text:
Need help with this word problem using hyperbolas and need the final answer in (x,y)

John was in the lead, Ed was 1.5 miles behind and Jeff was 2 miles behind John. Then they heard an explosion. John heard it first and Ed heard it a second later and Jeff heard it 1.5 seconds after John. First quad, find (x,y) of explosion. Using hyperbolas, and please explain. I need help finding the speed that the sound traveled and i'm getting stuck. We're not using the actual speed of sound so I think we may need to use a system of equations or something. I tried setting up 1 sec + j sec = distance but I can't find the distance.

Thank You!

• Welcome to stackexchange. "Needing help" is fine, but asking us to do the whole problem for you is not. Please edit the question to show us what you tried and where you are stuck. – Ethan Bolker Apr 19 at 1:34
• Sorry about that, i have showed what I tried – No me gusta matematicas Apr 19 at 1:38
• @Nomegustamatematicas is there any more information given? It would be impossible to determine an $(x,y)$ coordinate if the problem is only talking about one dimension. – rb612 Apr 19 at 1:40
• If you’re not given the speed of sound, you’re either expected to look it up for yourself or use a coordinate system in which the speed of sound equals $1$. – amd Apr 19 at 1:52
• @rb612 Certainly. A hyperbola is the locus of points with a constant difference in distance to a pair of fixed points. This problem involves finding the intersection of a pair of hyperbolas. – amd Apr 19 at 5:38

This becomes tricky given that you don't have knowledge of where the explosion happened relative to the 3 people. This means you have to examine 2 cases. You're right though, that you have to set up a system of equations.

Imagine John, Ed, and Jeff are on the number line. John is at mile $$2$$, Ed is at mile $$0.5$$, and Jeff is just starting, at mile $$0$$. This keeps the consistency of how far they are apart.

Now, let's let $$x$$ be the location of the explosion, and let $$y$$ be the velocity of the explosion. We have to set up some systems of equations.

We need to first figure out distance. We know that John heard the explosion first, so this means that it had to have happened closest to John, so at least at or after mile marker 1.25 (or else it would be closer to Ed). So since we don't know where this happens, we don't know if x is greater than or smaller than 2 (since it could've happened in front of John or close behind and he still would've heard it first!) But we need distance to be positive, so we have to add an absolute value sign.

So the distance John is away would be $$|x-2|$$, the distance Ed would be away is $$x-0.5$$ (since we know it happens in front of Ed), and the distance Jeff would be away is just $$x$$. Now you have that velocity is $$y$$.

So, since velocity is the distance over the time, you have $$\frac{|2-x|}{y}$$ is the time it takes for John to hear it, $$\frac{|x-0.5|}{y}$$ for Ed, and $$\frac{x}{y}$$ for Jeff.

Now you know that the time John heard it, plus 1 second, is the time Ed heard it. Similarly, time John heard plus 1.5 seconds is the time Jef heard it. Can you set up a system of equations and solve from here?


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# Electronic transport properties of the Al0.5TiZrPdCuNi alloy in the high-entropy alloy and metallic glass forms

## Abstract

High-entropy alloys (HEAs) are characterized by a simultaneous presence of a crystal lattice and an amorphous-type chemical (substitutional) disorder. In order to unravel the effect of crystal-glass duality on the electronic transport properties of HEAs, we performed a comparative study of the electronic transport coefficients of a 6-component alloy Al0.5TiZrPdCuNi that can be prepared either as a HEA or as a metallic glass (MG) at the same chemical composition. The HEA and the MG states of the Al0.5TiZrPdCuNi alloy both show large, negative-temperature-coefficient resistivity, positive thermopower, positive Hall coefficient and small thermal conductivity. The transport coefficients were reproduced analytically by the spectral conductivity model, using the Kubo-Greenwood formalism. For both modifications of the material (HEA and MG), contribution of phonons to the transport coefficients was found small, so that their temperature dependence originates predominantly from the temperature dependence of the Fermi–Dirac function and the variation of the spectral conductivity and the related electronic density of states with energy within the Fermi-level region. The very similar electronic transport coefficients of the HEA and the MG states point towards essential role of the immense chemical disorder.

## Introduction

High-entropy alloys (HEAs) are multi-component ($$N\ge$$ 5) crystalline solid solutions including random solutions and partially ordered ones1,2. HEAs possess enormous chemical (substitutional) disorder, similar to the one encountered in metallic glasses (MGs)3. Due to the simultaneous presence of a crystal lattice and an amorphous-type chemical disorder, HEA structures exhibit crystal-glass duality and can be conveniently termed as a “metallic glass on a crystal lattice”. Formation, stability, micro- and nano-structure and mechanical properties of HEAs have been widely investigated in the past4,5,6, while physical properties were less investigated, with the exception of magnetism and superconductivity7,8,9,10,11,12,13,14,15,16,17. A scarcely investigated topic of the HEAs is the electronic transport properties (electrical conductivity, thermoelectric power, electronic thermal conductivity and Hall coefficient), in relation to the crystal-glass duality. Crystallinity of the HEA structure introduces crystal-specific features like the electronic band structure with energy gaps between the bands, multi-branch Fermi surface and phononic dispersion relation, which largely determine the transport properties of crystalline solids. In MGs, there is no lattice and these features are absent. However, chemical disorder that is present to a similar extent in both the HEAs and the MGs also importantly influences the electronic transport properties. In this work, we address the role of the crystal-glass duality on the transport properties of HEAs by performing a comparative study of the transport coefficients of a 6-component metallic alloy Al0.5TiZrPdCuNi that can be prepared either as a HEA or as a MG at the same chemical composition. This has allowed us to unravel the effects of crystallinity and chemical disorder, in order to show how similar or different HEAs are from MGs.

There exists a rare class of near-equiatomic multicomponent alloys that can be prepared either in the crystalline state or as a bulk metallic glass (having the smallest dimension larger than 1 mm) at the same chemical composition. Such alloys are denoted as high-entropy bulk metallic glasses (HE-BMGs)18,19,20,21. The crystalline state of these alloys cannot be claimed to be a HEA, because the primary phase to form is always an intermetallic compound, rather than a solid solution phase. Upon crystallization, either during cooling or heating, an amorphous/intermetallic-compound composite is formed first, converting finally to fully intermetallic compounds (several of them) or perhaps a mixture of intermetallic compounds and solid solution phases. The Al0.5TiZrPdCuNi alloy can be also prepared in the crystalline or amorphous states at the same chemical composition22,23, but is rather unique in this class of alloys for two reasons. Firstly, the amorphous state can be prepared as ribbons of some 10-µm thickness by fast cooling of the spinning melt, which are too thin to be classified as a BMG, but conform to a MG. Secondly, the crystalline state of this alloy can indeed be prepared as a single-phase solid solution (HEA) with a bcc structure, either as melt-spun ribbons at a slower cooling rate or in the bulk rod form of 1.5 mm diameter by conventional Cu-mold casting. According to the available literature18,19,20,21,22,23, the Al0.5TiZrPdCuNi is the only composition that can form both the HEA and the MG states.

The single-phase bcc HEA state of the Al0.5TiZrPdCuNi is a quenched metastable state, because the equilibrium state of this system is a multi-phase crystalline state, composed of various intermetallic compounds. The multi-phase crystalline state as the equilibrium state can be expected on the basis of binary mixing enthalpies of the constituent elements (see Supplementary Table S1), which are highly negative (causing attraction) for most elemental pairs, with the most drastic examples of Pd-Zr ($$\Delta {H}_{mix}^{PdZr}=$$ − 91 kJ mol–1) and Pd-Ti ($$\Delta {H}_{mix}^{PdTi}=$$ – 65 kJ mol–1)24,25. The corresponding values of the total mixing enthalpy of this alloy $$\Delta {H}_{mix}=$$ – 46.7 kJ mol–1 and the atomic-size-difference (geometric) parameter $$\delta =$$ 8.8% are distinctly outside the ranges of crystalline solid solutions in the $$\Delta {H}_{mix}$$ vs. $$\delta$$ phase diagram22,23,26, indicating that the HEA state in the Al0.5TiZrPdCuNi alloy is of an unconventional type.

## Results

### Samples preparation and characterization

The HEA and the MG samples of the alloy with the nominal composition Al0.5TiZrPdCuNi were prepared along the steps described in Refs.22,23. The MG sample was a ribbon of 20 µm thickness and 1 mm width. Its XRD pattern (Fig. 1a) is typical of amorphous structures, showing a broad halo at $$2\theta \approx$$ 40°. The SEM EDS elemental maps (Fig. 2a) indicate a homogeneous distribution of the six constituting elements on the µm scale. The EDS-determined composition (in at.%) is Al7.0Ti21.1Zr15.3Pd14.0Cu21.3Ni21.3. The HEA sample was a rod of 1.3 mm2 cross section. Its XRD pattern (Fig. 1b) reveals the presence of two major phases and one minor phase. The two major phases are (1) a bcc with the unit cell parameter $$a=$$ 3.10 Å and (2) a cubic, type Pd2TiAl (Heusler alloy), space group Fm $$\overline{3 }$$ m, $$a=$$ 6.20 Å (this unit cell edge is exactly twice that of the bcc phase, so that some XRD peaks of the two phases overlap), whereas the minor phase is orthorhombic, type Ni10Zr7, space group Aea2 (No. 41), $$a=$$ 9.20, $$b=$$ 9.20 and $$c=$$ 12.3 Å. The SEM BSE image and the elemental maps are shown in Fig. 2b, where the two major phases (one dark and one bright) are clearly discerned, each composed of crystallites of µm dimensions. Some small black inclusions are also visible at the borders between the two phases. High solid solubility of the elements is evident from the elemental maps. Al is mostly concentrated in the dark phase, which contains also all other elements, with a slight enhancement of Pd. Its EDS composition is Al13.9Ti16.4Zr16.4Pd21.3Cu15.0Ni17.0. The dark phase corresponds to the Pd2TiAl-type cubic phase, which is quite far from the ideal, stoichiometric composition due to the high solubility of the elements. The bright phase is a solid solution of all elements except Al (a small amount of Al is still dispersed in this phase) and corresponds to the bcc phase. Its EDS composition is Al4.4Ti19.1Zr21.9Pd17.9Cu19.8Ni16.9. The minor phase (orthorhombic, Ni10Zr7-type) cannot be easily identified in the SEM BSE image, very likely due to the high solid solubility of the elements. Neglecting the minor phase (of molar fraction less than 10%), the bcc (bright) phase occupies about 70% of the sample’s volume, whereas the Pd2TiAl-type cubic phase (dark) occupies about 30%.

The bcc and the Pd2TiAl-type phases are structurally related. The latter is also based on bcc, but with a specific atomic ordering that can be described in an enlarged, 2 $$\times$$ 2 $$\times$$ 2 unit cell (comprising eight bcc cells). For the Pd2TiAl stoichiometric composition, Ti and Al replace each other alternatively in the corners of the bcc cells, whereas Pd occupies the bcc cell centers (see Supplementary Fig. S1). Due to the high solubility of the elements, other elements also partially occupy the corners and the cell centers. In principle, the entire lattice of the Al0.5TiZrPdCuNi HEA phase (neglecting the minor phase) can be loosely viewed as a bcc structure with a unit cell parameter $$a=$$ 3.10 Å, where a larger part of it is chemically disordered and a smaller part is partially ordered (approximating rather well a HEA). This is somewhat different from the previous report on the Al0.5TiZrPdCuNi HEA phase22, which could be, under specific conditions, synthesized as a single-phase, chemically disordered bcc solution with a slightly larger lattice parameter $$a=$$ 3.20 Å (perhaps due to a slightly different chemical composition) in the entirety of its volume.

### Experimental transport coefficients, specific heat and magnetic properties

The transport coefficients, the specific heat and the magnetic properties were measured comparatively for the HEA and MG samples under identical experimental conditions, described in the “Methods” section. The electrical resistivity between 400 and 2 K is shown in Fig. 3a. Both samples exhibit negative-temperature-coefficient (NTC) resistivity that increases upon cooling, but there exists a pronounced difference in the magnitude of the resistivity and the NTC. For the MG sample, the 400-K resistivity amounts to $${\rho }_{400K}^{MG}=$$ 277 µΩ cm, whereas the 2-K resistivity is $${\rho }_{2K}^{MG}=$$ 295 µΩ cm, yielding the increase by $${NTC}^{MG}=\left({\rho }_{2K}^{MG}-{\rho }_{400K}^{MG}\right)/{\rho }_{400K}^{MG}=$$ 7%. The resistivity of the HEA sample is considerably lower, with $${\rho }_{400K}^{HEA}=$$ 157 µΩcm, $${\rho }_{2K}^{HEA}=$$ 162 µΩcm and $${NTC}^{HEA}=$$ 3%. The thermopower (the Seebeck coefficient $$S$$) of both samples is positive (Fig. 3b), increasing roughly linearly with the temperature (slight nonlinearity, more pronounced for the HEA sample, can be observed at temperatures below 100 K). The thermopower of the HEA sample is a factor of about two larger than that of the MG sample, amounting at 380 K to $${S}_{380K}^{HEA}=$$ 7.3 µVK–1, as compared to $${S}_{380K}^{MG}=$$ 3.2 µV K–1 of the MG sample. The thermal conductivity in the temperature range 1.8–390 K is shown in Fig. 3c, where it is seen that $${\kappa }^{HEA}$$ of the HEA sample is a bit larger than $${\kappa }^{MG}$$ of the MG sample in most of the investigated temperature range, but due to a somewhat faster increase of $${\kappa }^{MG}$$ at higher temperatures, both samples reach the same 390-K value of $${\kappa }_{390K}^{MG}=$$ $${\kappa }_{390K}^{HEA}=$$ 8.2 W m–1 K–1. The Hall coefficient $${R}_{H}={E}_{y}/{j}_{x}{B}_{z}$$ of both samples, measured in a magnetic field range $${\mu }_{0}H=$$ ±9 T, is positive and temperature-independent (Fig. 3d), with the Hall coefficient of the MG sample, $${R}_{H}^{MG}=$$ 3.7 $$\times$$ 10–11 m3 C–1 only insignificantly larger than that of the HEA sample, $${R}_{H}^{HEA}=$$ 3.2 $$\times$$ 10–11 m3 C–1. The low-temperature specific heat $$C$$ between 1.8 and 4.5 K in a $$C/T$$ vs. $${T}^{2}$$ plot is shown in Fig. 4, where it is observed that $${C}^{HEA}$$ is a bit larger than $${C}^{MG}$$. The fit with the standard expression $$C/T=\gamma +\alpha {T}^{2}$$, where $$\gamma$$ and $$\alpha$$ are the electronic and lattice specific heat coefficients, respectively, has yielded $${\gamma }^{HEA}=$$ 2.91 mJ mol–1 K–2 and $${\gamma }^{MG}=$$ 2.28 mJ mol–1 K–2. The Debye temperatures, calculated from $$\alpha$$ are $${\theta }_{D}^{HEA}=$$ 251 ± 5 K and $${\theta }_{D}^{MG}=$$ 262 ± 5 K. The electronic specific heat coefficient is directly proportional to the electronic DOS at the Fermi energy $$g\left({\varepsilon }_{F}\right)$$ via the relation $$\gamma =\left({\pi }^{2}/3\right){k}_{B}^{2}g\left({\varepsilon }_{F}\right)$$, which can also be written as $$\gamma =$$ 2.358 $$g\left({\varepsilon }_{F}\right)$$, where $$\gamma$$ is given in units [mJ mol–1 K–2] and $$g\left({\varepsilon }_{F}\right)$$ is then obtained in units [states/(eV $$\cdot$$ atom)]27. From this relation we obtain $${g}^{HEA}\left({\varepsilon }_{F}\right)=$$ 1.23 states/(eV $$\cdot$$ atom) and $${g}^{MG}\left({\varepsilon }_{F}\right)=$$ 0.97 states/(eV $$\cdot$$ atom), with their ratio $${g}^{HEA}\left({\varepsilon }_{F}\right)/{g}^{MG}\left({\varepsilon }_{F}\right)=$$ 1.27. The field-cooled magnetic susceptibility $$\chi =M/H$$ of both samples, measured between 400 and 2 K in several magnetic fields between 1 and 7 T (Fig. 5) is positive, field-independent and also practically temperature independent (a small Curie upturn due to extrinsic magnetic impurities can be observed at low temperatures). The isothermal magnetization cures, $$M\left(H\right)$$, are shown in the inset of Fig. 5, exhibiting linear paramagnetic behavior up to the highest field of 7 T. These results demonstrate that Ni is in a non-magnetic state and both structural forms (HEA and MG) of the Al0.5TiZrPdCuNi alloy exhibit Pauli spin paramagnetism of the conduction electrons. The magnetic susceptibility is a sum of the Pauli paramagnetic susceptibility and the Larmor diamagnetic susceptibility of closed atomic shells, $$\chi ={\chi }_{P}+{\chi }_{L}$$ (the Landau orbital diamagnetic contribution can be neglected due to the enormous chemical and structural disorder in the samples), where $${\chi }_{P}$$ and $${\chi }_{L}$$ are of the same order of magnitude. An estimate of the Larmor susceptibility from literature tables28, using different ionization states of the elements yields the range $${\chi }_{L}=$$ − [0.127, 0.153]$$\times$$ 10–9 m3 mol–1, which is to be contrasted with the total susceptibility values (e.g., at 300 K) $${\chi }^{HEA}=$$ 0.98 $$\times$$ 10–9 m3 mol–1 and $${\chi }^{MG}=$$ 0.89 $$\times$$ 10–9 m3 mol–1. The comparison shows that $$\left|{\chi }_{L}\right|$$ accounts for about 15% of the total susceptibility, wherefrom we can make a rough estimate of the Pauli susceptibility ratio of the HEA and MG samples as $${\chi }_{P}^{HEA}/{\chi }_{P}^{MG}\approx$$ 1.1. The Pauli spin susceptibility has a simple relation to the electronic density of states at the Fermi energy, $$g\left({\varepsilon }_{F}\right)$$, via the relation $${\chi }_{P}={\mu }_{0}{\mu }_{B}^{2}g\left({\varepsilon }_{F}\right)$$, where $${\mu }_{0}$$ is the permeability of vacuum and $${\mu }_{B}$$ is the Bohr magneton. The susceptibility analysis then yields the ratio $${g}^{HEA}\left({\varepsilon }_{F}\right)/{g}^{MG}\left({\varepsilon }_{F}\right)\approx$$ 1.1, which is in fair agreement with the value 1.27 determined from the specific heat (where the latter should be considered as more precise).

### Theoretical analysis of the transport coefficients

The electronic transport coefficients (electrical conductivity $$\sigma =1/\rho$$, thermopower $$S$$, electronic thermal conductivity $${\kappa }_{el}$$ and Hall coefficient $${R}_{H}$$) were analyzed analytically by the spectral conductivity model, using Kubo–Greenwood formalism29,30 and the assumptions that (1) the HEA and MG structures are both spatially isotropic (for the HEA structure, this follows from the average cubic symmetry of the lattice), so that the tensorial transport coefficients reduce to scalars and (2) the effect of phonons is minor, so that the temperature dependence of the electronic transport coefficients originates predominantly from the temperature dependence of the Fermi–Dirac (FD) function. The second assumption follows from the fact that the immense chemical (substitutional) disorder and the associated topological disorder (lattice distortions) due to different atomic sizes in the HEA structure or complete absence of the lattice in the MG structure represent quenched defects, which break translational periodicity of the system. This causes elastic scattering of the conduction electrons at an extremely rapid rate at any temperature, providing the main scattering mechanism for the electronic transport phenomena.

Within the spectral conductivity model, the coefficients $$\sigma \left(T\right)$$, $$S\left(T\right)$$ and $${\kappa }_{el}\left(T\right)$$ can all be derived from a single, material-dependent quantity, the spectral conductivity $$\sigma \left(\varepsilon \right)$$, which is related to the electronic DOS $$g\left(\varepsilon \right)$$ via the Einstein relation

$$\sigma \left(\varepsilon \right)=\left({e}^{2}/V\right)g\left(\varepsilon \right)D\left(\varepsilon \right),$$
(1)

where $$D\left(\varepsilon \right)$$ is the electronic spectral diffusivity and $$V$$ is the sample volume. We shall further use the approximation that the energy dependence of $$D\left(\varepsilon \right)$$ can be neglected in the vicinity of the Fermi level $${\varepsilon }_{F}$$, by taking $$D\left(\varepsilon \right)\approx D\left({\varepsilon }_{F}\right)$$, so that the shape of $$\sigma \left(\varepsilon \right)$$ within the Fermi-level region is the same as the shape of the DOS $$g\left(\varepsilon \right)$$ (to a constant multiplicative factor).

Within the above model, the electrical conductivity $$\sigma \left(T\right)$$, the Seebeck coefficient $$S\left(T\right)$$ and the electronic thermal conductivity $${\kappa }_{el}\left(T\right)$$ are calculated from

$$\sigma \left(T\right)=\int d\varepsilon \sigma \left(\varepsilon \right) \left(-\partial f/\partial \varepsilon \right),$$
(2)
$$S\left(T\right)={\left[eT\sigma \left(T\right)\right]}^{-1}\int d\varepsilon \sigma \left(\varepsilon \right) \left(\varepsilon -\mu \right) \left(-\partial f/\partial \varepsilon \right),$$
(3)

and

$${\kappa }_{el}\left(T\right)=\left[{L}_{22}\left(T\right)/{e}^{2}T\right]-T\sigma \left(T\right){S}^{2}\left(T\right),$$
(4a)

with

$${L}_{22}\left(T\right)=\int d\varepsilon \sigma \left(\varepsilon \right) {\left(\varepsilon -\mu \right)}^{2}\left(-\partial f/\partial \varepsilon \right).$$
(4b)

The parameter $$e$$ in the above equations is the electric charge of the carriers (its sign distinguishes between electrons and holes), $$f={\left\{exp\left[\left(\varepsilon -\mu \right)/{k}_{B}T\right]+1\right\}}^{-1}$$ is the FD function and $$\mu$$ is the chemical potential. At low temperatures (such as room temperature), $$\mu$$ can be expressed as31

$$\mu \left(T\right)\approx {\varepsilon }_{F}-{\left({k}_{B}T\right)}^{2}\left({\pi }^{2}/6\right){\left[dln g\left(\varepsilon \right)/d\varepsilon \right]}_{{\varepsilon }_{F}}={\varepsilon }_{F}-\xi {T}^{2}.$$
(5)

Within the $$D\left(\varepsilon \right)\approx D\left({\varepsilon }_{F}\right)$$ approximation, we can replace $$g\left(\varepsilon \right)$$ in Eq. (5) by $$\sigma \left(\varepsilon \right)$$, which allows to relate the parameter $$\xi$$ to the Seebeck coefficient using Mott’s formula

$${S}^{Mott}\left(T\right)=\left({\pi }^{2}/3\right)\left({k}_{B}^{2}/e\right){\left[dln \sigma \left(\varepsilon \right)/d\varepsilon \right]}_{{\varepsilon }_{F}}T$$
(6a)

as

$$\xi =\left(e/2\right)\left[{S}^{Mott}\left(T\right)/T\right].$$
(6b)

The experimental thermopower $$S\left(T\right)$$ data can hence be used to set the starting value of the $$\xi$$ fit parameter. In practical terms, the theoretical analysis of the transport coefficients is done by assuming a specific functional form of the spectral conductivity $$\sigma \left(\varepsilon \right)$$ and then performing simultaneous fitting of the quantities $$\sigma \left(T\right)$$ and $$S\left(T\right)$$. After satisfactory fits are obtained, $${\kappa }_{el}\left(T\right)$$ is calculated theoretically, because this quantity is not available experimentally, but only the total thermal conductivity $$\kappa \left(T\right)$$ (that includes the phononic contribution) is measured. A comparison of the theoretical $${\kappa }_{el}\left(T\right)$$ and the experimental $$\kappa \left(T\right)$$ is then used to estimate the residual phononic thermal conductivity of the HEA and MG phases.

Regarding the Hall coefficient, the magneto-transport is less understood. A positive Hall coefficient, $${R}_{H}>0$$, has been reported for many amorphous transition metals and amorphous alloys containing transition metals32,33,34, which made the sign of $${R}_{H}$$ in disordered materials a well-known problem35,36,37. Within the Kubo formalism, the Hall coefficient in the weak-field limit is given by

$${R}_{H}=\left(1/B\right){\sigma }_{xy}\left(B\right)/{\sigma }_{xx}^{2},$$
(7)

where the magnetic field $$B$$ points along the $$z$$ direction, whereas the external electric field and the Hall field point along the $$x$$ and the negative $$y$$ direction, respectively. $${\sigma }_{xy}$$ denotes the off-diagonal element of the conductivity tensor and $${\sigma }_{xx}$$ is the diagonal element. Their analytical expressions for amorphous structures can be found e.g. in Ref.38. Both tensor elements are proportional to the square of the electric charge, $${\sigma }_{xx}, {\sigma }_{xy}$$ $$\propto {e}^{2}$$, and hence do not distinguish between electrons and holes, i.e. the sign of the charge carriers does not determine directly the sign of $${R}_{H}$$. It was then conjectured39 that the sign of the Hall coefficient is correlated with the negative sign of the derivative of the electronic DOS at the Fermi energy, $${R}_{H}\propto -{\left(dg/d\varepsilon \right)}_{{\varepsilon }_{F}}$$. This consideration was supported by other literature reports34,37,38, including a numerical re-examination of the problem40. In our analysis of the Hall coefficient of the HEA and MG samples, we shall check the validity of the $${R}_{H}\propto -{\left(dg/d\varepsilon \right)}_{{\varepsilon }_{F}}$$ relation, i.e. whether the sign of the Hall coefficient equals the sign of the negative DOS derivative within the Fermi level region. An independent check here is that the sign of $${R}_{H}$$ should be in agreement with the sign of the thermopower $$S$$, which also depends on the DOS derivative at $${\varepsilon }_{F}$$, in a product with the sign of the charge carriers $$e$$ (as can be seen from Eq. (6a)).

In a measurement of the electronic transport coefficients, the experimentally observable part of the spectral conductivity $$\sigma \left(\varepsilon \right)$$ is determined by the position and width of the symmetric, bell-shaped FD function derivative, $$-df/d\varepsilon$$, which is centered at the chemical potential $$\mu$$. Its full width at half maximum (FWHM) $${\Delta }_{f}=$$ 3.5 $${k}_{B}T$$ amounts at room temperature to $${\Delta }_{f}^{300K}=$$ 90 meV, whereas at $$T=$$ 2 K it narrows to $${\Delta }_{f}^{2K}=$$ 0.6 meV. In the $$T\to 0$$ limit, $$-df/d\varepsilon$$ converges to a delta function $$\delta \left(\varepsilon -{\varepsilon }_{F}\right)$$ and Eq. (2) yields the relation between the zero-temperature residual resistivity $$\rho \left(0\right)$$ and the spectral conductivity at $${\varepsilon }_{F}$$ as $$\rho \left(0\right)=1/\sigma \left({\varepsilon }_{F}\right)$$. Due to the temperature dependence of $$\mu$$ according to Eq. (5), the experimentally observable part of $$\sigma \left(\varepsilon \right)$$ is shifting with temperature on the energy axis, so that different parts of $$\sigma \left(\varepsilon \right)$$ and the DOS $$g\left(\varepsilon \right)$$ contribute to the integrals in Eqs. (2), (3), (4b) and (7). An estimate of $$\xi$$ from the thermopower data shown in Fig. 3b via Eq. (6b) gives for the HEA sample $${\xi }^{HEA}=$$ –9.6 $$\times$$ 10–9 eVK–2, yielding at 380 K a tiny shift of the chemical potential from $${\varepsilon }_{F}$$ by $${\mu }_{380K}^{HEA}-{\varepsilon }_{F}\approx$$ 1.4 meV. An estimate of $$\xi$$ for the MG sample gives $${\xi }^{MG}=$$ –4.2 $$\times$$ 10–9 eVK–2 and $${\mu }_{380K}^{MG}-{\varepsilon }_{F}\approx$$ 0.6 meV. The shift of $$\mu$$ on the energy scale is very small, so that the electronic transport coefficients are sensitive to a narrow portion of $$\sigma \left(\varepsilon \right)$$ (and the DOS $$g\left(\varepsilon \right)$$) in the energy interval of about ± 100 meV around $${\varepsilon }_{F}$$. Significant temperature dependence of the transport coefficients can be expected when the DOS changes with energy substantially on this scale.

Modeling the spectral conductivity $$\sigma \left(\varepsilon \right)$$ can be performed via different trial functions, a common property of which should be a substantial variation within the experimentally observable energy range. We shall use the model by Landauro and Solbrig, originally developed for quasicrystals41,42,43,44,45. A common feature that quasicrystals share with the HEAs and MGs is the low contribution of phonons to the transport coefficients, because phonon propagation in a non-periodic quasicrystalline structure is strongly hindered. Within this model, the spectral resistivity (the inverse spectral conductivity) $$\rho \left(\varepsilon \right)=1/\sigma \left(\varepsilon \right)$$ is constructed as a superposition of two Lorentzians (the details are given in the Supplementary Information). Choosing proper positions and widths of the Lorentzians on the energy axis within the Fermi-level region allows constructing $$\sigma \left(\varepsilon \right)$$ that exhibits either simple monotonous, positive- or negative-slope behavior, or a minimum (like the pseudogap in the DOS at $${\varepsilon }_{F}$$ in quasicrystals) or a maximum, with possible fine structure on the scale of a few meV. The actual shape of $$\sigma \left(\varepsilon \right)$$ is then adjusted by a simultaneously fitting of the $$\rho \left(T\right)$$ and $$S\left(T\right)$$ experimental data.

The fits of $$\sigma \left(T\right)$$ and $$S\left(T\right)$$ of the HEA and MG samples with Eqs. (2) and (3) are shown by solid curves in Fig. 3a,b. Excellent fits were obtained by the spectral conductivity functions presented in Fig. 6 (the sets of fit parameters determining $${\sigma }^{HEA}\left(\varepsilon \right)$$ and $${\sigma }^{MG}\left(\varepsilon \right)$$ are given in Supplementary Table S2). For converging results, the integrations had to be performed over an energy interval of ± 10 $${k}_{B}T$$ around $${\varepsilon }_{F}$$, which amounts to ± 0.17 eV at 400 K. This value defines the experimentally observed portions of the $$\sigma \left(\varepsilon \right)$$ and $$g\left(\varepsilon \right)$$ quantities. For both the HEA and the MG samples, the spectral conductivity is a negative-sloping function within the Fermi-level region, with a small dip at $${\varepsilon }_{F}$$. The $${\sigma }^{HEA}\left({\varepsilon }_{F}\right)$$ is by a factor of 1.8 larger than $${\sigma }^{MG}\left({\varepsilon }_{F}\right)$$, accounting for the fact that $${\rho }^{HEA}\left(0\right)$$ is smaller than $${\rho }^{MG}\left(0\right)$$ by the same factor. The final values of the fit parameters that shift the chemical potential on the temperature axis are $${\xi }^{HEA}=$$ –15.0 $$\times$$ 10–9 eVK–2 and $${\xi }^{MG}=$$ –4.6 $$\times$$ 10–9 eVK–2, very close to the starting values determined from the Mott formula of Eq. (6b). In Fig. 6, the function $$-\partial f/\partial \varepsilon$$ is also shown at the temperature of 400 K, where the shift of the maximum relative to $${\varepsilon }_{F}$$ due to the temperature-dependent chemical potential is unobservable on this energy scale.

The negative-sloping $$\sigma \left(\varepsilon \right)$$ within the Fermi-level region and the assumption of negatively charged carriers, $$e=-\left|e\right|$$ (electrons) reproduce quantitatively the NTC electrical resistivity and the positive thermopower with all their nonlinear features in the entire measured temperature range 2–400 K for both samples. Knowing $$\sigma \left(T\right)$$ and $$S\left(T\right)$$, the electronic thermal conductivity $${\kappa }_{el}\left(T\right)$$ was calculated from Eqs. (4a,b) and is presented by solid lines in Fig. 3c. It is observed that the calculated $${\kappa }_{el}^{HEA}\left(T\right)$$ and $${\kappa }_{el}^{MG}\left(T\right)$$ account for the majority part of the experimental total thermal conductivities $${\kappa }^{HEA}\left(T\right)$$ and $${\kappa }^{MG}\left(T\right)$$, supporting the assumption that the phononic thermal conductivity $${\kappa }_{ph}$$ is small for both samples. The negative derivative $$d\sigma /d\varepsilon <0$$ and the associated $$dg/d\varepsilon <0$$ within the Fermi-level region also consistently explain the positive sign of the Hall coefficient $${R}_{H}$$ and hence support the relation $${R}_{H}\propto -{\left(dg/d\varepsilon \right)}_{{\varepsilon }_{F}}$$ for the investigated chemically and topologically disordered HEA and MG samples.

## Discussion and conclusions

The electronic transport coefficients of the HEA and MG states of the Al0.5TiZrPdCuNi alloy reveal certain similarities as well as differences. The two states have in common the enormous chemical disorder, but differ in the presence/absence of a crystal lattice. In both cases, the contribution of phonons to the transport coefficients is small. While phonons cannot propagate in the amorphous state due to the absence of a crystal lattice, the lattice in the HEA state is unsuitable for the phonon propagation because of the topological distortion and random distribution of masses (different chemical elements) on the lattice sites. An exception are long-wavelength acoustic phonons, which “see” both structures as an elastic continuum and still contribute to the transport phenomena. Due to the weak phonon contribution, the temperature dependence of the electronic transport coefficients originates predominantly from the temperature dependence of the FD function and the variation of the spectral conductivity $$\sigma \left(\varepsilon \right)$$ and the related electronic DOS $$g\left(\varepsilon \right)$$ with energy within the Fermi-level region. For the Al0.5TiZrPdCuNi alloy, $${\sigma }^{HEA}\left(\varepsilon \right)$$ and $${\sigma }^{MG}\left(\varepsilon \right)$$ are both decreasing functions in that energy range, with some fine structure on the 10-meV scale very close to $${\varepsilon }_{F}$$. Under the realistic assumption that the spectral diffusivity $$D\left(\varepsilon \right)$$ does not change significantly within the Fermi-level region, the electronic DOSs $${g}^{HEA}\left(\varepsilon \right)$$ and $${g}^{MG}\left(\varepsilon \right)$$ show the same shape as their respective spectral conductivities. Knowing the DOS at $${\varepsilon }_{F}$$ values from the specific heat measurements (recall that their ratio is $${g}^{HEA}\left({\varepsilon }_{F}\right)/{g}^{MG}\left({\varepsilon }_{F}\right)=$$ 1.27), while the $$T\to$$ 0 resistivities yield $${\sigma }^{HEA}\left({\varepsilon }_{F}\right)/{\sigma }^{MG}\left({\varepsilon }_{F}\right)={\rho }^{MG}\left(0\right)/{\rho }^{HEA}\left(0\right)=$$ 1.8, the Einstein relation of Eq. (1) yields the ratio of the spectral diffusivities $${D}^{HEA}\left({\varepsilon }_{F}\right)/{D}^{MG}\left({\varepsilon }_{F}\right)=$$ 1.4. By assuming that the electronic diffusion constant can be described by the Einstein formula $$D={\mu }_{e}{k}_{B}T/e$$, valid for an electrically charged Brownian particle of charge $$e$$ in an electric field $$E$$ with mobility $${\mu }_{e}$$ (defined as the ratio of the electronic drift velocity $${v}_{d}$$ to the magnitude of the electric field, $${\mu }_{e}={v}_{d}/E$$), we can estimate the ratio of the electronic mobilities in the HEA and MG phases to be $${\mu }_{e}^{HEA}/{\mu }_{e}^{MG}\approx$$ 1.4, i.e. the mobilities are about the same. This number is independently confirmed from the Hall coefficient and the conductivity using expression $${\mu }_{e}=\sigma {R}_{H}$$, which yields a very similar value $${\mu }_{e}^{HEA}/{\mu }_{e}^{MG}=\left({\sigma }^{HEA}\left({\varepsilon }_{F}\right)/{\sigma }^{MG}\left({\varepsilon }_{F}\right)\right)\cdot \left({R}_{H}^{HEA}/{R}_{H}^{MG}\right)\approx$$ 1.5, pointing toward the essential role of chemical disorder in determining the electronic transport coefficients of both the HEA and the MG states, whereas the presence/absence of the (topologically distorted) crystal lattice is of minor importance, but still experimentally observable. The enormous chemical and topological disorders represent quenched defects in the structure, which scatter the electrons elastically at an extremely rapid rate at any temperature and strongly reduce the transport phenomena relative to the chemically ordered crystals. The order of magnitude of the relaxation time $${\tau }_{0}$$ (the mean time between two elastic scattering events) can be estimated from the Einstein conductivity of Eq. (1), by considering that the diffusion constant is given by $$D={l}_{0}^{2}/{\tau }_{0}$$, where $${l}_{0}$$ is the mean free path. Due to the immense disorder, it is reasonable to consider that the mean free path in the HEA lattice assumes its limiting (constant) value $${l}_{0}\approx a$$, where $$a=$$ 3.1 Å is the unit cell length. For the MG sample, $${l}_{0}$$ can be taken as the nearest-neighbor distance, again about 3 Å. Taking the experimental $$g\left({\varepsilon }_{F}\right)$$ and $${\rho }_{2K}$$ values, we obtain $${\tau }_{0}\sim$$ 10–15 s for both structural forms of the material. This is to be contrasted with pure metals near absolute zero, where $${\tau }_{0}\sim$$ 10–9 s is typical and $${l}_{0}$$ is on the order of 106 atomic distances (becoming $${\tau }_{0}\sim$$ 10–14 s and $${l}_{0}$$ about 100 atomic distances at room temperature). From the transport-properties point of view, HEAs appear very similar to MGs, but are quite different from translationally periodic, chemically ordered crystals.

## Methods

XRD patterns were recorded on a PANalytical X'Pert PRO MPD X-ray powder diffractometer using Cu Kα1 radiation ($$\lambda =$$ 1.54056 Å). SEM BSE imaging and EDS measurements of the chemical composition and elemental mapping were conducted on a focused ion beam scanning electron microscope FEI HeliosNanolab 650, equipped with EDS system from Oxford Instruments with X-max SDD detector. Electrical resistivity, thermoelectric power, thermal conductivity, Hall coefficient and specific heat were measured by a Quantum Design Physical Property Measurement System (PPMS 9 T). Magnetic properties were recorded on a Quantum Design MPMS3 SQUID magnetometer.

## Data availability

The datasets generated during and/or analyzed during the current study are available from the corresponding author on reasonable request.

## References

1. Yeh, J. W. et al. Nanostructured high-entropy alloys with multiple principal elements: Novel alloy design concepts and outcome. Adv. Eng. Mater. 6, 299–303 (2004).

2. Yeh, J. W. Recent progress in high-entropy alloys. Ann. Chim. Sci. Mater. 31, 633–648 (2006).

3. Takeuchi, A. & Inoue, A. Classification of bulk metallic glasses by atomic size difference, heat of mixing and period of constituent element and its application to characterization of the main alloying element. Mater. Trans. 46, 2817–2829 (2005).

4. Murthy, B. S., Yeh, J.-W., Ranganathan, S. & Bhattacharjee, P. P. High-Entropy Alloys, 2nd ed. (Elsevier, 2019).

5. Gao, M. C., Yeh, J.-W., Liaw, P. K. & Zhang., Y. (eds.). High-Entropy Alloys: Fundamentals and Applications (Springer International Publishing Switzerland, 2016). https://doi.org/10.1007/978-3-319-27013-5.

6. Zhang, Y. et al. Microstructures and properties of high-entropy alloys. Prog. Mater. Sci. 61, 1–93. https://doi.org/10.1016/j.pmatsci.2013.10.001 (2014).

7. Koželj, P. et al. Discovery of a superconducting high-entropy alloy. Phys. Rev. Lett. 113, 107001 (2014).

8. Vrtnik, S. et al. Superconductivity in thermally annealed Ta-Nb-Hf-Zr-Ti high-entropy alloys. J. Alloys Compd. 695, 3530–3540 (2017).

9. Sun, L. & Cava, R. J. High-entropy alloy superconductors: Status, opportunities and challenges. Phys. Rev. Mater. 3, 090301 (2019).

10. von Rohr, F. O., Winiarski, M. J., Tao, J., Klimczuk, T. & Cava, R. J. Effect of electron count and chemical complexity in the Ta-Nb-Hf-Zr-Ti high-entropy alloy superconductor. PNAS 113, E7144–E7150 (2016).

11. Jasiewicz, K., Wiendlocha, B., Korbeń, P., Kaprzyk, S. & Tobola, J. Superconductivity of Ta34Nb33Hf8Zr14Ti11 high entropy alloy from first principles calculations. Phys. Status Solidi Rapid Res. Lett. 10, 415–419 (2016).

12. von Rohr, F. O. & Cava, R. J. Isoelectronic substitutions and aluminium alloying in the Ta-Nb-Hf-Zr-Ti high-entropy alloy superconductor. Phys. Rev. Mater. 2, 034801 (2018).

13. Yuan, Y. et al. Superconducting Ti15Zr15Nb35Ta35 high-entropy alloy with intermediate electron-phonon coupling. Front. Mater. 5, 72. https://doi.org/10.3389/fmats.2018.00072 (2018).

14. Ishizu, N. & Kitagawa, J. New high-entropy alloy superconductor Hf21Nb25Ti15V15Zr24. Results Phys. 13, 102257 (2019).

15. Stolze, K., Cevallos, F. A., Kong, T. & Cava, R. J. High-entropy alloy superconductors on an α-Mn lattice. J. Mater. Chem. C 6, 10441 (2018).

16. Stolze, K., Tao, J., von Rohr, F. O., Kong, T. & Cava, R. J. Sc−Zr−Nb−Rh−Pd and Sc−Zr−Nb−Ta−Rh−Pd high-entropy alloy superconductors on a CsCl-type lattice. Chem. Mater. 30, 906–914 (2018).

17. Marik, S., Varghese, M., Sajilesh, K. P., Singh, D. & Singh, R. P. Superconductivity in equimolar Nb–Re–Hf–Zr–Ti high entropy alloy. J. Alloys Compd. 769, 1059–1063 (2018).

18. Ma, L., Wang, L., Zhang, T. & Inoue, A. Bulk glass formation of Ti–Zr–Hf–Cu–M (M=Fe Co, Ni) alloys. Mater. Trans. 43, 277–280 (2002).

19. Zhao, K., Xia, X. X., Bai, H. Y., Zhao, D. Q. & Wang, W. H. Room temperature homogeneous flow in a bulk metallic glass with low glass transition temperature. Appl. Phys. Lett. 98, 141913 (2011).

20. Takeuchi, A. et al. Pd20Pt20Cu20Ni20P20 high-entropy alloy as a bulk metallic glass in the centimeter. Intermetallics 19, 1546–1554 (2011).

21. Gao, X. Q. et al. High mixing entropy bulk metallic glasses. J. Non-Cryst. Solids 357, 3557–3560 (2011).

22. Takeuchi, A. et al. Al0.5TiZrPdCuNi high-entropy (H-E) alloy developed through Ti20Zr20Pd20Cu20Ni20 H-E glassy alloy comprising inter-transition metals. Mater. Trans. 54, 776–782 (2013).

23. Nagase, T., Takeuchi, A., Amiya, K. & Egami, T. Solid state amorphization of metastable Al0.5TiZrPdCuNi high entropy alloy investigated by high voltage electron microscopy. Mater. Chem. Phys. 210, 291–300 (2018).

24. de Boer, F. R., Boom, R., Mattens, W. C. M., Miedema, A. R. & Niessen, A. K. Cohesion in Metals: Transition Metal Alloys (Cohesion and Structure) (North Holland, 1988).

25. Takeuchi, A. & Inoue, A. Calculations of mixing enthalpy and mismatch entropy for ternary amorphous alloys. Mater. Trans. 41, 1372–1378 (2000).

26. Guo, S., Hu, Q., Ng, C. & Liu, C. T. More than entropy in high-entropy alloys: Forming solid solutions or amorphous phase. Intermetallics 41, 96–103 (2013).

27. Mizutani, U. Introduction to the Electron Theory of Metals 39 (Cambridge, 2001).

28. Bain, G. A. & Berry, J. F. Diamagnetic corrections and Pascal’s constants. J. Chem. Educ. 85, 532–536 (2008).

29. Kubo, R. Statistical-mechanical theory of irreversible processes. I. General theory and simple applications to magnetic and conduction problems. J. Phys. Soc. Jpn. 12, 570–586 (1957).

30. Greenwood, D. A. The Boltzmann equation in the theory of electrical conduction in metals. Proc. Phys. Soc. 71, 585–591 (1958).

31. Aschroft, N. W. & Mermin, N. D. Solid State Physics 46 (Saunders College Publishing, 1976).

32. Morgan, G. J. & Howson, M. A. Linear response formulae for the Hall coefficient. J. Phys. C Solid State Phys. 18, 4327–4334 (1985).

33. Howson, M. A. & Gallagher, B. L. The electron transport properties of metallic glasses. Phys. Rep. 170, 265–324 (1988).

34. Nguyen-Manh, D., Mayou, D., Morgan, G. J. & Pasturel, A. The Hall coefficient and the derivative of the density of states in amorphous metals. J. Phys. F 17, 999–1012 (1987).

35. Weir, G. F., Howson, M. A., Gallagher, B. L. & Morgan, G. J. Hybridization in amorphous metals. Philos. Mag. 47, 163–176 (1983).

36. Berger, L. Side-jump mechanism for the Hall effect of ferromagnets. Phys. Rev. B 2, 4559–4566 (1970).

37. Movaghar, B. The sign of the Hall effect in disordered materials. In Physics of Disordered Materials (eds. Adler, D., Fritzsche, H. & Ovshinsky, S. R.) 399–412 (Plenum Press, 1985).

38. Houari, A. & Harris, R. The Hall effect in simple glassy structures. J. Phys. Condens. Matter 3, 1505–1512 (1991).

39. Bush, G. & Güntherodt, H.-J. Electronic properties of liquid metals and alloys. In Solid State Physics 29 (eds. Ehrenreich, H., Seitz, F. & Turnbull, D.) 235–313 (Academic Press, 1974).

40. Houari, A., Mebrouki, M., Dib, A. F. R. & Ould-Kaddour, F. Numerical investigation of the Hall effect in disordered materials. Phys. B 291, 387–393 (2000).

41. Landauro, C. V. & Solbrig, H. Temperature dependence of the electronic transport in Al–Cu–Fe phases. Mater. Sci. Eng. A 294–296, 600–603 (2000).

42. Solbrig, H. & Landauro, C. V. Electronic transport parameters and spectral fine structure: From approximants to quasicrystals. In Quasicrystals, Structure and Physical Properties (ed. Trebin, H. R.) 254–271 (Wiley-VCH, 2003).

43. Landauro, C. V. & Solbrig, H. Modeling the electronic transport properties of Al–Cu–Fe phases. Phys. B 301, 267–275 (2001).

44. Dolinšek, J. et al. Intrinsic electrical, magnetic, and thermal properties of single-crystalline Al64Cu23Fe13 icosahedral quasicrystal: Experiment and modeling. Phys. Rev. B 76, 054201 (2007).

45. Bobnar, M. et al. Electrical, magnetic and thermal properties of single-grain Ag42In42Yb16 icosahedral quasicrystal: Experiment and modeling. Phys. Rev. B 84, 134205 (2011).

## Acknowledgements

The Slovenian authors acknowledge the financial support from the Slovenian Research Agency (research core funding No. P1-0125). Q. Hu acknowledges funding from the National Science Foundation of China (No. 52061016). A.J. acknowledges technical support from the CENN Nanocenter of the Jožef Stefan Institute, Ljubljana.

## Author information

Authors

### Contributions

Conceptualization of the work was done by M.W., Q.H., S.G. and J.D. The work was designed by M.W. and J.D. Transport properties and specific heat data acquisition and analysis were done by M.W., M.K., S.V., J.L. and D.G., whereas A.J. and A.M. performed XRD, electron microscopy imaging and EDS elemental mapping data acquisition and analysis. P.K. has performed magnetic data acquisition and analysis. C.W. and Q.H. performed materials synthesis. J.D. has performed data interpretation and wrote original draft manuscript, whereas review and editing of the manuscript was done by A.M., M.W., P.K., Q.H. and S.G. All authors discussed and commented on the manuscript.

### Corresponding authors

Correspondence to Qiang Hu or Janez Dolinšek.

## Ethics declarations

### Competing interests

The authors declare no competing interests.

### Publisher's note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

## Rights and permissions

Reprints and Permissions

Wencka, M., Krnel, M., Jelen, A. et al. Electronic transport properties of the Al0.5TiZrPdCuNi alloy in the high-entropy alloy and metallic glass forms. Sci Rep 12, 2271 (2022). https://doi.org/10.1038/s41598-022-06133-7

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# Math Help - How to evaluate this limit?

1. ## How to evaluate this limit?

I need to evaluate the lim x->0 tan^2(x)/x

I can use any method to do this. I know the limit is zero but I wasn't sure what method I could use to evaluate this.

It could be re-written as:
(sinx/cosx)(sinx/cosx)/x

The sin of 0 is 0 and therefore makes the whole function zero, would this be sufficient enough to evaluate it?

2. ## Re: How to evaluate this limit?

You could use L'Hôpital's rule to write the limit as:

$2\lim_{x\to0}\tan(x)\sec^2(x)=0$

3. ## Re: How to evaluate this limit?

Hi zzizi,

lim x->0 sin(x)/x =1,
lim x->0 tan(x)/x =1,
lim x->0 tan(x)/x *tan(x) = 1*0 = 0

4. ## Re: How to evaluate this limit?

Thank you both very much!

5. ## Re: How to evaluate this limit?

Originally Posted by zzizi
It could be re-written as:
(sinx/cosx)(sinx/cosx)/x

The sin of 0 is 0 and therefore makes the whole function zero, would this be sufficient enough to evaluate it?
Watch that last comment! Yes, sin(0) = 0, but x is in the denominator so x = 0 makes the expression indeterminate.

-Dan

6. ## Re: How to evaluate this limit?

Originally Posted by zzizi
I need to evaluate the lim x->0 tan^2(x)/x

I can use any method to do this. I know the limit is zero but I wasn't sure what method I could use to evaluate this.

It could be re-written as:
(sinx/cosx)(sinx/cosx)/x

The sin of 0 is 0 and therefore makes the whole function zero, would this be sufficient enough to evaluate it?
\displaystyle \begin{align*} \lim_{ x \to 0}\frac{\tan^2{x}}{x} &= \lim_{x \to 0}\frac{\sin{x}}{x} \cdot \lim_{x \to 0} \frac{\sin{x}}{\cos^2{x}} \\ &= 1 \cdot \frac{0}{1^2} \\ &= 0 \end{align*}

7. ## Re: How to evaluate this limit?

Originally Posted by topsquark
Watch that last comment! Yes, sin(0) = 0, but x is in the denominator so x = 0 makes the expression indeterminate.

-Dan
Yes, perhaps I was a bit hasty there.

8. ## Re: How to evaluate this limit?

Originally Posted by Prove It
\displaystyle \begin{align*} \lim_{ x \to 0}\frac{\tan^2{x}}{x} &= \lim_{x \to 0}\frac{\sin{x}}{x} \cdot \lim_{x \to 0} \frac{\sin{x}}{\cos^2{x}} \\ &= 1 \cdot \frac{0}{1^2} \\ &= 0 \end{align*}
Wonderful! thank you sir!


text:
# zbMATH — the first resource for mathematics

Tikhonov replacement functionals for iteratively solving nonlinear operator equations. (English) Zbl 1078.47030
The paper is concerned with an ill-posed nonlinear equation $$F(x)=y$$, where $$F: X \to Y$$ is a twice continuously differentiable operator between Hilbert spaces $$X$$ and $$Y$$. Let $$y^{\delta} \in Y$$ be an available approximation to $$y$$ with $$\| y^{\delta}-y \| \leq \delta$$. The authors suggest a two-level Tikhonov-type regularization method $$x_{k+1}=\text{argmin}_{x \in X}{\widetilde J}_{\alpha}(x,x_k)$$, $$k=0,1, \dots$$, $$\alpha \to 0$$, where $${\widetilde J}_{\alpha}(x,a)=J_{\alpha}(x)+ C\| x-a\|^2- \| F(x)-F(a)\|^2$$, $$C, \alpha>0$$, and $$J_{\alpha}(x)=\| y^{\delta}-F(x)\|^2+\alpha \| x-{\overline x}\|^2$$ is the classical Tikhonov functional. Let the derivative $$F^{\prime}(x)$$ be Lipschitz continuous on $$X$$ with a constant $$L$$. It is shown that the modified functional $${\widetilde J}_{\alpha}(x,x_k)$$ is strictly convex and that $$x_{k+1}$$ can be obtained by the fixed point iteration, provided that $$L \sqrt{J_{\alpha}(a)}<C+\alpha$$. Moreover, if $$\| F(x)-F({\widetilde x})-F^{\prime}({\widetilde x})(x-{\widetilde x})\| \leq \| F(x)-F({\widetilde x})\|$$ $$\forall x, {\widetilde x} \in X$$ and the solution $$x^*$$ to the original equation satisfies $$x^*-{\overline x} \in R(F^{\prime *}(x^*))$$, then the method with an appropriate stopping rule $$\alpha=\alpha(\delta)$$ generates approximations converging to $$x^*$$ as $$\delta \to 0$$.

##### MSC:
 47J06 Nonlinear ill-posed problems 47J25 Iterative procedures involving nonlinear operators
TIGRA
Full Text:


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## ROSENKREUTZ HOME NEWS

### The Pandemic 2020 New Class Announcement! Register Now!

by Eugene Passofaro -
Number of replies: 5

Dear Friends,

I am Beginning lectures this week for the New Online Class Series (opening today) The Pandemic 2020!!!

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by Laura Turner -
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by Eugene Passofaro -

About the middle is some interesting news about the peaceful demonstrations in Berlin of estimated 1 Million people who want to end the lock-downs and regain their freedom.

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Dr Simone Gold speaks out on remedies for Covid 2 and Hydroxychloroquine

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### Re: Where Did Covid-19 Originate?

by Laura Turner -
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14 words

### Where Did Covid-19 Originate?

by Eugene Passofaro -
Where Did Covid-19 Originate?

It is now a very important question for China is asking apologies from the USA President for claiming it came from their laboratory in Wuhan. Dr. Shi Zhengli is head of the laboratory. Had she possibly a hand in removing any evidence to connect the Chinese Laboratory she was working to create a Deadly Virus?  What is remarkable is that the Nobel Prize winning scientist Montagnier claims his testing of the virus showed a sequence of protein which was man-made and was in the Covid protein chain sequence revealing there was no other possible way it could have formed. Most certainly, to make it more effective the Bat virus was crucial to complete the sequences of the viral material. Here is a very important article below,  and the author claims to have done the mathematical formulas necessary to reveal if there is evidence the records of the protein genome had been intentionally changed after the virus broke out and spread in the wider world.  The article is presented for your own consideration and research.

Dr. Shi Zhengli, known as China’s batwoman, was worried.  Her creation, a highly infectious coronavirus, had leaked from her lab in October.  Early efforts to stop the widening circle of infections had failed.  It had spread like wildfire through the densely populated city of Wuhan.

The Communist authorities were also worried that the source of the epidemic would be traced back to the Wuhan lab.  To muddy the waters, on January 3rd,  China’s National Health Commission ordered all biolabs in China to destroy not only the samples of the SARS-CoV-2 that they had isolated from those infected, but also the genetic sequencing of the virus’ RNA strand.  Dr. Shi must have been only too happy to cover her own tracks—evidence suggests she had already started doing so–by destroying the evidence of the deadly virus’ actual origins in her lab.

It was the equivalent of trying to wipe the fingerprints from a gun that had just been used to commit murder.

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Dr. Shi registered her new virus on January 27th, 2020, with the National Center for Biotechnology Information (NCBI) of the U.S. National Institutes of Health, the customary repository for such information.  She called it RaTG-13, Ra for Rhinolophus affinis, the Latin name of the Intermediate Horseshoe Bat, and 13 for 2013, the year she supposedly “discovered” it.

A lot of people have been taken in by Dr. Shi’s clever “discovery,” which is looking more and more like a forgery. No one else has independently verified its existence.  No other lab has a sample of it, and no one else has ever sequenced it.  And of course, they likely never will, because more and more evidence suggests that it exists only in a string of letters on her computer.

It was a brilliant scheme, and it almost succeeded.  The Chinese virology community—following new, strict Party guidelines—has published a flurry of studies suggesting that the existence of RaTG-13 proves that SARS-CoV-2 came from nature.  They claim that other “first cousins” of the China Coronavirus will soon be found to exist in nature if we just keep looking.  They do genomic analyses showing that RaTG-13 and SARS-CoV-2 are 96% identical throughout the whole sequence of the viral genome.  They calculate that the two share a common ancestor a few decades back.

Actually, the only thing the two virus genomes have in common is Dr. Shi herself, who engineered the one and appears to have fabricated the other.

As clever as Batwoman is, however, she did not commit the perfect deception.  She left behind a few key clues that reveal, as clearly as fingerprints on a murder weapon, what she was up to.

One blogger, writing at Nerd Has Power, has brilliantly teased these findings out of the data. The blogger’s efforts can be read here if you have a free afternoon and a strong math background.  I offer here a summary of one of the blogger’s critiques, in the hope of making the blogger’s general line of argument accessible to the layman. Because he published his raw data, I and others have been able to check and verify his work.  I offer here a summary of one of the blogger’s critiques, in the hope of making the blogger’s general line of argument accessible to the layman.

An Impossible Ratio:  As viruses evolve, they mutate.  That is to say, one of the four nucleotides is randomly replaced by another.  Most of these random mutations do not produce changes in the amino acids that make up the protein.  Such mutations are called “synonymous”, since the three-nucleotide “codon” still codes for the same amino acid despite the change.  Like a synonym in a thesaurus, it “looks” different but “means”—in terms of the amino acid and resulting protein—the same thing.

But then there are “non-synonymous” mutations.  These are mutations that do change the resulting amino acid and hence the configuration of the resulting protein.  In nature, the ratio of synonymous to non-synonymous is approximately 5:1.

Here’s where Dr. Shi got into trouble.  When typing in the genomic sequence of her “discovery” she made way too many non-synonymous changes at the beginning.  Then, one-third of the way through the sequence, she apparently realized her error.  After that, she made way too few non-synonymous changes.  So while the entire genome has the expected 5:1 ratio, there are stretches where the ratio is closer to 2:1, and other long stretches where it is as high as 44:1.

Nature’s mutations are random.  Dr. Shi’s “mutations” are not. Dr. Lawrence Sellin has calculated that the odds that her “mutations” occurred naturally in just one area—the critical spike protein–at almost ten million to one.

If RaTG-13 is merely a strategic deception, as I believe it is, then what real coronavirus did Dr. Shi enhance in “Gain-of-Function” research to create the highly infectious SARS-CoV-2?  Evidence suggests that a coronavirus from a People’s Liberation Army biolab may have played this role.

If Dr. Shi’s research was in any way connected with the PLA’s bioweapons research program, this would go a long way towards explaining China’s bizarre behavior – the secrecy, the serial lying, the brutal quarantine, the persecution of whistle-blowers, the seeding of the virus around the world, and the furious rejection by Beijing of the idea of an international commission to investigate the origin and spread of the disease.  It all seems wildly overwrought, even by the standards of a conspiratorial Communist Party whose leaders suffer from a pathological paranoia.

Be that as it may, there is one thing about Dr. Shi’s research that we do know with absolute, ironclad certainty: The U.S. was helping to fund it.  From 2013 onward, the National Institute of Allergies and Infectious Diseases, run by the now-famous Dr. Anthony Fauci, gave $7.4 million for research involving gain-of-function work and collecting and studying bat coronaviruses,$600,000 of which went to the Wuhan Institute of Virology.  Much of the rest went to other labs in China.

If Dr. Fauci was as familiar with the Chinese Communist Party as he is with viruses, he would not have sent the Wuhan Institute of Virology one thin dime. He would have known that the PRC has a long-running program to develop bioweapons. He would have concluded that China’s only BSL-4 high containment lab was the obvious place to carry out at least some of this research.  And he definitely did know that the “Gain-of-Function” research he was funding had the potential to create “Pathogens of Pandemic Potential” that might not be used for peaceful ends.

Apparently none of these considerations ever crossed his mind.

Now that Western intelligence services have concluded that SARS-CoV-2 probably is a man-made chimera that leaked from a lab — a lab that he was funding — you might think that Dr. Fauci would be inclined to reevaluate the role he played in this global debacle.

Instead, he continues to tell us that there is “no scientific evidence the coronavirus was made in a Chinese lab” and that there is “no evidence” that it leaked from the lab.

It is a sad end to a career that was at least long, if not particularly illustrious.

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### Problems in BackPropagation and Gradient Descent

1. the gradient of reward signals given to the agent is realised many timesteps in the future. Questions above can be seem as Credit Assignment

2. there is the issue of being stuck in a local optimum.

### Pseudo code of Basic Evolution Strategy

1. Easier to scale in a distributed setting(easy to parallelize).
2. It does not suffer in settings with sparse rewards.
3. It has fewer hyperparameters.
4. It is effective at finding solutions for RL tasks.

### Improvement of Covariance Matrix Adaptive Evolution Strategy

We want to explore more and increase the standard deviation of our search space.
And there are times when we are confident we are close to good optima and just want to fine-tune the solution.

### Details of algorithm

1. Calculate the fitness score of each candidate solution in generation $(g)$.

2. Isolates the best 25% of the population in generation $(g)$, in purple.

1. Using only the best solutions, along with the mean $\mu^{(g)}$ of the current generation (the green dot), calculate the covariance matrix $C^{(g+1)}$ of the next generation.

1. Sample a new set of candidate solutions using the updated mean $\mu^{(g+1)}$ and covariance matrix $C^{(g+1)}$.

### OpenAI Evolution Strategy

In particular, $\sigma$ is fixed to a constant number, and only the $\mu$ parameter is updated at each generation.

Although its performance is not the best, it is possible to scale to over a thousand parallel workers.

0%


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Timezone: »

Poster
Approximating Orthogonal Matrices with Effective Givens Factorization
Thomas Frerix · Joan Bruna

Wed Jun 12 06:30 PM -- 09:00 PM (PDT) @ Pacific Ballroom #164

We analyze effective approximation of unitary matrices. In our formulation, a unitary matrix is represented as a product of rotations in two-dimensional subspaces, so-called Givens rotations. Instead of the quadratic dimension dependence when applying a dense matrix, applying such an approximation scales with the number factors, each of which can be implemented efficiently. Consequently, in settings where an approximation is once computed and then applied many times, such a representation becomes advantageous. Although effective Givens factorization is not possible for generic unitary operators, we show that minimizing a sparsity-inducing objective with a coordinate descent algorithm on the unitary group yields good factorizations for structured matrices. Canonical applications of such a setup are orthogonal basis transforms. We demonstrate numerical results of approximating the graph Fourier transform, which is the matrix obtained when diagonalizing a graph Laplacian.


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SEARCH HOME
 Math Central Quandaries & Queries
 Question from Iris, a student: Help please: me on this type of math problem I can get through step one after that I get lost. Maybe you can explain it to me in a easier way to get the correct results. I have read the examples, but I am still lost cannot move on until the next practice until I get this one Sample standard deviation The following are distances (in miles) traveled to the workplace by employees of a certain computer company. 15,29,32 28 18 28 Find the standard deviation of this sample of distances. Round your answer to at least two decimal places. (If necessary, consult a list of formulas.) The measurements given constitute a sample, and we're asked to compute the sample standard deviation. The sample standard deviation, denoted by , is a measure of the "spread" of the measurements in a sample about their mean . In order to find the standard deviation, we first find the mean of the sample: We then compute the standard deviation as follows: How do I get 232 from these numbers if you could break in down by each step, I know that you Multiply & then add, but what numbers do you Multiply and add together. I am not getting it. I need help (15-25)x2+(29-25)x2+(31-25)x2+(38-25)x2+(18-25)x2+(28-25)x2 over 6-1 I need a step by step hoe the answer is 232 i START LIKE THIS TRYING TO SOLVE (15-25)=-10 10X10=100 YHEN I GET LOST ON THE NEXT THINS TO DO, i DO THIS FOR EACH SET OF NUMBERS.

Iris,

You are correct that the mean of the sample is 25 miles.Your process for finding the sample standard deviation is correct but you need to be more careful with the numbers. You need to add

$(15 - 25)^2$
$(29 - 25)^2$
$(32 - 25)^2$
$(28 - 25)^2$
$(18 - 25)^2$ and
$(28 - 25)^2$

and then divide the sum by $6 - 1 = 5.$

Penny

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.


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# Refinement

From electron density to your finished model

Now that initial atom assignments have been made; the next stage is to refine the structure to see how well your assignments fit the data and improve the model.

## Initial Atom Type Assignment

1. Under the tab click on the arrow next to the Refine button so that it is pointing upwards and highlighted orange. The refinement options are now displayed.

2. Click on the Refine button and Olex2 starts the structure refinement. Alternatively, you can type >>refine or press CTRL+R.

During a refinement Olex2 is calculating structure factors based on your current structural model and comparing them to the experimentally measured values. Adjustments are then made to the atomic parameters to try and improve the fit.

3. If one of the atoms you assigned is now represented as a comparatively huge sphere (not just slightly larger than the others), then it is not real. Select it and press the DELETE key on your keyboard (or right-click and select Delete from the context menu). If you delete an atom by mistake, use CTRL+Z to undo.

4. If there are peaks larger than ~3 units, then there are missing C or O atoms. Select these peaks and assign them to either C or O and refine again. Repeat this step until there are no more large Q-peaks left.

5. At this point, you can hide all remaining Q-peaks: in click on the hide Q icon (or use CTRL+Q) to hide them.

6. Rotate your structure and examine the size of the atom spheres, they should all be approximately the same size. If this is not the case, you have assigned some of the atoms incorrectly. In this image, atoms 3 are correct, with similarly sized spheres. Atom 1 is too light (it should be heavier) and atom 2 is too heavy (it should be lighter).

The atoms are displayed as spheres when they are modelled isotropically (i.e. one parameter to define the atomic displacement parameter). The size of all atom spheres is approximately the same, but there are exceptions to the rule. For example, if there are long floppy chains (e.g. propyl or butyl) in your structure, the size of the spheres increases along the chain to the distal atom due to the increased libration. Atomic disorder can also cause changes in the visual representation of the spheres.

7. If a sphere is significantly smaller than the others, the atom type is too light e.g. assigned as a carbon atom, but it should be an oxygen atom. If it is distinctively larger, it is too heavy. Select the offending spheres and change their atom types. If the sphere representing the atom has become too small to see, draw a rectangle (SHIFT+LEFT MOUSE) around the atom to select it.

If an assigned atom type is too light, there is not enough electron density available to fit the experimental data, so the refinement pulls the electron density into a smaller volume to try and improve the fit to the experimental data. If the assigned atom type is too heavy, it has more electron density than expected and the refinement increases the volume over which the electron density is spread to improve the fit to the experimental data.

8. Carry out further refinement cycles (and adjust atom types) until all atoms are approximately represented by spheres of the same size. If the structure had become unrecognisable, you would need to click Solve to return to the original structure and start the solving and refining processes again.

At this stage, you should have R1 ~ 8–9%. The biggest Q peak should be ~ 1.2.

## Anisotropic Refinement

1. So far, we have limited the shape in which we can place electron density to a sphere. The next stage is to model the atoms as ellipsoids: using anisotropic displacement parameters, ADP, requires 6 parameters to define the volume occupied by the electrons during this anisotropic refinement, rather than the single one that was used to set the sphere during isotropic refinement.

2. Click on the rugby-ball shaped blue icon on at the top right. Olex2 will then automatically carry out an anisotropic refinement.

Atoms are now displayed more like rugby balls. If they appear like pancakes, tetrahedra or are comparatively much larger than other atoms in the structure, you must investigate. Common reasons for odd shaped ellipsoids include incorrect atom assignments, atomic disorder, erroneous space group assignment or poor data quality.

1. Again, the atoms should be roughly of similar size and shape in similar environments. If you have incorrect atom assignments at this stage, Olex2 might display tetrahedra instead of the expected ellipsoids. Such atoms have become non-positive definite, N.P.D, and are in reality heavier (i.e. O not C). Very large ellipsoids indicate the atom type should be lighter (i.e. C not O). In the illustration: atoms 3 are correctly assigned atom types with similar sized spheres. Atom 1 is non-positive definite (atom type is too light), and 2 is a large ellipsoid (atom type is too heavy).

1. The next stage is to add hydrogen atoms to your structure. In click on Add H in the top right-hand corner (or type >>hadd). Olex2 automatically adds hydrogen atoms and includes them in the subsequent refinement.

The hydrogen atoms are automatically added using a riding model with appropriate AFIX instructions. Please refer to page \pageref{hydrogen-atoms} for more details on this.

2. In , clicking on the H square cycles through Show H / Show H with H-bonds / Hide H (alternatively use CTRL+H). Stop at Show H with hydrogen bonds.

Hydrogen bonding interactions are shown as red dashed lines between a hydrogen atom and the acceptor atom.

3. Check that there have been no extra hydrogen atoms generated. If there are any, delete them. Select atoms by clicking on them with the LEFT MOUSE button and then press the DELETE key. Refine the structure.

If you select some hydrogen atoms and use the Delete All H button Sor type >>kill \$H in the command-line, only the selected atoms will be deleted.

4. Repeat hydrogen atom addition as before. This time Olex2 will only add the missing hydrogen atoms. If for any reason, they are still incorrect, you have to add them manually, which is what you also have to do if hydrogen atoms were missing after Add H. Refer to page \pageref{add-hydrogen-atoms} how to do this.

5. The first assignment of hydrogen atom positions is not always fully optimised, particularly for less well-defined groups such as –OH. If there are Q-peaks close to oxygen atoms that are not where the hydrogen is, the hydrogen position may need adjusting. The image shows an incorrectly positioned hydrogen atom 1, where the correct position is at Q-peak 2.

6. To optimise the hydrogen atom positions there are two options: multiple refinement cycles using the Refine button may see the hydrogen atom(s) rotate onto the Q-peak (sometimes updating the weighting scheme, see below, helps the refinement escape the false minimum and to rotate the H-atom(s) into the correct position). Alternatively, select a hydrogen atom and then in choose Fit option from the Select group or atom(s)and then… tool. Once in Fit mode, holding the SHIFT key and LEFT MOUSE button down whilst dragging the relevant hydrogen atom allows you to move the hydrogen atom into the desired position. Press ESC when finished. Repeat if more than one atoms needs to be relocated.

When in the Fit mode it is possible to rotate the structure to see if the relocated hydrogen atom is now on the Q-peak. Remove your finger from the mouse and SHIFT key, hold the LEFT MOUSE button down and move the mouse.

7. Refine the structure.

The refinement engine automatically adjusts the position to maintain the appropriate geometric constraints dictated by the AFIX instruction.

8. Rotate the structure and examine the hydrogen positions to make sure that the structure looks sensible. If required, repeat adjustments to the hydrogen atom positions.

9. Adjust the weighting scheme. Under the refinement options, there is a Weight option. Tick the box next to Auto Update Weights. Click Refine and repeat until the weights are no longer changing.

10. The structure is now solved and refined.

R1 should now be between 4–5%, wR2 around 10% and the largest Q peak about 0.3.


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## Friday, December 14, 2012

### Ratio, Proportions -- Beginning Problem Soving (SAT level)

Questions:
#1: At a county fair, if adults to kids ratio is 2 to 3 and there are 250 people at the fair. How many adults and how many kids?

Solution I :  Adults to Kids = 2 : 3 (given). Let there be 2x adults and 3x kids (only when two numbers have the same multiples can you simplify the two number to relatively prime.
2x + 3x = 250;  5x = 250; x = 50 so there are 2x or 2 * 50 = 100 adults and 3x or 3 * 50 = 150 kids.

Solution II:  From the given information, you know that every time when there are 5 parts (2 + 3 = 5), there
will be 2 parts for A, or $\frac{2}{5}}$  of the total, and 3 parts for B, or $\frac{3}{5}}$ of the total.
$\frac{2}{5}}$ * 250 = 100 adults and $\frac{3}{5}}$ * 250 = 150 kids

#2: In a mixture of peanuts and cashews, the ratio by weight of peanuts to cashews is 5 to 2. How many pounds of cashews will there be in 4 pounds of this mixture? (an actual SAT question)

Solution:
You can use either the first or second method (see above two solutions) but the second solution is much faster.
$\textcolor{red}{\frac{8}{7}}$ pounds

#3: Continue with question #1: How many more kids than adults go to the fair?

Solution I:
Use the method on #1 and then find the difference.
150 - 100 = 50 more kids.

Solution II:
Using the method for #1: solution II, you know  $\frac{2}{5}}$  of the all the people go to the fair are adults ,
and $\frac{3}{5}}$ of the total people go to the fair are kids.
$\frac{3}{5}}$ - $\frac{2}{5}}$  ) * 250 = 50 more kids.

#3: If the girls to boys ratio at an elementary school is 3 to 4 and there are 123 girls, how many boys are there at that elementary school?
Solution:
This one is easy, you set up the equation. Just make sure the numbers line up nicely.
Let there be "x" boys. $\frac{3}{4}$ = $\frac{123}{x}$ . Cross multiply to get x. Or since 123 is 41 * 3;
4 * 41 = 164 boys

#4 : If the girls to boys ratio at an elementary school is 2 to 5 and there are 78 more boys than girls, how many girls are there at the elementary school?

Solution I:
Again, you can use the algebra and let there be 2x girls and 5x boys.
According to the given, 5x - 2x = 3x = 78 so x = 26
Plug in and you get there are 2*26 = 52 girls and 5*26 = 130 boys.

Solution II:
If you keep expanding the ratio using the same multiples, 2 : 5 = 4 : 10 : 6 : 15...
Do you see that the difference of boys and girls are always multiples of 3 ( 5 - 2  = 3).
$\frac{78}{3}$ = 26 so there are 2 * 26 or 52 girls


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## General Chemistry: Principles and Modern Applications (10th Edition)

A saturated aqueous solution of $NaBr$ at 20$^{\circ}$C contains: $53.7g$ $NaBr$ $/ 100g$ $Solution$
1. The mass of the solution, in this is case, is the sum of the NaBr and the water mass. $mass(Solution) = 116g + 100g =216g$ 2. Now we know that there is 116g of NaBr for 216g of solution, and we just need to calculate this for 100 g of solution using a proportion. $\frac{116g}{216g} = \frac{x}{100g}$ $0.537 = \frac{x}{100g}$ $x = 0.537 \times 100g = 53.7g$


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# Proving sinxx=(1−x2π2)(1−x222π2)(1−x232π2)⋯\frac{\sin x}{x} =\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{2^2\pi^2}\right) \left(1-\frac{x^2}{3^2\pi^2}\right)\cdots

How to prove the following product?

Real analysis approach.

Let $\alpha\in(0,1)$, then define on the interval $[-\pi,\pi]$ the function $f(x)=\cos(\alpha x)$ and $2\pi$-periodically extended it the real line. It is straightforward to compute its Fourier series. Since $f$ is $2\pi$-periodic and continuous on $[-\pi,\pi]$, then its Fourier series converges pointwise to $f$ on $[-\pi,\pi]$:

Now take $x=\pi$, then we get

Fix $t\in(0,1)$. Note that for each $\alpha\in(0,t)$ we have $|(\alpha^2-n^2)^{-1}|\leq(n^2-t^2)^{-1}$ and the series $\sum_{n=1}^\infty(n^2-t^2)^{-1}$ is convergent. By Weierstrass $M$-test the series in the right hand side of $(2)$ is uniformly convergent for $\alpha\in(0,t)$. Hence we can integrate $(2)$ over the interval $[0,t]$. And we get

Finally, substitute $x=\pi t$, to obtain

Complex analysis approach

We will need the following theorem (due to Weierstrass).

Let $f$ be an entire function with infinite number of zeros $\{a_n:n\in\mathbb{N}\}$. Assume that $a_0=0$ is zero of order $r$ and $\lim\limits_{n\to\infty}a_n=\infty$, then

for some entire function $h$ and sequence of positive integers $\{p_n:n\in\mathbb{N}\}$. The sequence $\{p_n:n\in\mathbb{N}\}$ can be chosen arbitrary with only one requirement $-$ the series
is uniformly convergent on each compact $K\subset\mathbb{C}$.

Now we apply this theorem to the entire function $\sin z$. In this case we have $a_n=\pi n$ and $r=1$. Since the series

is uniformly convergent on each compact $K\subset \mathbb{C}$, then we may choose $p_n=1$. In this case we have

Let $K\subset\mathbb{C}$ be a compact which doesn’t contain zeros of $\sin z$. For all $z\in K$ we have

It is known that (here you can find the proof)

hence $h'(z)=0$ for all $z\in K$. Since $K$ is arbitrary then $h(z)=\mathrm{const}$. This means that

Since $\lim\limits_{z\to 0}z^{-1}\sin z=1$, then $C=1$. Finally,

This result is much more stronger because it holds for all complex numbers. But in this proof I cheated because series representation for $\cot z$ given above require additional efforts and use of Mittag-Leffler’s theorem.


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Address 5615 Beech Rd, Harbor Springs, MI 49740 (231) 242-4747

# error propagation capacitance Levering, Michigan

The question is to find the uncertainties in frequency by taking account of inductor being 5% accurate & capacitor being 8% accurate. 2. Log in with Facebook Log in with Twitter Your name or email address: Do you already have an account? The question is to find the uncertainties in frequency by taking account of inductor being 5% accurate & capacitor being 8% accurate. Fig. 2.

Would I just use the "2" ? reduce() in Java8 Stream API QED symbol after statements without proof Number of polynomials of degree less than 4 satisfying 5 points A piece of music that is almost identical to A link to the app was sent to your phone. Comment Stuck?

Express it in two forms, one for absolute error and one for relative (fractional) error. Would 50 have demonstrated the laws better? Yes it is indeed an LC circuit. share|cite|improve this answer answered Mar 30 '15 at 13:58 Rob Jeffries 40.3k368149 Yes, I also remarked that a correct formula includes correlations, so just the formula itself does not

units. You can do: $$\sqrt{x} \sqrt{x} = x^{\frac{1}{2}} x^{\frac{1}{2}} = x^{\frac{1}{2} + \frac{1}{2}} = x^1 = x$$ You can't do: $$\sqrt{x} \sqrt{y} = x^{\frac{1}{2}} y^{\frac{1}{2}} = xy^{\frac{1}{2} + \frac{1}{2}} = xy^1 = And I meant to have the " ]^2 " part at the very end to be INSIDE the square root. Aristotle, May 3, 2015 May 3, 2015 #9 Zondrina Homework Helper Aristotle said: ↑ Are you sure that you can just pull the sqrt C out like that? Make all the statements true How would you say "x says hi" in Japanese? You also need to plug in ##\sigma_C = 2## and ##\sigma_L = ?##. C1 and C2 are in parallel and added you should add the absolute uncertainties for those. Aristotle, May 3, 2015 May 3, 2015 #8 Aristotle When I did ∂F/∂L, I get: -[2π (LC)^1/2 ]^(-2) * ∂/∂L[(2π (LC)^(1/2)] = -[(2π(LC)^1/2 ]^(-2) * (π (LC)^(-1/2) * C ] **the The partial derivatives are also wrong. Discuss. (4) If you could have measured all of the resistors the factory manufactured, you could make some very definite statements about the average value and the distribution of the measured Consider one 20% resistor of size 20k and one 10% resistor of size 10k connected in series. We won’t spam you—promise. The percent uncertainty of C3 is 1% but the percent uncertainties of those are 10%. Is it Gaussian? However I thought if you multiplied two integers/variables, you add their exponents? I really appreciate your help Zondrina!! how to calculate the error propagation of dielectric constant K = Capacitance of dielelctric / Capacitance of air . What exactly is "absolute accuracy" and after that what would I have to do? But C3 in in series and there is multiplication and division involved so you need to use Percent Uncertainties. If you have a large bin of resistors, all manufactured to have the "same" resistance, they will, if measured precisely enough, have different resistance values. Was this enough to demonstrate the laws of error propagation? Voransicht des Buches » Was andere dazu sagen-Rezension schreibenEs wurden keine Rezensionen gefunden.Ausgewählte SeitenSeite 10TitelseiteInhaltsverzeichnisIndexVerweiseInhaltPart I Mathematical Methods1 Part II Computational Electromagnetics79 Part III Coupled Problems193 Part IV Circuit and Device What is the error limit of this result? Carbon resistors, used in all kinds of electrical and electronic circuits, are a good example. units for calculations, and omitting the units when you do the actual calculation. Aristotle, May 3, 2015 May 3, 2015 #14 Zondrina Homework Helper ##\sigma_L## and ##\sigma_C## represent the errors associated with measuring the inductance and capacitance of each component. Yes. How is the Heartbleed exploit even possible? asked 1 year ago viewed 5069 times active 5 months ago Get the weekly newsletter! Resistors connected in parallel. (2) The physical law for the effective resistance of resistors combined in parallel is: [2] 1 1 1 1 — = — + — + — + The results are not this simple when the two resistors have different uncertainty values. No, create an account now. We will not ask you to calculate or use standard deviations, though serious students might want to do this for practice. What is the error limit of this result? Comment Comments How do you figure the uncertainty is denominated by C3? Let us see if 2.7% is possible... –jaromrax Mar 30 '15 at 11:18 So, on my opinion, the result 1.78\Omega is not quite correct, I would rather prefer 1.40\Omega. So the absolute errors we know is that inductor is 5 percent and capacitor is 8 percent...& know that C= 25 F and L=L=1.12x10^(-25) H. Answer: 0.75k, which is 10% of 7.5k. (5) Were you surprised at these results? For example, we know if we measure the inductance, we may write it as:$$L = (1.12 \times 10^{-25} \pm \sigma_L) \space H$$If we measure the capacitance we may write The equivalent capacitance of a small capacitor in series with a large one is nearly that of the small one. The attempt at a solution Well taking the partial derivatives of f respect to L, I get: C sqrt(LC) / 2pi. Yes, my password is: Forgot your password? For example, we know if we measure the inductance, we may write it as:$$L = (1.12 \times 10^{-25} \pm \sigma_L) \space H If we measure the capacitance we may write Thank you!

Browse hundreds of Physics tutors. Please try the request again. Use the ohmmeter to measure the resistance of each pair of two connected in series (5 measurements). The answer that comes out is in ##Hz##, so make sure to convert to ##GHz##.

This is not an accident. Over 6 million trees planted Forums Search Forums Recent Posts Unanswered Threads Videos Search Media New Media Members Notable Members Current Visitors Recent Activity New Profile Posts Insights Search Log in Still you can draw some conclusions about the parent distribution on the basis of your sample. Its two probes are then connected to the lead wires of the resistor (or combination of resistors) and the resistance value appears on the digital display.

So, is the percentage error method wrong? The system returned: (22) Invalid argument The remote host or network may be down.


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# How to calculate species abundance and diversity

1. ### blicker

19
1. The problem statement, all variables and given/known data
This is the sample population taken along a 5m transect:
ants:9
grasshoppers:1
pill bugs:4
spiders:1

2. Relevant equations
What formulas would i use to calculate the species abundance and diversity?

3. The attempt at a solution
I only know this formula H'=-$\Sigma$ (n$_{}i$/N)xIn(n$_{}i$/N) and i dont know what to do with this formula.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. ### I like Serena

6,181
Hi blicker!

Can it be that your formula is actually:

where "ln" is the natural logarithm (and also a button on your calculator), and where ni is the number of individuals of species i?

See: http://en.wikipedia.org/wiki/Shannon_index

3. ### blicker

19
yes thats it! but i dont know what numbers go where or how to calculate it. like would i include the sigma symbol and the letters above and below it?

4. ### I like Serena

6,181
The sigma symbol indicates that you have to sum.
So:
$$H'=-\sum_{i=1}^S {n_i \over N} \ln {n_i \over N} = -({n_1 \over N} \ln {n_1 \over N} + {n_2 \over N} \ln {n_2 \over N} + {n_3 \over N} \ln {n_3 \over N} + {n_4 \over N} \ln {n_4 \over N})$$

From wiki:
* ni The number of individuals in species i; the abundance of species i.
* S The number of species. Also called species richness.
* N The total number of all individuals
* pi The relative abundance of each species, calculated as the proportion of individuals of a given species to the total number of individuals in the community: $n_i\over N$

5. ### blicker

19
so would the equation for ants look like this:
-(9/15)In(9/15)

6. ### I like Serena

6,181
Well, Shannon's index specifies to sum all the terms.
So what you mention is only part of H'.

Btw, it is $ln$ ("logarithmus naturalis").

But I can't really tell you what your problem asks.
I'm not studying whatever science you're studying - I'm a mathematician.

I did find this article about diversity: http://en.wikipedia.org/wiki/Diversity_Index
And apparently $n_i$ is the "species abundance".
Furthermore $p_i={n_i \over N}$ is the "relative species abundance".

7. ### blicker

19
well thanks for your help anyway.

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text:
# complementary subspace

## Primary tabs

Defines:
complementary, direct sum, decomposition, orthogonal complement
Synonym:
algebraic complement
Type of Math Object:
Definition
Major Section:
Reference
Groups audience:

## Mathematics Subject Classification

Regarding the remark following Proposition1: I don’t believe that, given $U$, subspace of $V$, that $U$ has a unique complement in $V$. E.g. any two lines in $\mathbb{R}^{2}$ are complementary. Did I misunderstand something here?


text:
## SPHERICAL MIRRORS ----  CURVED SURFACES  - KEY CONCEPTS

Let us look at the below diagram, reflection at plane surfaces i.e. at plane mirror. All the laws of reflection at plane surfaces are valid and applicable to spherical mirrors also.

### SPHERICAL MIRROR ?

The mirrors which have curved surface are known as Spherical mirrors, these mirrors are part of a hollow glass sphere.

## CONVEX MIRROR AND ITS TERMS

The spherical mirror is said to be convex if the bulged surface is made smooth and the inner surface is silvered.

## CONCAVE MIRROR AND ITS TERMS

The spherical mirror is said to be concave if the bulged surface is silvered and the inner surface is made smooth.

SPHERICAL MIRRORS AND TERMS RELATED :

APERTURE: The surface of the mirror from which the reflection take place is called its aperture.

POLE: The central point on the surface of the mirror is called its pole (P)

CENTRE OF CURVATURE:

The centre of the sphere of which the spherical mirror is a part is called the centre of curvature of the mirror.

The radius of the sphere from which the spherical mirror is a part is called the radius of curvature of the mirror.

PRINCIPAL AXIS:

The line joining the pole and the centre of curvature is called the principal axis of the mirror.

FOCUS: (F)

The point on the principal axis where rays incident parallel to the principal axis converge to reappear or to diverge from after reflection is called the Focus of the spherical mirror.

FOCAL LENGTH (f):

The distance of the Focus from the pole is called the focal length (f) of the spherical mirror.

### REAL AND VIRTUAL IMAGES:

An image formed by the actual intersection of light rays is called a real image. A real image is formed on the screen.

If the rays of light responsible for an image do not actually intersect, the image is called virtual image.

Virtual image cannot be formed on the screen, as no light reaches where the image appears to form.

Observe the following ray diagram when a parallel light incidents on a concave mirror and understand the reflection concepts........

IMAGES FORMED BY A CONVEX MIRROR:

Irrespective of the position of the object , The image formed by the convex mirror is always virtual and erect but when we compare the the size of the image, the image size  is always diminished.

PRINCIPAL FOCUS OF A CONVEX MIRROR
The principal focus of a convex mirror, which is divergent mirror,  is a point on the principal axis of the mirror, at which all the incident rays parallel to the principal axis appears to diverge after reflection from the mirror.

The principal focus of the convex mirror is virtual and always lies behind the convex mirror.

#### observe the above ray diagram when parallel rays incident on both concave mirror and convex mirror and observe the principal focus.

from the above ray diagrams, write down the characteristics of the image formed.

## MIRROR FORMULA:

1/v + 1/u = 1/f

From the mirror formula, if we observe, we can understand that how the object distance,u, and image distance,v, are related to the focal length.  The distances are measured from the pole only. Pole to object is u, and pole to image is v,

MAGNIFICATION BY SPHERICAL MIRRORS:
Generally, we can understand that the magnification is the increase in the image size with respect to the size of the object which is produced by spherical mirrors.  The magnification is calculated by taking the ratio of height of the image to the height of the object ---- hi/ho.  It can also be calculated by the negative ratio of the image distance to the object distance.

What can we understand from the magnification ?

*** when 'm' value is positive it tells us image is virtual.
*** when 'm' value is negative it tells us image is real.

logical question combining mirror formula and magnification

FIND OUT THE OBJECT DISTANCE (u), when the magnification produced is 'm' and focal length is 'f'.

solution:

mirror formula is :  1/f = 1/v + 1/u

magnification is m = -v/u   so  v=-mu

substituting the 'm' value in formula:

1/f = 1/-mu   + 1/u

1/f = 1/u(1/-m + 1/u)

simplifying,   u = f/m(m-1)

## critical thinking problems:

1. Suppose that you are given a concave mirror of focal length 15 cm.  Then,

#### a) To get the magnification m=1, where the object is needed to be placedb)  When the object is placed at 30 cm, write the characteristics of the image formed.c)  Your friend told that the image formed at infinity, where would be the object is placed.

2. When a student is performing experiment with concave mirror of focal length 20 cm, then he observed that the size of the image is reduced to 1/3rd of its size of the object.  Calculate the object distance in this situation.


text:
## Data

• Title: $$\mathrm{SL}_{2}$$ tilting modules in the mixed case
• Authors: Louise Sutton, Daniel Tubbenhauer, Paul Wedrich and Jieru Zhu
• Status: Preprint. Last update: Mon, 17 May 2021 10:52:03 UTC
• LaTex Beamer presentation: Slides, Slides2, Slides3, Slides4

## Abstract

Using the non-semisimple Temperley-Lieb calculus, we study the additive and monoidal structure of the category of tilting modules for $$\mathrm{SL}_{2}$$ in the mixed case. This simultaneously generalizes the semisimple situation, the case of the complex quantum group at a root of unity, and the algebraic group case in positive characteristic. We describe character formulas and give a presentation of the category of tilting modules as an additive category via a quiver with relations. Turning to the monoidal structure, we describe fusion rules and obtain an explicit recursive description of the appropriate analog of Jones-Wenzl projectors. We also discuss certain theta values, the tensor ideals, mixed Verlinde quotients and the non-degeneracy of the braiding.

## A few extra words

In this paper we study categories of representations of $$\mathrm{SL}_{2}(\mathbb{K})$$, or its associated quantum group quantized by using an invertible variable q, in the so-called mixed case or mixed characteristic. This can be thought of as the culmination of the following cases:

• The semisimple case where $$\mathbb{K}$$ is for example the field of complex numbers.
• The complex quantum group case (at a root of unity) where $$\mathbb{K}$$ is a field containing the complex numbers, one works with the quantum group and q is a root of unity in $$\mathbb{K}$$.
• Characteristic p where $$\mathbb{K}$$ is for example an algebraically closed field of prime characteristic.
• The “everything else” case, where one works with the quantum group, q is a root of unity in $$\mathbb{K}$$, and $$\mathbb{K}$$ is of prime characteristic.
The latter is an exciting case because of it fractal-type representation theory, a property that it shares with the prime characteristic case, and because it has a non-trivial braiding, a property that it shares with the complex quantum group case. The first aim of this paper is to give an overview of the state of the arts regarding $$\mathrm{SL}_{2}(\mathbb{K})$$ in the mixed case, and, at the same time, to put an eye on the categorical properties of the representation categories. In particular, we recall/deduce some character formulas and generalize the main result of click to mixed characteristic. Our second aim, and maybe the real meat, is to study the representation categories associated to $$\mathrm{SL}_{2}(\mathbb{K})$$ as (braided) monoidal categories, namely the fusion rules. Classically, fusion rules express the structure constants for the representation ring such into indecomposable modules. Here is an example of such a fusion graph (the fusion graph of tensoring with the vector representation)
On the categorified level one is interested in explicitly describing the projection and inclusion maps realizing such decompositions. In the Temperley-Lieb context this means decomposing the tensor products into idempotents that project onto the indecomposable summands predicted by the fusion rule. A famous example is the recursion for the classical Jones--Wenzl projectors
We establish decompositions analogous to this recursion in the mixed setting. As an example, we show an instance going beyond the Jones-Wenzl recursion with summands that need not be simple (here we use colored boxes to encode projectors corresponding to modules that need not be simple):
We also study analogs of Kauffman-Lins type formulas, tensor ideals, Verlinde quotients, modified traces and the Müger center.