Mathematics for chemistry/The Basics

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Numbers

Real numbers come in several varieties and forms, integer / decimal, rational / irrational. Integers are used for counting indivisable objects but they can also be directed, e.g. -4 or zero. Decimals can always be expressed as fractions e.g. 4.673 = 4673/1000. Numbers like $\pi$ and $\sqrt{2}$ are irrational.

Remember that in SI numbers do not have commas between the thousands, there are spaces. e.g. $18617132$, (an integer) or $1.8617132\mathrm{x}10^7$. Commas are used as decimal points in many countries.

The Properties of Graphs

The most basic relationship between two variables $x$ and $y$ is a straight line, a linear relationship. $y=\mathrm{m}x+\mathrm{c}$ The variable $m$ is the gradient and $c$ is a constant which gives the intercept.

The equations can be more complex than this including higher powers of $x$ such as $y=\mathrm{a}{x}^2+\mathrm{b}x+\mathrm{c}$ This is called a quadratic equation and it follows a shape called a parabola. High powers of $x$ can occur giving cubic, quartic and quintic equations etc. In general as you go to higher powers the line mapping the variables wiggles more, often cutting the $x$-axis several times.

Practice

Plot $x^2-1$ between -3 and +2 in units of 1.

Plot $x^2+3x$ between -4 and +1 in units of 1.

Plot $2x^3-5x^2-12x$ between -5 and +4 in units of 1.

The Quadratic Formula

For the solutions of: $a{x}^2+bx+c$ There is a formula: $x=\frac{-b\pm \sqrt{(b^2-4ac)}}{2a}$ (Notice the line over the square root has the same priority as a bracket. Of course we all know by now that $\sqrt{a+b}$ is not equal to $\sqrt{a}+\sqrt{b}$ but errors of priority are among the commonest algebra errors in practice.) There is a formula for a cubic equation but it is rather complicated. Cubic and higher equations occur often in chemistry but if they do not factorise they are usually solved by computer.

Solve:

$2x^2-14x+9$

$1.56(x^2+3.67x+0.014)$

Notice the scope or range of the bracket.

$2z^2-14z+9$

$2x^2-4x+2$

$-45.1(1.2[A{]}^2-57.9[A]+4.193)$

Notice here our variable is a concentration not the ubiquitous $x$.

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 these surds are in the trig functions of the important symmetrical angles, e.g. $\sin 60=\frac{\sqrt{3}}{2}$ and so they appear all the time in mathematical expressions to do with 3 dimensional space.

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, wheras by splitting it up into a sum, a sum of standard integrals is obtained. This is the principal chemical application of partial fractions.

Here follows an example:

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

In the above $x-2\mathrm{m}\mathrm{u}\mathrm{s}\mathrm{t}=A(x-1)+B(x+1)$ 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 putting a common denominator in.

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

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 3/2 = -A

therefore the expression becomes

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

Later you will have learned these integrate to simple $\ln$ expressions.

Partial Fractions - problems

$\mathrm{(}\mathrm{1}\mathrm{)}\frac{3}{x^2-1}\mathrm{(}\mathrm{2}\mathrm{)}\frac{4x-2}{x^2+2x}$ $\mathrm{(}\mathrm{3}\mathrm{)}\frac{4}{(2x+1)(x-3)}\mathrm{(}\mathrm{4}\mathrm{)}\frac{7x+6}{x^2+3x+2}$ $\mathrm{(}\mathrm{5}\mathrm{)}\frac{x+1}{2x^2-6x+4}$

Partial Fractions - answers

$\mathrm{(}\mathrm{1}\mathrm{)}\frac{3}{2(x-1)}-\frac{3}{2(x+1)}\mathrm{(}\mathrm{2}\mathrm{)}\frac{5}{(x+2)}-\frac{1}{x}$ $\mathrm{(}\mathrm{3}\mathrm{)}\frac{4}{7(x-3)}-\frac{8}{7(2x+1)}\mathrm{(}\mathrm{4}\mathrm{)}\frac{8}{x+2}-\frac{1}{x+1}$ $\mathrm{(}\mathrm{5}\mathrm{)}\frac{3}{2(x-2)}-\frac{2}{2(x-1)}$

Polynomial Division

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

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

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

So our equation becomes: $3x-7+\frac{1}{1+x}$ This can be easily differentiated, and integrated. If this is differentiated with the quotient formula it is considerably harder to reduce to the the same form. The same procedure can be applied to partial fractions.

Substitutions and Expansions

You can see the value of changing the variable here:

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

This simplifies to :

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

Where

$X=\frac{J(J+1)}{ϵ}$

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

Evaluating $\frac{\mathrm{d}y}{\mathrm{d}X}$ gives:

    2           3                              2
 3 X  + 2     (X  + 2 X + 1) X   2 X - 2     (X  - 2 X + 1) X
 -------- - 2 ---------------- - ------- + 2 ----------------
   2                2     2        2               2     2
  X  - 9          (X  - 9)        X  + 9         (X  + 9)

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

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

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