Question Number 208 by 02@>@0 last updated on 25/Jan/15
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Answered by 123456 last updated on 16/Dec/14
![Random things LOl I lets f:[0∣1]→R be a integable and continuos function define ∥f∥ be [0∣1]→R^+ where ∥f∥=(√(∫_0 ^1 f^2 dx)) wich we easy can check that ∥cf∥=∣c∣∥f∥ ∥f∥≥0 then if ∥f∥≠0, exist a vesor function given by f^ =(f/(∥f∥))](https://www.tinkutara.com/question/Q214.png)
$$\boldsymbol{\mathrm{R}}\mathrm{andom}\:\boldsymbol{\mathrm{things}}\:\mathbb{L}\mathscr{O}\mathfrak{l}\:\mathcal{I} \\ $$$$\mathrm{lets}\:\mathrm{f}:\left[\mathrm{0}\mid\mathrm{1}\right]\rightarrow\mathbb{R}\:\mathrm{be}\:\mathrm{a}\:\mathrm{integable}\:\mathrm{and}\:\mathrm{continuos}\:\mathrm{function} \\ $$$$\mathrm{define}\:\parallel{f}\parallel\:\mathrm{be}\:\left[\mathrm{0}\mid\mathrm{1}\right]\rightarrow\mathbb{R}^{+} \:\mathrm{where} \\ $$$$\parallel{f}\parallel=\sqrt{\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{f}^{\mathrm{2}} {dx}} \\ $$$$\mathrm{wich}\:\mathrm{we}\:\mathrm{easy}\:\mathrm{can}\:\mathrm{check}\:\mathrm{that} \\ $$$$\parallel{cf}\parallel=\mid{c}\mid\parallel{f}\parallel \\ $$$$\parallel{f}\parallel\geqslant\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{if}\:\parallel{f}\parallel\neq\mathrm{0},\:\mathrm{exist}\:\mathrm{a}\:\boldsymbol{\mathrm{vesor}}\:\boldsymbol{\mathrm{function}}\:\mathrm{given}\:\mathrm{by} \\ $$$$\hat {{f}}=\frac{{f}}{\parallel{f}\parallel} \\ $$
Answered by 123456 last updated on 24/Dec/14
$${f}_{{n}} \left({x}\right)=\frac{{x}\left({x}+\mathrm{1}\right)\centerdot\centerdot\centerdot\left({x}+{n}\right)}{{x}\left({x}−\mathrm{1}\right)\centerdot\centerdot\centerdot\left({x}−{n}\right)},{n}\in\mathbb{N} \\ $$$${f}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{f}_{{n}} \left({x}\right)=? \\ $$$$\mathbb{R}\rightarrow\mathbb{R} \\ $$$${z}=\left\{−\mathrm{1},−\mathrm{2},…,−{n}\right\} \\ $$$${p}=\left\{\mathrm{0},+\mathrm{1},+\mathrm{2},…,+{n}\right\} \\ $$$${f}:\mathbb{N}×\left(\mathbb{R}/{p}\right)\rightarrow\mathbb{R} \\ $$$$\mathrm{f}:\:{n}×\:\:\:\:\:\:{x}\:\:\:\:\rightarrow{f} \\ $$