In the first equation I can't figure out where the $1/x$ part comes from...
When using first-order taylor approximation arount $x_t = x$, I just get
$ln(1)*(x_t-x) = (x_t-x)$
Thank you for your help!
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Here is a derivation of the taylor expansion of $\ln(1+x)$: https://math.stackexchange.com/questions/878374/taylor-series-of-ln1x . Does this help? – Andreas Lenz Nov 15 '22 at 19:38
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No, I still don't get it... – Alexis Boudreau Nov 15 '22 at 20:47
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The Taylor series of $\ln(1+y)$ around $y=0$ is $\ln(1+y)=0+y+O(y^2)$, as you can see in the linked post. Now replace $y$ with $\frac{x_t-x}{x}$. – Andreas Lenz Nov 15 '22 at 20:58
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Yes, but $x_t = x$, so it becomes: $(x-x)/x$ ,no? – Alexis Boudreau Nov 15 '22 at 21:05
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You are correct, but this is what you should expect. The first-order Taylor series approximation of $f(y)$ around $a$ is $f(y)=f(a)+f'(a)(y-a)$. The second term will always be zero at $y=a$. When $y=a$, the equation just becomes $f(a)=f(a)$. (This applies to Taylor series approximations of any order.) – smcc Dec 01 '22 at 20:15