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--- analise2:fourier [2022/05/25 11:41] – tahzibi
+++ analise2:fourier [2022/05/27 08:44] (current) – external edit 127.0.0.1
@@ Line 2: / Line 2: @@
 Polinômios trigonométricos: Toda função escrita como:
 $$
- f(x) = a_0 + \sum_{n=1}^{m} a_n cos(nx) + b_n sin(nx)
+ f(x) = \frac{a_0}{2} + \sum_{n=1}^{m} a_n cos(nx) + b_n sin(nx)
 $$
 onde $a_i, b_i$ são números complexos.
@@ Line 35: / Line 35: @@
 A demonstração é um cálculo:
-Mostramos que $\int_{a}^{b} f \bar{t_n} = \int f \sum_{m=1}^{n} \bar{\gamma_m} \bar{\phi_m} = \sum_{m=1}^{n} \bar{\gamma_m} \bar{c_m} , \int_{a}^{b} |t_n|^2 = \sum_{m=1}^{n} |\gamma_m}|^2.$
+Mostramos que $\int_{a}^{b} f \bar{t_n} = \int_{a}^{b} f \sum_{m=1}^{n} \bar{\gamma_m} \bar{\phi_m} = \sum_{m=1}^{n} \bar{\gamma_m} \bar{c_m} , \int_{a}^{b} |t_n|^2 = \sum_{m=1}^{n}  | \gamma_{m}|^2.$
 Após manipulações chegamos
 $$
- \int_{a}^{b} |f-t_n|^2 dx = \int_{a}^{b} |f|^2 dx -\sum_{m=1}^{n} |c_m}|^2 + \sum_{m=1}^{n} |\gamma_m} - c_m|^2
+ \int_{a}^{b} |f-t_n|^2 dx = \int_{a}^{b} |f|^2 dx -\sum_{m=1}^{n} |c_{m}|^2 + \sum_{m=1}^{n} | \gamma_{m} - c_m|^2.
 $$
+Isto mostra o resultado e inclusive teremos igualdade se somente se $c_m = \gamma_m.$