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  1. MHB Apprentice

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    #1
    Regarding the post below in :

    There are just two things left I need to wrap my mind around, after that I think I will have comprehended the epsilon-delta concept.

    In example 3 in the document epsilon-delta1.pdf where the task is to show that $ \displaystyle \lim_{x \to 5} \, (x^2) = 25$, they assume that there exists an $ \displaystyle M$ such that $ \displaystyle |x + 5| \leqslant M$.

    (1) Is it not supposed to be a strict inequality i.e. $ \displaystyle |x+5| < M$ and not $ \displaystyle |x+5| \leqslant M$? Why would the eventual equality between $ \displaystyle M$ and $ \displaystyle |x+5|$ ever be interesting?

    They make the aforementioned requirement when one arrives at

    $ \displaystyle |x-5| < \frac {\epsilon}{|x+5|} \, .$

    We somehow, normally through algebraic manipulations, wish to arrive at $ \displaystyle |x-5| < \frac{\epsilon}{M}$ and in their procedure, they write

    $ \displaystyle |x-5||x+5| < \epsilon \iff |x-5|M < \epsilon \, .$

    (2) The steps above have overlooked something. Sure, I can buy that $ \displaystyle |x-5||x+5| < |x-5|M$ because we stipulated an upper bound for $ \displaystyle |x+5|$ but just because $ \displaystyle |x-5|M$ is greater than $ \displaystyle |x-5||x+5|$ does not mean that it also must be less than epsilon, right?

    Drawing a number line, one can readily conclude that having a < c and a < b does not imply b < c.

    What is going on?

  2. Indicium Physicus
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    #2
    Quote Originally Posted by sweatingbear View Post
    Regarding the post below in :



    There are just two things left I need to wrap my mind around, after that I think I will have comprehended the epsilon-delta concept.

    In example 3 in the document epsilon-delta1.pdf where the task is to show that $ \displaystyle \lim_{x \to 5} \, (x^2) = 25$, they assume that there exists an $ \displaystyle M$ such that $ \displaystyle |x + 5| \leqslant M$.

    (1) Is it not supposed to be a strict inequality i.e. $ \displaystyle |x+5| < M$ and not $ \displaystyle |x+5| \leqslant M$? Why would the eventual equality between $ \displaystyle M$ and $ \displaystyle |x+5|$ ever be interesting?
    You can just make them all strict. I don't think it matters a whole lot in these cases.

    Quote Quote:
    They make the aforementioned requirement when one arrives at

    $ \displaystyle |x-5| < \frac {\epsilon}{|x+5|} \, .$

    We somehow, normally through algebraic manipulations, wish to arrive at $ \displaystyle |x-5| < \frac{\epsilon}{M}$ and in their procedure, they write

    $ \displaystyle |x-5||x+5| < \epsilon \iff |x-5|M < \epsilon \, .$

    (2) The steps above have overlooked something. Sure, I can buy that $ \displaystyle |x-5||x+5| < |x-5|M$ because we stipulated an upper bound for $ \displaystyle |x+5|$ but just because $ \displaystyle |x-5|M$ is greater than $ \displaystyle |x-5||x+5|$ does not mean that it also must be less than epsilon, right?
    Well, I'm not sure I agree that the steps have overlooked something. You let $\delta= \min(1, \epsilon/11)$. Therefore, $\delta \le 1$, which implies that $|x+5|<11$. So you can control that piece. But, $\delta \le \epsilon/11$, which means that $|x-5|<\epsilon/11$. Therefore, the product
    $$|x^{2}-5| = |x-5| |x+5| \le 11( \epsilon/11) = \epsilon.$$
    So the answer to your question is that it does imply that $|x-5||x+5| \le \epsilon$, because of the way you chose your $\delta$.

    Remember that $\epsilon$ is always chosen first. Then your goal is to find a formula $\delta = \delta( \epsilon)$ so that the condition of the limit holds.

    Quote Quote:
    Drawing a number line, one can readily conclude that having a < c and a < b does not imply b < c.

    What is going on?
    I would agree that $a<c$ and $a<b$ does not imply that $b<c$. However, that's not what's going on here. Instead, you have something more like $0<a<b$ and $0<c<d$, and therefore $ac<bd$: that is true.

    By the way, I would like to take this moment to write that I really appreciate the fact that you take the time to write meticulously crafted questions! Your $\LaTeX$ is impeccable, and really adds to the clarity of your questions.

  3. MHB Apprentice

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    #3 Thread Author
    @Ackbach: First off: Thank you very much for your kind remarks and your helpful reply.

    Quote Originally Posted by Ackbach View Post
    You can just make them all strict. I don't think it matters a whole lot in these cases.
    All right, I understand.


    Quote Originally Posted by Ackbach View Post
    Well, I'm not sure I agree that the steps have overlooked something. You let $\delta= \min(1, \epsilon/11)$.
    This reminds me: I have always wondered why we want $ \displaystyle \delta$ to be smallest of those two quantities (i.e. the two upper bounds). What is the rationale behind that? My guess is that since we are making $ \displaystyle x$ very close to the point of interest, we thusly wish to choose a small distance as possible to the mentioned point of interest. Is this a valid conclussion?

    Quote Originally Posted by Ackbach View Post
    I would agree that $a<c$ and $a<b$ does not imply that $b<c$. However, that's not what's going on here. Instead, you have something more like $0<a<b$ and $0<c<d$, and therefore $ac<bd$: that is true.
    All right, I see that requiring the numbers to positive allows to conclude that $ \displaystyle ac < bd$. However, I do not think that I have seen this one before:

    $ \displaystyle (0 < a < b) \ \wedge \ (0 < c < d) \implies ac < bd \, .$

    Could you perhaps direct to me a source where I can read up on it and/or eventually find some kind of proof? I checked out this but to no avail.

  4. Indicium Physicus
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    #4
    Quote Originally Posted by sweatingbear View Post
    @Ackbach: First off: Thank you very much for your kind remarks and your helpful reply.
    You're very welcome.

    Quote Quote:
    This reminds me: I have always wondered why we want $ \displaystyle \delta$ to be smallest of those two quantities (i.e. the two upper bounds). What is the rationale behind that?
    The reason is that we're going to need both conditions met simultaneously (a logical AND operation); that is, we need BOTH the condition $|x+5|\le 11$ AND $|x-5|< \epsilon/11$ to be true simultaneously. If you defined $\delta= \max (1, \epsilon/11)$, one of those conditions might fail. So you're banking on the property of the minimum, that if $a= \min(b,c)$, then $a\le b$ AND $a\le c$.

    Quote Quote:
    My guess is that since we are making $ \displaystyle x$ very close to the point of interest, we thusly wish to choose a small distance as possible to the mentioned point of interest. Is this a valid conclussion?
    I'm interpreting your question as, "We thusly wish to choose as small a distance as possible to the mentioned point of interest. Is this a valid conclusion?" (emphasis added) The answer is that there is no smallest distance to the point of interest, if you are in the real number system. The idea of a limit is not that you get as close as possible to a point of interest, but that you get as close as you want.

    Quote Quote:
    All right, I see that requiring the numbers to positive allows to conclude that $ \displaystyle ac < bd$. However, I do not think that I have seen this one before:

    $ \displaystyle (0 < a < b) \ \wedge \ (0 < c < d) \implies ac < bd \, .$

    Could you perhaps direct to me a source where I can read up on it and/or eventually find some kind of proof? I checked out this but to no avail.
    From the wiki, you get that if $a<b$, and $c>0$, then $ac<bc$. So, apply this logic twice, along with the transitivity of inequalities.

    Theorem: If $0<a<b$ and $0<c<d$, then $ac<bd$.

    Proof:

    1. Assume that $0<a<b$, and that $0<c<d$.
    2. Because $c>0$, it follows that $ac<bc$.
    3. By the transitivity of inequalities, it must be that $b>0$.
    4. Therefore, applying our logic again, it is the case that $bc<bd$.
    5. Therefore, $ac<bc<bd$.
    6. By the transitivity of inequalities, it follows that $ac<bd$. QED.

    The same logic follows through if you use non-strict inequalities.

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    #5
    When I arrive at $ \displaystyle |x-5||x+5| < \epsilon $

    I usually do the following $ \displaystyle |x+5|= |x-5+10|\leq |x-5|+10$

    So we have the following

    $ \displaystyle |x-5||x+5| \leq |x-5| \, ( |x-5|+10 )$

    which can be made arbitrarily small .

  6. MHB Apprentice

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    #6 Thread Author
    Thanks again for such a thorough reply!

    Quote Originally Posted by Ackbach View Post
    If you defined $\delta= \max (1, \epsilon/11)$, one of those conditions might fail.
    Honestly, I do not see why/how it would fail; would you mind elaborating?

    Quote Originally Posted by Ackbach View Post
    So you're banking on the property of the minimum, that if $a= \min(b,c)$, then $a\le b$ AND $a\le c$.
    Intuitively, I cannot grasp that having $a$ equal the smallest of the quantities $b$ and $c$ allows $a$ to equal $b$ and $c$ at the same time. I do not see how that makes sense, neither for the $\min(b,c)$ nor why $\max(b,c)$ cannot allow $a$ to equal $b$ and $c$ simultaneously.

    Quote Originally Posted by Ackbach View Post
    The answer is that there is no smallest distance to the point of interest, if you are in the real number system. The idea of a limit is not that you get as close as possible to a point of interest, but that you get as close as you want.
    Ah! Thank you for that clarification, helped me see things differently.

    Quote Originally Posted by Ackbach View Post
    From the wiki, you get that if $a<b$, and $c>0$, then $ac<bc$. So, apply this logic twice, along with the transitivity of inequalities.

    Theorem: If $0<a<b$ and $0<c<d$, then $ac<bd$.

    Proof:

    1. Assume that $0<a<b$, and that $0<c<d$.
    2. Because $c>0$, it follows that $ac<bc$.
    3. By the transitivity of inequalities, it must be that $b>0$.
    4. Therefore, applying our logic again, it is the case that $bc<bd$.
    5. Therefore, $ac<bc<bd$.
    6. By the transitivity of inequalities, it follows that $ac<bd$. QED.

    The same logic follows through if you use non-strict inequalities.
    Thanks for that proof, I will have a thorough look at it very soon.

  7. Indicium Physicus
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    #7
    Quote Originally Posted by sweatingbear View Post
    Honestly, I do not see why/how it would fail; would you mind elaborating?
    Well, suppose $1>\epsilon/11$, and so you let $\delta= \max(1,\epsilon/11)$. Then you'd have $\delta=1$. But you could have a small epsilon, say, $\epsilon=1$. Then $x$ could range up to, say, $5.9$, and you'd have
    $$|x+5||x-5|=10.9(0.9)=9.81> \epsilon = 1.$$

    Quote Quote:
    Intuitively, I cannot grasp that having $a$ equal the smallest of the quantities $b$ and $c$ allows $a$ to equal $b$ and $c$ at the same time.
    It doesn't. It lets $a \le b$ and $a \le c$ at the same time. $a\le b$ is the same thing as saying $a$ is less than OR equal to $b$. Review your logic!

  8. MHB Apprentice

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    #8 Thread Author
    @Ackbach: I must thank you for your patience replying to my posts, much appreciated.

    Quote Originally Posted by Ackbach View Post
    Well, suppose $1>\epsilon/11$, and so you let $\delta= \max(1,\epsilon/11)$. Then you'd have $\delta=1$. But you could have a small epsilon, say, $\epsilon=1$. Then $x$ could range up to, say, $5.9$, and you'd have
    $$|x+5||x-5|=10.9(0.9)=9.81> \epsilon = 1.$$
    I do not see the "problem" or "contradiction"; sorry but my mind is unrelenting and not allowing me to wrap it around this concept. Do you care to elaborate further?

    Quote Originally Posted by Ackbach View Post
    It doesn't. It lets $a \le b$ and $a \le c$ at the same time. $a\le b$ is the same thing as saying $a$ is less than OR equal to $b$. Review your logic!
    What if for example $ \displaystyle b = 4$ and $ \displaystyle c = 1$? If we use the "AND" operator, then that means we have two requirements for $ \displaystyle a$ that must be met: $ \displaystyle a$ must be less than or equal to $ \displaystyle 4$ and, simultaneously, less than or equal to $ \displaystyle 1$. Effectively, this implies that a simply must be less than or equal to $ \displaystyle 1$.

    Hm, wait a minute... So effectively thanks to the "AND" operator, the upper bound for $ \displaystyle a$ became the smallest quantity of $ \displaystyle b$ and $ \displaystyle c$. So this is why we let $ \displaystyle \delta$ equal the minimum of the two different bounds and not the maximum?

    I think I am on the verge of having a Heureka-moment, come on, somebody please push me into goal!

  9. Indicium Physicus
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    #9
    Quote Originally Posted by sweatingbear View Post
    @Ackbach: I must thank you for your patience replying to my posts, much appreciated.
    You're quite welcome! I wouldn't say, though, that replying to your posts is requiring all that much patience. It's the bad attitudes of some posters I've dealt with in the past that are tiresome to work with.

    Quote Quote:
    I do not see the "problem" or "contradiction"; sorry but my mind is unrelenting and not allowing me to wrap it around this concept. Do you care to elaborate further?
    Well, we want to guarantee that if $\delta$ equals something as a function of $\epsilon$, then $|f(x)-L|<\epsilon$. But I just provided an example where $\delta$ was some function of $\epsilon$ (in this case, just the constant function $1$), and that did not imply that $|f(x)-L|<\epsilon$. So that $\delta$ didn't work.

    Quote Quote:
    What if for example $ \displaystyle b = 4$ and $ \displaystyle c = 1$? If we use the "AND" operator, then that means we have two requirements for $ \displaystyle a$ that must be met: $ \displaystyle a$ must be less than or equal to $ \displaystyle 4$ and, simultaneously, less than or equal to $ \displaystyle 1$. Effectively, this implies that a simply must be less than or equal to $ \displaystyle 1$.
    Precisely.

    Quote Quote:
    Hm, wait a minute... So effectively thanks to the "AND" operator, the upper bound for $ \displaystyle a$ became the smallest quantity of $ \displaystyle b$ and $ \displaystyle c$. So this is why we let $ \displaystyle \delta$ equal the minimum of the two different bounds and not the maximum?
    You've got it.

    Quote Quote:
    I think I am on the verge of having a Heureka-moment, come on, somebody please push me into goal!
    You've fallen off the Eureka cliff... er... maybe I need to come up with a better analogy.

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    #10
    Quote Originally Posted by Ackbach View Post
    You're quite welcome! I wouldn't say, though, that replying to your posts is requiring all that much patience. It's the bad attitudes of some posters I've dealt with in the past that are tiresome to work with.



    Well, we want to guarantee that if $\delta$ equals something as a function of $\epsilon$, then $|f(x)-L|<\epsilon$. But I just provided an example where $\delta$ was some function of $\epsilon$ (in this case, just the constant function $1$), and that did not imply that $|f(x)-L|<\epsilon$. So that $\delta$ didn't work.



    Precisely.



    You've got it.



    You've fallen off the Eureka cliff... er... maybe I need to come up with a better analogy.
    Maybe "You've climbed to the Eureka peak"? :P

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