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Informal Derivation of Taylor Series

We have a function $ f(x)$ and we want to approximate it using an $ n$th-order polynomial:

$\displaystyle f(x) = f_0 + f_1 x + f_2 x^2 + \cdots + f_n x^n + R_{n+1}(x)
$

where $ R_{n+1}(x)$, the approximation error, is called the remainder term. We may assume $ x$ and $ f(x)$ are real, but the following derivation generalizes unchanged to the complex case.

Our problem is to find fixed constants $ \{f_i\}_{i=0}^{n}$ so as to obtain the best approximation possible. Let's proceed optimistically as though the approximation will be perfect, and assume $ R_{n+1}(x)=0$ for all $ x$ ( $ R_{n+1}(x)\equiv0$), given the right values of $ f_i$. Then at $ x=0$ we must have

$\displaystyle f(0) = f_0
$

That's one constant down and $ n-1$ to go! Now let's look at the first derivative of $ f(x)$ with respect to $ x$, again assuming that $ R_{n+1}(x)\equiv0$:

$\displaystyle f^\prime(x) = 0 + f_1 + 2 f_2 x + 3 f_2 x^2 + \cdots + n f_n x^{n-1}
$

Evaluating this at $ x=0$ gives

$\displaystyle f^\prime(0) = f_1.
$

In the same way, we find

\begin{eqnarray*}
f^{\prime\prime}(0) &=& 2 \cdot f_2 \\
f^{\prime\prime\prime}...
...cdot 2 \cdot f_3 \\
& \cdots & \\
f^{(n)}(0) &=& n! \cdot f_n
\end{eqnarray*}

where $ f^{(n)}(0)$ denotes the $ n$th derivative of $ f(x)$ with respect to $ x$, evaluated at $ x=0$. Solving the above relations for the desired constants yields

\begin{eqnarray*}
f_0 &=& f(0) \\
f_1 &=& \frac{f^{\prime}(0)}{1} \\
f_2 &=& \...
...dot 2\cdot 1} \\
& \cdots & \\
f_n &=& \frac{f^{(n)}(0)}{n!}.
\end{eqnarray*}

Thus, defining $ 0!\isdef 1$ (as it always is), we have derived the following polynomial approximation:

$\displaystyle \zbox {f(x) \approx \sum_{k=0}^n \frac{f^{(k)}(0)}{k!} x^k}
$

This is the $ n$th-order Taylor series expansion of $ f(x)$ about the point $ x=0$. Its derivation was quite simple. The hard part is showing that the approximation error (remainder term $ R_{n+1}(x)$) is small over a wide interval of $ x$ values. Another ``math job'' is to determine the conditions under which the approximation error approaches zero for all $ x$ as the order $ n$ goes to infinity. The main point to note here is that the Taylor series itself is simple to derive.


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Previous: Taylor Series Expansions
Next: Taylor Series with Remainder

written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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