vuoi
o PayPal
tutte le volte che vuoi
Serie di Fourier
c0 + ∑k=1∞ (ak cos kx + bk sin kx) = g(x)
dove g(x) = f(x + T) = f(2π/T x)
Teorema: g(x + T) = g(x + T)
dim.:
g(x + T) = f(2π/T (x + T)) = f(2π/T + 2π)
= f(2π/T) = g(x)
Convergenza:
- Assoluta convergenza
∑k=1∞ |ak cos kx + bk sin kx| ≤ |ak cos kx| + |bk sin kx| = (|ak| |cos kx| + |bk| |sin kx|)
≤ (|ak| + |bk|)
se ∑k=1∞ |ak|, ∑k=1∞ |bk| converge. ass. => serie Fourier converge
- 2o criterio di Dirichlet
∑k=1∞ ak cos kx, ∑k=1∞ bk sin kx
ak ≠ 0, bk ≠ 0 => serie di Fourier converge in (0,2π)
3) 2o criterio di Dirichlet
\[\sum_{k=1}^{\infty}(-1)^{k}a_k\sin kx \quad \sum_{k=1}^{\infty}(-3)^{k}b_k\sin kx\]
\[a_k \to 0 \quad b_k \to 0\]
\[serie \ di \ fourier \ converge \ in \ \forall x \neq \frac{\pi}{\pi}\]
\[\sum_{k=1}^{\infty} (-1)^{k} (-3)^{k} a_k = \sum_{k=1}^{\infty} a_k\]
Trova la somma:
1)
\[\int_{-\pi}^{\pi} f(x)dx = \int_{-\pi}^{\pi} \Big[\frac{a_0}{2} + \sum_{k=1}^{\infty} [a_k \cos kx + b_k \sin kx]\Big]dx = \frac{a_0}{2} 2\pi + \sum_{k=1}^{\infty} a_k\]
\[\int_{-\pi}^{\pi} \cos kx dx + b_k \int_{-\pi}^{\pi} \sin kx dx = a_0 = \frac{a_0}{\pi}\]
\[\Rightarrow\]
\[a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) dx\]
2)
\[\int_{-\pi}^{\pi} f(x) \cos mx\ dx = \int_{-\pi}^{\pi} \Big[\frac{a_0}{2} \cos mx + \sum_{k=1}^{\infty} a_k\Big] \int_{-\pi}^{\pi} \cos kx \cos mx \ dx + b_k\]
\[\int_{-\pi}^{\pi} \sin kx \cos mx = a_k \int_{-\pi}^{\pi} \frac{1}{2} [\cos(kx+mx) + \cos(kx-mx)] dx =\]
\[= a_k \frac{1}{2} \cdot 2\pi \quad , \ per \ k = m \]
\[\Rightarrow\]
\[a_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos kx dx\]