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# London dispersion forces and distance relationship

Why is the London dispersion force inversely related to distance but not to the second power but the sixth power? Why is there not an inverse square relationship as there is for Coulombic forces?

Is there something non-Coulombic about London dispersion forces?

One somewhat simplified way of looking at it is that the London dispersion forces are a dipole-dipole interactions. The interaction between two dipoles depends upon the relative orientation of the two dipoles. Some positions are attractive and some are replusive, but averaged over all the positions (and considering that lower energy positions are favored by Boltzmann statistics), the interaction is attractive with potential energy having an 1/$r^6$ dependence.

See for example Stephan Franzen's lecture

and especially the corporate Intermolecular Van Der Waals page

as well as Fritz London's The General Theory of Molecular Forces Trans. Faraday Soc., 1937, vol. 33, pages 8-26.

for more on the 1/$r^6$ dependence of dipoles.

However, a proper quantum mechanical treatment shows that London dispersion interaction is not always proportional to 1/$r^6$

For example, between two hydrogen atoms in 1s states, the potential energy is proportional to 1/$r^6$ but

between a hydrogen atom in a 1s state and a hydrogen atom in a 2p state, energy is proportional to 1/$r^3$

See: Complement $C_{XI}$ of Quantum Mechanics vol. 2 by Cohen-Tannoudji et al.

Also, in 1948 Verwey and Overbeek demonstrated experimentally that the London dispersion interaction is even weaker than 1/$r^6$ at long distance (say hundreds of Angstroms or more).

Casimir and Polder soon thereafter explained with quantum electrodynamics (QED) that the dependence should be 1/$r^7$ at relatively long distances.

For a nice historical overview and QED prespective see Some QED vacuum effects: van der waals forces in The Quantum Vacuum by Peter W. Milonni.

• @DavePhD: "but averaged over all the positions (and considering that lower energy positions are favored by Boltzmann statistics), the interaction is attractive" why then might one of my profs ask questions starting with "Based on vdW repulsions, there are how many different bond angles in ...)? – Dissenter Jan 14 '15 at 21:35
• @Dissenter I think he/she is referring to the "12" part of the Lennard Jones 6-12 potential, which has nothing to do with dipole-dipole interaction, it is due to the Pauli exclusion principle (like electron degeneracy pressure). So once the molecules are close enough that electron wavefunctions overlap, the 1/r^6 attractive transitions to 1/r^12 repulsive – DavePhD Jan 14 '15 at 23:04

London dispersion forces (or van der Waals) can be rationalized (classically) as a relationship between induced dipole moments.

Some electrostatic (Coulombic) relationships:

• Two Point Charges: $V = \frac{q_1 q_2}{4\pi\epsilon_0 r}$ (i.e., $r^{-1}$)
• Point Charge and Permanent Dipole: $V = \frac{\mu_1 q_2}{4\pi\epsilon_0 r^2}$ (i.e., $r^{-2}$)
• Two Permanent Dipoles: $V = \frac{\mu_1 \mu_2}{4\pi\epsilon_0 r^3}(1 - 3 \cos^2 \theta)$ (i.e., $r^{-3}$)

The last assumes that the polar molecules are stationary with angle $\theta$ between the dipole moment vectors. If the molecules rotate then it falls off to $1/r^6$.

In general, the interaction of a $2^n$-pole (dipole, quadrupole, octupole, etc.) with $2^m$-pole is $V \propto \frac{1}{r^{n+m-1}}$.

So why the $1/r^6$ for dispersion? Here's a quick justification. Let's take the system of a permanent dipole that's creating a induced dipole moment in another molecule:

$\mu_2^* = \alpha_2 E$

where $\alpha_2$ is the polarizability in the second molecule and E is the electric field generated by the polar molecule:

$E = \mu \sqrt{1 + 3\cos^2 \theta} / 4\pi\epsilon_0r^3$

The potential energy is given by:

$V = \alpha_2 E^2 / 2$

$V \propto r^{-6}$

Since there are multiple dispersion interactions (two dipole moments averaged over all rotations, permanent dipole-induced dipole, hydrogen bonding, and induced dipole-induced dipole moment) the general potential energy is classically approximated as:

$V = \frac{C}{r^6}$

• what is "$\theta$" here? the third result seems wrong. – RE60K Jan 14 '15 at 14:26
• That's the general result regardless of the angle theta between two dipole vectors. – Geoff Hutchison Jan 14 '15 at 14:28
• As to the difference in constants with induced dipoles, I checked these with Atkins' textbook before posting. IIRC, it depends on the different ways of defining polarizability. I'll check with a colleague who breathes electrostatics. – Geoff Hutchison Jan 14 '15 at 14:31
• @GeoffHutchison For permanent dipole- induced dipole $E = \mu\sqrt {1 + 3 \cos^2 \theta}/4\pi\epsilon_0 r^3$ and $V=-\alpha E^2/2$ Then according to London 1937 (explaining Debye's theory), considering all orientations, the $\cos^2 \theta$ term is 1/3. – DavePhD Jan 14 '15 at 15:09
• @DavePhD if you see the still in my answer and see the surroundings, the illustration 27 says $E=kp/r^3 . \sqrt{(1+3\cos^2\theta)}$ – RE60K Jan 14 '15 at 15:15


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# American Institute of Mathematical Sciences

June  2013, 2(2): 301-318. doi: 10.3934/eect.2013.2.301

## Higher differentiability in the context of Besov spaces for a class of nonlocal functionals

 1 University of Nebraskat-Lincoln, Department of Mathematics, 203 Avery Hall, PO BOX 880130, Lincoln NE 68588-0130, United States, United States

Received  November 2012 Revised  January 2013 Published  March 2013

The aim of this paper is to contribute to the nonlocal theory within the calculus of variations by studying two classes of nonlocal functionals. Since the nonlocal theory is not quite as developed as the local theory, a proof for the existence and uniqueness of minimizers is provided. However, the main result within the paper establishes the higher differentiability, in the context of Besov spaces, for minimizers of nonlocal functionals. This result is obtained under quadratic growth assumptions via the difference quotient method.
Citation: Mikil Foss, Joe Geisbauer. Higher differentiability in the context of Besov spaces for a class of nonlocal functionals. Evolution Equations & Control Theory, 2013, 2 (2) : 301-318. doi: 10.3934/eect.2013.2.301
##### References:
 [1] Tsegaye G. Ayele and Abraham N. Abebe, Properties of iterated norms in Nikol'skii-Besov type spaces with generalized smoothness,, Eurasian Mathematics Journal, 1 (2010), 20.   Google Scholar [2] Viktor I. Burenkov, A theorem on iterated norms for Nikol'skii-Besov spaces and its application,, (Russian) Trudy Mat. Inst. Steklov., 181 (1988), 27.   Google Scholar [3] Viktor I. Burenkov, "Sobolev Spaces on Domains,", Teubner-Texte zur Mathematik [Teubner Texts in Mathematics], 137 (1998).   Google Scholar [4] Bernard Dacorogna, "Direct Methods in the Calculus of Variations,", Second edition, 78 (2008).   Google Scholar [5] Lawrence C. Evans, "Partial Differential Equations,", Graduate Studies in Mathematics, 19 (1998).   Google Scholar [6] Guy Gilboa and Stanley Osher, Nonlocal linear image regularization and supervised segmentation,, Multiscale Modeling & Simulation, 6 (2007), 595.  doi: 10.1137/060669358.  Google Scholar [7] Guy Gilboa and Stanley Osher, Nonlocal operators with applications to image processing,, Multiscale Modeling & Simulation, 7 (2008), 1005.  doi: 10.1137/070698592.  Google Scholar [8] Enrico Giusti, "Direct Methods in the Calculus of Variations,", World Scientific Publishing Co. Inc., (2003).  doi: 10.1142/9789812795557.  Google Scholar [9] Brittney Hinds and Petronela Radu, Dirichlet's principle and wellposedness of solutions for a nonlocal $p$-Laplacian system,, Applied Mathematics and Computation, 219 (2012), 1411.  doi: 10.1016/j.amc.2012.07.045.  Google Scholar [10] Giovanni Leoni, "A First Course in Sobolev Spaces,", Graduate Studies in Mathematics, 105 (2009).   Google Scholar

show all references

##### References:
 [1] Tsegaye G. Ayele and Abraham N. Abebe, Properties of iterated norms in Nikol'skii-Besov type spaces with generalized smoothness,, Eurasian Mathematics Journal, 1 (2010), 20.   Google Scholar [2] Viktor I. Burenkov, A theorem on iterated norms for Nikol'skii-Besov spaces and its application,, (Russian) Trudy Mat. Inst. Steklov., 181 (1988), 27.   Google Scholar [3] Viktor I. Burenkov, "Sobolev Spaces on Domains,", Teubner-Texte zur Mathematik [Teubner Texts in Mathematics], 137 (1998).   Google Scholar [4] Bernard Dacorogna, "Direct Methods in the Calculus of Variations,", Second edition, 78 (2008).   Google Scholar [5] Lawrence C. Evans, "Partial Differential Equations,", Graduate Studies in Mathematics, 19 (1998).   Google Scholar [6] Guy Gilboa and Stanley Osher, Nonlocal linear image regularization and supervised segmentation,, Multiscale Modeling & Simulation, 6 (2007), 595.  doi: 10.1137/060669358.  Google Scholar [7] Guy Gilboa and Stanley Osher, Nonlocal operators with applications to image processing,, Multiscale Modeling & Simulation, 7 (2008), 1005.  doi: 10.1137/070698592.  Google Scholar [8] Enrico Giusti, "Direct Methods in the Calculus of Variations,", World Scientific Publishing Co. Inc., (2003).  doi: 10.1142/9789812795557.  Google Scholar [9] Brittney Hinds and Petronela Radu, Dirichlet's principle and wellposedness of solutions for a nonlocal $p$-Laplacian system,, Applied Mathematics and Computation, 219 (2012), 1411.  doi: 10.1016/j.amc.2012.07.045.  Google Scholar [10] Giovanni Leoni, "A First Course in Sobolev Spaces,", Graduate Studies in Mathematics, 105 (2009).   Google Scholar
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# How long could someone with locked in syndrome survive in 1915 [closed]

Let's assume a modern for its day, 1915, hospital gets a patient with locked in syndrome: how would he fare?

We will think of someone who is just locked in. This means they don't have any problem breathing or any other lack of vital processes. They can't thought communicate in any way, or move including swallowing food. Wikipedia article describing the condition

I realize that by this point IVs are invented, but I don't know if proper nutrition could be fed to the person. This is also an age before antibiotics, and I am guessing infection would be a major threat to the patient.

Would this person make it days, weeks months, years?

## closed as off-topic by Ash, sphennings, JBH, Azuaron, SecespitusOct 3 '17 at 19:15

• This question does not appear to be about worldbuilding, within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

• Johnny Got His Gun? – Katamori Oct 3 '17 at 11:53
• @Katamori has given you the required reading. It is a story about a WW1 veteran who is in the scenario you describe. – Willk Oct 3 '17 at 12:08
• I'm voting to close this question as off-topic because it belongs on either the History.SE or Health.SE. It is also a primarily story/plot question, not a worldbuilding one. – Ash Oct 3 '17 at 14:37
• The survivability of someone who is catatonic in 1915 is not a world building question. – Mazura Oct 3 '17 at 15:34
• While others have already mentioned why this question might not fit the format, there is another thing to consider for now and future questions: People do not necessarily know what locked in means. When using medical- or technical terms always consider providing either a link for reading up or a short explanation that is clearly an explanation of the used term. – dot_Sp0T Oct 3 '17 at 15:49

You seem to take a big leap from 'can't swallow' to 'needs IV nutrition'.

Tube feeding is a viable option that is used a lot in modern medicine as well; in fact, it is preferred to IV nutrition if at all possible (i.e. when the bowel works as it should). This preference is because of the risk of infection inherent in using long-term IV access, and also because providing adequate nutrition that can be passed directly into the bloodstream is difficult and therefore costly. (While you can pass any sort of pureed meal through a feeding tube, any IV nutrition needs to be the pure nutrients (glucose, fatty acids, etc.) and absolutely free of any microbes and irritants.)

In using a feeding tube there are two main ways:

• using an existing opening to reach the stomach (most commonly the nose)

This is more commonly used for short term tube feeding, since the mucous membranes of the existing opening will get irritated and damaged by the tube after a while, leading to increased risk of infection (and also pain for the patient). This link suggests a maximum counted in weeks.

• surgically creating an opening through which the feeding tube should pass

This carries all the 'usual' risks of any surgical procedure (anesthesiologic risk, risk of infection), plus the risk of infection in the surgically created opening that is open to the outside world long term. However, since such an opening does not directly communicate with a main blood vessel, the risk of complications from infection are slightly lower than with long-term IV access.

As to the availability of these options to a patient in 1915: R. Skeeter provided a link to an article about surgically placed feeding tubes that mentions the practice appearing in mid to late 1800 - implying that non-surgical feeding tubes were available before that already. Another link provided by R. Skeeter illustrates a modern-day patient with a surgically placed feeding tube surviving for several years at least.

Without antibiotics I would expect some infection to set in within a few months, maybe a lucky patient with an otherwise strong constitution would be able to survive longer. However, once the immediate concern of nutrition has been taken care of, a patient that is bedridden long term will develop other problems - bedsores being one of them - and overall the body's health will deteriorate due to the inactivity.

Taking it all together, my expectation would be that several months is the maximum.

I have the feeling he won't last long, few months at best.

IV's were developed, yes, but Teflon needles weren't. This means lack of flexible needles and therefore higher chances of long term complications while taking IV's. Add to this the lack of antibiotics, and you see that infections are highly likely.

From that to the grave the step is short.

One aspect of this answer is the purely medical one which L.Dutch has already answered to. Even under the best of circumstances, survival would be hung on a very thin thread as IV needles of that time would leave arms and legs so bruised and swollen within weeks that setting up a new IV would become more critical with each treatment. Any infection which is not unlikely to occur in spite of increasing hygiene standards, would most likely lead to the person's death.

Another aspect is a social one. Why would you treat such a patient, unless the hospital is provided with immense funding to make up for such experimental treatment of a person without any hope of recovery from the point of view of doctor's and people of that day alike?

An additional complication to think of is that 1915 is after the start of WWI. Therefore, the chance of treating such a patient in a country/area where hundreds, even thousands of wounded soldiers require immediate treatment and ressources are getting more and more scarce with the progress of the war is definitely decreasing inspite of all funding. Sooner or later they will lack even the basic IV needles and medications and then the beds.

• WHy treat is an interesting question for me. Without EEG,and MRI you have just a tiny guess of the patient recovering. A locked in person would have reflexes, so I really want to know if they would try to keep them alive – Andrey Oct 3 '17 at 15:29
• I thought a possible reason would be that well-off relatives refuse to give up hope and insist that the patient is kept alive against all medical advice and that a hospital would only go along with such treatment motivated by huge donations. – Alex2006 Oct 3 '17 at 15:32

A few weeks seems likely, but a lot of artistic license can be allowed

If the person can breathe, she will not need an "iron lung", which did not come into widespread use until after 1928. However, if the person cannot swallow food, she cannot swallow saliva either, nor cough, which is another voluntary movement that locked-in people lose. Gradually, fluid and bits of food would fill the lungs and the patient develops pneumonia if she does not drown first.

Pneumonia can develop rapidly for some patients and our locked-in person may also suffer from complications related to blood-cloths and urinary tract infection, which are two conditions that may develop from the prolonged immobilization of locked-in persons.

Even today, locked-in syndrome is deadly. Acute LIS of a "vascular origin" has an almost 90% mortality in the first 4 months [2], although those who remain alive and learn to communicate has an 80% survival rate over the next 10 years. [3]

[1] As an aside, Pneumonia may also occur in a similar way in normal people who are taking cough-suppressant medicine, or where brain surgery has damaged the cough reflex.

Ugh, that's a bleak outlook. Days.

Another 50-70 years earlier, someone with locked-in syndrome would likely have benefitted from an awesome treatment called trepanation, in the early 20th century they might as well have enjoyed the benefits of electrocution electrostimulation in one of its forms (at that time likely alternating-current cranial electrotherapy, two decades later electroconvulsive therapy would be en vogue).
Which, given the state of technology, was not much better than trepanation. Neither treatment precisely increases the subject's lifetime (or does anything beneficial for that matter), so... that's that.

On a more serious note, pressure ulcer is a frequent complication, and a cause of infection and death even today. In the early 20th century, neither was there any understanding of the problem or its causes, nor were there any applicable preventive measures or cures (clinitron bed, vacuum-assisted closure). Antibiotics? Well, come back in three decades. Someone being paralyzed/atonic in an early-20th-century style of bed... I'd expect Bad Things™ happening within days.

Parenteral nutrition with the goal of mid-term to long-term survival (that is, more than a few days) is only possible since the 1960s with Steramin and Lipofundin being among the earliest, and rather insufficient, products available. It really only "works well" since the mid-1980s. In the early 20th century, expect more-or-less isotonic saline infusions so you don't die from dehydration right away, if you're lucky. But surely, not much more.

Surgically-placed feeding tubes have been available since the late 19th century, so there's that possibility. Rates of success (in particular long-term) were, on the other hand side, not so great, as indicated by the word "desastrous" in [Minard G 2006] for... rather obvious reasons. Trans-esophageal feeding tubes have been used for two thousand or so years, but the lack of understanding in nutrient requirements made sustaining a person for more than a few weeks a "challenge". For example, the notion of essential amino acids only became known in the mid-1930s at all.

Then of course, you may have read the stories of Edgar Allan Poe. Although not likely in case vertical eye movement ability is retained, nevertheless possibly, someone suffering from locked-in syndrome might just get buried!


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# Tag Info

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This is when the attempt at accuracy creates confusion and misunderstanding. Back in the day, growth models were not incorporating technological progress, and led to a long-run equilibrium characterized by constant per capita magnitudes. Verbally, the term "steady-state" seemed appropriate to describe such a situation. Then Romer and endogenous growth ...

9

Eq. (2.64) can be written as (at first order) $$k_{t+1} - k^* = \lambda (k_t - k^*) \tag{1a}$$ Define the quantity $\kappa_t$ as $$\kappa_t \stackrel{\rm def}{=} k_t - k^* \tag{2}$$ So that $$\kappa_{t+1} = \lambda \kappa_t \tag{1b}$$ This can be solved very easy by realizing that \begin{eqnarray} \kappa_1 &=&\lambda \kappa_0 \\ \...

9

There is quite a bit of work being done in that area. One very recent example is Straub and Werning's working paper "Positive Long Run Capital Taxation: Chamley-Juff Revisited." The point seems to be that we need to consider the rate of convergence to the steady state. Also, there is other literature that gives some competing results (e.g., "the new ...

6

Ljungqvist and Sargent (2004). Recursive macroeconomic theory 2n ed. (ch. 15) present and review the issue. In the Concluding Remarks section, they mention two environments, where the "zero-optimal-capital tax rate" does not hold: Aiyagari(1995) presents a model with heterogeneous agents, incomplete insurance markets and borrowing constraints (i.e. a "...

4

Following the conversation with user @denesp at the comments of my previous answer, I have to clarify the following: the usual graphical device we use related to the basic Solow growth model (see for example here, figure 2) is not a phase diagram, since reasonably we call "phase diagrams" those that contain zero-change loci, identify the crossing points of ...

4

One general issue I see is that you try to include uncertainty in a framework developed for a deterministic setup. What you do is to use expected income in the equation of motion for human capital. Let $I_{a,t}$ denote the indicator function for attack, taking the value $1$ when there is an attack, and the value $0$ when there isn't. Then, properly, $$\dot{... 4 Rearranging the steady state equation$$ \overline{p}^{\alpha}=\alpha y\overline{p}^{\alpha-1}- \alpha\overline{p}^{\alpha}-\frac{a+1}{\sigma} $$we get$$ (1 + \alpha)\overline{p}^{\alpha}=\alpha y\overline{p}^{\alpha-1}- \frac{a+1}{\sigma}. $$As \alpha \in [0,1], the left hand side of the equation is increasing in \overline{p} and the right hand side ... 4 A system is explosive if its coefficients are non-stationary. Stationary is an important property to have in dynamic models as it tells us that an equilibrium value is obtainable (which is important in finding the BGP). If your system is explosive no equilibrium (and BGP) exists. In your example where you have the equality:$$k_t-k^* \simeq \lambda^t (k_0-...

4

A system is said to be in steady state if certain variables do not change over time (and where "certain variables" depend on the context and ought to have been clearly specified by the writer). Say we define a system to be in steady state if inflation does not change over time. Then if inflation is constant at 2%, we say that the system is in steady state ...

3

As you say the first step is to take log of both sides after that you are just applying the rules for logarithms and rearrange. For example: $$\ln (XZ)=\ln X + \ln Z$$ $$\ln X/Z= \ln X - \ln Z$$ $$\ln X^a = a \ln X$$ $$\ln 1 = 0$$ Also an important approximations that hold close to zero are applied here as well these are: $\ln(1+x) \approx x$ for $x$ ...

3

An axis is characterized not by the variable it measures, but by the unit of measurement. So an axis can measure any number of variables, as long as they are measured in the same measurement unit. In your case, the unit of measurement is "units of output", and both $(n+\delta)k$ and $sf(k)$ are measured in it (capital is measured in units of output, in case ...

3

Your first question (regarding constraints on the parameters) can be answered through first and second derivative analysis. In order to satisfy strictly increasing, we need $u'>0$ and to satisfy strictly concave, we need $u''<0$. What does this actually mean? $$u'(\frac{}{})=\phi p\theta c_t^{p-1}e^{\theta c_t^p}>0$$ Since we know that $c_t^{p-1}e^{... 3 According to your calculations MPK is not increasing in$K$. The Solow model assumes$0< \alpha < 1$, thus$\alpha - 1 < 0$and$K^{\alpha - 1}$is decreasing in$K$. 3 If$Y = C \cdot X$where$C$is constant and$\frac{\dot{X}}{X} = g$then we can solve for$\frac{\dot{Y}}{Y}$as follows: $$\frac{d}{dt} Y = \frac{d}{dt} C \cdot X = C \cdot \frac{dX}{dt} = C \cdot \dot{X} \Rightarrow$$ $$\frac{\dot{Y}}{Y} = \frac{C \cdot \dot{X}}{ C \cdot X} = \frac{\dot{X}}{X} = g$$ Therefore, as you concluded, they both grow at rate$...

3

I'm not sure I follow the logic on that equation having infinitely many solutions and steady states. In any case, in what follows are some guidelines for equilibrium selection. It depends a lot on the context. Here are some criteria, in descending order of relevance Clean Ways Is any of the steady states unstable? If so, it's less likely to be the one ...

2

If you have are trying to discretize the continuous time model $$\dot{\textbf{x}} = A\textbf{x},$$ then in discrete time you will have $$\textbf{x}_{t+1} = B \textbf{x}_t$$ but $A\neq B$, since $A$ describes the change in $\textbf{x}$ while $B$ describes the next value of $\textbf{x}$, not just the change. However $$\Delta\textbf{x}_t = \textbf{x}_{t+1} ... 2 The steady-state value k^* must be a fixed point : (1+g)(1+n)k^* = s(k^*)^{\alpha} +(1 - \delta) k^* Taking the difference between this equation and the dynamic one : (1+g)(1+n)(k_{t+1} - k^*) = s((k_{t+1})^{\alpha} - (k^*)^{\alpha}) +(1 - \delta) (k_t - k^*) Now if you denote by d_t = k_t - k^* the distance to steady-state, this gives : (1+g)(... 2 In the model with technological progress the capital per effective worker remains constant, implies that capital per worker grows at the rate of exogenous rate of technological progress. See Barro and Martin book, Chapter 1. 2 Your first question, if it's literally correct, is easy: The only way for u' to be positive for c=0 is for p=1. if p =1 then sign(\phi)=sign(\theta) so that the product is positive. But, since exp()>0 and p=1, the first term of u'' cancels out and the only way for u'' to be negative is for \phi to be negative. Therefore \theta<0 too. ... 2 Smith pointed out that as wealth was growing in any nation, the rate of profit would tend to fall and investment opportunities would diminish. Source The first half of the sentence makes it clear that this is not about steady state economies, as wealth is still growing. my question can be rephrased in a simpler way: If the economy was/became a ... 1 Start by writing the Lagrangian (as it sounds like you have already done):$$L=\sum_{t=0}^{\infty} \{\beta^t(\ln C_t^y + \ln C_t^o) + \lambda_t (C_t^y +\frac{C_t^o}{1+n} +k_{t+1}(n+1)-k_t-f(k_t))\}.$$First order conditions are given by$$\frac{\partial L}{\partial C_t^y} = \frac{\beta^t}{C_t^y} +\lambda_t = 0\frac{\partial L}{\partial C_t^o} = \frac{...

1

Usually the term steady state is derived from the Solow Model and its derivatives that seek to explain long-term economic growth. The steady state is a state in which the growth rate of the economy is constant (but positive!). In the Solow model, the growth rate is more or less a function of the saving rate. An economy might deviate from this because it's ...

1

Let production function $$F(K_t,L_t)=K_t^aL_t^{1-a}$$ $$max[c^*=f(k^*)-(δ+n)k^*]$$ with respect to $k^*$ Then $MPK=a(k^*)^{a-1}$ So $k_G= (\frac{a}{(δ+n)})^{1/1-a}$ which is golden rule level of capital stock per capita. Now calculate consumption at golden rule level $$c_G=f(k_G)-(δ+n)k_G$$ $$c_G=(k_G)^a-(δ+n)k_G$$ $$c_G=(\frac{a}{(δ+n)})^{a/1-a}-(δ+... 1 Ok, first try for your first question. From u'>0\Rightarrow \phi\theta p>0 \Rightarrow \phi,\theta, p\ne 0. From u''<0\Leftrightarrow \phi \theta p c_t^{p-1}\exp(\theta c_t^p)(p-1)c_t^{-1}+\phi \theta p c_t^{p-1}c_t^{p-1}\exp(\theta c_t^p)c_t^{-1}p\theta<0\Leftrightarrow \phi \theta p c_t^{p-1}\exp(\theta c_t^p)(p-1)c_t^{-1}+\phi \theta p c_t^{... 1 To stick to the simpler deterministic case, it is perfect foresight and the Transvarsality condition (TVC)that guarantees that an optimizing agent will stay on the saddle-path, because all other paths lead to the violation of TVC (corner solutions violate the TVC). In a deterministic environment and with perfect foresight the agent knows this and so he ... 1 In general, a steady state in discrete time is that the first difference is zero at some t=t_0, and that it remains so:$$y_t-y_{t-1}=x_t-x_{t-1}=0$$for all t\geq t_0. Hence, in a steady state, y_t and x_t are constant at some values y and x. Thus, y=\beta(y+\gamma z), which implies$$y=\frac{\beta\gamma z}{1-\beta}, and $x=\rho x+y$, which ...

1

I found the error in my derivation: I mistakenly supposet that the steady state of $p_t+1$ equals $\bar{\rho}$. I did not recognize that p was different from $\rho$ because of the poor quality of my printout. Reading the paper again on a screen I immediately spotted the mistake.

1

Normally, in the Solow model, there are not increasing marginal returns because by construction, Solow model assumes that marginal return of capital is decreasing. In fact this feature ensures the catch up between countries? which is the main insight of this model. There is another branch of literature on endogeneous growth models which has taken into ...

1

Upgrading an exchange of comments, a critical point in the question is the expression \begin{align} \frac{\partial u(\tilde x)}{\partial \rho}=\frac{1}{(\rho + u'(\tilde x) + 1)^2} \end{align} which is mistaken , because from \begin{align} u(\tilde x) = \frac{\rho + u'(\tilde x)}{\rho + u'(\tilde x) + 1}. \end{align} we obtain \begin{align} \frac{\...

1

I find that the system is saddle-path stable. Setting $z\equiv 1/k$ we get \begin{eqnarray*} \dot{z} & = & -z^2\left(\frac{\rho}{1+\rho} - \frac{1}{1+\lambda}\right) \\ \\ \dot{\lambda} & = & \rho \lambda - z \end{eqnarray*} The fixed point is $E=\{z^*, \lambda^*\} = \{1, 1/\rho\}$ The Jacobian of this system evaluated at the steady ...

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text:
# Evaluate an Integral Using a Specific Hint

## Homework Statement

$$\int$$ dx/(x4 + 16)

## Homework Equations

Hint: With a>0, x4 + a2 = (x2 + $$\sqrt{}2a$$x + a)(x2 - $$\sqrt{}2a$$x + a)

## The Attempt at a Solution

I've plugged this into the equation, which leaves me with:

$$\int$$ dx/[(x2 + $$\sqrt{}2a$$x + a)(x2 - $$\sqrt{}2a$$x + a)]

rock.freak667
Homework Helper
You could try splitting it into partial fractions now, which may make it easier.

How would I do that?

rock.freak667
Homework Helper
Rewrite the 1/hint

as

$$\frac{1}{{(x^2+ \sqrt{2a}x + a}{( x^2 - \sqrt(2a) + a)}} = \frac{Ax+B}{(x^2+ \sqrt{2a}x + a} + \frac{Cx+D}{ x^2 - \sqrt(2a) + a}$$

then bring the right side to the same denominator and equate the numerators. Then choose values of x to get the constants A,B,C,D.

Yeah. This was my first approach but if you plug in the value of a = 4, it seems the two factors of the hints never equal 0... Maybe this is just me.

Char. Limit
Gold Member
Yeah. This was my first approach but if you plug in the value of a = 4, it seems the two factors of the hints never equal 0... Maybe this is just me.

If you can't pick values of x that decide certain values, there is another option, although it's not as fun: find the powers of x on each side and equate them to get a system of equations that you can solve.

rock.freak667
Homework Helper
Yeah. This was my first approach but if you plug in the value of a = 4, it seems the two factors of the hints never equal 0... Maybe this is just me.

They don't need to equal to zero necessarily, the equation would be true for all values of x.

Okay... I tried to solve for the constants but I am stuck... I get this equation:

1 = (Ax + B)(x2 - 2*sqrt(2)*x + 4) + (Cx + D)(x2 + 2*sqrt(2)*x + 4)

I can't see how to solve this either by finding an x where the factors are zero or by solving a system of equations... How can I turn this into a system of equations?

rock.freak667
Homework Helper
Okay... I tried to solve for the constants but I am stuck... I get this equation:

1 = (Ax + B)(x2 - 2*sqrt(2)*x + 4) + (Cx + D)(x2 + 2*sqrt(2)*x + 4)

I can't see how to solve this either by finding an x where the factors are zero or by solving a system of equations... How can I turn this into a system of equations?

Right, well you can multiply it out and equate coefficients OR you can put in values for x and get equations. For example, you can choose x=2,3,etc. and you will get equations to solve.

But I thought when you put in values of x it should make one of the factors be 0 so that you can solve for the other one?

rock.freak667
Homework Helper
But I thought when you put in values of x it should make one of the factors be 0 so that you can solve for the other one?

I did not mean it like that.

You have 4 unknowns and the equation is true for all values of x.

So you can put in 4 different values for x and get 4 equations.

I've tried solving the system of equations but cannot seem to get it not matter how hard I try... can someone please help me solve the system? After that I know how to do the integral.

rock.freak667
Homework Helper
What equations did you get?

Main equation: 1 = (Ax + B)(x2 - 2$$\sqrt{}2$$*x + 4) + (Cx + D)(x2 + 2$$\sqrt{}2$$*x + 4)

Set x = 0: 1/4 = B + D

Set x = sqrt(8): 1 = 8*sqrt(2)A + 8B + 40*sqrt(2)C + 20D

Set x = sqrt(2): 1 = -2*sqrt(2)A - 2B + 28*sqrt(2)C + 14D

Set x = 3*sqrt(2): 1 = 30*sqrt(2) + 10B + 102*sqrt(2) + 34D

Which by looking at the answers seems impossible plus possibly horribly wrong. Can someone tell me what they get please?!

After looking on the internet for a sample problem like this... I have found I am retarded... Figured out the problem: equated the coefficients wrong! Thank you all for your help! :P


text:
# Re: [O] Inline LaTeX fragments in emacs buffer

org-preview-latex-fragment is really great but what is even better is that
it also works outside org-mode! This makes my life a lot easier.

I set "C-c P" as a global key-binding for org-preview-latex-fragment and
"C-c p" to org-ctrl-c-ctrl-c. Then I can preview equations all equations in
ANY buffer with "C-u C-u C-c P". It's great to write equations in comments
of some programming buffer (no matter which programming language) and being
able to render the equations right there.

--
Darlan

At Sun, 13 May 2012 00:46:51 +0200,
Daimrod <daim...@gmail.com> wrote:
>
> Dov Grobgeld <dov.grobg...@gmail.com> writes:
>
> > Is there any way of getting org-mode to display inline LaTeX fragments
> > in the emacs buffer? E.g. I would like to be able to type:
> >
> > The size of the hypotenuse is $\sqrt{a^2+b^2}$
> >
> > and then the buffer immediately shows:
> >
> > The size of the hypotenuse is √a²+b²
> >
> > where √a²+b² is a png image rendered via dvipng.
> >
> > I assume a toggle would switch between either seeing the formula or
> > seeing the source of the formula, just like for the display of inline
> > images. Also, pressing a backspace when the cursor is at the end of
> > the formula should erase the trailing \$ sign, and turn off image
> > display of the formula, just like for [[links][name]].
> >
> > Has anything like this been implemented?
> >
> > Thanks!
> > Dov
>
> Yes, you can produce a preview image with C-c C-x C-l.
>
> See
> (info "(org) Previewing LaTeX fragments")
> and
> (info "(org) Embedded LaTeX")
>
>
>


text:
Situation 27.15.1. Here $S$ is a scheme, and $\mathcal{A}$ is a quasi-coherent graded $\mathcal{O}_ S$-algebra.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
# Parachutes for sale

### Help Support The Rocketry Forum:

#### kandsrockets

##### Well-Known Member
I have two parachutes for sale. Both of these chutes are rated for 8-14 lb rockets at a decent rate of 15-20 fps. They are made of .75 oz rip-stop nylon and sewn with T-70 mil spec nylon thread. All seams are double stitched and all edges are hemmed to protect from fraying. Cords are 550 mil spec Para cord. These are not cheap parachutes. The colors are Desert Tan and OD Green. I had made these for a V2 project that never got done. I already have 3 chutes this size and do not need more. These have never been used. They come with there own storage bags. The only difference between the two is one has black cords the other has white cords.

I am asking $50.00 ea plus$8.00 for shipping in the conus. If you are interested drop me a PM

#### kandsrockets

##### Well-Known Member
Here it is all packed in its bag

Also here is the style of chute. The one below is one that I have used more than 10 times. These are great chutes.

#### kandsrockets

##### Well-Known Member
Had a chance to take a picture of the chute for sale.


text:
# Expectation value of trials needed to get $k$ consecutive outcomes

Suppose that independent trials, each of which is equally likely to have any of $m$ possible outcome, are performed until the same outcome occurs $k$ consecutive times. If $N$ denotes the number of trials, show that $$E[N] = \frac{m^k-1}{m-1}$$

This is a homework question. I was trying to reverse engineer this into a GP but without success.

I also tried an induction approach to $k$. For getting one outcome, the number of trials needed is always $1$, so $k=1$ is a trivial case. Now assume that the expectation value of trials needed for $k$ consecutive outcomes is $E[N]$, then how do I find the update to $E[N]$ for $k+1$.

I figured it out.

Say you've got $k$ identical outcomes for the first time after $N_k$ trials. Now, if you want the $(k+1)^{th}$ outcome in the next turn, the probability is $\frac{1}{m}$ and the number of extra turns needed is $1$ so the expectation of $N_{k+1}$ conditionally on this is $N_k + \frac{1}{m}$.

Now what if you don't get the same outcome on this next try? You get a different outcome hence you start all over again and you need an extra number steps with mean $E[N_{k+1}]$.

To sum up, $E[N_{k+1} \mid N_k] = N_k + \frac{1}{m} +E[N_{k+1}](1-\frac{1}{m})$.

This recursion gives $E[N_{k+1}] = E[N_k] + \frac{1}{m} +E[N_{k+1}](1-\frac{1}{m})$, hence $E[N_{k+1}] = mE[N_k] + 1$, which, together with $E[N_1]=1$, yields the desired identity.

• "you need to get that different thing $k$ times now" Not necessarily "that different thing" but it is true that one starts anew hence your intuition is correct. +1. – Did Aug 12 '16 at 13:12
• +1 for not only answering your own question, but doing so in a much slicker way than Marko Reidel's method. – J.G. Aug 18 '18 at 19:49

We use the following generating function for $k\ge 2:$

$$G(z, u) = z^k\times \sum_{q\ge 0} u^q (z+z^2+\cdots+z^{k-1})^q \\ = z^k \sum_{q\ge 0} u^q z^q (1+z+\cdots z^{k-2})^q \\ = z^k \sum_{q\ge 0} u^q z^q \frac{(1-z^{k-1})^q}{(1-z)^q} \\ = z^k \frac{1}{1-uz(1-z^{k-1})/(1-z)} = z^k \frac{1-z}{1-z-uz(1-z^{k-1})}.$$

As a sanity check we should get one when we sum the probabilities of first getting $k$ consecutive outcomes after $n$ trials. This is

$$m \sum_{n\ge 0} m^{-n} [z^n] \left. z^k \frac{1-z}{1-z-uz(1-z^{k-1})} \right|_{u=m-1} \\ = m \sum_{n\ge 0} m^{-n} [z^n] z^k \frac{1-z}{1-z-(m-1)z(1-z^{k-1})} \\ = m \sum_{n\ge 0} m^{-n} [z^n] z^k \frac{1-z}{1-mz+(m-1)z^k} \\ = m \frac{1}{m^k} \frac{1-1/m}{(m-1)/m^k} = \frac{1}{m^k} \frac{m-1}{(m-1)/m^k} = 1$$

and the sanity check goes through. Now for the expectation we obtain

$$m \sum_{n\ge 1} n \frac{1}{m^n} [z^n] \left. z^k \frac{1-z}{1-z-uz(1-z^{k-1})} \right|_{u=m-1} \\ = \sum_{n\ge 1} n \frac{1}{m^{n-1}} [z^n] z^k \frac{1-z}{1-z-(m-1)z(1-z^{k-1})} \\ = \left. \left(z^k \frac{1-z}{1-mz+(m-1)z^{k}}\right)'\right|_{z=1/m} \\ = \left.\left(k z^{k-1} \frac{1-z}{1-mz+(m-1)z^{k}} - z^k \frac{1}{1-mz+(m-1)z^{k}} \\ - z^k \frac{1-z}{(1-mz+(m-1)z^{k})^2} (k(m-1)z^{k-1}-m) \right)\right|_{z=1/m}.$$

Using $\left.1-mz+(m-1)z^{k}\right|_{z=1/m} = (m-1)/m^{k}$ we get

$$\frac{k}{m^{k-1}} (1-1/m) \frac{m^{k}}{m-1} - \frac{1}{m^k} \frac{m^{k}}{m-1} - \frac{1}{m^k} (1-1/m) \frac{m^{2k}}{(m-1)^2} \left(\frac{k(m-1)}{m^{k-1}}-m\right) \\ = k - \frac{1}{m-1} - \frac{1}{m^{k+1}} \frac{m^{2k}}{m-1} \frac{k(m-1)}{m^{k-1}} + \frac{1}{m^{k+1}} \frac{m^{2k}}{m-1} m \\ = k - \frac{1}{m-1} - k + \frac{m^k}{m-1}.$$

We thus obtain the end result

$$\bbox[5px,border:2px solid #00A000]{ \frac{m^k-1}{m-1}}$$

for m>1 , since we have m outcomes done k times isn’t that m^k by multiplication principle


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# What is the value of 11^x-11^(x+2) where x is the largest

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What is the value of 11^x-11^(x+2) where x is the largest [#permalink]  25 Aug 2011, 12:34
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What is the value of 11^x-11^(x+2), where x is the largest integer such that 11^x is a factor of 30,030?

A. -1331
B. -1320
C. -121
D. -120
E. -1
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Jul 2012, 00:51, edited 2 times in total.
EDITED THE QUESTION.
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Re: Help needed.......Quant [#permalink]  25 Aug 2011, 12:46
siddharthvaid wrote:
Q. What is the value of 11x-11(x^+2) where x is the largest integer such that 11x is a factor of 30030?

a) -1331
b) -1320
c) -121
d) -120
e) -1

By 11(x^+2) do you mean 11* x*x or 11(x*x + 2) ?
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Re: Help needed.......Quant [#permalink]  25 Aug 2011, 16:10
I think there is some problem with either the values in the ques stem or the ans choices. something is wrong. What is the source of this ques?
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Re: Help needed.......Quant [#permalink]  25 Aug 2011, 21:06
i saw this questions on the kaplan problem set and am confused as well. am beginning to think that kaplan's questions are diminishing in quality.

so first step, I tried to solve for x. x=largest integer so that 11x is a factor of 30,030.

See if 30,030 is divisible by 11. IT IS! 11x2730=30,030.

Therefore, largest integer of x = 2730. since, 30,030 is a factor of itself.

11x-11x^2=11(2730)-11(2730^2) = some INCREDIBLY HUGE NUMBER... uhh. something is wrong here kaplan.
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Re: Help needed.......Quant [#permalink]  25 Aug 2011, 21:10
Might just be an odd one
Gr8 . one less question to worry about
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Re: Help needed.......Quant [#permalink]  25 Aug 2011, 22:53
siddharthvaid wrote:
Q. What is the value of 11x-11(x^+2) where x is the largest integer such that 11x is a factor of 30030?

a) -1331
b) -1320
c) -121
d) -120
e) -1

siddharthvaid:
 ! Please post PS questions in the PS sub-forum: gmat-problem-solving-ps-140/Please post DS questions in the DS sub-forum: gmat-data-sufficiency-ds-141/No posting of PS/DS questions is allowed in the main Math forum.

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Re: Help needed.......Quant [#permalink]  31 Aug 2011, 21:51
Expert's post
siddharthvaid wrote:
Q. What is the value of 11x-11(x^+2) where x is the largest integer such that 11x is a factor of 30030?

a) -1331
b) -1320
c) -121
d) -120
e) -1

The given expression is $$11x - 11x^2$$ = $$11x (1 - x)$$

11x needs to be factor of 30030 and x needs to be the largest integer possible. This means 11x needs to be the largest factor possible. The largest factor of a number is the number itself.
The largest factor of 30030 is 30030 = (11 * 2730)
x must be 2730

The value of 11x(1-x) = 30030*(-2729)
There is definitely something wrong in the expression, either in the book or in this particular reproduction.
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Re: Help needed.......Quant [#permalink]  11 Jul 2012, 19:53
Correct expression is 11^x - 11^(x+2).

x = 1 => answer is –1320.
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Re: What is the value of 11^x-11^(x+2) where x is the largest [#permalink]  12 Jul 2012, 00:55
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What is the value of 11^x-11^(x+2), where x is the largest integer such that 11^x is a factor of 30,030?

A. -1331
B. -1320
C. -121
D. -120
E. -1

Given that $$11^x$$ is a factor of $$30,030=2*3*5*7*11*13$$. Since $$x$$ is an integer then $$x=1$$.

$$11^x-11^{x+2}=11-11^3=11(1-11^2)=-11*120=-1320$$.

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Re: What is the value of 11^x-11^(x+2) where x is the largest [#permalink]  13 Jul 2012, 08:32
11^x - 11^(x+2 ) = 11^x(1-121)

=> 11^x (-120)

given 11^x is a factor of 30030.

the only option which satisfies is -1320 (11 * -120 =-1320)
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Re: What is the value of 11^x-11^(x+2) where x is the largest [#permalink]  13 Jul 2012, 23:53
how did we infer that x=1?
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Re: What is the value of 11^x-11^(x+2) where x is the largest [#permalink]  14 Jul 2012, 02:32
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Expert's post
pavanpuneet wrote:
how did we infer that x=1?

We have that $$30,030=2*3*5*7*11*13$$ is divisible by $$11^x$$ (where $$x$$ is an integer). Now, ask yourself what can be the largest integer value of $$x$$. Could it be 2 or more?
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Re: What is the value of 11^x-11^(x+2) where x is the largest [#permalink]  14 Jul 2012, 04:24
Should Be A, I didnt read the questions fully
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Re: What is the value of 11^x-11^(x+2) where x is the largest [#permalink]  14 Jul 2012, 04:26
Expert's post
asax wrote:
Should Be A, I didnt read the questions fully

OA for this question is B, not A. Check this: what-is-the-value-of-11-x-11-x-2-where-x-is-the-largest-119607.html#p1103719
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Re: What is the value of 11^x-11^(x+2) where x is the largest [#permalink]  01 Oct 2012, 05:21
We have 11 ^ x - 11 ^x+2

Simplifying : 11 ^ x - 11^x * 11^2

Taking 11^x as common we get : 11^x (1 - 11^2)

--> 11^x * -120

The question asks us to find the max. value for x so that 11^x is a factor of 30030

Prime Factors of 30030 are 5 x 3 x 2 x 11 x 7 x 13..

Because 11 appears only once , therefore the maximum value that x can have is 1 ....

Using the value of x as 1 and adding it the equation we simplified we get

11 x -120 = - 1320 (B)
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Re: What is the value of 11^x-11^(x+2) where x is the largest [#permalink]  05 Nov 2012, 03:53
Here 30030/11 gives 2730. This cannot be divided further and hence 11 can have the power of only 1. substituting in the given equation gives the ans -1320.
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Re: What is the value of 11^x-11^(x+2) where x is the largest [#permalink]  10 Dec 2012, 05:56
Given
x is the largest integer 30030

Find the Value of
11^x - 11^(x + 2)

Simplify 11^x - 11^(x + 2) we get
(11^x)(1-11^2) -> (11^x)(1-11)(1+11) -> (11^x)(-10)(12)

Elimination (11^x)(-10)(12)
1.From this we know that the value has to be -ve
2.The units digit must be a 0.

Hence eliminate options A,C,E. Now we are left with B and D

Look at the equation again (11^x)(-10)(12) -> (11^x)(-120)
Now in option D we have -120 this can possible only if x is the above equation is 0. However we are told that 11^X is a factor of 30030. So x has to be greater than 0. Which eliminates the option D.

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Re: What is the value of 11^x-11^(x+2) where x is the largest [#permalink]  10 Dec 2012, 06:21
Expert's post
Given
x is the largest integer 30030

Find the Value of
11^x - 11^(x + 2)

Simplify 11^x - 11^(x + 2) we get
(11^x)(1-11^2) -> (11^x)(1-11)(1+11) -> (11^x)(-10)(12)

Elimination (11^x)(-10)(12)
1.From this we know that the value has to be -ve
2.The units digit must be a 0.

Hence eliminate options A,C,E. Now we are left with B and D

Look at the equation again (11^x)(-10)(12) -> (11^x)(-120)
Now in option D we have -120 this can possible only if x is the above equation is 0. However we are told that 11^X is a factor of 30030. So x has to be greater than 0. Which eliminates the option D.

11^x would be a factor of 30,030 even if x=0, since 11^0=1 and 1 is a factor of every integer. The point is that we are looking for the largest possible value of integer x such that 11^x is a factor of 30,030, which is for x=1.

Hope it's clear.
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Re: What is the value of 11^x-11^(x+2) where x is the largest [#permalink]  10 Dec 2012, 07:09
Agree. So just to make sure we can see if 30030 is divisible by 11. If it is then we really dont care by how much because we know that x is not 0 now. So the only option we are left with is now B.
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Re: What is the value of 11^x-11^(x+2) where x is the largest [#permalink]  27 Feb 2014, 19:31
30,030 has 11 as a factor
So largest value of x possible = 1

Just place x = 1 in the equation

11 - 1331 = -1320 = Answer = B
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Re: What is the value of 11^x-11^(x+2) where x is the largest   [#permalink] 27 Feb 2014, 19:31

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text:
# What is the equation of the normal line of f(x)=(5+4x)^2 at x=7?

Nov 14, 2015

$y - 1089 = - \frac{1}{264} \left(x - 7\right)$

#### Explanation:

The normal line to a tangent is perpendicular to the tangent at a point. Therefore, we must first find the slope of the tangent using the derivative.

We can use the chain rule to differentiate ${\left(5 + 4 x\right)}^{2}$. By the chain rule, we know:
$f ' \left(x\right) = 2 \left(5 + 4 x\right) \cdot \frac{d}{\mathrm{dx}} \left[5 + 4 x\right]$
$f ' \left(x\right) = 2 \left(5 + 4 x\right) \cdot 4$
color(blue)(f'(x)=8(5+4x)

We can now find the slope of the tangent line at $x = 7$.
$f ' \left(7\right) = 8 \left(5 + 4 \left(7\right)\right) = \textcolor{red}{264}$

Now, since we want the normal line, which is perpendicular to the tangent line, we want the opposite reciprocal slope: color(indigo)(-1/264

We can use point-slope form to quickly find the equation of the normal line: $y - {y}_{1} = m \left(x - {x}_{1}\right)$

We will use the point $\left(7 , 1089\right)$, which can be determined by plugging $7$ into the original equation. Thus, the equation of the normal line is color(green)(y-1089=-1/264(x-7).

Aside: if you haven't yet learned the chain rule, or want another way to differentiate ${\left(5 + 4 x\right)}^{2}$, just rewrite it as $25 + 40 x + 16 {x}^{2}$, the derivative of which is $40 + 32 x = \textcolor{b l u e}{8 \left(5 + 4 x\right)}$.


text:
Describes how to value options on commodity futures contract using the Black formula in QuantLib Python

Visit here for other QuantLib Python examples. If you found these posts useful, please take a minute by providing some feedback.

The Black-Scholes equation is the well known model to price equity European options. In the case of equities, the spot price fluctuates and hence the intuition to model as a stochastic process makes sense. In the case of commodities, however, the spot price does not fluctuate as much, and hence cannot be modeled as a stochastic process to value options on commodities. In the 1976 paper [1], Fischer Black addressed this problem by modeling the futures price, which demonstrates fluctuations, as the lognormal stochastic process.

The futures price at time $t$, $F_t$ with a is modeled as:

\begin{eqnarray} dF_t &=& \sigma_t F_t dW \end{eqnarray}

where $\sigma_t$ is the volatility of the underlying, and $dW$ is the Weiner process.

The value of an option at strike $K$, maturing at $T$, risk free rate $r$ with volatility $\sigma$ of the underlying is given by the closed form expression:

\begin{eqnarray} c &=& e^{-r T} [FN(d_1) - KN(d_2)] \nonumber \\ p &=& e^{-r T} [KN(-d_2) - FN(-d_1)] \end{eqnarray}

where \begin{eqnarray} d_1 &=& \frac{\ln(F/K) + (\sigma^2/2)T}{\sigma\sqrt{T}} \nonumber\\ d_2 &=& \frac{\ln(F/K) - (\sigma^2/2)T}{\sigma\sqrt{T}} = d_1 - \sigma\sqrt{T} \end{eqnarray}

This formula can be easily used to price caps, swaptions, futures options contract. Here we will use QuantLib to price the options on commodity futures.

In [1]:
import QuantLib as ql
import math
In [2]:
calendar = ql.UnitedStates()
bussiness_convention = ql.ModifiedFollowing
settlement_days = 0
day_count = ql.ActualActual()

## Option on Treasury Futures Contract

Options on treasury futures (10 Yr Note TYF6C 119) can be valued using the Black formula. Let us value a Call option maturing on December 24, 2015, with a strike of 119. The current futures price is 126.95, and the volatility is 11.567%. The risk free rate as of December 1, 2015 is 0.105%. Let us value this Call option as of December 1, 2015.

In [3]:
interest_rate = 0.00105
calc_date = ql.Date(1,12,2015)
yield_curve = ql.FlatForward(calc_date,
interest_rate,
day_count,
ql.Compounded,
ql.Continuous)
In [4]:
ql.Settings.instance().evaluationDate = calc_date
option_maturity_date = ql.Date(24,12,2015)
strike = 119
spot = 126.953# futures price
volatility = 11.567/100.
flavor = ql.Option.Call

discount = yield_curve.discount(option_maturity_date)
strikepayoff = ql.PlainVanillaPayoff(flavor, strike)
T = yield_curve.dayCounter().yearFraction(calc_date,
option_maturity_date)
stddev = volatility*math.sqrt(T)

black = ql.BlackCalculator(strikepayoff,
spot,
stddev,
discount)
In [5]:
print "%-20s: %4.4f" %("Option Price", black.value() )
print "%-20s: %4.4f" %("Delta", black.delta(spot) )
print "%-20s: %4.4f" %("Gamma", black.gamma(spot) )
print "%-20s: %4.4f" %("Theta", black.theta(spot, T) )
print "%-20s: %4.4f" %("Vega", black.vega(T) )
print "%-20s: %4.4f" %("Rho", black.rho( T) )
Option Price        : 7.9686
Delta               : 0.9875
Gamma               : 0.0088
Theta               : -0.9356
Vega                : 1.0285
Rho                 : 7.3974

## Natural Gas Futures Option

I saw this question on quantlib users group. Thought I will add this example as well.

Call option with a 3.5 strike, spot 2.919, volatility 0.4251. The interest rate is 0.15%.

In [6]:
interest_rate = 0.0015
calc_date = ql.Date(23,9,2015)
yield_curve = ql.FlatForward(calc_date,
interest_rate,
day_count,
ql.Compounded,
ql.Continuous)
In [7]:
ql.Settings.instance().evaluationDate = calc_date
T = 96.12/365.

strike = 3.5
spot = 2.919
volatility = 0.4251
flavor = ql.Option.Call

discount = yield_curve.discount(T)
strikepayoff = ql.PlainVanillaPayoff(flavor, strike)
stddev = volatility*math.sqrt(T)

strikepayoff = ql.PlainVanillaPayoff(flavor, strike)
black = ql.BlackCalculator(strikepayoff, spot, stddev, discount)
In [8]:
print "%-20s: %4.4f" %("Option Price", black.value() )
print "%-20s: %4.4f" %("Delta", black.delta(spot) )
print "%-20s: %4.4f" %("Gamma", black.gamma(spot) )
print "%-20s: %4.4f" %("Theta", black.theta(spot, T) )
print "%-20s: %4.4f" %("Vega", black.vega(T) )
print "%-20s: %4.4f" %("Rho", black.rho( T) )
Option Price        : 0.0789
Delta               : 0.2347
Gamma               : 0.4822
Theta               : -0.3711
Vega                : 0.4600
Rho                 : 0.1597

## Conclusion

In this notebook, I demonstrated how Black formula can be used to value options on commodity futures. It is worth pointing out that different vendors usually have different scaling conventions when it comes to reporting greeks. One would needs to take that into account when comparing the values shown by QuantLib with that of other vendors

## References

[1] Fischer Black, The pricing of commodity contracts, Journal of Financial Economics, (3) 167-179 (1976)

Click here to download the ipython notebook on treasury futures.

quantlib   python   finance

Related Post

I am Goutham Balaraman, and I explore topics in quantitative finance, programming, and data science. You can follow me @gsbalaraman.

Checkout my book

Updated posts from this blog and transcripts of Luigi's screencasts on YouTube is compiled into QuantLib Python Cookbook .


text:
# Eigenvalue Analysis and Model Reduction in the Treatment of Disc Brake Squeal

Figure 1a. General view of FE car brake model: industrial model with adjacent components.
Disc brake squeal is a frequent and annoying phenomenon. It arises from self-excited vibrations caused by friction forces at the pad-rotor interface for an industrial brake model [1] (see Figure 1a and 1b). In order to satisfy customers, the automotive industry has been trying for decades to reduce squeal by changing the design of the brake and the disc. So far, it has found no satisfactory solutions that can be implemented in a systematic way. To improve the situation, several car manufacturers, suppliers, and software companies initiated a joint project, supported by the German Federal Ministry of Economics and Technology, which included two mechanical engineering groups at Technical University (TU) Berlin and TU Hamburg-Harburg, and the numerical analysis group at TU Berlin [1]. The goal of the project was to develop a mathematical model of a brake system with all effects that may cause squeal, to simulate the brake behavior for many different parameters, and to generate a small-scale reduced-order model that can be used for optimization.

The basic finite element (FE) model for analysis and numerical methods is expressed as  the macroscopic equation of motion

$M_\Omega\ddot{u} +D_\Omega\dot{u} +K_\Omega u=f, \quad (1)$

Figure 1b. General view of FE car brake model: simple academical model.

where $$u$$ contains the coordinates (in the FE basis) of the displacements and $$f$$ is an external force. $$M_\Omega$$, $$D_\Omega$$, and $$K_\Omega\in\mathbb{R}^{n,n}$$ are large, sparse, parameter–dependent coefficient matrices that collect terms proportional to acceleration, velocity, and displacement, respectively. Here $$M_\Omega$$ is a positive semi-definite mass matrix. The nonsymmetric matrix $$D_\Omega$$ collects damping and gyroscopic effects, and the nonsymmetric matrix $$K_\Omega$$ collects stiffness and circulatory effects. The parameter $$\Omega$$ denotes the rotational speed of the brake disc.

Other possible parameters include operating conditions (such as temperature and pad pressure) and material properties (friction coefficient, brake geometry and mass distribution, effects of wear, damping, etc.). The excitation of vibrational modes can be investigated by computing the eigenvalues with positive real parts (associated with the unstable behavior of the model) of the quadratic eigenvalue problem associated with (1). This is still a major computational challenge, particularly if solving for many parameter values.

The goal of the project’s numerical component was to produce from the large-scale system (1) a small-scale system that can be used for optimizing the original system with respect to parameter variations, and in which the system’s complete unstable behavior is present. A particular version of the proper orthogonal decomposition (POD) method was used to generate the reduced-order model. The method computes all eigenvectors associated with right-half-plane eigenvalues for several sample parameter values. This is done by companion linearization and a scaled version of the shift-and-invert Arnoldi method [3] with a tricky shift selection procedure.

The so-called sample matrix is then formed from all these eigenvectors and the equations of motion are projected to the subspace associated with the large singular values of this matrix. This reduces the size of the parametric FE model from one million to about fifty while retaining provably good accuracy for the whole parameter set.

We had access to three FE models: an academic, relatively low-dimensional model, depicted in Figure 1b, which was used to develop the method; and two large-scale models, actually used in industry. Figure 1a shows one of them. While applying the new technique to the industrial models, we noticed that the eigenvalue computation was severely ill-conditioned; actually, the eigenvalue problem was often close to a singular problem. An analysis of the FE models revealed that they contained highly stiff springs used to avoid rigid connections, a modeling trick that has also been observed in a different context [2]. The negative effects of this technique came as a surprise to the industrial partners, and a similar analysis of other car manufacturers’ models demonstrated that this technique is used commonly, although with significant differences in the chosen spring constants.

Figure 2. Relation of the various error metrics and the reduced dimension.

We devised a sensitivity estimation technique based on pseudo-spectra to automatically detect the presence of these artificial stiff springs and to properly treat the rigid connections associated with the eigenvalue infinity. In combination with effective scaling and shift-and-invert transformations, we implemented and delivered a Python program to the industrial partners. This code will not only be useful in brake squeal studies, but also in other parameter-dependent vibration problems, which have non-proportional damping and other nonsymmetric terms.

To observe the advantage of the new POD method, consider our first (ill-conditioned) industrial model of over 1.2 million degrees of freedom, where the parameter is the scaled rotational speed of the disc. The new method produces better accuracy at lower reduced dimensions when compared to the traditional modal truncation method, as seen in Figure 2. The traditional modal truncation method uses as projection space the eigenvectors associated with the smallest real eigenvalues of a simplified symmetric and definite linear eigenvalue problem obtained by omitting all damping, gyroscopic, and circulatory terms in (1).

Figure 3. Selected eigenvalues for the two different methods, POD and traditional modal truncation, color coded with their residuals (log scale).

Figure 3 shows a striking example observed for the (very ill-conditioned) second industrial model; the eigenvalue approximations of the two methods are often close together but occasionally noticeably distinct, and sometimes even in different half-planes! In this case the traditional modal truncation method would have missed an eigenvalue that leads to squeal. Moreover, the residuals (indicated by the color of the shown marks (o,+)) of the eigenvalues computed by the traditional method $$(\approx 10^{-7})$$ are much higher than those by POD $$(\approx 10^{-11})$$.

In conclusion, the new model reduction method, although more costly, produces much smaller reduced models with an accuracy that is far better than the traditional modal truncation method commonly used in industry.

Acknowledgments: The authors are grateful for financial support from the IGF project 16799N of the Research Association GFaI.

References

[1] Gräbner, N., Mehrmann, V., Quraishi, S., Schröder, C., & von Wagner, U. (2015). Numerical methods for parametric model reduction in the simulation of disc brake squeal. Institut für Mathematik, TU Berlin. Manuscript submitted for publication.

[2] Kannan, R., Hendry, S., Higham, N.J., & Tisseur, F. (2014). Detecting the causes of ill-conditioning in structural finite element models. Comput. Struct., 133, 79–89.

[3] Lehoucq, R.B., Maschhoff, K., Sorensen, D., & Yang, C. (1996). ARPACK Software Package. Rice University.

Volker Mehrmann is full professor at TU Berlin. He is a member of  the German Academy of Science and Engineering (aca-tech), a SIAM Fellow, and past president of GAMM, the International Association for Applied Mathematics and Mechanics. His research interests include numerical mathematics/scientific computing, applied and numerical linear algebra, control theory, and the theory and numerical solution of differential-algebraic equations. He is editor-in-chief of Linear Algebra and its Applications and Chair of the Research Center MATHEON in Berlin. Christian Schröder is a postdoctoral fellow at TU Berlin, where his research interests include numerical linear algebra and model order reduction.


text:
# How do you solve 2x^2 + 4x +10 = 0 using the quadratic formula?

May 11, 2016

The solutions for the equation are:
color(blue)( x = -1 + 2i

 color(blue)(x = -1 - 2i

#### Explanation:

$2 {x}^{2} + 4 x + 10 = 0$

The equation is of the form color(blue)(ax^2+bx+c=0 where:

$a = 2 , b = 4 , c = 10$

The Discriminant is given by:

$\Delta = {b}^{2} - 4 \cdot a \cdot c$

$= {\left(4\right)}^{2} - \left(4 \cdot 2 \cdot 10\right)$

$= 16 - 80 = - 64$

The solutions are normally found using the formula
$x = \frac{- b \pm \sqrt{\Delta}}{2 \cdot a}$

$x = \frac{\left(- 4\right) \pm \sqrt{- 64}}{2 \cdot 2} = \frac{- 4 \pm 8 i}{4}$

$= \frac{- 4 \pm 8 i}{4} = \frac{4 \left(- 1 \pm 2 i\right)}{4}$

$= \frac{\cancel{4} \left(- 1 \pm 2 i\right)}{\cancel{4}}$

$= - 1 \pm 2 i$

$x = - 1 + 2 i$

$x = - 1 - 2 i$


text:
## Synthesis Study Finds Ocean Acidification from CO2 Emissions Could Increase by 150% by 2050; Substantial Irreversible Damage to Ocean Ecosystems

##### 16 December 2009

Given increasing emissions of CO2 and the subsequent increased absorption by the oceans, ocean acidity could increase by 150% by 2050, according to a major new review and synthesis study released by the Secretariat of the Convention on Biological Diversity (CBD). This is an increase 100 times faster than any change in acidity experienced in the marine environment over the last 20 million years, giving little time for evolutionary adaptation within biological systems, the report says.

The launch of the study—Scientific Synthesis of the Impacts of Ocean Acidification on Marine Biological Diversity—was prepared in collaboration with the UNEP World Conservation Monitoring Centre (UNEP-WCMC) and marked Oceans Day during the current climate change negotiations in Copenhagen. The study presents a review and synthesis of existing literature and other scientific information on the potential impacts of ocean acidification on marine biodiversity.

The global atmospheric concentration of CO2 has increased from a pre-industrial value of 280 ppm to 384 ppm, leading to a 30% increase in the acidity of the oceans. This significant increase is 100 times faster than any change in acidity experienced in the marine environment for the last 20 million years and represents a rare geological event in the Earth’s history.

Atmospheric CO2 concentrations are predicted to increase throughout the 21st century and could exceed 800 ppm by 2100 if anthropogenic emissions continue along current trends. Ocean acidification is a direct consequence of increasing atmospheric CO2 concentrations that is occurring independently of climate change. It can be predicted with a high level of certainty based on the complex but predictable marine carbonate chemistry reactions and cycles of CO2 as it dissolves in seawater. Ocean acidification will follow the accelerating trend in world CO2 emissions, leading to a 150–185% increase in acidity by 2100, under current emission rates (a decrease of 0.4–0.45 pH units).

...Increasing ocean acidification reduces the availability of carbonate minerals in seawater, important building blocks for marine plants and animals. Carbonate ion concentrations are now lower than at any other time during the last 800,000 years. Furthermore, given current emission rates, it is predicted that the surface waters of the highly productive Arctic Ocean will become under-saturated with respect to essential carbonate minerals by the year 2032, and the Southern Ocean by 2050 with disruptions to large components of the marine food web. Seasonal fluctuations in carbonate mineral saturation in the Southern Ocean could mean that detrimental conditions for the continuing function of marine ecosystems, especially calcifying organisms, develop on much shorter timeframes.

—Scientific Synthesis of the Impacts of Ocean Acidification

According to the study, seas and oceans absorb approximately one quarter of the carbon dioxide emitted to the atmosphere from the burning of fossil fuels, deforestation, and other human activities. As more and more carbon dioxide has been emitted into the atmosphere, the oceans have absorbed greater amounts at increasingly rapid rates. Without this level of absorption by the oceans, atmospheric CO2 levels would be significantly higher than at present and the effects of global climate change would be more marked. However, the absorption of atmospheric CO2 results in changes to the chemical balance of the oceans, causing them to become more acidic.

Ocean acidification is irreversible on timescales of at least tens of thousands of years, and substantial damage to ocean ecosystems can only be avoided by urgent and rapid reductions in global emissions of CO2. Attention must be given for integration of this critical issue at the global climate change debate in Copenhagen.

This CBD study provides a valuable synthesis of scientific information on the impacts of ocean acidification, based on the analysis of more than 300 scientific literatures, and it describes an alarming picture of possible ecological scenarios and adverse impacts of ocean acidification on marine biodiversity.

—Ahmed Djoghlaf, Executive Secretary of the Convention

Among other findings, the study shows that increasing ocean acidification will mean that by 2100 some 70% of cold water corals, a key refuge and feeding ground for commercial fish species, will be exposed to corrosive waters.

In addition, given the current emission rates, it is predicted that the surface water of the highly productive Arctic Ocean will become under-saturated with respect to essential carbonate minerals by the year 2032, and the Southern Ocean by 2050 with disruptions to large components of the marine food source, in particular those calcifying species, such as foraminifera, pteropods, coccolithophores, mussels, oysters, shrimps, crabs and lobsters, which rely on calcium to grow and mature.

An emerging body of research suggests that many of the effects of ocean acidification on marine organisms and ecosystems will be variable and complex and will affect different species in different ways. Evidence from naturally acidified locations confirms, however, that although some species may benefit, biological communities under acidified seawater conditions are less diverse and calcifying (calcium-reliant) species absent.

Many questions remain regarding the biological and biogeochemical consequences of ocean acidification for marine biodiversity and ecosystems, and the impacts of these changes on oceanic ecosystems and the services they provide, for example, in fisheries, coastal protection, tourism, carbon sequestration and climate regulation.

Launched at the Earth Summit in Rio de Janeiro in 1992, the Convention on Biological Diversity is an international treaty for the conservation and sustainable use of biodiversity and the equitable sharing of the benefits from the utilization of genetic resources. The headquarters of the Secretariat of the Convention is located in Montreal.

Resources

"This is an increase 100 times faster than any change in acidity experienced in the marine environment over the last 20 million years, giving little time for evolutionary adaptation within biological systems"

This says it all. Goodbye seafood if we don't radically change our energy consumption patterns.

.

Could Increase by 150% by 2050...

LOL!!! Once again the key word "could" is thrown in (X years from now when proved wrong: "Look, you HATER, we said could... not WILL.") as well as some date so far off that it truly is a completely meaningless statement.

Who, exactly is FUNDING this "science?" What kinds of extreme profits/wages/retirement plans are being raked in by "studying" these issues and coming up with the "correct" result?

We need to take DRASTIC measures - MASSIVE tax increases and draconian loss of freedom because "The fact is that we can't account for the lack of warming at the moment and it is a travesty that we can't.... " - Kevin Trenberth, one of the leading authors of the IPCC report on climate change.

Science... LOL!!! Follow the \$

.

Again, who is funding the Goracle, stan peterson? Exxon Mobile, OPEC ?

Treehugger: Probbly the same peopel who're funding the IPCC's lobby groups. That would be Exxon-Mobil, Shell Oil, the WWF, the DOE, a myriad of investment companies, several investment banks, etc. You know, everyone who has a financial interest in screwing everyone else.

Who's funding YOU?

As for this study, I have a question... I thought CO2 levels were going to rise exponentially when the supposed warming heated up the oceans so they released more CO2? Now they're absorbing it instead? WTF? Which is it? You can't have both.

Wake up and smell the malarky, people. This isn't about the environment, it's about MONEY. Why do you think all your precious climate change activist groups are being shut out of the talks at COP15? Because it's the BANKERS and BUSINESS who are running the show, and NOT concern for nature and the environment.

Once again I ask: when are we going to stop playing around with this stupid CO2 distraction and start talking about REAL environmental problems?

Indeed, Goracle is just a mouthpiece for the carbon-lobby PR firms' talking points.

And the aggressive ignorance from the likes of Aaron Turpen doesn't help.

Collapse of ocean food chains may turn out to be the most catastrophic consequence of AGW. How many millions of humans depend on sea life for employment and/or food? I fear it's a grim world we are creating. Care for some jellyfish salad?

Like SciAm, Climate Progress, NY Times, etc., too many people are reading GCC. The quality of its posts and historical level of discussion means it has become a target.

The approach is obvious, if unable to debate the facts intelligently, then shout it down.

.

jcwinnie, You make VERY good points. The standard discussion of the Globalwarmists is to shout "the science is settled" then refuse to discuss the so-called science. "Science" that is based on data that the "scientists" refuse to release (peer reviewed?!? Not if it's not released.)

Yes, a discussion needs to take place with regard to climate, why data is not released, why scientists break the law by refusing to comply with FOIA requests, etc. There also needs to be a real discussion with regard to POLLUTION, not this silly CO2 topic. Most everyone is against pollution and are for energy independence (although many are against energy independence if it means using our own energy in the form of coal, oil, or natural gas). Why do the Globalewarmists refuse to discuss the "science?" The discussion is always focused on what to do about "science" that has very little basis in fact - it is based on models that DO NOT match today's conditions.

.

EP and JC,
Seems some think there is some mileage to be got as bandleader for the ignorant rowdy's.
I guess they get to play follow the leader as reward. The prerequisite display of ignorance is truly spectacular.

jcwinnis said;

"The approach is obvious, if unable to debate the facts intelligently, then shout it down."

Those shouting admit they have no real, defensible position. Goracle is a prime example...

.

Those shouting admit they have no real, defensible position.. Goracle is a prime example...

Nice shout, Will S!

LOL... Science!

,

As with anything involving climate is helps to look at the long term record to get some perspective.

According to the 2009 paper by Liu et al, they reconstructed the paleo-pH record of the past 7000 years in the South China Sea. The study shows a natural variation of pH between 8.3 and 7.90. The low arrived about 6000 years ago and the high about 1000 years ago. We are currently at a broad global average pH of 8.0-8.1 Clearly the low (less basic) of 6k years ago, was not caused by man-made CO2.

There is further evidence of pH variation coincident with the PDO. Does this mean we don't worry about dumping crap in the oceans? Or urban street runoff into oceans?? Hell no! Those are real pollution problems and addressing them should be a part of any industrialized nation's environmental protection plan.

Liu, Y., Liu, W., Peng, Z., Xiao, Y., Wei, G., Sun, W., He, J. Liu, G. and Chou, C.-L. 2009. Instability of seawater pH in the South China Sea during the mid-late Holocene: Evidence from boron isotopic composition of corals. Geochimica et Cosmochimica Acta 73: 1264-1272

Goracle,

Do you post under multiple names?

I live in Seattle. The shellfish industry in W. Wa. is experiencing for the first time in it's long history the effects of ocean ph eating their shellfish seed stock before it is old enough to develop a heavy enough shell to protect them. Don't tell me this is some B.S. science.
The people with their heads in the sand or else where need to wake up. Also if this last decade is again the warmest on record, then how is it unexplained cooling?
Is the north polar ice regrowing? Has the melting of the major ice caps reversed themselves? Those giant icebergs floating by New Zealand change course? Did the Mauna Loa Atmospheric Observatory suddenly see a decrease in Co2? And once the N.Pole is gone, whats to stop the permafrost from melting out and releasing bil. of tons of methane?

I do have to agree with some of Goracles first post. The public is getting tired of opinion and projection. Words are not pictures and we hear it all the time. The nightly news is not fact anymore it is opinion. it is entertainment. So much so that the average citizen has either become numb to the effects or drifts off into a haze with the passing words and grasps only the most exotic or radical crap flung at them. What we need are facts and then images to back it up. We need to see more comparative images of the polar ice and the sea level rise. We need to see pacific islanders moving from their traditional homelands. We need to see Northern indigenous peoples pointing out the melting of permafrost that has never melted before in written or oral history. We need to see images of receding ice pack that once inhabited the same regions years before where those same indigenous peoples used to hunt or where shore birds used to forage for food. Enough words let's see the bare facts in real terms.

jeffa,

thank you for being honest about your emotional attachment to the issues. That, however does not replace good science.

Cats and Dogs living together!!!! This two days after the Russians point out that the British MET office was Cherry picking termperature readings like Michael Mhann (sp) cherry picks tree rings.

I think in 30 years the last believer in AGW will be pulled out of a cave on some remote "dry" pacific island, like those Japanese soldiers that spent 30 years in the Jungle still believing the war was going on.

The fact that the basic facts of the post are being argued with shows that the sceptics don't care about the science. High school chemistry tells us that CO2 is absorbed by water to form a weak acid, carbonic acid. More CO2 inthe atmosphere equals lower ph. Measurement and observation tell us that it's happening

Are you planning to go and live in the Maldives Esabre? I think not. I live in Australia and we are being hit hard by climate change, living in a fool's paradise doesn't help anyone

The comments to this entry are closed.


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The Global Biodiversity Information Facility (GBIF)is a compilation of presence data for a broad range of species (absence data may be included). If you're interested in birds, data is funneled into GBIF from eBird, VertNet and other datasets. You can search widely across taxa, but you may be out of luck if you study viruses or microorganisms.

Accessing data from GBIF only takes creating an account and writing a few lines of code. You can download by species, geographical area or other query words. In the code below, I'll take you through downloading monk parakeet observations and filtering the data prior to mapping. To run the following code, you need to create your own account on GBIF, and install the rgbif package.

library(rgbif)

# obtain taxon key for species of interest
key <- name_suggest(q = "Myiopsitta monachus", rank = "species")$key[1] key # 2479407  Since we'll be making maps in later posts, it makes sense to ensure that we're downloading sufficient data. How many Myiopsitta monachus obsvervations have coordinates? occ_search(taxonKey = key, limit = 20, hasCoordinate = TRUE)  Now you can use the taxon key to download GBIF data. The following chunk of code may look funny, since the function names don't correspond to their actual functions. The logic is more straightforward than the functions' names. • once you start the download query, you can check its status • continue checking until the query status shows up as SUCCEEDED • then download the zip file pulled by the query • import the data to R You don't have to worry about saving the zip file. Once you import data to R, you can save it at any point as a .csv file in your working directory. # insert your GBIF account info to start a query mp <- occ_download('taxonKey = 2479407', user = "", pwd = "", email = "") # check query status occ_download_meta(mp) # when query is done running, download data mp <- occ_download_get(mp, overwrite = TRUE) # import into R and begin manipulating # ignore warning messages about changing class of certain variables suppressPackageStartupMessages(library("dplyr")) mp <- tbl_df(occ_download_import(mp)) # save data to Desktop to avoid downloading from GBIF again write.csv(mp, paste("Myiopsitta_monachus_GBIF_", Sys.Date(), ".csv", sep = ""))  The data is in your hands! It's time to familiarize yourself with the data and prune out any fishy observations. mp <- read.csv(paste("Myiopsitta_monachus_GBIF_", Sys.Date(), ".csv", sep = ""), header = TRUE)  Here are some basic functions to explore data. names(mp) # column names str(mp) # information about variable types and length head(mp) # display the head of the data frame View(mp) # viewing data frame in new RStudio tab  Some observations may belong to other species entirely, but that isn't the case here. unique(mp$vernacularName)

## [1] Monk Parakeet
## [2]
## [3] Monk Parakeet; Cockatoos; Parrots; Birds; vertebrates; chordates; animals
## [4] Monk Parakeet, Quaker Parrot
## [5] Monk Parakeet (Cliff)
## [6] MONK PARAKEET
## [7] munkpapagoi; Monk Parakeet
## 7 Levels:  Monk Parakeet MONK PARAKEET ... munkpapagoi; Monk Parakeet

You can use regular expression to remove any observations that may belong to a different species

table(mp$vernacularName)  ## ## ## 3524 ## Monk Parakeet ## 47267 ## MONK PARAKEET ## 1 ## Monk Parakeet (Cliff) ## 7 ## Monk Parakeet, Quaker Parrot ## 2 ## Monk Parakeet; Cockatoos; Parrots; Birds; vertebrates; chordates; animals ## 79 ## munkpapagoi; Monk Parakeet ## 1  mp1 <- mp[grep("luchsi", mp$scientificName, ignore.case = TRUE, invert = TRUE), ]
nrow(mp1)

## [1] 50839

Coordinate system, necessary to determine points are plotting correctly

table(mp1$verbatimCoordinateSystem)  ## ## decimal degrees deg. min. sec. ## 50573 16 26 ## Degrees Minutes Seconds UTM ## 5 219  Do observations without a common name have the correct scientific name? sn <- mp1[grep("monk parakeet", mp1$vernacularName, ignore.case = TRUE, invert = TRUE),
grep("scientificName$", names(mp1))] unique(sn)  ## [1] Myiopsitta monachus (Boddaert, 1783) ## [2] Myiopsitta monachus cotorra (Vieillot, 1818) ## [3] Myiopsitta monachus subsp. monachus ## [4] Myiopsitta monachus calita (Jardine & Selby, 1830) ## 6 Levels: Myiopsitta luchsi (Finsch, 1868) ...  Once you're sure all observations have at least been recorded as the species in question, make the data ready for mapping. # remove observations without coordinates (e.g., with NAs) mp2 <- mp1[which(!is.na(mp1$decimalLatitude) & !is.na(mp1$decimalLongitude)), ] # quality control # length(which(is.na(mp1$decimalLatitude) & is.na(mp1$decimalLongitude))) # nrow(mp1) - nrow(mp2) # remove sightings down by Antarctica mp3 <- mp2[!mp2$decimalLatitude < -50,]

# more quality control
# length(which(mp2$decimalLatitude < -50)) == nrow(mp2) - nrow(mp3)  Around fall 2014, GBIF uploaded a whole lot of absence data, which looks really strange when plotted on a map. I was assured by a GBIF data scientist that most absence data will have an absent tag associated with the occurrenceStatus variable, and should thus be easy to remove. table(mp3$occurrenceStatus)

##
##                                absent confirmed breeding
##              47334               2589                 11
##            present
##                108

# remove data that are actually absences
mp4 <- mp3[grep("absent", mp3$occurrenceStatus, invert = TRUE, ignore.case = TRUE), ] # quality control # length(grep("absent", mp3$occurrenceStatus, ignore.case = TRUE))
# nrow(mp3) - nrow(mp4)

# manipulate the data to retain columns useful for mapping
mp5 <- mp4[, grep(paste(c("decimalLatitude", "decimalLongitude", "^genus$", "speciesKey", "vernacularName" , "year", "taxonID", "scientificName$", "verbatimLocality", "stateProvince",
"gbifID", "verbatimCoordinateSystem", "occurrenceStatus"),
collapse = "|"), names(mp4))]

Use the basic functions mentioned above to examine the structure of mp6.

str(mp5)

## 'data.frame':    47453 obs. of  13 variables:
##  $gbifID : int 593683661 593689298 593701262 593713668 593722845 593748880 593748887 593752425 593755094 593760948 ... ##$ decimalLatitude         : num  25.7 29.7 32.8 41.2 28 ...
##  $decimalLongitude : num -80.3 -95.4 -96.7 -73.1 -82.7 ... ##$ genus                   : Factor w/ 1 level "Myiopsitta": 1 1 1 1 1 1 1 1 1 1 ...
##  $occurrenceStatus : Factor w/ 4 levels "","absent","confirmed breeding",..: 1 1 1 1 1 1 1 1 1 1 ... ##$ scientificName          : Factor w/ 6 levels "Myiopsitta luchsi (Finsch, 1868)",..: 2 2 2 2 2 2 2 2 2 2 ...
##  $stateProvince : Factor w/ 257 levels "","A Coruña",..: 94 233 233 71 94 233 233 169 233 149 ... ##$ taxonID                 : int  NA NA NA NA NA NA NA NA NA NA ...
##  $verbatimCoordinateSystem: Factor w/ 5 levels "","decimal degrees",..: 1 1 1 1 1 1 1 1 1 1 ... ##$ verbatimLocality        : Factor w/ 329 levels "","; ; ; ; ; ; ;",..: 1 1 1 1 1 1 1 1 1 1 ...
##  $vernacularName : Factor w/ 7 levels "","Monk Parakeet",..: 2 2 2 2 2 2 2 2 2 2 ... ##$ year                    : int  2011 2011 2011 2011 2011 2011 2011 2011 2011 2011 ...
##  \$ speciesKey              : int  2479407 2479407 2479407 2479407 2479407 2479407 2479407 2479407 2479407 2479407 ...

names(mp5)

##  [1] "gbifID"                   "decimalLatitude"
##  [3] "decimalLongitude"         "genus"
##  [5] "occurrenceStatus"         "scientificName"
##  [7] "stateProvince"            "taxonID"
##  [9] "verbatimCoordinateSystem" "verbatimLocality"
## [11] "vernacularName"           "year"
## [13] "speciesKey"

nrow(mp5)

## [1] 47453

We are done! After creating a .csv file in the working directory, we can move on to use the data for mapping purposes.

write.csv(mp5, paste("Myiopsitta_monachus_filtered_GBIF_", Sys.Date(), ".csv", sep = ""))


text:
The mole is a quantity of substance. The new prefix ‘guaca’ is defined suchthat one guacamole equals Avocado’s Number.

You are watching: How many atoms in a pound mole

—G. Byrne.

For much better or worse, there space a variety of unit systems that engineers have to compete with ~ above a constant basis.

The two equipment we will address most frequently are the following:

unit system

selected base units

SI (International device of Units)

$$\sim$$, $$\sikg$$, $$\siN$$, $$\sis$$

United says customary

$$\sift$$, $$\silb_f$$, $$\silb_m$$, $$\sis$$

## What is a pound?¶

Some resources use the term pound to median pound massive (lbm or $$\silb_m$$) andothers usage it to mean pound force (lbf or $$\silb_f$$).This is analogous come the difference in between mass and weight, i m sorry are additionally commonly confused.

Weight is the pressure exerted ~ above a massive by the regional gravitational field,$$f = m \cdot a = m \cdot g$$, wherein $$\sig$$ is the acceleration the gravity.Thus, $$\silb_f$$ = $$\silb_m$$ $$\cdot$$ $$\sig$$.

## What is a gmol?… a kgmol?… a lbmol?¶

First, what is a mole?

A mole is a certain number of molecules, atoms, electrons, etc.In the SI system, the mole has about $$\num6.022140d23$$ particles.This is correctly defined as a gram mole, though SI units officiallydesignate this together a mole.

Alternatively, a $$\sikgmol$$ (kilogram mole) is same to $$1000$$gram moles, and therefore can be believed of as having $$1000$$times $$\num6.022140d23$$ particles.

A $$\silbmol$$ (pound mole) has $$454$$ time $$\num6.022140d23$$particles and also is thus larger than a $$\sigmol$$.

Another way to look at the mole unit is together a unit offered to convert thenumber of moles to mass, using the molecular load ($$MW$$).

In summary

$$\sigmol$$ = (mass in grams) / $$MW$$

$$\silbmol$$ = (mass in pounds) / $$MW$$

or

mass in $$\sig$$ = $$MW$$ $$\cdot$$ $$\sigmol$$

mass in $$\silb_m$$ = $$MW$$ $$\cdot$$ $$\silbmol$$

## Converting in between units¶

Converting devices is a very essential skill for CBEs.

It helps united state to

use a consistent collection of devices in calculations

transfer results into and out of various other unit systems of interest

assure the equations space formulated correctly

Your textbook, Introduction to chemistry Engineering: tools for Today and also Tomorrowhas tables of common conversion determinants (page xix of the 5th edition):

Perry’s chemical Engineers’ Handbook (highly recommended) and theCRC Handbook that Chemistry and also Physics have many more conversion factors.

There space different, equivalent, ways that you will certainly see systems written.

For instance, the devices for the heat transfer coefficient, $$h$$,can it is in written plenty of ways:

 $$\mathrmW \cdot m^-2 \cdot K^-1$$ $$\mathrmW/m^2/K$$ $$\siW/m^2.K$$ $$\mathrmW/(m^2 K)$$ $$\mathrmW/(m^2 \cdot K)$$ $$\mathrmW m^-2 K^-1$$ $$\mathrmW/m^2 \cdot K$$ $$\frac\mathrmW\mathrmm^2 \cdot K$$

Example:

Convert $$\SI5W/m^2.\degree C$$to $$\siBtu/s.ft^2.\degree F$$

\<\SI5\frac\siW\sim^2.\degree C \cdot\left(\frac\SI3.412Btu/hr\SI1W\right)\left(\frac\SI1hr\SI3600s\right)\left(\frac\SI1m\SI3.28ft\right)^2\left(\frac\SI1\degree C\SI1.8\degree F\right)= \SI2.447d-4\frac\siBtu\sis.ft^2.\degree F\>

Exercise: Unit conversion

In their third most outstanding prank to date, CBE students totally drained Horsetooth reservoir.

As the water resource engineer in charge, it is your obligation to refill it. ☹

A pair of passionate interns working for girlfriend volunteer to walk to the store to choose up the required water in 1-liter bottles.

A 1-liter party of water is around $$\SI11inches$$ tall.

See more: What Is Everyday Motion Usually Relative To, Speed And Velocity

If the compelled bottles to be placed end to end, just how long would certainly the ridge be?

In almost right terms, would the length be the street from…

the soil floor to the optimal of the Petronas pair Towers ($$\approx \SI450meters$$)?

sea level to the optimal of mount Everest ($$\approx \SI8.9d3meters$$)?

Fort Collins come Denver ($$\approx \SI100km$$)?

the planet to the Moon ($$\approx \SI380d3km$$)?

the earth to Mars ($$\approx \SI55d6km$$ at your closest)?

the sun to Pluto ($$\approx \SI4.4d9km$$ at your closest)?

Exercise: Unit switch for tradition units

In physics, the first radiation constant, $$c_1$$,is approximately $$\SI3.741d-16W.m^2$$,where $$\siW$$ is watts ( $$\siJoules/s$$ ), a unit that power.

Being a clever engineering student, friend have occurred a brand-new system ofunits that you find more intuitive. The pertinent units and also conversionfactors in your mechanism are as follows: