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So I'm confused, here with what is highlighted. When the book says of "order $1/y_-$" you will reduce the displacement by a factor of $1/e$. Does of order mean when the time is equal to $1/y_-$, if that's the case (I've tried it with a few examples) you won't ever get exactly X - (1/e)X for the displacement, it will always be less than that from what I observed. So what do they mean? I feel like I'm overthinking this.

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  • $\begingroup$ "Reduced by a factor of $1/e$" doesn't mean "reduced from $X$ to $X - (1/e)X$." It means "reduced from $X$ to $(1/e)X$." Incidentally, the notation is not $y$, it's $\gamma$ (gamma). $\endgroup$
    – alephzero
    Sep 8, 2018 at 15:45
  • $\begingroup$ Hi, welcome to Physics SE! Can you convert the picture of the text into typed-out, formatted text? It makes the content index-able by search engines, and shows up better on different devices' displays. For formulae, try MathJax instead. $\endgroup$
    – user191954
    Sep 8, 2018 at 16:15
  • $\begingroup$ I can't I'm in the army doing all this on a phone I have no idea how to type it in with math text $\endgroup$ Sep 8, 2018 at 16:24

1 Answer 1

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Go with your feelings on this one: you are overthinking it.

"On the order of" does not mean "exactly", and "reduced by a factor of $1/e$" means:

$$ X\rightarrow X/e \ne X-X/e $$

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