1

I have tried this: http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-2-applications-of-differentiation/part-a-approximation-and-curve-sketching/problem-set-3/ and checked my solution of the problem 2A-6.

In the solutions they solve it this way: $$\tan\theta=\frac{\sin \theta}{\cos\theta}\approx \frac{\theta}{1-\theta^2/2}$$ Until this point I did it exactly this way but I don't get the next steps: $$\frac{\theta}{1-\theta^2/2} \approx \theta (1+\theta^2//2) \approx \theta $$

What has been done here? How does the minus sign change to a plus and what does the dubble-slash mean?

user50224
  • 966
  • 2
    One can also do this by using that $\tan x$ is odd, and so its Taylor series only has odd terms. Thus, its quadratic approximation agrees with its linear approximation, and so you only need to compute $\tan(0) = 0$ and $\left.\frac{d}{dx} \tan x\right\vert_{x = 0} = \sec^2 x \vert_{x = 0} = 1$. – Travis Willse Aug 31 '14 at 13:20
  • 1
    Normally you would rather write $\sim$ than $\approx$. – Krokop Aug 31 '14 at 14:09
  • @krokop If the only needed thing is an equivalent, I don't understand why the intermediate step would be useful. – Jean-Claude Arbaut Aug 31 '14 at 14:44

3 Answers3

2

Hint

Use the Taylor series:

$$(1-x)^\alpha=1-\alpha x+\cdots$$ with obviously $\alpha=-1$ in your case.

1

here is error or mistyping $$ \tan\theta=\frac{\sin \theta}{\cos\theta}\approx \frac{\theta}{1-\theta^2/2} $$ right way: $$\tan\theta=\frac{\sin \theta}{\cos\theta}=\frac{\sin \theta}{\sqrt{(1-\sin\theta^2)}}\approx \frac{\theta}{(1-\theta^2)^{1/2}} = \frac{\theta}{(1-\theta^2)^{1/2}} \approx \theta (1+\theta^2/2) \approx \theta $$ reason: linearixation of $$ (1+x)^\alpha=1+\alpha x+\cdots $$ so $$ \frac{1}{\sqrt{(1+x)}}=(1+x)^{-1/2}=1-{x/2} +\cdots $$

silent
  • 11
0

I don't know exactly what the double slash here means but this is simply a regular expansion of the kind

$1/(1-x) = 1 + x + x^2 + ... $

In this case you have $x=\theta^2 /2$

Avrham Aton
  • 445