Expressing Functions with Taylor Series¶

A sequence is an ordered set of numbers denoted by the list of numbers inside parentheses. For example, $s = (s_1, s_2, s_3, \cdots)$ means $s$ is the sequence $s_1, s_2, s_3, \cdots$ and so on. In this context, “ordered” means that $s_1$ comes before $s_2$, not that $s_1 < s_2$. Many sequences have a more complicated structure. For example, $s = (n^2, n\in N)$ is the sequence 0, 1, 4, 9, $\cdots$. A series is the sum of a sequence up to a certain element. An infinite sequence is a sequence with an infinite number of terms, and an infinite series is the sum of an infinite sequence.

A Taylor series expansion is a representation of a function by an infinite series of polynomials around a point. Mathematically, the Taylor series of a function, $f(x)$, is defined as:

\[ f(x) = \sum_{n = 0}^{\infty} \frac{f^{(n)}(a)(x-a)^n}{n!}, \]

where $f^{(n)}$ is the $n^\mathrm{th}$ derivative of $f$ and $f^{(0)}$ is the function $f$

TRY IT! Compute the Taylor series expansion for $f(x) = 5x^2 + 3x + 5$ around $a = 0$, and $a = 1$. Verify that $f$ and its Taylor series expansions are identical.

First compute derivatives analytically:

\[\begin{eqnarray*} f(x) &=& 5x^2 + 3x + 5\\ f^{\prime}(x) &=& 10x + 3\\ f''(x) &=& 10 \end{eqnarray*}\]

Around a = 0:

\[f(x) = \frac{5x^0}{0!} + \frac{3x^1}{1!} + \frac{10x^2}{2!} + 0 + 0 + \cdots = 5x^2 + 3x + 5 \]

Around a = 1:

\[\begin{eqnarray*} f(x) &=& \frac{13(x-1)^0}{0!} + \frac{13(x-1)^1}{1!} + \frac{10(x-1)^2}{2!} + 0 + \cdots\\ &=& 13 + 13x - 13 + 5x^2 - 10x + 5 = 5x^2 + 3x + 5 \end{eqnarray*}\]

Note: The Taylor series expansion of any polynomial has finite terms because the $n^\mathrm{th}$ derivative of any polynomial is 0 for $n$ large enough.

TRY IT! Write the Taylor series for $\sin(x)$ around the point $a = 0$.

Let $f(x) = \sin(x)$. Then according to the Taylor series expansion, $$f(x) = \frac{\sin(0)}{0!}x^0 + \frac{\cos(0)}{1!}x^1 + \frac{-\sin(0)}{2!}x^2 + \frac{-\cos(0)}{3!}x^3 + \frac{\sin(0)}{4!}x^4 + \frac{\cos(0)}{5!}x^5 + \cdots.$$

The expansion can be written compactly by the formula $$f(x) = \sum_{n = 0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}, $$ which ignores the terms that contain $\sin(0)$ (i.e., the even terms). However, because these terms are ignored, the terms in this series and the proper Taylor series expansion are off by a factor of $2n+1$; for example the $n = 0$ term in formula is the $n = 1$ term in the Taylor series, and the $n = 1$ term in the formula is the $n = 3$ term in the Taylor series.mm

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Python Numerical Methods

Expressing Functions with Taylor Series¶