Question

I'd like to request an explanation on the algebra in Unit 3 2A:6 (http://goo.gl/u3Ru4).
Though this question has already been taken up here (http://openstudy.com/study#/updates/4e52e8770b8b958804b69c23) I can't seem to duplicate the algebraic steps, so if someone could post them here that would be really helpful.

Answer 1

Are you asking for an algebraic explanation of the quadratic equation which can be found here: http://tiny.cc/6x3tkw
Or are you asking about the algebra involved in this particular application of the quadratic formula?

Answer 2

\[f(\theta ) ≈ f(\theta _{0})+ f'(\theta _{0})(\theta − \theta _{0})+ \frac{f''(\theta _{0})(\theta − \theta 0)}{2}\\if\\ (\theta ≈ \theta _{0})\]
So plugging in tangent and it's first and second derivative into the above equation we have:
\[f(\theta ) \approx tan(0) + sec^2(0)*(\theta -0) + \frac{tan(0)sec^2(0)*(\theta -0)}{2} \]

\[tan(0) = 0 , sec(0) = 1\]

So that all reduces to \[tan(\theta ) \approx \theta \]

Answer 3

Oh, but now I see the solution from the link you gave... you are asking how Gerrison went from \[\frac{\theta }{1- \frac{\theta}{2}} \approx \theta(1+\frac{\theta}{2})\]

Well, notice that it is not an equal sign, and that since this is a quadratic approximation, and theta is close to zero, any term that is higher than power 2, can be ignored, so:

\[\frac{\theta }{1- \frac{\theta}{2}} * \frac{1+ \frac{\theta^2}{2}}{1+ \frac{\theta^2}{2}} = \frac{\theta(1+\frac{\theta}{2})}{1- \frac{\theta^4}{4}} \approx \theta(1+\frac{\theta}{2})\]

Answer 4

Thats great! Thanks alot for the detailed explanation, and sorry if I didn't formulate my question in an easily understandable way. If you would allow one last question I'd like to know how Gerrison gets from \[\theta+\frac{\theta^2}{2}\approx \theta\]

He can't drop the \[\theta^2\] since its a quadratic expanion, right?

Answer 5

You are correct, that he could not ignore an order 2 term... but now I see that I had a typo in my work above. It should have read

\[\frac{\theta }{1- \frac{\theta^2}{2}} * \frac{1+ \frac{\theta^2}{2}}{1+ \frac{\theta^2}{2}} = \frac{\theta(1+\frac{\theta^2}{2})}{1- \frac{\theta^4}{4}} \approx \theta(1+\frac{\theta^2}{2})\]

So he is really only ignoring \[ \frac{\theta^3}{2}\]

Sorry for the typo. But thank you for the question. I think when I was working this problem I just skimmed over his answer and didn't even realize what he had done.

Answer 6

Wow, yeah, I should have checked that myself. Anyway thanks again for the great effort, really appreciate it, and think I've now fully understood this method.