12 dez 2015, 20:28
12 dez 2015, 20:51
13 dez 2015, 21:47
pedrodaniel10 Escreveu:Para \(n\geq 1\):
\(f(x)=1 \Rightarrow f'(x)=0
f(x)=x \Rightarrow f'(x)=1
f(x)=x^2 \Rightarrow f'(x)=2x
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f(x^{n-1})=x^2 \Rightarrow f'(x)=(n-1)x^{n-2}\)
Pelo que:
\(D_{\beta \rightarrow \beta }=\begin{bmatrix} 0 & 1 & 0 & . & 0\\ . & 0 & 2 & . & .\\ . & . & 0 & ... & .\\ . & . & . & . & n-1\\ 0 & 0 & 0 & . & 0 \end{bmatrix}\)