6.3 Taylor and maclaurin series  (Page 8/13)

[T] e ^x approximated by $1+x,a=0$

[T] $\text{sin}x$ approximated by x , $a=0$

[T] $\text{ln}x$ approximated by $x-1,a=1$

[T] $\text{cos}x$ approximated by $1,a=0$

$x^4$ at $a=\mathrm{-1}$

$1+x+x^2+x^3$ at $a=\mathrm{-1}$

$\text{sin}x$ at $a=\pi$

$\text{cos}x$ at $a=2\pi$

$\text{sin}x$ at $x=\frac{\pi }{2}$

$\text{cos}x$ at $x=\frac{\pi }{2}$

$e^x$ at $a=\mathrm{-1}$

$e^x$ at $a=1$

$\frac{1}{{\left(x-1\right)}^2}$ at $a=0$ ( Hint: Differentiate $\frac{1}{1-x}.)$

$\frac{1}{{\left(x-1\right)}^3}$ at $a=0$

$F\left(x\right)={\displaystyle {\int }_0^x\text{cos}\left(\sqrt{t}\right)dt};f\left(t\right)={\displaystyle \sum _{n=0}^{\infty }{\left(\mathrm{-1}\right)}^n\frac{t^n}{\left(2n\right)\text{!}}}$ at $a=0$ ( Note : $f$ is the Taylor series of $\text{cos}\left(\sqrt{t}\right).)$

$f\left(x\right)=2-x$

$f\left(x\right)=x^3$

$f\left(x\right)={\left(x-2\right)}^2$

$f\left(x\right)=\text{ln}x$

$f\left(x\right)=\frac{1}{x}$

$f\left(x\right)=\frac{1}{2x-x^2}$

$f\left(x\right)=\frac{x}{4x-2x^2-1}$

$f\left(x\right)=e^{\text{−}x}$

$f\left(x\right)=e^{2x}$

${\displaystyle \sum _{n=0}^{\infty }\frac{1}{n\text{!}}}$

${\displaystyle \sum _{n=0}^{\infty }\frac{2^n}{n\text{!}}}$

${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(\mathrm{-1}\right)}^n{\left(2\pi \right)}^{2n}}{\left(2n\right)\text{!}}}$

${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(\mathrm{-1}\right)}^n{\left(2\pi \right)}^{2n+1}}{\left(2n+1\right)\text{!}}}$

[T] Plot ${\text{sin}}^{2}x-{\left(S_5\left(x\right)\right)}^2$ on $[\text{−}\pi ,\pi ].$ Compare the maximum difference with the square of the Taylor remainder estimate for $\text{sin}x.$

[T] Plot ${\text{cos}}^{2}x-{\left(C_4\left(x\right)\right)}^2$ on $[\text{−}\pi ,\pi ].$ Compare the maximum difference with the square of the Taylor remainder estimate for $\text{cos}x.$

[T] Plot $\left|2S_5\left(x\right)C_4\left(x\right)-\text{sin}\left(2x\right)\right|$ on $[\text{−}\pi ,\pi ].$

[T] Compare $\frac{S_5\left(x\right)}{C_4\left(x\right)}$ on $\left[\mathrm{-1},1\right]$ to $\text{tan}x.$ Compare this with the Taylor remainder estimate for the approximation of $\text{tan}x$ by $x+\frac{x^3}{3}+\frac{2x^5}{15}.$

[T] Plot $e^x-e_4\left(x\right)$ where $e_4\left(x\right)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}$ on $[0,2].$ Compare the maximum error with the Taylor remainder estimate.

(Taylor approximations and root finding.) Recall that Newton’s method $x_{n+1}=x_n-\frac{f\left(x_n\right)}{f\prime \left(x_n\right)}$ approximates solutions of $f\left(x\right)=0$ near the input $x_0.$

1. If $f$ and $g$ are inverse functions, explain why a solution of $g\left(x\right)=a$ is the value $f\left(a\right)\text{of}f.$
2. Let $p_N\left(x\right)$ be the $N\text{th}$ degree Maclaurin polynomial of $e^x.$ Use Newton’s method to approximate solutions of $p_N\left(x\right)-2=0$ for $N=4,5,6.$
3. Explain why the approximate roots of $p_N\left(x\right)-2=0$ are approximate values of $\text{ln}\left(2\right).$

$\underset{x\to 0}{\text{lim}}\frac{\text{cos}x-1}{x^2}$

$\underset{x\to 0}{\text{lim}}\frac{\text{ln}\left(1-x^2\right)}{x^2}$

$\underset{x\to 0}{\text{lim}}\frac{e^{x^2}-x^2-1}{x^4}$

$\underset{x\to 0^{+}}{\text{lim}}\frac{\text{cos}\left(\sqrt{x}\right)-1}{2x}$