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Mathematics for Chemistry/Number theory

Numbers
Real numbers come in several varieties and forms;

 * Natural numbers are integers that are greater than or equal to zero.
 * Integers are whole numbers used for counting indivisible objects, together with negative equivalents and zero, e.g. 42, -7, 0
 * Rational numbers can always be expressed as fractions, e.g. 4.673 = 4673/1000.
 * Irrational numbers, unlike rational numbers, cannot be expressed as a fraction or as a definite decimal, e.g. $$\pi$$ and $$\sqrt 2$$

It is also worth noting that the imaginary unit and therefore complex numbers are used in chemistry, especially when dealing with equations concerning waves.

Surds
The origin of surds goes back to the Greek philosophers. It is relatively simple to prove that the square root of 2 cannot be a ratio of two integers, no matter how large the integers may become. In a rather Pythonesque incident the inventor of this proof was put to death for heresy by the other philosophers because they could not believe such a pure number as the root of 2 could have this impure property.

(The original use of quadratic equations is very old, Babylon many centuries BC.) This was to allocate land to farmers in the same quantity as traditionally held after the great floods on the Tigris and Euphrates had reshaped the fields. The mathematical technology became used for the same purpose in the Nile delta.

When you do trigonometry later you will see that surds are in the trigonometric functions of the important symmetrical angles, e.g. $$ \sin 60^\circ = \frac {\sqrt{3} } {2} $$ and so they appear frequently in mathematical expressions regarding 3 dimensional space.

Notation
The notation used for recording numbers in chemistry is the same as for other scientific disciplines, and appropriately called scientific notation, or standard form. It is a way of writing both very large and very small numbers in a shortened form compared to decimal notation. An example of a number written in scientific notation is

$$4.65 \times 10^6$$

with $$.65$$ being a coefficient termed the significand or the mantissa, and $$6$$ being an integer exponent. When written in decimal notation, the number becomes

$$4650000$$.

Numbers written in scientific notation are usually normalised, such that only one digit precedes the decimal point. This is to make order of magnitude comparisons easier, by simply comparing the exponents of two numbers written in scientific notation, but also to minimise transcription errors, as the decimal point has an assumed position after the first digit. In computing and on calculators, it is common for the $$\times 10$$ ("times ten to the power of") to be replaced with "E" (capital e). It is important not to confuse this "E" with the mathematical constant e.

Engineering notation is a special restriction of scientific notation where the exponent must be divisible by three. Therefore, engineering notation is not normalised, but can easily use SI prefixes for magnitude.

Remember that in SI, numbers do not have commas between the thousands, instead there are spaces, e.g. $$18~617~132$$, (an integer) or $$1.861~713~2 {\rm x} 10^{7}$$. Commas are used as decimal points in many countries.

Exponents
Consider a number $$x^n$$, where $$x$$ is the base and $$n$$ is the exponent. This is generally read as “$$x$$ to the $$n$$” or “$$x$$ to the power of $$n$$”. If $$n=2$$ then it is common to say “$$x$$ squared”, and if $$n=3$$ then “$$x$$ cubed”. Comparing powers (exponentiation) to multiplication for positive integer values of n $$(n = 1, 2, 3\dots)$$, it can be demonstrated that

$$4x = x+x+x+x$$, i.e. four lots of $$x$$ added together

$$x^4 = x \times x \times x \times x$$, i.e. $$x$$ multiplied by itself four times.

For $$x^1$$, the result is simply $$x$$. For $$x^0$$ the result is $$1$$.

$$x^n$$ can be reduced to multiplication like this if $$n$$ is an integer: $$x^n = \underbrace{x \times \cdots \times x}_n.$$

Order of operations
When an expression contains different operations, they must be evaluated in a certain order. Exponents are evaluated first. Then, multiplication and division are evaluated from left to right. Last, addition and subtraction are evaluated left to right. Parentheses or brackets take precedence over all operations. Anything within parentheses must be calculated first. A common acronym used to remember the order of operations is PEMDAS, for "Parentheses, Exponents, Multiplication, Division, Addition, Subtraction". Another way to remember this acronym is "Please Excuse My Dear Aunt Sally".

Keep in mind that negation is usually considered multiplication. So in the case of $$-x^2$$, the exponent would be evaluated first, then negated, resulting in a negative number.

Take note of this example:

$$3 + x \times 2, \text{ let } x=5.$$

If evaluated incorrectly (left-to-right, with no order of operations), the result would be 16. Three plus five gives eight, times two is 16. The correct answer should be 13. Five times two gives ten, plus three gives 13. This is because multiplication is solved before addition.

Partial fractions
Partial fractions are used in a few derivations in thermodynamics and they are good for practicing algebra and factorisation. It is possible to express quotients in more than one way. Of practical use is that they can be collected into one term or generated as several terms by the method of partial fractions. Integration of a complex single term quotient is often difficult, whereas by splitting it up into a sum, a sum of standard integrals is obtained. This is the principal chemical application of partial fractions.

An example is

$$ \frac {x-2} {(x+1) (x-1) } = \frac { A (x-1) + B (x+1) } {(x+1) (x-1) } = \frac {A} {(x+1) } + \frac {B} {(x-1) } $$

In the above $$ x-2 $$ must equal $$ A (x-1) + B (x+1) $$ since the denominators are equal. So we set $$x$$ first to +1 giving $$ -1 = 2B$$. Therefore B = -1/2. If we set $$x = -1$$ instead $$-3 = -2A$$, therefore $$A = 3/2$$. So

$$ \frac {x-2} {(x+1) (x-1) } =  \frac {3} {2(x+1) } - \frac {1} {2(x-1) } $$

We can reverse this process by use of a common denominator.

$$ \frac {3} {2(x+1) } - \frac {1} {2(x-1) } = \frac {3 (x-1) - (x+1)} {2(x+1)(x-1) } $$

The numerator is $$2x -4$$, so it becomes

$$ \frac {(x-2) } {(x+1)(x-1) } $$

which is what we started from.

So we can generate a single term by multiplying by the denominators to create a common denominator and then add up the numerator to simplify. A typical application might be to convert a term to partial fractions, do some calculus on the terms, and then regather into one quotient for display purposes. In a factorised single quotient it will be easier to see where numerators go to zero, giving solutions to $$f(x) = 0$$, and where denominators go to zero giving infinities.

A typical example of a meaningful infinity in chemistry might be an expression such as

$$ \frac A {{(E-E_a)}^2} $$

The variable is the energy E, so this function is small everywhere, except near $$E_a$$. Near $$E_a$$ a resonance  occurs and the expression becomes infinite when the two energies are precisely the same. A molecule which can be electronically excited by light has several of these resonances.

Here is another example. If we had to integrate the following expression we would first convert to partial fractions:

$$ \frac {3x} {2x^2 - 2x -4} = \frac {3x} {2 (x+1) (x-2)} = \frac A {x+1} + \frac B {x-2} $$

so

$$ \frac 3 2 x = A (x-2) + B (x-1) $$

let $$x = 2$$ then $$3 = B$$

let $$x = 1$$ then $$\frac{3}{2} = -A$$

therefore the expression becomes

$$ \frac 3 {x-2} - \frac 3 {2(x+1)} $$

Later you will learn that these expressions integrate to give simple expressions in terms of the natural logarithm.

Problems
$${\rm (1)}\frac {3} {x^2-1}{\rm (2)}\frac {4x-2} {x^2+2x}$$

$${\rm (3)}\frac {4} {(2x+1) (x-3)}{\rm (4)}\frac {7x+6} {x^2+3x+2}$$

$${\rm (5)}\frac {x+1} {2x^2-6x+4}$$

Answers
$${\rm (1)}\frac {3} {2(x-1) } - \frac {3} {2(x+1) }{\rm (2)}\frac {5} {(x+2) } - \frac {1} {x }$$

$${~\rm (3)}\frac {4} {7(x-3) } - \frac {8} {7(2x+1) }{\rm (4)}\frac {8} {x+2 } - \frac {1} {x+1 }$$

$${\rm (5)}\frac {3} {2(x-2) } - \frac {2} {2(x-1) }$$

Polynomial division
This is related to partial fractions in that its principal use is to facilitate integration.

Divide out

$$\frac { 3x^2 -4 x -6} { 1 + x}$$

like this

3x - 7 -      x + 1  ) 3x2  -4x   -6                3x2  +3x                -                 0   -7x   -6                     -7x   -7                            1

So our equation becomes $$3 x - 7 + \frac {1}{1+x}$$

This can be easily differentiated, and integrated. If this is differentiated with the quotient formula it is considerably harder to reduce to the same form. The same procedure can be applied to partial fractions.

Substitutions and expansions
You can see the value of changing the variable by simplifying

$$\frac {{\left(\frac { J (J+1)} {\epsilon} \right) }^3 + 2 {\left(\frac { J (J+1)} {\epsilon} \right) } +1 } {{\left(\frac { J (J+1)} {\epsilon} \right) }^2 -9} - \frac {{\left(\frac { J (J+1)} {\epsilon} \right) }^2 - 2 {\left(\frac { J (J+1)} {\epsilon} \right) } +1 } {{\left(\frac { J (J+1)} {\epsilon} \right) }^2 + 9}$$

to

$$\frac {x^3 + 2x + 1 } {x^2 - 9} - \frac {x^2 - 2x + 1 } {x^2 + 9}$$

where

$$x = \frac { J (J+1)} {\epsilon}$$

This is an example of simplification. It would actually be possible to differentiate this with respect to either $$J$$ or $$\epsilon$$ using only the techniques you have been shown. The algebraic manipulation involves differentiation of a quotient and the chain rule.

Evaluating $$\frac { {\rm d} y} { {\rm d} x}$$ gives

$$ \frac{3x^2 + 2}{x^2 - 9} -2 \frac{(x^3 + 2x + 1)x}{(x^2 - 9)^2} - \frac{2x - 2}{x^2 + 9} +2 \frac{(x^2 - 2x + 1)x}{(x^2 + 9)^2} $$

Expanding this out to the $$J$$s and $$\epsilon$$s would look ridiculous.

Substitutions like this are continually made for the purpose of having new, simpler expressions, to which the rules of calculus or identities are applied.

title: Mathematics for Chemistry/Number theory


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Talk:Esperanto/Word synthesis

Venturo or veturo?
It says here: ŝip- (ship) + ventur (travel) = ŝip(o)veturo (a journey by ship) This doesn't look correct. What is there correct translation of 'a journey' in Esperanto? --JorisvS 15:05, 4 August 2006 (UTC)

The noun form of "journey" is actually voyaĝo; "journey by ship" should be ŝipovoyaĝo. But I'm still an early Esperantist, so my opinion is worth mud. --74.130.244.188 (talk) 02:16, 28 October 2008 (UTC)

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: Talk:Esperanto/Word synthesis


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File:Answer-Keys 06.jpg

Licensing
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title: File:Answer-Keys 06.jpg


text: X86 Disassembly/Floating Point Examples

title: X86 Disassembly/Floating Point Examples


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Mathematics for Chemistry/Units and dimensions

Units, multipliers and prefixes
It is usually necessary in chemistry to be familiar with at least three systems of units, Le Système International d'Unités (SI), atomic units as used in theoretical calculations and the unit system used by the experimentalists. Thus if we are dealing with the ionization energy, the units involved will be the Joule (J), the Hartree (Eh, the atomic unit of energy), and the electron volt (eV).

These units all have their own advantages;

 * The SI unit should be understood by all scientists regardless of their field.
 * The atomic unit system is the natural unit for theory as most of the fundamental constants are unity and equations can be cast in dimensionless forms.
 * The electron volt comes from the operation of the ionization apparatus where individual electrons are accelerated between plates which have a potential difference in Volts.

An advantage of the SI system is that the dimensionality of each term is made clear as the fundamental constants have structure. This is a complicated way of saying that if you know the dimensionality of all the things you are working with you know an awful lot about the mathematics and properties such as scaling with size of your system. Also, the same system of units can describe both the output of a large power station (gigaJoules), or the interaction of two inert gas atoms, (a few kJ per mole or a very small number of Joules per molecule when it has been divided by Avogadro's number).

In SI the symbols for units are lower case unless derived from a person's name, e.g. ampere is A and kelvin is K.

Note the use of capitals and lower case and the increment on the exponent being factors of 3. Notice also centi and deci are supposed to disappear with time leaving only the powers of 1000.

Conversion factors
The $$\hat { } $$, (sometimes call caret or hat), sign is another notation for to the power of. E means times 10 to the power of, and is used a great deal in computer program output.

Energy
An approximation of how much of a chemical bond each energy corresponds to is placed next to each one. This indicates that light of energy 4 eV can break chemical bonds and possibly be dangerous to life, whereas infrared radiation of a few cm-1 is harmless.

 * 1 eV = 96.48530891 kJ mol-1 (Near infrared), approximately 0.26 chemical bonds
 * 1 kcal mol-1 = 4.184000000 kJ mol-1 (Near infrared), approximately 0.01 chemical bonds
 * 1 MHz = 0.399031E-06 kJ mol-1 (Radio waves), approximately 0.00 chemical bonds
 * 1 cm-1 = 0.01196265819 kJ mol-1 (Far infrared), approximately 0.00 chemical bonds

Wavelength, generally measure in nanometres and used in UV spectroscopy is defined as an inverse and so has a reciprocal relationship.

Length
There is the metre, the Angstrom (10-10 m), the micron (10-6 m), the astronomical unit (AU)and many old units such as feet, inches and light years.

Angles
The radian to degree conversion is 57.2957795130824, (i.e. a little bit less than 60, remember your equilateral triangle and radian sector).

Dipole moment
1 Debye = 3.335640035 Cm x 10-30     (coulomb metre)

Magnetic Susceptibility
1 cm3 mol-1 = 16.60540984 JT-2 x 1030   (joule tesla2)

Old units
Occasionally, knowledge of older units may be required. Imperial units, or convert energies from BTUs in a thermodynamics project, etc.

In university laboratory classes you would probably be given material on the Quantity Calculus notation and methodology which is to be highly recommended for scientific work.

A good reference for units, quantity calculus and SI is: I. Mills, T. Cuitas, K. Homann, N. Kallay, K. Kuchitsu, Quantities, units and symbols in physical chemistry, 2nd edition, (Oxford:Blackwell Scientific Publications,1993).

Unit labels
The labelling of tables and axes of graphs should be done so that the numbers are dimensionless, e.g.  temperature is $$T / K$$,

$$\ln (k/k_0)$$ and energy mol / kJ etc.

This can look a little strange at first. Examine good text books like Atkins Physical Chemistry which follow SI carefully to see this in action.

The hardest thing with conversion factors is to get them the right way round. A common error is to divide when you should be multiplying, also another common error is to fail to raise a conversion factor to a power.

$$ \text{Conversion factor} = \frac {\text{Units required}} {\text{Units given}}$$

1 eV = 96.48530891 kJ mol-1

1 cm-1 = 0.01196265819 kJ mol-1

To convert eV to cm-1, first convert to kJ per mole by multiplying by 96.48530891 / 1. Then convert to cm-1 by multiplying by 1 / 0.01196265819 giving 8065.540901. If we tried to go directly to the conversion factor in 1 step it is easy to get it upside down. However, common sense tell us that there are a lot of cm-1s in an eV so it should be obviously wrong.

1 inch = 2.54 centimetres. If there is a surface of nickel electrode of 2 * 1.5 square inches it must be 2 * 1.5 * 2.542 square centimetres.

To convert to square metres, the SI unit we must divide this by 100 * 100 not just 100.

Dimensional analysis
The technique of adding unit labels to numbers is especially useful, in that analysis of the units in an equation can be used to double-check the answer.

An aside on scaling
One of the reasons powers of variables are so important is because they relate to the way quantities scale. Physicists in particular are interested in the way variables scale in the limit of very large values. Take cooking the turkey for Christmas dinner. The amount of turkey you can afford is linear, (power 1), in your income. The size of an individual serving is quadratic, (power 2), in the radius of the plates being used. The cooking time will be something like cubic in the diameter of the turkey as it can be presumed to be linear in the mass.

(In the limit of a very large turkey, say one the diameter of the earth being heated up by a nearby star, the internal conductivity of the turkey would limit the cooking time and the time taken would be exponential. No power can go faster / steeper than exponential in the limit. The series expansion of $$e^x$$ goes on forever even though $$1 / n!$$ gets very small.)

Another example of this is why dinosaurs had fatter legs than modern lizards. If dinosaurs had legs in proportion to small lizards the mass to be supported rises as length to the power 3 but the strength of the legs only rises as the area of the cross section, power 2. Therefore the bigger the animal the more enormous the legs must become, which is why a rhino is a very chunky looking version of a pig.

There is a very good article on this in Cooper, Necia Grant; West, Geoffrey B., Particle Physics: A Los Alamos Primer, ISBN 0521347807


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Wikijunior talk:How Things Work/Microwave oven

Well, actually, not all molecules are resonating, just the ones of water. Otherwise you could not really cook with the thing; it would blow all food apart.
 * 2006-08-04: JH, R&D professional for White Goods Manufacturer

 * Thanks for the information, if you have anymore that could help expand the module that would be appreciated. --MJBT 06:26, 17 August 2006 (UTC)

 * Someone just added "I guess that is it". I reverted the page because such a comment does not add to the module and is more appropriate for the discussion page.--MJBT 06:26, 17 August 2006 (UTC)

Heating Food
Microwaves DO NOT cook food from the inside out. Microwaves absorb (and thus heat) from the outside in. Microwaves are generated by the magnetron, so they initially exist outside of the food. The microwaves are absorbed by the food, and the deeper (thicker) the food is, the less the microwaves penetrate to the inside. The absorption decays exponentially, thus they initially are absorbed most on the outside. But the microwaves may not be totally absorbed by the outer layers, so may penetrate all the way to the center of the food, thus the food is completely heated. The "heated from the inside" is a complete myth, and upon even the most rudimentary questioning fails the common sense test.


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Mathematics for Chemistry/Differentiation

Free web-based material from HEFCE
There is a DVD on differentiation at Math Tutor.

The basic polynomial
The most basic kind of differentiation is:

$$f (x) = x^n$$

$$f^{'} (x) = n x^{n-1}$$

There are two simple rules:

 * 1) The derivative of a function times a constant is just the same constant times the derivative.
 * 2) The derivative of a sum of functions is just the sum of the two derivatives.

To get higher derivatives such as the second derivative keep applying the same rules.

One of the big uses of differentiation is to find the stationary points of functions, the maxima and minima. If the function is smooth, (unlike a saw-tooth), these are easily located by solving equations where the first derivative is zero.

The chain rule
$$\frac {dy} {dx} = \frac {dy} {dw}. \frac {dw} {dx}$$

This is best illustrated by example: find $$\frac {dy} {dx}$$ given $$y = { (x^4 + 1)}^9$$

Let $$y = u^9$$ and $$u = x^4 + 1$$.

Now $$\frac {dy} {du} = 9 u^8$$ and $$\frac {du} {dx} = 4 x^3$$

So using the chain rule we have $$9 { (x^4 + 1)}^8 .4 x^3 = 36 x^3 { (x^4 + 1)}^8$$

Differentiating a product
$$\frac {d(u.v)} {dx} = v \frac {du} {dx} + u \frac {dv} {dx}$$

Notice when differentiating a product one generates two terms. (Terms are mathematical expression connected by a plus or minus.) An important point is that terms which represent physical quantities must have the same units and dimensions or must be pure dimensionless numbers. You cannot add 3 oranges to 2 pears to get 5 orangopears. Integration by parts also generates an extra term each time it is applied.

Differentiating a quotient
You use this to differentiate $$\tan$$.

$$ \frac {d} {dx} \left( \frac {u} {v}  \right) = \frac {  v \frac {du} {dx}  - u \frac {dv} {dx} } {v^2} $$

Problems
Differentiate with respect to $$x$$

$$ 3 x^2 - x ( 1 + x) (1-x) $$

Notice we have $$(a^2-b^2)$$. $$ 4 x^7 -3 x^2 $$

$$ {( 3 x + 2 )}2 + e^x $$

$$ 8 x^6 - 12 {\sqrt x} $$

$$ x {( x + 3)}2 $$

$$ x^2 ( 3 x - ( 2 + x ) (2 - x) ) $$

Evaluate the inner brackets first.

Evaluate

$$\frac {\rm d} { {\rm d} z} \left( e^{z} - \frac {z^9} {18} - \frac 3 {z^2} \right)$$

$$ \frac {\rm d} { {\rm d} c} \left( \sqrt { c^5} \right) $$

$$ \frac {\rm d} { {\rm d} \omega} \left( \frac 1 {\omega } - \frac 1 {\omega ^2} - \frac 1 {\omega ^3} \right) $$

$$ \frac {\rm d} { {\rm d} \phi} \left( e^\phi - \left( \frac 1 {\phi^2} + \frac 3 2  \phi ^ 2 \right) \right) $$

$$ \frac {\rm d} { {\rm d} r} \left( e^{5 r} - \left( \frac a {r^3} + \frac b {r^6} + \frac c {r^9} \right) \right) $$

a, b and c are constants. Differentiate with respect to $$r$$.

$$ 3 r e^{-3r} $$

$$ \frac {\rm d} { {\rm d} x} \left( x ^6 e^{x} \right) $$

Answers
$$ 3x^2 + 6x -1 $$

$$ 28x^6 -6x $$

$$ e^x + 18x + 12 $$

$$ 48x^5 - \frac 6 {\sqrt x} $$

$$ 3x^2 + 12x + 9 $$

$$ 4x^3 + 9x^2 - 8x $$

$$ e^z + \frac{z^8}{2} + \frac{6}{z^3} $$

$$ \frac 5 2 c ^{3/2} {\rm or} \frac 5 2 {\sqrt {c^3}} $$

$$ - \frac{1}{\omega ^2} + \frac 2 {\omega ^3} + \frac 3 {\omega ^4}$$

$$e^\phi + \frac 2 {\phi^3} - 3 \phi$$

$$5 e^{5 r} + \frac {3 a} {r^4} + \frac {6 b} {r^7} + \frac {9 c} {r^{10}}$$

$$e^{-3r} ( 3 - 9 r)$$

$$x^5 e^x (x+6)$$

Harder differentiation problems
Differentiate with respect to $$x$$:

$$x^4 {( x+9)}^5$$

$$5 x^2 {( x^2 -7x +9)}^5$$

Differentiate with respect to $$r$$

$${( r^2 + 3r -1)}^3 e^{-4r}$$

Differentiate with respect to $$\omega$$

$$\frac {1} { {\omega}^4} \left( { {\omega}^2 -3 \omega -19} \right)$$

$$\frac {e^{\omega}} { {\omega}^4}$$

Evaluate

$$\frac { {\rm d} } { {\rm d} z} \left( e^{-4z} \left( {z-1}^2 \right) \right)$$

$$\frac { {\rm d} } { {\rm d} \phi} \left( \phi ^2 e^{-2\phi} \left( 1 - \frac 1 {{\phi}^2} \right) \right)$$

Using differentiation to check turning points
$$\frac{dy}{dx}$$ is the tangent or gradient. At a minimum $$\frac{dy}{dx}$$ is zero. This is also true at a maximum or an inflection point. The second gradient gives us the nature of the point. If $$\frac{d^2y}{dx^2}$$ is positive the turning point is a minimum and if it is negative a maximum. Most of the time we are interested in minima except in transition state theory.

If the equation of $$ y = x^3$$ is plotted, is is possible to see that at $$x = 0$$ there is a third kind of point, an inflection point, where both $$\frac{dy}{dx}$$ and $$\frac {d^2y} {dx^2}$$ are zero.

$$ y = f(x) \frac{dy}{dx} = f^{'}(x) \frac{d^2y}{dx^2} = f^{''}(x)$$

Plot $$x^3+x^2-6x$$ between -4 and +3, in units of 1. (It will speed things up if you factorise it first. Then you will see there are 3 places where $$f(x)=0$$ so you only need calculate 5 points.) By factorising you can see that this equation has 3 roots. Find the 2 turning points. (Differentiate once and find the roots of the quadratic equation using $$x=-b\dots$$. This gives the position of the 2 turning points either side of zero. As the equation is only in $$x^3$$ it has 3 roots and 2 maxima / minima at the most therefore we have solved everything. Differentiate your quadratic again to get $$\frac{d^2y}{dx^2}$$. Notice that the turning point to the left of zero is a maximum i.e. $$\frac{d^2y}{dx^2} = -ve$$ and the other is a minimum i.e. $$\frac{d^2y}{dx^2} = +ve$$.

What is the solution and the turning point of $$y=x^3$$.

Solve $$x^3-x = 0$$, by factorisation.

(The 3 roots are -3,0 and +2. $$\frac {{\rm d}y} { {\rm d} x} = f^{'}(x) = 3x^2 + 2x-6$$

Solutions are $$1/3(\sqrt {19} -1)$$ and $$-1/3(\sqrt {19} +1)$$, i.e. -1.7863 and 1.1196.

$$\frac{d^2y}{dx^2} = f^{''}(x)= 6x+2$$

There are 3 coincident solutions at $$x=0$$, $$\frac{d^2y}{dx^2} = 0$$, at 0 so this is an inflection point.

The roots are 0, 1 and -1.


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User:Logtar

I am a Computer Science Graduate from DeVry.

I am a Windows corporate programming, and have been traveling the internet since 1995. I am a huge motorcycle enthusiast, but I also enjoy translation, interpretation, videogames, movies and books.

My blog http://blog.logtar.com is more visited than my actual homepage http://logtar.com, but I try to update them both.

title: User:Logtar


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Mac OS X Tiger/Appendix E

=Introduction= This Appendix contains the top 30 mistakes people ask when using Mac hardware and software. It's a good idea to read through this section before you use your Mac, as it will clarify some of the things you are likely to run into at first.

=The Mistakes=

Windows switcher mistakes

 * Closing the Red Dot on the top left and thinking the program has quit like it might have in Windows.

 * People using the control button on the keyboard like they would on a Windows operating system instead of Apple's alternative control button which is marked by the apple logo and is often referred to as the command button.

Hardware Mistakes

 * In Macintosh computers, the "Motherboard" is referred to as the "Logic Board"

Software Mistakes

 * Running programs directly from disk images without dragging them to the Applications folder. The disk image is just a virtual disk created by the .dmg, the program inside should be copied to your computer so you can eject the disk image and trash the .dmg file.

 * The Dock only has links to files (with the exception of Trash), the actual Applications are located in your Applications folder

Changes from Previous Versions

 * Spotlight isn't just good for finding lost files, it is often a quicker way to open Applications and documents when compared to using Finder to find the item.

title: Mac OS X Tiger/Appendix E


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Talk:Adventist Youth Honors Answer Book/Arts and Crafts/Airplane Modeling

Other Sources of Planes
Guillow's also makes a line of rubberband-powered balsa/tissue airplane kits. I've gotten them for as little as $10.00 each through Amazon. But I see the AMA Cub is even less than that! Wow. Good find! Jim Thomas 18:49, 4 August 2006 (UTC)