0: Mathematical preliminaries

Trigonometric identities

Trigonometry and the identities obeyed by trigonometric functions arise in many of the calculations we will be doing this semester. Hence, it is important for you to be familiar with these. Let us begin by recalling the definitions of the basic functions \(\cos(x)\) and \(\sin(x)\). Here, we will assume \(x \epsilon [0, 2\pi]\), although \since the functions are periodic, it does not matter if we extend the domain of \(x\). However, by restricting \(x\) in this way, we can make the convenient connection of \(x\) to one of the angles in a right triangle (see figure).

Figure 10:

Let \(c\) be the length of the hypotenuse of the triangle, and let \(a\) and \(b\) be the sides adjacent and opposite to the angle \(x\). Then, we have the definitions

From the pythagorean theorem \(a^2 + b^2 = c^2\), so that

The above relation gives us our first and most basic trigonometric identity satisfied by \(\cos(x)\) and \(\sin(x)\):

Many other identities follow straightforwardly from this one. For example, the other basic trigonometric functions \(tan(x)\), \(cot(x)\), \(sec(x)\) and \(csc(x)\), defined as

\[tan(x) = \frac{\sin(x)}{\cos(x)}, cot(x) = \frac{1}{tan(x)} = \frac{\cos(x)}{\sin(x)}\]

\[sec(x) = \frac{1}{\cos(x)}, csc(x) = \frac{1}{\sin(x)}\]

satisfy a number of related identities. If we divide our first identity through by \(\cos^{2}(x)\), for example, we obtain an identity satisfied by \(tan(x)\) and \(sec(x)\):

\[1+\frac{\sin^{2}(x)}{\cos^{2}(x)} = \frac{1}{\cos^{2}(x)}\]

\[1+tan^{2}(x) = sec^{2}(x)\]

Similary, if we divide through by \(\sin^{2}(x)\), we obtain

Note that the definitions of \(\cos(x)\) and \(\sin(x)\) imply that for certain values of \(x\), \(x = 0\), \(x = \pi/2\), \(x = \pi\), \(x = 3\pi/2\) and \(x = 2\pi\),

\[\cos(0) = 1, \sin(0) = 0\]

\[\cos(\pi/2)=0, \sin(\pi/2)=1\]

\[\cos(\pi) = -1, \sin(\pi)=0\]

\[\cos(3\pi/2)=0, \sin(3\pi/2)=-1\]

\[\cos(2\pi) = \cos(0) = 1, \sin(2\pi) = \sin(0) = 0\]

A number of other identities concern trigonometric functions whose arguments are sums or differences of angles \(x\pm y\):

\[\cos(x+y) = \cos(x)\cos(y)-\sin(x)\sin(y), \cos(x-y) = \cos(x)\cos(y)+\sin(x)\sin(y)\]

\[\sin(x+y) = \sin(x)\cos(y)+\cos(x)\sin(y), \sin(x-y) = \sin(x)\cos(y)-\cos(x)\sin(y)\]

In fact, the difference formulas can be derived from the sum formulas u\sing the simple symmetry relations \(\cos(-x)= \cos(x)\) and \(\sin(-x) = -\sin(x)\). In the next section, we will introduce a very simple technique for proving these relations.

Now, if we set \(x = y\), we can derive a useful identity for \(\cos^{2}(x)\). Setting \(x=y\) in the \(\cos(x+y)\) and \(\cos(x-y)\) identities, and adding the two identities together, we obtain

\[\cos(2x) = \cos^{2}(x)-\sin^{2}(x)\]

\[\cos(0) = 1 = \cos^{2}(x)+\sin^{2}(x)\]

\[1+\cos(2x) = 2\cos^{2}(x)\]

\[\cos^{2}(x) = \frac{1}{2}[1+\cos(2x)]\]

If we subtract the identities instead of adding them, we can derive a similar identity for \(\sin^{2}(x)\):

Finally, we note that \(\cos(x)\) and \(\sin(x)\) are expressible as infinite power series (also known as Taylor series) as

\[\cos(x)=1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+...=\sum_{k=0}^{\infty}(-1)^{k}\frac{1}{(2k)!}x^{2k}\]

\[\sin(x) = x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\frac{1}{7!}x^7+... = \sum_{k=0}^{\infty}(-1)^{k}\frac{1}{(2k+1)!}x^{2k+1}\]

These power series expressions are actually used in calculators and computer programs to evaluate \(\cos(x)\) and \(\sin(x)\).

Complex numbers

Complex numbers play a key role in quantum mechanics, even though anything that can be measured experimentally must ultimately be expressed as a real number. Without quantum theory, it is not possible to understand the nature of the chemical bond, and \since quantum theory is expressed in terms of complex numbers, it is important to review some of the basic facts about complex quantities. Complex numbers are so named because they are composed of two types of more fundamental numbers: real numbers and imaginary numbers. Real numbers are very familiar and are the numbers we deal with on a daily basis. However, when one embarks on the study of nonlinear algebraic equations, it rapidly becomes clear that some equations do not have solutions within the set of real numbers. Take the example of the equation

There are no real numbers \(x\) that satisfy this equation. The reason is clear if we just naively solve for \(x\) to find \(x=\pm\sqrt{-1}\). The quantity \(\sqrt{-1}\) is not defined within the real numbers. But mathematicians are never satisfied with the idea that something cannot be defined. In this case, the conundrum can be solved by simply extending the concept of a ``number'' to allow \(\sqrt{-1}\) to be defined. This involves inventing a new set of numbers, called ``imaginary numbers'' (for no better reason than ``imaginary'' is one of the opposites to ``real''), in which \(\sqrt{-1}\) has a definite value, denoted \(i\). Then, the solution to the above equation is simply \(x=\pm i\). The number \(i\) is the fundamental ``integer'' unit in the set of imaginary numbers just as \(1\) is the fundamental integer unit in the real numbers. So, just as any real number \(a\) can always be expressed as \(a\cdot 1\), any imaginary number can be expressed as a multiple of \(i\). If \(\beta\) is an imaginary number, then we can write \(\beta\) as \(b\cdot i\), where \(b\) is a real number. The number \(i\), itself, is really \(1\cdot i\), and clearly, \(1\) is real.

Given imaginary numbers, the most general kind of number we can now imagine has both real and imaginary parts, and such a number is known as a complex number. A complex number \(z\) is expressed as the sum of its real and imaginary parts. If the real part of \(z\) is \(a\) and the imaginary part of \(z\) is \(bi\), then we would write

Note that this is similar to expres\sing a two-dimensional vector \(v=x\hat{i}+y\hat{j}\). Indeed, we can exploit this analogy and use the ordered-pair notation to write a complex number as \(z=(a,bi)\) as an alternate notation. As we will see, there are many such analogies between complex numbers and two-dimensional vectors. For instance, we can think of a complex number as a point in a plane in which the horizontal axis corresponds to the real parts of all complex numbers and the vertical axis corresponds to the imaginary parts. The complex number \(z=a+bi\) would be the point \((a,b)\) in this plane. In order to extract the real and imaginary parts of \(z\), we introduce the functions \(Re\) and \(Im\) such that \(Re(z) = a\) and \(Im(z)=b\).

We noted above that experimental measurements ultimately yield real numbers, so we need a way of going from a complex number to a real one. For this, we exploit the analogy with two-dimensional vectors even more and define the ``measure'' or magnitude of a complex number \(z=a+ib\), which we denote as \(\left | z\right |\), as

which is the same as the magnitude of a two-dimensional vector \(v = (a,b) = a\hat{i}+b\hat{j}\). Another important definition is the so-called complex conjugate of \(z\), denoted \(z^{*}\), which is defined as follows: If \(z=a+ib\), then \(z^{*}=a-ib\). In general, all we have to do to obtain the complex conjugate of a number is replace \(i\) by \(-i\). Note that the magnitude of \(z\) is

\[\left | z \right |=\sqrt{z^{*}z}=\sqrt{(a-ib)(a+ib)}=\sqrt{a^2+iab-iab+b^2}=\sqrt{a^2+b^2}\]

where we have used the fact that \(i(-i)=-i^2=1\). Thus, a complex number times its complex conjugate is a real number.

What happens if we are given a complex number, but it is not expressed in the simple form \(z=a+ib\)? Can we manipulate it so that it finally does have this form? The answer is 'yes' with a little mathematical creativity. Consider, first, a simple product \(z=(u+iv)(x+iy)\). This can be expressed in the canonical form by simply multiplying it out algebraically:

\[z=ux+ivx+iuy-vy=(ux-vy)+i(vx+uy)\]

Next, consider a ratio \(z=(u+iv)/(x+iy)\). In order to simplify this, we use the fact that \((x+iy)(x-iy)\) is real. Thus, we multiply \(z\) by \(1\) in the form of \((x-iy)/(x-iy)\):

\[z=\frac{u+iv}{x+iy}\frac{x-iy}{x-iy}=\frac{(u+iv)(x-iy)}{(x+iy)(x-iy)}=\frac{ux+yv+i(vx-uy)}{x^2+y^2}=\frac{ux+yv}{x^2+y^2}+i\frac{vx-uy}{x^2+y^2}\]

One of the most frequently occuring complex number forms is

where \(\theta\) is a real number. The complex conjugate of \(z\) is obtained simply by replacing \(i\) by \(-i\):

and so that magnitude of \(z\) is

So, \(z\) is a complex number whose square magnitude is \(1\). On the other hand, if we seek to represent \(z\) in canonical form as \(z=a+ib\), then it follows that

\since \(a\) and \(b\) are actually functions of \(\theta\), we can ask what two functions of \(\theta\) are related by \(a^2+b^2=1\). We have already seen what these are - they are \(\cos(\theta)\) and \(\sin(\theta)\) \since \(\cos^{2}(\theta)+\sin^{2}(\theta)=1\). Thus, in canonical form, we can let \(a=\cos(\theta)\)and \(b=\sin(\theta)\) and write

which is known as Euler's formula. The above argument is not a very rigorous one, as there are many other functions \(a\) and \(b\) of \(\theta\) we could have chosen that satisfy \(a^2+b^2=1\). However, we can actually prove rigorously that \(a\) and \(b\) must be, respectively, \(\cos(\theta)\) and \(\sin(\theta)\) u\sing the fact that the function \(exp(x)\) has the following Taylor series expression:

Applying this to \(exp(i\theta)\), we obtain

\[e^{i\theta}=1+i\theta+\frac{(i\theta)^{2}}{2!}+\frac{(i\theta)^3}{3!}+\frac{(i\theta)^4}{4!}+\frac{(i\theta)^5}{5!}+...=\sum_{k=0}^{\infty}\frac{(i\theta)^k}{k!}\]

Now the even power terms are purely real, and the odd powers are purely imaginary, so if we wish to put the above power series into the canonical form, we obtain

\[e^{i\theta}=1-\frac{\theta^{2}}{2!}+\frac{\theta^{4}}{4!}-\frac{\theta^{6}}{6!}+...+i\left ( \theta-\frac{\theta^{3}}{3!}+\frac{\theta^{5}}{5!+...} \right )=\cos(\theta)+i \sin(\theta)\]

which proves Euler's formula.

Euler's formula can be used to simplify proofs of trigonometric identities. For example, we know that \(z^{*}z=1\). But from Euler's formula, \(z^{*}z=\cos^{2}(\theta)+\sin^{2}(\theta)\), so \(\cos^{2}(\theta)+\sin^{2}(\theta)=1\). If we use the fact that

Thus, adding \(z\) to \(z^{*}\), we obtain

Subtracting \(z\) from \(z^{*}\) yields

Now consider \(\cos^{2}(\theta)\). From the above formula, we have

\[=\frac{1}{4}\left ( 2+e^{2i\theta}+e^{-2i\theta}\right )=\frac{1}{2}+\frac{1}{4}\left ( e^{2i\theta}+e^{-2i\theta}\right )=\frac{1}{2}\left [1+\cos(2\theta)\right ]\]

Finally, note that if we multiply Euler's formula by a real number \(A\), we have a complex number of the form

In canonical form

and \(\left | z \right |^{2}=A exp(-i\theta)*A exp(i\theta)=A^2\), so \(\left | z \right | = A\). When we write \(z=A exp(i\theta)\), The angle \(\theta\) is called the phase of \(z\).

Derivative of a function

A line, described by the equation

has a constant slope \(m\) defined by

where the above equations means compute the change, \(\Delta y\) in \(y\) for a given change, \(\Delta x\) in \(x\) as shown in the figure:

Figure 11:

For a general function, such as that shown below:

Figure 12:

the slope is not a constant. However, it is possible to define the slope of the curve at each point by taking the slope of the line tangent to the curve at each point. The slope will also be a function of \(x\) and is called the derivative of the function \(f(x)\). To see how the slope is computed, consider a line connecting two points of the function corresponding to two points \(x\) and \(x+\Delta x\), where \(\Delta x\) is a small number:

Figure 13:

The slope of this line is

To find the slope at the \single point \(P\), we need to let \(\Delta x\) go to 0, so that the line intersects the curve at the point \(P\), where it will be tangent to the curve. Thus, by taking the limit as \(\Delta x \rightarrow 0\), we obtain the derivative of \(f(x)\) at the point \(P\). But \since the point \(P\) is arbitrary, we may perform this operation at any point and, thus, obtain the derivative of \(f(x)\) at any point. The general expression is

where \(f'(x)\) and \(df/dx\) are standard notations for the derivative.

As an example, consider the function \(f(x) = x^2\). According to the formula:

\[f'(x)=lim_{\Delta x \rightarrow 0}\frac{(x+\Delta x)^2-x^2}{\Delta x}=lim_{\Delta x \rightarrow 0}\frac{x^2+2x\Delta x+\Delta x^2 -x^2}{\Delta x}=lim_{\Delta x \rightarrow 0}(2x+\Delta x)=2x\]

Thus, the derivative of \(x^2\) is \(2x\). In fact, in general, if \(f(x)=x^n\), \(f'(x)=nx^{n-1}\). A few other functions and their derivatives are listed below:

\[f(x)=ln x, f'(x)=\frac{1}{x}\]

\[f(x)=\sin x, f'(x)=\cos x\]

\[f(x) = \cos x' f'(x) = -\sin x\]

Other important rules of differentiation are the product rule:

if \(h(x)=f(x)g(x)\) then \(h'(x)=f'(x)g(x)+f(x)g'(x)\)

and the chain rule:

if \(h(x)=f(g(x))\), then \(h'(x)=f'(g(x))g'(x)\)

Integral of a function

It is often necessary to calculate the area under a function, \(f(x)\) between two points \(x_1\) and \(x_2\). Such an operation is necessary, for example, in mechanics to compute the work performed by an agent exerting a force on an object. An analogous situation arises in thermodynamics in computing the work done by an agent exerting an external pressure on a system.

Consider the area under the curve designated below:

Figure 14:

One way to calculate the area approximately would be to cover the area with narrow rectangles or width \(\Delta x\):

Figure 15:

\(a\). Thus, we have

\[\int_{0}^{1}xe^{-ax}dx=-\frac{d}{da}\left [ \left . -\frac{1}{a}e^{-ax}\right |_{0}^{1} \right ]=-\frac{d}{da}\left [ \frac{1}{a}(1-e^{-a})\right ] = \frac{1}{a^2}(1-e^{-a})-\frac{1}{a}e^{-a}\]

In three dimensions, the point \(P\) is located by specifying the distance of \(P\) from the origin of a given coordinate system, the angle \(\theta\) of the vector from the origin to the point from the positive \(z\)-axis, and the angle \(\phi\) of the projection of this vector into the \(x\)-\(y\)plane from the positive \(x\)-axis. The transformation from \((r,\phi,\theta)\) to \((x,y,z)\) is

\[x=r\sin\theta \cos\phi\]

\[y=r\sin\theta \sin\phi\]

\[z=r\cos\theta\]

and the transformation from \((x,y,z)\) to \((r,\phi,\theta)\) is

\[r=\sqrt{x^2 +y^2 +z^2}\]

\[\phi=tan^{-1}\frac{y}{x}\]

\[\theta=tan^{-1}\frac{\sqrt{x^2 +y^2}}{z}\]

From the transformation equations, it can be shown that an integral of a function \(f(x,y,z)\)

\[I=\int \int \int f(x,y,z) \ dx \ dy \ dz\]

in spherical coordinates becomes

\[I=\int \int \int \tilde{f}(r,\phi,\theta) \ r^{2} \ dr \ \sin\theta \ d\theta \ d\phi\]

Here \(\tilde{f}(r,\phi,\theta)\) is the transformed form of the function expressed in spherical polar coordinates. If the integral is over all space, then the limits in Cartesian coordinates are

\[I=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x,y,z) \ dx \ dy \ dz\]

and in spherical polar coordinates, this becomes

\[I=\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}\tilde{f}(r,\phi,\theta) \ r^2 \ dr \ \sin\theta \ d\theta \ d\phi)\]

Note that \(\theta\) is integrated from \(0\) to \(\pi\) only, which covers only the positive \(z\)-axis, but \(\phi\) is integrated from \(0\) to \(2\pi\). Many of the functions \(\tilde{f}(r,\phi,\theta)\) we will have to deal with are simple products of functions of one variable, i.e. \[\tilde{f}(r,\phi,\theta)=g(r)h(\phi)v(\theta)\]

for this type of function, the three-dimensional integral in polar coordinates becomes a product of ordinary one-dimensional integrals:

\[I=\left [ \int_{0}^{2\pi}d\phi h(\phi)\right ] \left [ \int_{0}^{\pi}d\theta \sin\theta v(\theta) \right ] \left [ \int_{0}^{\infty}dr r^2 g(r) \right ]\]

Contributors and Attributions

Mark Tuckerman (New York University)