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## Taylor PolynomialsProbably the most important application of Taylor series is to
We look at various Taylor polynomials: T_1(x)&=&f(a) + f'(a) (x-a)\\ T_2(x)&=&f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2\\ T_3(x)&=&f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3\\ &\vdots&\\ T_n(x)&=&f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n \end{eqnarray}$ You may recognize the first Taylor polynomial above. (Does it look familiar? Take a moment to
think about this.) It is the Taylor polynomials extend the idea of
linearization. To approximate $f$ at a given
value of $x$, we will use $T_n(x)$ for a value of $n$ that gives a
good enough approximation. We see from $T_n(x)$ above that we
will need to find an $x$-value $a$ at which we can evaluate the
derivatives of $f$. Our
approximations get better the more terms we include,
and they also allow us to get further from $a$ and still have a good
approximation.
$\begin{array}{lll} If you know a series,
finding an $n^{th}$ degree Taylor polynomial is easy.
If you don't have the series at hand, you can compute the polynomial
as we just did for sine.My calculator approximates $\sin(.5)=.4794255386$, and $T_7(.5)=.5-\frac{.5^3}{6}+\frac{.5^5}{120}-\frac{.5^7}{5040}=.4794255332$.
This video will compute approximations of $e^x$ and $\ln(x)$ near $a=0$, and $\sin x$ near $a=\frac{\pi}{4}$. |