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Problem 398

# Consider the infinite trigonometric series $\left(a_{0} / 2\right)+{ }^{\infty} \Sigma_{n=1}\left(a_{n} \cos n x+b_{n} \sin n x\right)$ and assume that it converges uniformly for all $$\mathrm{x} \in(-\pi, \pi)$$. It can then be considered as a function $$f$$ of $$x$$ with period $$2 \pi$$, i.e. $\mathrm{f}(\mathrm{x})=\left(\mathrm{a}_{0} / 2\right)+{ }^{\infty} \sum_{\mathrm{n}=1} \mathrm{a}_{\mathrm{n}} \cos \mathrm{n} \mathrm{x}+\mathrm{b}_{\mathrm{n}} \sin \mathrm{n} \mathrm{x}$ Determine the values of $$a_{n}, b_{n}$$ in terms of $$f(x)$$

Expert verified
In summary, the values of $$a_n$$ and $$b_n$$ in terms of $$f(x)$$ can be determined using the following formulas: $$a_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x) dx$$ $$a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x) \cos(nx) dx$$ $$b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x) \sin(nx) dx$$ These expressions can be used to find the Fourier series coefficients for any given periodic function $$f(x)$$ with period $$2\pi$$.
See the step by step solution

## Step 1: Identify Fourier series coefficients formula

First of all, let's note down the Fourier series coefficients formula for $$a_n$$ and $$b_n$$: $$a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x) \cos(nx) dx$$ $$b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x) \sin(nx) dx$$ This is what we'll use to find the values of $$a_n$$ and $$b_n$$ for the given function $$f(x)$$.

## Step 2: Compute the coefficient a_0

We need to compute $$a_0$$ separately because the formula for it is slightly different. The formula for $$a_0$$ is: $$a_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x) dx$$ Now, we have the formula for $$a_0$$. Let's move forward and find the values of $$a_n$$ and $$b_n$$ in terms of $$f(x)$$.

## Step 3: Apply the formulas for a_n and b_n

Using the formulas from step 1, let's find the general expressions for coefficients $$a_n$$ and $$b_n$$. It's important to notice that we are not computing a specific value, but we are looking for expressions in terms of $$f(x)$$: $$a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x) \cos(nx) dx$$ $$b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x) \sin(nx) dx$$ We have obtained expressions for $$a_n$$ and $$b_n$$ in terms of $$f(x)$$. With these expressions in place, you can substitute any given function $$f(x)$$ to determine its Fourier series coefficients $$a_n$$ and $$b_n$$.

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