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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.

Henry Cai
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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 $g$.

Andrei
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  • I don’t quite familiar with the Taylor expansion, I only know some formulae for maclaurin expansion, but what you say really helps. – Henry Cai Feb 10 '20 at 22:53
  • To be honest sir, I don’t quite get why does this approximation of 1 + $\lambda$ to 1 would imply the exponential term would also be approximated to one. – Henry Cai Feb 10 '20 at 22:56
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    Taylor expansion around 0 = MacLaurin expansion – Andrei Feb 10 '20 at 22:57
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    In your expansion of the exponential, the first two terms are also $1+\lambda$, so there is no reason to say one of them is $1$ and the other $1+\lambda$. And if $\lambda$ is small, then $\lambda^2<<\lambda$ – Andrei Feb 10 '20 at 22:58
  • I see , thank you for your great help. May I ask one more thing: if I want to approximate $(1+ \lambda)^n e^{-n\lambda}$, can I just expand them and multiplying them. And is there any consistency in the terms that we should keep, for example can I keep the binomial term to the power of 2 and the exponential using maclaurin expansion term to the power of 3. I am quite confusing with the terms that are allowed to keep, is that arbitrary? Thank you very much. – Henry Cai Feb 10 '20 at 23:02
  • If you what to expand to power $p$, you need to keep all the terms in each expansion up to that power. The reason for that is that they can contribute to the $p$ term when multiplied with the $0^{th}$ term. So in the above example power $3$ comes from $(0,3)$, $(1,2)$, $(2,1)$, and $(3,0)$ terms. Note that if you decide to neglect power $4$ and above, $(3,1)$, $(3,2)$, $(2,2)$, ... are negligible. – Andrei Feb 11 '20 at 04:02