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Epsilon distance between two terms

  1. May 20, 2017 #1
    1. The problem statement, all variables and given/known data
    How close is [itex]x[/itex] to [itex]x_0[/itex] ([itex]x_0 \neq 0[/itex]) so that
    45680e6d259043cbafd126849b14f6cb.png

    2. Relevant equations



    3. The attempt at a solution
    I tried to use absolute value properties:[tex]- \epsilon \lt \frac{\sqrt{x_0^2+1}}{x_0^3} - \frac{\sqrt{x^2+1}}{x^3} \lt \epsilon[/tex]By adding in the three sides, we have:[tex]\frac{\sqrt{x^2+1}}{x^3} - \epsilon \lt \frac{\sqrt{x_0^2+1}}{x_0^3} \lt \frac{\sqrt{x^2+1}}{x^3} + \epsilon[/tex]But that doesn't say anything about the distance between the terms in the absolute value, and I don't know how to algebrically organize the inequality to achieve that. I'd greatly appreciate some help!
     
  2. jcsd
  3. May 20, 2017 #2

    Charles Link

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    A practical approach would be to write ## x=x_o+\delta ## and perform Taylor expansions on the functions to get a first order result for ## \delta ## as a function of ## \epsilon ##, dropping the higher order terms, ## \delta^2 ##, ## \delta^3 ##, etc...
     
  4. May 20, 2017 #3
    Unfortunately, I don't know what that is yet, or how to do it. :/
     
  5. May 20, 2017 #4

    Charles Link

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    For ## x^3 ##, it is simply a binomial expansion,(## (x_o+\delta)^3=x_o^3+3x_o^2 \delta+... ## ), and when you have it in the form ## \frac{1}{1+a \delta} ## it is equal to ## 1-a \delta ##. (approximately). The square term ## ( x_o+\delta)^2 =x_0^2+2 x_o \delta ## (approximately), and ## \sqrt{1+b \delta} =1+\frac{b}{2} \delta ## (approximately). With these equations, you can get an expression between ## \delta ## and ## \epsilon ## to first order. Otherwise, it looks somewhat difficult.
     
  6. May 20, 2017 #5

    PeroK

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    Hint: What about using ##y^2 - y_o^2 = (y-y_0)(y+y_0)##?
     
  7. May 20, 2017 #6

    PeroK

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    This is standard Real Analysis. Taylor Series can't often be used in these cases.
     
  8. May 20, 2017 #7

    Ray Vickson

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    That will give an approximation, but probably not a rigorous upper bound. Obtaining the latter seems much harder.
     
  9. May 20, 2017 #8

    PeroK

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    The question may not require the greatest possible value for ##\delta## - any value for ##\delta## should do, as long as it's rigorously proven to imply the inequality involving ##\epsilon##.
     
  10. May 20, 2017 #9
    I don't follow. Should I try transforming the expression inside the absolute value into a difference of squares?
     
  11. May 20, 2017 #10

    PeroK

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    No. Instead, let ##y= \frac{\sqrt{x^2+1}}{x^3}##
     
  12. May 20, 2017 #11
    If [itex]
    y= \frac{\sqrt{x^2+1}}{x^3}[/itex], shouldn't it be [itex] y_0 - y[/itex], considering the term with [itex]x_0[/itex] is the one from which
    [itex]\frac{\sqrt{x^2+1}}{x^3}[/itex] is being subtracted? Or should I switch their signs?
     
  13. May 20, 2017 #12

    PeroK

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    My hint will take you so far, but after that it's still very difficult. From your questions you need a lot more practice at limits before you tackle this one. Where did you get this problem? It seems quite far above your current level.

    To answer your question ##|a - b| = |b - a|##. So, it doesn't matter which way round the expression inside the modulus is.
     
  14. May 20, 2017 #13
    It's part of a list of exercises my professor assigned us. Should I leave it aside?
     
  15. May 20, 2017 #14

    PeroK

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    You could try without the numerator. Try with ##y = \frac{1}{x^3}##. Have you already done one like that?
     
  16. May 20, 2017 #15
    Does it relate to Bernoulli's inequality?
     
  17. May 20, 2017 #16

    PeroK

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    Not really. With these ##\epsilon-\delta## proofs, you are trying to show something different from the sort of approximations you get with the Binomial Theorem or Taylor Series.

    Let's take a simple example of the function ##x^2## and show that it is continuous at any point ##x_0 > 0##. It's relatively easy to take ##x_0 = 0## as a separate case. And, it's easy to show the case where ##x_0 < 0## once we have the case proved for ##x_0 > 0##.

    So, let ##x_0 > 0## and let ##\epsilon > 0##

    We start by looking at the difference ##|x^2 - x_0^2|##.

    The first idea is to rewrite this as:

    ##|x^2 - x_0^2|= |x - x_0||x + x_0|##.

    The next idea is to note that if we could find an upper bound for ##|x + x_0|## that would be very useful. The trick is to consider ##|x - x_0| < \frac{x_0}{2}##. (This is where we need ##x_0 > 0##.) Then we have:

    ##\frac{x_0}{2} < x < \frac{3x_0}{2}##

    Hence ##|x + x_0| = x + x_0 < \frac{5x_0}{2}##

    And ##|x^2 - x_0^2|= |x - x_0||x + x_0| < |x - x_0|\frac{5x_0}{2}##

    Finally, if we take ##|x - x_0| < \epsilon \frac{2}{5x_0}##, then:

    ##|x^2 - x_0^2| < \epsilon##

    Putting this altogether we have:

    ##|x - x_0| < min \lbrace \epsilon \frac{2}{5x_0}, \frac{x_0}{2} \rbrace \ \Rightarrow |x^2 - x_0^2| < \epsilon##

    And this shows that ##x^2## is continuous at ##x_0 > 0##. Note that I didn't actually use the symbol ##\delta## at all. Some people don't like this, but I don't see that the symbol ##\delta## is necessary outside of the definition. You may like to think about this yourself.

    Now, you may wish to try a function such as ##x^3##, ##\frac{1}{x^2}## or ##\frac{1}{x^3}##, which are gradually getting harder and trickier. I don't know what ones your professor has assigned you, but until you can do these ones, you will have to leave the harder ones. The one you started with is very hard - perhaps too hard.
     
  18. May 20, 2017 #17
    Is [itex]
    \delta = \mathrm{min}(1,\frac{\varepsilon}{2|x_0|+1})[/itex] incorrect for [itex]x^2[/itex]?
     
  19. May 21, 2017 #18

    PeroK

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    That looks like a simpler solution for ##x^2##. There are lots of possible ways to do these. I was trying to use some additional ideas that would work for things like ##\frac{1}{x^2}## so my solution was more complicated than it needed to be for the simple example.

    It looks like you have got the basic idea.

    Is the one you posted the first one you are stuck on?
     
  20. May 21, 2017 #19
    It's the first one I got stuck on because it's the first one I tried to do. So I started by trying to do your example to see if I could reach the same result as you did. But since you told me it's still correct, that's great!

    But indeed, [itex]x^3[/itex] is trickier, I don't know if I'm doing it right.
     
  21. May 21, 2017 #20

    PeroK

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    I'll tell you why I think the one you got stuck on it "too hard". Imagine, instead, you want to prove that ##\frac{\sqrt{x^2+1}}{x^3}## is continuous from first principles. Here's how I'd do it:

    First, I'd prove that if ##f## and ##g## are continuous, then:

    ##f+g, fg, f/g## and ##f ° g## are continuous, with the suitable constraints. To do this, you just need to show that an appropriate ##\delta## exists. You don't actually have to find it for every possible function!

    Then, I'd show that ##x^2 + 1, \sqrt{x}## and ##\frac{1}{x^3}## are continuous (using ##\epsilon-\delta##).

    Finally, I'd put the two together.

    In other words, the one you started with is so complicated that finding an actual ##\delta## is not very enlightening. It's enough to show from first principles that a ##\delta## exists.

    On the other hand, practising with, say, ##\frac{1}{x^3}## is enlightening, as you need to find some useful tricks. The more complicated one just uses the same tricks over and over, with things getting pointlessly complicated.

    That's one point of view, anyway!

    In any case, unless and until you can do the constituent functions: ##\sqrt{x^2 + 1}## and ##\frac{1}{x^3}##, it's pointless trying to do the whole thing. So, definitely do the easier ones first.
     
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