Can I prove the finiteness of an expression involving hyperreals?

[Buffu]

I need to prove whether the given expression is finite but not infinitesimal/infinite/infinitesimal.$$H-K\over H^2 + K^2$$ Where $H, K$ are +ve infinite numbers.

I did the following,

Let $H = \alpha K, \alpha \in R$
then,
$${\alpha K-K\over \alpha^2K^2 + K^2} \implies {\alpha - 1\over K(\alpha^2 + 1)}$$
Since ${\alpha - 1\over (\alpha^2 + 1)}$ is finite and $K$, infinite. Thus the given expression is infinitesimal.

I am not sure if this is correct.
Am I correct ?

 

[fresh_42]

It all depends on the assumption $H=\alpha K$ with $\alpha \in R$.
So a) does it mean $\alpha \in \mathbb{R}$ or $\alpha \in {}^*\mathbb{R}$ and b) this is meant by "+ve infinite"?

I'm not used to hyperreals, so maybe the question is silly. But I try to rule out cases like $H=K^L$ or $H=\varepsilon + K$ in which it becomes intricate. If you may assume $H= \text{ real }\cdot K$, then your argument seems to be correct.

 

[Buffu]

a) does it mean α∈Rα∈R\alpha \in \mathbb{R} or α∈∗R

I meant $\alpha \in \mathbb{R}$ ie set of reals.

"+ve infinite"?

If I am right then for a infinitesimal $\varepsilon$, there are two infinities $1/\varepsilon$ and $-1/\varepsilon$.
Former being larger than every real and latter being smaller than every real.

I'm not used to hyperreals, so maybe the question is silly. But I try to rule out cases like H=KLH=KLH=K^L or H=ε+KH=ε+KH=\varepsilon + K in which it becomes intricate. If you may assume H= real ⋅KH= real ⋅KH= \text{ real }\cdot K, then your argument seems to be correct.

My main problem is that I don't know if my method proves that the given expression is infinitesimal for all the possible relations between $H,K$, some which you mentioned,$H=K^L, H=\varepsilon + K...$

Do you know any other method to prove this ?

 

[fresh_42]

As far as I can see, the infinite hyperreals are basically defined by inverse infinitesimals.
So I would write $\varepsilon = \frac{1}{K}$ and $\eta = \frac{1}{H}$.
This leads to an expression $(\varepsilon - \eta)\frac{\varepsilon \cdot \eta}{\varepsilon^2 + \eta^2}$ which boils it down to the difference of two infinitesimals around zero: $\varepsilon - \eta$. Maybe one can formally work with $st$ from here on. It is in principle the same argument as yours, I simply found more sources dealing with infinitesimals than with infinites.

 

[Buffu]

This leads to an expression (ε−η)ε⋅ηε2+η2(ε−η)ε⋅ηε2+η2(\varepsilon - \eta)\frac{\varepsilon \cdot \eta}{\varepsilon^2 + \eta^2}

Sum/product of two infinitesimal is infinitesimal. But the quotient is indeterminate form.

So, $(\varepsilon - \eta)\frac{\varepsilon \cdot \eta}{\varepsilon^2 + \eta^2}$ =infinitesimal(infinitesimal/infinitesimal) = indeterminate.
How should I proceed now ?

which boils it down to the difference of two infinitesimals around zero: ε−ηε−η\varepsilon - \eta.

I don't get this part ??

 

[fresh_42]

I'm not sure, whether the above is a valid transformation or not and you should go on. $H+K$ or $\frac{\varepsilon}{\eta}$ are already indeterminate, i.e. the question itself has indeterminates. My thought has been, that $\frac{\varepsilon \eta}{\varepsilon^2 + \eta^2}=1$ (I think) and $\varepsilon - \eta$ is left. Then I would have proceeded like this:
$$\varepsilon - \eta = \varepsilon + (-\eta) = \varepsilon + \delta \text{ all infinitesimal }$$

 

[Buffu]

My thought has been, that εηε2+η2=1εηε2+η2=1\frac{\varepsilon \eta}{\varepsilon^2 + \eta^2}=1

$\frac{\varepsilon \eta}{\varepsilon^2 + \eta^2}=1$
Why do you think that ? I want to know. Most likely you are right, I don't know much non-standard analysis.

the question itself has indeterminates.

Most questions in the exercise were in indeterminate form, we need to transform them to determinate form using algebra, and then find there nature.

 

[Buffu]

@haruspex What do you think ?

 

[haruspex]

@haruspex What do you think ?

Not an area I've ever indulged in before, but from a bit of reading...
Seems to me we have to allow for H and K (or ε and η) being of different orders. So assuming a ratio which is real is not valid.
The sum of squares term will necessarily be dominated by the higher infinite order (lower infinitesimal order) if they are different. So $\frac{\epsilon\eta}{\epsilon^2+\eta^2}$ is either real or infinitesimal. it cannot be infinite. You can arrive at the same using H and K.

 

[mfb]

How can it be real? If they are of the same order, we have the right approach above. If one is of higher order, it should dominate, so we divide (lower order)/(higher order) - the result should still be infinitesimal.

 

[haruspex]

How can it be real? If they are of the same order, we have the right approach above. If one is of higher order, it should dominate, so we divide (lower order)/(higher order) - the result should still be infinitesimal.

Yes, my fault, I was just looking at the $\frac{\eta\epsilon}{\eta^2+\epsilon^2}$ term being discussed. Now I look back at the oroginal post I agree with you - must be infinitesimal.

 

[Buffu]

I will give a try to complete solution :-

Case 1: - $H > K$

$${H - K \over H^2 + K^2 } = {1/H - K/H^2 \over 1 + K^2/H^2 } = {\text{infinitesimal} - \text{infinitesimal} \over \text{finite} +\text{infinitesimal} } = {\text{infinitesimal} \over \text{finite}} = \text{infinitesimal}$$

Case 2:- $K > H$
$${H - K \over H^2 + K^2 } = -\left( { K - H \over H^2 + K^2 }\right)$$
Then we proceed as above to get negative infinitesimal.

Case 3:- $H = K$
$${H - K \over H^2 + K^2 } = 0$$
Which is a real infinitesimal.

@haruspex , @mfb , @fresh_42

Is this correct ?

 

[haruspex]

I will give a try to complete solution :-

Case 1: - $H > K$

$${H - K \over H^2 + K^2 } = {1/H - K/H^2 \over 1 + K^2/H^2 } = {\text{infinitesimal} - \text{infinitesimal} \over \text{finite} +\text{infinitesimal} } = {\text{infinitesimal} \over \text{finite}} = \text{infinitesimal}$$

Case 2:- $K > H$
$${H - K \over H^2 + K^2 } = -\left( { K - H \over H^2 + K^2 }\right)$$
Then we proceed as above to get negative infinitesimal.

Case 3:- $H = K$
$${H - K \over H^2 + K^2 } = 0$$
Which is a real infinitesimal.

@haruspex , @mfb , @fresh_42

Is this correct ?

Your argument does not work in case 3. H-K need not be zero. It could as much as the same infinity as H and K.
Use a method similar to the other cases.

 

[mfb]

H=K could be meant literally. But then K²/H² is "infinitesimal or finite".

 

[Buffu]

H=K could be meant literally. But then K²/H² is "infinitesimal or finite".

Finite but not infinitesimal, because then $K^2/H^2 = 1$, am I correct ?

 

[Buffu]

Your argument does not work in case 3. H-K need not be zero. It could as much as the same infinity as H and K.
Use a method similar to the other cases.

Why ? $H - K = H - H = 0$.

 

[mfb]

Finite but not infinitesimal, because then $K^2/H^2 = 1$, am I correct ?

That's not what I meant, but my post was a bit short:
If K=H means literally identical in case 3, then H>K includes every case where H>K, including (for example) H=w+1, K=w, in this case K²/H² is finite.

From H>K, you cannot conclude that K²/H² is infinitesimal. It could be, but you don't know, it could also be finite.

 

[Buffu]

That's not what I meant, but my post was a bit short:
If K=H means literally identical in case 3, then H>K includes every case where H>K, including (for example) H=w+1, K=w, in this case K²/H² is finite.

From H>K, you cannot conclude that K²/H² is infinitesimal. It could be, but you don't know, it could also be finite.

$H = w + 1, K = w$
$w = 1/\varepsilon$

$${H\over K} = { w + 1\over w } = {{ 1 + \varepsilon \over \varepsilon } \over 1/\varepsilon} = { 1 + \varepsilon} = \text{Finite}$$

yes your point is correct.
So how should I solve the question now.

 

[mfb]

The way you did, just replace "infinitesimal" by "finite or infinitesimal" at the place we discussed.

 

[Buffu]

The way you did, just replace "infinitesimal" by "finite or infinitesimal" at the place we discussed.

I did not get what you said , Should I replace "Infinitesimal" everywhere with "Infinitesimal or finite".

 

[mfb]

Not everywhere. Just for the ratio K^2/H^2.

 

[Buffu]

Not everywhere. Just for the ratio K^2/H^2.

Ok that worked.
We still get infinitesimal as the answer.
Thanks for the help. @mfb , @haruspex , @fresh_42.

 

[haruspex]

Why ? $H - K = H - H = 0$.

I think mfb has already clarified this, but just to confirm...
Because of the way your argument ran in the other cases (e,g, K²/H² infinitesimal) I had interpreted H>K etc. as referring to orders of infinity instead of taking it literally. Hence I did not take the = literally.

 

[alan2]

I know this is old but:

$$\frac{{H - K}}{{{H^2} + {K^2}}} = \frac{H}{{{H^2} + {K^2}}} - \frac{K}{{{H^2} + {K^2}}} = \varepsilon - \delta$$

which is infinitesimal. This is a homework problem from Keisler.