Complex Numbers

 

 

Strange as it may seem, we've got to put some time in on understanding sinusoids to be able to deal with the physics of blood flow. No doubt you are familiar with Fourier "analysis"; we can determine the extent to which various frequencies are present in a signal.  This is done quite rapidly every time you perform a Doppler study where the frequencies of reflected sound waves are analyzed for frequency content.  However engineers and mathematicians also use Fourier transform in quite a different way – to solve a wide range of mathematical problems. Unfortunately this article is largely mathematical, but the goal is to be able to reduce certain kinds of differential equations to algebra that we can all understand.

The Exponential Function

This may seem an odd way to start, but here goes.  Most everyone will have had to deal with the exponential function:

\(\Large y=A e^x\)

You supply this function with an input, the value of \(x\).  The function does something with this number and gives you back \(y\).  As a practical matter, this is a button on your calculator or a function on your spreadsheet that you can invoke whenever whenever forced.  Ever wonder what goes on in your calculator when you press the button?  How does it figure out the value of \(y\)?  Look it up in a table?  Kick the calculator elves in their collective butt to work it out for you?

Before I give you to answer, let me tell you the question.  The function \(y=A e^x\) is the solution of this differential equation:

\(\Large \frac{dy(x)}{dx}=y(x)\)

In English, this mathematical expression says that the slope of the function \(y(x)\) is equal to the value of the function itself.

 

A plot of a small section of \(y=e^x\) (plotted in blue) shows how the slope of the function is equal to the value of the function at selected values of \(x\). 

\(A\) can be any arbitrary constant and is chosen to meet some other criterion in the problem, e.g. the value of the function at some particular value of \(x\).  In English, we are looking for a function whose rate of change (slope) is equal to the value of the function itself.  That turns out to be a really important function that shows up everywhere! You could say that the function, \(y=e^x\), is defined as the solution of this differential equation.  Now, here's the answer:

\(\Large y(x)=e^x= \frac{1}{0!}+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+..+\frac{x^n}{n!}+..\)

The \(!\) is shorthand for the factorial function 

\(\Large n! \equiv n(n-1)(n-2)(n-3)..(2)(1)\)

\(0!\) is strangely equal to 1.  The expression for \( y=e^x\) can thus be written more compactly:

\(\Large y(x)=e^x=\Sigma_{n=0}^\infty \frac{x^n}{n!}\)

This is a formula, a recipe for obtaining \(y=e^x\).  It works very well for relatively small values of \(x\), but your computer or calculator may have more efficient ways of computing it for a large value of \(x\).  Let's give it a try and stick in a value for \(x\); how about 1.0?  

 The chart shows the first few terms of this infinite series with the partial sum ( sum of terms up to that that number of terms ) in the far right column.  For the particular value of \(x\) chosen, the terms get small pretty quick in the sequence and it doesn't take very many of them before we have a pretty accurate value for \(e^x\).  Perhaps you will recognize that the far right column converges on the value of the magical number \(e\), i.e. \(e^1\).  The reason this number is so special is that it falls out of the sky from a solution of the above differential equation.  Note that many functions are defined by your calculator or computer as a series of terms like this.  This is called a convergent series in that the sum adds up to a specific number.  We wouldn't have to look very far to find other series that add up to infinity; these aren't very useful.

The sequence of terms in this instance is called a Taylor series.  There's no guarantee that the solution of a differential equation can be represented by a Taylor series, but let's check to see if this one actually fills the bill – does it satisfy the original differential equation (or did I just lie to you)?  For this we have to know how to differentiate ( take the derivative ) of each of the terms.  Somewhere back in the math section we actually derived the following fact about the derivative of this kind of term.

\(\Large \frac{d}{dx} k x^n = k n x^{n-1}\)

In English this says that if we want to find the (first) derivative (the slope) of a function where \(x\) is raised to some power, \(n\), we simply multiply the function by the exponent (\(n\) ) and decrease the power of \(x\) by one; the constant multiplying the original function stays put. Now we apply this rule to the above series.  Differentiation is a linear operator; the derivative operator is applied to whole thing by applying it to each term in the sum:

\(\Large \frac{d}{dx}y(x)=\frac{d}{dx}[ \frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+..+\frac{x^n}{n!}+..]\)

 

\(\Large \frac{dy}{dx}=[ 0+1\frac{x^{1-1}}{1!}+2\frac{x^{2-1}}{2!}+3\frac{x^{3-1}}{3!}+4\frac{x^{4-1}}{4!}+5\frac{x^{5-1}}{5!}+..+n\frac{x^{n-1}}{n!}+..]\)

 

\(\Large \frac{dy}{dx}=[ \frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+..+\frac{x^{n-1}}{(n-1)!}+..]\)

 So we have differentiated the function \(y=e^x\) and obtained, \(e^x\)!  This is a function whose derivative ( rate of change ) is equal to the value of the function itself.  

This Taylor series can also be used for simple extensions of the definition of the function.  For example:

\(\Large y=e^{kx}= \frac{(kx)^0}{0!}+\frac{(kx)^1}{1!}+\frac{(kx)^2}{2!}+\frac{(kx)^3}{3!}+\frac{(kx)^4}{4!}+\frac{(kx)^5}{5!}+..+\frac{(kx)^n}{n!}+..\)

 And since k could certainly be equal to -1:

\(\Large y=e^{-x}= \frac{(-x)^0}{0!}+\frac{(-x)^1}{1!}+\frac{(-x)^2}{2!}+\frac{(-x)^3}{3!}+\frac{(-x)^4}{4!}+\frac{(-x)^5}{5!}+..+\frac{(-x)^n}{n!}+..\)

\(\Large y=e^{-x}= 1-\frac{x^1}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-\frac{x^5}{5!}+-+-..\)

Whether the term has a \(+\) or \(-\) sign relates only to whether there is an even or odd power of \(x\) in the term.

 

So here is the first result that we need to have on hand:

\(\Large \frac{d}{dx} e^{kx} = k e^{kx} \)

\(k\) can be any number, or even a function as long as it doesn't depend on \(x\). We could take the derivative of this function AGAIN:

\(\Large \frac{d^2}{dx^2} e^{kx} = \frac{d}{dx} ke^{kx} = k^2 e^{kx} \)

This is the second derivative of the first function and the usual notation is shown.  We could keep taking the derivative over and over again.  The \(n^{th}\) derivative is:

\(\Large \frac{d^n}{dx^n} e^{kx} = k^n e^{kx} \)


Sinusoidal Functions

We'll start right out with the question this time; what function(s) would satisfy the following relationship?

\(\Large \frac{d^2y(x)}{dx^2}+y(x)=0\)

And here is the answer:

\(\Large y_1(x) = A \cos(x), y_2(x) = B \sin(x)\)  

The solution(s) to this differential equation are the \(\cos\) and \(\sin\) functions. There is a good chance that you've always thought about these 2 functions in relation to trigonometry; Each is of course a ratio of length when \(x\) represents an angle in a right triangle.  The cosine function is the ratio of the "side adjacent" ( to the angle ) divided by the hypotenuse; the sine function is the ratio of the "side opposite" to the hypotenuse.  But these functions can be defined by the above linear differential equation.  This is a second-order differential equation and the solution is 2 independent functions. \(A\) and \(B\) are arbitrary constants that are chosen to fit the circumstances of the problem; any values for \(A\) and \(B\) fulfill the differential equation. We would also find that solution to the the differential equation

\(\Large \frac{d^2y(x)}{dx^2}+k^2 y(x)=0\)

consists of:

 

\(\Large y_1(x) = A \cos(kx), y_2(x) = B \sin(kx)\)  

Just as we saw for the exponential function, we can find a Taylor series for each of the sine and cosine functions that satisfies the differential equation and could be used by your calculator to determine what to spit out when you push the button.  Again these might not be the actual algorithms your calculator uses:

\(\Large y_1(x) = \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!}.. +.. - + (-1)^{n} \frac{x^{2n}}{(2n)!}\)

\(\Large y_1(x) = \Sigma_{n=0}^\infty  (-1)^{n} \frac{x^{2n}}{(2n)!}\)

and:

\(\Large y_2(x) = \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}.. +.. - + (-1)^{n} \frac{x^{2n+1}}{(2n+1)!}\)

\(\Large y_2(x) = \Sigma_{n=0}^\infty (-1)^{n} \frac{x^{2n+1}}{(2n+1)!}\)

If you look these up somewhere else, you may find formulas that look a little bit different depending on whether the summation starts at n=0 or n=1.  Looking closely at the formulas you'll see that \((-1)^n\) causes the sign of the term to alternate between plus and minus.  Check closely and you'll also see that \(2n\) recapitulates the exponent ( and factorial denominator ) in the series for the cosine function, \(2n+1\) works for the sine function.

Now for the derivative functions.  I'm just going to put down the answers this time although we will use the series solutions in a moment:

\(\Large \frac{d}{dx} \cos(kx) = -k\sin(kx) \)

\(\Large \frac{d}{dx} \sin(kx) =  k\cos(kx) \)

If you keep on differentiating a sinusoidal function, you eventually get back to the original function (multiplied by a constant):

\(\Large \frac{d}{dx} \cos(kx) = -k\sin(kx) \)

\(\Large \frac{d^2}{dx^2}  \cos(kx) = \frac{d}{dx} -k\sin(kx) = -k^2\cos(kx) \)

\(\Large \frac{d^3}{dx^3}  \cos(kx) = \frac{d}{dx} -k^2\cos(kx) = k^3\sin(kx) \)

\(\Large \frac{d^4}{dx^4}  \cos(kx) = \frac{d}{dx} k^3\sin(kx) = k^4\cos(kx) \)

 

 

 

 

 

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