Small angle approximation for (dθ/dt)^2=0

[V_Permendur]

Hey guys, I am looking for a textbook that I can cite as a source for a project, for which I am doing the math on.

I know that for a 22° approximation sinθ=θ and cosθ=1-\frac{θ^{2}}{2}

but for a 5° approximation sinθ=θ but now cosθ=1


and that's all fine and dandy, but I am looking through a paper on an inverted pendulum on a cart, and after solving their system lagrangian, which I have done, when it came down to linearize the final equations, they were saying that the \dot{\theta}²=0

and i believe this comes from the taylor series expansion. Unfortunately I am terrible at taylor series, and I want to know more about this, and what I really need is a textbook that has this information in there, that I can use as a reference source.

if anyone can name a book that will have this information (and hopefully the page that its on as well) if my university library doesn't have it, then hopefully I can find it in some library and if not, hopefully i can download a pdf of it somewhere.

Here is the paper from where I am getting most of this information. It seems to be the most complete. Scroll down to page 12 to see their approximation.
http://web.mit.edu/2.737/www/extra_files/andrew.pdf

thanks guys.

-Robby

 

[Simon Bridge]

You should find the Taylor Series in any entry-level (college) physics textbook.

I know that for a 22° approximation sinθ=θ and cosθ=θ-θ^2/2

That does not work ...
\theta=22^\circ \text{ is } 11\pi/90\text{rad}=0.38397\text{rad}
\theta -\theta^2/2 = 0.38397
\cos(\theta)=0.92718
... it's actually closer to 1. The approximation works better for \sin(\theta) - but \theta - \frac{1}{6}\theta^3 is better.

For any f(x) where you know its value at some point x=a then the values nearby, that is: x ≈ a, will be close to:
f(a)+\frac{f^\prime}{1!}(x-a)+\frac{f^{\prime\prime}}{2!}(x-a)^2+\frac{f^{\prime\prime\prime}}{3!}(x-a)^3 +\cdots
... the farther you go from f(a) the more terms you need in the approximation.
This series is called "The Taylor Series".

For the par-axial approximation you want to expand the functions around x=a=0
For f(x)=cos(x), f(x=0)=1, f'(x)=-sin(x), f''(x)=-cos(x) so just substitute into the formula for the series.

\cos(\theta) \approx 1 - 0 - \frac{1}{2}\theta^2 - 0 + \cdots and similarly,
\sin(\theta) \approx 0 + \theta - 0 - \frac{1}{6}\theta^3 + \cdots

Notice that the first term in the expansion for each dosn't actually help ... so you use at least the first two terms.
In practice this works for angles quite close to 0, the rule of thumb is "less than 22deg" but that's just a rule of thumb.
If you use the first three terms (say your angles are about 22deg) you get the sort of relationship that puzzled you.

You don't normally need to cite anything for the theory - just state that you are using the Taylor series expansion, and state the "order" of the expansion.

 

[V_Permendur]

You should find the Taylor Series in any entry-level (college) physics textbook.



That does not work ...
\theta=22^\circ \text{ is } 11\pi/90\text{rad}=0.38397\text{rad}
\theta -\theta^2/2 = 0.38397
\cos(\theta)=0.92718
... it's actually closer to 1. The approximation works better for \sin(\theta) - but \theta - \frac{1}{6}\theta^3 is better.

For any f(x) where you know its value at some point x=a then the values nearby, that is: x ≈ a, will be close to:
f(a)+\frac{f^\prime}{1!}(x-a)+\frac{f^{\prime\prime}}{2!}(x-a)^2+\frac{f^{\prime\prime\prime}}{3!}(x-a)^3 +\cdots
... the farther you go from f(a) the more terms you need in the approximation.
This series is called "The Taylor Series".

For the par-axial approximation you want to expand the functions around x=a=0
For f(x)=cos(x), f(x=0)=1, f'(x)=-sin(x), f''(x)=-cos(x) so just substitute into the formula for the series.

\cos(\theta) \approx 1 - 0 - \frac{1}{2}\theta^2 - 0 + \cdots and similarly,
\sin(\theta) \approx 0 + \theta - 0 - \frac{1}{6}\theta^3 + \cdots

Notice that the first term in the expansion for each dosn't actually help ... so you use at least the first two terms.
In practice this works for angles quite close to 0, the rule of thumb is "less than 22deg" but that's just a rule of thumb.
If you use the first three terms (say your angles are about 22deg) you get the sort of relationship that puzzled you.

You don't normally need to cite anything for the theory - just state that you are using the Taylor series expansion, and state the "order" of the expansion.

lol sorry i just changed it in my post! for 22 degree approx what i meant to say was cosθ=1-\frac{θ^{2}}{2}

the information i really need is how the heck they did the taylor series expansion with that derivative. i have my old math book and i can see how they got the taylor series for sine and cosine, but there's nothing in it similar to what they did in that paper. did you see the link?

 

[Simon Bridge]

On p12, eq2.26 is just the 1st order Taylor series expansion about \theta=0.
eq2.28 just applies this definition to the situation of small oscillations about one of the equilibrium points.
I'm afraid you'll have to be more specific about where you get lost.

 

[V_Permendur]

On p12, eq2.26 is just the 1st order Taylor series expansion about \theta=0.
eq2.28 just applies this definition to the situation of small oscillations about one of the equilibrium points.
I'm afraid you'll have to be more specific about where you get lost.

yes i probably should have listed the equation numbers.

im confused on 2.25-2.27

 

[Simon Bridge]

All right, but in what way?

1st step in scientific method is to identify the problem :)

For instance, perhaps you are confused by the way the epsilon vanishes?
2.25 defines \epsilon = \theta-\theta_0 where \theta_0 is some position of interest... in this case: an equilibrium position. Thus epsilon corresponds to a small angle off equilibrium.

To analyse the system you need to be able to know \sin(\epsilon) and so on.

The first condition considered is when \theta_0=0
So, from 2.25, what is epsilon?

... that help?