Why Does Taylor's Theorem Use +O(ε) Instead of -O(ε)?

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SUMMARY

Taylor's Theorem utilizes the notation +O(ε) to describe the magnitude of error as ε approaches 0, emphasizing that the sign of the error can vary with ε. The expression f = g + O(ε) indicates that there exists a positive constant M such that |f - g| ≤ M|ε|. Conversely, using -O(ε) would imply a transformation to g = f + O(ε), which is mathematically equivalent. Therefore, the convention of using +O(ε) is universally accepted in mathematical literature.

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gionole
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Help me with taylor's theorem
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Help me with taylor's theorem
I am trying to grasp how the last equation is derived. I understand everything, but the only thing problematic is why in the end, it's ##+O(\epsilon)## and not ##-O(\epsilon)##. It will be easier to directly attach the image, so please, see image attached.
 

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I am not accustomed the way of the text. What is the text you use ?
 
O-notation tells you about the magnitude of the error in some limit, in this case \epsilon \to 0; the sign of the error can depend on \epsilon, so it is conventional to use a plus sign.
 
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The expression, ##f=g+O(\epsilon)## means that there exists such positive number ##M## that ##|f-g| \leq M|\epsilon|##.
OTOH, ##f=g-O(\epsilon)## means ##g=f+O(\epsilon)##, which means that there exists such positive number ##M## that ##|g-f| \leq M|\epsilon|##. This is the same as above. So, one can always use ##+O(\epsilon)##.
 
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