Why Did Rudin Use Absolute Value When Calculating the Derivative?

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Discussion Overview

The discussion centers on the use of absolute value in the calculation of the derivative of a piecewise function as presented in Rudin's text. Participants explore the reasoning behind this choice, particularly in the context of limits and the behavior of the sine function.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant questions why absolute value is used in the limit calculation for the derivative of the function defined piecewise, specifically at the point where x equals zero.
  • Another participant suggests that the use of absolute value relates to the property of the sine function being odd, which could influence the non-negativity of the limit.
  • A different viewpoint mentions that checking for absolute convergence is more straightforward in this example, referencing concepts from earlier chapters in Rudin's book.
  • One participant notes that the Newton quotient at zero is bounded by a quantity that approaches zero as t approaches zero, implying a relationship to the limit process.

Areas of Agreement / Disagreement

Participants express differing perspectives on the reasoning behind the use of absolute value, indicating that multiple competing views remain regarding its necessity and implications in the limit calculation.

Contextual Notes

Some assumptions about the behavior of the sine function and the implications of absolute convergence are not fully explored, leaving room for further discussion on these points.

Who May Find This Useful

Readers interested in mathematical analysis, particularly those studying derivatives and limit processes in advanced calculus, may find this discussion relevant.

Bachelier
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On Page 106 in baby rudin (diff. chapter) when he tries to calculate the derivative of the fuction


$$f(x) = \begin{cases}
x^2 sin(\frac{1}{x}) & \textrm{ if }x ≠ 0 \\
0 & \textrm{ if }x = 0 \\
\end{cases}$$

rudin used the absolute value in trying to compute the limit as ##t → 0##

##i.e##

##\left|\frac{f(t) - f(0)}{t - 0}\right| = \left|t \ sin(\frac{1}{x})\right| ≤ |t|##

Why the abs. value?
 
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I think it has to see with the fact that sin(-x)=-sin(x) . The limit then will be non-negative. Since is negative
in the 4th quadrant.
 
Absolute convergence implies convergence as in chapter 3 of the same book. And it turns out in this example checking absolute convergence is more obvious.
 
It just shows that the Newton quotient at zero is squeezed by a number that goes to zero with t.
 
Thank you all
 

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