text:
# [Webkit-unassigned] [Bug 110070] Implement MathML linebreaking

bugzilla-daemon at webkit.org bugzilla-daemon at webkit.org
Sun Jan 15 23:21:26 PST 2017

https://bugs.webkit.org/show_bug.cgi?id=110070

BillyMorris <cs002238 at gmail.com> changed:

----------------------------------------------------------------------------
CC|                            |cs002238 at gmail.com

--- Comment #1 from BillyMorris <cs002238 at gmail.com> ---
You are right! Linebreaking is a  must for users, it is really convenient to track the process of execution if you have that bug. However I have recently addressed http://topessaywriter.org/ IT experts to get the bug:
<script type="text/x-mathjax-config">
MathJx.Hub.Config({
Tex: {
good_break: '\\mlToken{mo}[linebreak="good_break"]{}',
no_break: ['\\mlToken{mo}[linebreak="no_break"]{#1}',2],
invisibletimes: ['\\mmlToken{mo}{\u2062}']
}  } }); </script>

Everything was alright for more than a year and now I've got that type of problem with bug. Has anyone afced such problem before? I just do not want to address the services to bother experts with such silly question.

--
You are receiving this mail because:
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# Difference: StewartBoogertPhotometry20161021 (8 vs. 9)

#### Revision 921 Oct 2016 - AnishaVadher

Line: 1 to 1

 META TOPICPARENT name="StewartBoogertPhotometry2016"
-- LauraBevilacqua - 18 Oct 2016
Line: 19 to 19
Anisha:
Changed:
<
<
• Corrected histrogram of number of photoelectrons for a 0.1s exposure in the dark frame with 500 bins - logarithmicly normalized:
>
>
• Corrected histrogram of number of photoelectrons for a 0.1s exposure in the dark frame with 500 bins - logarithmicly normalized:

Changed:
<
<
• The median intensity as a function of exposure time.:
>
>
• The median intensity as a function of exposure time.:

• Line of best fit: y = 917.43 + 0.08x
Changed:
<
<
• Error as a function of exposure time:
>
>
• Error as a function of exposure time:

• Line of best fit: y = 15.76 + 0.73x
%BEGINLATEX% Mean number of effective photoelectrons : $$\nu = \nu_0 + \mu t.$$ Actual number of photoelectrons: $$n = n_0 + m,$$ where $n_0$ follows a Gaussian distribution and may come from electronic noise and $m$ follows a Poisson distribution as a function of $\mu t$. Error in $m \sigma_m$ come from $\sqrt{\mu t}$.
Line: 60 to 60

• Images scanned for stars and correct star count is returned
• Edge detection helps identify star position
Changed:
<
<
• The maximum pixel value was found to be 65195 from the fits file below, as opposed to Aaron's given value of 65535
• Threshold of 100 set so if pixels are above this value they are set as white background, if below they are coloured black (remember reverse grayscale):
• Edge detection:
>
>
• The maximum pixel value was found to be 65195 from the fits file below, as opposed to Aaron's given value of 65535
• Threshold of 100 set so if pixels are above this value they are set as white background, if below they are coloured black (remember reverse grayscale):
• Edge detection:

Changed:
<
<
• Maximum pixel value found to be 62673, compared to Aaron's 60000:
• Threshold set as 20 so that stars don't touch, achieves accurate count of 8:

>
>
• Maximum pixel value found to be 62673, compared to Aaron's 60000:
• Threshold set as 20 so that stars don't touch, achieves accurate count of 8:

 META FILEATTACHMENT attachment="SkewTest2.png" attr="" comment="Testing x and y correlation (P) for P=0, P=0.5, P=0.999" date="1476827216" name="SkewTest2.png" path="SkewTest2.png" size="19013" user="zyva010" version="1" attachment="Histogram_of_number_of_pixel_values_for_0.1s_exposure_500_bins.png" attr="" comment="Corrected histrogram of number of photoelectrons for a 0.1s exposure in the dark frame with 500 bins - logarithmicly normalized" date="1476982418" name="Histogram_of_number_of_pixel_values_for_0.1s_exposure_500_bins.png" path="Histogram_of_number_of_pixel_values_for_0.1s_exposure_500_bins.png" size="18847" user="zyva858" version="1"

Copyright © 2008-2021 by the contributing authors. All material on this collaboration platform is the property of the contributing authors.
Ideas, requests, problems regarding RHUL Physics Department TWiki? Send feedback

format: plain


text:
## How do I generate a bootstrap file on Ubuntu?

2

1

I would like to generate a Bitcoin and Dogecoin (qt) bootstrap file on my ubuntu machine. Is there any way to do this? I know it's possible with Windows but I'm on Ubuntu. Thanks.

possible duplicate: Creating my own bootstrap.dat?

– Murch – 2015-05-12T08:17:22.480

2

cd ~/.bitcoin/blocks

Or, to bootstrap testnet, find your local testnet blockchain:

cd ~/.bitcoin/testnet3/blocks

Concatenate all your blkxxxxx.dat files and place the result in bootstrap.dat:

cat blk* > bootstrap.dat

Warning: Using tar -cf (as instructed by this answer) did not work for me. I'm importing from bitcoind into btcd.

0

It should be as easy as

cd /home/USER/.bitcoin/blocks

tar -cf bootstrap.dat blk*

For dogecoin just change the directory to match

Pretty sure that just creates a tarfile with the blk files in it. – Nick ODell – 2014-12-16T00:38:57.530

ya bitcoind seems to recognize that format just fine. might not be a true bootstrap but i have used it successfully on older bitcoin clones. Might not work the same these days. Ill test tonight and update my post as needed – Mullick – 2014-12-21T04:15:38.853

@Mullick: Were you able to confirm your solution? – Murch – 2015-05-12T08:18:11.570


text:
# Illustration Parallelschaltung von drei Widerständen und deren Spannungen

Get illustration

Share — copy and redistribute the material in any medium or format

Adapt — remix, transform, and build upon the material for any purpose, even commercially.

Sharing and adapting of the illustration is allowed with indication of the link to the illustration.

Three electrical resistors $$R_1$$, $$R_2$$ and $$R_3$$ are connected parallel (side by side). that is they are connected so that different current flows through all three. The total current $$I$$ splits at the junctions so that current $$I_1$$ flows through the first resistor, current $$I_2$$ flows through the second resistor, and current $$I_3$$ flows through the third resistor. This is the characteristic of a parallel connection of resistors!

For the current to flow at all, a voltage source is connected to the resistors, which supplies the voltage $$U$$. The positive pole of the voltage source is connected to one end of all three resistors and the negative pole is connected to the other end of all three resistors.

When resistors are connected in parallel, the voltages across the resistors are equal to the applied voltage $$U$$:$U ~=~ U_1 ~+~ U_2 ~+~ U_3$


text:
# Strong Approximations of BSDEs in a domain

* Auteur correspondant
Abstract : We study the strong approximation of a Backward SDE with finite stopping time horizon, namely the first exit time of a forward SDE from a cylindrical domain. We use the Euler scheme approach of Bouchard and Touzi, Zhang 04}. When the domain is piecewise smooth and under a non-characteristic boundary condition, we show that the associated strong error is at most of order $h^{\frac14-\eps}$ where $h$ denotes the time step and $\eps$ is any positive parameter. This rate corresponds to the strong exit time approximation. It is improved to $h^{\frac12-\eps}$ when the exit time can be exactly simulated or for a weaker form of the approximation error. Importantly, these results are obtained without uniform ellipticity condition.
Mots-clés :
Type de document :
Article dans une revue
Bernoulli, Bernoulli Society for Mathematical Statistics and Probability, 2009, 15 (4), pp.1117-1147. <10.3150/08-BEJ181>
Domaine :

https://hal.archives-ouvertes.fr/hal-00177481
Contributeur : Stephane Menozzi <>
Soumis le : lundi 15 septembre 2008 - 11:39:15
Dernière modification le : lundi 29 mai 2017 - 14:22:50
Document(s) archivé(s) le : mardi 21 septembre 2010 - 17:13:27

### Fichiers

BMELECTRO.pdf
Fichiers produits par l'(les) auteur(s)

### Citation

Bruno Bouchard, Stephane Menozzi. Strong Approximations of BSDEs in a domain. Bernoulli, Bernoulli Society for Mathematical Statistics and Probability, 2009, 15 (4), pp.1117-1147. <10.3150/08-BEJ181>. <hal-00177481v2>

Consultations de
la notice

## 520

Téléchargements du document


text:
# Count arrays with size n, sum k and largest element m

I'm trying to solve pretty complex problem with combinatorics.

Namely, we have given three numbers N, K, M. Now we want to count how many different arrays of integers are there with length N, sum K and all the elements in the range [1, M]

Constraints:

• 1 <= N <= 100
• 1 <= K <= 100
• 1 <= M <= 100

Example

Let's say N = 2, K = 5, M = 3. This means that we want to count arrays of integers of size 2 with sum of all elements equal to 5 and elements in range [1, 3]. There are total of 2 arrays: {2, 3} and {3, 2}. Please note that the order of the elements also matters, {2, 3} is not equal to {3, 2}

Second example: N = 4, K = 7, M = 3. We want to count arrays of length 4, sum of 7 and elements in range [1, 3].

There are total of 16 possible way of arrays: (1,1,2,3), (1,1,3,2), (2,1,1,3), (3,1,1,2), (2,3,1,1), (3,2,1,1), (1,2,3,1), (1,3,2,1), (1,2,1,3), (1,3,1,2), (1,2,2,2), (2,1,2,2), (2,2,1,2), (2,2,2,1)

What I have tried

I know that one solution is to generate all possible arrays, but such algorithm in best case will work in complexity O(N!) which is far too big for N = 100. I started thinking about solving this with three-dimensional dynamical programming, but I cannot find the relations between the states.

I'm thinking about this way: Let f(i, j, l) be the number of arrays of length i, sum j, and largest element l. We can see that for i = 0 f(i,j,l) = 0, so this is I think the base case. Also f(1, 1, 1) = 1 is another base case.

Now I cannot find the relations between the states. Can you give me some hints how to find the relations between the states. Thanks in advance.

• Nice problem. Can you credit the source where you encountered the problem? – D.W. Jul 25 '17 at 16:11
• The problem is from one macedonian site for training, and the problem in original is in macedonian, but i can give you the test cases if you want them – someone12321 Jul 25 '17 at 16:30

You can use dynamic programming. For each $0 \leq i \leq N$ and $0 \leq s \leq K$, count the number of arrays of length $i$, consisting of numbers in the range $\{1,\ldots,M\}$, which sum to exactly $s$. The running time is $O(NK)$.

Explicitly, denoting the array by $a$, we have $a(0,0) = 1$, $a(0,s) = 0$ otherwise, and $$a(i,s) = \sum_{t=1}^M a(i-1,s-t),$$ where $a(i-1,r) = 0$ if $r < 0$.

As Hendrik Jan mentions in the comments, we can improve on this $O(NKM)$ algorithm by using the recursion $$a(i,s) = a(i,s-1) - a(i-1,s-1-M) + a(i-1,s-1),$$ with suitable base cases.

Alternatively, we can obtain an explicit expression: the answer is the coefficient of $x^K$ in the generating function $$(x + \cdots + x^M)^N = x^N \left(\frac{1-x^M}{1-x}\right)^N.$$ In other words, we are looking for the coefficient of $x^{K-N}$ in $$\left( \sum_{i=0}^N (-1)^i \binom{N}{i} x^{iM} \right) \left( \sum_{j=0}^\infty \binom{j+N-1}{N-1} x^j \right),$$ which has the closed form $$\sum_{i=0}^{\min(N,\lfloor (K-N)/M\rfloor)} (-1)^i \binom{N}{i} \binom{K-iM-1}{N-1}.$$

• The numbers $a(i,s)$ represent a matrix of solutions for "fixed" $M$. The numbers $a(i,s)$ are a "running sum" of $M$ consecutive elements from the previous row. So, can't the complexity be improved to $O(NK)$, as $a(i,s) = a(i,s-1) + a(i-1,s-1) - a(i-1,s-M-1)$? give or take typo's. – Hendrik Jan Jul 25 '17 at 1:22
• @HendrikJan Yes, that's right! I missed this somehow. – Yuval Filmus Jul 25 '17 at 5:12

You actually don't need the largest element l in your recursive function. It doesn't make such sense to use it, since the largest number in an array has no influence on the other numbers.

In most (all?) dynamic programming problems you have to think about the last step / the last part. For f(i, j) you have an array with i numbers. You can imagine, that in the last step you added the last number to the array. So before you added the last number, the array only consists of i-1 numbers.

Now, if the added number was 1, then there are f(i-1, j-1) many arrays. If the added number was 2, then there are f(i-1, j-2) many arrays. And so on...

Add all those possibilities up, and you end up with f(i, j).

I don't pretend this is the most efficient solution. Neither it has a strong theoretical merit, but you can still find it useful to verify your results. Here's a small program in Prolog, (using SWI Prolog for constraint satisfaction part), which can calculate or verify your guess:

% -*- mode: prolog; prolog-system: "swi" -*-
:- use_module(library(clpfd)).

sums(K, [], K).
sums(A, [X | Xs], K) :-
indomain(A),
indomain(X),
B is A + X,
sums(B, Xs, K).
sums([X | Xs], K) :- sums(X, Xs, K).

nmk_problem_helper(N, K, M, List) :-
indomain(N),
indomain(M),
indomain(K),
length(List, N),
List ins 1..M,
sums(List, K).

nmk_problem(N, K, M, Arrays) :-
findall(X, nmk_problem_helper(N, K, M, X), Arrays).

nmk_problem_all_helper(List) :-
[N, M, K] ins 1..100,
indomain(N),
indomain(M),
indomain(K),
length(List, N),
List ins 1..M,
sums(List, K).

nmk_problem_all(Arrays) :-
findall(X, nmk_problem_all_helper(X), Arrays).

Example usage:

?- nmk_problem(4, 7, 3, X).
X = [[1, 1, 2, 3], [1, 1, 3, 2], [1, 2, 1, 3], [1, 2, 2, 2], [1, 2, 3, 1], [1, 3, 1|...], [1, 3|...], [2|...], [...|...]|...].


text:
MathSciNet bibliographic data MR1287610 46-01 (47-01 65-01) Trenogin, V. A. \cyr Funktsionalʹnyĭ analiz. (Russian) [Functional analysis] Second edition. "Nauka'', Moscow, 1993. 440 pp. ISBN: 5-02-014891-1 Links to the journal or article are not yet available

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

 American Mathematical Society 201 Charles Street Providence, RI 02904-6248 USA


text:
# Can one prove $(g_1H)(g_2H) = (g_1g_2)H$, $g_1, g_2 \in G$ if and only if $gH = Hg$ for all $g \in G$?

Let $G$ be a group and $H$ a subgroup of $G$.

I have proven in an exercise that $gH = Hg$ for all $g \in G$ implies $(g_1H)(g_2H) = (g_1g_2)H$, $g_1, g_2 \in G$ where $(g_1H)(g_2H) = (xy | x\in g_1H, y\in g_2H)$.

Can one prove $(g_1H)(g_2H) = (g_1g_2)H$, $g_1, g_2 \in G$ if and only if $gH = Hg$ for all $g \in G$ ?

I have been sitting trying to come up with a counter example. Can anyone help ? Thanks in advance.

/Nicolas

Yes, it is. Put $g_2=g_1^{-1}$, then $g_1Hg_1^{-1}H=H$, hence $g_1Hg_1^{-1}\subseteq H$.

1) Putting $g_1=g_2=1$ we get $H$ is a subgroup (so the condition "$H$ is a subgroup" is unnecessary).
2) Put $g_2=g_1^{−1}$, then $g_1Hg_1^{−1}H=H$. Hence $g_1Hg_1^{−1}\subseteq H$, i.e. $g_1H\subseteq Hg_1$. From here Hg_1^{−1}\subseteq g_1^{−1}H. This is true for all $g_1$, so $xH\subseteq Hx$ and $Hx\subseteq xH$ for all $x$.
• Sorry, are you proving $(g_1H)(g_2H) = (g_1g_2)H$, $g_1, g_2 \in G$ implies $gH = Hg$ for all $g \in G$ ? - or is this some counter example ? – Shuzheng Oct 11 '13 at 20:48
• It follows from $gHg^{-1}\subseteq H$ that $gH\subseteq Hg$ and $Hg^{-1}\subseteq g^{-1}H$ for ALL $g\in G$. So $xH=Hx$ for all $x\in G$. – Boris Novikov Oct 11 '13 at 20:56
• But as this proof shows, $gHg^{-1} \leq G$ is just a subgroup in $G$ not neccesarily a subset of $H$ for arbitary $g \in G$ ? proofwiki.org/wiki/Conjugate_of_Subgroup_is_Subgroup – Shuzheng Oct 12 '13 at 9:11
• But your condition implies that $H$ is a subgroup: take $g_1=g_2=1$. – Boris Novikov Oct 12 '13 at 9:35
• Sorry, my mistake. I've edited. What i'm interested in is: $(g_1H)(g_2H)=(g_1g_2H)$ implies $\forall g \in G (gH = Hg)$, then I have proven both ways in the bicondition. Also, thanks for your help so far Boris. – Shuzheng Oct 14 '13 at 11:04


text:
# Bidirectional solid state switch for solenoid

I'm building a circuit to control a flip dot display, which is basically a matrix of solenoids which you control with bidirectional current across the 25 columns and 7 rows of the board.

The attached schematic shows the circuit so far - I'm using a bank of H-Bridges to control the columns, and they switch polarity depending whether I want to flip a particular dot one way or the other.

The challenge comes with controlling the rows, at the bottom of the image. If I'm targeting a particular row, the positive voltage has to go down one path, and the negative down another. So I imagine I'm looking for a bidirectional DC solid state switch that can handle a peak of about .5A.

Any suggestions on how I should implement that part of the circuit? Electromechanical relays are not good because these switches will be used thousands of times a day (in a clock). I also need to be able to handle up to 0.5A.

Edit: I've added a more accurate schematic for the flip dot board (25 columns, 7 rows)

update:

Thanks to all the ideas folks contributed to help me solve the problem. I finally finished the flip-dot clock project - you can check it out at http://www.dhenshaw.com/dottie

If there's interest I'll post a schematic of what I built (I'd also have to draw it first!)

• Both sides of the solenoids are H-Bridges. You already have the technology you need. – gbulmer Aug 18 '14 at 13:55
• So when all power is off, the flipped dots hold their positions whether "on" or "off"? And by extension, that means that only a pulse is needed to change the orientation of any one dot, with no requirement for sustained power to the solenoid? EDIT: will these solenoids even tolerate being energized full-time? – TDHofstetter Aug 26 '14 at 4:40
• Correct - a 100ms pulse flips a dot and it then stays in place. You could keep them energized all the time, but that would be a waste of electricity (9v @ 350mA) – David Aug 26 '14 at 18:12
• I'd be interested in your final schematic whenever you have a chance to draw it up! – cristoper Oct 25 '14 at 16:04
• ...did you manage the final schematics for Dottie? Would be very interesting for me, because I just got some 5x7 flip dot units. Cheers Udo Seligmann / Wuppertal, Germany p.s. is Dottie still running ok? – Udo Seligmann Jan 18 '17 at 16:52

## 2 Answers

David, while you may be able to find the appropriate switches and make your current approach work, I think there is a neater solution to your problem.

Each row and column in the matrix can be high, or low, or floating. This gives you random access to each solenoid, so you can run current through one solenoid in either direction, while no current is running through other solenoids.

I drew the half-bridges with MOSFETs, but they can be implemented in other ways too.

update:

The circuit in the dot board with the diodes seems a bit odd to me. It would still require nTransistors = 2*nRows + 2*nCols, but now the dot board has 2x connections per row instead of 1. I wonder what the rationale of the dot board designers was. I also wonder if I’m missing something.

Then again, we’re situated roughly in the same neck of the woods. Perhaps, I can physically look at your dot board.

• Thanks for your suggestion - it makes sense, except that the solenoids (which are on a complicated PCB with built-in diodes) aren't wired as simple as this... each solenoid has one line to control column, but two lines to control the row, which are dependent on voltage polarity - like my diagram above. I did try to wire it per your suggestion, but it didn't work (nothing flipped either way). I've manually wired up some pushbuttons & got the dots to flip either way, so I'm pretty sure my schematic depiction is accurate (but obviously not my electronics implementation!) – David Aug 19 '14 at 3:29
• @David It sounds like each one of your dots has 3 connections and 2 separate solenoids. But in that case, it seems odd to me that it has 2 solenoids per dot and you need to reverse polarity. Do you have a datasheet for your flip dots? – Nick Alexeev Aug 19 '14 at 3:40
• Sorry for the delay... so to be more specific, I uploaded a schematic that shows how there's one "column selector" line but two row selector lines. The PCB includes the diodes. – David Aug 26 '14 at 3:05
• @David I have updated my answer. Have a look. – Nick Alexeev Aug 31 '14 at 4:43
• In the first schematic in this answer, you'll find that when you try to activate L11, a parasitic current will also flow through the other three. The parasitic path has 3x resistance of L11, so you'll get 1/3 the current, but that still may be enough to flip a solenoid. I actually just built an array like this of 5x8 only to find that I was unintentionally actuating solenoids that share a row and column. It looks like the diodes in the second schematic save you from this issue – miles60 Mar 21 at 22:07

This one fell right off the bottom of the page & got forgotten.

David, if you already have your enableable H-bridge, then all you need to add to your original drawing is one pair of diodes per cell to ensure that negative voltage drives the cell to one state and positive voltage drives the cell to the other state.

The key will be "enableable". When you're not actually driving any cells to any new state, the whole H-bridge will need to drop power.


text:
## Seminar information archive

#### Algebraic Geometry Seminar

15:00-16:30   Room #122 (Graduate School of Math. Sci. Bldg.)
Thomas Ducat (University of Warwick)
Divisorial Extractions from Singular Curves in Smooth 3-Folds (ENGLISH)
[ Abstract ]
Consider a singular curve C contained in a smooth 3-fold X.
Assuming the existence of a Du Val general elephant S containing C,
I give a normal form for the equations of C in X and an outline of how to
construct a divisorial extraction from this curve. If the general S is
Du Val of type D_{2k}, E_6 or E_7 then I can give some explicit
conditions for the existence of a terminal extraction. A treatment of
the D_{2k+1} case should be possible by similar means.

### 2014/01/21

#### Tuesday Seminar of Analysis

16:30-18:00   Room #128 (Graduate School of Math. Sci. Bldg.)
Nao Hamamuki (Graduate shool of Mathematical Sciences, the University of Tokyo)
An improved level set method based on comparison with a signed distance function (JAPANESE)
[ Abstract ]
In the classical level set method, a slope of a solution to level set
equations can be close to zero as time develops even if the initial
slope is large, and this prevents one from computing interfaces given as
the level set of the solution. To overcome this issue we introduce an
improved equation by adding an extra term to the original equation.
Then, by applying a comparison principle to the signed distance function
to the interface, we prove that, globally in time, the slope of a
solution of the initial value problem is preserved near the zero level set.

#### Tuesday Seminar on Topology

16:30-17:30   Room #056 (Graduate School of Math. Sci. Bldg.)
Naohiko Kasuya (The University of Tokyo)
On contact submanifolds of the odd dimensional Euclidean space (JAPANESE)
[ Abstract ]
We prove that the Chern class of a closed contact manifold is an
obstruction for codimension two contact embeddings in the odd
dimensional Euclidean space.
By Gromov's h-principle,
for any closed contact $3$-manifold with trivial first Chern class,
there is a contact structure on $\\mathbb{R}^5$ which admits a contact
embedding.

#### Tuesday Seminar on Topology

17:30-18:30   Room #056 (Graduate School of Math. Sci. Bldg.)
Xiaolong Li (The University of Tokyo)
Weak eigenvalues in homoclinic classes: perturbations from saddles
with small angles (ENGLISH)
[ Abstract ]
For 3-dimensional homoclinic classes of saddles with index 2, a
new sufficient condition for creating weak contracting eigenvalues is
provided. Our perturbation makes use of small angles between stable and
unstable subspaces of saddles. In particular, by recovering the unstable
eigenvector, we can designate that the newly created weak eigenvalue is
contracting. As applications, we obtain C^1-generic non-trivial index-
intervals of homoclinic classes and the C^1-approximation of robust
heterodimensional cycles. In particular, this sufficient condition is
satisfied by a substantial class of saddles with homoclinic tangencies.

### 2014/01/20

#### Seminar on Geometric Complex Analysis

10:30-12:00   Room #126 (Graduate School of Math. Sci. Bldg.)
Hideki Miyachi (Osaka University)
タイヒミュラー距離の幾何学とその応用 (JAPANESE)

#### GCOE Seminars

16:00-17:00   Room #118 (Graduate School of Math. Sci. Bldg.)
Victor Isakov (The Wichita State University)
Increasing stability in the inverse problems for the Helmholtz type prposed equations (ENGLISH)
[ Abstract ]
We report on new stability estimates for recovery of the near field from the prposed scattering amplitude prposed and for Schroedinger potential from the Dirichlet-to Neumann map. In these prposed esrtimates prposed unstable (logarithmic part) goes to zero as the wave number grows. Proofs prposed are using prposed new bounds for Hankel functions and complex and real geometrical optics prposed solutions.

#### Algebraic Geometry Seminar

15:30-17:00   Room #126 (Graduate School of Math. Sci. Bldg.)
Taro Sano (University of Warwick)
Deforming elephants of Q-Fano 3-folds (ENGLISH)
[ Abstract ]
Shokurov and Reid proved that a Fano 3-fold with canonical
Gorenstein singularities has a Du Val elephant, that is,
a member of the anticanonical linear system with only Du Val singularities.
The classification of Fano 3-folds is based on this fact.
However, for a Fano 3-fold with non-Gorenstein terminal singularities,
the anticanonical system does not contain such a member in general.
Alt{\\i}nok--Brown--Reid conjectured that, if the anticanonical system is non-empty,
a Q-Fano 3-fold can be deformed to that with a Du Val elephant.
In this talk, I will explain how to deform an elephant with isolated
singularities to a Du Val elephant.

### 2014/01/15

#### FMSP Lectures

14:50-16:20   Room #056 (Graduate School of Math. Sci. Bldg.)
Rinat Kashaev (University of Geneva)
Lectures on quantum Teichmüller theory IV (ENGLISH)
[ Abstract ]
Quantum Teichmüller theory leads to a family of unitary infinite dimensional projective representations of the mapping class groups of punctured surfaces. One of the recent applications of this theory is the construction of state integral three-manifold invariants related with hyperbolic geometry.
In these lectures it is planned to address the following subjects:
1) Penner’s coordinates in the decorated Teichmüller space.
2) Ratio coordinates.
3) Quantization.
4) The length spectrum of simple closed curves.
[ Reference URL ]
http://fmsp.ms.u-tokyo.ac.jp/FMSPLectures_Kashaev.pdf

#### Operator Algebra Seminars

16:30-18:00   Room #118 (Graduate School of Math. Sci. Bldg.)
Yoh Tanimoto (Univ. Tokyo)
Wedge-local fields in integrable models with bound states (JAPANESE)

#### GCOE Seminars

16:00-17:00   Room #370 (Graduate School of Math. Sci. Bldg.)
Victor Isakov (The Wichita State University)
Increasing stability of the continuation for the Helmholtz type equations (ENGLISH)
[ Abstract ]
We derive conditional stability estimates for the Helmholtz type equations which are becoming of Lipschitz type for large frequencies/wave numbers. Proofs use splitting solutions into low and high frequencies parts where we use energy (in particular) Carleman estimates. We discuss numerical confirmation and open problems.

#### GCOE Seminars

17:00-18:00   Room #370 (Graduate School of Math. Sci. Bldg.)
Jin Cheng (Fudan University)
A numerical method for solving the inverse heat conduction problem without initial value (ENGLISH)
[ Abstract ]
In this talk, we will present some results for the inverse heat conduction problem for the heat equation of determining a boundary value at in an unreachable part of the boundary. The main difficulty for this problem is that the initial value is unknown by the practical reason. A new method is prposed to solve this problem and the nuemrical tests show the effective of this method. Some theoretic analysis will be presented. This is a joint work with J Nakagawa, YB Wang, M Yamamoto.

#### Number Theory Seminar

16:40-17:40   Room #056 (Graduate School of Math. Sci. Bldg.)
Stephen Lichtenbaum (Brown University)
Special values of zeta-functions of schemes (ENGLISH)
[ Abstract ]
We will give conjectured formulas giving the behavior of the
seta-function of regular schemes projective and flat over Spec Z at
non-positive integers in terms of Weil-etale cohomology. We will also
explain the conjectured relationship of Weil-etale cohomology to etale
cohomology, which makes it possible to express these formulas also in terms
of etale cohomology.

### 2014/01/14

#### FMSP Lectures

14:50-16:20   Room #056 (Graduate School of Math. Sci. Bldg.)
Rinat Kashaev (University of Geneva)
Lectures on quantum Teichmüller theory III (ENGLISH)
[ Abstract ]
Quantum Teichmüller theory leads to a family of unitary infinite dimensional projective representations of the mapping class groups of punctured surfaces. One of the recent applications of this theory is the construction of state integral three-manifold invariants related with hyperbolic geometry.
In these lectures it is planned to address the following subjects:
1) Penner’s coordinates in the decorated Teichmüller space.
2) Ratio coordinates.
3) Quantization.
4) The length spectrum of simple closed curves.
[ Reference URL ]
http://fmsp.ms.u-tokyo.ac.jp/FMSPLectures_Kashaev.pdf

#### Tuesday Seminar on Topology

17:00-18:00   Room #056 (Graduate School of Math. Sci. Bldg.)
Rinat Kashaev (University of Geneva)
State-integral partition functions on shaped triangulations (ENGLISH)
[ Abstract ]
Quantum Teichm\\"uller theory can be promoted to a
generalized TQFT within the combinatorial framework of shaped
triangulations with the tetrahedral weight functions given in
terms of the Weil-Gelfand-Zak transformation of Faddeev.FN"s
quantum dilogarithm. By using simple examples, I will
illustrate the connection of this theory with the hyperbolic
geometry in three dimensions.

#### Tuesday Seminar of Analysis

16:30-18:00   Room #128 (Graduate School of Math. Sci. Bldg.)
Kenichi Ito (University of Tsukuba)
Threshold properties for one-dimensional discrete Schr\\"odinger operators (JAPANESE)
[ Abstract ]
We study the relation between the generalized eigenspace and the asymptotic expansion of the resolvent around the threshold $0$ for the one-dimensional discrete Schr\\"odinger operator on $\\mathbb Z$. We decompose the generalized eigenspace into the subspaces corresponding to the eigenstates and the resonance states only by their asymptotics at infinity, and classify the coefficient operators of the singlar part of resolvent expansion completely in terms of these eigenspaces. Here the generalized eigenspace we consider is largest possible. For an explicit computation of the resolvent expansion we apply the expansion scheme of Jensen-Nenciu (2001). This talk is based on the recent joint work with Arne Jensen (Aalborg University).

#### Lie Groups and Representation Theory

16:30-18:00   Room #126 (Graduate School of Math. Sci. Bldg.)
Masaki Mori (the University of Tokyo)
A cellular classification of simple modules of the Hecke-Clifford
superalgebra (JAPANESE)
[ Abstract ]
The Hecke--Clifford superalgebra is a super version of
the Iwahori--Hecke algebra of type A. Its simple modules
are classified by Brundan, Kleshchev and Tsuchioka using
a method of categorification of affine Lie algebras.
However their constructions are too abstract to study in practice.
In this talk, we introduce a more concrete way to produce its
simple modules with a generalized theory of cellular algebras
which is originally developed by Graham and Lehrer.
In our construction the key is that there is a right action of
the Clifford superalgebra on the super-analogue of the Specht module.
With the help of the notion of the Morita context, a simple module
of the Hecke--Clifford superalgebra is made from that of
the Clifford superalgebra.

#### GCOE Seminars

15:00-16:00   Room #118 (Graduate School of Math. Sci. Bldg.)
Calderon problem for Maxwell's equations in cylindrical domain (ENGLISH)
[ Abstract ]
We prove some uniqueness results in determination of the conductivity, the permeability and the permittivity of Maxwell's equations from partial Dirichlet-to-Neumann map.

### 2014/01/11

#### Monthly Seminar on Arithmetic of Automorphic Forms

14:00-16:00   Room #123 (Graduate School of Math. Sci. Bldg.)
Dihua Jiang (School of Mathematics, University of Minnesota) 14:00-14:45
A product of tensor product L-functions of quasi-split classical groups of Hermitian type. (jointly with Lei Zhang) (ENGLISH)
Dihua Jiang (School of Mathematics, University of Minnesota) 15:00-15:45
A product of tensor product L-functions of quasi-split classical groups of Hermitian type, Part II. (jointly with Lei Zhang) (ENGLISH)

### 2014/01/10

#### FMSP Lectures

14:50-16:20   Room #056 (Graduate School of Math. Sci. Bldg.)
Rinat Kashaev (University of Geneva)
Lectures on quantum Teichmüller theory II (ENGLISH)
[ Abstract ]
Quantum Teichmüller theory leads to a family of unitary infinite dimensional projective representations of the mapping class groups of punctured surfaces. One of the recent applications of this theory is the construction of state integral three-manifold invariants related with hyperbolic geometry.
In these lectures it is planned to address the following subjects:
1) Penner’s coordinates in the decorated Teichmüller space.
2) Ratio coordinates.
3) Quantization.
4) The length spectrum of simple closed curves.
[ Reference URL ]
http://fmsp.ms.u-tokyo.ac.jp/FMSPLectures_Kashaev.pdf

### 2014/01/09

#### FMSP Lectures

14:50-16:20   Room #056 (Graduate School of Math. Sci. Bldg.)
Rinat Kashaev (University of Geneva)
Lectures on quantum Teichmüller theory I (ENGLISH)
[ Abstract ]
Quantum Teichmüller theory leads to a family of unitary infinite dimensional projective representations of the mapping class groups of punctured surfaces. One of the recent applications of this theory is the construction of state integral three-manifold invariants related with hyperbolic geometry.
In these lectures it is planned to address the following subjects:
1) Penner’s coordinates in the decorated Teichmüller space.
2) Ratio coordinates.
3) Quantization.
4) The length spectrum of simple closed curves.
[ Reference URL ]
http://fmsp.ms.u-tokyo.ac.jp/FMSPLectures_Kashaev.pdf

### 2014/01/08

#### Operator Algebra Seminars

16:30-18:00   Room #122 (Graduate School of Math. Sci. Bldg.)
Sakae Fuchino (Kobe University)
Dow's metrization theorem and beyond (JAPANESE)

#### Number Theory Seminar

16:40-17:40   Room #056 (Graduate School of Math. Sci. Bldg.)
Sho Yoshikawa (University of Tokyo)
Roots of the discriminant of an elliptic curves and its torsion points (JAPANESE)
[ Abstract ]
We give an explicit and intrinsic description of (the torsor defined by the 12th roots of) the discriminant of an elliptic curve using the group of its 12-torsion points and the Weil pairing. As an application, we extend a result of Coates (which deals with the characteristic 0 case) to the case where the characteristic of the base field is not 2 or 3. This is a joint work with Kohei Fukuda.

### 2013/12/26

#### Operator Algebra Seminars

16:30-18:00   Room #118 (Graduate School of Math. Sci. Bldg.)
Shuhei Masumoto (Univ. Tokyo)
Countable Chain Condition for $C^*$-algebras (ENGLISH)

### 2013/12/25

#### GCOE Seminars

16:00-17:00   Room #270 (Graduate School of Math. Sci. Bldg.)
Kazufumi Ito (North Carolina State University)
Nonsmooth Nonconvex Optimization Problems (ENGLISH)
[ Abstract ]
A general class of nonsmooth nonconvex optimization problems is discussed. A general existence theory of solutions, the Lagrange multiplier theory and sensitivity analysis of the optimal value function are developed. Concrete examples are presented to demonstrate the applicability of our approach.

### 2013/12/24

#### Tuesday Seminar on Topology

16:30-18:00   Room #056 (Graduate School of Math. Sci. Bldg.)
Tirasan Khandhawit (Kavli IPMU)
Stable homotopy type for monopole Floer homology (ENGLISH)
[ Abstract ]
In this talk, I will try to give an overview of the
construction of stable homotopy type for monopole Floer homology. The
construction associates a stable homotopy object to 3-manifolds, which
will recover the Floer groups by appropriate homology theory. The main
ingredients are finite dimensional approximation technique and Conley
index theory. In addition, I will demonstrate construction for certain
3-manifolds such as the 3-torus.


text:
# 6.6: Graph Quadratic Functions Using Properties

##### Learning Objectives

By the end of this section, you will be able to:

• Recognize the graph of a quadratic function
• Find the axis of symmetry and vertex of a parabola
• Find the intercepts of a parabola
• Graph quadratic functions using properties
• Solve maximum and minimum applications

Before you get started, take this readiness quiz.

1. Graph the function $$f(x)=x^{2}$$ by plotting points.
If you missed this problem, review Example 3.54.
2. Solve: $$2 x^{2}+3 x-2=0$$.
If you missed this problem, review Example 6.45.
3. Evaluate $$-\frac{b}{2 a}$$ when $$a=3$$ and $$b=-6$$.
If you missed this problem, review Example 1.21.

## Recognize the Graph of a Quadratic Function

Previously we very briefly looked at the function $$f(x)=x^{2}$$, which we called the square function. It was one of the first non-linear functions we looked at. Now we will graph functions of the form $$f(x)=a x^{2}+b x+c$$ if $$a \neq 0$$. We call this kind of function a quadratic function.

##### Definition $$\PageIndex{1}$$

A quadratic function, where $$a, b$$, and $$c$$ are real numbers and $$a≠0$$, is a function of the form

$$f(x)=a x^{2}+b x+c$$

We graphed the quadratic function $$f(x)=x^{2}$$ by plotting points.

Every quadratic function has a graph that looks like this. We call this figure a parabola. Let’s practice graphing a parabola by plotting a few points.

##### Example $$\PageIndex{1}$$

Graph: $$f(x)=x^{2}-1$$.

Solution:

We will graph the function by plotting points.

 Choose integer values for $$x$$, substitute them into the equation and simplify to find $$f(x)$$. Record the values of the ordered pairs in the chart. Plot the points, and then connect them with a smooth curve. The result will be the graph of the function $$f(x)=x^{2}-1$$.
##### Exercise $$\PageIndex{1}$$

Graph $$f(x)=-x^{2}$$.

##### Exercise $$\PageIndex{2}$$

Graph $$f(x)=x^{2}+1$$.

All graphs of quadratic functions of the form $$f(x)=a x^{2}+b x+c$$ are parabolas that open upward or downward. See Figure 9.6.6

Notice that the only difference in the two functions is the negative sign before the quadratic term ($$x^{2}$$ in the equation of the graph in Figure 9.6.6). When the quadratic term, is positive, the parabola opens upward, and when the quadratic term is negative, the parabola opens downward.

##### Definition $$\PageIndex{2}$$

Parabola Orientation

For the graph of the quadratic function $$f(x)=a x^{2}+b x+c$$, if

##### Example $$\PageIndex{2}$$

Determine whether each parabola opens upward or downward:

1. $$f(x)=-3 x^{2}+2 x-4$$
2. $$f(x)=6 x^{2}+7 x-9$$

Solution:

a. Find the value of $$a$$.

Since the $$a$$ is negative, the parabola will open downward.

b. Find the value of $$a$$.

Since the $$a$$ is positive, the parabola will open upward.

##### Exercise $$\PageIndex{3}$$

Determine whether the graph of each function is a parabola that opens upward or downward:

1. $$f(x)=2 x^{2}+5 x-2$$
2. $$f(x)=-3 x^{2}-4 x+7$$
1. up
2. down
##### Exercise $$\PageIndex{4}$$

Determine whether the graph of each function is a parabola that opens upward or downward:

1. $$f(x)=-2 x^{2}-2 x-3$$
2. $$f(x)=5 x^{2}-2 x-1$$
1. down
2. up

## Find the Axis of Symmetry and Vertex of a Parabola

Look again at Figure 9.6.10. Do you see that we could fold each parabola in half and then one side would lie on top of the other? The ‘fold line’ is a line of symmetry. We call it the axis of symmetry of the parabola.

We show the same two graphs again with the axis of symmetry.

The equation of the axis of symmetry can be derived by using the Quadratic Formula. We will omit the derivation here and proceed directly to using the result. The equation of the axis of symmetry of the graph of $$f(x)=a x^{2}+b x+c$$ is $$x=-\frac{b}{2 a}$$.

So to find the equation of symmetry of each of the parabolas we graphed above, we will substitute into the formula $$x=-\frac{b}{2 a}$$.

Notice that these are the equations of the dashed blue lines on the graphs.

The point on the parabola that is the lowest (parabola opens up), or the highest (parabola opens down), lies on the axis of symmetry. This point is called the vertex of the parabola.

We can easily find the coordinates of the vertex, because we know it is on the axis of symmetry. This means its
$$x$$-coordinate is $$-\frac{b}{2 a}$$. To find the $$y$$-coordinate of the vertex we substitute the value of the $$x$$-coordinate into the quadratic function.

## Axis of Symmetry and Vertex of a Parabola

The graph of the function $$f(x)=a x^{2}+b x+c$$ is a parabola where:

• the axis of symmetry is the vertical line $$x=-\frac{b}{2 a}$$.
• the vertex is a point on the axis of symmetry, so its $$x$$-coordinate is $$-\frac{b}{2 a}$$
• the $$y$$-coordinate of the vertex is found by substituting $$x=-\frac{b}{2 a}$$ into the quadratic equation.
##### Example $$\PageIndex{3}$$

For the graph of $$f(x)=3 x^{2}-6 x+2$$ find:

1. the axis of symmetry
2. the vertex

Solution:

a.

 The axis of symmetry is the vertical line $$x=-\frac{b}{2 a}$$. Substitute the values $$a,b$$ into the equation. $$x=-\frac{-6}{2 \cdot 3}$$ Simplify. $$x=1$$ The axis of symmetry is the line $$x=1$$.

b.

 The vertex is a point on the line of symmetry, so its $$x$$-coordinate will be $$x=1$$. Find $$f(1)$$. Simplify. The result is the $$y$$-coordinate. $$f(1)=-1$$ The vertex is $$(1,-1)$$.
##### Exercise $$\PageIndex{5}$$

For the graph of $$f(x)=2 x^{2}-8 x+1$$ find:

1. the axis of symmetry
2. the vertex
1. $$x=2$$
2. $$(2,-7)$$
##### Exercise $$\PageIndex{6}$$

For the graph of $$f(x)=2 x^{2}-4 x-3$$ find:

1. the axis of symmetry
2. the vertex
1. $$x=1$$
2. $$(1,-5)$$

## Find the Intercepts of a Parabola

When we graphed linear equations, we often used the $$x$$- and $$y$$-intercepts to help us graph the lines. Finding the coordinates of the intercepts will help us to graph parabolas, too.

Remember, at the $$y$$-intercept the value of $$x$$ is zero. So to find the $$y$$-intercept, we substitute $$x=0$$ into the function.

Let’s find the $$y$$-intercepts of the two parabolas shown in Figure 9.6.20.

An $$x$$-intercept results when the value of $$f(x)$$ is zero. To find an $$x$$-intercept, we let $$f(x)=0$$. In other words, we will need to solve the equation $$0=a x^{2}+b x+c$$ for $$x$$.

\begin{aligned} f(x) &=a x^{2}+b x+c \\ 0 &=a x^{2}+b x+c \end{aligned}

Solving quadratic equations like this is exactly what we have done earlier in this chapter!

We can now find the $$x$$-intercepts of the two parabolas we looked at. First we will find the $$x$$-intercepts of the parabola whose function is $$f(x)=x^{2}+4 x+3$$.

 $$f(x)=x^{2}+4 x+3$$ Let $$f(x)=0$$. $$\color{red}0\color{black}=x^{2}+4 x+3$$ Factor. $$0=(x+1)(x+3)$$ Use the Zero Product Property. $$x+1=0 \quad x+3=0$$ Solve. The $$x$$-intercepts are $$(-1,0)$$ and $$(-3,0)$$.

Now we will find the $$x$$-intercepts of the parabola whose function is $$f(x)=-x^{2}+4 x+3$$.

 $$f(x)=-x^{2}+4 x+3$$ Let $$f(x)=0$$. $$\color{red}0 \color{black}=-x^{2}+4 x+3$$ This quadratic does not factor, so we use the Quadratic Formula. $$a=-1, b=4, c=3$$ Simplify. $$x=\frac{-2(2 \pm \sqrt{7})}{-2}$$ $$x=2 \pm \sqrt{7}$$ The $$x$$-intercepts are $$(2+\sqrt{7}, 0)$$ and $$(2-\sqrt{7}, 0)$$.

We will use the decimal approximations of the $$x$$-intercepts, so that we can locate these points on the graph,

$$(2+\sqrt{7}, 0) \approx(4.6,0) \quad(2-\sqrt{7}, 0) \approx(-0.6,0)$$

Do these results agree with our graphs? See Figure 9.6.34

### Find the Intercepts of a Parabola

To find the intercepts of a parabola whose function is $$f(x)=a x^{2}+b x+c$$:

$$y$$-intercept

Let $$x=0$$ and solve for $$f(x)$$.

$$x$$-intercepts

Let $$f(x)=0$$ and solve for $$x$$

##### Example $$\PageIndex{4}$$

Find the intercepts of the parabola whose function is $$f(x)=x^{2}-2 x-8$$.

Solution:

 To find the $$y$$-intercept, let $$x=0$$ and solve for $$f(x)$$. $$f(0)=-8$$ When $$x=0$$, then $$f(0)=-8$$. The $$y$$-intercept is the point $$(0,-8)$$. To find the $$x$$-intercept, let $$f(x)=0$$ and solve for $$x$$. Solve by factoring. $$0=(x-4)(x+2)$$ $$4=x \quad-2=x$$ When $$f(x)=0$$, then $$x=4$$ or $$x=-2$$. The $$x$$-intercepts are the points $$(4,0)$$ and $$(-2,0)$$.
##### Exercise $$\PageIndex{7}$$

Find the intercepts of the parabola whose function is $$f(x)=x^{2}+2 x-8$$.

$$y$$ -intercept: $$(0,-8) x$$ -intercepts $$(-4,0),(2,0)$$

##### Exercise $$\PageIndex{8}$$

Find the intercepts of the parabola whose function is $$f(x)=x^{2}-4 x-12$$.

$$y$$ -intercept: $$(0,-12) x$$ -intercepts $$(-2,0),(6,0)$$

In this chapter, we have been solving quadratic equations of the form $$a x^{2}+b x+c=0$$. We solved for $$x$$ and the results were the solutions to the equation.

We are now looking at quadratic functions of the form $$f(x)=a x^{2}+b x+c$$. The graphs of these functions are parabolas. The $$x$$-intercepts of the parabolas occur where $$f(x)=0$$.

For example:

The solutions of the quadratic function are the $$x$$ values of the $$x$$-intercepts.

Earlier, we saw that quadratic equations have $$2, 1$$, or $$0$$ solutions. The graphs below show examples of parabolas for these three cases. Since the solutions of the functions give the $$x$$-intercepts of the graphs, the number of $$x$$-intercepts is the same as the number of solutions.

Previously, we used the discriminant to determine the number of solutions of a quadratic function of the form $$a x^{2}+b x+c=0$$. Now we can use the discriminant to tell us how many $$x$$-intercepts there are on the graph.

Before you to find the values of the $$x$$-intercepts, you may want to evaluate the discriminant so you know how many solutions to expect.

##### Example $$\PageIndex{5}$$

Find the intercepts of the parabola for the function $$f(x)=5 x^{2}+x+4$$.

Solution:

 To find the $$y$$-intercept, let $$x=0$$ and solve for $$f(x)$$. When $$x=0$$, then $$f(0)=4$$. The $$y$$-intercept is the point $$(0,4)$$. To find the $$x$$-intercept, let $$f(x)=0$$ and solve for $$x$$. Find the value of the discriminant to predict the number of solutions which is also the number of $$x$$-intercepts. Since the value of the discriminant is negative, there is no real solution to the equation. There are no $$x$$-intercepts.
##### Exercise $$\PageIndex{9}$$

Find the intercepts of the parabola whose function is $$f(x)=3 x^{2}+4 x+4$$.

$$y$$-intercept: $$(0,4)$$ no $$x$$-intercept

##### Exercise $$\PageIndex{10}$$

Find the intercepts of the parabola whose function is $$f(x)=x^{2}-4 x-5$$

$$y$$-intercept: $$(0,-5)$$ $$x$$-intercepts $$(-1,0),(5,0)$$

## Graph Quadratic Functions Using Properties

Now we have all the pieces we need in order to graph a quadratic function. We just need to put them together. In the next example we will see how to do this.

##### Example $$\PageIndex{6}$$ How to Graph a Quadratic Function Using Properties

Graph $$f(x)=x^{2}-6x+8$$ by using its properties.

Solution:

 Step 1: Determine whether the parabola opens upward or downward. Look $$a$$ in the equation $$f(x)=x^{2}-6x+8$$ Since $$a$$ is positive, the parabola opens upward. $$f(x)=x^{2}-6x+8$$ $$\color{red}{a=1, b=-6, c=8}$$ The parabola opens upward. Step 2: Find the axis of symmetry. $$f(x)=x^{2}-6x+8$$ The axis of symmetry is the line $$x=-\frac{b}{2 a}$$. Axis of Symmetry $$x=-\frac{b}{2 a}$$ $$x=-\frac{(-6)}{2 \cdot 1}$$ $$x=3$$ The axis of symmetry is the line $$x=3$$. Step 3: Find the vertex. The vertex is on the axis of symmetry. Substitute $$x=3$$ into the function. Vertex $$f(x)=x^{2}-6x+8$$ $$f(3)=(\color{red}{3}\color{black}{)}^{2}-6(\color{red}{3}\color{black}{)}+8$$ $$f(3)=-1$$ The vertex is $$(3,-1)$$. Step 4: Find the $$y$$-intercept. Find the point symmetric to the $$y$$-intercept across the axis of symmetry. We find $$f(0)$$. We use the axis of symmetry to find a point symmetric to the $$y$$-intercept. The $$y$$-intercept is $$3$$ units left of the axis of symmetry, $$x=3$$. A point $$3$$ units to the right of the axis of symmetry has $$x=6$$. $$y$$-intercept $$f(0)=(\color{red}{0}\color{black}{)}^{2}-6(\color{red}{0}\color{black}{)}+8$$ $$f(0)=8$$ The $$y$$-intercept is $$(0,8)$$. Point symmetric to $$y$$-intercept: The point is $$(6,8)$$. Step 5: Find the $$x$$-intercepts. Find additional points if needed. We solve $$f(x)=0$$. We can solve this quadratic equation by factoring. $$x$$-intercepts $$\color{red}{0}\color{black}{=}x^{2}-6x+8$$ $$\color{red}{0}\color{black}{=}(x-2)(x-4)$$ $$x=2 or x=4$$ The $$x$$-intercepts are $$(2,0)$$ and $$(4,0)$$. Step 6: Graph the parabola. We graph the vertex, intercepts, and the point symmetric to the $$y$$-intercept. We connect these $$5$$ points to sketch the parabola.
##### Exercise $$\PageIndex{11}$$

Graph $$f(x)=x^{2}+2x-8$$ by using its properties.

##### Exercise $$\PageIndex{12}$$

Graph $$f(x)=x^{2}-8x+12$$ by using its properties.

We list the steps to take in order to graph a quadratic function here.

### To Graph a Quadratic Function Using Properties

1. Determine whether the parabola opens upward or downward.
2. Find the equation of the axis of symmetry.
3. Find the vertex.
4. Find the $$y$$-intercept. Find the point symmetric to the $$y$$-intercept across the axis of symmetry.
5. Find the $$x$$-intercepts. Find additional points if needed.
6. Graph the parabola.

We were able to find the $$x$$-intercepts in the last example by factoring. We find the $$x$$-intercepts in the next example by factoring, too.

##### Example $$\PageIndex{7}$$

Graph $$f(x)=x^{2}+6 x-9$$ by using its properties.

Solution:

 Since $$a$$ is $$-1$$, the parabola opens downward. To find the equation of the axis of symmetry, use $$x=-\frac{b}{2 a}$$. $$x=-\frac{b}{2 a}$$ $$x=-\frac{6}{2(-1)}$$ $$x=3$$ The axis of symmetry is $$x=3$$. The vertex is on the line $$x=3$$. Find $$f(3)$$. $$f(x)=-x^{2}+6 x-9$$ $$f(3)=-9+18-9$$ $$f(3)=0$$ The vertex is $$(3,0)$$. The $$y$$-intercept occurs when $$x=0$$. Find $$f(0)$$. $$f(x)=-x^{2}+6 x-9$$ Substitute $$x=0$$. Simplify. $$f(0)=-9$$ The point $$(0,-9)$$ is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is $$(6,-9)$$. Point symmetric to the $$y$$-intercept is $$(6,-9)$$ The $$x$$-intercept occurs when $$f(x)=0$$. Find $$f(x)=0$$. Factor the GCF. Factor the trinomial. Solve for $$x$$. Connect the points to graph the parabola.
##### Exercise $$\PageIndex{13}$$

Graph $$f(x)=3 x^{2}+12 x-12$$ by using its properties.

##### Exercise $$\PageIndex{14}$$

Graph $$f(x)=4 x^{2}+24 x+36$$ by using its properties.

For the graph of $$f(x)=-x^{2}+6 x-9$$, the vertex and the $$x$$-intercept were the same point. Remember how the discriminant determines the number of solutions of a quadratic equation? The discriminant of the equation $$0=-x^{2}+6x-9$$ is $$0$$, so there is only one solution. That means there is only one $$x$$-intercept, and it is the vertex of the parabola.

How many $$x$$-intercepts would you expect to see on the graph of $$f(x)=x^{2}+4 x+5$$?

##### Example $$\PageIndex{8}$$

Graph $$f(x)=x^{2}+4 x+5$$ by using its properties.

Solution:

 Since $$a$$ is $$-1$$, the parabola opens downward. To find the equation of the axis of symmetry, use $$x=-\frac{b}{2 a}$$. The equation of the axis of symmetry is $$x=-2). The vertex is on the line \(x=-2$$. Find $$f(x)$$ when $$x=-2$$. The vertex is $$(-2,1)$$. The $$y$$-intercept occurs when $$x=0$$. Find $$f(0)$$. Simplify. The $$y$$-intercept is $$(0,5)$$. The point $$(-4,5)$$ is two units to the left of the line of symmetry. The point to units to the right of the line of symmetry is $$(0,5)\. Point symmetric to the \(y$$-intercept is $$(-4,5)$$. The $$x$$-intercept occurs when $$f(x)=0$$. Find $$f(x)=0$$. Test the discriminant. Since the value of the discriminant is negative, there is no real solution and so no $$x$$-intercept. Connect the points to graph the parabola. You may want to choose two more points for greater accuracy.
##### Exercise $$\PageIndex{15}$$

Graph $$f(x)=x^{2}-2 x+3$$ by using its properties.

##### Exercise $$\PageIndex{16}$$

Graph $$f(x)=-3x^{2}-6 x-4$$ by using its properties.

Finding the $$y$$-intercept by finding $$f(0)$$ is easy, isn't it? Sometimes we need to use the Quadratic Formula to find the $$x$$-intercepts.

##### Example $$\PageIndex{9}$$

Graph $$f(x)=2 x^{2}-4 x-3$$ by using its properties.

Solution:

 Since $$a$$ is $$2$$, the parabola opens upward. To find the equation of the axis of symmetry, use $$x=-\frac{b}{2 a}$$. $$x=-\frac{b}{2 a}$$ $$x=-\frac{-4}{2 \cdot 2}$$ $$x=1$$ The equation of the axis of symmetry is $$x=1$$. The vertex is on the line $$x=1$$. Find $$f(1)$$. $$f(1)=2-4-3$$ $$\f(1)=-5) The vertex is \((1,-5)$$. The $$y$$-intercept occurs when $$x=0$$. Find $$f(0)$$. Simplify. $$f(0)=-3$$ The $$y$$-intercept is $$(0,-3)$$. The point $$(0,-3)$$ is one unit to the left of the line of symmetry. Point symmetric to the $$y$$-intercept is $$(2,-3)$$ The point one unit to the right of the line of symmetry is $$(2,3)$$. The $$x$$-intercept occurs when $$y=0$$. Find $$f(x)=0$$. Use the Quadratic Formula. Substitute in the values of $$a,b$$ and $$c$$. $$x=\frac{-(-4) \pm \sqrt{(-4)^{2}-4(2)(3)}}{2(2)}$$ Simplify. Simplify inside the radical. $$x=\frac{4 \pm \sqrt{40}}{4}$$ Simplify the radical. $$x=\frac{4 \pm 2 \sqrt{10}}{4}$$ Factor the GCF. $$x=\frac{2(2 \pm \sqrt{10})}{4}$$ Remove common factors. $$x=\frac{2 \pm \sqrt{10}}{2}$$ Write as two equations. $$x=\frac{2+\sqrt{10}}{2}, \quad x=\frac{2-\sqrt{10}}{2}$$ Approximate the values. $$x \approx 2.5, \quad x \approx-0.6$$ The approximate values of the $$x$$-intercepts are $$(2.5,0)$$ and $$(-0.6,0)$$. Graph the parabola using the points found.
##### Exercise $$\PageIndex{17}$$

Graph $$f(x)=5 x^{2}+10 x+3$$ by using its properties.

##### Exercise $$\PageIndex{18}$$

Graph $$f(x)=-3 x^{2}-6 x+5$$ by using its properties.

## Solve Maximum and Minimum Applications

Knowing that the vertex of a parabola is the lowest or highest point of the parabola gives us an easy way to determine the minimum or maximum value of a quadratic function. The y-coordinate of the vertex is the minimum value of a parabola that opens upward. It is the maximum value of a parabola that opens downward. See Figure 9.6.124.

### Minimum or Maximum Values of a Quadratic Function

The y-coordinate of the vertex of the graph of a quadratic function is the

• minimum value of the quadratic equation if the parabola opens upward.
• maximum value of the quadratic equation if the parabola opens downward.
##### Example $$\PageIndex{10}$$

Find the minimum or maximum value of the quadratic function $$f(x)=x^{2}+2 x-8$$.

Solution:

 $$f(x)=x^{2}+2 x-8$$ Since $$a$$ is positive, the parabola opens upward. The quadratic equation has a minimum. Find the equation of the axis of symmetry. $$x=-\frac{b}{2 a}$$ $$x=-\frac{2}{2 \times 1}$$ $$x=-1$$ The equation of the axis of symmetry is $$x=-1$$. The vertex is on the line $$x=-1$$. $$f(x)=x^{2}+2 x-8$$ Find $$f(-1)$$. $$f(-1)=1-2-8$$ $$f(-1)=-9$$ The vertex is $$(-1,-9)$$. Since the parabola has a minimum, the $$y$$-coordinate of the vertex is the minimum $$y$$-value of the quadratic equation. The minimum value of the quadratic is $$-9$$ and it occurs when $$x=-1$$.

Show the graph to verify the result.

##### Exercise $$\PageIndex{19}$$

Find the maximum or minimum value of the quadratic function $$f(x)=x^{2}-8 x+12$$.

The minimum value of the quadratic function is $$−4$$ and it occurs when $$x=4$$.

##### Exercise $$\PageIndex{20}$$

Find the maximum or minimum value of the quadratic function $$f(x)=-4 x^{2}+16 x-11$$.

The maximum value of the quadratic function is $$5$$ and it occurs when $$x=2$$.

We have used the formula

to calculate the height in feet, $$h$$, of an object shot upwards into the air with initial velocity, $$v_{0}$$, after $$t$$ seconds .

This formula is a quadratic function, so its graph is a parabola. By solving for the coordinates of the vertex $$(t,h)$$, we can find how long it will take the object to reach its maximum height. Then we can calculate the maximum height.

Example $$\PageIndex{11}$$

The quadratic equation $$h(t)=-16 t^{2}+176 t+4$$ models the height of a volleyball hit straight upwards with velocity $$176$$ feet per second from a height of $$4$$ feet.

1. How many seconds will it take the volleyball to reach its maximum height?
2. Find the maximum height of the volleyball.

Solution:

Since $$a$$ is negative, the parabola opens downward. The quadratic function has a maximum.

a. Find the equation of the axis of symmetry.

$$\begin{array}{l}{t=-\frac{b}{2 a}} \\ {t=-\frac{176}{2(-16)}} \\ {t=5.5}\end{array}$$

The equation of the axis of symmetry is $$t=5.5$$.

The vertex is on the line $$t=5.5$$.

The maximum occurs when $$t=5.5$$ seconds.

b. Find $$h(5.5)$$.

Use a calculator to simplify.

$$h(t)=488$$

The vertex is $$(5.5,488)$$.

Since the parabola has a maximum, the $$h$$-coordinate of the vertex is the maximum value of the quadratic function.

The maximum value of the quadratic is $$488$$ feet and it occurs when $$t=5.5$$ seconds.

After $$5.5$$ seconds, the volleyball will reach its maximum height of $$488$$ feet.

##### Exercise $$\PageIndex{21}$$

Solve, rounding answers to the nearest tenth.

The quadratic function $$h(t)=-16 t^{2}+128 t+32$$ is used to find the height of a stone thrown upward from a height of $$32$$ feet at a rate of $$128$$ ft/sec. How long will it take for the stone to reach its maximum height? What is the maximum height?

It will take $$4$$ seconds for the stone to reach its maximum height of $$288$$ feet.

##### Exercise $$\PageIndex{22}$$

A path of a toy rocket thrown upward from the ground at a rate of $$208$$ ft/sec is modeled by the quadratic function of $$h(t)=-16 t^{2}+208 t$$. When will the rocket reach its maximum height? What will be the maximum height?

It will take $$6.5$$ seconds for the rocket to reach its maximum height of $$676$$ feet.

## Key Concepts

• Parabola Orientation
• For the graph of the quadratic function $$f(x)=a x^{2}+b x+c$$, if
• $$a>0$$, the parabola opens upward.
• $$a<0$$, the parabola opens downward.
• Axis of Symmetry and Vertex of a Parabola The graph of the function $$f(x)=a x^{2}+b x+c$$ is a parabola where:
• the axis of symmetry is the vertical line $$x=-\frac{b}{2 a}$$.
• the vertex is a point on the axis of symmetry, so its $$x$$-coordinate is $$-\frac{b}{2 a}$$.
• the $$y$$-coordinate of the vertex is found by substituting $$x=-\frac{b}{2 a}$$ into the quadratic equation.
• Find the Intercepts of a Parabola
• To find the intercepts of a parabola whose function is $$f(x)=a x^{2}+b x+c$$:
• $$y$$-intercept
• Let $$x=0$$ and solve for $$f(x)$$.
• $$x$$-intercepts
• Let $$f(x)=0$$ and solve for $$x$$.
• How to graph a quadratic function using properties.
1. Determine whether the parabola opens upward or downward.
2. Find the equation of the axis of symmetry.
3. Find the vertex.
4. Find the $$y$$-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
5. Find the $$x$$-intercepts. Find additional points if needed.
6. Graph the parabola.
• Minimum or Maximum Values of a Quadratic Equation
• The $$y$$-coordinate of the vertex of the graph of a quadratic equation is the
• minimum value of the quadratic equation if the parabola opens upward.
• maximum value of the quadratic equation if the parabola opens downward.

## Glossary

A quadratic function, where $$a, b$$, and $$c$$ are real numbers and $$a≠0$$, is a function of the form $$f(x)=ax^{2}+bx+c$$.


text:
# [CEA] The VIMOS Public Extragalactic Redshift Survey (VIPERS): $Ω_{\rm m_0}$ from the galaxy clustering ratio measured at $z \sim 1$

We use a sample of about 22,000 galaxies at $0.65<z<1.2$ from the VIPERS PDR-1 catalogue, to constrain the cosmological model through a measurement of the galaxy {\it clustering ratio} $\eta_{g,R}$. This statistic has favourable properties, being defined as the ratio of two quantities characterizing the smoothed density field in spheres of given radius $R$: the value of its correlation function on a multiple of this scale, $\xi(nR)$, and its variance $\sigma^2(R)$. For sufficiently large values of $R$, this is a universal number, capturing 2-point clustering information independently of the linear bias and linear redshift-space distortions of the specific galaxy tracers. In this paper we discuss in detail how to extend the application of $\eta_{g,R}$ to quasi-linear scales and how to control and remove observational selection effects which are typical of redshift surveys as VIPERS. We verify the accuracy and efficiency of these procedures using mock catalogues that match the survey selection process. These results evidence the robustness of $\eta_{g,R}$ to non-linearities and observational effects, which is related to its very definition as a ratio of quantities that are similarly affected.
We measure $\Omega_{m,0}=0.270_{-0.025}^{+0.029}$. In addition to the great precision achieved on our estimation of $\Omega_m$ using VIPERS PDR-1, this result is remarkable because it appears to be in good agreement with a recent estimate $z\simeq 0.3$, obtained applying the same technique to the SDSS-LRG catalogue. It, therefore, suports the robustness of the present analysis. Moreover, the combination of these two measurements at $z\sim 0.3$ and $z\sim 0.9$ provides us with a very precise estimate $\Omega_{m,0}=0.274\pm0.017$ which highlights the great consistency between our estimation and other cosmological probes such as BAOs, CMB and Supernovae.

Date added: Tue, 15 Oct 13


text:
## anonymous one year ago .

1. anonymous

|dw:1441573642078:dw|

2. anonymous

@freckles

3. iwillrektyou

n this case, you need the distance formula.

4. iwillrektyou

Those are ordered pairs, right? That show where each point is located?

5. anonymous

ya

6. iwillrektyou

Do you know the distance formula?

7. anonymous

sqrt of x2-x1 ^2+ y2-y2^2 pronounced y-1, not y ^2

8. iwillrektyou

Yes. So in order to find the area, you have to find the distance between the length and the width seperately. First, do (-5, 3), and (7, -6).

9. iwillrektyou

Just substitute and solve. Let me know if you need help.

10. anonymous

64+169

11. anonymous

want me to continue or something???

12. iwillrektyou

Yes. I will check your answers as you go, if you would like.

13. anonymous

233 now what???

14. iwillrektyou

You have to do the same thing with the points (-2, 7) and (-5, 3)

15. anonymous

81+64 145

16. iwillrektyou

In order to find the width. Then, you just multiply the 2. I will be checking your work,

17. anonymous

what do i do in order to find the width??

18. iwillrektyou

To find the width, use the distance formula for these 2 points: (-2, 7) and (-5, 3)

19. anonymous

yah wchich is 145

20. iwillrektyou

Yes. To find the area, just multiply the length and width together.

21. anonymous

um that's 33785 AAAANNNNDDD thats not an answer choice you are wrong

22. iwillrektyou

I am checking your work, and a step was done wrong.

23. iwillrektyou

For the width, it is 5.

24. iwillrektyou

And I am double checking my work for the length,

25. anonymous

how do you get 5??????

26. iwillrektyou

Ok so, Sqrt (-5 + 2)^2 + (3 - 7)^2 --> Sqrt (-3)^2 + (-4)^2 --> Sqrt (9 + 16) --> 5

27. anonymous

okay okaywhats the lenght

28. iwillrektyou

15

29. iwillrektyou

But do you understand the concept?

30. freckles

"Yes. So in order to find the area, you have to find the distance between the length and the width seperately. First, do (-5, 3), and (7, -6). " $\text{ distance between these two points is } \\ \sqrt{(-5-7)^2+(3-(-6))^2} \\ =\sqrt{(-5-7)^2+(3+6)^2} \\ =\sqrt{(-12)^2+(9)^2} .. \text{ so on ... }$

31. anonymous

yeah so 75

32. iwillrektyou

Yes. 75 is correct.

33. anonymous

can you help with one more?

34. iwillrektyou

I will try, but I am not very good at geometry, as I just started learning it. But shoot.

35. anonymous

|dw:1441574726192:dw| find the perimeter

36. anonymous

no i dont think so i think this has to do with the distance formula

37. iwillrektyou

Hmm..... I am not sure how to do this. Do they provide you with an ordered pair that substitutes for the variables?

38. anonymous

nope @freckles do you have any idea?

39. freckles

|dw:1441574975096:dw| the bottom we don't really have to use the distance formula for you know the distance from a to 0 is a and the distance from -a to 0 is a so the bottom length is just a+a=2a

40. freckles

|dw:1441575013692:dw| we have to use the distance formula for the other two sides though

41. anonymous

okay

42. anonymous

thank you guys so much for helping btw so what do we do now???

43. iwillrektyou

Sorry I couldn't help. I'm only in 8th grade. :(

44. anonymous

im in 8th too

45. iwillrektyou

46. freckles

distance formula for (-a,c) to (0,b) since that would be the length of that one side distance formula for (a,c) to (0,b) since that would be the length for that last side then just add all the lengths up to find the perimeter value

47. anonymous

can you help me in doing that?

48. freckles

what do you need help with exactly? entering in the values in to the formula?

49. anonymous

yeah just enter and then i'll solve

50. freckles

I will do one the distance from (-a,c) to (0,b) is: $\sqrt{(-a-0)^2+(c-b)^2} \\ \sqrt{(-a)^2+(c-b)^2} \\ \sqrt{a^2+(c-b)^2}$ you try finding the distance from (a,c) to (0,b) now..

51. anonymous

so would it be sqrt (a+a)^2+(c-c)^2?

52. freckles

are you using the points (a,c) and (0,b)?

53. freckles

the change of x is a-0 not a+a the change of y is c-b not c-c

54. anonymous

so sqrt (a-o)^2+(c-b)^2?

55. freckles

yes and a-0 is just a

56. freckles

|dw:1441575476337:dw| $\text{ so the distance between } (-a,c) \text{ and } (a,c) \text{ is } 2a \\ \text{ and the distance between } (a,c) \text{ and } (0,b) \text{ is } \sqrt{a^2+(c-b)^2} \\ \text{ and the distance between } (-a,c) \text{ and } (0,b) \text{ is } \sqrt{a^2+(c-b)^2}$ now just add up all 3 distances to find the perimeter

57. anonymous

so 2 sqrt (a^2+ (c-b)^2 +2a

58. freckles

yes: $2a+2 \sqrt{a^2+(c-b)^2}$

59. anonymous

60. anonymous

the answer choices are 2a sqrt a^2-(c-b)^2 or 2a sqrt a^2+(c+b)^2

61. freckles

$2a+2 \sqrt{a^2+(c-b)^2} \\ \text{ is the same as } 2a+2 \sqrt{a^2+(b-c)^2} \\ \text{ which is the same as } 2a+2\sqrt{(b-c)^2+a^2} \\ \text{ this is the same as } 2\sqrt{(b-c)^2+a^2}+2a \\ \text{ there is a lot of other ways \to express this }$

62. freckles

you can also factor out the 2... $2(\sqrt{(b-c)^2+a^2}+a)$

63. anonymous

but thats not even a choice

64. freckles

it is probably a choice but it is just written a bit different as I said there is a few different ways to express that same answer

65. anonymous

|dw:1441575733104:dw| these are my choices

66. freckles

67. anonymous

|dw:1441575827490:dw|

68. anonymous

that second one should be a plus 2a

69. freckles

yeah those answers are totally wrong I don't why they put the square root over everything like that

70. anonymous

it shouldn't be over the 2a sorry my bad

71. freckles

you do know when you asked me is this the answer I said yes... you wrote so 2 sqrt (a^2+ (c-b)^2 +2a and then I said "yes $2a+2 \sqrt{a^2+(c-b)^2}$" and I said yes assuming you meant to write a ) at the end of the (c-b)^2 there that is 2sqrt(a^2+(c-b)^2)+2a

72. freckles

addition is commutative 3+2 is the same as 2+3

73. freckles

$2a+2 \sqrt{a^2+(c-b)^2} \text{ is the same as } 2 \sqrt{a^2+(c-b)^2}+2a$

74. anonymous

sorry its late a night im really tire

75. anonymous

yah but thats still not an answer?

76. anonymous

im not talking about the 2a or the 2, look inside the sqrt. the answer has it as a^2-(c-b)^2 or a^2+(c+b)^2

77. freckles

both of those are wrong though

78. anonymous

well that's what they have as an answer....

79. freckles

well I don't know what you want me to tell you

80. freckles

recall the distance formula is: $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

81. freckles

do you see there is a plus between the two square parts?

82. freckles

and there is a minus in between the things inside the ( )

83. anonymous

yeah i get it that you are right i just don't know which one to pick

84. freckles

the ordered pairs were definitely (0,b) and (a,c) and (-a,c) right?

85. anonymous

yup

86. freckles

the only thing I can do is guess which one they meant as the right answer but either guess I make is going to be wrong whether it is marked right or not

87. anonymous

should i go with the plus or the minus?

88. freckles

let me confirm these are the choices: $2a+2 \sqrt{a^2-(c-b)^2} \\ 2a+2 \sqrt{a^2+(c+b)^2}$

89. freckles

there are no other choices?

90. freckles

like you can't pick a none of these?

91. anonymous

the only other choices are 6a and 9a

92. freckles

lol I don't know go with the -

93. freckles

94. anonymous

okay

95. anonymous

Anyone else have any ideas?

96. anonymous

@freckles i think i know what the answer is