# What is the mathematically correct way to write this summation?

1. Feb 18, 2012

### 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:
$$\displaystyle\int^{\theta_{2\pi}}_{\theta_{0}} \tau d\theta\approx(\theta_{2\pi}-\theta_{0})\frac{f(\theta_{0})+f(\theta_{2\pi})}{2}$$
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.?

2. Feb 18, 2012

### tiny-tim

Hi Kalus!
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(θi) notation is weird … it should be something like f(2πi/180) )

3. Feb 18, 2012

### 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 θi and θi+1 or θi and θi + 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.

4. Feb 18, 2012

### Kalus

Thank you for the replies.

Is this better now?

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

5. Feb 18, 2012

### 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?

$$\displaystyle\int^{\theta_{2\pi}}_{\theta_{0}} \tau d\theta\approx(\theta_{2\pi}-\theta_{0})\frac{f(\theta_{0})+f(\theta_{2\pi})}{2}$$

6. Feb 19, 2012

### 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

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

7. Feb 19, 2012

### 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?

8. Feb 19, 2012

### 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 θo to θ)

9. Feb 19, 2012

### Kalus

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

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