What formula should be used to find the Fourier series of an even function?

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To find the Fourier series of the even extension of the function f(x) = x defined on (0, π), it is correctly extended to f(x) = |x| on the interval (-π, π). The Fourier series for even functions only includes cosine terms, meaning the series consists of a_0 and a_n coefficients, while b_n terms are zero. The formulas to use are a_0 = (1/L) ∫ from -L to L of f(x) dx and a_n = (1/L) ∫ from -L to L of f(x) cos(nπx/L) dx. Clarification on the value of L is needed, but once established, the calculations can proceed.
J6204
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Homework Statement


In the following problem I am trying to extend the function $$f(x) = x $$ defined on the interval $$(0,\pi)$$ into the interval $$(-\pi,0)$$ as a even function. Then I need to find the Fourier series of this function.

Homework Equations

The Attempt at a Solution


So I believe I have extending the function onto the interval $(-\pi,0)$ correctly below.
$$f(x) = -x, (-\pi,0)$$
I am having a little trouble understanding the question. I believe I need to find the Fourier series of this function which is now -x. Since the function is now an even function the Fourier series should just consist of the terms a_0 and a_n since b_n has sin attached to it making it a odd function and therefore making it 0
However I am a little confused what formula now to use,
should I use the following formulas? and if so what is L? once I get these matters figured out I can proceed myself with the calculations, thanks!
$$f(x) = \frac{a_0}{2} + \sum^{\infty}_{n=0} a_n cos(nx)$$
$$a_0 = \frac{2}{L} \int^L_0f(x)dx$$
$$a_n = \frac{2}{L}\int^L_0f(x)cos(\frac{n\pi x}{L})dx$$​
 
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J6204 said:

Homework Statement


In the following problem I am trying to extend the function $$f(x) = x $$ defined on the interval $$(0,\pi)$$ into the interval $$(-\pi,0)$$ as a even function. Then I need to find the Fourier series of this function.

Homework Equations

The Attempt at a Solution


So I believe I have extending the function onto the interval $(-\pi,0)$ correctly below.
$$f(x) = -x, (-\pi,0)$$
I am having a little trouble understanding the question. I believe I need to find the Fourier series of this function which is now -x. Since the function is now an even function the Fourier series should just consist of the terms a_0 and a_n since b_n has sin attached to it making it a odd function and therefore making it 0
However I am a little confused what formula now to use,
should I use the following formulas? and if so what is L? once I get these matters figured out I can proceed myself with the calculations, thanks!
$$f(x) = \frac{a_0}{2} + \sum^{\infty}_{n=0} a_n cos(nx)$$
$$a_0 = \frac{2}{L} \int^L_0f(x)dx$$
$$a_n = \frac{2}{L}\int^L_0f(x)cos(\frac{n\pi x}{L})dx$$​
The function ##f(x) = x##, for ##x \in (0,\pi)##, extends as an even function to ##(-\pi,0)## just by using $$f(x) = |x|, \; x \in (-\pi,\pi).$$

The general formula is
$$a_n = \frac{1}{L} \int_{-L}^L f(x) \cos \left( \frac{n \pi x}{L} \right) \, dx,$$
but that becomes what you wrote above when ##f(x)## is an even function.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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