Click for printer friendely version of this HowTo Taylor Series Figure: Graphical Taylor analysis, the approximation includes only the linear term Often times you are working with awful functions as part of some analysis. They are impossible to integrate or differentiate or to find the roots of. As with many things, the trick is to find a reasonable approximation that is simple enough to work with. The Taylor Series allows you to do just that. With it you can decompose any complex function, $f(x)$, into an infinite series. Often, a truncated series will be used for for the analysis. For example, you might end up using only the first two terms of the series (the linear portion of the approximation). The difference between the value obtained from the infinite series and the truncated series can be called "error" or noise. The Taylor Series, or Taylor Expansion of any function, $f(x)$ is defined as: $$f(x) = f(a) + f'(a)(x-a) + f''(a)\frac{(x-a)^2}{2!}$$ $$\quad + \cdots + f^n(a)\frac{(x-a)^n}{n!} + \mathrm{remainder}.$$ (2.2.1) The first two terms are the most important because this is the linear approximation. It says that for any function, f(x), you can create an approximation of it around any point, $a$, by looking at the slope of the function at $f^{'}(a)$ and multiplying it by the distance between $a$ and another point you want to know something about. It is a straight line approximation, the value of the function at the point, $x=a$ added to the derivative $\frac{\Delta f}{\Delta x} \Delta x$. Approximate Exponentialno_title Example 2.2.0.1   Example 2.2.0.2 (Approximate Exponential)   To demonstrate the how the Taylor series can generate an approximation function for $f(x)$, we will define $f(x)$ as an exponential function and create an approximate function around zero. That is, we will let $f(x) = e^{-\lambda x}$ and $a = 0$. Since , $f'(x) = -\lambda e^{-\lambda x}$, the linear portion of the Taylor expansion is: $$f(x) = f(0) + f'(a)(x-0)$$ $$= 1 + (-\lambda)(x).$$ If we want to know $e^{-\lambda x}$ at $x = 0.1$, when $\lambda = 2$ then $$e^{-.1*2} = 1 + (-2)(0.1) = 1 - 0.2 = 0.8.$$ Compare this approximate answer to the correct answer to 6 decimal places, 0.818731. Note, if $-\lambda x$ is positive, then the series will diverge. However, when the exponent is negative, the series converges. That is, the signs of each term in the approximation will alternate. Thus, for some analyses, we can replace $e^{\lambda x}$ with $1 + \lambda x$ and continue the analysis. $\vert\boldsymbol{\vert}$ Click for printer friendely version of this HowTo

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