I Evaluating Taylor Series at the Mid-Point

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The discussion centers on understanding the error term of ##\mathcal{O}(\epsilon^{3})## in the Taylor Series expansion related to a specific equation. Participants seek clarification on why this particular error order is noted, as it differs from the more commonly encountered ##\mathcal{O}(\epsilon^{2})##. A reference to the midpoint method on Wikipedia is suggested for further insight into the topic. The conversation emphasizes the importance of recognizing the implications of the Taylor Series evaluation at the midpoint. Overall, the focus is on clarifying the mathematical reasoning behind the error classification.
thatboi
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Hi all,
I came across the following stackexchange post and was wondering if anyone could give any elaboration for why the answer claims that evaluating the Taylor Series resulted in ##\mathcal{O}(\epsilon^{3})## errors? I have not encountered such an expansion before.
EDIT: The equation at hand is:
$$f(x+\epsilon n, x) = f(x,x) + \epsilon n^{\mu}\frac{\partial f(x+\epsilon n,x)}{\partial n^{\mu}}\vert_{x+\frac{\epsilon}{2}n} + \mathcal{O}(\epsilon^{3}) $$.
 
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thatboi said:
Hi all,
I came across the following stackexchange post and was wondering if anyone could give any elaboration for why the answer claims that evaluating the Taylor Series resulted in ##\mathcal{O}(\epsilon^{3})## errors? I have not encountered such an expansion before.
To keep PF threads as self-contained as possible, could you post a summary here of the answer you're asking about?
 
renormalize said:
To keep PF threads as self-contained as possible, could you post a summary here of the answer you're asking about?
Ah that is true. I have edited the question.
 
thatboi said:
Ah that is true. I have edited the question.
Thanks. To see why the posted equation has errors of ##\mathcal{O}(\epsilon^{3})## rather than ##\mathcal{O}(\epsilon^{2})##, take a look at: https://en.wikipedia.org/wiki/Midpoint_method.
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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