24 Polinomio de Taylor

Veamos cuál es: dado un polinomio general de grado $\displaystyle n$,

1. 1.
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   $\displaystyle p(x_{0})=f(x_{0})$. Si evaluamos el polinomio, $\displaystyle p(x_{0})=a_{0}+a_{1}(x_{0}-x_{0})+\cdots a_{n}(x_{0}-x_{0})^{n}=a_{0}\Rightarrow a_{0}=f(x_{0})$.

2. 2.
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   $\displaystyle p^{\prime}(x_{0})=f^{\prime}(x_{0})\Rightarrow p^{\prime}(x)=0+a_{1}+a_{2}\cdot 2(x-x_{0})+\cdots+a_{n}n(x-x_{0})^{n-1}$. Como $\displaystyle p^{\prime}(x_{0})=f^{\prime}(x_{0})$, $\displaystyle p^{\prime}(x_{0})=a_{1}=f^{\prime}(x_{0})$.

3. 3.
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   $\displaystyle p^{\prime\prime}(x_{0})=f^{\prime\prime}$. $\displaystyle p^{\prime\prime}(x)=a_{2}2+a_{3}\cdot 3\cdot 2\cdot(x-x_{0})+\cdots+a_{n}\cdot n\cdot(n-1)\cdot(x-x_{0})^{n-2}$. Evaluando en $\displaystyle x_{0},$ $\displaystyle p^{\prime\prime}(x_{0})=a_{2}\cdot 2\Rightarrow a_{2}=f^{\prime\prime}(x_{0})/2$.

4. 4.
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   $\displaystyle p^{\prime\prime\prime}(x)=a_{3}\cdot 3\cdot 2+a_{4}\cdot 3\cdot 3\cdot 2(x-x_{0})+\cdots+a_{n}\cdot n\cdot(n-1)\cdot(n-2)\cdot(x-x_{0})^{n-3}\Rightarrow p^{\prime\prime\prime}(x_{0})=a_{3}\cdot 3\cdot 2$.

5. 5.
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   Lo hacemos hasta llegar a la derivada $\displaystyle n$-esima: $\displaystyle p^{(n)}(x_{0})=f^{(n)(x_{0})}\Rightarrow a_{n}\cdot n!=f^{(n)}(x_{0})\Rightarrow a_{n}=\frac{f^{(n)}(x_{0})}{n!}$.

Luego el polinomio de Taylor de grado $\displaystyle n$ de $\displaystyle f$ en $\displaystyle x_{0}$ es:

Por ejemplo, el polinomio de grado $\displaystyle 3$ de $\displaystyle f(x)=\sin x$ en el $\displaystyle 0$:

ya que $\displaystyle f(0)=\sin 0=0$, $\displaystyle f^{\prime}(x)=\cos x\Rightarrow f^{\prime}(0)=1$, $\displaystyle f^{\prime\prime}(x)=-\sin x\Rightarrow f^{\prime\prime}(0)=0$, $\displaystyle f^{\prime\prime\prime}(x)=-\cos x\Rightarrow f^{\prime\prime\prime}(0)=-1$.

Nos gustaria resolver $\displaystyle\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to x_{0}}\frac{f(x)\cdot h(x)}{g(x)}$, sabiendo que $\displaystyle f(x)\approx P_{n,x_{0},f}(x)=\frac{f^{(n)}(x_{0})}{n!}(x-x_{0})^{n}$.

Por ejemplo,

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  $\displaystyle\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to 0}\frac{\sin x}{x}=\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to 0}\frac{x-\frac{\sin(c)}{2!}x^{2}}{x}=\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to 0}1-\frac{\sin c}{2!}x=1$.

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  $\displaystyle\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to 0}\frac{\tan x-\sin x}{x^{3}}=\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to 0}\frac{\frac{3}{3!}x^{3}+\text{Error}}{x^{3}}=\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to 0}\frac{1}{2}+\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to 0}\frac{\text{Error}}{x^{3}}=\frac{1}{2}$ ya que $\displaystyle\tan x-\sin x\approx 3x^{3}$ (polinomio de Taylor de grado $\displaystyle 3$ en $\displaystyle 0$ ). $\displaystyle f(0)=\tan 0-\sin 0=0$, $\displaystyle f^{\prime}(x)=1+\tan^{2}x-\cos x\Rightarrow f^{\prime}(0)=1+0-1=0$, $\displaystyle f^{(2)}(x)=2\tan x(1+\tan^{2}x)+\sin x\Rightarrow f^{(2)}(0)=0$, $\displaystyle f^{(3)}(x)=2[(1+\tan^{2}x)(1+\tan^{2}x)+\tan x(2\tan x(1+\tan^{2}x))]+\cos x\Rightarrow f^{(3)}(0)=2+1\neq 0$. $\displaystyle\text{Error}=\frac{f^{(n+1)}(c)}{(x-x_{0})^{n+1}}=\frac{f^{(4)}(c)}{(x-x_{0})^{4}}$. Por otro lado, $\displaystyle\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to 0}\frac{\frac{f^{(4)}(c)}{4!}x^{4}}{x^{3}}=\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to 0}\frac{f^{(4)}(c)}{4!}\cdot x=0$ porque $\displaystyle c\in(0,x)$.

Entonces, siguiendo esta idea:

ya que $\displaystyle\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to x_{0}}\frac{f(x)}{(x-x_{0})^{n}}=\mathop{\operator@font l\acute{{\imath}}m}\limits_{x\to x_{0}}\frac{\frac{f^{(n)}(x_{0})}{n!}(x-x_{0})^{n}+\text{Error}}{(x-x_{0})^{n}}=\frac{f^{(n)}(x_{0})}{n!}$.

Decimos que $\displaystyle\frac{f^{(n)}(x_{0})}{n!}(x-x_{0})^{n}$ es el infinitesimo equivalente de $\displaystyle f$ cuando $\displaystyle x$ tiende a $\displaystyle x_{0}$.