Double Check My Work: Simplifying a Homework Statement

[zzmanzz]

Homework Statement

I got the problem down to:

$$\frac{2}{\pi} \left[ \int_{0}^{\pi/2} \frac{2}{\pi}xsin(nx) dx + \int_{\frac{\pi}{2}}^{\pi} (\frac{-2}{\pi}x+2)sin(nx) dx \right]$$

$$\frac{4}{\pi^2} \left[ \int_{0}^{\pi/2} xsin(nx) dx + \int_{pi/2}^{\pi} -xsin(nx) dx + \int_{pi/2}^{\pi} \pi sin(nx) dx \right]$$

$$\frac{4}{\pi^2} \left[ \left[ \frac{-x}{2n} cos(nx) +\frac{1}{n^2}sin(nx) \right]_{0}^{pi/2} + \left[ \frac{1}{n}xcos(nx)-\frac{1}{n^2}sin(nx) \right]_{\pi/2}^{\pi} + \left[\frac{-\pi}{n}cos(nx) \right]_{\pi/2}^{\pi} \right]$$

$$\frac{4}{\pi^2} \left[\left[\left[\frac{-\pi}{2n}cos(n\frac{\pi}{2}) +\frac{1}{n^2}sin(n\frac{\pi}{2}) \right] -\left[0\right]\right] + \left[ \left[\frac{\pi}{n}cos(n\pi)) - \frac{1}{n^2}sin(n\pi) \right] - \left[\frac{\pi}{2n}cos(n\frac{\pi}{2}) +\frac{1}{n^2}sin(n\frac{\pi}{2}) \right] \right] + \left[\frac{-\pi}{n}cos(n\frac{\pi}{2})+\frac{\pi}{n}cos(n\pi) \right] \right]$$

$$\frac{4}{\pi^2} \left[-\frac{\pi}{2n}cos(n\frac{\pi}{2}) +\frac{1}{n^2}sin(n\frac{\pi}{2}) + \frac{\pi}{n}cos(n\pi)) - \frac{1}{n^2}sin(n\pi) - \frac{\pi}{2n}cos(n\frac{\pi}{2}) -\frac{1}{n^2}sin(n\frac{\pi}{2}) - \frac{\pi}{n}cos(n\frac{\pi}{2})+\frac{\pi}{n}cos(n\pi) \right]$$

$$\frac{4}{\pi^2} \left[-\frac{\pi}{4n}cos(n\frac{\pi}{2}) + \frac{\pi}{n}cos(n\pi)) - \frac{1}{n^2}sin(n\pi) - \frac{\pi}{n}cos(n\frac{\pi}{2})+\frac{\pi}{n}cos(n\pi) \right]$$

I feel like i sccrewed something up to this point ... maybe turning some positive terms negative or vice versa? can someone just double check my work upto this point pleaseee.

Many thanks.

 

[Ray Vickson]

Homework Statement

I got the problem down to:

$$\frac{2}{\pi} \left[ \int_{0}^{\pi/2} \frac{2}{\pi}xsin(nx) dx + \int_{\frac{\pi}{2}}^{\pi} (\frac{-2}{\pi}x+2)sin(nx) dx \right]$$

$$\frac{4}{\pi^2} \left[ \int_{0}^{\pi/2} xsin(nx) dx + \int_{pi/2}^{\pi} -xsin(nx) dx + \int_{pi/2}^{\pi} \pi sin(nx) dx \right]$$

$$\frac{4}{\pi^2} \left[ \left[ \frac{-x}{2n} cos(nx) +\frac{1}{n^2}sin(nx) \right]_{0}^{pi/2} + \left[ \frac{1}{n}xcos(nx)-\frac{1}{n^2}sin(nx) \right]_{\pi/2}^{\pi} + \left[\frac{-\pi}{n}cos(nx) \right]_{\pi/2}^{\pi} \right]$$

$$\frac{4}{\pi^2} \left[\left[\left[\frac{-\pi}{2n}cos(n\frac{\pi}{2}) +\frac{1}{n^2}sin(n\frac{\pi}{2}) \right] -\left[0\right]\right] + \left[ \left[\frac{\pi}{n}cos(n\pi)) - \frac{1}{n^2}sin(n\pi) \right] - \left[\frac{\pi}{2n}cos(n\frac{\pi}{2}) +\frac{1}{n^2}sin(n\frac{\pi}{2}) \right] \right] + \left[\frac{-\pi}{n}cos(n\frac{\pi}{2})+\frac{\pi}{n}cos(n\pi) \right] \right]$$

$$\frac{4}{\pi^2} \left[-\frac{\pi}{2n}cos(n\frac{\pi}{2}) +\frac{1}{n^2}sin(n\frac{\pi}{2}) + \frac{\pi}{n}cos(n\pi)) - \frac{1}{n^2}sin(n\pi) - \frac{\pi}{2n}cos(n\frac{\pi}{2}) -\frac{1}{n^2}sin(n\frac{\pi}{2}) - \frac{\pi}{n}cos(n\frac{\pi}{2})+\frac{\pi}{n}cos(n\pi) \right]$$

$$\frac{4}{\pi^2} \left[-\frac{\pi}{4n}cos(n\frac{\pi}{2}) + \frac{\pi}{n}cos(n\pi)) - \frac{1}{n^2}sin(n\pi) - \frac{\pi}{n}cos(n\frac{\pi}{2})+\frac{\pi}{n}cos(n\pi) \right]$$

I feel like i sccrewed something up to this point ... maybe turning some positive terms negative or vice versa? can someone just double check my work upto this point pleaseee.

Many thanks.

Don't you remember the values for $\sin(n \pi)$ and $\cos(n \pi)$ for integer $n$? What about the values of $\cos(n \pi/2)$ for even $n$ and for odd $n$?

Also, when using LaTeX/TeX, use "\sin' and "\cos' instead of 'sin' and 'cos', because that will look much better and be much, much easier to read: you will get $\sin x, \: \cos x$ instead of $sin x, \; cos x$.

 

[Charles Link]

Just a quick look-over=in your integration by parts, first term, 3rd line, I don't get a "2" in the denominator.

 

[zzmanzz]

Just a quick look-over=in your integration by parts, first term, 3rd line, I don't get a "2" in the denominator.

you are right. thank you.. i was copying from my notes and might have caried that from the next step. good catch

 

[zzmanzz]

Don't you remember the values for $\sin(n \pi)$ and $\cos(n \pi)$ for integer $n$? What about the values of $\cos(n \pi/2)$ for even $n$ and for odd $n$?

Also, when using LaTeX/TeX, use "\sin' and "\cos' instead of 'sin' and 'cos', because that will look much better and be much, much easier to read: you will get $\sin x, \: \cos x$ instead of $sin x, \; cos x$.

Thanks for the reply.

So, looking back:$$\cos(n\pi) = (-1)^n$$

$$\sin(n\pi) = 0$$

$$\sin(\frac{n2}{\pi}) = (-1)^{((n-1)/2)} for n odd, 0 for even$$ the sin cancels out though and doesn't matter?

$$\cos(\frac{2n}{\pi}) = \frac{(1+(-1)^n)}{2*(-1)^{n/2}}$$