Is Calculating Only the Fourier Sine Series Sufficient for f(x) = x^3 on [-1,1]?

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Calculating only the Fourier sine series for the function f(x) = x^3 on the interval [-1, 1] is sufficient due to the odd nature of the function, which results in all cosine coefficients being zero. Since x^3 is odd, one can directly compute the sine series without needing to calculate the a0, an, or bn coefficients associated with the even terms. The discussion confirms that using the sine series provides a complete representation of the function. An explicit verification of the cosine terms shows they indeed evaluate to zero, reinforcing the validity of this approach. Thus, focusing solely on the Fourier sine series is an effective method for this function.
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Homework Statement



Determine a general Fourier series representation for f(x) = x^3 -1<x<1

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The Attempt at a Solution



May seem like a stupid Q, but would i have to calculate a0, an, bn or since i know that x^3 is an odd function, could jump straight into calculating the Fourier sine series for odd functions. Would that give me a general representation?
 
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looking at it one can directly jump to Fourier sine series
if you calculate all the terms initially assuming the complete Fourier representation you will find that coefficients associated with the even terms will go to zero.
so its more like using a known result.
 
Yes, you can restrict yourself to sines.

If you want you can explicitly check it, with an argument along the lines of
\int_{-1}^1 \cos(x) x^3 dx = \int_{-1}^0 \cos(x) x^3 dx + \int_0^1 \cos(x) x^3 dx = \int_0^1 \cos(-x) (-x)^3 dx + \int_0^1 \cos(x) x^3 dx = 0
because cos(-x) (-x)3 = - cos(x) x3
 
thanks.
 
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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