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

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The discussion centers on the use of +O(ε) in Taylor's Theorem, clarifying that it indicates the magnitude of error as ε approaches zero. The sign of the error can vary with ε, making the positive notation conventional. The expression f = g + O(ε) implies a bound on the difference |f - g|, while f = g - O(ε) can be rewritten as g = f + O(ε), showing that both forms convey similar information. Thus, using +O(ε) is standard practice in mathematical expressions. Understanding this notation is crucial for grasping the underlying concepts of error in limits.
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.
 
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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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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