-
Title: Orthonormalbasis of Trigonometric Functions
-
Series: Fourier Transform
-
Chapter: Fourier Series
-
YouTube-Title: Fourier Transform 4 | Orthonormalbasis of Trigonometric Functions
-
Bright video: https://youtu.be/6Z8EXPFH-pk
-
Dark video: https://youtu.be/0bhXht8mDUY
-
Ad-free video: Watch Vimeo video
-
Quiz: Test your knowledge
-
Dark-PDF: Download PDF version of the dark video
-
Print-PDF: Download printable PDF version
-
Thumbnail (bright): Download PNG
-
Thumbnail (dark): Download PNG
-
Subtitle on GitHub: ft04_sub_eng.srt missing
-
Timestamps (n/a)
-
Subtitle in English (n/a)
-
Quiz Content
Q1: Consider the trigonometric polynomial $f(x) = 1 + \sum_{k=1}^n a_k \cos(k x)$. How can the coefficients $a_k$ be calculated?
A1: $a_k = \frac{1}{\pi} \int_{-\pi}^{\pi} \cos(kx) f(x) , dx$
A2: $a_k = \int_{-\pi}^{\pi} \cos(kx) f(x) , dx$
A3: $a_k = \frac{1}{2 \pi} \int_{-\pi}^{\pi} \cos(kx) f(x) , dx$
A4: $a_k = -\frac{1}{2\pi} \int_{-\pi}^{\pi} \cos(kx) f(x) , dx$
Q2: Consider the trigonometric polynomial $f(x) = a_0 + \sum_{k=1}^n a_k \cos(k x)$. How can the coefficient $a_0$ be calculated?
A1: $a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) , dx$
A2: $a_0 = \frac{1}{\sqrt{2}\pi} \int_{-\pi}^{\pi} f(x) , dx$
A3: $a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) , dx$
A4: $a_0 = \frac{\sqrt{2}}{\pi} \int_{-\pi}^{\pi} f(x) , dx$
Q3: Consider the trigonometric polynomial $f(x) = a_0 + \sum_{k=1}^n a_k \cos(k x)+ \sum_{k=1}^n b_k \sin(k x)$. How can the coefficients $b_k$ be calculated?
A1: $b_k = \frac{1}{\pi} \int_{-\pi}^{\pi} \sin(kx) f(x) , dx$
A2: $b_k = \int_{-\pi}^{\pi} \sin(kx) f(x) , dx$
A3: $b_k = \frac{1}{2 \pi} \int_{-\pi}^{\pi} \sin(kx) f(x) , dx$
A4: $b_k = -\frac{1}{2\pi} \int_{-\pi}^{\pi} \sin(kx) f(x) , dx$
-
Last update: 2024-11
