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# E-Maths Note : Algebraic Expansion

This is a topic some of which is learnt ever since Primary 6.

Through my years of coaching, I notice quite a handful of students form blocks when they dealt with this topic that starts with "A" and ends with "a".

Haha! Yes! It's Alegbra! I was introduced of this thing in Primary 6. It was the ONLY method my dad taught me to solve word problems then. I didn't understand what he was telling me.... totally huh? huh? huh?

Well, I am glad I figure out the x-es & y-es eventually. Many students tend to see algebra to be abstract. In fact, it can be explained in terms of apples, oranges, pears .... ( This is my ONLY drawing subject- basic fruits) Those who have the "good fortune" will know the level of drawing :-)

Now I am going to share a few tips to make your learning of Algebra Expansion easier and simpler. One of the very powerful yet always underestimated strategies is the correct usage of brackets. Yes! Brackets are big deal okie!

3 golden algebra rules :

• $(A+B)^2 = (A)^2 +2(A)(B)+(B)^2$
• $(A-B)^2 = (A)^2 -2(A)(B)+(B)^2$
• $(A+B)(A-B)= (A)^2 -(B)^2$
• These 3 golden rules when applied correctly will definitely make your life easier. Now I'm going to share with you some common errors that students commit when they underestimate the importance of brackets

Qn: Expand $(3x+2y)^2$
Ans : $(3x+2y)^2= 3x^2+12xy+2y^2$ --> this is not right
Correct Ans : $(3x+2y)^2= (3x)^2+12xy+(2y)^2 = 9x^2+12xy+4y^2$

I hope this is a useful tip for you :)

### Related Post

Binomial Expansion Teaches how to choose the RIGHT partner (: I always like to share with my students what each Math topic has to do with their everyday life, particularly their future. Yesterday, I did Binomi...
E-Math Note : Usage of bracket in Expansion Another TIP for students taking Elementary Mathematics (E-Math) Qn: Expand and Simplify the following: (y+z)^2-(y-z)^2 Method 1 : Expansion u...

### Ai Ling Ong

Hi, I'm Ai Ling Ong. I enjoy coaching students who have challenges with understanding and scoring in 'O' Level A-Maths and E-Maths. I develop Math strategies, sometimes ridiculous ideas to help students in understanding abstract concepts the fast and memorable way. I write this blog to share with you the stuff I teach in my class, the common mistakes my students made, the 'way' to think, analyze... If you have found this blog post useful, please share it with your friends. I will really appreciate it! :)


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# Stat help. K2=95% confidence

by rowdy3
Tags: confidence, k295%, stat
 P: 31 To capture the actual Rent amount 95% of the time, predictions from the fitted equation should include a margin of error of about plus or minus ______ (round to the nearest ten dollars). $128$114 (answer) $57$228 Here's a pic of the table. How did they get the answer? http://picturetrail.com/sfx/album/view/24285022 I know k2= 95%. I know you do 2 * (standard error) but I don't see that on the table.

format: plain


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×

# Maths trick

Recently I discovered this video

In the video the man asks you to think of a three digit number, then through a series of steps, he makes your number into $$1089$$ every single time.

The purpose of this note is to explain how he does it.

First let's list the steps.

1. Think of a three digit number with unique digits (all of them must be different)
2. Find the reverse of that number (swap the first and last digits)
3. Take the largest of the two numbers and subtract the other one
4. Find the reverse of that number
5. Sum the reverse and the number you just worked out, you should get $$1089$$ as the answer

Now let's translate each step into an algebraic formula.

Step 1: A three digit number with unique digits is represented by the following notation:

$\overline{ABC} = 100A + 10B + C$

With each letter representing a different digit.

Step 2: The reverse of that first number can be represented using the same notation just with the letters flipped horizontally.

$\overline{CBA} = 100C + 10B + A$

Again with each letter representing a different digit.

Step 3: This step depends on which of the two numbers ($$\overline{ABC}$$ or $$\overline{CBA}$$) is larger. Either way will work but I'll show both anyway.

If $$\overline{ABC} > \overline{CBA}$$ (This also implies $$A > C$$)

$\overline{ABC} - \overline{CBA} = 99(A - C)$

If $$\overline{CBA} > \overline{ABC}$$ (This also implies $$A < C$$)

$\overline{CBA} - \overline{ABC} = 99(C - A)$

To make things easier we'l make $$x = A - C$$, this makes the equations read:

$\overline{ABC} - \overline{CBA} = 99x$

$\overline{CBA} - \overline{ABC} = -99x$

The second equation comes about because $$C - A = -(A - C) = -x$$. I'll use the first equation from now on and prove that the second equation works later.

We'll say for now that this new number we've gotten is $$\overline{DEF}$$, I'll show what each of those letters equals later on, we're just listing the steps as formula for now.

Step 4: The new number we have is equal to $$\overline{DEF}$$ so it's reverse is $$\overline{FED}$$.

Step 5: The final step can be represented by the equation below:

$\overline{DEF} + \overline{FED} = 1089$

Now that we have the steps sorted out, we find we only have two unknowns (technically three) out of all of the steps. The two unknowns are $$\overline{DEF}$$ and $$\overline{FED}$$, however we only need to find out one to figure out the other. It's not as easy as you might think.

We know $$\overline{DEF} = 99x$$ but that doesn't help us find the three digits we need in order to calculate $$\overline{FED}$$.

Let's start by taking a step back, what's $$99x$$ equal to?

$99x = \overline{ABC} - \overline{CBA}$

$99x = (100A + 10B + C) - (100C + 10B + A)$

We can use this to find the separate digits, since each digit in $$\overline{DEF}$$ is equal in some way to the corresponding digits in the two previous numbers. In other words:

$D = A - C$

$E = B - B = 0$

$F = C - A$

So now $$\overline{DEF} = 100(A-C) + 10(0) + (C-A)$$

We already said that $$x = A - C$$ so we'll substitute that in

$\overline{DEF} = 100x - x = 99x$

We've now just proved that we still have the same number, so we'll stop using $$x$$ for now and switch back to using the three digits.

$\overline{DEF} = 100(A - C) + 10(0) + (C - A)$

Now we have a problem, $$A > C$$ so $$C - A < 0$$, this doesn't make any sense since a digit can't be negative. To fix this problem we'll take a $$10$$ from the second digit and add it to the third digit.

$\overline{DEF} = 100(A - C) + 10(-1) + (10 + (C - A))$

We've encountered the same problem again so we'll take a $$100$$ from the first digit and add it to the second.

$\overline{DEF} = 100(A - C - 1) + 10(10 - 1) + (10 + (C - A))$

Now that all the digits are positive and below $$10$$, we can use the notation.

$\overline{DEF} = \overline{(A - C - 1)9(10 + (C - A))}$

Each set of brackets represent a digit in this case. We'll switch back to using $$x$$ to make it easier to calculate.

$\overline{DEF} = \overline{(x - 1)9(10 - x)}$

We can now calculate $$\overline{FED}$$

$\overline{FED} = \overline{(10 - x)9(x - 1)}$

We know $$\overline{DEF} = 99x$$ so we just need to calculate $$\overline{FED}$$ now

$\overline{FED} = 100(10 - x) + 10(9) + (x - 1)$

$\overline{FED} = 1000 - 100x + 90 + x - 1$

$\overline{FED} = 1089 - 99x$

Well we found the $$1089$$.

The last step involves summing the two opposites so we'll do that.

$\overline{DEF} + \overline{FED} = (99x) + (1089 - 99x)$

$\overline{DEF} + \overline{FED} = 1089 + (99x - 99x)$

$\overline{DEF} + \overline{FED} = 1089$

That shows what happens when $$\overline{ABC} > \overline{CBA}$$, so what happens when $$\overline{CBA} > \overline{ABC}$$.

The exact same thing happens, the only difference to the equations is that $$x = C - A$$.

Hope this wasn't too long for a note.

Note by Jack Rawlin
2 years, 1 month ago

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You have the general idea, though step 5 can be simplified much more. For example, there are only 10 possibilities, and it would be easier to check all the cases than work through your algebra. Think about how we can improve this proof further.

Staff - 2 years, 1 month ago

Does it work for the number 132 ??

- 2 years, 1 month ago

$$231 - 132 = 099$$

$$099 + 990 = 1089$$

- 2 years, 1 month ago

LOL thanx.No wonder i was getting it wrong.....

- 2 years, 1 month ago


text: # Data, Scripts, and Output for the use-case **BMI and Cortisol** This repository contains the information for the systematic review published by [Valk et al. (2021)][1]. The repository also contains the scripts and output for the simulation results re-enacting the screening phase using active learning in ASReview ([Van de Schoot et al., 2021][2]). ## Overview To increase the quality of a systematic review on Body Mass Index and cortisol, [Valk et al. (2021)][3] applied a 2-step screening. First, a search was performed identifying 996 records of which 93 were deemed as relevant. Then, a broader search was conducted identifying 3477 records. The 93 relevant records of the first search plus 93 irrelevant records were used as training data for the second search using ASReview. Then, 996 records were screened and the first 28 additionally found relevant records were inspected for full-text in detail and 3 additional papers appeared to be relevant. ## Search The search terms for the original search are available in the file 201116 Search query original search.docx and were used for the publication [Valk et al. (2021)][4]. A new search has been conducted with broader search terms as listed in the file 210111 Search query new search.docx. ## Raw data The data obtained in the first search are available in the file BMIandCortisol_original_data which includes (at least) the following columns: - *Included abstract* indicating which records have been included in the abstract inclusion phase (n_1=351); - *Included_fulltext* indicating which records have been included in the abstract inclusion phase (n_2=93); - *Title* containing the titles for the records; - *Abstract Note* containing the abstracts for the records. The second search resulted in an additional 3.477 records, see the file BMIandCortisol_newdata. A column titled *included* was added in the file titled BMIandCortisol_newdata_priorknowledge which containts prior knowledge. The 93 relevant records from BMIandCortisol_prior_inclusions.cs were added to the new data and labelled '1'. Also, 93 irrelevant records were selected, see BMIandCortisol_prior_exclusions.csv, added to the new data and labelled '0'. ## Data The raw data, BMIandCortisol_original_data.csv was split into two subsets  python scripts/split_data_with_multiple_labels.py BMIandCortisol_original_data.csv bmi_and_cortisol/data --split Included_abstract Included_fulltext  so that each subset contains only one column with labels denoting the relevant papers for the: 1. abstract inclusions: file name BMIandCortisol_original_data_Included_abstract.csv; 2. full text inclusions: file name BMIandCortisol_original_data_Included_fulltext.csv; A wordcloud of the words used in the abstracts of the 93 included papers was created using [asreview-wordcloud][5] (the script can be found in jobs.sh): ![enter image description here][6] ## Installation requirements For the installation requirements run the following code in the CLI:  pip install -r requirements.txt  ## Simulation The data files in the data-folder were used for running a simulation study. To run the simulation, run  sh jobs.sh  The results are stored in output/simulation. The simulation was conducted on the original data with 93 runs with each relevant record being a prior inclusion and 10 randomly chosen irrelevant records. In each run the same 10 irrelevant records have been used. To extract this information run  python scripts/get_prior_knowledge.py  The results are stored in output/tables. The dataset characteristics are obtained with  python scripts/merge_descriptives.py  and stored in output/tables. The metrics resulting from the simulation study per run, can be obtained with  python scripts/merge_metrics.py  and are stored in output/tables. ## Simulation Results The results are sumarized in this recall plot: ![enter][7] On average, after screening 56% of the records (n = 567), you would have found 95 % of all the relevant records (89 out of 93). If you would screen records in a random order, at this point you would have found 52 of the relevant records and finding 89 of the relevant records would take on average 972 records. In other words, the time that can be saved using active learning expressed as the percentage of records that do not have to be screened is 40% (sd=2.4), while still identifying 95% of the relevant records. This metric is also known as the Work Saved over random Sampling at 95% recall (WSS@95). Another way to interpret the results is the RRF metric. The RRF@10 is 39% (sd=1.76), meaning that after screening 10% of records, already 39% of the relevant records have been identified. The relevant record that was in row 44 of the dataset was the easiest to find over all trials (Title: *Hair Cortisol Concentrations in Overweight and Obese Children and Adolescents*). Discovering this record took screening 24 records on average, that is screening 2.35% of all records in the BMIandCortisol_original_data_Included_fulltext dataset. The record that was most difficult to find was in row 2 of the dataset, which was discovered after screening on average 702 records, 69.48% of all records in this dataset (title: *A double-edged sword: Relationship between full-range leadership behaviors and followers’ hair cortisol level*). ## ASReview Files The file BMIandCortisol_newdata_priorknowledge was used for screening in ASReview and contains (at least) the following colums: - title / abstract note of 3663 records, of which 186 labelled and 3477 unlabelled record; - included with 93 relevant and 93 irrelevant records. The file was uploaded in ASReview, the 186 labelled records were used as prior knowledge and the **996** records were screened via the default model settings in version 0.16 (see asreview-hairf-and-anthropometrics.asreview for the project files) of which **28** records were denoted as relevant, see the file asreview_result_asreview-hairf-and-anthropometrics.xlsx. The 996 was chosen to represent an equal amount of labour as was invested for the original search. Then, the first **28** additionally found relevant records were inspected in detail and 3 appeared to be relevant, see the file new_records_classified.xlsx. ## Contact For any questions or remarks, please send an email to m.mohseni@erasmusmc.nl . [1]: https://doi.org/10.1111/obr.13376 [2]: https://doi.org/10.1038/s42256-020-00287-7 [3]: https://doi.org/10.1111/obr.13376 [4]: https://doi.org/10.1111/obr.13376 [5]: https://zenodo.org/record/6625855#.Yu1bKnZBxPY://doi.org/10.1111/obr.13376 [6]: https://mfr.osf.io/export?url=https://osf.io/download/zj8d4/?direct=%26mode=render&format=2400x2400.jpeg [7]: https://mfr.osf.io/export?url=https://osf.io/download/2ufsv/?direct=%26mode=render&format=2400x2400.jpeg


text:
# Characterizing the Larkin-Ovchinnikov-Fulde-Ferrel phase induced by the chromomagnetic instability

Authors
Type
Published Article
Publication Date
Apr 04, 2006
Submission Date
Mar 26, 2006
Identifiers
DOI: 10.1103/PhysRevD.73.094016
Source
arXiv
We discuss possible destinations from the chromomagnetic instability in color superconductors with Fermi surface mismatch $\delta\mu$. In the two-flavor superconducting (2SC) phase we calculate the effective potential for color vector potentials $A_\alpha$ which are interpreted as the net momenta $q$ of pairing in the Larkin-Ovchinnikov-Fulde-Ferrel (LOFF) phase. When $1/\sqrt{2}<\delta\mu/\Delta<1$ where $\Delta$ is the gap energy, the effective potential suggests that the instability leads to a LOFF-like state which is characterized by color-rotated phase oscillations with small $q$. In the vicinity of $\delta\mu/\Delta=1/\sqrt{2}$ the magnitude of $q$ continuously increases from zero as the effective potential has negative larger curvature at vanishing $A_\alpha$ that is the Meissner mass squared. In the gapless 2SC (g2SC) phase, in contrast, the effective potential has a minimum at $gA_\alpha\sim\delta\mu\sim\Delta$ even when the negative Meissner mass squared is infinitesimally small. Our results imply that the chromomagnetic instability found in the gapless phase drives the system toward the LOFF state with $q\sim\delta\mu$.


text:
Dominic Steinitz dominic.steinitz at blueyonder.co.uk
Thu Mar 29 16:18:40 EDT 2007

> On 3/27/07, Dave at haskell.org <Dave at haskell.org> wrote:
> > Given the amount of material posted at haskell.org and elsewhere
> > explaining IO, monads and functors, has anyone considered publishing
> > a comprehensive book explaining those subjects?  (I am trying to
> > read all the material online, but books are easier to read and don't
> > require sitting in front of a computer to do so. Plus I can write in
> > books :-). )
>
> I've thought about writing extended tutorials on the relationship
> between Haskell programming and category theory but there's a problem
> I run into. It's tempting to make the identifications:
> types<->objects, Haskell function<->arrows, suitably polymorphic
> functions<->natural transformations, and so on. But the fact is, this
> doesn't really work in the obvious way even though it seems like it
> should at first (eg. Haskell functions aren't always total functions
> in the mathematical sense and if you allow partial functions you can
> do weird stuff). So either:
>
> (1) we need some technical work to patch up the differences (and
> unfortunately I don't know what the patch-up is),
> (2) we restrict ourselves to certain types of Haskell function for
> which the theory works or
> (3) we deliberately leave things a little vague.
>
> I usually tend to go for option (3), but that wouldn't be satisfactory
> for an extended treatment.
>
> Has anyone else given this subject much thought?
> --
> Dan
Dan,

It's a long time since I looked at this so this might be a complete red
herring but have you thought about using the category of \omega-CPOs. Types
are then objects (\omega-cpos) and functions are then arrows (continuous
functions in the \omega-cpo sense).

These papers may be of use:
1. Meijer http://research.microsoft.com/~emeijer/Papers/CWIReport.pdf
2. Mitchell Wand: Fixed-Point Constructions in Order-Enriched Categories
3. Freyd: Recursive Types Reduced to Inductive Types.

Dominic.


text:
### Home > CCAA > Chapter 3 Unit 2 > Lesson CC2: 3.2.4 > Problem3-77

3-77.

Simplify each multiplication problem below.

1. $\frac{5}{8}\cdot\frac{2}{3}$

First, multiply the numerators of the fractions.
$(5)(2)=10$

Then multiply the denominators of the fractions.
$(8)(3)$

Put the new numerator over the new denominator and then simplify.

$\frac{10}{24}=\frac{5}{12}$

1. $\frac{3}{4}\cdot\frac{2}{5}$

Use the same method as part (a).


text:
2012 CMS Winter Meeting

Fairmont Queen Elizabeth (Montreal), December 7 - 10, 2012

Operator Theory and Operator Algebras
Org: Ken Davidson, Evgenios Kakariadis and Laurent Marcoux (Waterloo)
[PDF]

SERBAN BELINSCHI, Queen's University
Using fully matricial functions to compute distributions of polynomials in unitarily invariant random matrices  [PDF]

In this talk we will indicate how fully matricial (or non-commutative) functions can be used to great effectiveness to find the asymptotic distributions of selfadjoint polynomials in unitarily invariant random matrices, as the size of the matrices tends to infinity. We will show that fully matricial functions remain somewhat useful also in studying some aspects of distributions of random matrices before letting the size of the matrix tend to infinity. The talk will be mostly a survey of recent and ongoing joint works with several co-authors.

GEORGE ELLIOTT, University of Toronto
What does classification mean?  [PDF]

From the point of view of C*-algebras, one should perhaps agree that classification means determining in some useful way the approximate unitary equivalence class of a morphism---as these form a category, which for separable C*-algebras distinguishes isomorphism classes. (An analogous situation holds for countable discrete groups.) For interesting classes of separable (amenable) C*-algebras, this has been achieved, by means of K-theoretical invariants. (On the other hand, no such generic invariants seem to work for countable groups.) A way of testing how difficult various cases are---the especially well behaved amenable case, in the sense of Toms and Winter, the general amenable case, or the general separable case---might be to look at the Borel complexity of the isomorphism relation (with respect to a natural Borel structure), in the sense of descriptive set theory. As it happens (work of many hands), except for the special case of AF algebras (dealt with by Bratteli and me forty years ago), which is simpler than the other cases, there would seem to be no noticeable variation at all in the Borel complexity (among all three of the cases mentioned above).

DOUGLAS FARENICK, University of Regina
Operator systems from discrete groups  [PDF]

If $\mathfrak u$ is a set of generators for a discrete group $G$, then the operator system $\mathcal S(\mathfrak u)$ generated by $\mathfrak u$ is the operator subsystem of the group C$^*$-algebra $C^*(G)$ spanned by the identity, the elements of $\mathfrak u$, and their inverses. This lecture will report on joint work with A. Kavruk, V. Paulsen, and I.G. Todorov concerning tensor products of such operator systems. A special case of interest occurs with the generators of a finitely generated free group.

MICHAEL HARTZ, University of Waterloo
Universal operator algebras associated to homogeneous varieties  [PDF]

In this talk, we will consider universal operator algebras generated by commuting row contractions satisfying homogeneous polynomial relations. These algebras can be realized as algebras of functions on the varieties defined by the relations. It turns out that their structure is closely related to the geometry of the associated algebraic varieties.

We will discuss the question of when two algebras of this are type isomorphic. In particular, we obtain a geometric condition for two such algebras being topologically isomorphic, thereby giving a positive answer to a conjecture of Davidson, Ramsey and Shalit.

ELIAS KATSOULIS, East Carolina University
C*-algebras and equivalences for C*-correspondences  [PDF]

We study several notions of shift equivalence for C*-correspondences and the effect that these equivalences have on the corresponding Pimsner dilations. Among others, we prove that non- degenerate, regular, full C*-correspondences which are shift equivalent have strong Morita equivalent Pimsner dilations. We also establish that the converse may not be true. These results settle open problems in the literature.

In the context of C*-algebras, we prove that if two non-degenerate, regular, full C*- correspondences are shift equivalent, then their corresponding Cuntz-Pimsner algebras are strong Morita equivalent. This generalizes results of Cuntz and Krieger and Muhly, Tomforde and Pask. As a consequence, if two subshifts of finite type are eventually conjugate, then their Cuntz-Krieger algebras are strong Morita equivalent. (Joint work with E. Kakariadis)

MATTHEW KENNEDY, Carleton University
A universal F. \& M. Riesz theorem  [PDF]

We generalize the F. \& M. Riesz theorem and Lebesgue's decomposition theorem to operator algebras generated by certain contractive tuples of operators. This class of algebras includes the multiplier algebra of every complete Nevanlinna-Pick space, for example the Hardy space, the Dirichlet space and the Drury-Arveson space. One consequence of our results is that these algebras have a unique Banach space predual. This is joint work with Dilian Yang.

DAVID KERR, Texas A\&M University
Combinatorial independence and sofic entropy  [PDF]

I will discuss some recent work with Hanfeng Li in which we initiate a study of combinatorial independence in the context of sofic entropy. The project involves, among other things, ergodic theory on ultraproducts and applications to the Fuglede-Kadison determinant in group von Neumann algebras.

MARCELO LACA, University of Victoria
Phase transitions from dilation matrices and self similar actions  [PDF]

I will talk about some recent computations of equilibrium states for the natural periodic dynamics of Cuntz-Pimsner algebras and their Toeplitz extensions. The main examples arise from integer dilation matrices, and, more generally, from self-similar group actions. This is joint work with Iain Raeburn, Jacqui Ramagge, and Michael Whittaker.

ZHIQIANG LI, University of Toronto
On simple inductive limits of splitting interval algebras with dimension drops  [PDF]

In this talk, we are going to investigate simple inductive limits built on splitting interval algebras with dimension drops, which are typcial subhomogeneous building blocks on the unit interval. Actually, a classification will be given in both unital and certain nonunital cases. In the nonunital cases, such inductive limits give us simple stably projectionless C*-algebras with nontrivial K-theory.

JAMES MINGO, Queen's University
Second Order Freeness and Orthogonal Random Matrices  [PDF]

Second order freeness is a property exhibited by many ensembles of random matrices. In recent work with Mihai Popa and Emily Redelmeier we have shown that independent ensembles of random matrices are real asymptotically free of second order if one of the ensembles is invariant under conjugation by an orthogonal matrix.

SARAH PLOSKER, University of Guelph
Private Quantum Codes  [PDF]

Private quantum codes are a basic tool in quantum key distribution and quantum cryptography. We define private quantum channels mathematically, and consider a general notion of private quantum codes wherein qubits are encoded into quantum subsystems. Private quantum channels, private subspaces, and a previously considered notion of private subsystems are all captured as special cases of this general phenomena. We provide a simple example that highlights the main differences between mappings on subsystems and subspaces and show that certain classes of channels can only be private in this subsystem setting. We also set out testable conditions for deciding when a code is private for a given channel and we discuss connections with quantum error correction. These conditions can be regarded as the private analogue of the Knill-Laflamme conditions for quantum error correction. Joint work with T. Jochym-O'Connor, D. W. Kribs, and R. Laflamme.

ALEXEY POPOV, University of Waterloo
Every operator has an almost-invariant subspace  [PDF]

We show that any bounded operator on a separable, reflexive, infinite-dimensional Banach space admits a rank-one perturbation which has an invariant subspace of infinite dimension and codimension. In the non-reflexive spaces, we show that the same is true for operators which have non-eigenvalues in the boundary of their spectrum. In the Hilbert space, our methods produce perturbations that are also small in norm.

This is a joint work with Adi Tcaciuc.

HEYDAR RADJAVI, University of waterloo
Semigroups of Partial Isometries  [PDF]

We present results on the spatial structure of semigroups of partial isometries on a Hilbert space. If these semigroups are also assumed to be selfadjoint, then they coincide precisely with faithful representations of abstract inverse semigroups. We obtain a generalized weighted-composition form for the members of such semigroups. In the presence of additional hypotheses, e.g., when the semigroup is finitely generated, we get a simpler, matricial form for it. This is joint work with Alexey Popov.

CHRIS RAMSEY, University of Waterloo
Triangular UHF algebras  [PDF]

Triangular UHF algebras are direct limits of full triangular matrix algebras. I will talk about their isometric automorphism groups and dilation theory. In particular, we know what is the C*-envelope of the semidirect product of one of these algebras by an automorphism.

PAUL SKOUFRANIS, UCLA
Normal Limits of Nilpotents in C*-Algebras  [PDF]

In 1974 Herrero completely classified when a bounded normal operator on a complex, infinite dimensional, separable Hilbert space was a norm limit of bounded nilpotent operators. In this brief talk I will examine the C$^*$-analog of this problem; that is, given a C$^*$-algebra $\mathfrak{A}$ and a normal operator $N \in \mathfrak{A}$, when is $N$ a norm limit of nilpotent operators from $\mathfrak{A}$? In particular, I will discuss this problem for von Neumann algebras (based on type decomposition), unital simple purely infinite C*-algebras (which include the Cuntz algebras), and approximately finite dimensional C$^*$-algebras. It is surprising that Herrero's result generalize to many of these C$^*$-algebras yet in others this classification problem is incredibly complex.

AHMED SOUROUR, University of Victoria
Semigroups of matrices generated by a unitary orbit  [PDF]

I will discuss some recent - and some not so recent - results about the structure of multiplicative semigroups of matrices generated by a unitary orbit or a similarity orbit of one matrix. The recent results are based on joint work with H. Radjavi.

ROLAND SPEICHER, Saarland University
Sharp bounds for sums associated to graphs of matrices  [PDF]

We provide a simple algorithm for finding the optimal upper bound for sums of products of matrix entries of the form $S(N) := \sum_{j_1, ..., j_{2m} = 1}^N t_{j_1 j_2} t_{j_3 j_4} \cdots t_{j_{2m-1} j_{2m}}$ where some of the summation indices are constrained to be equal. The upper bound is easily obtained from a graph associated to the constraints in the sum.

DILIAN YANG, University of Windsor
The structure of nonself-adjoint 2-graph algebras  [PDF]

A nonself-adoint 2-graph algebra is a unital \textsc{wot}-closed algebra generated by 2 isometric tuples satisfying some commutation relations. In this talk, we show that such an algebra has a $2\times 2$ lower triangular form, whose 1st column is a slice of its enveloping von Neumann algebra. We will also discuss that, in the case of atomic representations, one could decompose further to obtain a $3\times 3$ lower triangular form, whose $(3,3)$-entry is an analytic 2-graph algebra.

This talk is based on recent joint work with Adam Fuller.


text:
On different definitions of groups

Let a group $G$ be defined as a $4$-tuple $G=(|G|,\mu_G,i_g,e_G)$ satisfying the following axioms:

1) $\mu_G(\mu_G(x,y),z)=\mu_G(x,\mu_G(y,z))$

2) $\mu_G(x,e_G)=x=\mu_G(e_G,x)$

3) $\mu_G(x,i_G(x))=e_G=\mu_G(i_G(x),x)$

To each group $G$, we associate a pair $G'=(|G|,\delta_G)$, with $\delta_G(x,y):=\mu_G(x,i_G(y))$. This pair characterises the group, in the sense that, if $G_1,G_2$ are two groups with $G_1'=G_2'$ then $G_1=G_2$.

Now the question: Suppose $|X|$ is any set and $\delta\colon|X|\times|X| \to |X|$ any map. I want to write down a set of axioms for the pair $X = (|X|, \delta)$, which will be necessary and sufficient for it to arise from a group $G$ in the manner described above. That is, assuming $|X|$ and $\delta$ given, find convenient necessary and sufficient conditions for there to exist a group $G$ such that $G'=(|X|,\delta)$.

My work: since $\delta$ will have to be the $\delta_G$ of some group $G$, I tried to express the group multiplication using $\delta$:

$\mu_G(x,y)=\delta(x,\delta(\delta(y,y),y))$

Then I required for this multiplication the usual axioms identifying groups, which assume the following form:

associativity:

$\delta(\delta(x,\delta(\delta(y,y),y)), \delta(\delta(z,z),z)=\delta(x,\delta(\delta(\delta(y,\delta(\delta(z,z),z)),\delta(y,\delta(\delta(z,z),z))),\delta(y,\delta(\delta(z,z),z)))$

unit:

$\delta(x,\delta(\delta(\delta(x,x),\delta(x,x)),\delta(x,x)))=x=\delta(\delta(x,x),\delta(\delta(x,x),x))$

inverse:

$\delta(y,y)=\delta(x,x)$

First question: does it make sense?

Second question: is there a better choice of axioms?

• The question makes good sense, and I'm pretty sure I've seen answers that involved shorter axioms than the ones in the question, but I don't remember where I saw them. (That's why this is a comment and not an answer.) Nov 1, 2017 at 16:58

1 Answer

You don't need to include the $i_g$ and $e_G$ in $G$, they can be defined in terms of $\delta$ but you will need that $|G|$ is not empty. After axiom $\delta(x,x) = \delta(y,y)$, define $e_G =\delta(x,x)$ and $x^{-1}= \delta(e_G,x)$. Then the rest of the axioms become much shorter. For example the associativity axiom becomes $\delta (x,(\delta (y,z^{-1})^{-1}) =\delta (\delta(x,y^{-1}),z^{-1}).$


text:
QM-eigenfunction of kinetic energy

1. Nov 16, 2008

PullMeOut

I have to check if a function is an eigenfunction of an operator or not.the problem is attached

View attachment aaaaaaa.doc

Last edited: Nov 16, 2008
2. Nov 16, 2008

PullMeOut

well the problem is solved thanks anyway


text:
# Tagged: product of ideals

## Problem 623

Let $I=(x, 2)$ and $J=(x, 3)$ be ideal in the ring $\Z[x]$.

(a) Prove that $IJ=(x, 6)$.

(b) Prove that the element $x\in IJ$ cannot be written as $x=f(x)g(x)$, where $f(x)\in I$ and $g(x)\in J$.

## Problem 174

Let $R$ be a commutative ring and let $P$ be an ideal of $R$. Prove that the following statements are equivalent:

(a) The ideal $P$ is a prime ideal.

(b) For any two ideals $I$ and $J$, if $IJ \subset P$ then we have either $I \subset P$ or $J \subset P$.


text:
(In order to load the stylesheet of this notebook, execute the last code cell in this notebook)

# Clustering the data from the Mayor's 24-hour hotline in Boston

## Introduction

In this assignment, we will do a clustering analysis on the 311 calls that we used in the previous homework. We will focus on choosing a value for k, clustering based on multiple features and detecting outliers.

In [32]:

%matplotlib inline
import pandas as pd
import seaborn as sns
from matplotlib import pyplot as plt

In [31]:

## Step 1. Simple clustering

The first thing that we will try is to cluster the geolocation data of the calls. The geolocation of a record consists of a pair (latitude, longitude), which represents the position of the record on the map. Since the dataset that we have spans only a small geographic location, i.e. Boston, we will use the Euclidean distance when measuring how close two points are. Techinically, we should pick a projection in order to transform the data into 2d, and measure the Euclidean distance in the projected space.

Since we are dealing with numeric data, we might consider centering and normalizing them. In order to do that, check the StandardScaler from scikit-learn.

For exampe, the result of a clustering with 10 clusters is the following:

Filter the dataset to get the latitude and longitude, transform it as required and run the k-means clustering algorithm. Argue about the number of clusters you used. (pts 50)

In [31]:

## Step 2. Using multiple features

The next step in our analysis is to use more features than just the geolocation of a record (whose clustering only gives us information about the neighborhoods, which should not be a huge revelation!) Filter the dataset to get more features, transform them as required and compute another clustering. You should argue about the quality of your clustering and use domain knowledge (i.e. the fact that you are locals) to explore the clusters you produced and present them to the reader. (pts 30)

In [ ]:

## Step 3. Detecting outliers

Now that we have a clustering of the records in the dataset, we can detect the outlying ones. These are the ones who are the farthest from the centroids of their clusters. Track them down and describe any interesting observations that you can make. (pts 20)

In [ ]:

In [21]:

# Code for setting the style of the notebook
from IPython.core.display import HTML
def css_styling():
return HTML(styles)
css_styling()

Out[21]:

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font-family: "Computer Modern";
src: url('http://mirrors.ctan.org/fonts/cm-unicode/fonts/otf/cmunss.otf');
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font-family: 'EB Garamond', serif;
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font-family: 'EB Garamond',Computer Modern, "Helvetica Neue", Arial, Helvetica, Geneva, sans-serif;
line-height: 145%;
font-size: 140%;
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border-color: rgba(0,0,0,0.10) !important;
background: #fafafa;
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.CodeMirror {
font-family: "Source Code Pro";
font-size: 90%;
}
.prompt {
display: None;
}
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div.output_scroll {
width: inherit;
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font-weight: 200;
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display: block;
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font-weight: 300;
font-size: 16pt;
color: #CD2305;
font-style: italic;
margin-bottom: .5em;
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display: block;
}
.warning {
color: rgb( 240, 20, 20 )
}

MathJax.Hub.Config({
TeX: {
extensions: ["AMSmath.js"]
},
tex2jax: {
inlineMath: [ ['$','$'], ["\$","\$"] ],
displayMath: [ ['$$','$$'], ["\$","\$"] ]
},
displayAlign: 'center', // Change this to 'center' to center equations.
"HTML-CSS": {
styles: {'.MathJax_Display': {"margin": 4}}
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# Annihilation operator in harmonic oscillator

In Wikipedia's QHO page there is a moment when the following is stated:

I don't know why "the ground state in the position representation is determined by $a|0\rangle=0$". I would say that the position representation of the ground state is rather $\langle x|0\rangle$, isn't it?

• Why $\langle x|a|0\rangle=0$? I thought that the annihilation operator couldn't be applied to the ground state. Does it return a $0$ if one does that?
• Is it possible to get operators out of a bra and a ket? I mean, for any operator $\hat{A}$, is $\langle\phi|\hat{A}|\psi\rangle=\hat{A}\langle\phi|\psi\rangle$ true? In the first case I would be doing the inner product between a bra ($\langle\phi|$) and a ket ($\hat{A}|\psi\rangle$), but in the second case I'm applying the operator to a constant. So... that doesn't seem right to me, but I'd appreciate it if you told me.
• Related to the last item: what happens when an operator is applied to a constant? Do I get another operator?

• How does it jump from the second line to the third one (I mean from the one with the derivative in it to the one with the $\exp$ function)? I have absolutely no idea about that.

1. The annihilation operator is a linear operator. A linear operator can be applied to ANY state. And yes, it returns zero when applied to the ground state.

2. You can really take this as a definition. The definition of the momentum operator $\hat{p}$ is the operator such that $\langle x|\hat{p}|\psi\rangle=-i\hbar \psi'(x)$. One could write this as $\langle x|\hat{p}|\psi\rangle=-i\hbar \frac{d}{dx}\langle x| \psi\rangle$. "x" isn't a constant here, so you're not applying an operator to a constant.

3. Physicists love to abuse notation. If you formalize it, you can't apply operators (like $\hat{p}$ and $\hat{x}$ and $\hat{a}$ and $\hat{a}^\dagger$) to constants, you can only apply them to elements in your Hilbert space.

4. The jump from the second line to the third line is made by actually writing out the differential equation $\psi_0'(x)=x \alpha \psi_0(x)$ and solving it, by whatever method you want. (I prefer "by observation" :)

The phrasing of "the ground state in the position representation is determined by ..." is a bit weird, I agree. What is really meant is that the equation $a|0\rangle=0$, when written in the position basis, gives rise to a first order ordinary differential equation which can be solved pretty easily.

• Thanks for asnwering. In your example, you wrote $\langle x|\hat{p}|\psi\rangle$ and then "took out" the operator $\hat{p}$ from between those vectors. Is it always "legal" to extract an operator from between two vectors? – Tendero May 5 '16 at 1:14
• No. You're applying the definition of p. – user12029 May 5 '16 at 1:57

Concerning point 2:

Operators do not always come through each other cleanly, but there are some very basic rules that always apply, which can be turned into less tedious rules that apply in special cases. Often the latter are taught first, causing mass confusion.

General rule: Operators can be expressed as

(Sum over a in the set of eigenvectors ) |a > eigenvalue(a) < a|

If there are an infinite number of eigenvectors, then the eigenvalue should have a differential quantity in it like 'dx'. Example:

x operator = integral over x: |x > x dx < x|

If you're working in some other basis you can instead express it as (Sum over vectors a, b) |a > matrix-element[a, b] < b|

Also, always put all these integrals all the way out front. So,

< x | p |Ψ> = (Integral over p) < x|p > p dp < p | Ψ>

You are free to reorder < x|p >, p, dp, and < p|Ψ > as you wish since the two brackets are just scalars, and p isn't an operator, just a point in momentum space, and dp is a differential quantity of momentum space. If you have more than one vector quantity, then you're restricted in moving those vectors around in the same fashion as you would be in regular linear algebra (if you reorder a cross-product you need to negate it, for example). Just, things of the form < label 1|label 2 > are scalars.

Getting back to this particular case, < x|p > = (a const depending on the dimensionality) e^{i*(p/ℏ)*(dot)x}. Try taking the gradient of < x | Ψ > and compare, and you'll see that it comes out so that what you said up front is right.

The same general technique can be used to find how to swap other operators around. There are generally deep reasons that they should have these particular relationships. Like, momentum-in-dimension-1 is orthogonal to position-in-dimension-2, so there aren't any interactions between these terms, so those operators can flow past each other freely. But there are relationships between p and x in the same dimension, so they have to change something on the way past.

Concerning point 3:

If a and c are states and B is a scalar operator, then

< a| B |c > is a scalar (see above)

B < a|c > is an operator. This operator is equal to B times the scalar, < a|c >. This expression does not involve applying the operator to anything. It just gets scaled.


text:
# Tag Info

### Market Data Sources Bloomberg Vs Reuter

There are a few things to consider: Price On average Thomson Reuters is known to be less costly than Bloomberg. One thing to consider when looking to save money is that most vendors will use some ...
• 191

### Mapping symbols between tickers, Reuters RICs and Bloomberg tickers

Each vendor has their own symbology universe and each exchange, market, or country may have their own standard identifiers. To date the majority of identifiers have also been dynamic, i.e. when a ...
• 153
Accepted

### What structural model does Reuters use for default probability?

Reuters uses a proprietary model defined StarMine structural/SmartRatios Credit Risk model that has been developed by themselves and provided with the Reuters data service. It does not exist a formal ...
• 2,446
Accepted

### Is Reuters Instrument Code (RIC) case-sensitive?

Yes, Thomson Reuters RIC codes are case-sensitive. I believe the confusion comes from the fact that some frontends will 'assist' you if you misspell and automatically convert. That's just a trick. ...

### Database of ETF and underlying index from Thomson Reuters or Bloomberg

You can find the underlying data for the big indices, such as S&P 500, on our website www.simfin.com, freely available and instantly downloadable as excel. To this date, we have financial ratios, ...
• 81
Accepted

### Mapping I/B/E/S to Compustat via 6-digit CUSIP

The main problem of linking Compustat with IBES is not the fact that Compustat's cusip is 9 character, whereas IBES is 8-character. The main issue is that Compustat Cusip is header (most recent), ...
• 6,923

### Conversion stock symbols Google Finance vs. Reuters

There isn't a reliable mechanism. Some things to think about: It's common for venues to abuse the security reference of the primary venue to mean anywhere that this security is traded. So - for some ...
• 121
Accepted

### How does Reuters quote caps?

I don't know exactly about Reuters but often implied volas in the Black 76 world are quoted (forward) ATM. Thus the forward equals the strike and they dissappear from the formula:  C = E[(F-K)^+] = ...
• 13.3k

### Mapping I/B/E/S to Compustat via 6-digit CUSIP

Since I don't have SAS, I wrote a python script to create the mapping table between Compustat and IBES via CRSP. The code is available on my GitHub: https://github.com/snauhaus/link_compustat_ibes ...
• 227
Accepted

### Convert 3M rates to 6M rates using Basis Swaps (3M vs 6M)

Let's say 1yr semiannual rate versus 6m Libor is 2.00% and 1yr basis swap is 6m libor = 3m libor + 15bp. Then , to a first approximation 1yr rate versus 3m libor is 2.00-0.15= 1.85%. More ...
• 14.3k
1 vote

### Do companies' Reuters Instrument Codes (RICs) change over time?

You've stumbled upon one of the fundamental challenges to working with historical equity data. The quick answers are that (1) yes, identifiers can and do change over time; (2) there isn't a clear/...
• 1,608
1 vote

### Database of ETF and underlying index from Thomson Reuters or Bloomberg

So here's the answer for Bloomberg I suppose plus some extra advice when looking at the data you get from there. There is the function ETF that lets you screen for ETFs. Unfortunately, I have not ...
• 2,894
1 vote

### Vendor data aggregation for Options on Futures

I think the best you may be able to do here is creating your own unique identifier in your code based off underlying, maturity date, put or call, and strike price, and then match the two data sources ...
1 vote

### Market Data Sources Bloomberg Vs Reuter

Well for bond prcing, since there isn't a real market you'll need multiple data feeds anyway to create your bid/ask price. The question then becomes can either bloomberg or reuters. Both will allow ...
• 1,349
1 vote

### Bloomberg alternatives for fixed income data

https://www.money.net/ is what I like to use it can be used in your browser and they have a 14 day trial to see if you like it. Its relatively cheap and nice to use.
• 167
1 vote

### Bloomberg alternatives for fixed income data

The DTCC provides a free near-real-time swap activity here: https://rtdata.dtcc.com/gtr/dashboard.do You can subscribe to this service via RSS Feeds and mine it as needed. Swap trades must be ...
1 vote

### Mapping symbols between tickers, Reuters RICs and Bloomberg tickers

AAPL.O maps to AAPL US Equity (composite) or AAPL UW Equity ( Nasdaq NSM) thats stright common stock but what about preferred stock ? HWM_p.A maps to EP051790@SAM2 Pf or Unit trust ? PHGE_u.A maps ...
1 vote

### Mapping symbols between tickers, Reuters RICs and Bloomberg tickers

How to translate Eikon rics to Bloomberg identifier in Excel (works fine for different Equities from many stock exchanges): Cell A1. Write Eikon ric, for example: AAPL.O Cell B1 =TR(A1;"TR.SEDOL";;). ...
1 vote

### Mapping symbols between tickers, Reuters RICs and Bloomberg tickers

OpenFIGI is also a source of organizing and standardizing symbol tickers.
• 161
1 vote

### Mapping symbols between tickers, Reuters RICs and Bloomberg tickers

magtia.com We have a (cheap) commercial solution that addresses Symbol mapping from RIC to BBG to ISIN CUSIP activ ICE you name it We cover all markets and OTC. Problem is... the churn rate of new ...
1 vote

### Mapping symbols between tickers, Reuters RICs and Bloomberg tickers

For Europe, CBOE (batstrading) provides a csv list of symbols that trade on their exchange. Their file has a mapping between ISIN, Currency and Mic and Bloomberg Primary ticker: https://batstrading....

Only top scored, non community-wiki answers of a minimum length are eligible


text:
# DC Components and Frequency

I have started learning digital communication and was on line coding. I read there are certain characteristics that we need to keep in mind while choosing the right line coding method. One of these is considering DC component. I'm not quite able to understand what DC component really is, what and why does it poses a problem in digital communication?

To explain my doubt, I would like to tell two different definitions of DC component (at least they seem different to me) that I read from two different sources. First,

Signals creates very low frequencies when the voltage level in a digital signal is constant for a while. These frequencies around zero are called DC components, creates problems for a system that cannot pass low frequencies or system that uses electrical coupling.

Second,

Over one cycle period of a waveform, if all the positive voltages are cancelled by negative voltages, then, DC component of the waveform is 0. In line coding, the signal with the non-zero DC component is treated as a distorted one and it can create errors in received signals.

I don't seem to have any clue about what the first definition wants to say. As far as I know, the definition of frequency is number of cycles completed per unit time. Actually, how do you even measure the frequency of a digital signal? And how does it relate to the DC component?

I understand the second definition in itself pretty well but I don't understand how does it relates to the first definition. Also, what problem does it poses in transmission if the positive voltages do not cancel the negative ones?

P.S : A less mathematical reply would be highly appreciated since I'm new to this field.

The simplest waveform is one that encodes a bit $$1$$ with a pulse of positive amplitude, and a bit $$0$$ with a pulse of equal but negative amplitude. You may think that, for random bits, the negative pulses cancel out with the positive pulses and the average DC level of the signal is 0. This is in general not true over short time periods; even random bits will have sequences with many more $$1$$'s than $$0$$'s, or viceversa. This differential causes the signal's DC to grow, which causes distortion at the receiver end.
Line encoding can help with this problem. For example, the Alternate Mark Inversion line code encodes a bit $$0$$ to a pulse of amplitude zero, and a bit $$1$$ to pulses of positive and negative amplitude, alternating. So, every positive pulse is followed by a negative pulse, and the signal's DC never builds up.


text:
At the end of this step, you will understand the concepts of underfitting and overfitting, and you will be able to apply these ideas to make your models more accurate.

# Experimenting With Different Models¶

Now that you have a reliable way to measure model accuracy, you can experiment with alternative models and see which gives the best predictions. But what alternatives do you have for models?

You can see in scikit-learn's documentation that the decision tree model has many options (more than you'll want or need for a long time). The most important options determine the tree's depth. Recall from the first lesson in this micro-course that a tree's depth is a measure of how many splits it makes before coming to a prediction. This is a relatively shallow tree

In practice, it's not uncommon for a tree to have 10 splits between the top level (all houses) and a leaf. As the tree gets deeper, the dataset gets sliced up into leaves with fewer houses. If a tree only had 1 split, it divides the data into 2 groups. If each group is split again, we would get 4 groups of houses. Splitting each of those again would create 8 groups. If we keep doubling the number of groups by adding more splits at each level, we'll have $2^{10}$ groups of houses by the time we get to the 10th level. That's 1024 leaves.

When we divide the houses amongst many leaves, we also have fewer houses in each leaf. Leaves with very few houses will make predictions that are quite close to those homes' actual values, but they may make very unreliable predictions for new data (because each prediction is based on only a few houses).

This is a phenomenon called overfitting, where a model matches the training data almost perfectly, but does poorly in validation and other new data. On the flip side, if we make our tree very shallow, it doesn't divide up the houses into very distinct groups.

At an extreme, if a tree divides houses into only 2 or 4, each group still has a wide variety of houses. Resulting predictions may be far off for most houses, even in the training data (and it will be bad in validation too for the same reason). When a model fails to capture important distinctions and patterns in the data, so it performs poorly even in training data, that is called underfitting.

Since we care about accuracy on new data, which we estimate from our validation data, we want to find the sweet spot between underfitting and overfitting. Visually, we want the low point of the (red) validation curve in

# Example¶

There are a few alternatives for controlling the tree depth, and many allow for some routes through the tree to have greater depth than other routes. But the max_leaf_nodes argument provides a very sensible way to control overfitting vs underfitting. The more leaves we allow the model to make, the more we move from the underfitting area in the above graph to the overfitting area.

We can use a utility function to help compare MAE scores from different values for max_leaf_nodes:

In [1]:
from sklearn.metrics import mean_absolute_error
from sklearn.tree import DecisionTreeRegressor

def get_mae(max_leaf_nodes, train_X, val_X, train_y, val_y):
model = DecisionTreeRegressor(max_leaf_nodes=max_leaf_nodes, random_state=0)
model.fit(train_X, train_y)
preds_val = model.predict(val_X)
mae = mean_absolute_error(val_y, preds_val)
return(mae)

The data is loaded into train_X, val_X, train_y and val_y using the code you've already seen (and which you've already written).

In [2]:
# Data Loading Code Runs At This Point
import pandas as pd

melbourne_file_path = '../input/melbourne-housing-snapshot/melb_data.csv'
# Filter rows with missing values
filtered_melbourne_data = melbourne_data.dropna(axis=0)
# Choose target and features
y = filtered_melbourne_data.Price
melbourne_features = ['Rooms', 'Bathroom', 'Landsize', 'BuildingArea',
'YearBuilt', 'Lattitude', 'Longtitude']
X = filtered_melbourne_data[melbourne_features]

from sklearn.model_selection import train_test_split

# split data into training and validation data, for both features and target
train_X, val_X, train_y, val_y = train_test_split(X, y,random_state = 0)

We can use a for-loop to compare the accuracy of models built with different values for max_leaf_nodes.

In [3]:
# compare MAE with differing values of max_leaf_nodes
for max_leaf_nodes in [5, 50, 500, 5000]:
my_mae = get_mae(max_leaf_nodes, train_X, val_X, train_y, val_y)
print("Max leaf nodes: %d  \t\t Mean Absolute Error:  %d" %(max_leaf_nodes, my_mae))

Max leaf nodes: 5  		 Mean Absolute Error:  347380
Max leaf nodes: 50  		 Mean Absolute Error:  258171
Max leaf nodes: 500  		 Mean Absolute Error:  243495
Max leaf nodes: 5000  		 Mean Absolute Error:  254983

Of the options listed, 500 is the optimal number of leaves.

# Conclusion¶

Here's the takeaway: Models can suffer from either:

• Overfitting: capturing spurious patterns that won't recur in the future, leading to less accurate predictions, or
• Underfitting: failing to capture relevant patterns, again leading to less accurate predictions.

We use validation data, which isn't used in model training, to measure a candidate model's accuracy. This lets us try many candidate models and keep the best one.


text:
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# A retailer bought a shirt at wholesale and marked it up 80%

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Joined: 12 Oct 2011
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GMAT 1: 700 Q48 V37
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A retailer bought a shirt at wholesale and marked it up 80%  [#permalink]

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Updated on: 26 Sep 2019, 09:12
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A retailer bought a shirt at wholesale and marked it up 80% to its initial price of $45. By how many more dollars does he need to increase the price to achieve a 100% markup? A. 1 B. 2 C. 3 D. 4 E. 5 Kind of confused here. A 80% markup over$45 would be $81, a 100% markup would be$90, so in order to get a 100% markup he would need to increase the price by $9?! Either there's a mistake in the question or I misunderstood it. Originally posted by BN1989 on 16 Apr 2012, 05:04. Last edited by Abhishek009 on 26 Sep 2019, 09:12, edited 1 time in total. Used Spoiler ##### Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 59211 Re: A retailer bought a shirt at wholesale and marked it up 80% [#permalink] ### Show Tags 16 Apr 2012, 05:15 7 7 BN1989 wrote: A retailer bought a shirt at wholesale and marked it up 80% to its initial price of$45. By how many more dollars does he need to increase the price to achieve a 100% markup?

A. 1
B. 2
C. 3
D. 4
E. 5

Kind of confused here. A 80% markup over $45 would be$81, a 100% markup would be $90, so in order to get a 100% markup he would need to increase the price by$9?! Either there's a mistake in the question or I misunderstood it.

The point is that $45 is already a retail price, so a price after 80% mark up on wholesale price. Say wholesale price x, then 1.8x=45 --> x=25. In order to achieve a 100% markup a retail price should be 2*25=$50, so $5 more. Answer: E. _________________ ##### General Discussion Senior Manager Joined: 10 Jul 2013 Posts: 282 Re: A retailer bought a shirt at wholesale and marked it up 80% [#permalink] ### Show Tags 12 Aug 2013, 06:30 BN1989 wrote: A retailer bought a shirt at wholesale and marked it up 80% to its initial price of$45. By how many more dollars does he need to increase the price to achieve a 100% markup?

A. 1
B. 2
C. 3
D. 4
E. 5

Kind of confused here. A 80% markup over $45 would be$81, a 100% markup would be $90, so in order to get a 100% markup he would need to increase the price by$9?! Either there's a mistake in the question or I misunderstood it.

...
tricky question...especially the wording..... Answer is E
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Re: A retailer bought a shirt at wholesale and marked it up 80%  [#permalink]

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12 Aug 2013, 06:40
1
1
BN1989 wrote:
A retailer bought a shirt at wholesale and marked it up 80% to its initial price of $45. By how many more dollars does he need to increase the price to achieve a 100% markup? A. 1 B. 2 C. 3 D. 4 E. 5 Kind of confused here. A 80% markup over$45 would be $81, a 100% markup would be$90, so in order to get a 100% markup he would need to increase the price by $9?! Either there's a mistake in the question or I misunderstood it. For me the typical solution is like below: Attachments price.png [ 16.12 KiB | Viewed 10170 times ] _________________ Asif vai..... Intern Joined: 23 Apr 2012 Posts: 48 GMAT 1: 690 Q49 V34 Re: A retailer bought a shirt at wholesale and marked it up 80% [#permalink] ### Show Tags 26 Nov 2013, 08:54 Got the answer, but not sure if i understand the language of the question ! It says "marked it up 80% to initial price OF$45" -> shouldn't this imply that the initial price was $45 ? Math Expert Joined: 02 Sep 2009 Posts: 59211 Re: A retailer bought a shirt at wholesale and marked it up 80% [#permalink] ### Show Tags 26 Nov 2013, 08:59 surbhi87 wrote: Got the answer, but not sure if i understand the language of the question ! It says "marked it up 80% to initial price OF$45" -> shouldn't this imply that the initial price was $45 ? Wholesale price = x. Initial price =$45.
Mark up = 80%.

1.8x=45 --> x=25.

In order to achieve a 100% markup a retail price should be 2*25=$50, so$5 more.

Hope this helps.
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Re: A retailer bought a shirt at wholesale and marked it up 80%  [#permalink]

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26 Nov 2013, 11:39
Bunuel wrote:

Wholesale price = x.
Initial price = $45. Mark up = 80%. 1.8x=45 --> x=25. In order to achieve a 100% markup a retail price should be 2*25=$50, so $5 more. Hope this helps. Thanks Bunnel. I'm going to need a little more help here ! You said initial price was 45$. Then markup was 80% hence final price would be 45+((80/100)*45) . This would lead to 81.
100% markup from 45 should mean 90. hence 90-81= 9.
This was my initial approach. When i couldnt find 9 in the options, then i went on to do 1.8x=45
But i still dont get why the language of the question doesnt mean that the initial price was actually 45.
Per your equation , 1.8x=45 , it seems that 45 was the final price after the markup .
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Re: A retailer bought a shirt at wholesale and marked it up 80%  [#permalink]

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26 Nov 2013, 11:46
surbhi87 wrote:
Bunuel wrote:

Wholesale price = x.
Initial price = $45. Mark up = 80%. 1.8x=45 --> x=25. In order to achieve a 100% markup a retail price should be 2*25=$50, so $5 more. Hope this helps. Thanks Bunnel. I'm going to need a little more help here ! You said initial price was 45$. Then markup was 80% hence final price would be 45+((80/100)*45) . This would lead to 81.
100% markup from 45 should mean 90. hence 90-81= 9.
This was my initial approach. When i couldnt find 9 in the options, then i went on to do 1.8x=45
But i still dont get why the language of the question doesnt mean that the initial price was actually 45.
Per your equation , 1.8x=45 , it seems that 45 was the final price after the markup .

The point is that $45 is already a retail price (initial price), so a price after 80% mark up on wholesale price. Please check here: a-retailer-bought-a-shirt-at-wholesale-and-marked-it-up-130759.html#p1074292 _________________ Intern Status: Preparing for GMAT Joined: 10 Dec 2013 Posts: 13 Location: India Concentration: Marketing, Leadership GMAT 1: 530 Q46 V18 WE: Other (Entertainment and Sports) Re: A retailer bought a shirt at wholesale and marked it up 80% [#permalink] ### Show Tags 04 Aug 2014, 18:53 The words 80% of the initial price are so misleading! More often than not, many people would be thinking Oh! Initial price has to be the cost price. Why would Gmat do something like that? Manager Joined: 11 Jun 2014 Posts: 52 Concentration: Technology, Marketing GMAT 1: 770 Q50 V45 WE: Information Technology (Consulting) Re: A retailer bought a shirt at wholesale and marked it up 80% [#permalink] ### Show Tags 04 Aug 2014, 21:00 "A retailer bought a shirt at wholesale and marked it up 80% to its initial price of$45"

I think it sounds vague, but its actually correct, when the sentence says, "marked it up", it refers to 'wholesale price'.. marked it up to get the initial price of $45. Initial price is not wholesale price here.. its the price after mark up. BTW this can be solved without any effort. 80% is 4/5 of the wholesale price. so the current ratio of sale price to cost price is 1+4/5 = 9/5. When we see 9/5 and$45, its rings a bell.. so the wholesale price ( cost) must be 25 and to get a 100% markup, the sale price should be 50, so the price needs to be increased by 5.
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Re: A retailer bought a shirt at wholesale and marked it up 80%  [#permalink]

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06 Aug 2014, 03:13
Marked it up 80% to its initial price of $45 So cost in wholesale $$= 45*\frac{100}{180} = 25$$ 100% marked up means 25*2 = 50 Increase required = 50-45 = 5 Answer = E _________________ Kindly press "+1 Kudos" to appreciate Intern Joined: 23 Dec 2014 Posts: 47 Re: A retailer bought a shirt at wholesale and marked it up 80% [#permalink] ### Show Tags 11 Jan 2015, 15:57 surbhi87 wrote: Got the answer, but not sure if i understand the language of the question ! It says "marked it up 80% to initial price OF$45" -> shouldn't this imply that the initial price was $45 ? Yes, wording is really tricky or poor !!! I thought$45 was the retail price at the first sight.
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Re: A retailer bought a shirt at wholesale and marked it up 80%  [#permalink]

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11 Jan 2015, 16:50
Hi All,

The first sentence in this prompt clearly states the 'order' of events:

1) A retailer bought a shirt at wholesale (let's say X dollars)
2) Marked it up 80%.....1.8X dollars
3) To its initial price of $45 From this, we have an algebra equation: 1.8X = 45 From here, the math can be done in a couple of different ways.... You COULD solve for X then redo the entire calculation to figure out what the increase would need to be for the price to be marked up 100% OR..... You COULD set up a ratio.... Since we need a 100% mark-up, we're looking for the value of 2X. This means that we want to increase 1.8X by 0.2X.... 1.8X/0.2X = 45/Y where Y is the amount that the price of the item would need to be MARKED-UP to make the markup 100% 1.8X/.2X = 45/Y 9/1 = 45/Y 9Y = 45 Y = 5 Final Answer: GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at: Rich.C@empowergmat.com The Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★ Retired Moderator Joined: 19 Mar 2014 Posts: 917 Location: India Concentration: Finance, Entrepreneurship GPA: 3.5 Re: A retailer bought a shirt at wholesale and marked it up 80% [#permalink] ### Show Tags 26 Jun 2017, 10:30 A retailer bought a shirt at wholesale and marked it up 80% to its initial price of$45. By how many more dollars does he need to increase the price to achieve a 100% markup?

Let the wholesale price for this shirt be x

$$x + 0.8x = 45$$

$$1.8x = 45$$

$$x = \frac{45}{1.8}$$

$$x = 25$$

So, wholesale price = $25, 100% increase will make the price =$ 50

Diff: $$50 - 45 = 5$$

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Re: A retailer bought a shirt at wholesale and marked it up 80%  [#permalink]

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28 Jun 2017, 16:52
BN1989 wrote:
A retailer bought a shirt at wholesale and marked it up 80% to its initial price of $45. By how many more dollars does he need to increase the price to achieve a 100% markup? A. 1 B. 2 C. 3 D. 4 E. 5 We can let x = the increased price and create the following proportion: 1.8/45 = 2/x 1.8x = 90 x = 90/1.8 = 50 Thus, the item needs to be marked up by 5 more dollars. Answer: E _________________ # Scott Woodbury-Stewart Founder and CEO Scott@TargetTestPrep.com 122 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 09 Oct 2016 Posts: 32 Re: A retailer bought a shirt at wholesale and marked it up 80% [#permalink] ### Show Tags 22 Jul 2017, 07:21 BN1989 wrote: A retailer bought a shirt at wholesale and marked it up 80% to its initial price of$45. By how many more dollars does he need to increase the price to achieve a 100% markup?

A. 1
B. 2
C. 3
D. 4
E. 5

Kind of confused here. A 80% markup over $45 would be$81, a 100% markup would be $90, so in order to get a 100% markup he would need to increase the price by$9?! Either there's a mistake in the question or I misunderstood it.

We are asked increase the price to achieve a 100% markup.
From the question, we already increased to to 80% markup.
We need to increase the initial price by more 20% to achieve the 100%.
180% = 45
notice that both numbers (180 and 45) are divisible by 9
20% = 5
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Re: A retailer bought a shirt at wholesale and marked it up 80%  [#permalink]

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22 Jul 2017, 07:44
Stem states that 1.8x = 45
100% mark up is 2x, i.e an additional 0.2x

So, (0.2x/1.8x)*45 = 5
Answr:E
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Re: A retailer bought a shirt at wholesale and marked it up 80%  [#permalink]

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04 Aug 2018, 07:26
This is how i solved it:

C.P = 45
M.P = 45*1.8 = 81 (80% mark up)
In order to achieve a 100% mark up

(45+x) * 1.8 = 100
we need to find x

So solving the equation, we get :

x = 5
Answer = E which means if 45 is increased by 5 then it will be a 100 percent mark up.
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Re: A retailer bought a shirt at wholesale and marked it up 80%  [#permalink]

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26 Sep 2019, 09:11
BN1989 wrote:
A retailer bought a shirt at wholesale and marked it up 80% to its initial price of $45. By how many more dollars does he need to increase the price to achieve a 100% markup? A. 1 B. 2 C. 3 D. 4 E. 5 Kind of confused here. A 80% markup over$45 would be $81, a 100% markup would be$90, so in order to get a 100% markup he would need to increase the price by $9?! Either there's a mistake in the question or I misunderstood it. $$\frac{45}{180}*200 = 50$$ Thus, price must be increased by$ 5, Answer must be (E)
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Re: A retailer bought a shirt at wholesale and marked it up 80%  [#permalink]

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30 Sep 2019, 08:09

Solution

Given
• A retailer marked up the price of a shirt by 80% to 45.

To find
• By how many more dollars does he need to increase the price to achieve a 100% mark-up

Approach and Working out
• Price after 80% markup= 45
o So, 1.8* Cost price = 45
o Cost price = 45/1.8 = 450/18 = 50/2=25
• So, the price after 100% mark up = 50

Hence, the price of the shirt if increased by \$5 will give 100% mark up.

Thus, option E is the correct answer.

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Re: A retailer bought a shirt at wholesale and marked it up 80%   [#permalink] 30 Sep 2019, 08:09

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# A Group Homomorphism and an Abelian Group

## Problem 207

Let $G$ be a group. Define a map $f:G \to G$ by sending each element $g \in G$ to its inverse $g^{-1} \in G$.
Show that $G$ is an abelian group if and only if the map $f: G\to G$ is a group homomorphism.

## Proof.

### $(\implies)$ If $G$ is an abelian group, then $f$ is a homomorphism.

Suppose that $G$ is an abelian group. Then we have for any $g, h \in G$
\begin{align*}
f(gh)&=(gh)^{-1}=h^{-1}g^{-1}\\
&=g^{-1}h^{-1} \text{ since } G \text{ is abelian}\\
&=f(g)f(h).
\end{align*}
This implies that the map $f$ is a group homomorphism.

### $(\impliedby)$ If $f$ is a homomorphism, then $G$ is an abelian group.

Now we suppose that the map $f: G \to G$ is a group homomorphism.
Then for any $g, h \in G$, we have
$f(gh)=f(g)f(h) \tag{*}$ since $f$ is a group homomorphism.
The left hand side of (*) is
$f(gh)=(gh)^{-1}=h^{-1}g^{-1}.$ Thus we obtain from (*) that
$h^{-1}g^{-1}=g^{-1}h^{-1}.$ Taking the inverse of both sides, we have
$gh=hg$ for any $g, h \in G$.
It follows that $G$ is an abelian group.

## Related Question.

Another problem about the relation between an abelian group and a group homomorphism is:
A group is abelian if and only if squaring is a group homomorphism

### More from my site

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• Pullback Group of Two Group Homomorphisms into a Group Let $G_1, G_1$, and $H$ be groups. Let $f_1: G_1 \to H$ and $f_2: G_2 \to H$ be group homomorphisms. Define the subset $M$ of $G_1 \times G_2$ to be $M=\{(a_1, a_2) \in G_1\times G_2 \mid f_1(a_1)=f_2(a_2)\}.$ Prove that $M$ is a subgroup of $G_1 \times G_2$.   […]
• A Group is Abelian if and only if Squaring is a Group Homomorphism Let $G$ be a group and define a map $f:G\to G$ by $f(a)=a^2$ for each $a\in G$. Then prove that $G$ is an abelian group if and only if the map $f$ is a group homomorphism.   Proof. $(\implies)$ If $G$ is an abelian group, then $f$ is a homomorphism. Suppose that […]
• Group Homomorphism, Conjugate, Center, and Abelian group Let $G$ be a group. We fix an element $x$ of $G$ and define a map $\Psi_x: G\to G$ by mapping $g\in G$ to $xgx^{-1} \in G$. Then prove the followings. (a) The map $\Psi_x$ is a group homomorphism. (b) The map $\Psi_x=\id$ if and only if $x\in Z(G)$, where $Z(G)$ is the […]
• Abelian Groups and Surjective Group Homomorphism Let $G, G'$ be groups. Suppose that we have a surjective group homomorphism $f:G\to G'$. Show that if $G$ is an abelian group, then so is $G'$.   Definitions. Recall the relevant definitions. A group homomorphism $f:G\to G'$ is a map from $G$ to $G'$ […]
• A Homomorphism from the Additive Group of Integers to Itself Let $\Z$ be the additive group of integers. Let $f: \Z \to \Z$ be a group homomorphism. Then show that there exists an integer $a$ such that $f(n)=an$ for any integer $n$.   Hint. Let us first recall the definition of a group homomorphism. A group homomorphism from a […]
• Abelian Normal subgroup, Quotient Group, and Automorphism Group Let $G$ be a finite group and let $N$ be a normal abelian subgroup of $G$. Let $\Aut(N)$ be the group of automorphisms of $G$. Suppose that the orders of groups $G/N$ and $\Aut(N)$ are relatively prime. Then prove that $N$ is contained in the center of […]
• Surjective Group Homomorphism to $\Z$ and Direct Product of Abelian Groups Let $G$ be an abelian group and let $f: G\to \Z$ be a surjective group homomorphism. Prove that we have an isomorphism of groups: $G \cong \ker(f)\times \Z.$   Proof. Since $f:G\to \Z$ is surjective, there exists an element $a\in G$ such […]

### 1 Response

1. 03/05/2017

[…] Another problem about the relation between an abelian group and a group homomorphism is: A group homomorphism and an abelian group. […]

##### Order of the Product of Two Elements in an Abelian Group

Let $G$ be an abelian group with the identity element $1$. Let $a, b$ be elements of $G$ with order...

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text:
# In $(\mathbf Z/p^r \mathbf Z)^*$, finding an element with order $p-1$.

Let $p$ be an odd prime number.
I want to prove that $(\mathbf Z/p^r \mathbf Z)^*$ is an cyclic group.
I have known that $\overline {p-1} \in (\mathbf Z/p^r \mathbf Z)^*$ is of order $p^{r-1}$.
Since the order of the group is $(p-1)p^{r-1}$, if we find an element with order $p-1$, we conclude that $(\mathbf Z/p^r \mathbf Z)^*$ must have an element with order $(p-1)p^{r-1}$.

$\because$ Every finite commutive group has an element with order of its exponent.

So how to I find the existence of an element with order $p-1$?

I think it is possible by induction, for $r=1$ it is the classical result I assume to be true.

Take $0\leq x\leq p-1$ such that $x$ is of order $p-1$ in $\mathbb{Z}/p \mathbb{Z}^*$. We will be looking for an element of order $p-1$ in $\mathbb{Z}/p^2 \mathbb{Z}^*$ under the form :

$$x_2=x+\alpha p$$

Where $\alpha$ is to be determined, we have :

$$x_2^{p-1}=x^{p-1}+(p-1)\alpha x^{p-2} p\text{ mod } p^2$$ $$x_2^{p-1}=x^{p-1}-\alpha x^{p-2} p\text{ mod } p^2$$

Now because $x$ is of order $p-1$ we have $x^{p-1}=1+t_1p$ hence :

$$x_2^{p-1}=1+(t_1-\alpha x^{p-2}) p\text{ mod } p^2$$

Hence $\alpha$ must verify the equation $t_1-\alpha x^{p-2}=0$ mod $p$. This is clearly solvable because $x^{p-2}$ can be inversed mod $p$, hence one can construct $\alpha$ such that :

$$x_2:=x+\alpha p\text{ is of order dividing } p-1$$

Now $x_2=x$ mod $p$ so its order must be divided by the order of $x$ mod $p$ which is $p-1$ hence $x_2$ is exactly of order $p-1$.

Assume now that $x_{r-1}$ is of order $p-1$ in $\mathbb{Z}/p^{r-1} \mathbb{Z}^*$ then define :

$$x_r:=x_{r-1}+\alpha p^{r-1}$$

Likewise, this define an element in $\mathbb{Z}/p^{r} \mathbb{Z}^*$ and :

$$x_r^{p-1}=x_{r-1}^{p-1}+(p-1)\alpha x_{r-1}^{p-2} p^{r-1}\text{ mod } p^r$$

$$x_r^{p-1}=x_{r-1}^{p-1}-\alpha x_{r-1}^{p-2} p^{r-1}\text{ mod } p^r$$

But $x_{r-1}^{p-1}=1+t_{r-1}p^{r-1}$ hence we have that :

$$x_r^{p-1}=1+(t_{r-1}-\alpha x_{r-1}^{p-2}) p^{r-1}\text{ mod } p^r$$

Now the equation in $\alpha$ : $t_{r-1}-\alpha x_{r-1}^{p-2}=0$ mod $p$ is solvable hence one can find $\alpha$ such that $x_r$ is of order dividing $p-1$, now by construction $x_r$ mod $p^{r-1}$ is $x_{r-1}$ hence its order must be divided by $p-1$, finally $x_r$ is an element of order $p-1$ in $\mathbb{Z}/p^r \mathbb{Z}^*$.

• I appreciate your answer. It is very good. – Jeong Jun 12 '15 at 11:16


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Home > Standard Error > Interpretation Of Standard Error In Regression

# Interpretation Of Standard Error In Regression

## Contents

That assumption of normality, with the same variance (homoscedasticity) for each $\epsilon_i$, is important for all those lovely confidence intervals and significance tests to work. Standard error statistics are a class of statistics that are provided as output in many inferential statistics, but function as descriptive statistics. Why I Like the Standard Error of the Regression (S) In many cases, I prefer the standard error of the regression over R-squared. I don't question your knowledge, but it seems there is a serious lack of clarity in your exposition at this point.) –whuber♦ Dec 3 '14 at 20:54 @whuber For useful reference

McHugh. Standard error: meaning and interpretation. Fitting so many terms to so few data points will artificially inflate the R-squared. Thanks for writing!

## Standard Error Of Estimate Interpretation

In fact, the level of probability selected for the study (typically P < 0.05) is an estimate of the probability of the mean falling within that interval. It is an even more valuable statistic than the Pearson because it is a measure of the overlap, or association between the independent and dependent variables. (See Figure 3).     You may find this less reassuring once you remember that we only get to see one sample!

Lane PrerequisitesMeasures of Variability, Introduction to Simple Linear Regression, Partitioning Sums of Squares Learning Objectives Make judgments about the size of the standard error of the estimate from a scatter plot The residual standard deviation has nothing to do with the sampling distributions of your slopes. If the Pearson R value is below 0.30, then the relationship is weak no matter how significant the result. Standard Error Of Prediction Another use of the value, 1.96 ± SEM is to determine whether the population parameter is zero.

It is calculated by squaring the Pearson R. Standard Error Of Regression Formula If A sells 101 units per week and B sells 100.5 units per week, A sells more. That is, of the dispersion of means of samples if a large number of different samples had been drawn from the population.   Standard error of the mean The standard error Does he have any other options?Martha (Smith) on Should Jonah Lehrer be a junior Gladwell?

Further Reading Linear Regression 101 Stats topics Resources at the UCLA Statistical Computing Portal
© 2007 The Trustees of Princeton University. The Standard Error Of The Estimate Is A Measure Of Quizlet The larger the standard error of the coefficient estimate, the worse the signal-to-noise ratio--i.e., the less precise the measurement of the coefficient. Alas, you never know for sure whether you have identified the correct model for your data, although residual diagnostics help you rule out obviously incorrect ones. That is, the absolute change in Y is proportional to the absolute change in X1, with the coefficient b1 representing the constant of proportionality.

1. You bet!
2. Researchers typically draw only one sample.
3. Specifically, although a small number of samples may produce a non-normal distribution, as the number of samples increases (that is, as n increases), the shape of the distribution of sample means
4. Find the Centroid of a Polygon What would You-Know-Who want with Lily Potter?
5. Then you would just use the mean scores.
6. Then subtract the result from the sample mean to obtain the lower limit of the interval.
7. The point that "it is not credible that the observed population is a representative sample of the larger superpopulation" is important because this is probably always true in practice - how
8. Designed by Dalmario.
9. How to know if a meal was cooked with or contains alcohol?

## Standard Error Of Regression Formula

for 95% confidence, and one S.D. You may wonder whether it is valid to take the long-run view here: e.g., if I calculate 95% confidence intervals for "enough different things" from the same data, can I expect Standard Error Of Estimate Interpretation The smaller the standard error, the closer the sample statistic is to the population parameter. Standard Error Of Regression Coefficient Edit : This has been a great discussion and I'm going to digest some of the information before commenting further and deciding on an answer.

Now, the standard error of the regression may be considered to measure the overall amount of "noise" in the data, whereas the standard deviation of X measures the strength of the see here even if you have ‘population' data you can't assess the influence of wall color unless you take the randomness in student scores into account. In RegressIt you could create these variables by filling two new columns with 0's and then entering 1's in rows 23 and 59 and assigning variable names to those columns. You can see that in Graph A, the points are closer to the line than they are in Graph B. Linear Regression Standard Error

K? Can I visit Montenegro without visa? Name: Jim Frost • Monday, April 7, 2014 Hi Mukundraj, You can assess the S value in multiple regression without using the fitted line plot. this page If you know a little statistical theory, then that may not come as a surprise to you - even outside the context of regression, estimators have probability distributions because they are

Does he have any other options?zbicyclist on Some people are so easy to contact and some people aren't.Carol on Should Jonah Lehrer be a junior Gladwell? Standard Error Of Estimate Calculator However, in rare cases you may wish to exclude the constant from the model. Jim Name: Olivia • Saturday, September 6, 2014 Hi this is such a great resource I have stumbled upon :) I have a question though - when comparing different models from

## Formulas for a sample comparable to the ones for a population are shown below.

Imagine we have some values of a predictor or explanatory variable, $x_i$, and we observe the values of the response variable at those points, $y_i$. In multiple regression output, just look in the Summary of Model table that also contains R-squared. Intuitively, this is because highly correlated independent variables are explaining the same part of the variation in the dependent variable, so their explanatory power and the significance of their coefficients is What Is A Good Standard Error In fact, even with non-parametric correlation coefficients (i.e., effect size statistics), a rough estimate of the interval in which the population effect size will fall can be estimated through the same

Conveniently, it tells you how wrong the regression model is on average using the units of the response variable. In a scatterplot in which the S.E.est is small, one would therefore expect to see that most of the observed values cluster fairly closely to the regression line. Allen Mursau 4.924 προβολές 23:59 Standard Deviation vs Standard Error - Διάρκεια: 3:57. Get More Info Being out of school for "a few years", I find that I tend to read scholarly articles to keep up with the latest developments.

price, part 4: additional predictors · NC natural gas consumption vs.


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# Can Linear Constant coefficient Difference Equations always represent an LTI system?

Can we say that a Linear Constant Coefficient Difference Equation can always represent a Linear Shift Invarient system ? Are there any conditions which need to be satisfied additionally by these kind of equations to be able to do that?

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WHat about final rest condition? (i.e if x(t) = 0, for t > t0 then y(t) is also 0 for t > t0 ) whether now the differential equation satisfies LSI or not?

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Welcome to MSE! In its current form, this is probably better as a comment as opposed to an answer (I realize you don't yet have enough reputation). perhaps you can improve it to make it an answer. Regards –  Amzoti May 23 '13 at 13:09
Thank you.Initial rest condition is not necessary condition for linearity. But it is essential for Time invariance to satisfy.For linearity alone, zero initial condition like y(1) = 0 (which is not initial rest condition) is enough. –  Ram May 25 '13 at 12:58

http://ocw.mit.edu/resources/res-6-007-signals-and-systems-spring-2011/lecture-notes/MITRES_6_007S11_lec06.pdf

It is not necessary that a linear constant coefficient difference equation must represent an LTI system. It will represent an LTI system if and only if the solution satisfies the initial rest condition, namely if $x[n] = 0$ for $n<n_0$, then $y[n] = 0$ for $n<n_0$


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# A network has seven nodes and five independent loops. The number of branches in the network is

This question was previously asked in
TN TRB EC 2012 Official Paper
View all TN TRB EC Papers >
1. 13
2. 12
3. 11
4. 10

Option 3 : 11
Free
Network Theory
601
30 Questions 40 Marks 40 Mins

## Detailed Solution

Concept:

The number of meshes in a network is given as:

M = B – (N – 1)

B → number of branches

M → number of meshes

N → number of nodes

Analysis:

Given: N = 7, M = 5

The number of branches is calculated as:

B = M + (N - 1)

= 5 + 6 = 11

The Number of Node pair voltage = $$\frac{{N\left( {N - 1} \right)}}{2}$$


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# CCEA Double award unit 1 solutions

## Moments

Forces can cause an object to rotate and the turning effect of the force is called a moment. If a resultant force acts on an object about a fixed turning point called the pivot or fulcrum it will cause the object to rotate e.g. using a spanner, a screwdriver, opening a door hanging on hinges.

## Demo

In this tutorial you will learn how to calculate the moment in a real life example.

The equation for this calculation is written like this:

$M = { \text F \; \text x \; \text D}$

## Chilled practice question

A 5 N force is applied to a spanner 15 cm from the pivot. Calculate the moment.

## Frozen practice question

A screwdriver was used to open a tin of paint with a moment of 1.5 Nm exerted 25 cm from the pivot. What force was being applied ?

## Science in context

Forces can cause an object to rotate and the turning effect of the force is called a moment. If a resultant force acts on an object about a fixed turning point called the pivot or fulcrum it will cause the object to rotate e.g. using a spanner, a screwdriver, opening a door hanging on hinges. M = F x d, where M = the moment of a force (Nm), = force applied (N) and d (m) is the distance from the pivot or fulcrum.

## Millie’s Master Methods

### Millie’s Magic Triangle

The ability to rearrange equations is the first step to successfully solve Physics calculations. Millie’s Magic Triangle Teaches you to be confident when transposing an equation allowing …

### Calculation Master Method

Performing and mastering this routine will guarantee you maximum marks when solving Physics calculations. Calculation Master Class Top Mark Routine Applying this technique will ensure you gain …

## The Fridge Physics Store

### Teacher Fast Feedback

Feedback to students in seconds – Voice to label thermal bluetooth technology Click …

### Get Fridge Physics Merch

Why not buy a Fridge Physics baseball cap, woollen beanie, hoodie or polo shirt, all colours and sizes available. Free delivery to anywhere in the UK! Fridge …

## Pressure

Pressure can be calculated if you know the force applied to or by an object and the contact area. The units are newtons per square meter, or N/m2.

## Demo

In this tutorial you will learn how to calculate the pressure exerted by an object if you have been given the force applied to the object and given area.

The equation for this calculation is written like this:

$P = { \text F \; \text / \; \text A}$

## Chilled practice question

The base of a brief case has an area of 25 cm 2 and a mass of 5 Kg. Calculate the pressure exerted by the briefcase when placed on a surface. Take the force of Gravity as 10N/Kg.

## Frozen practice question

An elephant exerts a pressure of 53500 N/m2 one of its feet has an area of 0.15 m2. Calculate the mass of the elephant. Take the force of Gravity as 10N/Kg.

## Science in context

Pressure can be calculated if you know the force applied and the area using the physics equation, P = F/A. Pressure is equal to force divided by area, units are newtons per square meter, or N/m2.

## Millie’s Master Methods

### Millie’s Magic Triangle

The ability to rearrange equations is the first step to successfully solve Physics calculations. Millie’s Magic Triangle Teaches you to be confident when transposing an equation allowing …

### Calculation Master Method

Performing and mastering this routine will guarantee you maximum marks when solving Physics calculations. Calculation Master Class Top Mark Routine Applying this technique will ensure you gain …

## The Fridge Physics Store

### Teacher Fast Feedback

Feedback to students in seconds – Voice to label thermal bluetooth technology Click …

### Get Fridge Physics Merch

Why not buy a Fridge Physics baseball cap, woollen beanie, hoodie or polo shirt, all colours and sizes available. Free delivery to anywhere in the UK! Fridge …

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