iScience
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|1-0.[itex]\bar{9}[/itex]|=dx?
Is Understanding Differential Notation Necessary for Learning Advanced Math?
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No. First of all ##0.\bar{9} = 1##. Secondly, as you were told in the other thread, ##dx## is not a real number.
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No. We have that ##1=0.999...##. So ##|1-0.999...|=0##. On the other hand, ##dx## is not a real number.
Gold Member
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See https://www.physicsforums.com/showthread.php?t=507002 for a good explanation of why ##0.999...=1##
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iScience
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WannabeNewton said:
how does 0.999...=1?... ...how?...
just to be sure, I am not rounding..
that's like saying i have a piece wise function.. f(x)={2x , 0<x<5}
{3 , x=5 }
{2x , 5<x<∞}
the statement 4.999999...=5 is NOT even saying as x approaches 5... no. This is like saying at 4.999999... wer're already AT 5.
how does this make sense??
and if 4.99999... = 5 , then this must also be true 5.00000000...00001 = 5.
okay so.. now we have three "fives" on the number line... let's work our way outward shall we
4.999...[itex]\equiv[/itex]5 so now.. 4.999...9998 is also 5.. and 4.999...9997 is also 5...
5-[itex]\Sigma[/itex]dxi=5 no matter what the value of i is.. even as it approaches ∞ this is still five.
and now we have an infinite number of fives on the number line. the implications do not make sense to me.
if i am wrong with my analysis, i assume i am wrong from the point when i start to work my way outward after the point where 4.999... is said to be equal to five..
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HS-Scientist said:
That is actually the second part of the FAQ, so be sure to read the first part as well: https://www.physicsforums.com/showthread.php?t=507001
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Gold Member
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You really have to stop treating ##dx## like it's some kind of number. It's fine for hand wavy arguments in subpar physics books but not for what you're interested in here.
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If you want to be rigorous about it, then for any function ##f:\mathbb{R}\rightarrow \mathbb{R}##, you can define
[tex]df:\mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}: (p,h)\rightarrow f^\prime(p) h[/tex]
In particular, we have that
[tex]dx:\mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}: (p,h)\rightarrow h[/tex]
The main reason for this definition is to make the following true:
[tex]df = \frac{df}{dx} dx[/tex]
If you want to do physics, then infinitesimals come in very handy. Those infinitesimals are made rigorous by the hyperreal and surreal number system. But they are not denoted as ##dx##.
In short: you need to stop treating ##dx## as a number, or even worse: as a real number. It isn't, and it's very confusing.
[tex]df:\mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}: (p,h)\rightarrow f^\prime(p) h[/tex]
In particular, we have that
[tex]dx:\mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}: (p,h)\rightarrow h[/tex]
The main reason for this definition is to make the following true:
[tex]df = \frac{df}{dx} dx[/tex]
If you want to do physics, then infinitesimals come in very handy. Those infinitesimals are made rigorous by the hyperreal and surreal number system. But they are not denoted as ##dx##.
In short: you need to stop treating ##dx## as a number, or even worse: as a real number. It isn't, and it's very confusing.
mathsciguy
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About 0.999...=1, is representing it as a geometric infinite series a valid proof?
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Yes, it is probably the simplest strictly valid proof.
0.9999...= .9+ .09+ .009+ .0009+ ...= .9(1)+ .9(.1)+ .9(.01)+ .9(.001)+ ...= .9(1)+ .9(.1)+ .9(.1^2)+ .9(.1^3)+ ...
That is a greometric series of the form [itex]\sum ar^n[/itex] with a= .9, r= .1. A geometric series with |r|< 1 (which is the case here: |r|= .1< 1) converges to a/(1- r).
Here, a/(1- r)= .9/(1- .1)= .9/.9= 1.
0.9999...= .9+ .09+ .009+ .0009+ ...= .9(1)+ .9(.1)+ .9(.01)+ .9(.001)+ ...= .9(1)+ .9(.1)+ .9(.1^2)+ .9(.1^3)+ ...
That is a greometric series of the form [itex]\sum ar^n[/itex] with a= .9, r= .1. A geometric series with |r|< 1 (which is the case here: |r|= .1< 1) converges to a/(1- r).
Here, a/(1- r)= .9/(1- .1)= .9/.9= 1.
iScience
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guys what level of math are you all in? ie what's the highest math courses you've taken? i just finished diff eq and can't wait to get to where you guys are at.. things are so confusing for me right now :'(
Gold Member
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Don't worry, it will all come in due time. Most of the people answering you are doing their PhDs/masters or already have their PhDs so don't fret, you'll learn it all in sooner or later.iScience said:
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Yes, at PhD level, but the "geometric series" proof that 0.999...= 1 is one that I learned in PreCalculus.
Mandelbroth
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I think that ##|1-0.\bar{9}|=0##, which admittedly IS close to 0 if you want to think about it that way, seeing as the distance between the two is 0. 
For now, it's best to treat ##dx## and ##df## as variable holders for notation. Later, in differential geometry, you'll be introduced to the concept of the "exterior derivative," which extends the idea of the differential to a meaningful, mathematically rigorous context. For now, though, just treat it as notation.
It's not about your level of education in school. It's about how much you study by yourself. You can get a lot further by self study than you can by a formal college education, in my opinion.
For now, it's best to treat ##dx## and ##df## as variable holders for notation. Later, in differential geometry, you'll be introduced to the concept of the "exterior derivative," which extends the idea of the differential to a meaningful, mathematically rigorous context. For now, though, just treat it as notation.
Whereas you and micro seem to do better than most of us without a degree.WannabeNewton said:
It's not about your level of education in school. It's about how much you study by yourself. You can get a lot further by self study than you can by a formal college education, in my opinion.