Keisler Elementary Calculus: computing standard parts of expressions

[rstor]

TL;DR Summary

In Keisler Elementary Calculus page 39, example 4, the author makes the observation that both the numerator and denominator are nonzero infinitesimals. I am unsure of how they came to this conclusion (at the beginning of the example) for the expression in the denominator.

In Keisler Elementary Calculus page 39, example 4 it shows how to compute the standard parts of the following expression:

Example 4: If $\epsilon$ is infinitesimal but non zero, find the standard part of

$b=\frac {\epsilon} {5-\sqrt{25+ε}}$

Before calculating the standard parts the author makes the observation that "Both the numerator and denominator are nonzero infinitesimals"
I understand from the previous section that if this is the case then the quotient of two infinitesimals is in indeterminate form and therefore more work needs to be done.

I do not understand how we can (right away) come to the conclusion that the denominator is a nonzero infinitesimal. From what I understand the denominator is a sum of finites and therefore the result is finite though possibly infinitesimal (see page 31).

I had a similar question in this thread where the member Orodruin wrote in response to what I said about the expression $\sqrt{4 + \epsilon} ~~- 2$ that the expression can be:
"Finite, yes, but you cannot rule out infinitesimal. To do that you need to look at the square root and conclude that the finite but not infinitesimal parts of the square root and the -2 cancel and sum up to zero. What remains must therefore be infinitesimal (or zero) so you need to look at the infinitesimal part of the square root."

Why then in this case does the author conclude at the start that the expression in the denominator is a nonzero infinitesimal instead of saying that it is possibly infinitesimal?

 

[fresh_42]

$5-\sqrt{25+0}=0$ so the denominator is zero if $\varepsilon =0.$ Hence $5-\sqrt{25+\varepsilon }$ is infinitesimal, a slight variation of zero around zero.

 

[rstor]

$5-\sqrt{25+0}=0$ so the denominator is zero if $\varepsilon =0.$ Hence $5-\sqrt{25+\varepsilon }$ is infinitesimal, a slight variation of zero around zero.

I am confused. The author indicates that $\varepsilon ≠0.$ In the previous section (1.5) one might use algebraic manipulation to conclude if an expression is infinitesimal, finite but not infinitesimal, or infinite using the rules outlined on page 31. Also I understood from my previous post that if you had two positive finite numbers then the resulting sum could only be finite (not infinitesimal). However in this case we have a difference of finites in the denominator where it is not readily shown (at least to me) that the finite values cancel leaving only an infinitesimal. How can we therefore conclude that the denominator is infinitesimal?

 

[fresh_42]

Of course, we have $\varepsilon >0.$ It is only a heuristic to set $\varepsilon =0$ in order to see where the expression lies. We get in our example, that the denominator is zero; end of heuristic consideration.

Next, we get back to mathematics and consider $\varepsilon >0$ but infinitesimal small. However, we know from our heuristic consideration, that $5-\sqrt{25+\varepsilon }<0$ is surely infinitesimal, too.

Finally, we get to the third step that tells us that we have to apply further considerations since we are in the case $\frac{\text{infinitesimal}}{\text{infinitesimal}}.$

Allow me a personal question: Why in the world do you want to learn calculus via hyperreals?

 

[PeroK]

Why in the world do you want to learn calculus via hyperreals?

There are some who prefer hyperreals to real epsilons:

A major breakthrough in 19th mathematics was to develop calculus on a rigorous basis without recourse to "infinities" or "infinitesimals".

You may consider it a breakthrough. I consider it a step backwards that mathematical pedagogy has still not overcome.

 

[rstor]

Of course, we have $\varepsilon >0.$ It is only a heuristic to set $\varepsilon =0$ in order to see where the expression lies. We get in our example, that the denominator is zero; end of heuristic consideration.

Next, we get back to mathematics and consider $\varepsilon >0$ but infinitesimal small. However, we know from our heuristic consideration, that $5-\sqrt{25+\varepsilon }<0$ is surely infinitesimal, too.

Finally, we get to the third step that tells us that we have to apply further considerations since we are in the case $\frac{\text{infinitesimal}}{\text{infinitesimal}}.$

Allow me a personal question: Why in the world do you want to learn calculus via hyperreals?

Thank you for taking the time to elaborate. I can see how if we assume that $\epsilon=0$ that the denominator equals zero as you mentioned in your earlier post: $5-\sqrt{25+0}=0$

I suppose one could alternatively apply the standard value function to the denominator and also conclude that it equals zero this way. I am assuming from these methods one can see that the denominator is some infinitesimal value and since we have a quotient of infinitesimals further work is needed. I was trying to follow only the methods outlined in the book, thinking that the author would have intended the reader to use tools that were previously discussed to come to this conclusion (perhaps some kind of algebraic manipulation)...I also may have overlooked something that the author mentioned.

I enjoy electronics as a hobby and want to learn things at a deeper level. I am self studying calculus so I can learn physics for this reason. I started with KA Stroud Engineering Mathematics and have temporarily stopped at the point where it goes into calculus. I do not have an interest in learning proof based math at a very rigorous level and as Stroud is a math methods book I found it to be a good fit for me. One concern I had was that because it lacks rigor I might have issues when it comes time to apply what I learn to new situations wrt calculus. See preface of the book entitled "Infinitesimal Calculus" by Henle and Kleinberg (Amazon Look Inside) regarding such concerns. I found many posts on this forum and elsewhere suggesting Keisler's text. Here are some examples: [Ex1]PhysicsForum , [Ex2]PhysicsForum, [Ex3]PhysicsForum, [Ex4]PhysicsForumInsights, [Ex5]Quora and this.

Many of these posts helped me to decide to use hyperreal based text to learn calculus. I recently came across Calculus Set Free Infinitesimals to the Rescue by Bryan Dawson and a video series on Youtube that appears to follow his book. I might decide to switch to this book however I am unsure at this point.

 

[PeroK]

Many of these posts helped me to decide to use hyperreal based text to learn calculus. I recently came across Calculus Set Free Infinitesimals to the Rescue by Bryan Dawson and a video series on Youtube that appears to follow his book. I might decide to switch to this book however I am unsure at this point.

I have a degree in pure mathemantics and did a lot of analysis, functional analysis and measure theory, but I never studied the hyperreals. The physics books I've studied tend to use standard real analysis (not the hyperreals) - if at all. I would suggest that in terms of both mathematics and physics you need standard real analysis more than you need non-standard real analysis.

Non-standard analysis has its advocates, of course, but you should be aware that its not as widely taught and used as standard analysis.

 

[rstor]

Simple solution: $5-\sqrt{25+\epsilon}=5-5\sqrt{1+\frac{\epsilon}{25}}\approx -\frac{\epsilon}{50}$. Final answer is $b=-50$.

An earlier attempt at this question, I had reached to something similar to part of your solution as follows:
$b=\frac {\epsilon} {5-\sqrt{25+ε}}$

$b=\frac {\epsilon} {5-\sqrt{25(1+\frac{ε}{25})}}$

$b=\frac {\epsilon} {5-5\sqrt{1+\frac{ε}{25}}}$
At this point I was unsure of what could be done to identify the denominator as an infinitesimal. I also do not understand how you arrive at the following in the denominator:
$5-5\sqrt{1+\frac{\epsilon}{25}}\approx -\frac{\epsilon}{50}$

 

[Frabjous]

Infinitesimals are small

An earlier attempt at this question, I had reached to something similar to part of your solution as follows:
$b=\frac {\epsilon} {5-\sqrt{25+ε}}$

$b=\frac {\epsilon} {5-\sqrt{25(1+\frac{ε}{25})}}$

$b=\frac {\epsilon} {5-5\sqrt{1+\frac{ε}{25}}}$
At this point I was unsure of what could be done to identify the denominator as an infinitesimal. I also do not understand how you arrive at the following in the denominator:
$5-5\sqrt{1+\frac{\epsilon}{25}}\approx -\frac{\epsilon}{50}$

1) He’s taking a series expansion which is something that you probably do not know how to do yet.
2) To check if something is an infinitesimal, set the ε=0. If the thing is then equal to zero, it is then an infinitesimal. If it is finite, it is not. If you end up with something divided by zero, you have to investigate further.
3) To solve for the entire fraction, complete the square as was done in your other post.

edit: be careful about breaking things into parts

 

[fresh_42]

1) He’s taking a series expansion which is something that you probably do not know how to do.

You do not need Taylor, Bernoulli is sufficient:
https://en.wikipedia.org/wiki/Bernoulli's_inequality

But I get $b=-10$.

$$
5-\sqrt{25+\varepsilon }=5\left(1-\sqrt{1+\dfrac{\varepsilon }{25}}\right)\approx 5\left(1-\left(1+\dfrac{1}{2}\cdot \dfrac{\varepsilon }{25}\right)\right)=5\left(-\dfrac{\varepsilon }{50}\right)=-\dfrac{\varepsilon }{10}
$$

 

[Frabjous]

You do not need Taylor, Bernoulli is sufficient:
https://en.wikipedia.org/wiki/Bernoulli's_inequality

How do you know there is near equality? (I am not saying there isn’t)

 

[fresh_42]

How do you know there is near equality? (I am not saying there isn’t)

Right. I get $0>5-\sqrt{25+\varepsilon }\geq -\dfrac{\varepsilon }{10}.$

 

[rstor]

I have a degree in pure mathemantics and did a lot of analysis, functional analysis and measure theory, but I never studied the hyperreals. The physics books I've studied tend to use standard real analysis (not the hyperreals) - if at all. I would suggest that in terms of both mathematics and physics you need standard real analysis more than you need non-standard real analysis.

Non-standard analysis has its advocates, of course, but you should be aware that its not as widely taught and used as standard analysis.

I find that I have no interest in learning an epsilon delta method for calculus. It is my understanding that unless I am going into real analysis, omitting this is okay. Occasionally I ask those in other fields such as graduates of computer science, engineering, or those studying physics at university and they for example don't appear to have studied or remember what the epsilon delta definition of a limit is. I figure someone like myself, who is not going into mathematics as a major, and whose time is limited; that there might be other areas of study which would be of greater benefit (i.e. math methods, physics, electrical engineering). I was going to abandon the idea of studying calculus at any level deeper than what would be presented in a math methods book such as Stroud. It was when I found that there was an alternative to epsilon-delta based calculus that I thought it may warrant further investigation.

It does seem to be as you mentioned, not widely taught, and therefore I am concerned that if I need one on one in person help my options may be very limited.

 

[fresh_42]

It does seem to be as you mentioned, not widely taught, and therefore I am concerned that if I need one on one in person help my options may be very limited.

Well, you can ask your questions here, even though our knowledge of hyperreals is a bit thin. It is basically the decision whether you want to learn calculus pre-Cauchy or post-Cauchy (with a grain of salt). We are only a bit allergic against those heroes who treat $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ as if they were numbers. The epsilontic doesn't have those pits.

 

[Frabjous]

Right. I get $0>5-\sqrt{25+\varepsilon }\geq -\dfrac{\varepsilon }{10}.$

Does this prove that -epsilon/20 isn’t acceptable?

 

[fresh_42]

Does this prove that -epsilon/20 isn’t acceptable?

You won. I would need the derivative, which basically is Bernoulli as well as the recursion start of Taylor. But one derivative is sufficient.

 

[Frabjous]

You won.

Thanks. Feel free to edit this excursion out so that we do not confuse the OP.

 

[fresh_42]

Thanks. Feel free to edit this excursion out so that we do not confuse the OP.

No. It has value. Bernoulli and Taylor both start with the first derivative, best seen in my favorite Weierstraß formulation:
$$
f(x_0+ \varepsilon )=f(x_0) + \left. \dfrac{df(x)}{dx}\right|_{x=x_0}\cdot \varepsilon + r(\varepsilon )\, \text{ with } \,\dfrac{r(\varepsilon )}{\varepsilon }\stackrel{\varepsilon \to 0}{\longrightarrow }0
$$
This notation of a derivative is especially useful when dealing with hyperreals since it avoids limits.

 

[mathman]

An earlier attempt at this question, I had reached to something similar to part of your solution as follows:
$b=\frac {\epsilon} {5-\sqrt{25+ε}}$

$b=\frac {\epsilon} {5-\sqrt{25(1+\frac{ε}{25})}}$

$b=\frac {\epsilon} {5-5\sqrt{1+\frac{ε}{25}}}$
At this point I was unsure of what could be done to identify the denominator as an infinitesimal. I also do not understand how you arrive at the following in the denominator:
$5-5\sqrt{1+\frac{\epsilon}{25}}\approx -\frac{\epsilon}{50}$

Arith. error in post - corrected b=-10.

 

[mathman]

Simple solution: $5-\sqrt{25+\epsilon}=5-5\sqrt{1+\frac{\epsilon}{25}}\approx -\frac{\epsilon}{10}$. Final answer is $b=-10$

 

[fresh_42]

Simple solution: $5-\sqrt{25+\epsilon}=5-5\sqrt{1+\frac{\epsilon}{25}}\approx -\frac{\epsilon}{10}$. Final answer is $b=-10$

As you can see from the discussion above: the approximation isn't obvious. One needs a series expansion or at least the first derivative.

 

[PeroK]

I find that I have no interest in learning an epsilon delta method for calculus. It is my understanding that unless I am going into real analysis, omitting this is okay.

Yes, I wouldn't recommend going into the foundations of calculus at this stage. The reference I use for all things calculus is here:

https://tutorial.math.lamar.edu/

He only touches on the epsilon-delta now and again when he needs to. E.g.

https://tutorial.math.lamar.edu/classes/calcI/defnoflimit.aspx

 

[mathman]

As you can see from the discussion above: the approximation isn't obvious. One needs a series expansion or at least the first derivative.

Binomial expansion for (1+x)^(1/2) will do - no calculus.

 

[Frabjous]

Binomial expansion for (1+x)^(1/2) will do - no calculus.

In what class does one learn this?

 

[Office_Shredder]

I think the easiest way to get the infinitesimal part is to square is.

$$5-\sqrt{25+\epsilon}=b\epsilon $$
(Omitting the stuff smaller than the standard part of the infinitesimal)

$5-b\epsilon =\sqrt{25+\epsilon}$
$25-10b\epsilon+b^2\epsilon^2 =25+\epsilon$
Which gives $b=-1/10$

With a little more care you can make this fully rigorous (e.g. how do you know I didn't drop something bigger than $\epsilon$ on the right at the start)

 

[mathman]

In what class does one learn this?

I have no idea.

 

[fresh_42]

The binomial series (non-integer powers) are in line with our suggestions, Taylor series, or first derivative. It is calculus, not just a formula.

The most elementary approach is probably @Office_Shredder's in post #25.