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

  • #1
gionole
281
24
Homework Statement
Help me with taylor's theorem
Relevant Equations
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.
 

Attachments

  • Screenshot 2023-12-27 at 1.32.29 PM.png
    Screenshot 2023-12-27 at 1.32.29 PM.png
    8.6 KB · Views: 49
Physics news on Phys.org
  • #2
I am not accustomed the way of the text. What is the text you use ?
 
  • #3
O-notation tells you about the magnitude of the error in some limit, in this case [itex]\epsilon \to 0[/itex]; the sign of the error can depend on [itex]\epsilon[/itex], so it is conventional to use a plus sign.
 
  • Like
Likes gionole
  • #4
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)##.
 
  • Like
  • Love
Likes WWGD and gionole

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

Why does Taylor's theorem use +O(ε) instead of -O(ε)?

The notation +O(ε) is used in Taylor's theorem to indicate the order of the error term. The plus sign signifies that the error term is non-negative and provides an upper bound on the size of the error. The choice of +O(ε) over -O(ε) is a matter of convention and clarity, ensuring that the error term is understood to be an added, rather than subtracted, quantity.

What does the O(ε) term represent in Taylor's theorem?

The O(ε) term in Taylor's theorem represents the error term, which quantifies the difference between the actual function value and the value predicted by the Taylor polynomial. It indicates that the error is of the order of ε, meaning it becomes small as ε approaches zero. This term helps to approximate the function more accurately by accounting for higher-order terms that are not included in the polynomial.

Is the sign of the O(ε) term important in Taylor's theorem?

The sign of the O(ε) term is not inherently important; what matters is the magnitude and order of the error. The notation +O(ε) is used conventionally to indicate that the error term is added to the polynomial approximation. This helps to avoid confusion and ensures that the error is interpreted as an additional term that bounds the approximation error from above.

Can Taylor's theorem be stated with -O(ε) instead of +O(ε)?

While it is theoretically possible to state Taylor's theorem with -O(ε), it is unconventional and can lead to confusion. The standard practice is to use +O(ε) to clearly indicate that the error term is added to the polynomial approximation. Using -O(ε) could imply a different interpretation of the error term, which might not align with the conventional understanding of Taylor series approximations.

How does the use of +O(ε) affect the interpretation of Taylor's theorem?

The use of +O(ε) in Taylor's theorem affects the interpretation by clearly indicating that the error term is an additional, non-negative quantity that bounds the approximation error. This notation helps to communicate that the actual function value is within the range defined by the Taylor polynomial plus the error term, providing a more intuitive and standardized understanding of the theorem's accuracy and limitations.

Back
Top