I'm trying to follow some notes my supervisor has written and I've got the first three terms of a Taylor series $$1 - (-k \lambda) + (-k\lambda)^2 /2$$ becomes $$k\lambda +O(k\lambda^2)$$ Is this correct? What happened to the "1"?
$1 - (-k \lambda) + (k\lambda)^2 /2 = (k\lambda +O(k\lambda^2))$?
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taylor-expansion
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0You should as well ask where is the "$\frac{1}{2}$", they are all inside the $O$ term. – 2012-05-29
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0@Gigili can you give some more information, I don't have a grasp of whats going on yet. e.g. if the first four terms of the Taylor series were used what would that give. – 2012-05-29
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0There is a distinction between 'little-o' and 'big-o'. I am presuming you intended the latter? – 2012-05-29
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The Taylor polinomial of order $n$ of $e^{-x}$ is
$$e^{-x}=1-x+\frac{x^2}{2!}-\cdots+\frac{(-1)^nx^{n}}{n!}+o(x^{n})$$
Most probably, what he wrote is that
$$e^{-({-\lambda k})}=1-({-\lambda k})+\frac{(-\lambda k)^2}{2!}+o((-\lambda k)^{2})$$
$$e^{{\lambda k}}=1+\lambda k+\frac{\lambda^2 k^2}{2!}+o((\lambda k)^{2})$$
I don't see why $(a)$ he disregarded the $1$, $(b)$ he disregarded the term of degree $2$. Maybe he subtracted the series with that of $e^{-\lambda k}$?
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1The notation above was 'big-o'. – 2012-05-29
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0@copper.hat Yes, I see now. But since there was no $=$ signs I really didn't consider it, and since it was Taylor series, I assumed it was little-$o$. That confused me. – 2012-05-29
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0Depends on whether he really meant 'big-O$ or not. He tagged 'asymptotics', which is why I presumed it was... – 2012-05-29
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0Thanks, turns out there were a few typos. It was supposed to be expanding $1 - e^{-\lambda k}$... I really shouldn't go through this stuff at 2:15am. Un-accepted answer if you need to change. Ticked asymptotics because looking at what happens when $\lambda \rightarrow 0$ in the rest of the text. – 2012-05-29
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1My answer is incorrect then. Please take a little more care with your question. I have deleted my answer. – 2012-05-29
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0Sorry you spent the time on it, I tried to reply early earlier. :) – 2012-05-29
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0@copper.hat What was written out was how to get a close approximation to solve $e^{-k}(1 - e^{-\lambda k}) = \lambda/2 \,(1 + \lambda)^{-1} \,(1 + \lambda)^{\frac{-1}{\lambda}}$ as $\lambda \rightarrow 0$. And it was big O in the text... – 2012-05-29
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0The critical distinction is whether you were looking at $\lambda \to \infty$ or $\lambda \to 0$. You were looking at the latter, I presumed the former because you tagged 'asymptotics' and you used 'big-O' notation, which, in my experience, is more typically used for 'growth' rates. – 2012-05-29
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0@copper.hat Thanks again, in any case you've given me a few things to ask next time I have a meeting. – 2012-05-29
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0In this situation the 'big-O' notation is also correct. It was my inference that was wrong. – 2012-05-29