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Title: Integrable Functions
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Series: Fourier Transform
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Chapter: Fourier Series
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YouTube-Title: Fourier Transform 5 | Integrable Functions
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Bright video: https://youtu.be/2NUNRQEp7YE
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Dark video: https://youtu.be/LgRBO95N0xM
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Ad-free video: Watch Vimeo video
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Quiz: Test your knowledge
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Dark-PDF: Download PDF version of the dark video
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Print-PDF: Download printable PDF version
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Thumbnail (bright): Download PNG
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Thumbnail (dark): Download PNG
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Subtitle on GitHub: ft05_sub_eng.srt missing
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Timestamps (n/a)
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Subtitle in English (n/a)
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Quiz Content
Q1: Consider the function $f: [0,1] \rightarrow \mathbb{R}$ given by $f(0) = 0$ and $f(x) = \frac{1}{x}$ for $x > 0$. Is the function integrable?
A1: No, because $\int_0^1 | f(x) | , dx = \infty $.
A2: Yes, because $\int_0^1 | f(x) | , dx < \infty $.
A3: No, because $f$ is unbounded.
A4: Yes, because $f$ is bounded.
Q2: Consider the function $f: [0,1] \rightarrow \mathbb{R}$ given by $f(0) = 0$ and $f(x) = \frac{1}{\sqrt{x}}$ for $x > 0$. Is the function integrable?
A1: No, because $\int_0^1 | f(x) | , dx = \infty $.
A2: Yes, because $\int_0^1 | f(x) | , dx < \infty $.
A3: No, because $f$ is unbounded.
A4: Yes, because $f$ is bounded.
Q3: Consider the function $f: [0,1] \rightarrow \mathbb{R}$ given by $f(0) = 0$ and $f(x) = \frac{1}{\sqrt{x}}$ for $x > 0$. Is the function square-integrable?
A1: No, because $\int_0^1 | f(x) |^2 , dx = \infty $.
A2: Yes, because $\int_0^1 | f(x) |^2 , dx < \infty $.
A3: No, because $f$ is unbounded.
A4: Yes, because $f$ is bounded.
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Last update: 2024-11
