Math / Calculus

Question $f(x) = \begin{cases} 4 & -\pi \le x < 0 \\ 0 & 0 \le x < \pi \end{cases}$
A. $f(x) = 2 + \frac{8}{3\pi}\sin x + \frac{8}{\pi}\cos 3x + \dots$ B. $f(x) = 2 - \frac{8}{\pi}\cos x - \frac{8}{3\pi}\cos 3x - \dots$ C. $f(x) = 2 - \frac{8}{\pi}\sin x - \frac{8}{3\pi}\sin 3x - \dots$ D. $f(x) = 2 + \frac{8}{\pi}\cos x + \frac{8}{3\pi}\cos 3x + \dots - \frac{8}{\pi}\sin x - \frac{8}{3\pi}\sin 3x - \dots$
Which of the following graphs shows three periods of the function?

Studdy Solution

STEP 1

What is this asking? We need to find the first few terms of the Fourier series for a piecewise function and then pick the graph that shows three periods of this function. Watch out! The function is defined piecewise, so we need to be careful with the integration limits when calculating the Fourier coefficients.
Also, remember that the Fourier series approximates a periodic function, so we need to consider the periodic extension of the given function.

STEP 2

1. Define the function
2. Calculate $a_0$
3. Calculate $a_n$
4. Calculate $b_n$
5. Optimize the formula
6. Sketch the graph

STEP 3

Our function $f(x)$ is defined as: $f(x) = \begin{cases} 4, & -\pi \le x < 0 \\ 0, & 0 \le x < \pi \end{cases}$ This function has a period of $2\pi$ .

STEP 4

The formula for $a_0$ is: $a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx$

STEP 5

Since $f(x)$ is piecewise, we split the integral: $a_0 = \frac{1}{\pi} \left( \int_{-\pi}^{0} 4 \, dx + \int_{0}^{\pi} 0 \, dx \right)$

STEP 6

The second integral is zero, and the first integral is easy: $a_0 = \frac{1}{\pi} \left[ 4x \right]_{-\pi}^{0} = \frac{1}{\pi} (4 \cdot 0 - 4 \cdot (-\pi)) = \frac{4\pi}{\pi} = 4$ So, $a_0 = \textbf{4}$ .

STEP 7

The formula for $a_n$ is: $a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx$

STEP 8

Again, we split the integral: $a_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} 4 \cos(nx) \, dx + \int_{0}^{\pi} 0 \cdot \cos(nx) \, dx \right)$

STEP 9

The second integral is zero.
For the first integral: $a_n = \frac{4}{\pi} \left[ \frac{\sin(nx)}{n} \right]_{-\pi}^{0} = \frac{4}{n\pi} (\sin(0) - \sin(-n\pi)) = \frac{4}{n\pi} (0 - 0) = 0$ So, $a_n = \textbf{0}$ for all $n > 0$ .

STEP 10

The formula for $b_n$ is: $b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx$

STEP 11

Splitting the integral: $b_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} 4 \sin(nx) \, dx + \int_{0}^{\pi} 0 \cdot \sin(nx) \, dx \right)$

STEP 12

The second integral is zero.
For the first integral: $b_n = \frac{4}{\pi} \left[ -\frac{\cos(nx)}{n} \right]_{-\pi}^{0} = \frac{-4}{n\pi} (\cos(0) - \cos(-n\pi)) = \frac{-4}{n\pi} (1 - \cos(n\pi))$ Since $\cos(n\pi) = (-1)^n$ , we have $1 - \cos(n\pi) = 1 - (-1)^n$ .

STEP 13

Putting it all together, the Fourier series is: $f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx))$ $f(x) = \frac{4}{2} + \sum_{n=1}^{\infty} \left(0 \cdot \cos(nx) + \frac{-4(1 - (-1)^n)}{n\pi} \sin(nx) \right) = 2 - \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{1 - (-1)^n}{n} \sin(nx)$

STEP 14

The first three non-zero terms are: $f(x) \approx 2 - \frac{8}{\pi} \sin(x) - \frac{8}{3\pi} \sin(3x) - \dots$

STEP 15

The graph will be a square wave oscillating between 0 and 4, centered at 2, with a period of $2\pi$ .

STEP 16

The Fourier series is approximately $f(x) = 2 - \frac{8}{\pi}\sin(x) - \frac{8}{3\pi}\sin(3x) + \dots$ and the correct graph is C.

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Studdy Solution

STEP 1

STEP 2

1. Define the function
2. Calculate $a_0$
3. Calculate $a_n$
4. Calculate $b_n$
5. Optimize the formula
6. Sketch the graph

STEP 3

Our function $f(x)$ is defined as: $f(x) = \begin{cases} 4, & -\pi \le x < 0 \\ 0, & 0 \le x < \pi \end{cases}$ This function has a period of $2\pi$ .

STEP 4

The formula for $a_0$ is: $a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx$

STEP 5

Since $f(x)$ is piecewise, we split the integral: $a_0 = \frac{1}{\pi} \left( \int_{-\pi}^{0} 4 \, dx + \int_{0}^{\pi} 0 \, dx \right)$

STEP 6

The second integral is zero, and the first integral is easy: $a_0 = \frac{1}{\pi} \left[ 4x \right]_{-\pi}^{0} = \frac{1}{\pi} (4 \cdot 0 - 4 \cdot (-\pi)) = \frac{4\pi}{\pi} = 4$ So, $a_0 = \textbf{4}$ .

STEP 7

The formula for $a_n$ is: $a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx$

STEP 8

Again, we split the integral: $a_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} 4 \cos(nx) \, dx + \int_{0}^{\pi} 0 \cdot \cos(nx) \, dx \right)$

STEP 9

The second integral is zero.
For the first integral: $a_n = \frac{4}{\pi} \left[ \frac{\sin(nx)}{n} \right]_{-\pi}^{0} = \frac{4}{n\pi} (\sin(0) - \sin(-n\pi)) = \frac{4}{n\pi} (0 - 0) = 0$ So, $a_n = \textbf{0}$ for all $n > 0$ .

STEP 10

The formula for $b_n$ is: $b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx$

STEP 11

Splitting the integral: $b_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} 4 \sin(nx) \, dx + \int_{0}^{\pi} 0 \cdot \sin(nx) \, dx \right)$

STEP 12

The second integral is zero.
For the first integral: $b_n = \frac{4}{\pi} \left[ -\frac{\cos(nx)}{n} \right]_{-\pi}^{0} = \frac{-4}{n\pi} (\cos(0) - \cos(-n\pi)) = \frac{-4}{n\pi} (1 - \cos(n\pi))$ Since $\cos(n\pi) = (-1)^n$ , we have $1 - \cos(n\pi) = 1 - (-1)^n$ .

STEP 13

STEP 14

The first three non-zero terms are: $f(x) \approx 2 - \frac{8}{\pi} \sin(x) - \frac{8}{3\pi} \sin(3x) - \dots$

STEP 15

The graph will be a square wave oscillating between 0 and 4, centered at 2, with a period of $2\pi$ .

STEP 16

The Fourier series is approximately $f(x) = 2 - \frac{8}{\pi}\sin(x) - \frac{8}{3\pi}\sin(3x) + \dots$ and the correct graph is C.

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Studdy Solution

STEP 1

STEP 2

1. Define the function2. Calculate a0a_0a0​3. Calculate ana_nan​4. Calculate bnb_nbn​5. Optimize the formula6. Sketch the graph

STEP 3

Our function f(x)f(x)f(x) is defined as: f(x)={4,−π≤x<00,0≤x<π f(x) = \begin{cases} 4, & -\pi \le x < 0 \\ 0, & 0 \le x < \pi \end{cases} f(x)={4,0,​−π≤x<00≤x<π​ This function has a period of 2π2\pi2π.

STEP 4

The formula for a0a_0a0​ is: a0=1π∫−ππf(x) dx a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx a0​=π1​∫−ππ​f(x)dx

STEP 5

Since f(x)f(x)f(x) is piecewise, we split the integral: a0=1π(∫−π04 dx+∫0π0 dx) a_0 = \frac{1}{\pi} \left( \int_{-\pi}^{0} 4 \, dx + \int_{0}^{\pi} 0 \, dx \right) a0​=π1​(∫−π0​4dx+∫0π​0dx)

STEP 6

STEP 7

The formula for ana_nan​ is: an=1π∫−ππf(x)cos⁡(nx) dx a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx an​=π1​∫−ππ​f(x)cos(nx)dx

STEP 8

Again, we split the integral: an=1π(∫−π04cos⁡(nx) dx+∫0π0⋅cos⁡(nx) dx) a_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} 4 \cos(nx) \, dx + \int_{0}^{\pi} 0 \cdot \cos(nx) \, dx \right) an​=π1​(∫−π0​4cos(nx)dx+∫0π​0⋅cos(nx)dx)

STEP 9

STEP 10

The formula for bnb_nbn​ is: bn=1π∫−ππf(x)sin⁡(nx) dx b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx bn​=π1​∫−ππ​f(x)sin(nx)dx

STEP 11

Splitting the integral: bn=1π(∫−π04sin⁡(nx) dx+∫0π0⋅sin⁡(nx) dx) b_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} 4 \sin(nx) \, dx + \int_{0}^{\pi} 0 \cdot \sin(nx) \, dx \right) bn​=π1​(∫−π0​4sin(nx)dx+∫0π​0⋅sin(nx)dx)

STEP 12

STEP 13

STEP 14

The first three non-zero terms are: f(x)≈2−8πsin⁡(x)−83πsin⁡(3x)−… f(x) \approx 2 - \frac{8}{\pi} \sin(x) - \frac{8}{3\pi} \sin(3x) - \dots f(x)≈2−π8​sin(x)−3π8​sin(3x)−…

STEP 15

The graph will be a square wave oscillating between 0 and 4, centered at 2, with a period of 2π2\pi2π.

STEP 16

The Fourier series is approximately f(x)=2−8πsin⁡(x)−83πsin⁡(3x)+…f(x) = 2 - \frac{8}{\pi}\sin(x) - \frac{8}{3\pi}\sin(3x) + \dotsf(x)=2−π8​sin(x)−3π8​sin(3x)+… and the correct graph is **C**.

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1. Define the function
2. Calculate $a_0$
3. Calculate $a_n$
4. Calculate $b_n$
5. Optimize the formula
6. Sketch the graph

Our function $f(x)$ is defined as: $f(x) = \begin{cases} 4, & -\pi \le x < 0 \\ 0, & 0 \le x < \pi \end{cases}$ This function has a period of $2\pi$ .

The formula for $a_0$ is: $a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx$

Since $f(x)$ is piecewise, we split the integral: $a_0 = \frac{1}{\pi} \left( \int_{-\pi}^{0} 4 \, dx + \int_{0}^{\pi} 0 \, dx \right)$

The formula for $a_n$ is: $a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx$

Again, we split the integral: $a_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} 4 \cos(nx) \, dx + \int_{0}^{\pi} 0 \cdot \cos(nx) \, dx \right)$

The formula for $b_n$ is: $b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx$

Splitting the integral: $b_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} 4 \sin(nx) \, dx + \int_{0}^{\pi} 0 \cdot \sin(nx) \, dx \right)$

The first three non-zero terms are: $f(x) \approx 2 - \frac{8}{\pi} \sin(x) - \frac{8}{3\pi} \sin(3x) - \dots$

The graph will be a square wave oscillating between 0 and 4, centered at 2, with a period of $2\pi$ .

The Fourier series is approximately $f(x) = 2 - \frac{8}{\pi}\sin(x) - \frac{8}{3\pi}\sin(3x) + \dots$ and the correct graph is C.