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# Dendrite (metal)

A silver crystal, electrolytically refined with visible dendritic structures
A pure copper crystal with dendritic structure, electrolytic made.
Dendritic crystallization after melting inside sealed ampules of rubidium and caesium metal

A dendrite in metallurgy is a characteristic tree-like structure of crystals growing as molten metal solidifies, the shape produced by faster growth along energetically favourable crystallographic directions. This dendritic growth has large consequences in regard to material properties.

Dendrites form in unary (one-component) systems as well as multi-component systems. The requirement is that the liquid (the molten material) be undercooled, aka supercooled, below the freezing point of the solid. Initially, a spherical solid nucleus grows in the undercooled melt. As the sphere grows, the spherical morphology becomes unstable and its shape becomes perturbed. The solid shape begins to express the preferred growth directions of the crystal. This growth direction may be due to anisotropy in the surface energy of the solid–liquid interface, or to the ease of attachment of atoms to the interface on different crystallographic planes, or both (for an example of the latter, see hopper crystal). In metallic systems, interface attachment kinetics is usually negligible (for non-negligible cases, see dendrite (crystal)). In metallic systems, the solid then attempts to minimize the area of those surfaces with the highest surface energy. The dendrite thus exhibits a sharper and sharper tip as it grows. If the anisotropy is large enough, the dendrite may present a faceted morphology. The microstructural length scale is determined by the interplay or balance between the surface energy and the temperature gradient (which drives the heat/solute diffusion) in the liquid at the interface.[1]

As solidification proceeds, an increasing number of atoms lose their kinetic energy, making the process exothermic. For a pure material, latent heat is released at the solid–liquid interface so that the temperature remains constant until the melt has completely solidified. The growth rate of the resultant crystalline substance will depend on how fast this latent heat can be conducted away. A dendrite growing in an undercooled melt can be approximated as a parabolic needle-like crystal that grows in a shape-preserving manner at constant velocity. Nucleation and growth determine the grain size in equiaxed solidification while the competition between adjacent dendrites decides the primary spacing in columnar growth. Generally, if the melt is cooled slowly, nucleation of new crystals will be less than at large undercooling. The dendritic growth will result in dendrites of a large size. Conversely, a rapid cooling cycle with a large undercooling will increase the number of nuclei and thus reduce the size of the resulting dendrites (and often lead to small grains).

Smaller dendrites generally lead to higher ductility of the product. One application where dendritic growth and resulting material properties can be seen is the process of welding. The dendrites are also common in cast products, where they may become visible by etching of a polished specimen.

As dendrites develop further into the liquid metal, they get hotter because they continue to extract heat. If they get too hot, they will remelt. This remelting of the dendrites is called recalescence. Dendrites usually form under non-equilibrium conditions.

An application of dendritic growth in directional solidification is gas turbine engine blades which are used at high temperatures and must handle high stresses along the major axes. At high temperatures, grain boundaries are weaker than grains. In order to minimize the effect on properties, grain boundaries are aligned parallel to the dendrites. The first alloy used in this application was a nickel-based alloy (MAR M-200) with 12.5% tungsten, which accumulated in the dendrites during solidification. This resulted in blades with high strength and creep resistance extending along the length of the casting, giving improved properties compared to the traditionally-cast equivalent.[2]

## Computational Modeling

Phase-field simulation of dendritic solidification of a pure material using the model developed by Kobayashi with six-fold anisotropy. The white region represents solid ${\displaystyle (\phi =1)}$ and the blue region represents liquid ${\displaystyle (\phi =0)}$.

The first computational model of dendritic solidification was published by Kobayashi,[3] who used a phase-field model to solve two coupled partial differential equations describing the evolution of the phase-field, ${\displaystyle \phi }$ (with ${\displaystyle \phi =0}$ in the liquid phase and ${\displaystyle \phi =1}$ in the solid phase), and the temperature field, ${\displaystyle T}$, for a pure material in two dimensions:

${\displaystyle \tau {\frac {\partial \phi }{\partial t}}=-{\frac {\partial }{\partial x}}\left(\epsilon {\frac {\partial \epsilon }{\partial \theta }}{\frac {\partial \phi }{\partial y}}\right)+{\frac {\partial }{\partial y}}\left(\epsilon {\frac {\partial \epsilon }{\partial \theta }}{\frac {\partial \phi }{\partial x}}\right)+\nabla \cdot \left(\epsilon ^{2}\nabla \phi \right)+\phi (1-\phi )\left(\phi -{\frac {1}{2}}+m+a\chi \right)}$

which is an Allen-Cahn equation with an anisotropic gradient energy coefficient:

${\displaystyle \epsilon (\theta )={\bar {\epsilon }}\left[1+\delta \cos(j\theta )\right]}$

where ${\displaystyle {\bar {\epsilon }}}$ is an average value of ${\displaystyle \epsilon }$, ${\displaystyle \theta }$ is the angle between the interface normal and the x-axis, and ${\displaystyle \delta }$ and ${\displaystyle j}$ are constants representing the strength and mode of anisotropy, respectively.

The parameter ${\displaystyle m}$ describes the thermodynamic driving force for solidification, which Kobayashi defines for a supercooled melt as:

${\displaystyle m(T)={\frac {\alpha }{\pi }}\tan ^{-1}\left[\gamma (T_{e}-T)\right]}$

where ${\displaystyle \alpha }$ is a constant between 0 and 1, ${\displaystyle \gamma }$ is a positive constant, and ${\displaystyle T_{e}}$ is the dimensionless equilibrium temperature. The temperature has been non-dimensionalized such that the equilibrium temperature is ${\displaystyle T_{e}=1}$ and the initial temperature of the undercooled melt is ${\displaystyle T=0}$.

The evolution equation for the temperature field is given by

${\displaystyle {\frac {\partial T}{\partial t}}=\nabla ^{2}T+K{\frac {\partial \phi }{\partial t}}}$

and is simply the heat equation with a source term due to the evolution of latent heat upon solidification, where ${\displaystyle K}$ is a constant representing the latent heat normalized by the strength of the cooling.

When this system is numerically evolved, random noise representing thermal fluctuations is introduced to the interface via the ${\displaystyle a\chi }$ term, where ${\displaystyle a}$ is the magnitude of the noise and ${\displaystyle \chi }$ is a random number distributed uniformly on ${\displaystyle [-0.5,0.5]}$.


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### Particle, Astrophysics, and Nuclear Physics Seminar

Monday, October 14, 2019
2:00pm – 3:00pm

Storrs Campus
GS-413.E

Dr. Masakaki Tomii, Department of Physics, University of Connecticut Non-perturbative matching of Wilson coefficients from four-flavor to three-flavor theories using lattice QCD

The first lattice result for direct CP violation in $$K \to \pi\pi$$ decay in the Standard Model published by RBC/UKQCD in 2015 is desired to be more precise to see the deviation from the experimental value more clearly. In this talk, I will overview RBC/UKQCD’s calculation of $$K \to \pi\pi$$ matrix elements mentioning a particular error source: Since RBC/UKQCD calculated the matrix elements with the four-quark operators without the charm quark, we need the corresponding Wilson coefficients in the three-flavor theory, which can be obtained by converting from the four-flavor theory. Perturbative matching may result in a significant uncertainty since it needs to be done below charm threshold. To resolve this problem, we are trying to do this matching in a non-perturbative prescription using lattice QCD. I will explain the strategy of the matching and show some results of an exploratory calculation.

Contact:

Prof. L. Jin

Physics Department (primary), College of Liberal Arts and Sciences, UConn Master Calendar

format: plain


text:
# [Tex/LaTex] How to install titlesec and secsty packages?

installingmiktexpackages

Possible Duplicate:
How do I update my TeX distribution?
How can I change the package repository?

Please help me to install packages. When I install the options for random repository package is not shown.


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## July 20, 2014

### The Place of Diversity in Pure Mathematics

#### Posted by Tom Leinster

Nope, this isn’t about gender or social balance in math departments, important as those are. On Friday, Glasgow’s interdisciplinary Boyd Orr Centre for Population and Ecosystem Health — named after the whirlwind of Nobel-Peace-Prize-winning scientific energy that was John Boyd Orr — held a day of conference on diversity in multiple biological senses, from the large scale of rainforest ecosystems right down to the microscopic scale of pathogens in your blood.

I used my talk (slides here) to argue that the concept of diversity is fundamentally a mathematical one, and that, moreover, it is closely related to core mathematical quantities that have been studied continuously since the time of Euclid.

In a sense, there’s nothing new here: I’ve probably written about all the mathematical content at least once before on this blog. But in another sense, it was a really new talk. I had to think very hard about how to present this material for a mixed group of ecologists, botanists, epidemiologists, mathematical modellers, and so on, all of whom are active professional scientists but some of whom haven’t studied mathematics since high school. That’s why I began the talk with an explanation of how pure mathematics looks these days.

I presented two pieces of evidence that diversity is intimately connected to ancient, fundamental mathematical concepts.

The first piece of evidence is a connection at one remove, and schematically looks like this:

maximum diversity $\leftrightarrow$ magnitude $\leftrightarrow$ intrinsic volumes

The left leg is a theorem asserting that when you have a collection of species and some notion of inter-species distance (e.g. genetic distance), the maximum diversity over all possible abundance distributions is closely related to the magnitude of the metric space that the species form.

The right leg is a conjecture by Simon Willerton and me. It states that for convex subsets of $\mathbb{R}^n$, magnitude is closely related to perimeter, volume, surface area, and so on. When I mentioned “quantities that have been studied continuously since the time of Euclid”, that’s what I had in mind. The full-strength conjecture requires you to know about “intrinsic volumes”, which are the higher-dimensional versions of these quantities. But the 2-dimensional conjecture is very elementary, and described here.

The second piece of evidence was a very brief account of a theorem of Mark Meckes, concerning fractional dimension of subsets $X$ of $\mathbb{R}^n$ (slide 15, and Corollary 7.4 here). One of the standard notions of fractional dimension is Minkowski dimension (also known by other names such as Kolmogorov or box-counting dimension). On the other hand, the rate of growth of the magnitude function $t \mapsto \left| t X \right|$ is also a decent notion of dimension. Mark showed that they are, in fact, the same. Thus, for any compact $X \subseteq \mathbb{R}^n$ with a well-defined Minkowski dimension $dim X$, there are positive constants $c$ and $C$ such that

$c t^{dim X} \leq \left| t X \right| \leq C t^{dim X}$

for all $t \gg 0$.

One remarkable feature of the proof is that it makes essential use of the concept of maximum diversity, where diversity is measured in precisely the way that Christina Cobbold and I came up with for use in ecology.

So, work on diversity has already got to the stage where application-driven problems are enabling advances in pure mathematics. This is a familiar dynamic in older fields of application such as physics, but I think the fact that this is already happening in the relatively new field of diversity theory is a promising sign. It suggests that aside from all the applications, the mathematics of diversity has a lot to give pure mathematics itself.

Next April, John Baez and friends are running a three-day investigative workshop on Entropy and information in biological systems at the National Institute for Mathematical and Biological Synthesis in Knoxville, Tennessee. I hope this will provide a good opportunity for deepening our understanding of the interplay between mathematics and diversity (which is closely related to entropy and information). If you’re interested in coming, you can apply online.

Posted at July 20, 2014 2:10 PM UTC

TrackBack URL for this Entry:   https://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/2755

### Re: The Place of Diversity in Pure Mathematics

Like John, I had the urge to rewrite my blog entry. I think it’s better now.

Posted by: Tom Leinster on July 20, 2014 11:14 PM | Permalink | Reply to this

### Re: The Place of Diversity in Pure Mathematics

Thanks for mentioning that Banff workshop on information and entropy in biological systems!

Anyone who wants more details can check out my blog article on Azimuth. With a little luck Tobias Fritz, Tom Leinster and Christina Cobbold will be there, as well as other people whose names aren’t so familiar on the n-Café: biologists and information theorists.

Posted by: John Baez on July 21, 2014 7:45 AM | Permalink | Reply to this

### Re: The Place of Diversity in Pure Mathematics

Did you really mean Banff, or were you mixing up the Knoxville workshop with this one?

Posted by: Mark Meckes on July 21, 2014 8:06 AM | Permalink | Reply to this

### Re: The Place of Diversity in Pure Mathematics

It was a nice talk. For those of you who are slightly worried about Tom’s use of genus rather than Euler characteristic on slide 6, he did say, probably just for my benefit, that he really meant Euler characteristic, but that was harder to draw and there was probably only two people in the room who appreciated the difference :-)

I think that the generic biologist finds our desire for precision and process of abstraction very foreign. Afterwards, at dinner, the biologist without a maths degree that I was with said at one point to me “You speak a lot like Tom.” He was referring to my use of language, perhaps one might say, to my pedantry in choice and interpretation of words.

Posted by: Simon Willerton on July 21, 2014 6:15 PM | Permalink | Reply to this

### Diversity in Pure Mathematics

Interesting work. This metatheorem also applied to dimensionality problems in Physics, Complex Systems and Logical and Physical Power Systems.

Posted by: Jonah Lissner on August 20, 2014 6:47 PM | Permalink | Reply to this

### Re: The Place of Diversity in Pure Mathematics

Tom, have you looked into notions of diversity in finance (eg the notion of a diversified portfolio)? I wonder whether mathematicians can learn anything about diversity from financiers/finance.

Posted by: trent on August 21, 2014 10:50 PM | Permalink | Reply to this

### Re: The Place of Diversity in Pure Mathematics

Someone mentioned that to me before (maybe you?), but I can’t say I’ve looked into it properly. On the other hand, I have learned some things about diversity from economists.

Posted by: Tom Leinster on August 23, 2014 3:22 PM | Permalink | Reply to this

### Re: The Place of Diversity in Pure Mathematics

Wasn’t me. Curious to hear what you’ve learned from economists though.

Posted by: trent on August 23, 2014 9:43 PM | Permalink | Reply to this

Post a New Comment


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# Risk Neutral Evaluation - Exchange/Spread Options

I have two assets, $S_1$ and $S_2$, which follow geometric Brownian motion processes. This implies that both $S_1$ and $S_2$ have a lognormal distribution.

I'm trying to get the exchange option price formula through the risk neutral valuation, or, in other words $$C(t,S_1,S_2)=e^{-r(T-t)}E(\max\{S_1-S_2,0\})$$.

How do I calculate the expected value $E(\max\{S_1-S_2,0\})$??

-

Your formula, as it stands, is incorrect, at least is if $E$ means the "expected value under real-world probabilities".

I wrote a blog post explaining the basic rationale behind risk-neutral pricing where you will see that if the Fundamental Theorem of Asset Pricing theorem holds, you can write:

Let $X_t=S_{1,t}-S_{2,t}$

$$e^{-rt} X_t = \mathbb{E}_\mathbb{Q}[e^{-rT} \max(X_T,0) | \mathcal{F}_t]$$

where $\mathbb{E}_\mathbb{Q}$ stands for "expectation under the risk-neutral probability".

From there, what you have to do is to define the risk-neutral dynamics of $S_1$ and $S_2$. When you say that the assets are following a lognormal distribution, I'll assume that they follow are independent Geometric Brownian Motion. In this case, their risk neutral dynamics are well know (see here):

$$dS_{1,t}= r S_{1,t} dt + \sigma_1 S_{1,t} dW^{\mathbb{Q}}_{1,t}$$ $$dS_{2,t}= r S_{2,t} dt + \sigma_2 S_{2,t} dW^{\mathbb{Q}}_{2,t}$$

where is the risk-free rate, and $dW^{\mathbb{Q}}_{i,t}=dW_{i,t}+\frac{\mu_i-r}{\sigma_i}dt$.

So, you have:

$$dX_t=dS_{1,t}-dS_{2,t}=r (S_{1,t} -S_{2,t})dt + + \sigma_1 S_{1,t} dW^{\mathbb{Q}}_{1,t} - dS_{2,t}= r S_{2,t} dt - \sigma_2 S_{2,t} dW^{\mathbb{Q}}_{2,t}$$

and:

$$X_t = e^{-r(T-t)} \mathbb{E}_\mathbb{Q}[\max(X_T,0) | \mathcal{F}_t]= e^{-r(T-t)} \mathbb{E}_\mathbb{Q}[\mathbb{1}_{X_T>0}X_T | \mathcal{F}_t]$$

Then you carry on manipulation the equation, I would suggest changing the numeraire...

-

Let's generalize to the slightly more common case of an option on a spread of correlated assets $S_{1,2}$ with correlation coefficient $\rho$. That is to say, write $s = S_1-S_2$, and you want

$$e^{-rT}\mathbb{E}\left[ (s_T-K)^+ \right]$$

If $S_{1,2}$ follow geometric brownian motions (GBM) then the terminal distribution for $s_T$ is not in any simple form. Let me note, however, that it is common among professionals to simply assume $s$ follows an ordinary brownian motion with first two moments that match the corresponding moments of the "true" distribution, in which case the terminal distribution is a gaussian, and we can derive the option value

$$V=e^{-rT}(S_2-S_1) N\left( \frac{e^{-rT}(S_2-S_1)-K e^{-rT}}{v} \right) + v \cdot n\left( \frac{e^{-rT}(S_2-S_1)-K e^{-rT}}{v} \right)$$

where

$$v := e^{rT}\sqrt{ S_1^2(e^{\sigma_1^2 T}-1)-2S_1S_2(e^{\rho \sigma_1\sigma_2 T}-1) + S_2^2 (e^{\sigma_2^2 T}-1)}$$

To get an "exact" solution, you can instead observe that we write this as

$$\frac{dS_1}{S_1} = r dt + \sigma_1 dW^{(1)} \\ \frac{dS_2}{S_2} = r dt + \sigma_2 \left( \rho dW^{(1)} + \sqrt{1-\rho^2} dW^{(2)}\right)$$

in which case the expectation integral becomes

$$e^{-rT}\int_{z_1=-\infty}^{\infty} \int_{z_2=-\infty}^{\infty} (S_1(z_1)-S_2(z_1,z_2)-K)^+ n(z_1) n(z_2) dz_1 dz_2$$

which we can change to

$$e^{-rT}\int_{z_1=S_1^{-1}(K)}^{\infty} P( S_2^0, S_1(z_1)-K, r^\prime(z_1), \sigma^\prime(z_1) ) n(z_1) dz_1$$

for adjusted drift $r^\prime(z_1)$ and adjusted vol $\sigma^\prime(z_1)$ in the Black-Scholes put option pricing formula $P(\cdot)$. Here for clarity $S_2^0$ indicates the initial value of $S_2$, while $S_{1,2}(\cdot)$ indicate the usual terminal value functions for GBMs using the customary volatilities and drifts.

This integral is one-dimensional and can be efficiently evaluated using numerical quadrature, such as a trapezoid rule scheme.

-

The solution can be found in, say, Bjork's book. Assume that the two Brownian motions are uncorrelated. I'm also swapping the symbols $S_{1,T}$ and $S_{2,T}$ in the payoff function.

We use the stock $S_1$ as numeraire. Note that $$\frac{V_t}{S_{1,t}} = \mathbb{E}^{Q^1} \left[ \frac{S_{1,T} \max \left(S_{2,T}/S_{1,T} - 1 \right)}{S_{1,T}} \middle\vert \mathcal{F}_t \right] = \mathbb{E}^{Q^1} \left[ \max \left(\frac{S_{2,T}}{S_{1,T}} - 1 \right) \middle\vert \mathcal{F}_t \right] \; .$$ Let $Z_t \equiv S_{2,t}/S_{1,t}$. Since $Z_t$ is a martingale under the $Q^1$ measure by definition, $$dZ_t = Z_t (\sigma_2 - \sigma_1) \; dW^{Q^1} \; .$$ We are essentially pricing a European call with strike $K=1$ for a stock $Z_t$ that has no drift but has volatility $\sigma = | \sigma_2 - \sigma_1 |$. The price is $$V_t = S_{1,t} \left\{ Z_t N(d_1) - N(d_2) \right\} = S_{2,t} N(d_1) - S_{1,t} N(d_2) \; ,$$ where $d_1 = \frac{\log Z_t + \frac{1}{2} \sigma^2 (T-t)} {\sigma \sqrt{T-t}}$ and $d_2 = d_1 - \sigma \sqrt{T-t}$.

See also Margrabe's formula [ http://en.wikipedia.org/wiki/Margrabe's_formula ].

-


text:
org.apache.commons.math3.analysis.interpolation

Interface MultivariateInterpolator

• All Known Implementing Classes:
MicrosphereInterpolator

public interface MultivariateInterpolator
Interface representing a univariate real interpolating function.
Since:
2.1
• Method Summary

Methods
Modifier and Type Method and Description
MultivariateFunction interpolate(double[][] xval, double[] yval)
Computes an interpolating function for the data set.
• Method Detail

• interpolate

MultivariateFunction interpolate(double[][] xval,
double[] yval)
throws MathIllegalArgumentException,
DimensionMismatchException,
NoDataException,
NullArgumentException
Computes an interpolating function for the data set.
Parameters:
xval - the arguments for the interpolation points. xval[i][0] is the first component of interpolation point i, xval[i][1] is the second component, and so on until xval[i][d-1], the last component of that interpolation point (where d is thus the dimension of the space).
yval - the values for the interpolation points
Returns:
a function which interpolates the data set
Throws:
MathIllegalArgumentException - if the arguments violate assumptions made by the interpolation algorithm.
DimensionMismatchException - when the array dimensions are not consistent.
NoDataException - if an array has zero-length.
NullArgumentException - if the arguments are null.


text:
# Create all numbers from 1-100 using 1,3,3,6

Create all the numbers from $$1$$ to $$100$$ using the numbers $$1$$,$$3$$,$$3$$, and $$6$$.

• You can only use each number once, except for the $$3$$, of which you have two.
• You can use addition ($$x+y$$), subtraction ($$x-y$$), division ($$\frac{x}{y}$$), multiplication ($$x\times y$$), exponentiation ($$x^y$$) and roots ($$\sqrt[\leftroot{-2}\uproot{2}x]{y}$$).
• You can combine numbers like $$1$$ and $$3$$ to $$13$$ etc.
• You must use all numbers. EDIT: no factorials, in squareroots 2 is hidden, no combining results of operations, you can use parentheses and start with negative numbers, no rounding and no decimal points. Good Luck
• Can we combine the results of operations? For example, is $(1+3) | 36 = 436$ (where | indicates concatenation) – user46002 Jan 27 '19 at 22:14
• If we need to take a square root, is the two implied? – user46002 Jan 27 '19 at 22:14
• can we use factorial? – Omega Krypton Jan 27 '19 at 23:23
• Factorials are most likely out, but what about parentheses, unary minus (like starting with -1) and decimal points? – Bass Jan 28 '19 at 0:25
• If decimal is allowed then round would probably valid too? – Mukyuu Jan 28 '19 at 3:31

These get harder with larger numbers, but here are the first 40 (and a couple of the easier ones after that) with the digits in order:

1 to 10

1: $$1 + 3 + 3 - 6$$
2: $$(1 + 3) \times 3 / 6$$
3: $$1*3 * 3 - 6$$
4: $$13 - 3 - 6$$
5: $$-1^{33} +6$$
6: $$1\times3-3+6$$
7: $$1 + 3 -3 +6$$
8: $$1+3/3 + 6$$
9: $$1^3 \times (3+6)$$
10: $$1 + \sqrt[3]3^6$$

11 to 20

11: $$\sqrt{1+3}+3+6$$
12: $$1\times 3 + 3 + 6$$
13: $$1 + 3+3+6$$
14: $$-1 + 3\times 3+6$$
15: $$-1\times3 + 3\times 6$$
16: $$1 - 3 + 3 \times 6$$
17: $$-1^3 +3\times 6$$
18: $$(1+3)*3+6$$
19: $$13 + \sqrt{36}$$
20: $$-1 + 3^3 - 6$$

21 to 30

21: $$1 * 3^3 - 6$$
22: $$13 + 3 + 6$$
23: $$-13+36$$
24: $$(1+3)\times\sqrt{36}$$
25: $$1 - 3 + \sqrt3^6$$
26: $$-1+33-6$$
27: $$1*33-6$$
28: $$1+33-6$$
29: $$-1 + 3 + \sqrt3^6$$
30: $$(-1+3+3)\times 6$$

31 to 40

31: $$13+3*6$$
32: $$-1+3^3+6$$
33: $$13*3-6$$
34: $$1+3^3+6$$
35: $$-1+(3+3)\times6$$
36: $$1\times(3+3)\times 6$$
37: $$1^3+36$$
38: $$\sqrt{1+3}+36$$
39: $$1\times3 + 36$$
40: $$1+33+6$$

41 to 50 (getting much harder now, so from now on, only the easier ones)

41:
42: $$(1+3+3)\times 6$$
43:
44:
45: $$13\times3+6$$
46:
47:
48: $$(-1 + 3 \times 3) \times 6$$
49: $$13+36$$
50:

51 to 60

51:
52:
53: $$-1 +3 \times 3 \times 6$$
54: $$1\times 3 \times 3 \times 6$$
55: $$1 + 3 \times 3 \times 6$$
56:
57:
58: $$(1+3)^3-6$$
59:
60: $$(1+3\times3)\times6$$

61 to 70

61:
62:
63:
64: $$(1+3/3)^6$$
65:
66:
67:
68:
69:
70: $$(1+3)^3+6$$

71 to 80

71:
72: $$(1+3)\times 3 \times 6$$
73:
74:
75:
76:
77:
78: $$13 \times \sqrt{36}$$
79:
80: $$-1 + 3\times\sqrt3^6$$

81 to 90

81: $$1\times3\times\sqrt3^6$$
82: $$1+3\times\sqrt3^6$$
83:
84:
85:
86:
87:
88:
89:
90:

91 to 100

91:
92:
93:
94:
95:
96: $$(13+3)\times6$$
97:
98:
99:
100:

• Base64 because rot13 doesn't work with numbers: U29tZSBvdGhlcnM6CjQxID0gMzYgKyA2IC0gMQphbmQKNDMgPSAoM14zKSArIDE2CmFuZAo1NSA9IDYxIC0gc3FydCgzNikKYW5kCjY3ID0gNjEgLSBzcXJ0KDM2KQphbmQKNzYgPSA2MyArIDEzCmFuZAo5NCA9IDYxICsgMzM= – Outman Jan 28 '19 at 11:19

### Here's a solution for most of them. The remaining 21 are impossible.

• Most of them only use simple arithmetic.
• Some of them use exponentiation.
• Some use square roots.

Edit: Using only the binary operators (including digit concatenation), the number of possible combinations is pretty small (arrange three binary operators, then fill in the four numbers in some order) and I double-checked all of them with a computer. Most numbers are solvable with basic arithmetic, some require exponentiation, some require negation or square roots, and the rest are apparently impossible to solve without some extra operation such as rounding.

Accordingly, it is provably impossible to construct the following twenty-one numbers:

41, 44, 46, 47, 56, 68, 69, 74, 77, 79, 83, 85, 86, 89, 90, 91, 92, 95, 97, 98, 100.

The rest can be constructed as follows:

1-25 (complete)

1. (6+1)-(3+3)
2. (3*3)-(6+1)
3. (3*3)-(6*1)
4. 13-(6+3)
5. 36-31
6. 3*(3+1)-6
7. (6+1)+(3-3)
8. (6-3)3-1
9. 6
(3-1)-3
10. 3+13-6
11. 6+3+3-1
12. (6+3+3)*1
13. 6+3+3+1
14. (6*3)-(3+1)
15. 13 + (6/3)
16. 13 + (6-3)
17. 33-16
18. (1+3)*3 + 6
19. (6 + 1/3)*3
20. (3*6)+(3-1)
21. (13-6)*3
22. 13+3+6
23. 36-13
24. (6+1)*3 +3
25. 16+(3*3)

26-50 (except 41, 44, 46, 47)

26. 33 - (6+1)
27. 33 - (6*1)
28. 61 - 33
29. 31 - (6/3)
30. (6+3+1)*3
31. 13 + (6*3)
32. 63-31
33. 13*3 - 6
34. 31+(6-3)
35. 6*(3+3) -1
36. (6+3)(3+1)
37. 6
(3+3) + 1
38. 33+(6-1)
39. 33+(6*1)
40. 31+3+6
41. round[36 + √(31)]
42. (3+3+1)*6
43. 16 + (3^3)
44. round[(3-√3)^16]
45. (3*13)+6
46. round[3 * (13 + √6)]
47. round[31√3] - 6
48. (3*3 - 1) *6
49. 36+13
50. 63-13

51-75 (except 56, 68, 69, 74)

51. 16*3 + 3
52. 61 - (3*3)
53. (3*3*6)-1
54. 1*3*3*6
55. 61-(3+3)
56. √(3136)     (!)
57. (13+6)*3
58. (3+1)^3 - 6
59. 63 - (3+1)
60. (63*1) - 3
61. 61 + (3 - 3)
62. 61 + (3/3)
63. (6+1)*(3*3)
64. 63 + (1^3)
65. 63 + (3-1)
66. 63 + (3*1)
67. 36 + 31
68. round[63 + √31]
69. round[6 √133]
70. 61+(3*3)
71. (6^3)/3 - 1
72. 36 * (3-1)
73. (6^3)/3 + 1
74. round[3 √613]
75. 13*6 - 3

76-100 (except: 77, 79, 83, 85*, 86, 89,90,91,92, 95, 97, 98, 100.)

76. 63+13
77. round[61 * (3-√3)]
78. 13 * √(36)
79. round[6*13 + √3]
80. √(3^6) * 3 - 1
81. 13*6 + 3
82. √(3^6) * 3 + 1
83. round[(3+31)*√6]
84. 3^√(16) + 3
85. -----
86. round[√(6+√3) * 31]
87. (31*3) - 6
88. 61 + (3^3)
89. round[63 * √(3-1)]
90. round[13 * √(6/3)]
91. round[16 * √33]
92. round[3^√(3*6-1)]
93. 31 * (6-3)
94. 63+31
95. round[3*31 + √6]
96. (13+3)*6
97. round[(√3)^6 * √13]
98. round[36 * (1+√3)]
99. (3*31)+6
100. round[3*(31+√6)]

If you're allowed to concatenate the results of operations (e.g. (3+1)|5 = 45 ) then a solution for 85 is :

85. (3*3)|1 - 6

Python:

from itertools import permutations
from math import sqrt, floor, ceil

concat_literal_numbers_only = True

ops = { "+" : lambda a,b: a+b,
"-" : lambda a,b: a - b,
"/" : lambda a,b : a/float(b),
"*" : lambda a,b : a*b,
"^" : lambda a,b : a**b,
"C" : lambda a,b : float(str(a) + str(b)),
"n" : lambda a : -a,
"s" : lambda a : sqrt(a),
#"f" : lambda a : floor(a)
}

arity = {"+" : 2,
"-" : 2,
"/" : 2,
"*" : 2,
"^" : 2,
"C" : 2,
"n" : 1,
"s" : 1,
"f" : 1,
}

# print ops["/"](1,3)

# args: number of open args available
# nums: available digits to be used
# ops : tuple indicating commands used so far

def evaluate(cmds) :
"""Consume the list of commands in prefix notation, producing a pair (ans, unconsumed_symbols)"""

x = cmds.pop(0)
if not ops.get(x) :
return (x, cmds)
else :
args = []
for y in range(arity[x]) :
try :
(a, cmds) = evaluate(cmds)
args += [a]
except OverflowError :
return (None, None)
return (ops.get(x)(*args), cmds)

def score(ops):
ret = 0
ret += ops.count("+")
ret += 1.1*ops.count("-")
ret += 2 * ops.count("*")
ret += 3 * ops.count("/")
ret += 3 * ops.count("n")
ret += 4 * ops.count("^")
ret += 4 * ops.count("s")
ret += 4 * ops.count("f")
ret += 4 * ops.count("w")

# ret += 4 * ops.count("fs")
# ret += 4 * ops.count("cs")
return ret

agenda = [{"args" : 1, "nums" : [1,3,3,6], "ops" : []}]
seen = {}

only_search_for = None
ret = []

def finish(ops) :
global ret
global seen

ops_tmp = ops[:]
try :
n,_ = evaluate(ops_tmp)
except :
n = None

if n is None or not (0  score(ops) :
seen[n] = ops

print ops,"\t",n

while agenda :
x = agenda.pop(0)

if not x["nums"] and not x["args"] : # finished: used up all numbers; no open spaces.
finish(x["ops"])

if len(x["nums"]) == x["args"] : # fill in numbers only
for nums in set(permutations(x["nums"])) :
finish(x["ops"] + list(nums))

# print {"args" : 0,
#        "nums" : [],
#        "ops" : x["ops"] + list(nums)}

elif len(x["nums"]) > x["args"] :
# add new operators

for op in ops.keys() :
if arity[op] == 1 and x["ops"] and x["ops"][-1] == op :
continue # limit repeated unary operations
if arity[op] == 1 and x["ops"] and arity.get(x["ops"][-1]) == 1 :
continue # limit repeated unary operations

if (concat_literal_numbers_only and x["ops"] and (x["ops"][-1] == "C" or (len(x["ops"])>1 and x["ops"][-2] == "C")) and op != "C") :
continue

new_x = {"args" : x["args"] + arity[op] - 1,
"nums" : x["nums"],
"ops" : x["ops"] + [op]}
agenda = [new_x] + agenda

if x["args"] == 1 :
continue

for n in set(x["nums"]) :
new_nums = x["nums"][:]
new_nums.remove(n)
new_x = {"args" : x["args"] - 1,
"nums" : new_nums,
"ops" : x["ops"] + [n]}

agenda = [new_x] + agenda

# SHOW HOW TO MAKE ALL OF THE NUMBERS
miss = []
for i in range(0+1,100+1) :
if not seen.get(i) :
miss += [i]
print i, "\t", seen.get(i, "---")

# SHOW WHICH NUMBERS WERE MISSED
print "missed: ", miss

# IF YOU'RE LOOKING FOR ALL POSSIBLE WAYS TO MAKE SOMETHING, SHOW THEM HERE.
if only_search_for is not None :
ret = sorted(ret, key=score)
for x in ret:
print x

• Although it's unlikely that you'd find the missing solutions there, looks like the code ignores Nth roots entirely. – Bass Jan 29 '19 at 9:47
• @Bass . Thanks, yes ---- when posting the code I took out some operators that didn't add to the number of solutions. – user326210 Jan 30 '19 at 1:00

We can generate any integer using only $$1$$, $$3$$, $$3$$ and $$6$$ with the introduction of one special function.

The function in question is the Logarithm to an arbitrary base $$b$$ , or $$\log _{b} (x)$$.

To begin, let's discuss square root stacking.
$$\sqrt{\sqrt{a}}$$ is equivalent to $$\sqrt[4]{a}$$, and $$\sqrt{\sqrt{\sqrt{a}}}$$ is equivalent to $$\sqrt[8]{a}$$, which can be rewritten as $$a^\frac{1}{8}$$. This pattern continues indefinitely; $$a$$ with $$n$$ square roots stacked to it is equal to $$a^\frac{1}{2^n}$$.

The laws of logarithms state that $$\log _{b} (x^a) = a \cdot \log _{b} (x)$$. If we take the logarithm to base $$b$$ or our previous square root stack, we get $$\log _{b} (a^\frac{1}{2^n})$$, or $$\frac{1}{2^n} \cdot \log _{b}a$$. Setting $$a$$ and $$b$$ as 3 means that $$log _{3} (\sqrt{\sqrt[...]{3}})$$, with $$n$$ square roots, is equal to $$\frac{1}{2^n} \cdot \log _{3}3$$, or $$2^{-n}$$. $$(\frac{1}{a^x} = a^{-x})$$

$$\sqrt{\sqrt{16}} = 2$$, and $$\log _{b}(b^a) = a$$. As such,

$$0 = -\log _{\sqrt{\sqrt{16}}}(\log_{3}(3))$$
$$1 = -\log _{\sqrt{\sqrt{16}}}(\log_{3}(\sqrt{3}))$$
$$2 = -\log _{\sqrt{\sqrt{16}}}(\log_{3}(\sqrt{\sqrt{3}}))$$
$$3 = -\log _{\sqrt{\sqrt{16}}}(\log_{3}(\sqrt{\sqrt{\sqrt{3}}}))$$
$$4 = -\log _{\sqrt{\sqrt{16}}}(\log_{3}(\sqrt{\sqrt{\sqrt{\sqrt{3}}}}))$$
And so on, such that the amount of square roots is equal to your desired number.

This works for all integers; for negative numbers simply remove the $$-$$ at the start.

This was inspired by Numberphile's video on the four 4s.

Here are some:

1: $$3 + 3 - 6 + 1$$
2: $$3 * 3 - (6 + 1)$$
3: $$3 * 3 * 1 - 6$$
4: $$3 * 3 - 6 + 1$$
5: $$(3 * 6) / 3 - 1$$
6: $$(3 * 6) / 3 * 1$$
7: $$(3 * 6) / 3 + 1$$
8: $$3 * 3 - 1 ^ 6$$
9: $$(3 * 6) / (3 - 1)$$
10: $$3 * 3 + 1 ^ 6$$
11: $$36 / 3 - 1$$
12: $$36 / 3 * 1$$
13: $$36 / 3 + 1$$
14: $$3 * 6 - (3 + 1)$$
15: $$3 * 6 - (3 * 1)$$
16: $$3 * 6 - (3 - 1)$$
17: $$3 * 6 - 1 ^ 3$$
18: $$3 * 6 * 1 ^ 3$$
19: $$3 * 6 + 1 ^ 3$$
20: $$3 * 6 + 3 - 1$$
22: (omega kyrpton did some) $$3 * 6 + 3 + 1$$

I will do more later.

1-20: (Credits to @YoutRied)

1: $$3 + 3 - 6 + 1$$
2: $$3 * 3 - (6 + 1)$$
3: $$3 * 3 * 1 - 6$$
4: $$3 * 3 - 6 + 1$$
5: $$(3 * 6) / 3 - 1$$
6: $$(3 * 6) / 3 * 1$$
7: $$(3 * 6) / 3 + 1$$
8: $$3 * 3 - 1 ^ 6$$
9: $$(3 * 6) / (3 - 1)$$
10: $$3 * 3 + 1 ^ 6$$
11: $$36 / 3 - 1$$
12: $$36 / 3 * 1$$
13: $$36 / 3 + 1$$
14: $$3 * 6 - (3 + 1)$$
15: $$3 * 6 - (3 * 1)$$
16: $$3 * 6 - (3 - 1)$$
17: $$3 * 6 - 1 ^ 3$$
18: $$3 * 6 * 1 ^ 3$$
19: $$3 * 6 + 1 ^ 3$$
20: $$3 * 6 + 3 - 1$$

21-29

21: $$3 * 6 + 3 * 1$$
22: $$( 1 + 3 ) ! - ( 6 / 3 )$$
23: $$( 1 + 3 ) ! - ( 6 - 3 )$$
24: $$( 6 - 3 / 3 - 1 ) !$$
25: $$1 * 3 ^ 3 - floor(\sqrt{6})$$
26: $$( 6 - 3 ) ^ 3 - 1$$
27: $$( 6 - 3 ) ^ 3 * 1$$
28: $$( 6 - 3 ) ^ 3 + 1$$
29: $$31 - 6 / 3$$

41-50: (Credits to @Bass for 42, 45, 49)

41: $$(-1+3!)+36$$
42: $$(1+3+3)\times 6$$
43: $$31 + 6 * ceil(\sqrt{3})$$
44: $$floor( 1 * 3 * \sqrt{6 ^ 3})$$
45: $$13\times3+6$$
46: $$ceil(\sqrt{6 ^ 3} + 31)$$
47: $$floor(\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{31!}}}}})+36$$
48: $$6 * ( 3 * 3 - 1 )$$
49: $$13+36$$
50: $$(6+1)^2 + 3 - 3$$

51-60:

51: $$( 3 * 6 - 1 ) * 3$$
52: $$( 3 + 3 + 1 ) * ceil(\sqrt{6})$$
53: $$-1+( 3 * 3 * 6 )$$
54: $$1*3 * 3 * 6$$
55: $$1+3*3*6$$
56: $$61-3!+floor(\sqrt{3})$$
57: $$1*63-3!$$
58: $$1+63-3!$$
59: $$floor(\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{6^{(3-1)}}}}}}*3)$$
60: $$(1+3*3)*6$$

61-70:

61: $$63-3+1$$
62: $$63+1-ceil(\sqrt{3})$$
63: $$63-floor(\sqrt{3})+1$$
64: $$63+ceil(\sqrt{3})-1$$
65: $$63+3-1$$
66: $$63+3*1$$
67: $$63+3+1$$
68: $$61+3!+floor(\sqrt{3})$$
69: $$61+3!+ceil(\sqrt{3})$$
70: $$61+3*3$$

71-80

71: $$(3+1)!*3-floor(\sqrt{\sqrt{6}})$$
72: $$(3+1)*3*6$$
73: $$(3+1)!*3+floor(\sqrt{\sqrt{6}})$$
74: $$(3+1)!*3+floor(\sqrt{6})$$
75: $$(3+1)!*3+ceil(\sqrt{6})$$
76: $$ceil(\sqrt{\sqrt{\sqrt{ceil(\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{ceil(\sqrt{3!!})!}}}}})}}})*(3*6+1)$$
76:(much simpler) $$13*6-ceil(\sqrt{3})$$
77: $$13*6-floor(\sqrt{3})$$
78: $$13*floor(\sqrt{3})*6$$
79: $$13*6+floor(\sqrt{3})$$
80: $$13*6+ceil(\sqrt{3})$$

81-90:

81: $$(6+3)^{(3-1)}$$
82: $$(6+3)^{ceil(\sqrt{3})}+1$$

• Who said you could use factorials? – Yout Ried Jan 28 '19 at 0:20
• What are number 23 (plus you probably can't use factorials and 24? I don't get them. – Yout Ried Jan 28 '19 at 1:08
• Oops forgot a ")" and maybe you're missing a factorial for number 24 – Yout Ried Jan 28 '19 at 1:16

Expanding on Bass's answer, I added some new numbers.

(I lost track on which numbers I added, though 1-40 is all Bass)

1: $$1 + 3 + 3 - 6$$
2: $$(1 + 3) \times 3 / 6$$
3: $$1*3 * 3 - 6$$
4: $$13 - 3 - 6$$
5: $$-1^{33} +6$$
6: $$1\times3-3+6$$
7: $$1 + 3 -3 +6$$
8: $$1+3/3 + 6$$
9: $$1^3 \times (3+6)$$
10: $$1 + \sqrt[3]3^6$$ 11: $$\sqrt{1+3}+3+6$$
12: $$1\times 3 + 3 + 6$$
13: $$1 + 3+3+6$$
14: $$-1 + 3\times 3+6$$
15: $$-1\times3 + 3\times 6$$
16: $$1 - 3 + 3 \times 6$$
17: $$-1^3 +3\times 6$$
18: $$(1+3)*3+6$$
19: $$13 + \sqrt{36}$$
20: $$-1 + 3^3 - 6$$
21: $$1 * 3^3 - 6$$
22: $$13 + 3 + 6$$
23: $$-13+36$$
24: $$(1+3)\times\sqrt{36}$$
25: $$1 - 3 + \sqrt3^6$$
26: $$-1+33-6$$
27: $$1*33-6$$
28: $$1+33-6$$
29: $$-1 + 3 + \sqrt3^6$$
30: $$(-1+3+3)\times 6$$
31: $$13+3*6$$
32: $$-1+3^3+6$$
33: $$13*3-6$$
34: $$1+3^3+6$$
35: $$-1+(3+3)\times6$$
36: $$1\times(3+3)\times 6$$
37: $$1^3+36$$
38: $$\sqrt{1+3}+36$$
39: $$1\times3 + 36$$
40: $$1+33+6$$
41:
42: $$(1+3+3)\times 6$$
43: $$3^3 + 16$$
44:
45: $$13\times3+6$$
46:
47:
48: $$(-1 + 3 \times 3) \times 6$$
49: $$13+36$$
50:
51: $$16*3+3$$
52: $$61 - 3\times3$$
53: $$-1 +3 \times 3 \times 6$$
54: $$1\times 3 \times 3 \times 6$$
55: $$1 + 3 \times 3 \times 6$$
56:
57: $$(6\times3+1)\times3$$
58: $$(1+3)^3-6$$
59: $$63 - 3 - 1$$
60: $$(1+3\times3)\times6$$
61: $$63 - 3 + 1$$
62: $$63 - 1^3$$
63: $$63 * 1^3$$
64: $$(1+3/3)^6$$
65: $$63 + 3 - 1$$
66: $$1 \times (63 + 3)$$
67: $$63 + 3 + 1$$
68:
69:
70: $$(1+3)^3+6$$ 71: $$6^3 / 3 - 1$$
72: $$(1+3)\times 3 \times 6$$
73: $$6^3 / 3 + 1$$
74:
75: $$3^(3+1) - 6$$
76: $$63+13$$
77:
78: $$13 \times \sqrt{36}$$
79:
80: $$-1 + 3\times\sqrt3^6$$
81: $$1\times3\times\sqrt3^6$$
82: $$1+3\times\sqrt3^6$$
83:
84:
85:
86:
87: $$3^(3+1) + 6$$
88: $$61 + 3^3$$
89:
90:
91:
92:
93:
94: $$33 + 61$$
95:
96: $$(13+3)\times6$$
97:
98:
99: $$31\times3+6$$
100:

Only need 41, 44, 46, 47, 50, 56, 68, 69, 74, 77, 79, 83, 84, 85, 86, 89, 90, 91, 92, 93, 94, 95, 97, 98, and 100 now!

Partial answer 1-50 (w/e 41,47):

$$1= 1+3+3-6$$
$$2= 1 + (\frac{6}{(3+3)})$$
$$3= 1^3+(\frac{6}{3})$$
$$4= (\frac{6}{3})+3-1$$
$$5= (\frac{6}{3})+3^1$$
$$6= 6^1+3-3$$
$$7= 6+1-3+3$$
$$8= 6 + 3 - 1^3$$
$$9= 1^3 * (3+6)$$
$$10= 1^3 +3+6$$
$$11= 13 - (\frac{6}{3})$$
$$12= 6+3+3^1$$
$$13= 6+3+3+1$$
$$14= 6*3 - 3 - 1$$
$$15= 6*3 - 3^1$$
$$16= 16 + 3 - 3$$
$$17= 16 + (\frac{3}{3})$$
$$18= (\frac{6*3}{1^3})$$
$$19= 6*3+1^3$$
$$20= 6*3+3-1$$
$$21= 6*3+3^1$$
$$22= 6*3+3+1$$
$$23= 36-13$$
$$24= 6*(3+1^3)$$
$$25= 16+(3*3)$$
$$26= 13*(\frac{6}{3})$$
$$27= 33-6^1$$
$$28= 33-6+1$$
$$29= 31-(\frac{6}{3})$$
$$30= 6*(3+3-1)$$
$$31= 13+3*6$$
$$32= 3^3+6-1$$
$$33= (\frac{33}{1^6})$$
$$34= 33+1^6$$
$$35= (3+3)*6-1$$
$$36= (3+3)^1*6$$
$$37= 1+(3+3)*6$$
$$38= 33+6-1$$
$$39= 33+6^1$$
$$40= 1+33+6$$
$$41=$$
$$42= (1+3+3)*6$$
$$43= 16+3^3$$
$$44= round(\sqrt{6^3}*3^1)$$
$$45= 3*3*(6-1)$$
$$46= ceil(\sqrt{6^3}*3)+1$$
$$47=$$
$$48= ((3*3)-1)*6$$
$$49= 16+33$$
$$50= 63-13$$

Alrighty, I'm piggybacking off of @OmegaKrypton else and adding some of my own.

1 to 10

1: $$1 + 3 + 3 - 6$$
2: $$(1 + 3) \times 3 / 6$$
3: $$1^3 +3/6$$
4: $$13 - 3 - 6$$
5: $$-1^{33} +6$$
6: $$1\times3-3+6$$
7: $$1 + 3 -3 +6$$
8: $$1+3/3 + 6$$
9: $$1^3 \times (3+6)$$
10: $$1^3 + 3+6$$

11 to 20

11: $$\sqrt{1+3}+3+6$$
12: $$1\times 3 + 3 + 6$$
13: $$1 + 3+3+6$$
14: $$-1 + 3\times 3+6$$
15: $$-1\times3 + 3\times 6$$
16: $$1 - 3 + 3 \times 6$$
17: $$-1^3 +3\times 6$$
18: $$(1+3)*3+6$$
19: $$13 + \sqrt{36}$$
20: $$-1 + 3^3 - 6$$

21 to 30

21: $$1 * 3^3 - 6$$
22: $$13 + 3 + 6$$
23: $$-13+36$$
24: $$(1+3)\times\sqrt{36}$$
25: $$3*6+3!-1$$ or $$(6-1)^(3!/3)$$
26: $$-1+33-6$$
27: $$1*33-6$$
28: $$1+33-6$$
29: $$36-3!-1$$
30: $$(-1+3+3)\times 6$$

31 to 40

31: $$13+3*6$$
32: $$-1+3^3+6$$
33: $$13*3-6$$
34: $$1+3^3+6$$
35: $$-1+(3+3)\times6$$
36: $$1\times(3+3)\times 6$$
37: $$1^3+36$$
38: $$\sqrt{1+3}+36$$
39: $$1\times3 + 36$$
40: $$1+33+6$$

41 to 50

41:
42: $$(1+3+3)\times 6$$
43: $$3^3+16$$
44:
45: $$13\times3+6$$
46:
47:
48: $$16*(3!-3)$$
49: $$13+36$$
50: $$63-13$$

I added a few. It's getting late here; will come back tomorrow.


text:
MathCS Seminar

Title: The density of squarefree values taken by a polynomial
Colloquium: N/A
Speaker: Manjul Bhargava of Princeton
Contact: John Duncan, john.duncan@emory.edu
Date: 2017-10-20 at 4:00PM
Venue: W201
Abstract:
It is well known that the density of integers that are squarefree is $6/\pi^2$, giving one of the more intriguing occurrences of $\pi$ where one might not a priori expect it! A natural next problem that has played an important role in number theory is that of understanding the density of squarefree values taken by an integer polynomial. We survey a number of recent results on this problem for various types of polynomials - some of which use the ABC Conjecture'' and some of which do not.


text:
# Calendar Math, Days, Weeks Years

On the Time Math page, it was explained how there are 24 hours that make 1 whole day.

A day starts and begins at midnight.

## Today, Yesterday, Tomorrow

The current day we are in at any time, is called today.

The day that has just passed is called yesterday, and the day coming up after today is called tomorrow.

## Names of Days

A period of 7 days make up 1 week.

Each of these 7 different days in a week have their own name.

MONDAY , TUESDAY , WEDNESDAY , THURSDAY , FRIDAY , SATURDAY , SUNDAY

There can be a difference of views on whether a week begins on a Monday or a Sunday, neither is strictly right or wrong.

But for clarity on this page, we'll go with the view that a week begins on a Monday, and ends on a Sunday.

So with this thinking, 3 weeks in a row looks like:

Saturday and Sunday together are referred to as the "weekend", as Saturday is the beginning of the end of a week.
Regardless of whether the thinking is that a week starts on a Monday or a Sunday.

It's also the case that the other 5 days are referred to as "weekdays".

A period of 1 week doesn't necessarily have to be from Monday to Sunday.

Any 7 day period is a week.

For example one Wednesday to the next Wednesday.

## Months and Years

52 weeks and 1 day, make up 1 whole year.

Years we'll come to, but what's important to understand first are months.

One month is quite a bit longer than a week. Roughly just over 4 times longer in general.

There are 12 months in one year, and like with days, months have their own names also.

The 12 months of a year in order, from first to last, are:

JANUARY , FEBRUARY , MARCH , APRIL , MAY , JUNE

JULY , AUGUST , SEPTEMBER , OCTOBER , NOVEMBER , DECEMBER

Unlike weeks, not all months are the same length.

- April , June , September , November have 30 days.

- January , March , May , July , August , October , December have 31 days.

The month of February is slightly different to the other 11 months of a year.

February usually has only 28 days, but every 4 years February has 29 days.

The years when February has an extra day are called "Leap years". Which is explained more in the next section.

The image below shows the calendar for the whole of the year 2016, which was a Leap year.

## Leap Years

So how does the Leap Year system work, and why does the month of February have 28 days for 3 years, and then 29 on the fourth year?

As mentioned, there are 52 weeks and 1 day in a year, which is 365 days.

But this is actually only for most years, and not every year.

One year is classed as how long it takes our planet Earth to orbit around the Sun.

This is roughly 365 days, but not exactly.

It really takes our planet Earth very close to 365 days and  $\dpi{100}&space;\fn_jvn&space;\tfrac{1}{4}$  of a day to orbit the Sun.

365.25 days to 2 decimal places.

This is why we have a leap year every 4 years.

The four extra  $\dpi{100}&space;\fn_jvn&space;\tfrac{1}{4}$  days of the previous 3 years get added together with the  $\dpi{100}&space;\fn_jvn&space;\tfrac{1}{4}$  day from the fourth year, and made into one whole extra day in this fourth year, February 29.

The last Leap Year we had was 2016.

The next will be 2020, then 2024, then 2028 and so on.

However, every 100 years when we expect that there would be a Leap Year, we don't have one.

Except when its's every 400 years.

For example, the year 1500 was NOT a leap year, but the year 1600 was.

The years 1700, 1800 and 1900 were NOT leap years, but the year 2000, as it was another 400 years after the year 1600, year 2000 was a leap year.

This can sound rather complicated at first reading.

More specific information on leap years can be seen in this Leap Year Rules section on the Skywise Unlimited website.

Which expands on some of the information on leap years mentioned on this page.

## Year Numbers

So a year is when the Earth makes one journey fully around the sun, and it keeps going round and round again, year after year.

Each new year has a number to identify it, the year 2018 is followed by 2019, then 2020 and so on.

Where do the numbers come from?

The years as we count them at present, are numbered starting from the birth of Jesus, over 2000 years ago.

These years after Jesus birth, are formally regarded as  2018 AD  or  2020 AD etc.

The birth of Jesus is recognized as the year  1 AD.

AD is short for "Anno Domini", which is latin and means "year of our lord".

Though in another discussion, there is still some debate on the official date and year of Jesus birth. But as things stand just now, we are in the year 2018.

The years that came before Jesus birth, have BC after them.

The BC stands for "Before Christ".

So 300 years before Jesus was born, is referred to as  300 BC.

Some specific groups of years have their own individual names that are commonly used.

The most common terms are DECADE, CENTURY and MILLENNIUM.

1 Decade = 10 Years   ,   1 Century = 100 Years   ,   1 Millennium = 1000 Years

## Calendar Math, Dates

The date on any given day can be represented in by numbers, which represent the day, month and year.

DAY / MONTH / YEAR

However, in America, it's more common to start with the month, rather than day.

MONTH / DAY / YEAR

June is the sixth month of a year, and as such, the fourth day in June in 2018 numerically would be:

04 / 06 / 2018     or     06 / 04 / 2018

›   ›  Calendar Math


text:
14 videos

1 videos

28 videos

4 videos

# Collection Conference on Arithmetic Geometry in honor of Luc Illusie

Organizer(s) Lei Fu (Tsinghua University), Fabrice Orogozo (CNRS&IMJ-PRG), Weizhe Zheng (MCM, AMSS, CAS)
Date(s) 6/7/21 - 6/11/21
00:00:00 / 00:00:00
8 22

## Motivic connections over a finite field

Work in progress with Michael Groechenig.

If $X$ is a smooth variety over a perfect field $k$ of characteristic $p>0$ which lifts to $W_2(k)$, its de Rham complex up to a certain level splits. It is the spectacular theorem of Deligne-Illusie from 1987, which led to manifold developments. It enabled Ogus-Vologodsky to develop a version of the Simpson correspondence, and later to Lan-Sheng-Zuo to define when $X$ is projective Higgs-de Rham flows, which over finite fields for flat connections with vanishing Chern classes, are preperiodic.

On the other hand, over the field of complex numbers, Brunebarbe-Klingler-Totaro proved, in answer to a question I had posed, that if $X$ smooth projective has no degree $\geq 1$ symmetric differential forms, then all flat connections are motivic, in fact they have finite monodromy. Their proof is transcendental.

We investigate a version of this theorem over a finite field. Under Deligne-Illusie $W_2(\mathbb{F}_q)$ assumption, we suspect that precisely the same vanishing condition on global symmetric differential forms forces flat connections with vanishing Chern classes to have finite monodromy. As of today, we can prove it in rank $\leq 3$.

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• 01:06:10
published on June 7, 2021

## On the derived Lusztig correspondence

By Gérard Laumon

01:12:10
published on June 7, 2021

## Cohomology of stacks of shtukas

By Cong Xue

01:07:56
published on June 7, 2021

## Grothendieck–Serre in the quasi-split unramified case

By Kęstutis Česnavičius

01:02:23
published on June 7, 2021

## Prisms, prismatic neighborhoods, and p-de Rham cohomology

By Arthur Ogus

01:02:07
published on June 7, 2021

## Logarithmic abelian varieties

By Kazuya Kato

01:09:35
published on June 8, 2021

## Micro support of a constructible sheaf in mixed characteristic

By Takeshi Saito

01:14:41
published on June 8, 2021

## Logarithmic geometry and resolution of singularities

By Michael Temkin

59:03
published on June 8, 2021

## Motivic connections over a finite field

By Hélène Esnault

53:55
published on June 8, 2021

## Exponential sums and finite groups

By Nicholas Katz

01:10:20
published on June 8, 2021

## Representability results for flat cohomology

By Martin Olsson

01:06:51
published on June 9, 2021

## Cohomology of prismatic crystals

By Yichao Tian

01:03:01
published on June 9, 2021

## Basic reductions of abelian varieties

By Yunqing Tang

01:13:53
published on June 9, 2021

## Intersection cohomology of Shimura varieties and pizza

By Sophie Morel

01:10:48
published on June 9, 2021

## Height pairing and vanishing cycles

By Alexander Beilinson

01:13:42
published on June 10, 2021

## Hodge theory for non-Archimedean analytic spaces

55:31
published on June 10, 2021

By Miaofen Chen

55:27
published on June 10, 2021

## On a generalization of Colmez’s functor

By Yongquan Hu

01:08:37
published on June 10, 2021

## The absolute prismatic site

By Bhargav Bhatt

01:14:28
published on June 10, 2021

## Remarks on p-adic logarithmic cohomology theories

By Akhil Mathew

01:07:42
published on June 11, 2021

## The relative Hodge-Tate spectral sequence

By Ahmed Abbes

01:05:40
published on June 11, 2021

## Comparison of oriented products and rigid toposes

By Ofer Gabber

53:39
published on June 11, 2021

## Jet bundles and differential calculus

By Bao Chau Ngo

## Register

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text:
# Imaginary integration?

1. Jul 8, 2010

### Unit

How is this possible?

$$\int_{i\infty}^\pi e^{ix} dx = i$$

I mean, I understand that the integral of exp(ix) is -i exp(ix) and then you evaluate that from π to i∞ — but that's exactly it, how does one "draw a line" from (π, 0) on the Argand plane to (0, ∞)? (assuming Argand plane tuples (a, b) ↔ a + bi)

EDIT: fixed the integrand, thanks Mute

Last edited: Jul 8, 2010
2. Jul 8, 2010

### Mute

(slight typo in your original post - the integrand in the latex reads $e^{i\pi}$, not $e^{ix}$!)

Pick any contour that connects the two points.

e.g., a contour that runs in a straight line from $x = 0 + i\infty$ to $x = 0 + i0$ (i.e., down the imaginary axis), then from there to $x = \pi + i0$ (along the real axis). This gives

$$\int_{i\infty}^\pi dx~e^{ix} = \int_{i\infty}^{i0}dx~e^{ix} + \int_0^\pi dx~e^{ix} = -i\int_{-\infty}^0 dy~e^y + \int_0^\pi dx~e^{ix} = -i + 2i = i$$
where I changed variables in the first integral $y = ix$.

To see that this has to be the value for any contour, consider this contour plus another arbitrary contour that runs from the point (pi,0) back to (0,infinity), forming a closed loop. The total closed contour contains no poles, so the closed contour integral has to be zero. Thus, the arbitrary contour running from (pi,0) to (0,infinity) must exactly cancel out the original contour. Because the value of a contour running from point A to point B is minus the value of the contour running from point B to point A, this establishes that any arbitrary contour from (0,infinity) to (pi,0) of exp(ix) gives the value i.

Last edited: Jul 8, 2010
3. Jul 8, 2010

### Unit

I have never heard of a "contour" before, but this makes a lot of sense! Thank you! I was trying to imagine one straight line, clearly impossibly.

Just a few questions: What do you mean by "the total closed contour contains no poles"? And, I don't understand how this integral represents the "area" under the curve. Is that solely a Cartesian xy concept?

4. Jul 8, 2010

### Mute

If you haven't studied any complex analysis before my explanation may be a bit hard to understand. In the complex plane one typically does integration on contours (curves in the complex plane). This is basically summing the value of the function to be integrated at each point along the curve, and so has no interpretation of an area in the argand plane; however, if you were to parametrize the curve by the arc length traveled from some starting point, then I suppose the contour integral would be the area under the integrand as a function of the traversed arc length.

Anywho, one of the basic results of complex analysis is that any function that doesn't have singularities (poles) will give zero when integrand around a closed curve. This is what I used to determine that your integral had to give i for any curve.

See: http://en.wikipedia.org/wiki/Line_integral (talks about both vector line integrals and complex line (contour) integrals)


text:
# Trigonometry Identities.

1. Jul 26, 2006

i need to prove the following identity

$$\sin^4 - \cos^4 = 1 - 2\cos^2$$

i have tried approaching this identity by solving the RHS (right hand side) of the equation but this lead no where. however i am unsure on how to maniplulate the LHS of this equation because of the power of 4. would somebody please be able to give me a slight hint in which direction to head to prove this?
many thanks,

2. Jul 26, 2006

Do you know how to factorize $$a^2-b^2$$? How can you use this to help you factorize $$a^4-b^4$$?

Also, the identity should have a variable (i.e. instead of $$sin^4$$, it should be $$sin^4x$$).

All the best!

3. Jul 26, 2006

oh, my mistake i forgot the $\theta$ sign after the trigonometric functions. i am not sure on how top factorise, sorry :(

4. Jul 26, 2006

### HallsofIvy

Long, long, long, before you learn about trigonometry, you should have learned that a2- b2= (a- b)(a+ b). And if we replace a by x2 and b by y2, we get that
x4- y4= (x2- y2)(x2+ y2). And what is x2+ y2 in this case?

5. Jul 26, 2006

can $x^2 + y^2$ be factorised? ive tried $(x + y)(x + y)$, $(x - y)(x - y)$, but neither have worked. am i completely missing the point here? its been a while since i have had to factorise. thanks

6. Jul 26, 2006

### BobG

In this case you don't have to, since $x^2=sin^2\theta$ and $y^2=cos^2\theta$.

7. Jul 26, 2006

### sdekivit

we don't need to factorize. Just use:

$$sin^{4} \theta = sin^{2} \theta \cdot sin^{2} \theta$$

$$sin^{2} \theta = 1 - cos^{2} \theta$$

$$(1-cos^{2} \theta) \cdot (1-cos^{2} \theta) = 1 - 2\cdot cos^{2} \theta + cos^{4} \theta$$

Now substract $cos^{4} \theta$ and you got your proof.

8. Jul 26, 2006

okay thanks for the help people, sorry to be a little slow

9. Jul 26, 2006

### hmm?

In addition to what these people said: you should try to remember identities that will serve as shortcuts when proving identities.

like: Sin^2 + Cos^2=1 <= this may be the most important one

Sin^2= 1 - Cos^2

Cos^2= 1 - Sin^2

10. Jul 26, 2006

### VietDao29

You're missing an angle here.
It should read sin2x, not just sin2 :)

11. Jul 26, 2006

### hmm?

You're absolutely right--all those -1 from omitted thetas and xs from test questions still haven't sunk in :/...haha.

12. Jul 26, 2006

### HallsofIvy

Unfortunately, I have had students declare that
$$\frac{sin x}{sin y}= \frac{x}{y}$$!

13. Jul 27, 2006

### Office_Shredder

Staff Emeritus
There aren't any parentheses, so it could just be that s, i, and n are all variables

14. Jul 27, 2006

### VietDao29

Well, I used to think $$\frac{sin x}{x} = sin$$ but then I found out it's worse than that. It's not just any sin, it's an X-rated video filmed in someone's kitchen sinc.


text:
Math Help - needing help on how to solve equations

1. needing help on how to solve equations

my last math class was in 2005 here it is 5 years later and i am still struggling with the same stuff basically i need to know the steps to solve equations i will set some examples

3(x - 1) - 12 = 0
-1(y + 3) = 10
-4x = 44
7 - z = 15
-7c + 1 = -20
5(3y - 2) = 16y

2. Originally Posted by Russell010887
my last math class was in 2005 here it is 5 years later and i am still struggling with the same stuff basically i need to know the steps to solve equations i will set some examples

3(x - 1) - 12 = 0
-1(y + 3) = 10
-4x = 44
7 - z = 15
-7c + 1 = -20
5(3y - 2) = 16y

$3(x-1)-12=0\Rightarrow 3x-3-12=0 \Rightarrow3x-15=0 \Rightarrow 3x=15 \Rightarrow x=\frac{15}{3} \Rightarrow x= 5$

$-4x=44 \Rightarrow 4x=-44 \Rightarrow x=\frac{-44}{4} \Rightarrow x=-11$

$5(3y-2)=16y \Rightarrow 15y-10 = 16y \Rightarrow -10 = y \Rightarrow y=-10$

and so on

3. All of the questions are similar. First you expand the brackets and then get the letter on its own side.

Remember that:
$a(b+c)=ab+ac$

4. Originally Posted by Russell010887
my last math class was in 2005 here it is 5 years later and i am still struggling with the same stuff basically i need to know the steps to solve equations i will set some examples

Whatever is done to the unknown value, undo it. You undo adding something by subtracting that same thing. You undo multiplying by division, etc. And, of course, you "undo" step by step in the opposite order in which it was "done". And whatever you do to one side of the equation you do to the other side as well.

3(x- 1)- 12= 0

If you were to start from "x" and calculate this you would: subtract 1, multiply by 3, then subtract 12. To "undo" that, you first add 12, then divide by 3 then add 1:"
3(x- 1)- 12+ 12= 0+ 12
3(x- 1)= 12

3(x- 1)/3= 12/3
x- 1= 4

x- 1+ 1= 4+ 1
x= 5

3(x - 1) - 12 = 0
Here, 1 is subtracted from x, then it is multiplied by 3, then 12 is subtracted. To "undo", add 12, then divide by 3, then add 1:
3(x- 1)- 12+ 12= 0+ 12
3(x- 1)= 12

3(x- 1)/3= 12/3
x- 1= 4

x-1+ 1= 4+ 1
x= 5

Try that on the rest of these

-1(y + 3) = 10
-4x = 44
7 - z = 15
-7c + 1 = -20
5(3y - 2) = 16y

5. Math.com Algebra Worksheet Generator
This is a big help. You can generate algrebra equations from simple one step equations to much harder equations.
You can also create an exam if you wish by putting a number of different equations to build up to getting harder as they go.

Also, If you struggle to work one out or want to know how to solve it. You can use this website to enter an equation that will solve it but also show you step by step how to do it.
Algebra.Help -- Equation Calculator

I'm currently learning to advance my own skills in math so these are helping a lot for me.


text:
$\cos(t)$ as a function of (t-2)?

As the title suggests, I'm doing a Laplace transform problem using the $t$-shift theorem, and I've almost got it, I just can't work out how to transform $\cos(t)$ into a function of $(t-2)$, I saw a trig identity

$$\cos(x)\cos(y) = \frac{1}{2} [\cos(x - y) + \cos(x + y)]$$

But I can't quite work out how to re-arrange it, any help much appreciated.

Thanks.

-
$\cos(t)=f(t-2)$ if $f(t)=\cos(t+2)$. Is that what you're looking for? –  Jonas Meyer May 21 '11 at 8:32
Thanks but I'm looking to get f(t) = cos(t) but all terms that include t are in the form (t-2). eg something like; cos(t) = f(t) = A.sin(t-2) + B.cos(t-2) –  tomatosource May 21 '11 at 10:46

$\cos t = \cos((t-2)+2)$ is a function of $t-2$. If you prefer to express it as a linear combination of $\sin (t-2)$ and $\cos (t-2)$, you can indeed use a trig identity:
$\cos (x+y) = \cos x \cos y - \sin x \sin y$ will work (with $x=t-2$ and $y=2$).
@tomatosource, I think the point is to use the identity for $\cos(x+y)$ with $x=t-2$ and $y=2$. –  Gerry Myerson May 21 '11 at 11:00


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# Is There A Way To Fix Windows Update Error 80200053?

Asked By 20 points N/A Posted on -

Hey, I was trying to update my windows when suddenly; I encountered Windows Update error 80200053. What does it mean exactly and how can I fix it my problem?

SHARE
Answered By 0 points N/A #330933

## Is There A Way To Fix Windows Update Error 80200053?

To fix Windows Update error 80200053, you need to download and install the Windows Update agent to fix this problem.


text:
# natcap.invest.routing package¶

## natcap.invest.routing.delineateit module¶

DelineateIt entry point for exposing pygeoprocessing’s watershed delineation routine to a UI.

natcap.invest.routing.delineateit.execute(args)

Delineateit: Watershed Delineation.

This ‘model’ provides an InVEST-based wrapper around the pygeoprocessing routing API for watershed delineation.

Upon successful completion, the following files are written to the output workspace:

• snapped_outlets.shp - an ESRI shapefile with the points snapped to a nearby stream.
• watersheds.shp - an ESRI shapefile of watersheds determined by the d-infinity routing algorithm.
• stream.tif - a GeoTiff representing detected streams based on the provided flow_threshold parameter. Values of 1 are streams, values of 0 are not.
Parameters: workspace_dir (string) – The selected folder is used as the workspace all intermediate and output files will be written.If the selected folder does not exist, it will be created. If datasets already exist in the selected folder, they will be overwritten. (required) results_suffix (string) – This text will be appended to the end of output files to help separate multiple runs. (optional) dem_uri (string) – A GDAL-supported raster file with an elevation for each cell. Make sure the DEM is corrected by filling in sinks, and if necessary burning hydrographic features into the elevation model (recommended when unusual streams are observed.) See the ‘Working with the DEM’ section of the InVEST User’s Guide for more information. (required) outlet_shapefile_uri (string) – This is a vector of points representing points that the watersheds should be built around. (required) flow_threshold (int) – The number of upstream cells that must into a cell before it’s considered part of a stream such that retention stops and the remaining export is exported to the stream. Used to define streams from the DEM. (required) snap_distance (int) – Pixel Distance to Snap Outlet Points (required) None

## natcap.invest.routing.routedem module¶

RouteDEM entry point for exposing the natcap.invest’s routing package to a UI.

natcap.invest.routing.routedem.execute(args)

RouteDEM: D-Infinity Routing.

This model exposes the pygeoprocessing d-infinity routing functionality in the InVEST model API.

Parameters: workspace_dir (string) – The selected folder is used as the workspace where all intermediate and output files will be written. If the selected folder does not exist, it will be created. If datasets already exist in the selected folder, they will be overwritten. (required) dem_uri (string) – A GDAL-supported raster file containing a base Digital Elevation Model to execute the routing functionality across. (required) pit_filled_filename (string) – The filename of the output raster with pits filled in. It will go in the project workspace. (required) flow_direction_filename (string) – The filename of the flow direction raster. It will go in the project workspace. (required) flow_accumulation_filename (string) – The filename of the flow accumulation raster. It will go in the project workspace. (required) threshold_flow_accumulation (int) – The number of upstream cells that must flow into a cell before it’s classified as a stream. (required) multiple_stream_thresholds (bool) – Set to True to calculate multiple maps. If enabled, set stream threshold to the lowest amount, then set upper and step size thresholds. (optional) threshold_flow_accumulation_upper (int) – The number of upstream pixels that must flow into a cell before it’s classified as a stream. (required) threshold_flow_accumulation_stepsize (int) – The number of cells to step up from lower to upper threshold range. (required) calculate_slope (bool) – Set to True to output a slope raster. (optional) slope_filename (string) – The filename of the output slope raster. This will go in the project workspace. (required) calculate_downstream_distance (bool) – Select to calculate a distance stream raster, based on uppper threshold limit. (optional) downstream_distance_filename (string) – The filename of the output raster. It will go in the project workspace. (required) None


text:
Why $n^\mu T_{\mu\nu}$ is called the pressure?

1. Aug 11, 2014

ccnu

I read in the textbook which says that, according to the usual definition the absolute value of $n^\mu T_{\mu\nu}$ is just the pressure. $T_{\mu\nu}$ is the energy-momentum tensor and $n^\mu$ is a four dimensional normal vector.

2. Aug 12, 2014

ChrisVer

As given I don't see that this holds in general... For example take a perfect fluid for which it can be written in rest frame:
$T_{\mu \nu} = diag( \rho, p , p , p )$
Then:
$n^{\mu}T_{\mu \nu} = n^{0}T_{00} + n^{ii}T_{ii} = n^{0} \rho + (n^{11}+n^{22}+n^{33}) p$

Which doesn't have to be equal to the pressure... If though you choose $n^{0}=0$ (so it's not just any 4-dim unit vector) things can get better.

Last edited: Aug 12, 2014
3. Aug 12, 2014

ccnu

Oh~~ yes, I forgot to say that $n^0 = 0$, and $n^\mu$ is pure spatial. Thanks a lot!

4. Aug 13, 2014

Orodruin

Staff Emeritus
Chris, your LHS has a free index and your RHS does not.

In a general frame you would have
$$n^\mu T_{\mu\nu} = n^\mu \left((\rho+p) u_\mu u_\nu - p g_{\mu\nu}\right) = (\rho+p) u_\nu (n \cdot u) - p n_\nu,$$
where u is the 4-velocity of the fluid.

5. Aug 13, 2014

Meir Achuz

The association of $n^\mu T_{\mu\nu}$ with the pressure at a point follows from the fact that its integral over a closed surface equals the rate of change of momentum within the closed surface.
There are simple examples where $n^\mu T_{\mu\nu}$ is NOT the pressure at a point on the surface. For instance the pressure on a dielectric slab due to a point charge a distance d from the slab is not $n^\mu T_{\mu\nu}$.

6. Aug 13, 2014

ChrisVer

oops sorry... yes the correct form would have to be:
$n^{i} T_{i \mu} \equiv n_{i} p$

7. Aug 13, 2014

Orodruin

Staff Emeritus
$n^{i} T_{i \mu} \equiv n_{\color{Red}\mu} p$

Assuming $n^0 = 0$ and that we are in the rest frame of the fluid.

8. Aug 13, 2014

ChrisVer

the n0 is zero
and also mu=i for the expression not to be zero...

For the rest frame yes, I just corrected the expression I gave in my previous post

9. Aug 13, 2014

Orodruin

Staff Emeritus
Even if the expression is non-zero only for spatial $\mu$, you must still have the same free indices on both sides of your equality. In your case you have i as a summation index on one side and as a free index on the other. I can guess what you mean because I know what you are aiming for, but formally it does not make sense.


text:
# Regex matching "\text{Hom}(.+,.+)" does not work

Here's a screenshot showing everything:

As you can see my template_regex is:

\text\{Hom\}$$.+,.+$$

It's not matching either of the nodes shown in the lower right, but also shown in my front end (which I'm kind of showing off) here:

Those nodes exist in the Neo4j database, however, they're not being matched by this very basic regular expression.

def escape_regex_str(string):
string = string.replace('(', r'$$') string = string.replace(')', r'$$')

The top line of the above function solved the issue. Might be with t since escaping that is usually the tab character, so we need to double escape some of the backslashes. :)


text:
Sciencemadness Discussion Board » Fundamentals » Chemistry in General » Removing trace impurities from H2O2 Select A Forum Fundamentals   » Chemistry in General   » Organic Chemistry   » Reagents and Apparatus Acquisition   » Beginnings   » Responsible Practices   » Miscellaneous   » The Wiki Special topics   » Technochemistry   » Energetic Materials   » Biochemistry   » Radiochemistry   » Computational Models and Techniques   » Prepublication Non-chemistry   » Forum Matters   » Legal and Societal Issues   » Detritus   » Test Forum

Pages:  1
Author: Subject: Removing trace impurities from H2O2
VeritasC&E
Harmless

Posts: 25
Registered: 29-1-2018
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Removing trace impurities from H2O2

Hello,

I have some H2O2 at 9% concentration and I'd like to use it in small quantities for applications with puriss. reagents I would not like to contaminate.

How would you go about purifying it (in the sense of removing impurities other than water and H2O2 down to maybe <1ppm for any single impurity)?

I'm predominantly concerned about traces of metal ions and maybe stabilizer additives from the pre-dilution H2O2 solution it was probably made from.

I looked it up in a lab chemicals purification book but the process there was for concentrating H2O2, not riding it from impurities.
morganbw
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I do not want to come off as being rude but sometimes if you need the good stuff you need to buy. Of course you may figure it out, I am just pointing out my thought process.
Metacelsus
International Hazard

Posts: 2423
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Location: Cambridge, MA
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Mood: Double, double, toil and trouble

What are the impurities? An ion-exchange resin might work.

As below, so above.
VeritasC&E
Harmless

Posts: 25
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Mood: No Mood

Is there is no reasonable way to purify it and is it just better to produce it pure from scratch? Or do you mean in terms of time saving?

I like to purify stuff myself when possible. A point here is that the price per liter literally varies from ~$2 to ~$20 going between probably fairly clean H2O2 solution and puriss. H2O2. I can afford to do chemistry only at fairly priced consumables.
VeritasC&E
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Registered: 29-1-2018
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I'm not sure what they are. I know sometimes stabilizers are added to the more concentrated H2O2 it might have been made from, and other likely impurities would be ions from the water (say it's made from RO water for economical reasons), and maybe from the apparatus that was used to produce it (Fe, Ni, Cr for instance if stainless steel was used).

Do you know which stabilizers are commonly used in commercial H2O2 solutions?
DraconicAcid
International Hazard

Posts: 3404
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Location: The tiniest college campus ever....
Member Is Online

Mood: Semi-victorious.

I suspect hydrogen peroxide would be too prone to decomposition during purification to bother trying.

Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
VeritasC&E
Harmless

Posts: 25
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So we just have to rely on the pricing of big companies if we absolutely want it pure?

[Edited on 2-9-2020 by VeritasC&E]
unionised
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I can't see any theoretical objections to sub distillation at home. as long as you have the kit.
VeritasC&E
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 Quote: Originally posted by unionised I can't see any theoretical objections to sub distillation at home. as long as you have the kit.

Should I vacuum distill the solution below boiling point with the goal in mind to get a much purer (albeit initially less concentrated) H2O2 solution in the receiver? And in a second step boil away enough H2O to reconcentrate the purified solution to the initial 9%?

teodor
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As far as I remember from this video - https://www.youtube.com/watch?v=nVAe__ToAOY
you can distil without vacuum H2O2 as 25-30% mixture with water. I didn't try it by myself and if I would do I would take a lot of precautions, there are some safety issues especially if you have little experience in this.

[Edited on 2-9-2020 by teodor]
unionised
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I think you need to start by looking up "sub distillation" i.e. "sub boiling distillation".
zed
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H2O2 is easy to concentrate. Problem is, it usually contains stabilizers that are easily concentrated too.

Easily concentrated, and hard to eliminate, when present in your final product. In years past, acetanilide was a common stabilizer. Might still be. I don't know. But, in some applications it presents a real problem.

The solution to this problem, is to start with H2O2 that doesn't contain stabilizing agents, that are gonna mess things up.

Oh yeah, and apparently H2O2 may be able to "etch" ordinary glasses. Thereby contaminating itself. Grrrr.

Gotta find a way to obtain "better" peroxide, if your work is critical.

An ancient paper on the subject, yet still so contemporary!
There may be numerous contaminants in commercial peroxide, and they may have to be contended with under some circumstances. Perhaps the author's difficulties, will provide illumination.
https://archive.org/details/interactionofpla00shaf/page/n1/m...

[Edited on 2-9-2020 by zed]
VeritasC&E
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 Quote: Originally posted by unionised I think you need to start by looking up "sub distillation" i.e. "sub boiling distillation".

Isn't this just slow distillation under BP? Woudn't it be enhanced by vacuum? (which also also would allow a significantly reduced rate of decomposition)

Your suggestion is good but I just want to be sure I correctly understand the process to follow and how efficient that would be. How much H2O2 can I expect to carry away with the water vapour into the receiver flask (i.e. what concentration can I expect in my receiver starting with my 9% material)?

[Edited on 2-9-2020 by VeritasC&E]
VeritasC&E
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In theory, separation by freezing could be a route (assuming H2O2 is a bad solvent)

In practice however the process looks it could be very tedious (very close MPs), and hard to perform without contaminating the solution that is supposedly being purified.

In theory:

Freeze a portion of the solution > Pass through pre-cleaned Frit > Dilute again with Distilled Water > Mix Well > Repeat

The main problems I could see:

1) How to minimize H2O2 loss in the ice?
2) Contamination during filtration
3) Energy & Time Cost

[Edited on 2-9-2020 by VeritasC&E]
unionised
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Quote: Originally posted by VeritasC&E
 Quote: Originally posted by unionised I think you need to start by looking up "sub distillation" i.e. "sub boiling distillation".

Isn't this just slow distillation under BP? Woudn't it be enhanced by vacuum? (which also also would allow a significantly reduced rate of decomposition)

Your suggestion is good but I just want to be sure I correctly understand the process to follow and how efficient that would be. How much H2O2 can I expect to carry away with the water vapour into the receiver flask (i.e. what concentration can I expect in my receiver starting with my 9% material)?

[Edited on 2-9-2020 by VeritasC&E]

The important aspect is that you don't boil the material.
That way, you don't generate "spray" which carries over involatiles like stabilisers.
In principle, starting with 9% you end up with 9%.
But without any involatile materials.
VeritasC&E
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Quote: Originally posted by unionised
Quote: Originally posted by VeritasC&E
 Quote: Originally posted by unionised I think you need to start by looking up "sub distillation" i.e. "sub boiling distillation".

Isn't this just slow distillation under BP? Woudn't it be enhanced by vacuum? (which also also would allow a significantly reduced rate of decomposition)

Your suggestion is good but I just want to be sure I correctly understand the process to follow and how efficient that would be. How much H2O2 can I expect to carry away with the water vapour into the receiver flask (i.e. what concentration can I expect in my receiver starting with my 9% material)?

[Edited on 2-9-2020 by VeritasC&E]

The important aspect is that you don't boil the material.
That way, you don't generate "spray" which carries over involatiles like stabilisers.
In principle, starting with 9% you end up with 9%.
But without any involatile materials.

But isn't there a significant BP difference between H2O and H2O2? Do they form an azeotrope? In which case, would you know within which proportions and parameters?

[Edited on 2-9-2020 by VeritasC&E]
unionised
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They might form an azeotrope, I think so, but I'd need to look it up.

However, an azeotrope is only relevant if you are boiling something and...
VeritasC&E
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 Quote: Originally posted by unionised They might form an azeotrope, I think so, but I'd need to look it up. However, an azeotrope is only relevant if you are boiling something and...

Yes, we are not boiling.

What I mean is that unless they behave like such, I don't understand quite why I would get in the receiver flask the same concentration as in the original concentration when the BP of the two miscibles are so far appart. Per my limited understanding, I wouldn't expect that to happen. I'm not sure how to use their relative vapour pressures but I'd expect that they could be used to indicate what the concentration in the receiver might look like (I would expect it to be lower, or even somewhat higher, but not the same as in the initial solution).

[Edited on 3-9-2020 by VeritasC&E]
unionised
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Because, in principle, you evaporate all of the contents of the distillation flask and condense all of it in the receiver.
VeritasC&E
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 Quote: Originally posted by unionised Because, in principle, you evaporate all of the contents of the distillation flask and condense all of it in the receiver.

Is there no risk of overconcentrating the H2O2 in either the initial portion of the receiver or (more likely) in the very late portion of the source flask?

When only half of the solution has been carried over, do you know if the receiver will have an increased or rather a decreased H2O2 concentration?

I know 9% H2O2 is very safe, but I also know H2O2 vapours or more concentrated solutions can be dangerous, so it's important that I can visualize the dynamics of the process to know where H2O2 can get concentrated in case I need to do something to avoid it.

Thanks a lot for your contributions by the way!

[Edited on 3-9-2020 by VeritasC&E]
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Could the method of purification by crystallisation with pure urea or sodium carbonate be as good as distillation?

Also, you will unable to remove some usual tap water organic or gases like NH3 with a distillation. By the way, the presence of organic matter and small quantities of resulting organic peroxides probably is something which should be considered when performing distillation, what do you think?

[Edited on 3-9-2020 by teodor]

[Edited on 3-9-2020 by teodor]
unionised
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Quote: Originally posted by VeritasC&E
 Quote: Originally posted by unionised Because, in principle, you evaporate all of the contents of the distillation flask and condense all of it in the receiver.

Is there no risk of overconcentrating the H2O2 in either the initial portion of the receiver or (more likely) in the very late portion of the source flask?

When only half of the solution has been carried over, do you know if the receiver will have an increased or rather a decreased H2O2 concentration?

I know 9% H2O2 is very safe, but I also know H2O2 vapours or more concentrated solutions can be dangerous, so it's important that I can visualize the dynamics of the process to know where H2O2 can get concentrated in case I need to do something to avoid it.

Thanks a lot for your contributions by the way!

[Edited on 3-9-2020 by VeritasC&E]

When you are half way through, one container or the other will hold more than 9% H2O2
My guess is that water will evaporate preferentially at first leaving more concentrated H2O2 in the distillation flask and a more dilute solution in the receiver.
VeritasC&E
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Quote: Originally posted by unionised
Quote: Originally posted by VeritasC&E
 Quote: Originally posted by unionised Because, in principle, you evaporate all of the contents of the distillation flask and condense all of it in the receiver.

Is there no risk of overconcentrating the H2O2 in either the initial portion of the receiver or (more likely) in the very late portion of the source flask?

When only half of the solution has been carried over, do you know if the receiver will have an increased or rather a decreased H2O2 concentration?

I know 9% H2O2 is very safe, but I also know H2O2 vapours or more concentrated solutions can be dangerous, so it's important that I can visualize the dynamics of the process to know where H2O2 can get concentrated in case I need to do something to avoid it.

Thanks a lot for your contributions by the way!

[Edited on 3-9-2020 by VeritasC&E]

When you are half way through, one container or the other will hold more than 9% H2O2
My guess is that water will evaporate preferentially at first leaving more concentrated H2O2 in the distillation flask and a more dilute solution in the receiver.

Is there no danger towards the end of the process?
VeritasC&E
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 Quote: Originally posted by teodor Could the method of purification by crystallisation with pure urea or sodium carbonate be as good as distillation? Also, you will unable to remove some usual tap water organic or gases like NH3 with a distillation. By the way, the presence of organic matter and small quantities of resulting organic peroxides probably is something which should be considered when performing distillation, what do you think? [Edited on 3-9-2020 by teodor]

Hello! Could you describe that method?

[Edited on 3-9-2020 by VeritasC&E]
Fyndium
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I'm under impression H2O2 can be safely vacuum distilled to a high concentration.
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text:
# Magnetically deliverable calcium phosphate nanoparticles for localized gene expression

Puddu, Michela; Broguiere, Nicolas; Mohn, Dirk; Zenobi-Wong, Marcy; Stark, Wendelin J; Grass, Robert N (2015). Magnetically deliverable calcium phosphate nanoparticles for localized gene expression. RSC Advances, 5(13):9997-10004.

## Abstract

Iron oxide doped tricalcium phosphate (Fe$_2$O$_3$@TCP) nanoparticles were designed as transfection vehicles and prepared by flame spray synthesis. Both components are known to be non-toxic and biocompatible. Calcium phosphate (CaP) facilitates DNA entry into cells without the need for toxic cationic mediators, while magnetic iron oxide allows for particle localization at a target site. Flame spray synthesis ensures easy and low-cost nanoparticle production in a reproducible way. Fe$_2$O$_3$@TCP nanoparticles, exhibiting DNA-binding capacity in the presence of CaCl$_2$, were tested for transfection of a green fluorescent protein (GFP) encoding plasmid with Human Embryonic Kidney 293 (HEK 293) cells. Commercial magnetic agents, polyethylenimine (PEI) and standard calcium phosphate-mediated transfection were used for comparison. Transfection efficiency was estimated by GFP expression detected by fluorescence microscopy, while hoechst/ethidium homodimer-1 staining allowed the evaluation of method toxicity. We were able to efficiently transfect HEK 293 cells, and showed that Fe$_2$O$_3$@TCP particles and bound DNA can be concentrated in specific sites in a culture plate through the application of a magnetic field gradient to achieve localized transfection. While the commercial magnetic controls strongly affected cell growth and morphology, Fe$_2$O$_3$@TCP particles did not show marked toxicity and had only limited effects on cell proliferation. Overall performance in terms of transfection efficiency, cell proliferation and viability, were comparable to that of CaP and PEI, which lack magnetic targeting capability. The newly synthetized Fe$_2$O$_3$@TCP are, therefore, improved tools to deliver nucleic acids into cells and achieve spatial control of transfection.

## Abstract

Iron oxide doped tricalcium phosphate (Fe$_2$O$_3$@TCP) nanoparticles were designed as transfection vehicles and prepared by flame spray synthesis. Both components are known to be non-toxic and biocompatible. Calcium phosphate (CaP) facilitates DNA entry into cells without the need for toxic cationic mediators, while magnetic iron oxide allows for particle localization at a target site. Flame spray synthesis ensures easy and low-cost nanoparticle production in a reproducible way. Fe$_2$O$_3$@TCP nanoparticles, exhibiting DNA-binding capacity in the presence of CaCl$_2$, were tested for transfection of a green fluorescent protein (GFP) encoding plasmid with Human Embryonic Kidney 293 (HEK 293) cells. Commercial magnetic agents, polyethylenimine (PEI) and standard calcium phosphate-mediated transfection were used for comparison. Transfection efficiency was estimated by GFP expression detected by fluorescence microscopy, while hoechst/ethidium homodimer-1 staining allowed the evaluation of method toxicity. We were able to efficiently transfect HEK 293 cells, and showed that Fe$_2$O$_3$@TCP particles and bound DNA can be concentrated in specific sites in a culture plate through the application of a magnetic field gradient to achieve localized transfection. While the commercial magnetic controls strongly affected cell growth and morphology, Fe$_2$O$_3$@TCP particles did not show marked toxicity and had only limited effects on cell proliferation. Overall performance in terms of transfection efficiency, cell proliferation and viability, were comparable to that of CaP and PEI, which lack magnetic targeting capability. The newly synthetized Fe$_2$O$_3$@TCP are, therefore, improved tools to deliver nucleic acids into cells and achieve spatial control of transfection.

## Statistics

### Citations

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Item Type: Journal Article, refereed, original work 04 Faculty of Medicine > Center for Dental Medicine > Clinic for Preventive Dentistry, Periodontology and Cariology 610 Medicine & health English 2015 15 Oct 2015 13:54 05 Apr 2016 19:27 RSC Publishing 2046-2069 Publisher DOI. An embargo period may apply. https://doi.org/10.1039/C4RA13413C

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text:
# Dean Number

Written by Jerry Ratzlaff on . Posted in Dimensionless Numbers

Dean number, abbreviated De, is a dimensionless number used in momentum transfer for the flow in curved pipes and channels. The equation and calculation is shown below.

 $$\large{ De = \sqrt { \frac {d}{2r} } \frac {\rho v d}{ \mu } = \sqrt { \frac { d } { 2 r } } Re }$$ Where: $$\large{ De }$$ = Dean number $$\large{ \rho }$$  (Greek symbol rho) = density of the fluid $$\large{ d }$$ = diameter $$\large{ \mu }$$  (Greek symbol mu) = dynamic viscosity $$\large{ r }$$ = radius of curviture of the path of channel $$\large{ Re }$$ = Reynolds number $$\large{ v }$$ = axial velocity scale


text:
# maxentropy

0th

Percentile

##### Maximum entropy sampling

Starting point is a network $A[F]$ with $nf$ points. Now one has to select $ns$ points of a set of candidate sites to augment the existing network. The aim of maximum entropy sampling is to select a feasible D-optimal design that maximizes the logarithm of the determinant of all principal submatrices of $A$ arising by this expansion.

This algorithm is based on the interlacing property of eigenvalues. It starts with an initial solution given directly or provided by the greedy or dual-greedy approach. It uses a branch-and-bound strategy to calculate an optimal solution.

It is also possible to construct a completely new network, that means $nf=0$.

Keywords
spatial , design
##### Usage
maxentropy(A,nf,ns,method="d",S.start=NULL,rtol=1e-6,mattest=TRUE,etol=0,verbose=FALSE)
##### Arguments
A
Spatial covariance matrix $A$.
nf
Number of stations are forced into every feasible solution.
ns
Number of stations have to be added to the existing network.
method
Method to determine the initial solution: "d"=dual-greedy algorithm, "g"=greedy algorithm, "dc"=dual-greedy + interchange algorithm, "gc"=greedy + interchange algorithm, "c"=interchange algorithm + directly given initial solution. Otherwise this algorithm has to be started with an directly given initial solution.
S.start
Vector that gives the $ns$ indices contained in the initial solution of dimension $dim(A)[1]-nf$ that should to be improved.
rtol
The algorithm terminates if the optimal solution is obtained with a tolerance of $rtol$.
mattest
Logical, if TRUE a tes for for symmetry and positive definiteness of the matrix $A$ isperformed (default is TRUE).
etol
Tolerance for checking positve definiteness (default 0).
verbose
Logical, if TRUE some information is printed per iteration (default is FALSE).
##### Details

$A[F]$ denotes the principal submatrix of $A$ having rows and columns indexed by $1..nf$.

##### Value

A object of class monet containing the following elements:
S
Vector containing the indices of the added sites in the initial solution or 0 for the other sites.
det.start
Determinant of the principal submatrix indexed by the initial solution.
det
Determinant of the principal submatrix indexed by the optimal solution.
maxcount
Maximum of active subproblems.
iter
Number of iterations.

##### References

Ko, Lee, Queyranne, An exact algorithm for maximum entropy sampling, Operations Research 43 (1995), 684-691.

Gebhardt, C.: Bayessche Methoden in der geostatistischen Versuchsplanung. PhD Thesis, Univ. Klagenfurt, Austria, 2003

O.P. Baume, A. Gebhardt, C. Gebhardt, G.B.M. Heuvelink and J. Pilz: Network optimization algorithms and scenarios in the context of automatic mapping. Computers & Geosciences 37 (2011) 3, 289-294

greedy, dualgreedy, interchange

• maxentropy
##### Examples
x <- c(0.97900601,0.82658702,0.53105628,0.91420190,0.35304969,
0.14768239,0.58000004,0.60690101,0.36289026,0.82022147,
0.95290664,0.07928365,0.04833764,0.55631735,0.06427738,
0.31216689,0.43851418,0.34433556,0.77699357,0.84097327)
y <- c(0.36545512,0.72144122,0.95688671,0.25422154,0.48199229,
0.43874199,0.90166634,0.60898628,0.82634713,0.29670695,
0.86879093,0.45277452,0.09386800,0.04788365,0.20557817,
0.61149264,0.94643855,0.78219937,0.53946353,0.70946842)
A <- outer(x, x, "-")^2 + outer(y, y, "-")^2
A <- (2 - A)/10
diag(A) <- 0
diag(A) <- 1/20 + apply(A, 2, sum)

S.entrp<-c(0,7,0,9,0,11,0,13,14,0,0,0,0,0,0)
maxentropy(A,5,5,S.start=S.entrp)
maxentropy(A,5,5,method="g")
maxentropy(A,5,5)

Documentation reproduced from package edesign, version 1.0-13, License: GPL (>= 2.0)

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