„Analízis (MSc) típusfeladatok” változatai közötti eltérés
| 300. sor: | 300. sor: | ||
<big>b)</big> <math>\int_R \psi_n(x)dx = \int_0^\infty -(\frac{x^n}{n!} e^{-x})' dx = -\left[\frac{x^n}{n!} e^{-x}\right]_0^\infty = 0</math> | <big>b)</big> <math>\int_R \psi_n(x)dx = \int_0^\infty -(\frac{x^n}{n!} e^{-x})' dx = -\left[\frac{x^n}{n!} e^{-x}\right]_0^\infty = 0</math> | ||
<big>c)</big> <math>C_{\psi_n} = 2 \pi \int_{-\infty}^\infty \frac{\left| \hat{\psi} \right|^2}{|y|} dy</math> | |||
Először számoljuk ki a wavelet Fourier trafóját (felhasználom, hogy <math>\mathcal{F}(-f') = -iy\mathcal{F}(f)</math>): | |||
<math>\hat{\psi} = \mathcal{F}(-(\frac{x^n}{n!} e^{-x})' \cdot H(x)) = -\frac{iy}{n!} \mathcal{F}(x^n e^{-x}H(x)) = -\frac{iy}{n!} i^n \mathcal{F}(e^{-x}H(x))^{(n)} = -\frac{iy}{n!} i^n \left(\frac{1}{\sqrt{2\pi}} \frac{1}{1+iy}\right)^{(n)} =</math> | |||
<math>= -\frac{iy}{n!} i^n i^n (-1)(-2) \dots(-n) \frac{1}{\sqrt{2\pi}} \frac{1}{(1+iy)^{n+1}} = -iy \frac{n!}{n!} (-1)^n (-1)^n \frac{1}{\sqrt{2\pi}} \frac{1}{(1+iy)^{n+1}} = -iy \frac{1}{\sqrt{2\pi}} \frac{1}{(1+iy)^{n+1}}</math> | |||
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