I am trying to solve this problem:
Give that $p=(1 + \lambda) e^{-\lambda}$, and $q=1-p$.
Show that, when $\lambda$ is sufficiently small:
$q=\frac{1}{2}\lambda^2$.
So, what I did is $(1 + \lambda) \approx 1$, and $e^{-\lambda} \approx 1 - \lambda + \frac{1}{2}\lambda^2 \approx 1 + \lambda^2$. Then, Substituting q into $1-q=p$. We would get the answer.
However, I looked at the mark scheme after finishing this question. What the mark scheme did is that:
$(1+\lambda)e^{-\lambda^2} \approx (1+\lambda)(1 - \lambda +\frac{1}{2}\lambda^2 +...)\approx(1 + \frac{1}{2}\lambda^2+..)$. Here, they didn’t approximate $(1+\lambda)$ to 1(What they did, is they approximate the cubic term of $\lambda$ to 0). So, May I know, which way is the appropriate to do it, or both are correct.
Following this, if my way of doing is correct, does it mean that, once I approximate $(1+\lambda)$ to 1 once, I should do that in the following approximation, is there a such consistency? For example, can I do $(1+\lambda)e^{-\lambda} \approx (1)(1 + \lambda + \frac{1}{2} \lambda ^2)$ instead of $(1 + \lambda)(e^{-\lambda}) \approx (1)(1 + \frac{1}{2}\lambda^2)$
I don’t quite get the concept of approximation, and how to utilize it. Just like the question said ‘sufficiently small’, I don’t what how small is sufficient, does it have certain conditions?
Thank you very much for you guy’s replies, thank you very much for your help and sorry for any inconvenience caused.
No, you should never use different approximations when you multiply. Since you said that $(1+\lambda)\approx 1$, then immediately $e^{-\lambda}$ would also be approximately $1$. Are you familiar with the concept of Taylor expansion? For $x$ close to $0$ you have: $$f(x)\approx f(0)+f'(0)x+\frac12 f''(0)x^2+...\\g(x)\approx g(0)+g'(0)x+\frac12 g''(0)x^2+...$$ Let's assume that we want to keep terms up to power $2$ in $x$ in the approximation for $f(x)g(x)$. That means we have for $0^{th}$ power $$f(0)g(0)$$, for the first power $$(f(0)g'(0)+f'(0)g(0))x$$ and for the second power $$x^3(f(0)\frac 12 g''(0)+f'(0)g'(0)+\frac 12 f''(0)g(0))$$ Notice that for the third power of $x$ you would need to keep up to the third derivative of each $f$ and $g$.