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Thus, for Find all the "b" according to the graph Solve: $f(x)=\left\{\begin{a...

Aug 1, 2024

Solution by Steps

step 1

We need to find the Fourier coefficients

b_n

for the given piecewise function

f(x)

. The function is defined as:

f(x) = \begin{cases} 1 &amp; \text{if } x &gt; 0 \\ -1 &amp; \text{if } x &lt; 0 \end{cases}

The Fourier coefficient

b_n

is given by:

b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx

step 2

We split the integral into two parts:

b_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} (-1) \sin(nx) \, dx + \int_{0}^{\pi} 1 \cdot \sin(nx) \, dx \right)

step 3

Evaluate the first integral:

\int_{-\pi}^{0} (-\sin(nx)) \, dx = -\left[ \frac{-\cos(nx)}{n} \right]_{-\pi}^{0} = \frac{\cos(0) - \cos(-n\pi)}{n} = \frac{1 - \cos(n\pi)}{n}

step 4

Evaluate the second integral:

\int_{0}^{\pi} \sin(nx) \, dx = \left[ -\frac{\cos(nx)}{n} \right]_{0}^{\pi} = \frac{\cos(0) - \cos(n\pi)}{n} = \frac{1 - \cos(n\pi)}{n}

step 5

Combine the results:

b_n = \frac{1}{\pi} \left( \frac{1 - \cos(n\pi)}{n} + \frac{1 - \cos(n\pi)}{n} \right) = \frac{2}{\pi} \cdot \frac{1 - \cos(n\pi)}{n}

step 6

Simplify the expression:

b_n = \frac{2(1 - \cos(n\pi))}{n\pi}

Since

\cos(n\pi)

(-1)^n

, we have:

b_n = \frac{2(1 - (-1)^n)}{n\pi}

Thus,

b_n

is:

b_n = \begin{cases} \frac{4}{n\pi} &amp; \text{if } n \text{ is odd} \\ 0 &amp; \text{if } n \text{ is even} \end{cases}

Answer

b_n = \frac{4}{n\pi}

for odd

n

, and

b_n = 0

for even

n

Key Concept

Fourier Coefficients for Piecewise Functions

Explanation

The Fourier coefficients

b_n

for a piecewise function can be found by integrating the product of the function and

\sin(nx)

over the given interval, and then simplifying the result.