⧼exchistory⧽
13 exercise(s) shown, 0 hidden
BBy Bot
Nov 03'24
[math]
\newcommand{\ex}[1]{\item }
\newcommand{\sx}{\item}
\newcommand{\x}{\sx}
\newcommand{\sxlab}[1]{}
\newcommand{\xlab}{\sxlab}
\newcommand{\prov}[1] {\quad #1}
\newcommand{\provx}[1] {\quad \mbox{#1}}
\newcommand{\intext}[1]{\quad \mbox{#1} \quad}
\newcommand{\R}{\mathrm{\bf R}}
\newcommand{\Q}{\mathrm{\bf Q}}
\newcommand{\Z}{\mathrm{\bf Z}}
\newcommand{\C}{\mathrm{\bf C}}
\newcommand{\dt}{\textbf}
\newcommand{\goesto}{\rightarrow}
\newcommand{\ddxof}[1]{\frac{d #1}{d x}}
\newcommand{\ddx}{\frac{d}{dx}}
\newcommand{\ddt}{\frac{d}{dt}}
\newcommand{\dydx}{\ddxof y}
\newcommand{\nxder}[3]{\frac{d^{#1}{#2}}{d{#3}^{#1}}}
\newcommand{\deriv}[2]{\frac{d^{#1}{#2}}{dx^{#1}}}
\newcommand{\dist}{\mathrm{distance}}
\newcommand{\arccot}{\mathrm{arccot\:}}
\newcommand{\arccsc}{\mathrm{arccsc\:}}
\newcommand{\arcsec}{\mathrm{arcsec\:}}
\newcommand{\arctanh}{\mathrm{arctanh\:}}
\newcommand{\arcsinh}{\mathrm{arcsinh\:}}
\newcommand{\arccosh}{\mathrm{arccosh\:}}
\newcommand{\sech}{\mathrm{sech\:}}
\newcommand{\csch}{\mathrm{csch\:}}
\newcommand{\conj}[1]{\overline{#1}}
\newcommand{\mathds}{\mathbb}
[/math]
What cubic polynomial best approximates [math]x^4-2x^3+3x-3[/math] near [math]x=2[/math]?
BBy Bot
Nov 03'24
[math]
\newcommand{\ex}[1]{\item }
\newcommand{\sx}{\item}
\newcommand{\x}{\sx}
\newcommand{\sxlab}[1]{}
\newcommand{\xlab}{\sxlab}
\newcommand{\prov}[1] {\quad #1}
\newcommand{\provx}[1] {\quad \mbox{#1}}
\newcommand{\intext}[1]{\quad \mbox{#1} \quad}
\newcommand{\R}{\mathrm{\bf R}}
\newcommand{\Q}{\mathrm{\bf Q}}
\newcommand{\Z}{\mathrm{\bf Z}}
\newcommand{\C}{\mathrm{\bf C}}
\newcommand{\dt}{\textbf}
\newcommand{\goesto}{\rightarrow}
\newcommand{\ddxof}[1]{\frac{d #1}{d x}}
\newcommand{\ddx}{\frac{d}{dx}}
\newcommand{\ddt}{\frac{d}{dt}}
\newcommand{\dydx}{\ddxof y}
\newcommand{\nxder}[3]{\frac{d^{#1}{#2}}{d{#3}^{#1}}}
\newcommand{\deriv}[2]{\frac{d^{#1}{#2}}{dx^{#1}}}
\newcommand{\dist}{\mathrm{distance}}
\newcommand{\arccot}{\mathrm{arccot\:}}
\newcommand{\arccsc}{\mathrm{arccsc\:}}
\newcommand{\arcsec}{\mathrm{arcsec\:}}
\newcommand{\arctanh}{\mathrm{arctanh\:}}
\newcommand{\arcsinh}{\mathrm{arcsinh\:}}
\newcommand{\arccosh}{\mathrm{arccosh\:}}
\newcommand{\sech}{\mathrm{sech\:}}
\newcommand{\csch}{\mathrm{csch\:}}
\newcommand{\conj}[1]{\overline{#1}}
\newcommand{\mathds}{\mathbb}
[/math]
Another statement of Taylor's Theorem which gives a different form for the remainder is the following: Let [math]f[/math] be a function with continuous [math](n+1)\mathrm{st''[/math] derivative at every point of the interval [math][a,b][/math]. Then
[[math]]
f(b) = f(a) + f^\prime(a)(b-a) + \cdots +
\frac1{n!} f^{(n)} (a)(b-a)^n
[[/math]]
[[math]]
+ \int_a^b \frac{(b-t)^n}{n!} f^{(n+1)}
(t) \; dt
.
[[/math]]
}
- lab{9.8.12a}
Using integration by parts, show that
[[math]] \int_a^b \frac{(b-t)^n}{n!} f^{(n+1)} (t) \; dt [[/math]][[math]] = - \frac1{n!} f^{(n)} (a)(b-a)^n + \int_a^b \frac{(b-t)^{n-1}}{(n-1)!} f^{(n)} (t) \; dt . [[/math]]
- Using induction on [math]n[/math] and the result of part \ref{ex9.8.12a}, prove the above form of Taylor's Theorem in which the remainder appears as an integral.
BBy Bot
Nov 03'24
[math]
\newcommand{\ex}[1]{\item }
\newcommand{\sx}{\item}
\newcommand{\x}{\sx}
\newcommand{\sxlab}[1]{}
\newcommand{\xlab}{\sxlab}
\newcommand{\prov}[1] {\quad #1}
\newcommand{\provx}[1] {\quad \mbox{#1}}
\newcommand{\intext}[1]{\quad \mbox{#1} \quad}
\newcommand{\R}{\mathrm{\bf R}}
\newcommand{\Q}{\mathrm{\bf Q}}
\newcommand{\Z}{\mathrm{\bf Z}}
\newcommand{\C}{\mathrm{\bf C}}
\newcommand{\dt}{\textbf}
\newcommand{\goesto}{\rightarrow}
\newcommand{\ddxof}[1]{\frac{d #1}{d x}}
\newcommand{\ddx}{\frac{d}{dx}}
\newcommand{\ddt}{\frac{d}{dt}}
\newcommand{\dydx}{\ddxof y}
\newcommand{\nxder}[3]{\frac{d^{#1}{#2}}{d{#3}^{#1}}}
\newcommand{\deriv}[2]{\frac{d^{#1}{#2}}{dx^{#1}}}
\newcommand{\dist}{\mathrm{distance}}
\newcommand{\arccot}{\mathrm{arccot\:}}
\newcommand{\arccsc}{\mathrm{arccsc\:}}
\newcommand{\arcsec}{\mathrm{arcsec\:}}
\newcommand{\arctanh}{\mathrm{arctanh\:}}
\newcommand{\arcsinh}{\mathrm{arcsinh\:}}
\newcommand{\arccosh}{\mathrm{arccosh\:}}
\newcommand{\sech}{\mathrm{sech\:}}
\newcommand{\csch}{\mathrm{csch\:}}
\newcommand{\conj}[1]{\overline{#1}}
\newcommand{\mathds}{\mathbb}
[/math]
Let [math]f[/math] have a continuous second derivative at every point of an interval containing the number [math]a[/math] in its interior, and let [math]f^\prime(a) = 0[/math]. Show that [math]f[/math] has a local maximum value at [math]a[/math] if [math]f^{\prime\prime} (a) \lt 0[/math], and a local minimum value at [math]a[/math] if [math]f^{\prime\prime} (a) \gt 0[/math]. [Hint: Use the Taylor Formula [math]f(x) = T_1(x) + R_1[/math] and the fact that, if a continuous function is positive (or negative) at [math]a[/math], then it is positive (or negative) near [math]a[/math].]