What is the mathematically correct way to write this summation?

[Kalus]

How can I make this mathematically correct? I hope you see what I'm trying to do?...

If you have a graph where:
W=\displaystyle\int^{\theta_{2\pi}}_{\theta_{0}} \tau d\theta
Then the estimated area with the trapesium rule:
<br /> \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}
Since one cycle is split into 360 uniformly spaced points, or 359 segments, with 1 degree spacing, this gives a step size of h=\frac{2\pi}{360}=\frac{\pi}{180}.
Then the integration is:
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}))

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.?

 

[tiny-tim]

Hi Kalus!

... 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.?

Yes, the ∑ always goes up in values of one, eg \sum\limits_{i=1}^n, you can't put "2π" on top.

(btw, your f(θ_i) notation is weird … it should be something like f(2πi/180) )

 

[mathman]

The upper limit of i should be 359, since you have divided the domain into 359 intervals, and you started with i=0. It is not a matter of degrees or radians. Also the function points are θ_i and θ_i+1 or θ_i and θ_i + h.

Your sub-scripting for θ is also messy. For the sum you are using index subscripts while for the integral you are sub-scripting by angle values.

 

[Kalus]

Thank you for the replies.

Is this better now?

W=\displaystyle\int^{\theta_{2\pi}}_{\theta_{0}} \tau d\theta\approx \frac{h}{2}\displaystyle\sum\limits_{k=0}^{359} (f(\frac{k\pi}{180})+f(\frac{(k+1)\pi}{180}))

 

[Kalus]

or would this be more appropriate?

W=\displaystyle\int^{\theta_{2\pi}}_{\theta_{0}} \tau d\theta\approx \frac{\pi}{360}\displaystyle\sum\limits_{\theta=0}^{359} (f(\theta)+f(\theta+1))

How could I adjust this to follow the same formatting as above?

\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}

 

[tiny-tim]

Hi Kalus!

(just got up :zzz: …)

I'm not sure what the relation is between your τ and your f

I would expect to see something like

W=\displaystyle\int^{2\pi}_0 f(\theta) d\theta\approx \frac{1}{360}\displaystyle\sum\limits_{k=0}^{359} (f(k\pi/180)+f((k+1)\pi/180))

 

[Kalus]

Hi tiny-tim,

τ, the torque is a function of θ. I guess I should avoid using f at all?

I've changed it to this now:

The trapezoidal rule, which says:
\displaystyle\int^{b}_{a}f(x) dx\approx(b-a)\frac{f(a)+f(b)}{2}
Applied to the work integration:
\displaystyle\int^{\theta_{2\pi}}_{\theta_{0}} \tau d\theta\approx(2\pi)\frac{\tau(0)+\tau(2\pi)}{2}
Since one cycle is split into 360 uniformly spaced points, or 359 segments, with $1^{\circ}$ spacing, this gives a step size of h=\frac{2\pi}{360}=\frac{\pi}{180}
The integration can then be found with:

W=\displaystyle\int^{\theta_{2\pi}}_{\theta_{0}} \tau d\theta\approx \frac{\pi}{360}\displaystyle\sum\limits_{\theta=0}^{359} (\tau(\theta)+\tau(\theta+1))

Since the graph is in graph and τ is calculated in degrees it would be better for the summation to be in degrees. I have expressed the 1 degree step size in radians (pi/180), which becomes (pi/360) at the start when divided through by 2 following the formula.

Is this the best solution?

 

[tiny-tim]

Yes, that's fine.

(except your \displaystyle\int^{\theta_{2\pi}}_{\theta_{0}} d\theta should be \displaystyle\int^{2\pi}_0 d\theta …

θ goes from 0 to 2π, not from θ_o to θ_2π)

 

[Kalus]

Thanks :) I've adjusted that and it looks great now.