Suppose that \(f(x) = \sum_{n=-\infty}^{\infty} \frac{e^{5ni}}{1 + n^2} e^{nix}\), \(x \in [0, 2\pi]\). We can write \(f(x)\) in the form \(f(x) = \sum_{n=1}^{\infty} (A_n \sin(nx)... Suppose that \(f(x) = \sum_{n=-\infty}^{\infty} \frac{e^{5ni}}{1 + n^2} e^{nix}\), \(x \in [0, 2\pi]\). We can write \(f(x)\) in the form \(f(x) = \sum_{n=1}^{\infty} (A_n \sin(nx) + B_n \cos(nx)) + B_0\). Find \(A_n\), \(B_n\) (with \(n \neq 0\)), and \(B_0\).
Understand the Problem
The question asks us to rewrite a function (f(x)) given as a complex Fourier series into its real trigonometric Fourier series representation. This involves finding the coefficients (A_n), (B_n), and (B_0) in terms of the original complex Fourier coefficients.
Answer
$A_n = \frac{-2\sin(5n)}{1+ n^2}$ $B_n = \frac{2\cos(5n)}{1+ n^2}$ $B_0 = 1$
Answer for screen readers
$A_n = \frac{-2\sin(5n)}{1+ n^2}$
$B_n = \frac{2\cos(5n)}{1+ n^2}$
$B_0 = 1$
Steps to Solve
- Analyze the given complex Fourier series
We are given the complex Fourier series (f(x) = \sum_{-\infty}^{\infty} c_n e^{inx}), where (c_n = \frac{e^{5ni}}{1+ n^2}). We need to find the real Fourier series representation in terms of sine and cosine functions.
- Relate complex Fourier coefficients to real Fourier coefficients
The complex Fourier series can be converted to a real Fourier series using the following relationships: $A_n = i(c_n - c_{-n})$ $B_n = c_n + c_{-n}$ $B_0 = c_0$
- Calculate $A_n$
Using the formula $A_n = i(c_n - c_{-n})$, we substitute $c_n = \frac{e^{5ni}}{1+ n^2}$: $A_n = i\left(\frac{e^{5ni}}{1+ n^2} - \frac{e^{-5ni}}{1+ (-n)^2}\right) = i\left(\frac{e^{5ni} - e^{-5ni}}{1+ n^2}\right)$.
Recall Euler's formula: $e^{ix} = \cos(x) + i\sin(x)$. Therefore, $e^{5ni} = \cos(5n) + i\sin(5n)$ and $e^{-5ni} = \cos(-5n) + i\sin(-5n) = \cos(5n) - i\sin(5n)$. $A_n = i\left(\frac{\cos(5n) + i\sin(5n) - (\cos(5n) - i\sin(5n))}{1+ n^2}\right) = i\left(\frac{2i\sin(5n)}{1+ n^2}\right) = \frac{-2\sin(5n)}{1+ n^2}$.
- Calculate $B_n$
Using the formula $B_n = c_n + c_{-n}$, we substitute $c_n = \frac{e^{5ni}}{1+ n^2}$: $B_n = \frac{e^{5ni}}{1+ n^2} + \frac{e^{-5ni}}{1+ (-n)^2} = \frac{e^{5ni} + e^{-5ni}}{1+ n^2}$.
Using Euler's formula again: $B_n = \frac{\cos(5n) + i\sin(5n) + \cos(5n) - i\sin(5n)}{1+ n^2} = \frac{2\cos(5n)}{1+ n^2}$.
- Calculate $B_0$
Since $B_0 = c_0$, we substitute $n=0$ into $c_n = \frac{e^{5ni}}{1+ n^2}$: $B_0 = c_0 = \frac{e^{5(0)i}}{1+ 0^2} = \frac{e^0}{1} = \frac{1}{1} = 1$.
$A_n = \frac{-2\sin(5n)}{1+ n^2}$
$B_n = \frac{2\cos(5n)}{1+ n^2}$
$B_0 = 1$
More Information
The real Fourier coefficients are derived from the complex Fourier coefficients using Euler's formula to extract the real and imaginary parts, which correspond to the cosine and sine terms, respectively.
Tips
A common mistake is to forget to use Euler's formula or to incorrectly apply it when converting the complex exponentials into sine and cosine functions. Also, there might be sign errors when calculating $A_n$ and $B_n$. For $B_0$, forgetting to substitute n=0 into $c_n$ is another common mistake.
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