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I am computing the Fourier series expansion of the given signal x(t). In that I am having difficulties in calculating range of function $ \frac{1}{n\pi}(1-\cos(n\pi))\sin(\frac{n\pi}{2})$ (or) I am not able to get how can I arrive at equation (2) from equation (1)

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I have done some back calculation(and would like to know the forward way of calculating it)

For example, for the function to have a value of $ \frac{2n}{\pi} $

$$ \sin(\frac{n\pi}{2})=1 \,\, \text{and} \,\, \cos(n\pi)=-1$$

$$\implies \sin(\frac{n\pi}{2})=\sin((4k+1)\frac{\pi}{2}) \, and \, \cos(n\pi)=\cos(2k+1)\pi$$ $$\implies n=(4k+1) \, \text{and} \, n =(2k+1).$$

Which is a contradiction and doesn't match with the book.(Is this the case that among $\implies n=(4k+1) \, \text{and} \, n =(2k+1)$ ,we have to choose $n=(4k+1)$?)

And how have we arrived at 2nd part of equation (2)?

Please help in this regard.

Also clarify another doubt. How have we arrived at equation (3) from (2). It is confusing because it involves determination of the terms based on n which in turn is dependent on k.

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    HInT:take $x=\dfrac{n\pi}{2}$ so $$\frac{1}{n\pi}(1-cos(n\pi))sin(\frac{n\pi}{2})=\\\dfrac1{2a}(1-\cos 2a)\sin a =\\ \dfrac1{2a}(2\sin^2 a)\sin a =\\ \dfrac1{a}\sin^3 a $$ – 2017-02-19
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    Let $n=4k$ in (1), what do you get? – 2017-02-19
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    @MyGlasses : $\frac{1}{4k\pi}(1-cos(2k\pi)sin(\frac{4k\pi}{2})=0$ – 2017-02-19
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    For $n=4k+1$ what you get? – 2017-02-19
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    @MyGlasses: $$=\frac{1}{n\pi}(1-cos(4k+1)\pi)sin((4k+1)\frac{\pi}{2})$$ $$=\frac{1}{n\pi}(1-cos(4k\pi+\pi))sin((\frac{4k\pi}{2}+\frac{\pi}{2}))$$ $$=\frac{1}{n\pi}(1+cos(4k\pi))cos(\frac{4k\pi}{2})$$ $$=\frac{1}{n\pi}(2)(1)=\frac{2}{n\pi}$$ – 2017-02-19
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    Very good, for $n=4k+2$ and $n=4k+3$ do like that. got it.? – 2017-02-19
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    But sir I can do this if we know the answer i.e by back calculation. What is the forward way of calculating it? – 2017-02-19
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    No. This is forward way. we assume $n$ with four values $4k$, $4k+1$, , $4k+2$, $4k+3$ and decide what occur in these cases. – 2017-02-19
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    Sir why do we assume 4k, why not 2k or any other number? – 2017-02-19
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    Since periods of $\cos$ and $\sin$ is $4k$ here. – 2017-02-19
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    @MyGlasses Sir, fundamental period of $cos(n\pi)$ is 2 and fundamental period of $sin(\frac{n\pi}{2})$ is 4. Are we choosing 4k because it is the greater of the two? – 2017-02-19
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    You are right, I wanted to show that values of $\sin\dfrac{n\pi}{2}$ and $\cos n\pi$ change in a **step** and it is $4$. It is for $\sin$ function: $\dfrac{n\pi}{2}=2k\pi\implies n=4k$. but you said better. – 2017-02-19
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    @MyGlasses Sir, in this function there is a multiplication between sine and cos.Is this the reason we have considered the greater of the two fundamental periods? If it had been addition of the two functions, the fundamental period would have been the LCM(Least Common Multiple) of the individual periods i.e LCM(2,4)=4, which happens to be the same as the fundamental period in case of multiplication. But if the function had been $sin(\frac{2\pi n}{3})+cos(\frac{2\pi n}{5})$ ,then the fundamental period would have been LCM(3,5)=15. In that case should have we considered 15k? – 2017-02-19
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    First question is right. The second is right also. – 2017-02-19
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    @MyGlasses Sir why is it that in case of multiplication of signals,the fundamental period is the greater of the individual fundamental periods? – 2017-02-19
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    Could you please, give an example or a reference of this case to me to understand you question. – 2017-02-19

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