text:
# Horizontal line in LATEX

I want to draw a horizontal line in a LATEX file. The starting point is at the middle of the page and the length is the half of the paperwidth.

How can this be done? Thanks for your help.

This revised version works with documentclasses that have symmetric or skewed margins. I interpreted "middle of the page" to mean "middle of the paper", and not "middle of the textblock".

\documentclass{book}
\usepackage{lipsum}
\begin{document}
\noindent\kern.5\paperwidth\kern-1in\kern-\hoffset\kern-\oddsidemargin%
\makebox[0pt][l]{\makebox[.5\paperwidth]{\hrulefill}}\par

\lipsum[3]
\end{document}

BOOK:

ARTICLE:

• (My understanding of the question) Wouldn't \centerline{\hspace{0.5\textwidth}\makebox[.5\paperwidth]{\hrulefill}} be closer to the OP's request ? – BambOo Aug 21 '18 at 13:13
• Thank you for your reply but i want the starting point of the line to be at center of the page. Obviously in your solution the starting point of the line is not at the center of the page. – Apostolos K Aug 21 '18 at 13:14
• @ApostolosK Sorry. I will revise. – Steven B. Segletes Aug 21 '18 at 13:15
• @ApostolosK Please see revision. – Steven B. Segletes Aug 21 '18 at 13:21
• @ApostolosK That is not the question you asked. Which it is why it is the site policy that your question should be accompanied by a small self-contained code (starting with \documentclass, etc.) that demonstrates the problem. – Steven B. Segletes Aug 21 '18 at 13:34
\documentclass[twocolumn]{article}
\usepackage{tikz,lipsum}
\begin{document}

\lipsum[1-4]
\noindent%
\tikz[remember picture,overlay]{\draw(0,0)coordinate(o)--(o -| current page.east);}

New line here

\end{document}


text:
# Simulating from empirical copula density estimate

I went through these excellent slides by Prof. Arthur Charpentier on the use of Kernel density estimators (KDEs) with boundary correction to estimate Copulas in a nonparametric fashion.

As explained in the presentation, it is necessary to use KDEs with boundary correction because Copulas have bounded support $([0,1]^2)$ and as is well known, KDEs are biased in the corners (this leads to density underestimation). There are several methods to correct for boundary bias, but here I am interested in transformed Kernels.

In the univariate case:

• the objective is to estimate $f_{X}$ (i.e., get $\hat{f_{X} }$) via KDE
• problem: the observations $X$ can only take on values within a certain interval, (i.e., the support of $f_{X}$ is bounded), let it be $[0,1]$. Because the support is bounded we cannot use KDE
• to address this issue we use a transformation $T: [0,1] \to \mathbb{R}$ such that $Y=T(X)$ takes on values in the entire real line and so $f_{Y}$ has unbounded support. $T$ needs to be continuously differentiable, strictly increasing, bijective, and to have an inverse continuously differentiable. Such a function can be for example the quantile function (inverse CDF) of the Normal or Student's t distributions
• thanks to this trick, we can estimate $f_{Y}$ with a KDE without worrying about border bias
• then, the link between the KDE $\hat{f_{Y} }$ of $f_{Y}$ and the estimate $\hat{f_{X} }$ of $f_{X}$ (what we're interested in in the first place) is given by:

$\hat{f_{X} }(x) = \frac{\hat{f_{Y} }(y)}{\mid\frac{d}{dy}T^{-1}(y)\mid}\bigg|_{y=T(x)}$

(a good post summarizing this process with links to additional resources can be found here).

In the bivariate case, based on page 31 of the abovelinked presentation and on Prof. Charpentier's R code:

mat[i,j]<-su[Xi,Yj]/(dnorm(qnorm(s[i]))*dnorm(qnorm(s[j])))

this translates to:

$\hat{f_{X_{1}, X_{2}} }(x_{1},x_{2}) = \frac{\hat{f_{Y_{1},Y_{2}} }(y_{1},y_{2})}{\mid\frac{d}{dy_{1}}T^{-1}(y_{1})*\frac{d}{dy_{2}}T^{-1}(y_{2})\mid}\bigg|_{y_{1}=T(x_{1}),y_{2}=T(x_{2})}$

where $\hat{f_{X_{1}, X_{2}} }(x_{1},x_{2})$ is simply the Copula density estimate, and $\frac{d}{dy}T^{-1}$ is simply equal to a PDF $t$, because $T^{-1}$ (inverse of a quantile function) is a CDF and the derivative of a CDF is its PDF (hence the dnorm, and qnorm in the code above).

• How would one use a smoothed bootstrap scheme to sample from $\hat{f_{X_{1}, X_{2}} }(x_{1},x_{2})$? We basically have a product of two normal densities in the denominator and a bivariate Gaussian KDE in the numerator. Thus, should we just separately draw values with replacement from $T(X_{1})$ and $T(X_{2})$, add Gaussian noise (numerator) via rnorm in R for instance, and divide by a product of two Gaussian noises (denominator)? How would we ensure that the values generated are in $[0,1]$, and how do we account for the bandwidth of the bivariate KDE?
• Brownie points: in the equation of the Copula density estimate, look at the denominator: why is what should be equal to the derivative of some bivariate function (let it be the bivariate normal CDF) equal to the product of the derivatives of two univariate functions (the two dnorm's in the code chunk above)? I guess it comes from some basic property that I've forgotten?

format: plain


text:
I will now review some differential geometry. Namely, I’ll recall what it means to have a connection in a complex vector bundle ${E}$, and construct its curvature as an ${E}$-valued global 2-form.

Now there is a fancy, clean approach to the theory of connections and curvature on principal bundles over a group (and a vector bundle basically corresponds to one such over $GL_n$). This approach is awesomely slick and highly polished: basically, it axiomatizes the intuitive idea that a connection is a way of identifying different fibers of a vector bundle (via parallel transport). So what is a connection on a principal bundle over a manifold? It’s a compatible system of defining whether tangent vectors are horizontal: the horizontal curves are those that correspond to a parallel transport. Then all the comparatively ugly index-filled results in the classical approaches get transformed into elegant, short results about Lie-algebra valued differential forms.

In fact, the whole Chern-Weil business can be developed using this formalism, and it becomes very slick. But I would like to do it in a slightly less fancy way, using the Cartan formalism: this essentially amounts to working in frames systematically. Here a frame is a local system of sections which is a basis for a vector bundle, and constitutes a generalized form of local coordinates. We can formulate the notions of connections and curvature in terms of frames (they’re systems of forms associated to each frame that transform in a certain way).

The theory has a super-optimal amount of index-pushing to it, but nonetheless, it is one I would like to gain comfort with, e.g. because Griffiths-Harris use it in their book. When one wants to actually prove concrete, specific results about certain types of manifolds (e.g. Kahler manifolds), it may be helpful to use local coordinates. An analogy: the theory of derived categories replaces the Grothendieck spectral sequence with the statement that the derived functor of the composite is the composite of the derived functors. But for concrete instances, the spectral sequence is still huge. (more…)

Let ${M}$ be a Riemannian manifold with metric ${g}$, Levi-Civita connection ${\nabla}$, and curvature tensor ${X,Y,Z \rightarrow R(X,Y)Z}$. Define$\displaystyle R(X,Y,Z,W) := g\left( R(X,Y)Z, W\right).$ Apparently people sometimes refer to this as the curvature tensor, though it is probably not too confusing.

Some algebra

Recall also the following three identities, proved here:

• (Skew-symmetry) ${R(X,Y,Z,W) = -R(Y,X,Z,W)}$
• (Skew-symmetry) ${R(X,Y,Z,W) = -R(X,Y,W,Z)}$
• (Bianchi identity) ${R(X,Y,Z,W) + R(Z,X , Y, W) + R(Y, Z, X, W) = 0}$

I claim now that there is a type of symmetry:

$\displaystyle \boxed{ R(X,Y,Z,W) = R(Z,W,X,Y).}$ This is in fact a general algebraic lemma.

Lemma 1 Let ${V}$ be a real vector space and let ${R:V \times V \times V \times V \rightarrow \mathbb{R}}$ be a quadrilinear map satisfying the three bulleted identities. Then it satisfies the boxed one.

The proof is some slightly messy algebra, which I’ll only sketch. There is a geometrical way of thinking about this that Milnor presents in Morse Theory.

Let the Bianchi identity as written above be denoted by ${Rel_{X,Y,Z,W}}$. If we add ${Rel_{X,Y,Z,W}, Rel_{W,Y,Z,X}, Rel_{X,W,Z,Y}, Rel_{X,Y,W,Z}}$ and use skew-symmetry several times, we obtain

$\displaystyle R(W,Y,Z,X) + R(Z,W,Y,X) + R(X,W,Z,Y) = 0.$

Now using ${Rel_{W,Y,Z,X}}$ gives

$\displaystyle R(X,W,Z,Y) = R(Y,Z,W,X)$

and suitably interchanging all the variables gives the result.

Sectional curvature

Notation as above, for a 2-dimensional subspace ${S \subset T_p(M)}$, define the sectional curvature ${K(S)}$ as

$\displaystyle K(S) := R( E, F, F, E)$

if ${(E,F)}$ form an orthonormal basis for ${S}$.I now claim this is well-defined.

Proposition 2 If ${X,Y \in T_p(M)}$ span ${S}$, then$\displaystyle K(S) = \frac{ R(X,Y,Y,X) }{(X,X)(Y,Y) - (X,Y)^2}.$ In particular, ${K(S)}$ depends only on ${S}$ and is well-defined.

Indeed, write ${X = x_1 E + x_2 F, Y = y_1 E + y_2 F}$. Then this is a computation depending on the previous identities already proved. (Messy algebra and blogging do not mix well.)

Sectional curvature determines ${R}$

The sectional curvature actually encodes all the information contained in ${R}$. Indeed, if we had two Riemannian metrics on the same manifold with curvature tensors ${R,R'}$ with the same sectional curvature in all two-dimensional planes, then

$\displaystyle R'(X,Y,Y,X) = R(X,Y,Y,X)$ for all ${X,Y \in T_p(M)}$.

Consider the difference ${R' - R}$. Then it satisfies all the four identities in the first section of this post, along with ${B(X,Y,X,Y) = 0}$ (by skew-symmetry again). Also by the fourth identity, ${B(X,Y,X,Y')}$ is symmetric in ${Y,Y'}$, so the skew-symmetry just proved implies

$\displaystyle B(X,Y,X,Y') \equiv 0.$ In particular, we get another skew-symmetric identity:

$\displaystyle R(X,Y,Z,W) = -R(Z,Y,X,W)=R(Y,Z,X,W) .$

Applying this again gives

$\displaystyle R(X,Y,Z,W) = R(Z,X,Y,W) ,$ and the Bianchi identity clearly gives that ${B \equiv 0}$.

I should say something about the geometric interpretation about all this.  If $S \subset T_p(M)$, then $\exp_p(S \cap U)$ for $U$ a small neighborhood of $0 \in T_p(M)$ is a surface in $M$.  Then the Gauss curvature of that surface is $K(S)$.

It turns out that the curvature tensor associated to the connection from a Riemannian pseudo-metric ${g}$ has to satisfy certain conditions.  (As usual, we denote by $\nabla$ the Levi-Civita connection associated to $g$, and we assume the ground manifold is smooth.)

First of all, we have skew-symmetry

$\displaystyle R(X,Y)Z = -R(Y,X)Z.$

This is immediate from the definition.

Next, we have another variant of skew-symmetry:

Proposition 1 $\displaystyle g( R(X,Y) Z, W) = -g( R(X,Y) W, Z)$  (more…)


text:
# On two instances of spectral rigidity

 On some classes of semi-discrete darboux integrable equations Bilen, Ergün; Zheltukhın, Kostyantyn; Department of Mathematics (2017) In this thesis we consider Darboux integrable semi-discrete hyperbolic equations of the form $t_{1x} = f(t,t_{1}, t_{x}), \frac{\partial f}{\partial t_{x}} \neq 0.$ We use the notion of characteristic Lie ring for a classification problem based on dimensions of characteristic $x$- and $n$-rings. Let $A = (a_{ij})_{N\times N}$ be a $N\times N$ matrix. We also consider semi-discrete hyperbolic equations of exponential type \$u_{1,x}^{i} - u_{x}^{i} = e^{\sum a_{ij}^{+}u_{1}^{j} + \sum a_{ij}^{-}u^{j}}, i,j = 1...


text:
## GPU renderer Battle

Post and discuss your competition entries here!

Forum rules
1. All competition entries must be a new original work unless otherwise specified during a competition.
2. All submissions must be uploaded to the corresponding competition gallery no later than the date specified in the competition post.

### Re: GPU renderer Battle

When you post the latest version of the scene, I can run it with the under development micro-kernel edition of BIASPATHOCL. I have a GTX980 and a R9 290X where I can run the test.

Posts: 8404
Joined: Sat Apr 19, 2008 6:04 pm
Location: Italy

### Re: GPU renderer Battle

here's the scene
FStormSceneLux.7z

short note:
There's just too much difference between the two engines and the scene: i.e. FStorm is RGB, few lights (in the scene) don't contribute to GI (direct only), have different falloff (?), scene is lit by HDRI with portal on the balcony (why upper part/ceiling is lit orange and lower blueish), different shading system(?), more/better(?) postFX, tonemapping options (which consequently make 'the look' pop up)... otherwise are very close considering render times (yes, it all comes down to the time artist has to invest, so it really matters to have efficient workflow without tinkering, waiting for trial&error results (compiling on nvidia ), clicking/seeking all over the menus to get to proper settings in question).
... for now am done messing with my head and vision... sounds appearing

& here are raw render comparisons (1024samples)
burnin

Posts: 287
Joined: Mon May 03, 2010 8:04 pm

### Re: GPU renderer Battle

Ok, I did some tests with the GTX980 (because it is comparable to your Quadro):

- PATHOCL => 8:48secs

- New BIASPATHOCL with micro-kernels => 6:18secs

- I have then tuned some parameter. New BIASPATHOCL with

Code: Select all
accelerator.type = BVHaccelerator.instances.enable = 0

because you should use MBVH and true instance support only if you run out of memory => 5:20secs

- Cut path depths (i.e. reduce a bit the quality of rendering) with

Code: Select all
biaspath.pathdepth.total = 4biaspath.pathdepth.diffuse = 3biaspath.pathdepth.glossy = 2biaspath.pathdepth.specular = 2

result => 4:01secs

And I could continue removing volumes and replacing (arch)glasses with transparencies, etc.

P.S. this is one of the reason why render engine comparisons are so hard (if not always unfair), I can cut here, push there and pretty much get any time you want.

Posts: 8404
Joined: Sat Apr 19, 2008 6:04 pm
Location: Italy

### Re: GPU renderer Battle

Nice

Yup, my bad for leaving volumes untouched.
Archglass was intentionally there (wanted to see how it feels with interior scene, cuz in preview is brighter than background - but it's fast and subtle reflections help with the look)
- New BIASPATHOCL with micro-kernels => 6:18secs

Yes!
Is it coming soon?
- I have then tuned some parameter. New BIASPATHOCL with... because you should use MBVH and true instance support only if you run out of memory

didn't know that, thanks
- Cut path depths (i.e. reduce a bit the quality of rendering) with

Done that too, but wasn't satisfied
P.S. this is one of the reason why render engine comparisons are so hard (if not always unfair), I can cut here, push there and pretty much get any time you want.

Same thoughts here
Also FStorm uses post on benchmark scene, which makes it even harder to reproduce in detail.
Thank you, for everything.

BTW, did you experience any lag while compiling?
burnin

Posts: 287
Joined: Mon May 03, 2010 8:04 pm

### Re: GPU renderer Battle

burnin wrote:Nice

Yup, my bad for leaving volumes untouched.
Archglass was intentionally there (wanted to see how it feels with interior scene, cuz in preview is brighter than background - but it's fast and subtle reflections help with the look)
- New BIASPATHOCL with micro-kernels => 6:18secs

Yes!
Is it coming soon?

I've quickly made an experimental build for your convenience (viewtopic.php?f=30&t=13054&p=123962#p123962), but I can't gurantee that I've built it properly. It's marked as purely for testing since this is not yet completed by the developer and even if I did the compiling part right, things may not work as expected (or at all).

If testing with BIASPATHOCL, make sure to use sampler.type = "BIASPATHSAMPLER" or you will get an error message telling you to set it.

In fact, I'm slightly curious about these stats:

Tonga is a 'bit' more powerful (r9 380 vs r7 260x) and also working on a faster PCI-E lane.

Here are the stats when using PathOCL:

Here the contributions seem relative to the hardware power (similar with sobol as well). This may be just how biaspath works and expected, but I'd like someone more knowledgable to estimate.

biaspathocl Results after 500 sec/8.3 min (manually stopped, thus the 'box' parts):
Attachments
kalel

Posts: 258
Joined: Sun Jul 17, 2016 6:35 am

### Re: GPU renderer Battle

kalel wrote:Here the contributions seem relative to the hardware power (similar with sobol as well). This may be just how biaspath works and expected, but I'd like someone more knowledgable to estimate.

BIASPATHOCL performance is all about using a tile size large enough to keep the GPU busy, what have you used ? I have used a 1024x1024 tile with 1 sample per pixel in my tests.

Posts: 8404
Joined: Sat Apr 19, 2008 6:04 pm
Location: Italy

### Re: GPU renderer Battle

kalel wrote:Here the contributions seem relative to the hardware power (similar with sobol as well). This may be just how biaspath works and expected, but I'd like someone more knowledgable to estimate.

BIASPATHOCL performance is all about using a tile size large enough to keep the GPU busy, what have you used ? I have used a 1024x1024 tile with 1 sample per pixel in my tests.

Thanks, you mentioned something important to try out. Normally I try various tile sizes when setting up scenes to see what works best, but this time I was using whatever was set in the blend file (16 I guess), in order to quickly start rendering.

After changing the AA size to 1, I was able to not get out of memory errors when using 1024 tile size. The stats in this case are much better for every device, with Bonaire still contributing a somewhat larger percentage than with PathCL. CPU performance improved significantly as well.

The only issue with 1024 (considering that means the entire image) is that we're not getting some benefits from biaspath (making more samples on those tiles that are more noisy), but perhaps it's necessary to do this in order to utilize the GPU properly.
Attachments
Stats
kalel

Posts: 258
Joined: Sun Jul 17, 2016 6:35 am

### Re: GPU renderer Battle

Thanks for the build, kalel.
Gloomy sunday here, so testing mode on...
burnin

Posts: 287
Joined: Mon May 03, 2010 8:04 pm

### Re: GPU renderer Battle

burnin wrote:Thanks for the build, kalel.
Gloomy sunday here, so testing mode on...

You're welcome. Will be happy to see what you do. Sunny sunday here, but I'm not feeling well enough to use it properly.
kalel

Posts: 258
Joined: Sun Jul 17, 2016 6:35 am

### Re: GPU renderer Battle

Im kind of late trying out this render test scene.
Render time = 36 minutes
Using 2 x RX 480 + i7-6700K 4.0GHz CPU
Using LuxRender 1.7 x64 OpenCL Setup
Here is my result:
Attachments
david57

Posts: 208
Joined: Sun Aug 23, 2009 3:33 pm

Previous


text:
# Construct a triangle ABC in which BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct another triangle whose sides are3/4 times the corresponding sides of ΔABC. - Mathematics

Construct a triangle ABC in which BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct another triangle whose sides are3/4 times the corresponding sides of ΔABC.

#### Solution

Following steps are involved in the construction of a ΔA'BC' whose sides are3/4 of the corresponding sides of ΔABC:

Step 1

Draw a ΔABC with sides BC = 6 cm and AB = 5 cm and ∠ABC = 60°.

Step 2

Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

Step 3

Locate 4 points (as 4 is greater in 3 and 4) B1, B2, B3 and B4 on line segment BX.

Step 4

Join B4C and draw a line through B3 parallel to B4C intersecting BC at C'.

Step 5

Draw a line through C' parallel to AC intersecting AB at A'. ΔA'BC' is the required triangle.

Justification

The construction can be justified by proving

A'B=3/4AB, BC'=3/4BC, A'C'=3/4AC

In ΔA'BC' and ΔABC,

∠A'C'B = ∠ACB (Corresponding angles)

∠A'BC' = ∠ABC (Common)

∴ ΔA'BC' ∼ ΔABC (AA similarity criterion)

=>(A'B)/(AB)=(BC')/(BC)=(A'C')/(AC)

In ΔBB3C' and ΔBB4C,

∠B3BC' = ∠B4BC  (Common)

∠BB3C' = ∠BB4C (Corresponding angles)

∴ ΔBB3C' ∼ ΔBB4C (AA similarity criterion)

=>(BC')/(BC)=(BB_3)/(BB_4)

=>(BC')/(BC)=3/4

From (1) and (2), we obtain

(A'B)/(AB)=(BC')/(BC)=(A'C')/(AC)=3/4

=>A'B=3/4AB, BC'=3/4BC, A'C'=3/4AC

This justifies the construction.

Concept: Division of a Line Segment
Is there an error in this question or solution?


text:
# Properties

 Label 1216.1.g.a.417.2 Level $1216$ Weight $1$ Character 1216.417 Analytic conductor $0.607$ Analytic rank $0$ Dimension $2$ Projective image $D_{2}$ CM/RM discs -8, -19, 152 Inner twists $8$

# Related objects

## Newspace parameters

 Level: $$N$$ $$=$$ $$1216 = 2^{6} \cdot 19$$ Weight: $$k$$ $$=$$ $$1$$ Character orbit: $$[\chi]$$ $$=$$ 1216.g (of order $$2$$, degree $$1$$, minimal)

## Newform invariants

 Self dual: no Analytic conductor: $$0.606863055362$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(i)$$ Defining polynomial: $$x^{2} + 1$$ Coefficient ring: $$\Z[a_1, \ldots, a_{19}]$$ Coefficient ring index: $$1$$ Twist minimal: yes Projective image $$D_{2}$$ Projective field Galois closure of $$\Q(\sqrt{-2}, \sqrt{-19})$$ Artin image $D_4:C_2$ Artin field Galois closure of 8.0.378535936.1

## Embedding invariants

 Embedding label 417.2 Root $$-1.00000i$$ of defining polynomial Character $$\chi$$ $$=$$ 1216.417 Dual form 1216.1.g.a.417.1

## $q$-expansion

 $$f(q)$$ $$=$$ $$q-1.00000 q^{9} +O(q^{10})$$ $$q-1.00000 q^{9} +2.00000i q^{11} +2.00000 q^{17} +1.00000i q^{19} +1.00000 q^{25} -2.00000i q^{43} -1.00000 q^{49} -2.00000 q^{73} +1.00000 q^{81} +2.00000i q^{83} -2.00000i q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2q - 2q^{9} + O(q^{10})$$ $$2q - 2q^{9} + 4q^{17} + 2q^{25} - 2q^{49} - 4q^{73} + 2q^{81} + O(q^{100})$$

## Character values

We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/1216\mathbb{Z}\right)^\times$$.

 $$n$$ $$191$$ $$705$$ $$837$$ $$\chi(n)$$ $$1$$ $$-1$$ $$-1$$

## Coefficient data

For each $$n$$ we display the coefficients of the $$q$$-expansion $$a_n$$, the Satake parameters $$\alpha_p$$, and the Satake angles $$\theta_p = \textrm{Arg}(\alpha_p)$$.

Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
$$n$$ $$a_n$$ $$a_n / n^{(k-1)/2}$$ $$\alpha_n$$ $$\theta_n$$
$$p$$ $$a_p$$ $$a_p / p^{(k-1)/2}$$ $$\alpha_p$$ $$\theta_p$$
$$2$$ 0 0
$$3$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$4$$ 0 0
$$5$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$6$$ 0 0
$$7$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$8$$ 0 0
$$9$$ −1.00000 −1.00000
$$10$$ 0 0
$$11$$ 2.00000i 2.00000i 1.00000i $$0.5\pi$$
1.00000i $$0.5\pi$$
$$12$$ 0 0
$$13$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$14$$ 0 0
$$15$$ 0 0
$$16$$ 0 0
$$17$$ 2.00000 2.00000 1.00000 $$0$$
1.00000 $$0$$
$$18$$ 0 0
$$19$$ 1.00000i 1.00000i
$$20$$ 0 0
$$21$$ 0 0
$$22$$ 0 0
$$23$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$24$$ 0 0
$$25$$ 1.00000 1.00000
$$26$$ 0 0
$$27$$ 0 0
$$28$$ 0 0
$$29$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$30$$ 0 0
$$31$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$32$$ 0 0
$$33$$ 0 0
$$34$$ 0 0
$$35$$ 0 0
$$36$$ 0 0
$$37$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$38$$ 0 0
$$39$$ 0 0
$$40$$ 0 0
$$41$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$42$$ 0 0
$$43$$ − 2.00000i − 2.00000i 1.00000i $$-0.5\pi$$
1.00000i $$-0.5\pi$$
$$44$$ 0 0
$$45$$ 0 0
$$46$$ 0 0
$$47$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$48$$ 0 0
$$49$$ −1.00000 −1.00000
$$50$$ 0 0
$$51$$ 0 0
$$52$$ 0 0
$$53$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$54$$ 0 0
$$55$$ 0 0
$$56$$ 0 0
$$57$$ 0 0
$$58$$ 0 0
$$59$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$60$$ 0 0
$$61$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$62$$ 0 0
$$63$$ 0 0
$$64$$ 0 0
$$65$$ 0 0
$$66$$ 0 0
$$67$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$68$$ 0 0
$$69$$ 0 0
$$70$$ 0 0
$$71$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$72$$ 0 0
$$73$$ −2.00000 −2.00000 −1.00000 $$\pi$$
−1.00000 $$\pi$$
$$74$$ 0 0
$$75$$ 0 0
$$76$$ 0 0
$$77$$ 0 0
$$78$$ 0 0
$$79$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$80$$ 0 0
$$81$$ 1.00000 1.00000
$$82$$ 0 0
$$83$$ 2.00000i 2.00000i 1.00000i $$0.5\pi$$
1.00000i $$0.5\pi$$
$$84$$ 0 0
$$85$$ 0 0
$$86$$ 0 0
$$87$$ 0 0
$$88$$ 0 0
$$89$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$90$$ 0 0
$$91$$ 0 0
$$92$$ 0 0
$$93$$ 0 0
$$94$$ 0 0
$$95$$ 0 0
$$96$$ 0 0
$$97$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$98$$ 0 0
$$99$$ − 2.00000i − 2.00000i
$$100$$ 0 0
$$101$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$102$$ 0 0
$$103$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$104$$ 0 0
$$105$$ 0 0
$$106$$ 0 0
$$107$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$108$$ 0 0
$$109$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$110$$ 0 0
$$111$$ 0 0
$$112$$ 0 0
$$113$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$114$$ 0 0
$$115$$ 0 0
$$116$$ 0 0
$$117$$ 0 0
$$118$$ 0 0
$$119$$ 0 0
$$120$$ 0 0
$$121$$ −3.00000 −3.00000
$$122$$ 0 0
$$123$$ 0 0
$$124$$ 0 0
$$125$$ 0 0
$$126$$ 0 0
$$127$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$128$$ 0 0
$$129$$ 0 0
$$130$$ 0 0
$$131$$ − 2.00000i − 2.00000i 1.00000i $$-0.5\pi$$
1.00000i $$-0.5\pi$$
$$132$$ 0 0
$$133$$ 0 0
$$134$$ 0 0
$$135$$ 0 0
$$136$$ 0 0
$$137$$ 2.00000 2.00000 1.00000 $$0$$
1.00000 $$0$$
$$138$$ 0 0
$$139$$ − 2.00000i − 2.00000i 1.00000i $$-0.5\pi$$
1.00000i $$-0.5\pi$$
$$140$$ 0 0
$$141$$ 0 0
$$142$$ 0 0
$$143$$ 0 0
$$144$$ 0 0
$$145$$ 0 0
$$146$$ 0 0
$$147$$ 0 0
$$148$$ 0 0
$$149$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$150$$ 0 0
$$151$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$152$$ 0 0
$$153$$ −2.00000 −2.00000
$$154$$ 0 0
$$155$$ 0 0
$$156$$ 0 0
$$157$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$158$$ 0 0
$$159$$ 0 0
$$160$$ 0 0
$$161$$ 0 0
$$162$$ 0 0
$$163$$ − 2.00000i − 2.00000i 1.00000i $$-0.5\pi$$
1.00000i $$-0.5\pi$$
$$164$$ 0 0
$$165$$ 0 0
$$166$$ 0 0
$$167$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$168$$ 0 0
$$169$$ −1.00000 −1.00000
$$170$$ 0 0
$$171$$ − 1.00000i − 1.00000i
$$172$$ 0 0
$$173$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$174$$ 0 0
$$175$$ 0 0
$$176$$ 0 0
$$177$$ 0 0
$$178$$ 0 0
$$179$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$180$$ 0 0
$$181$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$182$$ 0 0
$$183$$ 0 0
$$184$$ 0 0
$$185$$ 0 0
$$186$$ 0 0
$$187$$ 4.00000i 4.00000i
$$188$$ 0 0
$$189$$ 0 0
$$190$$ 0 0
$$191$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$192$$ 0 0
$$193$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$194$$ 0 0
$$195$$ 0 0
$$196$$ 0 0
$$197$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$198$$ 0 0
$$199$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$200$$ 0 0
$$201$$ 0 0
$$202$$ 0 0
$$203$$ 0 0
$$204$$ 0 0
$$205$$ 0 0
$$206$$ 0 0
$$207$$ 0 0
$$208$$ 0 0
$$209$$ −2.00000 −2.00000
$$210$$ 0 0
$$211$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$212$$ 0 0
$$213$$ 0 0
$$214$$ 0 0
$$215$$ 0 0
$$216$$ 0 0
$$217$$ 0 0
$$218$$ 0 0
$$219$$ 0 0
$$220$$ 0 0
$$221$$ 0 0
$$222$$ 0 0
$$223$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$224$$ 0 0
$$225$$ −1.00000 −1.00000
$$226$$ 0 0
$$227$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$228$$ 0 0
$$229$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$230$$ 0 0
$$231$$ 0 0
$$232$$ 0 0
$$233$$ 2.00000 2.00000 1.00000 $$0$$
1.00000 $$0$$
$$234$$ 0 0
$$235$$ 0 0
$$236$$ 0 0
$$237$$ 0 0
$$238$$ 0 0
$$239$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$240$$ 0 0
$$241$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$242$$ 0 0
$$243$$ 0 0
$$244$$ 0 0
$$245$$ 0 0
$$246$$ 0 0
$$247$$ 0 0
$$248$$ 0 0
$$249$$ 0 0
$$250$$ 0 0
$$251$$ − 2.00000i − 2.00000i 1.00000i $$-0.5\pi$$
1.00000i $$-0.5\pi$$
$$252$$ 0 0
$$253$$ 0 0
$$254$$ 0 0
$$255$$ 0 0
$$256$$ 0 0
$$257$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$258$$ 0 0
$$259$$ 0 0
$$260$$ 0 0
$$261$$ 0 0
$$262$$ 0 0
$$263$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$264$$ 0 0
$$265$$ 0 0
$$266$$ 0 0
$$267$$ 0 0
$$268$$ 0 0
$$269$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$270$$ 0 0
$$271$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$272$$ 0 0
$$273$$ 0 0
$$274$$ 0 0
$$275$$ 2.00000i 2.00000i
$$276$$ 0 0
$$277$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$278$$ 0 0
$$279$$ 0 0
$$280$$ 0 0
$$281$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$282$$ 0 0
$$283$$ − 2.00000i − 2.00000i 1.00000i $$-0.5\pi$$
1.00000i $$-0.5\pi$$
$$284$$ 0 0
$$285$$ 0 0
$$286$$ 0 0
$$287$$ 0 0
$$288$$ 0 0
$$289$$ 3.00000 3.00000
$$290$$ 0 0
$$291$$ 0 0
$$292$$ 0 0
$$293$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$294$$ 0 0
$$295$$ 0 0
$$296$$ 0 0
$$297$$ 0 0
$$298$$ 0 0
$$299$$ 0 0
$$300$$ 0 0
$$301$$ 0 0
$$302$$ 0 0
$$303$$ 0 0
$$304$$ 0 0
$$305$$ 0 0
$$306$$ 0 0
$$307$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$308$$ 0 0
$$309$$ 0 0
$$310$$ 0 0
$$311$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$312$$ 0 0
$$313$$ 2.00000 2.00000 1.00000 $$0$$
1.00000 $$0$$
$$314$$ 0 0
$$315$$ 0 0
$$316$$ 0 0
$$317$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$318$$ 0 0
$$319$$ 0 0
$$320$$ 0 0
$$321$$ 0 0
$$322$$ 0 0
$$323$$ 2.00000i 2.00000i
$$324$$ 0 0
$$325$$ 0 0
$$326$$ 0 0
$$327$$ 0 0
$$328$$ 0 0
$$329$$ 0 0
$$330$$ 0 0
$$331$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$332$$ 0 0
$$333$$ 0 0
$$334$$ 0 0
$$335$$ 0 0
$$336$$ 0 0
$$337$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$338$$ 0 0
$$339$$ 0 0
$$340$$ 0 0
$$341$$ 0 0
$$342$$ 0 0
$$343$$ 0 0
$$344$$ 0 0
$$345$$ 0 0
$$346$$ 0 0
$$347$$ 2.00000i 2.00000i 1.00000i $$0.5\pi$$
1.00000i $$0.5\pi$$
$$348$$ 0 0
$$349$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$350$$ 0 0
$$351$$ 0 0
$$352$$ 0 0
$$353$$ −2.00000 −2.00000 −1.00000 $$\pi$$
−1.00000 $$\pi$$
$$354$$ 0 0
$$355$$ 0 0
$$356$$ 0 0
$$357$$ 0 0
$$358$$ 0 0
$$359$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$360$$ 0 0
$$361$$ −1.00000 −1.00000
$$362$$ 0 0
$$363$$ 0 0
$$364$$ 0 0
$$365$$ 0 0
$$366$$ 0 0
$$367$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$368$$ 0 0
$$369$$ 0 0
$$370$$ 0 0
$$371$$ 0 0
$$372$$ 0 0
$$373$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$374$$ 0 0
$$375$$ 0 0
$$376$$ 0 0
$$377$$ 0 0
$$378$$ 0 0
$$379$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$380$$ 0 0
$$381$$ 0 0
$$382$$ 0 0
$$383$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$384$$ 0 0
$$385$$ 0 0
$$386$$ 0 0
$$387$$ 2.00000i 2.00000i
$$388$$ 0 0
$$389$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$390$$ 0 0
$$391$$ 0 0
$$392$$ 0 0
$$393$$ 0 0
$$394$$ 0 0
$$395$$ 0 0
$$396$$ 0 0
$$397$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$398$$ 0 0
$$399$$ 0 0
$$400$$ 0 0
$$401$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$402$$ 0 0
$$403$$ 0 0
$$404$$ 0 0
$$405$$ 0 0
$$406$$ 0 0
$$407$$ 0 0
$$408$$ 0 0
$$409$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$410$$ 0 0
$$411$$ 0 0
$$412$$ 0 0
$$413$$ 0 0
$$414$$ 0 0
$$415$$ 0 0
$$416$$ 0 0
$$417$$ 0 0
$$418$$ 0 0
$$419$$ − 2.00000i − 2.00000i 1.00000i $$-0.5\pi$$
1.00000i $$-0.5\pi$$
$$420$$ 0 0
$$421$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$422$$ 0 0
$$423$$ 0 0
$$424$$ 0 0
$$425$$ 2.00000 2.00000
$$426$$ 0 0
$$427$$ 0 0
$$428$$ 0 0
$$429$$ 0 0
$$430$$ 0 0
$$431$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$432$$ 0 0
$$433$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$434$$ 0 0
$$435$$ 0 0
$$436$$ 0 0
$$437$$ 0 0
$$438$$ 0 0
$$439$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$440$$ 0 0
$$441$$ 1.00000 1.00000
$$442$$ 0 0
$$443$$ 2.00000i 2.00000i 1.00000i $$0.5\pi$$
1.00000i $$0.5\pi$$
$$444$$ 0 0
$$445$$ 0 0
$$446$$ 0 0
$$447$$ 0 0
$$448$$ 0 0
$$449$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$450$$ 0 0
$$451$$ 0 0
$$452$$ 0 0
$$453$$ 0 0
$$454$$ 0 0
$$455$$ 0 0
$$456$$ 0 0
$$457$$ 2.00000 2.00000 1.00000 $$0$$
1.00000 $$0$$
$$458$$ 0 0
$$459$$ 0 0
$$460$$ 0 0
$$461$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$462$$ 0 0
$$463$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$464$$ 0 0
$$465$$ 0 0
$$466$$ 0 0
$$467$$ 2.00000i 2.00000i 1.00000i $$0.5\pi$$
1.00000i $$0.5\pi$$
$$468$$ 0 0
$$469$$ 0 0
$$470$$ 0 0
$$471$$ 0 0
$$472$$ 0 0
$$473$$ 4.00000 4.00000
$$474$$ 0 0
$$475$$ 1.00000i 1.00000i
$$476$$ 0 0
$$477$$ 0 0
$$478$$ 0 0
$$479$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$480$$ 0 0
$$481$$ 0 0
$$482$$ 0 0
$$483$$ 0 0
$$484$$ 0 0
$$485$$ 0 0
$$486$$ 0 0
$$487$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$488$$ 0 0
$$489$$ 0 0
$$490$$ 0 0
$$491$$ 2.00000i 2.00000i 1.00000i $$0.5\pi$$
1.00000i $$0.5\pi$$
$$492$$ 0 0
$$493$$ 0 0
$$494$$ 0 0
$$495$$ 0 0
$$496$$ 0 0
$$497$$ 0 0
$$498$$ 0 0
$$499$$ − 2.00000i − 2.00000i 1.00000i $$-0.5\pi$$
1.00000i $$-0.5\pi$$
$$500$$ 0 0
$$501$$ 0 0
$$502$$ 0 0
$$503$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$504$$ 0 0
$$505$$ 0 0
$$506$$ 0 0
$$507$$ 0 0
$$508$$ 0 0
$$509$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$510$$ 0 0
$$511$$ 0 0
$$512$$ 0 0
$$513$$ 0 0
$$514$$ 0 0
$$515$$ 0 0
$$516$$ 0 0
$$517$$ 0 0
$$518$$ 0 0
$$519$$ 0 0
$$520$$ 0 0
$$521$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$522$$ 0 0
$$523$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$524$$ 0 0
$$525$$ 0 0
$$526$$ 0 0
$$527$$ 0 0
$$528$$ 0 0
$$529$$ −1.00000 −1.00000
$$530$$ 0 0
$$531$$ 0 0
$$532$$ 0 0
$$533$$ 0 0
$$534$$ 0 0
$$535$$ 0 0
$$536$$ 0 0
$$537$$ 0 0
$$538$$ 0 0
$$539$$ − 2.00000i − 2.00000i
$$540$$ 0 0
$$541$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$542$$ 0 0
$$543$$ 0 0
$$544$$ 0 0
$$545$$ 0 0
$$546$$ 0 0
$$547$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$548$$ 0 0
$$549$$ 0 0
$$550$$ 0 0
$$551$$ 0 0
$$552$$ 0 0
$$553$$ 0 0
$$554$$ 0 0
$$555$$ 0 0
$$556$$ 0 0
$$557$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$558$$ 0 0
$$559$$ 0 0
$$560$$ 0 0
$$561$$ 0 0
$$562$$ 0 0
$$563$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$564$$ 0 0
$$565$$ 0 0
$$566$$ 0 0
$$567$$ 0 0
$$568$$ 0 0
$$569$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$570$$ 0 0
$$571$$ 2.00000i 2.00000i 1.00000i $$0.5\pi$$
1.00000i $$0.5\pi$$
$$572$$ 0 0
$$573$$ 0 0
$$574$$ 0 0
$$575$$ 0 0
$$576$$ 0 0
$$577$$ −2.00000 −2.00000 −1.00000 $$\pi$$
−1.00000 $$\pi$$
$$578$$ 0 0
$$579$$ 0 0
$$580$$ 0 0
$$581$$ 0 0
$$582$$ 0 0
$$583$$ 0 0
$$584$$ 0 0
$$585$$ 0 0
$$586$$ 0 0
$$587$$ − 2.00000i − 2.00000i 1.00000i $$-0.5\pi$$
1.00000i $$-0.5\pi$$
$$588$$ 0 0
$$589$$ 0 0
$$590$$ 0 0
$$591$$ 0 0
$$592$$ 0 0
$$593$$ −2.00000 −2.00000 −1.00000 $$\pi$$
−1.00000 $$\pi$$
$$594$$ 0 0
$$595$$ 0 0
$$596$$ 0 0
$$597$$ 0 0
$$598$$ 0 0
$$599$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$600$$ 0 0
$$601$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$602$$ 0 0
$$603$$ 0 0
$$604$$ 0 0
$$605$$ 0 0
$$606$$ 0 0
$$607$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$608$$ 0 0
$$609$$ 0 0
$$610$$ 0 0
$$611$$ 0 0
$$612$$ 0 0
$$613$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$614$$ 0 0
$$615$$ 0 0
$$616$$ 0 0
$$617$$ −2.00000 −2.00000 −1.00000 $$\pi$$
−1.00000 $$\pi$$
$$618$$ 0 0
$$619$$ 2.00000i 2.00000i 1.00000i $$0.5\pi$$
1.00000i $$0.5\pi$$
$$620$$ 0 0
$$621$$ 0 0
$$622$$ 0 0
$$623$$ 0 0
$$624$$ 0 0
$$625$$ 1.00000 1.00000
$$626$$ 0 0
$$627$$ 0 0
$$628$$ 0 0
$$629$$ 0 0
$$630$$ 0 0
$$631$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$632$$ 0 0
$$633$$ 0 0
$$634$$ 0 0
$$635$$ 0 0
$$636$$ 0 0
$$637$$ 0 0
$$638$$ 0 0
$$639$$ 0 0
$$640$$ 0 0
$$641$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$642$$ 0 0
$$643$$ − 2.00000i − 2.00000i 1.00000i $$-0.5\pi$$
1.00000i $$-0.5\pi$$
$$644$$ 0 0
$$645$$ 0 0
$$646$$ 0 0
$$647$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$648$$ 0 0
$$649$$ 0 0
$$650$$ 0 0
$$651$$ 0 0
$$652$$ 0 0
$$653$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$654$$ 0 0
$$655$$ 0 0
$$656$$ 0 0
$$657$$ 2.00000 2.00000
$$658$$ 0 0
$$659$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$660$$ 0 0
$$661$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$662$$ 0 0
$$663$$ 0 0
$$664$$ 0 0
$$665$$ 0 0
$$666$$ 0 0
$$667$$ 0 0
$$668$$ 0 0
$$669$$ 0 0
$$670$$ 0 0
$$671$$ 0 0
$$672$$ 0 0
$$673$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$674$$ 0 0
$$675$$ 0 0
$$676$$ 0 0
$$677$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$678$$ 0 0
$$679$$ 0 0
$$680$$ 0 0
$$681$$ 0 0
$$682$$ 0 0
$$683$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$684$$ 0 0
$$685$$ 0 0
$$686$$ 0 0
$$687$$ 0 0
$$688$$ 0 0
$$689$$ 0 0
$$690$$ 0 0
$$691$$ 2.00000i 2.00000i 1.00000i $$0.5\pi$$
1.00000i $$0.5\pi$$
$$692$$ 0 0
$$693$$ 0 0
$$694$$ 0 0
$$695$$ 0 0
$$696$$ 0 0
$$697$$ 0 0
$$698$$ 0 0
$$699$$ 0 0
$$700$$ 0 0
$$701$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$702$$ 0 0
$$703$$ 0 0
$$704$$ 0 0
$$705$$ 0 0
$$706$$ 0 0
$$707$$ 0 0
$$708$$ 0 0
$$709$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$710$$ 0 0
$$711$$ 0 0
$$712$$ 0 0
$$713$$ 0 0
$$714$$ 0 0
$$715$$ 0 0
$$716$$ 0 0
$$717$$ 0 0
$$718$$ 0 0
$$719$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$720$$ 0 0
$$721$$ 0 0
$$722$$ 0 0
$$723$$ 0 0
$$724$$ 0 0
$$725$$ 0 0
$$726$$ 0 0
$$727$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$728$$ 0 0
$$729$$ −1.00000 −1.00000
$$730$$ 0 0
$$731$$ − 4.00000i − 4.00000i
$$732$$ 0 0
$$733$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$734$$ 0 0
$$735$$ 0 0
$$736$$ 0 0
$$737$$ 0 0
$$738$$ 0 0
$$739$$ 2.00000i 2.00000i 1.00000i $$0.5\pi$$
1.00000i $$0.5\pi$$
$$740$$ 0 0
$$741$$ 0 0
$$742$$ 0 0
$$743$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$744$$ 0 0
$$745$$ 0 0
$$746$$ 0 0
$$747$$ − 2.00000i − 2.00000i
$$748$$ 0 0
$$749$$ 0 0
$$750$$ 0 0
$$751$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$752$$ 0 0
$$753$$ 0 0
$$754$$ 0 0
$$755$$ 0 0
$$756$$ 0 0
$$757$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$758$$ 0 0
$$759$$ 0 0
$$760$$ 0 0
$$761$$ 2.00000 2.00000 1.00000 $$0$$
1.00000 $$0$$
$$762$$ 0 0
$$763$$ 0 0
$$764$$ 0 0
$$765$$ 0 0
$$766$$ 0 0
$$767$$ 0 0
$$768$$ 0 0
$$769$$ −2.00000 −2.00000 −1.00000 $$\pi$$
−1.00000 $$\pi$$
$$770$$ 0 0
$$771$$ 0 0
$$772$$ 0 0
$$773$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$774$$ 0 0
$$775$$ 0 0
$$776$$ 0 0
$$777$$ 0 0
$$778$$ 0 0
$$779$$ 0 0
$$780$$ 0 0
$$781$$ 0 0
$$782$$ 0 0
$$783$$ 0 0
$$784$$ 0 0
$$785$$ 0 0
$$786$$ 0 0
$$787$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$788$$ 0 0
$$789$$ 0 0
$$790$$ 0 0
$$791$$ 0 0
$$792$$ 0 0
$$793$$ 0 0
$$794$$ 0 0
$$795$$ 0 0
$$796$$ 0 0
$$797$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$798$$ 0 0
$$799$$ 0 0
$$800$$ 0 0
$$801$$ 0 0
$$802$$ 0 0
$$803$$ − 4.00000i − 4.00000i
$$804$$ 0 0
$$805$$ 0 0
$$806$$ 0 0
$$807$$ 0 0
$$808$$ 0 0
$$809$$ −2.00000 −2.00000 −1.00000 $$\pi$$
−1.00000 $$\pi$$
$$810$$ 0 0
$$811$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$812$$ 0 0
$$813$$ 0 0
$$814$$ 0 0
$$815$$ 0 0
$$816$$ 0 0
$$817$$ 2.00000 2.00000
$$818$$ 0 0
$$819$$ 0 0
$$820$$ 0 0
$$821$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$822$$ 0 0
$$823$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$824$$ 0 0
$$825$$ 0 0
$$826$$ 0 0
$$827$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$828$$ 0 0
$$829$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$830$$ 0 0
$$831$$ 0 0
$$832$$ 0 0
$$833$$ −2.00000 −2.00000
$$834$$ 0 0
$$835$$ 0 0
$$836$$ 0 0
$$837$$ 0 0
$$838$$ 0 0
$$839$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$840$$ 0 0
$$841$$ −1.00000 −1.00000
$$842$$ 0 0
$$843$$ 0 0
$$844$$ 0 0
$$845$$ 0 0
$$846$$ 0 0
$$847$$ 0 0
$$848$$ 0 0
$$849$$ 0 0
$$850$$ 0 0
$$851$$ 0 0
$$852$$ 0 0
$$853$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$854$$ 0 0
$$855$$ 0 0
$$856$$ 0 0
$$857$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$858$$ 0 0
$$859$$ − 2.00000i − 2.00000i 1.00000i $$-0.5\pi$$
1.00000i $$-0.5\pi$$
$$860$$ 0 0
$$861$$ 0 0
$$862$$ 0 0
$$863$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$864$$ 0 0
$$865$$ 0 0
$$866$$ 0 0
$$867$$ 0 0
$$868$$ 0 0
$$869$$ 0 0
$$870$$ 0 0
$$871$$ 0 0
$$872$$ 0 0
$$873$$ 0 0
$$874$$ 0 0
$$875$$ 0 0
$$876$$ 0 0
$$877$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$878$$ 0 0
$$879$$ 0 0
$$880$$ 0 0
$$881$$ −2.00000 −2.00000 −1.00000 $$\pi$$
−1.00000 $$\pi$$
$$882$$ 0 0
$$883$$ 2.00000i 2.00000i 1.00000i $$0.5\pi$$
1.00000i $$0.5\pi$$
$$884$$ 0 0
$$885$$ 0 0
$$886$$ 0 0
$$887$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$888$$ 0 0
$$889$$ 0 0
$$890$$ 0 0
$$891$$ 2.00000i 2.00000i
$$892$$ 0 0
$$893$$ 0 0
$$894$$ 0 0
$$895$$ 0 0
$$896$$ 0 0
$$897$$ 0 0
$$898$$ 0 0
$$899$$ 0 0
$$900$$ 0 0
$$901$$ 0 0
$$902$$ 0 0
$$903$$ 0 0
$$904$$ 0 0
$$905$$ 0 0
$$906$$ 0 0
$$907$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$908$$ 0 0
$$909$$ 0 0
$$910$$ 0 0
$$911$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$912$$ 0 0
$$913$$ −4.00000 −4.00000
$$914$$ 0 0
$$915$$ 0 0
$$916$$ 0 0
$$917$$ 0 0
$$918$$ 0 0
$$919$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$920$$ 0 0
$$921$$ 0 0
$$922$$ 0 0
$$923$$ 0 0
$$924$$ 0 0
$$925$$ 0 0
$$926$$ 0 0
$$927$$ 0 0
$$928$$ 0 0
$$929$$ −2.00000 −2.00000 −1.00000 $$\pi$$
−1.00000 $$\pi$$
$$930$$ 0 0
$$931$$ − 1.00000i − 1.00000i
$$932$$ 0 0
$$933$$ 0 0
$$934$$ 0 0
$$935$$ 0 0
$$936$$ 0 0
$$937$$ 2.00000 2.00000 1.00000 $$0$$
1.00000 $$0$$
$$938$$ 0 0
$$939$$ 0 0
$$940$$ 0 0
$$941$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$942$$ 0 0
$$943$$ 0 0
$$944$$ 0 0
$$945$$ 0 0
$$946$$ 0 0
$$947$$ − 2.00000i − 2.00000i 1.00000i $$-0.5\pi$$
1.00000i $$-0.5\pi$$
$$948$$ 0 0
$$949$$ 0 0
$$950$$ 0 0
$$951$$ 0 0
$$952$$ 0 0
$$953$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$954$$ 0 0
$$955$$ 0 0
$$956$$ 0 0
$$957$$ 0 0
$$958$$ 0 0
$$959$$ 0 0
$$960$$ 0 0
$$961$$ 1.00000 1.00000
$$962$$ 0 0
$$963$$ 0 0
$$964$$ 0 0
$$965$$ 0 0
$$966$$ 0 0
$$967$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$968$$ 0 0
$$969$$ 0 0
$$970$$ 0 0
$$971$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$972$$ 0 0
$$973$$ 0 0
$$974$$ 0 0
$$975$$ 0 0
$$976$$ 0 0
$$977$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$978$$ 0 0
$$979$$ 0 0
$$980$$ 0 0
$$981$$ 0 0
$$982$$ 0 0
$$983$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$984$$ 0 0
$$985$$ 0 0
$$986$$ 0 0
$$987$$ 0 0
$$988$$ 0 0
$$989$$ 0 0
$$990$$ 0 0
$$991$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$992$$ 0 0
$$993$$ 0 0
$$994$$ 0 0
$$995$$ 0 0
$$996$$ 0 0
$$997$$ 0 0 1.00000 $$0$$
−1.00000 $$\pi$$
$$998$$ 0 0
$$999$$ 0 0
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000

## Twists

By twisting character
Char Parity Ord Type Twist Min Dim
1.1 even 1 trivial 1216.1.g.a.417.2 yes 2
4.3 odd 2 inner 1216.1.g.a.417.1 2
8.3 odd 2 CM 1216.1.g.a.417.2 yes 2
8.5 even 2 inner 1216.1.g.a.417.1 2
19.18 odd 2 CM 1216.1.g.a.417.2 yes 2
76.75 even 2 inner 1216.1.g.a.417.1 2
152.37 odd 2 inner 1216.1.g.a.417.1 2
152.75 even 2 RM 1216.1.g.a.417.2 yes 2

By twisted newform
Twist Min Dim Char Parity Ord Type
1216.1.g.a.417.1 2 4.3 odd 2 inner
1216.1.g.a.417.1 2 8.5 even 2 inner
1216.1.g.a.417.1 2 76.75 even 2 inner
1216.1.g.a.417.1 2 152.37 odd 2 inner
1216.1.g.a.417.2 yes 2 1.1 even 1 trivial
1216.1.g.a.417.2 yes 2 8.3 odd 2 CM
1216.1.g.a.417.2 yes 2 19.18 odd 2 CM
1216.1.g.a.417.2 yes 2 152.75 even 2 RM


text:
# Entropy generation in laminar boundary layers of non-ideal fluid flows

*Corresponding author for this work

Research output: Chapter in Book/Conference proceedings/Edited volumeConference contributionScientificpeer-review

3 Citations (Scopus)

## Abstract

This paper documents a numerical study on entropy generation in zero-pressure gradient, laminar boundary layers of adiabatic non-ideal compressible fluid flows. The entropy generation is expressed in terms of dissipation coefficient $$C:\mathrm {d}$$ and its dependency on free-stream Mach number, fluid molecular complexity, and flow non-ideality is investigated systematically by means of a boundary layer code extended to treat fluids modeled with arbitrary equations of state. The results of the study show that the trend of dissipation coefficient follows that of an incompressible flow for complex fluid molecules like siloxanes in all thermodynamic and flow conditions. For simpler fluids like CO$$:2$$ the trend becomes inversely proportional to the free-stream Mach number and the $$C:\mathrm {d}$$ value can significantly reduce in the non-ideal flow regime, where strong thermo-physical property gradients occur near the wall.

Original language English Non-Ideal Compressible Fluid Dynamics for Propulsion and Power - Selected Contributions from the 2nd International Seminar on Non-Ideal Compressible Fluid Dynamics for Propulsion and Power, NICFD 2018 Francesca di Mare, Andrea Spinelli, Matteo Pini SpringerOpen 104-117 14 9783030496258 https://doi.org/10.1007/978-3-030-49626-5_8 Published - 2020 2nd International Seminar on Non-Ideal Compressible Fluid Dynamics for Propulsion and Power - Bochum, GermanyDuration: 4 Oct 2018 → 5 Oct 2018Conference number: 2

### Publication series

Name Lecture Notes in Mechanical Engineering 2195-4356 2195-4364

### Conference

Conference 2nd International Seminar on Non-Ideal Compressible Fluid Dynamics for Propulsion and Power NICFD 2018 Germany Bochum 4/10/18 → 5/10/18

### Bibliographical note

Green Open Access added to TU Delft Institutional Repository ‘You share, we take care!’ – Taverne project https://www.openaccess.nl/en/you-share-we-take-care
Otherwise as indicated in the copyright section: the publisher is the copyright holder of this work and the author uses the Dutch legislation to make this work public.

## Keywords

• Boundary layer
• Dissipation coefficient
• Non-Ideal Compressible Fluid Dynamics
• Organic Rankine cycle
• Supercritical carbon-dioxide

## Fingerprint

Dive into the research topics of 'Entropy generation in laminar boundary layers of non-ideal fluid flows'. Together they form a unique fingerprint.


text:
# Can this transistor (2N2222) take 6 V on emitter-base? Am I reading the datasheet incorrectly?

I have created the following circuit to better understand how to use a transistor as a switch*.

I've been struggling with understanding datasheets for transistors.

simulate this circuit – Schematic created using CircuitLab

Datasheet states EB maximum voltage

According to the datasheet the emitter-base maximum voltage is 6.0 V. In my circuit I am only applying 5 V. From reading the datasheet (below) would you expect that to be too much?

Also, I see that the secondary (output) circuit of collector-emitter can supposedly take up to 40 V(?). Does that mean I should be able to drive a circuit on the collector-emitter side that has up to 40 V on it?

Am I reading the datasheet properly?

I'm wondering because in my other (referenced question) circuit the transistor became very hot with only 5 V -- but that may have been due to being wired improperly, I'm not sure.

You can see the complete datasheet at:

https://www.onsemi.com/pub/Collateral/P2N2222A-D.PDF

* Note This question is directly related to Is it possible to use a NPN BJT as switch, from single power source?. However, I had the original circuit built improperly there, and I'm now wondering if the 2N2222 can take 5 V on its base pin (according to the datasheet).

• Visualize a diode where the arrow is in a BJT. The transistor will inherently constrain the base voltage to be at most one diode voltage (0.7V) above the emitter. Here you're trying to apply 5V. To limit the current, add a base resistor. The resistor will be 5V on one side and 0.7V on the other. – BB ON Apr 1 at 18:35
• Everyone "always mentions 0.7V" because the base-emitter junction is a PN junction exactly like a diode. It becomes forward biased at that voltage and there it is essentially at maximum conductivity. Applying 5V can't "turn it on" more. The base resistor defines the current allowed to enter the base. By applying a small base current, you get a large collector-emitter current gain. Please read BJT theory. – BB ON Apr 1 at 18:46
• No problem. A quick solution is to just add a resistor between SW1 and the base. Try 1K. – BB ON Apr 1 at 18:57
• You accepted my answer to your previous question, which explained why you need a resistor in series with the base. In your comment you said you had built my proposed circuit and it worked. Why are you trying without a base resistor again? – Peter Bennett Apr 1 at 20:18
• @raddevus 0.7V is about the amount of "bias voltage" must exist across a silicon diode before it stops blocking and starts passing normal current. It's due to the material used to make the semiconductor. For a diode made out of germanium, this bias voltage would be about 0.3V. I'm learning, too. – Suncat2000 Apr 2 at 15:30

Bi-Polar transistors multiply current. A small current from the Base to Emitter causes a large current from the Collector to Emitter. On the ON Semi Datasheet, see hfe on page 2 for the gain, which is between 50 and 375 for this transistor. Realistically it is probably ~ 200. This means that if you have a current flow of 1 mA in the base-emitter, there will be a current flow of 200 mA through the collector - emitter.

The Base Emitter junction is a small diode with a reverse breakdown voltage specified at 6 V. This circuit has no reverse voltage, BUT with 5 V on the Base and ~ 0.7 V drop across the B-E junction, you will have 4.3 V into a short circuit (~ 0.05 Ohms for wiring) which is 56 Amps / 369 Watts! Neither the transistor or the power supply will last more than a microsecond.

That is why you MUST PUT a current limiting RESISTOR in series with the BASE. A value of 430 Ohms would be ~ 10 mA Base current, but.... Use a standard 1.5k (1.47k) Ohm which will allow 2.67 mA Base current which means the Ic current will be ~ 534 mA (with Hfe =200). This is less than the 600 mA Ic allowed by the data sheet. The 200 Ohm resistor will limit the LED-C-E current to 14.5 mA.

Hope all is correct, the last time I did these calcs was over 40 years ago.

• Thanks very much, this definitely adds to my understanding and may be the most appropriate answer to this question since it references the datasheet -- I was also curious about that 600mA max rating that I saw there and you explained that too. – raddevus Apr 2 at 11:56

This is your circuit drawn as a schematic to read for understanding rather than as a wiring diagram (which is more about getting everything connected and not so much for understanding it.)

simulate this circuit – Schematic created using CircuitLab

The idea illustrated in your diagram, where the LED and a current-limiting resistor are placed in series in the collector circuit, is a common (and reasonable) approach. The BJT is operating as a "semiconductor switch" and this is one of several approaches for that behavior. So far, so good.

But the idea of directly tying $$\+5\:\text{V}\$$ to the base, when the emitter is also nailed down to ground, is not common and it's not good. This directly places a full, forward-biased $$\5\:\text{V}\$$ across the base-emitter diode. Keep in mind:

1. You only need from as little as $$\600\:\text{mV}\$$ to perhaps as much as $$\900\:\text{mV}\$$ (in most cases) to use the BJT as a switch.
2. For each additional $$\60\:\text{mV}\$$ (typically) you will get 10 times as much collector current (if permitted by the circuit portions connected to the collector) and 10 times as much base current (always possible.) Broadly speaking, the base current will be exponentially related to the applied forward-biasing voltage across the base and emitter.

You were applying $$\5\:\text{V}\$$!! This is way, way, way above what you should have been using. So the BJT was being literally flooded with base current. Of course it was getting hot! It was dissipating serious power. Might have even damaged the device (I'd probably throw the part away, in fact, after doing something like that.)

This is why a resistor is often applied to the base circuit.

simulate this circuit

The base resistor's voltage drop is a simple linear relationship to the current passing through it. The BJT's base-emitter junction current is an exponential relationship. So as the BJT's base-emitter diode junction tries to rapidly increase it's current, the resistor in series with it opposes this rapid change by dropping voltage. Very quickly, it will turn out that the resistor drops enough voltage so that the base-emitter junction's voltage is close to where it should be.

By using a resistor, you permit the base voltage to "find a stable and reasonable voltage drop" for its operation.

As others have pointed out, the Maximum Ratings section also specifies an absolute worst case reverse-biased voltage for the base-emitter. This is because the base-emitter PN junction diode can't handle a lot of reverse-bias voltage in a typical BJT. Diodes used in bridge rectifiers can often handle very large reverse-bias voltages across them. But not so much with BJTs. They aren't designed to handle much of that kind of stress. Instead, they just break down and avalanche. So the ratings there tell you what to watch out for. Often, people will add a separate diode (oriented opposite to the forward direction of the base-emitter junction of the BJT) going from base to ground in a case like this to protect the BJT ... just in case.

• Fantastic explanation. Very clear and helped me understand a lot. Thanks. – raddevus Apr 1 at 19:35
• @raddevus Thanks for the kind words and I'm glad it helped out. – jonk Apr 1 at 19:38
• Note that that 600-900mV would be doubled in the case of a Darlington transistor. (of course, the semiconductor used also changes the voltage, but silicon is the only one that matters in modern electronics. That may change, silicon carbide is becoming popular, but for now it's all silicon.) – Hearth Apr 2 at 2:45
• @Hearth If the OP had mentioned a Darlington configuration, or Szlikai, I suppose I would have said something more, including slower response times, and so on. But without a hint otherwise, I might have just confused the OP by listing out a table of possibilities that made no sense to the OP and couldn't be put into useful context for the question. I'm kind of glad I didn't mention it, to be honest. ;) – jonk Apr 2 at 3:19
• A key distinction between BJTs and FETs is that BJTs are thought of as being current-controlled devices, FETs are voltage-controlled. The transistor gain is one of the most important parameters when dealing with BJTs (i.e., i_c = i_b * beta, where beta is is the gain; if the base resistor were 0 ohms, just a wire, you'd draw max current...). That transistor gain also changes with temperature... – RichS Apr 2 at 5:22

$$\V_{EB}=V_E-V_B\$$.

The datasheet is saying the emitter can be (up to) 6 V above the base, not that the base can be 6 V above the emitter.

With $$\V_{BE}\$$ ($$\V_B-V_E\$$) at 6 V, an absurdly large current will flow into the base and burn out your transistor very quickly.

• @raddevus: To be clear - $V_{EB}$ is the maximum reverse voltage the base-emitter junction can withstand (where forward is the normal base current direction). – Transistor Apr 1 at 18:10


text:
Summation problem (first N positive integers)

1. Oct 5, 2013

GeorgeCostanz

1. The problem statement, all variables and given/known data

2. Relevant equations

so i kno the formula for the for the sum of the first N positive integers

when i = 1

3. The attempt at a solution

i kno the answer = n^2(n+1)/2

but could someone explain step by step how you reduce it to get the final answer? as if i'm in kindergarten? i'm slow, thanks.

my work:

following the formula: n(n+1)/2 * n(n+1)/2 * n(n+1)/2 ?

i don't understand how to reduce the terms to get the final answer provided

Last edited: Oct 5, 2013
2. Oct 5, 2013

Ray Vickson

PF rules require you to show us your work.

3. Oct 5, 2013

oh okay sure

4. Oct 5, 2013

rcgldr

Are you sure the upper bound on the second summation is "i" and not "n"? Note that this is a summation of the value "1", not the variable "i".

5. Oct 5, 2013

Dick

If the upper bound on the second summation were "n" then the sum would be n^3. You are only going to get the correct answer if the upper bound is "i". Start with the innermost sum and work your way out. It's not hard.


text:
# PHP and Oracle Session Login

Here it's I have a problem with my PHP Code + Oracle Login form.
In this PHP file, I make login function. But I have an error like this :

    Warning: oci_num_rows() expects parameter 1 to be resource, string given in C:\xampp\htdocs\developers\it\session.php on line 12
Wrong

-

<?php
session_start();
include ("config.php");
$username =$_POST['username'];
$password =$_POST['password'];
$do =$_GET['do'];
if($do=="login") {$cek = "SELECT PASSWORD, USER_LEVEL FROM T_USERS WHERE USERNAME='$username' AND PASSWORD='$password'";
$result = oci_parse($conn, $cek); oci_execute($result);
if(oci_num_rows($cek)==1) {$c = oci_fetch_array($result);$_SESSION['username'] = $c['username']; ociresult($c,"USERNAME");
$_SESSION['USER_LEVEL'] =$c['USER_LEVEL']; ociresult($c,"USER_LEVEL"); if($c['USER_LEVEL']=="ADMINISTRATOR")
{
}
else if($c['user_level']=="User") { header("location:user.php"); } else if($c['user_level']=="Root")
{
}
else if($c['user_level']=="Manager") { header("location:manager.php"); } else if($c['user_level']=="Admin")
{
}
else if($c['user_level']=="Director") { header("location:director.php"); } } else { echo "Wrong"; } } ?>  I have tried to search in google, but still don't find anything. Someone knows, what's the problem ? Thanks for advance. - ## 2 Answers According to your script instead of if(oci_num_rows($cek)==1)

you should call

if(oci_num_rows($result)==1)  - for wrong login, that's true. But when I try to login using correct login, it still give an message "Wrong". Any Help again ? – David Jul 30 '12 at 10:55 var_dump(oci_fetch_array($result)) and check –  Vitaliy Zurian Jul 30 '12 at 10:57
it show error like this : Parse error: syntax error, unexpected T_VARIABLE in C:\xampp\htdocs\developers\it\session.php on line 16 –  David Jul 30 '12 at 11:03
Sorry, man, but you should at least be familiar with PHP syntax... –  Vitaliy Zurian Jul 30 '12 at 11:04

You probably want to use $result and not $cek when you're asking for the number of rows returned from oci_num_rows(). However, you really want to avoid using $username and $password directly in the string like that. It'll make you wide open for SQL injection attacks, so look into using oci_parse together with oci_bind_by_name.

After that you should also always call exit() after the sequence of redirects, as the script will continue running if you don't (and that might be a security issue other places).

-