009 Posts: 539
Andrew,
I am impressed with your line of thought on this thread. I am very certain that you laid a golden egg on the forum.
I completely see where the use of the middle wire capacitance component was killing the charge in mario's system.
The conversation between Jbignes5 and yourself easily conveys the idea behind Chris Carson's device. His Device was operating on a negative phase shift where each capacitance was on the opposite half cycle producing an AC signal of "energy synthesis" from the dielectric field.
Thanks for pointing me to this thread. You are certainly looking at things clearly.
Dave
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#52
05-29-2011, 06:53 AM
Aaron Co-Founder & Moderator Join Date: Feb 2007 Location: Washington State Posts: 10,035
opposite half cycle AC energy synthesis
Quote:
Originally Posted by Web000x the idea behind Chris Carson's device. His Device was operating on a negative phase shift where each capacitance was on the opposite half cycle producing an AC signal of "energy synthesis" from the dielectric field.
Dave,
Would you mind elaborating on that?
Are you talking about the potential difference between two negatives
that alternate?
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Sincerely,
Aaron Murakami
#53
05-29-2011, 07:40 AM
Web000x Silver Member Join Date: Apr 2009 Posts: 539
Quote:
Originally Posted by Aaron Dave, Would you mind elaborating on that? Are you talking about the potential difference between two negatives that alternate?
Sure,
Eric Dollard mentioned the importance of inductance/capacitance parameter variation with respect to time for the "synthesis of energy". In Mr. Dollard's book titled "Symbolic Representation of the Generalized Wave", Read Pg 57 where at the bottom of the page it lists the quantity called -HB; H is receptance or negative resistance, and B = mhos in farads per second.
That quantity suggests energy production from the dielectric field by a varying capacity. I have yet to verify this experimentally, but this idea is due to standing on the shoulders of giants if you will.
Armagdn03 spoke about the exchange of charge between a dielectric parameter variation which stood the test of equations for realistic and theoretical application.
Peter L Quote:
Quote:
The sixth picture shows another, lower angle shot of the whole machine. It also shows how one section is out of phase with the other.
If you were to have two capacitors that vary out of phase like Chris Carson had, you would have a setup that could move a maximum amount of charge by the equation Ψ=eC (Ψ = Total Dielectric Induction in Coulomb, e = electrostatic potential in Volts, and C = Capacitance in Farads) between capacitor plates. The displacement current due to the potential differences can then run through a load while the capacitors are moving towards equilibrium. From my understanding, the Ψ=eC stays in balance.
Armagdn03's description of using conservation of charge (variable capacitance) vs. time manipulation has me convinced you can indeed change the energy of a system and collect a bit extra.
Dave
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Last edited by Web000x; 12-12-2011 at 03:37 AM.
#54
05-30-2011, 08:53 PM
Armagdn03 Silver Member Join Date: Oct 2007 Posts: 902
Quote:
Originally Posted by Web000x Andrew, I am impressed with your line of thought on this thread. I am very certain that you laid a golden egg on the forum. I completely see where the use of the middle wire capacitance component was killing the charge in mario's system. The conversation between Jbignes5 and yourself easily conveys the idea behind Chris Carson's device. His Device was operating on a negative phase shift where each capacitance was on the opposite half cycle producing an AC signal of "energy synthesis" from the dielectric field. Thanks for pointing me to this thread. You are certainly looking at things clearly. Dave
Thank you!
I also gave one really great solution to the switching problem encountered with the Hidink patent a couple posts back. There was the need for complicated switching schemes, which no longer exist with the method of accomplishment I showed. This device is in itself over-unity. I know personally of at least one person who has this down pat however out of respect to the original inventor, (which may still be alive) he has kept quiet.
Chris Carsons device was a super simple demonstration, but a mechanical implementation for energy production on that scale is a little silly when you consider the limitations of the system! 10,000 rpm on a spinning multi plate cap??? Wow talk about model building difficulty. Using phase changes of ionized gas, OR using uv light to excite a layer of phosphorus as depicted in these patents.
Keep in mind for this system the critical parameters for power generation.
1)Capacitance of the capacitors (total charge held)
2)Starting voltage across caps (Total charge held)
3)Impedance of the load (Time constants, dictates coulombs per second or watts)
4)Switching speed (directly related to time constants)
5)Limit on how quickly one can switch between between conductive plasma phase and non conductive phase. User Tecstatic has released some information on this in the private part of Heretical builders forum, and I will ask permission to re post here.
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#55
05-30-2011, 09:03 PM
Armagdn03 Silver Member Join Date: Oct 2007 Posts: 902
For those who enjoy the work of Eric Dollard this is a direct application of his theory. I will be posting with reference to four quadrant theory for a bit, but am a bit predisposed at the moment.
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#56
05-30-2011, 11:46 PM
7imix Senior Member Join Date: Sep 2010 Location: Planet Earth Posts: 380
Great thread, thanks for pointing me at it.
Also, which one is the hidink patent you mentioned?
I like the line of thought about using plasma tubes to control the switching, but I'm not clear on what that would look like. I have spent a lot of time studying the schematic of the voltage converter Eric uses in his car and trying to understand how I could create a solid state embodiment of that, but all the ideas I came up with are far too complex. Using plasma tubes is a great suggestion.
What kind of tubes are available? Would transient voltage suppressors be suitable at all? I have had a difficult time designing a circuit that uses them in a manner similar to how don smith uses them, but I have not spent too much time trying.
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#57
05-31-2011, 02:38 AM
Raui Senior Member Join Date: Dec 2008 Posts: 284
Quote:
Originally Posted by 7imix Great thread, thanks for pointing me at it. The two google patent links you just posted are 404. Can you re link them? Also, which one is the hidink patent you mentioned? I like the line of thought about using plasma tubes to control the switching, but I'm not clear on what that would look like. I have spent a lot of time studying the schematic of the voltage converter Eric uses in his car and trying to understand how I could create a solid state embodiment of that, but all the ideas I came up with are far too complex. Using plasma tubes is a great suggestion. What kind of tubes are available? Would transient voltage suppressors be suitable at all? I have had a difficult time designing a circuit that uses them in a manner similar to how don smith uses them, but I have not spent too much time trying.
Here are the patents your after;
Louis O'Hare Patent 1
Louis O'Hare Patent 2
Joseph Hiddink
Also just a note on Dollard's voltage converter. It may have been modified from that schematic I posted a while back since logic tells us that it will loose charge in the same way as Mario's device did earlier in the topic. I don't question any of his work in any way shape or form but I don't think this device is worth bothering about too much for reasons Andrew posted earlier. I had a friend replicate the circuit solid state and he said that it operated in just the same way as a boost converter does and was quite inefficient.
I don't wish to deter you from your study but I thought I'd put that out there.
Raui
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#58
05-31-2011, 04:22 AM
7imix Senior Member Join Date: Sep 2010 Location: Planet Earth Posts: 380
Yes, I am aware that the voltage converter is a common device and does not produce large amounts of excess energy... But I still want to study it so I understand how it works and can build something like it. This thread is helping with that a lot.
Also, there has to be SOME reason Dollard talks about this device. I would like to know what it is he is trying to teach us by talking about it.
When I build "normal" electrical devices, any misconceptions I may have about theory are easily dispelled since I get direct experience with measuring the device.
I am very curious if there is a good way to use transistors to charge capacitors in parallel and discharge in series, or vice versa. What I came up with seems horribly complicated. I have not been able to find any schematics or even learn about any common devices that do this.
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#59
05-31-2011, 06:12 AM
SuperCaviTationIstic Senior Member Join Date: Jan 2010 Posts: 331
Quote:
Originally Posted by Armagdn03 I also gave one really great solution to the switching problem encountered with the Hidink patent a couple posts back. There was the need for complicated switching schemes, which no longer exist with the method of accomplishment I showed. This device is in itself over-unity. I know personally of at least one person who has this down pat however out of respect to the original inventor, (which may still be alive) he has kept quiet.
I personally have contacted Joseph Hiddink. Facebook! He's alive and well, but VERY saddened and obsessed with his discovery/invention because no one (governments, manufacturing, r&d) has taken him seriously and helped him develop it.
Aaron, would you think it's ok for me to post his Facebook id? He seems to not be able to accept instant messages, and it sometimes takes him a while to respond to messages, but we have conversed a few times.
I really believe that if he saw a great interest in this, he would love to help out.... He does want to develop it commecially, but it's so simple and he doesn't seem to mind, and actually encourages its development.
And one more very interesting side note:
RogerArrick.com Roger Arrick Mystery Orb
more pictures are on the page
A metal sphere was found in a rive and x-rays were taken. It is 2 concentric spheres with a gas discharge tube in the middle surrounded by a metal cylinder. A wire runs to each sphere, the cylinder, and each end of the discharge tube. I've downloaded and backed up these pictures because it's vital PROOF that the government/militarily/machine uses this technology, and poor Joseph doesn't get a dime.
That's right, this not only produces high voltage, but the negative pulse has a matter repulsion effect and a positive pulse has an attraction effect.
Incase you didn't connect the dots....it's a flying saucer propulsion device
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#60
05-31-2011, 06:35 AM
SuperCaviTationIstic Senior Member Join Date: Jan 2010 Posts: 331
and if you think that's weird, it get WAY WAY WAY WEIRDER
Tom Swift and his Polar-Ray Dynasphere
I was doing a google search and I forget the term I was looking for, but I think I was trying to find out if anyone had made a piezo-electric concentric-spheres-capacitor oscillator out of quartz.
Guess what, young Tom Swift did! To power his Polar-Ray
from the text:
["Well, as you know, that double-walled crystal sphere at the stern is my electrostatic-field device. I've decided to call the gadget a 'Dynasphere' -- or 'Polar-Ray Dynasphere,']
also very weird that Keely invented a thing with that name, but that's completely different. (Or is it?)
Tom Swift's 'fictional' device is VERY similar to hiddink's device, yet acting as an osccilator with the medium of the highest Q known (correct me if I'm wrong) QUARTZ
And the whole electrostatic field shaping of his electro-propulsion ship is very clear and accurate.
Tell me I'm wrong.
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Associated Topics || Dr. Math Home || Search Dr. Math
### Deriving the Normal From the Binomial
```
Date: 07/11/98 at 22:45:41
From: Jason
Subject: Derivation of normal distribution
Please tell me how the normal distribution is derived from the
binomial distribution. Can it be proven that the normal distribution
approximates the binomial distribution as the number of trials goes to
infinity? Or can you please tell me a Web site or a book I should
consult?
Thank you very much for your time.
```
```
Date: 07/12/98 at 07:25:51
From: Doctor Anthony
Subject: Re: Derivation of normal distribution
Derivation of the Normal distribution from the Binomial distribution
---------------------------------------------------------------------
Let a variate take values 0, k, 2k, 3k, ..., nk with probabilities
given by successive terms of (q + p)^n.
Then the mean m = npk and the variance s^2 = npqk^2.
Suppose:
y = probability of occurrence of rk = C(n,r) p^r q^(n-r)
Also let:
y' = probability of occurrence (r+1)k = C(n,r+1)p^(r+1) q^(n-r-1)
Then:
y' - y = C(n,r+1)p^(r+1) q^(n-r-1) - C(n,r)p^r q^(n-r)
n!p^r q^(n-r-1)
= ---------------[(n-r)p - (r+1)q]
(r+1)! (n-r)!
And:
y' - y 1 1
------ = ------[np - r(p+q) - q] = ------[np - r - q] (Equation 1)
y (r+1)q (r+1)q
Let:
x = rk - npk, so that x is now the variate measured from the mean.
Then:
r = x/k + np and r+1 = x/k + np + 1
Thus:
k(r+1) = x + k + npk
k^2 (r+1)q = (x + k + npk)qk
Multiply top and bottom of the righthand side of Equation 1 by k^2.
Then:
y' - y [(np-r)k - qk]k
------ = --------------- [note that (np-r)k = -x]
y [x + k + npk]qk
(-x - kq)k
= ---------------- (Equation 2)
npqk^2 + (x+k)qk
We now let k = dx, so that y' - y = dy and let n ->infinity in such a
way that nk^2 is finite. Equation 2 can then be written as:
dy (-x - q dx)dx
---- = ----------------
y s^2 + (x+dx)q dx
As dx -> 0 this becomes:
dy -x dx
---- = ------
y s^2
Integrating, we get:
ln(y) = - x^2/(2s^2) + constant
y = e^(-x^2/(2s^2) + constant)
y = A e^(-x^2/(2s^2))
If we integrate this from -infinity to +infinity we get the area under
the curve, and we choose the constant A to make this area equal to 1.
To see how this integral is obtained, see:
Algebraic Integration of Standard Normal Distribution Function
http://mathforum.org/library/drmath/view/53628.html
We find that:
1
A = -----------
s.sqrt(2.pi)
So the formula for the Normal pdf curve is:
1
----------- e^(-x^2/(2s^2))
s sqrt(2pi)
- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
```Date: 06/21/2003 at 04:27:27
From: Jeff
Subject: Deriving the Normal From the Binomial
I'm having trouble following one of the steps in the derivation:
| y' - y (-x - kq)k
| ------ = ----------------
| y npqk^2 + (x+k)qk
|
|
| We now let k = dx, so that y' - y = dy and let n ->infinity
| in such a way that nk^2 is finite. The equationcan then be
| written as:
|
| dy (-x - q dx)dx
| ---- = ----------------
| y s^2 + (x+dx)q dx
|
| As dx -> 0 this becomes:
|
| dy -x dx
| ---- = ------
| y s^2
It seems to me that if we simplify the above expression just before
the dx -> 0 we get
-x dx - q dx^2
-----------------------
s^2 + x q dx + q dx^2
The terms with dx^2 will go to 0 very quickly and will contribute
very little, leaving
-x dx
--------------
s^2 + x q dx
What am I missing?
-Jeff
```
```Date: 06/30/2003 at 12:30:25
From: Doctor Anthony
Subject: Re: Deriving the Normal From the Binomial
Hi Jeff,
As dx -> 0, xq dx is zero _compared_ to s^2, which is why
-x dx
--------------
s^2 + x q dx
reduces to
-x dx
---------
s^2
In the numerator, we are _multiplying_ by dx and this cannot be ignored;
but in the denominator we are _adding_ a term with dx and this _can_ be ignored.
If we derive the formula for dy/dx in a simple case such as y = x^2 from
first principles we can reason as follows:
If (x+dx, y+dy) is a point on y = x^2 then
y+dy = (x+dx)^2
y+dy = x^2 + 2x dx + (dx)^2 Cancel y and x^2
dy = 2x dx + (dx)^2 Now divide through by dx
dy/dx = 2x + dx
Now let dx -> 0, so dy -> 0 also; and the left hand side -> 0/0
while the right hand side becomes 2x. (We can now ignore dx since it is
Now 0/0 is NOT equal to 0 or 1 but tends to different things in different
situations. (In this case it tends to 2x.)
For example
17 * 0 = 0
so dividing through by 0 (assuming this is allowed) you get
17 -> 0/0
Clearly we could also get
213 -> 0/0
or
-78 -> 0/0
or
anything you care to think of -> 0/0
In fact, the various values for dy/dx are all values we obtain for 0/0.
When you see an expression like dy/dx = 2x you can think of an
infinitely small increment in y divided by an infinitely small
increment in x (in effect 0/0) has the value 2x.
- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Probability
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# Worksheet on Arithmetic Mean
Practice the word problems on average given in the worksheet on arithmetic mean. Here the questions will help us to find the mean of the given numbers.
1. Find the arithmetic mean of the first ten natural numbers
2. Find the arithmetic mean of the first eight odd numbers
3. Find the arithmetic mean of the first seven multiples of 5
4. Find the arithmetic mean of all the factors of 20
5. Find the arithmetic mean of the first ten prime numbers
6. Find the arithmetic mean of the first eleven even numbers
7. The number of members in 10 families of a society are
2, 4, 3, 4, 2, 2, 3, 5, 1, 4.
Find the mean number of members per family.
8. The following are the number of math books issued in a school library during a week:
222, 120, 201, 322, 167, 273, 406 and 345
Find the average number of math books issued per day.
9. If the arithmetic mean of 7, 9, 11, 13, x, 21 is 13, find the value of x.
10. The daily minimum temperature recorded (in degree F) at a place during a week was as under:
Find the mean temperature.
11. The mean of 24 numbers is 35. If 3 is added to each number, what will be the new mean?
12. The mean of 20 numbers is 43. If 6 is subtracted from each of the numbers, what will be the new mean?
13. The mean of 15 numbers is 27. If each number is multiplied by 5, what will be the mean of the new numbers?
14. The mean of 15 numbers is 63. If each number is divided by 9, what will be the mean of the new numbers?
15. The percentages of marks obtained by 12 students of a class in mathematics are
36, 64, 47, 43, 50, 39, 81, 93, 72, 35, 53, 41.
Find the mean percentage of marks.
16. The mean of 20 numbers is 18. If 3 is added to each of the first ten numbers, find the mean of the new set of 20 numbers.
17. The mean weight of 8 girls in a group is 35 kg. The individual weights of seven of them are 35 kg, 28 kg, 43 kg, 25 kg, 33 kg, 51 kg and 22 kg. Find the weight of the eighth girls.
18. The mean of the marks scored by 50 students was found to be 39. Later on it was discovered that a score of 75 was misread as 25. Find the correct mean.
19. The mean of 100 items was found to be 64. Later on it was discovered that two items were misread as 26 and 9 instead of 36 and 90 respectively. Find the correct mean.
20. The mean of six numbers is 23. If one of the numbers is excluded, the mean of the remaining numbers is 20. Find the excluded number.
Answers for the worksheet on arithmetic mean are given below to check the exact of the above questions on mean (average).
1. 5.5
2. 8
3. 20
4. 7
5. 12.9
6. 12
7. 3
8. 257
9. x = 17
10. 35.4 °F
11. 38
12. 37
13. 135
14. 7
15. 54.5
16. 19.5
17. 43
18. 40
19. 64.91
20. 38
Math Homework Sheets
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
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Lesson Plan:
# Beat the Drum
no ratings yet
Subject
September 3, 2015
## Learning Objectives
Students will be able to count to 10 using one-to-one correspondence. Students will practice their fine motor skills.
## Lesson
### Introduction (5 minutes)
• Begin the lesson by explaining to your students that they will play a game about counting cubes, or determining how many cubes there are.
• Tell your students that they will use tongs to put one-inch cubes into a container. After putting all 10 cubes into the container, inform them that they can strike the drum with a drumstick.
• Inform your students that they will then switch positions with another student so that everyone gets a chance to count and strike the drum.
### Explicit Instruction/Teacher Modeling (5 minutes)
• Separate the class into two equal groups, and have each group sit around one container.
• Place 10 cubes and one pair of tongs by each container. Put the drum and drumsticks between the groups.
• Have one student in front of each container at a time.
• Demonstrate how to hold and use the tongs to pick up one cube.
• Discuss whether or not the piles have the same amount of cubes, or if they are equal. Potential discussion questions include: Do the piles look equal? How can we check to make sure that there are the same amount of cubes in each pile?
• Have someone from each group count the number of cubes. Suggest that since both groups have 10 cubes, the groups are, indeed, equal.
• Instruct your students to place the cubes into the container with the tongs when you give a signal.
• Ask them to count using one-to-one correspondence while placing their cubes in the container. For example, if a student is putting her third cube into the container, make sure she says "three" while doing so.
• Direct each student to strike the drum when he has put in all 10 cubes.
### Guided Practice/Interactive Modeling (5 minutes)
• Practice the game as a class for two more turns.
• Encourage your students to continue counting the cubes individually as they place them in containers.
• Count with your class if support is needed. Have the entire group count with that student.
### Independent Working Time (20 minutes)
• Have your students continue to take turns in their groups.
• Ask them to identify which group has more, less, or equal amounts of cubes in the container.
• Encourage good sportsmanship during play.
## Extend
### Differentiation
• Enrichment: Place 15-20 cubes in front of each student. Have students count either up to the number of cubes, or have them count by two's instead of using direct one-to-one correspondence.
• Support: Count and model for students having difficulty. Decrease the pile to five one-inch cubes, and allow students to use their fingers to place the cubes into the container. This will let them focus on counting.
## Review
### Assessment (5 minutes)
• To assess student understanding, notice if your students are counting with one-to-one correspondence.
• Note if your students are demonstrating understanding of more, less, and equal. Assess this by listening to the students' answers as they compare quantities.
• Note if your students are using a firm grasp on the tongs. If their grip is very loose, encourage them to grip the tongs tighter so that the cubes do not fall out.
### Review and Closing (5 minutes)
• Ask your students, as a group, to count to 10 aloud.
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### Diophantine N-tuples
Can you explain why a sequence of operations always gives you perfect squares?
### DOTS Division
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.
### Sixational
The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. Prove that all terms of the sequence are divisible by 6.
# Pair Products
##### Age 14 to 16 Challenge Level:
Choose four consecutive whole numbers.
Multiply the first and last numbers together.
Multiply the middle pair together.
Choose several different sets of four consecutive whole numbers and do the same.
What do you notice?
Can you explain what you have noticed? Will it always happen?
Click below to see how Charlie and Alison explained what they noticed.
Charlie said:
I noticed that the product of the outer pair was always $2$ less than the product of the inner pair.
I can explain this by labelling the four consecutive numbers $n, n+1, n+2, n+3$.
Outer pair: $n(n+3) = n^2 + 3n$
Inner pair: $(n+1)(n+2) = n^2 + 3n + 2$
Alison said:
I drew a diagram, in which the product of each pair is represented by the area of a rectangle:
The outer pair is represented by the red rectangle.
The inner pair is represented by the blue rectangle.
The purple area is common to both.
The area of the red strip will always be two units less than the area of the blue strip.
Therefore, the product of the outer pair is always two less than the product of the inner pair.
Instead of doing lots of calculations, can you use these representations to compare the product of the first and last numbers with the product of the second and penultimate numbers, when you have:
• $5$ consecutive whole numbers
• $6, 7, 8, \ldots x$ consecutive whole numbers
• $4$ consecutive even numbers
• $4$ consecutive odd numbers
• $5, 6, 7, 8, \ldots x$ consecutive even or odd numbers
• $4$ consecutive multiples of $3, 4, 5 \ldots$
• Decimals that differ by $1$, such as $1.2, 2.2, 3.2, 4.2$
• Four numbers going up in $3$s, such as $2, 5, 8, 11$
• Four numbers going up in $\frac{1}{2}$s, such as $4, 4\frac{1}{2}, 5, 5\frac{1}{2}$
Make up a few similar questions of your own. Impress your friends by giving them a calculator and 'predicting' what will happen!
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > eupth2lem3lem3 Structured version Visualization version GIF version
Theorem eupth2lem3lem3 28015
Description: Lemma for eupth2lem3 28021, formerly part of proof of eupth2lem3 28021: If a loop {(𝑃‘𝑁), (𝑃‘(𝑁 + 1))} is added to a trail, the degree of the vertices with odd degree remains odd (regarding the subgraphs induced by the involved trails). (Contributed by Mario Carneiro, 8-Apr-2015.) (Revised by AV, 21-Feb-2021.)
Hypotheses
Ref Expression
trlsegvdeg.v 𝑉 = (Vtx‘𝐺)
trlsegvdeg.i 𝐼 = (iEdg‘𝐺)
trlsegvdeg.f (𝜑 → Fun 𝐼)
trlsegvdeg.n (𝜑𝑁 ∈ (0..^(♯‘𝐹)))
trlsegvdeg.u (𝜑𝑈𝑉)
trlsegvdeg.w (𝜑𝐹(Trails‘𝐺)𝑃)
trlsegvdeg.vx (𝜑 → (Vtx‘𝑋) = 𝑉)
trlsegvdeg.vy (𝜑 → (Vtx‘𝑌) = 𝑉)
trlsegvdeg.vz (𝜑 → (Vtx‘𝑍) = 𝑉)
trlsegvdeg.ix (𝜑 → (iEdg‘𝑋) = (𝐼 ↾ (𝐹 “ (0..^𝑁))))
trlsegvdeg.iy (𝜑 → (iEdg‘𝑌) = {⟨(𝐹𝑁), (𝐼‘(𝐹𝑁))⟩})
trlsegvdeg.iz (𝜑 → (iEdg‘𝑍) = (𝐼 ↾ (𝐹 “ (0...𝑁))))
eupth2lem3.o (𝜑 → {𝑥𝑉 ∣ ¬ 2 ∥ ((VtxDeg‘𝑋)‘𝑥)} = if((𝑃‘0) = (𝑃𝑁), ∅, {(𝑃‘0), (𝑃𝑁)}))
eupth2lem3lem3.e (𝜑 → if-((𝑃𝑁) = (𝑃‘(𝑁 + 1)), (𝐼‘(𝐹𝑁)) = {(𝑃𝑁)}, {(𝑃𝑁), (𝑃‘(𝑁 + 1))} ⊆ (𝐼‘(𝐹𝑁))))
Assertion
Ref Expression
eupth2lem3lem3 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → (¬ 2 ∥ (((VtxDeg‘𝑋)‘𝑈) + ((VtxDeg‘𝑌)‘𝑈)) ↔ 𝑈 ∈ if((𝑃‘0) = (𝑃‘(𝑁 + 1)), ∅, {(𝑃‘0), (𝑃‘(𝑁 + 1))})))
Distinct variable groups: 𝑥,𝑈 𝑥,𝑉 𝑥,𝑋
Allowed substitution hints: 𝜑(𝑥) 𝑃(𝑥) 𝐹(𝑥) 𝐺(𝑥) 𝐼(𝑥) 𝑁(𝑥) 𝑌(𝑥) 𝑍(𝑥)
Proof of Theorem eupth2lem3lem3
StepHypRef Expression
1 trlsegvdeg.u . . . . 5 (𝜑𝑈𝑉)
2 fveq2 6645 . . . . . . . 8 (𝑥 = 𝑈 → ((VtxDeg‘𝑋)‘𝑥) = ((VtxDeg‘𝑋)‘𝑈))
32breq2d 5042 . . . . . . 7 (𝑥 = 𝑈 → (2 ∥ ((VtxDeg‘𝑋)‘𝑥) ↔ 2 ∥ ((VtxDeg‘𝑋)‘𝑈)))
43notbid 321 . . . . . 6 (𝑥 = 𝑈 → (¬ 2 ∥ ((VtxDeg‘𝑋)‘𝑥) ↔ ¬ 2 ∥ ((VtxDeg‘𝑋)‘𝑈)))
54elrab3 3629 . . . . 5 (𝑈𝑉 → (𝑈 ∈ {𝑥𝑉 ∣ ¬ 2 ∥ ((VtxDeg‘𝑋)‘𝑥)} ↔ ¬ 2 ∥ ((VtxDeg‘𝑋)‘𝑈)))
61, 5syl 17 . . . 4 (𝜑 → (𝑈 ∈ {𝑥𝑉 ∣ ¬ 2 ∥ ((VtxDeg‘𝑋)‘𝑥)} ↔ ¬ 2 ∥ ((VtxDeg‘𝑋)‘𝑈)))
7 eupth2lem3.o . . . . 5 (𝜑 → {𝑥𝑉 ∣ ¬ 2 ∥ ((VtxDeg‘𝑋)‘𝑥)} = if((𝑃‘0) = (𝑃𝑁), ∅, {(𝑃‘0), (𝑃𝑁)}))
87eleq2d 2875 . . . 4 (𝜑 → (𝑈 ∈ {𝑥𝑉 ∣ ¬ 2 ∥ ((VtxDeg‘𝑋)‘𝑥)} ↔ 𝑈 ∈ if((𝑃‘0) = (𝑃𝑁), ∅, {(𝑃‘0), (𝑃𝑁)})))
96, 8bitr3d 284 . . 3 (𝜑 → (¬ 2 ∥ ((VtxDeg‘𝑋)‘𝑈) ↔ 𝑈 ∈ if((𝑃‘0) = (𝑃𝑁), ∅, {(𝑃‘0), (𝑃𝑁)})))
109adantr 484 . 2 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → (¬ 2 ∥ ((VtxDeg‘𝑋)‘𝑈) ↔ 𝑈 ∈ if((𝑃‘0) = (𝑃𝑁), ∅, {(𝑃‘0), (𝑃𝑁)})))
11 2z 12002 . . . . . 6 2 ∈ ℤ
1211a1i 11 . . . . 5 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → 2 ∈ ℤ)
13 trlsegvdeg.v . . . . . . . 8 𝑉 = (Vtx‘𝐺)
14 trlsegvdeg.i . . . . . . . 8 𝐼 = (iEdg‘𝐺)
15 trlsegvdeg.f . . . . . . . 8 (𝜑 → Fun 𝐼)
16 trlsegvdeg.n . . . . . . . 8 (𝜑𝑁 ∈ (0..^(♯‘𝐹)))
17 trlsegvdeg.w . . . . . . . 8 (𝜑𝐹(Trails‘𝐺)𝑃)
18 trlsegvdeg.vx . . . . . . . 8 (𝜑 → (Vtx‘𝑋) = 𝑉)
19 trlsegvdeg.vy . . . . . . . 8 (𝜑 → (Vtx‘𝑌) = 𝑉)
20 trlsegvdeg.vz . . . . . . . 8 (𝜑 → (Vtx‘𝑍) = 𝑉)
21 trlsegvdeg.ix . . . . . . . 8 (𝜑 → (iEdg‘𝑋) = (𝐼 ↾ (𝐹 “ (0..^𝑁))))
22 trlsegvdeg.iy . . . . . . . 8 (𝜑 → (iEdg‘𝑌) = {⟨(𝐹𝑁), (𝐼‘(𝐹𝑁))⟩})
23 trlsegvdeg.iz . . . . . . . 8 (𝜑 → (iEdg‘𝑍) = (𝐼 ↾ (𝐹 “ (0...𝑁))))
2413, 14, 15, 16, 1, 17, 18, 19, 20, 21, 22, 23eupth2lem3lem1 28013 . . . . . . 7 (𝜑 → ((VtxDeg‘𝑋)‘𝑈) ∈ ℕ0)
2524nn0zd 12073 . . . . . 6 (𝜑 → ((VtxDeg‘𝑋)‘𝑈) ∈ ℤ)
2625adantr 484 . . . . 5 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → ((VtxDeg‘𝑋)‘𝑈) ∈ ℤ)
2713, 14, 15, 16, 1, 17, 18, 19, 20, 21, 22, 23eupth2lem3lem2 28014 . . . . . . 7 (𝜑 → ((VtxDeg‘𝑌)‘𝑈) ∈ ℕ0)
2827nn0zd 12073 . . . . . 6 (𝜑 → ((VtxDeg‘𝑌)‘𝑈) ∈ ℤ)
2928adantr 484 . . . . 5 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → ((VtxDeg‘𝑌)‘𝑈) ∈ ℤ)
30 z2even 15711 . . . . . . 7 2 ∥ 2
3119ad2antrr 725 . . . . . . . 8 (((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) ∧ 𝑈 = (𝑃𝑁)) → (Vtx‘𝑌) = 𝑉)
32 fvexd 6660 . . . . . . . 8 (((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) ∧ 𝑈 = (𝑃𝑁)) → (𝐹𝑁) ∈ V)
331ad2antrr 725 . . . . . . . 8 (((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) ∧ 𝑈 = (𝑃𝑁)) → 𝑈𝑉)
3422ad2antrr 725 . . . . . . . . 9 (((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) ∧ 𝑈 = (𝑃𝑁)) → (iEdg‘𝑌) = {⟨(𝐹𝑁), (𝐼‘(𝐹𝑁))⟩})
35 eupth2lem3lem3.e . . . . . . . . . . . . . 14 (𝜑 → if-((𝑃𝑁) = (𝑃‘(𝑁 + 1)), (𝐼‘(𝐹𝑁)) = {(𝑃𝑁)}, {(𝑃𝑁), (𝑃‘(𝑁 + 1))} ⊆ (𝐼‘(𝐹𝑁))))
3635adantr 484 . . . . . . . . . . . . 13 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → if-((𝑃𝑁) = (𝑃‘(𝑁 + 1)), (𝐼‘(𝐹𝑁)) = {(𝑃𝑁)}, {(𝑃𝑁), (𝑃‘(𝑁 + 1))} ⊆ (𝐼‘(𝐹𝑁))))
37 ifptru 1071 . . . . . . . . . . . . . 14 ((𝑃𝑁) = (𝑃‘(𝑁 + 1)) → (if-((𝑃𝑁) = (𝑃‘(𝑁 + 1)), (𝐼‘(𝐹𝑁)) = {(𝑃𝑁)}, {(𝑃𝑁), (𝑃‘(𝑁 + 1))} ⊆ (𝐼‘(𝐹𝑁))) ↔ (𝐼‘(𝐹𝑁)) = {(𝑃𝑁)}))
3837adantl 485 . . . . . . . . . . . . 13 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → (if-((𝑃𝑁) = (𝑃‘(𝑁 + 1)), (𝐼‘(𝐹𝑁)) = {(𝑃𝑁)}, {(𝑃𝑁), (𝑃‘(𝑁 + 1))} ⊆ (𝐼‘(𝐹𝑁))) ↔ (𝐼‘(𝐹𝑁)) = {(𝑃𝑁)}))
3936, 38mpbid 235 . . . . . . . . . . . 12 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → (𝐼‘(𝐹𝑁)) = {(𝑃𝑁)})
40 sneq 4535 . . . . . . . . . . . . 13 ((𝑃𝑁) = 𝑈 → {(𝑃𝑁)} = {𝑈})
4140eqcoms 2806 . . . . . . . . . . . 12 (𝑈 = (𝑃𝑁) → {(𝑃𝑁)} = {𝑈})
4239, 41sylan9eq 2853 . . . . . . . . . . 11 (((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) ∧ 𝑈 = (𝑃𝑁)) → (𝐼‘(𝐹𝑁)) = {𝑈})
4342opeq2d 4772 . . . . . . . . . 10 (((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) ∧ 𝑈 = (𝑃𝑁)) → ⟨(𝐹𝑁), (𝐼‘(𝐹𝑁))⟩ = ⟨(𝐹𝑁), {𝑈}⟩)
4443sneqd 4537 . . . . . . . . 9 (((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) ∧ 𝑈 = (𝑃𝑁)) → {⟨(𝐹𝑁), (𝐼‘(𝐹𝑁))⟩} = {⟨(𝐹𝑁), {𝑈}⟩})
4534, 44eqtrd 2833 . . . . . . . 8 (((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) ∧ 𝑈 = (𝑃𝑁)) → (iEdg‘𝑌) = {⟨(𝐹𝑁), {𝑈}⟩})
4631, 32, 33, 451loopgrvd2 27293 . . . . . . 7 (((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) ∧ 𝑈 = (𝑃𝑁)) → ((VtxDeg‘𝑌)‘𝑈) = 2)
4730, 46breqtrrid 5068 . . . . . 6 (((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) ∧ 𝑈 = (𝑃𝑁)) → 2 ∥ ((VtxDeg‘𝑌)‘𝑈))
48 z0even 15708 . . . . . . 7 2 ∥ 0
4919ad2antrr 725 . . . . . . . 8 (((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) ∧ 𝑈 ≠ (𝑃𝑁)) → (Vtx‘𝑌) = 𝑉)
50 fvexd 6660 . . . . . . . 8 (((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) ∧ 𝑈 ≠ (𝑃𝑁)) → (𝐹𝑁) ∈ V)
5113, 14, 15, 16, 1, 17trlsegvdeglem1 28005 . . . . . . . . . 10 (𝜑 → ((𝑃𝑁) ∈ 𝑉 ∧ (𝑃‘(𝑁 + 1)) ∈ 𝑉))
5251simpld 498 . . . . . . . . 9 (𝜑 → (𝑃𝑁) ∈ 𝑉)
5352ad2antrr 725 . . . . . . . 8 (((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) ∧ 𝑈 ≠ (𝑃𝑁)) → (𝑃𝑁) ∈ 𝑉)
5422adantr 484 . . . . . . . . . 10 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → (iEdg‘𝑌) = {⟨(𝐹𝑁), (𝐼‘(𝐹𝑁))⟩})
5539opeq2d 4772 . . . . . . . . . . 11 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → ⟨(𝐹𝑁), (𝐼‘(𝐹𝑁))⟩ = ⟨(𝐹𝑁), {(𝑃𝑁)}⟩)
5655sneqd 4537 . . . . . . . . . 10 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → {⟨(𝐹𝑁), (𝐼‘(𝐹𝑁))⟩} = {⟨(𝐹𝑁), {(𝑃𝑁)}⟩})
5754, 56eqtrd 2833 . . . . . . . . 9 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → (iEdg‘𝑌) = {⟨(𝐹𝑁), {(𝑃𝑁)}⟩})
5857adantr 484 . . . . . . . 8 (((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) ∧ 𝑈 ≠ (𝑃𝑁)) → (iEdg‘𝑌) = {⟨(𝐹𝑁), {(𝑃𝑁)}⟩})
591adantr 484 . . . . . . . . . 10 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → 𝑈𝑉)
6059anim1i 617 . . . . . . . . 9 (((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) ∧ 𝑈 ≠ (𝑃𝑁)) → (𝑈𝑉𝑈 ≠ (𝑃𝑁)))
61 eldifsn 4680 . . . . . . . . 9 (𝑈 ∈ (𝑉 ∖ {(𝑃𝑁)}) ↔ (𝑈𝑉𝑈 ≠ (𝑃𝑁)))
6260, 61sylibr 237 . . . . . . . 8 (((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) ∧ 𝑈 ≠ (𝑃𝑁)) → 𝑈 ∈ (𝑉 ∖ {(𝑃𝑁)}))
6349, 50, 53, 58, 621loopgrvd0 27294 . . . . . . 7 (((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) ∧ 𝑈 ≠ (𝑃𝑁)) → ((VtxDeg‘𝑌)‘𝑈) = 0)
6448, 63breqtrrid 5068 . . . . . 6 (((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) ∧ 𝑈 ≠ (𝑃𝑁)) → 2 ∥ ((VtxDeg‘𝑌)‘𝑈))
6547, 64pm2.61dane 3074 . . . . 5 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → 2 ∥ ((VtxDeg‘𝑌)‘𝑈))
66 dvdsadd2b 15648 . . . . 5 ((2 ∈ ℤ ∧ ((VtxDeg‘𝑋)‘𝑈) ∈ ℤ ∧ (((VtxDeg‘𝑌)‘𝑈) ∈ ℤ ∧ 2 ∥ ((VtxDeg‘𝑌)‘𝑈))) → (2 ∥ ((VtxDeg‘𝑋)‘𝑈) ↔ 2 ∥ (((VtxDeg‘𝑌)‘𝑈) + ((VtxDeg‘𝑋)‘𝑈))))
6712, 26, 29, 65, 66syl112anc 1371 . . . 4 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → (2 ∥ ((VtxDeg‘𝑋)‘𝑈) ↔ 2 ∥ (((VtxDeg‘𝑌)‘𝑈) + ((VtxDeg‘𝑋)‘𝑈))))
6827nn0cnd 11945 . . . . . . 7 (𝜑 → ((VtxDeg‘𝑌)‘𝑈) ∈ ℂ)
6924nn0cnd 11945 . . . . . . 7 (𝜑 → ((VtxDeg‘𝑋)‘𝑈) ∈ ℂ)
7068, 69addcomd 10831 . . . . . 6 (𝜑 → (((VtxDeg‘𝑌)‘𝑈) + ((VtxDeg‘𝑋)‘𝑈)) = (((VtxDeg‘𝑋)‘𝑈) + ((VtxDeg‘𝑌)‘𝑈)))
7170breq2d 5042 . . . . 5 (𝜑 → (2 ∥ (((VtxDeg‘𝑌)‘𝑈) + ((VtxDeg‘𝑋)‘𝑈)) ↔ 2 ∥ (((VtxDeg‘𝑋)‘𝑈) + ((VtxDeg‘𝑌)‘𝑈))))
7271adantr 484 . . . 4 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → (2 ∥ (((VtxDeg‘𝑌)‘𝑈) + ((VtxDeg‘𝑋)‘𝑈)) ↔ 2 ∥ (((VtxDeg‘𝑋)‘𝑈) + ((VtxDeg‘𝑌)‘𝑈))))
7367, 72bitrd 282 . . 3 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → (2 ∥ ((VtxDeg‘𝑋)‘𝑈) ↔ 2 ∥ (((VtxDeg‘𝑋)‘𝑈) + ((VtxDeg‘𝑌)‘𝑈))))
7473notbid 321 . 2 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → (¬ 2 ∥ ((VtxDeg‘𝑋)‘𝑈) ↔ ¬ 2 ∥ (((VtxDeg‘𝑋)‘𝑈) + ((VtxDeg‘𝑌)‘𝑈))))
75 simpr 488 . . . . 5 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → (𝑃𝑁) = (𝑃‘(𝑁 + 1)))
7675eqeq2d 2809 . . . 4 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → ((𝑃‘0) = (𝑃𝑁) ↔ (𝑃‘0) = (𝑃‘(𝑁 + 1))))
7775preq2d 4636 . . . 4 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → {(𝑃‘0), (𝑃𝑁)} = {(𝑃‘0), (𝑃‘(𝑁 + 1))})
7876, 77ifbieq2d 4450 . . 3 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → if((𝑃‘0) = (𝑃𝑁), ∅, {(𝑃‘0), (𝑃𝑁)}) = if((𝑃‘0) = (𝑃‘(𝑁 + 1)), ∅, {(𝑃‘0), (𝑃‘(𝑁 + 1))}))
7978eleq2d 2875 . 2 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → (𝑈 ∈ if((𝑃‘0) = (𝑃𝑁), ∅, {(𝑃‘0), (𝑃𝑁)}) ↔ 𝑈 ∈ if((𝑃‘0) = (𝑃‘(𝑁 + 1)), ∅, {(𝑃‘0), (𝑃‘(𝑁 + 1))})))
8010, 74, 793bitr3d 312 1 ((𝜑 ∧ (𝑃𝑁) = (𝑃‘(𝑁 + 1))) → (¬ 2 ∥ (((VtxDeg‘𝑋)‘𝑈) + ((VtxDeg‘𝑌)‘𝑈)) ↔ 𝑈 ∈ if((𝑃‘0) = (𝑃‘(𝑁 + 1)), ∅, {(𝑃‘0), (𝑃‘(𝑁 + 1))})))
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ↔ wb 209 ∧ wa 399 if-wif 1058 = wceq 1538 ∈ wcel 2111 ≠ wne 2987 {crab 3110 Vcvv 3441 ∖ cdif 3878 ⊆ wss 3881 ∅c0 4243 ifcif 4425 {csn 4525 {cpr 4527 ⟨cop 4531 class class class wbr 5030 ↾ cres 5521 “ cima 5522 Fun wfun 6318 ‘cfv 6324 (class class class)co 7135 0cc0 10526 1c1 10527 + caddc 10529 2c2 11680 ℤcz 11969 ...cfz 12885 ..^cfzo 13028 ♯chash 13686 ∥ cdvds 15599 Vtxcvtx 26789 iEdgciedg 26790 VtxDegcvtxdg 27255 Trailsctrls 27480 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2113 ax-9 2121 ax-10 2142 ax-11 2158 ax-12 2175 ax-ext 2770 ax-rep 5154 ax-sep 5167 ax-nul 5174 ax-pow 5231 ax-pr 5295 ax-un 7441 ax-cnex 10582 ax-resscn 10583 ax-1cn 10584 ax-icn 10585 ax-addcl 10586 ax-addrcl 10587 ax-mulcl 10588 ax-mulrcl 10589 ax-mulcom 10590 ax-addass 10591 ax-mulass 10592 ax-distr 10593 ax-i2m1 10594 ax-1ne0 10595 ax-1rid 10596 ax-rnegex 10597 ax-rrecex 10598 ax-cnre 10599 ax-pre-lttri 10600 ax-pre-lttrn 10601 ax-pre-ltadd 10602 ax-pre-mulgt0 10603 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-ifp 1059 df-3or 1085 df-3an 1086 df-tru 1541 df-ex 1782 df-nf 1786 df-sb 2070 df-mo 2598 df-eu 2629 df-clab 2777 df-cleq 2791 df-clel 2870 df-nfc 2938 df-ne 2988 df-nel 3092 df-ral 3111 df-rex 3112 df-reu 3113 df-rab 3115 df-v 3443 df-sbc 3721 df-csb 3829 df-dif 3884 df-un 3886 df-in 3888 df-ss 3898 df-pss 3900 df-nul 4244 df-if 4426 df-pw 4499 df-sn 4526 df-pr 4528 df-tp 4530 df-op 4532 df-uni 4801 df-int 4839 df-iun 4883 df-br 5031 df-opab 5093 df-mpt 5111 df-tr 5137 df-id 5425 df-eprel 5430 df-po 5438 df-so 5439 df-fr 5478 df-we 5480 df-xp 5525 df-rel 5526 df-cnv 5527 df-co 5528 df-dm 5529 df-rn 5530 df-res 5531 df-ima 5532 df-pred 6116 df-ord 6162 df-on 6163 df-lim 6164 df-suc 6165 df-iota 6283 df-fun 6326 df-fn 6327 df-f 6328 df-f1 6329 df-fo 6330 df-f1o 6331 df-fv 6332 df-riota 7093 df-ov 7138 df-oprab 7139 df-mpo 7140 df-om 7561 df-1st 7671 df-2nd 7672 df-wrecs 7930 df-recs 7991 df-rdg 8029 df-1o 8085 df-oadd 8089 df-er 8272 df-map 8391 df-en 8493 df-dom 8494 df-sdom 8495 df-fin 8496 df-dju 9314 df-card 9352 df-pnf 10666 df-mnf 10667 df-xr 10668 df-ltxr 10669 df-le 10670 df-sub 10861 df-neg 10862 df-nn 11626 df-2 11688 df-n0 11886 df-xnn0 11956 df-z 11970 df-uz 12232 df-xadd 12496 df-fz 12886 df-fzo 13029 df-hash 13687 df-word 13858 df-dvds 15600 df-edg 26841 df-uhgr 26851 df-ushgr 26852 df-uspgr 26943 df-vtxdg 27256 df-wlks 27389 df-trls 27482 This theorem is referenced by: eupth2lem3lem7 28019
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# Nonstandard Models of Arithmetic 27
MW: Enayat’s second major result is:
Theorem 7: Every countable recursively saturated model of PA+ΦT is a T-standard model of PA.
Recall the definition of ΦT: the set of formulas of the form
{φ → Con(Tn) : φ∈L(PA), n∈ℕ}
where Tn is the first n statements of T, and φ is φ∈L(PA) translated into a formula of L(ZF).
Notice the countability condition, which was not needed for the reverse result, Proposition 6. Enayat mentions that the quickest way to prove the theorem is via a replendence argument. Someday I may do a post about this, but the proof he gives is worth our attention. It falls into four parts. Let N be a countable recursively saturated model of PA+ΦT. Here’s a quick preview—details to come, so don’t worry if not all of it makes sense yet.
1. Using the Arithmetized Completeness Theorem, plus recursive saturation, he obtains a model U of T whose ω is both an end-extension of N and elementarily equivalent to it: ωUeN and Th(ωU)=Th(N). (Recall that Th(N) is the set of all sentences of L(PA) that hold in N, and likewise for Th(ωU).)
2. Next he observes that Proposition 6 applies to ωU, so it’s recursively saturated.
3. Since ωUeN, their standard systems are the same: SSy(ωU)=SSy(N). The standard system of a nonstandard model of PA is the family of subsets of ℕ that can be coded by elements of the model.
4. Finally, he notes that any two countable recursively saturated models of PA are isomorphic if they have the same first-order theory and the same standard system. That is, if Th(M)=Th(N) and SSy(M)=SSy(N) and M and N are countable and recursively saturated, then MN. So N is isomorphic to the ω of a model of T. But that’s the definition of T-standard. QED
The rest of this post expands on (1), with (2)–(4) relegated to later posts. Let’s start with the Arithmetized Completeness Theorem (ACT). I wrote a Topics post about this, but it’s pretty long so I’ll just cherry-pick the info we need.
The Completeness Theorem says that a consistent theory T has a model U. The ACT says that a consistent arithmetic (aka representable) theory has an arithmetic model. Arithmetic here means definable via formulas of L(PA). Thus, the set of Gödel numbers of the axioms of T is definable by a formula of L(PA): φ∈T iff θ(⌜φ⌝) holds, for some θ(x)∈L(PA). Likewise, the domain, relations, constants, and functions of U are all specified by formulas in L(PA).
Moreover, the ACT says this can all be proved inside PA. The proof of the ACT shows how, starting with a formula defining a consistent T, we can whip up formulas for the all the semantic aspects of U. Foremost among these is the truth predicate SatU: SatU:(⌜φ⌝) for a sentence φ expresses the fact that U satisfies φ.
It’s convenient to gloss over the distinction between a formula and its Gödel number, so I’ll mostly omit the brackets ⌜⌝ below. Imagine that formulas are numbers.
Because all this is happening inside PA, we can apply it to any model N. For nonstandard N, matters take a curious turn. First off, we’ll have nonstandard formulas, including some of nonstandard length. (If Formula(x) is the formalization of “x is a formula”, then any nonstandard d satisfying Formula(d) represents a nonstandard formula.)
The truth predicate SatU defines satisfaction for all sentences in the language of T, including nonstandard ones. The proof of the ACT shows that SatU extends ordinary standard satisfaction: if φ is standard, then SatU(φ) iff U⊧φ. To avoid confusion, I will use the symbol ⊧ only for satisfaction in the standard sense.
The notion of consistency also demands a second look. Con(T), a sentence in the language of arithmetic, says that there is no proof of a contradiction, not just no standard proof. An instructive example: by Gödel’s Second Incompleteness Theorem, the theory PA+¬Con(PA) is consistent. Suppose N is a model of it. Then N⊧¬Con(PA). So in N, there is a proof of a contradiction, but no standard proof.
The ACT demands that N⊧Con(T) before it goes to work. Showing that T is consistent in the standard sense doesn’t cut it.
OK, let’s look at Enayat’s step (1). We need to construct a model of T, starting with a model of PA. Not only that, but ΦT is knocking on the door of Con(T). This suggests using the ACT. Larry Manevitz observed (in the 70s) something similar. Let N be a model of PA. Then
If N⊧Con(ZF), then there is a model of ZF whose ω is an end extension of N.
If N⊧Con(ZF), the ACT immediately hands us a model U of ZF. Manevitz’s result falls out from looking at the recursion that imbeds ℕ as an initial segment of any model (say M) of PA. You recursively define a map nnM with Successor(n)↦SuccessorM(nM). A straightforward induction shows that the image is an initial segment. This can all be formalized in PA, so it works for any model N of PA. In particular, N an be imbedded as an initial segment of ωU. That’s Manevitz’s result. It works without change for any arithmetic extension T of ZF: if N⊧Con(T), then there is a model of T whose ω is an end extension of N.
We just need to deal with two issues to get Enayat’s step (1):
• ΦT gives us Con(Tn) for any n∈ℕ. (Just let φ ≡ 1=1.) This is not quite the same as Con(T).
• We also need Th(ωU)=Th(N).
For the first bullet point, we might think we’re in the clear since Con(Tn) for all n∈ℕ implies that T is consistent. But we saw above that N⊧Con(T) is stronger. Overspill comes to the rescue. Since N⊧Con(Tn) for all n∈ℕ, we have N⊧Con(Td) for some nonstandard d. We can apply the ACT to Td to get a model U of it. But Td includes all the standard formulas of T, so UT with T interpreted in the standard sense.
For the second bullet point, we appeal to recursive saturation. The trick here: code Th(N) via a nonstandard element of N. Any element v of N can be regarded as a bitstring; let’s write vi for the i-th bit of v. Also write φi for the sentence with Gödel number i. If v is to code Th(N), we need vi=1 when N⊧φi and vi=0 when N⊧¬φi. In other words,
N⊧(φivi=1).
Now, the set {φivi=1 : i∈ℕ} is a recursive set of formulas with the free variable v. It is obviously finitely satisfiable, so it’s a recursive type. Recursive saturation says there is an element aN realizing this type:
N⊧(φiai=1), for all i∈ℕ.
Next step is to whip up a theory combining T and Th(N). Once again we do this first for finite fragments of the theory. Let
Γ(n) = Tn+{φi : i<n and ai=1}
N⊧ΦT tells us that N⊧Con(Γ(n)) for all n∈ℕ. By overspill, N⊧Con(Γ(d)) for some nonstandard d. Apply the ACT to Γ(d). It hands us a U satisfying Γ(d). A fortiori U is a model of T (standardly speaking) and of exactly those φi that hold in N. For the record, note that N “believes” there are nonstandard sentences φe, but if e>d then Γ(d) says nothing about them. Fortunately, we care about φi only for standard i. Summing up, Th(ωU)=Th(N), and by the Manevitz argument, ωUeN. (1) is done!
Finally, a point to ponder. It looks as though the ACT-provided truth predicate SatU offers a way around Tarski’s Undefinability Theorem. Namely, N⊧φ iff SatU). What gives?
The answer lies buried in the details of Topics 4 (the Addendum) and Topics 10 (a remark at the end of the Proof Sketch). I’ll explain next time.
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# Lesson: Investigating Torque
### Summary
Students learn about torsion as a force acting upon structures and have the opportunity to design something to withstand this force.
### Engineering Connection
Understanding how torsion affects objects helps engineers design products and structures (from bicycles to bridges) that are safe and sound. For civil and mechanical engineers, evaluation of the effect of torsional forces on objects, such as supporting beams in buildings or machine parts, is critical to making sure that structures and machines do not fail.
### Educational Standards
• International Technology and Engineering Educators Association: Technology
• Massachusetts: Science
### Learning Objectives
• Students learn the concept of a moment (torque) of a force and learn how to calculate moments.
• Students learn how moments ("turning forces") create bending and torsion loads on structures.
• Students understand the effects of bending and torsion loads.
• Students gain an appreciation of how engineers can design a structure to resist bending and torsion.
### Introduction/Motivation
Introduce students to all the keywords and recap the concepts from Fairly Fundamental Facts about Forces and Structures lesson.
### Lesson Background and Concepts for Teachers
Students should have a basic understanding of tension, compression, shear, bending, torsion and concept of a moment (torque). Review Lesson 1: Fairly Fundamental Facts about Forces and Structures, and complete the Introduction to Loads Acting on Structures lesson before beginning this lesson.
Moment and torque can be use interchangeably, physicists tend to use the word torque and engineers tend to use moment when referring to forces that cause rotation. The ability of any beam or structural member to resist bending and torsion, depends on the following factors (variables):
Material:: Every material has a different yield strength, tensile strength, and shear strength, which ultimately determine the load that a material can withstand and the amount of deformation (stretching, bending, twisting) that accompanies a given load.
Size: Engineers calculate the moment of inertia of a beam or column, which is a measure of the size and shape of its cross-sectional area, and how far away the area is from the center of the beam. The greater the moment of inertia, the greater the load that can be carried by the structural member. This means that increasing the cross-sectional area of a beam or taking a certain amount of area and spreading it out farther from the center, will increase the strength and stiffness of the beam (see Figure A).
It might be instructive for students to draw on graph paper different designs for beams, showing how the cross-sectional area, or the distribution of area can increase to make a stronger, stiffer beam. Have them try to draw two beam cross-sections that have the same areas, but different moments of inertia (meaning that the area of one beam is spread out farther away from the center, and the area of the other is more concentrated around the center).
Reinforcement / Composite Structure: Many structural members are composite materials, which means that they are made from two or more different materials bonded together. Foam board is an example of a composite material; it is a layer of foam sandwiched between two layers of paper. Reinforced concrete has steel rods (called rebars, short for reinforcing bars) that are placed inside the form before the concrete is poured. Concrete is a material that is very strong in compression, but very weak in tension; the steel rebars can take great tensile loads and thus they overcome the weakness of the concrete and make the composite material much stronger. Fiberglass, which is used to make canoes, is mostly a plastic epoxy resin; the epoxy resin by itself would not be that strong, however, it is reinforced by glass fibers inside that are very strong in tension.
Structural Bracing: Any structural members that help a structure to resist bending and/or torsion. Examples: wire cables (called guy wires) bracing a tower; truss bracing in bridges, towers and skyscrapers (a truss structure is a triangular formation of long, thin bars pinned together at the ends); brackets and braces such as those used to hold up book shelves and store signs, and strengthen table legs and dump truck bodies.
### Associated Activities
• Wimpy Radar Antenna - Students reinforce an antenna tower made from foam insulation so that it can withstand specified bending and twisting moments (torques) with minimal deflection. They discuss the problem, run initial tests and graph the results. Then they design, construct and test sturdier towers, and graph the results.
### Assessment
Assess students' understanding, individually or as a group, using the Investigating Questions provided in the associated activity.
### Contributors
Douglas Prime, Center for Engineering Educational Outreach, Tufts University
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# How to calculate the interior angles of a convex, equilateral heptagon given some of the interior angles of the heptagon. [closed]
I cannot find an easy-that of which holds true for any knowledge of mathematics-way to explain, to a peer, the solution, and the reason the solution is true, to the following problem:
Let H be a convex, equilateral heptagon whose angles measure (in degrees) 168°, 108°, 168°, 108°, x°, y°, and z° in clockwise order. Computer the number y°.
Will someone please not only show the solution, but put the steps to solve it, and why those steps work, in simple terms so that I may expound it to my peer in a rudimentary fashion?
## closed as unclear what you're asking by Aretino, Yanior Weg, Lord Shark the Unknown, Shailesh, YuiTo ChengMay 11 at 4:04
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
• Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. – saulspatz May 10 at 15:02
• I clearly stated in the problem everything that is necessary to solve the problem, and I also explained that I can solve the problem, as it is not, to me, a challenging problem, but I require simple steps to reach the solution, and a simple way to explain the, for a peer. – Evan Merrill May 10 at 15:09
• Then perhaps you ought to ask this question on matheducators.stackexchange.com – saulspatz May 10 at 15:16
• If you know the solution you are kindly requested to explain it in your question. – Aretino May 10 at 15:19
• Thank you, I had hitherto not been aware of that site’s existance. I, however, until I find necssecary upon precedement, will keep question here because I have observed questions such as these on this site. – Evan Merrill May 10 at 15:20
There is no answer. I constructed it in GeoGebra and the two points are further that $$2$$ units apart. You need to reach from $$A'$$ to $$B''$$ with two segments of length $$1$$, but they are over $$2$$ apart in $$y$$ alone.
• @EvanMerrill I checked the answer by hand, the way Ross Millikan indicated in one comment to your question. I used antother coordinate system than them. The distance between the 2 endpoints of the sequence of angles you give in the question is $\approx 2.6$ times the side length, both in my and their calculation. Please check that you got the angles and their order correct! – Ingix May 10 at 17:16
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# Show that the Diffusion Eqn after substitution gives the Helmholtz Eqn
• jkthejetplane
In summary: algebraic manipulation means for example that you ll transfer the denominator ##\psi(r)## on the right hand side, then you also may transfer ##\kappa## on the right hand side and then transfer all the right hand side...
jkthejetplane
Homework Statement
Question is shown below. Basically i am wondering if i am over thinking it or am i just plugging it in? I do not see where psi or theta are possible given to do any further work. Same would go for part b.
Relevant Equations
Helmholtz as defined in my text given below as well
Question:
Helmholtz as defined in text:
My attempt so far
vanhees71 and Delta2
So you conclude that each side of your last equation must be equal to some constant $C$, which means that $\psi$ satisfies...
And then you need to justify why $C \leq 0$.
(The constant $k$ which appears in the Helmholtz equation is not necessarily equal to the diffusivity $\kappa$ which appears in the diffusion equation. You need to clearly distinguish between these in your answer.)
hutchphd, vanhees71 and Delta2
pasmith said:
So you conclude that each side of your last equation must be equal to some constant $C$, which means that $\psi$ satisfies...
And then you need to justify why $C \leq 0$.
(The constant $k$ which appears in the Helmholtz equation is not necessarily equal to the diffusivity $\kappa$ which appears in the diffusion equation. You need to clearly distinguish between these in your answer.)
Ok so that makes sense as that is how i have started similar problems in the pas that i had boundary conditions for.
I got Θ(t) = Θ(0)eCt and for the other ψ(r) = sin(ψC/$\kappa$) + cos(ψC/$\kappa$)
Then i do the same for Helmholtz eq? OR
Do i set the C/$\kappa$ = $k$? Then plug back in my general solution of ψ(r) and the second derivative of it to the Helmholtz eq hoping to yield 0? Showing that it is only true when $C \leq 0$ ?
No , i believe its not the easiest path to solve for ##\psi(\mathbf{r})## and then show that this ##\psi## satisfies Helmholtz. It is also wrong that you decide to replace the operator ##\nabla^2## with ##\frac{\partial^2}{\partial r^2}## cause we might be in 2 or 3 dimensions where the ##\nabla^2## operator has different expression. You have shown that $$\kappa\frac{\nabla^2\psi(\mathbf{r})}{\psi(\mathbf{r})}=C$$
From the above equation , how can you manipulate it algebraically to show that it is equivalent to Helmholtz equation ,I repeat again, without solving that equation for ##\psi##, just some algebraic manipulation is all that is needed and replacement of ##\frac{C}{\kappa}## with a proper constant so that we have the Helmholtz equation at the end of the algebraic manipulation.
Last edited:
vanhees71
Delta2 said:
No , i believe its not the easiest path to solve for ##\psi(\mathbf{r})## and then show that this ##\psi## satisfies Helmholtz. It is also wrong that you decide to replace the operator ##\nabla^2## with ##\frac{\partial^2}{\partial r^2}## cause we might be in 2 or 3 dimensions where the ##\nabla^2## operator has different expression. You have shown that $$\kappa\frac{\nabla^2\psi(\mathbf{r})}{\psi(\mathbf{r})}=C$$
From the above equation , how can you manipulate it algebraically to show that it is equivalent to Helmholtz equation ,I repeat again, without solving that equation for ##\psi##, just some algebraic manipulation is all that is needed and replacement of ##\frac{C}{\kappa}## with a proper constant so that we have the Helmholtz equation at the end of the algebraic manipulation.
So I am just a bit confused still. I need to solve the diff equation for ψ, then just manipulate it to look like Helmholtz?
jkthejetplane said:
So I am just a bit confused still. I need to solve the diff equation for ψ, then just manipulate it to look like Helmholtz?
If you really want to solve a differential equation, you can manipulate it to look like Helmholtz, because it is easy to solve Helmholtz equation (Your teacher told you the formal solution, didn't he?). This also applies to other diff equations that are easy to solve. We always do this when solving Schroedinger equation.
pai535 said:
If you really want to solve a differential equation, you can manipulate it to look like Helmholtz, because it is easy to solve Helmholtz equation (Your teacher told you the formal solution, didn't he?). This also applies to other diff equations that are easy to solve. We always do this when solving Schroedinger equation.
No he hasn't. This is an upper level class but i had taken a 5 year break from school so i think he assumes we remember everything.
jkthejetplane said:
No he hasn't. This is an upper level class but i had taken a 5 year break from school so i think he assumes we remember everything.
For this question, you only need to give the equation (and maybe don't need to solve it).
For Helmholtz equation, you can solve it by separation of variables.
I am not really getting anywhere. Do you have to be super vague on here? I feel like no one gives me direct answers haha. I just want to understand the problem
No you don't have to solve any equation for ##\psi## (I repeat myself for 3rd time and others have said this as well).
Just algebraic manipulation of this equation $$\kappa\frac{\nabla^2\psi(\mathbf{r})}{\psi(\mathbf{r})}=C$$ and you ll have Helmholtz equation.
Algebraic manipulation means for example that you ll transfer the denominator ##\psi(r)## on the right hand side, then you also may transfer ##\kappa## on the right hand side and then transfer all the right hand side to the left hand side so you ll have equation of "something"=0. That something will look like Helmholtz equation.
Delta2 said:
$$\kappa\frac{\nabla^2\psi(\mathbf{r})}{\psi(\mathbf{r})}=C$$
From the above equation , how can you manipulate it algebraically to show that it is equivalent to Helmholtz equation ,I repeat again, without solving that equation for ##\psi##, just some algebraic manipulation is all that is needed and replacement of ##\frac{C}{\kappa}## with a proper constant so that we have the Helmholtz equation at the end of the algebraic manipulation.
If you want to know how to get Helmholtz from diffusion equation, this is what you want. You only need to figure out how you can transform the above equation into the form of Helmholtz.
If you want to solve the Helmholtz equation, I will give a hint in Cartesian coordinates.
Do separation of variables: ##\psi(\vec{r}) = X(x)Y(y)Z(z)##
Then: $$YZ\frac{d^2 X}{dx^2} + ZX\frac{d^2 Y}{dy^2} + XY\frac{d^2 Z}{dz^2} + k^2XYZ = 0.$$
Divide ##XYZ## on both sides: $$\frac{X''}{X}+\frac{Y''}{Y}+\frac{Z''}{Z} + k^2 = 0.$$
You can try to solve it now.
Delta2
You can't solve the $\psi$ equation; you don't know what the boundary is or what condition you have to apply there.
You can solve the $\theta$ equation, which will tell you what values of $C$ are physically realistic (because you can rule out values of $C$ which lead to solutions which grow exponentially with time...).
Last edited:
pai535 and Delta2
I think this discussion overcomplicates things. You just make an separation ansatz
$$T(t,\vec{x})=\Theta(t) \psi(\vec{r})$$
as in the text shown in the scan in #1. Plugging this into the diffusion equation then leads to the Helmholtz equation for ##\psi(\vec{r})## by just defining ##k## in the right way.
Delta2
## 1. What is the Diffusion Equation?
The Diffusion Equation is a mathematical equation that describes the process of diffusion, which is the movement of particles from an area of high concentration to an area of low concentration. It is often used in physics, chemistry, and other scientific fields to model the spread of substances through a medium.
## 2. What is the Helmholtz Equation?
The Helmholtz Equation is a mathematical equation that describes the behavior of waves in a medium. It is used in many areas of science and engineering, such as acoustics, electromagnetics, and fluid dynamics.
## 3. How are the Diffusion and Helmholtz Equations related?
The Diffusion and Helmholtz Equations are related through substitution. By substituting the diffusion coefficient and the diffusion flux into the Helmholtz Equation, we can show that the resulting equation is equivalent to the Diffusion Equation.
## 4. Why is it important to show that the Diffusion Equation after substitution gives the Helmholtz Equation?
Showing that the Diffusion Equation after substitution gives the Helmholtz Equation is important because it helps us understand the underlying principles of diffusion and wave behavior. It also allows us to apply the same mathematical techniques and solutions to both equations, making it easier to solve complex problems in various fields of science and engineering.
## 5. Can you provide an example of how the Diffusion Equation and Helmholtz Equation are used in real-world applications?
One example is in the study of heat transfer. The Diffusion Equation is used to model the diffusion of heat through a material, while the Helmholtz Equation is used to study the propagation of thermal waves. By understanding these equations and their relationship, we can better predict and control heat transfer in various systems, such as in building insulation or electronic devices.
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# It’s A Dog’s World - PowerPoint PPT Presentation
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It’s A Dog’s World. Student’s Name Class Date. Farmer Jones raises dogs. He needs to feed his puppies. There are 12 puppies in his barn. One bowl will feed 4 puppies. How many bowls of food does Farmer Jones need to fix?.
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It’s A Dog’s World
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## It’s A Dog’s World
Student’s Name
Class
Date
Farmer Jones raises dogs. He needs to feed his puppies. There are 12 puppies in his barn. One bowl will feed 4 puppies. How many bowls of food does Farmer Jones need to fix?
Solve the problem and then click on the right arrow buttons to see the author’s solution.
There are 12 dogs in the barn.
• Click once on the screen to see 1 group of ten.
• Click a second time on the screen to see 2 ones.
One bowl will feed 4 dogs.
Click once on the screen to see the dogs at their bowl.
12 – 4
Use one more bowl to feed 4 more dogs.
Click once on the screen to see these dogs at their bowl.
12 - 4 - 4
Use one more bowl to feed 4 more dogs.
Click once on the screen to see these dogs at their bowl.
12 - 4 - 4 - 4
There are no dogs left in the barn. We subtracted 3 groups of 4 dogs.
12 - 4 - 4 - 4 = 0
Farmer Jones will need to use 3 bowls to feed his dogs.
12 - 4 - 4 - 4 = 0
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# 【状元之路】2015-2016高中英语 Unit 4 Astronomy the science of the stars单元综合测评 新人教版必修3
Unit 4
Astronomy: the science of the stars
(时间：90 分钟 满分：110 分)
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A
B
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D 答案与解析： B 推理判断题。通读全文可知 Andromeda 是一个星座，像太阳系； Upsilon Andromeda 是一颗恒星，像太阳；three planets 是它的三个卫星，因此可推出 B 项正确。 2．According to Debra Fischer，in the new solar system, ________. A．the farther a planet is from the star, the longer it takes to go around the star B．the larger a planet is，the shorter it takes to go around the star C．it takes all the planets around three and a half to four years to go around the star D．the planets move around the star at the same(均匀) speed 答案与解析：A 推理判断题。根据天文常识及本文所述，任何行星的卫星的运转周期不可能是 一致的，越是远离行星，一般运转周期越长。 3．What led to Marcy's discovery of the planets? A．That he found a dog owner was pulled by his dog. B．The tremble of Upsilon Andromeda. C．That any planet has gravity. D．That he believes any star has its planets. 答案与解析：B 细节理解题。由倒数第二段可知。 4．We can infer that the scientists are ________. A．to find out whether the planets can support life B．to find means to communicate with the living beings on the planets C．to find means to travel to the planets D．soon to be able to answer the question of whether there is life on these planets 答案与解析：A 推理判断题。其他几项提到的不是本文重点或没提及。 B The earth is the only planet that scientists think has life. Why does the earth have life while the other planets ( 行星) don't? For one thing, the earth is just the right temperature. As the_third_planet from the sun, the earth seems to be just the right distance away. The planets which are closer to the sun are hot that their surfaces bake (烘烤) in the sun. The farthest planets are cold balls. When the earth developed — which scientists believe may have happened about 4 billion years ago — many gases covered the earth. The gases caused the earth to be hot. But something
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wonderful happened. The temperature was just right for thick clouds to form. It rained very hard for a very long time. This gave the earth its oceans. Water made it possible for plants to grow. The plants created oxygen in the atmosphere. Oxygen is the gas that humans and animals breathe. Only one other planet in the solar system seems to be something like the earth. That planet is Mars (火星). Mars is smaller than the earth, and it is quite a bit cooler. But it is not too cold for humans. On some days, the temperatures are as low as a winter day in the northern United States. If you wore a special space suit, you could walk around on Mars. You would have to bring your own air to breathe, the air on Mars is too thin to breathe. Mars has the largest volcano (火山) in the solar system. It is sixteen miles high. The highest volcano on the earth is five miles high. The most unexpected (未预测到的) sight on Mars is dried up river beds. Scientists believe that Mars was once much better than it is now. Does this mean there could have been living things on Mars? Scientists are not sure, but there has been no sign so far. 5．Which of the following is NOT the reason that the earth has life? A. The earth is just the right temperature. B. The volcanos on earth are lower than those on Mars. C. There is oxygen in the atmosphere of the earth. D. The earth seems to be just the right distance away from the sun. 答案与解析：B 细节理解题。文中提到地球是可证实的唯一有生命存在的太阳系行星，其原因 包括地球合适的温度、距太阳距离适中和大气里含有氧气等。B 项火山与此无关。 6 ． The underlined phrase “the third planet” in the first paragraph ________. A. the Sun C. the Earth B. the Moon D. the Mars refers to
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have compiled some of the popular ways by which time travel is made possible in fiction books： ●Through the wormhole( 虫 孔 )— It is a shortcut between points in space － time. According to Einstein, an object can pass through this if it can travel at the speed of light with infinite mass. An example of a wormhole is described in Riddle of the Red Bible. ●Through a black hole — A black hole is a vacuum in space where light cannot even escape or pass through. In some sci－fi movies, like Star Trek, black holes become the means to travel through time. ●Through time machines — In fiction, time machines are complex vehicles that can travel faster than light. A time machine can be a strange vehicle like TARDIS in Doctor
Who or a specially modified car like the one used in Back to the Future.
●Through a parallel(平行的) universe — Another popular way to travel through time is to be able to slip or slide into a parallel universe where one can go back to a point in time and see a different reality. Though we know time travel may not be possible, it is still an entertaining subject and a heated topic for most of us. 语篇解读 本文作者在文中介绍了科幻小说里讲述时间旅行的几种方式。 8．The underlined word “enthralled” in the first paragraph can be replaced “________”． A．thrilled C．shocked B．confused D．upset by
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parents. Greet the
children warmly and let them know you are glad to see them. Children learn a lot from the
space. Her school had an unused __26__， and Katie thought it was the perfect spot. The school liked the garden idea, and all the students were __27__ to help. But Katie didn't __28__ there. She worked with local __29__who gave her plots (小块土地) of land for additional (额外的) gardens. Katie hoped to grow all kinds of crops, __30__ she didn't know how. She asked for help from an expert gardener. Together, they __31__ the types of vegetables that grow best in the area. A seed company then gave them all the plants. Katie's __32__ grew even bigger than her cabbage. Her group, Katie's Krops, now has seven gardens. All the land __33__ that they can grow a lot of food for the needy. Encouraged by Katie, lots of kids and adults __34__ to plant and water. Katie's Krops has __35__ soup kitchens over 5,000 pounds of vegetables so far. 16．A. studied C. planted 17．A. though C. until 18．A. fresh C. sweet 19．A. build C. leave 20．A. sent C. told 21．A. houses C. clothes 22．A. kindness C. protection 23．A. recognize C. feed 24．A. silence C. trust 25．A. forgot C. doubted 26．A. desk C. field 27．A. excited C. brave 28．A. stop C. live 29．A. students B. cut D. washed B. if D. because B. hard D. big B. accept D. find B. returned D. lent B. jobs D. meals B. happiness D. question B. hire D. save B. luck D. satisfaction B. promised D. decided B. room D. book B. hurry D. possible B. lie D. help B. farmers
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C. teachers 30．A. so C. for 31．A. cooked C. hid 32．A. trouble C. picture 33．A. means C. feels 34．A. hate C. refuse 35．A. sold C. shown 答案与解析：
D. workers B. or D. but B. chose D. changed B. idea D. mistake B. hopes D. advises B. happen D. volunteer B. bought D. given
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Krops，所以 Katie 用蔬菜救助人们的“想法(idea)”变得更大了。 33．A 所有这些土地“意味着(means)”他们能为生活贫困的人提供更多食物。 34．D 受 Katie 的鼓励，很多小孩和成人都“自愿(volunteer)”来干活。 35．D 由上文可知 Katie 送给施食处蔬菜。所以在此用 given。 第二节 语法填空(共 10 小题；每小题 1.5 分，满分 15 分) 阅读下面材料，在空白处填入适当的内容(1 个单词)或括号内单词的正确形式。 Once a man was walking along a beach. The sun was shining and it was a beautiful day. In the distance he could see a person 36.________(go) back and forth between the surf's edge 37.________ the beach. As the man approached, he could see there were hundred of starfish stranded(搁浅的) on the sand as the result of the natural 38.________(behave) of the tide. The man 39.________(stick) by the apparent uselessness of the task. There were far too many starfish. Many of them were sure to die. 40.________ he approached, the person continued the task of picking up the starfish one by one and throwing them into the ocean. As he came up to the person, he said, “You must be crazy. There are 41.________(thousand) of miles of beach 42.________(cover) with starfish. You can't 43.________(possible) make a difference.” The person looked at the man. He then bent down and picked up one 44.________(many) starfish and threw it back into the ocean. He turned back to the man and said, “It surely made a difference 45.________ that one！” 答 案 ： 36.going 37.and 38.behavior 39.was stuck 40.As/When 41.thousands 42.covered 43.possibly 44.more 45.to 第三部分 写作(共两节，满分 35 分) 第一节 短文改错(每小题 1 分，满分 10 分) 英语课上，老师要求同桌同学相互修改作文。假设以下短文为你同桌所写，请你对其进行修改。 文中共有 10 处错误，每句中最多有两处。错误涉及一个单词的增加、删除或修改。 增加：在缺词处加一个漏词符号(∧)，并在此符号下面写出该加的词。 删除：把多余的词用斜线(\)划掉。 修改：在错的词下画一横线，并在该词下面写出修改后的词。 注意：1.每处错误及其修改均仅限一词。 2．只允许修改 10 处，多者(从第 11 处起)不计分。 3．必须按答题要求做题，否则不给分。 My mother died months before and my father was often out on the business, so I felt very lonely. I missed my old life then my mother was still alive and we were happy. Laura was my best friends and cared about me very much. One Friday when I thought I will have to spend the weekend alone, I couldn't help cry. She sat with me, not saying something while I cried until I felt calmly. From that day on she often encouraged me face my life bravely. Gradually I began to consider the reality calmly. Under her help I was able to deal with my feelings better. 答案：
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3．开头和结尾已经写出，不计入总词数。 Dear Mr. Smith， I am writing to request the valuable chance to be one of the homestay families for the American __________________________________________ Yours sincerely， Li Hua 范文： Dear_Mr._Smith， I_am_writing_to_request_the_valuable_chance_to_be_one_of_the_homestay_families_for _the_American_students. As an active girl in senior one, Guangming School, I'm quite good at English. My parents, who are kind and friendly, are for my idea that I host an American student. We have a clean and comfortable house, which lies in a beautiful area of the city. Luckily, we have extra rooms for guests. My mother cooks well so the student will be able to enjoy delicious Chinese food. What's more, my father is an excellent driver and we will show the foreign student around some local places of interest in our area. I'd greatly appreciate it if I could have the chance to make friends with her/him. I'm looking forward to your reply. Yours_sincerely， Li_Hua students._________________________________ ______________________________________________________________________________________
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### ...学年高中英语 Unit 4 Astronomy the science of the...
2015-2016学年高中英语 Unit 4 Astronomy the science of the stars单元词汇句子练习 新人教版必修3_英语_高中教育_教育专区。高中英语 ...
### ...3 Unit 4 Astronomy:the science of the stars》文...
【人教版】2016高三英语一轮复习《Book 3 Unit 4 Astronomy:the science of the stars》文档(含解析)_英语_高中教育_教育专区。1.system n.系统;体系;制度 ...
### ...学年高中英语 Unit 4 Astronomy the science of the...
2014-2015学年高中英语 Unit 4 Astronomy the science of the stars单元综合测试(I)新人教版必修3_高三英语_英语_高中教育_教育专区。2014-2015 学年高中英语 ...
### ...3 Unit 4 Astronomy:the science of the stars》文...
【人教版】2016高三英语大一轮复习《Book 3 Unit 4 Astronomy:the science of the stars》文档(含解析)_英语_高中教育_教育专区。1.system n.系统;体系;制度 ...
### ...学年高中英语 Unit 4 Astronomy the science of the...
2015-2016学年高中英语 Unit 4 Astronomy the science of the stars单元词汇句子练习 新人教版必修3_英语_高中教育_教育专区。Unit 4 Astronomy: the science of...
### 2015年高中英语 Unit4 Astronomy the science of the s...
2015高中英语 Unit4 Astronomy the science of the stars单元测试1 新人教版必修3_英语_高中教育_教育专区。必修三 Unit 4Astronomy: the science of the ...
### ...必修三Unit 4 Astronomythe science of the stars
2015届高考英语一轮单元测试:人教版必修三Unit 4 Astronomythe science of the stars_英语_高中教育_教育专区。A卷 基础知识检测 一、用所给单词的适当形式填空 ...
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# psychology
posted by .
It's for a psychology experiment, i'm comparing the reaction times of people to the number of hours of sleep they get per night on average. Please give me some ideas on how to carry out this experiment. For example, like what will be a good way to select subjects, just male or both female and male, age, and so on. Any help will be greatly appreciate! Thanks!!!
• psychology -
It seems like you are looking to correlate sleep times to reaction times. If you will actually be doing the experiment rather than just proposing one, it might be convenient to choose college peers. (I am assuming you are in college.)
You might want to do a pilot study with one gender, just to get the "bugs" out. Be aware that reaction times can vary depending on the hour of the day.
What type of task will you be using to measure reaction times?
Are you going to rely on self-reporting for hours of sleep?
Will the subjects be measured over a period of days, say a week, to get an average?
You have many such questions to answer in creating your experiment. Good luck!
• psychology -
For the samples, i was thinking about recruiting 10 females age range from 19-25. Yes, i am pretty much will have a survey about general backgrounds, and number of hours of sleep on average per night. I will give the survey to the participants before testing the reaction time.
To test the reaction time, the method i'm thinking about doing is: I hold the ruler. Line the subject's hand up with the bottom. Drop the ruler and ask the subject to grabs it. Where they grab it will be shown by a measurement on the ruler. That will be a measure of reaction time. Like you said, since reaction times can vary depending on the hour of the day, the time of day will be the same for everybody. ( in the morning 10-12 maybe?) i only have two week to carry out the experiment so the subjects will just be measured over a period of days.
• psychology -
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# 2nd order ODE
#### ssonia
Given y''(t)+7x'(t)= 0 with x'(0)=a constant U
Solve for y'
Can l integrate the above ODE directly with respect to a variable t? hence,
y'(t)=-7x(t)
Thanks
#### romsek
MHF Helper
Given y''(t)+7x'(t)= 0 with x'(0)=a constant U
Solve for y'
Can l integrate the above ODE directly with respect to a variable t? hence,
y'(t)=-7x(t)
Thanks
yes but there will be a constant of integration
$y^\prime(t) = -7x(t) + C$
#### ssonia
And if y'(t) = a constant V, then C=v+7x(t) ?
Yes.
#### ssonia
Thanks Archie....
#### HallsofIvy
MHF Helper
And if y'(t) = a constant V, then C=v+7x(t) ?
Yes, but how do you know that y' is a constant? Do you mean y'(0)= V? (If so there is no need to say "a constant".)
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2009hw4solution
2009hw4solution - Name NetID HADM 2222 Fall 2009 Prof Q Ma...
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Name: ___________________________________ NetID: _______________________ Prof. Q. Ma, HADM 2222 Fall 2009 1/6 HADM 2222 Fall 2009, Prof. Q. Ma Homework assignment #4 [Due 11:59 a.m. Friday, October 16, 2009, Statler 435 drop box] 1. What is the IRR of the following set of cash flows? Year Cash Flow 0 -18,000 1 9,800 2 7,500 3 7,300 Solution: The IRR is the interest rate that makes the NPV of the project equal to zero. So, the equation that defines the IRR for this project is: 0 = –\$18,000 + \$9,800/(1+IRR) + \$7,500/(1+IRR) 2 + \$7,300/(1+IRR) 3 Using a spreadsheet, financial calculator, or trial and error to find the root of the equation, we find that: IRR = 18.49% 2. Bumble’s Bees, Inc., has identified the following two mutually exclusive projects: Year Cash Flow A Cash Flow B 0 -37,000 -37,000 1 19,000 6,000 2 14,500 12,500 3 12,000 19,000 4 9,000 23,000 a. What is the IRR for each of these projects? Using the IRR decision rule, which project should the company accept? Is this decision necessarily correct? Solution: The IRR is the interest rate that makes the NPV of the project equal to zero. The equation for the IRR of Project A is: 0 = –\$37,000 + \$19,000/(1+IRR) + \$14,500/(1+IRR) 2 + \$12,000/(1+IRR) 3 + \$9,000/(1+IRR) 4 Using a spreadsheet, financial calculator, or trial and error to find the root of the equation, we find that: IRR = 20.30% The equation for the IRR of Project B is: 0 = –\$37,000 + \$6,000/(1+IRR) + \$12,500/(1+IRR) 2 + \$19,000/(1+IRR) 3 + \$23,000/(1+IRR) 4 Using a spreadsheet, financial calculator, or trial and error to find the root of the equation, we find that: IRR = 18.55% Examining the IRRs of the projects, we see that the IRRA is greater than the IRRB, so IRR decision rule implies accepting project A. This may not be a correct decision; however, because the IRR criterion has a ranking problem for mutually exclusive projects. To see if the IRR decision rule is correct or not, we need to evaluate the project NPVs.
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Name: ___________________________________ NetID: _______________________ Prof. Q. Ma, HADM 2222 Fall 2009 2/6 b. If the required return is 11 percent, what is the NPV for each of these projects? Which project will the company choose if it applies the NPV decision rule? Solution: The NPV of Project A is: NPVA = \$6,588.52 NPVA = –\$37,000 + \$19,000/1.11+ \$14,500/1.11 2 + \$12,000/1.11 3 + \$9,000/1.11 4 And the NPV of Project B is: NPVB = \$7,594.13 NPVB = –\$37,000 + \$6,000/1.11 + \$12,500/1.11 2 + \$19,000/1.11 3 + \$23,000/1.11 4 The NPVB is greater than the NPVA, so we should accept Project B. c. Over what range of discount rates would the company choose project A? Project B? At what discount rate would the company be indifferent between these two projects? Explain. Solution:
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# Questions tagged [pitch]
Pitch of an audio signal representing a musical note, measured in units of octave and relative to a pitch standard such as A440, is the base-2 logarithm of the frequency ratio of the fundamental frequency of the note to the frequency of the pitch standard.
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### Calculating the fundamental frequency using Autocorrelation often gives half the expected value
I'm currently writing a mobile app which needs to analyse musical notes and find the fundamental frequency to determine the pitch. To do this I'm reading in audio data, taking an FFT, taking the auto-...
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1 vote
2k views
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PDA
View Full Version : Due date field
ABC123
Stewart Spiers
01-30-2005, 03:26 AM
After entering a date in a character field is it possible to write a script to display a due date in another field, next month for example. If so how to I do this??
Stewart Spiers
Tom Cone Jr
01-30-2005, 03:55 AM
Read the manual for information on date arithmetic. You could define a calc field in field rules for the due date using an expression that adds a fixed number of days to the other date field value.
-- t
Jay Talbott
01-30-2005, 04:55 AM
In my humble opinion, you are asking for trouble using a character field to gather date information. If the user type Febary 1, 2005, or 2/31/05, for example, how will your program know what date the user wants.
You should use a date field for dates.
I hope this helps.
Regards,
Jay Talbott
Bill Warner
01-30-2005, 05:15 AM
I agree wholeheartedly with Jay. Use date fields!
Here are some examples of date math:
DATE2 = DATE1 + 15
will add 15 days to DATE1 to give DATE2
will add one month to DATE1 (i.e., the same day of the month - Jan 27 to Feb 27). Be careful with this when dealing with February!
Tom Cone Jr
01-30-2005, 01:13 PM
I agree. In my previous I missed the fact that a character type field is being used. A bad business, in my opinion. Date arithmetic, obviously, requires date type fields.
-- t
Stewart Spiers
01-31-2005, 12:54 AM
I have made 2 new Date fields, Date1 and Date 2
I have tried entering your suggestions, Date2 = Addmonths(1,1) or Date2 = Date1+15 but each time I am told “Expression does not evaluate to a numeric value".
Stewart Spiers
Tom Cone Jr
01-31-2005, 03:09 AM
Stewart,
when someone offers a suggestion to you and it involves a function you're not familiar with it's wise to read the docs to "decode" what the suggestion is doing. If you review the help file info on the AddMonths() function you'll quickly see that the first argument that is passed to the function must be a date. AddMonths(1,1) is invalid since the first "1" is not a date.
The attachment is a simple example I whipped up for you. It shows how to do this in field rules for a table. Unzip the attachment to an empty folder. enter a few records, change a few. Study the field rules. Lookup the pieces you don't understand. Then, finally, integrate it into your app.
-- tom
Stewart Spiers
02-01-2005, 12:57 AM
Thanks for trying to help me.
Unfortunately when trying to run your example it says it was written with the Professional Edition of Alpha5 and cannot open it as I use the Home Edition of Alpha5
Is there a way around that problem
Stewart Spiers
Tom Cone Jr
02-01-2005, 02:14 AM
Next time please mention that you're using the Home Edition up front.
The fix is easy.
Use Windows Explorer to copy the Test table and its dictionaries to a new folder.
Use the home edition to create a new database in that same new folder and then "Add" the Test table to it.
Use the default form or browse to enter and change a few records.
Then study the field rules for the Test table.
-- t
Tom Cone Jr
02-01-2005, 02:20 AM
An even easier fix would be to simple open the home editioin and create a new database in the current folder (where you unzipped the original download). then add the Test table to that database.
-- t
Stewart Spiers
02-02-2005, 12:20 AM
t,
Thanks very much now it all works.
I would love to be able to study the field rules but cannot find any article in the instruction books relating to dates which says anything about how to write the program you have sent me.
Is this because I only have the home edition of Alpha5
Where should I look
Stewart Spiers
Tom Cone Jr
02-02-2005, 12:51 AM
The best source of technical "help" is the compiled help file (CHM) called "Alpha Five Help", available as a free,large, download from the Learning Center. The link to the LC is on the Alpha Software home page.
I recommend you buy and study all 3 of Dr. Wayne's books if you'd like to learn how to use Alpha Five more effectively.
If you haven't worked through the free tutorials, that's higly recommended, also.
In the file I sent you I created a single field rule, for the Due_Date field. It uses a single expression to calculate the value to store in Due_Date, by adding one month to the Bill_Date field. If you haven't found the field rules for the table you should work through the user's guide before doing anything else.
-- t
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DÄ internationalArchive44/2010Linear Regression Analysis
# Linear Regression Analysis
## Part 14 of a Series on Evaluation of Scientific Publications
Dtsch Arztebl Int 2010; 107(44): 776-82. DOI: 10.3238/arztebl.2010.0776
#### Schneider, A; Hommel, G; Blettner, M
Background: Regression analysis is an important statistical method for the analysis of medical data. It enables the identification and characterization of relationships among multiple factors. It also enables the identification of prognostically relevant risk factors and the calculation of risk scores for individual prognostication.
Results: After a brief introduction of the uni- and multivariable regression models, illustrative examples are given to explain what the important considerations are before a regression analysis is performed, and how the results should be interpreted. The reader should then be able to judge whether the method has been used correctly and interpret the results appropriately.
Conclusion: The performance and interpretation of linear regression analysis are subject to a variety of pitfalls, which are discussed here in detail. The reader is made aware of common errors of interpretation through practical examples. Both the opportunities for applying linear regression analysis and its limitations are presented.
The purpose of statistical evaluation of medical data is often to describe relationships between two variables or among several variables. For example, one would like to know not just whether patients have high blood pressure, but also whether the likelihood of having high blood pressure is influenced by factors such as age and weight. The variable to be explained (blood pressure) is called the dependent variable, or, alternatively, the response variable; the variables that explain it (age, weight) are called independent variables or predictor variables. Measures of association provide an initial impression of the extent of statistical dependence between variables. If the dependent and independent variables are continuous, as is the case for blood pressure and weight, then a correlation coefficient can be calculated as a measure of the strength of the relationship between them (Box 1 gif ppt).
Regression analysis is a type of statistical evaluation that enables three things:
• Description: Relationships among the dependent variables and the independent variables can be statistically described by means of regression analysis.
• Estimation: The values of the dependent variables can be estimated from the observed values of the independent variables.
• Prognostication: Risk factors that influence the outcome can be identified, and individual prognoses can be determined.
Regression analysis employs a model that describes the relationships between the dependent variables and the independent variables in a simplified mathematical form. There may be biological reasons to expect a priori that a certain type of mathematical function will best describe such a relationship, or simple assumptions have to be made that this is the case (e.g., that blood pressure rises linearly with age). The best-known types of regression analysis are the following (Table 1 gif ppt):
• Linear regression,
• Logistic regression, and
• Cox regression.
The goal of this article is to introduce the reader to linear regression. The theory is briefly explained, and the interpretation of statistical parameters is illustrated with examples. The methods of regression analysis are comprehensively discussed in many standard textbooks (13).
Cox regression will be discussed in a later article in this journal.
Methods
Linear regression is used to study the linear relationship between a dependent variable Y (blood pressure) and one or more independent variables X (age, weight, sex).
The dependent variable Y must be continuous, while the independent variables may be either continuous (age), binary (sex), or categorical (social status). The initial judgment of a possible relationship between two continuous variables should always be made on the basis of a scatter plot (scatter graph). This type of plot will show whether the relationship is linear (Figure 1 gif ppt) or nonlinear (Figure 2 gif ppt).
Performing a linear regression makes sense only if the relationship is linear. Other methods must be used to study nonlinear relationships. The variable transformations and other, more complex techniques that can be used for this purpose will not be discussed in this article.
Univariable linear regression
Univariable linear regression studies the linear relationship between the dependent variable Y and a single independent variable X. The linear regression model describes the dependent variable with a straight line that is defined by the equation Y = a + b × X, where a is the y-intersect of the line, and b is its slope. First, the parameters a and b of the regression line are estimated from the values of the dependent variable Y and the independent variable X with the aid of statistical methods. The regression line enables one to predict the value of the dependent variable Y from that of the independent variable X. Thus, for example, after a linear regression has been performed, one would be able to estimate a person’s weight (dependent variable) from his or her height (independent variable) (Figure 3 gif ppt).
The slope b of the regression line is called the regression coefficient. It provides a measure of the contribution of the independent variable X toward explaining the dependent variable Y. If the independent variable is continuous (e.g., body height in centimeters), then the regression coefficient represents the change in the dependent variable (body weight in kilograms) per unit of change in the independent variable (body height in centimeters). The proper interpretation of the regression coefficient thus requires attention to the units of measurement. The following example should make this relationship clear:
In a fictitious study, data were obtained from 135 women and men aged 18 to 27. Their height ranged from 1.59 to 1.93 meters. The relationship between height and weight was studied: weight in kilograms was the dependent variable that was to be estimated from the independent variable, height in centimeters. On the basis of the data, the following regression line was determined: Y= –133.18 + 1.16 × X, where X is height in centimeters and Y is weight in kilograms. The y-intersect a = –133.18 is the value of the dependent variable when X = 0, but X cannot possibly take on the value 0 in this study (one obviously cannot expect a person of height 0 centimeters to weigh negative 133.18 kilograms). Therefore, interpretation of the constant is often not useful. In general, only values within the range of observations of the independent variables should be used in a linear regression model; prediction of the value of the dependent variable becomes increasingly inaccurate the further one goes outside this range.
The regression coefficient of 1.16 means that, in this model, a person’s weight increases by 1.16 kg with each additional centimeter of height. If height had been measured in meters, rather than in centimeters, the regression coefficient b would have been 115.91 instead. The constant a, in contrast, is independent of the unit chosen to express the independent variables. Proper interpretation thus requires that the regression coefficient should be considered together with the units of all of the involved variables. Special attention to this issue is needed when publications from different countries use different units to express the same variables (e.g., feet and inches vs. centimeters, or pounds vs. kilograms).
Figure 3 shows the regression line that represents the linear relationship between height and weight.
For a person whose height is 1.74 m, the predicted weight is 68.50 kg (y = –133.18 + 115.91 × 1.74 m). The data set contains 6 persons whose height is 1.74 m, and their weights vary from 63 to 75 kg.
Linear regression can be used to estimate the weight of any persons whose height lies within the observed range (1.59 m to 1.93 m). The data set need not include any person with this precise height. Mathematically it is possible to estimate the weight of a person whose height is outside the range of values observed in the study. However, such an extrapolation is generally not useful.
If the independent variables are categorical or binary, then the regression coefficient must be interpreted in reference to the numerical encoding of these variables. Binary variables should generally be encoded with two consecutive whole numbers (usually 0/1 or 1/2). In interpreting the regression coefficient, one should recall which category of the independent variable is represented by the higher number (e.g., 2, when the encoding is 1/2). The regression coefficient reflects the change in the dependent variable that corresponds to a change in the independent variable from 1 to 2.
For example, if one studies the relationship between sex and weight, one obtains the regression line Y = 47.64 + 14.93 × X, where X = sex (1 = female, 2 = male). The regression coefficient of 14.93 reflects the fact that men are an average of 14.93 kg heavier than women.
When categorical variables are used, the reference category should be defined first, and all other categories are to be considered in relation to this category.
The coefficient of determination, r2, is a measure of how well the regression model describes the observed data (Box 2 gif ppt). In univariable regression analysis, r2 is simply the square of Pearson’s correlation coefficient. In the particular fictitious case that is described above, the coefficient of determination for the relationship between height and weight is 0.785. This means that 78.5% of the variance in weight is due to height. The remaining 21.5% is due to individual variation and might be explained by other factors that were not taken into account in the analysis, such as eating habits, exercise, sex, or age.
In formal terms, the null hypothesis, which is the hypothesis that b = 0 (no relationship between variables, the regression coefficient is therefore 0), can be tested with a t-test. One can also compute the 95% confidence interval for the regression coefficient (4).
Multivariable linear regression
In many cases, the contribution of a single independent variable does not alone suffice to explain the dependent variable Y. If this is so, one can perform a multivariable linear regression to study the effect of multiple variables on the dependent variable.
In the multivariable regression model, the dependent variable is described as a linear function of the independent variables Xi, as follows: Y = a + b1 × X1 + b2 × X2 +…+ bn × Xn . The model permits the computation of a regression coefficient bi for each independent variable Xi (Box 3 gif ppt).
Just as in univariable regression, the coefficient of determination describes the overall relationship between the independent variables Xi (weight, age, body-mass index) and the dependent variable Y (blood pressure). It corresponds to the square of the multiple correlation coefficient, which is the correlation between Y and b1 × X1 + ... + bn × Xn.
It is better practice, however, to give the corrected coefficient of determination, as discussed in Box 2. Each of the coefficients bi reflects the effect of the corresponding individual independent variable Xi on Y, where the potential influences of the remaining independent variables on Xi have been taken into account, i.e., eliminated by an additional computation. Thus, in a multiple regression analysis with age and sex as independent variables and weight as the dependent variable, the adjusted regression coefficient for sex represents the amount of variation in weight that is due to sex alone, after age has been taken into account. This is done by a computation that adjusts for age, so that the effect of sex is not confounded by a simultaneously operative age effect (Box 4 gif ppt).
In this way, multivariable regression analysis permits the study of multiple independent variables at the same time, with adjustment of their regression coefficients for possible confounding effects between variables.
Multivariable analysis does more than describe a statistical relationship; it also permits individual prognostication and the evaluation of the state of health of a given patient. A linear regression model can be used, for instance, to determine the optimal values for respiratory function tests depending on a person’s age, body-mass index (BMI), and sex. Comparing a patient’s measured respiratory function with these computed optimal values yields a measure of his or her state of health.
Medical questions often involve the effect of a very large number of factors (independent variables). The goal of statistical analysis is to find out which of these factors truly have an effect on the dependent variable. The art of statistical evaluation lies in finding the variables that best explain the dependent variable.
One way to carry out a multivariable regression is to include all potentially relevant independent variables in the model (complete model). The problem with this method is that the number of observations that can practically be made is often less than the model requires. In general, the number of observations should be at least 20 times greater than the number of variables under study.
Moreover, if too many irrelevant variables are included in the model, overadjustment is likely to be the result: that is, some of the irrelevant independent variables will be found to have an apparent effect, purely by chance. The inclusion of irrelevant independent variables in the model will indeed allow a better fit with the data set under study, but, because of random effects, the findings will not generally be applicable outside of this data set (1). The inclusion of irrelevant independent variables also strongly distorts the determination coefficient, so that it no longer provides a useful index of the quality of fit between the model and the data (Box 2).
In the following sections, we will discuss how these problems can be circumvented.
The selection of variables
For the regression model to be robust and to explain Y as well as possible, it should include only independent variables that explain a large portion of the variance in Y. Variable selection can be performed so that only such independent variables are included (1).
Variable selection should be carried out on the basis of medical expert knowledge and a good understanding of biometrics. This is optimally done as a collaborative effort of the physician-researcher and the statistician. There are various methods of selecting variables:
Forward selection
Forward selection is a stepwise procedure that includes variables in the model as long as they make an additional contribution toward explaining Y. This is done iteratively until there are no variables left that make any appreciable contribution to Y.
Backward selection
Backward selection, on the other hand, starts with a model that contains all potentially relevant independent variables. The variable whose removal worsens the prediction of the independent variable of the overall set of independent variables to the least extent is then removed from the model. This procedure is iterated until no dependent variables are left that can be removed without markedly worsening the prediction of the independent variable.
Stepwise selection
Stepwise selection combines certain aspects of forward and backward selection. Like forward selection, it begins with a null model, adds the single independent variable that makes the greatest contribution toward explaining the dependent variable, and then iterates the process. Additionally, a check is performed after each such step to see whether one of the variables has now become irrelevant because of its relationship to the other variables. If so, this variable is removed.
Block inclusion
There are often variables that should be included in the model in any case—for example, the effect of a certain form of treatment, or independent variables that have already been found to be relevant in prior studies. One way of taking such variables into account is their block inclusion into the model. In this way, one can combine the forced inclusion of some variables with the selective inclusion of further independent variables that turn out to be relevant to the explanation of variation in the dependent variable.
The evaluation of a regression model requires the performance of both forward and backward selection of variables. If these two procedures result in the selection of the same set of variables, then the model can be considered robust. If not, a statistician should be consulted for further advice.
Discussion
The study of relationships between variables and the generation of risk scores are very important elements of medical research. The proper performance of regression analysis requires that a number of important factors should be considered and tested:
1. Causality
Before a regression analysis is performed, the causal relationships among the variables to be considered must be examined from the point of view of their content and/or temporal relationship. The fact that an independent variable turns out to be significant says nothing about causality. This is an especially relevant point with respect to observational studies (5).
2. Planning of sample size
The number of cases needed for a regression analysis depends on the number of independent variables and of their expected effects (strength of relationships). If the sample is too small, only very strong relationships will be demonstrable. The sample size can be planned in the light of the researchers’ expectations regarding the coefficient of determination (r2) and the regression coefficient (b). Furthermore, at least 20 times as many observations should be made as there are independent variables to be studied; thus, if one wants to study 2 independent variables, one should make at least 40 observations.
3. Missing values
Missing values are a common problem in medical data. Whenever the value of either a dependent or an independent variable is missing, this particular observation has to be excluded from the regression analysis. If many values are missing from the dataset, the effective sample size will be appreciably diminished, and the sample may then turn out to be too small to yield significant findings, despite seemingly adequate advance planning. If this happens, real relationships can be overlooked, and the study findings may not be generally applicable. Moreover, selection effects can be expected in such cases. There are a number of ways to deal with the problem of missing values (6).
4. The data sample
A further important point to be considered is the composition of the study population. If there are subpopulations within it that behave differently with respect to the independent variables in question, then a real effect (or the lack of an effect) may be masked from the analysis and remain undetected. Suppose, for instance, that one wishes to study the effect of sex on weight, in a study population consisting half of children under age 8 and half of adults. Linear regression analysis over the entire population reveals an effect of sex on weight. If, however, a subgroup analysis is performed in which children and adults are considered separately, an effect of sex on weight is seen only in adults, and not in children. Subgroup analysis should only be performed if the subgroups have been predefined, and the questions already formulated, before the data analysis begins; furthermore, multiple testing should be taken into account (7, 8).
5. The selection of variables
If multiple independent variables are considered in a multivariable regression, some of these may turn out to be interdependent. An independent variable that would be found to have a strong effect in a univariable regression model might not turn out to have any appreciable effect in a multivariable regression with variable selection. This will happen if this particular variable itself depends so strongly on the other independent variables that it makes no additional contribution toward explaining the dependent variable. For related reasons, when the independent variables are mutually dependent, different independent variables might end up being included in the model depending on the particular technique that is used for variable selection.
Overview
Linear regression is an important tool for statistical analysis. Its broad spectrum of uses includes relationship description, estimation, and prognostication. The technique has many applications, but it also has prerequisites and limitations that must always be considered in the interpretation of findings (Box 5 gif ppt).
Conflict of interest statement
The authors declare that they have no conflict of interest as defined by the guidelines of the International Committee of Medical Journal Editors.
Manuscript submitted on 11 May 2010, revised version accepted on 14 July 2010.
Translated from the original German by Ethan Taub, MD
Corresponding author
Prof. Dr. rer. nat. Maria Blettner
Department of Medical Biometrics, Epidemiology, and Computer Sciences
Johannes Gutenberg University
Obere Zahlbacher Str. 69
55131 Mainz
Germany
1.
Fahrmeir L, Kneib T, Lang S: Regression – Modelle, Methoden und Anwendungen. 2nd edition. Berlin, Heidelberg: Springer 2009
2.
Bortz J: Statistik für Human-und Sozialwissenschaftler. 6th edition. Heidelberg: Springer 2004.
3.
Selvin S: Epidemiologic Analysis. Oxford University Press 2001.
4.
Bender R, Lange S: Was ist ein Konfidenzintervall? Dtsch Med Wschr 2001; 126: T41.
5.
Sir Bradford Hill A: The environment and disease: Association or Causation? Proc R Soc Med 1965; 58: 295–300. MEDLINE
6.
Carpenter JR, Kenward MG: Missing Data in Randomised Controlled Trials: A practical guide. Birmingham, Alabama: National Institute for Health Research; 2008. Publication RM03/JH17/MK.
8.
Horn M, Vollandt R: Multiple Tests und Auswahlverfahren. Stuttgart: Gustav Fischer Verlag 1995.
Departrment of Medical Biometrics, Epidemiology, and Computer Sciences, Johannes Gutenberg University, Mainz, Germany: Dipl. Math. Schneider, Prof. Dr. rer. nat. Hommel,
Prof. Dr. rer. nat. Blettner
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Table 1
1 Fahrmeir L, Kneib T, Lang S: Regression – Modelle, Methoden und Anwendungen. 2nd edition. Berlin, Heidelberg: Springer 2009 2 Bortz J: Statistik für Human-und Sozialwissenschaftler. 6th edition. Heidelberg: Springer 2004. 3 Selvin S: Epidemiologic Analysis. Oxford University Press 2001. 4 Bender R, Lange S: Was ist ein Konfidenzintervall? Dtsch Med Wschr 2001; 126: T41. 5 Sir Bradford Hill A: The environment and disease: Association or Causation? Proc R Soc Med 1965; 58: 295–300. MEDLINE 6 Carpenter JR, Kenward MG: Missing Data in Randomised Controlled Trials: A practical guide. Birmingham, Alabama: National Institute for Health Research; 2008. Publication RM03/JH17/MK. 7 EMEA: Poiints to consider on multiplicity issues in clinical trials. 8 Horn M, Vollandt R: Multiple Tests und Auswahlverfahren. Stuttgart: Gustav Fischer Verlag 1995.
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# Quantitative analysis: households with vehicles
Introduction
This paper conducts quantitative analysis of the car ownership by the state in the years 2015 and 2016. The analysis conducted includes descriptive statics; mean, median, mode, variance, minimum, maximum and standard deviation. Also, graphical analysis such frequency histogram, box plot and scatter plots are conducted. The data for 2015 and 2016 are then compared, and their correlation determined.
1. The arrangement of the 2016 Vehicles per Household data in descending order generates the table below
In this data set, Jurupa Valley, Moreno Valley, California shares an average of 2.32, Corona and Simi Valley, California shares an average of 2.29, Fontana, Norwalk, and Pomona shares the average of 2.27 while Santa Ana, California and Temecula, California Shares an average of 2.25. This result in 4 classes as observed in the histogram.
The plotting of the histogram in excel for the averages generated the histogram chart below
Figure 1 2016 Histogram
• Relative Frequency Graph
The frequency plot for the data set 2016 percentage of households without vehicles was very unusual. The graph did not indicate any normal distribution pattern for the data set. The graph is presented in the chart below due to some errors in the research model or data collection.
Figure 2 2016 frequency histogram
• Box plot
In order to construct the box plot, the minimum, first quartile, third quartile, median and maximum values are determined. These values are then used to obtain the P-values as indicated performed in excel and displayed in the second column of the table.
The box chat is displayed in figure 3.
Table one
Table 1 Box Plot Values
Figure 3 2016 box plot
Figure 4 box chat
• Numerical Descriptive Statistics for the “2016 Vehicles per Household
The descriptive analysis for the 2016 vehicles per household generated the result displayed in the table below
Table 2Figure 5 2016 numerical (descriptive) analysis
The average value for the 2016 vehicles per household is 1.7154 (correct to 4 decimal places). The maximum value is 2.36 while the minimum value is 0.63. The minimum and the maximum values are located around the mean. These three values can, therefore, be used as a measure of central tendency. The mode is 1.69, and the median is 1.72 these values to illustrate the central tendency or the distribution of cars per household. The standard deviation value is 0.299 correct to three decimal places. The low standard deviation indicates that the number of cars per household is concentrated around the mean. The variance also emphasizes this statistical analysis given that it is lower than the standard deviation; the variance is 0.0894 correct to four decimal places.
• Outliers in 2016 Percentage Of Households Without Vehicle
In the determination of outliers, the first quartile, third quartile and inter-quartile ranges were determined. These values were then used to obtain the lower and the upper bound values. The values were then used to extrapolate the outliers in the excel as displayed in the excel format. The result for the analysis generated an above 95% true for the outliers. This indicates that the data set was greatly biased or skewed. The impact of these is that the data is unreliable. Outliers affect the quality of dataset and should be dealt with in case they were a result of experimental error.
Table 3 5. Outliers in 2016 Percentage of Households without Vehicle
• Comparing 2016 and 2015 percentages for households without vehicles.
• Descriptive analysis
The descriptive analysis for the data set generated the following results
Table 4 2015 Descriptive Analysis Result
The values generated in the 2016 calculations were not percentages as this are. Therefore these are two standards which make their comparison impossibility unless they were of the same unit.
• Outliers.
The values used in the determination of the outliers are as indicated below. The outliers indicated false for all the values as contained in the excel. This indicates that the data collected in the section of the data were accurately collected which justify the result of the study.
Table 5 2015 outlier values
• Frequency Histogram.
The frequency plot for the result of 2015 was similar to the plot for 2016. There was no normal distribution curve for the data set. The plot is displayed in the chart below and also contained in the excel
Figure 5 frequency histogram for the 2015 dataset
• box plot
The box plot for the data set in 2015 is similar to that of 2016. The plot is contained in the below. The third series lies before 10.00% while the fourth series greatly before 10.00% with a small portion spilling over to the second range of 10-20%.
Figure 6 2015 box plot
• Scatter plot for the 2015 and 2016 the average number of households with vehicles
The scatter plot displays a rising linear representation which is an indicator of a positive correlation. Thus there is a correlation between the data sets.
Figure 8 scatter plot for the 2015 and 2016 data on average households with vehicles
Conclusion
The quantitative analysis indicates the correlation between the data samples for 2015 and 2016. However, there are differences as represented in the outlier analysis whereby 2016 had a higher percentage of almost 100% outliers while the 2015 dataset had no outlier. This indicates that the research procedure for the 2016 analysis could have been falsified, or the results were accurate but there is a confounding variable.
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illustrates graphs and information of conic sections - Maple Programming Help
Home : Support : Online Help : Education : Student Package : Precalculus : Interactive : Student/Precalculus/ConicsTutor
Student[Precalculus][ConicsTutor] - illustrates graphs and information of conic sections
Calling Sequence ConicsTutor() ConicsTutor(f)
Parameters
f - (optional) input equation of the conic section in cartesian xy-coordinates or polar rt-coordinates
Description
• The ConicsTutor command launches a tutor interface that illustrates the graph and provides information about the related conic section.
• The equation f can be one of the following forms:
– $\mathrm{expr1}=\mathrm{expr2}$ where expr1 and $\mathrm{expr2}$ are of degree at most 2, in terms of x and y.
– $\mathrm{expr}$ where $\mathrm{expr}$ is of degree at most $2$ in terms of x and y.
– $r=g\left(t\right)$ where $g\left(t\right)$ is the polar form of an equation for a conic section, in terms of the variable t.
– $g\left(t\right)$ where $g\left(t\right)$ is the polar form of an equation for a conic section, in terms of the variable t.
– Note that $g\left(t\right)$ can either be a constant or be in the form $\frac{a}{b+cd\left(t\right)}$ , where a, b, and c are real numbers, and d is the sine or cosine function.
• If f is not specified, ConicsTutor uses a default function.
Examples
> $\mathrm{with}\left(\mathrm{Student}[\mathrm{Precalculus}]\right):$
> $\mathrm{ConicsTutor}\left(\right)$
> $\mathrm{ConicsTutor}\left({x}^{2}+2xy+{y}^{2}+2x-2y+4\right)$
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# Vinai Datta Thatiparthi
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@RungeBot f(x)=(sin(x))/x
.@jskherman Here's f(x)=(sin(x))/x interpolated using 20 equally spaced points (blue) and 20 Chebyshev points (red). For your function, Runge's phenomenon is not a problem.
.@jskherman Here's f(x)=((e^x)(sin(abs(x))))/(sin(x)cos(x)+e^x) interpolated using 20 equally spaced points (blue) and 20 Chebyshev points (red). For your function, Runge's phenomenon is quite bad.
QOTO: Question Others to Teach Ourselves
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All cultures welcome.
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https://www.neetprep.com/questions/3893-Physics/693-Current-Electricity?courseId=3786&testId=1061969-Past-Year----MCQs&subtopicId=412-Wheatstone-Bridge
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The resistances of the four arms $$P,Q,R~\text{and}~S$$ in a Wheatstone’s bridge are $$10~\Omega,30~\Omega,30~\Omega$$ and $$90~\Omega$$ respectively. The emf and internal resistance of the cell are $$7~\text{volt}$$ and $$5~\Omega$$ respectively. If the galvanometer resistance is $$50~\Omega$$ the current drawn from the cell will be:
1. $$0.2~\text{A}$$
2. $$0.1~\text{A}$$
3. $$2.0~\text{A}$$
4. $$1.0~\text{A}$$
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Three resistances $$\mathrm P$$, $$\mathrm Q$$, and $$\mathrm R$$, each of $$2~\Omega$$ and an unknown resistance $$\mathrm{S}$$ form the four arms of a Wheatstone bridge circuit. When the resistance of $$6~\Omega$$ is connected in parallel to $$\mathrm{S}$$, the bridge gets balanced. What is the value of $$\mathrm{S}$$?
1 $$2~\Omega$$ 2 $$3~\Omega$$ 3 $$6~\Omega$$ 4 $$1~\Omega$$
Subtopic: Wheatstone Bridge |
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## Scattering Parameters
Have you ever finished a big project--and then the next morning wished you could have added just one more feature? Well, thanks to the miracle of the internet, I get to keep writing all the articles I want to supplement the information in my latest book, High-Speed Signal Propagation. This article on Scattering Parameters (S-parameters) relates the transmission matrices described in that book to the S-parameter matrices provided by a network analyzer.
## Scattering Parameters
by Dr. Howard Johnson
A scattering matrix (S-parameter matrix) is one way to describe the operation of a linear, time-invariant two-port circuit. A two-port network is defined as any linear device where a signal goes in one side and comes out the other.
The S-parameter matrix is rapidly becoming very popular as a way to characterize connectors and cables for high-speed applications above 1 Gb/s.
The measurement setup associated with S-parameters is as follows (Figure 1).
From the test equipment, two cables having characteristic impedance Z0 lead to the left and right sides, respectively, of the device under test (DUT).
Using the first (left-side) cable, inject a sinusoidal signal (in1) of unit amplitude into the DUT. The test equipment records the amplitude and phase of the signal (out1) reflected back onto the first cable from the DUT, and also the amplitude and phase of the signal (out2) conveyed through the DUT to the second cable on the other side.
In a separate experiement, using the second (right-side) cable, inject a sinusoidal signal (in2) of unit amplitude into the DUT. The test equipment records the amplitude and phase of the signal (out2) reflected from the right side of the DUT, and the amplitude and phase of the signal (out1) conveyed through the DUT to the other (left) side. The complete S-parameter matrix is a combination of these four basic measurements.
The four elements of an S-parameter matrix may be reported as complex numbers (with real and imaginary parts) or in logarithmic units (as dB magnitude and phase).
NOTE: This procedure as I described it measures circuit performance at only one single frequency. The entire procedure is usually performed on a dense grid of frequencies spanning the range of interest, such that the parameters s11, etc., are all functions of frequency.
Provided that the reflection coefficients s11 and s22 are relatively small, you may estimate the effect of cascading several two-port networks by merely multiplying the s21 coefficients of the individual components (or, if they are in logarithmic units, by adding the dB values of the s21 coefficients). Such a calculation determines, TO FIRST ORDER, the magnitude and phase of a signal that propagates straight through the cascade proceeding from left to right through each component. This is the beauty of S-parameter analysis, and one key reason it is used in the design of highly cascaded systems like radio receivers and chains of linear amplifiers.
Unfortunately, the overall transfer function of a highly cascaded system equals the product of the s21 terms if and only if the reflections are all negligible. If the reflections are significant, the gain does not equal the product of s21 terms.
If you want to model a system with reflections, a more sophisticated analysis is required. That's the purpose of the Transmission Matrix, also called the Transfer Matrix, or A-parameters.
The measurement setup associated with transmission parameters is defined as follows (Figure 2).
The four transmission parameters are measured by first stimulating the circuit on the left side (V1 and I1) while holding the right side open-circuited. This condition ensures I2=0, under which condition you may easily determine a11 = V1/V2 and a21 = I1/V2.
Then the right side is shorted to ground, ensuring V2=0, while you measure a12 = V1/I2 and a22 = I1/I2.
The measurements are repeated on a dense grid of frequencies spanning the range of interest, such that the parameters a11, etc., all become functions of frequency.
One difficulty associated with transmission parameters is that in a practical high-frequency circuit it is extremely difficult to obtain the perfect open-circuit and short-circuit conditions required to make good measurements. For example, the open-circuit measurement is always corrupted by parasitic capacitance shunting port 2. The short-circuit measurement is corrupted by parasitic inductance in series with port 2.
The S-parameter method circumvents the open/short difficulties by always connecting both ports to transmission lines with stable, well-defined impedances. To the extent that such a test setup better represents the actual working conditions of your circuit, the S-parameter method generates a more accurate model. Note, however, that the S-parameter method requires that the measurements be made at some particular impedance Z0. For best results, this impedance should be reasonably close to the impedance under which your circuit element will actually operate.
To make clear the importance of the test-circuit impedance, consider that a good 50-ohm connector, if implemented with proper vias, and when measured with Z0=50 ohms may exhibit very small reflection coefficients s11 and s22. The same connector, if measured with a test impedance of Z0=75 ohms might produce terrible reflection coefficients. The values of the S-parameter matrix are thus quite sensitive to the impedances used during the test. The subject of correcting an S-parameter matrix to compute the result you would have gotten at a different level of test impedance is considered at the end of this article.
The transmission parameters are the parameters best suited for calculating the performance of cascaded systems, taking into account reflections.
Given a set of system components represented by transmission matrices A, B,...Z, the transmission matrix representing the cascaded combination of all three pieces is constructed by simply forming the matrix-multiplication product of all the pieces: AB...Z. The resulting combined transmission matrix properly represents all the transmission gains and all the internal reflections associated with the combination of system components.
The transmission-matrix description excels at modeling systems that incorporate multiple cascaded sections with noticeable reflections, such as a chip I/O driver followed by a chip package, a pcb trace, a connector, another pcb trace, another chip package, and a parasitic load at the receiver.
Supposing that you have an S-parameter description of a connector, and an S-parameter description of a backplane, how do you combine these pieces to produce an S-parameter description of the whole system?
This is done by first converting each S-parameter description to a transmission-matrix description and then multiplying together the transmission matrices corresponding to the system components you intend to cascade. The result is a transmission-matrix description of the whole system.
To facilitate your work, here are the conversions from S-matrix format to transmission-matrix (A-matrix) format. These formulas assume that Z0 is purely real and the same on both sides of the S-parameter test setup. If that isn't true, more complicated formulas apply (see for example, Chan Chan and Chan, "Analysis of Linear Networks and Systems", Addison Wesley, 1972, Lib. Congress Cat. Card No. 70-156589).
NOTE: The conversion calculations must be repeated done for each frequency on your dense grid. That makes a lot of computation, but why else do we have computers?
Given the source and load impedances in the network you may extract the transmission gain of the system (including the effects of internal reflections) directly from the combined transmission matrix (see Appendix C of High-Speed Signal Propagation).
Alternately, if you have access to an S-parameter simulator you can convert the combined transmission matrix back to a single S-parameter matrix representing the whole system, and then use it. You'll get the same answer either way.
Now let's look at one of the very interesting uses of the transmission matrix: scaling the length of a transmission line.
Sometimes it happens that you have measured the S-parameters of a piece of cable, or a backplane trace, and wish to have an S-parameter description of a longer (or shorter) length of the same cable. The solution to this problem is to first convert the S-parameter description to a transmission matrix, then scale the length, and finally convert back to S-parameters.
To see how length-scaling works with transmission parameters, first consider the simple case of doubling the length of the transmission medium. This effectively cascades two identical sections of transmission line. If the transmission-matrix for a single section of line is A, then the result for a double-length section would equal A2 (using matrix multiplication). Further integer-length extensions are provided by compounding higher and higher powers of the matrix A, such that the complete model for a section n-times longer than the basic unit section equals An.
An even more interesting fact (which I don't have space to prove here) is that the power-of-A formula works even for non-integer n, and also for fractional n. Therefore if you want to model a section of transmission line whole length is, say, .316 times the length of your basic unit section you simply form the matrix A0.316.
If you haven't previously encountered the idea of raising a matrix to a fractional power, let me tell you a little about it. The concept is that you first decompose the matrix using an eigenvalue decomposition. Routines for eigenvalue decomposition exist in all the major mathematical-spreadsheet packages (MatLab, Mathematica, and my favorite, MathCad). The eigenvalue decomposition produces three matrices, U, D, and V, which when multiplied together (UDV) equal your original matrix A. Without going into a lot of detail, the matrix D is always a diagonal matrix. The diagonal elements of D are called the eigenvalues of A (see any book on linear algebra, for example Gilbert Strang, "Linear Algebra and Its Applications", Academic Press, 1976, ISBN 0-12-673650-2). To form the matrix Ax for any value of x, form a new matrix E by raising each element of the diagonal matrix D to the power of x, and then use your new E to form a new matrix UEV, which becomes your answer. Convert UEV from transmission-matrix format back to S-parameter format and you now have a length-scaled S-parameter model of the whole system.
Let me advise you that such manipulations work best on systems, such as long uniform transmission lines, that incorporate few if any internal reflections. It's OK if you have reflections at the front and back ends of the line, as those will properly scale in time, but if you have internal reflections (such as vias) internal to your unit-length standard measurement then when you scale that measurement to, say, half length the equations will produce some kind of system with different internal reflections, such that when you cascade two half-length systems in series you will get the original reflection pattern observed during the test. That answer is mathematically correct, but it probably isn't what you want. The original unit-length S-parameter setup should incorporate only the main body of the transmission structure, not any imperfections along the way. Any mismatches at the ends of the transmission structure are properly handled by this technique.
My final subject has to do with correcting the value of Z0 used during S-parameter characterization.
The conversion from S-parameter to transmission-matrix format depends in each direction upon an assumed value for the characteristic impedance Z0 of the test setup. If you have made S-parameter measurements at one level of impedance and wish to see what S-parameter matrix you would have gotten with a different level of impedance Z1, then convert from S-parameters to transmission-parameters using Z0, and then go back the other way using Z1.
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### Email Archive: hol-info (read-only)
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[Hol-info] Simple Mathematical induction From: Tony Johnson - 2009-11-11 22:29 Hi, How would I go about proving something simple like this? 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2} Thanks,
Re: [Hol-info] Simple Mathematical induction From: Michael Norrish - 2009-11-11 23:27 Tony Johnson wrote: > How would I go about proving something simple like this? > 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2} First you need to define your summation. Life is usually easiest if you define things by pattern-matching recursion: val sum_def = Define (sum 0 = 0) /\ (sum (SUC n) = SUC n + sum n) ; Such a definition is a prime candidate for automatic rewriting, so I would then add the definition to the built-in simpset val _ = export_rewrites ["sum_def"] Then you get to state your theorem. Proving things with DIV is usually a pain, so I recast by multiplying both sides by two: val sum_form = store_thm( "sum_form", 2 * sum n = n * (n + 1), ) What's the ? First an induction Induct_on n THEN ... Following induction steps by rewriting is usually safe, so Induct_on n THEN SRW_TAC [][] This eliminates the base case automatically and gives the goal 2 * (SUC n + sum n) = SUC n * (SUC n + 1) ------------------------------------------- 2 * sum n = n * (n + 1) where the formula below the line is the inductive hypothesis. The IH would apply as a rewrite if the multiplication had been pushed into the addition. This is the theorem arithmeticTheory.LEFT_ADD_DISTRIB. Let's assume you have done an open arithmeticTheory so that we don't have to use fully-qualified names everywhere. Then, we can modify our tactic to be Induct_on n THEN SRW_TAC [][LEFT_ADD_DISTRIB] THEN ... This gives 2 * SUC n + (n * n + n) = SUC n * SUC n + SUC n ------------------------------------------------- 2 * sum n = n * (n + 1) The IH has been applied and now our goal is pure arithmetic. As it stands it is not a useful instance of a Presburger formula because of the n * n and the SUC n * SUC n. But the latter can be expanded to a formula including the former with the use of the standard MULT_CLAUSES, which includes the conjuncts |- SUC n * m = n * m + m |- n * SUC m = n + n * m So, we modify the tactic to be Induct_on n THEN SRW_TAC [][LEFT_ADD_DISTRIB, MULT_CLAUSES] THEN ... This gives a goal of 2 + 2 * n + (n * n + n) = n + n * n + (SUC n + 1) --------------------------------------------------- 2 * sum n = n * (n + 1) This is now a good instance of a Presburger formula (with n*n substituted out for an arbitrary variable), meaning that DECIDE_TAC will solve it. So our final tactic is Induct_on n THEN SRW_TAC [][LEFT_ADD_DISTRIB, MULT_CLAUSES] THEN DECIDE_TAC and the final proof looks like val sum_form = store_thm( "sum_form", 2 * sum n = n * (n + 1), Induct_on n THEN SRW_TAC [][LEFT_ADD_DISTRIB, MULT_CLAUSES] THEN DECIDE_TAC) The form with division can subsequently be done (in a slightly hacky way) as val sum_form_div = store_thm( "sum_form_div", sum n = n * (n + 1) DIV 2, SRW_TAC [][Once EQ_SYM_EQ] THEN MATCH_MP_TAC DIV_UNIQUE THEN Q.EXISTS_TAC 0 THEN SRW_TAC [][GSYM sum_form, MULT_COMM]); Michael.
[Hol-info] pred_set question From: Michael Norrish - 2009-11-11 22:31 Hi Jianjun, > I have difficulty proving something like > !x y z. ((x UNION y = z) /\ DISJOINT x y) ==> (x = z DIFF y). > Any help please? Simple equalities on sets usually fall over if you attack them with extensionality and METIS_TAC. In this case, the tactic SRW_TAC [][DISJOINT_DEF, EXTENSION] THEN METIS_TAC [] does the job (assuming that you have pred_setTheory open to pick up the theorems DISJOINT_DEF and EXTENSION). Michael.
[Hol-info] Call for Bids (ITP 2011) From: Lawrence Paulson - 2009-11-11 15:29 It is time to begin the process of selecting a host for ITP 2011, the International Conference on Interactive Theorem Proving. Following tradition from TPHOLs, the hosts of the previous conference (ITP 2010) are running the process. There are two phases: solicitation of bids and voting. This message concerns the first phase. A long-standing TPHOLs convention is that the conference should be held in a continent different from the location of the previous meeting, and therefore no bids to host ITP 2011 in Europe will be accepted. Based on ITP and TPHOLs history, ITP 2011 will likely be held in July, August or September. (The ACL2 Workshop has taken place at various times of year.) Bids should be sent to itp10@... and should include at least the following information: - name and email address of a contact person - names of other people involved - address of website for the bid - approximate dates of the conference - structure (e.g., k workshop days and n days of presentations followed by excursion...) - advantages of the proposed venue An example of a previous winning bid is here: http://web2.comlab.ox.ac.uk/oucl/conferences/TPHOLs2005/bid.html Deadline for all bids is Monday, 4 January 2010. Shortly after that, the bids will be made public and the voting phase will take place. The people eligible to vote are those who are seriously thinking of attending ITP 2011. The voting system used will be Single Transferable Vote between all received bids. (Note: ACL2 papers will be welcome at ITP 2011, regardless of whether or not there is a separate ACL2 workshop in 2011.) Matt Kaufmann and Larry Paulson
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# Download St 314 – Day #13 Notes A. Normal Approximation to the Binomial
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```St 314 – Day #13 Notes
A. Normal Approximation to the Binomial
Recall the binomial distribution:
P(Y = y) = p(y) = (n y )py (1-p)n-y
y = 0, 1, …
which gives the probability of y successes out of n independent trials, where the
probability of success on any one trial is p (0 < p < 1).
If n is large this becomes difficult to compute. However, we can use the Central Limit
Theorem (CLT) to approximate the r.v. Y with a normal r.v. Y* where Y and Y* have the
same mean and variance:
E(Y* ) = E(Y) = np
Var(Y* ) = Var(Y) = np(1 – p) = npq
Note: For this to work, we need n large enough so that both np and nq are at least five
(and 10 is suggested)
One problem: For the binomial r.v. Y:
P(Y = a) is a positive number for a = 0, 1, …
But for the normal r.v. Y* :
P(Y* = a) =
for all a (including 0, 1, …)
The fix is discussed in the text (p. 133) and suggested by the diagram below:
Binomial Distribution
probability
0.15
Event prob.,Trials
0.01,1000
0.12
0.09
0.06
0.03
0
0
2
4
6
8
10
12
-1-
14
16
18
20
Result: we use a continuity correction and replace P(Y* = a) – which equals zero – with:
P(a – 0.5 ≤ Y* ≤ a + 0.5)
which more accurately reflects the probability function p(y) of the binomial r.v. Y.
Note: Other results from using this correction are given in the text (p. 133). Some
examples:
P(Y ≤ 10) →
P(Y > 12) →
B. HTs and CIs for p
Suppose we have a random sample of size n from a binomial distribution.
Let Y be the number of successes.
Obvious estimator of p is the sample proportion:
Y can be approximated by a normal r.v. with
mean =
variance =
Consequently, the sampling distribution of the estimator
by a normal distribution with:
can be approximated
The hypothesis testing procedure then becomes:
Step 1: Let p0 be the nominal value of the proportion p. Then our null hypothesis is:
and our alternative hypothesis is one of:
depending on what is of concern in the problem.
-2-
Step 2: If the null hypothesis is true, then
is:
. Let q0 = 1 – p0 . Then the test statistic
which is approximately distributed as
Note: As before, n should satisfy:
and ≥ 10 is recommended.
Step 3: The critical region depends on Ha:
If Ha: p < p0 then we use a lower one-tailed test and thus reject H0 if:
If Ha: p > p0 then we use an upper one-tailed test and thus reject H0 if:
If Ha: p ≠ p0 then we use a two-tailed test and thus reject H0 if:
Step 4: Perform the n trials, and say we observe y successes. Then let
and compute Z.
Step 5: Reach conclusion and state it in words.
Example 4.10 (p. 180): This deal with the breaking strength of carbon fibers. Historically
10% are non-conforming. Concern is with developing a monitoring procedure, so
changes from the 10% level in either direction are of interest.
Step 1:
H0 :
Ha:
Step 2: Test statistic is:
Z=
We need:
and
Thus, n ≥
n∗p0 ≥ 5
n∗q0 ≥ 5
⇒
⇒
n≥
n≥
.
Step 3: Choice of α =
we reject H0 if:
(see text for discussion) and two-sided alternative means
-3-
Step 4: From the data: n =
and y =
so
Step 5: Since Z =
we do not reject the null hypothesis H0 .
In words: there is not sufficient evidence from the data to reject the claim that the true
proportion of non-conforming fibers is different than the nominal value of 10%.
As described in the text (p. 183), the appropriate CI for the proportion p is:
Note: this uses an estimated standard error of:
Example 4.11 (p. 184): The carbon fibers problem again. With α = 0.01 we have Zα/2 =
so a 99% CI for p is:
Finally, the CI result can be used to determine the sample size n that would be necessary
if we wanted to produce an interval with half-width no greater than some pre-determined
constant B.
Setting:
Results in:
Note: in the above, p0 is either:
(a) a reasonable guess of p if we have one, or
(b) 0.5 if we don’t have such a guess.
-4-
```
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mechmat competition1998
Competition for University Students
of Mechanics and Mathematics Faculty
of Kyiv State Taras Shevchenko University.
1. See William Lowell Putnam Math. Competition, 1996, B1.
2. See William Lowell Putnam Math. Competition, 1989, A4.
3. See William Lowell Putnam Math. Competition, 1997, B6.
4. Let q ∈ C, q = 1. Prove that for every non-singular matrix A ∈ M atn×n (C) there exists
non-singular matrix B ∈ M atn×n (C) such that
AB − qBA = I.
(V. Mazorchuk)
5. See William Lowell Putnam Math. Competition, 1992, B6.
6. See William Lowell Putnam Math. Competition, 1989, A6.
7. See William Lowell Putnam Math. Competition, 1997, B2.
8. Does there exist a function f ∈ C(R) such that for every real number x we have
1
f (x + t)dt = arctan x?
0
9. See William Lowell Putnam Math. Competition, 1997, A4.
10. The sequence {xn , n ≥ 1} ⊂ R is defined as follows:
1
x1 = 1, xn+1 =
+ { n}, n ≥ 1,
2 + xn
where {a} denotes fractional part of a. Find the limit
N
1
x2k .
lim
N →∞ N
k=1
(A. Kukush)
(A. Dorogovtsev, Jr.)
11. See William Lowell Putnam Math. Competition, 1995, A5.
12. Let B be complex Banach space and linear operators A, C ∈ L(B) be such that
σ(AC 2 ) {x + iy|x + y = 1} = ∅.
Prove that
σ(CAC) {x + iy|x + y = 1} = ∅.
(A. Dorogovtsev)
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### Home > APCALC > Chapter 4 > Lesson 4.4.2 > Problem4-151
4-151.
The annual cycle of daily high temperatures in Cabanaville is shown in the graph below from January 1st to December 31st. The $x$-axis is marked in months and the $y$-axis represents temperature in $^\circ\text{F}$.
1. Approximately when is Cabanaville at its hottest? The coldest? How can you tell?
This is a temperature graph. Change in temperature is a rate. How is rate shown on a temperature graph? How about the 'fastest' rate?
2. When is the temperature changing the fastest? What is the name for this type point on
the graph?
This is a temperature graph. During what month was temperature the highest?
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HOME WISE Uranium Project > Calculators >
# Uranium Mine Feasibility Calculator
(last updated Dec. 6, 2022)
Requires Netscape 3.0, Internet Explorer 3.0 or higher. JavaScript must be enabled.
For educational purposes only. No warranty.
Investors: please take note of our Precautionary Notice
Calculate the economic feasibility of a uranium mine and mill. HELP
1. Select a predefined parameter set, or enter and/or change parameters at your discretion,
2. enter a value into any one field of the Lifetime Material Balance (if none was preset), and
3. click the "Calculate" button.
"t" = metric tonne, "Mt" = million t, "Mlbs" = million lbs, "M\$" = \$ million, \$ = US\$.
> See also: Uranium Mine Feasibility Overview
Your browser does not support JavaScript, or you have disabled it.
The Calculator won't work.
Enable JavaScript (Preferences Menu) and reload the page.
Waste generatedReserve/Product
AmountAmountUranium contentsValue
[Mlbs U3O8][t U3O8][t U][M\$]
Ore deposit Mt Ore in situ
Mine Mt Waste Rock Mt Ore mined
Ore Sorting Mt Ore Reject Mt Ore post sorting
Mill Mt Mill Tailings t Ore Concentrate
Lifetime of operation: years
Capital
Cost
Nominal
Oper. Cost
Subtotal
(Nominal Cost)
Long-Term
Mgmt. Cost
Total
(True Cost)
Mining M\$ M\$ M\$ M\$ M\$
Ore Sorting M\$ M\$ M\$ M\$ M\$
Milling M\$ M\$ M\$ M\$ M\$
Gen. & Admin. M\$ M\$ M\$ M\$ M\$
All
stages
Total M\$
(=\$/lb U3O8)
M\$
(=\$/lb U3O8)
M\$
(=\$/lb U3O8)
M\$
(=\$/lb U3O8)
M\$
(=\$/lb U3O8)
Share (nom.) % % %
Share (true) % % % % %
Profit / Loss at
Nominal Cost
Profit / Loss at
True Cost
(Ore Concentrate Value ./. Cost) M\$
(=\$/lb U3O8)
(=%)
M\$
(=\$/lb U3O8)
(=%)
Please be aware that in most cases the figures shown do not comprise further cost components, such as:
acquistion cost of the property/deposit, exploration cost, sales cost, taxes and royalties, among others.
## Specific Results
Resulting Specific Operating Cost [\$/lb U3O8 in Ore Concentrate]
Nominal OperatingLong-Term Management
Mining \$/lb U3O8 \$/lb U3O8
Ore Sorting \$/lb U3O8 \$/lb U3O8
Milling \$/lb U3O8 \$/lb U3O8
Gen. & Admin. \$/lb U3O8 \$/lb U3O8
Total \$/lb U3O8 \$/lb U3O8
Resulting Uranium Concentrations
in Waste generatedin Reserve/Product
Ore deposit Uranium Ore in situ: wt_% U
MineWaste Rock: (conc. not specified)Uranium Ore mined: wt_% U
Ore SortingOre sorting reject: wt_% UUranium Ore post sorting: wt_% U
MillMill Tailings: wt_% UUranium Ore Concentrate: wt_% U
Resulting Mass Flow
Mine Actual Mine Ore production rate: Mt/a
Actual Mine capacity usage: %
Ore Sorting Percentage of uranium contents in mined ore rejected by sorting: wt_%
Mill Actual Mill Ore processing rate: Mt/a
Actual Mill capacity usage: %
Actual production rate of uranium in ore concentrate:
Mlbs/a U3O8 = t/a U3O8 = t/a U
General Overall uranium recovery from Ore in situ to Ore concentrate: wt_%
## Parameter Entry
Select Parameter Set:
OP = open pit · UG = underground · HL = heap leach · ISL = in situ leach
Notes:
Process Parameters
Mine Grade of In Situ Ore: wt_% U = wt_% U3O8 (optional)
Waste/Ore Ratio:
Ore Production Capacity: Mt/a (optional)
Ore Extraction Losses: wt_%
Grade of Ore mined: wt_% U = wt_% U3O8
Ore Sorting Ore rejected: wt_%
Relative Ore grade improvement by sorting: %
Mill Ore Processing Capacity: Mt/a
Design Life: years (optional)
Extraction Losses: wt_% (auto estimate)
Uranium concentration in Ore Concentrate: wt_% U
Cost Parameters
Capital CostNominal Operating CostLong-Term Management Cost
Mining \$ per t/a mined
\$ million
\$ per t ore mined
\$ per t waste rock
\$ per t ore mined
\$ per t waste rock
Ore Sorting \$ per t/a mined
\$ million
\$ per t ore mined
\$ per t ore reject
\$ per t ore mined
\$ per t ore reject
Milling \$ per t/a milled
\$ million
\$ per t ore milled
\$ per lb U3O8 produced
\$ per t tailings
\$ per t ore milled
\$ per lb U3O8 produced
\$ per t tailings
General and
\$ per t/a mined
\$ million
\$ per t ore mined
\$ per lb U3O8 produced
\$ per t ore mined
\$ per lb U3O8 produced
Year of assessment: Inflation: % take inflation into account
Price Parameters
Market Price
Natural Uranium in
Ore Concentrate
\$ per lb U3O8 = \$ per kg U
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Playing Keno the Appropriate Way
Keno is a form of lottery-like betting game available mainly in some country casinos and also sold as a progressive game at some national state lotteries. The origin of this name Keno is somewhat vague, however, most experts agree that it came in the Japanese pronunciation of the word”koku” which means fortune. Although this seems to be an unlikely origin for the title, the game did achieve popularity in Japan where it was allegedly first played in the 16th Century. A version of Keno was also developed in Great Britain, probably since an advancement in the Japanese machine.
The game of Keno involves playing with a mathematical grid comprising a million positions where each level seems for the whole session. To place a wager, the player must identify the amount which lies between every pair of adjacent numbers. The player could call the amount Keno (so”ten”), and mark that the numbers by composing them out, or in many instances – typing them in a computer program. After the player wins a wager, they have to remove one of their winning combinations by the gridnevertheless, they're not required to remove all their winning combinations. Should they choose to accomplish this, the winnings will then be multiplied by ten.
1 intriguing facet of this Keno system is that there are certain kinds of balls used in the sport. 먹튀검증 One kind of ball is called the”coolcat” that is the smallest ball used at the sport. Another sort of ball is that the”red flag” that is nearly identical to your hot dog. Thenthere are smaller chunks called”hot potato” that are employed in place of other chunks. These balls are not used in any official games of this Keno system but instead are utilized to refer to these tiny reddish stripes that gamers use to indicate their winning combinations within the grid.
Besides the chunks used from the Keno method, players can also opt to buy exclusive die cuts. This is fundamentally a plastic piece with all the numbers on it cut in half. Players may then quickly set their bet with the die cut because the whole ball or cross is marked with the winning amount. This allows players who'd rather not utilize the Keno platform to still play keno at an actual casino game.
Though this kind of gambling game is relatively new and does not yet have a big following as the world wide web does, it's quickly gaining acceptance. One reason for its increase in popularity is because of the simplicity of working with each of the necessary equipment. By way of example, all a player needs is a pair of basic playing cardsa standard deck of cards and a pencil to begin with Keno.
If you're simply looking for a fun new way to spend your Sunday afternoon, Keno may be the ideal game that you play with. The attractiveness of playing Keno without investing any money into the true game is that this form of entertainment can be enjoyed any time of the week. It's possible for you to enjoy Keno with your family and friends as readily as you would enjoy a night at the casino. Another benefit of Keno is you can play the game in literally any place. However, for optimal game play along with a truly challenging experience, it is very important to not forget that there are certain wager sizes along with other aspects that needs to be taken under account before you start betting.
Ordinarily, Keno games demand a minimum of 3 decks of cards and the bet is placed on the previous ten figures on the respective winning hand. The first ten numbers will remain specific for Keno matches and will probably be ordered by the random number generator (RNG). The player may call any number and the trader will browse the cards to determine which ten numbers will be the upcoming ones to be predicted.
The objective of this strategy is to try and generate a number selection which provides the best chance at hitting the jackpot. The randomness of the amounts generated is a part of the pleasure of enjoying Keno, but it is also very important to think about how likely the chosen numbers are in the winning amounts. It is often times more likely that an experienced Keno player can successfully predict the amount combinations employed from the random number generator than it is to just choose any random number from the box. In the event you decide to play Keno online casinothen you will most likely wish to stay with regular balls. Standard sized Keno sets do not have the utmost amount of twist, which means you should normally select amounts which are not as likely to cause you to be known out.
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# Maximum PC
It is currently Sun Apr 19, 2015 3:22 am
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Post subject: Memory addressing in a 32-bit OS (AKA: Where's My RAM?)Posted: Thu Dec 27, 2007 4:14 pm
Million Club [PC]*
Joined: Sun Feb 12, 2006 5:29 pm
Posts: 4914
Location: Motor City, folding for Mom
The basics of memory addressing in XP or Vista 32-bit:
The total number of system address space happens to exactly equal the amount of address space needed by 4GB of physical memory.
However, onboard devices / resources (NB/SB chips, parallel/serial ports, NIC, etc.) also need address space, as do PCI/PCIe/AGP cards. Devices & resources are allocated address space from the top of the pool, down. This works out to (depending on the specific system) to (typically) 3.12~3.5GB worth of space left. 3.5 is a rare high number; some system configs put that number as low as 2.9GB.
And, the gotcha: onboard memory on a videocard also needs address space. This is also allocated from top-down, after system resources.
So: 4GB (equivalent) address space - resources = 3.12GB available - vidcard memory (lets say 256MB, unless he's got a 512MB frame buffer) = 2.87GB available space. Increase the frame buffer (or, especially, run SLI/Crossfire), and the number shrinks further. Sucks for the people with a pair of 768MB buffers, as they can't even address 2GB RAM.
The math: each bit of memory on that stick requires an address. 8 bits/byte, 1,024 bytes/KB, and so on.
But, so does each bit of memory on that videocard.
32 bit = 34,359,738,368 addresses.
64 bit = 1,099,511,627,776 addresses.
chart to explain it better:
See that yellow area? As you add more onboard devices (or, more videocard RAM), it grows. Down.
As it grows, the grey area (RAM address pool) gets smaller. Smaller it gets, the less RAM you can address.
Sources:
http://www.codinghorror.com/blog/archives/000811.html
And, good read (with layman's terms!), found by Nastyman
*Edit: with 4GB RAM installed, Vista x86 SP1 will show all four gigs in System Properties; however, Task Manager (as well as most system-monitor utilities) will continue to show only that RAM that's actually addressable by the system. So, if the BIOS and System Properties show 4GB, but Task Manager only reports, say, 2.8GB, that's normal.
(updated 5 July 2008)
*Edit (16 Sep 09):
Sovereign installed Win7 (x86) on one of his machines, which has 4GB RAM installed.
The Resource Monitor (in Task Manager) gives a very good graphical representation of where, exactly, The Missing RAM went:
Soveriegn wrote:
The "Hardware Reserved" shows the exact amount that was mapped to Memory-Mapped I/O devices (like ATI HyperMemory, PCI devices and the like).
Top
Post subject: Re: Memory addressing in a 32-bit OS (AKA: Where's My RAM?)Posted: Sun Nov 28, 2010 12:11 am
Bitchin' Fast 3D Z8000*
Joined: Tue Jun 29, 2004 11:32 pm
Posts: 2555
Location: Somewhere between compilation and linking
cup wrote:
Sucks for the people with a pair of 768MB buffers, as they can't even address 2GB RAM.
Sure they can address those 2GB of ram. Presumably, they could fill the buffer with an image, make a system call and see the image on their monitor. Honestly, I don't see the big deal. You have the same MMIO regardless of whether you have 3GB or 4GB of memory in the system. Given that most desktop computers now have either a 256MB or 512MB video card, you're not wasting all that much memory... probably less than \$10 worth!
cup wrote:
The math: each bit of memory on that stick requires an address.
Wrong. I don't believe there has ever been a "bit" addressable computer and modern computers certainly aren't addressed in this manner. If you look at the ISA for most processors, you'll see that not only can you not load/store a single bit, but typically, you also have a variety of "boundary" conditions (ie 4 byte, 8 byte, even 16 and 32 byte boundaries depending on the instruction).
cup wrote:
http://www.codinghorror.com/blog/archives/000811.html
*Edit: with 4GB RAM installed, Vista x86 SP1 will show all four gigs in System Properties; however, Task Manager (as well as most system-monitor utilities) will continue to show only that RAM that's actually addressable by the system. So, if the BIOS and System Properties show 4GB, but Task Manager only reports, say, 2.8GB, that's normal.
(updated 5 July 2008)
*Edit (16 Sep 09):
Sovereign installed Win7 (x86) on one of his machines, which has 4GB RAM installed.
The Resource Monitor (in Task Manager) gives a very good graphical representation of where, exactly, The Missing RAM went:
Soveriegn wrote:
The "Hardware Reserved" shows the exact amount that was mapped to Memory-Mapped I/O devices (like ATI HyperMemory, PCI devices and the like).
[/quote]
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Reconstruction of the
2001 Pegsdon formation
1. Draw a circle and with it, the horizontal and vertical centerlines.
2. Copy this circle with its center to the lower end of the vertical centerline.
3. Notice that, since both circles are equal and pass through each other's center, they form the Vesica Pisces symbol. Therefore, both intersecting points together with the upper end of the vertical centerline form an equilateral triangle.
Draw this triangle.
4. Construct the inscribed circle of this triangle.
5. Copy this circle four times, to both ends of the horizontal centerline and to both intersections of circle 4 with the vertical centerline.
6. Construct the inscribed hexagon (regular 6-sided polygon) of circle 1, pointing up.
7. In turn, construct the inscribed circle of hexagon 6.
8. The next circle to construct is one that fits between circles 7 and 4. To do so, first, copy circle 4 two times, one to the intersection of the horizontal centerline with circle 7 (on the right), the other to the intersection with the circle just copied (on the left), see figure.
9. The line connecting the intersecting points of the rightmost circle 5 with the latter circle 8 divides the part of the horizontal centerline between circles 7 and 4 into two equal parts. Construct this line.
10. Construct the circle tangent to circles 7 and 4, centered at the horizontal centerline.
11. Copy this circle to the center of the formation.
12. The vertical row of circles 4 and 5, again, form Vesica Pisces symbols. And again, an equilateral triangle can be constructed. Do so.
13. Construct the inscribed circle of this triangle.
14. Elongate the vertical centerline until it intersects circle 2 at its lowest point. Copy circle 13 to this point.
15. Circles 1, 4, 5, 7, 11 and 14 together form the basic parts for the final reconstruction.
16. Deleting all parts of circles not used, and making black relevant areas, finishes the reconstruction of the 2001 Pegsdon formation. Black denotes flattened crop.
17.
courtesy The Crop Circle Connector
photo by: Andrew King
This is how the match with the aerial image looks like.
Copyright © 2001, Zef Damen, The Netherlands
Personal use only, commercial use prohibited.
Since 1-February-2005
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# 9.8: Digraph Spellings of Long ‘oo’
Difficulty Level: At Grade Created by: CK-12
## Digraph Spellings of Long ‘oo’
1. You have seen that the long ‘oo’ sound, [u¯]\begin{align*}[\bar{\mathrm{u}}]\end{align*}, is often spelled <u>\begin{align*}<\mathrm{u}>\end{align*} or <o>. It is also often spelled with combinations of two vowel letters. When two vowel letters work together as a team to spell a single vowel sound, they are called a digraph. In all but three of the following words [u¯]\begin{align*}[\bar{\mathrm{u}}]\end{align*} is spelled with vowel digraphs. Underline the letters that spell [u¯]\begin{align*}[\bar{\mathrm{u}}]\end{align*}:
ch\underline{oo}ses\underline{u}icidec\underline{ou}pong\underline{oo}sech\underline{ew}sthr\underline{ou}ght\underline{oo}br\underline{ui}segr\underline{ou}psn\underline{u}isancel\underline{oo}sey\underline{ou}thr\underline{ew}n\underline{oo}dlessm\underline{oo}thj\underline{ui}ces\underline{ui}taven\underline{u}ecr\underline{ui}sec\underline{ou}garkn\underline{ew}m\underline{oo}dl\underline{o}sepr\underline{oo}fj\underline{ew}elp\underline{oo}dleb\underline{oo}tsd\underline{ew}r\underline{ou}tinebr\underline{oo}d\begin{align*}& \text{ch\underline{oo}se} && \text{thr\underline{ou}gh} && \text{l\underline{oo}se} && \text{j\underline{ui}ce} && \text{kn\underline{ew}}&& \text{p\underline{oo}dle}\\ & \text{s\underline{u}icide} && \text{t\underline{oo}} && \text{y\underline{ou}} && \text{s\underline{ui}t} && \text{m\underline{oo}d}&& \text{b\underline{oo}ts}\\ & \text{c\underline{ou}pon} && \text{br\underline{ui}se} && \text{thr\underline{ew}} && \text{aven\underline{u}e} && \text{l\underline{o}se} && \text{d\underline{ew}}\\ & \text{g\underline{oo}se} && \text{gr\underline{ou}ps} && \text{n\underline{oo}dles} && \text{cr\underline{ui}se} && \text{pr\underline{oo}f} && \text{r\underline{ou}tine}\\ & \text{ch\underline{ew}s}&& \text{n\underline{u}isance} && \text{sm\underline{oo}th} && \text{c\underline{ou}gar}&& \text{j\underline{ew}el}&& \text{br\underline{oo}d}\end{align*}
2. Sort the words into these groups:
Words in which [u¯]\begin{align*}[\bar{\mathrm{u}}]\end{align*} is not spelled with a digraph...
suicideavenuelose\begin{align*}& suicide && avenue && lose\end{align*}
Words in which
<oo> <ou>
choose mood coupon
goose proof through
too poodle groups
loose boots you
noodles brood cougar
smooth routine
Words in which
<ui> <ew>
bruise chews
nuisance threw
juice knew
suit jewel
cruise dew
3. You have worked with six ways of spelling [u¯]\begin{align*}[\bar{\mathrm{u}}]\end{align*}. Write them below and give at least one word that contains each spelling:
Spellings of [u¯]\begin{align*}[\bar{\mathrm{u}}]\end{align*} Example Words
<u>\begin{align*}\end{align*} junior, rumor, ruble, . . .
<o> shoe, who, prove, . . .
<oo> choose, loose, noodles, . . .
<ou> cougar, coupon, group, . . .
<ui> bruise, juice, nuisance, . . .
<ew> chew, threw, jewel, . . .
4. You have learned eight patterns, like VCC and VCV, for marking long and short vowels. Unfortunately, although these patterns are very useful when vowels are spelled by single letters, they are not useful when vowels are spelled with vowel digraphs. So vowel patterns like VCC and VCV cannot help when you are spelling vowel sounds with digraphs. But there are other kinds of patterns that can help, as we'll see in the next lesson.
Word Venn. All of the following words contain the sound [u¯]\begin{align*}[\bar{\mathrm{u}}]\end{align*}. Into circle A put only those words that contain a digraph spelling of [u¯]\begin{align*}[\bar{\mathrm{u}}]\end{align*}. Into circle B put only those words that contain an instance of final <e> deletion. Inside the rectangle but outside the circles put any other of the words in the list:
approvalassumedbruisingchoosyconsumercougarcouponcruisergluedimproveincludingjewelryjuicyjuniorknewloosennuisancerublerumorshoe\begin{align*} & \text{approval} && \text{cougar} && \text{including} && \text{loosen}\\ & \text{assumed} && \text{coupon} && \text{jewelry} && \text{nuisance}\\ & \text{bruising} && \text{cruiser} && \text{juicy} && \text{ruble}\\ & \text{choosy} && \text{glued} && \text{junior} && \text{rumor}\\ & \text{consumer} && \text{improve} && \text{knew} && \text{shoe}\end{align*}
Teaching Notes.
Item 1. The word through raises the complexities posed by the consonant digraph <gh>. Old English had consonant sounds that linguists call velar fricatives, which means that they were pronounced back in the mouth at the velum and they were pronounced with a hissing or friction. In Old English one of these was spelled <g> and the other <h>. Over the centuries the two converged and came to be spelled <gh>. Thus, hundreds of years ago <gh> spelled a velar fricative sound like that at the end of the Scottish pronunciation of loch or the German pronunciation of Bach. Over time that sound dropped out of English, but the <gh> usually stayed in the written words, with a new pronunciation. After short vowels spelled with a digraph it came to be pronounced [f], as in laugh, tough, cough..
The complexities arise with words like brought, freight, straight, and tight and like weigh, though, and through. In the first group, with the cluster <ght>, we treat the <gh> as part of the spelling of the sound [t]. Thus, in such words [t] is said to be spelled <ght>, due to a simplifying of earlier pronunciation with no concomitant change in spelling. However, after long vowels the <gh> (with no following <t>) is no longer pronounced, as in the second group of words: weigh, though, through. To say that in weigh, for instance, [a¯]\begin{align*}[\bar{\mathrm{a}}]\end{align*} is spelled <eigh> blurs the consonant-vowel distinction. It seems better to treat the <gh> in such words as a diacritic, marking a preceding long vowel, much like silent final <e>. Thus we would say, for instance, that [a¯]\begin{align*}[\bar{\mathrm{a}}]\end{align*} is sometimes spelled <ei> before <gh> (weigh) or <ght> (weight).
(At the front of words <gh> is pronounced [g], as in ghost and ghastly, ghetto, ghoul. It is also pronounced [g] inside recent adoptions from Italian, like spaghetti. This <gh> does not come from the earlier sound in loch and Bach. For more on <gh> = [g] see AES, pp. 209-10, 352.)
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You are here: / / / DC Brush Motor
# DC Brush Motor
Views: 29 Author: Site Editor Publish Time: 2018-01-10 Origin: Site
DC Brush Motor
Principles of operation
Reference to the chart reveals useful performance information valid for all fulling servomotors.
It shows speed n, current I, output power P and efficiency ¦Ç plotted against torque M for a given supply voltage U. Torque M is a function of the current I and the torque constant k (expressed in Nm/A). The motor develops its maximum torque Ms at stall (n=0), when the current is maximum and determined only by the supply voltage U and the rotor resistance R:
With increasing speed, an increasing back EMF E is induced in the armature which tends to reduce the current:
The value of E is the product of angular speed ¦Ø (expressed in rad/s) and the torque constant (expressed in V/rad/s=Vs=Nm/A):
E = k¦Ø
Thus, the supply voltage splits into two parts: RI, necessary to establish the current I in the armature, which generates the torque
M, and k¦Ø to overcome the induced voltage, in order to generate the speed¦Ø:
U = RI + k¦Ø
No-load speed no is a function of the supply voltage and is reached when E becomes almost equal to U; no-load current Io is a function of friction torque:
Power output P is the product of angular speed¦Ø and torque M (P = M.¦Ø); for a given voltage it reaches its maximum Pmax at half the stall torque Ms, where efficiency is close to 50%. The maximum continuous output power is defined by an hyperbola delimiting the continuous and intermittent operation ranges.
Efficiency¦Çis the mechanical to electrical power ratio (¦Ç= Pm / Pel). Maximum efficiency¦Çmax occurs at relatively high speed. Its value depends upon the ratio of stall torque and friction torque and thus is a function of the supply voltage:
The maximum continuous torque depends upon dissipated power (I2R), its maximum value is determined by:
Where Tmax is the maximum tolerated armature temperature, Tamb is the ambient temperature, Rmax is the rotor resistance at temperature Tmax and Rth is the total thermal resistance (rotor-body-ambient).
At a given torque M, increasing or decreasing the supply voltage will increase or decrease the speed. The speed-torque function varies proportionally to the supply voltage U.
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Time remaining:
##### i dont nderstand i need to write a system of equations for it
Mathematics Tutor: None Selected Time limit: 1 Day
a family member has some five dollar bills and one dollar bills in her wallet . altogether she has 18 bills and a total of \$62. how many of each bill does she have?
Mar 27th, 2015
5x+y=62 (1)
x+y = 18 (2)
(1)-(2),
4x = 44
x = 11,
y = 18-x =7
Mar 26th, 2015
...
Mar 27th, 2015
...
Mar 27th, 2015
Dec 3rd, 2016
check_circle
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# Coordinate geometry question
• Nov 13th 2007, 10:23 AM
hymnseeker
Coordinate geometry question
Just a really quick one, suppose I have a line of equation: y=1/2x+5/2
The next part of the question asks for this in the form of ax+by+c=0
How do I convert from y=mx+c to ax+by+c=0?
Thank youuu! :)
• Nov 13th 2007, 10:24 AM
Jhevon
Quote:
Originally Posted by hymnseeker
Just a really quick one, suppose I have a line of equation: y=1/2x+5/2
The next part of the question asks for this in the form of ax+by+c=0
How do I convert from y=mx+c to ax+by+c=0?
Thank youuu! :)
bring everything to one side and leave 0 on the other side and (in this case) multiply through by 2 when done
• Nov 13th 2007, 10:26 AM
hymnseeker
Thanks for the reply, so just let y=0?
1/2x+5/2=0
Does this satisfy what the question wants in asking for ax+by+c=0?
• Nov 13th 2007, 10:29 AM
Jhevon
Quote:
Originally Posted by hymnseeker
Thanks for the reply, so just let y=0?
1/2x+5/2=0
Does this satisfy what the question wants in asking for ax+by+c=0?
did you read what i said? i said bring everything over to one side, how does that translate to equating y to zero?
$y = \frac 12x + \frac 52$ ............subtract y from both sides
$\Rightarrow 0 = \frac 12x - y + \frac 52$ ...............multiply through by 2
$\Rightarrow 0 = x - 2y + 5$
which is the same thing as
$x - 2y + 5 = 0$
here a = 1, b = -2 and c = 5
• Nov 13th 2007, 10:35 AM
hymnseeker
Thank you for your help. Needed to see it visually - and in answer to your question (although I suspect it was rhetorical), yes, I did read what you said first off.
• Nov 13th 2007, 11:03 AM
Jhevon
Quote:
Originally Posted by hymnseeker
Thank you for your help. Needed to see it visually - and in answer to your question (although I suspect it was rhetorical), yes, I did read what you said first off.
yes, it was rhetorical. you're welcome, and good luck with your class
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# Is it possible to create an infinitely differentiable piecewise function?
The following piecewise function is continuous, but not differentiable: $$f(x) = \begin{cases}x, & x\leq0 \\ -x, & x\geq 0\end{cases}$$
This piecewise function is differentiable, but has a discontinuity in its second derivative: $$g(x) = \begin{cases} 0, & x\leq0\\ x^2, & x\geq0\end{cases}$$
It seems that no matter how "smooth" a piecewise function appears to be, there will always be some discontinuity in one of its derivatives. My guess is that this is because one of the sides will drop to zero before the other, and you even end up with $0 = \text{some nonzero constant}$. (Like in my second example; the second derivative is $0$ and $2$.) If the two have the same magnitude and drop to zero at the same time, either they have a different constant $x^0$ term, meaning they are not continuous; they have a different $x^n$ term, which will eventually drop to $x^0$, resulting in the previous case; or they have no differences and it isn't really a piecewise function at all.
However, this is only valid for finite polynomials.
Is it possible to create a piecewise function so that all of its derivatives are continuous? If not, like I suspect, what are some proofs of this fact other than the outline I gave? And what does this mean for the nature of functions and continuity -- it almost seems to imply that a very small section of a function can "force" the rest of the curve into a single, locked state.
• In the complex plane things are close to what you are imagining. If a function is differentiable once it is differentiable infinitely many times, it is equal to its Taylor series out to the radius of convergence, etc. In the reals, all bets are off, as the classic bump function demonstrates May 10 '17 at 4:53
The canonical example is the function $$f(x) = \begin{cases}e^{-\frac{1}{x}}, & x>0 \\ 0, & x\leq 0\end{cases}$$
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# Water pressure and flow restricton - does size matter?
• R_Rose
In summary, the conversation discusses the process of filtering liquids using a bucket filled with filter medium and various drain hole arrangements. The reduction of flow was unexpected and there may be additional problems caused by varying levels of clogging. It is suggested to use a uniform mesh bottom plate and a layer of coarse granular material for better flow. The idea of not having a bottom at all is also considered, but the need for structural support and the presence of extra forces are taken into account. A pre-filter is also mentioned as a useful addition to improve the main filter's performance.
R_Rose
I'm trying to figure out how to best filter some liquids. A 5 gal bucket is filled with some filter medium (say sand on top 1/2 and fine charcoal on bottom 1/2). The liquid is slowly added at the top and gravity is used and vacuum may be used at some point with a complete seal along the bottom of the bucket.
Now what I'm wondering is about the number and size of holes in the bottom of the bucket. I've done this with 2 smaller holes (diameter of a pencil) and there was a steady drip from each hole, about 1 a second. I used a different bucket and moved the holes from the extreme edges to 1/3 of the way from the edge so there were 2 holes directly across from each other - same size as before and another in the center and then 3 small slits about 1/2 inch long by 2mm across around the edges. This time flow was about 1/5 the rate as before.
The reduction of flow was the opposite of what I was expecting and was wondering if there may be some reason for this which I didn't understand such as there being the same pressure in the vessel but spread out over more holes which would decrease the pressure at each hole. Since the surface of the filter material wants to hold onto the water, this decrease of pressure makes a big difference in flow.
Is that a reasonable conclusion or am I experiencing something else I have to figure out.
You are forcing the draining liquid to take paths which are mostly sideways through the lower layers of filtering material rather than straight down and out .
Depending on the placement of your limited number of drain holes these paths could be of different lengths and present different levels of flow resistance .
There may be additional problems caused by flow being active in some holes and sluggish in others thus causing varying levels of clogging .
In any case your total drain hole flow area is probably insufficient .
Bottom plate of a filter should ideally be a uniform mesh of many generous sized holes .
Often useful to have a layer of very coarse granular material at lowest level of a filter .
Last edited:
sophiecentaur and anorlunda
Try sketching the likely flow paths for different arrangements of drain holes .
Ideally, there would be no bottom at all in the bucket so that each part of the filter material would work with the same flow rate. Something at the bottom is needed, of course, which has to be strong enough to hold the load and the holes need to be small enough so that the filter medium can support itself. That's all obvious stuff and if you look at colanders and sieves, used in cooking, their construction is mostly holes. So why not copy existing designs? Is there some extra force or load involved that isn't there in the kitchen?
Nidum said:
Often useful to have a layer of very coarse granular material at lowest level of a filter
A good idea and that's how commercial, high volume filters work.
Also, a coarse, pre-filter can allow the main filter to do its job better by allowing better flow of the liquid into it.
## 1. How does size affect water pressure and flow restriction?
The size of a pipe or restriction in a water system can have a significant impact on the pressure and flow of the water. A smaller size will result in higher pressure but lower flow, while a larger size will result in lower pressure but higher flow. This is due to the principles of fluid dynamics, where a smaller area for the water to flow through creates more resistance and therefore higher pressure.
## 2. Can a smaller size restriction increase water pressure?
Yes, a smaller restriction can increase water pressure, as mentioned above. However, this increase in pressure may not always be desirable, as it can also lead to potential damage or leaks in the system.
## 3. How can I determine the optimal size for a water restriction?
The optimal size for a water restriction depends on various factors, such as the desired pressure and flow rate, the volume of water being transported, and the size of the pipes in the system. It is best to consult with a professional or use mathematical calculations to determine the most suitable size for your specific water system.
## 4. Does the material of the restriction affect water pressure and flow?
The material of the restriction can also impact water pressure and flow. Some materials, such as rougher or more rigid materials, can create more resistance and therefore affect the pressure and flow. It is important to consider both the size and material when designing a water system.
## 5. Are there any other factors besides size that can affect water pressure and flow restriction?
Yes, there are other factors that can affect water pressure and flow restriction, such as the elevation of the system, the temperature of the water, and the presence of any clogs or debris in the pipes. It is essential to regularly maintain and monitor a water system to ensure optimal pressure and flow.
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#### Abstract
Two kinds of series representations, referred to as the Engel series and the Cohen-Egyptian fraction expansions, of elements in two different fields, namely, the real number and the discrete-valued non-archimedean fields are constructed. Both representations are shown to be identical in all cases except the case of real rational numbers.
#### 1. Introduction
It is well known [1] that each nonzero real number can be uniquely written as an Engel series expansion, or ES expansion for short, and an ES expansion represents a rational number if and only if each digit in such expansion is identical from certain point onward. In 1973, Cohen [2] devised an algorithm to uniquely represent each nonzero real number as a sum of Egyptian fractions, which we refer to as its Cohen-Egyptian fraction (or CEF) expansion. Cohen also characterized the real rational numbers as those with finite CEF expansions. At a glance, the shapes of both expansions seem quite similar. This naturally leads to the question whether the two expansions are related. We answer this question affirmatively for elements in two different fields. In Section 2, we treat the case of real numbers and show that for irrational numbers both kinds of expansion are identical, while for rational numbers, their ES expansions are infinite, periodic of period , but their CEF expansions always terminate. In Section 3, we treat the case of a discrete-valued non-archimedean field. After devising ES and CEF expansions for nonzero elements in this field, we see immediately that both expansions are identical. In Section 4, we characterize rational elements in three different non-archimedean fields.
#### 2. The Case of Real Numbers
Recall the following result, see, for example, Kapitel IV of [1], which asserts that each nonzero real number can be uniquely represented as an infinite ES expansion and rational numbers have periodic ES expansions of period .
Theorem 2.1. Each is uniquely representable as an infinite series expansion, called its Engel series (ES) expansion, of the form where Moreover, if and only if for all sufficiently large .
Proof. Define then . If is already defined, put Observe that is the least integer and We now prove the folowing.Claim 2. We have .Proof. First, we show that for all by induction. If , we have seen that . Assume now that for . By (2.3), we see that . Since and we have . If there exists such that , then and so , contradicting the minimal property of and the Claim is proved.From the Claim and (2.3), we deduce that and . Iterating (2.4), we get To establish convergence, let Since and for all , the sequence of real numbers is increasing and bounded above by . Thus, exists and so By the Claim, showing that any real number has an ES expansion. To prove uniqueness, assume that we have two infinite such expansions such that with the restrictions and the same restrictions also for the 's. From the restrictions, we note that If , then by (2.12) we also have , forcing . If , then (2.12) shows that , forcing again . In either case, cancelling out the terms in (2.12) we get Since , then so . But there is exactly one integer satisfying these restrictions. Thus, . Cancelling out the terms and in (2.14) and repeating the arguments we see that for all .
Concerning the rationality characterization, if its ES expansion is infinite periodic of period , it clearly represents a rational number. To prove its converse, let . Since
we see that is a rational number in the interval whose denominator is . In general, from (2.4), we deduce that each is a rational number in the interval whose denominator is . But the number of rational numbers in the interval whose denominator is is finite implying that there are two least suffixes such that . Thus, by (2.3), we have . From (2.2), we know that the sequence is increasing. We must then have and the assertion follows.
Remark 2 s. In passing, we make the following observations. (a)For , we have (b)If , then and so for all . (c)If , then its ES expansion is
To construct a Cohen-Egyptian fraction expansion, we proceed as in [2] making use of the following lemma.
Lemma 2.2. For any , there exist a unique integer and a unique such that
Proof. Let and . Put and so To prove uniqueness, assume so that Since there is only one integer with this property, we deduce and consequently, proving the lemma.
Theorem 2.3. Each is uniquely representable as a CEF expansion of the form subject to the condition and no term of the sequence appears infinitely often. Moreover, each CEF expansion terminates if and only if it represents a rational number.
Proof. To construct a CEF expansion for , define If , then the process stops and we write . If , by Lemma 2.2, there are unique and such that Thus, If then the process stops and we write . If , by Lemma 2.2, there are and such that the last inequality being followed from and . Observe also that Continuing this process, we get with If some , then the process stops, otherwise the series convergence follows at once from
To prove uniqueness, let
with the restrictions (2.23) on both digits and . Now It is clear that the restrictions (2.23) imply the strict inequality in (2.33). This also applies to the right-hand sum in (2.32). Equating integer and fractional parts in (2.32), we get Since , then so . But there is exactly one integer satisfying these restrictions. Then and Proceeding in the same manner, we conclude that for all .
Finally, we look at its rationality characterization. If , then , say where . From (2.30), we see that each is a rational number whose denominator is . Using this fact and the second inequality condition in (2.30), we deduce that for some , that is, the expansion terminates. On the other hand, it is clear that each terminating CEF expansion represents a rational number. Now suppose that is irrational and there is a and integer such that for all . Then
Since , it follows that is rational, which is impossible.
The connection and distinction between ES and CEF expansions of a real number are described in the next theorem.
Theorem 2.4. Let and the notation be as set out in Theorems 2.1 and 2.3.
(i) If , then its ES expansion is infinite periodic of period , while its CEF expansion is finite. More precisely, for , let its ES and CEF expansions be, respectively,
If is the least positive integer such that , then and the digits terminate at .
(ii) If , then its ES and its CEF expansions are identical.
Proof. Both assertions follow mostly from Theorems 2.1, 2.3, and Remark (b) except for the result related to the expansions in (2.38) which we show now.
Let and let be the least positive integer such that . We treat two seperate cases.
Case 1 (). In this case, we have and . Since , we get and so . We have , and so the CEF expansion terminates. On the other hand, by Remark (a) after Theorem 2.1, we have .Case 2 (). Thus, and . By Lemma 2.2, we have . For , assume that and . Then Since , again by Lemma 2.2, . This shows that Since , we have and thus From the construction of CEF, we know that . Thus, showing that . Furthermore, implying that the CEF terminates at , and by Remark (a) after Theorem 2.1, .
#### 3. The Non-Archimedean Case
We recapitulate some facts about discrete-valued non-archimedean fields taken from [3, Chapter ]. Let be a field complete with respect to a discrete non-archimedean valuation and its ring of integers. The set is an ideal in which is both a maximal ideal and a principal ideal generated by a prime element . The quotient ring is a field, called the residue class field. Let be a set of representatives of . Every is uniquely of the shape
for some , and define the order of by , with . The head part of is defined as the finite series
Denote the set of all head parts by
The Knopfmachers' series expansion algorithm for series expansions in [4] proceeds as follows. For , let Define If is already defined, put
if , where and which may depend on . Then for
The process ends in a finite expansion if some . If some , then is not defined. To take care of this difficulty, we impose the condition
Thus
When , the algorithm yields a well-defined (with respect to the valuation) and unique series expansion, termed non-archimedean Engel series expansion. Summing up, we have the following.
Theorem 3.1. Every has a finite or an infinite convergent non-archimedean ES expansion of the form where the digits are subject to the restrictions
Now we turn to the construction of a non-archimedean Cohen-Egyptian fraction expansion, in the same spirit as that of the real numbers, that is, by way of Lemma 2.2. To this end, we start with the following lemma.
Lemma 3.2. For any such that , there exist a unique such that and a unique such that
Proof. Let . Then Putting , we show now that . Since , we have where , and so Thus To prove the uniqueness, assume that there exist such that and such that From , we get . If , since we have . Using , we deduce that which is a contradiction. Thus, and so .
For a non-archimedean CEF expansion, we now prove the following.
Theorem 3.3. Each has a non-archimedean CEF expansion of the form where This series representation is unique subject to the digit condition (3.19).
Proof. Define and . Then . If , the process stops and we write . If , by Lemma 3.2, there are and such that where and So If , the process stops and we write . If , by Lemma 3.2, there are and such that where and So Continuing the process, in general, where Thus, We observe that the process terminates if . Next, we show that . By construction, we have . Assume that , then Regarding convergence, consider It remains to prove the uniqueness. Suppose that has two such expansions Since and , we have . Similarly, yielding by uniqueness and . Putting we have and so By Lemma 3.2, since is the unique element in with such property, we deduce . Continuing in the same manner, we conclude that the two expansions are identical.
It is clear that the construction of non-archimedean ES and CEF expansions is identical which implies at once that the two representations are exactly the same in the non-archimedean case.
#### 4. Rationality Characterization in the Non-Archimedean Case
In the case of real numbers, we have seen that both ES and CEF expansions can be used to characterize rational numbers with quite different outcomes. In the non-archimedean situation, though ES and CEF expansions are identical, their use to characterize rational elements depend significantly on the underlying nature of each specific field. We end this paper by providing information on the rationality characterization in three different non-archimedean fields, namely, the field of -adic numbers and the two function fields, one completed with respect to the degree valuation and the other with respect to a prime-adic valuation.
The following characterization of rational numbers by -adic ES expansions is due to Grabner and Knopfmacher [5].
Theorem 4.1. Let . Then is rational, , if and only if either the -adic ES expansion of is finite, or there exist an and an such that where .
Now for function fields, we need more terminology. Let denote a field and an irreducible polynomial of degree over . There are two types of valuation in the field of rational functions , namely, the -adic valuation , and the degree valuation defined as follows. From the unique representation in ,
set Let and be the completions of , with respect to the -adic and the degree valuations, respectively. The extension of the valuations to and is also denoted by and .
For a characterization of rational elements, we prove the following.
Theorem 4.2. The CEF of or in terminates if and only if .
Proof. Although the assertions in both fields and are the same, their respective proofs are different. In fact, when the field has finite characteristic, both results have already been shown in [6] and the proof given here is basically the same.
We use the notation of the last section with added subscripts or to distinguish their corresponding meanings.
If the CEF of in either field is finite, then is clearly rational. It remains to prove the converse and we begin with the field . Assume that . By construction, each and so can be uniquely represented in the form
where with gcd. Since and , it is of the form where are polynomials over , not all , of degree and . Thus, yielding By construction, we have Substituting (4.4) and (4.5) into (4.8) and using lead to Since gcd, it follows that , and so successively, we have which together with (4.9) yield Using (3.19) and (4.7), we consequently have This shows that for all large which implies that from some onwards, , and so , that is, the expansion terminates.
Finally for the field , assume that . Without loss of generality, assume . By the Euclidean algorithm, we have
where From the Euclidean algorithm, which is, in the terminology of Lemma 3.2, Again, from the Euclidean algorithm, which is, in the terminology of Lemma 3.2, Proceeding in the same manner, in general we have There must then exist such that , that is, . Thus, the CEF of is where , which is a terminating CEF.
#### Acknowledgments
This work was supported by the Commission on Higher Education and the Thailand Research Fund RTA5180005 and by the Centre of Excellence In Mathematics, the Commission on Higher Education.
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My Math Forum Stress strain analysis
Real Analysis Real Analysis Math Forum
October 27th, 2017, 06:18 AM #1 Member Joined: Jan 2017 From: Huddersfield Posts: 36 Thanks: 0 Stress strain analysis I have been given a question on stress and strain which I have completed using 3 methods and come to the correct answer with each method, however I must then go into detail about the advantages and disadvantages of all 3, to which I am struggling, if anybody could give a quick summary of each that would be great help (I am a numbers man and not a great writer) The 3 methods used were : Hand calculations (pen paper and calculator) Spreadsheet template (Microsoft excel) Solid modelling (solidworks analysis) What are the advantages/disadvantages of a simple model/complicated model? Speed, repeatability, flexibility and accuracy? Any help is greatly appreciated
October 27th, 2017, 07:54 AM #2 Senior Member Joined: Jun 2015 From: England Posts: 853 Thanks: 258 Well you must have experiences of your own since you have worked it out three ways. But I assume that you did not develop the spreadsheet or solid model software yourself? The spreadsheet you might have done yourself or used your college's own but I assume the solid model is a commercial package So I will confine my remarks to somethings you probably haven't thought of. Cost increases in the order 1 < 2 < 3 Probably convenience goes in the same order. But there is a trade off between effort and results. I haven't counted the effort involved in learning the software and setting it up. So if you only want to do the calculation once this trade off is quite different from if you want to perform many similar calculations. Hope this helps.
October 27th, 2017, 09:24 AM #3 Member Joined: Jan 2017 From: Huddersfield Posts: 36 Thanks: 0 Yes that was a great help No I performed the calculations myself on all 3, I just struggled to think of reasons as to disadvantages to the spreadsheet and solid modelling, learning the software is a cracking one, would never have thought of that Cheers
October 28th, 2017, 06:00 PM #4 Senior Member Joined: Sep 2016 From: USA Posts: 438 Thanks: 249 Math Focus: Dynamical systems, analytic function theory, numerics A few more comments which might be helpful. 1. Pencil and paper results are only valid for the model which is being implemented. Models are naturally a simplified representation for a system. They can do a good job of failthfully reproducing the behavior or a poor job. The pencil and paper results apply directly to the model but don't necessarily apply to the real problem. 2. Excel is dogshit and forcing people to use it should be a war crime. 3. Solidworks applies numerical methods to solve problems and thus has the opposite problem as in (1). Namely, you get approximate solutions whose accuracy may depend heavily on the problem, the numerical method, or both. In particular, finite element methods have the fantastic ability to solve differential equations even when the solutions are not smooth enough. However, this means it is possible (though unlikely in a real world application) to produce a solution which doesn't actually exist.
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# Strain in different direction
1. Oct 20, 2016
1. The problem statement, all variables and given/known data
In this question , I am not convinced that δAD = δAC cos theta
2. Relevant equations
3. The attempt at a solution
, i think it should be
δAC = δAD cos theta , am i right ?
I think so because we normally get the non-vertical and non -horizontal line and then cos the angle or sin the angle to get the length in x and y direction , right ?
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2. Oct 20, 2016
### Staff: Mentor
No. You're not right. Call L the distance between C and D. Does this distance change?
3. Oct 20, 2016
no
no , the distance doesnt chnage , so ?
4. Oct 20, 2016
### Staff: Mentor
Call x the distance from A to C. Using the Pythagorean theorem, in terms of L and x, what is the distance between A and D?
5. Oct 20, 2016
AD = sqrt ( L^2 + x^2 )
6. Oct 20, 2016
### Staff: Mentor
Good. Now, if the distance between A and C changes by dx, in terms of L, x, and dx, by how much does the distance between A and D change?
7. Oct 20, 2016
can you
can you give some hint , i have no idea
8. Oct 20, 2016
### Staff: Mentor
9. Oct 20, 2016
yes , and ?
10. Oct 20, 2016
### Staff: Mentor
Apparently, that wasn't a big enough hint. What is the derivative with respect to x of $\sqrt{L^2+x^2}$?
11. Oct 20, 2016
0.5(2x) (1/ $\ sqrt{L^2+x^2}$ )
12. Oct 20, 2016
### Staff: Mentor
What is 0.5(2x) equal to?
13. Oct 20, 2016
X
14. Oct 20, 2016
### Staff: Mentor
You really need to learn to use LaTex to display your equations. The current way you do it is totally unacceptable. Here is the link: https://www.physicsforums.com/help/latexhelp/
Use LaTex from now on.
As far as the derivative of $\sqrt{L^2+x^2}$ is concerned, you obtained:$$\frac{x}{\sqrt{L^2+x^2}}$$Is that what you meant?
15. Oct 20, 2016
Yes, so , what are you trying to say?
16. Oct 20, 2016
### Staff: Mentor
You still don't see it, huh?
Well, if dx is the change in the length of AC, then the change in length of AD is $\frac{x}{\sqrt{L^2+x^2}}dx$.
Or equivalently, if $\delta_{AC}$ is the change in length of AC, then the change in length of AD is $\frac{AC}{AD}\delta_{AC}$:
$$\delta_{AD}=\frac{AC}{AD}\delta_{AC}$$
But,$$\frac{AC}{AD}=\cos{\theta}$$
Therefore,$$\delta_{AD}=\delta_{AC}\cos{\theta}$$
17. Oct 20, 2016
Do you mean AC (del AC ) = xdx ?
18. Oct 20, 2016
ok , but , i still dun know why Ac = Ad cos theta , where theta is between AC and AD ..... because normally , we will make the 'slanted line ' to get the horizontal x -direction and vertical (y-direction ) , am i right ?
Last edited: Oct 20, 2016
19. Oct 20, 2016
one more thing, how could AD = AC cos theta , where cos theta = 4/5 ,
it's given that AD = 2 , AC = 1.6
AC cos theta = 1.6(4/5) = 1.28 which is not = AD (2m)
20. Oct 20, 2016
i dont understand this part , we know that δ = PL / AE , we also know that δAD = (4/5)δAC , so , shouldnt (4/5)δAC = (4/5)(1.6) / AE ?
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## Conversion formula
The conversion factor from decimeters to millimeters is 100, which means that 1 decimeter is equal to 100 millimeters:
1 dm = 100 mm
To convert 211.3 decimeters into millimeters we have to multiply 211.3 by the conversion factor in order to get the length amount from decimeters to millimeters. We can also form a simple proportion to calculate the result:
1 dm → 100 mm
211.3 dm → L(mm)
Solve the above proportion to obtain the length L in millimeters:
L(mm) = 211.3 dm × 100 mm
L(mm) = 21130 mm
The final result is:
211.3 dm → 21130 mm
We conclude that 211.3 decimeters is equivalent to 21130 millimeters:
211.3 decimeters = 21130 millimeters
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 millimeter is equal to 4.7326076668244E-5 × 211.3 decimeters.
Another way is saying that 211.3 decimeters is equal to 1 ÷ 4.7326076668244E-5 millimeters.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that two hundred eleven point three decimeters is approximately twenty-one thousand one hundred thirty millimeters:
211.3 dm ≅ 21130 mm
An alternative is also that one millimeter is approximately zero times two hundred eleven point three decimeters.
## Conversion table
### decimeters to millimeters chart
For quick reference purposes, below is the conversion table you can use to convert from decimeters to millimeters
decimeters (dm) millimeters (mm)
212.3 decimeters 21230 millimeters
213.3 decimeters 21330 millimeters
214.3 decimeters 21430 millimeters
215.3 decimeters 21530 millimeters
216.3 decimeters 21630 millimeters
217.3 decimeters 21730 millimeters
218.3 decimeters 21830 millimeters
219.3 decimeters 21930 millimeters
220.3 decimeters 22030 millimeters
221.3 decimeters 22130 millimeters
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# Find diagonal
Find diagonal of cuboid with length=20m width=25m height=150m
Result
d = 153.379 m
#### Solution:
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#### To solve this example are needed these knowledge from mathematics:
Pythagorean theorem is the base for the right triangle calculator. See also our trigonometric triangle calculator.
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FACTOID # 21: 15% of Army recruits from South Dakota are Native American, which is roughly the same percentage for female Army recruits in the state.
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Encyclopedia > Return on capital
Return on capital, also known as Return On Invested Capital (ROIC) is defined as
` NOPLAT / Invested Capital `
usually expressed as a percentage. In accounting, NOPLAT is an acronym for: Net Operating Profit Less Adjusted Taxes. ... A percentage is a way of expressing a proportion, a ratio or a fraction as a whole number, by using 100 as the denominator. ...
NOPLAT = Net Operating Profit Less Adjusted Tax - used to normalise effects of company's capital structure. It's the net profit with a few costs backed out, cost of interest and depreciation (accrual accounting of capital expenditures). Capital expenditures (CAPEX) are expenditures used by a company to acquire or upgrade physical assets such as equipment, property, industrial buildings. ...
When the ROIC is greater than the cost of capital (usually measured as weighted average cost of capital), the company is creating value. When it is less than the cost of capital, value is destroyed. The cost of capital is just one of many costs in a company, so a company that has a profit on its income statement must by definition be "creating value". The cost of capital for a firm is a weighted sum of the cost of equity and the cost of debt (see the financing decision). ... The weighted average cost of capital (WACC) is used in finance to measure a firms cost of capital. ...
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Return on Invested Capital - ROIC (378 words) Return on Invested Capital or ROIC calculation points out the profitability of a company’s equity and debt. ROIC is a better indicator of stock quality and strength than Return on Equity because it takes a company’s debt into account. That’s a three percentage points increase in total return on investment coming from Company B. Remember when you own equity in a firm, you are also responsible for any incurred debt.
Value Investing Encyclopedia: Return on Capital (466 words) In most cases, return on capital (however computed) is a good indicator of the relative profitability of various firms. Return on capital is also known as return on invested capital, and abbreviated as either ROC or ROIC. The fact that a source uses the term “return on capital” (or ROC) rather than “return on invested capital” (or ROIC) is not a good indication that cash assets are being counted as capital.
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Fraction as Percentage - Students Explore
# Fraction as Percentage
Fraction can be convert into percentage with various methods, but in this post we will show two methods to convert fraction as percentage. First we will discuss the easy method, and we recommend you to follow first method because it is easy as compared to second method which we will show you in this post.
You May Also Like: How to Convert Percentage into Fraction
## How to Convert Fraction into Percentage (Method 1)
Basically there are two steps in this method to convert fraction into percentage, as follow:
1. First divide the fraction to a decimal.
2. Then in second step multiply the decimal by 100 and put symbol of percentage (%) to get percent value.
It seems so easy, but still students get confused. It is because they often do not know how to convert fraction into decimal.
## How to Convert Fraction into Decimal
Conversion of fraction into decimal is so easy for example:
1/5 = 1 ÷ 5 = 0.2
Here we had a fraction of 1/5, we converted it into decimal as 0.2 by dividing 1 by 5. Now move to the second step, that is multiplication of decimal with 100. Lets see.
0.2 × 100 = 20%.
You May Also Like: How to Convert Percentage into Fraction
## Fraction as Percentage
Look at another example, Convert 6/12 into percentage.
Solution:
Step 1: divide 6 by 12 as, 6÷12 = 1/2 = 0.5 ( here we get decimal of fraction)
Step 2: multiply the decimal by 100. 0.5 x 100 = 50%. ( 50% is the answer of 6\12).
Example 2: Convert 7/4 into percentage
Solution:
Step 1: divide 7 by 4 = 7\4 = 1.75
Step 2: Multiply 1.75 by 100 = 175%.
Therefore, the fraction 74 is equivalent to 175%.
## Conversion of Fraction as Percentage (Method 2)
In this method we first check what is the value at the denominator, if it is 100 then ok, if it is not 100 then we must have to be convert it into 100 by multiplying the number which exact divide the 100. Then we also multiply that number with the numerator, more you may understand by following example:
### Convert 3/5 into percentage?
Solution:
Here we can see the there is no 100 at the denominator but it is 5 at the denominator, so we have to convert it into 100.
Now if you we multiply 5 by 20 i.e. 5×20 = 100, it will become 100, we must also have to multiply 20 by numerator i.e. 3×20 = 60.
3/5 = 3×20\5×20 = 60\100 = 60%.
I hope you understand this method, if you still have any query then please do ask in the comment section given below this post. I have some homework for you to solve this at comment section too and also share this post with your friends, classmates and colleagues.
You May Also Like: How to Convert Percentage into Fraction
• 3\6
• 1\100
• 7\8
• 8\7
### 2 Responses
1. March 23, 2022
[…] How to convert Fraction into Percentage […]
2. March 26, 2022
[…] You May Also Like: What is Percentage and Why we Use it in Maths What is Concept of Average and it Uses Is Median the Average ? Hence Proved – Maths Fraction as Percentage […]
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# Suppose H () = (Sx-4)4 Find two functions f and g such that (fog) () = H () Neither function
###### Question:
Suppose H () = (Sx-4)4 Find two functions f and g such that (fog) () = H () Neither function can be the identity function. (There may be more than one correct answer) 0 g ()
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## Differential and Integral Equations
### Existence of global solutions to the Cauchy problem for some reaction-diffusion system
Munemitsu Hirose
#### Abstract
We consider the Cauchy problem for the following reaction-diffusion system: $$\begin{cases} \displaystyle \frac{\partial u_i}{\partial t} = \Delta u_i +g_i(x,t) \prod_{j=1}^m {u_j}^{p_{ij}}, & x \in {\bf R}^n, \ t > 0, \ i=1,2,\cdots,m, \\ u_i(x,0)=f_i(x) \geq 0, \ \not\equiv 0, & x \in {\bf R}^n, \ i=1,2,\cdots,m, \end{cases}$$ where $n \geq 3$, $m \geq 2$, $p_{ij} \geq 0$ $( 1 \leq i, j \leq m ),$ $\prod_{j=1}^m {u_j}^{p_{ij}} = {u_1}^{p_{i1}} {u_2}^{p_{i2}} \cdots$ ${u_m}^{p_{im}} ,$ $(i=1,2,\cdots,m)$ and $f_i(x)$ ($i=1,2,\cdots,m$) is a non-negative, bounded and continuous function in ${\bf R}^n$. In this paper, we show the existence of non-negative and global solutions $u_i(x,t)$ ($i=1,2,\cdots,m$) for the above Cauchy problem when $g_i(x,t)$ ($i=1,2,\cdots,m$) and $p_{ij} \geq 0$ ($1 \leq i, j \leq m$) satisfy some conditions.
#### Article information
Source
Differential Integral Equations, Volume 23, Number 7/8 (2010), 671-684.
Dates
First available in Project Euclid: 20 December 2012
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# Matrix equation notation
I have been reading through a paper and saw a notation that i did not know about. It says that we have to solve the matrix equation $\Delta \mathrm{h}'=\mathrm{Lh}'$. That it clear to me, but what does the following mean: Image of equation.
I was not able to find any information on what that means. Can anybody tell man or atleast give me some keywords to help me.
• Depends on the context. Need more information, e.g., which paper? Commented Oct 26, 2016 at 20:31
I will start with the left-most matrix, as I think this is what you will be struggling the most. This defines a $(k-b+1)\times k$ matrix. It means that the first $k$ rows and $k$ columns of the matrix consist of the $k\times k$ matrix $L$. Then, below this, we have an identity matrix in the bottom right corner, of size $(k-b+1)\times (k-b+1)$ Then, the gaps on the left hand side are filled by zeros.
This may be better explained by example. Let's take $k=5$ and $b=3$ and let $$L=\begin{pmatrix} 1&3&2&1&8\\ 9&7&0&1&3\\ 2&3&1&1&4\\ 7&0&1&3&10\\ 8&1&9&3&2 \end{pmatrix}.$$
Then the matrix described by your picture is
$$\begin{pmatrix} 1&3&2&1&8\\ 9&7&0&1&3\\ 2&3&1&1&4\\ 7&0&1&3&10\\ 8&1&9&3&2\\ 0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1 \end{pmatrix}$$
If, instead, we let $b=2$ then we would obtain the following matrix.
$$\begin{pmatrix} 1&3&2&1&8\\ 9&7&0&1&3\\ 2&3&1&1&4\\ 7&0&1&3&10\\ 8&1&9&3&2\\ 0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1 \end{pmatrix}$$
Then, according to your picture, we are multiplying this matrix by another matrix $h'$ and obtaining another matrix consisting of $(\Delta h)'$ at the top and 0's underneath (to make up the required dimension of the matrix).
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StudentShare solutions
# Finite element analysis2 - Essay Example
## Extract of sample Finite element analysis2
It can be used to analyze either small or large-scale deflection under loading or applied displacement. It and can analyze determine elastic deformation, or "permanently bent out of shape"permanent plastic deformation. The Widespread use of personal and mainframe computers in the second half of the 20th century catalyzed applications involving FEA is required because ofdue to the astronomical number of calculations needed to analyze analyse a large structure. Today, FEA is an integral part of nearly all design and development projects.
As mentioned, thisThis report has been complied compiled to document the results of of tensile testing on a buckle connection simulated by the Finite finite Element element Analysis analysisof a buckle connection. The function objective of the FE Analysisanalysis was to obtain determine the maximum stress max of the buckle connection, which is based must be designed to resist withstand a 450 Kg kg of tensile loadon, as written in assignment.
It is also important to note that although the method of finite elements is a powerful analysis tool that is widely applied, it is still simply an approximate numerical estimate. In other words, although results accurate enough for engineering purposes are obtainable via precise modelling and adequate numbers of elements, exact solutions are highly unlikely.The buckle connection was designed to be made from Stainless Steel plate. A comprehensive set of properties for the specified material is shown in table 1
Material properties of Stainless Steel
The next section discusses the assumptions made in preparing for the analysis; then the modelling is described. The subsequent sections include the finite element analysis calculations, a discussion and then conclusions and recommendations are provided.
ASSUMPTIONSAssumptions:
As in any engineering problem, the first step is to develop a list of assumptions which will be used as guidelines to help bound the physical problem. Assumptions necessary for finite element analysis are divided into four categories: geometry, properties, mesh, and boundary conditions.
In the begging of doing this report in FEA Package there are numbers of assumptions were made which give create more imaginations as key in testing any design before test it in real life. First, in terms of the geometry, it is assumed that the drawings provided are accurate representations of the buckle and also that the determination of pass/fail judgement for the part is based simply on the tensile test of the flat plate of the buckle and that no other variations in the standard belt assembly come into play, e.g. length of the belt, etc. Shell elements are utilized and it is assumed that they can adequately model the part. Thickness of the buckle was assumed to be constant across the part and the element type chosen was selected in part because its input data allows incorporation of thickness values. ly
Material properties were the next consideration whereby it was assumed that all material properties supplied provided by the designerin the problem statement were correct accurate and provided a true representation of the material specifiedto be analysed. . Ambient temperature and pressure are assumed to have no effect on the analysis results. Another basic assumption made was that the material was homogeneous and isotropic and had no discontinuities or residual ...Show more
## Summary
The Finite finite Element element Analysis method was first introduced in the 1950's and has been continually developed ever sincesince then for solving complicated mathematical problems. Its applications range from stress analysis and heat conduction to fluid flow, lubrication, electric and magnetic fields…
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Natural Number Multiplication is Commutative/Euclid's Proof
Theorem
In the words of Euclid:
If two numbers by multiplying one another make certain numbers, the numbers so produced will be equal to one another.
Proof
Let $A, B$ be two (natural) numbers, and let $A$ by multiplying $B$ make $C$, and $B$ by multiplying $A$ make $D$.
We need to show that $C = D$.
We have that $A \times B = C$.
So $B$ measures $C$ according to the units of $A$.
But the unit $E$ also measures $A$ according to the units in it.
So $E$ measures $A$ the same number of times that $B$ measures $C$.
Therefore from Proposition $15$ of Book $\text{VII}$: Alternate Ratios of Multiples‎ $E$ measures $B$ the same number of times that $A$ measures $C$.
We also have that $A$ measures $D$ according to the units of $B$ since $B \times A = D$.
But the unit $E$ also measures $B$ according to the units in it.
Therefore $E$ measures $B$ the same number of times that $A$ measures $D$.
But we also have that $E$ measures $B$ the same number of times that $A$ measures $C$.
So $A$ measures $C$ and $D$ the same number of times.
Therefore $C = D$.
$\blacksquare$
Historical Note
This proof is Proposition $16$ of Book $\text{VII}$ of Euclid's The Elements.
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## Glue Factory Stakes
Logic Grid puzzles come with a handy interactive grid that will help you solve the puzzle based on the given clues.
Puzzle ID: #33598 Fun: (2.86) Difficulty: (3.26) Category: Logic-Grid Submitted By: cnmne
Five horses have just competed in the annual "Glue Factory Stakes". With the given clues, determine for each horse: its color, its number, its jockey, and where it finished.
Finish: Win, Place, Show, 4th, 5th
Horses: Ghoulish Pleasure, Infirm, Roughage, Seattle Slow, Secret Harriet
Colors: Black, Red, Yellow, Blue, White
Numbers: #18, #29, #37, #46, #50
Jockeys: Allen, Jerry, Luke, Nathan, Tim
1) The 5 horses are: Infirm, the horse ridden by Tim, the horse that wore black, #50, and the horse that Placed.
2) The 5 jockeys are: Allen, the one that rode Seattle Slow, the one that rode the horse that wore red, the one that rode #18, and the one that rode the horse that Showed.
3) Secret Harriet finished ahead of the horse that wore blue, which finished ahead of the horse ridden by Luke.
4) #50 finished ahead of the horse that wore white, but finished behind Roughage.
5) Jerry finished next to #37.
6) Allen finished next to #29.
7) The horse that wore blue finished two positions away from #50.
8) The horse that wore yellow finished two positions away from #46.
9) Luke finished at least two positions away from #46.
10) Nathan finished at least two positions away from #50.
11) Luke rode a horse that wore an even number.
12) Only one of the horses that wore a primary color also wore an even number.
## What Next?
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# Gluing the ends of a cylinder. Can we get other than a torus?
Let $X=S^1 \times I$ be a cylinder, where $S^1$ is the 1-dimensional circle. If we glue the "bottom" boundary $S^1 \times 0$ and the "top" boundary $S^1\times 1$ by a homeomorphism sending $x\times 0$ to $x\times 1$, the resulting manifold is homeomorphic to a torus.
If we chose another homeomorphism to glue boundaries, can we get a manifold that is not homeomorphic to a torus? (I only consider oriented manifolds so Klein bottle are omitted.)
Or no matter what homeomorphism we choose, is the resulting manifold is homeomorphic to a torus?
Consider the orientations on the circles $S^1 \times 0$ and $S^1 \times 1$ coming from a choice of orientation on $S^1$. If the gluing map $f : S^1 \times 1 \to S^1 \times 0$ preserves orientation, the quotient is homeomorphic to a torus. If not, it is homeomorphic to a Klein bottle. The proof requires one to know that two self-homeomorphisms of $S^1$ are isotopic if they both preserve orientation or if they both reverse orientation. The homeomorphism between the quotient spaces of $S^1 \times [0,1]$ may then be constructed by "absorbing" the isotopy into the product structure on $S^1 \times [0,1]$.
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# A mixture of gases contains 319 torr N2, 273 torr O2, and 285 torr Ar. What is the total pressure of the mixture?
Mar 28, 2015
The total pressure of the mixture will be $\text{877 torr}$.
When dealing with mixtures of gases, you must keep in mind that the total pressure of the mixture can be broken down as a sum of the partial pressures of all the individual gases that make up that mixture - this is known as Dalton's law.
In other words, each gas that's part of the mixture adds its partial pressure to the total pressure exercited in the container.
Imagine that you start with an empty flask. You put the nitrogen in first and record its pressure - 319 torr. Then you get it out, put in the oxygen, and record its pressure - 273 torr.
Finally, you get the oxygen out, put in the argon, and record its pressure - 285 torr.
Now, it only makes sense that when you put all three gases together in the same flask, the total pressure will be the sum of their individual pressures. When in the mixture, these individual pressures will become their partial pressures.
Therefore,
${P}_{\text{total") = P_("nitrogen") + P_("oxygen") + P_("argon}}$
${P}_{\text{total") = "319 torr" + "273 torr" + "285 torr" = color(red)("877 torr}}$
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# loop over matrix: subscript out of bounds
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## loop over matrix: subscript out of bounds
I have a basic for loop with a simple matrix. The code is doing what it is supposed to do, but I'm still wondering the error "subscript out of bounds". What would be a smoother way to code such a basic for loop? myMatrix <- matrix(0,5,12) for(i in 1:nrow(myMatrix)) { for(i in 1:ncol(myMatrix)) { myMatrix[i,i] = -1 myMatrix[i,i+1] = 1 }} print(myMatrix) Thanks in advance! [[alternative HTML version deleted]] ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: loop over matrix: subscript out of bounds
Both loops are on 'i', which is a bad idea. :-) Also myMatrix[i,i+1] will be out-of-bounds if i = ncol(myMatrix) On Mon, Aug 6, 2018 at 12:02 PM, Maija Sirkjärvi <[hidden email]> wrote: > I have a basic for loop with a simple matrix. The code is doing what it is > supposed to do, but I'm still wondering the error "subscript out of > bounds". What would be a smoother way to code such a basic for loop? > > myMatrix <- matrix(0,5,12) > for(i in 1:nrow(myMatrix)) { > for(i in 1:ncol(myMatrix)) { > myMatrix[i,i] = -1 > myMatrix[i,i+1] = 1 > }} > print(myMatrix) > > Thanks in advance! > > [[alternative HTML version deleted]] > > ______________________________________________ > [hidden email] mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help> PLEASE do read the posting guide http://www.R-project.org/> posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > [[alternative HTML version deleted]] ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: loop over matrix: subscript out of bounds
Hello, Eric is right but... You have two assignments. The second sets a value that will be overwritten is the next iteration by myMatrix[i,i] = -1 when 'i' becomes the next value. If you fix the second index and use 'j', you might as well do myMatrix[] = -1 myMatrix[, ncol(myMatrix)] = 1 Hope this helps, Rui Barradas Às 10:24 de 06/08/2018, Eric Berger escreveu: > Both loops are on 'i', which is a bad idea. :-) > Also myMatrix[i,i+1] will be out-of-bounds if i = ncol(myMatrix) > > > On Mon, Aug 6, 2018 at 12:02 PM, Maija Sirkjärvi <[hidden email]> > wrote: > >> I have a basic for loop with a simple matrix. The code is doing what it is >> supposed to do, but I'm still wondering the error "subscript out of >> bounds". What would be a smoother way to code such a basic for loop? >> >> myMatrix <- matrix(0,5,12) >> for(i in 1:nrow(myMatrix)) { >> for(i in 1:ncol(myMatrix)) { >> myMatrix[i,i] = -1 >> myMatrix[i,i+1] = 1 >> }} >> print(myMatrix) >> >> Thanks in advance! >> >> [[alternative HTML version deleted]] >> >> ______________________________________________ >> [hidden email] mailing list -- To UNSUBSCRIBE and more, see >> https://stat.ethz.ch/mailman/listinfo/r-help>> PLEASE do read the posting guide http://www.R-project.org/>> posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> > > [[alternative HTML version deleted]] > > ______________________________________________ > [hidden email] mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html> and provide commented, minimal, self-contained, reproducible code. > ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: loop over matrix: subscript out of bounds
In reply to this post by Maija Sirkjärvi Quoting Maija Sirkjärvi <[hidden email]>: > I have a basic for loop with a simple matrix. The code is doing what it is > supposed to do, but I'm still wondering the error "subscript out of > bounds". What would be a smoother way to code such a basic for loop? > > myMatrix <- matrix(0,5,12) > for(i in 1:nrow(myMatrix)) { > for(i in 1:ncol(myMatrix)) { > myMatrix[i,i] = -1 > myMatrix[i,i+1] = 1 > }} > print(myMatrix) > > Thanks in advance! > Perhaps you do not need loops at all? myMatrix <- matrix(0, 5, 12) diag(myMatrix) <- -1 diag(myMatrix[, -1]) <- 1 -- Enrico Schumann Lucerne, Switzerland http://enricoschumann.net______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: loop over matrix: subscript out of bounds
In reply to this post by Rui Barradas Thanks for help! However, changing the index from i to j for the column vector changes the output. I would like the matrix to be the following: -1 1 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 -1 1 0 0 0 ..... etc. How to code it? Best, Maija >> myMatrix <- matrix(0,5,12) >> for(i in 1:nrow(myMatrix)) { >> for(i in 1:ncol(myMatrix)) { >> myMatrix[i,i] = -1 >> myMatrix[i,i+1] = 1 >> }} >> print(myMatrix) ma 6. elok. 2018 klo 13.58 Rui Barradas ([hidden email]) kirjoitti: > Hello, > > Eric is right but... > > You have two assignments. The second sets a value that will be > overwritten is the next iteration by myMatrix[i,i] = -1 when 'i' becomes > the next value. > > If you fix the second index and use 'j', you might as well do > > myMatrix[] = -1 > myMatrix[, ncol(myMatrix)] = 1 > > Hope this helps, > > Rui Barradas > > Às 10:24 de 06/08/2018, Eric Berger escreveu: > > Both loops are on 'i', which is a bad idea. :-) > > Also myMatrix[i,i+1] will be out-of-bounds if i = ncol(myMatrix) > > > > > > On Mon, Aug 6, 2018 at 12:02 PM, Maija Sirkjärvi < > [hidden email]> > > wrote: > > > >> I have a basic for loop with a simple matrix. The code is doing what it > is > >> supposed to do, but I'm still wondering the error "subscript out of > >> bounds". What would be a smoother way to code such a basic for loop? > >> > >> myMatrix <- matrix(0,5,12) > >> for(i in 1:nrow(myMatrix)) { > >> for(i in 1:ncol(myMatrix)) { > >> myMatrix[i,i] = -1 > >> myMatrix[i,i+1] = 1 > >> }} > >> print(myMatrix) > >> > >> Thanks in advance! > >> > >> [[alternative HTML version deleted]] > >> > >> ______________________________________________ > >> [hidden email] mailing list -- To UNSUBSCRIBE and more, see > >> https://stat.ethz.ch/mailman/listinfo/r-help> >> PLEASE do read the posting guide http://www.R-project.org/> >> posting-guide.html > >> and provide commented, minimal, self-contained, reproducible code. > >> > > > > [[alternative HTML version deleted]] > > > > ______________________________________________ > > [hidden email] mailing list -- To UNSUBSCRIBE and more, see > > https://stat.ethz.ch/mailman/listinfo/r-help> > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html> > and provide commented, minimal, self-contained, reproducible code. > > > [[alternative HTML version deleted]] ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: loop over matrix: subscript out of bounds
> Thanks for help! > However, changing the index from i to j for the column vector changes the > output. I would like the matrix to be the following: > -1 1 0 0 0 0 0 > 0 -1 1 0 0 0 0 > 0 0 -1 1 0 0 0 > ..... > etc. > How to code it? as Enrico Schumann showed you: Without any loop, a very nice R-ish way (see his message)! Martin > Best, > Maija ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: loop over matrix: subscript out of bounds
Thanks, but I didn't quite get it. And I don't get it running as it should. ti 7. elok. 2018 klo 10.47 Martin Maechler ([hidden email]) kirjoitti: > > > Thanks for help! > > However, changing the index from i to j for the column vector changes the > > output. I would like the matrix to be the following: > > > -1 1 0 0 0 0 0 > > 0 -1 1 0 0 0 0 > > 0 0 -1 1 0 0 0 > > ..... > > etc. > > > How to code it? > > as Enrico Schumann showed you: Without any loop, a very nice > R-ish way (see his message)! > > Martin > > > Best, > > Maija > > [[alternative HTML version deleted]] ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: loop over matrix: subscript out of bounds
Hello, If it is not running as you want it, you should say what went wrong. Post the code that you have tried and the expected output, please. (In fact, the lack of expected output was the reason why my suggestion was completely off target.) Rui Barradas On 07/08/2018 09:20, Maija Sirkjärvi wrote: > Thanks, but I didn't quite get it. And I don't get it running as it should. > > ti 7. elok. 2018 klo 10.47 Martin Maechler ([hidden email] > ) kirjoitti: > > > > Thanks for help! > > However, changing the index from i to j for the column vector > changes the > > output. I would like the matrix to be the following: > > > -1 1 0 0 0 0 0 > > 0 -1 1 0 0 0 0 > > 0 0 -1 1 0 0 0 > > ..... > > etc. > > > How to code it? > > as Enrico Schumann showed you: Without any loop, a very nice > R-ish way (see his message)! > > Martin > > > Best, > > Maija > ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: loop over matrix: subscript out of bounds
Thanks! If I do it like this: myMatrix <- matrix(0,5,5*2-3) print(myMatrix) for(i in 2:nrow(myMatrix)) for(j in 2:ncol(myMatrix)) myMatrix[i-1,j-1] = -1 myMatrix[i-1,j] = 1 print(myMatrix) I get the following result: [,1] [,2] [,3] [,4] [,5] [,6] [,7] [1,] -1 -1 -1 -1 -1 -1 0 [2,] -1 -1 -1 -1 -1 -1 0 [3,] -1 -1 -1 -1 -1 -1 0 [4,] -1 -1 -1 -1 -1 -1 1 [5,] 0 0 0 0 0 0 0 However. The result that I would need to get would be like this: [,1] [,2] [,3] [,4] [,5] [,6] [,7] [1,] -1 1 0 0 0 0 0 [2,] 0 -1 1 0 0 0 0 [3,] 0 0 -1 1 0 0 0 [4,] 0 0 0 -1 1 0 0 [5,] 0 0 0 0 -1 1 0 I'd rather not create symmetric matrices as I would really like to learn how to do this thing "the hard way" as I find matrix iteration to be quite a basic procedure in everything I'm trying to do. Thanks again! Maija ti 7. elok. 2018 klo 17.37 Rui Barradas ([hidden email]) kirjoitti: > Hello, > > If it is not running as you want it, you should say what went wrong. > Post the code that you have tried and the expected output, please. > (In fact, the lack of expected output was the reason why my suggestion > was completely off target.) > > Rui Barradas > > On 07/08/2018 09:20, Maija Sirkjärvi wrote: > > Thanks, but I didn't quite get it. And I don't get it running as it > should. > > > > ti 7. elok. 2018 klo 10.47 Martin Maechler ([hidden email] > > ) kirjoitti: > > > > > > > Thanks for help! > > > However, changing the index from i to j for the column vector > > changes the > > > output. I would like the matrix to be the following: > > > > > -1 1 0 0 0 0 0 > > > 0 -1 1 0 0 0 0 > > > 0 0 -1 1 0 0 0 > > > ..... > > > etc. > > > > > How to code it? > > > > as Enrico Schumann showed you: Without any loop, a very nice > > R-ish way (see his message)! > > > > Martin > > > > > Best, > > > Maija > > > [[alternative HTML version deleted]] ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: loop over matrix: subscript out of bounds
You only need one "for loop" for(i in 2:nrow(myMatrix)) { myMatrix[i-1,i-1] = -1 myMatrix[i-1,i] = 1 } HTH, Eric On Wed, Aug 8, 2018 at 12:40 PM, Maija Sirkjärvi <[hidden email]> wrote: > Thanks! > > If I do it like this: > > myMatrix <- matrix(0,5,5*2-3) > print(myMatrix) > for(i in 2:nrow(myMatrix)) > for(j in 2:ncol(myMatrix)) > myMatrix[i-1,j-1] = -1 > myMatrix[i-1,j] = 1 > print(myMatrix) > > I get the following result: > > [,1] [,2] [,3] [,4] [,5] [,6] [,7] > [1,] -1 -1 -1 -1 -1 -1 0 > [2,] -1 -1 -1 -1 -1 -1 0 > [3,] -1 -1 -1 -1 -1 -1 0 > [4,] -1 -1 -1 -1 -1 -1 1 > [5,] 0 0 0 0 0 0 0 > > However. The result that I would need to get would be like this: > > [,1] [,2] [,3] [,4] [,5] [,6] [,7] > [1,] -1 1 0 0 0 0 0 > [2,] 0 -1 1 0 0 0 0 > [3,] 0 0 -1 1 0 0 0 > [4,] 0 0 0 -1 1 0 0 > [5,] 0 0 0 0 -1 1 0 > > I'd rather not create symmetric matrices as I would really like to learn > how to do this thing "the hard way" as I find matrix iteration to be quite > a basic procedure in everything I'm trying to do. > > Thanks again! > Maija > > > > > ti 7. elok. 2018 klo 17.37 Rui Barradas ([hidden email]) kirjoitti: > > > Hello, > > > > If it is not running as you want it, you should say what went wrong. > > Post the code that you have tried and the expected output, please. > > (In fact, the lack of expected output was the reason why my suggestion > > was completely off target.) > > > > Rui Barradas > > > > On 07/08/2018 09:20, Maija Sirkjärvi wrote: > > > Thanks, but I didn't quite get it. And I don't get it running as it > > should. > > > > > > ti 7. elok. 2018 klo 10.47 Martin Maechler ([hidden email] > > > ) kirjoitti: > > > > > > > > > > Thanks for help! > > > > However, changing the index from i to j for the column vector > > > changes the > > > > output. I would like the matrix to be the following: > > > > > > > -1 1 0 0 0 0 0 > > > > 0 -1 1 0 0 0 0 > > > > 0 0 -1 1 0 0 0 > > > > ..... > > > > etc. > > > > > > > How to code it? > > > > > > as Enrico Schumann showed you: Without any loop, a very nice > > > R-ish way (see his message)! > > > > > > Martin > > > > > > > Best, > > > > Maija > > > > > > > [[alternative HTML version deleted]] > > ______________________________________________ > [hidden email] mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help> PLEASE do read the posting guide http://www.R-project.org/> posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > [[alternative HTML version deleted]] ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: loop over matrix: subscript out of bounds
Thanks a lot ! That's it! Maija ke 8. elok. 2018 klo 12.53 Eric Berger ([hidden email]) kirjoitti: > You only need one "for loop" > > for(i in 2:nrow(myMatrix)) { > myMatrix[i-1,i-1] = -1 > myMatrix[i-1,i] = 1 > } > > HTH, > Eric > > > On Wed, Aug 8, 2018 at 12:40 PM, Maija Sirkjärvi < > [hidden email]> wrote: > >> Thanks! >> >> If I do it like this: >> >> myMatrix <- matrix(0,5,5*2-3) >> print(myMatrix) >> for(i in 2:nrow(myMatrix)) >> for(j in 2:ncol(myMatrix)) >> myMatrix[i-1,j-1] = -1 >> myMatrix[i-1,j] = 1 >> print(myMatrix) >> >> I get the following result: >> >> [,1] [,2] [,3] [,4] [,5] [,6] [,7] >> [1,] -1 -1 -1 -1 -1 -1 0 >> [2,] -1 -1 -1 -1 -1 -1 0 >> [3,] -1 -1 -1 -1 -1 -1 0 >> [4,] -1 -1 -1 -1 -1 -1 1 >> [5,] 0 0 0 0 0 0 0 >> >> However. The result that I would need to get would be like this: >> >> [,1] [,2] [,3] [,4] [,5] [,6] [,7] >> [1,] -1 1 0 0 0 0 0 >> [2,] 0 -1 1 0 0 0 0 >> [3,] 0 0 -1 1 0 0 0 >> [4,] 0 0 0 -1 1 0 0 >> [5,] 0 0 0 0 -1 1 0 >> >> I'd rather not create symmetric matrices as I would really like to learn >> how to do this thing "the hard way" as I find matrix iteration to be quite >> a basic procedure in everything I'm trying to do. >> >> Thanks again! >> Maija >> >> >> >> >> ti 7. elok. 2018 klo 17.37 Rui Barradas ([hidden email]) kirjoitti: >> >> > Hello, >> > >> > If it is not running as you want it, you should say what went wrong. >> > Post the code that you have tried and the expected output, please. >> > (In fact, the lack of expected output was the reason why my suggestion >> > was completely off target.) >> > >> > Rui Barradas >> > >> > On 07/08/2018 09:20, Maija Sirkjärvi wrote: >> > > Thanks, but I didn't quite get it. And I don't get it running as it >> > should. >> > > >> > > ti 7. elok. 2018 klo 10.47 Martin Maechler ( >> [hidden email] >> > > ) kirjoitti: >> > > >> > > >> > > > Thanks for help! >> > > > However, changing the index from i to j for the column vector >> > > changes the >> > > > output. I would like the matrix to be the following: >> > > >> > > > -1 1 0 0 0 0 0 >> > > > 0 -1 1 0 0 0 0 >> > > > 0 0 -1 1 0 0 0 >> > > > ..... >> > > > etc. >> > > >> > > > How to code it? >> > > >> > > as Enrico Schumann showed you: Without any loop, a very nice >> > > R-ish way (see his message)! >> > > >> > > Martin >> > > >> > > > Best, >> > > > Maija >> > > >> > >> >> [[alternative HTML version deleted]] >> >> ______________________________________________ >> [hidden email] mailing list -- To UNSUBSCRIBE and more, see >> https://stat.ethz.ch/mailman/listinfo/r-help>> PLEASE do read the posting guide >> http://www.R-project.org/posting-guide.html>> and provide commented, minimal, self-contained, reproducible code. >> > > [[alternative HTML version deleted]] ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: loop over matrix: subscript out of bounds
In reply to this post by Eric Berger >>>>> Eric Berger on Wed, 8 Aug 2018 12:53:32 +0300 writes: > You only need one "for loop" > for(i in 2:nrow(myMatrix)) { > myMatrix[i-1,i-1] = -1 > myMatrix[i-1,i] = 1 > } > > HTH, > Eric and why are you not using Enrico Schumann's even nicer solution (from August 6) that I had mentioned too ? Here's the link to it in the (official) R-help archives: https://stat.ethz.ch/pipermail/r-help/2018-August/455673.htmlMaija said > Thanks, but I didn't quite get it. And I don't get it running as it should. and actually she is right that that version does not work for all dimensions of 'myMatrix' -- it does need ncol(.) >= 3 but neither does the above solution -- it only works for nrow(.) >= 2 Here's a function version of Enrico's that does work in all cases(!) without a for loop -- including examples (as comments) mkMat <- function(n=5, m=7) { M <- matrix(0, n,m) diag(M) <- -1 ## this fails when m == ncol(M) <= 2, and ', drop=FALSE' does *not* help : ## diag(M[, -1]) <- 1 ## diag(M[, -1, drop=FALSE]) <- 1 ## This *does* work: M[col(M) - row(M) == 1L] <- 1 M } mkMat() ## [,1] [,2] [,3] [,4] [,5] [,6] [,7] ## [1,] -1 1 0 0 0 0 0 ## [2,] 0 -1 1 0 0 0 0 ## [3,] 0 0 -1 1 0 0 0 ## [4,] 0 0 0 -1 1 0 0 ## [5,] 0 0 0 0 -1 1 0 mkMat(3,5) ## [,1] [,2] [,3] [,4] [,5] ## [1,] -1 1 0 0 0 ## [2,] 0 -1 1 0 0 ## [3,] 0 0 -1 1 0 mkMat(5,3) ## [,1] [,2] [,3] ## [1,] -1 1 0 ## [2,] 0 -1 1 ## [3,] 0 0 -1 ## [4,] 0 0 0 ## [5,] 0 0 0 ## Show that all small (m,n) work: for(m in 0:3) for(n in 0:3) { cat(sprintf("(%d,%d):\n", n,m)); print(mkMat(n,m)) } ## (output not shown here) > > On Wed, Aug 8, 2018 at 12:40 PM, Maija Sirkjärvi <[hidden email]> > wrote: > > > [.............] > > [.............] > > However. The result that I would need to get would be like this: > > > > [,1] [,2] [,3] [,4] [,5] [,6] [,7] > > [1,] -1 1 0 0 0 0 0 > > [2,] 0 -1 1 0 0 0 0 > > [3,] 0 0 -1 1 0 0 0 > > [4,] 0 0 0 -1 1 0 0 > > [5,] 0 0 0 0 -1 1 0 ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: loop over matrix: subscript out of bounds
> >>>>> Eric Berger on Wed, 8 Aug 2018 12:53:32 +0300 writes: > > > You only need one "for loop" > > for(i in 2:nrow(myMatrix)) { > > myMatrix[i-1,i-1] = -1 > > myMatrix[i-1,i] = 1 > > } Or none, with matrix-based array indexing and explicit control of the indices to prevent overrun in : mkMat <- function(n=5, m=7) { M <- matrix(0, n,m) i <- 1:min(n,m) j <- i[i
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## Re: loop over matrix: subscript out of bounds
Hello, There are now three solutions to the OP's problem. I have timed them and the results depend on the matrix size. The solution I thought would be better, Enrico's diag(), is in fact the slowest. As for the other two, Eric's for loop is 50% fastest than the matrix index for small matrices but its relative performance degrades as the matrix becomes bigger and bigger. library(microbenchmark) #Enrico Schumann mkMat_diag <- function(nr = 5, nc = 7) { M <- matrix(0, nr, nc) diag(M) <- -1 diag(M[, -1]) <- 1 M } #Eric Berger mkMat_loop <- function(nr = 5, nc = 7) { M <- matrix(0, nr, nc) for(i in 2:nrow(M)) { M[i - 1, i - 1] <- -1 M[i - 1, i] <- 1 } M } #S.Ellison mkMat_index <- function(nr = 5, nc = 7) { M <- matrix(0, nr, nc) i <- 1:min(nr, nc) j <- i[i < nc] M[ cbind(i, i) ] <- -1 M[ cbind(j, j + 1) ] <- 1 M } microbenchmark( loop = mkMat_loop(), index = mkMat_index(), diag = mkMat_diag(), times = 1e3 ) microbenchmark( loop = mkMat_loop(50, 70), index = mkMat_index(50, 70), diag = mkMat_diag(50, 70) ) microbenchmark( loop = mkMat_loop(500, 700), index = mkMat_index(500, 700), diag = mkMat_diag(500, 700) ) Hope this helps, Rui Barradas On 08/08/2018 12:59, S Ellison wrote: > >>>>>>> Eric Berger on Wed, 8 Aug 2018 12:53:32 +0300 writes: >> >>> You only need one "for loop" >>> for(i in 2:nrow(myMatrix)) { >>> myMatrix[i-1,i-1] = -1 >>> myMatrix[i-1,i] = 1 >>> } > > Or none, with matrix-based array indexing and explicit control of the indices to prevent overrun in : > > mkMat <- function(n=5, m=7) { > M <- matrix(0, n,m) > i <- 1:min(n,m) > j <- i[i M[ cbind(i,i) ] <- -1 > M[ cbind(j, j+1) ] <- 1 > M > } > > > > > ******************************************************************* > This email and any attachments are confidential. Any use...{{dropped:8}} > > ______________________________________________ > [hidden email] mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html> and provide commented, minimal, self-contained, reproducible code. > ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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# Activities On Completing Flow Chart Solving Quadratic Equations By The Square Flowchart
By Liam Simcha at October 11 2018 16:43:18
Here are the types of evidence you can use: Images ; A photo is often a great way to show a point. You can use a photo in three ways: * Literally: If you're talking about a piece of equipment, show a photo of it rather than describe its specifications in bulleted text. You can use callouts that point to the various features and label them. * Metaphorically: Sometimes a point you're making is a concept, rather than a fact. For example, you may be talking about tough times ahead, so you could show a photo of a rocky road or a steep staircase. * Schematically: If you're talking about a process, you can show it with a diagram or add arrows to point out parts of a photo.
The key to process improvement is to clearly communicate process definitions (the way in which the company wants the processes to be carried out) to the people in charge of their execution (through training, process descriptions publication, etc...). The better process participants understand the process definition, the higher the probability that the process is carried out according to it. They are better implemented through obtaining buy-in than through imposing directives.
## Gallery of Activities On Completing A Flow Chart
How to create a flowchart When you're working on a complex project, creating the flowchart itself may be a time-consuming task. Here are six simple steps you can follow to create even complex flowcharts: Start by defining the end result of the process or project. The end result could be anything such as completing a user manual, writing a complex software process, installing a new part, or performing a test. List the various steps involved to achieve the end result. This will take some research. In complex processes, each step could have a series of sub steps. The steps involved to create a user manual could be: a. Meet with SME ; b. Research existing documentation ; c. Videotape the procedure ; d. Take photographs ; e. Create illustrations ; f. Develop the user guide ; g. Test the user guide ; h. Make changes/adjustments ; i. Deliver final product.
For the production of rice, the energy consuming equipments used include blowers, elevators, motors, boilers and steam distribution and many more. The efficiency of the product (rice) depends on the utilities maintained by the rice production mills such as electricity, air, water, labour, etc. Many of the rice plants in India and also in the different parts of the world adhere to several procedures for rice processing such as drying of rice, cleaning of rice, milling, whitening, polishing, grading, blending, sorting and packaging. There are different types of sorters and separators used for the rice to be free from any type of dust, fungal infections, plastic granules, unwanted grain, etc.
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# The Grey Labyrinth is a collection of puzzles, riddles, mind games, paradoxes and other intellectually challenging diversions. Related topics: puzzle games, logic puzzles, lateral thinking puzzles, philosophy, mind benders, brain teasers, word problems, conundrums, 3d puzzles, spatial reasoning, intelligence tests, mathematical diversions, paradoxes, physics problems, reasoning, math, science.
Author Message
ctrlaltdel
Member of the Daedalians
Posted: Sat Apr 06, 2002 12:10 pm Post subject: 1 i wasn't sure where to post this, because they are really chestnuty puzzles, so as an explanation and an excuse i give you the big picture: i am going to organize a sort of treasure hunt game in my city where on different places the players will have to solve something and move on to the next stop. i need to know the TIMES it would take them to finish the puzzles, because it will help me organize the whole thing. knowing all the answers its hard for me to judge the times. so more important to me than solutions (which no doubt you will find) are the times it took you to solve them. if you can put those times in your posts pls. if you know these than don't bother to post stuff like: 0.02 seconds. oh and one thing... this is NOT a contest!!! so if you put the times (please do!), be sincere about it, anyway i need to know the average not the genius times. enough chat, here goes: 1. you have 3 matches in one row, 4 matches in the second, and 5 in the third row. two players take turns taking matches. you can take any number of matches from any one row. the player who takes the last match loses. if you are the first to go, what is your winning move? 2. you have 4 beer bottles. arrange them on a table so that each of the corks would be the same distance from any other of the four. 3. 149, 162, 536, 496, x 4. give me one word for each group of colors: red, yellow, green black, red white, black white, red, black, yellow white, blue, red (leave this out - slovak flag) red, white red, orange, yellow, green, blue, violet ...ok, thats enuff - the rest is too local/language/culture based...
Tahnan
Daedalian Member
Posted: Sun Apr 07, 2002 6:11 am Post subject: 2 I decided not to try to solve these, mostly because 1. Man, it's Nim again. I can never, ever remember the strategy for Nim. 2. Do you mean "...from any of the other three?" This was enough of a chestnut that, assuming the answer is , it took me no time at all. 3. No real reason not to try to solve this one. 4. I'm not sure what you mean "one word for." I assume the last one is , but I don't know a single word for "red and white." ? ? ? But I figured the feedback was necessary.
ctrlaltdel
Member of the Daedalians
Posted: Sun Apr 07, 2002 7:34 am Post subject: 3 well thnx, yes i welcome any form of feedback ok, so you don't play nim, didn't bother with the number sequence, solved the bottles in no time (hey that's good to know), and you were not sure about the 'one word for' - ok i might think about rephrasing that... anyway the last one was correct, and for the red and white the answer is the third of your choices. if you got the rest of them too, how long did it take you?
Tahnan
Daedalian Member
Posted: Sun Apr 07, 2002 5:49 pm Post subject: 4 I actually didn't try the others, because as I said, I had no idea if this is what you meant. Black/red is immediately a , r/y/g is--now that I'm thinking--immediately a . I suppose w/b is a , but the other rings no bells (and note that I got 'red and white' on my third try).
ctrlaltdel
Member of the Daedalians
Posted: Sun Apr 07, 2002 10:51 pm Post subject: 5 well in the original 'version' they know the number of letters for each word. you got those fine except for black/red - for this the word is 5 letters. and the last one is 5 letters too. or it can be 6 letters if you use plural (so obviously the potential 6th is 's') thnx for trying at least! edit: add the 6th letter... [This message has been edited by ctrlaltdel (edited 04-07-2002 07:02 PM).]
Death Mage
Raving Lunatic
Posted: Sun Apr 07, 2002 11:17 pm Post subject: 6 1) Punch the other player in the face and force him to make only moves you want him to. 4) Oh that's easy. One word for all of them, "colors" :P
CrystyB
Misunderstood Guy
Posted: Sun Apr 07, 2002 11:32 pm Post subject: 7 Nim? Lovely game. The strategy for it however would not be a chestnut. It's pretty damn hard to get to it if you don't know a thing about it. In general, it goes like this: Covert the three to base 2 and modify one of these numbers so that the XOR-sum of them would be 0. e.g. for 15, 19 and 7, 01111+10011+00111=11011, so take 11 from the second one, giving 01111+01000+00111=00000. In the 3-4-5 case, take two from the first. On reviewing the problems, neither are chestnuts!!! (IMO i mean)
ctrlaltdel
Member of the Daedalians
Posted: Mon Apr 08, 2002 7:44 am Post subject: 8 well, nim rules might be hard to break in general if you don't know anything about them... however, im not asking for this at all - luckily for the players i just want to know the solution to this case - which they can gain in any way they see fit. they can sit there and try a frew dozen combinations for all i care. and the answer you gave is one of the correct ones.
ctrlaltdel
Member of the Daedalians
Posted: Wed Apr 10, 2002 9:46 pm Post subject: 9 -bump- yea yea yea, im bumping my own thread. i really need to find out about those times.
Alter Ego
Daedalian Member
Posted: Wed Apr 10, 2002 10:07 pm Post subject: 10 Just passing through - busy lately - but here's the answer to 3) 481. Took me about 20 seconds. But I guess I was looking for a trick.
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Message Subject Math: 6÷2(1+2) = ?
Poster Handle Anonymous Coward
Post Content
the 6/2 would need distributed into the (2+1) if you want to do distribution.
As seen on every math website I have found.
Quoting: Forgotten
Those websites are wrong, OR the (6/2) was in parentheses. Any fraction using the '/', used as a coefficient, MUST be in parentheses, or else, the entire bottom becomes the denominator. I proved this with the identity law, and that isn't good enough, here is an online algebra lesson:
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# Math Challenge 2019
Discussion in 'Mathematics' started by Shin0bi3Z, Jul 2, 2019.
1. ### Shin0bi3ZNew commenter
OK, I think doing that helps to see the whole number that you will end up with a lot faster.
2. ### sammiloczyNew commenter
27+48, I'd borrow 2 from 27 and place on the 48 to make the calculation 25+50
minstrel likes this.
I assume the post refers to what processes are being done for the calculation.
You are confusing efficiency with a particular algorithm.
Efficiency is something like take 3 from the 48 and add it to the 27 to give 30 + 45 which has an eviqualent total to the original calculation.
Whilst notionally you might think want to say that there is only one calculation, this is not what you are actually doing.
For example.
This ignores the process of increasing 48 to 50 and registering the difference is 2. Which is then used in the second step.
You choice is the addition by column algorithm. Or a form of chunking where you add to make one or both of the values 'round' (multiples of 10) and then correct.
In terms of efficience the measure of the efficiency changes upon how many addends one needs to add up.
In terms of mental speed I find the chunking quicker and somewhat automatic.
Why not 20 + 40 = 60, 8+7 =15, 60 +15 =75? If you going to separate the units and the tens that is?
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# Funny Numbers
9 replies to this topic
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Posted 23 April 2013 - 04:56 AM
The first funny number: If a 1 is placed after this five-digit number, the result is triple the number you'd get by putting the 1 in front. The second funny number: It's the second-smallest odd number greater than 1 that's a perfect cube and also a perfect square. The third funny number is the smallest with the property that if any prime between 10 and 20 is divided into it, the remainder is 1. What's the exact product of the three funny numbers?
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### #2 BobbyGo
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Posted 23 April 2013 - 04:40 PM
Spoiler for
Edited by BobbyGo, 23 April 2013 - 04:40 PM.
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### #3 BobbyGo
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Posted 23 April 2013 - 05:42 PM
*Tried to edit original answer to fix mistake, but was too late.
Spoiler for
Edited by BobbyGo, 23 April 2013 - 05:43 PM.
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### #4 superprismatic
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Posted 23 April 2013 - 06:03 PM Best Answer
Spoiler for I get
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### #5 bhramarraj
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Posted 23 April 2013 - 06:07 PM
Spoiler for
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### #6 k-man
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Posted 23 April 2013 - 06:36 PM
Spoiler for my take
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### #7 bonanova
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Posted 23 April 2013 - 06:52 PM
Spoiler for my take
" by " is not " into "
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Vidi vici veni.
### #8 k-man
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Posted 23 April 2013 - 07:08 PM
Spoiler for my take
" by " is not " into "
It isn't indeed. And English is my second language, so all this time I've thought that it meant the same thing when applied to division.
So, BobbyGo has it then...
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### #9 BobbyGo
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Posted 23 April 2013 - 07:23 PM
It isn't indeed. And English is my second language, so all this time I've thought that it meant the same thing when applied to division.
So, BobbyGo has it then...
Unless superprismatic is right. I didn't even consider that the third number might be less than the primes instead of greater.
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Posted 23 April 2013 - 08:01 PM
sorry but 1 is what i was looking for.
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# Precalculus
posted by on .
To estimate the height of a building, two students find the angle of elevation from a point (at ground level) down the street from the building to the top of the building is 33 degrees. From a point that is 300 feet closer to the building, the angle of elevation (at ground level) to the top of the building is 49 degrees. If we assume that the street is level, use this information to estimate the height of the building.
The height of the building is_____
• Precalculus - ,
h/tan 49° = h/tan 33° - 300
solve for h
### Answer This Question
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# Morphisms $\mathrm{SL}_n(\mathbb Z) \to \mathrm{SL}_m(\mathbb Z)$
From a result Obtained by O. Schreier and B. L. van der Waerden [Math. Sem. Univ. Hamburg 6, 303- 322 (1928)], one can show that for two fields $\mathbb F$ and $\mathbb G$, and integers $n>m>2$, the only group homomorphism $\mathrm{SL}_n(\mathbb F) \to \mathrm{SL}_m(\mathbb G)$ is the trivial homomorphism.
My question is: Is there a non trivial group homomorphism $\mathrm{SL}_n(\mathbb Z) \to \mathrm{SL}_m(\mathbb Z)$ with $n>m>2$?
Thanks.
-
For $n$ at least 7 it has to be trivial since such homomorphism has to kill the alternating group $A_n$. Something similar probably also works for the smaller numbers. – Misha Mar 30 '13 at 2:41
See mathoverflow.net/questions/70311/… for a related question. – Guntram Mar 30 '13 at 4:36
## 2 Answers
I'll try to flesh out the first comment to give an answer: for all $n\ge 4$ there is a representation $\rho:A_{n+1}\to \rm SL_n(\bf Z)$ whose image normally generates $\rm SL_n(\bf Z)$ (because it injects into all $\mathrm{PSL}_{n-1}(\mathbf{Z} / N)$ for $N\ge 1$ and $\rm SL_n(\bf Z)$ has CSP). But for $n\ge 5$ there is no nontrivial complex representation of $A_{n+1}$ in dimension less than $n$, so the image of $\rho$ must be contained in the kernel of any morphism $\rm SL_n(\bf Z)\to\rm SL_m(\bf Z)$ if $m < n$, and so this morphism must be trivial.
This leaves the case $n=4$ when there is a real representation of dimension 3 ; however (as pointed in the comment below) it is not rational, so the argument above works as well.
-
For $n=4$, there are complex representations, just not rational, of dim $3$, because $A_5$ is the symmetry group of an oriented dodecahedron. – Will Sawin Mar 30 '13 at 16:17
Right. I have edited accordingly, thanks for pointing that out. – Jean Raimbault Mar 31 '13 at 3:24
$SL_n(\mathbb Z)$ is generated by elementary matrices. In general let $E_n(R)$ be the subgroup of $SL_n(R)$ generated by elementary matrices. Let's show that if $m\lt n$ and $n\ge 3$ then the only homomorphism $E_n(R)\to GL_m(F)$ is the trivial one, if $F$ is a field. Wlog $F$ is algebraically closed.
Let $T_n(R)\subset E_n(R)$ be the group of strictly upper-triangular matrices. It is generated by elementary matrices $e_{ij}(a)$ where $1\le i\lt j\le n$ and $a\in R$. Let's first show that every homomorphism $h:T_n(R)\to GL_m(R)$ takes $e_{1n}(a)$ to a scalar matrix. This is by induction on $n$. The element $e_{1n}(a)$ is central in $T_n(R)$, so the image of $h$ is contained in the centralizer of $h(e_{1n}(a))$. If $h(e_{1n}(a))$ is not a scalar, then by conjugating we can assume the image of $h$ is contained in $GL_{m_1}(F)\times GL_{m_2}(F)$ for some positive $m_1$ and $m_2$ adding up to $m$. (EDIT: I have changed the next few sentences to correct an error.) Or rather, that the image of $h$ is contained in the subgroup of $GL_m(F)$ preserving $F^{m_1}\times 0\subset F^{m_1}\times F^{m_2}=F^m$. By induction on $n$ the images of $e_{1,n-1}(a)$ under the resulting homomorphisms to $GL_{m_1}(F)$ and $GL_{m_2}(F)$ are central. (To begin the induction, if $n=3$ then $m_1$ and $m_2$ are both $1$ so this step still is OK.) So $h$ takes $T_n(R)$ into some conjugate of $T_m(F)$. Since $n>m$, it follows that $e_{1n}(a)$, being an $(n-1)$-fold commutator in $T_n(R)$, to the identity after all.
In particular a homomorphism $E_n(R)\to GL_m(F)$ must take $e_{1n}(a)$ to a scalar matrix. Of course more generally it must take $e_{ij}(a)$ to a scalar matrix. Since $n\ge 3$, we can express $e_{ij}(a)$ as the commutator of $e_{ik}(a)$ and $e_{kj}(1)$, and it follows that $e_{ij}(a)$ goes to the identity.
-
That's a great argument, both more elementary and more easily generalized than competing arguments. However, to nitpick: you seem to assume that all matrices are diagonalizable when you talk of a product of $GL$. – Ben Wieland Apr 9 '13 at 20:17
Good point. I'm sure that I never in my life thought that every matrix over an algebraically closed field is diagonalizable, but I did mess up somehow. I've corrected some typos, and will try to corret the thinko soon. – Tom Goodwillie Apr 9 '13 at 23:28
I don't follow the step "If h(e1n(a)) is not a scalar, then by conjugating we can assume the image of h is contained in GLm1(F)×GLm2(F) for some positive m1 and m2 adding up to m." What about the case that n=3, m=2 and h(e13(a)) = [[1,a],[0,1]]? – Guntram Apr 10 '13 at 4:41
That step is wrong. I'll fix the proof. – Tom Goodwillie Apr 10 '13 at 13:55
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## Conversion formula
The conversion factor from feet to millimeters is 304.8, which means that 1 foot is equal to 304.8 millimeters:
1 ft = 304.8 mm
To convert 25.8 feet into millimeters we have to multiply 25.8 by the conversion factor in order to get the length amount from feet to millimeters. We can also form a simple proportion to calculate the result:
1 ft → 304.8 mm
25.8 ft → L(mm)
Solve the above proportion to obtain the length L in millimeters:
L(mm) = 25.8 ft × 304.8 mm
L(mm) = 7863.84 mm
The final result is:
25.8 ft → 7863.84 mm
We conclude that 25.8 feet is equivalent to 7863.84 millimeters:
25.8 feet = 7863.84 millimeters
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 millimeter is equal to 0.00012716433701601 × 25.8 feet.
Another way is saying that 25.8 feet is equal to 1 ÷ 0.00012716433701601 millimeters.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that twenty-five point eight feet is approximately seven thousand eight hundred sixty-three point eight four millimeters:
25.8 ft ≅ 7863.84 mm
An alternative is also that one millimeter is approximately zero times twenty-five point eight feet.
## Conversion table
### feet to millimeters chart
For quick reference purposes, below is the conversion table you can use to convert from feet to millimeters
feet (ft) millimeters (mm)
26.8 feet 8168.64 millimeters
27.8 feet 8473.44 millimeters
28.8 feet 8778.24 millimeters
29.8 feet 9083.04 millimeters
30.8 feet 9387.84 millimeters
31.8 feet 9692.64 millimeters
32.8 feet 9997.44 millimeters
33.8 feet 10302.24 millimeters
34.8 feet 10607.04 millimeters
35.8 feet 10911.84 millimeters
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Idiom - Equally well : GMAT Verbal Section
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# Idiom - Equally well
Author Message
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### Show Tags
05 Jul 2007, 11:12
Can someone explain the idiom equally well?
and also the use of equal rather than equivalent.
If you have any questions
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Re: Idiom - Equally well [#permalink]
### Show Tags
06 Jul 2007, 02:35
bmwhype2 wrote:
Can someone explain the idiom equally well?
Let me try this one...
Well (adjective) - used to refer to health
E.g.
I am feeling equally well
E.g.
Chorbakr is an equally well established community.
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Re: Idiom - Equally well [#permalink]
### Show Tags
06 Jul 2007, 02:48
bmwhype2 wrote:
and also the use of equal rather than equivalent.
equal = equal
equivalent = nearly equal
Re: Idiom - Equally well [#permalink] 06 Jul 2007, 02:48
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NAG Library Manual, Mark 28.5
Interfaces: FL CL CPP AD
```/* nag_sparse_direct_real_gen_norm (f11mlc) Example Program.
*
* Copyright 2022 Numerical Algorithms Group.
*
* Mark 28.5, 2022.
*/
#include <nag.h>
#include <stdio.h>
int main(void) {
double anorm;
Integer exit_status = 0, i, n, nnz;
double *a = 0;
Integer *icolzp = 0, *irowix = 0;
/* Nag types */
Nag_NormType norm;
NagError fail;
INIT_FAIL(fail);
printf(
"nag_sparse_direct_real_gen_norm (f11mlc) Example Program Results\n\n");
/* Skip heading in data file */
scanf("%*[^\n] ");
/* Read order of matrix and number of right hand sides */
scanf("%" NAG_IFMT "%*[^\n] ", &n);
/* Read the matrix A */
if (!(icolzp = NAG_ALLOC(n + 1, Integer))) {
printf("Allocation failure\n");
exit_status = -1;
goto END;
}
for (i = 1; i <= n + 1; ++i)
scanf("%" NAG_IFMT "%*[^\n] ", &icolzp[i - 1]);
nnz = icolzp[n] - 1;
/* Allocate memory */
if (!(a = NAG_ALLOC(nnz, double)) || !(irowix = NAG_ALLOC(nnz, Integer))) {
printf("Allocation failure\n");
exit_status = -1;
goto END;
}
for (i = 1; i <= nnz; ++i)
scanf("%lf%" NAG_IFMT "%*[^\n] ", &a[i - 1], &irowix[i - 1]);
/* Calculate 1-norm */
norm = Nag_RealOneNorm;
/* nag_sparse_direct_real_gen_norm (f11mlc).
* 1-norm, infinity-norm, largest absolute element, real
* general matrix
*/
nag_sparse_direct_real_gen_norm(norm, &anorm, n, icolzp, irowix, a, &fail);
if (fail.code != NE_NOERROR) {
printf("Error from nag_sparse_direct_real_gen_norm (f11mlc).\n%s\n",
fail.message);
exit_status = 1;
goto END;
}
/* Output norm */
printf("%s\n%7.3f\n", "One-norm", anorm);
/* Calculate M-norm */
norm = Nag_RealMaxNorm;
/* nag_sparse_direct_real_gen_norm (f11mlc), see above. */
nag_sparse_direct_real_gen_norm(norm, &anorm, n, icolzp, irowix, a, &fail);
if (fail.code != NE_NOERROR) {
printf("Error from nag_sparse_direct_real_gen_norm (f11mlc).\n%s\n",
fail.message);
exit_status = 1;
goto END;
}
/* Output norm */
printf("\n");
printf("%s\n%7.3f\n", "Max", anorm);
/* Calculate I-norm */
norm = Nag_RealInfNorm;
/* nag_sparse_direct_real_gen_norm (f11mlc), see above. */
nag_sparse_direct_real_gen_norm(norm, &anorm, n, icolzp, irowix, a, &fail);
if (fail.code != NE_NOERROR) {
printf("Error from nag_sparse_direct_real_gen_norm (f11mlc).\n%s\n",
fail.message);
exit_status = 1;
goto END;
}
/* Output norm */
printf("\n");
printf("%s\n%7.3f\n", "Infinity-norm", anorm);
END:
NAG_FREE(a);
NAG_FREE(icolzp);
NAG_FREE(irowix);
return exit_status;
}
```
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# Copyright 2011 Casa Software Ltd.
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## Transcription
1 Table of Contents Variable Forces and Differential Equations... 2 Differential Equations... 3 Second Order Linear Differential Equations with Constant Coefficients... 6 Reduction of Differential Equations to Standard Forms by Substitution Simple Harmonic Motion The Simple Pendulum Solving Problems using Simple Harmonic Motion Circular Motion Mathematical Background Polar Coordinate System Polar Coordinates and Motion Examples of Circular Motion
2 Variable Forces and Differential Equations A spring balance measures the weight for a range of items by exerting an equal and opposite force to the gravitational force acting on a mass attached to the hook. The spring balance is therefore capable of applying a variable force, the source of which is the material properties of a spring. When in equilibrium, the spring balance and mass attached to the hook causes the spring to extend from an initial position until the resultant force is zero. Provided the structure of the spring is unaltered by these forces, the tension in the spring is proportional to the extension of the spring from the natural length of the spring. The tension due to the spring is an example of a force which is a function of displacement: Hooke s Law, empirically determined (determined by experiment), states for a spring of natural length when extended beyond the natural length exerts a tension proportional to the extension. Introducing the constant known as modulus of elasticity for a particular spring (or extensible string), the tension due to the extension of the spring is given by: The term natural length means the length of a spring before any external forces act to stretch or compress the spring. If a particle is attached to a light spring and the spring is stretched to produce a displacement from the natural length of the spring, then the force acting upon the particle due to the spring is given by Applying Newton s second law of motion, where the equation can be written in terms of and derivatives of as follows. Equation (1) is a second order linear differential equation, the solution of which provides the displacement as a function of time in the form. Differential equations are often 2
3 encountered when studying dynamics, therefore before returning to problems relating to the motion of particles attached to elastic strings and springs the technical aspects of differential equations will be considered. Differential Equations Ordinary differential equations involve a function and derivative of the function with respect to an independent variable. For example the displacement from an origin of a particle travelling in a straight line might be expressed in the form of a differential equation for the displacement in the form A differential equation is a prescription for how a function and functions obtained by differentiating the function can be combined to produce a specific function, in this case. Whenever the derivative of a function is involved, a certain amount of information is lost. The integral of a derivative of a function is the function plus an arbitrary constant. The arbitrary constant represents the lost information resulting from when the derivative is calculated. For example, The two functions and both have the same derivative therefore if presented with the derivative alone, the precise nature of the function is unknown; hence the use of a constant of integration whenever a function is integrated. Combining derivatives to form a differential equation for a function also means information about the function is missing within the definition and for this reason the solution to a differential equation must be expressed as a family of solutions corresponding to constants introduced to accommodate the potential loss of information associated with the derivatives. A general solution to Equation (2) is and are constants yet to be determined. Both and are solutions to the differential equation as are any number of other choices for the values of and. For a given problem, if at a given time the position and the derivative of position are known, then a specific solution from the set of solutions represented by Equation (3) can be obtained. The method used to establish solutions to equations of the standard form, of which Equation (2) is an example, will be discussed in detail later. Solving general differential equations is a large subject, so for sixth form mechanics the types of differential equations considered are limited to a subset of equations which fit standard forms. Equations (1) and (2) are linear second order differential equations with constant coefficients. To begin with, solutions for certain standard forms of first order differential equations will be considered. 3
4 The differential equations used to model the vertical motion of a particle with air resistance prescribe the rate of change of velocity in terms of velocity: Or depending on the model used for the resistance force, Equations (3) and (4) are first order differential equations specifying the velocity as a function of time. Equation (3) is a linear first order differential equation since and appear in the equation without products such as, or. Equation (4) is nonlinear because appears in the equation. These first order differential equations (3) and (4) are also in a standard form, namely, The key point being the derivative can be expressed as the product of two function where one function expresses a relationship between the dependent variable while the other only involves the independent variable. For equation (4) and. The solution for the standard form (5) is obtained by assuming The solution relies on the separation of the variables. For Equation (3),, therefore the solution can be obtained as follows: and If the is particle initially released from rest, then when, therefore, hence The same procedure could be used to find a solution for the nonlinear differential equation (4). Equation (3) represents a first order linear differential equation for which two standard forms can apply. In addition to being open to direct integration using (5) and (6), Equation (3) is of the form: Differential equations of the form (7) can be solved by determining a, so called, integrating factor such that the differential equation can be reduced to an equivalent equation: 4
5 If, then by the rule for differentiating products If Equation (7) is multiplied throughout by the integrating factor Equation (9) will reduce to Equation (8) provided And Equation (10) is valid provided Or Applying the solution based on separation of variables yields Equation (9) can now be written in the form Therefore if (11) is obtained as follows is used in Equation (8), an equivalent differential equation to Equation Since Equation (3) can be written in the standard form defined by Equation (7), namely, 5
6 We can therefore identify the following functions and, therefore the solution requires an integrating factor of, therefore Applying the same initial conditions as before, namely, when yields resulting in the same answer as before Two different methods applied to a single problem leading to the same conclusion provide a sense of reassurance. An alternative to explicitly solving a differential equation is to calculate the solution using numerical methods. It is important to realise, however, that even when an expression or a numerical solution is produced, there is the possibility an assumption used in the solution is invalid and therefore the solution is only valid for a limited range of the independent variable. An equation of the form requires the condition since the solution involves. The importance of such restrictions can be nicely illustrated by the follow sequence of algebraic steps applied to any number contradiction. leading to a So far so good, but attempting to divide by leads to In terms of manipulation of numbers, these steps appear fine but for the step in which is eliminated. Dividing by zero is clearly shown to produce an incorrect answer. Differential equations may have conditions leading to similar issues, but for now it is sufficient to understand the solution techniques for differential equations and defer these problematic considerations for those studying mathematics at a higher level than this text. Second Order Linear Differential Equations with Constant Coefficients Dynamics problems involving Newton s second law of motion often involve second order linear differential equations as illustrated in the derivation of Equation (1) for a particle attached to a light spring. For an understanding of simple harmonic motion it is sufficient to investigate the solution of differential equations with constant coefficients: 6
7 That is, equations of the form (12) for which, and are all constant. The equation of motion for a particle attached to a light spring is of the form (12) where,, and. Apart from being important mathematical methods for mechanics in their own right, solutions of first order differential equations play a role in solving equations of the form (12). Before writing down the solution for Equation (12), first the solution for the equation must be established. While is a function obtained from the function, the act of differentiating could be defined in the sense that Similarly, the second derivative of might be expressed as Using these alternative forms for the first and second derivative of expressed as Equation (14) could be It might seem reasonable to think of these operations expressed by Equation (15) in an equivalent form using the analogy for factorising a quadratic equation as If Equations (15) and (16) are equivalent, then the solution obtained from the first order differential equation might reasonably be expected to be 7
8 Applying separation of variables Thus a solution to Equation (16) obtain from the methods above is. Since the roots for the quadratic polynomial are also interchangeable when Equation (17) was chosen, it might also be reasonable to assume is also a function which satisfies Equation (14). Since Therefore substituting into Equation (14) And since the value is a root of, hence is a solution of the differential equation (14). Similarly, must be a solution and since, is also a solution of the differential equation (14). Equation (18) is consistent with the previous discussion about potential loss of information resulting from differentiating a function, namely, the second derivative of a function potentially needs two constants of integration to allow for a class of functions all of which have the same second derivative. The introduction of two constants in the solution serves to introduce the necessary generality needed to accommodate the range of functions satisfying Equation (14). Repeated Root for The generality of the solution (18) runs into problems if the quadratic equation has repeated roots, in which case Equation (18) reduces to Namely only a single constant and function appear in the solution. It becomes necessary to look for a further solution before all the possible solutions to the differential equation are obtained. It can be shown that if is a solution of (14), then is also a solution of (14). The 8
9 fact that a second solution is required and the method for constructing the second solution are both consequences of theory beyond the scope of this text, so simply showing that is a solution of (14) will suffice. Substituting into the left-hand side of (14) If is a repeated root of then so and For repeated roots of the auxiliary equation, the general solution of (14) is Complex Roots for The motivation for considering differential equations was the equation of motion for a particle attached to a light spring. The resulting differential equation is written in the form of a second order differential equation with constant coefficients: The auxiliary equation is therefore This quadratic equation has no real roots, however the complex roots are, where and. The solution (18) still applies in the sense and Using Euler s formula, the complex solution to Equation (19) is While expressed as a complex valued function of a real variable, the Equation (20) suggests and are solutions of Equation (19). 9
10 First consider Copyright 2011 Casa Software Ltd. Therefore is indeed a solution of (19). Similarly is another solution. The real valued general solution of (19) is therefore of the form Defining the alternative constants and as follows: The solution to equation (19) can be written as follows: Using the solution can be expressed in the form The solution (22) is an alternative formulation of solution (21), in which the constants and can be interpreted as the amplitude or maximum displacement from the centre for the oscillation of a particle attached to a spring and as defining the initial displacement of the particle at the time. Equation (22) is the more common form used when analysing dynamics problems described as simple harmonic motion, of which a particle on a spring is one example of this type of motion. More generally, the auxiliary equation has complex roots of the form and whenever the and. Under these circumstances the solution as prescribed by Equation (18) takes the form: Following a similar analysis used to obtain Equation (20) the complex valued solution is of the form 10
11 Since the differential equation (14) has real coefficients and equates to zero, it might be reasonable to assume both real and imaginary part of the complex solution must be solutions of the differential equation (14). A solution of the form is therefore a nature first choice to test by substitution into the differential equation. Substituting into Therefore, Since the complex roots of the auxiliary equation are obtained from it follows that and, therefore Similarly Thus, is a solution of whenever the auxiliary equation has complex roots and, with. 11
12 A similar argument shows Copyright 2011 Casa Software Ltd. solution of can be expressed in the form: is also a solution of (14) and therefore a real valued To solve a second-order homogeneous linear differential equation of the form: Determine the root and of the auxiliary equation The general solution for the differential equation is then one of the following three options: Roots of Auxiliary Equation and General Solution Real roots: Real roots: Complex Roots: There are problems in mechanics for which the homogeneous differential equation is replaced by an equation of the form: For example, the motion of a particle of mass force in the direction of the x-axis given by attached to a spring with a constant additional results in the equation of motion: Or The general solution for these types of problems reduces to three steps: 1. Find the general solution for the complementary homogenous equation 2. Find any function not part of the complementary solution which satisfies 12
13 3. Add the two solutions together to form the general solution for (24) The function referred to as is known as a particular integral, while is the complementary function. Since the complementary function is established using the table above, the problem is therefore to construct a particular integral for equations of the form (24). Consider the equation of motion: If then the solution for has already been established to be The problem is to find the simplest function such that Since the function appears on the left-hand-side if is used as a solution where is a constant, the first and second derivatives of are zero and therefore The general solution is therefore For these specific types of second order differential equations it is possible to find many different particular integrals, however it can be shown that the general solution constructed from the complementary function and any one of these particular integrals result in the same answer when boundary conditions are applied of the form and. Methods for determining particular integral for differential equations of the form (24) are again beyond the scope of this text, so a limited number of special cases will be tabulated with their use illustrated by example. 13
14 Form for f(t) Form for Particular Integral To illustrate the use of these particular integrals, consider the problem of a particle attached to a spring, where instead of a constant force the disturbing force varies with time according to. Newton s second law of motion yields the equation Or if Solving Equation (25) involves determining the complementary function equation for the homogeneous and finding an appropriate particular integral for the function. Since the form for the function matches where, the particular integral must be of the form, where both and must be determined by substituting into the identity Given the form And Therefore Equating coefficients for sine and cosine yields 14
15 Therefore for the particular integral is And the general solution is Example: Given the boundary conditions and at and the differential equation Find as a function of. Solution: The first step is to calculate the general solution to the homogenous differential equation It is important to determine the complementary function first since there is always the possibility the standard option for the particular integral corresponding to is included in the two functions used to construct the complementary function. Obtaining the complement function allows an informed decision to be made when selected the form for the particular integral. The auxiliary equation corresponding to is The complementary function is therefore constructed using the form for two distinct real roots: Since cannot be constructed from the particular integral can be chosen to be Substituting into 15
16 The particular integral must satisfy the differential equation Therefore The general solution for the differential equation is Applying the boundary conditions and at Therefore at results in the equation for the constants and Solving the simultaneous equations (a) and (b) yields and. The boundary conditions result is the particular solution Example: A particle of mass attached to a spring is subject to three forces: i. A tension force ii. A damping force proportional to the velocity iii. A disturbing force By applying Newton s second law of motion, express the displacement in terms of time as a differential equation and solve the differential equation for the general solution. Solution: Newton s second law of motion allows these three forces to be combined in the form 16
17 Equation (a) is a second order linear differential equation with constant coefficients. The solution is therefore of the form. The complementary function is obtained from the general solution for the corresponding homogeneous equation The auxiliary equation for Equation (b) is Since the roots for the auxiliary equation are complex and therefore the complementary function is of the form Where and. The particular integral for the function. Since the form for the function matches where and, the particular integral must be of the form, where both and must be determined by substituting into the identity Given the form And Therefore Collecting coefficients of and : 17
18 Coefficient of : Copyright 2011 Casa Software Ltd. Coefficient of : The particular integral for Equation (a) is therefore With general solution Reduction of Differential Equations to Standard Forms by Substitution The discussions above are concerned with finding solutions to a select group of differential equations appearing in standard forms, namely, 1. first order differential equations where the variables can be separated to allow direct integration, 2. first order linear differential equations by determining an integrating factor and 3. second order linear differential equations with constant coefficients. These standard forms can also be useful for problems where a change of variable transforms a differential equation from one form to a standard form for which a solution can be found. By way of example, consider the second order differential equation for a particle attached to a light spring. Rather than treating the problem as a second order differential equation, using Therefore displacement : is can be expressed as a first order differential equation involving velocity and 18
19 Substituting implies, therefore the equation is transformed to a first order linear differential equation Using direct integration Since If the constant of integration is rewritten in the for (imposing which is require by Equation (26) relates velocity to displacement and since equation for. is also a first order differential Using separation of variables Making the substitution Hence Since, 19
20 These two solutions are of the form That is, the same solution for the original differential equation is recovered by direct integration as by applying the theory for the second order linear differential equation to the displacement as a function of time. Equation (22) and Equation (27) are identical. 20
21 Simple Harmonic Motion While the introductory problems in mechanics involving the motion of a particle are often concerned with moving a particle from one place to another, there is an important class of problems where a particle goes through a motion, but at some point in the trajectory the particle returns to the initial position. An obvious example of repetitious motion is a Formula 1 race car which must execute a sequence of laps of a race circuit. Other examples might be the hands of an analogue clock or the vibrations in a tuning fork. The key characteristic for all these motions is that after a time period, the particle or particles retraces over ground previously encountered. While periodic motion is often complex in nature, many problems can be reduced by approximation to a more simple form known as Simple Harmonic Motion (SHM). An example of such an approximation is a simple pendulum, where for small oscillations the motion can be approximated to simple harmonic motion. Simple Harmonic Motion is an oscillation of a particle in a straight line. The motion is characterised by a centre of oscillation, acceleration for the particle which is always directed towards the centre of oscillation, and the acceleration is proportional to the displacement of the particle from the centre of oscillation. These statements are encapsulated in the differential equation. Simple Harmonic Motion For constant motion the equation of has solution Such a linear motion is precisely the motion of a particle of mass attached to a spring of natural length when moving on a smooth horizontal surface after being displaced a distance from the natural length of the spring before being released. When approximating a motion as simple harmonic, the problem is reduced to that of a straight line trajectory for a particle corresponding to the x-coordinate of an equivalent particle moving in a circle of radius with a constant speed. This motion of a particle in a circle provides a geometric perspective for simple harmonic motion expressed in the solution. 21
22 The trajectory for a particle undergoing Simple Harmonic Motion is described by a cosine functions in terms of time, hence the name for the motion. sine or Simple Harmonic behaviour: More complex oscillation can be analysed in terms of combinations of sine and cosine functions, so understanding the more fundamental problem of simple harmonic motion provides the basis for understanding these more general problems. More complicated periodic motion can be recreated by combining these three sinusoidal motions representing simple harmonic oscillations. For example the motion of a piano string and be synthetically modelled from a number of sinusoidal motions. A technologically significant problem is that of interpreting infrared spectra, which can be understood in terms of oscillations associated with molecular bonds. A molecular bond is modelled as springs connecting two masses, hence the relevance of SHM. Infrared spectra are used to 22
23 characterise materials for medical science and other key areas of technology. The following Fourier Transform Infrared (FTIR) spectrum illustrates numerous oscillations in intensity which can be traced back to vibrations associated with carbon-hydrogen bonds in polystyrene (PS). FTIR Spectra from polystyrene. The Simple Pendulum A simple pendulum consists of a particle of mass attached to one end of a light inextensible string of length where the other end of the string is attached to a fixed point. The particle when at rest hangs vertically below the fixed point. The particle and string when displaced from the equilibrium position oscillate in a circular arc in the same vertical plane. This physical description suggests the particle moves in 2D and therefore the motion will not behave like simple harmonic motion. The value in studying the simple pendulum lies in observing the types of approximation and restrictions to the motion of the particle that allow a description in terms of simple harmonic motion. A diagram for the simple pendulum showing the forces acting on the particle of mass helps to write down the equations of motion using the unit vectors and, to express the displacement from the origin in terms of the tension exerted by the inextensible string and weight : and In general, these equations for the simple pendulum do not match the equation for simple harmonic motion, however if the length of the string is large compared with the vertical displacement then, therefore 23
24 The assumption that is large compared with also suggests that the acceleration in the direction is small too, therefore the component for the equations of motion yields Applying these approximation to the equation of motion for the direction then becomes on replacing and where Thus, the motion of a simple pendulum for which the length of the string is large compared to the vertical displacement of the mass reduces under approximation to simple harmonic motion. Note the condition that is large compared with is equivalent to stating the maximum angle for the oscillations is small. Also, the assertion that is geometrically equivalent to observing for large and small the trajectory of the particle for small angles is almost without curve, that is, can be approximated by a straight line. The reason for analysing a mechanical system such as the simple pendulum is to extract useful information. Historically, a pendulum offered a means of measuring time, the method being to count the number of complete oscillations. Once the motion of a pendulum is characterised in terms of simple harmonic motion, the mathematics of the solution provides the means of calculating such useful parameters. Solving Problems using Simple Harmonic Motion Simple harmonic motion is referred to a periodic because after a time interval or period, the same trajectory for the particle begins afresh. This statement is mathematically described by the displacement for the particle must be the same at two times and : The shortest time is called the period and is therefore. For a simple pendulum of length, the time period is determined by and is therefore 24
25 In general, if a problem can be expressed in the form of simple harmonic motion, that is, the equation of motion is of the form then the time for one complete oscillation is given by Comparing solution to Implies Note: the time period for a particle moving under simple harmonic motion is independent of the maximum displacement from the centre of oscillation known as the amplitude. The velocity for the particle does depend on the amplitude. Given the displacement for SHM, Eliminating from these two equations by multiplying Equation (1) by the resulting equations yields before squaring and adding Using 25
26 Equation (3) shows the velocity for a particle moving in SHM is a maximum when and zero when the displacement of the particle is at either of the extreme positions from the centre of oscillation, namely,. Example The port at Teignmouth is in the Teign estuary. A sand bar at the mouth of the river Teign prevents ships from entering the port apart from when the tides raise the water level sufficiently to allow ships to pass over the sand bar and into the port. The minimum and maximum water level due to tidal influences for a certain day is known to be at hours and at hours, respectively. By modelling the water level at the sand bar as varying according to simple harmonic motion, estimate the earliest time after low tide a ship requiring a depth of can cross the sand bar and enter the port. Solution While the high and low water depth will vary on a daily basis, for the time interval between low water at and high water at, the variation in tidal depths is sufficiently small to allow these variations to be approximated by a single sinusoidal function, hence the application of simple harmonic motion to the changes in water depth. Over a longer time interval, the approximation would breakdown, but for the problem as stated a reasonable estimate for the time at which a ship requiring of water to pass safely can be calculated using simple harmonic motion. This example states that simple harmonic motion can be used to approximate the water depth. The problem therefore does not involve showing that simple harmonic motion is appropriate, but simply requires the application of the solution to SHM to the conditions given in the question. The maximum and minimum depths are effectively boundary conditions for the SHM solution: where the simple harmonic oscillations occur about the mean depth for the water, namely, The actual depth of water is given by 26
27 The centre of oscillation for the SHM is, and the maximum displacement of the water from the mean depth is the amplitude for the SHM. These boundary conditions for the simple harmonic motions can be expressed in terms of displacement from the centre of oscillation for the water where corresponds to low water and time is expressed in minutes. Since low water occurs at, high water occurs after low water or after low water, thus and A complete oscillation would cause the water to change for low water to high water and back to low water again, therefore the time period for the SHM will be twice the time to go from low water to high water. The time period for the SHM is. The relationship between and the time period is, therefore. The SHM solutions will be completely determined once the phase shift establishes the displacement when and since is fixed. The phase shift Therefore The time at which a ship requiring equation of water to pass over the sand bank is obtained from the 27
28 Since time is measured from low tide at, the ship must wait at least before crossing the sand bar. The earliest time the ship should attempt to enter the river is estimated to be. Circular Motion Newton s First Law of Motion states: Every body remains stationary or in uniform motion in a straight line unless it is made to change that state by external forces. Thus, unless an external force acts on a particle, the path of the particle is that of a straight line. Whenever a particle deviates from moving in a straight line a force must act upon the particle therefore a particle moving along the circumference of a circle must have a force acting causing the circular motion. Mathematical Background The following topics are relevant to the mechanics of a particle moving in a plane. These subjects represent the technical aspects of mathematics which help an understanding of the physics of a moving particle. It is therefore useful to refresh these subjects to aid the mechanics discussion which follows. Polar Coordinate System Contrary to popular belief, mathematics is designed to make problems easier to solve. The mathematical techniques introduced in this section are only introduced to achieve the goal of simplifying mechanics problems. For example, the Cartesian equation for a circle of radius is When the Cartesian coordinate system is replaced by 2D polar coordinates, the same equation is simply Studying polar coordinates is therefore an appropriate subject for anyone considering the motion of a particle following a circular path. 28
29 Cartesian Coordinates 2D Cartesian coordinates are defined with respect to two directions, the x-axis and the y-axis, specified to be at right-angles to each other such that the y-axis is obtained from the x-axis by rotating the x-axis through in an anti-clockwise direction. A pair of numbers is used to identify a position in a plane representing the distance from the origin in the direction of the x- axis coupled with a distance from the origin in the y-axis direction. The intersection of lines parallel to the axes and located by these distances from the origin define the position specified by the Cartesian coordinates. The same location in a 2D plane can be specified relative to the same origin by defining an initial line and a point at the origin known as the pole from which the initial line emanates in the direction specified by the x-axis in the Cartesian coordinate system. The same position with Cartesian coordinates is defined by specifying two polar coordinates with respect to the pole and initial line as distance of the point from the pole and the angle between a line drawn from the pole to the point makes and the initial line measured in an anti-clockwise direction. Polar Coordinates The relationship between the Cartesian coordinates and the polar coordinates are derived from the geometry of a right-angled triangle, namely, 29
30 and Copyright 2011 Casa Software Ltd. With these relationships between Cartesian and polar coordinate systems the equation for a circle of radius in Cartesian coordinates and can be expressed in terms of polar coordinates and as follows: Equation of circle Substitute and Using and the simplest of expressions for a circle of radius is A particle mass attached to an end of a light inextensible string of length moving on a smooth horizontal surface constrained by the string attached to a fixed point follows a path described in polar coordinates by the equation. The pole is taken as the point at which the string is attached to the surface. A particle mass moving in a vertical plane constrained by a light rod of length to a fixed point follows a circular path is also described in polar coordinates by the equation. The difference between horizontal motion in a circle and vertical motion in a circle is gravity acts in the same vertical plane as the motion of the particle. Gravity therefore performs work on the particle with time. 30
31 Polar Coordinates and Motion A particle moving in a plane is described using Cartesian coordinates by two functions of time, and. Similarly, when using a polar coordinate system the same location for a particle in a plane is described by functions of time specifying how the polar coordinates change with time, namely, and. The position vector for a particle changing with time is either or placing the pole at the centre of the circle and the initial line in the direction of the unit vector and using and The components of the position vector for a particle is therefore expressed as a function times a function of a function, that is, of the form and as a result, when differentiating the components of the position vector to obtain the velocity vector, the derivative is obtained by applying the rules for differentiating products and the chain rule and Thus, when expressed in polar coordinates the position vector for a particle yields the velocity of the particle on differentiation with respect to time as follows. Using Newton s notation, For motion in a circle of radius,, therefore for circular motion 31
32 The quantity Copyright 2011 Casa Software Ltd. is the rate of change of angle with time and is referred to as the angular speed of the particle. Thus for circular motion with radius, the speed of the particle is. The direction for the velocity of a particle moving in a circle might reasonably be expected to be in the direction of the tangent to the circle. Since the tangent line to a circle is at right-angles to the diameter line passing through the point of intersection with the tangent line, the position vector of the particle for the origin placed at the centre of the circle should be perpendicular to the velocity vector. The vector product or dot product of two vectors and is defined as Two non-zero vectors are orthogonal, that is at right-angles, if and only if. Therefore the dot product of the position vector relative to the centre of motion and the velocity vector should be zero. Since and Therefore the direction of the velocity vector is at right angles to the line connecting the particle to the centre of rotation. The acceleration of a particle moving in a circle of radius is obtained by differentiating the velocity vector. Again using the product and chain-rule for differentiation: Since the unit vectors in the direction of the position vector and the velocity vectors are the acceleration can be expressed using these two orthogonal components as 32
33 The velocity unit vector is in the direction of the tangent to the circle traced out by the motion of a particle, and the magnitude for the component of acceleration in the direction of the tangent is. The component of acceleration in the direction of the position vector for the particle (unit vector ) shows that an acceleration of magnitude must act on the particle for the motion to trace a circle of radius. The minus sign indicates the acceleration responsible for the circular motion acts towards the centre of the circle. For a particle of mass attached by a string causing the particle to move in a circular path of radius with angular speed radians per second, in the absence of an external force, the tension Newton in the string must be Since the speed on the particle is given by, therefore For circular motion in a horizontal plane, the resultant force in the vertical direction must be zero. The only force acting on the particle is the force causing the circular motion, which acts at rightangles to the direction of motion. Since the line of action of the force and the direction of motion are at right-angles to each other the force does no work and therefore a particle moving in a horizontal plane experiences no change to the speed of the particle. These statements can be expressed mathematically using Newton s second law of motion: 33
34 For circular motion in the horizontal plane: Resolving vertically: Resolving radially: Applying Newton s second law: Since and are perpendicular unit vectors Newton s second law dictates and Assuming and, this implies. Thus, for circular motion in the absence of an external force, the rate of change of angle with time is constant. Since the speed of the particle is given by, the speed is therefore constant too. For circular motion in the vertical plane, gravity acts as an external force to the circular motion of the particle. Circular motion in the vertical plane is a closed system only when gravity is included and within this closed system energy is conserved, therefore the motion in a vertical plane gains in kinetic energy of the particle are achieved through work done by gravity. For motion in a conservative force field, of which gravity is an example: For horizontal circular motion with no external forces there is no change in potential energy for the particle, hence One further point regarding a particle moving in a circular path without external forces, the force required to cause the circular motion is always directed towards the same point and therefore taking moments about the centre of rotation yields zero moment (the line of action of the force passes through the centre of rotation and therefore the distance from the centre to the line of action of the force is zero, hence the moment about the centre is zero). The force constraining the particle to move in a circle does not cause the particle to change the angular speed of the particle about the centre of rotation. This geometric observation about moments is algebraically stated above in the expression derived for motion in a horizontal circle. 34
35 Examples of Circular Motion While a string is an obvious means of constraining a particle to move in a circular path there are many examples of technological importance in which circular motion is performed. Not least is the near circular motion of the Earth around the Sun or a geostationary satellite carefully positioned in orbit around the Earth so television signal can be beamed to fix locations at the planet surface; both trajectories are the result of matching the gravitational force to the force required for circular motion. Force due to radial electrostatic field strength acting on a charged particle with charge is Force for a circular path of radius is. Electrostatic Sector Only particles with speed and mass such that exit the electrostatic sector apertures. A double focusing magnetic sector mass spectrometer used by the semiconductor industry works based on balancing forces acting on charged particles causing curved motion for charged particles. In simple terms, only those charge particles with precise characteristics are allowed to move in circular paths. These circular paths are constrained by forces first from an electrostatic force field followed by a magnetic force field. Together these two circular motions allow only certain mass of a particle to reach the detector. The motion of ions in a double focusing mass spectrometers used in practice are more involved than circular paths, but as a basic model for the apparatus circular paths illustrate the principle. When a car or a bicycle follows a circular track, the force allowing the circular motion is that of friction between the wheels and the track. While a string attached to a fixed point and a particle provides a physical connection between the force and the resulting circular path, in the case of friction the force is localized at the point of contact between the road and a wheel, but logically the motion is constrained to a circle of radius determined by the balancing of the frictional force to the centripetal force required for a circular trajectory. 35
36 Example Copyright 2011 Casa Software Ltd. Two particles and are connected by a light inextensible string which is threaded through a smooth tube of length. The tube is fixed to a smooth horizontal table with a hole positioned so that the tube is perpendicular to the table surface and particle hangs vertically beneath the table. Particle A is set in motion following a circular path on the horizontal table with constant angular speed such that particle is in a state of equilibrium. The mass of particle is and the mass of particle is. 1. Assuming the string is sufficiently long to avoid touching the table bottom, show that particle remains in contact with the table provided. 2. Show that at the point particle is about to leave the table the radius of the circular path is given by. The smooth tube can be modelled as a ring fixed at a position through which the string passes. Given that particle is suspended below the table in equilibrium, the forces acting on particle must sum to zero. The two forces acting on particle are the tension from the inextensible string and gravity, thus resolving vertically for particle : : From the geometry for the string 36
37 Since the string passes over a smooth ring the tension acting on particle acting on particle. is the same as the tension While particle is in motion, the motion is only in the horizontal plane and therefore the component of force in the vertical direction must be zero. Resolving forces acting on particle the vertical direction yields: in : Since particle moves with constant angular speed, the component of force acting towards the centre of motion is constant in magnitude and equal to, where is the radius for the circular motion. Resolving in the radial direction with respect to the centre of rotation : Equation (2) can be expressed as Dividing Equation (3) by Equation (4) yields and since For particle to be in contact with the horizontal surface, therefore Using the equilibrium state of particle, Equation (2) is rewritten using Equation (1) in the form Thus when, the point at which particle is about to leave the horizontal table surface, 37
38 Since, Example: Motion in a Vertical Circle A particle of mass is attached to a light inextensible string of length. The other end of the string is fixed to a point. The particle is held at the same height as the point with the string held taut before an impulse causes motion for the particle in a vertical plane with initial speed. Determine an expression for the velocity of the particle in terms of the initial speed, the acceleration due to gravity, the string length and the angle between the string and the horizontal line passing through and the initial position for the particle. Since the particle is moving in the vertical plane, the angular speed is not constant and the acceleration when resolved radially and tangentially is of the form Resolving in the two perpendicular directions and allows the problem to be addressed using a natural pair of orthogonal directions for circular motion. Such a choice is analogous to resolving perpendicular and parallel to an inclined plane. Applying conservation of energy the change in kinetic energy must be equal to the change in potential energy. where is the vertical displacement of the particle during the motion to a point making and angle with the horizontal. The negative sign for the term indicates kinetic energy is lost for positive vertical displacements. Since 38
39 As an alternative approach, the work done can be calculated using the following argument, the merit of which is the use of a negative sign for the work done is automatically included. Consider the motion of the particle in a circle in terms of radial and tangential directions. These two directions are perpendicular therefore the work done by the forces acting on the particle is obtained by summing the product of the component forces in these directions with the displacement in these directions. The advantage of choosing the radial and tangential directions lies in observing for circular motion. That is, the particle always remains the same distance from the pole positioned at. Since the displacement in the radial direction is zero, the work done in the radial direction is zero too, so all that remains is to calculate the work done in the tangential direction. The component of force in the tangential direction is and, for small changes in angle, the displacement in the direction is, thus If these products of force times small steps are summed to approximate the work done between and an angle, moving to the limit as we obtain the integral Thus, the same result is obtained, but by applying integration techniques for polar coordinates from FP3, the sign in the energy equation is recovered from the mathematics. 39
40 To further illustrate the uses of FP2 and differential equations, the same problem can be solved directly from Newton s second law of motion applied to the two orthogonal directions and. Resolving the forces: The component of force in radial direction is The component of force in tangential direction is Applying Newton s second law of motion Therefore two equations are obtained and Let, then Therefore the differential equation is transformed to a variable separable equation as follows Now, therefore applying the initial condition, the constant is determined as follows. Therefore substituting for C and the solution previously obtained is recovered, namely, 40
41 41
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### KEANSBURG SCHOOL DISTRICT KEANSBURG HIGH SCHOOL Mathematics Department. HSPA 10 Curriculum. September 2007
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# Taxation for Caregivers
Home Fairmark Forum Other Tax Topics Taxation for Caregivers
Viewing 3 posts - 1 through 3 (of 3 total)
• Author
Posts
• #5004
todd
Participant
Each month my brother receives a \$2000 check from my elderly walker-bound mother. This check is for:
Caregiving: \$1500
Handyman/Gardening/Housekeeping: \$300
Rent assistance: \$200
The caregiving requires that he stay at our mom’s house 4 days/week. It includes ordering & dispensing meds, assisting mom to get in and out of bed, grooming, light cooking, transporting mom to appointments, encouraging and leading physical therapy exercises and fall prevention/quick recovery.
My brother also does some gardening/handyman work for a few other people, from which he averages \$150/month. He has no other income.
Question 1: Which 1040 line should my brother put his monthly earned income of \$1950 (\$1500 + \$300 + \$150)?
Question 2: Can my mom deduct her \$1500/month caregiving expenses on her return?
Question 3: Is there anything else we should know?
#5033
Florida
Participant
This is my take on question 1:
He will pay both self-employment tax (15.3%) and income tax (10%) on the \$1950 per month, \$23,400 per year.
Self-employment tax (medicare + social security) is 15.3% or (.153 x \$23,400) = \$3580.
His income tax is calculated after deducting his standard deduction and deducting 50% of the amount of his self-employment tax. Income of \$23,400 – \$1,790 (1/2 SE tax) – \$12,200 (standard deduction) = \$9,410. His tax rate is 10% so he will owe \$941 in income tax.
Total tax will be \$3588 + \$941 = \$4,529
He will use Schedule C-EZ and Schedule SE to document his income and calculate his self-employment tax.
His self-employment income from Schedule C-EZ will flow back to line 3 on Schedule 1 and then to line 7a on Form 1040. The deductible part of his SE tax (50% of the total SE tax) will flow from Schedule SE to line 14 in the Adjustments to Income area of Schedule 1 and then to line 8a on his Form 1040.
The amount of his self-employment tax from Schedule C-EZ will flow to Schedule 2 line 4 to line 15 on Form 1040. He will put his income tax amount on line 12a of Form 1040.
Looks like he will need to fill out:
Form 1040
Schedule 1
Schedule 2
Schedule C-EZ
Schedule SE
Hopefully, someone will point out any errors or you can double-check by looking at the forms. I always use tax software so never really look at forms but they are available on the IRS website with a quick search.
The good news is he is building social security credits.The bad news is he has a tax bill due in April 2020 if he hasn’t been making estimated payments during the year…which he probably needs to do in 2020.
#5038
todd
Participant
Thank you so much Florida. Very helpful!
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# Thread: The Covariance of Two Random Variables
1. ## The Covariance of Two Random Variables
It is known that when random variables X and Y independent, cov(X,Y)=0. However, inverse is not true. Show that, by an example, that we can have cov(X,Y)=0 and X and Y are not independent.
2. Originally Posted by atalay
It is known that when random variables X and Y independent, cov(X,Y)=0. However, inverse is not true. Show that, by an example, that we can have cov(X,Y)=0 and X and Y are not independent.
Here is one...
$X$ takes values $\{-1,0,1\}$ with probabilities $\{\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\}.$
Let $Y = X^2.$
$Cov(X,Y) = EX^3 - EXEX^2 = 0$
Since, $EX = EX^3 = 0.$
But clearly $X$ and $Y$ are dependent.
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# q5pf09 - 3. 16.6 kJ 4.-16.6 kJ 4. Consider the reaction...
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1. Assuming the apparatus itself absorbs no heat, what will be the final temperature of a bomb calorimeter's heat sink consisting of 100 mL of water at 15 °C if the reaction releases 6.276 kJ of heat? 1. 30 K 2. 303 °C 3. 30 °C 4. 0 °C 5. -303 °C 6. 273 K 2. Calculate the change in enthalpy for the reaction below based on the provided data. N 2 H 4 (l) + H 2 (g) 2NH 3 (g) ∆H = -201.1 kJ∙mol -1 N 2 (g) + 3H 2 (g) 2NH 3 (g) ∆H = -91.8 kJ∙mol -1 CH 3 OH(l) CH 2 O(g) + H 2 (g) ∆H = 85.2 kJ∙mol -1 -------------------------------------------------------------------------- CH 2 O(g) + N 2 (g) + 3H 2 (g) N 2 H 4 (l) + CH 3 OH(l) ∆H = ? 1. ∆H = -28.4 kJ∙mol -1 2. ∆H = -207.7 kJ∙mol -1 3. ∆H = 194.5 kJ∙mol -1 4. ∆H = -378.1 kJ∙mol -1 5. ∆H = 24.1 KJ∙mol -1 3. Calculation the work (w) for the following reaction conducted at 1000 °C: SF 6 (g) + O 3 (g) SO 3 (g) + 3 F 2 (g) 1. 21.2 kJ 2. -21.2 kJ
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Unformatted text preview: 3. 16.6 kJ 4.-16.6 kJ 4. Consider the reaction below. 2 S 2 O 2 (g) + 4 F 2 (g) → 2 S 2 (g) + 4 OF 2 (g) Its change in entropy would likely be (positive/negative/either) and (large/small) 1. either, small 2. negative, small 3. positive, small 4. either, large 5. negative, large 6. positive, large 5. Which of the follwing reactions would spontaneous at some temperatures and non-spontaneous at other temperatures? rxn ∆S rxn (J∙mol-1 ∙K-1 ) ∆H rxn (kJ∙mol-1 ) I-25.20 2.45 II 1.15 879.23 III 13.93-367.10 IV-4.76-98.04 1. I ans II 2. I and III 3. I and IV 4. II and III 5. II and IV 6. III and IV 6. What would be the total energy associated with the motion of a gaseous system composed of 1 mole each of CO 2 , O 2 and O 3 ? 1. 12 RT 2. 24 RT 3. 18RT 4. 9RT 5. 6RT 6. 15RT...
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## This note was uploaded on 03/04/2011 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.
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# [NumPy-Tickets] [NumPy] #1982: In-place arithmetic operations: wrong calculation order?
NumPy Trac numpy-tickets@scipy....
Fri Nov 18 12:23:38 CST 2011
```#1982: In-place arithmetic operations: wrong calculation order?
------------------------+---------------------------------------------------
Reporter: ling | Owner: somebody
Type: defect | Status: new
Priority: normal | Milestone: Unscheduled
Component: numpy.core | Version: 1.6.0
Keywords: |
------------------------+---------------------------------------------------
In-place operation like "subtract(b[:-1], b[1:], b[:-1])" used to work
fine in Numpy 1.4.1, but sometimes fails with Numpy 1.6.0. See the
example below (the problematic output is Out[6]):
{{{
In [1]: import numpy; numpy.__version__
Out[1]: '1.6.0'
In [2]: a = numpy.arange(12).reshape((3, 4))
In [3]: b = a[::-1].copy(); numpy.subtract(b[:-1], b[1:])
Out[3]:
array([[4, 4, 4, 4],
[4, 4, 4, 4]])
In [4]: numpy.subtract(b[:-1], b[1:], b[:-1])
Out[4]:
array([[4, 4, 4, 4],
[4, 4, 4, 4]])
In [5]: b = a[::-1]; numpy.subtract(b[:-1], b[1:])
Out[5]:
array([[4, 4, 4, 4],
[4, 4, 4, 4]])
In [6]: numpy.subtract(b[:-1], b[1:], b[:-1])
Out[6]:
array([[4, 5, 6, 7],
[4, 4, 4, 4]])
}}}
Similar problem happens to other in-place operations (e.g., multiply,
divide).
--
Ticket URL: <http://projects.scipy.org/numpy/ticket/1982>
NumPy <http://projects.scipy.org/numpy>
My example project
```
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## Thursday, March 28, 2013
### Counting Descriptions- a new'' complexity class
Let nσ(w) is the number of σ's in w.
We often ask our students about languages like { w | na(w) = 2nb(w) } (CFL but not REG). Lets formally define languages that are like that.
A COUNTING DESCRIPTION is a boolean combination of linear equations and inequalities involving nσ(w). For example
(na(w) ≤ 2nb(w)+3) AND NOT( nc(w) = nb(w) ).
We denote a counting description by E. Let L(E) = { w : E(w) is true } A lang L is CD if there exists an E such that L=L(E). What to make of the class CD?
The following items came out of emails between myself and Eric Allender, Dave Barrington, and Neil Immerman.
1. CD is contained in 1-way log space and in uniform TC0.
2. CD is incomparable to REG since (1) from above we see there are langs in CD that are not REG, and (2) ba* is not in CD.
3. CD intersect REG is in uniform AC0
4. Parikh's Theorem yields a large class of CD's that produce context free languages.
5. If E is a Boolean combination of threshold and mod statement, each about a single nσ(w), then L(E) is regular
6. The following papers may help answer some of the questions one could ask: here and here.
Are the following decidable:
1. Given a counting description E, is L(E) regular?
2. Given a counting description E, is L(E) context free?
Richard Beigel has shown these problems, with an unbounded alphabet size, are NP-hard. see here. I am very curious about the case where the alphabet size is bounded. Are there other (NATURAL!) classes one could ask about? NOT context sensitive since CSL contain log space. NOT the class DSPACE((log n)1/2) since that's not natural. Maybe some sublinear class would be interesting. We can also ask variants of these questions involving any combinations of the following variants:
1. Do not allow negation.
2. Do not allow intersection.
3. Do not allow inequalities.
4. Do not allow additive constants.
5. Do not allow multiplicative constants.
6. Only allow a bounded alphabet size.
7. Allow other types of equations, for example n_a(w) = n_b(w)2.
8. Allow other primitives such ast n_a(w) == n_b(w) mod 9.
9. Have a non-uniform version of Counting Descriptions and bound the size of the formula as a function of n.
In problem 7 if you allow any poly then the problem is undecidable by the solution to Hilbert's tenth problem.
#### 4 comments:
1. A trivial remark after item 2.(2): Your languages are closed under permutation. The class CD may then be compared to other classes closed under permutation. I am not at all a specialist of these things but I suspect such classes have been studied before!
2. sounds somewhat similar to timed automata. maybe some natural bridge thms possible?
3. The concept of semilinearity is very relevant here -- so all MCFGs/LCFRSs are all semilinear too (permutation equivalent to a regular language.)
You might also look at the use of string kernels to generalise beyond n_a(w) to n_{abc}(w) -- i.e. look st linear constraints on continuous or discontinuous substrings.
4. Look at "Parikh automata".
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Cody
# maeffchen
Rank
Score
1 – 50 of 251
#### Problem 44945. Calculate BMI
Created by: Debanjana Mukherjee
#### Problem 44946. Solve a System of Linear Equations
Created by: Pooja Lalan
#### Problem 44944. Convert from Fahrenheit to Celsius
Created by: Debanjana Mukherjee
#### Problem 44943. Calculate Amount of Cake Frosting
Created by: Debanjana Mukherjee
#### Problem 186. The Tower of Hanoi
Created by: eric landiech
Tags hanoi
#### Problem 2869. There are 10 types of people in the world
Created by: James
#### Problem 1227. Generate a random matrix A of (1,-1)
Created by: Alex P.
Tags random, matrix, +-1
#### Problem 15. Find the longest sequence of 1's in a binary sequence.
Created by: Cody Team
#### Problem 2891. Binary code (array)
Created by: Yannick
#### Problem 1547. Relative ratio of "1" in binary number
Created by: Marek Kuklis
Tags matlab, binary
#### Problem 43977. Converting binary to decimals
Tags binary, basics
#### Problem 1295. Bit Reversal
Created by: @bmtran (Bryant Tran)
#### Problem 108. Given an unsigned integer x, find the largest y by rearranging the bits in x
Created by: AMITAVA BISWAS
Tags binary
#### Problem 2522. Convert given decimal number to binary number.
Created by: Pritesh Shah
Tags numbers
#### Problem 34. Binary numbers
Created by: Cody Team
Tags matlab
#### Problem 2678. Find out sum and carry of Binary adder
Created by: Pritesh Shah
Tags binary
#### Problem 1036. Cell Counting: How Many Draws?
Created by: Ned Gulley
#### Problem 41. Cell joiner
Created by: Cody Team
Tags matlab, strings
#### Problem 953. Pi Estimate 1
Created by: Ed Hall
Tags for loop
Created by: goc3
#### Problem 44660. Perimeter of a semicircle
Created by: Srishti Saha
#### Problem 44060. Volume Pillar
Created by: Erik Luiten
Tags volume, pillar, fun
#### Problem 43215. Convert radians to degrees
Created by: Jamil Kasan
#### Problem 44273. Given a square and a circle, please decide whether the square covers more area.
Created by: AMITAVA BISWAS
#### Problem 43298. Calculate area of sector
Created by: Jang geun Choi
Tags easy, matlab, simple
#### Problem 120. radius of a spherical planet
Created by: AMITAVA BISWAS
#### Problem 1658. Simple equation: Annual salary
Created by: matlab.zyante.com
Tags easy, salary, zyante
#### Problem 1066. Multiples of a Number in a Given Range
Created by: @bmtran (Bryant Tran)
#### Problem 1925. Smith numbers
Created by: Sky Sartorius
Tags factors, sum
#### Problem 83. Prime factor digits
Created by: Cody Team
#### Problem 148. Factorize THIS, buddy
Created by: the cyclist
#### Problem 1590. find the maximum element of the matrix
Created by: Przemyslaw
Tags max matrix
Created by: Cem
#### Problem 109. Check if sorted
Created by: AMITAVA BISWAS
#### Problem 43016. Find the next Fibonacci number
Created by: Steven Van Vaerenbergh
Tags fibonacci
#### Problem 44311. Number of Even Elements in Fibonacci Sequence
Created by: Mehmet OZC
#### Problem 876. Sum the entries of each column of a matrix which satisfy a logical condition.
Created by: Ken Deeley
#### Problem 1179. Knights and Knaves (part 1)
Created by: Alfonso Nieto-Castanon
Tags puzzles, logic
#### Problem 3092. Return fibonacci sequence do not use loop and condition
Created by: Binbin Qi
#### Problem 713. Find the maximum number of decimal places in a set of numbers
Created by: Vincent
1 – 50 of 251
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£
# £1 in 1915 → £0.89 in 1914
### UK Inflation Rate, £1 in 1915 to 1914
According to the Office for National Statistics composite price index, prices in 1914 are 10.91% lower than average prices throughout 1915. The pound experienced an average inflation rate of 12.24% per year during this period, meaning the real value of a dollar decreased.
In other words, £1 in 1915 is equivalent in purchasing power to about £0.89 in 1914.
The 1914 inflation rate was 0.00%. The inflation rate in 1915 was 12.24%. The 1915 inflation rate is higher compared to the average inflation rate of 4.55% per year between 1915 and 2019.
Average inflation rate 12.24% Converted amount (£1 base) £0.89 Price difference (£1 base) £-0.11 CPI in 1915 11.000 CPI in 1914 9.800 Inflation in 1914 0.00% Inflation in 1915 12.24%
### How to Calculate Inflation Rate for £1, 1914 to 1915
This inflation calculator uses the following inflation rate formula:
CPI in 1914CPI in 1915
×
1915 GBP value
=
1914 GBP value
Then plug in historical CPI values. The UK CPI was 11 in the year 1915 and 9.8 in 1914:
9.811
×
£1
=
£0.89
£1 in 1915 has the same "purchasing power" or "buying power" as £0.89 in 1914.
To get the total inflation rate for the 1 years between 1914 and 1915, we use the following formula:
CPI in 1914 - CPI in 1915CPI in 1915
×
100
=
Cumulative inflation rate (1 years)
Plugging in the values to this equation, we get:
9.8 - 1111
×
100
=
-11%
Politics and news often influence economic performance. Here's what was happening at the time:
• French troops force Germans back into the Champagne region, suffering losses of 50,000 men and gaining several hundred yards.
• Germany uses poison gas for the first time in World War I.
• Almost 430,000 French, British, and Germans are killed in the Battle of Loos.
• The one-millionth Model T is produced by Ford Motor Company.
### Data Source & Citation
Raw data for these calculations comes from the composite price index published by the UK Office for National Statistics (ONS). A composite index is created by combining price data from several different published sources, both official and unofficial. The Consumer Price Index, normally used to compute inflation, has only been tracked since 1988. All inflation calculations after 1988 use the Office for National Statistics' Consumer Price Index, except for 2017, which is based on The Bank of England's forecast.
You may use the following MLA citation for this page: “£1 in 1915 → 1914 | UK Inflation Calculator.” U.S. Official Inflation Data, Alioth Finance, 20 Jul. 2019, https://www.officialdata.org/1915-GBP-in-1914?amount=1.
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# Need some help understanding this problem about maximizing graph connectivity
I was wondering if someone could help me understand this problem. I prepared a small diagram because it is much easier to explain it visually.
Problem I am trying to solve:
1. Constructing the dependency graph Given the connectivity of the graph and a metric that determines how well a node depends on the other, order the dependencies. For instance, I could put in a few rules saying that
• node 3 depends on node 4
• node 2 depends on node 3
• node 3 depends on node 5
But because the final rule is not "valuable" (again based on the same metric), I will not add the rule to my system.
2. Execute the request order Once I built a dependency graph, execute the list in an order that maximizes the final connectivity. I am not sure if this is a really a problem but I somehow have a feeling that there might exist more than one order in which case, it is required to choose the best order.
First and foremost, I am wondering if I constructed the problem correctly and if I should be aware of any corner cases. Secondly, is there a closely related algorithm that I can look at? Currently, I am thinking of something like Feedback Arc Set or the Secretary Problem but I am a little confused at the moment. Any suggestions?
PS: I am a little confused about the problem myself so please don't flame on me for that. If any clarifications are needed, I will try to update the question.
-
It looks like you are trying to determine an ordering on requests you send to nodes with dependencies (or "partial ordering" for google) between nodes.
If you google "partial order dependency graph", you get a link to here, which should give you enough information to figure out a good solution.
In general, you want to sort the nodes in such a way that nodes come after their dependencies; AKA topological sort.
-
Thank you for the input. I think I am looking for a topological sort. The only next step is to figure out a good way to construct a DAG out of the graph that I have. Thanks once again. – Legend May 11 '10 at 23:47
I'm a bit confused by your ordering constraints vs. the graphs that you picture: nothing matches up. That said, it sounds like you have soft ordering constraints (A should come before B, but doesn't have to) with costs for violating the constraint. An optimal algorithm for scheduling that is NP-hard, but I bet you could get a pretty good schedule using a DFS biased towards large-weight edges, then deleting all the back edges.
-
Thanks for your input. Sorry about the confusion. I did mention this point in my question. I am still in the stage of forming the question so pardon me for the mess up :) – Legend May 11 '10 at 23:46
If you know in advance the dependencies of each node, you can easily build layers.
It's amusing, but I faced the very same problem when organizing... the compilation of the different modules of my application :)
The idea is simple:
``````def buildLayers(nodes):
layers = []
n = nodes[:] # copy the list
while not len(n) == 0:
layer = _buildRec(layers, n)
if len(layer) == 0: raise RuntimeError('Cyclic Dependency')
for l in layer: n.remove(l)
layers.append(layer)
return layers
def _buildRec(layers, nodes):
"""Build the next layer by selecting nodes whose dependencies
"""
result = []
for n in nodes:
if n.dependencies in flatten(layers): result.append(n) # not truly python
return result
``````
Then you can pop the layers one at a time, and each time you'll be able to send the request to each of the nodes of this layer in parallel.
If you keep a set of the already selected nodes and the dependencies are also represented as a set the check is more efficient. Other implementations would use event propagations to avoid all those nested loops...
Notice in the worst case you have O(n3), but I only had some thirty components and there are not THAT related :p
-
Thanks for your time. I guess both of us are solving the same problem in two different domains :) My problem has one more constraint: I have not decided on a proper metric yet to decide how a node depends on the other. – Legend May 11 '10 at 23:48
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# August 22, 2014 in roman numeral
Here you will see step by step solution to convert August 22, 2014 date to roman numeral. How to write August 22, 2014 as a roman numeral? August 22, 2014 as a roman numeral written as VIII.XXII.MMXIV (MM.DD.YYYY), please check the explanation that how to convert 22 August, 2014 in roman number.
## Answer: August 22, 2014 in roman numeral
VIII.XXII.MMXIV
### How to convert August 22, 2014 in roman number?
To convert the August 22, 2014 in roman number simply expand the each number from month, date and year from hindu-arabic number to roman numerals, then replace the all numbers of expanded form of date, month and year with respective roman numerals.
#### Solution for August 22, 2014 to roman numeral
Given date is => 22-08-2014
After expanding number from Month, Date and year, this table provides a simple and explanation of how to convert the date 'August 22, 2014' to its Roman number:
MonthDayYear
Date 08 [Aug] 22 2014
Expanded Number Values 5 + 1 + 1 + 1 10 + 10 + 1 + 1 1000 + 1000 + 10 + 4
Roman Numeral Values V + I + I + I X + X + I + I M + M + X + IV
Result = VIII XXII MMXIV
Hence, in order to correct roman numerals date combination of Month, Day and Year of August 22, 2014 is written as VIII.XXII.MMXIV or in 22/August/2014 [DD/MM/YYYY] format it is written as XXII/VIII/MMXIV.
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# 653 UK tablespoons in imperial cups
## Conversion
653 UK tablespoons is equivalent to 34.473580593433 imperial cups.[1]
## Conversion formula How to convert 653 UK tablespoons to imperial cups?
We know (by definition) that: $1\mathrm{brtablespoon}\approx 0.0527926195917811\mathrm{brcup}$
We can set up a proportion to solve for the number of imperial cups.
$1 brtablespoon 653 brtablespoon ≈ 0.0527926195917811 brcup x brcup$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{brcup}\approx \frac{653\mathrm{brtablespoon}}{1\mathrm{brtablespoon}}*0.0527926195917811\mathrm{brcup}\to x\mathrm{brcup}\approx 34.47358059343306\mathrm{brcup}$
Conclusion: $653 brtablespoon ≈ 34.47358059343306 brcup$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 imperial cup is equal to 0.0290077207759061 times 653 UK tablespoons.
It can also be expressed as: 653 UK tablespoons is equal to $\frac{1}{\mathrm{0.0290077207759061}}$ imperial cups.
## Approximation
An approximate numerical result would be: six hundred and fifty-three UK tablespoons is about thirty-four point four six imperial cups, or alternatively, a imperial cup is about zero point zero three times six hundred and fifty-three UK tablespoons.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic.
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# Integer to English Words
Convert a non-negative integer to its english words representation. Given input is guaranteed to be less than 231 – 1.
1. Create a list of numbers for which you have to know the text. For example, there is no way you could come up with text for 1-9, 10-20, 30, 40… without knowing them!
2. Create a subroutine that solves this problem for 3 digit numbers (up-to 999). If you find the number in the above map return string, else build it. For eg. 123
```1 2 0 3 - - - | - - - | - - - ^ Hundred ^ Twenty ^ There```
3. Use the above routine to convert digits in batches of 3. Depending on the batch, add Thousand, Million and/or Billion to the converted batch of 3 digits. E.g. to convert 2,190,876,001
```0 0 2 1 9 0 8 7 6 0 0 1 - - - | - - - | - - - | - - - ^ Billion ^ Million ^ Thousand```
Gotcha’s
• If num is 0 output “Zero”. This is the only case you will use zero.
• If any of the 3 digits in the batches are 0, output nothing! E.g. 1000123, make sure you don’t output an extra Thousand anywhere, meaning something like “One Million Thousand One Hundred Twenty Three”.
```class Solution { Map<Integer, String> num2wordsMap = new HashMap<Integer, String>() {{ put(1, "One"); put(2, "Two"); put(3, "Three"); put(4, "Four"); put(5, "Five"); put(6, "Six"); put(7, "Seven"); put(8, "Eight"); put(9, "Nine"); put(10, "Ten"); put(11, "Eleven"); put(12, "Twelve"); put(13, "Thirteen"); put(14, "Fourteen"); put(15, "Fifteen"); put(16, "Sixteen"); put(17, "Seventeen"); put(18, "Eighteen"); put(19, "Nineteen"); put(20, "Twenty"); put(30, "Thirty"); put(40, "Forty"); put(50, "Fifty"); put(60, "Sixty"); put(70, "Seventy"); put(80, "Eighty"); put(90, "Ninety"); put(100, "One Hundred"); put(200, "Two Hundred"); put(300, "Three Hundred"); put(400, "Four Hundred"); put(500, "Five Hundred"); put(600, "Six Hundred"); put(700, "Seven Hundred"); put(800, "Eight Hundred"); put(900, "Nine Hundred"); }}; public String numberToWords(int num) { if (num == 0) return "Zero"; String ret = ""; int index = 0; while ( num > 0 ) { int lastThree = 0; lastThree += (num % 10); num /= 10; lastThree += ((num % 10)*10); num /= 10; lastThree += ((num % 10)*100); num /= 10; if (lastThree==0) { index++; continue; } String threeDigitRet = threeDigitNumberToWords(lastThree); if ( index == 0 ) { ret = threeDigitRet; } else if ( index == 1 ) { ret = threeDigitRet + " Thousand " + ret; } else if ( index == 2 ) { ret = threeDigitRet + " Million " + ret; } else if ( index == 3 ) { ret = threeDigitRet + " Billion " + ret; } index++; } return ret.trim(); } /** * convert 3 numbers to text */ public String threeDigitNumberToWords(int num) { int units = num % 10; int tens = (num/10) % 10; int hundreds = (num/100) % 10; int tensNum = tens*10 + units; String ret = ""; if (!numberToWordsMap(num).equals("")) { ret += numberToWordsMap(num); } else { ret += hundreds != 0 ? numberToWordsMap(hundreds) + " Hundred " : ""; if (!numberToWordsMap(tensNum).equals("")) { ret += numberToWordsMap(tensNum); } else { ret += (tens != 0 ? numberToWordsMap(tens*10) + " " : ""); ret += numberToWordsMap(units); } } return ret; } public String numberToWordsMap(int n) { return num2wordsMap.getOrDefault(n, ""); } }```
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[RESOLVED] How to solve arithmetical problems?
CodeGuru Home VC++ / MFC / C++ .NET / C# Visual Basic VB Forums Developer.com
# Thread: [RESOLVED] How to solve arithmetical problems?
1. Junior Member
Join Date
Mar 2014
Posts
1
## [RESOLVED] How to solve arithmetical problems?
Here is the problem I need to solve:
Here is the code I created in order to solve it:
Code:
```double numx, numz, result;
numx = Convert.ToDouble(textBox1.Text);
numz = Convert.ToDouble(textBox2.Text);
result = Math.Pow(numx, 2) + Math.Pow(numz, 2) / 1 - Math.Pow(numx, 2) - Math.Pow(numz, 2) / 2;
label1.Text = result.ToString();```
But when solving the problem manually I get different answer, so what I'm doing wrong? A help would be appreciated and thanks for your time in advance.
Kind regards,
Ron
2. Member
Join Date
Jun 2011
Location
Buenos Aires, Argentina
Posts
130
## Re: How to solve arithmetical problems?
You are grouping wrong. Multiplication and division precede addition and substraction.
Code:
`result = (Math.Pow(numx, 2) + Math.Pow(numz, 2)) / (1 - (Math.Pow(numx, 2) - Math.Pow(numz, 2)) / 2);`
When having doubts with this sort of problems, reduce your expression to chunks of the ecuation.
Code:
```A = X^2 + Z^2
B = X^2 - Z^2
C = B / 2
D = 1 - C
E = A / D
result = E```
Hope it helps. Good luck!
Last edited by Nikel; March 28th, 2014 at 09:53 AM. Reason: Just formatting.
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# Solving a trigonometric equation
I'm solving this equation:
$$\sin(3x) = 0$$
The angle is equal to 0, therefore:
$$3x=0+2k\pi \space\vee\space3x= (\pi-0)+2k\pi$$
$$x = \frac {2}{3}k\pi \space \vee \space x = \frac {\pi + 2k\pi}{3}$$
$$x = k\frac {\pi}{3}$$
It looks like the two trigonometric equations have been combined into one. I must have made a mistake. Any hints?
• Notice that for $\sin x = 0$, the solutions $x = 0 + 2k\pi \; \vee \; x = \pi + 2k\pi$ can be combined as $x = k\pi$ (where $k \in \mathbb{Z}$). Draw the solutions and realise that you're not "missing" anything: both ways of writing down the solutions contain the exact same angles. Commented Feb 12, 2016 at 13:07
• Thanks a lot @StackTD. This makes a lot of sense. Feel free to post your comment as an answer and I'll be glad to accept it. Commented Feb 12, 2016 at 13:14
• Thanks; I elaborated a bit in the answer. Commented Feb 12, 2016 at 13:18
• Notice that:$$\left\{\frac23k\pi:k\in\Bbb Z\right\}=\left\{\dots,-\frac43,-\frac23,\frac23,\frac43,\dots\right\}$$and:$$\{ \frac{\pi+2k\pi}3:k\in\Bbb Z\} =\\\left\{\dots,-\frac33\pi,-\frac13\pi,\frac13\pi, \frac33\pi,\dots\right\}$$ Commented Feb 12, 2016 at 13:30
All we need is for $3x$ to be an integer multiple of $\pi$. In other words
$$3x = k\pi \Rightarrow x = \frac{k\pi}{3}$$
You're not missing anything: sometimes it's possible to efficiently combine sets of solutions.
Notice that for $\sin x=0$, the solutions $x=0+2k\pi$ ($0$ and then 'adding full circles') $\vee \; x=\pi+2k\pi$ ($\pi$ and then 'adding full circles') can be combined as $x=k\pi$ ($0$ and 'adding half circles'), where always $k \in \mathbb{Z}$.
Draw the solutions and realise that you're not 'missing' anything: both ways of writing down the solutions contain the exact same angles; you 'run through' the same angles.
This is not always possible for equations of the form $\sin x = c$ (only if $c=k\pi$) or $\cos x = c$ (only if $c=\pi/2+k\pi$), but it is always possible for $\tan x = c$ since the solutions $$x = \arctan c + 2k\pi \, \vee x = \pi + \arctan c + 2k\pi$$can alwayes be combined as $$x = \arctan c + k\pi$$ You can easily see this by drawing a trigonometric circle and visualising the solutions.
Your solutions are: $$x = \frac{\pi}{3}(2k),\phantom{NNNNNNNN} x = \frac{\pi}{3}(2k + 1).$$ So, you've shown that $x$ is either $\frac{\pi}{3}$ times an even integer, or else $\frac{\pi}{3}$ times an odd integer. In other words, $x$ is $\frac{\pi}{3}$ times an integer.
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# Grade Curve Array Program Problem
• (2 Pages)
• 1
• 2
## 25 Replies - 5467 Views - Last Post: 21 April 2010 - 06:05 PMRate Topic: //<![CDATA[ rating = new ipb.rating( 'topic_rate_', { url: 'https://www.dreamincode.net/forums/index.php?app=forums&module=ajax§ion=topics&do=rateTopic&t=168322&s=f45eafa77baf3606304f2b58f6e3f896&md5check=' + ipb.vars['secure_hash'], cur_rating: 0, rated: 0, allow_rate: 0, multi_rate: 1, show_rate_text: true } ); //]]>
### #1 JavaBunny
Reputation: 0
• Posts: 32
• Joined: 21-February 10
# Grade Curve Array Program Problem
Posted 11 April 2010 - 11:51 AM
My homework assignment is to create a program using an array so I decided to create a program that creates a grade curve but I'm stuck. I guess time off for Spring Break got to me. Here is what I have so far. Can anyone give me some hints to get me back on track? Any help would be greatly appreciated!
```//GradeCurveArray.java
{
public status void main (String[] args)
int [] gradesArray = {95, 79, 89, 78, 82};
for (int student = 0; student < grades.length; student++)
system.out.printf("Student %2d", student+1);
//call method getMaximum
}//end main
public static int getMaximum(int grades [])
{
{
{
}//end inner for
}//end outer for
}//end Method getMaximum
modifyArray (array); //pass array reference
public static void modifyArray( int [] arrayC)
{
for (int counter=0; counter<arrayC.length; counter++)
arrayC[counter]+=C;
}//end method modifyArray
```
Is This A Good Question/Topic? 0
## Replies To: Grade Curve Array Program Problem
### #2 japanir
• jaVanir
Reputation: 1014
• Posts: 3,025
• Joined: 20-August 09
## Re: Grade Curve Array Program Problem
Posted 11 April 2010 - 11:58 AM
the main method should be declared static:
```public static void main(String[] args)
```
also, you are not using the for loop correctly here:
``` for (int grade ; studentGrades)
```
should be:
``` for (int grade : studentGrades)
```
you call this method:
```modifyArray (array); //pass array reference
```
outside any method block.
perhaps it should be inside the main method?
what is C?
```int curveGrade = C
```
you have a parameter passed to that method called arrayC, but it is an array of int values, so you cannot pass it to an int variable.
did you declare the outputGrades method? you call it in the main method, but i see no implementation for it.
### #3 JavaBunny
Reputation: 0
• Posts: 32
• Joined: 21-February 10
## Re: Grade Curve Array Program Problem
Posted 11 April 2010 - 12:12 PM
Thanks. I have fixed the static method, and loop. I moved the modifyArray line here in the main method.
```system.out.printf("Student %2d", student+1);
modifyArray (array);//pass array reference
//call method getMaximum
```
Do you think that will be alright? For the question about C. I was trying to figure out how to write the code that will take the highest grade, subtract it from 100 and then add that number to all of the array values and then display them as the final grades but I just couldn't figure out how to do it for some reason. I have tried to write this program by piecing together a bunch of stuff from my book and I think that it part of what is causing my problems. Any suggestions?
japanir, on 11 April 2010 - 11:58 AM, said:
the main method should be declared static:
```public static void main(String[] args)
```
also, you are not using the for loop correctly here:
``` for (int grade ; studentGrades)
```
should be:
``` for (int grade : studentGrades)
```
you call this method:
```modifyArray (array); //pass array reference
```
outside any method block.
perhaps it should be inside the main method?
what is C?
```int curveGrade = C
```
you have a parameter passed to that method called arrayC, but it is an array of int values, so you cannot pass it to an int variable.
did you declare the outputGrades method? you call it in the main method, but i see no implementation for it.
This post has been edited by Theresa Schultz: 11 April 2010 - 06:39 PM
### #4 Luckless
• </luck>
Reputation: 293
• Posts: 1,146
• Joined: 31-August 09
## Re: Grade Curve Array Program Problem
Posted 11 April 2010 - 12:42 PM
Quote
``` for (int student = 0; student < grades.length; student++)
```
i believe this should be:
``` for (int student = 0; student < gradesArray.length; student++)
```
This post has been edited by Luckless: 11 April 2010 - 12:44 PM
### #5 JavaBunny
Reputation: 0
• Posts: 32
• Joined: 21-February 10
## Re: Grade Curve Array Program Problem
Posted 11 April 2010 - 12:48 PM
Thanks! I have fixed that in the code.
[quote name='Luckless' date='11 April 2010 - 12:42 PM' timestamp='1271014940' post='986543']
### #6 Luckless
• </luck>
Reputation: 293
• Posts: 1,146
• Joined: 31-August 09
## Re: Grade Curve Array Program Problem
Posted 11 April 2010 - 12:52 PM
So what errors remain?
### #7 JavaBunny
Reputation: 0
• Posts: 32
• Joined: 21-February 10
## Re: Grade Curve Array Program Problem
Posted 11 April 2010 - 01:01 PM
Unfortunately I haven't even gotten to the point of errors yet because I got stuck. The biggest block right now is the last part (modifyArray) where I am trying to figure out how to take the highGrade and subtract it from 100 and then add the difference to all of the other grades in the array and then print that list. How do I do that? I know what I have on there now is not right.
Luckless, on 11 April 2010 - 12:52 PM, said:
So what errors remain?
### #8 Luckless
• </luck>
Reputation: 293
• Posts: 1,146
• Joined: 31-August 09
## Re: Grade Curve Array Program Problem
Posted 11 April 2010 - 01:10 PM
```int amountToAdd = 100- getMaximum(gradesArray) ;
for(int i = 0; i < arrayC.length; i ++){
}
```
this takes the highest grade and subtracts it from 100 then adds amountToAdd to each index location in arrayC[](which is really gradesArray[]). Hope this helps you.
This post has been edited by Luckless: 11 April 2010 - 01:12 PM
### #9 JavaBunny
Reputation: 0
• Posts: 32
• Joined: 21-February 10
## Re: Grade Curve Array Program Problem
Posted 11 April 2010 - 06:12 PM
Luckless, on 11 April 2010 - 01:10 PM, said:
```int amountToAdd = 100- getMaximum(gradesArray) ;
for(int i = 0; i < arrayC.length; i ++){
}
```
this takes the highest grade and subtracts it from 100 then adds amountToAdd to each index location in arrayC[](which is really gradesArray[]). Hope this helps you.
### #10 JavaBunny
Reputation: 0
• Posts: 32
• Joined: 21-February 10
## Re: Grade Curve Array Program Problem
Posted 11 April 2010 - 06:25 PM
I am getting a compile error on my code. It says "Java49 reached end of file while parsing.}//end class GradeCurveArray" Please let me know what I need to fix to get this to work correctly.
```//GradeCurveArray.java
{
public static void main (String[] args)
int [] gradesArray = {95, 79, 89, 78, 82};
for (int student = 0; student < gradesArray.length; student++)
system.out.printf("Student %2d", student+1);
modifyArray (array);//pass array reference
//call method getMaximum
}//end main
public static int getMaximum(int grades [])
{
{
{
}//end inner for
}//end outer for
}//end Method getMaximum
public static void modifyArray( int [] arrayC)
{
for (int i=0; i < arrayC.length; i ++)
{
}//end method modifyArray
for (int student = 0; student < gradesArray.length; student++)
system.out.printf("Student %2d", student+1);
}//end main
```
[quote name='Theresa Schultz' date='11 April 2010 - 06:12 PM' timestamp='1271034762' post='986790']
### #11 citus
Reputation: 16
• Posts: 140
• Joined: 28-March 10
## Re: Grade Curve Array Program Problem
Posted 12 April 2010 - 02:21 AM
You have a number of problems.
1. You have a bunch of words spelled incorrectly. Java is a CaSe-SENsiTiVe language. Also, you left out one curly brace "}" at the bottom of your code.
``` system.out.printf("Student %2d", student+1);
//.......
system.out.printf("Student %2d", student+1);
```
Both of those should be "System" instead of "system".
2.
``` outputGrades(gradesArray);
```
Is this method (outputGrades(int[])) declared and defined in another piece of code? I don't see any declaration for it.
3.
``` modifyArray (array);//pass array reference
```
You do not define a variable named "array". Did you mean to use modifyArray(gradesArray)?
4.
``` for (int[] studentGrades : grades)
{
{
}//end inner for
}//end outer for
```
You are not using the for-each loop properly. You are trying to use an array (int[] studentGrades) which is already declared, to iterate through an array of ints (int[] grades) which is passed into the method as a parameter. Try using the below code.
``` public static int getMaximum(int grades [])
{
{
{
}
}//end outer for
}//end Method getMaximum
```
EDIT the forum messed up the last post I was trying to edit so I will try this again.
I don't know what you are trying to do with printf, but you may as well just simplify things and write a method like printMyStudentArray(int[]) which will handle printing out the numbers for you. Also please read this link for more information on the printf() method.
Lastly.
``` //Add Grade Curve amount to array elements
public static void modifyArray( int [] arrayC)
{
for (int i=0; i < arrayC.length; i ++)
{
}//end method modifyArray
//If you are trying to end the method here then you are
//missing a "}". If you do end the method here then all
//of the code below this point is outside of any method
//declared in your class. You will either need to remove
//the code or move it inside of a method for execution
for (int student = 0; student < gradesArray.length; student++)
system.out.printf("Student %2d", student+1);
}//end main
```
This post has been edited by citus: 12 April 2010 - 02:50 AM
### #12 JavaBunny
Reputation: 0
• Posts: 32
• Joined: 21-February 10
## Re: Grade Curve Array Program Problem
Posted 15 April 2010 - 07:52 PM
I am trying to write a program that takes an array of 5 grades, finds the highest grade, subtracts the highest grade from 100 and adds the difference to the rest of the values in the array and redisplays them as the grade curve. Here is the code that I have but I get an error that says java:60: class, interface, or enum expected }//end class GradeCurveArray. Please help me fix the program so it will work.
```//GradeCurveArray.java
{
public static void main (String[] args)
int [] gradesArray = {95, 79, 89, 78, 82};
for (int student = 0; student < gradesArray.length; student++)
System.out.printf("Student %2d", student+1);
modifyArray (array);//pass array reference
}//end main
public static int getMaximum(int grades [])
{//assume first element of grades array is largest
{
}//end outer for
}//end Method getMaximum
public static void modifyArray( int [] gradesArray)
{
for (int i=0; i < gradesArray.length; i++)
}//end method modifyArray
{
for (int student = 0; student < gradesArray.length; student++)
System.out.printf("Student %2d", student+1);
}//end main
```
### #13 pbl
• There is nothing you can't do with a JTable
Reputation: 8378
• Posts: 31,956
• Joined: 06-March 08
## Re: Grade Curve Array Program Problem
Posted 15 April 2010 - 09:51 PM
modifyArray (array);//pass array reference
should probably be
should probably be
### #14 Dogstopper
Reputation: 2965
• Posts: 11,222
• Joined: 15-July 08
## Re: Grade Curve Array Program Problem
Posted 16 April 2010 - 03:36 AM
Ypu also have one too many ending braces at the end. Delete the one that says "end main" (At the VERY end), as you have already ended main.
### #15 JavaBunny
Reputation: 0
• Posts: 32
• Joined: 21-February 10
## Re: Grade Curve Array Program Problem
Posted 16 April 2010 - 08:14 AM
``` //GradeCurveArray.java
{
public static void main (String[] args)
int [] gradesArray = {95, 79, 89, 78, 82};
for (int student = 0; student < gradesArray.length; student++)
System.out.printf("Student %2d", student+1);
}//end main
public static int getMaximum(int grades [])
{//assume first element of grades array is largest
{
}//end outer for
}//end Method getMaximum
public static void modifyArray( int [] gradesArray)
{
for (int i=0; i < gradesArray.length; i++)
}//end method modifyArray
{
for (int student = 0; student < gradesArray.length; student++)
System.out.printf("Student %2d", student+1);
| 3,449
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| 2.640625
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latest
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en
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| 9,438,429
| 7,458
|
Re: Mathematica Weirdness
• To: mathgroup at smc.vnet.net
• Subject: [mg116705] Re: Mathematica Weirdness
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Thu, 24 Feb 2011 06:24:30 -0500 (EST)
You need to use more precision
N[Integrate[1000000*Exp[x^2 - 12*x]*x^14, {x, 0, 1}], 20]
2.7482124237345664276
NIntegrate[1000000*Exp[x^2 - 12*x]*x^14, {x, 0, 1}]
2.74821
Bob Hanlon
---- Steve Heston <sheston at rhsmith.umd.edu> wrote:
=============
My question is why I get a negative integral of a positive
function?
Integrate[1000000*Exp[x^2-12*x]*x^14,{x,0,1}]//N
Integrate[1000000*Exp[x^2-12*x]*x^14,{x,0.,1}]//N
NIntegrate[1000000*Exp[x^2-12*x]*x^14,{x,0,1}]
The first line gives a negative answer, while the second two lines give
identical positive answers. Something is strange here.
Steve
Steven L. Heston
Associate Professor
Finance Department
Robert H. Smith School of Business
4447 Van Munching Hall Van Munching Hall
University of Maryland
College Park, MD 20742-1815
301-405-9686 TEL
301-405-0359 FAX
sheston at rhsmith.umd.edu
http://www.rhsmith.umd.edu
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# Six Functions of a Dollar. Made Easy! Business Statistics AJ Nelson 8/27/2011 1
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## Transcription
1 Six Functions of a Dollar Made Easy! Business Statistics AJ Nelson 8/27/2011 1
2 Six Functions of a Dollar Here's a list. Simple Interest Future Value using Compound Interest Present Value Future Value of an Annuity Present Value of an Annuity Amortization of Future Value (aka Sinking Fund Factor ) Amortization of Present Value (aka Partial Payment Factor) Business Statistics AJ Nelson 8/27/2011 2
3 Wait a minute!!!!!!! That's Seven Functions, not Six. Easy Explanation: Simple Interest is not considered one of the six functions. Then why is it listed!!!!!!! Most college finance textbooks begin with Simple Interest as a starting point for teaching Future Value where interest is compounded. Are we good now? O.K. Rock On!!!!!!! Business Statistics AJ Nelson 8/27/2011 3
4 Simple Interest Future Value Present Value Interest Rate (eg Monthly or Annual) Number of periods (eg months or years) Notice with Simple Interest no interest is earned on prior year interest. Interest is earned only on the original principal. ALWAYS be sure to match the interest rate period to that used in the number of periods. In other words, do not use an annual interest rate and a period expressed in months. EXCEL CLASS EXAMPLE: Use Excel to calculate the following future values: Initial Investment \$20,000 Monthly Interest Rate is 0.3% Compare periods of 6 months, 12 months, 24 months, 60 months, & 240 months Business Statistics AJ Nelson 8/27/2011 4
5 Future Value Using Compound Interest Future Value Present Value Interest Rate (eg Monthly or Annual) Number of periods (eg months or years) Notice with Compound Interest interest is earned on prior year interest. Interest is earned not only on the original principal but also on interest earned during the course of the investment. ALWAYS be sure to match the interest rate period to that used in the number of periods. In other words, do not use an annual interest rate and a period expressed in months. EXCEL CLASS EXAMPLE: Use Excel to calculate the following future values: Initial Investment \$20,000 Monthly Interest Rate is 0.3% Compare periods of 6 months, 12 months, 24 months, 60 months, & 240 months THEN compare the results with your earlier Excel simple interest calculations. Business Statistics AJ Nelson 8/27/2011 5
6 DERIVATION: Present Value Using Compound Interest Start With Future Value Formula Present Value Formula EXCEL CLASS EXAMPLE: Use Excel to calculate the following present values: Desired Future Value \$50,000 Annual Interest Rate is 5% Compare periods of 1 year, 2 years, 5 years, 10 years, & 20 years Business Statistics AJ Nelson 8/27/2011 6
7 Compounding Vs. Discounting Compound Factor Compounding Present Value Future Value Discounting Discount Factor Business Statistics AJ Nelson 8/27/2011 7
8 Notice that the first two functions of a dollar translate: A lump sum future value to present value OR A lump sum present value to a future value. The last four functions consider annuities. Annuities are a series of payments. In the case of the functions of a dollar, we consider a series of all equal payments. And Now the Formulas!!!!! Business Statistics AJ Nelson 8/27/2011 8
9 Future Value Of An Annuity Future Value Payment Interest Rate (eg Monthly or Annual) Number of periods (eg months or years) EXCEL CLASS EXAMPLE: Use Excel to calculate the following future values: Compare saving \$2,000 per year, \$5,000 per year, and \$10,000 per year Annual Interest Rate is 6% Consider a periods of 30 years. Business Statistics AJ Nelson 8/27/2011 9
10 Present Value Of An Annuity Future Value Payment Interest Rate (eg Monthly or Annual) Number of periods (eg months or years) Notice that if you divide the formula for Future Value of an Annuity by the discount factor ( ), one derives the formula for Present Value of an Annuity. ALWAYS be sure to match the interest rate period, and payment period with that used in the number of periods. In other words, do not use an annual interest rate with a monthly payment. EXCEL CLASS EXAMPLE: Use Excel to calculate the following present values: Compare a loan with an annual payment of \$2,000 per year, \$5,000 per year, and \$10,000 per year. Annual Interest Rate is 6% Consider a loan term of 30 years. THEN Confirm the Future Value of the Annuities (see previous page) divided by the discount factor equals the Present Value of the Annuities (calculated above), Business Statistics AJ Nelson 8/27/
11 Amortization Of Future Value AKA: Sinking Fund Factor Start With Future Value of Annuity Formula Multiply by reciprocal of ( ) Amortization of Future Value Formula EXCEL CLASS EXAMPLE: Use Excel to calculate the following amortization of future values (aka sinking fund factor): Start with a saving goal (future value) of \$100,000 Annual Interest Rate is 6% Compare results for periods of 5 years, 10 years, and 15 years. Business Statistics AJ Nelson 8/27/
12 Amortization Of Present Value AKA: Partial Payment Factor Start With Present Value of Annuity Formula Multiply by reciprocal of ( ) Amortization of Future Value Formula EXCEL CLASS EXAMPLE: Use Excel to calculate the following amortization of present values (aka partial payment factor): Start with a loan amount (present value) of \$300,000 Annual Interest Rate is 6% Compare results for periods of 10 years,15 years and 30 years. Business Statistics AJ Nelson 8/27/
13 Summary of Formulas Six Functions of a Dollar Present Value Future Value Payment Interest Rate (eg Monthly or Annual) Number of periods (eg months or years) Future Value using Compound Interest Present Value Future Value of an Annuity Present Value of an Annuity Amortization of Future Value (aka Sinking Fund Factor ) Amortization of Present Value (aka Partial Payment Factor) Business Statistics AJ Nelson 8/27/
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### Solutions to Problems: Chapter 5
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Associated Topics || Dr. Math Home || Search Dr. Math
### Conversion of Time Range into Minutes
```Date: 11/13/2002 at 17:24:03
From: David Bird
Subject: Conversion of time range into minutes
Hello,
I'm currently relieving a position that contains a quality improvement
role. One report is related to utilisation of time, requiring me to
report on 'time available' and 'actual time used' within operating
theatres.
Let's say the "time available" is from 0830 (start) until 1600
(finish), and the "time used" is 0835 - 1704.
What equation/conversion can I use to find the number of minutes
between the start and end times for both 'available' and 'actual' time
on a 24-hour clock?
Having an equation/conversion would save literally hours of currently
manual calculations.
Regards,
David.
```
```
Date: 11/14/2002 at 23:10:36
From: Doctor Ian
Subject: Re: Conversion of time range into minutes
Hi David,
I'm not entirely sure I understand what you're asking. To avoid doing
the calculations 'manually', do you intend to program a computer or
equivalent can subtract times directly.
To subtract by hand, there are two cases. First, the number of minutes
in the end time is greater than in the start time. In this case, you
can just subtract the times as if they were integers:
1744
- 0835
------
909
The second case is _not_ like this:
1704
- 0835
------
????
In this case, you can still do something like an integer subtraction,
instead of borrowing a 1 from the third column (which would represent
100 minutes), you borrow a 6 (which represents 60 minutes):
6 6
|
17 04
- 08 35
-------
8 29
^ ^
| |
| +------ this is (60+4) minus 35
|
+--------- this is (17-1) minus 8
That's really the only difference. If the interval spans midnight, you
can borrow 24 hours,
24
1 6
|
02 04
- 16 35
-------
9 29
^ ^
| |
| +------ this is (60+4) minus 35
|
+--------- this is (2-1+24) minus 16
Does this help?
- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
Middle School Calendars/Dates/Time
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# quarter to ell conversion
Conversion number between quarter and ell [ell] is 0.2. This means, that quarter is smaller unit than ell.
### Contents [show][hide]
Switch to reverse conversion:
from ell to quarter conversion
### Enter the number in quarter:
Decimal Fraction Exponential Expression
quarter
eg.: 10.12345 or 1.123e5
Result in ell
?
precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential:
### Calculation process of conversion value
• 1 quarter = (exactly) (0.2286) / (1.143) = 0.2 ell
• 1 ell = (exactly) (1.143) / (0.2286) = 5 quarter
• ? quarter × (0.2286 ("m"/"quarter")) / (1.143 ("m"/"ell")) = ? ell
### High precision conversion
If conversion between quarter to metre and metre to ell is exactly definied, high precision conversion from quarter to ell is enabled.
Decimal places: (0-800)
quarter
Result in ell:
?
### quarter to ell conversion chart
Start value: [quarter] Step size [quarter] How many lines? (max 100)
visual:
quarterell
00
102
204
306
408
5010
6012
7014
8016
9018
10020
11022
Copy to Excel
## Multiple conversion
Enter numbers in quarter and click convert button.
One number per line.
Converted numbers in ell:
Click to select all
## Details about quarter and ell units:
Convert Quarter to other unit:
### quarter
Definition of quarter unit: ≡ 1⁄4 yd .
Convert Ell to other unit:
### ell
Definition of ell unit: ≡ 45 in (In England usually) . of chiefly historical interest
← Back to Length units
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HONORS MTH 152 Math for Liberal Arts II Mathematics at
math home | courses | prerequisites | faculty | FAQ | Master Math
The honors section of Math 152 is offered during the Spring semester on Monday afternoons 4:00 – 6:40 on the Loudoun campus.You may qualify for the course in one of two ways: your algebra score on the math placement test must be at least 40 OR you are recommended by a NVCC math instructor The honors section of Math 152 covers all the required topics that are presented in the regular 152 class (modular arithmetic, graphing, linear programming, probability, combinations and permutations, and statistics) but also covers four additional units. These extra topics are purely for enrichment and are not included as test material. Instead, students are expected to submit solutions to a series of problems on each of the topics. Classic math problems will be introduced with some possible solutions. For example, the first honors project deals with the problem of apportionment. In the House of Representatives, there are 435 members. Each state is allocated a certain number of representatives based on their population. If a state has ten percent of the nation's population, that would entitle the state to 43.5 representatives. What is done with the ".5"? Do we round up? Then we will total more than 435 members. The apportionment problem was studied by George Washington, John Adams, Thomas Jefferson, and other founding fathers. We will look at the various methods used throughout the nation's history. You will be asked to allocate representatives to a legislature in a fictional country using each of the methods studied. The format of the class will be to cover the required curriculum during the first part of the class, and then proceed to a class discussion of the enrichment topic. Other honors projects will include the topics of modular arithmetic, probability, and statistics. Handouts are provided with explanations for each of the four topics. You will have three to four weeks to complete each assignment. There will be time at the end of each class to discuss any difficulties that arise on the assignments. The class is small to encourage cooperation among the students in completing the projects. The focus of the honors section is to introduce students to new ideas in mathematical problem solving that might not arise in other courses. You will not be able to register by telephone or on NovaConnect for this course. Honors students can register in person at the Student Services Center. If you have any questions regarding the Math 152 honors course, please contact Karen Barr in LR 212 or call 450 - 2611 or by e-mail at kbarr@nvcc.edu.
Top of page | NVCC home NVCC-Loudoun home
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## Correlational Analysis
Before beginning this assignment, please watch the following videos on correlation:
Correlation – The Basic Idea Explained (Links to an external site.)
Correlation Basics (Links to an external site.)
Study Description: A school educator is interested in determining the potential relationship between grade point average (GPA) and IQ scores among ninth graders. The educator takes a random sample of 30 ninth graders aged 14 years old and administers the Wechsler Intelligence Scale for Children-Fourth Edition (WISC-IV). The WISC-IV includes a Full Scale IQ (FSIQ; however, for this assignment we will just call it IQ).
Output file: See Week 5 SPSS Output.pdf file (Links to an external site.).
1. Hypotheses – Formulate null and alternative hypotheses. What do you think is the relationship between IQ scores and GPA?
2. Variables – Describe the scale of measurement (nominal, ordinal, interval, or ratio) for each of the variables.
3. Correlation – Write an overview of the results of the correlation (at least two paragraphs), including the appropriate and necessary statistical results within sentences and in proper APA formatting. Be sure to provide sufficient explanation for any numbers presented. Consider the following in your overview and conclusions:
• Is there a significant correlation between IQ scores and GPA? If so, what does a significant correlation mean?
• Using the correlation table and scatterplot, explain whether the relationship is positive, negative, or no correlation.
• Describe the strength of the relationship (e.g. very strong, moderate, weak, etc.).
• What conclusions can we draw from these results? What conclusions can we NOT make using these results?
Write a total of 400-700 words in response to these questions
### Discussion Forum: Course Reflection
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SOCW 6311 wk 10 Assignment: Designing a Plan for Outcome Evaluation Social workers can apply knowledge and skills learned from conducting one type of evaluation to others. Moreover, evaluations themselves can….
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Re: Fit rectangle to set of points
• To: mathgroup at smc.vnet.net
• Subject: [mg69343] Re: Fit rectangle to set of points
• From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
• Date: Thu, 7 Sep 2006 23:58:17 -0400 (EDT)
• Organization: The Open University, Milton Keynes, UK
• References: <edom0n\$hu2\$1@smc.vnet.net>
```buttgereit at biomediclab.de wrote:
> Dear Group,
>
> I have a set of 2D points given for which I would like to find the
> smallest rectangle circumscribing all the points.
>
> Doing this with a circle instead of a rectangle was easy --- for the
> rectangle I don't see the trick.
>
> Any hints?
>
> TIA + Best Regards,
>
> Peter
>
HI Peter,
Do you mean something like that:
In[1]:=
pts = Table[Random[], {10}, {2}]
Out[1]=
{{0.830387, 0.0810041}, {0.123819, 0.121942},
{0.0578849, 0.938962}, {0.773195, 0.00453692},
{0.858906, 0.0116332}, {0.678336, 0.100916},
{0.862732, 0.780806}, {0.283974, 0.056643},
{0.600934, 0.766803}, {0.242359, 0.669791}}
In[2]:=
xleft = Min[pts[[All,1]]]
Out[2]=
0.0578849
In[3]:=
xright = Max[pts[[All,1]]]
Out[3]=
0.862732
In[4]:=
ybottom = Min[pts[[All,2]]]
Out[4]=
0.00453692
In[5]:=
ytop = Max[pts[[All,2]]]
Out[5]=
0.938962
In[6]:=
ListPlot[pts, Epilog ->
{Line[{{xleft, ybottom}, {xleft, ytop},
{xright, ytop}, {xright, ybottom},
{xleft, ybottom}}]}];
Regards,
Jean-Marc
```
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| 583
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| 3.296875
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www.surfhydrodynamics.com
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1. #### THE OUTLINE: (THE TEMPLATE, THE PLAN VIEW)
The outline of a surfboard represents the shape of the perimeter of the board, the equivalent terms are planform and template. It is defined by :
• The length of the board
• The maximum width.
We then give as information
• The position of this maximum width (wide point) in relation to the center of the length of the board. Two other dimensions end up describing the outline,
• The width of the nose
• The tail width taken respectively is one foot (30.48 cm) from the front end of the board and one foot from the rear end of the board..
the dimensions defining the outline
THE OUTLINE SURFACE : The surface of the outline partly determines the waterline volume (the other determining factor being the thickness of the board)
A large surface therefore allows a large flotation. The drag force opposing the movement of the rower is made up of 2 main components: frictional drag and shape drag. (See drag chapter) At train speeds, frictional drag remains negligible and shape drag is predominant. It is caused by changes in direction and speed, constrained by shapes not parallel to the displacement. These non-parallel shapes constitute the frontal surface (submerged surface projected perpendicular to the displacement).
The shape hydrodynamic drag force is proportional to the frontal area and the square of the speed. Rowing with a surfboard, whose buoyancy volume keeps the surfer's entire body out of the water, is therefore easier than with a small board that lets part of the surfer's body sink, because the frontal surface is less important.
Floating therefore allows you to better use your rowing energy and to reach higher speeds. It will thus be easier to travel long distances or to catch up with the wave of the wave in order to go on it. The low shape drag of a high flotation board will thus allow you to be propelled by a wave with a very low slope, or to take the wave ahead of its peak, because the slightest component of the force of gravity resulting from a slight slope, will quickly exceed the drag. The surfer will therefore be able to take the necessary speed to go into planing and perform his take off.
Gravity drag comparative diagram following slope: If P (thrust resulting from the gravity projected on the slope), is greater than T (the drag of resistance to progress), the board accelerates, the take off is then possible. The surface also determines planing lift: A large surface makes it possible to reach or maintain planing with less speed, the board with a large surface sinks and brakes less in the maneuvering phases where the speed drops.
#### The length of the outline: The advantages / disadvantages of length :
Flotation: The length therefore brings the advantages of the buoyancy volume, but for the same submerged volume, the frontal surface of a long board is less than that of a wide board, so we will gain significantly in form drag when paddling (especially if the surface is choppy) and will increase the theoretical hull speed limit. In planing, length is no longer an advantage, as we will see in the chapter dealing with friction drag and the Lindsay Lord experiments.
submersion of the rocker curve not adapted to the wave curve: A long board will find its place in a big wave and/or a weak curvature, but it will swallow if the wave is hollow and of small size.
: wave length and curve
Considering only the dynamic aspect of inertia, without going into hydrodynamics, a long board requires more energy to be put into rotation because it is heavier but also because of the distance between the pivot point (which we assume here at the level of the fins) and the center of inertia of the board. The maneuvering effort is determined by this distance.
To give board A a speed of 1 revolution/sec around its center of gravity, a force of 3.8 kg would have to be applied at 1 meter from the center of rotation for one second. If we place the pivot point 1.3m from the center of gravity of board A, this force increases from 3.8 kg to 12 kg To give board B a speed of 1 revolution/sec around its center of gravity, we would have to apply a force of 0.3 kg at 1 meter from the center of rotation for one second. If we place the pivot point 0.6m from the center of gravity of board B, this force increases from 0.3 kg to 1.2 kg.
The variation of the distance between the center of inertia of the board and the pivot point is therefore already of a very great influence on the maneuverability in terms of inertia, we will see that this pivot point is also very influential on the 'hydrodynamic. It therefore seems wise to take care of the placement of this pivot before worrying about a difference of 0.01 kg between carbon or classic technology in the choice of fins.
It should be noted that a forward position of the pivot point reduces the torque to be produced and therefore improves maneuverability in Yaw. This is what happens when the daggerboards are moved forward or when the daggerboards have an automatic orientation system which modifies the position of the center of the curve. (See ADAC system)
• Sequence of maneuvers :
The position of the surfer on a long board must be moved from the back, during the tight turns phases, to the front to seek acceleration. This movement can be used to perform dance steps over the length, but the intuitiveness of the maneuvers is greatly affected. Shortboards allow sequences of maneuvers without necessarily moving.
• Duck bar passage :
A long board can be impossible to submerge when ducking the bar, and it can become very energy-consuming to pass the bar without taking advantage of this maneuver. All that remains is to turn the board into a barrel and squeeze it with all your strength so as not to send it over the other surfers in the spot or let go of the board if the wave is too powerful.
To sum up, the advantage of a short board lies mainly in the maneuverability without moving the feet and the passage of the bar in duck. The advantage of a long board is mainly in the ease of catching more waves and stability. One method of choice is to get the board as long as possible, to catch as many waves as possible depending on the physical and wave conditions, but short enough to allow room for the maneuvers the surfer desires to produce. Choosing a board that is too short to have great maneuverability is nonsense if you miss 50% of the waves of the day...
#### The width of the outline :
When paddling, for an equivalent flotation, a wide board has a larger frontal surface, and therefore generates more shape drag than a long board. When the speed increases and you enter planing, the shape drag becomes less important because a large part of the submerged frontal surface disappears. Lindsay Lord's tests described in a previous chapter, giving the drag as a function of the Width/Length ratio, tell us that the optimal ratio is around 0.5, and that above this ratio, the width becomes penalizing when the surface of the the water is irregular. This is related to the frontal area hitting the chop, which a long, narrow board cuts through better. One of the biggest advantages of the width is the ability to maintain flotation and planing even in maneuvers where you rely solely on the tail (rear) of the board to lift the front of the board. plank and turn radically, as in the "tick-tock" maneuvers in skateboarding. In these phases of tight turns, at reduced speed, in support on the back, the board floats more if it is wide. This avoids sinking and blocking the board.
Yaw maneuverability gain : Another advantage in terms of maneuverability of the boards with the large width, is the possibility (with equal outline surface) of creating a stronger outline curve.
One can easily imagine that it is easier to spin a floating disc than a long rectangular board. We will therefore gain in “yaw” maneuverability on a wide outline with a tight radius of curvature.
However, the rounded outline at the front of the board tends to push the rail out of the water as the speed increases. This can make the rail difficult to submerge, resulting in a board that slips forward. Without rail attachment at the front, it is impossible to take advanced positions generating speed. We will favor forms of parallel rails for speed.
hydrodynamic thrust forces on the front part of the rail. The front curve of the outline tends to eject the rail from the water. The hydrostatic thrust (floating of the submerged volume) also tends to push the submerged part out of the water in a direction perpendicular to the surface.
Loss of maneuverability inRoll : A large width increases the rocking force in "roll" necessary to sink the rail, by increasing the volume to be submerged and the distance between the center of the board and the rail. It is therefore more difficult to push the rail deep into the wave to exploit accelerating carved turns and/or to quickly switch from rail to rail. Wide boards are therefore naturally more stable in Roll, they sail flatter and more difficult to exploit the effects of cut curves that the rails combined with the rocker curve can provide. We will see in the optimizations how a “hydroactive” type rail can free us from this width problem. The term “clunky” designates this difficulty of passing from rail to rail, by the sound that the board makes when it slams when passing from its stable flat position. A basic rule indicates that if the foot must be brought closer to the rail to obtain the grip, it is that the board is too wide, or that the fins are badly implanted because they generate cant (see chapter fins). Couple diagram of Roll/Roll wide board narrow board:
Outline : Influence of the position of the wide point poition
The position of the wide part of the board determines the area where the lift will be the most important. The further this position is from the center, the more we mark the style of surfing that we can practice with.
• Placed on the back of the center of the board, the wide part will allow the surfer a more rearward position. It will therefore limit sinking and braking for a surfing style favoring tricks and maneuvers using support on the rear foot to turn. This type of surfing on the back favors the demonstrative figures appreciated by the competitors.
• Placed on the front of the center of the board, the wide part allows more forward support and faster surfing. The maneuvers exploit the shapes: By pressing on the rail and the curve of the rocker, the surfer produces propulsive, incisive and precise cut turns. This type of surfing is more in tune with the shapes and hydrodynamic effects of the board and exploits the fluidity of the wave in a more instinctive surf than demonstration.
• By mixing the two, we produce a versatile surf..
The outline therefore determines the type of wave and surf for which the board is intended. Here are some basic types of boards whose outline varies according to the style of surfing you want to practice: :
###### OUTLINE gun :
Big wave surfing, the origin of its name “Gun” would be due to a reflection made in 1956 by a shaper working on a form intended for big waves: “You don't hunt the elephant (big waves) with a gun for children, we use an elephant gun! "Elephant Gun" remained the name Gun. So this elephant gun is very thin at the back and narrow, this small width offers very little frontal surface, this allows you to reach high speeds. The narrow outline makes it easy to switch from rail to rail (Roll/Roulis) and to sink the rail deeply. Rarely equipped with side fins close to the rail because they induce cant. A single central fin provides a centered support increasing the sinking of the rail and the exploitation of the rocker in the cut curves. Its narrow Pintail with low lift, sinks and digs a furrow which stabilizes the trajectory by maintaining the rear. The rearward position of the fin further increases this directionality. This type of board can reach more than 3 meters to hunt mammoths!
OUTLINE long bord : (ou Malibu)
the surf of the origins, imported by Duke in California with the surf, it was abandoned during the shortboard revolution, but has come back in force since 1990 with a lightness and hydrodynamic qualities that make it easier to surf than at the origins. A large rounded nose clearly distinguishes it from the Gun. Its width provides the flotation and the planing surface giving it, respectively, a faster oar, and the possibility of surfing low slopes. Its flotation prohibits the duck, but allows it to take the waves ahead of the peak with little effort. The number of waves that it becomes possible to surf in a session therefore becomes greater than with a shortboard. The inertia of this type of board needs to be tamed and used, the longboarder moves from the front to the back of the board depending on the speed or maneuverability he is looking for. It allows you to go from fairly big conditions to small wave conditions where tricks and various steps on the board are an art in itself. Shapers' longboards can be very light and the levels of tricks performed by pro longboarders no longer have much to envy to shortboards!
OUTLINE mini Malibu : wider and thicker than a short board, it allows better paddling thanks to its flotation. It thus allows older surfers to keep a chance to catch waves in territories populated by young shortboarders, while keeping more radical maneuvers than a Malibu (longboard)! An identical tail and nose width, but a wide point (beautiful master) far enough back to maintain a fairly radical turning ability at the rear.
OUTLINE Egg : a mini Malibu with a more advanced wide point, so possible support further forward without burying. This board is therefore intended for faster but less radical surfing than the Malibu.
OUTLINE hybrid :Board that connects to the mains to go up to the peak thanks to its electric motor. It recharges its battery when its propeller spins upside down as it rolls down the wave. Has a citrus press and coffee grinder that makes very good cappuccinos (optional). More seriously, this type of board offers a compromise between Fish Egg shortboard/mini Malibu. Less radical than a shortboard, it keeps a nose suitable for ducks and a rear wide enough for maneuvers in back support, but allows advanced support from a wide point in front and a fairly wide nose. It is comparable to a larger Fish with a narrower tail. It offers lift and maneuverability and allows you to surf more types of waves than the Fish.
OUTLINE short Board :
between 1.5 and 2.2 meters, it is the most widespread board since the shortboard revolution initiated in 1966 by the radical figures of Nat Young at the world championships. The shapers linked to this revolution are the Australian Bob McTavish, and the Californian George Greenought. More difficult to row because of its low buoyancy, its low inertia and its small surface make it very maneuverable without making any movement on the board. The ducks are easy to dive thanks to a pointed nose and a reduced volume, but its low lift implies a constant search for speed and very fast sequences so that it does not sink into the dead spots (zone where the board is almost stopped). The term "engagement" defines the revolutionary style of shortboard surfing in comparison to pre-existing surfing, where the "hang ten" or "hang five" were the most extreme figures.
Radical shortboard maneuvers tend to slow and sink the board in skid-type turns. The stimulus will therefore have to be produced by pumping. The contribution of drive, maneuverability and pump acceleration generated by dynamic fins (see ADAC system), lend themselves particularly well to this type of surfing..
OUTLINE Fish :
It is a wide shortboard with a swallow tail tail. Ideal for beach break and small fast waves. Thick and wide, the Fish has a flotation that makes paddling easier. Its advanced "wide point" (master bau) allows it to get support providing speed without burying, its tailcoat extends the rail and its drive effect, while offering width to avoid sinking the rear in low turns. speeds. The low rocker is matched to wave curvature to limit frontal area and drag. The fins are symmetrical profiles and parallel to the axis of the board to generate a minimum of drag. The curvature and wave speed necessary to propel this small surface in planing make it difficult to use it in the less pronounced troughs of medium-sized waves..
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2nd Grade Math Worksheets: Number Worksheets For Preschool Math Worksheet Pdf Do Make Flyer Using Microsoft Word Grade Addition Free One Step Equations With And Subtraction Children Activity To Print Kids Problems. Free Printable Worksheets For Middle School Social Studies Ordinal Numbers Activities For Kindergarten Z Pictures For Kindergarten | Ukulele-roysakuma
# Number Worksheets For Preschool Math Worksheet Pdf Do Make Flyer Using Microsoft Word Grade Addition Free One Step Equations With And Subtraction Children Activity To Print Kids Problems
Published at Friday, 17 January 2020. 2nd Grade Math Worksheets. By .
How many little masterful works of art do you currently have in your home? Whether your child attends preschool or kindergarten or is home schooled, arts and crafts are a natural part of any curriculum. What do you do with these masterpieces? Most are proudly displayed on refrigerator doors. But you can take it a step further and create clever displays for your child has art work. The best part of course is that your child can help you with this special little project in your home.
Academically, parents can use preschool worksheets to help teach their children some of the basic skills they will need for kindergarten and school. This will include counting to ten, recognizing shapes and colors, being able to hold a pencil or crayon properly, and coloring in without scribbling. Basic math concepts such as recognizing patterns, understanding quantity and some simple addition and subtraction will be useful. By the time your child is ready for kindergarten or school, they should be able to recognize their own name and other simple written words. The sounds of each letter of the alphabet should be familiar to your child, and they should understand the principle of reading from left to right, which way to hold a book, and possibly even be starting to read three and four-letter words.
Show a magazine or picture book to children. Ask them to identify all instances of the given letter in any page. Hand out letter cards to all children. Call out a letter. The child with that card has to come in front of the class and display the letter. Divide the class into two groups. Give one group letter cards. Give other group various objects. The first group will hold up a letter. The second group should hold up an object that starts with that letter. Where can we get kindergarten abc worksheets?You can design them yourself and print them out. Many printed workbooks are available. A good alternative is to download printable kindergarten abc worksheets. This is advantageous because you can then usually print out any sheet any number of times.
File name:
### Number Worksheets For Preschool Math Worksheet Pdf Do Make Flyer Using Microsoft Word Grade Addition Free One Step Equations With And Subtraction Children Activity To Print Kids Problems
Image Size: 960 x 1243 Pixels
File Type: Image/jpg
Total Gallery: 22 Pictures
File Size: 140 kb
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# TOJ 4109 Cyuunibyou's problem
I make mistakes on this problem's meaning.So I get WA many times.So sad.
The portal:http://acm.tju.edu.cn/toj/showp4109.html
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
int even_tot,odd_tot;
void Deal_with(){
int n;
while(scanf("%d",&n),n){
int temp;
even_tot = 0;
odd_tot = 0;
for(int i=0;i<n;i++){
scanf("%d",&temp);
if(temp&1){
odd_tot ++;
}
else {
even_tot ++;
}
}
if(odd_tot > even_tot && n % 2 == 1){
puts("Justice will prevail over evil.");
}
else if(even_tot > odd_tot && n % 2 == 0){
puts("While the priest climbs a post, the devil climbs ten.");
}
else {
puts("I love this world.");
}
}
}
int main(void){
freopen("a.in","r",stdin);
Deal_with();
return 0;
}
#### Instrction Arrangement （hdu 4109 差分约束）
2015-08-25 23:17:12
#### hdu4109拓扑
2015-08-19 10:29:03
#### hdu4109(拓扑求关键路径)
2015-08-17 20:12:13
#### TOJ-ACM
2013-12-16 21:58:37
#### TOJ 题目分类
2010-11-03 22:39:00
#### TOJ 3772.Rupxup's Math Problem
2017-07-31 20:48:50
#### Codeforces-954F：Runner's Problem（矩阵快速幂+离散化）
2018-03-25 22:15:19
#### found pre-existing rpmdb problem
2016-01-21 08:54:09
#### 【模拟】【HDU1443】 Joseph
2014-10-01 20:10:35
#### TOJ1065Factorial
2017-10-10 14:25:30
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# combustion of fuel problem
## Recommended Posts
octane is burnt with 80% air
if it is theoritical air ,the equation wiill be
C8H18+a(O2+3.76N2)------>xCO2+yH2O+a*(3.76N2)
and form that equation we can find the coefficient a,x,y.
for x=8;
2y=18, so y=9;
2a=2x+y, so 2a=2*8+9, a=12.5
but when it comes to less than theoritical air there should be co present ,shouldn't it?
then what will be the equation
c8h8+a(o2+3.76n2)----->xco2+yh2o+a*(3.76n2)+co
if i give some coefficient to co my question is how i can find the value coefficient of co here ?
##### Share on other sites
why don't you do it like the first one?
##### Share on other sites
i tried .lets see...
if we declare another coefficient d for co then
c8h8+a(o2+3.76n2)----->xco2+yh2o+a*(3.76n2)+dco
8=x+d;..........(1)
2y=18, so y=9;
2a=2x+y+d
=>2x+d=2a-9;...........(2)
we get 3 variable and two equation.but we need 3 equations to solve for 3 variable...
##### Share on other sites
if you're sure with the equations then let's try...
First having equations less than the number of variables doesn't mean there is no solution.
I guess you should write d or x interms of 'a' then insert any number that satisfy the equation.
##### Share on other sites
then how can i do it
i cant proceed any further than the equations i wrote above
## Create an account
Register a new account
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• #### Activity
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• Create New...
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## BIM-MPS Details, Part IV
### The MPS on the BIM (BIMMPS)
As much as viewing the MPS on a universal grid is absolutely amazing in its combination of simple geometry to visually reveal the basic and fundamental relationships in the form of AREAS -- areas that add and subtract to inform all the AREAS that define all three generations within-- it is truly when we examine the BIMMPS that the numerical, quantity based relationships of these AREAS are unfolded in an in-depth and repeating NPS of their own that the profound connections are realized!
As before, the NPS of the BIMMPS at first looks formidable as a group. Taking a few examples of the smaller MPS and drawing a clear NPS picture will establish a fractal-template that will be repeated with each and every MPS thereafter -- only the numbers will change.
### Background
Taken from the BIM-MPS_Details-I.html page back in Book I. The central x+y column will be the focus of this section. While x may seem, and indeed is, quite simple and straightforward, y reveals all the Butterfly Fractal 1 pattern parts previous to x. Just follow the examples and the whole secret to the Perfect Numbers, Odd Complements and the resulting Mersenne Prime Squares will be revealed. And such a simple, repetitive pattern it is!
As one can see, the color bands that make up the PN can form all three "generations" withing the MPS space. They do so in a completely fractal manner, based on the exponential power of 2. They do so in a repeating NPS template that is exactly the same for each MPS, with only the relative numbers being different.
BIM-MPS-PN-1--1.mp4
BIM-MPS-PN-1--1.gif
As an overview, let's look at a half a dozen examples before diving in to a more detailed look. To note: only the y values listed are illustrated within the much larger x+y=z central column shown.
BIM+BF-8.gif
BFsums-x=2_p=2
BFsums-x=4_p=3
BFsums-x=8_p=4
BFsums-x=16_p=5
BFsums-x=32_p=6
BFsums-x=64_p=7
Looking at BFsums-x=16_p=5MPS-0-22 in detail:
BFsums-x=16_p=5MPS-0
BFsums-x=16_p=5MPS-1
BFsums-x=16_p=5MPS-2
BFsums-x=16_p=5MPS-3
BFsums-x=16_p=5MPS-4
BFsums-x=16_p=5MPS-5
BFsums-x=16_p=5MPS-6
BFsums-x=16_p=5MPS-7
BFsums-x=16_p=5MPS-8
BFsums-x=16_p=5MPS-9
BFsums-x=16_p=5MPS-10
BFsums-x=16_p=5MPS-11
BFsums-x=16_p=5MPS-12
BFsums-x=16_p=5MPS-13
BFsums-x=16_p=5MPS-14
BFsums-x=16_p=5MPS-15
BFsums-x=16_p=5MPS-16
BFsums-x=16_p=5MPS-17
BFsums-x=16_p=5MPS-18
BFsums-x=16_p=5MPS-19
BFsums-x=16_p=5MPS-20
BFsums-x=16_p=5MPS-21
BFsums-x=16_p=5MPS-22
And finally, looking atBFsums-x=64_p=7MPS-0-22 in detail:
MPS-0-22/BFsums-x=64_p=7_0
MPS-0-22/BFsums-x=64_p=7_1
MPS-0-22/BFsums-x=64_p=7_2
MPS-0-22/BFsums-x=64_p=7_3
MPS-0-22/BFsums-x=64_p=7_4
MPS-0-22/BFsums-x=64_p=7_5
MPS-0-22/BFsums-x=64_p=7_6
MPS-0-22/BFsums-x=64_p=7_7
MPS-0-22/BFsums-x=64_p=7_8
MPS-0-22/BFsums-x=64_p=7_9
MPS-0-22/BFsums-x=64_p=7_10
MPS-0-22/BFsums-x=64_p=7_11
MPS-0-22/BFsums-x=64_p=7_12
MPS-0-22/BFsums-x=64_p=7_13
MPS-0-22/BFsums-x=64_p=7_14
MPS-0-22/BFsums-x=64_p=7_15
MPS-0-22/BFsums-x=64_p=7_16
MPS-0-22/BFsums-x=64_p=7_17
MPS-0-22/BFsums-x=64_p=7_18
MPS-0-22/BFsums-x=64_p=7_19
MPS-0-22/BFsums-x=64_p=7_20
MPS-0-22/BFsums-x=64_p=7_21
MPS-0-22/BFsums-x=64_p=7_22
### Yet another advanced section. And it will be addressed in the Appendix I in Book VI. Here is just a little bit. The MPS on the BIM (BIMMPS)
###### Bringing it all together: Parts I - II - III - IV
Let's repeat a key concept covered in Part II and III:
The Perfect Numbers may be expressed as the sequential running sums () of the cubes of the sequential ODD numbers as 1³ + 3³ + 5³ ... and we have the genesis of the Butterfly Fractal~3~.
In the Butterfly Fractal~1~ and Butterfly Fractal~2~, we find the MPS parameters based around the EVEN numbers, especially within the exponential power of 2. When we look at their summations, we find it delivers the ODDs we are looking for that inform the MPS directly.
This gave the x and z=Mp directly and the y, xz=PN and xy=CR indirectly by calculation.
Now, in the Butterfly Fractal~3~, we find the MPS parameters based around the ODD numbers, especially the Running Sums (∑) of the ODD numbers sequence of 1,3,5,… giving the MPS and the ODD³ sequence of 1³, 3³, 5³,… giving the PN!
of Numbers (Part I)
### NEXT: BOOK III: An Artist, Mathematician and Choreographer...Oceans of Numbers (Part II)
~~~~~~~~
~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~
Mersenne Prime Squares (Part I) the Introductory white paper.
Mersenne Prime Squares (Part II) 3 Simple Intros.
Mersenne Prime Squares (Part III) the Advanced white paper.
The MPS Project The Mersenne Prime Squares Project.
MathspeedST: TPISC Media Center
MathspeedST: eBook (free)
Apple Books
Artist Link in iTunes Apple Books Store: Reginald Brooks
### Back to Part III of the BIM-Goldbach_Conjecture.
BACK: ---> PRIMES Index on a separate White Paper
BACK: ---> Periodic Table Of PRIMES (PTOP) and the Goldbach Conjecture on a separate White Paper (REFERENCES found here.)
BACK: ---> Periodic Table Of PRIMES (PTOP) - Goldbach Conjecture ebook on a separate White Paper
BACK: ---> Simple Path BIM to PRIMES on a separate White Paper
BACK: ---> PRIMES vs NO-PRIMES on a separate White Paper
BACK: ---> TPISC_IV: Details_BIM+PTs+PRIMES on a separate White Paper
BACK: ---> PRIME GAPS on a separate White Paper
Reginald Brooks
Brooks Design
Portland, OR
brooksdesign-ps.net
KEYWORDS TAGS: BMP, BIM-MersenneSquare-PerfectNumber, Mersenne Primes, Mersenne Prime Square, Perfect Number, Perfect Square, MPS, Mp, PTOP, Periodic Table Of PRIMES, PRIMES vs NO-PRIMES, algebraic geometry, BIM, TPISC, The Pythagorean - Inverse Square Connections, Pythagorean Triangles, DNA, Zika virus, pentagon, decagon, double pentagon, composite axial DNA double-helix, Pythagorean Triples, primitive Pythagorean Triples, non-primitive Pythagorean Triples, Pythagorean Theorem, Pythagorus Theorem, The Dickson Method, BBS-ISL Matrix, Expanded Dickson Method, r-sets, s-set, t-sets, Pair-sets, geometric proofs, MathspeedST, leapfrogging LightspeedST FASTER than the speed of light, Brooks (Base) Square- Inverse Square Law (ISL), BBS-ISL Matrix grid, The Architecture Of SpaceTime (TAOST), The Conspicuous Absence Of Primes (TCAOP), A Fresh Piece Of Pi(e), AFPOP, Numbers of Inevitability, LightspeedST, Teachers, Educators and Students (TES), number theory, ubiquitous information, FASTER than the speed of light, primes, prime numbers, fractals, mathematics, Universe, cosmos, patterns in number, DSEQEC, Double-Slit Experiment-Quantum Entanglement Conjecture, CaCost, Creation and Conservation of SpaceTime.
Art Theory 101 / White Papers Index
PIN: Pattern in Number...from primes to DNA. | PIN: Butterfly Primes...let the beauty seep in. | PIN: Butterfly Prime Directive...metamorphosis. | PIN: Butterfly Prime Determinant Number Array (DNA) ~conspicuous abstinence~. | GoDNA: the Geometry of DNA (axial view) revealed. | SCoDNA: the Structure and Chemistry of DNA (axial view). | The LUFE Matrix | The LUFE Matrix Supplement | The LUFE Matrix: Infinite Dimensions | The LUFE Matrix: E=mc2 | Dark Matter=Dark Energy | The History of the Universe in Scalar Graphics | The History of the Universe_update: The Big Void | Quantum Gravity ...by the book | The Conservation of SpaceTime | LUFE: The Layman's Unified Field Expose` | GoMAS: The Geometry of Music, Art and Structure ...linking science, art and esthetics. Part I | Brooks (Base) Square (BBS): The Architecture of Space-Time (TAOST) and The Conspicuous Absence of Primes (TCAOP) - a brief introduction to the series | more White Papers... net.art index | netart01: RealSurReal...aClone, 2001 | netart02: Hey!Ufunk'n with my DNA? | netart03: 9-11_remembered | netart04: Naughty Physics (a.k.a. The LUFE Matrix) | netart05: Your sFace or Mine? | netart06: Butterfly Primes | netart07: Music-Color-ISL | netart08: BBS-ISL matrix | netart09: BBS-interactive (I) | netart10: Sunspots and Solar Flares | netart11: Music-Color-ISL (II-III) | Art Theory 101: PIN, DNA, LUFE Matrix, GoMAS, BBS index | home Copyright 2021-23, Reginald Brooks, Brooks Design. All rights reserved. iTunes, iTunes Store, Apple Books, iBooks, iBooks Store, iBooks Author, Mac OS are registered® trademarks of Apple Inc. and their use on this webpage does not reflect endorsement by Apple Inc.
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# Evaluate $\lim\limits_{n \to \infty}\frac{\mathrm{sgn}(n^2-3n+2)}{e^{n+1}}$
Solve : $$\lim\limits_{n \to \infty}\frac{\mathrm{sgn}(n^2-3n+2)}{e^{n+1}}$$ Now, i started by applying the quotent rule $$\frac{\lim\limits_{n \to \infty} \mathrm{sgn}(n^2-3n+2)}{\lim\limits_{n \to \infty}e^{n+1}}$$The second limit gives out infinity . I do have a problem with the first limit : $$\mathrm{sgn}(\lim\limits_{n \to \infty}(n^2-3n+2))$$ and then $$\mathrm{sgn}(\infty)$$ which i am not sure how to evaluate . Wolfram says the answer is 0 .I just want an explanation for the first limit ( the one with the sign function ).
• for the signum part, you have to know, that this part is bounded by $1$. Mar 17, 2020 at 10:41
$$0\xleftarrow[n\to\infty]{}\frac{-1}{e^{n+1}}\le\frac{\text{sgn}(n^2-3n+2)}{e^{n+1}}\le\frac1{e^{n+1}}\xrightarrow[n\to\infty]{}0$$
• Do you think we can take $(-1)^n$ instead of that $\text{sgn}$? Mar 17, 2020 at 10:51
When $$n \to \infty, n^2-3n+2 \sim n^2,~ so ~ \mathrm{sgn}(n^2-3n+2)=+1$$ So the required limit is $$L=\lim_{n \to \infty} \frac{\mathrm{sgn}(n^2-3n+2)}{e^{n+1}}=\frac{1}{\infty}=0$$
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http://www.mywordsolution.com/question/explain-how-will-the-mix-of-inputs-in-shenzen/919575
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+1-415-315-9853
info@mywordsolution.com
## Economics
Basic Economics Macroeconomics Microeconomics Business Economics Econometrics International Economics Managerial Economics Game Theory Public Economics
Determination of production function with the given input and output data.
You have been hired as a plant manager for a firm that produces widgets (Q) in Angola, Indiana. Widget production requires machine time (K) and labor time (L). The factory works three eight-hour shifts each day. Below is input and output data from the last twenty-one shifts:
Shift Widgets Machine Hours Labor Hours 1 4067 2540 1029 2 4313 2786 1042 3 4493 2757 1165 4 4503 2397 1421 5 4066 2754 920 6 4641 2364 1556 7 3929 2604 916 8 4222 2530 1131 9 4272 2852 986 10 4728 2637 1399 11 4310 2724 1073 12 3980 2430 1039 13 5011 2773 1499 14 4495 2492 1341 15 3971 2433 1032 16 4597 2652 1298 17 4662 2392 1548 18 4533 2296 1532 19 4400 2722 1128 20 4392 2829 1065 21 4715 2511 1487
1.Your engineers tell you that the marginal product of both labor and machine time diminishes as the volume of production expands during any one shift. What is your production function? You should round all parameters to exactly two decimal places.The chief operating officer of the company asks you to meet a production goal of 4800 widgets during the next shift.
2.In a table, find out how many machine hours are required to meet the Q = 4800 production goal if L=300, L=400, L=500, L=600, L=700, L=800, L=900, and L=1000. Graph the isoquant that these calculations imply. describe in very clear and complete terms why the isoquant has the shape that you observe.
3.The unit depreciation expense of one machine hour is \$130 and the expense of one labor hour is \$65. What combination of machine and labor hours minimizes the cost of producing 4800 widgets? Do you believe that the firm should make production at Angola more or less capital intensive? describe your reasoning in full detail.
4.The firm is considering a movement of the plant to Shenzen, China where labor is cheaper. The same mathematical relationship between inputs and outputs will hold. A new plant in Shenzen would enjoy a one hour expense for labor of \$40. Because of a scarcity of modern equipment, the expense of one machine hour would jump to \$320. Can the move to China be financially justified? Theoretically, describe how will the mix of inputs in Shenzen compare to the mix of inputs in Angola?
For political reasons, the plant is not moved to China. You receive news that certain machinery is unavailable for the foreseeable future. Due to unscheduled changeover work, use of machinery is constrained to 1500 hours each shift. This news comes at a bad time. The COO tells you that market demand for output has dramatically grown. The equilibrium price for a widget has risen to \$160.
5.You are told to produce a quantity that maximizes profit. How many units do you produce and what is your profit? How many machine and labor hours are used in production.
6.At this level of profit-maximizing output, what is your marginal cost, average variable cost, and average fixed cost of production?
7.The COO notifies all plant managers in the company of a simplified bonus program. A plant manager's bonus will be tied to one measurement - average production cost. The lower a plant's average production cost, the higher the bonus. Is this type of bonus structure in the interest of the company? Use theoretical and graphical insights from chapter five of the textbook to describe your reasoning.
• Reference No.:- M919575
Have any Question?
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Quadtrees are data structures used to store digital images. For our purposes, the images will be simple bitmaps, where every pixel is either a 1 (black) or a 0 (white). A quadtree representation of a bitmap is obtained as follows: first, associate the root node with the entire image. If the entire image is either all 1’s or all 0’s, then store that value in the node and you’re done. Otherwise divide the region into four equal size quadrants, add four children to the root, and assign each child one of the four regions in the following order: the first child gets the upper left quadrant, the second the upper right, the third the lower left and the fourth the lower right. Then recursively apply the above rules to each of the children. For example, the 4x4 image on the left would be represented by the quadtree on the right:
Note that this procedure only works as stated if the image is a square and has a side length equal to a power of 2. For those images which do not meet those requirements, we pad the rows and columns with 0’s (on the right and on the bottom, respectively) until we have a bitmap of the appropriate size. For example, a 5x13 image would be converted to a 16x16 bitmap (with the original image residing in the upper left portion, and the remainder of the image filled with 0’s).
While quadtrees can result in a significant savings in space over the original image, even more savings can be achieved if we identify repeated subtrees. For example, in the tree above, the first and third subtrees of the root are identical, so we could replace the root of the third subtree with a reference to the first subtree, obtaining something that symbolically looks like the following:
We will call these compressed quadtrees super quadtrees, or squadtrees. For our purposes the use of a reference saves space only when the tree it replaces has height of at least 1. Thus, while we could replace 5 of the nodes which contain a B with references to the first node with a B, this would not in practice save any space in the compression. Using this rule, our squadtree contains only 12 nodes, as opposed to 17 in the original quadtree. Your job for this problem is to take a set of images and determine the number of nodes in both the resulting quadtrees and squadtrees.
### 输入格式
Input will consist of multiple problem instances. Each instance will start with a single line containing two integers n and m, indicating the number of rows and columns in the image. The maximum values for these integers is 128. The next n lines will each contain m characters representing the image to process. A black pixel will be represented by a ‘1’, and a white pixel will be represented by a ‘0’. The input line 0 0 will terminate input and should not be processed.
### 输出格式
For each problem instance, output two integers separated by a single space. The first value is the number of nodes in the quadtree for the problem instance, and the second is the number of nodes in the squadtree.
### 样例
Input
4 4
1011
0111
1010
0111
6 7
1110111
1010101
0000000
0100010
1011101
1010101
0 0
Output
17 12
61 24
0 人解决，1 人已尝试。
0 份提交通过，共有 1 份提交。
9.9 EMB 奖励。
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Since a sample is used to infer the population, the sample must be a representative sample. Read on to see what we mean by that!
In other words, it can’t be biased. Instead, it must represent the entire population as accurately as possible.
Among other things, statisticians could use the following guidelines to ensure that the sample is a representative sample.
First, they can make sure that the survey is not a survey self-selected survey.
In a self-selected survey, the participants volunteer to answer the questions.
When people volunteer, they usually have a strong opinion on the matter.
For example, suppose Candidates A, B, and C are applying for the position of superintendent in a school district.
A sample of 500 people is selected to answer questions about the candidates.
What if 100 of these participants are closely related to Candidate B and volunteer to take the survey?
It is likely that the poll is biased and does not represent the population correctly.
In addition, the sample can not be too small. Imagine that you select a sample of 200 participants from 10,000,000 residents.
It is likely that this sample does not represent the population correctly.
The sample not only has to be big enough, it also has to be Response rate can not be too low.
For example, a sample of 5000 people was selected from a population of 100,000.
The sample size looks really good. However, if only 30 people took the survey, the response rate is too low.
Sometimes the wrong people are chosen to represent the population. There are many ways this can happen.
• Choose just New York State to find the median income of all Americans.
• Using just the height of professional basketball players to estimate the average height of all men.
• ask just Mothers who don’t believe in breastfeeding when breastfeeding is important.
Regardless of whether the sample is representative of the population or not, different studies can still produce different results.
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# Inertial frame of reference
In physics, an inertial frame of reference, or inertial frame for short, is a frame of reference in which the observers move without the influence of any accelerating or decelerating force.
If you can find an inertial frame of reference for a given situation, then it can always be transformed by a change of coordinates into one in which the observers do not move at all.
The transformation from one inertial frame of reference to another is done using Lorentz transformations, or, at speeds considerably below the speed of light, Galilean transformations.
An inertial reference frame is a space-time coordinate system that neither rotates nor accelerates. In real life, such frames of reference are purely theoretical, because gravitational force (and thus acceleration) exists everywhere in the known universe. However, they may be approximated very well in intergalactic space, or to a lesser extent within the confines of a coasting spacecraft.
Inertial frames of reference appear prominently in both Newtonian relativity and Einstein's special theory of relativity.
In Newtonian mechanics, any mass viewed from an inertial reference frame will appear either to be stationary or to be moving at constant speed in a straight line, if and only if the sum of forces acting upon that mass is zero. (This is also known as Newton's first law of motion.)
Different inertial reference frames may have different origins at any given moment in time, and their respective origins may be moving at constant speed and direction relative to each other. An object which is stationary in one inertial frame will also appear to be stationary in another inertial frame.
non-inertial frame of reference is a coordinate system which is accelerating.
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X
•
## How do I determine which size Chessboard will best match my House of Staunton Chess set?
The minimum square size for a set of properly proportioned Staunton Pattern Chessmen is such that the width of the base of the King should be 78% of the width of a square. In other words, you should divide the King's base diameter by 0.78 to determine the ideal square size for the Chessboard. Let's look at an example - the Reykjavik II Chessmen. The Reykjavik II has a King base diameter of 1.75", which is the norm for a 3.75" King. If you enter 1.75 / 0.78 into a calculator, the result is 2.243, which is rounded to 2.25". Hence, the minimum Chessboard square size for the Reykjavik II Chessmen would be a 2.25".
If you purchase a Chessboard that has a smaller square size than was is considered the minimum, the pieces will be crowded together on the Chessboard. This will significantly increase the likelihood of Chess pieces being knocked over during the heat of battle.
Please keep in mind that this formula will determine the minimum square size for the Chessboard. Depending on availability, you can always purchase a slightly larger size Chessboard than what is considered the minimum and it would work just as well. Using the same Reykjavik II Chessmen example, either a 2.25" or 2.375" square Chessboard would be considered ideal. The chart below details the most common King base diameters and its corresponding ideal square sizes.
King Base Diameter Ideal Square Size 1.5" 1.875" - 2.0" Squares 1.75" 2.25" - 2.375" Squares 1.875" 2.375" - 2.5" Squares 2.0" 2.5" - 2.75" Squares
•
## How do I take care of my new House of Staunton Chess set and/or Chessboard?
Your House of Staunton Chessmen should be lightly waxed and buffed either once or twice per year, using only a high quality paste wax with a cotton cloth or cheesecloth. We recommend Liberon's Paste Wax. Please make sure that you carefully follow the manufacturer's instructions on the paste wax carefully, as improper waxing could ruin the finish on your Chessmen. Above all else - use the wax sparingly and allow the wax to dry before buffing. You should NEVER use a liquid polish.
•
## Why are some sets (and boards) so much more expensive?
Many items determine the cost of a chess set. Ebony is more expensive than rosewood, which is, in turn, more expensive that boxwood, etc.
The quality and detail of the Knights is another price factor. A set of finely carved knights can represent up to 50% of the total cost of a chessmen.
Then, the quality of the turnings, the finish and the uniformity also add to cost. Most Indian-made sets are relatively inexpensive. They are almost always of very poor quality. They will have burn and tear marks, burrs, poor finishes, etc. The Indian manufacturers also have no regard for U.S. and International copyright laws, and will make counterfeit copies of most anything on the market. Why buy an inferior copy of something genuine that can last many generations?.
Finally, look at the storage box. A very high quality box can have a value of several hundred dollars or more. Most Indian-made sets come in very inexpensive boxes, hence can be marketed significantly below a high quality chess set housed in a fine box.
•
## Does the House of Staunton have replacement Chess pieces should I lose or break one?
Absolutely! The House of Staunton maintains a very large inventory of replacement Chess pieces, so spares are always available.
Because our Chess sets are handcrafted out of natural materials, there can be some variation in the grain/color/cutting of the Chess pieces, so we strongly encourage those customers interested in a replacement Chess piece to send in an original for matching purposes. This is especially important in the case of Rosewood, Golden Rosewood and Blood Rosewoods, which can vary significantly in color.
If you do not want to send back an original piece, we can not guarantee that replacement pieces will perfectly match with the rest of the Chess set. The price of replacement Chess pieces will vary, depending on the Chess set, wood selection and piece type. Please Contact Us for pricing information.
•
## Do you repair chess pieces and boards?
The House of Staunton is happy to endorse Ron Fromkin, owner of the Japanese Repository Inc. for his expert chess restoration services. We have been using Ron for many years and have always found his work to be top quality and fairly priced.
Ron has been restoring chess and other artworks for the last 24 years. He works in ivory, horn, bone, antler, and other natural materials. He repairs chess boards, and also does lathe turning. Contact him directly at ivoryrepair@yahoo.com. For additional information see his website at www.ivoryrepair.com.
•
## How do I place an order from the House of Staunton?
Via fax You can fax us the items you wish to order, your shipping and payment details, and your contact information (daytime telephone number and email address). We can only accept credit card orders via fax.
Via email: You can email us your order by sending us the items you wish to order, your shipping and payment details, and your contact information (daytime telephone number) to customerservice@houseofstaunton.com. We recommend that you break your order up into multiple emails for security purposes.
Via mail order: You can mail us the items you wish to order, your shipping and payment details, and your contact information (daytime telephone number and email address). We accept credit card, money order and personal check payments via mail order.
We accept no responsibility for those customers who mail us cash. All payments must be made to the order of The House of Staunton, Inc. If you are paying via money order or personal check, please Contact Us to determine the shipping charges.
•
## What forms of payment does The House of Staunton accept?
We accept VISA, MasterCard, Discover, American Express, PayPal (id: sales@houseofstaunton.com), Money Orders (in US Dollars), Personal Checks and Money Wires (on purchases over \$200.00.) Please Contact Us should you have any questions.
•
## What is the best way to insert the cross (finial) into the King?
Generally, you twist and seat the cross. However, the "wings" of the cross can break off rather easily if you twist too hard. You can use soap or wax to lubricate the shank if it does not seat easily. If it is too loose, a small bit of paper can be used to help wedge it tight. It is better not to glue it in, because then it is very hard to fix if it breaks. And it is almost always better to replace a broken cross, rather than the entire piece..
•
## What is the difference between Ebonized and Genuine Ebony Chess pieces?
Ebonization, the process in which Ebonized Chessmen are created,. In this process, Boxwood (the wood used to create the white pieces) chess pieces are treated with a special stain to turn them jet black in color. When done properly, brand new Ebonized Chessmen are nearly indistinguishable from Genuine Ebony Chessmen.
Ebonized Chess pieces are often referred to as faux Ebony, because it has the look and feel of Genuine Ebony at a fraction of the cost.
Over time, the Ebonizing can wear off on the sharp corners and details of the Chessmen, but it is easily restored using a black marker.
•
## Which House of Staunton Chess sets are "triple-weighted"?
Unlike other Chess set manufacturers, The House of Staunton does not describe its Chess sets as triple weighted because this term has no meaning today.
The term was originally coined by the Drueke Company to describe how many metal slugs were inside their Chessmen. Drueke used the term triple weighted to describe Chess Sets that had 3 metal slugs. Since then, many Chess manufacturers have bastardized the term and it is now used to describe any Chess set with any type of metal weighting. .
Rather than use a meaningless marketing terms to describe our Chessmen, we tell you the exact weight of our Chessmen in ounces so you know exactly what you are getting. Please do not fall for such marketing trickery, as we have seen some Chess sets advertised as triple weighted that weighed less than 30 ounces. If the House of Staunton were to use such standards, our Weighted Plastic Collector Chessmen would be septuple (7x) weighted!
•
•
## Where can I found the House of Staunton Return Policy?
The House of Staunton Return Policy can be found here.
•
## What do I do if the item I received was damaged in transit?
The House of Staunton uses the finest packaging materials to protect its packages and adheres to the strictest packing guidelines to minimize the likelihood of your package being damaged in transit.. We understand how frustrating it can be to receive an item that is damaged.
That being said, in the unlikely event that the item that you received is damaged, please rest assured that the House of Staunton is committed to providing the finest customer service and our professional staff will work quickly to resolve the situation and get you a replacement item at absolutely no cost to you.
The House of Staunton Damaged Item Policy is discussed in the Return Policy. If you have any questions about your item, please do not hesitate to contact us
•
While we do not have gift wrapping available, we do have full-color customizable gift cards that can be inserted into the package at no cost. Please let us know at the time of purchase and we'd be happy to help you. We have cards of all types available, including Birthday, Graduation, Hanukkah and All Occasion Gift Cards.
•
The House of Staunton offers a wide selection of custom-made Products that are available for purchase. This includes our Signature Contemporary-line of Chessboards and our 40,000 year-old Mammoth Ivory Chessmen. Given the nature of these products, these are not items that are typically carried by the House of Staunton and all sales are handled on a "custom made" basis. At the time that an order is placed, we require a deposit of 50% of the full purchase price. We do not begin production on the custom item until your deposit has been received.
If you decide that you do not want to purchase the custom-made item, you have the option of converting your deposit into a store credit that can redeemed any other House of Staunton-brand Chess product. The credit will be in the full amount of the deposit.
We are sorry, but no cash refunds will be made for custom items
•
## What is your policy on discounts and combining discounts?
Coupons can NOT be combined with
Any other discounts, such as other coupons and price matching
Any promotional offers, such as free gifts, discounted additional purchases and free shipping.
Any purchases made through 3rd parties, such as House of Staunton-authorized resellers.
Any other discounts - such as other coupons and price matching
Any promotions - such as free gifts or free shipping.
Any purchases made through 3rd parties, such as eBay, Amazon.com and House of Staunton resellers.
Thank you for understanding.
•
## What is the difference between Lacquered and Polished Sets?
The lacquered Chessmen have a high gloss finish, compared to the other Chessmen:
•
## What is the difference between the Collector Chessmen and the Imperial Collector Chessmen?
The difference between the Collector Chessmen and the Imperial Collector Chessmen are the Knights included in the set.
The Collector Chessmen feature the legendary Jaques of London Knight design, which is modeled after the noble steeds from the Greek Parthenon (Elgin Marbles) and are unsurpassed in beauty and craftsmanship.
The Imperial Collector Chessmen feature our Imperial Knight design, which features a down turned head, flowing mane and offers breathtaking level of detail.
The Knights in the image below are of our Ivory Collector Chessmen and are crafted from 40,000 year old Mammoth Ivory.
•
## What are your policies regarding International Shipping?
Please be advised that we will only ship Internationally via trackable methods.
We WILL NOT ship via any other less expensive and untrackable means, such as US Postal Airmail or US Postal Surface/Economy. This is necessary to meet the standard credit card requirement that all sellers be able to provide PROOF OF DELIVERY on all shipments.
When shipping internationally, delays or additional fees assessed by your country's customs department are beyond our control. Our customers are strongly encouraged to research their country's importing rules and if any additional fees, such as GST, duties or VAT are applicable. These fees are not included in the shipping costs and are,regrettably, your responsibility.
If you have any questions, we encourage you to contact your local Customs Officer to determine if any such fees are applicable to your purchase. We have no access to that information and will not be able to assist you in this matter.
•
## What is the House of Staunton Collector's Club?
The House of Staunton Collector's Club offers special savings for individuals interested in collecting House of Staunton-brand Chess Products. For a one-time fee of \$50.00 USD, the individual will receive a lifetime membership to the House of Staunton Collector's Club.
Benefits include:
Discounts on House of Staunton-brand Chess equipment. With a minimum discount of 10% off MSRP, the savings can equal or exceed the membership fee on your first purchase!
Significant discounts on close-out House of Staunton-brand Chess Equipment.
Early notice on sets about to be retired.
Early notice on sets about to be introduced, including ourlimited-edition chessmen we introduce every year.
•
## I need to have a package shipped the same day that I order it. Is that possible?
If you are in a hurry and need your package to be shipped the same day that you order it, add rush processing to your order. Orders placed before 12 PM (US Central time) with rush shipping will ship out the same day.
This option is for customers who want special delivery service, or just want regular shipping - but want the order shipped out ASAP (and do not want to pay for a shipping option with higher rates). To expedite delivery of your order, add RUSH PROCESSING to your order now. Rush orders received by 12:00 pm (US Central Time) on a business day will ship same day. Otherwise they will ship on the following business day.
Please note that all bespoke and custom-made items are shipped directly from the manufacturer or crafted or demand, and same day shipping is not available.
•
## What is the warranty on House of Staunton-brand Chess sets and Chess boards?
From the date the item is purchased, the House of Staunton will warranty that its branded products will be free of manufacturing defects for a period of one year and will replace any items that exhibit manufacturing defects at absolutely no charge to the customer.
Examples of defects that ARE COVERED under this warranty:
If Hairline cracks develop in the base of Wooden Chessmen.
If the metal weight inside become loose and moves around.
Examples of defects that ARE NOT COVERED under this warranty:
Any physical defects that were caused by improper use of the product, such as being dropped on the ground, being chewed by the dog, pieces that are improperly stored.
Any pieces that become lost.
Any pieces that are damaged due to improper waxing.
The above examples do not represent every possible defect; all warranty claims are handled on a case-by-case basis.
This warranty only applies to customers who have purchased a brand new House of Staunton brand chess board or chess set directly from the House of Staunton, or one of our authorized retailers.
Items purchased from non-authoirzed retailers, or in used condition through venues such as eBay or Craigslist are considered to be sold in AS-IS condition, and is not eligible for warranty coverage.
•
## Does the House of Staunton allow Payment Plans / Lay Away on Purchases?
Absolutely! The House of Staunton offers a very flexible payment program (commonly referred to as Lay Away) that allows our customers to make payments on purchases over time.
This program is ideal for customers interested in a limited edition Chess set but lacks the financial resources to complete the purchase immediately. Once we receive the initial payment (10% of the purchase price / Minimum \$50.00), that Chess set will be removed from our inventory and a Production Number (if applicable) is generated.
Once the initial payment has been received, we strongly recommend our customers leave a credit card on file with us and a specific amount will be withdrawn each month. For example, a \$699 Ebony Collector would require a \$70.00 and 9 monthly payments of \$70.00 USD to be paid in full. Once the final payment is received, the Chess set will ship. There are no credit checks or personal information required to participate.
We do require that a payment be made each month and that the purchase be completed within one calendar year. That being said, multiple payments each month are fine and you can pay it off as soon as you want.
•
•
## My purchase is a gift that is being delivered directly to the recipient. Can I include a personalized card in the package so they know who it is from?
Absolutely! The House of Staunton would be happy to include a personalized card in the package for FREE. We include an actual greeting card, complete with sealed envelope!
During the checkout process, you will have the opportunity to enter the personalized message that you want to include on the card. It is located at the bottom of the information entry screen, where you enter your mailing and billing address.
Please be as specific as you can with the message. If the card is for a particular reason, such as Birthday, Graduation, Wedding or Christmas/Hannukah, please specify that and we'll include your message on an appropriate card.
•
## What kind of oil or wax can be used to maintain the wooden chessmen in top condition?
NEVER use an oil on wooden chessmen. It will darken the finish and can cause the finish to become "gummy."
We recommend wiping the chessmen after each use with cloth from an old T-Shirt. Pieces should only be waxed about once a year or less with a high quality paste wax, such as MinWax or Lebrons (Clear color). The finish must be applied very sparingly, and only applied to the areas which are actually handled, never the detailed areas such as the mane, eyes and mouths of the Knights, for example.
Wax should be removed with a piece of a T-Shirt (This same piece of cloth should be kept to wipe off the chessmen after each use). A pointed tooth pick can be used to remove wax residue from the detailed areas if needed.
You can tell if the wax is ready for removal if it offers some resistance to the motion of the cloth. The wax must be removed between 15 and 30 minutes after applying. Never let the wax sit for more than 30 minutes or you will not be able to remove it without destroying the finish.
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