- #1
Kalus
- 37
- 0
How can I make this mathematically correct? I hope you see what I'm trying to do?...
If you have a graph where:
[tex]W=\displaystyle\int^{\theta_{2\pi}}_{\theta_{0}} \tau d\theta[/tex]
Then the estimated area with the trapesium rule:
[tex]
\displaystyle\int^{\theta_{2\pi}}_{\theta_{0}} \tau d\theta\approx(\theta_{2\pi}-\theta_{0})\frac{f(\theta_{0})+f(\theta_{2\pi})}{2}[/tex]
Since one cycle is split into 360 uniformly spaced points, or 359 segments, with 1 degree spacing, this gives a step size of [tex]h=\frac{2\pi}{360}=\frac{\pi}{180}[/tex].
Then the integration is:
[tex]W=\displaystyle\int^{\theta_{2\pi}}_{\theta_{0}} \tau d\theta\approx \frac{h}{2}\displaystyle\sum\limits_{i=0}^{2\pi} (f(\theta_i)+f(\theta_{i+h}))[/tex]
Now, I seem to be mixing degrees and radians... when you use the summation symbol, is it always assumed that the summation goes up in values of one, i.e, if you have summation between k=1 and k=10 that this will go up 1,2,3,4 etc.?
If you have a graph where:
[tex]W=\displaystyle\int^{\theta_{2\pi}}_{\theta_{0}} \tau d\theta[/tex]
Then the estimated area with the trapesium rule:
[tex]
\displaystyle\int^{\theta_{2\pi}}_{\theta_{0}} \tau d\theta\approx(\theta_{2\pi}-\theta_{0})\frac{f(\theta_{0})+f(\theta_{2\pi})}{2}[/tex]
Since one cycle is split into 360 uniformly spaced points, or 359 segments, with 1 degree spacing, this gives a step size of [tex]h=\frac{2\pi}{360}=\frac{\pi}{180}[/tex].
Then the integration is:
[tex]W=\displaystyle\int^{\theta_{2\pi}}_{\theta_{0}} \tau d\theta\approx \frac{h}{2}\displaystyle\sum\limits_{i=0}^{2\pi} (f(\theta_i)+f(\theta_{i+h}))[/tex]
Now, I seem to be mixing degrees and radians... when you use the summation symbol, is it always assumed that the summation goes up in values of one, i.e, if you have summation between k=1 and k=10 that this will go up 1,2,3,4 etc.?