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# Do supernovae produce powerful gravitational waves?

I would think that the amount of energy produced by one supernova is much more than a pair of colliding black holes, no? Wouldn't that much material and energy output from the explosion generate a powerful gravitational wave?

• Define "powerful". – Dmitry Grigoryev Sep 14 '16 at 16:04

Actually the peak luminosity for the first binary black hole merger LIGO detected was $$3.6 \times 10^{49} \,\text{Watts}$$ (source). Whereas the brightest supernova ever recorded was ASASSN-15lh which had a peak luminosity of $$2 \times 10^{38} \,\text{Watts}$$ (source). In fact the energy radiated by the binary black hole merger during the $$0.2 \,\text{s}$$ it took to merge was over 10 times the combined luminosity of the observable universe (source).

This translates to the total energy emitted to be $$5.4 \times 10^{47} \,\text{J}$$ for the black hole merger and $$(1.1 \pm 0.2) \times 10^{45} \,\text{J}$$ for the ASASSN-15lh supernova. So as you can see, the binary black hole merger released around 100 times more energy than the brightest known supernova.

In relation to the gravitational waves emitted by supernovae, their wave-forms are not as well understood as the "chirps" you get from merging black hole binaries, however they are still expected to be strong enough to be detected in the local universe.

Gravitational waves are produced by aspherical accelerating systems. A common example of a system that would produce them is two stars orbiting each other; an example of a system that would not produce them is a totally stationary, non-rotating spherical star. (More specifically, the production of radiation depends on the mass quadrupole moment and its time derivatives.

Supernovae are, on the whole, not spherically symmetric. We can see a simple example of this in pulsar kicks, where a pulsar resulting from a supernova is expelled at an enormous speed (generally several hundred km/s) that cannot be accounted for by the motion of the originals star alone. There must be some drastic asymmetry responsible.

What precisely would cause the emission of gravitational waves in a supernova? There are quite a few potential culprits (see Ott et al. (2012) and Kotake et al. (2005)):

• Asymmetric core bounce, possibly linked to progenitor rotation
• Convection inside the resulting neutron star
• Interactions between the neutron star and the shock wave
• Asymmetric neutrino emission
• Aspherical gas outflows

On the whole, rotation will have a large impact, especially when it comes to core bounce and convection. This should be obvious; it creates an asymmetry before the collapse.

Will we be able to detect those waves? It seems likely. Figure 132 of Kotake et al. indicates that some of our present detectors should be sensitive enough to detect neutrino- and matter- based gravitational waves from close sources ($\sim10\text{ kpc}$):

So far, LIGO hasn't been successful in this area, but it's quite possible for it to detect gravitational waves from supernovae.

If you're interested, I know of at least one site that has openly available data from theoretical simulations that you can compare to detection capabilities.

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# 3.8.3: Kelvin waves

As coastal engineers we are interested in the tidal propagation and the tidal range along the boundaries of oceans and seas. To understand more about the propagation along closed coastal boundaries we need a further understanding of the rotary wave forming the amphidromic system. These waves depend on the existence of a closed boundary. A definition sketch is given in Fig. 3.32. Wave propagation is in the positive $$y$$-direction in the Northern Hemisphere (NH), hence $$c$$ is positive. Such a wave would deflect from this eastern boundary in the Southern Hemisphere (SH) because of Coriolis. Therefore, for the SH we expect a wave propagating in the negative $$y$$-direction ($$c$$ negative).

As before, we assume a more or less horizontal bottom slope and a small tidal amplitude compared to the water depth. Balancing inertia, Coriolis, the pressure gradient and bed friction leads to the following reduced shallow water equations in two horizontal dimensions:

$\dfrac{\partial u}{\partial t} - fv = -g \dfrac{\partial \eta}{\partial x} - \dfrac{\tau_{b,x}}{\rho h}\label{eq3.8.3.1}$

$\dfrac{\partial v}{\partial t} + fu = -g \dfrac{\partial \eta}{\partial y} - \dfrac{\tau_{b,y}}{\rho h}\label{eq3.8.3.2}$

where

 $$u, v$$ velocity in $$x,y$$-direction $$m/s$$ $$f$$ $$2 \omega_e \sin \varphi$$ is the Coriolis parameter with the earth's angular velocity $$\omega_e = 72.9 \times 10^{-6}\ rad/s$$ and the latitude $$\varphi$$ positive in the NH and negative in the SH $$1/s$$ $$\eta$$ oscillatory water level variation $$m$$ $$\tau_{b, x}$$ bottom shear stress in $$x$$ direction $$N/m^2$$ $$\tau_{b, y}$$ bottom shear stress in $$x$$ direction $$N/m^2$$ $$\rho$$ water density $$kg/m^3$$ $$h$$ water depth $$m$$

The Coriolis acceleration has been introduced in the above equations of motion in such a way that it makes a right angle with the particle velocity and acts towards the right (to starboard side) in the NH and towards the left (to port) in the SH (see Intermezzo 3.1). Check for yourself the signs of the Coriolis terms in Eqs. $$\ref{eq3.8.3.1}$$ and $$\ref{eq3.8.3.2}$$. In doing so, note that the latitude $$\varphi$$ and hence the Coriolis parameter $$f$$ are assumed to have positive values in the NH and negative values in the SH.

Continuity requires:

$\dfrac{\partial \eta}{\partial t} + h \left (\dfrac{\partial u}{\partial x} + \dfrac{\partial v}{\partial y} \right ) = 0$

If we neglect friction and realise that the velocity $$u$$ at the closed boundary (the coast- line) is zero the balance equations reduce to:

$-fv = -g \dfrac{\partial \eta}{\partial x}\label{eq3.8.3.3}$

$\dfrac{\partial v}{\partial t} = -g \dfrac{\partial \eta }{\partial y} \label{eq3.8.3.4}$

$h \dfrac{\partial v}{\partial y} = -\dfrac{\partial \eta}{\partial t}\label{eq3.8.3.5}$

Notice that the cross-shore ($$x$$) momentum balance (Eq. $$\ref{eq3.8.3.3}$$) is geostrophic: there is a balance between Coriolis force and the pressure gradient due to water level differences. This is comparable to the influence of Coriolis on the flow in a confined channel (see Intermezzo 3.6). The alongshore momentum balance, Eq. $$\ref{eq3.8.3.4}$$, is the same as that for shallow water gravity waves (the progressive tidal waves from Intermezzo 3.5). Equation $$\ref{eq3.8.3.4}$$ combined with the reduced continuity equation, Eq. $$\ref{eq3.8.3.5}$$ yields a set of equations that it equivalent to Eqs. 3.8.1.2 and 3.8.1.3 and thus has an equivalent solution. Now substitute this solution $$\eta = \hat{\eta} \cos (\omega t - ky)$$ and $$v = g/c \eta$$ with $$c = \pm \sqrt{gh}$$ in the geostrophic flow equation, Eq. $$\ref{eq3.8.3.3}$$. The resulting equation now gives a solution known as a Kelvin wave:

$\eta (x, y, t) = \eta_0 e^{(\tfrac{fx}{c})} \cos (\omega t - ky)$

$v(x, y, t) = \tfrac{c}{h} \eta_0 e^{(\tfrac{fx}{c})} \cos (\omega t - ky)$

The Kelvin wave propagates along the coast (in the $$y$$-direction) at the shallow water speed. The alongshore velocity is in phase with the water level (as for the propagating wave of Intermezzo 3.5). The amplitude is maximum at the coast ($$\eta_0$$) and then decays with distance from the coast (in the negative $$x$$-direction). The scale of the decay is $$c/f$$ in which the variables are the latitude and the water depth:

$\dfrac{c}{f} = \sqrt{gh} \dfrac{1}{1.46 \times 10^{-4} \times \sin \varphi} = 21453 \dfrac{\sqrt{h}}{\sin \varphi}$

At 45° latitude this amounts to about 1900 km for the deep oceans with an average water depth of 4000 m and to a good 200 km for a shallow sea with a typical water depth of 50 m.

The value of $$c/f$$ cannot be negative, since this would make the sea level grow exponentially offshore. This implies, as expected, that the Kelvin wave propagates in the positive $$y$$-direction in the NH (where $$f$$ is positive) and in the negative $$y$$-direction in the SH (where $$f$$ is negative).

## Intermezzo 3.6 Geostrophic momentum balance

If the flow takes place in a confined conduit or channel that prevents a deviation of the course (i.e. a steady current), the Coriolis acceleration causes a pressure gradient across the conduit:

$\dfrac{1}{\rho} \dfrac{\partial p}{\partial n} = 2 \omega_e V \sin \varphi$

where

 $$\rho$$ water density $$kg/m^3$$ $$\rho$$ water pressure $$N/m^2$$ $$n$$ normal to and directed to starboard of the current $$V$$ -

In open channel flow, the pressure gradient becomes visible as a gradient of the water surface:

$\dfrac{1}{\rho} \dfrac{\partial p}{\partial n} = g \dfrac{\partial \eta}{\partial n}$

Note: upon comparing the above two equations (Eqs. 3.45 and 3.46) with the full depth-integrated shallow water equations it can be seen that only the pressure gradient due to the water level surface and the Coriolis acceleration are retained.

As an example, we compute the sea level difference across the Strait of Florida. The Florida Current is located at latitude 26°N; the current velocity is about 1 m/s; the width of the Strait of Florida is about 80 km.

$\dfrac{1}{\rho} \dfrac{\partial p}{\partial n} = 2 \cdot 0.729 \times 10^4 \cdot \sin 26^{\circ} \cdot 1 = 6.4 \times 10^{-5} m/s^2$

The elevation difference over 80 km is computed as follows:

$\Delta \eta = \dfrac{1}{\rho} \dfrac{\partial p}{\partial n} \dfrac{\Delta x}{g} = \dfrac{6.4 \times 10^{-5}}{9.81} \cdot 80 \times 10^3 = 0.52m$

The observed value is 0.45 m, which is close to our estimate. (Similar computations can be made for e.g. the Western Scheldt, the British Channel or the Dutch Texel Inlet).

The Kelvin wave is a coastally trapped wave; it needs a coastline. In the NH, the wave propagates poleward along an eastern boundary and equatorward along a western boundary with its maximum amplitude at the boundary. It thus forms a wave trapped to the boundary and rotating counter-clockwise around an amphidromic point (like a standing wave, but now rotating). The rotation is clockwise in the SH.

In the North Sea multiple amphidromic points can be observed (see Fig. 3.31). The Kelvin wave enters the North Sea basin from the north. Some of the energy is dissipated in the basin and some is reflected from the shallow areas in the southern part of the North Sea. This reflected wave forms its own amphidromic system.

For the ideal Kelvin wave, friction is not taken into account. Neglecting friction can be a good approximation for deeper water. Near the coast inertia is relatively unimportant but bottom friction needs to be taken into account. The velocity is then governed by the balance between the bottom friction and the alongshore water level gradient. This will be treated in more detail in Ch. 5.

This page titled 3.8.3: Kelvin waves is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Judith Bosboom & Marcel J.F. Stive (TU Delft Open) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.


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# Tag Info

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As someone who has now taught CS for many years, let me first reassure you that teaching clear, clean coding is entirely worthwhile. I think this is especially true for those who will not go off into software professionally, because editing, and being able to express ideas in a clean, clear manner is a skill that serves people well in any field. (I've ...

6

I emphasize readability and maintenance of code. Therefore, I will assign a project and later turn around and require another student to complete the second half of it. Everyone ends up maintaining someone else's code as a result. I think you should continue teaching your good habits and thinking styles. That said, you see a student doing Olympiad problems ...

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Back off on clean coding until they can appreciate it. Take named constants (const int maxScore=14) as an example. Showing const along with regular variables will only confuse students -- why is there a special word to make a variable useless? Later using it to "fix" if(s1>14) will probably still confuse them -- it's more typing and more ...

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A famous example is Strassen's matrix multiplication algorithm, whose running time satisfies the recurrence $T(n) = 7T(n/2) + O(n^2)$. A similar example is Karatsuba's algorithm for fast integer multiplication, whose running time satisfies the recurrence $T(n) = 3T(n/2) + O(n)$. Other algorithms for multiplying matrices or integers supply even more examples ...

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I'll give you a straight, simple, and practical answer. Teach clean coding and simple effective thinking. Because in the practical world, the codes are big and have a lot of integration. The reality is that a single coder does not code all of it. As a teacher, you should teach something that is practically feasible. Teach a student to write such a code that ...

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Learning is like searching for a treasure hidden in the forest. The teacher can teach to use tools that make the task easier, also provides some tips in order to deal with implementation problems realted to the tool's maturity. The teacher should avoid re-invent the wheel, althougth must speak about the history of how it was invented.

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Teaching introduces a strong two-way dependency. As a student, one needs a certain amount of trust in the teacher - otherwise, if they already know better, then why on Earth would they need the teacher in the first place. So, if the student is there of their own volition, this already established a need: they need help. The student needs to understand that ...

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Programming vs language This is the difference between learning to program, and learning a language. It will have analogies in human languages (English, French, etc.) One can know a language, but is one understood; is one poetic; is one concise, is one un-ambiguous. Does it scale Un-clean code does not scale. It is like goto and global valuables, they are ...

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Explain and demonstrate the purpose of functions, in context. Putting everything in main() might work for toy programs, but in real programming you want to use functions for a specific reason: you want your code to be able to repeatedly do perform a particular task, producing a particular kind of output when given a particular kind of input. Depending on the ...

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There will come a day when a problem is too hard or too large for him to solve on his own (of course). He is being academically lazy and it will handicap him. Thats a common pitfall for bright kids. It's totally OK to experiment and 'go your own way' when learning for the first time. Learning should be fun. It will help him relate to new information he ...

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What do you mean your students are not going to become professional coders? Is your education only meant to become a hobby for them? You mentioning coding competitions seems to indicate that this is the case. I think what you really meant is that these people will not be full-time programmers but that they will program as part of their jobs. People in this ...

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I'll probably get some hate for this, but I personally hate coding in classes and Java in general, it's just not my cup of tea. But I can appreciate good, clean code. I program mostly in php and this makes the use of proper classes less important, as you can just physically separate it into separate pages. But I understand when classes are necessary/use-full....

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Sciencemadness Discussion Board » Fundamentals » Chemistry in General » Making Potassium metal by distillation with Al and KOH Select A Forum Fundamentals   » Chemistry in General   » Organic Chemistry   » Reagents and Apparatus Acquisition   » Beginnings   » Responsible Practices   » Miscellaneous   » The Wiki Special topics   » Technochemistry   » Energetic Materials   » Biochemistry   » Radiochemistry   » Computational Models and Techniques   » Prepublication Non-chemistry   » Forum Matters   » Legal and Societal Issues   » Detritus   » Test Forum

Author: Subject: Making Potassium metal by distillation with Al and KOH
Yb(NO3)3
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Making Potassium metal by distillation with Al and KOH

I recently came across a video of distilling potassium by reducing KOH with Al powder, I was not aware that it was possible to use aluminium powder to do this, especially since magnesium did not work for codys lab, I, however also got this method to work very well with little difficulty, same as the video I used a CO2 cartridge with a copper tube for my reaction vessel, added 10g of KOH, dried with a gas torch and put in 3.5g of 200mesh Al powder (cheap one not flake or dark), my gas torch proved to be not big and hot enough so I put it in a simple blast furnace, this worked quickly and the potassium distilled over into a test-tube, the potassium seemed pure, no contamination aside from oxidation which is unavoidable without a inert gas, I got about 2g maybe. It passed all my tests: shiny and metallic when fresh, melts about 60*C, ignites and burns the recognizable potassium purple in water then explodes.
As I said I was not aware this reaction worked so any ideas are welcome, this seems like a very easy way to make potassium, easier than the catalyzed magnesium reduction (in my opinion), and also cheaper and more pure than this method but with sodium and KCl

The video came from Огненное ТВ and was called PREPARATION OF POTASSIUM AND SODIUM IN HOME CONDITIONS

The picture is not that great if it came through, the test-tube had potassium coating the inside, and I have not had time to clean it up much.

j_sum1

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Well, I am impressed. This is a good result.
I have seen this video before -- I subscribe to this channel. But I watch few videos since I don't speak Russian.

It sounds like it is worth giving this a crack myself.
Thanks for highlighting this method for us.

AJKOER
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Possible reaction path:

3 [ 2 KOH + Heat --> K2O + H2O ]

Al + 3 H2O + Heat --> Al(OH)3 + 3/2 H2

3/2 [ H2 + K2O = 2 K (g) + H2O ]

3/2 [ K2O + H2O --> 2 KOH ]

Implied net reaction for the above:

6 KOH + Al + Heat --> Al(OH)3 + 3 K (g) + 3 KOH

or simply:

3 KOH + Al + Heat --> Al(OH)3 + 3 K (g)

[Edited on 1-5-2018 by AJKOER]
Sigmatropic
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Good speculation AKJOER, except that Al(OH)3 will dehydrate into aluminium oxide, releasing water. Which will feed back into the reaction, chewing up more Al under the formation of hydrogen. Also why exactly do you think hydrogen reacts with potassium oxide to form water and elemental potassium. It is not only strongly endothermic but also uphill with regard to the oxidation potential.
I would speculate this reaction is better described by:
6 KOH + 2 Al --> K2Al2O4 + 2 K2O + 3 H2
Seeing that K2Al2O4 is comparable to the complex K2O*Al2O3 I would argue the equations below are equivalent.
2KOH - - > K2O + H2O
3 H2O + 2 Al --> Al2O3 + 3 H2

To arrive at potassium metal a thermitic reaction between K2O and Al. Probably also ending in K2Al2O4.
4K2O + Al --> 2 K + K2Al2O4
VSEPR_VOID
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Are there any high temperature desiccants that can be used to absorb any water produced by the thermal decomposition of the aluminium hydroxide?

I am not on a network that permits me to go to ebay and see, but could you use your method to produce cheap potassium metal and sell it? KOH and Al powder are dirt cheap.

[Edited on 1-5-2018 by VSEPR_VOID]

AJKOER
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I did some research, and per a source on the action of H2 on NaOH (which I think may parallel KOH), we have:

H2 + NaOH = NaH + H2O (see Eq 1 on page 2192)

and at 360 C :

NaH + Heat --> Na + 1/2 H2

Source: See 'Revisiting the Hydrogen Storage Behavior of the Na-O-H System' by Jianfeng Mao , et al., in Materials 2015, 8, 2191-2203; doi:10.3390/ma8052191 . Link to download file: https://www.mdpi.com/1996-1944/8/5/2191/pdf . To quote:

"The DTA profile for NaH shows one endothermic peak at ca. 360 °C, which can be assigned to the decomposition of NaH to Na metal and hydrogen."
--------------------------------------------------

Actually, in my opinion, the actual chemistry is likely a function of the highly anodic metal (like Mg, Zn, Al, Fe, Sn,.., see http://www.zygology.com/cms/upload_area/pdf/Zyg-Anodic-Index... where Zn and Al are the frequent cited sources of so called 'nascent hydrogen' with HCl) reacting with water as follows:

H2O = H+ + OH-

Al --> Al3+ + 3 e-

e- + H+ = .H = 1/2 H2 (as .H + .H = H2)

Hence, the cited Equation (1) above could actually be the net of the following reactions:

.H + OH- = e- + H2O = e-(aq) (see Eq 37 and Table 2 at https://pdfs.semanticscholar.org/d696/b35956e38351dd2eae6706... )

e- + .H = H- (see http://pubs.rsc.org/-/content/articlelanding/1966/tf/tf96662... )

Net of the two above:

2 .H + OH- = H- + H2O

or, seemingly (but not so at very high temperatures as H2 + Heat = .H + .H):

H2 + OH- = H- + H2O

Further:

H- + Heat --> .H + e- = 1/2 H2 + e- (as .H + .H = H2)

and:

e- + Na+ = Na

My point is if one is going to give credit to the hydrogen, and ignore the role of the highly anodic metal, it is really the mono-atomic hydrogen radical (.H formed courtesy of the anodic metal or high temperatures), or as historically sometimes referred to, in my estimation, as 'nascent' hydrogen. Here is a recent account of the surface chemistry of adsorbed .H on aluminum and applications, see https://books.google.com/books?id=1etfSdk55SYC&pg=PA817&... . This last reference per Equation (5) appears to support my reaction presentation in terms of 2 .H (which is adsorbed on an aluminum surface) and not H2, to quote:

" PbS + 2 .H = Pb + H2S (5) "

[Edited on 3-5-2018 by AJKOER]

[Edited on 3-5-2018 by AJKOER]
Yb(NO3)3
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Thanks for the input, to add to it in practice not much water is produced, just slight condensation in the test-tube, so that's not a problem

VSEPR_VOID It is very cheap but laws and regulations in my country make it almost imossible to buy potassium let alone sell it, I had to wait 2 years for Al powder to be available.
Foeskes
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Cody used KCl not KOH. I'm pretty sure using KOH with magnesium would work but would be really violent and self sustaining. Like the Mg and NaOH reaction.
Diachrynic
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This is great! How do you think it got? More than 1000 °C?

we apologize for the inconvenience
Yb(NO3)3
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Diachrynic The potassium boils at 759*C so it would not be necessary to get to 1000*, also I used copper tubing to distill out the potassium so I couldn't let it get that hot, unfortunately when I attempted cesium a while ago the copper tubing did melt, just when I had something I couldn't afford to fail it did.

Just a photo of the reaction vessel, cost me about $4, no welding or brazing just a friction fit, so pretty much anyone could do it. Yb(NO3)3 Harmless Posts: 6 Registered: 30-4-2018 Member Is Offline Mood: No Mood The photos didn't come through so here they are: Yb(NO3)3 Harmless Posts: 6 Registered: 30-4-2018 Member Is Offline Mood: No Mood I just made a basic video on the process so thought I might mention it here, I have got almost exactly the same results twice in a row so it doesn't seem hard. video: https://youtu.be/2ORTcQB8kOc j_sum1 Administrator Posts: 5174 Registered: 4-10-2014 Location: Oz Member Is Offline Mood: Metastable, and that's good enough. Nice! Well done. This looks like a reliable, efficient setup. VSEPR_VOID International Hazard Posts: 719 Registered: 1-9-2017 Member Is Offline Mood: Fullerenes What heat source did you use to get it to 1000C? Within cells interlinked Within cells interlinked Within cells interlinked Yb(NO3)3 Harmless Posts: 6 Registered: 30-4-2018 Member Is Offline Mood: No Mood I used wood as fuel for the furnace, it does burn through it very quickly but coal is too hot and can melt the copper tubing, I think charcoal briquettes would be probably be better. Foeskes Hazard to Others Posts: 156 Registered: 25-2-2017 Member Is Offline Mood: No Mood I decided to try it out and it worked! (I used a small butane brazing torch by kovea in a flower pot lined with sand and plaster.$
I got 3 spheres of potassium and when added to water it burns with a nice purple flame.
I used a co2 canister with a high pressure rated brass tube (just pushed in no screws)
For some reason the zinc on the receiving end of the apparatus disappeared leaving only copper.
And sone wired black chunks were in the oil with the K, it reacts with water violently while sinking and seems to produce CuO and is brittle. I assume it's a mix of potassium and copper oxide or some kind of cuperate from burning potassium.
The main problem I have is before the potassium drops into oil it would stick to the tube and oxidize.
I put some butane in the container so ignore the bubbles.

aga
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Sweet !

metalresearcher
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I tried, but to no avail

I used a welded steel retort (22mm OD) attached with a gas pipe coupling to a 12mm copper tube ending in a test tube with lamp (mineral) oil.
I first heated the KOH to 600 C to dry it and then let the lye (which is actually molten KOH) freeze and cool down and added the Al powder and tightened the screws very tightly. Then I heated the retort to 950 C (measured with an infrared contactless pyrometer), but it only bubbled slightly for less than a minute and after that nothing happened, even no bubbling. A few times I turned off the burner to check for leaks (which then should appear as white smoke) but no smoke was detected and when the flame was on, no purple/magenta coloring (the K+ ion color) of the flame took place.

But adding water in the opened retort afterwards resulted in a small explosion (so actually a very few K metal was in it), but why did it not evaporate, despite the retort temperature is well above the bp of K (768 C) ?

elementcollector1
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Improper mixing, maybe? Though the aluminum would also be molten at that point...

What was your stoichiometry? Maybe you lost most of your K to aluminate and need more starting Al.

Elements Collected:52/87
Latest Acquired: Cl
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metalresearcher
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I used stabilized Al powder (3.5g) and 11g KOH of which is I guess 7-8 g real KOH, rest water.
Today I did another try, now with Mg turnings.
I let it run for about 20 minutes, as long as it bubbled in the test tube with the oil.

Here a video of yesterdays attempt with Al powder and after it today's attempt.

https://www.metallab.net/jwplayer/video.php?v=L2NsaXBzL0tPSC...

Afterwards there was some tiny amount of K in the retort, because when watering it, flames appeared (see video).

[Edited on 2019-10-15 by metalresearcher]

draculic acid69
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Koh is about 10%water until it's dried so 11g would be more like 9.9g koh
7-8 is a bit on the low side.

 Sciencemadness Discussion Board » Fundamentals » Chemistry in General » Making Potassium metal by distillation with Al and KOH Select A Forum Fundamentals   » Chemistry in General   » Organic Chemistry   » Reagents and Apparatus Acquisition   » Beginnings   » Responsible Practices   » Miscellaneous   » The Wiki Special topics   » Technochemistry   » Energetic Materials   » Biochemistry   » Radiochemistry   » Computational Models and Techniques   » Prepublication Non-chemistry   » Forum Matters   » Legal and Societal Issues   » Detritus   » Test Forum


text:
## swiftskier96 Group Title Help!?!?!?!? 16. What is the equation, in standard form, of a parabola that models the values in the table? x - [-2, 0, 4] f(x) - [4, -6, 70] y = 3x² + 4x – 6 y = 4x² + 3x – 6 y = 6x² + 3x – 4 y = –4x² – 3x + 6 one year ago one year ago

1. ZeHanz Group Title

Standard form is y=ax²+bx+c. You've got these points on it: (-2, 4), (0,-6) and (4,70). Just substitute these for x and f(x) in the standard formula. Hint: begin with (0,-6) for a quick start!

2. swiftskier96 Group Title

So id begin with something like -6 = 3(0)^2 + 4(0) - 6 and if i got -6, the equation would be right?

3. swiftskier96 Group Title

4. phi Group Title

yes, that is the easy way. notice that the last 2 equations will fail

5. phi Group Title

because with x=0, just ignore any term with x in it.

6. swiftskier96 Group Title

So you're saying it has to be either A or B?

7. phi Group Title

y = 3x² + 4x – 6 y = 4x² + 3x – 6 y = 6x² + 3x – 4 y = –4x² – 3x + 6 with x= 0 these turn into y= -6 y= -6 y= -4 y= +6 only the first two have y=-6.

8. phi Group Title

are you sure you don't have a typo in f(x) values?

9. swiftskier96 Group Title

Ok. How do i figure out which ones the right answer? No, those are the right values.

10. phi Group Title

test y = 3x² + 4x – 6 replace x with -2 y = 3* -2*-2 + 4*-2 + -6 what is that?

11. phi Group Title

Do I have to multiply and add little numbers for you ?

12. swiftskier96 Group Title

I got -38. I think thats wrong... Hold on. Ok seriously? Im trying to do it. Im not as stupid as you think i am. Just give me a second.

13. phi Group Title

I think you are smart, but don't like to do this stuff

14. phi Group Title

always do it in small methodical steps y = 3* -2*-2 + 4*-2 + -6 3 * (-2*-2) is 3*4 = 12 4*-2 is -8 -6 is -6 add up 12-8-6 = 12 -14 = -2

15. ZeHanz Group Title

y=ax²+bx+c. Point (0,-6), so x = 0, y = -6: $-6=a \cdot 0^2 + b \cdot 0 + c \Leftrightarrow -6 = c$ So now you know c = -6. Now try the next point, (-2, 4), so set x = -2 and y = 4 in y=ax²+bx-6:$4=a \cdot (-2)^2 + b \cdot -2 -6 \Leftrightarrow 4a-2b-6=4 \Leftrightarrow 4a-2b=10$Last point, (4, 70), so set x = 4 and y = 70:$70=a \cdot 4^2 + b \cdot 4 -6 \Leftrightarrow 16a+4b-6=70 \Leftrightarrow 16a+4b=76$So now you have to solve the system: 4a-2b=10 16a+4b=76

16. phi Group Title

which does not match 4 . so it looks like equation 2 is the answer we can check: y = 4x² + 3x – 6 replace x with -2: y = 4* -2*-2 +3 * -2 -6 type this into google 4* -2*-2 +3 * -2 -6=

17. swiftskier96 Group Title

I got 4.

18. phi Group Title

so it must be equation 2 if we really wanted to check we could replace x with 4, and we will get 70

19. ZeHanz Group Title

So now you have to solve the system: 4a-2b=10 multiply with 2 16a+4b=76 leave the same Then it is: 8a-4b=20 16a+4b=76 ----------- + 24a = 96, so a = 96/24 = 4 Plug a=4 in 4a-2b=10, this gives: 16-2b=10, so 2b=6 and b =3 Formula: y=ax²+bx+c, with a=4, b=3, c=-6: Answer: y = 4x²+3x-6, that's answer B.


text:
### Advances in Logic Locking: Past, Present, and Prospects

Hadi Mardani Kamali, Kimia Zamiri Azar, Farimah Farahmandi, and Mark Tehranipoor

##### Abstract

Logic locking is a design concealment mechanism for protecting the IPs integrated into modern System-on-Chip (SoC) architectures from a wide range of hardware security threats at the IC manufacturing supply chain. Logic locking primarily helps the designer to protect the IPs against reverse engineering, IP piracy, overproduction, and unauthorized activation. For more than a decade, the research studies that carried out on this paradigm has been immense, in which the applicability, feasibility, and efficacy of the logic locking have been investigated, including metrics to assess the efficacy, impact of locking in different levels of abstraction, threat model definition, resiliency against physical attacks, tampering, and the application of machine learning. However, the security and strength of existing logic locking techniques have been constantly questioned by sophisticated logical and physical attacks that evolve in sophistication at the same rate as logic locking countermeasure approaches. By providing a comprehensive definition regarding the metrics, assumptions, and principles of logic locking, in this survey paper, we categorize the existing defenses and attacks to capture the most benefit from the logic locking techniques for IP protection, and illuminating the need for and giving direction to future research studies in this topic. This survey paper serves as a guide to quickly navigate and identify the state-of-the-art that should be considered and investigated for further studies on logic locking techniques, helping IP vendors, SoC designers, and researchers to be informed of the principles, fundamentals, and properties of logic locking.

Available format(s)
Publication info
Preprint. MINOR revision.
Keywords
Logic Locking
Contact author(s)
h mardanikamali @ ufl edu
History
Short URL
https://ia.cr/2022/260

CC BY

BibTeX

@misc{cryptoeprint:2022/260,
author = {Hadi Mardani Kamali and Kimia Zamiri Azar and Farimah Farahmandi and Mark Tehranipoor},
title = {Advances in Logic Locking: Past, Present, and Prospects},
howpublished = {Cryptology ePrint Archive, Paper 2022/260},
year = {2022},
note = {\url{https://eprint.iacr.org/2022/260}},
url = {https://eprint.iacr.org/2022/260}
}

Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.


text:
# Triangular waveform to square waveform circuit

I am working on a frequency counter project and I am in the midst of thinking about how to go about doing a triangular wave to square wave converter. I have done some research but couldn't find any circuit diagram that could help. It was mostly just triangular wave to sine wave. I would appreciate if anyone could help. Thanks.

• So I'm guessing the triangle-to-sine converter circuits you found were probably all integrators? What do you think you'd get if you fed a triangle into a differentiator? – brhans Apr 3 at 11:51
• Or a comparator? – Andy aka Apr 3 at 11:53
• The right answer depends a bit on whether you need it to work over a wide range of amplitudes and/or frequencies, and whether the triangle wave has noise on it that has to be rejected. – Spehro Pefhany Apr 3 at 12:36
• Differentiate it. – Chu Apr 3 at 18:20

A comparator is used to produce square waves:

simulate this circuit – Schematic created using CircuitLab

At high frequencies ($$\ > \frac{1}{2\pi R C}\$$), the R-C filter passes only the DC offset of the input signal, and the output switches at the zero-crossings relative to this level.

At low frequencies, the R-C filter simply produces a time lag and the output switches based on the slope of the input signal.

Something like the following would work, change the ratio of r2 to r3 to change the duty cycle

simulate this circuit – Schematic created using CircuitLab

The LM393 is a comparator, when the signal at the + input is greater the signal at the - input the output is pulled high via R1, when the signal at - is greater than the signal at + the output is driven low. The potential divider R2 and R3 defines the set point at which the transition from high to low happens.


text:
# Proof of Peano's existence theorem for ODEs

Theorem: Given a continuous function $f:(a,b)\times (c,d)\to \mathbb R$, the problem $$\begin{cases} y'(t)=f(t,y(t))\\ y(t_0)=y_0\end{cases}$$ has a local solution in some neighbourhood of a given point $t_0$.

A "standard" proof of Peano existence theorem makes use of the Ascoli-Arzelà theorem. I suspect that the following proof, which doesn't, is therefore wrong. But where?

Consider, instead of the open rectangle $(a,b)\times (c,d)$, a closed rectangle contained in it (since the thesis is local, this makes no difference), say $[a',b']\times[c',d']$. Consider polynomials $P_n(t,y)$ which converge uniformly to $f$ on the closed rectangle. These exist because of the Weiertrass theorem. Now the problems $$\begin{cases} y_n'(t)=P_n(t,y_n(t))\\ y_n(t_0)=y_0\end{cases}$$ have a unique global solution in $[a,b]$, since $P_n$ is a polynomial, hence globally Lipschitz on the rectangle (recall that globally Lipschitz implies global existence by gluing local solutions).

Now we say that the function $P_n$ has an uniform limit $f$, and therefore the functions $t\to y_n'(t)$ have an uniform limit, which we shall denote by $g(t)$. (This is the point which I am less sure of.)

Now we have a sequence of functions $y_n$ which take the same value in $t_0$, which are $C^1([a',b;])$ and whose derivatives converge uniformly to a function $g$. Therefore a limit for $y_n$ exists, is differentiable in a neighbourhood of $t_0$ and its derivative is $g$. (This is a theorem of elementary analysis: $f_n(a)\to l$ for a given $a$, $f'_n\to g$ uniformly, then $f_n$ has a limit $f$ which satisfies $f(a)=l$, $f'=g$.)

Thank you in advance for helping me finding the (possible) error.

I am not sure the proof is wrong or correct; however here are is one unwarranted conclusions that is drawn.

You say that the problems have a unique global solution in $[a,b]$. I don't see that hold. They have a unique ${\bf local}$ solution, but the interval where the uniqueness holds may depend on $n$ and may well shrink to 0. To check that this does or does not happen you need to unpack the proof of existence and uniqueness and see if you can guarantee a lower bound to the width of the existence-uniqueness intervals.

The argument you use that "globally Lipshitz etc" is not quite correct. Your polynomial approximates $f$ in the given rectangle. The solution may leave that rectangle. Moreover the Lipshitz constants for your polynomials $P_n$ is not uniformly bounded with respect to $n$ (in general).

As an example consider the simple equation

$y' =y^2,\ \ \ y(0)=1$. The solution is $y(t)= \frac {1}{1-t}$ which has no global existence on [0,2] (for example), although the ODE is nice and polynomial (and even time-independent).

In your ode you should have $y_n'(t)=P_n(t,y_n(t))$. And you have to extract from the sequence of $y_n$ a converging subsequence. For this Ascoli helps a lot. After that your argument works to show that the limit of the subsequence verifies the ode.

• Thank you. So there is need to extract also to say that the $y_n'$ converge? Uniform convergence of the $P_n$ (as two-variable polynomia) doesn't suffice, right? – W. Rether Sep 1 '16 at 19:18
• The convergence of $y_n'=P_n(t,y_n(t))$ (and the fact that the limit is continuous) comes for free when $y_n$ and $P_n$ converges uniformly. Luckily enough because you can't get that from Ascoli. – H. H. Rugh Sep 1 '16 at 20:10

I think that one of the major problems proof-verifying your question is that there are missing details. I don't know if you agree with me in this point but it was the feeling I got first time I read your question. In particular, the objections made by @user490850 made sense to me at that time and I won't repeat them here.

But the basic idea of using Weierstrass' approximation is good and it dates backs according to my research to Constantin Corduneanu who developed this proof in his article "Sur les inégalités différentielles" (it wasn't the main theme of the article). I'm pretty sure that he quoted Differential Equations by Birkhoff and Rota in its edition of the 30s.

Your theorem is a particular case. You could very well take $\mathbb{R}^n$ instead of $\mathbb{R}$ as in Proving Peano's Existence Theorem by approximating with $C^{\infty}$ functions using Weierstrass' Theorem. I haven't considered the latter as a duplicate for this reason. You can check my answer there as well as the linked document that contains the notations that may be strange and other details. On the other hand, the tough part is the $\mathbb{R}$ case...


text:
How to compute this integral?

I'm trying to calculate this integral :

$$I(z,k,a)= \displaystyle\int_{1}^{\infty} t^2\, \text{ArcTanh} \left(\sqrt{\frac{t^2-1}{t^2}} \dfrac{k}{z}\right)\, e^{-a\,t} \, dt$$

Where :

• $\textrm{ArcTanh(...)}$ : is the inverse hyperbolic tangent function.

• $z=x+i\,y$ : is a complex with the assymption $y<<<x$.

• $k$ and $a$ are reals $>0$.

Mathematica didn't calculate it. Maybe it is too complicated to be done.

In Mathematica input form :

Assuming[{a>0&&k>0},Integrate[t^2 (ArcTanh[Sqrt[(t^2-1)/t^2](k/z)])Exp[-a t],{t,1,Infinity}]]

If I simply enter that into Mathematica, it instantly returns the same expression.

Please, how do I go about this ? Is there any tricks that can be applied ?

Is there a way to get the symbolic result ?

Thank's.

• i like to start by considering a particular case, say a=k=z=1. In this case even that does not yield a closed form result, so I'd call to prospect of obtaining a result for the general case doubtful. May 9 '16 at 19:26
• Is a symbolic solution absolutely necessary, or could you get away with calculating values of this integral numerically after assigning specific values to the parameters? May 9 '16 at 20:40
• @MarcoB Numerical solutions are already calculated, but I would like to calculate the symbolic solution for solving a non linear equation and discuss the analytical solutions. May 9 '16 at 20:53
• @Betatron Then this may not be a Mathematica problem, but a mathematical one, i.e. the existence of a closed-form antiderivative is doubtful. Notice also that, for what it's worth, your Assuming[...] expression evaluates for close to 30 s then returns the expression unevaluated in MMA 10.4.0. May 9 '16 at 22:32
• If you are sure that analytic solution exists please provide some background and preceding steps of your derivation. May 10 '16 at 7:45

2 Answers

Preliminary post

For k/z = 1 there is a closed form result:

I[1,1,a]=((4 + a^2) BesselK[0, a] + a (4 BesselK[1, a] + a BesselK[2, a]))/(2 a^3)

The derivation and the extension to k/z != 1 requires some manual interaction to help Mathematica which we will show in the following.

Solution

Summary

As the integral to be calculated is returned unevaluated by Mathematica we show here to get useful partial results by using a method I like to call "man-machine-interaction". This means we carry out a joint work with paper and pencil on one side and Mathematica on the other side.

The case m = k/z == 1 can be solved symbolically. If m < 1 we show how to get the symbolic soluton to aribtrary precision as a power series in (1-m).

Preliminary remarks

1) The integral to be calculated is

fI := Integrate[
t^2 (ArcTanh[m Sqrt[(t^2 - 1)/t^2]]) Exp[-a t], {t, 1, Infinity},
Assumptions -> {a > 0, 0 < m < 1}]

The numerical integral presents, however, no problem

fN[a_, m_] :=
NIntegrate[t^2 (ArcTanh[m Sqrt[(t^2 - 1)/t^2]]) Exp[-a t], {t, 1, Infinity}]

2) Here we have set m = k/z, because k and z enter only as a quotient.

3) Also we have introduced natural restrictions on the parameters "a" and "m". The restriction for m comes from the requirement that fI be a real quantity. This in turn requires the argument of the ArcTanh to be between 0 and 1 if we keep the upper limit of integration of t at infinity.

4) The factor t^2 in the integrand can be generated by differentiating twice with respect to "a". Hence it is sufficient to study the simpler intergral without the factor t^2.

Hence the integrand to be studied becomes

fi = Exp[-a t] ArcTanh[m Sqrt[(-1 + t^2)/t^2]];

Step by step solution

The first step will be a partial integration. This step is not done automatically by Mathematica, hence we do it manually.

fi can be written as

fi = fid + fi1

where

fid1 = -(1/a) Exp[-a t] ArcTanh[m Sqrt[(-1 + t^2)/t^2]];

fid = D[fid1, t];

and

fi1 = Simplify[ (E^(-a t) m (2/t - (2 (-1 + t^2))/t^3))/(
2 a Sqrt[(-1 + t^2)/t^2] (1 - (m^2 (-1 + t^2))/t^2)), {a > 0, t > 0,
0 < m < 1}]

(*
Out[99]= (E^(-a t) m)/(a Sqrt[-1 + t^2] (t^2 - m^2 (-1 + t^2)))
*)

Indeed

Simplify[fi == fid + fi1, {a > 0, t > 0, 0 < m < 1}]

(* Out[101]= True *)

Because

fid1 /. t -> 1

(* Out[102]= 0 *)

Limit[fid1, t -> \[Infinity], Assumptions -> a > 0]

(* Out[103]= 0 *)

the integral becomes

fI2 := Integrate[(E^(-a t) m)/(
a Sqrt[t^2 - 1] (t^2 + m^2 (1 - t^2))), {t, 1, \[Infinity]},
Assumptions -> {a > 0, 0 < m < 1}]

This integral is also returned unevaluated by Mathematica. Hence we look first at a special case.

The case m = 1

If m = 1 the integrand simplifies and the integral can be done expicitly

(E^(-a t) m)/(a Sqrt[t^2 - 1] (t^2 + m^2 (1 - t^2))) /. m -> 1

(* Out[125]= E^(-a t)/(a Sqrt[-1 + t^2]) *)

fI3 = Integrate[%, {t, 1, \[Infinity]}, Assumptions -> a > 0]

(* Out[126]= BesselK[0, a]/a *)

In order to find the original integral we need to differentiate this expression twice

fI4 = D[fI3, {a, 2}] // Simplify

(*
Out[127]= ((4 + a^2) BesselK[0, a] + a (4 BesselK[1, a] + a BesselK[2, a]))/(2 a^3)
*)

The result in the original designation

$$I(1,1,a)=\frac{\left(a^2+4\right) K_0(a)+a (4 K_1(a)+a K_2(a))}{2 a^3}$$

The numerical agreement is excellent:

Plot[fN[a, 1]/fI4, {a, 0.1, 2}, PlotRange -> {{0, 2}, {0.99, 1.01}},
PlotLabel ->
"Integral for m = 1\nComparison of symbolical and numerical solution",
AxesLabel -> {"a", "fN/fI4"}]

Series expansion

Now we can proceed further developing the integrand of fi1 into a power series about m = 1.

Remark: Although it leads to cumbersome expressions I took four terms in order to possibly find a rule in the result (but, alas, I didn't)

Series[fi1, {m, 1, 4}] // Normal

(*
Out[122]= E^(-a t)/(a Sqrt[-1 + t^2]) + (E^(-a t) (-1 + m) (-1 + 2 t^2))/(
a Sqrt[-1 + t^2]) + (E^(-a t) (-1 + m)^2 (1 - 5 t^2 + 4 t^4))/(
a Sqrt[-1 + t^2]) + (
E^(-a t) (-1 + m)^4 (-1 + t^2)^(3/2) (1 - 12 t^2 + 16 t^4))/a + (
E^(-a t) (-1 + m)^3 (-1 + 9 t^2 - 16 t^4 + 8 t^6))/(a Sqrt[-1 + t^2])
*)

fis = List @@ %;

Timing[Integrate[fis, {t, 1, \[Infinity]}, Assumptions -> {a > 0}]]

(*
{1233.4843069, {BesselK[0, a]/a, ((-1 + m) BesselK[2, a])/a, (
3 (-1 + m)^2 (4 a BesselK[0, a] + (8 + a^2) BesselK[1, a]))/a^4, (
15 (-1 + m)^4 (a (112 + a^2) BesselK[2, a] +
4 (168 + 5 a^2) BesselK[3, a]))/
a^6, ((-1 +
m)^3 (24 a (20 + a^2) BesselK[0, a] + (960 + 168 a^2 + a^4) BesselK[1,
a]))/a^6}}
*)

Finally, we need the second derivative with respect to a:

D[%[[2]], {a, 2}] // Simplify

(*
{((4 + a^2) BesselK[0, a] + a (4 BesselK[1, a] + a BesselK[2, a]))/(
2 a^3), ((-1 + m) (a^2 BesselK[0, a] + 4 a BesselK[1, a] +
8 BesselK[2, a] + 2 a^2 BesselK[2, a] + 4 a BesselK[3, a] +
a^2 BesselK[4, a]))/(4 a^3), (
3 (-1 + m)^2 (16 a (20 + a^2) BesselK[0,
a] + (640 + 144 a^2 + 3 a^4) BesselK[1, a] +
a (8 + a^2) (16 BesselK[2, a] + a BesselK[3, a])))/(4 a^6), (1/(
4 a^8))15 (-1 + m)^4 (a^3 (112 + a^2) BesselK[0, a] +
32 a^2 (91 + a^2) BesselK[1, a] + 29568 a BesselK[2, a] +
592 a^3 BesselK[2, a] + 2 a^5 BesselK[2, a] +
112896 BesselK[3, a] + 5184 a^2 BesselK[3, a] +
52 a^4 BesselK[3, a] + 16128 a BesselK[4, a] +
432 a^3 BesselK[4, a] + a^5 BesselK[4, a] +
672 a^2 BesselK[5, a] + 20 a^4 BesselK[5, a]), (1/(
4 a^8))(-1 + m)^3 (8 a (10080 + 600 a^2 + 7 a^4) BesselK[0, a] +
3 (53760 + 11840 a^2 + 368 a^4 + a^6) BesselK[1, a] +
a (8 (2880 + 456 a^2 + 7 a^4) BesselK[2, a] +
a (960 + 168 a^2 + a^4) BesselK[3, a]))}
*)

Observation

Related simpler example where Mathematica needs some help

This integral is returned unevaluated

Integrate[Exp[-a Cosh[u]], {u, 0, \[Infinity]}, Assumptions -> a > 0]

(*
Out[124]= Integrate[E^(-a Cosh[u]), {u, 0, \[Infinity]}, Assumptions -> a > 0] *)

But with the substitution Cosh[u] -> t we obtain an eqivalent expression of the integral which now is deone by Mathematica:

Integrate[Exp[-a t]/Sqrt[t^2 - 1], {t, 1, \[Infinity]}, Assumptions -> a > 0]

(* Out[142]= BesselK[0, a] *)

• "[...] I'll show it later." -- Looking forward to see it! :) May 15 '16 at 17:58
• @Dr. Wolfgang Hintze, If I simply enter into Mathematica I[1,1,a] it instantly returns the same expression. Please, How did you get the closed form I[1,1,a]=((4 + a^2) BesselK[0, a] + a (4 BesselK[1, a] + a BesselK[2, a]))/(2 a^3) ? Thanks May 15 '16 at 20:30

For k/z = 1, and integrating by parts:

ClearAll["Global*"]
Iv[t_] := Exp[-a*t]*t^2;
u[t_] := ArcTanh[Sqrt[(t^2 - 1)/t^2]];
v = Integrate[Iv[t], t];
Du = Simplify@D[u[t], t];
Int == u[t]*v - Integrate[Du*v, t]

HoldForm[Integrate[Exp[-a*t]*t^2*ArcTanh[Sqrt[(t^2 - 1)/t^2]], {t, 1, Infinity}] ==
Limit[-((E^(-a t) (2 + 2 a t + a^2 t^2) ArcTanh[Sqrt[(-1 + t^2)/
t^2]])/a^3), t -> Infinity] -
Limit[-((
E^(-a t) (2 + 2 a t + a^2 t^2) ArcTanh[Sqrt[(-1 + t^2)/t^2]])/
a^3), t -> 1] +
Integrate[(E^(-a t) (2 + 2 a t + a^2 t^2))/(
Sqrt[1 - 1/t^2] t), {t, 1, Infinity}]/a^3]

$\int_1^{\infty } \exp (-a t) t^2 \tanh ^{-1}\left(\sqrt{\frac{t^2-1}{t^2}}\right) \, dt=\lim_{t\to \infty } \, -\frac{e^{-a t} \left(2+2 a t+a^2 t^2\right) \tanh ^{-1}\left(\sqrt{\frac{-1+t^2}{t^2}}\right)}{a^3}-\lim_{t\to 1} \, -\frac{e^{-a t} \left(2+2 a t+a^2 t^2\right) \tanh ^{-1}\left(\sqrt{\frac{-1+t^2}{t^2}}\right)}{a^3}+\frac{\int_1^{\infty } \frac{e^{-a t} \left(2+2 a t+a^2 t^2\right)}{\sqrt{1-\frac{1}{t^2}} t} \, dt}{a^3}$

Limit[u[t]*v, t -> Infinity, Assumptions -> a > 0] -
Limit[u[t]*v, t -> 1, Assumptions -> a > 0]

$0$

ExpandAll@Simplify[Du*v, t > 0]

in1 = Assuming[{a > 0, t > 0}, Integrate[-((2 E^(-a t))/(a^3 Sqrt[-1 + t^2])), {t, 1, Infinity}]];
in2 = Assuming[{a > 0, t > 0}, Integrate[-((2 E^(-a t) t)/(a^2 Sqrt[-1 + t^2])), {t, 1,
Infinity}]];
in3 = Assuming[{a > 0, t > 0}, Integrate[(-E^(-a t) t^2)/(a Sqrt[-1 + t^2]), {t, 1, Infinity}]];
HoldForm[Integrate[Exp[-a*t]*t^2*ArcTanh[Sqrt[(t^2 - 1)/t^2]], {t, 1,
Infinity}]] == -(in1 + in2 + in3)

$\int_1^{\infty } \exp (-a t) t^2 \tanh ^{-1}\left(\sqrt{\frac{t^2-1}{t^2}}\right) \, dt=\frac{2 K_0(a)}{a^3}+\frac{2 K_1(a)}{a^2}+\frac{a K_0(a)+K_1(a)}{a^2}$

where: $K_0(a)$ and $K_1(a)$ the modified Bessel function of the second kind.


text:
Home > Mean Square > Squared-error

# Squared-error

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text:
# Math Help - conditional variance of stochastic integral

1. ## conditional variance of stochastic integral

Hallo!
I've problems to compute the conditional variance of a Brownian Integral.

I've written my problem in an extra file in the attachment.

I appreciate any help.

2. By using Itô's formula I think I can show that
$var(\int\limits_0^Tf(u)dB_u|\mathcal{F}_t)=\int\li mits_t^Tf^2(u)du.$

Does anyone agree?


text:
# Derive the time equation for the collision of two masses due to Newtonian gravity

alan123hk
Assume m2 masses a significant percentage of Earth, like the moon say.
We have a controlled Earth (not spinning, a perfect sphere, no air) all alone in space (so the sun isn't accelerating it for instance) and I have a stationary moon-size rigid balloon at the same altitude as the orbit of the moon. We let go of it and time how long it takes to hit. Now we let go of something the mass of the moon (and stronger to avoid Roche effects) from the same altitude and it will hit the ground in significantly less time. It accelerates more or less at the same rate at first, but gains speed quicker because the Earth is accelerating up towards the massive moon but not the balloon moon. Each travels a different distance because the Earth is not sitting still.

Since seeing this thread yesterday, I have been trying to derive the time equation for the collision of two masses due to Newtonian gravity. Unfortunately, this seems to be much more difficult than I thought before, so I haven't made much progress yet. 🤔

Last edited by a moderator:

Richard R Richard
The weight at the height of the Earth's surface is Mg and the aerodynamic drag of an object is proportional to its speed squared. Setting up Newton's equations

$$mg-kv ^ 2 = ma$$

When the dimensions of two test objects are equal ## k ## is a constant.

When the equilibrium of forces is reached, the acceleration ## to ## becomes zero or it descends with a constant velocity, whose value is obtained by solving

$$mg-kv ^ 2 = 0$$

$$mg = kv ^ 2$$

$$v = \sqrt {\dfrac {mg}{k}}$$

According to this, the more massive, the higher terminal velocity, then it falls faster.

But as I said this is only true if k is constant (see drag coefficient),

Massive objects do not always fall first,

• A 747 airplane has more mass than a car, however both launched from a certain height and with horizontal speed, the car will fall first.

• An empty balloon falls before the same inflated balloon of the same mass ...

• A satellite with a certain tangential velocity does not fall, no matter how much its mass is ...

Since seeing this thread yesterday, I have been trying to derive the time equation for the collision of two masses due to Newtonian gravity. Unfortunately, this seems to be much more difficult than I thought before, so I haven't made much progress yet. 🤔

To know the time of fall of an object, taking into account that gravity varies with height and in the absence of friction (this almost leaves the thread) is

We already know the modulus of acceleration of gravity falls with the square of the distance to the center.

$$a_g = \dfrac {GM_T}{r ^ 2}$$

It is logical to think that the relative acceleration with respect to the surface for an object at a higher altitude is less than one at a low altitude. these types of corrections are made by means of differential equations.
Let's see the simple calculation of frictionless free fall, An object of mass ## m ## is dropped without initial velocity from a height ## r_o ## with respect to a massive body of mass ## M ##

Using the conservation of energy you have a first order differential equation that is relatively easy to integrate.

$$\displaystyle \frac {1}{2} v ^ 2-G \frac {M}{r} = - G \frac {M}{r_0}$$

from where

$$v = - \displaystyle \frac {dr}{dt} = \sqrt {2 GM} \sqrt {\frac {1}{r} - \frac {1}{r_0}} = \sqrt {\frac { 2 GM}{r_0}} \sqrt {\frac {r_0-r}{r}}$$

if we try to integrate

$$\displaystyle- \int_{r_0} ^ r {\sqrt {\frac {r}{r_0-r}} dr} = \sqrt {\frac {2 GM}{r_0}} \int_0 ^ t {dt}$$

The integral on the left is a bit heavy to perform. My source, friend from another forum has made, the change of variable $$u ^ 2 = r / (r_0-r)$$.

Finally

$$\displaystyle- \int_{r_0} ^ r {\sqrt {\frac {r}{r_0-r}} dr} = r_0 \left (\frac {\pi}{2} - \arctan \sqrt {\frac {r}{r_0-r}} + \frac {\sqrt {r} \sqrt {r_0-r}}{r_0} \right)$$

then

$$t (r) = \displaystyle \sqrt {\frac {r_0 ^ 3}{2GM}} \left (\frac {\pi}{2} - \arctan \sqrt {\frac {r}{r_0-r }} + \frac {\sqrt {r} \sqrt {r_0-r}}{r_0} \right)$$

(this makes sense as long as ## r ## is greater to the radius of the earth).

alan123hk
alan123hk
To know the time of fall of an object, taking into account that gravity varies with height and in the absence of friction (this almost leaves the thread) is
We already know the modulus of acceleration of gravity falls with the square of the distance to the center.

But what I mean is that in the absence of air resistance and initial speed, these two objects will accelerate toward each other at the same time due to the action of Newtonian gravity. I believe that the impact time will depend on the mass of the two objects (M1, M2), the initial distance between them (R0), and may also include the radii of the two objects (D1, D2).

But what I mean is that in the absence of air resistance and initial speed, these two objects will accelerate toward each other at the same time due to the action of Newtonian gravity. I believe that the impact time will depend on the mass of the two objects (M1, M2), the initial distance between them (R0), and may also include the radii of the two objects (D1, D2).
This has been discussed before here and is not trivial. One trick was to treat the trajectory as a degenerated ellipse and appply Kepler's laws.

Nugatory and bhobba
Mentor
This has been discussed before here and is not trivial. One trick was to treat the trajectory as a degenerated ellipse and appply Kepler's laws.
Exactly. It is one of those problems easy to state but hard to do. As an aside, if you were to include air resistance, then the problem is very tough because you have turbulence and must solve the Navier-Stokes Equation. Even showing you can do it will not only possibly get you a Nobel, maybe even a Fields Medal, but also a million big ones from the Clay Institute - it is one of the Millenium prizes. By making some approximations, it, while still hard, can be done using the drag equation:
https://physics.info/drag/

Thanks
Bill

Richard R Richard
alan123hk
Exactly. It is one of those problems easy to state but hard to do. As an aside, if you were to include air resistance, then the problem is very tough because you have turbulence and must solve the Navier-Stokes Equation. Even showing you can do it will not only possibly get you a Nobel, maybe even a Fields Medal, but also a million big ones from the Clay Institute - it is one of the Millenium prizes. By making some approximations, it, while still hard, can be done using the drag equation

I am very happy that this discussion thread has reopened, and I agree with you.

I searched the Internet for a long time and finally found an equation that can be easily derived under a special situation (including no air resistance, no initial velocity in any direction,..). This derivation method cleverly uses the concept of conservation of momentum.

So far, in the derivation process of the equation shown below, I have not found anything unreasonable.

$$\textrm{Energy conservation (KE = PE): } \frac{p^2}{2}\left( \frac{1}{m} + \frac{1}{M} \right) = GMm\left(\frac{1}{r} - \frac{1}{r_0}\right)$$
$$\frac{dr}{dt} = -(v + V) = -p\left( \frac{1}{m} + \frac{1}{M} \right)$$
$$\int_0^T dt = -\int_{r_0}^0 dr \sqrt{\frac{rr_0}{2G(M+m)(r_0-r)}} = \frac{\pi}{2\sqrt{2}}\frac{r_0^{3/2}}{\sqrt{G(M+m)}}$$

where p is the combined momentum of the two bodies.

bhobba
Homework Helper
Gold Member
2022 Award
alan123hk and bhobba
Homework Helper
2022 Award
So far, in the derivation process of the equation shown below, I have not found anything unreasonable.
I note that this is exactly the result obtained by @Richard R Richard #2 with the recognition that the M is the total system mass.

$$t (r) = \displaystyle \sqrt {\frac {r_0 ^ 3}{2GM}} \left (\frac {\pi}{2} - \arctan \sqrt {\frac {r}{r_0-r }} + \frac {\sqrt {r} \sqrt {r_0-r}}{r_0} \right)$$

Just plug in r=0 and use the total mass ##M=m_1 +m_2##

$$t (0) = \displaystyle \sqrt {\frac {r_0 ^ 3}{2G(m_1+m_2)}} \left (\frac {\pi}{2} \right)$$

bhobba
Richard R Richard
##r = 0## is an idealization, any real massive object has a surface with a radius r, which cannot be traversed without impact. On Earth for example ##r = R_T = 6354km## It should also be noted that if we idealize and think that there is a hole that allows us to reach ##r = 0##, we must change the equations because the mathematical function of gravity inside the Earth is not the same as outside.

bhobba
Homework Helper
2022 Award
Yes I was assuming point masses. The generalizations at that point are easy...you did the hard parts

bhobba
Mentor
This is really good. When something is hard, figuring it out for yourself, even with the help of the internet, is a real eye-opener. That is how I learned some of the most 'most satisfying' bits of knowledge. It's not they are unknown or anything like that - it's simply they are not often discussed. One I figured out by myself was what Newtons Laws really mean. You can do a search on this forum for the answer. I am about to do a post on 0 to the power 0. Good work, guys. This is what this forum is all about.

Thanks
Bill

alan123hk, Richard R Richard and hutchphd
alan123hk
Now that I have an equation for collision time, I try to use the principles of conservation of momentum and energy to derive approximate equations for collision position and collision velocity.

collision position : -

$$r_1 = \frac {(r_0-d)m_2} {m_1+m_2} ~~~~~~~~~~~~~ r_2 = \frac {(r_0-d)m_1} {m_1+m_2}$$
collision velocity : -
$$V_1 = \sqrt { \frac {2G M_1 M_2 ( \frac 1 d -\frac 1 r_0 )} { 1+ \frac {M_1} {M_2} } } ~~~~~~~~~~~ V_2 = \sqrt { \frac {2G M_1 M_2 ( \frac 1 d -\frac 1 r_0 )} { 1+ \frac {M_2} {M_1} } }$$

where
d = diameter of m1 and m2, and r0 = r1+r2+d
r
1
and r2 = moving distance of M1 and M2

Last edited:
Richard R Richard
I detect an incompatibility of proposals between what I explained and what you conclude. Not because it is wrong but because they are two different problems. In my proposal the fall time is for a particle of negligible mass compared to the gravitational mass ## M ##. Your case deals with a problem of two bodies of masses of the same order of magnitude And only in the case that the system does not have angular momentum, and in the absence of other external forces, the collision position will be exactly at the Center of mass. As you also affirm that each mass has the same diameter, the collision occurs when their centers are one diameter apart, the most massive body will be closer to the MC in the same proportion as the mass ratio. The velocity relationship is also derived from the conservation of energy, it is advisable to use a reference system originating in the MC

alan123hk
I detect an incompatibility of proposals between what I explained and what you conclude. Not because it is wrong but because they are two different problems. In my proposal the fall time is for a particle of negligible mass compared to the gravitational mass M. Your case deals with a problem of two bodies of masses of the same order of magnitude And only in the case that the system does not have angular momentum, and in the absence of other external forces, the collision position will be exactly at the Center of mass. As you also affirm that each mass has the same diameter, the collision occurs when their centers are one diameter apart, the most massive body will be closer to the MC in the same proportion as the mass ratio. The velocity relationship is also derived from the conservation of energy, it is advisable to use a reference system originating in the MC
Thanks for your advice. I have been engaged in engineering work for decades since leaving school, and now I am trying to regain my interest in physics after retirement. Most of the basics of mathematics and physics that I have learned before are now forgotten, maybe now I have to spend some time to relearn and improve.

Homework Helper
Gold Member
2022 Award
Now that I have an equation for collision time, I try to use the principles of conservation of momentum and energy to derive approximate equations for collision position and collision velocity.
The energy (and hence relative velocity) at the collision can be simply calculated from conservation of energy. And the motion of the two masses relative to the initial rest frame is, by conservation of momentum, in inverse proportion to their masses.

The time taken is the tricky thing to calculate. You don't need the time to calculate the other quantities.

alan123hk
The time taken is the tricky thing to calculate. You don't need the time to calculate the other quantities.

Yes, I did not use an equation about time. I mean, after obtaining the time equation, I naturally want to know more information, so I try to derive the collision velocity and collision position equations.

Now I am thinking about more complicated things. For example, when a small asteroid approaches the Earth, how do scientists calculate the elliptical trajectory of its fall, the time and location of its impact on the earth’s atmosphere. As for entering the earth’s atmosphere, due to the huge air resistance, I believe that only with the help of supercomputers can we accurately calculate the time and location of the impact on the ground.

Speaking of this, I'm afraid I 'm a little off topic again.

Last edited:


text:
# On the Structure of the Completion of a Normed Lattice

On the Structure of the Completion of a Normed Lattice Let X be a normed lattice and Y be the norm completion of X with a natural embedding π : X → Y . By the Kawai- Luxemburg theorem, X is embedded as an order dense set and π preserves all suprema and infima iff X satisfies the condition (A o ) (i.e., the norm has pseudo σ-Lebesgue property). Let X o be the largest ideal in X having the condition (A o ); let Y (o) be the band in Y generated by π X o and Y (1) be the complementary band to Y (o) . The structure of Y and, in particular, of the bands Y (o) and Y (1) are studied. The conditions for Y (o) to be a projection band and π X o to be topologically dense in Y (o) are obtained. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Positivity Springer Journals

# On the Structure of the Completion of a Normed Lattice

, Volume 11 (1) – Oct 13, 2006
19 pages

/lp/springer_journal/on-the-structure-of-the-completion-of-a-normed-lattice-wNEGE8E2bF
Publisher
Springer Journals
Subject
Mathematics; Fourier Analysis; Operator Theory; Potential Theory; Calculus of Variations and Optimal Control; Optimization; Econometrics
ISSN
1385-1292
eISSN
1572-9281
D.O.I.
10.1007/s11117-006-2020-3
Publisher site
See Article on Publisher Site

### Abstract

Let X be a normed lattice and Y be the norm completion of X with a natural embedding π : X → Y . By the Kawai- Luxemburg theorem, X is embedded as an order dense set and π preserves all suprema and infima iff X satisfies the condition (A o ) (i.e., the norm has pseudo σ-Lebesgue property). Let X o be the largest ideal in X having the condition (A o ); let Y (o) be the band in Y generated by π X o and Y (1) be the complementary band to Y (o) . The structure of Y and, in particular, of the bands Y (o) and Y (1) are studied. The conditions for Y (o) to be a projection band and π X o to be topologically dense in Y (o) are obtained.

### Journal

PositivitySpringer Journals

Published: Oct 13, 2006

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text:
Hey guys, take a look at my work for a stationary integral?

1. Nov 13, 2012

Kidphysics

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

Path between O=(0,0) and P=(0,4) for which the integral

int. O to P (x (1-y'^2))^(1/2) dx

becomes stationary.

2. Relevant equations

http://en.wikipedia.org/wiki/Calculus_of_variations#Example

3. The attempt at a solution

Okay guys,

so mimicking the example from wiki but with L= (x)^1/2 (1-y'^2)^1/2

dL/dy=0 and dL/dy'= -(x)^1/2 y' / (1-y'^2)^1/2 and then d/dx dL/df' = - y'/[(2x^1/2)(1-y'^2)^1/2]

and since d/dx dL/df' =0 this implies y'=0 and we integrate to get y=c, adhering to boundary conditions y must equal 0... was that the correct way sorry for not writing in latex if I must I will redo the post

2. Nov 13, 2012

haruspex

That last expression looks too simple. There should be a y" in there, and two different powers of x. Remember it's not a partial derivative this time.

3. Nov 13, 2012

Kidphysics

ahhhh you are correct! y is a function of x..

4. Nov 13, 2012

Kidphysics

well.. I used the product rule and the chain rule and now I am looking at an unbelievable

2 x^1/2 y'^2 y'' - 1/2 x^-1/2 y' - x^1/2 y'' + 1/2 x^-1/2 y'^3 =0 .

I believe I took the derivative correctly this time but now I am lost as what to do with this..

5. Nov 13, 2012

Kidphysics

well.. I used the product rule and the chain rule and now I am looking at an unbelievable

2 x^1/2 y'^2 y'' - 1/2 x^-1/2 y' - x^1/2 y'' + 1/2 x^-1/2 y'^3 =0 .

I believe I took the derivative correctly this time but now I am lost as what to do with this.

6. Nov 13, 2012

haruspex

Allow me to put that in LaTex:

$$2 x^{1/2} y'^2 y'' - (1/2)x^{-1/2} y' - x^{1/2} y'' + (1/2) x^{-1/2} y'^3 =0$$
Simplifying:
$$4 x y'^2 y'' - y' - 2x y'' + y'^3 =0$$
But that's not quite what I get. Writing K = (1-y'2)1/2:
L = x1/2K
∂K/∂y' = -y'/K
∂L/∂y' = -x1/2y'/K
$$0 = -\frac{d(∂L/∂y')}{dx} = \frac{x^{-1/2}y'}{2K} + \frac{x^{1/2}y''}{K} - \frac{x^{1/2}y'\frac{∂K}{∂y'}y''}{K^2}$$
$$0 = y'+ 2xy'' - 2\frac{xy'\frac{∂K}{∂y'}y''}{K}$$
$$0 = y'+ 2xy'' + 2\frac{xy'^2y''}{K^2}$$
$$0 = (y'+ 2xy'')(1-y'^2) + 2xy'^2y''$$
$$0 = y'+ 2xy'' - y'^3$$
Is that right or did I get sign wrong somewhere?

7. Nov 13, 2012

Kidphysics

hey thanks, jumped out of bed to respond.. I'm sure you have the math correct but I can't see how this helps us get a solution the the original problem. At this step we have a nonlinear ode so I'm a little frightened we are on the wrong track? Thank you for helping thus far

8. Nov 13, 2012

haruspex

It's not hard from there. There are no y terms, so it's first order, and you can get it into the form f(x) = g(t)t' where t = y', so then it's just an integration to get an equation relating x to y'.


text:
### Guestblogging on AstroBites

Some of the graduate students at the Harvard Center for Astrophysics recently started a blog called AstroBites. Authors on the blog, usually grad students, provide easily digestible reviews of recent astronomy papers appearing on the pre-print archive (aka astro-ph). The blog provides a nice intermediate between higher-level reviews of papers (and the papers themselves), and intro astronomy text books, providing a thoroughly modern view of a wide range of astronomy topics.

In addition to the review articles, there are regular features on career development and general insights into modern astronomy research, from a grad student perspective. For example, this post on what it's like to observe for the first time on a major telescope, or this post on data mining.

After a recent visit to the CfA one of the blog's founders, Nathan, contacted me and invited me to write a guest post on a topic of my choosing. The post Zen and the Art of Astronomy Research appeared today. Go check it out, and browse the AstroBites archive while you're there.

Next up: my long-promised guest post on AstroBetter (sorry for the delay, Kelle!)

blissful_e said…
Great article - I think anyone in a challenging degree programme would benefit from that advice.

The part that made me laugh out loud was when you talked about ending sentences with prepositions. :) Ah, the horror of writing like we talk.
mama mia said…
Did I ever tell you John, that I love my son-in-law? You rock! I read that article and I could hear your voice as if you were speaking to students face to face...
JohnJohn said…
Awwww, thanks Nonna! :)

E: I wish I could that the ending-prepositions-with thing was original. But I agree, pretty dang funny!

### On the Height of J.J. Barea

Dallas Mavericks point guard J.J. Barea standing between two very tall people (from: Picassa user photoasisphoto).

Congrats to the Dallas Mavericks, who beat the Miami Heat tonight in game six to win the NBA championship.

Okay, with that out of the way, just how tall is the busy-footed Maverick point guard J.J. Barea? He's listed as 6-foot on NBA.com, but no one, not even the sports casters, believes that he can possibly be that tall. He looks like a super-fast Hobbit out there. But could that just be relative scaling, with him standing next to a bunch of extremely tall people? People on Yahoo! Answers think so---I know because I've been Google searching "J.J. Barea Height" for the past 15 minutes.

So I decided to find a photo and settle the issue once and for all.

I then used the basketball as my metric. Wikipedia states that an NBA basketball is 29.5 inches in circumfe…

### Finding Blissful Clarity by Tuning Out

It's been a minute since I've posted here. My last post was back in April, so it has actually been something like 193,000 minutes, but I like how the kids say "it's been a minute," so I'll stick with that.
As I've said before, I use this space to work out the truths in my life. Writing is a valuable way of taking the non-linear jumble of thoughts in my head and linearizing them by putting them down on the page. In short, writing helps me figure things out. However, logical thinking is not the only way of knowing the world. Another way is to recognize, listen to, and trust one's emotions. Yes, emotions are important for figuring things out.
Back in April, when I last posted here, my emotions were largely characterized by fear, sadness, anger, frustration, confusion and despair. I say largely, because this is what I was feeling on large scales; the world outside of my immediate influence. On smaller scales, where my wife, children and friends reside, I…

### The Force is strong with this one...

Last night we were reviewing multiplication tables with Owen. The family fired off doublets of numbers and Owen confidently multiplied away. In the middle of the review Owen stopped and said, "I noticed something. 2 times 2 is 4. If you subtract 1 it's 3. That's equal to taking 2 and adding 1, and then taking 2 and subtracting 1, and multiplying. So 1 times 3 is 2 times 2 minus 1."

I have to admit, that I didn't quite get it at first. I asked him to repeat with another number and he did with six: "6 times 6 is 36. 36 minus 1 is 35. That's the same as 6-1 times 6+1, which is 35."

Ummmmm....wait. Huh? Lemme see...oh. OH! WOW! Owen figured out

x^2 - 1 = (x - 1) (x +1)

So $6 \times 8 = 7 \times 7 - 1 = (7-1) (7+1) = 48$. That's actually pretty handy!

You can see it in the image above. Look at the elements perpendicular to the diagonal. There's 48 bracketing 49, 35 bracketing 36, etc... After a bit more thought we…


text:
# 13.1 Core file: accessing external memory on ti tms320c54x

 Page 1 / 1
Large amounts of data can be stored in external memory. The READPROG and WRITPROG macros copy data between internal memory and external memory.

## Introduction

The TI DSP evaluation boards you use have a large amount of memory; in addition to the 32K words internal to the DSP,there are another 256K words of memory installed on the EVM board. For many exercises, the data sets are small, and youworked with only the on-chip memory of the DSP and were not expected to consider how the use of memory impactedperformance. However, the large delays often required in audio processing, for example, require that many thousands ofsamples be stored in memory. There is not enough memory on the DSP microprocessor itself to store a second or more of samplesat a 44.1 kHz sample rate, so the off-chip memory must be used.

## Evm memory maps

As you have seen, the TI TMS320C54x DSP has two separate memory spaces, called Program and Data. Usually, Programcontains your assembled program, and Data contains data, but sometimes it may be convenient or more efficient to violatethis convention. (For instance, the firs instruction requires filter coefficients in the Program address space.) The Data space is 64K long and is accessedusing the 16-bit auxiliary registers ( ARx ). Although the Program space is normally accessed using 16-bit literals stored in your program code,the Program space is, in fact, significantly larger than 64K. Using special "extended addressing" instructions, the TI DSPcan access up to 8192K-words of memory in the Program space. The extended addressing instructions include far calls andjumps that reset the full 23-bit program counter, as well as accumulator-addressed data-transfer instructions.

## Internal and external memory

In many exercises, it is possible to store program instructions and data entirely in the DSP's on-chip("internal") memory. This internal memory has several advantages over off-chip ("external") memory: it is muchfaster (data stored can be accessed without delay), and multiple reads and writes can access the DSP's on-chipmemory simultaneously. However, many applications (including the audio delay effect of Using External Memory ) require a data buffer too large to fit into the on-chip memory. For these large buffers, wemust use the larger but slower external memory.

When writing programs that require large amounts of memory, use the internal memory to hold your code, filtercoefficients, and any small buffers you need. External memory should be used for large buffers that you only accessa few times per sample, like the delay buffer described in Audio Effects: Using External Memory .

## Tms320c549x dsp evm memory maps

As these memory maps show, the EVM's Data address space is addressed fully by the 16-bit auxiliary registers( ARx ) and address-extension words and the mapping of Data memory is not affected by the OVLY bit. However, because the Program memory space is much larger than can be addressed by the 16-bitaddressing register or the 16-bit literals stored in the program, it is split up into 64K (16-bit) pages by thehardware. Normal instructions, such as call , firs , and mvpd accept only 16-bit addresses, and can therefore only address the current "page"(usually address in the form 00xxxxh , which corresponds to the addresses the linker uses for yourprogram's code). To access the full 23-bit address space, the DSP offers special accumulator-addressed load, store,and jump instructions.

Further complicating matters is the fact that the OVLY bit affects the mapping of the Program memory space. If you remember, before we load our DSPprogram, we have to change the PMST to FFE0h . We do this to set the OVLY bit in the PMST , which maps the internal memory into both the Program and Data spaces of the DSP. If OVLY is 1, the internal memory appears in both the Data and Program memory address space at locations 0080h to 07FFFh . Therefore, with OVLY set, anything written into Data memory below 07FFFh will overwrite a program stored in the same location.

This is why the memory allocated for your program - 6000h - 7FFFh - does not overlap with any of the space allocated for the datasegments.
In addition, copies of the internal memory also appear in the extended Program address space, occupyinglocations 0080h - 7FFFh of each page. Therefore, with OVLY set, any addresses to Program memory locations in the form of xx0000h - xx7FFFh reference internal memory.

When OVLY is zero, internal memory is not mapped into the Program space at all; in this case, theProgram space includes only external memory. In this mode, all 192K words of external Program RAM are accessible,although several wait states will be required for accessing each item of memory. In the overlay mode, only addresses inthe ranges of 08000h - 0FF00h , 1800h - 1FFFFh , and 28000h - 2FFFFh are available to store your data buffers; the remaining addresses areunmapped or map to the on-chip RAM.

To escape this confusion and allow the full 192K-words of external Program RAM to be used for your data buffers, thecore file provides mechanisms for manipulating the PMST indirectly. Instead of accessing the external Program RAM directly, we can use the special macrosto access the RAM that is normally "hidden" by the internal memory. This allows us to use the full range of externalmemory available: addresses 000000h - 00FF00h and 010000h - 02FFFF . However, since addresses 00FF00h - 00FFFFh are reserved by the core file, you must be careful not to writeto addresses in this range.

## Accessing extended program ram

The core file provides two macros for accessing data stored in the external Program RAM: READPROG and WRITPROG . These macros allow the processor to copy data between data memory and external Programmemory. Both macros address external Program memory using the value in accumulator A . READPROG reads data from the external Program memory location pointed to by A and writes it to the data memory location pointed toby AR1 . WRITPROG reads data from the memory location pointed to by AR1 and writes it to the location in external Program RAM specified byaccumulator A . Both macros take one parameter, a count; specifying 1 reads or writes one word from externalmemory, and specifying some other number $n$ transfers $n$ words starting at the locations pointed to by A and AR1 . AR1 is left pointing at the word after the last word read or written; noother registers are modified.

For instance, the following code fragment loads the value contained in memory location 023456h into the location 0064h in data memory using the READPROG macro:

1 stm #64h,AR1 ; load 64h into AR1 2 ld #02h,16,A ; load 02h in high part of A 3 adds #3456h,A ; fill in low part of A 4 ; A contains 023456h 5 READPROG 1 ; read from 023456h in external Program RAM 6 ; into *AR1 in Data RAM

The WRITPROG macro can be used similarly to write into extended Program RAM:

1 stm #64h,AR1 ; load 64h into AR1 2 ld #02h,16,A ; load 02h in high part of A 3 adds #3456h,A ; fill in low part of A 4 ; A contains 023456h 5 WRITPROG 1 ; write from *AR1 in Data RAM to 6 ; 023456h in external Program RAM

Note that Code Composer will not display or allow you to change the contents of the external Program RAM on thememory-dump or disassembly screen, though you can view or change it indirectly by watching the effects of the READPROG and WRITPROG macros on data memory.

where we get a research paper on Nano chemistry....?
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
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text:
Remark 93.16.3. If the morphism $f : \mathcal{S}_ U \to \mathcal{X}$ of Lemma 93.16.2 is only assumed surjective, flat and locally of finite presentation, then it will still be the case that $f_{can} : [U/R] \to \mathcal{X}$ is an equivalence. In this case the morphisms $s$, $t$ will be flat and locally of finite presentation, but of course not smooth in general.

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text:
# With excess of $Cl_2$ ammonia forms

$\begin{array}{1 1}(a)\;NH_4Cl&(b)\;N_2\\(c)\;NCl_3&(d)\;NClNH_3\end{array}$

With excess of $Cl_2$ ammonia forms $NCl_3$
Hence (c) is the correct answer.


text:
## Diagonalizable Linear Transformation

My own question. I am sure it has an answer. But my Linear Algebra is so embarrasingly awful I am not sure. I have a dislike to the Linear Algebra involving matrix computations and applications of that such as linear mappings from $\mathbb{R}^n\mapsto\mathbb{R}^m$. However, there is a side of Linear Algebra that I like, it is about general vector spaces over fields (and therefore no matrices). I was thinking whether there is a generalization of "diagonalization" in general:

Let $V,U$ be (finite) vector spaces over a field $F$. Let $\phi: V\mapsto U$ be a linear transformation. Is there such a notion as " $\phi$ is a diagonalizable linear transformation".