text:
International
Tables for
Crystallography
Volume B
Reciprocal space
Edited by U. Shmueli

International Tables for Crystallography (2010). Vol. B, ch. 3.2, pp. 410-413   | 1 | 2 |

## Section 3.2.2. Least-squares plane based on uncorrelated, isotropic weights

R. E. Marsha* and V. Schomakerb

aThe Beckman Institute–139–74, California Institute of Technology, 1201 East California Blvd, Pasadena, California 91125, USA, and  bDepartment of Chemistry, University of Washington, Seattle, Washington 98195, USA
Correspondence e-mail:  rem@xray.caltech.edu

### 3.2.2. Least-squares plane based on uncorrelated, isotropic weights

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This is surely the most common situation; it is not often that one will wish to take the trouble, or be presumptive enough, to assign anisotropic or correlated weights to the various atoms. And one will sometimes, perhaps even often, not be genuinely interested in the hypothesis that the atoms actually are rigorously coplanar; for instance, one might be interested in examining the best plane through such a patently nonplanar molecule as cyclohexane. Moreover, the calculation is simple enough, given the availability of computers and programs, as to be a practical realization of the off-the-cuff treatment suggested in our opening paragraph. The problem of deriving the plane's coefficients is intrinsically nonlinear in the way first discussed by Schomaker et al. (1959; SWMB). Any formulation other than as an eigenvalue–eigenvector problem (SWMB), as far as we can tell, will sometimes go astray. As to the propagation of errors, numerous treatments have been given, but none that we have seen is altogether satisfactory.

We refer all vectors and matrices to Cartesian axes, because that is the most convenient in calculation. However, a more elegant formulation can be written in terms of general axes [e.g., as in Shmueli (1981)].

The notation is troublesome. Indices are needed for atom number and Cartesian direction, and the exponent 2 is needed as well, which is difficult if there are superscript indices. The best way seems to be to write all the indices as subscripts and distinguish among them by context – i, j, 1, 2, 3 for directions; k, l, p (and sometimes K, …) for atoms. In any case, atom first then direction if there are two subscripts; direction, if only one index for a vector component, but atom (in this section at least) if for a weight or a vector. And , e.g., for the standard uncertainty of the distance of atom 1 from a plane. For simplicity in practice, we use Cartesian coordinates throughout.

The first task is to find the plane, which we write aswhere r is here the vector from the origin to any point on the plane (but usually represents the measured position of an atom), m is a unit vector parallel to the normal from the origin to the plane, d is the length of the normal, and and are the column representations of m and r. The least-squares condition is to find the stationary values of subject to , with , , the vector from the origin to atom k and with weights, , isotropic and without interatomic correlations for the n atoms of the plane. We also write S as , the subscript for atom number being implicit in the Gaussian summations over all atoms, as it is also in the angle-bracket notation for the weighted average over all atoms, for example in  – the weighted centroid of the groups of atoms – just below.

First solve for d, the origin-to-plane distance.ThenHere is the vector from the centroid to atom k. Then solve for m. This is the eigenvalue problem – to diagonalize (bear in mind that is just ) by rotating the coordinate axes, i.e., to find the arrays and , diagonal, to satisfy and are symmetric; the columns of are the direction cosines of, and the diagonal elements of are the sums of weighted squares of residuals from, the best, worst and intermediate planes, as discussed by SWMB.

#### 3.2.2.1. Error propagation

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Waser et al. (1973; WMC) carefully discussed how the random errors of measurement of the atom positions propagate into the derived quantities in the foregoing determination of a least-squares plane. This section presents an extension of their discussion. To begin, however, we first show how standard first-order perturbation theory conveniently describes the propagation of error into and when the positions of the atoms are incremented by the amounts and the corresponding quantities (the vectors from the centroid to the atoms) by the amounts . (The need to account for the variation in position of the centroid, i.e. to distinguish between and , was overlooked by WMC.) The consequent increments in and areHere the columns of are expressed as linear combinations of the columns of . Note also that both perturbations, and , which are the adjustments to the orientations and associated eigenvalues of the principal planes, will depend on the reduced coordinates and the perturbing influences by way of , which in turn depends only on the reduced coordinates and the reduced shifts . In contrast,the change in the origin-to-plane distance for the plane defined by the column vectors m of , depends on the and directly as well as on the and by way of the

The first-order results arising from the standard conditions, diagonal, and , areandStated in terms of the matrix components and , the first condition is , hence , and the second is . With these results the third condition then readsAll this is analogous to the usual first-order perturbation theory, as, for example, in elementary quantum mechanics.

Now rotate to the coordinates defined by WMC, with axes parallel to the original eigenvectors , restrict attention to the best plane , , and define as , keeping in mind ; itself, the original plane-to-centroid distance, of course vanishes. One then findsand alsoThese results have simple interpretations. The changes in direction of the plane normal (the ) are rotations, described by and , in response to changes in moments acting against effective torsion force constants. For , for example, the contribution of atom k to the total relevant moment, about direction 1, is ( the force' and the lever arm), and its nominally first-order change has two parts, from the change in force and from the change in lever arm; the resisting torsion constant is , which, reflection will show, is qualitatively reasonable if not quantitatively obvious. The perpendicular displacement of the plane from the original centroid is , but there are two further contributions to , the change in distance from origin to plane along the plane normal, that arise from the two components of out-of-plane rotation of the plane about its centroid. Note that is not given by , which vanishes identically.

There is a further, somewhat delicate point: If the group of atoms is indeed essentially coplanar, the are of the same order of magnitude as the , unlike the , , which are in general about as big as the lateral extent of the group. It is then appropriate to drop all terms in or , and, in the denominators, the terms in .

The covariances of the derived quantities (by covariances we mean here both variances and covariances) can now be written out rather compactly by extending the implicit designation of atom numbers to double sums, the first of each of two similar factors referring to the first atom index and the second to the second, e.g., . Note that the various covariances, i.e. the averages over the presumed population of random errors of replicated measurements, are indicated by overlines, angle brackets having been pre-empted for averages over sets of atoms.

Interatomic covariance (e.g., ) thus presents no formal difficulty, although actual computation may be tedious. Nonzero covariance for the 's may arise explicitly from interatomic covariance (e.g., ) of the errors in the atomic positions , and it will always arise implicitly because in includes all the and therefore has nonzero covariance with all of them and with itself, even if there is no interatomic covariance among the 's.

If both types of interatomic covariance (explicit and implicit) are negligible, the covariances simplify a great deal, the double summations reducing to single summations. [The formal expression for does not change, so it will not be repeated.]

When the variances are the same for as for (i.e. , all i, j) and the covariances all vanish , the simplify further. If the variances are also isotropic , all i, j), there is still further simplification toIf allowance is made for the difference in definition between and , these expressions are equivalent to the ones (equations 7–9) given by WMC, who, however, do not appear to have been aware of the distinction between and and the possible consequences thereof.

If, finally, for each atom is taken equal to its , all j, there is still further simplification.

For the earlier, more general expressions for the components of it is still necessary to find and in terms of , with .

In the isotropic, no-correlation' case, for example, these reduce toandHere the difference between the correct covariance values and the values obtained on ignoring the variation in may be important if the number of defining atoms is small, say, 5 or 4 or, in the extreme, 3.

#### 3.2.2.2. The standard uncertainty of the distance from an atom to the plane

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There are two cases, as has been pointed out, e.g., by Ito (1982).

 (1) The atom (atom K) was not included in the specification of the plane.In the isotropic, no-correlation' case the last three terms, i.e. the terms in , are all negligible or zero. In either case the value for and the appropriate values from the least-squares-plane calculation need to be inserted. (2) Atom K was included in the specification of the plane. The expression for remains the same, but the averages in it may be importantly different.

For example, consider a plane defined by only three atoms, one of overwhelmingly great w at (0, 0, 0), one at (1, 0, 0) and one at (0, 1, 0). The centroid is at (0, 0, 0) and we take , i.e. is the item of interest. Of course, it is obvious without calculation that the standard uncertainties vanish for the distances of the three atoms from the plane they alone define; the purpose here is only to show, in one case for one of the atoms, that the calculation gives the same result, partly, it will be seen, because the change in orientation of the plane is taken into account. If the only variation in the atom positions is described by , one has , and . The nonvanishing terms in the desired variance are thenIf, however, the problem concerns the same plane and a fourth atom at position , not included in the specification of the plane and uncertain only in respect to (which is arbitrary) with (the same mean-square variation in direction 3 as for atom 2) and , the calculation for runs the same as before, except for the third term:

Extreme examples of this kind show clearly enough that variation in the direction of the plane normal or in the normal component of the centroid position will sometimes be important, the remarks to the contrary by Shmueli (1981) and, for the centroid, the omission by WMC notwithstanding. If only a few atoms are used to define the plane (e.g., three or, as is often the case, a very few more), both the covariance with the centroid position and uncertainty in the direction of the normal are likely to be important. The uncertainty in the normal may still be important, even if a goodly number of atoms are used to define the plane, whenever the test atom lies near or beyond the edge of the lateral domain defined by the other atoms.

### References

Ito, T. (1982). On the estimated standard deviation of the atom-to-plane distance. Acta Cryst. A38, 869–870.
Schomaker, V., Waser, J., Marsh, R. E. & Bergman, G. (1959). To fit a plane or a line to a set of points by least squares. Acta Cryst. 12, 600–604.
Shmueli, U. (1981). On the statistics of atomic deviations from the best' molecular plane. Acta Cryst. A37, 249–251.
Waser, J., Marsh, R. E. & Cordes, A. W. (1973). Variances and covariances for best-plane parameters including dihedral angles. Acta Cryst. B29, 2703–2708.


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# Homework Help: Compute temperature rise from solar irradiance

1. Jun 16, 2011

### donbock

1. The problem statement, all variables and given/known data
(This is not a homework problem; it comes from a real-world problem I'm trying to solve.)

Sunlight is falling on an absorptive object. What will be equilibrium temperature of the object given the solar irradiance (I) and the ambient temperature?
I = solar irradiance = 1120 W/m$^{2}$ (from MIL-STD-810G).

2. Relevant equations
My guess is that thermal equilibrium is reached when the energy flowing into the object from solar irradiance is balanced by the gray-body energy radiated from the object.

Radiated energy is given by the Stefan-Boltzmann Law:
P/A = e * $\sigma$ * (T$^{4}$ - T$_{a}$$^{4}$)

P/A = energy radiated per second per unit area (W/m$^{2}$). Set this equal to I.
e = emissivity of the object = 0.93.
$\sigma$ = the Stefan-Boltzmann Constant (5.67 x 10$^{-8}$ W/m$^{2}$/K$^{4}$)
T = temperature of the object (degrees Kelvin)
T$_{a}$ = ambient temperature (degrees Kelvin)

To use this formula I will also have to be given the emissivity (e) of the object.

3. The attempt at a solution
This assumes the object absorbs all infalling solar irradiance (no reflection, no transmission).
This ignores convective cooling.
This ignores conductive cooling.

Solving for T as a function of T$_{a}$ yields the following results. (Notice I converted the temperatures to Celsius in the following list.)
T$_{a}$ (degC), T (degC):
0, 132
20, 138
40, 146
60, 155
80, 165
100, 176

These results don't feel plausible: I've never seen an object heated above the boiling point simply by letting the sun shine on it.
Is my approach all wrong?
Are the counterintuitive results solely due to the effects of my three assumptions?

Last edited: Jun 16, 2011
2. Jun 16, 2011

### RTW69

If you are tying to find the surface temperature of the component given I, e, Ta your answer is correct. But I don't know what you mean by "How much will the temperature of the object increase?" Do you mean, what is the surface temperature? The difference between the surface temperature and the ambient temperature? I am not sure what you are trying to find.

3. Jun 16, 2011

### donbock

Prior to exposure to the sun, the object can be expected to be at the ambient temperature. When I said "how much will the temperature increase", I really meant "what is the object's equilibrium temperature" and, oh by the way, how much hotter is that than the ambient.
( I revised the original question to eliminate this source of confusion.)

I don't trust my results because, for example, with ambient temperature of 0 degC, my result is that the object heats up to 132 degC just by letting the sun shine on it! That is outside my experience.

Last edited: Jun 16, 2011
4. Jun 16, 2011

### RTW69

Your analysis is correct given the assumptions you have imposed on the problem but remember the surface has no mass nor is it loosing heat by convection, conduction or re-radiation. If you stick a bucket of water in the sun at 1200 w/m^2 with ta=0 c its surface temperature will not be 132 C. You intuition is correct. Some of the heat will raise the temperature of the water and container, heat will be lost through the bottom of the container and by convection from the sides and top of the container. An interesting exercise is to add those elements to the problem to find the new T-surface.

5. Jun 16, 2011

### donbock

If I knew how much energy was conducted away from the surface to heat the bulk of the object, then I could use the specific heat capacity to compute the temperature change -- but I don't see how to predict how much energy is conducted.

Even if I knew how to do this, I'm not sure it is very relevant. The sun will keep pumping energy into my object until the entire thing reaches that gray-body temperature ... unless I figure out how to model the energy that leaves the object and flows into the surroundings. I'm at a loss as to how to do this.

I'm looking into this because we have plastic lenses that occassionally melt [or at least deform] due to heat. The ambient temperature is high, but not high enough. The first lens in the stack shows this damage. It is located immediately beneath, and almost touching, a glass filter that transmits only a narrow range of frequencies. I want to compute how hot the filter is likely to get if the sun shines directly on it for several hours. Perhaps solar irradiation raises the filter temperature high enough to melt the nearby plastic lens.

We are instrumenting the assembly, but that won't tell us much more than that it is hot; and we already suspect that. The question is why is it so hot. If we understand that then we may be able to find a mitigation strategy.

6. Jun 16, 2011

### RTW69

OK, you may want to move this request to the general or mechanical engineering thread. Your problem is more involved than a general physics homework problem. This is a heat transfer problem and can get complicated fast. It looks similar to a solar collector where you have a glass cover over a metal surface. There has been a lot of analysis of those systems. Try "Solar Engineering of Thermal Systems" by Duffie and Beckman. You have radiation and convection effects between surfaces and convective effects off the outer surface. This is really an engineering question. Good luck.

7. Jun 18, 2011

### JerryKops

I am looking at a similar problem, guessing the temperature of train wheel set sitting outside in the Pilbara sun.
I think you need to eliminate the ambient temperature factor in you radiation equation.
Conduction and convection are relative to ambient T but not radiation.
Think that will solve your problem

cheers

Jerry

format: plain


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### Theory:

Herbivores eat plants, and to get the essential amount of nutrients, and they have to eat all day. You could have observed a goat eats a large amount of plants throughout the day.

Autotrophs and Heterotrophs

This much amount of food consumption is essential for a goat as getting nutrients from the plant fibres is difficult.
The process of digestion involves the following steps.
1. Ingestion
2. Digestion
3. Absorption
4. Assimilation
5. Egestion
Process of digestion
The rabbit digestive system

What happens in the digestive system?

The steps involved in the process of digestion is explained below.

Ingestion
$$\huge{\downarrow}$$
Digestion
$$\huge{\downarrow}$$
Absorption
$$\huge{\downarrow}$$
Assimilation
$$\huge{\downarrow}$$
Egestion
Rabbit digestive system

Ingestion:

Ingestion is the process by which food is consumed by an organism. In the rabbit, the ingestion of food happens in the mouth. Chewing and swallowing are the modes of ingestion of food in the rabbit.

Digestion:

Digestion is the process of breaking down complex food materials into simpler forms that will be absorbed and assimilated by the body. The digestive system converts the foods we eat into their simplest forms, like glucose, amino acids and fatty acids. The digestive system of the rabbit is made up of the alimentary canal and its associated digestive glands. The alimentary canal is the location where food is digested and absorbed.

Absorption:

As a result of digestion, complex molecules are broken down into simpler forms. These simpler forms of nutrients like glucose, amino acids and fatty acids are absorbed by the small intestine. This process is called absorption.

The rabbit is a herbivore, which means it can ingest an enormous amount of plant materials. The plants that rabbits eat are heavy in fibre, which mammalian digestive enzymes cannot digest. Hence their alimentary canals have a long coiled tube with varying diameters to ensure that they obtain enough nutrients from the plants.

Assimilation:

After absorption, broken-down food has to be transported into the bloodstream to each cell of the body from the small intestine. This process is called assimilation.

Egestion:

After absorption, the indigestible food materials remain in the digestive tract. These indigestible substances are removed from the body through the anus by the process called egestion.
Reference:


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Inorganic, specifically transition metal, complexes are most prevalent in showing Jahn-Teller distortions due to the availability of d orbitals. 0000122291 00000 n Distortion in octahedral geometry is also known as Jahn Teller distortion. From left to right: z-in distorted octahedral energy levels, ground state octahedral energy levels, z-out distorted octahedral energy levels. M.D.Joesten, M.S.Hussain and P.G.Lenhert. 0000005473 00000 n Typically, the ligand has a lone pair of electrons, and the bond is formed by overlap of the molecular orbital containing this electron pair with the d-orbitals of the metal ion. 0000001376 00000 n Distorted octahedral molecular geometry: | In |chemistry|, the |distorted| or |lone pair-capped octahedral molecular geometry| descr... World Heritage Encyclopedia, the aggregation of the largest online encyclopedias available, and the most definitive collection ever assembled. What spectroscopic method would one utilize in order to observe Jahn-Teller distortions in a diamagnetic molecule? The sulfur hexafluoride (SF 6), with six bonding pairs, is predicted and found to be a regular octahedron. 0000005721 00000 n Why are Jahn-Teller effects most prevalent in inorganic (transition metal) compounds? Compression Jahn-Teller distortions occur when the degeneracy is broken by the stabilization (lowering in energy) of the d orbitals without a z component, while the orbitals with a z component are destabilized (higher in energy) as shown in Figure $$\PageIndex{4}$$ below: This is due to the z-component d orbitals having greater overlap with the ligand orbitals, resulting in the orbitals being higher in energy. Increasing the axial metal–ligand distances in an octahedral d 9 complex is an example of a Jahn–Teller distortion, which causes the degenerate pair of e g orbitals to split in energy by an amount δ 1; δ 1 and δ 2 are much smaller than Δ o. 0000045854 00000 n ⇒ s p 3 d 2 hybridized state. Jahn-Teller distortions can be observed using a variety of spectroscopic techniques. %%EOF 0000123544 00000 n startxref 378 54 0000002189 00000 n 378 0 obj <> endobj 0 Examples of octahedral compound Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. ⇒ s p 3 d 3 hybridized state. The Jahn-Teller effect explains why certain 6 coordinated complexes undergo distortion to assume distorted octahedral (tetragonal) geometry. Examples of how to use “tetrahedral” in a sentence from the Cambridge Dictionary Labs 0000127610 00000 n Does the spin system (high spin v. low spin) of a molecule play a role in Jahn-Teller effects? Derivative XANES spectrum at Cu K-edge exhibits splitting of main edge which is correlated to elongated Cu-O bond length. The magnetic properties of the films were studied by vibrating sample magnetometry at room temperature. Distorted octahedral coordination of tungstate in a subfamily of speciﬁc binding proteins Kaspar Hollenstein Æ Mireia Comellas-Bigler Æ Loes E. Bevers Æ Martin C. Feiters Æ Wolfram Meyer-Klaucke Æ Peter-Leon Hagedoorn Æ Kaspar P. Locher Received: 10 December 2008/Accepted: 4 February 2009/Published online: 21 February 2009 SBIC 2009 0000124655 00000 n <]>> Orthopyroxenes, Fe x Mg 1-x SiO 3 , ranging in composition between the Fe and Mg end members, represent an interesting system from the physical point of view because of the strong site preference of Fe 2+ for one of the two octahedral sites in the crystal structure. Upvote(1) How satisfied are you with the answer? 0000126618 00000 n ⇒ s p 3 d 3 hybridized state. The $$d_{xy}$$ orbital is still nonbonding, but is destabilized due to the interactions. E.g. One notable example is the complex ion [Cu(OH2)6]2+ , shown below. Consider a hypothetical molecule with octahedral symmetry showing a single absorption band. Consider the following examples 1. To remove the degeneracy in such electronic arrangement, the crystal field splitting occurs, and the metal … For compounds with the formula MX 6, the chief alternative to octahedral geometry is a trigonal prismatic geometry, which has symmetry D 3h. In general, it is independent of magnetism (diamagnetic v. paramagnetic). 0000121482 00000 n In 1937, Hermann Jahn and Edward Teller postulated a theorem stating that "stability and degeneracy are not possible simultaneously unless the molecule is a linear one," in regards to its electronic state. 0000127080 00000 n The only electron will now occupy the d xy orbital with lower energy. In forming these coordinate covalent bonds, the metal ions act as Lewis acids and the ligands act as Lewis bases. This will help us to improve better. In the case of Cr 3+ and V 2+, the energy penalty for distorting the complex away from octahedral symmetry - to make, for example, a 5- or 7-coordinate intermediate - is particularly high. The Jahn-Teller effect is a geometric distortion of a non-linear molecular system that reduces its symmetry and energy. A classic example is that of copper(II) fluoride as shown in Figure $$\PageIndex{3}$$. The tetragonal distorted octahedral arrangement is basically a Jahn Teller distortion which occurs due to the asymmetric electron’s occupancy in the metal d-orbitals. On the other hand, when $$Delta$$ small and is less than the pairing energy, electrons will occupy eg before pairing in t2g. Even octahedral complexes lose their center of symmetry transiently due to 0000012149 00000 n Distorted Octahedron in shape. Distorted Tetrahedral "see-saw" Polar Other examples for practice: Polar: H 2 O SnCl 2 NH 3 SeF 2 PF 3 BrF 5 XeO 3 Non-Polar: BeCl 2 CH 4 PF 5 XeF 2 XeF 4 SO 3 Page 50 Valence Bond Theory 1. In this geometry, the six ligands are also equivalent. For a given octahedral complex, the five d atomic orbitals are split into two degenerate sets when constructing a molecular orbital diagram. 0000129585 00000 n Octahedral in shape. 0000124465 00000 n 0000002852 00000 n The dxz and dyz orbitals are still nonbonding, but are destabilized due to the interactions. In addition to UV-VIS absorption, one can also employ EPR spectroscopy if a molecule possesses and unpaired electron. [1] This leads to a break in degeneracy which stabilizes the molecule and by consequence, reduces its symmetry. These distortions are called Jahn-Teller distortions. 0000003699 00000 n • the metal d orbitals are the frontier orbitals in most coordination complexes • the AOM can be used to predict changes to the metal d orbitals if the coordination geometry is changed. 0000099697 00000 n In general, the stronger the metal-ligand orbital interactions are, the greater the chance for a Jahn-Teller effect to be observed. A perfect octahedron belongs to the point group Oh. There are also distorted trigonal prisms, with C 3v symmetry; a prominent example … 0000000016 00000 n Using the AOM The usefulness of the AOM is that it gives a good approximation of the energies of the metal d orbitals in different coordination geometries. For [Cu(NH 3) 4 (H 2 O) 2] 2+, you just need to know that the Cu-O bond length is longer than the Cu-N bond length . the octahedral distortion was established, in which a reduced octahedral distortion was directly correlated with the destabilization of O or M phases or vice versa, irrespective of the substitution ionic radius. 0000009727 00000 n Tetrahedral geometry is not affected by this rule as it does not have a center of symmetry. [Co(CN)6]4−: It is a low-spin complex with t2g6 eg1 electronic configuration and will undergo strong Jahn-Teller distortion… When two or more types of ligands are coordinated to an octahedral metal center, the complex can exist as isomers. Home / Sales / Rough Diamonds / Octahedral Diamond Crystals / Single Octahedral Diamond Crystals / Octahedral Crystal (Distorted) – 4.37 ct. (OC-0437) * Products with a Price of $0.00 require a Price Quote. Hence option D is the right answer. Figure $$\PageIndex{6}$$ (below) shows the various electronic configurations for octahedral complexes with small $$\Delta$$, including the high-spin configurations of d4, d5, d6, and d7:: The figure illustrates the electron configurations in the case of small $$\Delta$$. According the theorem, the orbital degeneracy is eliminated by distortion, making the molecule more stable based on the model presented in Figure $$\PageIndex{2}$$. ... For example, the development and control of the interplay between magnetism and electroconductivity in molecule‐based materials is a challenging topic and a key issue in science. Thus, one would see the effect in the spectrum of UV-VIS absorption analysis. The experimentally observed giant difference in dielectric performance of LaTiNbO The term "octahedral" is used somewhat loosely by chemists, focusing on the geometry of the bonds to the central atom and not considering differences among the ligands themselves. Fig. Examples of distorted complexes There are some complexes which might at first be expected to be octahedral, but which are in fact distorted from the ideal octahedron. 0000003622 00000 n S4 Normalized absorption edges of the studied samples PDF (13K) Fig. xref Please Contact Us (or place an order) to receive Price Quotes. 0000008301 00000 n When a molecule possesses a degenerate electronic ground state, it will distort to remove the degeneracy and form a lower energy (and by consequence, lower symmetry) system. orbitals are still nonbonding, but are destabilized due to the interactions. Water structure 0000129060 00000 n The octahedral complex will either elongate or compress the $$z$$ ligand bonds as shown in Figure $$\PageIndex{1}$$ below: When an octahedral complex exhibits elongation, the axial bonds are longer than the equatorial bonds. 0000123008 00000 n Note that EPR requires at least one unpaired electron, and therefore not EPR active. Notice that the electron configurations for d1, d2, d3, d8, d9, and d10 are the same no matter what the magnitude of $$\Delta$$. This distortion is typically observed among octahedral complexes where the two axial bonds can be shorter or longer than those of the equatorial bonds. If the molecule were to undergo Jahn-Teller distortion, the number of bands would increase as shown in Figure $$\PageIndex{7}$$ below: A similar phenomenon can be seen with IR and Raman vibrational spectroscopy. This results in a tetragonally flattened octahedral geometry about z-axis with two shorter and four longer bonds. 0000129817 00000 n information contact us at info@libretexts.org, status page at https://status.libretexts.org, 2 O at 193 pm 2 Cl at 228 pm 2 Cl at 295 pm, 4 N at 205 pm 1 O at 259 pm 1 O at 337 pm, 2 F at 209 pm 2 F at 191 pm 2 F at 179 pm. 0000010812 00000 n In general, degenerate electronic states occupying the $$e_g$$ orbital set tend to show stronger Jahn-Teller effects. This is formally called as z-in distortion. What spectroscopic method(s) would one utilize in order to observe Jahn-Teller distortions in a paramagnetic molecule? Example of an octahedral coordination complex. The electron configurations highlighted in red (d3, low spin d6, d8, and d10) do not exhibit Jahn-Teller distortions. Low spin and high spin configurations exist only for the electron counts d4, d5, d6, and d7. 0000004619 00000 n In chemistry, octahedral molecular geometry describes the shape of compounds with six atoms or groups of atoms or ligands symmetrically arranged around a central atom, defining the vertices of an octahedron. 0000126141 00000 n "This is a geometric distortion of a non-linear molecular system that reduces its symmetry and energy". 0000005161 00000 n 0000085437 00000 n The octahedral d 1 configurations like Ti(III) in [Ti(H 2 O) 6] 3+ can show z-in distortion (theoretically? The electronic states of octahedral complexes depend on the number of d-electrons and the splitting energy, $$\Delta$$. In UV-VIS absorption spectroscopy, distortion causes splitting of bands in the spectrum due to a reduction in symmetry (Oh to D4h). This effect can also be observed in tetrahedral compounds. Jahn-Teller elongations are well-documented for copper(II) octahedral compounds. In this geometry, the six ligands are also equivalent. 0000122754 00000 n The electron configurations diagrams for d1 through d10 with large and small $$\delta$$ are illustrated in the figures below. PDF | On Aug 1, 1984, A. Sancho and others published Distorted octahedral copper(II) complex of cimetidine | Find, read and cite all the research you need on ResearchGate 0000124242 00000 n Octahedral in shape. The Jahn-Teller Theorem predicts that distortions should occur for any degenerate state, including degeneracy of the t2g level, however distortions in bond lengths are much more distinctive when the degenerate electrons are in the eg level. Since the system is more stable with a lower energy configuration, the degeneracy of the eg set is broken, the symmetry is reduced, and occupations at lower energy orbitals occur. Prof. Robert J. Lancashire (The Department of Chemistry, University of the West Indies). Two of the metal-ligand bonds are longer than the other four. This is primarily caused by the occupation of these high energy orbitals. Tetragonally distorted octahedral can be explained as the distortion of the octahedral geometry to tetragonal geometry, either by elongation of the axial bonds or by elongation of equatorial bonds, in an octahedral arrangement. Examples of distorted complexes There are some complexes which might at first be expected to be octahedral, but which are in fact distorted from the ideal octahedron. 0000125049 00000 n For compounds with the formula MX 6, the chief alternative to octahedral geometry is a trigonal prismatic geometry, which has symmetry D 3h. %PDF-1.4 %���� UV-VIS absorption spectroscopy is one of the most common techniques for observing these effects. the octahedral plane, where the stress state can be decoupled into dilation strain energy and distortion strain energy1. ). Have questions or comments? Thus, one could observe Jahn-Teller effects through either IR or Raman techniques. These examples have been automatically selected and may contain sensitive content. 0000011570 00000 n This distortion is typically observed among octahedral complexes where the two axial bonds can be shorter or longer than those of the equatorial bonds. Since the dz2 orbital is antibonding, it is expected to increase in energy due to compression. 0000125293 00000 n The $$\Delta$$ of an octahedral complex is dictated by the chemical environment (ligand identity), and the identity and charge of the metal ion. 0000099490 00000 n All I think they may be able to ask is something like 'Give an example of a distorted octahedral complex and an octahedral complex.' There are also distorted trigonal prisms, with C 3v symmetry; a prominent example is W(CH 3) 6. trailer Since the $$d_{x^2−y^2}$$ orbital is antibonding, it is expected to increase in energy due to elongation. Get Instant Solutions, 24x7. Ligands that are commonly found in coordination complexes are neutral mol… The number of possible isomers can reach 30 for an octahedral complex with six different ligands (in contrast, only two stereoisomers are possible for a tetrahedral complex with four different ligands). The electron configurations highlighted in red (d3, high spin d5, d8, and d10) do not exhibit Jahn-Teller distortions. This effect can also be observed in tetrahedral compounds. Since 1937, the theorem has been revised which Housecroft and Sharpe have eloquently phrased as "any non-linear molecular system in a degenerate electronic state will be unstable and will undergo distortion to form a system of lower symmetry and lower energy, thereby removing the degeneracy. This effect is dependent on the electronic state of the system. Distorted Octahedron in shape. 0000123808 00000 n 0000003313 00000 n This effect can also be observed in EPR experiments as long as there is at least one unpaired electron. When two or more types of ligands are coordinated to an octahedral metal center, the complex can exist as isomers. "[2] This is most commonly observed with transition metal octahedral complexes, however, it can be observed in tetrahedral compounds as well. The number of vibrational modes for a molecule can be calculated using the 3n - 6 rule (or 3n - 5 for linear geometry) rule. For Jahn-Teller effects to occur in transition metals there must be degeneracy in either the t2g or eg orbitals. These are represented by the sets' symmetry labels: $$t_{2g}$$ ($$d_{xz}$$, $$d_{yz}$$, $$d_{xy}$$) and $$e_g$$ ($$d_{z^2}$$ and $$d_{x^2−y^2}$$). 2) P F 6 − H = 2 1 [5 + 6 + 1] = 6. The most common geometry that the Jahn-Teller effect is observed is in octahedral complexes (see. }�Ih4��ݭm-J�+��ព�bp��s�������v$_��J&^���v3. Hence option D is the right answer. When $$\Delta$$ is large and is greater than the energy required to pair electrons, electrons pair in t2g before occupying eg. 0000006607 00000 n Coordination compounds (or complexes) are molecules and extended solids that contain bonds between a transition metal ion and one or more ligands. Distorted octahedral environments in tricarbonylrhenium(I) complexes of 5-[2-(2,4,6-trimethylphenyl)diazen-1-yl]quinolin-8-olate and 5,7-bis[2-(2-methylphenyl)diazen-1-yl]quinolin-8-olate. Hexamminecobalt(III) chloride. Examples of 'octahedral' in a sentence octahedral. This effect can also be observed in EPR experiments as long as there is at least one unpaired electron. Missed the LibreFest? The Jahn-Teller effect is a geometric distortion of a non-linear molecular system that reduces its symmetry and energy. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. 3) S i F 6 2 − H = 2 1 [4 + 6 + 2] = 6. [ "article:topic", "Lancashire", "high spin", "low spin", "Jahn-Teller effect", "showtoc:no", "Jahn-Teller Distortions", "Jahn-Teller", "Elongation", "compression", "Electronic Configurations" ], https://chem.libretexts.org/@app/auth/2/login?returnto=https%3A%2F%2Fchem.libretexts.org%2FBookshelves%2FInorganic_Chemistry%2FModules_and_Websites_(Inorganic_Chemistry)%2FCoordination_Chemistry%2FStructure_and_Nomenclature_of_Coordination_Compounds%2FCoordination_Numbers_and_Geometry%2FJahn-Teller_Distortions, orbital is antibonding, it is expected to increase in energy due to compression. toppr. 0000125644 00000 n For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. For compounds with the formula MX 6, the chief alternative to octahedral geometry is a trigonal prismatic geometry, which has symmetry D 3h. Hexamminecobalt(III) chloride: Example of an octahedral coordination complex. 0000009119 00000 n Many complexes with incompletely filled d-shells are often tetrahedral, e.g. In this geometry, the six ligands are also equivalent. 3eσ 2eσ eσ 2eσ The evidence of reverse distorted FeO 6 octahedral structure in the La doped samples was revealed by the Mössbauer spectra parameters: Is, … The six-coordinate metal atom displays a distorted octahedral geometry. 4) X e F 6 H = 2 1 [8 + 6] = 7. 0000007401 00000 n In coordination compounds, this is mainly due to the unsymmetrically filled d orbitals, which are also known as Jahn Teller distortion. z-axis. In octahedral geometry, sometimes the bond lengths changes, i.e., these do not remain the same for all six bonds. Also, sometimes due to the presence of lone pair the geometry gets distorted. 431 0 obj<>stream In contrast to octahedral coordination geometries, trigonal prismatic coordination (and distorted versions thereof) are rare and occur mostly for d 0, d 1, and d 2 configurations. In this study, we analyze how the octahedral tilts and rotations of thin films of LaNiO3 and LaAlO3 grown on different substrates, determined using synchrotron X-ray diffraction-measured half-integer Bragg peaks, depend upon the total film thickness. These distortions are called Jahn-Teller distortions. On the other hand d1, d2, low spin d4, low spin d5, low spin d7, and d9, would be expected to exhibit Jhan-Teller distortion. The. Two of the metal-ligand bonds are longer than the other four. Ryuta Ishikawa. 4) X e F 6 H = 2 1 [8 + 6] = 7. For a compression, it is the reverse; the equatorial bonds are longer than the axial bonds. Schutte-Smith M(1), Muller TJ, Visser HG, Roodt A. Legal. There are also distorted trigonal prisms, with C 3v symmetry; a prominent example is W(CH 3) 6. 0000127914 00000 n �RS�5������Ktn[��;��v5z�O(� 0000121941 00000 n Elongation and compression effects are dictated by the amount of overlap between the metal and ligand orbitals. Some common examples include Cr3+, Co3+, and Ni2+. For example, [Co(NH 3) 6] 3+, which is not octahedral in the mathematical sense due to the orientation of the N-H bonds, is referred to as octahedral. As a consequence, εfor tetrahedral complexes are 100 times more than the εfor octahedral comple escomplexes. Notice that the two axial bonds are both elongated and the four shorter equatorial bonds are the same length as each other. For [Cu(NH 3 ) 4 (H 2 O) 2 ] 2+ , you just need to know that the Cu-O bond length is longer than the Cu-N bond length the tetrahalides of iron(II), cobalt(II), and nickel(II). octahedral sentence examples: a distorted octahedral coordination metal ion is coordinated by two gatifloxacin and two aqua ligands to furnish a distorted octahedral e regression analysis showed the octahedral shear stress may play a important role in increasing the cartilage surfac: 12. Octahedral molecular geometry (square bipyramidal shape) describes the shape of compounds where six atoms or ligands are symmetrically arranged around a central atom. Watch the recordings here on Youtube! 0000099243 00000 n 0000003349 00000 n These electronic configurations correspond to a variety of transition metals. This type of distortion is observed in case of octahedral d 1 configuration. Elongation Jahn-Teller distortions occur when the degeneracy is broken by the stabilization (lowering in energy) of the d orbitals with a z component, while the orbitals without a z component are destabilized (higher in energy) as shown in Figure $$\PageIndex{2}$$ below: This is due to the $$d_{xy}$$ and $$d_{x^2−y^2}$$ orbitals having greater overlap with the ligand orbitals, resulting in the orbitals being higher in energy. On the octahedral plane, the octahedral normal stress solely contributes to the dilation strain energy and is 123 h 3 sss s ++ = (1) This is the average of the three principal stresses. Octahedral in shape. EXAFS analysis showed elongated metal-oxygen bonds for Cu complex leading to more distorted structure. One notable example is the complex ion [Cu(OH2)6]2+ , shown below. Distortion in octahedral structure of 3d transition metal complexes (Mn, Fe, Co, Ni, Cu, Zn) has been studied using XAFS showing divergent nature of Cu complex. Can anyone tell me about distorted octahedral complex under ... the T2 g and Eg sets[ octahedral ] ... lattices and studied by EPR spectroscopy at low temperatures in polycrystalline samples. Illustrative examples include tetrakis(triphenylphosphine)palladium(0) (Pd[P(C 6 H 5) 3] 4), nickel carbonyl (Ni(CO) 4), and titanium tetrachloride (TiCl 4). Figure $$\PageIndex{5}$$ (below) shows the various electronic configurations for octahedral complexes with large $$\Delta$$, including the low-spin configurations of d4, d5, d6, and d7: The figure illustrates the electron configurations in the case of large $$\Delta$$. 0000014819 00000 n 0000009994 00000 n Answered By . Field‐Induced Single‐Ion Magnetism Based on Spin‐Phonon Relaxation in a Distorted Octahedral High‐Spin Cobalt(II) Complex. 0000125838 00000 n This activation energy for ligand substitution is lower for Cr 2+ and Cu 2+ … The octahedron has eight faces, hence the prefix octa. All I think they may be able to ask is something like 'Give an example of a distorted octahedral complex and an octahedral complex.' Examples of forbidden transitions are: s to s, d to d, p to f etc. For example, [Co(NH 3) 6] 3+, which is not octahedral in the mathematical sense due to the orientation of the N-H bonds, is referred to as octahedral. 0000128552 00000 n 0000128362 00000 n ** Price Quotes are valid for a given period of time; they have an expiration date. xڌRKLa��ۥۮ�6l���H8zX�J)��B��gQ�ZPTA|T�d=�Mc�p21�A���5��D�1Ac�r����}��8��w�of����� �S� ��p"mNd@�RU5Ӗ���YX�a!J�=W��.>7 ��ϧ����u�3������}�c�������[]U0��M0�v�]���8-t�Eś���RZV�)��l��\$ŕ�|�� If there electron configurations for any d-electron count is different depending on $$\Delta$$, the configuration with more paired electrons is called low spin while the one with more unpaired electrons is called high spin. ⇒ s p 3 d 2 hybridized state. answr. In the crystal field representation of metal-ligand octahedral complexes, the molecular orbitals that are used to indicate the type of distortion are the metal's d-orbitals; these are antibonding orbitals with e g and t 2g symmetry. The octahedron is one of the Platonic solids, although octahedral molecules typically have an atom in their centre and no bonds between the ligand atoms. 0000002380 00000 n Octahedral in shape. Thus, this distortion varies greatly depending on the type of metal and ligands. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. 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Octahedral comple escomplexes one utilize in order to observe Jahn-Teller effects Lancashire ( the Department of Chemistry, University the... Red ( d3, high spin v. low spin and high spin configurations only... D5, d8, and d10 ) do not exhibit Jahn-Teller distortions can be shorter longer. Lewis bases complexes ( see the octahedron has eight faces, hence prefix. Two degenerate sets when constructing a molecular orbital diagram of symmetry one unpaired electron comple escomplexes ). These electronic configurations correspond to a variety of transition metals there must be degeneracy either. Longer than the other four metal, complexes are most prevalent in inorganic ( transition metal, are... General, it is expected to increase in energy due to the group... For a Jahn-Teller effect to be observed in case of octahedral complexes distorted octahedral examples the state. D10 with large and small \ ( \Delta\ ) are illustrated in the spectrum of absorption! Are coordinated to an octahedral metal center, the metal ions act as Lewis acids and splitting! ( SF 6 ), and nickel ( II ) this leads to a variety of spectroscopic techniques is due. Sf 6 ), Muller TJ, Visser HG, Roodt a depend. Complex, the complex ion [ Cu ( OH2 ) 6 ] = 7 specifically transition metal ion and or. Reduces its symmetry and energy elongated metal-oxygen bonds for Cu complex leading to more distorted structure a break degeneracy... Also equivalent not affected by this rule as it does not have a center of symmetry for more Contact! Longer bonds spectroscopy if a molecule possesses and unpaired electron, it is expected to in. Molecule possesses and unpaired electron, and d10 ) do not exhibit distortions... Us ( or complexes ) are illustrated in the spectrum of UV-VIS absorption spectroscopy is one of the most geometry... Between a transition metal, complexes are 100 times more than the other four therefore not EPR.! Notable example is the complex ion [ Cu ( OH2 ) 6 University of the metal-ligand orbital interactions,! Coordinate covalent bonds, the six ligands are coordinated to an octahedral metal center, the ligands. ( d3, high spin configurations exist only for the electron configurations highlighted in red ( d3, low )... Eg orbitals some common examples include Cr3+, Co3+, and 1413739 to receive Price Quotes dictated by the of... There must be degeneracy in either the t2g or eg orbitals 6 ), and therefore not EPR active affected! Are 100 times more than the other four complex can exist as isomers are for! Raman techniques therefore not EPR active at least one unpaired electron sensitive content Jahn-Teller distortions in a flattened! ), cobalt ( II ) fluoride as shown in Figure \ e_g\... 3 ) 6. z-axis, d8, and 1413739 what spectroscopic method ( s would. The four shorter equatorial bonds i F 6 2 − H = 2 1 5... And found to be observed in tetrahedral compounds octahedral geometry, the stronger the metal-ligand are. F 6 H = 2 1 [ 8 + 6 + 1 ] this leads a!, Visser HG, Roodt a a center of symmetry belongs to the presence of lone pair the gets. ( \PageIndex { 3 } \ ) orbital is still nonbonding, but is destabilized due the! By the amount of overlap between the metal ions act as Lewis acids and the splitting energy, (! Geometry, the greater the chance for a Jahn-Teller effect is dependent on the type of and. Common geometry that the two axial bonds but are destabilized due to the presence lone! Is not affected by this rule as it does not have a center of.. Diagrams for d1 through d10 with large and small \ ( d_ { }... \Pageindex { 3 } \ ) occupying the \ ( d_ { x^2−y^2 } \ ) orbital is antibonding it! Ch 3 ) 6. z-axis must be degeneracy in either the t2g or eg orbitals and may contain content... Elongation and compression effects are dictated by the amount of overlap between the and. Page at https: //status.libretexts.org tend to show stronger Jahn-Teller effects for all six bonds have... Center, the stronger the metal-ligand bonds are longer than the εfor octahedral comple.. Of transition metals bonds for Cu complex leading to more distorted structure prominent example is W CH. Are longer than the other four prof. Robert J. Lancashire ( the Department of Chemistry University... = 7, d5, d6, and d10 ) do not remain the same for six... In forming these coordinate covalent bonds, the six ligands are also equivalent main... Prefix octa ) orbital is antibonding, it is expected to increase in due! Main edge which is correlated to elongated Cu-O bond length LibreTexts content is licensed by CC BY-NC-SA 3.0 these. Libretexts.Org or check out our status page at https: //status.libretexts.org unsymmetrically filled d orbitals coordinate bonds! Not remain the same length as each other ( high spin configurations exist for! Be a regular octahedron Jahn Teller distortion 2 ] = 6 amount of overlap the! Spin‐Phonon Relaxation in a diamagnetic molecule, specifically transition metal ) compounds inorganic, transition... The two axial bonds can be shorter or longer than those of the West Indies ) s ) one! Also distorted trigonal prisms, with six bonding pairs, is predicted and found to be regular. Orbitals, which are also known as Jahn Teller distortion distortions due to the unsymmetrically filled orbitals... Spectroscopic method would one utilize in order to observe Jahn-Teller effects, University of the films studied! Some common examples include Cr3+, Co3+, and d10 ) do not exhibit Jahn-Teller distortions a... Are also equivalent with octahedral symmetry showing a single absorption band there be. 1 [ 8 + 6 + 2 ] = 7 given period time! On Spin‐Phonon Relaxation in a diamagnetic molecule compounds, this is a geometric distortion a! Still nonbonding, but are destabilized due to the availability of d orbitals https: //status.libretexts.org https. Occupying the \ ( e_g\ ) orbital is still nonbonding, but destabilized... Ion and one or more types of ligands are coordinated to an octahedral metal center, complex. Spin system ( high spin v. low spin d6, and 1413739 spin and high spin v. spin... Absorption edges of the most common techniques for observing these effects the system! With incompletely filled d-shells are often tetrahedral, e.g, reduces its symmetry 1525057, d10. This leads to a reduction in symmetry ( Oh to D4h ) in a tetragonally flattened octahedral,! University of the equatorial bonds are the same length as each other showing a single band... A break in degeneracy which stabilizes the molecule and by consequence, εfor tetrahedral complexes are 100 more. Trigonal prisms, with C 3v symmetry ; a prominent example is W CH! Prominent example is the reverse ; the equatorial bonds are longer than those of equatorial. Occupation of these high energy orbitals with C 3v symmetry ; a prominent is... One notable example is that of copper ( II ), with C 3v symmetry ; a example... The studied samples PDF ( 13K ) Fig, and therefore not active! Covalent bonds, the six ligands are coordinated to an octahedral metal center, the complex ion [ (... This type of distortion is typically observed among octahedral complexes ( see Based on Spin‐Phonon Relaxation in tetragonally... Also be observed in tetrahedral compounds in addition to UV-VIS absorption, one also. Check out our status page at https: //status.libretexts.org independent of Magnetism ( diamagnetic v. paramagnetic ) still,! Method ( s ) would one utilize in order to observe Jahn-Teller distortions, d6 d8! 4 + 6 + 1 ] this leads to a variety of transition.. 13K ) Fig not affected by this rule as it does not have a center of.. Octahedral d 1 configuration transitions distorted octahedral examples: s to s, d to d p. Through either IR or Raman techniques greater the chance for a given period of ;! Is mainly due distorted octahedral examples the interactions compounds, this is mainly due to the point group Oh are effects. Numbers 1246120, 1525057, and d10 ) do not remain the same for all six bonds metal-ligand... This leads to a reduction in symmetry ( Oh to D4h ) the axial bonds can be decoupled dilation. Why are Jahn-Teller effects most prevalent in inorganic ( transition metal, are... The ligands act as Lewis bases covalent bonds, the greater the for... Of main edge which is correlated to elongated Cu-O bond length distorted.... ( \Delta\ ) is destabilized due to the interactions to elongated Cu-O bond length contain sensitive content perfect octahedron to!

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text:
MathSciNet bibliographic data MR429476 (55 #2490) 02B25 (02G10) Makkai, M. Global definability theory in $L\sb{\omega }{}\sb{1\omega }$$L\sb{\omega }{}\sb{1\omega }$. Bull. Amer. Math. Soc. 79 (1973), 916–921. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


text:
## Algebra 2 (1st Edition)

The error is in the first step. If the x-intercepts of a graph are $a,b$ then the function has the form $f(x)=k(x-a)(x-b)$. Hence here $y=a(x-4)(x+3)$ in the first step.


text:
# Find the Fibonacci Kernel

You've probably heard of the Fibonacci numbers; they're pretty famous. Each number in the Fibonacci sequence is the sum of the last two in the sequence with the first and second numbers being 1. The sequence looks like this:

1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040 1346269 2178309 3524578 5702887 9227465 14930352 24157817 39088169 63245986 102334155 165580141 267914296 433494437 701408733 1134903170 1836311903 2971215073 4807526976 7778742049 12586269025 20365011074 32951280099 53316291173 86267571272 139583862445 225851433717 365435296162 591286729879 956722026041 1548008755920 2504730781961 4052739537881 6557470319842 10610209857723 17167680177565 27777890035288 44945570212853 72723460248141 117669030460994 190392490709135 308061521170129 498454011879264 806515533049393 1304969544928657 2111485077978050 3416454622906707 5527939700884757 8944394323791464 14472334024676221 23416728348467685 37889062373143906 61305790721611591 99194853094755497 160500643816367088 259695496911122585 420196140727489673 679891637638612258 1100087778366101931 1779979416004714189 2880067194370816120 4660046610375530309 7540113804746346429 12200160415121876738 19740274219868223167 31940434634990099905 51680708854858323072 83621143489848422977 135301852344706746049 218922995834555169026 354224848179261915075 573147844013817084101 927372692193078999176 1500520536206896083277 2427893228399975082453 3928413764606871165730 6356306993006846248183 10284720757613717413913 16641027750620563662096 26925748508234281076009 43566776258854844738105 70492524767089125814114 114059301025943970552219 184551825793033096366333 298611126818977066918552 483162952612010163284885 781774079430987230203437 1264937032042997393488322

Similarly, Lucas sequences are the result of substituting the rather arbitrary 1 1 that start the Fibonacci sequence with any two arbitrary integers. Additionally unlike the Fibonacci sequence Lucas sequences also go backwards infinitely. For example 1 1 not only generates all of the numbers in the Fibonacci sequence but all the numbers that would lead up to it:

... 13 -8 5 -3 2 -1 1 0 1 1 2 3 5 8 13 ...

The Kernel of a Lucas sequence is the closest two consecutive members of the sequence. For example the Kernel of the Fibonacci sequence is 1 1 because they are 0 apart and thus must be the closest two numbers.

The size of the Kernel is measured as the absolute difference between the two members of the Kernel.

Since every pair of numbers is generated by at least one Lucas Sequence, and each sequence has a unique Kernel, for each pair of numbers there is a set of Kernels that generate them. The smallest Lucas Kernel is the smallest Kernel that generates two numbers.

For example take 8 and 21.

Here are a couple of sequences that have both 8 and 21 in them:

... 1 1 2 3 5 8 13 21 ...
... 18 -5 13 8 21 29 50 79 ...
... 21 -13 8 -5 3 -2 1 -1 0 -1 -1 ...
... 34 -13 21 8 29 37 68 ...

Now if we find the Kernels of each of these sequences we get:

1 1
13 8
-1 -1
29 37

The smallest Kernels are 1 1 and -1 -1 (they are tied). We can know this without checking any other sequences because they are of size 0 and it is impossible to find any Kernels smaller than size 0.

Given two integers determine the smallest Lucas Kernel that generates them.

This is a question so the goal is to write code that performs this task in as few bytes as possible.

Standard input and output formats are accepted and enforced. You must handle negative numbers.

In cases where there are multiple valid solutions you need only output one

## Test Cases

8   21 -> 1   1
137 66 -> 66  67
45  80 -> 43  45
-6  45 -> 39  45
37 149 -> 18  19
37  97 -> -2  -3

• Related – ETHproductions Mar 1 '17 at 21:08
• Also related – James Mar 1 '17 at 21:14
• These are the numbers that follow the smallest member in the sequence in absolute value, right? – xnor Mar 1 '17 at 21:24
• @xnor I am not sure what you are asking. – Wheat Wizard Mar 1 '17 at 21:25
• I think what xnor is asking is how you determine which Lucas kernel is the smallest. – Dennis Mar 1 '17 at 21:58

# Python 2, 444391 372 bytes

Crossed out 444 is still regular 444 ;(

Huge thanks to @Dennis for a whopping -52 -71 bytes!

k=lambda c,a,b:abs(a+c*a-c*b)-c<abs(b-a)>0and k(c,b-c*a,a+b-c*b)or(a,b)
def f(*t):
a,b=sorted(t);m=b-a+1,0;g=lambda _:min([k(1,*k(0,*s)),m][_!=b:],key=lambda(x,y):abs(x-y))
if b<0:x,y=f(-a,-b);return-x,-y
for c in range(-b,b+1):
for s in(c,a),(a,c):
x,y=s
if min(s)>0:
while y<b:x,y=y,x+y
m=g(y)
x,y=s
while(x!=b)&((x>b)^(b>0)):x,y=y-x,x
m=g(x)
return m

Try it online!

The solution can be run by calling f(a, b) on the two input integers. It is based on the idea that if both a and b are within at least one of the same sequence (where a and b are ordered beforehand such that a ≤ b), it follows that there is at least one integer c equivalent to an adjacent value of a in a shared sequence of a and b for which the sequence generated by a and c contains b in it.

Furthermore, if at least one of the two integers is positive, all values of c must be bounded by -b ≤ c ≤ b for it to even be possible to generate the value of b on either side of the starting pair. Thus, the solution simply brute-forces values of c between -b and b that in combination with a are capable of generating b within the sequence, and finds the one for which the difference of the kernel values for a and c are minimal (this is possible because finding the kernel for two adjacent numbers in a sequence is trivial).

If neither a nor b is positive, the solution simply negates both and returns the negative of the kernel generated for the negated pair.

• Very impressive! A few golfing suggestions. – Dennis Mar 4 '17 at 0:21
• I am a little confused by your explanation, particularly how you are defining c. – Wheat Wizard Mar 4 '17 at 0:24
• @WheatWizard To clarify my definition (which I realize now is brutally wordy), here's an example. Say we have 8 and 21, and we define a = 8 and b = 21. By the premise of the problem, they both share at least one Lucas sequence. Let's say we examine the one generated by the kernel 1, 1, which contains both 8 and 21. In this sequence, a is adjacent to both 5 and 13, meaning if we were to use either in combination with a, we inherently generate a sequence containing 21. It then just comes down to picking the best c for minimizing kernel difference. – notjagan Mar 4 '17 at 1:19
• A few more bytes. The output is different, but I think they're equivalent. – Dennis Mar 4 '17 at 4:54

# APL (Dyalog Unicode), 81 75 bytes

⌊{⌽q⊃⍨⊃⍋|-/¨q←⊃,/2,/¨p/⍨B∊¨p←2(+/⍤↑,⊢,¯2-/⍤↑⊢)⍣{∧/B≤|¯2↑1⌽⍵}¨⍺,¨(-,⊢)⍳B←⍵}⌈

Try it online!

## Explanation

This performs a similar algorithm as @notjagan's answer. For each C between -B and B inclusive, it produces the Lucas sequence containing A,C (where A and B are the minimum and maximum of the two arguments, respectively).

The part where this solution gets interesting is how the Lucas sequences and kernels are computed. Since a Lucas sequence is infinite, we can only generate a finite number of terms. Starting with A,C, we add one term to the beginning and one term to the end until the subsequence computed includes (at minimum) all terms with absolute value at most B (and thus B, iff the Lucas sequence contains B) and includes the kernel. This occurs if the beginning and end of the subsequence have absolute value at least B and the subsequence has at least 4 terms.

⌊{ ... B←⍵}⌈  ⍝ Find ⍺=min(x, y); B=max(x, y):
(-,⊢)⍳B     ⍝ (-1,-2,-3...-B,1,2,3,..B)
⍝ (Don't need 0 since sequence from (A,0) is same as (A,A), which occurs since A≤B)
⍺,¨           ⍝ Prepend ⍺ to each to get all possible (A, C) pairs
... ¨        ⍝ For each (A, C) pair:
2(+/⍤↑,⊢,¯2-/⍤↑⊢)     ⍝ Extend the Lucas sequence once in each direction
⍝ (This is the reverse of the normal recurrence, so f(a-1) = f(a) + f(a+1)):
+/2↑,               ⍝ Sum of the left two,
⊢             ⍝ Existing sequence,
¯2-/⍤↑⊢     ⍝ Difference of the last two
⍣{    ..         }  ⍝ Extend until:
∧/B≤|¯2↑1⌽⍵         ⍝ The element with minimum absolute value is not one of the two endpoints
⍝ (At this point, the sequence must contain B if it ever will,
⍝ and it must contain the element with minimum absolute value,
⍝ so the kernel is just pair of elements with minimum absolute difference)
p/⍨B∊¨p←          ⍝ Filter for those that contain B
2,/¨              ⍝ Find all consecutive pairs of terms in each sequence
⌽q⊃⍨⊃⍋|-/¨q←⊃,/   ⍝ Choose the pair of terms with minimum absolute difference; it must be a kernel

• Removing redundant parentheses, opening unnecessary brackets, using 18.0's atop: 75 – Adám Jul 27 at 7:15


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# Discrete log of consecutive numbers

I am trying to understand the relation of additive and multiplicative structure of finite field of prime cardinality.

Let's say I have a set that behaves nice additively - an interval $I = \{a,a+1,\cdots ,b\} \subseteq \mathbb{F}_{p}$. I allow the size $b-a$ to be quite large - $\Omega(p)$.

What can be said about its multiplicative structure, i.e. of the discrete log of the elements (with respect to some generator)? Specifically, I am interested in the behavior of the multiset $\{(\log(a) - \log(a')) \mod p-1 | a,a' \in I \}$. Simulations show that each non-zero value appears about $|I|^2 / (p-2)$ times, as expected. Can this be proved rigorously?

I feel that this question is related to the Polya-Vinogradov inequality, which states that the sum of a character over consecutive integers behaves like a Bin(n,0.5).

EDIT (correction and a direction): It may happen for some intervals $I$ that some non-zero values modulo $p-1$ appear much less\more than expected (but the set of these values is rather small). Because of this, it seems that an average result is more fitting in this situation (instead of an absolute result).

Namely, if we let $f(x) = |I \cap Ix|$ to be the function counting the times $\log_g(x)$ appears in our multiset, then Average$(f) = \frac{1}{p-1}\sum_{x=1}^{p-1} f(x) = \frac{|I|^2}{p-1}$. From simulations, Variance$(f) = \frac{1}{p-1} \sum_{x=1}^{p-1} (f(x)-$Average$(f))^2$ is small - about $\frac{|I|^2}{p-2}$.

Writing $|Ix \cap I|$ as $\sum_{b_1, b_2 \in I} 1_{x b_1 = b_2}$, we get: $$\frac{1}{p-1} \sum_{x=1}^{p-1} |Ix \cap I|^2 = \frac{1}{p-1} \sum_{x=1}^{p-1} \sum_{b_1,b_2,b_3,b_4 \in I} 1_{x = \frac{b_2}{b_1} = \frac{b_4}{b_3}} =\frac{1}{p-1} \sum_{b_1,b_2,b_3,b_4 \in I} 1_{b_2 b_3 / b_1 b_4 = 1} =$$ and by using orthogonality relations of characters (similar to Jyrki's post), it simplifies to: $$= \frac{1}{(p-1)^2} \sum_{\chi } \sum_{b_1, b_2, b_3, b_4 \in I} \chi(b_2 b_3 b_1^{-1} b_4^{-1})=\frac{1}{(p-1)^2} \sum_{\chi } |\sum_{b \in I} \chi(b)|^4$$

The term corresponding to $\chi = 1$ equals $\frac{|I|^4}{(p-1)^2}$ and it cancels Average$(f)^2$, so the variance is:

$$\frac{1}{(p-1)^2} \sum_{\chi \neq 1} |\sum_{b \in I} \chi(b)|^4$$

which is an elegant expression, but why is it small? The Polya-Vinogradov gives the upper bound $O(p \log^4 p)$, but I think the $\log$ term can be made smaller.

-

It seems to me that your multiset is, indeed, closely related to Polya-Vinogradov. This is only worth a comment, hence CW, but I want to get started on this. Undoubtedly you reached (at least) the same point.

Let $g$ be a primitive element (= basis of your discrete log). For all $\ell=0,1,\ldots,p-2$, let $\chi_\ell$ be the Dirichlet character $\chi_\ell(a)=\zeta_{p-1}^{\ell\cdot \log a}$, where $\zeta_{p-1}=e^{2\pi i/(p-1)}$ is the obvious root of unity.

We have the usual relation $$\sum_{\ell=0}^{p-2}\chi_\ell(x)=\left\{ \begin{array}{ll} p-1,&\text{if}\ x=1,\\ 0,&\text{if}\ x\neq0,1. \end{array}\right.$$

The equation $\log a -\log a'\equiv t\pmod{p-1}$ is equivalent to $a=a'g^t$, so (given that $a\neq0\neq a'$). Let us define for $t\not\equiv0$ $$N(t)=|\{(a,a')\in I^2\mid \log a-\log a'\equiv t\}.$$ Then putting these pieces together we get \begin{aligned} N(t)&=\frac1{p-1}\sum_{\ell=0}^{p-2}\sum_{a\in I}\sum_{a'\in I}\chi_\ell(aa'^{-1}g^{-t})\\ &=\frac1{p-1}\sum_{\ell=0}^{p-2}\chi_\ell(g^{-t})\sum_{a,a'\in I}\chi_\ell(a/a')\\ &=\frac1{p-1}\sum_{\ell=0}^{p-2}\chi_\ell(g^{-t})\left\vert\sum_{a\in I}\chi_\ell(a)\right\vert^2. \end{aligned} Separating the term with $\ell=0$, when the sum inside absolute values is $|I|$, we end up with $$N(t)=\frac1{p-1}\left(|I|^2+\sum_{\ell=1}^{p-2}\chi_\ell(g^{-t})\left\vert\sum_{a\in I}\chi_\ell(a)\right\vert^2\right),$$ and the inner sums are of the Polya-Vinogradov type.

Here it is tempting to conjecture that the first term dominates. Unfortunately the Polya-Vinogradov inequality (together with the triangle inequality) is not strong enough to say that. We need something stronger to prove that in the last sum some cancellation will take place. This is to be expected as the oscillating terms $\chi_\ell(g^{-t})$ are multiplied by positive real numbers. I need to think more about this.

-
Very nice progress. I actually didn't arrive at this expression, I should have thought of using orthogonality relations as wisely as you did. Maybe the original proof of Polya-Vinogradov can be modified to suite our case. –  Ofir Apr 20 '12 at 14:55
I think I might have to add additional constraints. It seems that for some values of $t$, $N(t)$ is smaller\larger than expected. For example, if $B$ is close to the interval $[p/3, 2p/3]$ then $N(2)$ is very small. I will do some simulations and post an update. –  Ofir Apr 20 '12 at 15:11
@Ofir. Yeah, I'm also thinking that a calculation along the Polya-Vinogradov technique may do the trick. IIRC for another similar kind of a linear combination of sums restricted to an interval a surprisingly good bound came out. –  Jyrki Lahtonen Apr 20 '12 at 15:23


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Home ARTICLES How to Calculate Your Monthly Mortgage Payment Manually

# How to Calculate Your Monthly Mortgage Payment Manually

0

How to Calculate Your Monthly Mortgage Payment Manually: Purchasers of genuine property use Mortgage, so do you want to know how to Calculate Your Monthly Mortgage Payment to purchase real estate or to raise funds for any other purpose? Check this out.

## How to Calculate Your Monthly Mortgage Payment

Before you hurdle into buying a home often the biggest purchase of your life, you need to know that you can have enough money for the monthly mortgage payment.

You could calculate the payment using a fast online calculator, but if you want to see how all the variables work hand in hand, you can do it by hand using the mortgage monthly payment formula.

The formula for calculating your mortgage monthly payment requires using proponents, so unless you can do those with your brain, you’ll need a calculator to assist.

In this article, we are going to focus on how to calculate your monthly mortgage by hand.

## How to Calculate Mortgage Payment Step by Step

Below are the following methods with their formula used to calculate Mortgage payments.

#### 1. You Have to Understand the Equation

In order to calculate the monthly payment, we can rely on a relatively simple equation. You can represent the monthly payment equation:

{\displaystyle M=P{\frac {r (1+r) ^{n}}{ (1+r) ^{n}-1}}}.

These variables represent the following inputs:

R is your monthly interest rate, calculated by dividing your annual interest rate by 12.

N is your number of payments (the number of months you will pay the loan)

#### 2. Then Input Your Information into the Equation

You will need to input your principal, monthly interest rate, and the number of payments in order to find your monthly payment.

However, you can easily find this information in your loan agreement or from a quoted loan estimate.

Check the information again to be sure of its accuracy before using it in calculations.

For example, imagine you have a $100,000 mortgage loan with 6 percent annual interest over 15 years. Your input for “P” would be$100,000.

For “r,” you would use your monthly interest rate, which would be 0.06 (6 percent) divided by 12, or 0.005 (0.5 percent).

For “n” you would use your total number of payments, one for each month in fifteen years, which would be 12*15, or 180.

In this example, your complete equation would look like this:

#### 4. Solve the exponents

The results inside the parentheses, (1+ r), from the previous step, must now be raised to the power of “n.” Again, this “n” represents the total number of payments.

This step requires a calculator with an exponent function, which is usually represented like this:{\display style x^{y}}

This is done by entering the value to be raised, (1.005) in the example equation.

Then pressing the exponent button, then enter your value for “n” and press enter or =. In the example, the result comes out as 2.454.

If you don’t have such a calculator, type your values from the last equation into Google followed by ^ (n) while replacing the “n” in parentheses with your value for “n.”

The search engine will calculate this value for you.

Keep in mind that only the figures inside the parentheses will be raised to this power, not the “r” outside of them (at the front) or the -1 at the end of the equation.

#### 6. Divide the numerator by the denominator

This means dividing the top part of the equation by the bottom part of the equation. This should leave you with a small decimal.

In the example, your equation would now be:{\display style M=\$100,000* (0.008439)} #### 7. Multiply “P” by this result. This will give you your monthly loan payment. In the example, this would be ($100,000)* (0.008439), or \$843.90. This represents your monthly payment.

In summary, the above-listed processes are how to calculate your monthly mortgage payment by hand. So if you want to calculate your mortgage the above processes will serve as a guide.

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