Monday, January 28, 2013

Clarifying the Taylor Series

A friendly warning, the following is bit esoteric, (a fancy word for hazy), and is still very much a work in progress.  I'm having fun with it though, and I think it's an insightful review of the Taylor series in any event, so if you're interested, read on.

My previous post was about a new way to memorize the Taylor series.  It emphasized what the Taylor series did rather than how it was derived.  I hadn't completely thought things through and had introduced a new term, "the inverse chain rule".  +John Baez pointed out that the explanation as it stood sounded rather mysterious, and he was right.  His comment made me rethink the entire Taylor series process again, and I think I, (hopefully), have a much more clear explanation now!

My premise is that the easy way to memorize the Taylor series is to understand what is going on in each term.  Hopefully, the same simple thing will be happening in each term making the whole process easy to remember and apply.

First, what does a Taylor series do?  It provides an approximation for a function near a point where we already know the function's value.  Let's take a look at the first two terms of a Taylor series centered at the point x equals 0:

The first term is the value of the function we already know at x equals 0.  To find the value of our function at a different location, as a first approximation, we can try to multiply the distance to our new location times the slope of our function at x equals zero.  In other words, we find out how our function changes with respect to x and then multiply that change by x and add that change to the known value of our function to get the new value of our function at x.

Let's add one more term to the series


The x squared term looks very similar to the x term.  It's tempting to wonder if the coefficient in front of the x squared term is the amount our function varies with respect to x squared, (the slope of our function with respect to x squared).  As it turns out, that's exactly what it is.  We now have an easy way to remember the Taylor series.  Each term multiplies [the slope of the function with respect to the nth power of x] times [the distance to our new point raised to the nth power]


The Second Glance
This explanation is easier for me to remember because I only have to remember what one term does.  I don't have to think about a series of terms, or how to operate on that series to make the terms I'm not interested in go away.  Every term does the same operation for a new power of x.  That operation is:  

multiply [the slope of the function with respect to x to the n] by [x to the n].  

This interpretation seemed obvious at first glance since the nth derivative of the function is right there in the definition of each term.  The derivative was actually what pointed me towards the idea of associating each term with the slope of the function with respect to the associated power of x in the first place.  

When I took a second look though, there was a problem.  If the nth derivative was actually finding the dependence of the function with respect to x to the n, then why was there a factor of n factorial in each term?  It turns out that the nth derivative does not in fact find the slope of a function with respect to [x to the nth power].

An example helped me see what was going on.  Suppose someone asked me to take the slope of the function x cubed with respect to x cubed.  Suspecting that I might not know how to take the derivative of a function with respect to x cubed, I'd try a graphical method first.  I'd use the values of x cubed as the x axis of my graph and plot x cubed with respect to them getting what turns out to be a straight line with slope of one.


It's an intuitive result.  Of course, x cubed changes as a straight line with respect to itself.  It's like asking what the slope of the line y equals x is with respect to x.

Now, suppose that instead of using a graph I'd decided to use derivatives.  My first guess would be that to find the slope of x cubed with respect to x I'd take the third derivative.  After all, there's an x cubed in the denominator of the derivative

.

I'd get the wrong answer though because the third derivative is not the same operation as finding the slope of a function with respect to x cubed.  After taking the derivative of x cubed three times, I'd get six, not one, the correct answer I'd found by doing the problem graphically.


So, that didn't work but it's still far easier to take a derivative than it is to draw a graph in most cases.  So, if I could somehow still use the derivative idea it would be great.  It turns out that it's easy to patch up the answer I got from taking the third derivative.  If I just divide the third derivative by the factors that resulted from doing the three derivatives (three times two times one), I'll get the correct answer of one.

Although I can't prove it yet, I claim that this is a general result and so, if I want to get the slope of  function with respect to x to the nth power, I use the following formula


The right hand side is identical to the value of the coefficient of the nth term of a Taylor series bring us back to my initial premise and, (for me anyway), easier way of memorizing the Taylor series:

The nth term of the Taylor series is 

[the slope of the function with respect to x to the n] times [the distance to the point where we want to approximate the function raised to the nth power]. 

In other words, we're just applying the same simple slope fix we saw in the term where n equals one over and over to new values of n.

Picture of the Day:
From 1/28/13




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