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# Generalized least squares

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In statistics, generalized least squares (GLS) is a technique for estimating the unknown parameters in a linear regression model when there is a certain degree of correlation between the residuals in a regression model. In these cases, ordinary least squares and weighted least squares can be statistically inefficient, or even give misleading inferences. GLS was first described by Alexander Aitken in 1936. [1]

## Method outline

In standard linear regression models we observe data ${\displaystyle \{y_{i},x_{ij}\}_{i=1,\dots ,n,j=2,\dots ,k}}$ on n statistical units. The response values are placed in a vector ${\displaystyle \mathbf {y} =\left(y_{1},\dots ,y_{n}\right)^{\mathsf {T}}}$, and the predictor values are placed in the design matrix ${\displaystyle \mathbf {X} =\left(\mathbf {x} _{1}^{\mathsf {T}},\dots ,\mathbf {x} _{n}^{\mathsf {T}}\right)^{\mathsf {T}}}$, where ${\displaystyle \mathbf {x} _{i}=\left(1,x_{i2},\dots ,x_{ik}\right)}$ is a vector of the k predictor variables (including a constant) for the ith unit. The model forces the conditional mean of ${\displaystyle \mathbf {y} }$ given ${\displaystyle \mathbf {X} }$ to be a linear function of ${\displaystyle \mathbf {X} }$, and assumes the conditional variance of the error term given ${\displaystyle \mathbf {X} }$ is a known nonsingular covariance matrix ${\displaystyle \mathbf {\Omega } }$. This is usually written as

${\displaystyle \mathbf {y} =\mathbf {X} \mathbf {\beta } +\mathbf {\varepsilon } ,\qquad \operatorname {E} [\varepsilon \mid \mathbf {X} ]=0,\ \operatorname {Cov} [\varepsilon \mid \mathbf {X} ]=\mathbf {\Omega } .}$

Here ${\displaystyle \beta \in \mathbb {R} ^{k}}$ is a vector of unknown constants (known as “regression coefficients”) that must be estimated from the data.

Suppose ${\displaystyle \mathbf {b} }$ is a candidate estimate for ${\displaystyle \mathbf {\beta } }$. Then the residual vector for ${\displaystyle \mathbf {b} }$ will be ${\displaystyle \mathbf {y} -\mathbf {X} \mathbf {b} }$. The generalized least squares method estimates ${\displaystyle \mathbf {\beta } }$ by minimizing the squared Mahalanobis length of this residual vector:

${\displaystyle \mathbf {\hat {\beta }} ={\underset {b}{\operatorname {argmin} }}\,(\mathbf {y} -\mathbf {X} \mathbf {b} )^{\mathsf {T}}\mathbf {\Omega } ^{-1}(\mathbf {y} -\mathbf {X} \mathbf {b} )={\underset {b}{\operatorname {argmin} }}\,\mathbf {y} ^{\mathsf {T}}\,\mathbf {\Omega } ^{-1}\mathbf {y} +(\mathbf {X} \mathbf {b} )^{\mathsf {T}}\mathbf {\Omega } ^{-1}\mathbf {X} \mathbf {b} -\mathbf {y} ^{\mathsf {T}}\mathbf {\Omega } ^{-1}\mathbf {X} \mathbf {b} -(\mathbf {X} \mathbf {b} )^{\mathsf {T}}\mathbf {\Omega } ^{-1}\mathbf {y} \,,}$

where the last two terms evaluate to scalars, resulting in

${\displaystyle \mathbf {\hat {\beta }} ={\underset {b}{\operatorname {argmin} }}\,\mathbf {y} ^{\mathsf {T}}\,\mathbf {\Omega } ^{-1}\mathbf {y} +\mathbf {b} ^{\mathsf {T}}\mathbf {X} ^{\mathsf {T}}\mathbf {\Omega } ^{-1}\mathbf {X} \mathbf {b} -2\mathbf {b} ^{\mathsf {T}}\mathbf {X} ^{\mathsf {T}}\mathbf {\Omega } ^{-1}\mathbf {y} \,.}$

This objective is a quadratic form in ${\displaystyle \mathbf {b} }$.

Taking the gradient of this quadratic form with respect to ${\displaystyle \mathbf {b} }$ and equating it to zero (when ${\displaystyle \mathbf {b} ={\hat {\beta }}}$) gives

${\displaystyle 2\mathbf {X} ^{\mathsf {T}}\mathbf {\Omega } ^{-1}\mathbf {X} {\hat {\beta }}-2\mathbf {X} ^{\mathsf {T}}\mathbf {\Omega } ^{-1}\mathbf {y} =0}$

Therefore, the minimum of the objective function can be computed yielding the explicit formula:

${\displaystyle \mathbf {\hat {\beta }} =\left(\mathbf {X} ^{\mathsf {T}}\mathbf {\Omega } ^{-1}\mathbf {X} \right)^{-1}\mathbf {X} ^{\mathsf {T}}\mathbf {\Omega } ^{-1}\mathbf {y} .}$

### Properties

The GLS estimator is unbiased, consistent, efficient, and asymptotically normal with ${\displaystyle \operatorname {E} [{\hat {\beta }}\mid \mathbf {X} ]=\beta }$ and ${\displaystyle \operatorname {Cov} [{\hat {\beta }}\mid \mathbf {X} ]=(\mathbf {X} ^{\mathsf {T}}\Omega ^{-1}\mathbf {X} )^{-1}}$. GLS is equivalent to applying ordinary least squares to a linearly transformed version of the data. To see this, factor ${\displaystyle \mathbf {\Omega } =\mathbf {C} \mathbf {C} ^{\mathsf {T}}}$, for instance using the Cholesky decomposition. Then if we pre-multiply both sides of the equation ${\displaystyle \mathbf {y} =\mathbf {X} \mathbf {\beta } +\mathbf {\varepsilon } }$ by ${\displaystyle \mathbf {C} ^{-1}}$, we get an equivalent linear model ${\displaystyle \mathbf {y} ^{*}=\mathbf {X} ^{*}\mathbf {\beta } +\mathbf {\varepsilon } ^{*}}$ where ${\displaystyle \mathbf {y} ^{*}=\mathbf {C} ^{-1}\mathbf {y} }$, ${\displaystyle \mathbf {X} ^{*}=\mathbf {C} ^{-1}\mathbf {X} }$, and ${\displaystyle \mathbf {\varepsilon } ^{*}=\mathbf {C} ^{-1}\mathbf {\varepsilon } }$. In this model ${\displaystyle \operatorname {Var} [\varepsilon ^{*}\mid \mathbf {X} ]=\mathbf {C} ^{-1}\mathbf {\Omega } \left(\mathbf {C} ^{-1}\right)^{\mathsf {T}}=\mathbf {I} }$, where ${\displaystyle \mathbf {I} }$ is the identity matrix. Thus we can efficiently estimate ${\displaystyle \mathbf {\beta } }$ by applying Ordinary least squares (OLS) to the transformed data, which requires minimizing

${\displaystyle \left(\mathbf {y} ^{*}-\mathbf {X} ^{*}\mathbf {\beta } \right)^{\mathsf {T}}(\mathbf {y} ^{*}-\mathbf {X} ^{*}\mathbf {\beta } )=(\mathbf {y} -\mathbf {X} \mathbf {b} )^{\mathsf {T}}\,\mathbf {\Omega } ^{-1}(\mathbf {y} -\mathbf {X} \mathbf {b} ).}$

This has the effect of standardizing the scale of the errors and “de-correlating” them. Since OLS is applied to data with homoscedastic errors, the Gauss–Markov theorem applies, and therefore the GLS estimate is the best linear unbiased estimator for β.

## Weighted least squares

A special case of GLS called weighted least squares (WLS) occurs when all the off-diagonal entries of Ω are 0. This situation arises when the variances of the observed values are unequal (i.e.  heteroscedasticity is present), but where no correlations exist among the observed variances. The weight for unit i is proportional to the reciprocal of the variance of the response for unit i. [2]

## Feasible generalized least squares

If the covariance of the errors ${\displaystyle \Omega }$ is unknown, one can get a consistent estimate of ${\displaystyle \Omega }$, say ${\displaystyle {\widehat {\Omega }}}$, [3] using an implementable version of GLS known as the feasible generalized least squares (FGLS) estimator. In FGLS, modeling proceeds in two stages: (1) the model is estimated by OLS or another consistent (but inefficient) estimator, and the residuals are used to build a consistent estimator of the errors covariance matrix (to do so, one often needs to examine the model adding additional constraints, for example if the errors follow a time series process, a statistician generally needs some theoretical assumptions on this process to ensure that a consistent estimator is available); and (2) using the consistent estimator of the covariance matrix of the errors, one can implement GLS ideas.

Whereas GLS is more efficient than OLS under heteroscedasticity or autocorrelation, this is not true for FGLS. The feasible estimator is, provided the errors covariance matrix is consistently estimated, asymptotically more efficient, but for a small or medium size sample, it can be actually less efficient than OLS. This is why, some authors prefer to use OLS, and reformulate their inferences by simply considering an alternative estimator for the variance of the estimator robust to heteroscedasticity or serial autocorrelation. But for large samples FGLS is preferred over OLS under heteroskedasticity or serial correlation. [3] [4] A cautionary note is that the FGLS estimator is not always consistent. One case in which FGLS might be inconsistent is if there are individual specific fixed effects. [5]

In general this estimator has different properties than GLS. For large samples (i.e., asymptotically) all properties are (under appropriate conditions) common with respect to GLS, but for finite samples the properties of FGLS estimators are unknown: they vary dramatically with each particular model, and as a general rule their exact distributions cannot be derived analytically. For finite samples, FGLS may be even less efficient than OLS in some cases. Thus, while GLS can be made feasible, it is not always wise to apply this method when the sample is small. A method sometimes used to improve the accuracy of the estimators in finite samples is to iterate, i.e. taking the residuals from FGLS to update the errors covariance estimator, and then updating the FGLS estimation, applying the same idea iteratively until the estimators vary less than some tolerance. But this method does not necessarily improve the efficiency of the estimator very much if the original sample was small. A reasonable option when samples are not too large is to apply OLS, but throwing away the classical variance estimator

${\displaystyle \sigma ^{2}*(X'X)^{-1}}$

(which is inconsistent in this framework) and using a HAC (Heteroskedasticity and Autocorrelation Consistent) estimator. For example, in autocorrelation context we can use the Bartlett estimator (often known as Newey-West estimator since these authors popularized the use of this estimator among econometricians in their 1987 Econometrica article), and in heteroskedastic context we can use the Eicker–White estimator. This approach is much safer, and it is the appropriate path to take unless the sample is large, and "large" is sometimes a slippery issue (e.g. if the errors distribution is asymmetric the required sample would be much larger).

The ordinary least squares (OLS) estimator is calculated as usual by

${\displaystyle {\widehat {\beta }}_{\text{OLS}}=(X'X)^{-1}X'y}$

and estimates of the residuals ${\displaystyle {\widehat {u}}_{j}=(Y-X{\widehat {\beta }}_{\text{OLS}})_{j}}$ are constructed.

For simplicity consider the model for heteroskedastic errors. Assume that the variance-covariance matrix ${\displaystyle \Omega }$ of the error vector is diagonal, or equivalently that errors from distinct observations are uncorrelated. Then each diagonal entry may be estimated by the fitted residuals ${\displaystyle {\widehat {u}}_{j}}$ so ${\displaystyle {\widehat {\Omega }}_{OLS}}$ may be constructed by

${\displaystyle {\widehat {\Omega }}_{\text{OLS}}=\operatorname {diag} ({\widehat {\sigma }}_{1}^{2},{\widehat {\sigma }}_{2}^{2},\dots ,{\widehat {\sigma }}_{n}^{2}).}$

It is important to notice that the squared residuals cannot be used in the previous expression; we need an estimator of the errors variances. To do so, we can use a parametric heteroskedasticity model, or a nonparametric estimator. Once this step is fulfilled, we can proceed:

Estimate ${\displaystyle \beta _{FGLS1}}$ using ${\displaystyle {\widehat {\Omega }}_{\text{OLS}}}$ using [4] weighted least squares

${\displaystyle {\widehat {\beta }}_{FGLS1}=(X'{\widehat {\Omega }}_{\text{OLS}}^{-1}X)^{-1}X'{\widehat {\Omega }}_{\text{OLS}}^{-1}y}$

The procedure can be iterated. The first iteration is given by

${\displaystyle {\widehat {u}}_{FGLS1}=Y-X{\widehat {\beta }}_{FGLS1}}$
${\displaystyle {\widehat {\Omega }}_{FGLS1}=\operatorname {diag} ({\widehat {\sigma }}_{FGLS1,1}^{2},{\widehat {\sigma }}_{FGLS1,2}^{2},\dots ,{\widehat {\sigma }}_{FGLS1,n}^{2})}$
${\displaystyle {\widehat {\beta }}_{FGLS2}=(X'{\widehat {\Omega }}_{FGLS1}^{-1}X)^{-1}X'{\widehat {\Omega }}_{FGLS1}^{-1}y}$

This estimation of ${\displaystyle {\widehat {\Omega }}}$ can be iterated to convergence.

Under regularity conditions any of the FGLS estimator (or that of any of its iterations, if we iterate a finite number of times) is asymptotically distributed as

${\displaystyle {\sqrt {n}}({\hat {\beta }}_{FGLS}-\beta )\ {\xrightarrow {d}}\ {\mathcal {N}}\!\left(0,\,V\right).}$

where n is the sample size and

${\displaystyle V=\operatorname {p-lim} (X'\Omega ^{-1}X/T)}$

here p-lim means limit in probability

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## References

1. Aitken, A. C. (1936). "On Least-squares and Linear Combinations of Observations". Proceedings of the Royal Society of Edinburgh. 55: 42–48.
2. Strutz, T. (2016). Data Fitting and Uncertainty (A practical introduction to weighted least squares and beyond). Springer Vieweg. ISBN   978-3-658-11455-8., chapter 3
3. Baltagi, B. H. (2008). Econometrics (4th ed.). New York: Springer.
4. Greene, W. H. (2003). Econometric Analysis (5th ed.). Upper Saddle River, NJ: Prentice Hall.
5. Hansen, Christian B. (2007). "Generalized Least Squares Inference in Panel and Multilevel Models with Serial Correlation and Fixed Effects". Journal of Econometrics . 140 (2): 670–694. doi:10.1016/j.jeconom.2006.07.011.


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# A Question about Irreducible Polynomials [closed]

I am doing some self-study in the area of Cryptography. I am using the Third Edition of the book "Cryptography Theory and Practice" by Douglas R. Stinson. Based upon the information on page 105, in the field $F_{2^8}$, I believe the following two polynomials are multiplicative inverses.

$x^6 + x^4 + x + 1$
$x^7 + x^6 + x^3 + x$

If they are and I multiply them together, I should get the irreducible polynomial for the field right? Here is what I get when I multiply the two polynomials together in $F_{2^8}$. \begin{eqnarray*} (x^6 + x^4 + x + 1)(x^7 + x^6 + x^3 + x) &=& (x^6)(x^7 + x^6 + x^3 + x) + \\ && (x^4)(x^7 + x^6 + x^3 + x) \\ &+& (x + 1)(x^7 + x^6 + x^3 + x) \\ (x^6 + x^4 + x + 1)(x^7 + x^6 + x^3 + x) &=& (x^{13} + x^{12} + x^9 + x^7) + \\ && (x^{11} + x^{10} + x^7 + x^5) \\ &+& (x + 1)(x^7 + x^6 + x^3 + x) \\ (x^6 + x^4 + x + 1)(x^7 + x^6 + x^3 + x) &=& (x^{13} + x^{12} + x^9 + x^7) + \\ && (x^{11} + x^{10} + x^7 + x^5) + \\ && (x^8 + x^7 + x^4 + x^2) + (x^7 + x^6 + x^3 + x) \\ (x^6 + x^4 + x + 1)(x^7 + x^6 + x^3 + x) &=& x^{13} + x^{12} + x^{11} + x^9 + x^{10} + x^5 + \\ && x^8 + x^7 + x^4 + x^2 + x^7 + x^6 + x^3 + x \\ (x^6 + x^4 + x + 1)(x^7 + x^6 + x^3 + x) &=& x^{13} + x^{12} + x^{11} + \\ && x^{10} + x^9 + x^8 + x^6 + x^5 + x^3 + x \\ (x^6 + x^4 + x + 1)(x^7 + x^6 + x^3 + x) &=& x^{5} + x^{4} + x^{3} + x^{2} + x^9 + x^8 + x^6 + x^5 + x^3 + x \\ (x^6 + x^4 + x + 1)(x^7 + x^6 + x^3 + x) &=& x^{4} + x^{3} + x^{2} + x^9 + x^8 + x^6 + x^3 + x \\ (x^6 + x^4 + x + 1)(x^7 + x^6 + x^3 + x) &=& x^{4} + x^{3} + x^{2} + x + x^8 + x^6 + x^3 + x \\ (x^6 + x^4 + x + 1)(x^7 + x^6 + x^3 + x) &=& x^8 + x^6 + x^4 + x^2 \\ \end{eqnarray*} This is not the irreducible polynomial. What am I missing?

Thanks

Bob

• You are not correctly reducing the polynomial after multiplication. To get a polynomial in $\mathbb{F}_{2^8}$ from one with terms of degree of eight or more you have to divide it by the irreducible polynomial (you can't just mod 8 the terms degrees). – puzzlepalace Mar 20 '17 at 21:17
• You have reduce the polynomial in step 6 using "long division modulo 2" as shown here. – Occams_Trimmer Mar 20 '17 at 21:29
• "If they are and I multiply them together, I should get the irreducible polynomial for the field right?" Wrong. – fkraiem Mar 21 '17 at 7:00
• I'm voting to close this question as off-topic because it is about general mathematics. – fkraiem Mar 21 '17 at 7:01

In the page $104$ you can see: $$F_{2^8}=\frac{Z_2[x]}{x^8+x^4+x^3+x+1}.$$ this means that irreducible polynomial that generate the $F_{2^8}$ is $x^8+x^4+x^3+x+1$. so, in multiplications, you should note that $x^8=-x^4-x^3-x-1=x^4+x^3+x+1$. Therefore $x^{8+i}=x^8\cdot x^i=x^{4+i}+x^{3+i}+x^{1+i}+x^i$.

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# What is Arnold Hotel Font™?

This is in the spirit of the What is a Word/Phrase™ series started by JLee with a special brand of Phrase™ and Word™ puzzles.

If a font shares a specific commonality with another font, it's an Arnold Hotel Font™.

Use the following examples below to find the commonality. Clarification: This is asking specifically about the typeface being used, not the name of the typeface/letters contained within.

And, if you want to analyze, here is a CSV version:

Arnold Hotel Font™, Non-Arnold Hotel Font™
GARAMOND,CALIBRI
ANTONIO,VERDANA
UNIVERS,CAMBRIA
GOTHAM,BODONI
BEMBO,GILL SANS
AVENIR,LUCIDA
CHAMPION,PERPETUA
FORMATA,ROCKWELL
GEORGIA,TAHOMA
VITESSE,ONYX
CALIFORNIAN,PAPYRUS

Note: There are other Arnold Hotel Fonts™ than the ones on the list.

Hint 1

(((((((((7+2)-1)+1)+1)-1)+1)+1)+1)+2)

Hint 2

Brownie birth Home

Hint 3

The person in charge, alternatively, the place they're in charge of:
MPIUTRCDHUDEANIELS
MIINCDHIAAENLAMCROBBIE
WMAALRLYALCAENLDOH
MOIHCIHOASETLAVTDERAKE
EPREINCNJSBTAARTREON
RROUBTEGRETRLSBARCHI
RIOLBLEIRNTOJIJSONES
BIROUWCAESHTAARTREELD
JMOIANNNGEASBOETLA
RNOENBNRIAESDKGAREEN
JNLOARNTDHIWSEMSATRETRINN
RWEIBSECCOCNASBILNANK
SMAIMCUHEILGLASNTSATNALTEEY
MMAIRCKHSISGCAHNLISSEL

For @hdsdv (and anyone else following along at home)

For your spoiler block #1, take a look at the second piece of documentation(s) you mentioned.
For your spoiler block #2: Who are some famous/well-known Arnolds?
For your spoiler block #3: refer back to my hint for your spoiler block #1

• Would you like to clarify whether this is actually about the names or the typefaces themselves? (Of course the ambiguity may be entirely deliberate and I'm not suggesting there's anything wrong with that.) – Gareth McCaughan Oct 22 at 9:47
• Good comment! Let me clarify. – Gunslinger711 Oct 22 at 12:12
• @GarethMcCaughan and therefore there's no word-prop tag ;) – Omega Krypton Oct 22 at 12:27
• There are many typefaces called Garamond, some of which are very different from one another. Several of the things listed here are actually families with many members (including e.g. roman and italic variants which may have very different letterforms). Again, any ambiguity here may be very much intended, but would you like to offer any clarification on how to identify e.g. a particular version of Garamond or a particular weight of Univers if we need to? (Again: absolutely fine not to.) – Gareth McCaughan Oct 22 at 16:20
• @Deusovi In addition to the typefaces' names and the letterforms themselves, there are also things like their designers, the foundries that sold them, the dates when they were created, what if anything they were named after, what other typefaces they were inspired by (or shamelessly copied from), etc. The names are more useful than the letterforms for trying to find such things. – Gareth McCaughan Oct 24 at 0:57

That's quite a big hint. It looks like the Arnold Hotel fonts

are or have been used in the logos or official branding guides of Big Ten universities.
Iowa - Antonio
Michigan - Univers
Illinois - Univers
Ohio State - Univers (deprecated, replaced by Proxima Nova and Capita)
Michigan State - Gotham
Maryland - Bembo
Penn State - Avenir
Purdue - Champion
Rutgers - Formata
Indiana - Georgia
Wisconsin - Vitesse

Still working on these:

Northwestern - Georgia (I had to find their Word document template to see this, and if this is the intended mapping I find it somewhat disappointing - their official fonts are Periódico, Campton, and Akkurat)
Minnesota - ??? Currently Neutraface, cannot find any references to any of the listed fonts.

Oh also,

I have no idea how to connect the title. Trying to incorporate the hint, I looked for broadly famous Arnolds (Palmer, Schwarzenegger, and there are very few others). Arnold Palmer went to Wake Forest, an ACC school. Arnold Schwarzenegger graduated from University of Wisconsin - Superior. I suppose this could be the connection but it seems very tenuous, since this association is not well-known or often-discussed. There are some Big Ten athletes named Arnold (Arnie Risen, Kenny Arnold) but again they don't seem broadly famous enough to justify the title. I feel like I'm still missing something here.

As for the missing one,

Californian is UC Berkeley's logo. Not sure how that fits with all the Big Ten schools.

• You're most of the way there! I'll add a few hints up in my original post that will hopefully help you (or someone else) wrap it up. – Gunslinger711 Oct 25 at 16:32
• I was able to find rot13(Trbetvn va Abegujrfgrea'f qbjaybnqnoyr Jbeq qbphzrag grzcyngr. Guvf frrzf yvxr n fgergpu, ohg bu jryy. Nf sbe Zvaarfbgn, abar bs gurve qbphzrag grzcyngrf ner ninvynoyr sbe qbjaybnq jvgubhg ybttvat va gb gurve flfgrz. Gur Zvaarfbgn jbeqznex ybbxf xvaq bs yvxr Pnyvsbeavna, ohg vg vfa'g (lbh pna frr zvabe qvssreraprf va gur G naq S, sbe rknzcyr.) – hdsdv Oct 25 at 17:32


text: Formula, Derivation, Calculate the area, Area using diameter and circumference at BYJUâS. This is illustrated below in brown. Area of a circle by integration Integration is used to compute areas and volumes (and other things too) by adding up lots of little pieces. A circle is a shape with all points at the boundary having the same distance to the centre. Investigate the cases when circle center is on the â¦ Equation of circle- $2x^2+ \lambda xy+2y^2+( \lambda -4)x+6y-5=0$ find area of the circle. This is illustrated below in brown. If a circle touches both the axes, then consider the center of the circle to be $$(r, r)$$, where $$r$$ is the The equation of a circle can be found using the centre and radius. The area of a circle is number of square units inside the circle. Geometrical shapes are fun to play with. The diameter of a circle calculator uses the following equation: Area of a circle = Ï * (d/2) 2 Where: Ï is approximately equal to 3.14. This page looks to give a general run through of how the formula for the area of a circle can be derived. Watch Area of a Circle Equation Definition Proof Examples - WORLD ENTERTAINMENT on Dailymotion Find the equation of the tangent line. Semicircle Definition A semicircle is a half circle. Attempt- For converting the equation from second degree to first degree $\lambda xy=0$. Plug the circumference into the formula. Area is defined as the region occupied by a two-dimensional shape. Circle A is given by the equation (x â 4) 2 + (y + 3) 2 = 29. Standard formula to calculate the area of a circle is: A=Ïr². By finding the area of go shopping with my old papa he is still alive!And i The surface area of circle is 4Ïr 2 .The diameter is the distance across two extreme ends of a circle passing through the center. For the equation of a line, you need a point (you have it) and the lineâs slope. Then, its radius is doubled. Look at the graph below, can you express the equation of the circle in standard form? Explanation: The general equation of a circle is (x â h) 2 + (y â k) 2 = r 2, where (h, k) represents the location of the circle's center, and r represents the length of its radius. The greeks considered the Egyptians as the inventors of geometry. Equation of a Circle: The equation of the circle can be formed using the Center-Radius Formula. You can ask me questions directly at https://www.simor.org/user/1723/saul-remi_hernandez/rooms Equation for the area of a circle is derived with integration And so: All points are the same distance from the center. We suggest to use the same formula to find the area of a circle which uses You might think that means it will have half the perimeter of a circle, but that is not true. Q. Whenever a math problem tells you the circumference C of a circle, you can use this equation to find the radius r.Replace C in the equation with the circumference of the circle in your problem: The following formulas are used for circle calculations. Ï is defined as the ratio of the circumference of a circle to its diameter.. Calculates the area and circumference of a circle given the radius or diameter. Area A = pi x radius x radius A = 3.14 x 1 x 1 A = 3.14 x 1 = 3.14 sq.cm Notice that the calculated area of the circle is same in both the methods. Show Solutions Since the radius of this this circle is 2, and its center is (3,1) , this circle's equation is. The distance between any point of the circle and the centre is called the radius. Derive Area of Circle Formula The Area of Circle formula is: AREA = Ï × radius 2 But how does that come about? 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### Match the vector equations in Exercises 1-8 with …

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University of Central Arkansas
Problem 2

# Match the vector equations in Exercises 1-8 with the graphs (a)-(h) given here.$$\mathbf{r}(t)=\mathbf{i}+\mathbf{j}+t \mathbf{k}, \quad-1 \leq t \leq 1$$

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## Video Transcript

okay. Were given a vector equation. Are, uh, T is equal to, um t squared minus one. Um, j plus two tea. Kay whips. I went too far out there. Um, okay. And t goes from negative One toe, one inclusive. Okay. And we want to kind of pair it up with a crap. Let's go ahead and develop our own graph. Let's see if we can do this and then look at our options. Okay? So, Z, why is right there in the ex is coming out of the board. But I don't even have to worry about the x axis because Jay is the Y. Playing Kay is a seaplane, so I know I only have something in the Y, um Z playing. Okay, so if I start right off at Ti umm is zero, then my wife value is out here. Negative one. If I have a, um t of negative one. Um, I have nothing on the why. It's gonna be in my Z. So this will be down here at negative too. And the same thing with a positive one. I will be up here at Ah, positive, too. So I'm gonna start with negative one and go to positive, too, in that direction. And so if I look at my choices, this is gonna be choice F.

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Chapter 3: – Chemical Bonding
B.sc 1st year Book
(Page 1)

# CHEMICAL BONDING

A chemical bond is defined as the force of attraction between two or more atoms or Ions or molecules that holds various constituent particles”. The combination of these particles (atoms, ions, or molecules) is called chemical bonding; the followings are the reasons for their combination.
Concept of decrease of energy of combining atoms: When the two atoms form a bond, their potential energy decreases, i.e. bonded atoms have lower energy than non-bonded atoms and are more stable due to chemical bonding.

## Electronic theory of valency or octet rule :

In 1916 the scientist’s Lewis, Kossel, and Langmuir proposed that while acquiring noble gases configuration that is He(Z=2), Ne: 2, 8; Ar: 2,8, 8; Kr:2,8,18,8 and so on. The elements of noble gases have 8 electrons in their outermost orbit(except He) and do not react to form a chemical bond hence the other atoms also try to acquire 8 electrons In their outermost orbit and become most stable. This is called the ‘octet rule, But helium has only two electrons called ‘duplet’ and is also as stable as an octet. The octet rule is followed by an atom or Ion by forming a chemical bond between the atoms either by transfer of electrons in case of ionic bonds or by sharing of electrons between the atoms, forming covalent bonds or by donation of a lone pair of electrons between the atoms forming a coordinate bond to complete their octet.
“The tendency of the atoms to complete 8 electrons in the outermost shell is called “octet rule” or “electronic theory of valency“.

### The main features of the electronic theory of valency are as follows.

(i) When an atom adopts 8 electrons in its outermost orbit ( 2 in the case of He) by transferring, gaining, or sharing it becomes chemically stable and does not take part in chemical combination.
(ii) The atoms having less than 8 electrons in their outermost shell are chemically active and have a tendency to combine with other atoms. If the atoms have less than 4 electrons in their outermost shell, has a tendency to lose them and if they have more than 4 electrons in the outermost shell, tend to gain electrons during the chemical combination to attain a stable configuration.
(iii) The two atoms chemically combine together either by the transfer of an electron from the outermost shell of one atom to another or by sharing of 1,2,3 or 4 electrons between the valance shell of both combining atoms. Such a chemical combination produces a stable configuration of the rule of the octet.
(iv) The tendency of an atom to transfer or share its electron is a measure of its chemical reactivity.

## lonic/or Electrovalent bond:

The ionic/or electrovalent bond is formed by the loss of one or more electrons from the valence shell of an atom that is highly electropositive to the valence shell of a highly electronegative atom. The atom which loses electrons forms a cation and the atom which accepts electrons forms an anion. There is an electrostatic force of attraction between the opposite Ions resulting in the formation of an ionic bond. During the formation of an Ionic bond, energy is released. The cations and anions generated by the transference of electrons acquire inert gas configuration (ns 2 or ns2np6). Examples of the formation of ionic bonds In some electrovalent compounds are given below.
$$\left(a\right)\:Na\:\left(2,\:8,\:1\right)\frac{-e^-}{ }Na^+\left(2,\:8\right)$$
$$Cl\:\left(2,\:8,\:7\right)\frac{-e^-}{ }Cl^-\left(2,\:8,\:8\right)$$
Na+ + Cl →  NaCl (Due to electrostatic force of attraction)
$$\left(b\right)\:Mg\left(2,\:8,\:2\right)\:\frac{-2e^-}{ }\:Mg^{2+}\left(2,\:8\right)$$
$$O\:\left(2,\:6\right)\:\frac{+2e^-}{ }O^{2-}\left(2,\:8\right)$$
Mg2+ + O2+ → MgO
$$\left(c\right)\:Al\left(2,\:8,\:3\right)\frac{-3e^-}{ }Al^{3+}\left(2,\:8\right)$$
$$3\:Cl\left(2,\:8,\:7\right)\:\frac{+3e^-}{ }3Cl^-\left(2,\:8,\:8\right)$$
Al3+ + 3Cl → AlCl3

### Characteristics of Ionic compounds :

1. They are formed by the combination of metal and non-metals.
2. Ionic bonds are formed between cations and anions by the electrostatic force of attraction.
3. These compounds are highly soluble in polar solvents like H2O, NH3, Liq. SO2 etc and insoluble in nonpolar solvents like CCl4, C6H6, and Liq. CS2 etc.
4. They ionize in solution or in a fused state.
5. They are good conductors of heat and electricity in solution or in a molten state. But in a crystalline state, they are bad conductors of electricity because ions are not movable.
6. They show high values of melting and boiling points.
7. They possess a non-directional polar linkage. Hence they are incapable of exhibiting any type of isomerism.
8. They undergo fast reactions.
9. They have high density.
10. They are highly brittle:
11. They have regular symmetry.

### Factors Influencing the Formation of Ionic bond:

In the formation of an ionic bond between A and X, the following steps will occur.

(i) Atom ‘A’ releases an electron by absorbing energy equal to its ionization energy (I.E.) and converted into cation c+.

A(g) + energy absorbed (I.E) → c+(g) + e

(ii) In this step atom Xaccepts the released electron from atom A and converted it into an anion,
X(g) + e → a(g) + Energy released (EA)

(iii) Lastly cations c+ and anion a- combine together by an electrostatic force of attraction to give a stable ionic crystal c+a- In this step, energy equal to lattice energy of the Ionic compound c+a- Is released, which is represented as
$$\Delta \:H_{Lattice}=-\frac{e^2}{r_c^++r^-_a}$$
where the negative sign indicates the energy is released; a represents ionic charges on c+ and aions; rc+ andra are ionic radii of cation and anion respectively. In other words, the greater the value of lattice energy of the resulting ionic compound, the greater will be the ease of its formation
$$c^+_{\left(g\right)}+a_{\left(g\right)}^-\rightarrow c^+a^-\left(s\right)\:+\:Energy\:released\:\left(\Delta H_{lattice}\right)$$

Thus, the overall energy change (E-Ionic ) in the formation of ionic crystal c+a−(s) is given by :
$$E_{Ionic}=\left(I.P.\right)_A-\left(EA\right)_X-\frac{e^2}{r^+_c+r^-_a}$$

If c+a is stable, the ionic energy (Elonic) is negative.
1,0.
i.e $$\frac{e^2}{r^+_c+r^-_a}\:>\:\left[\left(IP\right)_A-\left(EA\right)_X\right]$$

Thus, the different steps discussed above indicate that the following factors facilitate the formation of an ionic crystal by a metal and a non-metal.
(i) Atom ‘A‘ must-have low ionization energy.
(ii) Atom ‘ X ‘ should have a high value of electron affinity.
(iii) The value of lattice energy of c+a- ionic crystal must be high.
(iv) There should be a large electronegativity difference (>1.80) between the atoms ‘ A ‘ and ‘ X ‘ 1. e. A should be highly electropositive and X should be highly electronegative. For example, the energy change involved in the formation of ionic bonds in NaCl is discussed in the Born-Haber cycle. (ref, fig, 2.08).

## Variable electrovalency (oxidation states):

The elements which change their valency and show two or more valencies are said to possess variable valency. The metals belonging to p and d-blocks show variable electrovalency. The reason behind this behavior is either an inert pair effect in p-block or incomplete (n−1)d-orbitals in the case of transitional elements.

### (a) p-block elements :

 Heavier elements of group IIIA :Ga(+3, +1); In(+3, +1); Ti(+3, +1); Heavier elements of group IVA :Ge(+4, +2); Sn(+4, +2); Pb(+4, +2); Heavier elements of group VA :Sb(+5, +3); Bi(+5, +3);
These metals show greater stability in a lower oxidation state as compared to a higher one. This is due to the inert pair effect.

### (b) Transitional elements :

The transitional elements having the electronic configuration (π−1)s2,p6, d1 to 10ns1 to 2 form cations with variable positive electrovalency. For example, Fo shows +2,+3, and titanium show +2,+3, and +4oxidation states due to loss of 3 d and 4 s electrons respectively. Because of the fact that the (n−1)d and ns orbitals in the atoms of these elements have almost the same energy, the removal of electrons from (n−1) d-orbitals takes place as easily as electrons of ns-orbitals.
When all the valence electrons (electrons of incomplete shell) are removed from an atom, the residual part is called the ‘core’. The core may have a stable configuration similar to inert gases (more common in p-block elements) or pseudo-inert gas configuration as observed in transitional elements and heavier p-block elements. These are the two reasons for the variable electrovalency of the elements.

### (c) Pseudo-inert gas configuration :

Those elements have 18 electrons in their outermost orbits such as Cu, Ag, Au of IB, and Zn, Cd, and Hgof IIB groups.
 IB Cu=3s23p63d10s1 Cu+=3s23p63d10 Cu2+=3s23p63d9 Ag=4s24p64d105s1 Ag+=4s24p64d10 Ag2+=4s24p64d9 Au=5s25p65d106s1 Au+=5s25p65d10 Au3+=5s25p65d8 IIB Zn=3s23p64d105s2 Zn2+=3s23p64d10 Cd=4s24p64d105s2 Cd2+=4s24p64d10 Hg=5s25p85d106s2 Hg2+=5s25p85d10
Monovalent cations of the IB group are not stable because of the presence of unstable (n−1) s2p6d10 configuration. Consequently, they are converted into Cu2+, Ag2+, and Au3++ which are relatively more stable and have 18 electrons in the Kernel, and are held more strongly with increased nuclear charge.
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# 5.2 Unit circle: sine and cosine functions  (Page 3/12)

 Page 3 / 12

If $\mathrm{cos}\left(t\right)=\frac{24}{25}$ and $t$ is in the fourth quadrant, find $\text{sin}\left(t\right).$

$\mathrm{sin}\left(t\right)=-\frac{7}{25}$

## Finding sines and cosines of special angles

We have already learned some properties of the special angles, such as the conversion from radians to degrees. We can also calculate sines and cosines of the special angles using the Pythagorean Identity    and our knowledge of triangles.

## Finding sines and cosines of 45° angles

First, we will look at angles of $\text{\hspace{0.17em}}45°\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}\frac{\pi }{4},$ as shown in [link] . A $\text{\hspace{0.17em}}45°–45°–90°\text{\hspace{0.17em}}$ triangle is an isosceles triangle, so the x- and y -coordinates of the corresponding point on the circle are the same. Because the x- and y -values are the same, the sine and cosine values will also be equal.

At $\text{\hspace{0.17em}}t=\frac{\pi }{4}$ , which is 45 degrees, the radius of the unit circle bisects the first quadrantal angle    . This means the radius lies along the line $\text{\hspace{0.17em}}y=x.\text{\hspace{0.17em}}$ A unit circle has a radius equal to 1. So, the right triangle formed below the line $\text{\hspace{0.17em}}y=x\text{\hspace{0.17em}}$ has sides $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and and a radius = 1. See [link] .

From the Pythagorean Theorem we get

${x}^{2}+{y}^{2}=1$

Substituting $\text{\hspace{0.17em}}y=x,$ we get

${x}^{2}+{x}^{2}=1$

Combining like terms we get

$2{x}^{2}=1$

And solving for $\text{\hspace{0.17em}}x,$ we get

In quadrant I, $\text{\hspace{0.17em}}x=\frac{1}{\sqrt{2}}.\text{\hspace{0.17em}}$

At $\text{\hspace{0.17em}}t=\frac{\pi }{4}\text{\hspace{0.17em}}$ or 45 degrees,

$\begin{array}{l}\left(x,y\right)=\left(x,x\right)=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\hfill \\ x=\frac{1}{\sqrt{2}},y=\frac{1}{\sqrt{2}}\hfill \\ \mathrm{cos}\text{\hspace{0.17em}}t=\frac{1}{\sqrt{2}},\mathrm{sin}\text{\hspace{0.17em}}t=\frac{1}{\sqrt{2}}\hfill \end{array}$

If we then rationalize the denominators, we get

$\begin{array}{l}\mathrm{cos}\text{\hspace{0.17em}}t=\frac{1}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{2}}{2}\hfill \\ \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t=\frac{1}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{2}}{2}\hfill \end{array}$

Therefore, the $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ coordinates of a point on a circle of radius $\text{\hspace{0.17em}}1\text{\hspace{0.17em}}$ at an angle of $\text{\hspace{0.17em}}45°\text{\hspace{0.17em}}$ are $\text{\hspace{0.17em}}\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right).\text{\hspace{0.17em}}$

## Finding sines and cosines of 30° and 60° angles

Next, we will find the cosine and sine at an angle of $\text{\hspace{0.17em}}30°,$ or $\text{\hspace{0.17em}}\frac{\pi }{6}\text{\hspace{0.17em}}$ . First, we will draw a triangle inside a circle with one side at an angle of $\text{\hspace{0.17em}}30°,$ and another at an angle of $\text{\hspace{0.17em}}-30°,$ as shown in [link] . If the resulting two right triangles are combined into one large triangle, notice that all three angles of this larger triangle will be $\text{\hspace{0.17em}}60°,$ as shown in [link] .

Because all the angles are equal, the sides are also equal. The vertical line has length $\text{\hspace{0.17em}}2y,$ and since the sides are all equal, we can also conclude that $\text{\hspace{0.17em}}r=2y\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}y=\frac{1}{2}r.\text{\hspace{0.17em}}$ Since $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t=y\text{\hspace{0.17em}}$ ,

$\mathrm{sin}\left(\frac{\pi }{6}\right)=\frac{1}{2}r$

And since $\text{\hspace{0.17em}}r=1\text{\hspace{0.17em}}$ in our unit circle    ,

Using the Pythagorean Identity, we can find the cosine value.

The $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ coordinates for the point on a circle of radius $\text{\hspace{0.17em}}1\text{\hspace{0.17em}}$ at an angle of $\text{\hspace{0.17em}}30°\text{\hspace{0.17em}}$ are $\text{\hspace{0.17em}}\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right).\text{\hspace{0.17em}}$ At $t=\frac{\pi }{3}$ (60°), the radius of the unit circle, 1, serves as the hypotenuse of a 30-60-90 degree right triangle, $\text{\hspace{0.17em}}BAD,$ as shown in [link] . Angle $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ has measure $\text{\hspace{0.17em}}60°.\text{\hspace{0.17em}}$ At point $\text{\hspace{0.17em}}B,$ we draw an angle $\text{\hspace{0.17em}}ABC\text{\hspace{0.17em}}$ with measure of $\text{\hspace{0.17em}}60°.\text{\hspace{0.17em}}$ We know the angles in a triangle sum to $\text{\hspace{0.17em}}180°,$ so the measure of angle $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ is also $\text{\hspace{0.17em}}60°.\text{\hspace{0.17em}}$ Now we have an equilateral triangle. Because each side of the equilateral triangle $\text{\hspace{0.17em}}ABC\text{\hspace{0.17em}}$ is the same length, and we know one side is the radius of the unit circle, all sides must be of length 1.

The measure of angle $\text{\hspace{0.17em}}ABD\text{\hspace{0.17em}}$ is 30°. So, if double, angle $\text{\hspace{0.17em}}ABC\text{\hspace{0.17em}}$ is 60°. $\text{\hspace{0.17em}}BD\text{\hspace{0.17em}}$ is the perpendicular bisector of $\text{\hspace{0.17em}}AC,$ so it cuts $\text{\hspace{0.17em}}AC\text{\hspace{0.17em}}$ in half. This means that $\text{\hspace{0.17em}}AD\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}\frac{1}{2}\text{\hspace{0.17em}}$ the radius, or $\text{\hspace{0.17em}}\frac{1}{2}.\text{\hspace{0.17em}}$ Notice that $\text{\hspace{0.17em}}AD\text{\hspace{0.17em}}$ is the x -coordinate of point $\text{\hspace{0.17em}}B,$ which is at the intersection of the 60° angle and the unit circle. This gives us a triangle $\text{\hspace{0.17em}}BAD\text{\hspace{0.17em}}$ with hypotenuse of 1 and side $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ of length $\text{\hspace{0.17em}}\frac{1}{2}.\text{\hspace{0.17em}}$

#### Questions & Answers

The average annual population increase of a pack of wolves is 25.
Brittany Reply
how do you find the period of a sine graph
Imani Reply
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
Jhon Reply
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
Baptiste Reply
the sum of any two linear polynomial is what
Esther Reply
divide simplify each answer 3/2÷5/4
Momo Reply
divide simplify each answer 25/3÷5/12
Momo
how can are find the domain and range of a relations
austin Reply
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
Diddy Reply
6000
Robert
more than 6000
Robert
can I see the picture
Zairen Reply
How would you find if a radical function is one to one?
Peighton Reply
how to understand calculus?
Jenica Reply
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
rachel Reply
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
Reena Reply
what is foci?
Reena Reply
This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
Chris
how to determine the vertex,focus,directrix and axis of symmetry of the parabola by equations
Bryssen Reply

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## Sunday, July 28, 2013

### Crowdsourcing and information theory: The disconnect

In crowdsourcing, redundancy is a common approach to ensure quality. One of the questions that arises in this setting is the question of equivalence. Let's assume that a worker has a known probability $q$ of giving a correct answer, when presented with a choice of $n$ possible answers. If I want to simulate one high-quality worker workers of quality $q$, how many workers of quality $q' < q$ do we need?

Information Theory and the Noisy Channel Theorem

Information theory, and the noisy channel theorem, can give an answer to the problem: Treat each worker as a noisy channel, and measure the "capacity" of each user. Then, the sum of the capacities of the different workers should give us the equivalent capacity of a high-quality worker.

We have that the capacity $C(q,n)$ of a worker with quality $q$, who returns the correct answer with probability $q$, when presented with $n$ choices, is:

$C(q,n) = H(\frac{1}{n}, n) - H(q, n)$

where $H(q,n) = -q \cdot \log(q) - (1-q) \cdot \log(\frac{1-q}{n-1})$ is the entropy (aka uncertainty) of the worker.

Examples

The value $H(\frac{1}{n}, n) = \log(n)$ is the initial entropy that we have for the question, when no answers are given. Intuitively, when $n=2$, the initial uncertainty is equal to $\log(2)=1$ bit, since we need one bit to describe the correct answer out of the 2 available. When $n=4$, the uncertainty is equal to $\log(4)=2$ bits, as we need 2 bits to describe the correct answer out of the 4 available.

Therefore, a perfect worker, with quality $q=1$ will have $H(1,n)=0$ entropy, and therefore the capacity of a perfect worker is $\log(n)$.

Can Two Imperfect Workers Simulate a Perfect One?

Now, here comes the puzzle. Assume that we have $n=4$, and the workers have to choose among 4 possible answers. We also have also two workers with $q=0.85$, that select with 85% probability the correct answer out of 4 available. These workers have each capacity equal to $C(0.85, 4) = 1.15$ bits. At the same time, we have one perfect worker with $q=1$. This worker has a capacity of $C(1,4)=2$ bits. So, in principle, the two noisy workers are sufficient to simulate a perfect worker (and would leave a remaining 0.3 bits to use :-)

What am I missing?

My problem is that I do not get how to reach this theoretical limit. I cannot figure out how to use these two workers with $q=0.85$, in order to reconstruct the correct answer. Asking two workers to work in parallel will not cut it (still possible for both workers to agree and be incorrect). Sequential processing (get first a worker to select two out of the four answers, then the second one pick the correct out of the two) seems more powerful, but again I do not understand how to operationalize this.

According to information theory, these two  $q=0.85$ workers are equivalent, on average, with one perfect $q=1.0$ worker. (Actually, they seem to carry more information). And even if we avoid perfection, and we set target quality at $q=0.99$,  $C(0.99,4)=1.9$. I still cannot see how I can combine two workers with 85% accuracy to simulate a 99% accurate worker.

• Update 1 (thanks to Michael Nielsen): Information theory operates over a large amount of transmitted information, so posing the question as "answering a single question" makes it sound more impossible than it should.

We need 2 bits of information to transfer the answer for a multiple choice question with n=4 choices. Say that we have a total of N such questions. So, we need 2N bits to transfer perfectly ann the answers. If we have perfect workers, with $q=1$, we have that $C(1,4)=2$, and we need 2N bits / 2 bits/answer = N answers, from these workers.

Now, let's say that we have workers with $q'=0.85$. In that case $C(0.85, 4) = 1.15$ bits per answer. Therefore, we need 2N bits / 1.15 bits/answer = 1.74N answers from these  85% accurate workers in order to perfectly reconstruct the answers for these N questions.

So, if we get from these 85% workers a total of 100 answers (each one 85% correct), we should be able to reconstruct the 100% correct answer for ~57 (=100/1.74) questions.

Of course we should be intelligent of what exactly to ask and get these 100 answers.

I see in Wikipedia, in the article about the noisy channel theorem, that "Simple schemes such as 'send the message 3 times and use a best 2 out of 3 voting scheme if the copies differ' are inefficient error-correction methods" and that "Advanced techniques such as Reed–Solomon codes and, more recently, turbo codes come much closer to reaching the theoretical Shannon limit". Unfortunately, my familiarity with such coding schemes is minimal (i.e., I have no clue), so I cannot understand their applicability in a crowdsourcing setting.

So, here is my question: What coding schemes should we use in crowdsourcing in order to get closer to the theoretical limits given by Shannon? Or what is the fundamental thing that I miss? Because I do feel that I am missing something...

Any help would be appreciated.

• Update 2 (thanks to the comments by stucchio and syrnik): Information theory predicts that we can always recover the perfect answers from noisy workers, given sufficient worker capacity. For anyone that has worked in crowdsourcing, this sounds very strange, and seems practically infeasible. The problem does not seem to be in the assumptions of the analysis; instead it seems to rely on the feasibility of implementing a proper encoding scheme on top of human computation.

The key concept in information theory is the coding scheme that is used to encode the information, to make the transmission of information robust to errors. Information theory does not say how we can recover this perfect information using a noisy channel. Over time, researchers came up with appropriate encoding schemes that approach very closely the theoretical maximum (see above, Reed-Solomon codes, turbo codes, etc). However, it is not clear whether these schemes are translatable into a human computation setting.

Consider this gross simplification (which, I think, is good enough to illustrate the concept): In telecommunications, we put a "checksum" together with each message, to capture cases of incorrect information transmission. When the message gets transmitted erroneously, the checksum does not match the message content. This may be the result of corruption in the message content, or the result of corruption in the checksum (or both). In such cases, we re-transmit the message. Based on the noise characteristics of the channel, we can decide how long the message should be, how long the checksum should be, etc., to achieve maximum communication efficiency.

For example, consider using a parity bit, the simplest possible checksum computation. We count the number of 1 bits in the message: if the number of 1's is odd, we set the parity bit to 1, if the number of 1's is even, we set the parity bit to 0. The extra parity bit increases the size of the message but can be used to detect errors when the message gets transmitted over a noisy channel, and reduce the error rate. By increasing the number of parity bits we can reduce the error rate to arbitrarily low levels.

In a human computation setting, computing such a checksum is highly non-trivial. Since we do not really know the original message, we cannot compute at the source an error-free checksum. We can of course try to create "meta"-questions that will try to compute the "checksum" or even try to modify all the original questions to have an error-correcting component in them.

See now the key difference: In information theory, we have computed error-free the message to be transmitted with built-in error-correction. Consider now the same implementation in a human computation setting: We ask the user to inspect the previous k questions, and report some statistics about the previously submitted answers. The user now operates on the noisy message (i.e., the given, noisy answers), therefore even the error-free computation of the checksum is going to be noisy, defeating the purpose of an error-correcting code.

Alternatively, we can try to take the original questions, and try to ask them in a way that enforces some error-correcting capability. However, it is not clear that these "meta-questions" are going to have the same level of complexity for the humans, even if in the information-theoretic sense, they carry the same amount of information.

It seems that in a human computation setting we have noisy computation, as opposed to noisy transmission. Since the computation is noisy, there is a good chance that the computation these "checksums" is going to be correlated with the original errors. Therefore, it is not clear whether we can actually implement the proper encoding schemes on top of human computers, to achieve the theoretical maximums predicted by information theory.

Or, at least, this seems like a very interesting, and challenging research problem.