Series to represent alternate between 1 and -1

foo9008 · May 8, 2016

1. The problem statement, all variables and given/known data
we know that a2 , a4 ,a6 (even number ) = 0 , but when a1 , a3 , a5 (odd numbers) , the answer of an alternate between positive and negative .... in the second circle , the author represent it with (-1)^(n+1) , i dont think this is correct , this is because when n=3 , an = -2/ 3pi when n=3 , [ (-1) ^(3+1 ) ] = positive 1 , not negative 1 !!! can someone explain on this ?

2. Relevant equations

3. The attempt at a solution

stevendaryl · May 8, 2016

I think that the author redefined [itex]n[/itex]. To make it clear, you start off assuming that

[itex]f(x) = a_0 + \sum_n a_n cos(nx)[/itex]

Then, for this particular problem, you find that for [itex]n > 0[/itex], then [itex]a_n = 0[/itex] unless [itex]n[/itex] is odd. If [itex]n[/itex] is odd, then that means that [itex]n[/itex] can be written as:

[itex]n = 2n'-1[/itex]

where [itex]n' = 1, 2, 3, ...[/itex]

So the term [itex]-\frac{2}{\pi} \frac{cos(3x)}{1}[/itex] corresponds to [itex]n=3[/itex], but it corresponds to [itex]n' = 2[/itex]. In terms of [itex]n'[/itex], the general term is

[itex]\frac{2}{\pi} \frac{cos((2n'-1)x)}{2n'-1} (-1)^{n'+1}[/itex]

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Series to represent alternate between 1 and -1

foo9008

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