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# half equations for a lead-acid cell

we are learning about galvanic cells. I would like to know the half equations and overall equation that occurs in this lead acid cell. so far my understanding for the oxidisation section is that in a lead acid battery oxidisation occurs in lead which has the equation Pb →Pb2+. The equation for the overall reaction at the anode is Pb + SO42- → PbSO4 + 2e-. is this correct because i am confused about the sulphuric acid as i thought it was H2SO4 but a lot of sites have it as SO4 or HSO4.

• – Simon-Nail-It May 6 '19 at 12:50
• Remember that both $\ce{SO4^{2-}}$ and $\ce{HSO4^{-}}$ are from $\ce{H2SO4}$ – Simon-Nail-It May 6 '19 at 12:52


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# Call for Proposals: 2022 HSCI Cancer Program Pilot Grants

February 2, 2022

AWARD AMOUNT
Up to 100,000 for entire project period Project period not to exceed 2 years POSTED DATE Wednesday, February 2, 2022 DEADLINE FOR APPLICATION Friday, March 4, 2022 ANTICIPATED AWARD DATE April 1, 2022 The Harvard Stem Cell Institute (HSCI) Cancer Program invites applications for pilot grants to be awarded for research in 2022. The intent is to provide resources for innovative projects dissecting stem/developmental programs in solid tumors that would help foster collaborations within the HSCI research community. This funding is for innovative, high-impact research projects. Multi-investigator, multi-institution proposals are encouraged; of specific interest to this funding cycle are proposals that dissect mechanisms that enable malignant stem cells to evade the immune system. Additionally, the applications should promote the Program mission: To identify critical genes and pathways that sufficiently distinguish cancer from normal stem cells and hence may serve as candidate targets for therapy. Special consideration will be given to proposals that could lead to further funding, such as a program project grant. The HSCI Cancer Program intends to award up to two grants in response to this call for proposals, contingent upon the number and quality of the proposals received, as well as the availability of HSCI Cancer Program funds. All proposals will be reviewed by the HSCI Cancer Program and the Executive Committee. The HSCI reserves the right to not award any seed grants in this cycle. It is understood that any tool or reagent generated as part of an HSCI Cancer Program Pilot Grant will be available for sharing with the wider HSCI community. Any pilot grants awarded during this cycle will be funded at up to100,000 in total costs (including indirect costs, maximum 20% of total direct costs-basis), for the total project period not to exceed two years. For example, $100,000 for a one-year project, or$50,000 each year for a two-year project.

Application Due Date: Friday, March 4, 2022

## Eligibility criteria

All principal investigators working in the field of cancer, stem cell, and cellular therapy  and affiliated with Harvard, are welcome to apply.  Junior investigators are especially encouraged to apply. In applications from two collaborating labs, at least one of the labs must be affiliated with the HSCI and both labs must be affiliated with Harvard. In order to be awarded an HSCI pilot grant, both investigators must meet all of the following criteria (as verified by a research/finance administrator at each investigator’s home institution):

• Have Principal Investigator status at their home institution
• Have independent laboratory space assigned to them in their own name (i.e. not assigned space via another investigator)
• Have independent funding sources in their own name (i.e. start-up funds or sponsored awards)
• Have any necessary approvals for proposed research in place no later than April 1, 2022.

## Application requirements and guidelines

All proposals submitted in response to this call for proposals must contain the following components, each of which must include the enumerated elements:

1. Proposal Title & Abstract (1 pg. max.)

1.1. Proposal title, PI(s) name(s), PI(s) home institution(s)

1.2. Abstract (lay-oriented language, 2,500 characters, incl. spaces, max.)

2. Research Proposal (3pgs. max., with pages numbered)

2.1. Background/Rationale

2.2. Specific research aims

2.3. Preliminary studies (if applicable)

2.4. Anticipated research milestones

• Pages listing references do NOT count against 3 pg. limit

3. Annual Budget Pages

3.1. Total annual costs may not exceed $100,000, including indirect costs for up to two years 3.2. Indirect costs may not exceed 20% of total direct costs 3.3. Submit separate budgets which show the amount of funding called for each PI, whether PIs have different home institutions or not • Equipment is not allowed on HSCI seed grants • Travel may not exceed$1k

• HSCI budget template strongly encouraged (template available upon request)

4. Budget Justification (1 pg. max.)

4.1. Justify all proposed budget items at the line-item level of detail

4.2. Describe effort committed by all personnel on budget, even if no salary called

4.3. PI is required to commit effort on the project

5. PI Biosketches (4 pgs. max. per biosketch)

5.1. NIH format

5.2. List recent financial support, indicate if directly applicable to the proposed project

5.3. May include up to 10 listings of recent/relevant publications

6. Letter of Institutional Review & Approval

6.1. Must be signed by an authorized institutional representative (as determined by investigator’s home institution’s policies for research proposal submission) as confirmation that the home institution has reviewed and approved the research and budget

• For a proposal in which PIs have different home institutions, a letter from each institution is required

7. Appendices (optional)

7.1. One copy each of no more than two relevant publications may be attached to the proposal

## Submission process

The 2022 HSCI Cancer Program Pilot Grant application can be found at: https://hsci.formstack.com/forms/2022_cancer_program_pilot

Applicants must complete the online application by the submission deadline.

• Upload all proposal documents in PDF format
• Proposal components should be ordered according to the order listed above
• Proposals must be submitted by 11:59 PM Eastern Standard Time on Friday, March 4, 2022.

Incomplete applications will not be considered.

Applications will be reviewed by the HSCI Cancer Program and Executive Committee in March, and all applicants will be notified of the outcome of the review process by end of March 2022. Successful applicants will be issued a pilot grant with a funding start date of April 1, 2022. Project period is for April 1, 2022 to March 31, 2022  for one year duration and April 1, 2022 to March 31, 2023  for two year durations (total funds available remain the same).

## Review criteria

The goal of the HSCI Cancer Program is to identify critical genes and pathways that sufficiently distinguish cancer from normal stem cells and hence may serve as candidate targets for therapy. Factors for reviewers’ consideration include: scientific quality, relevance to the HSCI Cancer Program mission, career development and recruitment/retention, value for money, potential to promote collaboration within the HSCI community, and regulatory issues.

The HSCI Executive Committee will evaluate proposals primarily for their potential to advance these goals. However, the HSCI will also consider the potential of the proposal to secure funding from other sources as-is, and/or the potential of the proposed research, if successful, to secure follow-on funding from public, philanthropic, or commercial sources.

## Inquiries

Questions about the HSCI Cancer Program may be addressed to Robert Perez, HSCI Grants Officer, at robert_perez@harvard.edu.

Thank you for your interest in the HSCI Cancer Program.

format: plain


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# For Y~B(11, 0.3) what is P(|Y-5| >= 3)?

I feel this is a simple problem, yet I cannot seem to solve it. Any help is greatly appreciated.

• What does "B(11,0.3)" mean? Would it refer to a Binomial distribution, a Beta distribution, or perhaps something else? (The nature of the question suggests it's not a Beta, but even so you should explain your notation.)
– whuber
Oct 22, 2014 at 14:28
• The binomial-tag strongly hints at $B$ meaning binomial, but it would still be better to explain the notation in the question. Oct 22, 2014 at 14:32

In this case, there are few enough possible values that $Y$ may take, that it is feasible to simply enumerate all possible values of $Y$ for which $|Y-5|\ge3$ and sum the respective probabilities.

• Note that to get a correct answer it is crucial to distinguish "$\ge$" (as stated in the question) from "$\gt$" (as given in this answer).
– whuber
Oct 22, 2014 at 14:29
• Whoops. Edited to match the question. Oct 22, 2014 at 14:31

Note that $P(|Y-5|\geq 3)=1-P(2<Y<8)=1-P(Y<8)+P(Y\leq2)$. Now you have $Y\sim B(11,0.3)$. Therefore we have $P(|Y-5|\geq 3)=1-F_{B(11,0.3)}(8)+F_{B(11,0.3)}(2)$. Where $F_{B(11,0.3)}$ is the CDF of $B(11,0.3)$. Let me know if this helps.

• Small R code assuming that $B(11,0.3)$ means binomial with 11 trials and success probability $0.3$. 1-(dbinom(8,11,0.3)+dbinom(3,11,0.3)) Oct 22, 2014 at 16:08

Since $Y \sim B(n,p)$ is a discrete variable with distribution $\binom{n}{k}p^k(1-p)^{n-k}$ for $k \in \{0,\ldots,n\}$ you have that $$P( | Y-5| \geq 3) = 1 - \sum_{k=3}^7 P(Y = k) = 1 - \sum_{k=3}^7 \binom{n}{k}p^k(1-p)^{n-k}.$$ Computing the above with $p=0.3$ and $n=11$ you get $$P( | Y-5| \geq 3) \approx 0.317$$

Code in matlab/octave for the above computation is:

k = [3:7]; p = 0.3; n = 11;
1 - sum(bincoeff(n,k).*p.^k.*(1-p).^(n-k))

• Welcome to Cross Validated! Please note this is a self-study question. The other answers are deliberately incomplete. Oct 22, 2014 at 12:48


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# Generating hooked curve through “n” number of points with python

It is possible to generate a curve in python through n number of points and the curve to be hooked to the points? Extrapolating I want to be able to generate curve through center of cubes curve that will stay hooked with the cubes so any update to the cubes will update the curve.

import bpy
from mathutils import Vector
cubes = [ob for ob in bpy.context.scene.objects if ob.name.startswith('Cube')]
spline = bpy.context.active_object
length = len(spline.data.splines[0].points)
for i in range(length):
bpy.ops.curve.select_all(action='DESELECT')
point = spline.data.splines[0].points[i]
cube = cubes[i]
point.select = True
point.co = Vector((cube.location.x, cube.location.y, cube.location.z, 1))
cube.select_set(True)


text:
# Multiplicity function for tensor powers of modules of the An algebra

@article{Kulish2012MultiplicityFF,
title={Multiplicity function for tensor powers of modules of the An algebra},
author={Petr P. Kulish and Vladimir D. Lyakhovsky and O. V. Postnova},
journal={Theoretical and Mathematical Physics},
year={2012},
volume={171},
pages={666-674}
}
We consider the decomposition of the pth tensor power of the module $$L^{\omega _1 }$$ over the algebra An into irreducible modules, $$(L^{\omega _1 } )^{ \otimes p} = \sum\nolimits_v {m(v,p)L^v }$$. This problem occurs, for example, in finding the spectrum of an invariant Hamiltonian of a spin chain with p nodes. To solve the problem, we propose using the Weyl symmetry properties. For constructing the coefficients m(ν, p) as functions of p, we develop an algorithm applicable to powers of an… CONTINUE READING

#### Citations

##### Publications citing this paper.
SHOWING 1-6 OF 6 CITATIONS

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CITES BACKGROUND

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CITES BACKGROUND


text:
## Feb. 16, 2015--Feb. 20, 2015

Invited Speakers:

Dima Arinkin (Winsconsin-Madison), Pierre Baumann (Strasbourg), Gwyn Bellamy (Glasgow), Nicolas Guay (Alberta), Tatsuyuki Hikita (Kyoto), Daniel Juteau (Caen), Tina Kanstrup (Aarhus), Yoshiyuki Kimura (RIMS), Ryosuke Kodera (RIMS), Martina Lanini (RIMS, Erlangen), Kevin McGerty (Oxford), Anne Moreau (Poitiers), Katsuyuki Naoi (Tokyo University of Agriculture), Alexei Oblomkov (Massachusetts), Yoshiki Oshima (IPMU), Simon Riche (Clermont-Ferrand II), Alexey Sevastyanov (Aberdeen), Shunsuke Tsuchioka (Tokyo), Shintaro Yanagida (RIMS), Kentaro Wada (Sinsyu)

Place: Irako View Hotel (map, another map in Japanese), tel: 0531-35-6111

How to go: We will pick up invited speakers who are staying at the Comfort Hotel Chubu International Airport at the hotel in the morning of February 16.
If you are coming on your own there will be a free bus from West Exist of Toyohashi station to Irako View Hotel at 12:40 on Monday, February 16. It will take about 2 hours from Toyohashi station to the hotel. You can also take a public bus from # 1 bust stop of East Exist of Toyohashi station to Irako Misaki (the final bus stop), see the table table for weekdays (only in Japansese). It will cost ¥1,490. They will pick you up if you call the hotel from the final bus stop, Irako Misaki (from 8:00 to 18:30 only).
For returen there will be a free bus from the hotel to Toyohashi station at 14:00 on Friday, February 20.

Equipment: whitebords, projector

Program :

15:00-15:30 15:30--16:30 16:45--17:45 18:00--19:00 19:00-19:30
Mon, Feb. 16 registration Bellamy Yanagida McGerty Free discussion

9:00--10:00 10:15--10:45 11:00--12:00 12:00-13:00 13:00-14:30 14:30--15:00 15:15--16:15 16:30--17:00 17:00-19:00
Tue, Feb. 17 Baumann Tsuchioka Riche Lunch Free discussion Lanini Hikita Oshima Free discussion
Wed, Feb. 18 Sevastyanov Naoi Moreau Lunch Free discussion Free discussion Free discussion Free discussion Free discussion
Thu, Feb. 19 Oblomkov Kimura Arinkin Lunch Free discussion Kodera Guay Wada Free discussion

9:00--10:00 10:00--10:30 10:30--11:30 11:30--12:30 12:30--14:00
Fri, Feb. 20 Kanstrup break Juteau Lunch Free discussion

Talks

Dima Arinkin, Smooth categorical representations of reductive groups.

Let G be a complex reductive group. Consider categories equipped with smooth (sometimes called strong) action of G. Natural and important examples of such categories arise from geometry: if X is a variety equipped with an action of G (for instance, X=G itself, or X=G/B is the flag space of G), then the category of D-modules on X carries a smooth action of G. We view such categories as (smooth) categorical representations of G.
The theory of smooth categorical representations of G is similar to the representation theory of a reductive group over a finite field, I will discuss this similarity in my talk. The surprising twist (and the main result of this talk) is that the theory of smooth categorical representations is simpler than its classical counterpart: there are no cuspidal representations!

Pierre Baumann, Mirković-Vilonen bases and tensor products.

(This is work in progress joint with Gaussent and Littelmann.)
Let G be a connected reductive algebraic group over the field of complex numbers. The geometric Satake correspondence (due to Lusztig, Beilinson-Drinfeld, Ginzburg, and Mirković-Vilonen) provides a tensor equivalence between a category of perverse sheaves on the affine Grassmannian $Gr$ of the Langlands dual of $G$, endowed with a convolution product, and the category of finite-dimensional representations of $G$. Specifically, given a dominant weight $\lambda$ for $G$, the intersection homology of the closure of the Schubert cell $Gr^\lambda\subset Gr$ can be endowed with the structure of a $G$-module of highest weight $\lambda$. Further, one can define cycles in $\overline{Gr^\lambda}$, the so-called Mirković-Vilonen cycles, whose fundamental classes form a basis in the intersection homology. A similar construction allows to construct geometrically the tensor products of simple $G$-modules: the Schubert varieties $\overline{Gr^\lambda}$ must be replaced by convolution varieties and the Mirković-Vilonen cycles must be replaced by cycles defined by Goncharov and Shen. We will study the relationship between the basis formed by Goncharov and Shen's cycles and the basis given by tensor products of Mirković-Vilonen cycles.

Gwyn Bellamy, Counting resolutions of symplectic quotient singularities.

If V is a symplectic vector space and G a finite subgroup of Sp(V), then the quotient singularity V/G is a very interesting object to study, both from the geometric and representation-theoretic point of view. One of the motivational problems in trying to understand the singularities of V/G is that of deciding whether V/G admits a symplectic resolution or not. More generally, one can ask how many symplectic resolutions it admits. The goal of this talk is to explain how one can count the number of symplectic resolutions of V/G. I'll present an explicit formula for this number in terms of the dimension of a certain Orlik-Solomon algebra. The key to deriving this formula is to relate the resolutions of V/G to the Calogero-Moser deformations, where one can use the representation theory of symplectic reflection algebras.

Nicolas Guay, Deformed double current algebras associated to complex semisimple Lie algebras.

Almost ten years ago, deformed double current algebras for $\mathfrak{sl}_n$ appeared in the study of Schur-Weyl duality for rational Cherednik algebras of the symmetric group. For an arbitrary complex semisimple Lie algebra, they were defined a couple of years later as degenerate forms of affine Yangians. They have appeared recently in on-going work of V. Toledano-Laredo and Y. Yang on the elliptic Casimir connection. I will give an overview of a few results known about deformed double current algebras and I will discuss some possible directions for studying their representations, based in part on known results about quantum toroidal algebras and Yangians.

Tatsuyuki Hikita, An algebro-geometric realization of the cohomology ring of Hilbert scheme of points in the affine plane.

We show that the cohomology ring of Hilbert scheme of $n$-points in the affine plane is isomorphic to the coordinate ring of $\mathbb{G}_{m}$-fixed point scheme of the $n$-th symmetric product of $\mathbb{C}^{2}$ for a natural $\mathbb{G}_{m}$-action on it. This result can be seen as an analogue of a theorem of DeConcini, Procesi and Tanisaki on a description of the cohomology ring of Springer fiber of type A.

Daniel Juteau, Parabolic degeneration of rational Cherednik algebras.

This is a report on joint work with Armin Gusenbauer, Stephen Griffeth, and Martina Lanini. We give necessary conditions for finite dimensionality of a simple lowest weight module for the rational Cherednik algebra of a complex reflection group, and for the existence of a non-zero map between standard modules. The latter criterion involves a generalization of the combinatorics of standard Young tableaux to arbitrary complex reflections groups. This uses a parabolic degeneration procedure for Cherednik algebra modules, alternative to Bezrukavnikov-Etingof parabolic restriction.

Tina Kanstrup, Quasi-coherent Hecke categories and affine braid group actions.

We propose a geometric setting leading to categorical braid group actions.
First we consider the quasi-coherent Hecke category QCHecke(G,B) for a reductive group G with a Borel subgroup B. We show that a monoidal action of QCHecke(G,B) on a triangulated category gives rise to a categorification of degenerate Hecke algebra representation known as Demazure Descent Data.
Next we replace the group G by the derived group scheme LG of topological loops with values in G and consider QCHecke(LG,LB). A monoidal action of the category QCHecke(LG,LB) gives rise to a categorical action of the affine Braid group.
Trying to avoid heavy derived algebraic geometry methods, we present an example of the construction above in classical algebro-geometric terms. For a G-variety $X$, we construct a Braid group action on a category of equivariant matrix factorizations on the product of $T^*X$ and the Grothendieck variety for the Lie algebra of G. The potential for matrix factorizations is provided by the moment map.

Yoshiyuki Kimura, Remarks on quantum unipotent subgroups and the dual canonical basis.

Quantum unipotent subgroup is the quantum coordinate ring of the unipotent subgroup associated with a Weyl group element of a symmetrizable Kac-Moody group. By the works of Geiss-Leclerc-Schroer and Goodearl-Yakimov, it is known that it has a quantum cluster algebra structure. In this talk, I explain the opposite pro-unipotent subgroup associated with a given Weyl group element and its compatibility with the dual canonical basis and the multiplicative property between the dual canonical basis element in the quantum unipotent subgroups and the one in the opposite.

Ryosuke Kodera, Affine Yangian action on the Fock space.

We compare actions of Yangians of finite and affine type A on the Fock space. The former is constructed by Uglov using Drinfeld correspondence, while the latter is by Varagnolo via quiver variety.

Martina Lanini, Degenerate flags and Schubert varieties.

Introduced in 2010 by E. Feigin, degenerate flag varieties are degenerations of flag manifolds. It has been proven that, in type A and C, they share many properties with Schubert varieties. In this talk I will discuss joint work with Cerulli Irelli, where we prove a surprising fact about degenerate flags.

Kevin McGerty, KN stratifications, D-modules and a modified hyper-Kahler Kirwan map.

We will discuss a recollement for D-modules on stacks equipped with a certain kind of stratification which arises naturally in GIT. Applications to the cohomology of hyper-Kahler quotients such as quiver varieties will be given. This is joint work with Tom Nevins.

Anne Moreau, Motivic stringy invariants for spherical varieties and applications.

In this talk, we will be interested in the motivic integral over the arc space of a complex Q-Gorensein spherical G-variety X where G is a reductive connected group. We give a formula for the stringy E-function of X in term of its colored fan. As applications, we obtain a smoothness criterion for locally factorial horospherical varieties and we conjecture that this criterion still holds for any locally factorial spherical variety. We also compute the Poincaré series of the weighted graded Stanley-Reisner ring for a Q-Gorenstein horospherical variety $X$, which allows to describe the cohomology ring $H^*(X,C)$ is some particular cases. All this is based on joint (past and current) works with Victor Batyrev.

Katsuyuki Naoi, Minimal affinizations and their graded limits.

Minimal affinizations are a class of finite-dimensional simple modules over a quantum loop algebra. In this talk, I will introduce a Jacobi-Trudi type formula for the characters of minimal affinizations in classical types. This formula is proved by studying the graded limits (variants of classical limits) of minimal affinizations. If time permits, I would also like to give a sketch of the proof.

Alexei Oblomkov, Cohomology of the homogeneous Hitchin fibers and representation theory of Cherednik algebras.

The talk is based on the joint work with Z. Yun.
I will explain a geometric construction of the action of the rational Cherednik algebra on the cohomology of the homogeneous Hitchin fiber for the Hitchin system on the weighted projective line. The perverse and cohomological filtrations on the cohomology have a natural representation theoretic interpretation which will be explained. In type $A$ case, the ring structure of the cohomology will be presented.

Yoshiki Oshima, Howe duality and holomorphic representations.

The restriction of the oscillator representation of the metaplectic group to a reductive dual pair gives a correspondence between representations of different reductive groups, which is called the Howe duality. We study the representations of real reductive groups which correspond to holomorphic representations by the Howe duality using D-module realization.

Simon Riche, Exotic tilting sheaves, parity sheaves on affine Grassmannians, and the Mirkovic-Vilonen conjecture.

In this talk I'll explain the construction (obtained in a joint work with Carl Mautner) of an equivalence of categories relating tilting objects in the heart of Bezrukavnikov's "exotic t-structure" on the derived category of equivariant coherent sheaves on the Springer resolution of a connected reductive group over a field k of good characteristic, and Iwahori-constructible parity sheaves on the affine Grassmannian Gr of the Langlands dual group, with coefficients in k. This construction allows in particular to obtain a modular version of an equivalence due to Arkhipov-Bezrukavnikov-Ginzburg in characteristic 0, and to prove the last missing piece for the proof of the Mirkovic-Vilonen conjecture on stalks of standard spherical perverse sheaves on Gr in the expected generality.

Alexey Sevastyanov, A proof of De Concini-Kac-Procesi conjecture and Lusztig's partition.

In 1992 De Concini, Kac and Procesi observed that isomorphism classes of irreducible representations of a quantum group at odd primitive root of unity m are parameterized by conjugacy classes in the corresponding algebraic group G. They also conjectured that the dimensions of irreducible representations corresponding to a given conjugacy class $O$ are divisible by $m^{1/2 dim O}$. In this talk I shall outline a proof of an improved version of this conjecture and derive some important consequences of it related to q-W algebras.

A key ingredient of the proof are transversal slices S to the set of conjugacy classes in G. Namely, for every conjugacy class O in G one can find a special transversal slice S such that O intersects S and dim O=codim S. The construction of the slice utilizes some new combinatorics related to invariant planes for the action of Weyl group elements in the real reflection representation. The condition dim O=codim S is checked using some new mysterious results by Lusztig on intersection of conjugacy classes in algebraic groups with Bruhat cells.

Shunsuke Tsuchioka, On graded Cartan invariants of symmetric groups and Hecke algebras.

We consider graded Cartan matrices of the symmetric groups and the Iwahori-Hecke algebras. These matrices may also be interpreted as Gram matrices of the Shapovalov form on sums of weight spaces of a basic representation of an affine quantum group. We present a conjecture predicting the invariant factors of these matrices and proving that it gives the correct invariants when one specializes or localizes the ring $\mathbb{Z}[v,v^{-1}]$ in certain ways. This generalizes Evseev's theorem which settled affirmatively the K\"ulshammer-Olsson-Robinson conjecture that predicts the Cartan invariants of the Iwahori-Hecke algebras. This is a joint work with Anton Evseev.

Shintaro Yanagida, Quantum toroidal algebras and motivic Hall algebras of elliptic singular fibers.

We consider the Ringel-Hall algebra of coherent sheaves over the trees or cycles of projective lines. The Drinfeld double of the composition subalgebra of this Hall algebra inherits automorphisms induced from equivalences of the associated derived category. We construct non-trivial automoprhisms of the Drinfeld double in the case of Kodaira singular fibers of elliptic surfaces using the relative Fourier-Mukai transforms and the motivic version of the Hall algebra. These non-trivial automorphisms coincide with the one constructed by Miki for the quantum toroidal algebras in the case of type A.

Kentaro Wada, New realization of cyclotomic q-Schur algebras.

A cyclotomic $q$-Schur algebra $\mathscr{S}_{n,r}$, which has parameters $q, Q_0,Q_1,\dots,Q_{r-1}$, is a quasi-hereditary cover of the cyclotomic Hecke algebra of type $G(r,1,n)$ (i.e. Ariki-Koike algebra) in the sense of Rouquier. In the case where $r=1$, $\mathscr{S}_{n,1}$ is a classical $q$-Schur algebra, and it is known that $\mathscr{S}_{n,1}$ is a quotient of the quantum group $U_q (\mathfrak{gl}_m)$. However, in the case where $r>1$, $\mathscr{S}_{n,r}$ is not a quotient of a quantum group.

In this talk, we introduce a Lie algebra $\mathfrak{g}$ with parameters $Q_1,\dots, Q_{r-1}$ and an associative algebra $\mathcal{U}$ with parameters $q, Q_1,\dots,Q_{r-1}$. Then, $\mathscr{S}_{n,r}$, in the case where $q$ is generic (resp. $q=1$), is realized as a quotient of $\mathcal{U}$(resp. $U(\mathfrak{g})$). We see that $\mathfrak{g}$ is a deformation of the current Lie algebra of $\mathfrak{gl}_m$, and we see that $\mathcal{U}$ is "a $q$-analogue" of $U(\mathfrak{g})$. We also discuss some properties of them.

Acknowledgments: supported by RIMS Project 2014 Geometric Representation Theory.

Organizing committee: Hiraku Nakajima, Syu Kato, Tomoyuki Arakawa

Contact: arakawa at kurims.kyoto-u.ac.jp


text:
# Problem to solve simultaneous equations

Posted 14 days ago
247 Views
|
4 Replies
|
2 Total Likes
|
 Hi, I have problems to solve simultaneous equations! The NSolve command also does not lead to a solution. Mathematica runs the whole night, but does not give any solutions. Have anybody an idea? Thanks for your help! Attachments:
4 Replies
Sort By:
Posted 14 days ago
 After some simplifications by hand I get the expression below which one wants to set to 0 to solve for f. 16 Ds f^3 n p W^2 - k^3 \[Pi]^2 Q Coth[(H k \[Pi])/(4 f W)] Sec[\[CurlyPhi]] Without numeric values for the various parameters this is unlikely to have a "useful" solution.
Posted 13 days ago
 Thank you for your help first. Is there no general solution for this system of equations without using values for the parameters? Even if I use values for the parameters and set the equation 0, the Solve command does not get the f out of the Coth. Attachments:
 I look carefully at this Plot Plot[16*80*f^3*3.67*1*108.29^2-1^3/\[Pi]^2*210890* Coth[(20*1/\[Pi])/(4*f*108.29)]*Sec[0.445457],{f,-.2,.2}] to estimate locations of solutions.I then try FindRoot[16*80*f^3*3.67*1*108.29^2-1^3/\[Pi]^2*210890* Coth[(20*1/\[Pi])/(4*f*108.29)]*Sec[0.445457]==0,{f,0.175}] which returns {f->0.171224}


text:
Internet ad revenue may have reached $17 billion in the first half of the year, according to the Interactive Advertising Bureau, but the rate of growth declined between 2011 and 2012. In a Thursday report, the industry association said online ad revenue in 2012 climbed 14 percent, from$14.9 billion in the first six months of 2011. But between the first half 2010 and the first half of 2011, revenue increased 23 percent, indicating softening growth.

When asked about the slowdown, research executives at the IAB and PwC (which conducted the study for the association) mostly attributed it to the overall macroeconomic picture. In 2010, the economy was particularly weak, making growth in 2011 look especially strong. And uncertainty in the current economy could be dragging down spending this year.

David Silverman, a partner at PwC US, also said the appearance of a declining rate of growth could be a function of the growing industry.

“As your base gets bigger, your dollar growth can be higher but as a percentage it looks smaller. That’s just going to be a natural phenomenon that will happen over time as the base continues to grow,” he said.

According to the report, mobile continues to be a big gainer, with spending nearly doubling from $636 million in the first half of 2011 to$1.2 billion in the same period this year.  But though it’s seen as a new revenue source and diver of overall growth it remains a relatively small piece (7 percent) of the overall online ad spending pie. Among all categories of online advertising, search leads with 48 percent of the spending ($8.1 billion), followed by display with 33 percent ($5.6 billion).

Rich media saw a 32 percent decrease, the researchers said, but explained that spending in that category is moving to digital video, which rose 18 percent but remained flat year over year.

Another interesting finding from the IAB report was that brand dollars are moving online at a slightly slower pace compared to previous half-year reports. But Sherill Mane, the IAB’s SVP for research, analytics and measurement, said that while performance-based advertising (in which payment is based on a direct response) seems to be currently outpacing brand advertising (which is based on impressions), dollars for the latter type of advertising are continuing to come online.

While the IAB continues to tout the health of the online ad market – and cite the strength of its growth relative to that of other mediums – the majority of ad dollars are still spent offline. Online publishers and platforms are making headway with new ad formats that deliver increasingly premium experiences, but they’re just beginning to show that they can offer measurable results.

As for advertiser categories, the report said spending by pharma and healthcare was up 81 percent to \$1.1 billion in 2012 and automotive climbed 29 percent. Consumer packaged goods grew by just 4 percent but though that number seems small relative to the big gains in other categories, the researchers said it’s a solid number for a tough economy.

Image by  Tribalium via Shutterstock.


text:
# Conditional probability, uniform distributions

Problem: Suppose that the random variable X is uniformly distributed symmetrically around zero, but in such a way that the parameter is uniform on (0, 1); that is, suppose that $X\mid (A=a) \sim U(-a,a)$ with $A \sim U(0,1)$. Find the distribution of X.

My attemt: $f_{A,X}(a,x)=f_{X\mid A=a}(x) \cdot f_{A} (a)$.

$f_{A}(a)=1$ for $0<a<1$ and $f_{X,A=a}(x)=\frac{1}{2a}$ for $-a<x<a$ which means that $f_{A,X}(a,x)=\frac{1}{2a}$ for $0<a<1$ and $-a<x<a$. Next I integrate the joint distribution over a to find $f_{X}(x)$.

$f_{X}(x)=\int_0^1 f_{A,X}(a,x) da=\int_0^1 \frac{1}{2a}da$, but this integral is not convergent. Where do I go wrong?

Thanks.

-
I think the standard symbol where you use $\in$ would be $\sim$. –  joriki Mar 19 '13 at 2:04

You were correct up until setting up the integral for the marginal density: $$f_X(x) = \int_0^1 f_{A,X}(a,x) \mathrm{d}a = \int_0^1 \frac{1}{2a} [-a<x<a] \mathrm{d}x$$ For a given $-1<x<1$, the indicator function $[-a<x<a]$ is non-zero for $|x|<a<1$, therefore $$f_X(x) = \int_{|x|}^a \frac{\mathrm{d}a}{2a} = \frac{1}{2} \ln \frac{1}{|x|}$$


text:
# Binary to Decimal, Hex & Octal Conversion

Binary Value
Decimal  =  10
Hexa Decimal  =  A
Octal  =  12
GENERATE WORK

## Binary to Decimal, Hex & Octal - work with steps

Input Data :
Binary Value = 1010

Objective :
Find what is hex, decimal & octal Value from given binary value?

Solution :
Binary to Decimal
1010_2 = (1 \times 2^3) + (0 \times 2^2) + (1 \times 2^1) + (0 \times 2^0)

1010_2 = 8 + 0 + 2 + 0

1010_2 = 10_10
Binary to Hexa Decimal
Split the binary number from left to right each group 4 bits
1010
A

1010_2 = A_16

Binary to Octal
Split the binary number from left to right each group 3 bits
001010
12

1010_2 = 12_8

Binary to Decimal, Hex & Octal Converter - step by step conversion & solved example problems to perform binaty decimal, binary to hexadecimal & binary to octal conversion. The binary number system is the most influential & popular number system in the context of digital computations, electronics engineering & tele communications etc. It's also called base-2 number system only uses 0 and 1 to represent any sort of information. Generally 0 represents LOW or OFF state and 1 represents HIGH or ON state. In the context of digital computations, it's frequently required to convert binary numbers into other popular representations like in decimal, hexadecimal & octal number systems, where this Binary to Decimal, Hex & Octal Converter can be employed to perform such conversions easily. Besides, the step by step calculation along with solved example problem let the users easily understand how manually perform such conversions.

## Binary to Decimal Conversion

The binary number system is not the usual number system to easily understand & to read for humans, therefore, it is often required to convert the binary number into its equivalent decimal number to make it easily readable. The below step by step conversion & solved example let the users to understand how to convert binary to decimal number. The rightmost digit of the binary number has the weightage of 20 and the power of 2 will increase by 1 for each successive digit from right to left (see the solved example below). It's also called as the place value of binary digits. The sum of products of binary digits & place value provides its equivalent decimal value.

Step by step conversion:
step 1: Multiply the binary digit with place value for each digit. For example, for the binary number 1010, it would be like 0 x 20 = 0, 1 x 21 = 1, 0 x 22
= 0 & 1 x 23 = 8.
step 2: Sum all the product values provides an equivalent decimal.

Solved Example:
Convert the binary 100101 to its decimal equivalent

[(1) x 25] + [(0) x 24] + [(0) x 23] + [(1) x 22] + [(0) x 21] + [(1) x 20]
[1 x 32] + [0 x 16] + [0 x 8] + [1 x 4] + [0 x 2] + [1 x 1] = 37
The decimal equivalent for binary 100101 is 37.

Binary Fraction to Decimal Conversion
Solved Example:
The below solved example let the users to know how to convert fractional binary number 10101.0101 to its equivalent decimal form
[1 x 24] + [0 x 23] + [1 x 22] + [0 x 21] + [1 x 20] . 0 x 2-1 + 1 x 2-2 + 0 x 2-3 + 1 x 2-4
(16 + 0 + 4 + 0 + 1) . (0 + 0.25 + 0 + 0.0625)
10101.01012 = 21.3125

### Binary to Hex Conversion

Binary to Hex conversion can be done by divide the bits into groups from right to left side, each containing 4 bits. If the group is lack of 4 bits then add 0 or 0s to the left hand side to make sure each group containing 4 bits. The extra bits of 0 at the left side are called padding. The below solved example let the users to understand how to convert binary to decimal number.

Step by step conversion:
step 1: Split the given binary number into groups from right, each containing 4 bits.
step 2: Add 0 or 0s to the left side if any group is lack of 4 bits.
step 3: Find the Hex equivalent for each group.
step 4: Form the each group Hex number together in the same order.

Solved Example:
Convert the binary 1011010 to Hex number

10100102 => 0101 1010 grouped with padding
=> 5 A
= 5A16

### Binary to Octal Conversion

Binary to Octal conversion can be done by divide the bits into groups from right to left side, each containing 3 bits. If the group is lack of 3 bits then add 0 or 0s to the left hand side to make sure each group containing 3 bits. The extra bits of 0 at the left side are called padding. The below solved example let the users to understand how to convert binary to octal number.

Step by step conversion:
step 1: Split the given binary number into groups from right, each containing 3 bits.
step 2: Add 0 or 0s to the left side if any group is lack of 3 bits.
step 3: Find the Octal equivalent for each group.
step 4: Form the each group Octal number together in the same order.

Solved Example:
Convert the binary 10110 to Octal number

101102 => 010 110 grouped with padding
=> 2 6
101102 = 268
In number conversion, is being used in various general & digital applications, therefore, sometimes it's important to convert binary numbers into various representations & number systems. Here, the above step by step calculation & solved examples may useful for users to understand how the values are being used in binary conversions, however, when it comes to online for quick calculations, this binary to decimal, hex & octal converter helps the user to perform & verify such calculations as quick & easy as possible.


text:
+3 votes
395 views

For each element in a set of size $2n$, an unbiased coin is tossed. The $2n$ coin tosses are independent. An element is chosen if the corresponding coin toss was a head. The probability that exactly $n$ elements are chosen is

1. $\frac{^{2n}\mathrm{C}_n}{4^n}$
2. $\frac{^{2n}\mathrm{C}_n}{2^n}$
3. $\frac{1}{^{2n}\mathrm{C}_n}$
4. $\frac{1}{2}$
asked | 395 views

## 3 Answers

+9 votes
Best answer

answer - A

ways of getting n heads out of 2n tries = 2nCn

probability of getting exactly n heads and n tails = (1/2n)(1/2n)

number of ways = 2nCn/4n

answered by Boss (8.9k points)
selected by
what if we do as below:
no. of ways to get n heads = 2nCn
no. of total outcome = 2n
therefore, probability of getting exactly n heads out of 2n tosses = 2nCn /2n

what does it calculate?
How is it saying n heads out of 2n?
+1 vote

Answer :Option A

Here is the link for theory- http://stattrek.com/probability-distributions/binomial.aspx

answered by Active (1.6k points)
0 votes

Required Probability=$\frac{No. of Favourable Ways }{Total no. of Ways}$

No. of favourable ways= 2nC (bcoz select exactly n heads out of 2n tosses )

Total no. of ways= 2n tosses and each have 2 possibilities either H or T so total=$2^{2n}$=$4^n$ possibilities

So Ans is $\frac{\binom{2n}{n} }{4^{n}}$  which fits option A.

answered by Veteran (14.6k points)
edited
Answer:

+5 votes
1 answer
1
+14 votes
3 answers
2
+4 votes
2 answers
3


text:
# Publications by Katherine Blundell

## Uncovering the orbital dynamics of stars hidden inside their powerful winds: application to $η$ Carinae and RMC 140

Monthly Notices of the Royal Astronomical Society Oxford University Press 494 (2020) 17-35

D Grant, K Blundell, J Matthews

Determining accurate orbits of binary stars with powerful winds is challenging. The dense outflows increase the effective photospheric radius, precluding direct observation of the Keplerian motion; instead the observables are broad lines emitted over large radii in the stellar wind. Our analysis reveals strong, systematic discrepancies between the radial velocities extracted from different spectral lines: the more extended a line's emission region, the greater the departure from the true orbital motion. To overcome these challenges, we formulate a novel semi-analytical model which encapsulates both the star's orbital motion and the propagation of the wind. The model encodes the integrated velocity field of the out-flowing gas in terms of a convolution of past motion due to the finite flow speed of the wind. We test this model on two binary systems. (1), for the extreme case $\eta$ Carinae, in which the effects are most prominent, we are able to fit the model to 10 Balmer lines from H-alpha to H-kappa concurrently with a single set of orbital parameters: time of periastron $T_{0}=2454848$ (JD), eccentricity $e=0.91$, semi-amplitude $k=69$ km/s and longitude of periastron $\omega=241^\circ$. (2) for a more typical case, the Wolf-Rayet star in RMC 140, we demonstrate that for commonly used lines, such as He II and N III/IV/V, we expect deviations between the Keplerian orbit and the predicted radial velocities. Our study indicates that corrective modelling, such as presented here, is necessary in order to identify a consistent set of orbital parameters, independent of the emission line used, especially for future high accuracy work.

Show full publication list


text:
12.4 The regression equation  (Page 2/2)

 Page 2 / 2

$\epsilon$ = the Greek letter epsilon

For each data point, you can calculate the residuals or errors, ${y}_{i}-{\stackrel{^}{y}}_{i}={\epsilon }_{i}$ for $i=\text{1, 2, 3, ..., 11}$ .

Each $|\epsilon |$ is a vertical distance.

For the example about the third exam scores and the final exam scores for the 11 statistics students, there are 11 data points. Therefore, there are 11 $\epsilon$ values. If you square each $\epsilon$ and add, you get

$\left({\epsilon }_{1}{\right)}^{2}+\left({\epsilon }_{2}{\right)}^{2}+\text{...}+\left({\epsilon }_{11}{\right)}^{2}=\stackrel{11}{\underset{\text{i = 1}}{\Sigma }}{\epsilon }^{2}$

This is called the Sum of Squared Errors (SSE) .

Using calculus, you can determine the values of $a$ and $b$ that make the SSE a minimum. When you make the SSE a minimum, you have determined the points that are on the line of best fit. It turns out thatthe line of best fit has the equation:

$\stackrel{^}{y}=a+\text{bx}$

where $a=\overline{y}-b\cdot \overline{x}$ and $b=\frac{\Sigma \left(x-\overline{x}\right)\cdot \left(y-\overline{y}\right)}{{\Sigma \left(x-\overline{x}\right)}^{2}}$ .

$\overline{x}$ and $\overline{y}$ are the sample means of the $x$ values and the $y$ values, respectively. The best fit line always passes through the point $\left(\overline{x},\overline{y}\right)$ .

The slope $b$ can be written as $b=r\cdot \left(\frac{{s}_{y}}{{s}_{x}}\right)$ where ${s}_{y}$ = the standard deviation of the $y$ values and ${s}_{x}$ = the standard deviation of the $x$ values. $r$ is the correlation coefficient which is discussed in the next section.

Least squares criteria for best fit

The process of fitting the best fit line is called linear regression . The idea behind finding the best fit line is based on the assumption that the data are scattered about a straight line. The criteria for the best fit line is that the sum of the squared errors (SSE) is minimized, that is made as small as possible. Any other line you might choose would have a higher SSE than the best fit line. This best fit line is called the least squares regression line .

Computer spreadsheets, statistical software, and many calculators can quickly calculate the best fit line and create the graphs. The calculations tend to be tedious if done by hand. Instructions to use the TI-83, TI-83+, and TI-84+ calculators to find the best fit line and create a scatterplot are shown at the end of this section.

Third exam vs final exam example:

The graph of the line of best fit for the third exam/final exam example is shown below:

The least squares regression line (best fit line) for the third exam/final exam example has the equation:

$\stackrel{^}{y}=-173.51+\text{4.83x}\phantom{\rule{20pt}{0ex}}$
• Remember, it is always important to plot a scatter diagram first. If the scatter plot indicates that there is a linear relationship betweenthe variables, then it is reasonable to use a best fit line to make predictions for $y$ given $x$ within the domain of $x$ -values in the sample data, but not necessarily for $x$ -values outside that domain.
• You could use the line to predict the final exam score for a student who earned a grade of 73 on the third exam.
• You should NOT use the line to predict the final exam score for a student who earned a grade of 50 on the third exam, because 50 is not within the domain of the x-values in the sample data, which are between 65 and 75.

Understanding slope

The slope of the line, b, describes how changes in the variables are related. It is important to interpret the slope of the line in the context of the situation represented by the data. You should be able to write a sentence interpreting the slope in plain English.

INTERPRETATION OF THE SLOPE: The slope of the best fit line tells us how the dependent variable (y) changes for every one unit increase in the independent (x) variable, on average.

Third exam vs final exam example

• Slope: The slope of the line is b = 4.83.
• Interpretation: For a one point increase in the score on the third exam, the final exam score increases by 4.83 points, on average.

Using the linear regression t test: linregttest

1. In the STAT list editor, enter the X data in list L1 and the Y data in list L2, paired so that the corresponding (x,y) values are next to each other in the lists. (If a particular pair of values is repeated, enter it as many times as it appears in the data.)
2. On the STAT TESTS menu, scroll down with the cursor to select the LinRegTTest. (Be careful to select LinRegTTest as some calculators may also have a different item called LinRegTInt.)
3. On the LinRegTTest input screen enter: Xlist: L1 ; Ylist: L2 ; Freq: 1
4. On the next line, at the prompt β or ρ, highlight "≠ 0" and press ENTER
5. Leave the line for "RegEq:" blank
6. Highlight Calculate and press ENTER.

The output screen contains a lot of information. For now we will focus on a few items from the output, and will return later to the other items.

• The second line says y=a+bx. Scroll down to find the values a=-173.513, and b=4.8273 ; the equation of the best fit line is $\stackrel{^}{y}=-173.51+\text{4.83}x\phantom{\rule{20pt}{0ex}}$
• The two items at the bottom are $r^{2}$ = .43969 and $r$ =.663. For now, just note where to find these values; we will discuss them in the next two sections.

Graphing the scatterplot and regression line

1. We are assuming your X data is already entered in list L1 and your Y data is in list L2
2. Press 2nd STATPLOT ENTER to use Plot 1
3. On the input screen for PLOT 1, highlight On and press ENTER
4. For TYPE: highlight the very first icon which is the scatterplot and press ENTER
5. Indicate Xlist: L1 and Ylist: L2
6. For Mark: it does not matter which symbol you highlight.
7. Press the ZOOM key and then the number 9 (for menu item "ZoomStat") ; the calculator will fit the window to the data
8. To graph the best fit line, press the "Y=" key and type the equation -173.5+4.83X into equation Y1. (The X key is immediately left of the STAT key). Press ZOOM 9 again to graph it.
9. Optional: If you want to change the viewing window, press the WINDOW key. Enter your desired window using Xmin, Xmax, Ymin, Ymax

**With contributions from Roberta Bloom

where we get a research paper on Nano chemistry....?
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
1 It is estimated that 30% of all drivers have some kind of medical aid in South Africa. What is the probability that in a sample of 10 drivers: 3.1.1 Exactly 4 will have a medical aid. (8) 3.1.2 At least 2 will have a medical aid. (8) 3.1.3 More than 9 will have a medical aid.


text:
# Checking a panel unit root test in R done manually

I have spent much time looking for a special package that could run the Pesaran(2007) unit root test (which assumes cross-sectional dependence unlike most others) and I have found none. So, I decided to do it manually; however, I don't know where I'm going wrong, because my results are very different from Microsoft Excel's results (in which it is done very easily).

My data frame is made up of 22 countries with 506 observations of daily price indices. Following is the model to run using the Pesaran(2007) unit root test:

(i) With an intercept only

where $\overline{Y}$ is the cross-section average of the observations across countries at each time $t$ and $b$ is the coefficient of interest to us because it will allow us to compute the ADF test statistic and then determine whether the process is stationary or not.

I constructed each of these variables in the following way:

(Delta)Y(t)

dif.yt = diff(yt)
## yt is the object containing all the observations for a specific country
## (e.g. Australia)

Y(t-1)

yt.lag.1 = lag(yt, -1)

Y(bar)(t-1)

ybar.lag.1 = lag(c(rowMeans(x)), -1)
## x is the object containing my entire data frame

(Delta)Y(bar)(t-1)

dif.ybar.lag.1 = diff(ybar.lag.1)

(Delta)Y(bar)(t-2)

dif.ybar.lag.2 = diff(lag(c(rowMeans(x)), -2))

(Delta)Y(t-1)

dif.yt.lag.1 = diff(yt.lag.1)

(Delta)Y(t-2)

dif.yt.lag.2 = diff(lag(yt, -2)

After constructing each variable individually, I then run the linear regression

reg = lm(dif.yt ~ yt.lag.1[-1] + ybar.lag.1[-1] + dif.ybar.lag.1 +
dif.ybar.lag.2 + dif.yt.lag.1 + dif.yt.lag.2)
summary(reg)

It is obvious that the explanatory variables in my regression equation differ in length, so I'd like to know whether there is a way in R to make all the variables of equal length (perhaps with a function).

Also, I'd like to know whether the procedure I used was correct and if there are more optimal ways.

-
From ?lm, a warning on using with time series: Even if the time series attributes are retained, they are not used to line up series, so that the time shift of a lagged or differenced regressor would be ignored. –  Matthew Lundberg Apr 3 '12 at 3:33
@EricD.Brean That's absolutely right! –  SavedByJESUS Apr 3 '12 at 3:43
@MatthewLundberg Alright then. I take it that I will always have to manually adjust the length of the series as in "yt.lag.1[-1]" in these cases. But do you have an idea why the coefficients estimates are so different? Thank you. –  SavedByJESUS Apr 3 '12 at 3:45
Looks to be the case, Saved. I'm curious as to the outcome. –  Matthew Lundberg Apr 3 '12 at 3:47

 library(dynlm)

#object class is a zoo or ts

reg =   dynlm(d(yt) ~ (L(yt, 1) + L(ybar,1) + d(L(ybar,1) + ....
data = ~yourdata, start = .... other args)

summary(reg)

More details about the package: http://cran.r-project.org/web/packages/dynlm/dynlm.pdf

-
Thank you so much! It was very well actually! I have a few questions though. Does it mean that "d()" is the same as "diff()"? And is "L(xt, 1)" the same as "lag(xt, -1)"? Thank you once again. –  SavedByJESUS Apr 3 '12 at 23:49
I actually have a table with all the critical values for this test. So, testing hypotheses won't be a problem once I know the formula to calculate the ADF t statistic (coef/se(coef)). Thank you very much! –  SavedByJESUS Apr 4 '12 at 1:10


text:
## What are bitcoin miners really solving?

152

91

As with mining, what are the bitcoin miners really solving? I read they are solving hashes, but what does that really mean. Can we see what they are solving? Can someone give an example of what a bitcoin mining machine sees to solve?

2

– Stephen Gornick – 2013-02-28T23:52:30.587

Ok but WHAT data are we mining! Nobody seems to know. I don't care how it works I want to know where the data is coming from that bit mining is decryption or encrypting. – None – 2013-12-05T20:18:23.387

44

They try to find a random nonce (a little random data) that goes into a block and makes the block have a (SHA256) hash that (in binary) starts with a certain amount of 0's. The more zeroes the more rare hash is. A good hash' outcome is not predictable, and so you have to try a lot of times to find a good nonce.

The amount of zeroes are based on how difficult it is supposed to be to find a block. In Bitcoin it adjusts to have a new block every 10 minutes (on average, given the rate at which previous blocks are found).

Interesting: because the hashes are unpredictable it doesn't matter how the nonce changes! Most of the time it's just a number counting upwards from 0!

152

Here is an extremely simplified sketch of the problem, but it should give a pretty good idea of what the problem is.

The data:

This is the hash of the lastest block (shortened to 30 characters):

00000000000001adf44c7d69767585

These are the hashes of a few valid transactions waiting for inclusion (shortened).

5572eca4dd4
db7d0c0b845

And this the hash of one special transaction that you just crafted, which gives 25BTC (the current reward) to yourself:

916d849af76

Building the next block:

Now, let's use a gross approximation of what a new block might look like (the real one uses binary format). It contains the hash of the previous block and the hashes of those 3 transactions:

00000000000001adf44c7d69767585--5572eca4dd4-db7d0c0b845-916d849af76--

Now let's do mining by hand! Our goal is to complete this block with a nonce (a piece of garbage) such that the hash of the new block starts with 13 zeros (considering the previous hash, it seems that 13 zeroes is the current difficulty!).

Mining (trying to finalize this block):

Let's try with nonce=1, and compute the hash of the block (I'm using the md5 hash algorithm, but Bitcoin uses double sha256):

> echo "00000000000001adf44c7d69767585--5572eca4dd4-db7d0c0b845-916d849af76--1" | md5sum
8b9b994dcf57f8f90194d82e234b72ac

No luck, the hash does not start with a 0… Let's try with nonce=2

> echo "00000000000001adf44c7d69767585--5572eca4dd4-db7d0c0b845-916d849af76--2" | md5sum
5b7ce5bcc07a2822f227fcae7792fd90

No luck…

If we pursue until nonce=16, we get our first leading zero.

> echo "00000000000001adf44c7d69767585--5572eca4dd4-db7d0c0b845-916d849af76--16" | md5sum
03b80c7a34b060b33dd8fbbece79cee3

For nonce=208, we get two leading zeroes!

> echo "00000000000001adf44c7d69767585--5572eca4dd4-db7d0c0b845-916d849af76--208" | md5sum
0055e55df5758517c9bed0981b52ce4a

Continue like this… If you finally find a hash that has 13 leading zeroes… you're a winner! Other miners will now build upon your block, you've just got 25BTC.

But you'll have to be fast!

Back to step 1…

If someone manages to build a block before you do, you'll have to start again from the beginning with the new block's hash (the one of the winner).

Note that you're not actually "counting zeroes". In reality approximately the first four non zero numbers are being compared with something called a "target". This is stored in compact form in each block in a field called "bits". – halfbit – 2013-03-01T18:50:18.043

9Also your example is conceptually heading in the right direction, but the real success is when the sha256 hash of the header is less than the target. Example target: 00000000000001ae00000000000000 is greater than 00000000000001adf44c7d69767585 <-- this would be a valid hash. – halfbit – 2013-03-01T18:54:20.690

"And this the hash of one special transaction that you just crafted and which gives 25BTC (the current reward) to yourself: 916d849af76" How is this transaction crafted? – Leonardo Marques – 2013-07-01T17:05:07.803

@Reonarudo: the simplest is to use one of your addresses to produce a single standard output to be spent, but you can forge any kind of custom transaction with custom scripts as well if it pleases you (you are allowed a +25 BTC total balance for this transaction, additionnal inputs were not allowed some time ago, but maybe they are now). – Stéphane Gimenez – 2013-07-01T20:33:13.513

So normally I would just append my wallet address in the end between - and --? – Leonardo Marques – 2013-07-01T20:35:40.780

3@Reonarudo, not exactly, this is a very simplified sketch of what is actally done. Transactions are made with scripts which are often made from adresses, you can find more info on the Bitcoin wiki. – Stéphane Gimenez – 2013-07-01T20:39:13.757

3This basically sounds like a nice (simplified) summary, but at what point does the bitcoin network accept this as the next valid block, and what happens in the (unlikely) event that two different miners managed to submit a valid block almost simultaneously? – Tobias Kienzler – 2013-12-02T09:13:02.080

1Wow, great answer! There're a couple of things I still couldn't get my head around though: Let's say you've been searching for a solution for 6 minutes and there comes a new transaction. 1) what happens now? if I have to start over, then that sounds like everyone has to start over. Then how can blockchain guarantee to find a block in ~10 minutes? 2) How does that transaction come to me anyway? 3) What if I find a solution for this block and then a new transaction comes to me from some other node who accepted it for the block I just closed? – keremispirli – 2014-04-28T17:24:18.810

2>

• Yes everyone starts over, if you build on top of an old block there's little chance your blocks will be part of the longer chain and they will be disregarded by the rest of the network. There is no guaranty that a block is find within 10 minutes, it's just an average statistical outcome. 2) What transaction? Blocks are broadcast by whoever mined them and relayed by other nodes. 3) Nothing forces you to include all pending transactions, but the more transactions you include the more fees you can collect.
• < – Stéphane Gimenez – 2014-04-30T10:47:50.153

start="2">

• The transaction to be added to current block, the one that caused me and all(?) others to restart the calculation. How does it come to me? And also, 3) Why would I accept it for 0.001 btc if I am to find a new block and get 25 btc?
• Thanks Stephane! – keremispirli – 2014-05-01T12:39:53.640

67

The following is a description of the global, statistical gamble which is played every 10 or so minutes. The interval of the game is controlled by the difficulty which says how many "hashes" are needed per interval.

In other words, the difficulty and target define the "odds of the house" against your chance of getting a winning SHA hash. The nonce is the "scratch off" area.

Each hash consumes electricity, and emits heat, which requires additional cooling.

This is what is done with each hash:

Step 1

At a high level, the miner software takes a list of active transactions, and then groups them together in something called a "block".

Or more accurately stated: The miner software coverts all the transactions into a summary view called a "merkle root", and hashes it, which is representative of the transactions.

Step 2

Then mining software converts this to into a binary format called a Block Header, which also references the previous blocks (also called a chain).

Field           Purpose                          Updated when...               Size (Bytes)
Version         Block version number             You upgrade the software and   4
it specifies a new version

hashPrevBlock   256-bit hash of the previous     A new block comes in          32
hashMerkleRoot  256-bit hash based on all        A transaction is accepted     32
the transactions in the block

Time            Current timestamp as seconds     Every few seconds              4
since 1970-01-01T00:00 UTC

Bits            Current target in compact format   The difficulty is adjusted   4

Nonce           32-bit number (starts at 0)       A hash is tried (increments)  4

Step 3:

The miner hardware changes a small portion of this block called a "nonce".

Step 4:

The block header is hashed and compared to the Target as if it were simply a large number like 10,000,000 > 7,000,000 (the real numbers are much bigger, and in hex). The target is compressed and stored in each block in a field called bits.

An expanded target looks like this:

  Target   0000000000000083ef00000000000000000000000000000000000000000000000

And the goal is to make sure the SHA256 hash of the block is less than this value. In the example below "83ee" is smaller than "83ef"

To simplify this concept, you can ballpark the target by counting the leading zeros (as the other answer here explains). Here is an example:

Here is a sample block with transactions you can view on BlockChain.info. Look in the upper right hand corner of the webpage for this hash:

   Hash 0000000000000083ee9371ddff055eed7f02348e4eda36c741a2fc62c85bc5cf

That previous hash was from today and has 14 leading zeroes. Let's compare that to what was needed 3 years ago with block 100 which has 8 leading zeros.

   Hash 00000000a8ed5e960dccdf309f2ee2132badcc9247755c32a4b7081422d51899

Summary

So at the end of the day, all a miner does is:

1. Take a block header as input
2. Change the Nonce
3. Test if the Block Header hash is less than the Target. If it is, you win.
4. Go to step 2 (or go to step 1 if someone else won the block)

Want to see what Bitcoin-QT does when it finds a block?... I posted it here.. The information in this post will help you understand what happened.

Isn't the nonce (32 bits) far too small to be able to find a hash that must start with 14 leading zeroes (14x4 = 56 bits) ? Finding such a hash would require on average 2^56 / 2 attempts and with a 32 bit nonce only 2^32 attempts are possible before all possible nonce values are exhausted. Edit: with the time changing every second, I guess this only applies if hash rate exceeds 4 GH/sec (which I think it does...) – john16384 – 2017-09-06T21:09:14.910

Yup, you're right, ...from the URL on the Block header: Whenever Nonce overflows (which it does frequently), the extraNonce portion of the generation transaction is incremented, which changes the Merkle root. – halfbit – 2017-09-06T22:54:39.623

13

Mining provides a way to reach consensus on what the transaction ledger should look like and know that nobody is cheating.

That's the non-technical definition of mining.

The "authority" for double spending is the blockchain. The blockchain consists of the history of all blocks in the blockchain plus the next block of transactions. The reward subsidy currently is 25 BTC to the party that submits the next block. But hey ...you would like that 25 BTC (worth currently about \$825) as would I as would everyone else. So how do you make it so that I can't cheat and claim the block myself?

Well, you put in a system that you and I have to compete. That's what the proof of work does -- it makes it so that when I claim the reward it is easy to prove that I really did the work involved. So for me to have a 2% chance of solving a block I need to put in 2% of of the mining work. There's no way for me to put in less than 2% of all the work and still solve blocks at least 2% of the time (on average).

Thus as a result, when a transaction block is submitted, all the peers verify that there were no double spends, that the right amount of subsidy was claimed, and that the submitter truly expended the work necessary for that solution. With those three rules, then there doesn't not need to be a central authority managing the process or able to control the outcome.


text:
# ML Work-Flow (Part 5) – Feature Preprocessing

We already discussed first four steps of ML work-flow. So far, we preprocessed crude data by DICTR (Discretization, Integration, Cleaning, Transformation, Reduction), then applied a way of feature extraction procedure to convert data into machine understandable representation, and finally divided data into different bunches like train and test sets . Now, it is time to preprocess feature values and make them ready for the state of art ML model ;).

We need Feature Preprocessing in order to:

1. Evade scale differences between dimensions.
2. Convey instances into a bounded region in the space.
3. Remove correlations between different dimensions.

1. Evading scale differences reduces unit differences between particular feature dimensions. Think about Age and Height of your customers. Age is scaled in years and Height is scaled in cm's. Therefore, these two dimension values are distributed in different manners. We need to resolve this and convert data into a scale invariant representation before training your ML algorithm, especially if you are using one of the linear models like Logistic Regression or SVM (Tree based models are more robust to scale differences).
2. Conveying instances into a bounded region in the space resolves the representation biases between instances. For instance, if you work on a document classification problem with bag of words representation then you should care about document length since longer documents include more words which result in more crowded feature histograms. One of the reasonable ways to solve this issue is to divide each word frequency by the total word frequency in the document so that we can convert each histogram value into a probability of seeing that word in the document. As a result, document is represented with a feature vector that is 1 in total of its elements. This new space is called vector space model in the literature.
3. Removing correlations between dimensions cleans your data from redundant information exposed by multiple feature dimensions. Hence data is projected into a new space where each dimension explains something independently important from the other feature dimensions.

Okay, I hope now we are clear why we are concerned about these. Henceforth, I'll try to emphasis some basic stuff in our toolkit for feature preprocessing.

Standardization

• Can be applied to both feature dimensions or data instances.
• If we apply to dimensions, it reduces unit effect and if we apply to instances then we solve instance biases as in the case of the document classification problem.
• The result of standardization is that each feature dimension (instance) is scaled into defined mean and variance so that we fix the unit differences between dimensions.
• $z = (x-mu)/alpha$  : for each dimension (instance),  subtract the mean and divide by the variance of that dimension (instance) so that each dimension is kept inside a mean = 0 , variance = 1 curve.

Min Max Scaling

• Personally, I've not applied Min-Max Scaling to instances,
• It is still useful for unit difference problem.
• Instead of distributional consideration, it hinges the values in the range  [0,1].
• $x_{norm} = (x - x_{min})/(x_{max} - x_{min})$ :  Find max and min values of the feature dimension and apply the formula.

Caveat 1: One common problem of Scaling and Standardization is you need to keep min and max for Scaling, mean and variance values for Standardization for the novel data and the test time. We estimate these values from only the training data and assume that these are still valid for the test and real world data. This assumption might be true for small problems but especially for online environment this caveat should be dealt with a great importance.

Sigmoid Functions

• Sigmoid function naturally fetches given values into a [0, 1] range
• Does not need any assumption about the data like mean and variance
• It penalizes large values  more than the small ones.
• You can use other activation functions like tanh.

Caveat 2: How to choose and what to choose are very problem dependent questions. However, if you have a clustering problem then standardization seems more reasonable for better similarity measure between instance and if you intend to use Neural Networks then some particular kind of NN demands [0,1] scaled data (or even more interesting scale ranges for better gradient propagation on the NN model). Also, I personally use sigmoid function for simple problems in order to get fast result by SVM without complex investigation.

Zero Phase Component Analysis (ZCA Whitening)

• As I explained before, whitening is a process to reduce redundant information by decorrelating data with a final diagonal correlation matrix with preferable all diagonals are one.
• It has especially very important implications in Image Recognition and Feature Learning  so as to make visual cues more concrete on images.
• Instead of formula, it is more intuitive to wire some code

I tried to touch some methods and common concerns of feature preprocessing, by no means  complete. Nevertheless, a couple of takeaways from this post are; do not ignore normalizing your feature values before going into training phase and choose the correct method by investigating the values painstakingly.

PS: I actually promised to write a post per week but I am as busy as a bee right now and I barely find some time to write a new stuff. Sorry about it 🙁

# Normalize Your Database with NF rules.

The Normal Form rules are the basic rules that are created by the Relational Database concept creater Edgar F. Codd. The main purpose of these is to create database tables that do not have redundancy for the data inside. There are basically three NF variations 1NF, 2NF, 3NF (Actually now on it goes up to 6NF by other theoreticians but basic structure is based on 1NF 2NF 3NF) .

1NF points out:
- Table should not have columns that are based on same data
- Create separate tables for each group of related data and identify each row with a unique column (the primary key).

Example:
(Manager, Employee1, Employee2 ) are the columns but there are two Employee column that violate rule one.

(Manager, Employee) is good solution but it violates still second rule.

(Manager, EmployeeID) is intuitive solution since each Employee has only one Manager. Now our table is proper for 1NF.

2NF points out:
- Remove subset of data that causes multiple rows includes same information and create seperate table for them
-Use foreign keys to relate the tables that are newly created and previous one

EXAMPLE:
(EmpID,FirstName, LastName, City, Zipcode, ) is our table but if there are two employee with same city so zipcodem it makes two row with redundant data. So it is more appropriate to use another table that includes cities and correspondence zipcodes. In addition new new structure makes easier to add, remove or update address information. Instead of updating all the rows that have same city and zipcode, we just need to update one row in the new table.


text:
# Ind the cuical valuels rejecbon region(s) for a oft-tailed chi-square tost with sample size n = 19 and level ol

###### Question:

ind the cuical valuels rejecbon region(s) for a oft-tailed chi-square tost with sample size n = 19 and level ol signifcance a = 0.10. 0 cick the Icon t vew the Chi-Square Distrbution Table: Find tne critical Ivalue(s): (Round " three decimal places. needed. Use comma aepargtu neam needed ) Find the relection rgion(s) Choose the correct answer belox:

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# Average Velocity

## Displacement divided by time.

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Average Velocity

Students will learn the meaning of speed, velocity and average velocity.

### Key Equations

Speed = distance/time

Average Velocity

Guidance
Speed is the distance traveled divided by the time it took to travel that distance. Velocity is the instantaneous speed and direction. Average velocity is the displacement divided by the time.

#### Example 1

Pacific loggerhead sea turtles migrate over 7,500 miles (12,000 km) between nesting beaches in Japan and feeding grounds off the coast of Mexico. If the average speed of a loggerhead is about 45 km/day, how long does it take for it to complete the distance of a one-way migration?

Question: \begin{align*}t = ?\end{align*} [days]

Given: \begin{align*}d = 12,000 \ km\end{align*}

\begin{align*}{\;} \qquad v_{avg} = 45 \ km/day\end{align*}

Equation: \begin{align*}v_{avg} = \frac{d}{t}\end{align*} therefore \begin{align*}t = \frac{d}{v_{avg}}\end{align*}

Plug n’ Chug: \begin{align*}t = \frac{d}{v_{avg}} = \frac{12,000 \ km}{45 \ km/day} = 267 \ days\end{align*}

### Explore More

1. Two cars are heading right towards each other, but are 12 km apart. One car is going 70 km/hr and the other is going 50 km/hr. How much time do they have before they collide head on?
2. You drive the 10 miles to work at an average speed of 40 mph. On the way home you hit severe traffic and drive at an average speed of 10 mph. What is your average speed for the trip?
3. The following data represent the first 30 seconds of actor Crispin Glover’s drive to work.
Time (s) Position (m) Distance (m)
0 0 0
5 10 10
10 30 30
15 30 30
20 20 40
25 50 70
30 80 120

Sketch the graphs of position vs. time and distance vs. time. Label your \begin{align*}x\end{align*} and \begin{align*}y\end{align*} axes appropriately.

1. Why is there a discrepancy between the distance covered and the change in position during the time period between \begin{align*}t = 25 \;\mathrm{s}\end{align*} and \begin{align*}t = 30 \;\mathrm{s}\end{align*}?
2. What do you think is going on between \begin{align*}t = 10 \;\mathrm{s}\end{align*} and \begin{align*}t = 15 \;\mathrm{s}\end{align*}?
3. What is the displacement between \begin{align*}t = 10 \;\mathrm{s}\end{align*} and \begin{align*}t = 25 \;\mathrm{s}\end{align*}?
4. What is the distance covered between \begin{align*}t = 10 \;\mathrm{s}\end{align*} and \begin{align*}t = 25 \;\mathrm{s}\end{align*}?
5. What is the average velocity during the first \begin{align*}30\end{align*} seconds of the trip?
6. What is the average velocity between the times \begin{align*}t = 20 \;\mathrm{s}\end{align*} and \begin{align*}t = 30 \;\mathrm{s}\end{align*}?
7. During which time interval(s) was the velocity negative?
8. Sketch the velocity vs. time and speed vs. time graphs. Label your \begin{align*}x\end{align*} and \begin{align*}y\end{align*} axes appropriately.

1. 0.1 hours = 6 minutes

2. 16 mph

3. Discuss in class


text:
# BJT Transistor Collector Current

As seen from the question above, I'm required to calculate $I_C$ for both cases. Why am I not able to get the answer by using the formula $$I_C = I_S e^{V_{BE}/V_T}$$ ?

I was only able to get the answer through finding $V_B = 2.5\text{V}$ and assuming $V_{BE} = 0.7\text{V}$. $V_E$ is then $2.5-0.7 = 1.8\text{V}$, and with this value I can easily calculate $I_E$ and subsequently $I_C$.

But using the method above, I'm not able to get the answer for part (b), as for the case of $V_B$, $R_1$ is still equal to $R_2$, resulting in $V_B = 2.5\text{V}$. I'll still get the same answer.

I recall my lecturer saying that the equation $$I_C = I_S e^{V_{BE}/V_T}$$ can only be used when $V_{BE}$ is to be found through iteration. I can't remember the reasoning behind though.

Hope you guys could clarify on this.

The problem in part (b) is that the current through the voltage divider is comparable to the base current, and the voltage divider is significantly loaded down by said base current.

In part (a), the current $I$ through the (unloaded) voltage divider is $250\mu$A. The base current $I_B = I_C/\beta = 10\mu$A, so $I >> I_B$ and $I_B$ is negligible (i.e. the voltage divider isn't loaded down).

But in part (b), the current $I$ through the (unloaded) voltage divider would be $2.5\mu$A. Now, assuming $I_C = 100$mA, $I < I_B$ and the voltage divider is loaded down by the transistor's base current -- all of the current (and then some) would flow into the transistor rather than $R_2$. This means $V_B \neq 2.5$V.

I'm not sure what you've learned yet so I'm not sure the best way for you to calculate $I_C$ in part (b) -- perhaps your lecturer has given you an iterative procedure, based on your comment about finding $V_{BE}$. But I can tell you that your error is in assuming that $V_B = 2.5$V.

What this exercise is hinting at is that you can't make the bias resistors $R_1$ and $R_2$ arbitrarily high. The rule of thumb is that the unloaded voltage divider current should be $10$x the base current. If the voltage divider current is too low you will not get the desired $V_B$, and if it's much higher then you're just wasting current.

Here is a simplified schematic (you don't need the capacitors or input source to calculate the bias point) which you can simulate in CircuitLab:

simulate this circuit – Schematic created using CircuitLab

The simulation result shows $V_B \approx 1.2$V and $I_C \approx 275\mu$A. What's happening is that you have a $2.5$V Thevenin equivalent source, but $R_{\text{TH}} = 500\text{k}\Omega$, so even a small $I_B$ results in a significant voltage drop across $R_{\text{TH}}$, dropping $V_B$ well below $2.5$V. With $I_C \approx 275\mu$A, $I_B \approx 2.75\mu$A and the voltage drop across $R_{\text{TH}}$ is $I_B \times R_{\text{TH}} \approx 1.3$V. That means $V_B \approx 1.2$V.

To calculate $I_C$, note that $$V_B = V_{\text{TH}} - I_BR_{\text{TH}}\tag1$$ Also note that $$V_B = V_{BE} + I_ER_E \tag2$$ Since $I_E = (\beta + 1)I_B = 101 \times I_B$, you can substitute into $(2)$ for $I_B$, then equate $(1)$ and $(2)$.

That gives you $$V_{\text{TH}} - I_BR_{\text{TH}} = V_{BE} + \frac{(\beta + 1)I_B}{R_E}$$

I guessed $V_{BE} = 0.7$V and solved for $I_B \approx 3.6\mu$A. That means $I_C = \beta I_B \approx 360\mu$A, which is close to the answer.

• Based on your comment , the current I through the voltage divider is 250µA , but isn't that the current that flows into the base of the transistor? In that case shouldn't IB was well? Also , the way that I was taught to do this is to take the thevenin voltage equivalent to get Vb , in so for part a , the thevenin resistance would be 5k ohms and the thevenin voltage would be 2.5V. Could I take 2.5/5000 to get my Ib? Lastly , how were you able to calculate IB based on the current through the voltage divider?
– John
Mar 10 '15 at 19:27
• $250\mu$A is the current through the unloaded voltage divider (i.e. with the transistor disconnected from it). With the transistor connected the current through $R_1$ is $250\mu$A $+ I_B$. $250\mu$A flows through $R_2$ and the rest flows into the transistor base. When $I = 250\mu$A you can ignore the loading effect of the transistor, but not when the unloaded voltage divider current is only $2.5\mu$A.
– Null
Mar 10 '15 at 19:33
• I'm calculating $I_B$ based on the given answers for $I_C$ (working backwards). It's a sanity check to make sure that your assumption that $V_B = 2.5$V is valid, which is the case for part (a) but not part (b).
– Null
Mar 10 '15 at 19:36
• Ok I sort of get what you're trying to say. But then for (b) , how would I be able to calculate Ic then? The method that I've been taught was to convert R1 and R2 into a thevenin equivalent circuit , but even then I'm unable to the answer.
– John
Mar 10 '15 at 19:46