User:Mmitchell10/sandbox

Everyday uses of negative numbers

• Goal difference in association football and hockey; points difference in rugby football; net run rate in cricket; golf scores relative to par.

• British football clubs are deducted points if they enter administration, and thus have a negative points total until they have earned at least that many points that season.

• Bank account balances which are overdrawn.

• Refunds to a credit card or debit card are a negative debit.

• Temperatures which are colder than 0°C or 0°F.

• Storeys in a building below the ground floor.

Leading-order

The leading-order term within a mathematical equation or expression is the term with the biggest numerical value (irrespective of whether it is positive or negative). It is so-called because it has the largest order of magnitude. An equation or expression will typically contain terms of various different sizes, and if it has several terms which are approximately the same size and significantly bigger than all the rest, these are collectively referred to as the leading-order terms. The other lower-order terms might be able to be regarded as negligible.

A simple example

Leading-order terms for y=x³+5x+0.1

x	0.001	0.1	0.5	2	10
x3	0.000000001	0.001	0.125	8	1000
5x	0.005	0.5	2.5	10	50
0.1	0.1	0.1	0.1	0.1	0.1
y	0.105000001	0.601	2.725	18.1	1050.1

Within a given equation, the terms which are the leading-order terms may change as the variables in the equation change. For example, consider the equation y=x³+5x+0.1. For five different values of x, the table shows the sizes of the four terms in this equation, and which ones are leading-order.

As x increases further, the leading-order terms will stay as x³ and y, but as x decreases and then becomes more and more negative, which terms are leading-order will again change.

Two terms that are within a factor of 10 of each other should be regarded as of the same order, or magnitude, and two terms that are not within a factor of 100 of each other should not. However, in between is a grey area, and there is no strict cut-off for when two terms should or should not be regarded as approximately the same order. So there are no fixed boundaries where some terms are to be regarded as leading-order and others not. Instead the terms fade in and out, and deciding whether particular terms in a model are approximately leading-order or not, and if not, whether they are small enough to be regarded as unimportant and negligible, (two different questions), is often a matter of investigation and judgement, and will depend on the context. For example, if faced with terms with values 5, 30 and 100, then 5 and 30 are about the same order, and so are 30 and 100, but 5 and 100 are not really. So whether 5 should be regarded as one of the 'leading-order' terms, along with 30 and 100, is a matter of opinion.

Leading-order balance

Equations with only one leading-order term are possible, but rare. For example, the equation 100 = 1 + 1 + 1 + ... + 1, (where the right hand side comprises one hundred 1's). Usually an equation will contain at least two leading-order terms. In this case, on the assumption that the lower-order terms, and the parts of the leading-order terms that are the same size as the lower-order terms (perhaps the second significant figure onwards), are negligible, a new equation may be formed by dropping all these lower-order terms and parts of the leading-order terms. The remaining terms provide the leading-order balance, or dominant balance, and creating a new equation just involving these terms is known as taking an equation to leading-order. Analysing the behaviour given by this equation gives the leading-order behaviour of the model, which is the main behaviour - the true behaviour is only small deviations away from this.

Suppose we want to understand the leading-order behaviour of the example above. When x=0.001, the x³ and 5x terms may be regarded as negligible, and dropped, along with any values in the third decimal places onwards in the two remaining terms. This gives the leading-order balance y=0.1. Thus the leading-order behaviour of this equation at x=0.001 is that y is constant. Similarly, when x=0.5, the x³ and 0.1 terms may be regarded as negligible, and dropped, along with any values in the second decimal places onwards in the two remaining terms. This gives the leading-order balance y=5x. Thus the leading-order behaviour of this equation at x=0.5 is that y increases linearly with x. The leading-order behaviour may thus be investigated at any value of x. When the behaviour is more complicated ...leading-order balance between cubic and linear

Technical

writing each term as the sum of

In advanced mathematics, as well as colloquially in popular culture, especially geek culture, the phrase "order of magnitude" is used to denote a change in a numeric quantity, usually a measurement, by a factor of 10; that is, the moving of the decimal point in a number one way or the other, possibly with the addition of significant zeros.

big O notation - informally means the size of the term

examples of where it's useful, eg Couette flow, Hagen–Poiseuille equation, method of matches asy exps

Other

The equilibrium constant for this reaction, K_a1, at 25 °C, has been put at: 2.5×10⁻⁴ mol/litre (pK_a1 = 3.6); 1.72×10⁻⁴ mol/litre

$\epsilon$

$\infty$

Matching

We use matching to find the value of the constant $B$. The idea of matching is that the inner and outer solutions should agree for values of $t$ near the edge of the boundary layer. We need the outer limit of the inner solution to match the inner limit of the outer solution, ie.

$\lim _{\tau \rightarrow \infty }y_{I}=\lim _{t\to 0}y_{O},\,$

which gives $B=e$.

Solution valid for all $t$

To obtain our final, matched solution, valid on the whole domain, one popular method is the uniform method. In this method, we add the inner and outer approximations and subtract their overlapping value, $\,y_{overlap}$. In the boundary layer, we expect the outer solution to be approximate to the overlap, $y_{O}\,\sim \,y_{overlap}$. Far from the boundary layer, the inner solution should approximate it, $y_{I}\,\sim \,y_{overlap}$. Hence, we want to eliminate this value from the final solution. In our example, $y_{overlap}\,\sim \,e$. Therefore, the final solution is,

$y(t)=y_{I}+y_{O}-e=e\left({1-e^{-t/\epsilon }}\right)+e^{1-t}-e=e\left({e^{-t}-e^{-t/\epsilon }}\right).\,$

A Bjerrum plot is a graph of the equilibrium concentrations (or ratios of equilibrium concentrations) of the different species of a polyprotic acid in a solution, as functions of the solution's pH .

Normally the carbonate system is plotted, where the polyprotic acid is carbonic acid (a diprotic acid), and the different species of dissolved inorganic carbon (DIC) are carbonic acid, carbon dioxide, bicarbonate, and carbonate. In acidic conditions, the dominant form of DIC is CO₂; in basic (alkalinic) conditions, the dominant form is CO₃²⁻; and in between, the dominant form is HCO₃⁻. At every pH, the concentration of carbonic acid is

About 30-40% of the carbon dioxide released by humans into the atmosphere dissolves into the oceans, rivers and lakes . To maintain chemical equilibrium, some of it reacts with the water to form carbonic acid. Some of these extra carbonic acid molecules split up to give a carbonate ion and two hydrogen ions, thus increasing the ocean’s "acidity" (H⁺ ion concentration). This increasing acidity is thought to have a range of direct undesirable consequences such as depressing metabolic rates in jumbo squid and depressing the immune responses of blue mussels. (These chemical reactions also happen in the atmosphere, and as about 20% of anthropogenic CO₂ emissions are absorbed by the terrestrial biosphere, also in the ground soils between absorbed CO₂ and soil moisture. Thus anthropogenic CO₂ emissions to the atmosphere can increase the acidity of land, sea and air.)

Other chemical reactions are also triggered which result in an actual net decrease in the amount of carbonate ions available. In the oceans, this makes it more difficult for marine calcifying organisms, such as coral and some plankton, to form biogenic calcium carbonate, and existing such structures become vulnerable to dissolution. Thus, ongoing acidification of the oceans also poses a threat to the food chains connected with the oceans.

Acidification

Dissolving CO
₂ in seawater increases the hydrogen ion (H⁺
) concentration in the ocean, and thus decreases ocean pH, by the following chemical reactions:

CO
_2(aq) + H
₂O ⇌ H
₂CO
₃

H
₂CO
₃ ⇌ H⁺ + HCO₃⁻

HCO₃⁻ ⇌ H⁺ + CO₃²⁻.

Other biological impacts

Aside from calcification, organisms may suffer other adverse effects, either indirectly through negative impacts on food resources, or directly as reproductive or physiological effects. For example, the elevated levels of CO₂ may produce CO
₂-induced acidification of body fluids, known as hypercapnia. Also, increasing acidity is believed to:

It has even been suggested that ocean acidification will alter the acoustic properties of seawater, allowing sound to propagate further, increasing ocean noise and impacting animals that use sound for echolocation or communication.

However, as with calcification, as yet there is not a full understanding of these processes in marine organisms or ecosystems.

Bjerrum plot

In acidic conditions, the dominant form of the CO₂ compounds is CO₂, in basic conditions, the dominant form is CO₃²⁻, and in between, the dominant form is HCO₃⁻.

Bjerrum plot equations

Example Bjerrum plot: Change in carbonate system of seawater from ocean acidification.

If carbon dioxide, hydrogen ions, bicarbonate and carbonate ions are all dissolved in water, and at chemical equilibrium, their equilibrium concentrations are often assumed to be given by:

$[{\textrm {CO}}_{2}]_{eq}={\frac {[{\textrm {H}}^{+}]_{eq}^{2}}{[{\textrm {H}}^{+}]_{eq}^{2}+K_{1}[{\textrm {H}}^{+}]_{eq}+K_{1}K_{2}}}{\textrm {Tot}}[{\textrm {CO}}_{2}],$

$[{\textrm {HCO}}_{3}^{-}]_{eq}={\frac {K_{1}[{\textrm {H}}^{+}]_{eq}}{[{\textrm {H}}^{+}]_{eq}^{2}+K_{1}[{\textrm {H}}^{+}]_{eq}+K_{1}K_{2}}}{\textrm {Tot}}[{\textrm {CO}}_{2}],$

$[{\textrm {CO}}_{3}^{2-}]_{eq}={\frac {K_{1}K_{2}}{[{\textrm {H}}^{+}]_{eq}^{2}+K_{1}[{\textrm {H}}^{+}]_{eq}+K_{1}K_{2}}}{\textrm {Tot}}[{\textrm {CO}}_{2}],$

where the subscript 'eq' denotes that these are equilibrium concentrations, K₁ $K_{1}$ is the equilibrium constant for the reaction CO
₂ + H
₂O ⇌ H⁺ + HCO₃⁻, $K_{2}$ is the equilibrium constant for the reaction HCO₃⁻ ⇌ H⁺ + CO₃²⁻, and Tot[CO₂] is the (unchanging) total concentration of CO₂ compounds in the system, i.e. [CO₂] + [HCO⁻
₃] + [CO²⁻
₃].

A Bjerrum plot consists of these three species plotted against pH = –log₁₀[H⁺]. The fractions in these equations give the three species' relative proportions, and so if Tot[CO₂] is unknown, or the actual concentrations are unimportant, these proportions may be plotted instead.

5.8×10⁷ kg

Chemical and mathematical derivation of equations

Suppose that the interactions of carbon dioxide, hydrogen ions, bicarbonate and carbonate ions, all dissolved in water, are as follows:

CO
₂ + H
₂O ⇌ H⁺ + HCO₃⁻               (1)

      HCO₃⁻ ⇌ H⁺ + CO₃²⁻.               (2)

(Note that reaction (1) is actually the combination of two elementary reactions: CO
₂ + H
₂O ⇌ H
₂CO
₃ ⇌ H⁺ + HCO₃⁻.)

Assuming the mass action law applies to these two reactions, that water is abundant, and that the different chemical species are always well-mixed, their rate equations are:

${\frac {{\textrm {d}}[{\textrm {CO}}_{2}]}{{\textrm {d}}t}}=-k_{1}[{\textrm {CO}}_{2}]+k_{-1}[{\textrm {H}}^{+}][{\textrm {HCO}}_{3}^{-}],$

${\frac {{\textrm {d}}[{\textrm {H}}^{+}]}{{\textrm {d}}t}}=k_{1}[{\textrm {CO}}_{2}]-k_{-1}[{\textrm {H}}^{+}][{\textrm {HCO}}_{3}^{-}]+k_{2}[{\textrm {HCO}}_{3}^{-}]-k_{-2}[{\textrm {H}}^{+}][{\textrm {CO}}_{3}^{2-}],$

${\frac {{\textrm {d}}[{\textrm {HCO}}_{3}^{-}]}{{\textrm {d}}t}}=k_{1}[{\textrm {CO}}_{2}]-k_{-1}[{\textrm {H}}^{+}][{\textrm {HCO}}_{3}^{-}]-k_{2}[{\textrm {HCO}}_{3}^{-}]+k_{-2}[{\textrm {H}}^{+}][{\textrm {CO}}_{3}^{2-}],$

${\frac {{\textrm {d}}[{\textrm {CO}}_{3}^{2-}]}{{\textrm {d}}t}}=k_{2}[{\textrm {HCO}}_{3}^{-}]-k_{-2}[{\textrm {H}}^{+}][{\textrm {CO}}_{3}^{2-}],$

where [ ] denotes concentration, t is time, and $k_{1}$ and $k_{-1}$ are appropriate proportionality constants for reaction (1), called respectively the forwards and reverse rate constants for this reaction. (Similarly $k_{2}$ and $k_{-2}$ for reaction (2).)

 

At any equilibrium, the concentrations are unchanging, hence the left hand sides of these equations are zero. Then, from the first of these four equations, the ratio of rate constants equals the ratio of equilibrium concentrations, and this ratio, called $K_{1}$, is the equilibrium constant for reaction (1), i.e. (and similarly from the fourth equation for the equilibrium constant $K_{2}$ for reaction (2)),

$K_{1}={\frac {k_{1}}{k_{-1}}}={\frac {[{\textrm {H}}^{+}]_{eq}[{\textrm {HCO}}_{3}^{-}]_{eq}}{[{\textrm {CO}}_{2}]_{eq}}},$         (3)        

$K_{2}={\frac {k_{2}}{k_{-2}}}={\frac {[{\textrm {H}}^{+}]_{eq}[{\textrm {CO}}_{3}^{2-}]_{eq}}{[{\textrm {HCO}}_{3}^{-}]_{eq}}},$           (4)

where the subscript 'eq' denotes that these are equilibrium concentrations.

Rearranging (3) gives

$[{\textrm {HCO}}_{3}^{-}]_{eq}={\frac {K_{1}[{\textrm {CO}}_{2}]_{eq}}{[{\textrm {H}}^{+}]_{eq}}},$         (5)

and rearranging (4), then substituting in (5), gives

$[{\textrm {CO}}_{3}^{2-}]_{eq}={\frac {K_{2}[{\textrm {HCO}}_{3}^{-}]_{eq}}{[{\textrm {H}}^{+}]_{eq}}}={\frac {K_{1}K_{2}[{\textrm {CO}}_{2}]_{eq}}{[{\textrm {H}}^{+}]_{eq}^{2}}}.$         (6)

 

The total concentration of CO
₂ compounds in the system is given by

${\textrm {Tot}}[{\textrm {CO}}_{2}]=[{\textrm {CO}}_{2}]+[{\textrm {HCO}}_{3}^{-}]+[{\textrm {CO}}_{3}^{2-}]$

                    $=[{\textrm {CO}}_{2}]_{eq}\left(1+{\frac {K_{1}}{[{\textrm {H}}^{+}]_{eq}}}+{\frac {K_{1}K_{2}}{[{\textrm {H}}^{+}]_{eq}^{2}}}\right)$                 substituting in (5) and (6)

                    $=[{\textrm {CO}}_{2}]_{eq}\left({\frac {[{\textrm {H}}^{+}]_{eq}^{2}+K_{1}[{\textrm {H}}^{+}]_{eq}+K_{1}K_{2}}{[{\textrm {H}}^{+}]_{eq}^{2}}}\right).$

 

This gives the equation for $[{\textrm {CO}}_{2}]_{eq}$. The equations for $[{\textrm {HCO}}_{3}^{-}]_{eq}$ and $[{\textrm {CO}}_{3}^{2-}]_{eq}$ are obtained by substituting this into (5) and (6).