University of Florida/Egm6321/f09.team1.gzc/Mtg6

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Proof of Taylor series continued Since $\underset{x_0}{\overset{x}{\int }}{f}^1(t)dt=[f^0(t){]}_{t=x_0}^{t=x}$

(1)

$\Rightarrow (e)p.5-3$

Int. by parts (1)

$(2)\{\begin{array}{cc}& \underset{x_0}{\overset{x}{\int }}\underset{u^{\text{'}}}{\underbrace{1}}\underset{v}{\underbrace{f^{(1)}(t)}}dt={[uv]}_{x_0}^x-\underset{}{\overset{}{\int }}{uv}^{\text{'}}\\ & ={[t{f}^{(1)}t]}_{t=x_0}^{t=x}-\underset{\alpha }{\underbrace{\underset{x_0}{\overset{x}{\int }}t{f}^{(2)}dt}}\\ & =x{f}^1(x)-x_0{f}^{(1)}(x)0)-\alpha \\ & +x{f}^{(1)}(x_0)-x{f}^{(1)}(x_0)\\ & =\underset{\underset{x_0}{\overset{x}{\int }}x{f}^{(2)}(t)dt=:\beta }{\underbrace{[x{f}^{(}1)(x)-x{f}^{(}1)(x_0)]}}+(x-x_0)f^{(1)}(x_0)-\alpha \end{array}$

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Combine [ + β - α] into a single int. use (2) p.6-1 in (2) p.5-3:

$f(x)=f(x_0)+\frac{{(x-x_0)}^1}{1!}{f}^{(1)}(x_0)+\underset{\beta -\alpha }{\underbrace{\underset{x_0}{\overset{x}{\int }}(x-t)f^{(2)}(t)dt}}$

(1)

HW*2.1: 1) Do integration by parts on last term (integration) of (1) to reveal 3 more terms in Taylor series, i.e. , $\frac{{x-x_0}^2}{2!}{f}^{(2)}(x_0)+\frac{{x-x_0}^3}{3!}{f}^{(3)}(x_0)+\frac{{x-x_0}^4}{4!}{f}^{(4)}(x_0)$

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to verifyTemplate:Font Template:Font for (n+1) eTemplate:Fontwith R_(n+2)(x)

4) UseTemplate:Font onTemplate:Font Template:Fontto showTemplate:FontTemplate:Font

Template:Font$\underset{a}{\overset{b}{\int }}w(x)g(x)dx\stackrel{(1)}{=}g(\xi )\underset{a}{\overset{b}{\int }}w(x)dx$ $\xi \in [a,b]$

Template:Font$f(x)=sinx,x\in [0,\pi ]$ Constrast Taylor Series of f(.) around

$\{\begin{array}{cc}& x_0=\frac{\pi }{4}S10\\ & x_0=\frac{3\pi }{4}S11\end{array}$ for n = 0,1,2,...,10

Plot these series (for each n) Find (estimate) max $|R_{n+1}(x=\frac{3\pi }{4})|$

$(5)p.3-3\Rightarrow |R_{n+1}(x=\frac{3\pi }{4})|\leqq \frac{{x-x_0}^{n+1}}{(n+1)!}(\alpha )$

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$\alpha =max|f^{(n+1)}(t)|$

$\underset{\leqq 1(f(x)=sinx)}{\underbrace{t\in [x_0,x]}}$

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File:Mtg6.pg5.fig1.svg