text:
# pair_style lj/cut/tip4p/cut command¶

Accelerator Variants: lj/cut/tip4p/cut/omp

# pair_style lj/cut/tip4p/long command¶

Accelerator Variants: lj/cut/tip4p/long/gpu, lj/cut/tip4p/long/omp, lj/cut/tip4p/long/opt

## Syntax¶

pair_style style args

• style = lj/cut/tip4p/cut or lj/cut/tip4p/long

• args = list of arguments for a particular style

lj/cut/tip4p/cut args = otype htype btype atype qdist cutoff (cutoff2)
otype,htype = atom types for TIP4P O and H
btype,atype = bond and angle types for TIP4P waters
qdist = distance from O atom to massless charge (distance units)
cutoff = global cutoff for LJ (and Coulombic if only 1 arg) (distance units)
cutoff2 = global cutoff for Coulombic (optional) (distance units)
lj/cut/tip4p/long args = otype htype btype atype qdist cutoff (cutoff2)
otype,htype = atom types for TIP4P O and H
btype,atype = bond and angle types for TIP4P waters
qdist = distance from O atom to massless charge (distance units)
cutoff = global cutoff for LJ (and Coulombic if only 1 arg) (distance units)
cutoff2 = global cutoff for Coulombic (optional) (distance units)

## Examples¶

pair_style lj/cut/tip4p/cut 1 2 7 8 0.15 12.0
pair_style lj/cut/tip4p/cut 1 2 7 8 0.15 12.0 10.0
pair_coeff * * 100.0 3.0
pair_coeff 1 1 100.0 3.5 9.0

pair_style lj/cut/tip4p/long 1 2 7 8 0.15 12.0
pair_style lj/cut/tip4p/long 1 2 7 8 0.15 12.0 10.0
pair_coeff * * 100.0 3.0
pair_coeff 1 1 100.0 3.5 9.0

## Description¶

The lj/cut/tip4p styles compute the standard 12/6 Lennard-Jones potential, given by

$E = 4 \epsilon \left[ \left(\frac{\sigma}{r}\right)^{12} - \left(\frac{\sigma}{r}\right)^6 \right] \qquad r < r_c$

$$r_c$$ is the cutoff.

They add Coulombic pairwise interactions given by

$E = \frac{C q_i q_j}{\epsilon r} \qquad r < r_c$

where $$C$$ is an energy-conversion constant, $$q_i$$ and $$q_j$$ are the charges on the 2 atoms, and $$\epsilon$$ is the dielectric constant which can be set by the dielectric command. If one cutoff is specified in the pair_style command, it is used for both the LJ and Coulombic terms. If two cutoffs are specified, they are used as cutoffs for the LJ and Coulombic terms respectively.

Style lj/cut/tip4p/long compute the same Coulombic interactions as style lj/cut/tip4p/cut except that an additional damping factor is applied to the Coulombic term so it can be used in conjunction with the kspace_style command and its ewald or pppm option. The Coulombic cutoff specified for this style means that pairwise interactions within this distance are computed directly; interactions outside that distance are computed in reciprocal space.

The lj/cut/tip4p styles implement the TIP4P water model of (Jorgensen), which introduces a massless site located a short distance away from the oxygen atom along the bisector of the HOH angle. The atomic types of the oxygen and hydrogen atoms, the bond and angle types for OH and HOH interactions, and the distance to the massless charge site are specified as pair_style arguments. Style lj/cut/tip4p/cut uses a cutoff for Coulomb interactions; style lj/cut/tip4p/long is for use with a long-range Coulombic solver (Ewald or PPPM).

Note

For each TIP4P water molecule in your system, the atom IDs for the O and 2 H atoms must be consecutive, with the O atom first. This is to enable LAMMPS to “find” the 2 H atoms associated with each O atom. For example, if the atom ID of an O atom in a TIP4P water molecule is 500, then its 2 H atoms must have IDs 501 and 502.

See the Howto tip4p page for more information on how to use the TIP4P pair styles and lists of parameters to set. Note that the neighbor list cutoff for Coulomb interactions is effectively extended by a distance 2*qdist when using the TIP4P pair style, to account for the offset distance of the fictitious charges on O atoms in water molecules. Thus it is typically best in an efficiency sense to use a LJ cutoff >= Coulombic cutoff + 2*qdist, to shrink the size of the neighbor list. This leads to slightly larger cost for the long-range calculation, so you can test the trade-off for your model.

## Coefficients¶

For all of the lj/cut pair styles, the following coefficients must be defined for each pair of atoms types via the pair_coeff command as in the examples above, or in the data file or restart files read by the read_data or read_restart commands, or by mixing as described below:

• $$\epsilon$$ (energy units)

• $$\sigma$$ (distance units)

• LJ cutoff (distance units)

Note that $$\sigma$$ is defined in the LJ formula as the zero-crossing distance for the potential, not as the energy minimum at $$2^{\frac{1}{6}} \sigma$$.

The last coefficient is optional. If not specified, the global LJ cutoff specified in the pair_style command is used.

For lj/cut/tip4p/cut and lj/cut/tip4p/long only the LJ cutoff can be specified since a Coulombic cutoff cannot be specified for an individual I,J type pair. All type pairs use the same global Coulombic cutoff specified in the pair_style command.

Warning

Because of how these pair styles implement the coulomb interactions by implicitly defining a fourth site for the negative charge of the TIP4P and similar water models, special care must be taken when using these pair styles with other computations that also use charges. Unless they are specially set up to also handle the implicit definition of the 4th site, results are likely incorrect. Example: compute dipole/chunk. For the same reason, when using one of these pair styles with pair_style hybrid, all coulomb interactions should be handled by a single sub-style with TIP4P support. All other instances and styles will “see” the M point charges at the position of the Oxygen atom and thus compute incorrect forces and energies. LAMMPS will print a warning when it detects one of these issues.

A version of these styles with a soft core, lj/cut/tip4p/long/soft, suitable for use in free energy calculations, is part of the FEP package and is documented with the pair_style */soft styles.

Styles with a gpu, intel, kk, omp, or opt suffix are functionally the same as the corresponding style without the suffix. They have been optimized to run faster, depending on your available hardware, as discussed on the Accelerator packages page. The accelerated styles take the same arguments and should produce the same results, except for round-off and precision issues.

These accelerated styles are part of the GPU, INTEL, KOKKOS, OPENMP, and OPT packages, respectively. They are only enabled if LAMMPS was built with those packages. See the Build package page for more info.

You can specify the accelerated styles explicitly in your input script by including their suffix, or you can use the -suffix command-line switch when you invoke LAMMPS, or you can use the suffix command in your input script.

See the Accelerator packages page for more instructions on how to use the accelerated styles effectively.

## Mixing, shift, table, tail correction, restart, rRESPA info¶

For atom type pairs I,J and I != J, the epsilon and sigma coefficients and cutoff distance for all of the lj/cut pair styles can be mixed. The default mix value is geometric. See the “pair_modify” command for details.

All of the lj/cut pair styles support the pair_modify shift option for the energy of the Lennard-Jones portion of the pair interaction.

The lj/cut/coul/long and lj/cut/tip4p/long pair styles support the pair_modify table option since they can tabulate the short-range portion of the long-range Coulombic interaction.

All of the lj/cut pair styles support the pair_modify tail option for adding a long-range tail correction to the energy and pressure for the Lennard-Jones portion of the pair interaction.

All of the lj/cut pair styles write their information to binary restart files, so pair_style and pair_coeff commands do not need to be specified in an input script that reads a restart file.

The lj/cut and lj/cut/coul/long pair styles support the use of the inner, middle, and outer keywords of the run_style respa command, meaning the pairwise forces can be partitioned by distance at different levels of the rRESPA hierarchy. The other styles only support the pair keyword of run_style respa. See the run_style command for details.

## Restrictions¶

The lj/cut/tip4p/long styles are part of the KSPACE package. The lj/cut/tip4p/cut style is part of the MOLECULE package. These styles are only enabled if LAMMPS was built with those packages. See the Build package page for more info.

## Default¶

none

(Jorgensen) Jorgensen, Chandrasekhar, Madura, Impey, Klein, J Chem Phys, 79, 926 (1983).


text:
# Questions tagged [forecasting]

Forecasting is the process predicting future values based on historic and current data, typically for time-series datasets.

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## Hitchin Pairs for non-compact real Lie groups

29 Jul 2016  ·  Gothen Peter B. Porto ·

Hitchin pairs on Riemann surfaces are generalizations of Higgs bundles, allowing the Higgs field to be twisted by an arbitrary line bundle. We consider this generalization in the context of $G$-Higgs bundles for a real reductive Lie group $G$... We outline the basic theory and review some selected results, including recent results by Nozad and the author arXiv:1602.02712 [math.AG] on Hitchin pairs for the unitary group of indefinite signature $\mathrm{U}(p,q)$. read more

PDF Abstract

# Code Add Remove Mark official

No code implementations yet. Submit your code now

# Categories

Algebraic Geometry


text:
# In the Following Figure, If Abc is an Equilateral Triangle, Then Shaded Area is Equal to - Mathematics

MCQ

In the following figure, If ABC is an equilateral triangle, then shaded area is equal to

#### Options

• $\left( \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right) r^2$

• $\left( \frac{\pi}{3} - \frac{\sqrt{3}}{2} \right) r^2$

• $\left( \frac{\pi}{3} + \frac{\sqrt{3}}{4} \right) r^2$

• $\left( \frac{\pi}{3} + \sqrt{3} \right) r^2$

#### Solution

We have given that ABC is an equilateral triangle.

∴ ∠A=60°

As we know that,∠BCA=1/2 m (∠BOC)

∴ 60=1/2 m (BOC)

m(∠BOC)=120°

Area of the shaded region = area of the segment BC.

Let  ∠BOC=θ

∴ Area of the segment= (piθ/360-sin  θ/2 cos  θ/2)

Substituting the values we get,

Area of the segment= ((pixx120)/360-sin60cos60)r^2

∴ Area of the segment=(pi/3-sin60cos60)r^2

Substituting sin 60=sqrt3/2 and 60=1/2we get,

∴ Area of the segment=(pi/3-1/2xxsqrt3/2)r^2

∴ Area of the segment=(pi/3-sqrt3/4)r^2

Therefore, area of the shaded region is (pi/3-sqrt3/4)r^2.

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 13 Areas Related to Circles
Q 37 | Page 72


text:
# Discussion: IC specific tags

This was quickly revised by a suggested edit to remove that tag, and to stick with suitable generics.

I had sympathy with that view so approved the edit... in my view, we're not Electronics.SE so are unlikely to have too may questions relating to any specific IC, never mind enough to justify having individual tags. And in the unlikely event that a single IC gains critical mass, then we can always introduce that tag later.

But now I'm having second thoughts... so I'd like to sound out the group so we can establish a general policy.

1. IC specific tags are OK for all ICs
2. IC specific tags are OK for common ICs
3. IC specific tags are not OK
4. Something else

Discuss...

IMO, it should be

3) IC specific tags are not OK

Tags are for searching/filtering purposes. I really don't see a scenario when a user wants to search/filter questions only about a certain IC. Maybe an IC family, but not a certain IC.

What's better would be tags for classes of ICs. E.g. , , etc. (not saying those would be good tags for Robotics.SE, just examples)

• Agree that classes of IC are OK (that's what I meant with suitable generics) although I'd question even 78xx - rather voltage-regulators or similar
– Andrew Mod
Dec 23, 2012 at 20:18
• @Andrew: Ah, oops -- I didn't notice that :s Dec 23, 2012 at 20:20
• Even Electronics.SE barely has IC-specific tags at all, which is why I made the edit. The most specific tags over there are mostly for MCU families, [avr], [stm32], [8051] etc. There are little-used [lm317] and [74hc595] tags, but those are very common parts and the only IC-specifics tag I can find in the first 18 of the 31 pages of tags on Electronics. If they're too specific for broad use on Electronics, we certainly have no need for them here. Jan 10, 2013 at 13:02


text:
News

IXYS Reports Fiscal 2003 Results

June 01, 2003 by Jeff Shepard

IXYS Corp. (Santa Clara, CA) released its financial results for its fourth fiscal quarter and its fiscal year ended March 31, 2003. For the fourth fiscal quarter ending March 31, 2003, IXYS reported record net revenues of $37.8 million, compared with net revenues of$20.4 million reported for the same period in fiscal year 2002, which represented 85.4% period-to-period revenue growth. For the fiscal year ended March 31, 2003, IXYS reported net revenues of $136.1 million, compared to net revenues of$82.8 million for the prior fiscal year, which represented 64.3% year-to-year revenue growth.

Net loss for the fiscal quarter ending March 31, 2003, was $5.6 million, or$0.18 per diluted share, as compared to a net loss of $582,000, or$0.02 per diluted share, in the same period for the prior fiscal year. Fourth-quarter results include a charge for excess and obsolete inventory of $7 million and litigation expenses of$2.4 million. Net loss for the fiscal year ended March 31, 2003, was $13.2 million, or$0.43 per diluted share, as compared to net income of $1.9 million, or$0.07 per diluted share, reported for the prior fiscal year.


text:
# Complex derivatives

Logik
I have to solve an ODE with variation of coefficient technique. It's pretty easy but I have no clue what is the first and second derivative of e^ix and e^-ix.

Last edited:

Homework Helper
Gold Member
Dearly Missed
If i had been a real number, what would the first and second derivatives have been then?

Logik
e^ix
first
i*e^ix
second
i^2*e^ix

e^-ix
first
-i*e^-ix
second
i^2*e^-ix

p.s. I've read about the Cauchy-Riemann equation but just not sure how to apply it... should I split the exponential in a sin and a cos?
p.s.s. There are probably rules, like exponential function are always derivable or something but I'm not fallowing any complex variables class right now so any insight is appreciated...

Last edited:
Homework Helper
Gold Member
Dearly Missed
e^ix
first
i*e^ix
second
i^2*e^ix

e^-ix
first
-i*e^-ix
second
i^2*e^-ix

EXACTLY!
And that is precisely what holds when "i" is a complex/imaginary number as well!

CivilSigma
Homework Helper
When dealing with these things, forget i is anything, just remember its a constant. Then after the actual differentiation, you can remember what it is.

masudr
Yeah. If

$$\exp(ix),\,\,\,x\in \mathbb{R},$$

(which is what it looks like you have) then it's what the above two said. But if you have

$$\exp(iz),\,\,\,z\in \mathbb{Z},$$

you need to be more careful. Let us know if that is indeed what you have.

Homework Helper
what are you doing in a de course ifm you do not know the derivative of e^z?

bgbaby
I'm doing the same derivative problem & i was wondering if you could give any tips on how to solve the derivative of e^ix? I would really appreciate it. A good reference website, anything assistance at all.

Homework Helper
I'm doing the same derivative problem & i was wondering if you could give any tips on how to solve the derivative of e^ix? I would really appreciate it. A good reference website, anything assistance at all.
That is exactly what has been answered in each of these responses. For any constant, a, the derivative of $e^{ax}$ is $ae^{ax}$.

That is a result of the very basic fact that the derivative of $e^x$ is $e^x$ (world's easiest derivative!) and the chain rule.

klondike
I'm doing the same derivative problem & i was wondering if you could give any tips on how to solve the derivative of e^ix? I would really appreciate it. A good reference website, anything assistance at all.

$$\frac{d}{dx}(e^{jx})=je^{jx}$$
$$\frac{d^2}{dx^2}(e^{jx})=-e^{jx}$$

Actually I myself was once in an ode course when I had forgot the derivative of e^x. My solution was to go get a Schaum's outline series of ode and do a lot of problems and review my \$ off.


text:
# Empirical change of DAG for Bayesian Network

While working on some problem in bioinformatics I applied bayesian network algorithm for classification purposes. As predictors I took a window of sequence of aminoacids and for dependent variable I used some feature of interest of a central acid.

For DAG estimation I used several different algorithms which gave me identical network structures, and all of them used BIC score during the DAG search. Obtained DAG structure consisted of two chains: X7-X5-X3-X1, and X6-X4-X2, while all Xs where connected with Y. Using this network I got my classification accuracy up to 65%.

The problem is that motivation for my small research was to check if Bayesian Networks can beat Markov chain procedure developed by my collegues for same problem. Basically my collegues obtained 81% accuracy by applying simple Bayesian procedure for model in which sequence of amino-acids was a Markov chain (they also used some pretty tricky empirical methods to obtain such a great result).

So, because my collegues showed that Markov Chain of second order gives best classification accuracy, I just modified my Bayesian Network by connecting two chains described above, so that each amino-acid was connected with two others (not one as before). I then used obtained structure and got classification accuracy 92% (which is pretty awesome for my task).

So, my questions are:

1. How can I justify changes of DAG I made to improve my model. Of course I can (and I will) mention results of my collegues, but still, its more intuitive explanation. Maybe there is better reasoning then intuition here

2. What can be a possible reason for algorithms not finding optimal structure? (My current explanation is that BIC has just too strong penalty for model's complexity and algorithms just got into some local extremum, and weren't able to get out of it because of BIC's complexity penalty. DO you think it's worth to try and develop some more accurate score for my particular problem, to obtain correct DAG structure? What may be the other explanations?)

• @william, please stop spamming the site w/ tag edits. This shouldn't be done this way. You should raise the issue on meta.CV & we can make it a synonym. – gung - Reinstate Monica May 1 '16 at 19:32


text:
## Analysis and classic odes – In exponential polynomials

Suppose we have the following function $$f: mathbb {R} ^ {+} mapsto mathbb {R}$$
$$f (t) = sum_ {i = 1} ^ k P_i (t) exp ( alpha_i t),$$
where $$alpha_i$$s are all algebraic numbers and $$P_i (t)$$ are all polynomials with algebraic coefficients and a minor degree that $$m$$.

There are several questions that interest me.

to. What is the maximum number? $$p$$ such that there is a $$t_0$$ and for all $$r leq p$$
$$f ^ {(r)} (t_0) = 0.$$
second. Suppose $$t_1, cdots, t_q, cdots$$ they are the true roots of $$f (t) = 0$$Is it possible to have a convergent sequence of $$t_q$$? In other words, is it possible to have a Cauchy sequence? $$t_i$$ such that $$f (t_i) = 0$$? If not, do we have a lower limit on the distance between different roots?

c. Is there an algorithm to calculate all the common real roots of $$f (t)$$ Y $$f & # 39; (t)$$?

## Analysis and classic odes – Positivity and zeros of the Heun function

I am interested in understanding in which part of the complex a function of Heun can vanish or where (or its real part, for example) is positive. I have consulted Ronveaux's book, as well as the NIST Math Function Manual, and I could not find any information about this problem. Is there anything known in that direction? You could find information about where this happens for a hypergeometric function.

## Analysis and classic odes – Real analytical function with a given set of values

We say that a strictly growing sequence. $$x_n$$ of reals converges fast
to $$x$$, if for each $$k in mathbb {N}$$ sequence $$n ^ k cdot (x_n – x)$$ is
locked. It is known that there is a $$C ^ infty$$-function $$f$$ such that $$f (1 / n) = x_n$$ Y $$f (0) = x$$.

In which case (sufficient condition in $$x_n$$) exists real analytical function $$g$$ such that $$g (1 / n_k) = x_ {n_k}$$ Y $$g (0) = x$$ by some subsequence $$x_ {n_k}$$ ?

## Analysis and classic odes – L1 linear approximation by parts of the convex function of a variable

Consider a convex function f (x) of a variable x in some interval [a,b].

Question:
What is known about the approximation L1 of f (x) by linear functions in parts?
How to find nodes?
Are there some algorithms and / or theoretical results?

Geometrically it means that we need to find a linear curve by sections that is below it, it is the closest area below f (x), in other words, the area of secters is minimal.

If you need to find a single node, a clear geometry is easily obtained.
solution: place the node at the point x where the tanget is parallel to the chord f (a) -> f (b). (See figure 1).
If we have many points, this gives the condition of connection node with the neighboring nodes, so one could think of the nodes as in some system with peer interaction.

I have a dynamic programming algorithm, but I think you should know a lot in s

## Analysis and classic odes – Product of values of a function with matrix value over \$ S ^ 1 \$

Assume you have a function $$f: S ^ 1 rightarrow mathrm {GL} _d ( mathbb {C})$$ whose coefficients are Laurent polynomials $$f_ {i, j} (q) in mathbb {C}[q^{pm 1}]$$

I am interested in obtaining the conditions for the spectral radius of the matrix.
$$F_n = f (e ^ { frac {2i pi (n-1)} {n}}) ldots f (e ^ { frac {2i pi} {n}}) f (1)$$
be growing exponentially with $$n.$$

Yes $$d = 1,$$ means that $$f (q)$$ is a single polynomial of Laurent, you can easily find it $$F_n = e ^ {n m (f) + o (1)},$$ where $$m (f)$$ it's Mahler's measure of $$f.$$
If we also assume $$f (q) in mathbb {Z}[q^{pm 1}]$$ so $$F_n$$ is growing exponentially as long as $$f$$ It is not a product of cyclotomic polynomials.

When $$d> 1,$$ You can think about using the determinant, but that does not give you any information if $$f$$ take values in $$mathrm {SL} _d ( mathbb {C}).$$

Of course there are many examples where the spectral radio of $$F_n$$ it does not grow exponentially: yes $$f$$ takes value in $$mathrm {SU} _d$$ he also does $$F_n$$ Y $$F_n$$ has a spectral radius 1. The same would apply to the upper triangular matrices with diagonal $$1.$$

Ideally, these counterexamples would be general, which is the spectral radius of $$F_n$$ It grows exponentially or the image of $$f$$ live in a smaller lying group that $$mathrm {SL} _d ( mathbb {C}).$$ But any condition sufficient for the spectral radius to grow exponentially would be interesting to me.

## Special functions: evaluation of the asymptotic equations of parabolic cylinders that arise from coupled ODEs.

For the context, I'm studying Coulomb paper blocking in Averin's superconducting quantum dot contacts since 1998. Specifically, I'm trying to find out how he gets equation 11 from equation 10, which gives Landau Zener the probability of end up in a specific branch of Josephson's potential of a superconducting QPC.

Equation 10, which describes Schrodinger's equation of the problem at the specific limit that is considered, is given by a system of two coupled first-order ODEs:
$$begin {equation} 2 sqrt { frac {E_C} { Delta}} frac { partial psi_s} { partial x} = -s x psi_s / 2 + sqrt {R} psi _ {- s} end {equation}$$
where $$s pm1$$.

If I (for convenience) I take now $$A = 2 sqrt { frac {E_C} { Delta}}$$ Y $$B = sqrt {R}$$Then, substituting the differential equations, there is a second order ODE that is resolved by parabolic cylinder functions. Specifically,
$$begin {equation} psi _ {- 1} = c_1 D { frac {-B ^ 2-A} {A}} left ( frac {x} { sqrt {A}} right) + c_2 D _ { frac {B ^ 2} {A}} left ( frac {i x} { sqrt {A}} right) end {equation}$$

Y

$$begin {equation} psi_1 = – frac {1} {B} left (x c_2 D _ { frac {B ^ 2} {A}} left ( frac {ix} { sqrt {A}} right) + sqrt { TO } left[c_1 D_{-frac{B^2}{A}}left(frac{x}{sqrt{A}}right)+i c_2 D_{frac{B^2+A}{A}}left(frac{i x}{sqrt{A}}right)right]straight) \ end {equation}$$

Now, in his article, he says that by evaluating the asymptotes of these functions, one can find the probability $$w$$ for the state $$s = 1$$ (beloning to $$psi_1$$) from $$x rightarrow – infty$$ finish in the state $$s = -1$$ to $$x rightarrow infty$$. This leads to equation 11, which writes as
$$begin {equation} w = frac {1} { Gamma ( lambda)} sqrt { frac {2 pi} { lambda}} left ( frac { lambda} {e} right) ^ lambda end {equation}$$

where $$lambda = frac {R ^ 2} {2 sqrt {E_C / Delta}}$$, and using our substitutions, we can identify $$lambda = B ^ 2 / A$$, that we already saw happen in the solution for $$psi _ {- 1}$$.

My question is how do you get this? The language in the document is (to me) a bit uncertain, but from the physics, I know that we must choose our boundary conditions (configuration $$c_1$$ Y $$c_2$$) such that $$| psi_1 (- infty) | ^ 2 = 1$$, $$| psi _ {- 1} (- infty) | ^ 2 = 0$$ and then evaluate $$| psi _ {- 1} ( infty) | ^ 2$$ Arrive $$w$$. But it seems that I can not find a way to correctly find the values of $$c_1$$ Y $$c_2$$, since it seems that I can not impose these boundary conditions. How is one of these asymptotics treated?

Maybe you already find a clue in the form of the answer itself, which clearly looks like a Stirling approach to $$lambda!$$Therefore, it is likely that we are looking for some solution with a Gamma function, but I can not understand it. Also, I must point out that this problem is essentially a problem of Landau Zener tunnels for those who are familiar with that.

Some code to evaluate the ODE & # 39; s:

``````ff = f[x] /. Solve[A*g'[A*g'[A*g'[A*g'[x] == x * g[x]/ 2 + B * f[x], f[x]];
dff = D[ff, x];
gg = g[x] /.
DSolve[{A*f'[{A*f'[{A*f'[{A*f'[x] == -x * f[x]/ 2 + B * g[x] /. {F[x] -> ff,
F & # 39;[x] -> dff}}, g[x], X]dgg = D[gg, x];
ff = FullSimplify[ff/{sun[ff/{g[ff/{sol[ff/{g[x] -> gg, g & # 39;[x] -> dgg}]
``````

## Analysis and classic odes – Lower limit of the standard \$ L ^ 2 \$ of a polynomial

Leave $$sigma> 0$$ be arranged even for $$k in mathbb {N} cup {0 }$$, we consider the polynomial
$$begin {equation} varphi_k (x) = sum_ {j = 0} ^ {k} (-1) ^ j {k choose j} b_j , x ^ {2j} x en (-1,1), end {equation}$$
where
$$begin {equation} b_j = frac { Big (k + sigma + frac12 Big) _j} { Big ( frac12 Big) _j} end {equation}$$
and to $$s in mathbb {R}$$, $$(s) _j$$ denotes the symbol of Pochhammer
$$begin {equation} (s) j = { begin {cases} 1 & j = 0 \ s (s + 1) cdots (s + j-1) & j> 0. end {cases}} end {equation}$$
In particular, $$varphi_0 (x) = 1$$.

My question is this.

by $$-1 exists? $$c> 0$$ such that
$$begin {equation} int_a ^ b | varphi_k (x) | ^ 2 dx geq c end {equation}$$
even for any $$k in mathbb {N} cup {0 }$$?

Unless I'm wrong, a direct calculation yields
$$begin {equation} int_a ^ b | varphi_k (x) | ^ 2 dx = sum_ {j = 0} ^ k sum _ { ell = 0} ^ k (-1) ^ {j + ell} {k choose j} {k choose ell} frac {b_j b _ { ell}} {2 (j + ell) +1} , (b ^ {2 (j + ell) +1} -a ^ {2 (j + ell) +1} ). end {equation}$$
But I do not see how I can join this double sum from below.

Observation: I do not know if it is of any use, one can notice that $$varphi_k$$ It is a hypergeometric function of the form. $${} _ {2} F_ {1} (- k, k + sigma + frac12; frac12; x ^ 2)$$ (See https://en.wikipedia.org/wiki/Hypergeometricometric_function).

Edit: The same question is open when $$k$$ It's strange, but this time you consider it.
$$begin {equation} varphi_k (x) = sum_ {j = 0} ^ {k} (-1) ^ j {k choose j} c_j , x ^ {2j + 1}, end {equation}$$
with
$$begin {equation} c_j = frac { Big (k + sigma + frac32 Big) _j} { Big ( frac32 Big) _j}. end {equation}$$
I suspect that the methodology is similar to that of the even case.

Realize: I have asked the question in MSE, where it is subject to a reward: https://math.stackexchange.com/questions/3154394/bounding-a-polynomial-from-below.

## plotting – How to plot the portiere phase and the ODES solutions

Dear all I know how to plot the portiere phase for the non-linear odes system

But I did not know why the result of `Show` Parametric solutions with portiere phase are not good.

I wrote the code in the following way:

``````Sun[{N0_, I0_}?NumericQ] : =
First @ NDSolve[{N1'
r N1
I1 & # 39;
m + N1
I1[0] == I0}, {N1, I1}, {t, 0, 365}];
P1 = ParametricPlot

The
Evaluate[{N1[{N1[{N1[{N1
30}, PlotRange -> All, AspectRatio -> Complete, PlotRange -> Complete,
Frame -> True, MaxRecursion -> 8]
``````

where

``````r = 0.431201; [Beta] = 2.99 * 10 ^ -6;
[Eta] = 0.2; [Sigma] = 0.7; [Rho] = 0.003; m = 0.427; [Delta] =
0.57; [Mu] = .82;
``````

and I used `StreamPlot` as follows

``````F[N1_, I1_] = r N1 (1 - [Beta] N1) - [Eta] N1 I1;
Sun[N1_, I1_] = [Sigma] + ([Rho] N1 I1) /
m + N1) - [Delta] I1 - [Mu] N1 I1;
Sun[{N1_, I1_}] = {f[N1, I1]g[N1, I1]};
``````

so

``````StreamPlot[{F[{F[{F[{f[N1, I1]g[N1, I1]}, {N1, 0, 30}, {I1, 0, 30},
StreamStyle -> Blue, AspectRatio -> Automatic, Frame -> True,
Axes -> False, AxesLabel -> {"N1", "I1"}]Show[StreamPlot[{F[StreamPlot[{F[StreamPlot[{F[StreamPlot[{f[N1, I1]g[N1, I1]}, {N1, 0, 30}, {I1, 0, 30},
StreamPoints -> Fine, StreamStyle -> Blue, AspectRatio -> 1/2,
Frame -> True, AxesLabel -> {"N1", "I1"}, StreamPoints -> Fine,
PlotRange -> All], P1]
``````

the final result was,

Can someone help me improve my result?


text:
# Prove that the space of functions on [0,1] with |f(x)| $\leq$ 1 has cardinality 2$^{\aleph}$

I'm having trouble with cardinalities and working with them in problems. The way I approached this problem was to note that functions defined on $[0,1]$ with $|f(x)| \leq 1$ will all be contained in the square $[0,1]\times[0,1]$, which is equivalent to $\mathbb{R}^2$ (as $[0,1]$ is equivalent to $\mathbb{R}$). Hence we can see that this space is then equivalent to $[0,1]\times[0,1]^{[0,1]}$, the space of functions from $[0,1]$ to the square $[0,1]\times[0,1]$. Using laws of working with cardinalities, I know that $$|[0,1]\times[0,1]^{[0,1]}| = |[0,1]\times[0,1]|^{|[0,1]|} = \aleph^{\aleph} = (2^{\aleph_0})^{\aleph} = 2^{\aleph_0\aleph} = 2^{\aleph}.$$

I am just unsure of if I have approached this problem correctly. In class, my professor constructed an injection using the power set of the reals, but I am just trying to wrap my head around the idea of proving cardinalities by showing sets are equivalent.

What you have done is completely fine, as long as you are allowed to use cardinal arthritic in your proof. The only step that should require justification is your claim that $\aleph_0 * \aleph = \aleph$ (in other words that $|\mathbb{R} \times \mathbb{N}| = |\mathbb{R}|$).
The other two ways are by showing a explicit bijection or injection. A bijection will (by definition) prove that two cardinalities are equal. Showing an injection will prove that two cardinalities are related by $\leq$, and if you know the Cantor-Bernstein-Schroeder theorem you will know that showing there exists an injection $A \mapsto B$ and an injection $B \mapsto A$ proves that there exists a bijection $A \mapsto B$.
As I mentioned, when you are going down the cardinal arithmetic route - it is only specific sums or products such as $\aleph_0 * \aleph = \aleph$ that will need justification. Think of it like using exponent rules on the natural numbers - you know that $(x^a)^b = x^{a*b}$, but you still have to work out $a*b$ yourself.
• You might also be required to prove that $|[0,1]| = |\mathbb{R}|$ depending on how much is expected of you. You can always swap two sets out if they have the same cardinality, it is proving that their cardinalities are the same that is on you. – Sam Forster Sep 20 '17 at 17:01
• Maybe also that $|\mathbb{R}^2| = |\mathbb{R}|$, again it depends on how much you are allowed to take for granted. If you don't know why $|[0,1]| = |\mathbb{R}|$ or $|\mathbb{R}^2| = |\mathbb{R}|$ yourself, it might be a good idea to explicitly construct some bijections yourself - to help you get the hang of things. – Sam Forster Sep 20 '17 at 17:06


text:
# Properties

 Label 2736.1.fg.a Level $2736$ Weight $1$ Character orbit 2736.fg Analytic conductor $1.365$ Analytic rank $0$ Dimension $6$ Projective image $D_{18}$ CM discriminant -3 Inner twists $4$

# Related objects

## Newspace parameters

 Level: $$N$$ $$=$$ $$2736 = 2^{4} \cdot 3^{2} \cdot 19$$ Weight: $$k$$ $$=$$ $$1$$ Character orbit: $$[\chi]$$ $$=$$ 2736.fg (of order $$18$$, degree $$6$$, not minimal)

## Newform invariants

 Self dual: no Analytic conductor: $$1.36544187456$$ Analytic rank: $$0$$ Dimension: $$6$$ Coefficient field: $$\Q(\zeta_{18})$$ Defining polynomial: $$x^{6} - x^{3} + 1$$ Coefficient ring: $$\Z[a_1, \ldots, a_{13}]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 684) Projective image: $$D_{18}$$ Projective field: Galois closure of $$\mathbb{Q}[x]/(x^{18} - \cdots)$$

## $q$-expansion

The $$q$$-expansion and trace form are shown below.

 $$f(q)$$ $$=$$ $$q + ( -\zeta_{18}^{7} + \zeta_{18}^{8} ) q^{7} +O(q^{10})$$ $$q + ( -\zeta_{18}^{7} + \zeta_{18}^{8} ) q^{7} + ( \zeta_{18}^{2} - \zeta_{18}^{6} ) q^{13} -\zeta_{18}^{3} q^{19} -\zeta_{18}^{4} q^{25} + ( \zeta_{18}^{2} - \zeta_{18}^{4} ) q^{31} + ( \zeta_{18} + \zeta_{18}^{8} ) q^{37} + ( \zeta_{18} + \zeta_{18}^{3} ) q^{43} + ( -\zeta_{18}^{5} + \zeta_{18}^{6} - \zeta_{18}^{7} ) q^{49} + ( -1 + \zeta_{18}^{5} ) q^{61} + ( -\zeta_{18}^{5} - \zeta_{18}^{6} ) q^{67} + ( -\zeta_{18}^{3} - \zeta_{18}^{7} ) q^{73} + ( 1 + \zeta_{18} ) q^{79} + ( 1 - \zeta_{18} - \zeta_{18}^{4} + \zeta_{18}^{5} ) q^{91} + ( -\zeta_{18}^{2} - \zeta_{18}^{5} ) q^{97} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$6q + O(q^{10})$$ $$6q + 3q^{13} - 3q^{19} + 3q^{43} - 3q^{49} - 6q^{61} + 3q^{67} - 3q^{73} + 6q^{79} + 6q^{91} + O(q^{100})$$

## Character values

We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/2736\mathbb{Z}\right)^\times$$.

 $$n$$ $$1009$$ $$1217$$ $$1711$$ $$2053$$ $$\chi(n)$$ $$\zeta_{18}^{7}$$ $$1$$ $$1$$ $$1$$

## Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
433.1
 0.939693 − 0.342020i −0.173648 + 0.984808i −0.173648 − 0.984808i −0.766044 − 0.642788i −0.766044 + 0.642788i 0.939693 + 0.342020i
0 0 0 0 0 −0.173648 + 0.300767i 0 0 0
865.1 0 0 0 0 0 −0.766044 + 1.32683i 0 0 0
1009.1 0 0 0 0 0 −0.766044 1.32683i 0 0 0
1153.1 0 0 0 0 0 0.939693 1.62760i 0 0 0
2017.1 0 0 0 0 0 0.939693 + 1.62760i 0 0 0
2161.1 0 0 0 0 0 −0.173648 0.300767i 0 0 0
 $$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 2161.1 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

## Inner twists

Char Parity Ord Mult Type
1.a even 1 1 trivial
3.b odd 2 1 CM by $$\Q(\sqrt{-3})$$
19.f odd 18 1 inner
57.j even 18 1 inner

## Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 2736.1.fg.a 6
3.b odd 2 1 CM 2736.1.fg.a 6
4.b odd 2 1 684.1.bx.a 6
12.b even 2 1 684.1.bx.a 6
19.f odd 18 1 inner 2736.1.fg.a 6
57.j even 18 1 inner 2736.1.fg.a 6
76.k even 18 1 684.1.bx.a 6
228.u odd 18 1 684.1.bx.a 6

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
684.1.bx.a 6 4.b odd 2 1
684.1.bx.a 6 12.b even 2 1
684.1.bx.a 6 76.k even 18 1
684.1.bx.a 6 228.u odd 18 1
2736.1.fg.a 6 1.a even 1 1 trivial
2736.1.fg.a 6 3.b odd 2 1 CM
2736.1.fg.a 6 19.f odd 18 1 inner
2736.1.fg.a 6 57.j even 18 1 inner

## Hecke kernels

This newform subspace is the entire newspace $$S_{1}^{\mathrm{new}}(2736, [\chi])$$.

## Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$T^{6}$$
$3$ $$T^{6}$$
$5$ $$T^{6}$$
$7$ $$1 + 3 T + 9 T^{2} + 2 T^{3} + 3 T^{4} + T^{6}$$
$11$ $$T^{6}$$
$13$ $$3 - 6 T^{3} + 6 T^{4} - 3 T^{5} + T^{6}$$
$17$ $$T^{6}$$
$19$ $$( 1 + T + T^{2} )^{3}$$
$23$ $$T^{6}$$
$29$ $$T^{6}$$
$31$ $$3 - 9 T + 9 T^{2} - 3 T^{4} + T^{6}$$
$37$ $$3 + 9 T^{2} + 6 T^{4} + T^{6}$$
$41$ $$T^{6}$$
$43$ $$1 - 6 T + 12 T^{2} - 8 T^{3} + 6 T^{4} - 3 T^{5} + T^{6}$$
$47$ $$T^{6}$$
$53$ $$T^{6}$$
$59$ $$T^{6}$$
$61$ $$1 + 3 T + 12 T^{2} + 19 T^{3} + 15 T^{4} + 6 T^{5} + T^{6}$$
$67$ $$3 - 6 T^{3} + 6 T^{4} - 3 T^{5} + T^{6}$$
$71$ $$T^{6}$$
$73$ $$1 + 6 T + 12 T^{2} + 8 T^{3} + 6 T^{4} + 3 T^{5} + T^{6}$$
$79$ $$3 - 9 T + 18 T^{2} - 21 T^{3} + 15 T^{4} - 6 T^{5} + T^{6}$$
$83$ $$T^{6}$$
$89$ $$T^{6}$$
$97$ $$27 - 9 T^{3} + T^{6}$$


text:
# How to Customize the Command Prompt

I’m a big fan of the terminal: whether you’re leveraging a handful of commands (or more!) to improve your development process, or just using it to quickly move around your drives and open files and folders, the command line is an awesome tool. However, if you use it often, you’ll want to customize it to your needs. I'll show you how to do that today!

I'm often asked, "How did you get your command prompt to look like that?" Well, in this tutorial, I’ll show you exactly how to do it. It’s pretty simple, and won’t require too much of your time.

I should note that what I’m showing you is specifically for the bash shell; this is the default shell on Mac and most Linux systems. If you’d like a bash prompt on Windows, you might want to check out Cygwin.

Before we get started, let’s talk for a minute about how you customize your bash prompt. It’s not quite the same as you’re average application: there’s no preferences panel. Your customizations are stored in a file. If you’re on Linux (or using Cygwin), that will be your .bashrc file; on Mac, that’s your .bash_profile file. In both cases, this file is kept in your home directory (if you aren’t sure where that is for a Cygwin install, run the command echo $HOME). Note that I’ll only refer to the .bashrc file from here on out, but use the .bash_profile if you’re on a Mac. Note that on Macs (and possibly Linux machines; I’m not sure), files that begin with a period are hidden by default. To show them, run these two lines in the terminal  1 defaults write com.apple.finder AppleShowAllFiles TRUE  2 killall Finder  So, what goes in that .bashrc file? Each line is actually a command that you could run on the command line. In fact, that’s how these config files work: When you open the console, all the commands you’ve written in the config file are run, setting up your environment. So, if you just want to try out some of what I’ll show below, just type it on the command line itself. The simplicity here is beautiful. ## Customizing PS1 Let’s start with a definition. The prompt is what you see at the beginning of the line, each time you hit enter on the command line. Here’s what the default settings are for the Mac: In this case, the prompt is andrews-macbook:~ screencast$. There’s a few variables here: andrew-macbook is the name of this computer, ~ is the current directory (the home directory) and screencast is the username. Let’s customize this a bit.

Open up either your .bashrc file. The way we set what information is displayed in the prompt is with the PS1 variable. Add this to the file:

 1 PS1='->'

Notice that I don’t put spaces on either side of the equals sign; that’s necessary. Save this file in your home directory and re-open a terminal window. Now, you should have a prompt that looks like this:

I’ll note here that if you find it tedious to close and re-open your terminal each time you make a change to your .bashrc or .bash_profile, there’s a bit of a shortcut: You can load any bash customization file with the source command. Run this in your terminal:

 1 source ~/.bashrc

Still too long? Well, a single period (.) is an alias for source. Happy now? If you’re quick, you’ll realize that we can use the source command to include other files within our .bashrc file, if you want to split it up to keep in under control.

Let’s customize our prompt a bit more. We can use built-in variables in the string we assign to PS1 to include helpful information in the prompt; here’s a few useful one:

• \d: Date
• \h: Host
• \n: Newline
• \t: Time
• \u: Username
• \W: Current working directory
• \w: Full path to current directory

So, if you set your prompt to this:

 1 PS1='\n\W\n[\h][\u]->'

you should see something like this:

Notice a few things here: firstly, we’re using a bunch of the variables shown above to give us more information. But secondly, we’re including a few newlines in there, and getting a more interesting prompt: we have the current directory on one line, and then the actual prompt on the next line. I prefer my prompt this way, because I always have the same amount of space to write my commands, no matter how long the path to the current directory is. However, there’s a better way to do this, so let’s look at that now.

## Customizing PROMPT_COMMAND

The better way to do this is the use the PROMPT_COMMAND variable; the contents of this variable isn’t just a string, like with PS1. It’s actually a command that executed before bash displays the prompt. To give this a try, let’s add this to our .bashrc:

 1 PROMPT_COMMAND='echo "comes before the prompt"'

We’re using the echo command here; if you aren’t familiar with it, you just pass it a string, and it will write it to the terminal. By itself, it’s not incredibly useful (although you can use it to view variables: echo $PS1), but it’s great when used with other commands, so display their output. If you added the line above, you should see this: Let’s do something more useful here. Let’s write a bash function that we will assign to PROMPT_COMMAND. Try this:  1 print_before_the_prompt () {  2  echo "comes before the prompt"  3 }  4 5 PROMPT_COMMAND=print_before_the_prompt  If you use this, you shouldn’t see a difference in your prompt from what we have above. Now, let’s make this useful.  1 print_before_the_prompt () {  2  echo "$USER: $PWD"  3 }  4 5 PROMPT_COMMAND=print_before_the_prompt  6 7 PS1='->'  Here’s what you’ll get: That’s a good start, but I want to do a bit more. I’m going to use the printf command instead of echo because it makes including newlines and variables a bit easier. A quick background on the printf command: it takes several paramters, the first being a kind of template for the string that will be outputted. The other parameters are values that will be substituted into the template string where appropriate; we’ll see how this works. So let’s do this:  1 print_before_the_prompt () {  2  printf "\n%s: %s\n" "$USER" "$PWD"  3 }  See those %s parts in there? That means “interpret the value for this spot as a string”; for context, we could also use %d to format the value as a decimal number. As you can see, we have two %ss in the “template” string, and two other parameters. These will be placed into the string where the %ss are. Also, notice the newlines at the beginning and end: the first just gives the terminal some breathing room. The last one makes sure that the prompt (PS1) will be printed on the next line, and not on the same line as PROMPT_COMMAND. You should get a terminal like this: ## Adding Some Color Looking good! But let’s take it one step farther. Let’s add some color to this. We can use some special codes to change the color of the text in the terminal. It can be rather daunting to use the actual code, so I like to copy this list of variables for the color and add it at the top of my .bashrc file:  1 txtblk='\e[0;30m' # Black - Regular  2 txtred='\e[0;31m' # Red  3 txtgrn='\e[0;32m' # Green  4 txtylw='\e[0;33m' # Yellow  5 txtblu='\e[0;34m' # Blue  6 txtpur='\e[0;35m' # Purple  7 txtcyn='\e[0;36m' # Cyan  8 txtwht='\e[0;37m' # White  9 10 bldblk='\e[1;30m' # Black - Bold  11 bldred='\e[1;31m' # Red  12 bldgrn='\e[1;32m' # Green  13 bldylw='\e[1;33m' # Yellow  14 bldblu='\e[1;34m' # Blue  15 bldpur='\e[1;35m' # Purple  16 bldcyn='\e[1;36m' # Cyan  17 bldwht='\e[1;37m' # White  18 19 unkblk='\e[4;30m' # Black - Underline  20 undred='\e[4;31m' # Red  21 undgrn='\e[4;32m' # Green  22 undylw='\e[4;33m' # Yellow  23 undblu='\e[4;34m' # Blue  24 undpur='\e[4;35m' # Purple  25 undcyn='\e[4;36m' # Cyan  26 undwht='\e[4;37m' # White  27 28 bakblk='\e[40m' # Black - Background  29 bakred='\e[41m' # Red  30 badgrn='\e[42m' # Green  31 bakylw='\e[43m' # Yellow  32 bakblu='\e[44m' # Blue  33 bakpur='\e[45m' # Purple  34 bakcyn='\e[46m' # Cyan  35 bakwht='\e[47m' # White  36 37 txtrst='\e[0m' # Text Reset  There’s some method to this madness: The first set are turn on normal coloring. The second set turn on bold coloring. The third set turn on underlined coloring. And that fourth set turn on background coloring. That last one resets the coloring to normal. So, let’s use these!  1 print_before_the_prompt () {  2  printf "\n$txtred%s: $bldgrn%s \n$txtrst" "$USER" "$PWD"  3 }

Here, I’ve added $txtred before the first %s, and $bldgrn before the second %s; then, at the end, I’ve reset the text color. You have to do this because once you set a color, it will hold until you either use a new color or reset the coloring. You’ll also notice that when setting a variable, we don’t prefix it with a dollar sign; but we do use the dollar sign when using the variable: that’s the way bash variables work. This gives us the following:

If you’ve seen the screencasts that come with my book Getting Good with Git, you might remember that I have some version control information showing in my prompt. I got this idea from the excellent PeepCode “Advanced Command Line” screencast, which share this, as well as many other great tips.

To do this, we’re going to need to download and build the script that finds this information. Head on over to the repository for vcprompt, a script that outputs the version control information. If you’re familiar with the Mercurial version control system, you can use that to get the repo, but you’ll most likely want to hit that ‘zip’ link to download the script code as a zip file. Once you unzip it, you’ll have to build the script. To do this, just cd into the unzipped script folder and run the command make. Once this command runs, you should see a file named ‘vcprompt’ in the folder. This is the executable script.

So, how do we use this in our prompt? Well, this brings up an important rabbit-trail: how do we “install” a script (like this one) so that we can use it in the terminal? All the commands that you can run on the terminal are found in a defined array of folders; this array is the PATH variable. You can see a list of the folders currently in your PATH by running echo $PATH. It might look something like this:  1 /usr/bin:/bin:/usr/sbin:/sbin:/usr/local/bin:/usr/X11/bin  What we need to do is put the executable script vcprompt in a folder that’s in our path. What I like to do (and yes, I learned this trick from that PeepCode screencast, too) is create a folder called ‘bin’ (short for ‘binary’) in my home directory and add that folder to my PATH. Add this to your .bashrc:  1 export PATH=~/bin:$PATH

This sets PATH to ~/bin, plus whatever was already in the PATH variable. If we now put that vcprompt script into ~/bin, we will be able to execute it in any folder on the terminal.

So, now let’s add this to our prompt:

 1 print_before_the_prompt () {  2  printf "\n $txtred%s:$bldgrn%s $txtpur%s\n$txtrst" "$USER" "$PWD" "$(vcprompt)"  3 }  I’ve added $txtpur %s to the “template” string, and added the fourth parameter"$(vcprompt)". Using the dollar sign and parenthesis will execute the script and return the output. Now, you’ll get this: Notice that if the folder doesn’t use some kind of version control, nothing shows. But, if we’re in a repository, we get the version control system that’s being used (Git, in my case) and the branch name. You can customize this output a bit, if you’d like: check the Readme file that you downloaded with the source code for the vcprompt script. ## Moving On! Here’s our complete .bashrc or .bash_profile file:  1 export PATH=~/bin:$PATH  2 3 txtblk='\e[0;30m' # Black - Regular  4 txtred='\e[0;31m' # Red  5 txtgrn='\e[0;32m' # Green  6 txtylw='\e[0;33m' # Yellow  7 txtblu='\e[0;34m' # Blue  8 txtpur='\e[0;35m' # Purple  9 txtcyn='\e[0;36m' # Cyan  10 txtwht='\e[0;37m' # White  11 bldblk='\e[1;30m' # Black - Bold  12 bldred='\e[1;31m' # Red  13 bldgrn='\e[1;32m' # Green  14 bldylw='\e[1;33m' # Yellow  15 bldblu='\e[1;34m' # Blue  16 bldpur='\e[1;35m' # Purple  17 bldcyn='\e[1;36m' # Cyan  18 bldwht='\e[1;37m' # White  19 unkblk='\e[4;30m' # Black - Underline  20 undred='\e[4;31m' # Red  21 undgrn='\e[4;32m' # Green  22 undylw='\e[4;33m' # Yellow  23 undblu='\e[4;34m' # Blue  24 undpur='\e[4;35m' # Purple  25 undcyn='\e[4;36m' # Cyan  26 undwht='\e[4;37m' # White  27 bakblk='\e[40m' # Black - Background  28 bakred='\e[41m' # Red  29 badgrn='\e[42m' # Green  30 bakylw='\e[43m' # Yellow  31 bakblu='\e[44m' # Blue  32 bakpur='\e[45m' # Purple  33 bakcyn='\e[46m' # Cyan  34 bakwht='\e[47m' # White  35 txtrst='\e[0m' # Text Reset  36 37 print_before_the_prompt () {  38  printf "\n $txtred%s:$bldgrn%s $txtpur%s\n$txtrst" "$USER" "$PWD" "\$(vcprompt)"  39 }  40 41 PROMPT_COMMAND=print_before_the_prompt  42 PS1='->'

Well, that’s a crash-course on customizing your bash prompt. If you have any questions, be sure to drop them in the comments!


text: Radius of a Circumcircle Given a triangle with vertices A, B and C, find a formula for the radius R of the circumcircle. Every triangle has one (and only one) circumcircle, but most other polygons do not. The circumcircle is a triangle's circumscribed circle, i.e., the unique circle that passes through each of the triangle's three vertices. Given a triangle with known sides a, b and c; the task is to find the area of its circumcircle. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. The radius of the incircle is related to the area of the triangle. So, circumradius = Now, inradius of the equilateral triangle = Triangles. paulsrinjoy2234 is waiting for your help. For a triangle, it is the measure of the radius of the circle that circumscribes the triangle. This site is using cookies under cookie policy. So let me try to draw it. 1/2 units . • Given the side lengths of the triangle, it is possible to determine the radius of the circle. For a triangle, it is the measure of the radius of the circle that circumscribes the triangle. The radius of the circumcircle of a right angled triangle is 15 cm and the radius of its inscribed circle is 6 cm. Let A B C a triangle with acute angle γ, let M be the center of the circumcircle and r be its radius. Two pipes can fill a tank in 6 hours and 8 hours respectively. triangle area St . If point P is equidistant from A(1,3),B(-3,5) and C(5, -1), then find AP.ANSWER THIS QUESTION THEN I WILL MARK THEM AS BRAINLISTPLEASE ANSWER MY QUESTIONS..., 14. First, draw three radius segments, originating from each triangle vertex (A, B, C). Let the side be a . Below image shows an equilateral triangle with circumcircle: The formula used to calculate the area of circumscribed circle is: (π*a 2)/3. The circumcircle of a triangle is also known as circumscribed circle. FAQ. In this diagram I must find AC ( which is$\sqrt21$) in order to find the radius of the circle. Hence, its semiperimeter = 3a/2 . A regular polygon's radius is also the radius of the circumcircle. If 40% of an amount is 250, what will be 60% of that amount?O 500O None of these250O 375, If √(1-sin^2x=-cosx and √1-cos^2x=sinx then x lies in. Examples: Input: a = 2, b = 2, c = 3 Output: 7.17714 Input: a = 4, b = 5, c = 3 Output: 19.625 Approach: For a triangle with side lengths a, b, and c, Radius of the circumcircle: where A = √(s*(s-a)*(s-b)*(s-c)) and s = (a+b+c)/2 is the semiperimeter. The center of the circumcircle is called the circumcenter, which may lie outside the triangle. The circumcircle of a triangle or other polygon is the circle which passes through all of its vertices (if such a circle exists). The radius of the circle on which lie a set of points is, by definition, the radius of the circumcircle of any triangle with vertices at any three of those points. A triangle's three perpendicular bisectors,, and meet (Casey 1888, p. 9) at (Durell 1928). If we construct a circle that has a center at O and whose radius is this orange distance, whose radius is any of these distances over here, we'll have a circle that goes through all of the vertices of our triangle centered at O. Customer Voice. Given – In the figure ABC is a triangle in which ∠A = 30° To prove – BC is the radius of circumcircle of ∆ABC whose centre is O. A circle that passes through all vertices (corner points) of a polygon. The law of sines establishes a common ratio for all the lengths of the sides of a triangle to the sines of its angles. 1/2 units, And P is a point that is equidistance from vertices A, B and C of the triangle ABC then, Thus, AP will be nothing but the radius of the circumcircle, Therefore, the distance AP will be 1/2 units or 0.5 units. If the radius of the circumcircle of the triangle is 5/10 i.e. That ratio is the diameter (2 x radius) of the unique circle in which that triangle can be inscribed (circumscribed circle of the triangle). The circumradius of a cyclic polygon is the radius of the circumscribed circle of that polygon. If the radius of the circumcircle of the triangle is 5/10 i.e. Cloudflare Ray ID: 6172116b0e582a33 A third pipe can empty the same tank in 12 hours. 1 Formula for a Triangle; 2 Proof; 3 Formula for Circumradius; 4 Euler's Theorem for a Triangle; 5 Proof; 6 Right triangles. Performance & security by Cloudflare, Please complete the security check to access. \qquad (1) (2 sin A d 1 ) = 1 2. Thus, AP will be nothing but the radius of the circumcircle. (2) \left( \dfrac{d_2}{2 \sin (180^\circ - A)} \right)=25. Whats the radius of its circumcircle? The radius of the circumcircle of a triangle $$\Delta ABC$$ is generally denoted as R.Recall how we can construct the circumcircle, by first determining its center as the point of concurrency of the perpendicular bisectors of the sides of the triangle. Calculates the radius and area of the circumcircle of a triangle given the three sides. I know that the radius can be found by dividing the . Stack Exchange Network. The circumcircle of a triangle can be explained as the circle that passes through 3 vertices of a given triangle. Find the sides of the triangle. If the pairwise distances among three of the points are a, b, and c, then the circle's radius can be calculated by the sides a,b and c of the triangle. Triangles, rectangles, regular polygons and some other shapes have an incircle, but not all polygons. The radius of circle can be an edge or side of polygon inside the circle which is known as circumradius and center of circle is known as circumcenter. If the radius of the circumcircle of the triangle is 5/10 i.e. The center of the incircle is a triangle center called the triangle's incenter. …, gether, how long it will take to fillthe tank?None of these4.8 Hrs9 Hrs10 Hrs, Show that x = -1 and y = -2 is not a solution of the system of linearequationx + y = -3x - y = 4, If you want to show BOOBSJoin Id - 3863516002pas - yyou, 7V = 98 56 cm cube. Answers and Replies Related Introductory Physics Homework Help News on Phys.org. Please enable Cookies and reload the page. Since every triangle is cyclic, every triangle has a circumscribed circle, or a circumcircle. The radius of the circumcircle of a right angled triangle is 15 cm and the radius of its inscribed circle is 6 cm. Problem. Click hereto get an answer to your question  Find the area of a right triangle, if the radius of its circumcircle is 2.5 cm and the altitude drawn to the hypotenuse is 2 cm long. The incircle of triangle ABC has radius equal to 2 and the circumcircle of triangle ABC has radius equal to 6 … Thread starter primarygun; Start date Feb 8, 2005; Feb 8, 2005 #1 primarygun. Let's see if we can somehow relate some of these things with the area to the radius of the triangle's circumscribed circle. For a triangle, 3 sides are given. And P is a point that is equidistance from vertices A, B and C of the triangle ABC then. Now for an equilateral triangle, sides are equal. The center of the circumcircle of a triangle is located at the intersection of the perpendicular bisectors of the triangle. If all the pipes start working to Now, radius of incircle of a triangle = where, s = semiperimeter. If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. You can specify conditions of storing and accessing cookies in your browser. In this video we look at the derivation of a formula that compares the area of a triangle and the radius of its circumscribed circle. \qquad (2) (2 sin (1 8 0 ∘ − A) d 2 ) = 2 5. Q: What is the radius of the circumcircle of a triangle whose sides are 30 cm 36 cm and 30 cm ? The center of the circumcircle is called the circumcenter, and the circle's radius is called the circumradius. 1/2 units And P is a point that is equidistance from vertices A, B and C of the triangle ABC then P will be the circumcentre Thus, AP will be nothing but the radius of the circumcircle In the case of an equilateral triangle, where all three sides (a,b,c) are have the same length, the radius of the circumcircle is given by the formula: where s is the length of a side of … Property - 4: Circumcircle, Incircle, Excircle relations. Radius of circumcircle of a triangle ABC is5/10 units. ABC is a triangle with area equal to 20 . If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. In geometry, the circumscribed circle or circumcircle of a polygon is a circle which passes through all the vertices of the polygon. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, ... Circumcircle of an isosceles triangle … Your IP: 64.111.111.112 It can be inside or outside of the circle . All triangles are cyclic; that is, every triangle has a circumscribed circle. For a triangle, 3 sides are given. Circumcircle of a Triangle Calculator The circumcircle of a triangle can be explained as the circle that passes through 3 vertices of a given triangle. Published: 24 June 2019 Last Updated: 18 July 2019 , , - sides of a triangle - semiperimeter - circumcenter . Details Written by Administrator. Find the sides of the triangle. The center of this circle is called the circumcenter and its … Given a triangle with known sides a, b and c; the task is to find the area of its circumcircle. Contents. Calculate the radius of the circumcircle of an isosceles triangle if given sides ( R ) : Radius of the circumscribed circle of an isosceles triangle : = Digit 2 1 2 4 6 10 F • Calculates the radius and area of the circumcircle of a triangle given the three sides. Below is the circumcircle of a triangle (try dragging the points): Regular polygons do have circumcircles. Are we able to get it without using cosine law or sine law or heron formula? You may need to download version 2.0 now from the Chrome Web Store. 5. Q: If the radius of a circumcircle of an equilateral triangle is 6 cm ,then find the radius of its incircle. The formulas of circumcircle of a triangle is given below: Add your answer and earn points. Radius of circumcircle of a triangle = Where, a, b and c are sides of the triangle. Radius of the circumcircle of a triangle . By moving C on the arc over A B the angle γ doesn't change; move it to the intersection C ′ of A M and the circle. (1) The radius of the circumcircle of triangle A C D ACD A C D is (d 2 2 sin  (18 0 ∘ − A)) = 25. In the figure on the right, the red circle is the circumcircle of the triangle. and area = √3/4 × a². diarmeter, 28cm find the slant height of the cone, I am a number tell my identitiy take me two times overand add a thirty six to reach a centuryyou still need four , any b00bs Caring girljoin i want your Help Id - 3863516002pas - you, 12. Examples: Input: a = 2, b = 2, c = 3 Output: 7.17714 Input: a = 4, b = 5, c = 3 Output: 19.625 Approach: For a triangle with side lengths a, b, and c, Given below is the figure of Circumcircle of an Equilateral triangle. circumcircle area Sc . The radius of a circumcircle of an equilateral triangle is equal to (a / √3), where ‘a’ is the length of the side of equilateral triangle. DEF is the triangle formed by joining the points of contact of the incircle with the sides of the Δ A B C, prove that its sides are 2r cos 2 A , 2r cos 2 B , and 2r 2r cos 2 C , where r is the radius of incircle of Δ A B C. and its angles are (2 π − 2 A ), (2 π − 2 B ) and (2 π − 2 C ). Circumcircle of a triangle This online calculator determines the radius and area of the circumcircle of a triangle given the three sides person_outline Timur schedule 2011-06-24 21:15:14 Are we able to get it without using cosine law or sine law or heron formula? Calculate the radius of the circumcircle of an isosceles triangle if given sides ( R ) : Radius of the circumscribed circle of an isosceles triangle : = Digit 2 1 2 4 6 10 F What's the radius of its circumcircle? So the circumscribed circle is a circle that passes through all of the vertices of the triangle and every triangle has a circumscribed circle. 30, 24, 25 24, 36, 30 30, 40, 41 This is the hard part, right over here so it might look something like this That's fair enough. The radius of the circumcircle of triangle A B D ABD A B D is (d 1 2 sin  A) = 12.5. P will be the circumcentre . Questionnaire. In the case of a triangle, there is always a circumcircle possible, no matter what shape the triangle is. side a: side b: side c : circumradius r . Let A, B and C be the the vertices of a triangle. Since every triangle is cyclic, every triangle has a circumscribed circle, or a circumcircle. ‹ Derivation of Formula for Radius of Circumcircle up Derivation of Heron's / Hero's Formula for Area of Triangle › Log in or register to post comments 54292 reads Those quadrilaterals with circumcircles form a special class, known as cyclic quadrilaterals. Circumcircle of a triangle. The circumcircle of a triangle is also known as circumscribed circle. Construction – join OB and OC. (1) \left( \dfrac{d_1}{2 \sin A} \right)=12.5. Radius can be found like this: where S, area of triangle, can be found using Hero's formula Calculator determines radius, and having radius, area of circumcircle, area of triangle and area ratio - just for reference Circumcircle of a triangle 232 0. Therefore, the distance AP will be 1/2 units or … The formula for the radius of the circle circumscribed about a triangle (circumcircle) is given by R = a b c 4 A t where A t is the area of the inscribed triangle. 30, 24, 25 24, 36, 30 There exist three consecutive vertices A, B, C in every convex n-gon with n≥3, such that the circumcircle of triangle ABC covers the whole n-gon 1 Proving circumcenter lies on altitude Proof : ∠BOC = 2∠BAC = 2 × 30°= 60° Now in ∆OBC, OB = OC [Radii of the same circle] ∠OBC = ∠OCB But, in ΔBOC, From Thales the triangle A B C ′ is right-angled with a right angle at B. In this video we look at the derivation of a formula that compares the area of a triangle and the radius of its circumscribed circle. The ratio of the area of the incircle to the area of the triangle is less than or equal to {\displaystyle {\tfrac {\pi } {3 {\sqrt {3}}}}}, with equality holding only for equilateral triangles. diameter φ . area ratio Sc/St . So our circle would look something like this, my best attempt to draw it. Another way to prevent getting this page in the future is to use Privacy Pass. The polygon 's incenter, 25 24, 36, 30 property - 4: circumcircle incircle... A: side C: circumradius r triangles are cyclic ; that is, every triangle is 5/10 i.e radius... The circumcircle of an equilateral triangle is also known as circumscribed circle is circle... Circle 's radius is also the radius of its incircle of circumcircle of the of! ; that is, every triangle has a circumscribed circle 's incenter another way to prevent getting page! Incircle of a triangle, it is the radius of its incircle date Feb 8, 2005 Feb! Vertices of a given triangle the intersection of the circle that passes through 3 vertices of the circumscribed circle,. Perpendicular bisectors,, and meet ( Casey 1888, p. 9 ) at ( Durell 1928.! The hard part, right over here so it might look something like this that fair! You may need to download version 2.0 now from the Chrome web Store need download. Fill a tank in 12 hours not all polygons 2019 Last Updated: 18 July 2019,, the... Q: if the radius of the incircle is a point that is, triangle! Sides are equal - a ) d 2 ) ( 2 sin ( 1 ) \left \dfrac! 8 hours respectively know that the radius of incircle of a triangle = Please enable Cookies reload... { d_1 } { 2 \sin ( 180^\circ - a ) } \right =12.5... 1 ) = 2 5 dividing the to download version 2.0 now from the web! ∘ − a ) } \right ) =25 so it might look something like this that 's fair enough (... 1 8 0 ∘ − a ) } \right ) =12.5 circumcircles form a special radius of circumcircle of a triangle! Attempt to draw it 's radius is called the circumcenter, which lie. Vertices a, B, C ) radius can be explained as the circle that through... Or heron formula page in the case of a triangle now, inradius of the circumcircle of an triangle... ( corner points ) of a triangle, it is the measure of the is... To download version 2.0 now from the Chrome web Store sin a d ). Can be explained as the circle two pipes can fill a tank in 6 hours and 8 respectively!, incircle, Excircle relations the intersection of the circumcircle of a triangle with area equal to 20,. The circumcircle of a triangle is 5/10 i.e find the area of its incircle regular polygon 's is..., sides are 30 cm 36 cm and 30 cm is equidistance from a. May need to download version 2.0 now from the Chrome web Store July 2019,! ) d 2 ) radius of circumcircle of a triangle 1 2 and area of the circumcircle of a,... This page in the figure of circumcircle of the circumcircle of a given triangle will... Through all of the triangle is cyclic, every triangle has one ( and only ). You are a radius of circumcircle of a triangle and gives you temporary access to the web property circle 's radius called... Case of a triangle with known sides a, B, C ), three. Triangle ABC is5/10 units, known as circumscribed circle the hard part right... As cyclic quadrilaterals the the vertices of the polygon center of the vertices the. You temporary access to the web property sides a, B, C ) 30 cm cm! Since every triangle has a circumscribed circle with area equal to 20,,! Three sides all vertices ( corner points ) of a cyclic polygon is the figure circumcircle. Is 6 cm, then find the area of the incircle is to... ) of a triangle ABC then passes through all vertices ( corner points ) of a polygon corner )! Conditions of storing and accessing Cookies in Your browser 2019 Last Updated 18... Can specify conditions of storing and accessing Cookies in Your browser most other polygons do.! Please enable Cookies and reload the page it is possible to determine the radius the. Can fill a tank in 6 hours and 8 hours respectively the intersection of the circumcircle is called circumcenter... { 2 \sin ( 180^\circ - a ) } \right ) =25 accessing! Thus, AP will be nothing but the radius of the triangle 's three perpendicular bisectors,, sides.


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# Thread: trig proof

1. ## trig proof

In triangle ABC, angle C is 90 degrees. show that;

$\displaystyle sin2A = \frac{2ab}{c^{2}}$

$\displaystyle sin\frac{1}{2}A = \sqrt{\frac{c-b}{2c}}$

Can someone please explain what to do, as I dont understand the question very well.

Thanks in advance.

2. Originally Posted by Tweety
In triangle ABC, angle C is 90 degrees. show that;

$\displaystyle sin2A = \frac{2ab}{c^{2}}$

$\displaystyle sin\frac{1}{2}A = \sqrt{\frac{c-b}{2c}}$

Can someone please explain what to do, as I dont understand the question very well.

Thanks in advance.
use the fact that $\displaystyle \sin(2A)=2\sin(A)\cos(A)$

3. and remember,
$\displaystyle \sin(A)=\frac {\textrm{opposite}} {\textrm{hypotenuse}}$
$\displaystyle \cos(A)=\frac {\textrm{adjacent}} {\textrm{hypotenuse}}$
it's up to you to see what is $\displaystyle a$,$\displaystyle b$ and $\displaystyle c$.

4. Originally Posted by Raoh
and remember,
$\displaystyle \sin(A)=\frac {\textrm{opposite}} {\textrm{hypotenuse}}$
$\displaystyle \cos(A)=\frac {\textrm{adjacent}} {\textrm{hypotenuse}}$
it's up to you to see what is $\displaystyle a$,$\displaystyle b$ and $\displaystyle c$.

Oh so the sides are a,b and c? So do I draw a right angled triangle and and label the hypotenuse c?

5. Originally Posted by Tweety
Oh so the sides are a,b and c? So do I draw a right angled triangle and and label the hypotenuse c?
yes,the hypotenuse is c.


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Percentages, Fractions, and Decimal Values

# What fraction is equivalent to .25?

###### Wiki User

That fraction is one-fourth.

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## Related Questions

0.25 is a fraction. It is a fraction in decimal form rather than in the form of a ratio. However, that does not stop it being a fraction. Its equivalent, in rational form, is 25/100. You can simplify this rational fraction if required.

fraction equivalent for 0.36 = 36/100 or 9/25

An equivalent fraction is when one fraction can is equal to another. Example: 25/100=1/4

-1/25 is the fraction in lowest terms that is equivalent to -0.04

0.28 is equivalent to 28/100 = 7/25 as a fraction. Therefore the given number is equivalent to 7 + 7/25 = 182/25.

0.25 is a fraction. It is a fraction in decimal form rather than in the form of a ratio. However, that does not stop it being a fraction. Its equivalent, in rational form, is 25/100. You can simplify this rational fraction if required.

The equivalent fraction of 6.25 is 25/4 which is NOT a proper fraction.6.25 = 625/100 or 25/4 in improper fraction

It is: 0.52 = 13/25 as a fraction in its lowest terms

One possible equivalent fraction is 15125/1.

The equivalent fraction for a fifth is 2/10 3/15 4/20 5/25 ect.

Take a fraction, and multiply (or divide) the numerator and the denominator by the same non-zero integer, and you will have a fraction with the same value as the original (equivalent fraction). Example: 75/100; divide numerator and denominator by 25: 75 Ã· 25 = 3. 100 Ã· 25 = 4, so the equivalent fraction is 3/4.

5.08 is the decimal. The equivalent fraction is 127/25

If you simplify 25/120, you get 5/24. So that is one equivalent fraction.

0.64 is equivalent to 64/100 = 16/25 as a fraction.


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## Cryptology ePrint Archive: Report 2005/008

Comments on Distributed Symmetric Key Management for Mobile Ad hoc Networks" from INFOCOM 2004

J. Wu and R. Wei

Abstract: In IEEE INFOCOM 2004, Chan proposed a distributed key management scheme for mobile ad hoc networks, and deduced the condition under which the key sets distributed to the network nodes can form a cover-free family (CFF), which is the precondition that the scheme can work. In this paper, we indicate that the condition is falsely deduced. Furthermore, we discuss whether CFF is capable for key distributions in ad hoc networks.

Category / Keywords: cryptographic protocols / Key management

Date: received 5 Jan 2005, last revised 5 May 2005

Contact author: jwu1 at lakeheadu ca

Available format(s): Postscript (PS) | Compressed Postscript (PS.GZ) | PDF | BibTeX Citation

Short URL: ia.cr/2005/008

[ Cryptology ePrint archive ]


text:
# Why does the domain and range of $\sqrt x$ contain only positive real numbers?

Consider the function $f(x)=\sqrt x$.

Pretty much every resource I can find gives its domain as $\{x\in \mathbb{R} \mid x \ge 0\}$ and its range as $\{y\in \mathbb{R} \mid y \ge 0\}$

I don't see why the domain needs to be restricted to positive numbers. Isn't the square root of a negative number defined? That is, the square root of a negative number is a complex number of the form $a + bi$.

So why isn't the domain of $f(x)=\sqrt x$ instead $\{x\in \mathbb{R}\}$ and its range $\{y\in \mathbb{C} \mid y \ge 0\}$?

What am I missing?

• You can't write $y\geq 0$ when $y\in\mathbb{C},$ because there is no order on $\mathbb{C}.$ – Balloon Nov 26 '15 at 17:41
• This is relevant: en.wikipedia.org/wiki/Principal_branch – Irregular User Nov 26 '15 at 17:41
• If $y$ is an element of the complex number system, it makes no sense to state the requirement $y>0$ Complex numbers are not positive nor negative. They cannot be ordered in that way – imranfat Nov 26 '15 at 17:41
• The range in this case "is" $\{ y\in\mathbb{C}\;\colon\; \text{Re}(y)\geq 0\text{ and Im}(y)\geq 0\}$ – Szmagpie Nov 26 '15 at 17:45
• "Isn't the square root of a negative number defined?". No, in $\mathbb R$ it isn't. – Yves Daoust Nov 26 '15 at 18:40

The square root of a negative number is defined when discussing complex numbers, but when only real numbers are discussed (as is usual with most of calculus) there is no advantage to using the complex numbers as square roots.

Not an answer, but something heavy I want to get of my chest here: Some of you make statements as "The square root of a negative number is complex" That is from a mathematical standpoint a very poor way of "introducing" complex numbers. It falls in the same category as stating $i=\sqrt{-1}$ whereas $i^2=-1$ is a far better concept. A Squareroot of negative numbers are pretty meaningless in the real word as well as in the complex world, however, an equation like $x^2=-9$ has exactly two (complex) solutions by rewriting $-9=9i^2$ and then take square roots (which is NOT the same thing as saying +/-$\sqrt{-9}$) If we truly want to consider the complex number set as a field (which complex analysts want), we need to step away from this "squareroot of a negative number is complex" notion. More on this here: https://en.wikipedia.org/wiki/Complex_number

Because in these resources $f$ is defined as $$f:A\to B$$ where $A,B\subset \mathbb R$.

Of course square root function $g(z)=\sqrt z$ is defined for every complex argument, moreover - it is multivalued. It means that for example $g(-4)$ may be equal to $2i$ or $-2i$.

The same situation may be noticed on the real line; $x^2=4$ have two roots: $2$ and $-2$, but the function $f$ is single-valued. That's because it is only a branch of square root function, its principal value.

This set does not have a meaning: $\{y\in \mathbb{C} ~|~ y \ge 0\}$.
Complex numbers cannot be so simply compared with $\ge$
(unless they are real numbers i.e. unless their imaginary part is zero).

These texts you're reading, they assume the reader does not know yet, or does not want to know yet about complex numbers. Now, if the reader only knows real numbers, then the number x must be non-negative as square root is not defined on negative real numbers.

Also, if you look at the square root of x when x is a complex number, you will notice that one number has more than one complex square roots. This means that square root is not really a function, at least not a function $\mathbb{C} \to \mathbb{C}$

The real-valued function defined by $f(x)=\sqrt x$ has domain $\{x\in\Bbb R:x\ge0\}$. The complex-valued function defined by $g(x)=\sqrt x$ has domain $\Bbb C$. These are different functions (since they have different domains).

It is a question of which range is acceptable for your specific application. If you only want real numbers as results you have to set the domain to the non-negative real numbers. If you are ok with complex numbers you can also set the domain to $\mathbb{C}$.

Let me make this clear to you. See the basic rule of a 2-D function like $y=\sqrt{x}$. In mathematics, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. A function can always be plotted on a graph. How can you plot the graph of a complex number in real number graph? That isn't possible. Complex numbers do not exist at all. Everything is real. In mathematics we use complex numbers in order to make problems easier. That doesn't mean they start existing. That's why in order to obtain a real number graph the domain is set of positive numbers(including 0).

This is the definition of the real square root function, from $\mathbb R$ to $\mathbb R$.

You can also define a complex square root function, from $\mathbb C$ to $\mathbb C$.

A mixed function, from $\mathbb R$ to $\mathbb C$ is of little interest.