Non-standard calculus (infinitesimals)

[danne89]

Compute the standard part of this, please:
$$\frac{ \sqrt{H+1}}{ \sqrt{2H} + \sqrt{H-1}}$$, where H is positive infinite.

It probably should be some algebra trick I'm not familar with.

 

[vincentchan]

infinite can't divided by infinite... your question doesn't make sense at all
I think what you meant was...
$$\lim_{H \rightarrow \infty} \frac{\sqrt{H+1}}{\sqrt{2H}+\sqrt{H-1}}$$

hints:
for a very large $H$, you can assume $\sqrt{H+1}= \sqrt{H}$

 

[HallsofIvy]

No, vincent, he meant what he said. As the title said, this is "non-standard analysis" in which we have both infinite numbers and infinitesmals. (H is a positive infinite number.)

However, the result will be exactly the same as the lim as H-> infinity.

And, yes, it is true that for "infinite" H, H+1= H, H-1= H. This is exactly the same as
$$\frac{\sqrt{H}}{\sqrt{2H}+\sqrt{H}}= \frac{\sqrt{H}}{(\sqrt{2}+1)\sqrt{H}}= \frac{1}{\sqrt{2}+1}$$

 

[Hurkyl]

One algebra trick, it so happens, is exactly the same as standard analysis: divide the numerator and denominator by the "highest power" of H... in this case, it's 1/2.

And, yes, it is true that for "infinite" H, H+1= H, H-1= H.

That's incorrect: H+1 is never equal to H for any hyperreal, even the unlimited ones. (I think unlimited is the preferred term, over infinite or transfinite)

The unlimited hyperreals don't act like cardinal numbers -- they act like real numbers, in a very real sense. (Pun intended)


However, H/(H+1) would be a limited (aka finite) number with standard part 1, so that could be used fruitfully in this example, by replacing (H+1) with H * ((H+1) / H).

 

[danne89]

Ahh I see! It's really just a beginner calculus book, although it used infinitesimals.

 

[HallsofIvy]

Ah! Thanks for clearing up my error, Hurkyl. An expert on non-standard analysis I'm not. (I'm barely competent on STANDARD analysis!)