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Exercise 3.F.7

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Proof. We have that $\frac{d^{j}}{d x^{j}} x^{k} = \frac{k!}{(k - j)!} x^{k - j}$ for $k \geq j$ (this can easily be proven by induction on $j$ , but I don’t think that’s really the point here).

Thus, if $k = j$ , we have

\begin{aligned} φ_{j} (x^{k}) & = φ_{j} (x^{j}) \\ = \frac{\frac{d^{j}}{d x^{j}} x^{j} |_{x = 0}}{j!} \\ = \frac{\frac{j!}{(j - j)!} x^{j - j} |_{x = 0}}{j!} \\ = \frac{j!}{j!} \\ = 1 \end{aligned}

And if $k > j$ , we have

\begin{aligned} φ_{j} (x^{k}) & = \frac{\frac{d^{j}}{d x^{j}} x^{k} |_{x = 0}}{j!} \\ = \frac{\frac{k!}{(k - j)!} x^{k - j} |_{x = 0}}{j!} \\ = 0 \end{aligned}

Clearly, if $k < j$ then $\frac{d^{j}}{d x^{j}} x^{k} = 0$ . Therefore

φ_{j} (x^{k}) = {\begin{matrix} 1, & if j = k \\ 0, & if j \neq k \end{matrix}

Which is the definition of dual basis. □

celiopassos

2017-10-06 00:00

Exercise 3.F.7

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