User:Jkc0113

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The archaic lower case "s": ſ

The total area:

$\frac {\sqrt {3}}{4} + \sum_{n=2}^ \infty [\frac {\sqrt {3}}{4} (3) (\frac {1}{3})^{n-1} (4)^{n-2} ]\,$

The total perimeter:

$3 + 3 \sum_{n=1}^ \infty [(\frac{1}{3})^n (4)^{n-1}] \,$

I don't know how to solve them, but it's a start.

This is currently serving as my notepad for various physics formulas and stuff like that.

$a=\frac{8eV_{dc}}{mr^2 \omega^2}$

Q = 4 e V r f (m r 2 ω 2)

When a charged particle moves through a magnetic field B, it feels a force F given by the cross product:

$\vec{F} = q \vec{v} \times \vec{B}$

where $q\,$ is the electric charge of the particle, $\vec{v} \,$ is the velocity vector of the particle, and $\vec{B} \,$ is the magnetic field.

$F = q v B \sin\theta\,$

where $\theta \,$ is the angle between the $\vec{v} \,$ and $\vec{B} \,$ vectors.

Scalar form of Coulomb's Law

If one is interested only in the magnitude of the force, and not in its direction, it may be easiest to consider a simplified, scalar version of the law:

$F = k_C \frac{|q_1| |q_2|}{r^2}$

where:

$F \$ is the magnitude of the force exerted,

$q_1 \$ is the charge on one body,

$q_2 \$ is the charge on the other body,

$r \$ is the distance between them,

$k_C = \frac{1}{4 \pi \epsilon_0} \approx$ 8.988×10⁹ N m² C^-2 (also m F^-1) is the electrostatic constant or Coulomb force constant, and

$\epsilon_0 \approx$ 8.854×10⁻¹² C² N^-1 m^-2 (also F m^-1) is the permittivity of free space, also called electric constant, an important physical constant.

In cgs units, the unit charge, esu of charge or statcoulomb, is defined so that this Coulomb force constant is 1.

Table of derived quantities

Particle property

Relationship

Field property

Vector quantity

Force (on 1 by 2)

$\mathbf{F}_{12}= {1 \over 4\pi\epsilon_0}{q_1 q_2 \over r^2}\mathbf{\hat{r}}_{21} \$

$\mathbf{F}_{12}= q_1 \mathbf{E}_{12}$

Electric field (at 1 by 2)

$\mathbf{E}_{12}= {1 \over 4\pi\epsilon_0}{q_2 \over r^2}\mathbf{\hat{r}}_{21} \$

Relationship

$\mathbf{F}_{12}=-\mathbf{\nabla}U_{12}$

$\mathbf{E}_{12}=-\mathbf{\nabla}V_{12}$

Scalar quantity

Potential energy (at 1 by 2)

$U_{12}={1 \over 4\pi\epsilon_0}{q_1 q_2 \over r} \$

$U_{12}=q_1 V_{12} \$

Potential (at 1 by 2)

$V_{12}={1 \over 4\pi\epsilon_0}{q_2 \over r}$

Theory behind a Sector Instrument

The behavior of ions in a homogeous, linear, static electric or magnetic field (separately) as is found in a sector instrument is simple. The physics are described by a single equation called the Lorentz force law. This equation is the fundamental equation of all mass specrometric techniques and applies in non-linear, non-homogeneous cases too and is an important equation in the field of electrodynamics generally.

$\mathbf{F} = q (\mathbf{E} + \mathbf{v} \times \mathbf{B}),$

where E is the electric field strength, B is the magnetic field induction, q is the charge of the particle, v is its current velocity (expressed as a vector), and × is the cross product.

So the force on an ion in a linear homogeous electric field (an electric sector) is:

$F=qE\,$ ,

in the direction of the electric field, with positive ions and opposite that with negative ions.

Electric sector from a Finnigan MAT mass spectrometer (vacuum chamber housing removed)

The force is only dependent on the charge and electric field strength. The lighter ions will be deflected more and heavier ions less due to the difference in inertia and the ions will physically separate from each other in space into distinct beams of ions as they exit the electric sector.

And the force on an ion in a linear homogeneous magnetic field (a magnetic sector) is:

$F=qvB\,$ ,

perpendicular to both the magnetic field and the velocity vector of the ion itself, in the direction determined by the right-hand rule of cross products and the sign of the charge.

The force in the magnetic sector is complicated by the velocity dependence but with the right conditions (uniform velocity for example) ions of different masses will separate physically in space into different beams as with the electric sector.

Mattauch-Herzog

The Mattauch-Herzog geometry consists of a 31.82° electric sector, a drift length which is followed by a 90° magnetic sector of opposite curvature direction. The entry of the ions sorted primarily by charge into the magnetic field produces an energy focussing effect and much higher transmission than a standard energy filter. This geometry is often used in applications with a high energy spread in the ions produced where sensitivity is nonetheless required, such as spark source mass spectrometry (SSMS) and secondary ion mass spectrometry (SIMS).

Electronics formulas

$I=\frac{dQ}{dt}\,$

$I=\frac{V}{R} Amps \,$

$Q=It Coulombs \,$

P = V I w a t t s

P = I 2 R w a t t s

$P=\frac{V^2}{R} watts$

E = P t J o u l e s (o r k W h)

$R=\frac{\rho l}{\alpha}$

R θ = R 0 (1 + α 0 θ)

$\frac{R_1}{R_2}=\frac{1+\alpha_0 \theta_1}{1+\alpha_0 \theta_2}$

V = E - I r

$r = \frac{E - V}{I}$

Part 5: Computation of Equilibrium Values

Problem 5.1

Below is the multiplication tree.

[[[[[INSERT PIC]]]]]

The probability that the dominant A allele will be transmitted to the next generation is represented below in terms of $p\,$ , $q_1\,$ , $q_2\,$ , $q_3\,$ .

$P_A=p^2q_1+p(1-p)q_2\,$

Problem 5.2

The probability that the recessive a allele will be transmitted to the next generation is represented below in terms of $p\,$ , $q_1\,$ , $q_2\,$ , $q_3\,$ .

$P_a=p(1-p)q_2+(1-p)^2q_3\,$

Problem 5.3

The proportion of dominant A alleles in the next generation, represented by $f(p)\,$ , is displayed below.

$f(p)= \frac {P_A}{P_A + P_a} = {p^2q_1+p(1-p)q_2}{p^2q_1+2p(1-p)q_2+(1-p)^2q_3}\,$

Problem 5.4

Below, we find the three solutions for $p\,$ for

$f(p) = p\,$ ,

the three values of $p\,$ at which the Hardy-Weinberg Law is satisfied.

$f(p) = p\,$

$\frac{p^2q_1 + p(1-p)q_2}{p^2q_1 + 2p(1-p)q_2 + (1-p)^2q_3} = p\,$

$p^2q_1 + p(1-p)q_2 = p^3q_1 + 2p^2(1-p)q_2 + p(1-p)^2q_3\,$

$p^2q_1 + pq_2-p^2q_2 = p^3q_1 + 2p^2q_2 - 2p^3q_2 + pq_3 - 2p^2 + p^3q_3\,$

$(p^3-p^2)q_1 + (2p^2-2p^3 + p - p^2)q_2 + (p-2p^2+p^3)q_3 = 0\,$

$p(p-1)[pq_1 + (-2p+1)q_2 + (p-1)q_3] = 0 \,$

$pq_1+(-2p+1)q_2 +(p-1)q_3=0\,$

$pq_1-2pq_2+q_2 +(pq_3-q_3=0\,$

$p(q_1-2q_2+q_3)=q_3-q_2\,$

$p^* = \frac{q_3 -q_2}{q_1-2q_2+q_3}\,$

Problem 5.5 [[[[[NEED VERBAL EXPLANATION]]]]]

a.)

$p^* = \frac{0.25 - 0.75}{0.5-2(0.75)+0.25}\,$

$p^* = \frac {2}{3}\,$

$(AA,Aa,aa)=(\frac {2}{3}^2(0.5), 2(\frac {2}{3})(1-\frac {2}{3})(0.75), (1-\frac {2}{3})^2(0.25))\,$

$(AA,Aa,aa)=(\frac {2}{9},\frac {3}{18}, \frac {1}{36})\,$

b.)

$p^* = \frac{0.25 - 0.125}{0.5-2(0.125)+0.25}\,$

$p^* = \frac {1}{4}\,$

$(AA,Aa,aa)=(\frac {1}{4}^2(0.5), 2(\frac {1}{4})(1-\frac {1}{4})(0.125), (1-\frac {1}{4})^2(0.25))\,$

$(AA,Aa,aa)=(\frac{1}{32}, \frac {3}{128}, \frac {9}{64})\,$

c.)

$p^* = \frac{0.25 - \frac {1}{3}}{0.5-2(\frac {2}{3})+(0.25)}\,$

$p^* = -1\,$

$(AA,Aa,aa)\,$ cannot be determined, because $p^* = -1\,$ is not a valid value for

p

d.)

$p^* = \frac{0.5 - (\frac {1}{3})}{0.25-2(\frac {1}{3})+0.5}\,$

$p^* = 2\,$

$(AA,Aa,aa)\,$ cannot be determined, because $p^* = 2\,$ is not a valid value for

p

Problem 5.6

Suppose that $q_3-q_2>0\,$ and $q_1-q_2<0\,$ .

Because $q_1\,$ , $q_2\,$ , and $q_3\,$ are all positive real numbers less than 1,

$0<|q_3 - q_2|<1 \,$ and $0<|q_1 - q_2|<1\,$ .

$p^*= \frac {q_3-q_2}{q_1-2q_2+q_3} = \frac {q_3-q_2}{(q_1-q_2)+(q_3-q_2)}\,$

It follows that

$p^*=\frac {(+)}{(-)+(+)}\,$

The denominator must be less than $q_3-q_2\,$ (possibly negative), so either $p^*>1\,$ or $p^*<0\,$ . Therefore $0<p^*<1\,$ is not true so $p=p^*\,$ is unphysical.

Problem 5.7

Suppose that $q_3-q_2<0\,$ and $q_1-q_2>0\,$ .

Because $q_1\,$ , $q_2\,$ , and $q_3\,$ are all positive real numbers less than 1,

$0<|q_3 - q_2|<1 \,$ and $0<|q_1 - q_2|<1\,$ .

$p^*= \frac {q_3-q_2}{q_1-2q_2+q_3} = \frac {q_3-q_2}{(q_1-q_2)+(q_3-q_2)}\,$

It follows that

$p^*=\frac {(-)}{(+)+(-)}\,$

The denominator must be less than $q_3-q_2\,$ (possibly negative), so $|p^*|>1\,$ . Therefore $0<p^*<1\,$ is not true and $p=p^*\,$ is unphysical.

Problem 5.8

Suppose that $q_3-q_2>0\,$ and $q_1-q_2>0\,$ .

Because $q_1\,$ , $q_2\,$ , and $q_3\,$ are all positive real numbers less than 1,

$0<|q_3 - q_2|<1 \,$ and $0<|q_1 - q_2|<1\,$ .

$p^*= \frac {q_3-q_2}{q_1-2q_2+q_3} = \frac {q_3-q_2}{(q_1-q_2)+(q_3-q_2)}\,$

It follows that

$p^*=\frac {(+)}{(+)+(+)}\,$

The denominator must be greater than $q_3-q_2\,$ , and the expression must be positive, so $0<p^*<1\,$ . Therefore it is possible that $p=p^*\,$ and that this is a possible occurrence in nature.

Problem 5.9

Suppose that $q_3-q_2<0\,$ and $q_1-q_2<0\,$ .

Because $q_1\,$ , $q_2\,$ , and $q_3\,$ are all positive real numbers less than 1,

$0<|q_3 - q_2|<1 \,$ and $0<|q_1 - q_2|<1\,$ .

$p^*= \frac {q_3-q_2}{q_1-2q_2+q_3} = \frac {q_3-q_2}{(q_1-q_2)+(q_3-q_2)}\,$

It follows that

$p^*=\frac {(-)}{(-)+(-)}\,$

The expression must be positive and less than one, because the absolute value of the denominator is greater than the absolute value of the numerator. Because $0<p^*<1\,$ ,

$p=p^*\,$ is a possible natural occurrence.

Part 6: Stability of Equilibrium: Numerical Analysis

[[[[[NEED EXPLANATIONS]]]]]

To find $f(p)\,$ , the proportion of dominant A alleles present in the next generation, we use the formula

$f(p)=\frac {p^2q_1 +p(1-p)q_2}{p^2q_1 + 2p(1-p)q_2 + (1-p)^2q_3},$ .

Problem 6.1

Let $q_1=\frac {1}{2}\,$ , $q_2=\frac{3}{4}\,$ , and $q_3=\frac{1}{4}\,$ .

$p^*=\frac{0.25-0.75}{0.5-2(0.75)+(0.25)}\,$

$p^*=\frac {2}{3}\,$

(a) Let $p=0.01\,$

(a) 
(b) 
(c)

The proportion of dominant A alleles moves away from $p^*\,$ with each generation and 0.01 is representative of 0. Therefore, $p=0\,$ is an UNSTABLE equilibrium value.

(b) Let $p=0.99\,$

(a) 
(b) 
(c)

The proportion of dominant A alleles moves away from $p^*\,$ with each generation and 0.99 is representative of 1. Therefore, $p=1\,$ is an UNSTABLE equilibrium value.

(c) Let $p=p^*\,$ to the nearest tent.

$p=0.7\,$

(a) 
(b) 
(c)

The proportion of dominant A alleles moves away from $p^*\,$ with each generation and 0.7 is representative of $p *$ . Therefore, $p*\,$ is an UNSTABLE equilibrium value.

Problem 6.2

Let $q_1=\frac{1}{2}\,$ , $q_2=\frac{1}{8}\,$ , and $q_3=\frac{1}{4}\,$ .

$p^*=\frac{0.25-0.125}{0.5-2(0.125)+0.25}\,$

$p^*=0.25\,$ .

(a) Let $p=0.01\,$ .

(a) 
(b) 
(c)

The proportion of dominant A alleles moves away from $p^*\,$ with each generation and 0.01 is representative of 0. Therefore, $p=0\,$ is an UNSTABLE equilibrium value.

(b) Let $p=0.99\,$

(a) 
(b) 
(c)

The proportion of dominant A alleles moves closer to $p^*\,$ with each generation and 0.99 is representative of 1. Therefore, $p=1\,$ is a STABLE equilibrium value.

(c) Let $p=p^*\,$ to the nearest tenth.

$p=0.3\,$

(a) 
(b) 
(c)

The proportion of dominant A alleles moves away from $p^*\,$ with each generation and 0.3 is representative of $p *$ . Therefore, $p*\,$ is an UNSTABLE equilibrium value.

Problem 6.3

Let $q_1=\frac {1}{2}\,$ , $q_2=\frac{1}{3}\,$ , and $q_3=\frac{1}{4}\,$ .

$p^*=\frac{0.25-\frac{1}{3}}{0.5-2(\frac{1}{3})+0.25}\,$

$p^*=-1\,$

(a) Let $p=0.01\,$

(a) 
(b) 
(c)

$p *$ does not lie between 0 and 1, so $p *$ does not represent a physical equilibrium value, so the population's stability with respect to its polymorphic equilibrium value can not be evaluated.

(b) Let $p=0.99\,$

(a) 
(b) 
(c)

$p *$ does not lie between 0 and 1, so $p *$ does not represent a physical equilibrium value, so the population's stability with respect to its polymorphic equilibrium value can not be evaluated.

(c) Let $p=p^*\,$ to the nearest tent.

$p=-1\,$

$p *$ does not lie between 0 and 1, so $p *$ does not represent a physical equilibrium value, the population's stability with respect to its polymorphic equilibrium value can not be evaluated. Also, -1 is not a valid physical proportion of alleles, so the value of $f(p)\,$ has no meaning.

Problem 6.4

Let $q_1=\frac{1}{4}\,$ , $q_2=\frac{1}{3}\,$ , and $q_3=\frac{1}{2}\,$ .

$p^*=\frac{0.5-\frac{1}{3}}{0.25-2(\frac{1}{3})+0.5}\,$ .

$p^*=2\,$ .

(a) Let $p=0.01\,$

(a) 
(b) 
(c)

$p *$ does not lie between 0 and 1, so $p *$ does not represent a physical equilibrium value, so the population's stability with respect to its polymorphic equilibrium value can not be evaluated.

(b) Let $p=0.99\,$

(a) 
(b) 
(c)

$p *$ does not lie between 0 and 1, so $p *$ does not represent a physical equilibrium value, so the population's stability with respect to its polymorphic equilibrium value can not be evaluated.

(c) Let $p=p^*\,$ to the nearest tent.

$p=2\,$

$p *$ does not lie between 0 and 1, so $p *$ does not represent a physical equilibrium value, the population's stability with respect to its polymorphic equilibrium value can not be evaluated. Also, 2 is not a valid physical proportion of alleles, so the value of $f(p)\,$ has no meaning.

Part 7: Stability of Equilibrium: Analysis Using Calculus

[[[[[GIVE VERBAL EXPANATIONS]]]]]

Problem 7.1

$L(p)=f(a)+f'(a)(p-a).\,$

First we must find the derivative, $f'(p)\,$ .

By the Quotient rule,

$f'(p)=\frac {[2(q_1-q_2)p + q_2][(q_1-2q_2+q_3)p^2 - 2(q_3-q_2)p+q3]-[2(q_1-2q_2+q_3)p-2(q_3-q_2)][(q_1+q_2)p^2+pq_2]}{[(q_1-2q_2+q_3)p^2-2(q_3-q_2)p+q_3]^2}\,$

$f'(0)=\frac {q_2p}{q_3}\,$ .

Problem 7.2

Problem 7.3

$p=0\,$ is a stable equilibrium if and only if $f'(0)<1\,$ , because if the rate that $f(p)\,$ changed were greater than 1, $f(p)\,$ would increase and more dominant A alleles would be present, causing $p\,$ to be greater than 0.

Problem 7.4

$f'(1)=\frac{[(2q_1-2q_2)+q_2][(q_1-2q_2+q_3)-2q_3+2q_2+q_3]-[(2q_1-4q_2+2q_3)-2q_3+2q_2][(q_1+q_2)+q_2]}{[q_1-2q_2+q_3-2q_3+2q_2+q_3]^2}\,$

$f'(1)=\frac{(2q_1-q_2)q_1-(2q_1-2q_2)(q_1+2q_2}{q_1^2}\,$

$f'(1)=\frac{q_2(4q_2-3q_1)}{q_1^2}\,$

Problem 7.5

[[[[[SEPARATE PAGE]]]]]

Problem 7.6

Suppose that $q_2>q_1\,$ and $q_2>q_3\,$ . We will determine the stability of this polymorphic state.

$\frac{q_1q_2-2q_1q_3+q_2q_3}{q_2^2-q_1q_3} + 1\,$

$\frac{q_1q_2-2q_1q_3+q_2q_3-q_2^2+q_1q_3}{q_2^2-q_1q_3}\,$

$\frac{q_1q_2-q_1q_3+q_2q_3-q_2^2}{q_2^2-q_1q_3}\,$

$\frac{(q_1-q_2)(q_2-q_3)}{q_2^2-q_1q_3}\,$

$q_2>q_1\,$ and $q_2>q_3\,$ , so $(q_1-q_2)\,$ must be negative, $(q_2-q_3)\,$ must be positive, and $(q_2^2-q_1q_3)\,$ must be positive, so

$\frac{(q_1-q_2)(q_2-q_3)}{q_2^2-q_1q_3}=\frac {(-)(+)}{(+)}= (+)\,$ .

Therefore $f'(p^*)-1\,$ is negative, and therefore stable because $f'(p^*)<1\,$ .

Problem 7.7

Problem 7.8

Part 8: Estimating $p\,$ in the Hardy-Weinberg Equilibrium Law

Problem 8.1

$p^2=13/81\,$

$p=0.40062\,$

$(1-p)^2=26/81\,$

$p=0.43344\,$

Problem 8.2

In section 4, we demonstrated that $P(a,b,c)=\frac{n!}{a!b!c!}r^as^bt^c\,$ .

We can substitute $X_1\,$ for $a\,$ , $X_2\,$ for $b\,$ , $X_3\,$ for $c\,$ , the probability $p^2\,$ that $X_1\,$ will occur for $r\,$ , the probability $2p(1-p)\,$ that $X_2\,$ will occur for $s\,$ , and the probability $(1-p)^2\,$ that $X_3\,$ will occur for $t\,$ .

By substitution, we now have the formula for the most likely value of $p\,$ .

$L(p)=\frac{n!}{X_1X_2X_3}\,(p^2)^{X_1}(2p[1-p])^{X_2}([1-p]^2)^{X_3}$ .

Problem 8.3

a.)

$L(p)=\frac{81!}{13!42!26!}\,$ $(p^2)^{13}(2p[1-p])^{42}([1-p]^2)^{26}\,$

$L(p)=(1.64299*10^{33})\,$ $(p^2)^{13}(2p[1-p])^{42}([1-p]^2)^{26}\,$

To maximize $L(p)\,$ , we will find the critical point that yields the highest value of $L(p)\,$ .

$L(p)=(1.64399*10^{33})(2^48)(p^26)(p^42)(1-p)^{42}(1-p)^{52}\,$

$L(p)=(2^{42})(1.64399*10^{33})(p^{68})(1-p)^{94}\,$

Taking the derivative and setting the derivative equal to 0,

$L'(p)=0=(2^{42})(1.64399*10^{33})[-94(1-p)^{93}p^{68}+68p^{67}(1-p)^{94}]\,$

$0=(68p^{67}(1-p)^{94}\,$

$(68p^{67}(1-p)^{94}=94(1-p)^{93}p^{68}\,$

$\frac{68}{94}(1-p)=p\,$

$\frac{37}{47}=\frac{37}{47}p=p\,$

$\frac{37}{47}=\frac{81}{47}p\,$

$p=\frac{34}{81}\,$

$L(0)=0\,$

$L(1)=0\,$

$L(p)=(2^{42})(1.64399*10^{33})(0.41975)^{68}(1-0.41975)^{94}\,$

: $L(p)=0.643474\,$ .

Therefore, the value for $p\,$ that maximizes $L(p)\,$ is 0.41975.

b.)

$(AA,Aa,aa)=(p^2,2p[1-p],[1-p]^2)\,$

$(AA,Aa,aa)=(0.643474^2,2[0.643474][1-0.643474],[1-0.643474]^2)\,$

$(AA,Aa,aa)=(0.414059,0.45883,0.127111)\,$

$(E_1,E_2,E_3)=n*(AA,Aa,aa)\,$

$(E_1,E_2,E_3)=81*(0.414059,0.45883,0.127111)\,$

$(E_1,E_2,E_3)=(33.5388,37.1653,10.296)\,$ .

Problem 8.4

Though there is not a dominant/recessive relationship between the A and G alleles, we will assign for the sake of convenience the probabilities $p^2, 2p(1-p), (1-p)^2\,$ to the occurrences of the AA, AG, and GG combinations respectively.

a.)

$L(p)=\frac {42!}{11!23!8!}(p^2)^{11}(2p[1-p])^{23}([1-p]^2)^8\,$

$L(p)=(3.37682*10^{16})(2^{23})(p^{45})(1-p)^{39}\,$

$L(p)=(2.83268*10^{23})(p^{45})(1-p)^{39}\,$

To maximize $L(p)\,$ , we will find the critical point that yields the highest value of $L(p)\,$ .

$L'(p)=[(2.83268*10^{23})][45p^{44}(1-p)^{39}-39(1-p)^{38}p^{45}]\,$

$0=[(2.83268*10^{23})][45p^{44}(1-p)^{39}-39(1-p)^{38}p^{45}]\,$

$39(1-p)^{38}p^{45}=45p^{44}(1-p)^{39}\,$

$p=\frac{45}{39}(1-p)\,$

$p=\frac{45}{39}-\frac{45}{39}p\,$

$\frac{28}{13}p=\frac{45}{39}\,$

$p=\frac{15}{28}\,$

L(0)=0 L(1)=0 L(0.535714)=0.018148

Therefore, the value for $p\,$ that maximizes $L(p)\,$ is 0.535714.

b.)

$(AA,GA,GG)=(p^2,2p[1-p],[1-p]^2)\,$

$(AA,GA,GG)=(0.535714^2,2[0.535714][1-0.535714],[1-0.535714]^2)\,$

$(AA,GA,GG)=(0.286989,0.497449,0.215561)\,$

$(E_1,E_2,E_3)=n*(AA,GA,GG)\,$

$(E_1,E_2,E_3)=42*(0.286989,0.497449,0.215561)\,$

$(E_1,E_2,E_3)=(12.0536,20.8929,9.05358)\,$ .

Part 9: Consistency with the Hardy-Weinberg Equilibrium Law

$\int_0^t \frac {e^{-x/2}}{\sqrt {2 \pi x}}\,dx$

Problem 9.1

say which for]]]]]

$t=\frac{(X_1-E_1)^2}{E_1} + \frac {(X_2-E_2)^2}{E_2} + \frac {(X_3-E_3)^2}{E_3}.\,$

$t=\frac{(13-20.25)^2}{20.25} + \frac {(42-40.5)^2}{40.5} + \frac {(26-20.25)^2}{20.25}\,$

$t=4.28395\,$

Problem 9.2

The integral $\int_0^t \frac {e^{-x/2}}{\sqrt {2 \pi x}}\,dx$ cannot be directly solved by the Trapezoid Rule or Simpson's Rule because the function $\frac {e^{-x/2}}{\sqrt {2 \pi x}}\,$ does not have a defined value at $x = 0$ (the denominator would equal 0 at that point).

Problem 9.3

We must use a u-substitution to make the integral differentiable by Simpson's Rule.

$\int_0^t \frac {e^{-x/2}}{\sqrt {2 \pi x}}\,dx$

$\frac{1}{\sqrt{2\pi x}}\int_0^t \frac {e^{-x/2}}{\sqrt x}\,dx$

Let

$\frac{1}{\sqrt{2\pi x}}\int_0^\sqrt t \frac {e^{-u^2/2}}{\sqrt u}\,2udu$

$\frac{2}{\sqrt{2\pi x}}\int_0^\sqrt t e^{-u^2/2}\,du$

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My name is Jimbo Wales. I founded Wikipedia in 2001, and since 2006 have been Chair Emeritus of the Wikimedia Foundation, which I founded in 2003. Along with Angela Beesley, I was the co-founder of Wikia.

Talk · Statement of principles · Travel · Barnstars · Pictures · Funny pictures · Wikia · Contributions · All subpages

Wikia is a completely separate organization. Wikia is working on a search engine project unrelated to Wikipedia and the Wikimedia Foundation.

Contacting me

Press inquiries: If your press inquiry is for my personal views on some general topic, contact my assistant at sierra (at) wikia (dot) com. She knows how to find me as quickly as possible. If your press inquiry is about Wikia (a completely separate company from Wikimedia/Wikipedia, extending the Wikipedia social model to become "the rest of the library"), also contact Sierra at sierra (at) wikia (dot) com. If your press inquiry is strictly about Wikipedia, you can contact me directly by email or you can call the Foundation office and speak to our communications person Sandra, at +1 727 231 0101.

General Wikipedia questions: You will probably be satisfied by contacting the help desk. Remember, if you're with the press, please follow the instructions above.

Complaints: The best thing to do, if you have a complaint, is to start with the help desk. Ask a short, friendly question, and Wikipedians will love to help you. Contacting me directly with a complaint should be reserved for after you have exhausted all other remedies.

Invitations: If you want to invite me to speak at a conference, please send an email to wikispeaker (at) gmail (dot) com or call +1 (847) 380-1794. If you call this number with complaints about Wikipedia, they will not know what you are talking about, and you will not receive any help at all.

Other inquiries (related to me or my position) can be sent by e-mail to jwales (at) wikia (dot) com. (Press inquiries by e-mail are also always welcome.) To make sure I see your e-mail, the best way to slip it by my spam filters is to mention Wikipedia in the subject or body of the e-mail.

Statement of principles

As we move forward with software and social changes, I think it is imperative that I state clearly and forcefully my views on openness and the license. This page, like all Wikipedia pages, is a living, dynamic document, which I will update and clarify as legitimate questions arise.

I should point out that these are my principles, such that I am the final judge of them. This does not mean that I will not listen to you, but it does mean that at some ultimate, fundamental level, this is how Wikipedia will be run.

(But have no fear, as you will see, below.)

1.	Wikipedia's success to date is entirely a function of our open community. This community will continue to live and breathe and grow only so long as those of us who participate in it continue to Do The Right Thing. Doing The Right Thing takes many forms, but perhaps most central is the preservation of our shared vision for the NPOV and for a culture of thoughtful, diplomatic honesty.
2.	Newcomers are always to be welcomed. There must be no cabal, there must be no elites, there must be no hierarchy or structure which gets in the way of this openness to newcomers. Any security measures to be implemented to protect the community against real vandals (and there are real vandals, who are already starting to affect us), should be implemented on the model of "strict scrutiny". "Strict scrutiny" means that any measures instituted for security must address a compelling community interest, and must be narrowly tailored to achieve that objective and no other. For example: rather than trust humans to correctly identify "regulars", we must use a simple, transparent, and open algorithm, so that people are automatically given full privileges once they have been around the community for a very short period of time. The process should be virtually invisible for newcomers, so that they do not have to do anything to start contributing to the community.
3.	*"You* can edit this page right now" is a core guiding check on everything that we do.** We must respect this principle as sacred.
4.	Any changes to the software must be gradual and reversible. We need to make sure that any changes contribute positively to the community, as ultimately determined by me, in full consultation with the community consensus.
5.	*The GNU FDL license, the openness and viral* nature of it, are fundamental to the long-term success of the site.** Anyone who wants to use our content in a closed, proprietary manner must be challenged. We must adhere very strictly to both the letter and spirit of the license.
6.	*The mailing list will remain open, well-advertised, and will be regarded as the* place for meta-discussions about the nature of Wikipedia.** Very limited meta-discussion of the nature of the Wikipedia should be placed on the site itself. Wikipedia is an encyclopedia. The topic of Wikipedia articles should always look outward, not inward at the Wikipedia itself.
7.	Anyone with a complaint should be treated with the utmost respect and dignity. They should be encouraged constantly to present their problems in a constructive way in the open forum of the mailing list. Anyone who just complains without foundation, refusing to join the discussion, I am afraid I must simply reject and ignore. Consensus is a partnership between interested parties working positively for a common goal. I must not let the "squeaky wheel" be greased just for being a jerk.
8.	Diplomacy consists of combining honesty and politeness. Both are objectively valuable moral principles. Be honest with me, but don't be mean to me. Don't misrepresent my views for your own political ends, and I'll treat you the same way.

The original version of this Statement of Principles was first published on Wikipedia on 27 October 2001.

en	This user is a native speaker of English.

de-2

Dieser Benutzer hat fortgeschrittene Deutschkenntnisse.

I am learning German - Ich lerne Deutsch

I like getting simple messages from people in German, but unfortunately I'm not quite ready yet to conduct real conversations in German.

My name in other languages

You see, Wikipedia is not only a marvelous project — it is also a marvelous interlingual project. For those who speak languages other than English, here's my name translated or transliterated to many languages...

Hello you, speaking any language different from those in the above page — just pop in, and write your own!

Quotations

Wikiquote has a collection of quotations related to:

Jimmy Wales

"We make the Internet not suck." – Discussing Wikipedia on C-SPAN

"Imagine a world in which every single person on the planet is given free access to the sum of all human knowledge. That’s what we’re doing." – Slashdot interview

"...I advise the world to relax and chill a notch or two." – Slashdot posting

"Here, we are polite, thoughtful, smart, geeky people, trying only to do something which is undoubtedly good in the world: write and give away a free encyclopedia." – Talk:Jyllands-Posten Muhammad cartoons controversy

"Freedom of speech is critical for all cultures" – BBC News Website

"To me the key thing is getting it right. And if a person's really smart and they're doing fantastic work, I don't care if they're a high school kid or a Harvard professor; it's the work that matters."

"[Wikipedia is] like a sausage: you might like the taste of it, but you don't necessarily want to see how it's made."

Greater involvement by scientists would lead to a "multiplier effect", says Wales. Most entries are edited by enthusiasts, and the addition of a researcher can boost article quality hugely. "Experts can help write specifics in a nuanced way," he says. Nature special report

"...I trust you...See that link up there 'edit this page'? Go for it. It's a wiki world!"

My travel itinerary

These are the voyages of Jimbo Wales. I'm often traveling, but you can see where I'm voyaging next. Here is an incomplete list of Wikipedia meetups (which are awesome!) I have attended:

When you absolutely, positively need Jimbo overnight, fire him as a human cannonball!

You may edit this page!

Really, you can! Please feel free to! After all, that's what Wikipedia is about!

Now, you all know that this is my user page. I like to keep it a certain way, but the thing is, I trust you. Yes, I really do. I trust that you'll add something here that makes me smile :), that informs me, or that helps to inform others. If I have things in a certain format, I trust that you will respect that format. Actually, scratch that. Since this page is just so simple and plain, my ultimate dream is that some person who thinks it is fun would come along and make it look perfect, or close to perfect. See that link up there that says 'edit this page' ? Go for it. It's a "wiki world"! – Jimbo

N.B.: Many Wikipedians watch over my user page and will edit mercilessly or even remove altogether any bad faith alterations made. This is a wiki, after all.

Comments on how this page currently looks should be directed to my talk page.


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User:Jkc0113

From Wikipedia, the free encyclopedia

Scalar form of Coulomb's Law

Table of derived quantities

Theory behind a Sector Instrument

Mattauch-Herzog

Electronics formulas

Part 5: Computation of Equilibrium Values

Part 6: Stability of Equilibrium: Numerical Analysis

Part 7: Stability of Equilibrium: Analysis Using Calculus

Part 8: Estimating $p\,$ in the Hardy-Weinberg Equilibrium Law

Part 9: Consistency with the Hardy-Weinberg Equilibrium Law

When The Sanbox was created

userboxes were already there

Contacting me

Statement of principles

I am learning German - Ich lerne Deutsch

My name in other languages

Quotations

My travel itinerary

You may edit this page!

Views

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User:Jkc0113

From Wikipedia, the free encyclopedia

Scalar form of Coulomb's Law

Table of derived quantities

Theory behind a Sector Instrument

Mattauch-Herzog

Electronics formulas

Part 5: Computation of Equilibrium Values

Part 6: Stability of Equilibrium: Numerical Analysis

Part 7: Stability of Equilibrium: Analysis Using Calculus

Part 8: Estimating in the Hardy-Weinberg Equilibrium Law

Part 9: Consistency with the Hardy-Weinberg Equilibrium Law

When The Sanbox was created

userboxes were already there

Contacting me

Statement of principles

I am learning German - Ich lerne Deutsch

My name in other languages

Quotations

My travel itinerary

You may edit this page!

Views

Navigation

Interaction

Search

Languages

Part 8: Estimating $p\,$ in the Hardy-Weinberg Equilibrium Law