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	<title>Comments on: How can one do calculus with (nilpotent) infinitesimals?: An Introduction to Smooth Infinitesimal Analysis</title>
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	<link>http://xorshammer.com/2008/08/11/smooth-infinitesimal-analysis/</link>
	<description>Some things in mathematical logic that I find interesting</description>
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		<title>By: Todd Olson</title>
		<link>http://xorshammer.com/2008/08/11/smooth-infinitesimal-analysis/#comment-262</link>
		<dc:creator>Todd Olson</dc:creator>
		<pubDate>Fri, 05 Mar 2010 02:36:50 +0000</pubDate>
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		<description>Have you seen Keisler text book
Elementary Calculus: An Infinitesimal Approach
1976, 1986 available on line free at

http://www.math.wisc.edu/~keisler/calc.html</description>
		<content:encoded><![CDATA[<p>Have you seen Keisler text book<br />
Elementary Calculus: An Infinitesimal Approach<br />
1976, 1986 available on line free at</p>
<p><a href="http://www.math.wisc.edu/~keisler/calc.html" rel="nofollow">http://www.math.wisc.edu/~keisler/calc.html</a></p>
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		<title>By: Todd Trimble</title>
		<link>http://xorshammer.com/2008/08/11/smooth-infinitesimal-analysis/#comment-160</link>
		<dc:creator>Todd Trimble</dc:creator>
		<pubDate>Wed, 01 Apr 2009 15:43:11 +0000</pubDate>
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		<description>Robert, I wouldn&#039;t call myself an expert, but yes: the fact that smooth infinitesimal analysis is modeled in a &lt;i&gt;topos&lt;/i&gt; means &quot;smooth spaces&quot; include function spaces, where the calculus of variations naturally lives. For instance, one has a smooth space of smooth paths between two points of a manifold, and it is an easy proposition that tangent vectors in that smooth space are equivalent to vector fields along a chosen path. One can go on to perform analysis on smooth functionals on such function spaces within this setting in a very intuitive fashion.</description>
		<content:encoded><![CDATA[<p>Robert, I wouldn&#8217;t call myself an expert, but yes: the fact that smooth infinitesimal analysis is modeled in a <i>topos</i> means &#8220;smooth spaces&#8221; include function spaces, where the calculus of variations naturally lives. For instance, one has a smooth space of smooth paths between two points of a manifold, and it is an easy proposition that tangent vectors in that smooth space are equivalent to vector fields along a chosen path. One can go on to perform analysis on smooth functionals on such function spaces within this setting in a very intuitive fashion.</p>
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		<title>By: Robert</title>
		<link>http://xorshammer.com/2008/08/11/smooth-infinitesimal-analysis/#comment-154</link>
		<dc:creator>Robert</dc:creator>
		<pubDate>Thu, 12 Mar 2009 03:12:39 +0000</pubDate>
		<guid isPermaLink="false">http://xorshammer.wordpress.com/?p=3#comment-154</guid>
		<description>Can these notions be extended to the calculus of variations?</description>
		<content:encoded><![CDATA[<p>Can these notions be extended to the calculus of variations?</p>
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		<title>By: mkoconnor</title>
		<link>http://xorshammer.com/2008/08/11/smooth-infinitesimal-analysis/#comment-68</link>
		<dc:creator>mkoconnor</dc:creator>
		<pubDate>Mon, 15 Sep 2008 02:05:11 +0000</pubDate>
		<guid isPermaLink="false">http://xorshammer.wordpress.com/?p=3#comment-68</guid>
		<description>I definitely agree that the proofs in smooth infinitesimal analysis are much nicer than in classical calculus; that&#039;s the main reason why I like it.

You define the ``area under a curve&#039;&#039; function in a similar way to arclength as follows: You show that, given $latex f$, there is a unique function $latex g(a,b)$ (to be interpreted as the area under the curve $latex y = f(x)$ from $latex x = a$ to $latex x = b$ such that: for all $latex a,b,c$, $latex g(a,b) + g(b,c) = g(a,c)$ and, if $latex f$ is linear on $latex \lbrack a,b\rbrack$, then $latex g(a,b)$ is the appropriate value (i.e., the formula for the area of a trapezoid: $latex (f(a) + f(b)/2)\cdot (b-a)$.  You can then prove that there is a unique function $latex g(a,b)$ with these properties, and it&#039;s $latex \int_a^b f(t)\,dt$.

The way area is dealt with in smooth infinitesimal analysis (i.e., isolate some conditions that an area function must have, then prove there is a unique function satisfying those conditions) is not so different from what is done in the classical case.  The problem is that it&#039;s done over again for each problem type: that is, you use it once to determine the area of a cone, then once again to prove the fundamental theorem of calculus, etc. 

If I understand your question right, you&#039;re wondering if you can do it once and for all.  That is, can you prove that there is a unique function assigning to subsets of, say, $latex R^2$ an element of $latex R$ satisfying the appropriate conditions.  I don&#039;t know, but I would guess that the answer is &quot;no.&quot;  I would have to think some more about that (and probably learn some more first!).  If you have any thoughts, let me know.</description>
		<content:encoded><![CDATA[<p>I definitely agree that the proofs in smooth infinitesimal analysis are much nicer than in classical calculus; that&#8217;s the main reason why I like it.</p>
<p>You define the &#8220;area under a curve&#8221; function in a similar way to arclength as follows: You show that, given <img src='http://l.wordpress.com/latex.php?latex=f&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='f' title='f' class='latex' />, there is a unique function <img src='http://l.wordpress.com/latex.php?latex=g%28a%2Cb%29&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='g(a,b)' title='g(a,b)' class='latex' /> (to be interpreted as the area under the curve <img src='http://l.wordpress.com/latex.php?latex=y+%3D+f%28x%29&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='y = f(x)' title='y = f(x)' class='latex' /> from <img src='http://l.wordpress.com/latex.php?latex=x+%3D+a&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='x = a' title='x = a' class='latex' /> to <img src='http://l.wordpress.com/latex.php?latex=x+%3D+b&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='x = b' title='x = b' class='latex' /> such that: for all <img src='http://l.wordpress.com/latex.php?latex=a%2Cb%2Cc&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='a,b,c' title='a,b,c' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=g%28a%2Cb%29+%2B+g%28b%2Cc%29+%3D+g%28a%2Cc%29&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='g(a,b) + g(b,c) = g(a,c)' title='g(a,b) + g(b,c) = g(a,c)' class='latex' /> and, if <img src='http://l.wordpress.com/latex.php?latex=f&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='f' title='f' class='latex' /> is linear on <img src='http://l.wordpress.com/latex.php?latex=%5Clbrack+a%2Cb%5Crbrack&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='\lbrack a,b\rbrack' title='\lbrack a,b\rbrack' class='latex' />, then <img src='http://l.wordpress.com/latex.php?latex=g%28a%2Cb%29&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='g(a,b)' title='g(a,b)' class='latex' /> is the appropriate value (i.e., the formula for the area of a trapezoid: <img src='http://l.wordpress.com/latex.php?latex=%28f%28a%29+%2B+f%28b%29%2F2%29%5Ccdot+%28b-a%29&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='(f(a) + f(b)/2)\cdot (b-a)' title='(f(a) + f(b)/2)\cdot (b-a)' class='latex' />.  You can then prove that there is a unique function <img src='http://l.wordpress.com/latex.php?latex=g%28a%2Cb%29&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='g(a,b)' title='g(a,b)' class='latex' /> with these properties, and it&#8217;s <img src='http://l.wordpress.com/latex.php?latex=%5Cint_a%5Eb+f%28t%29%5C%2Cdt&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='\int_a^b f(t)\,dt' title='\int_a^b f(t)\,dt' class='latex' />.</p>
<p>The way area is dealt with in smooth infinitesimal analysis (i.e., isolate some conditions that an area function must have, then prove there is a unique function satisfying those conditions) is not so different from what is done in the classical case.  The problem is that it&#8217;s done over again for each problem type: that is, you use it once to determine the area of a cone, then once again to prove the fundamental theorem of calculus, etc. </p>
<p>If I understand your question right, you&#8217;re wondering if you can do it once and for all.  That is, can you prove that there is a unique function assigning to subsets of, say, <img src='http://l.wordpress.com/latex.php?latex=R%5E2&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='R^2' title='R^2' class='latex' /> an element of <img src='http://l.wordpress.com/latex.php?latex=R&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='R' title='R' class='latex' /> satisfying the appropriate conditions.  I don&#8217;t know, but I would guess that the answer is &#8220;no.&#8221;  I would have to think some more about that (and probably learn some more first!).  If you have any thoughts, let me know.</p>
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		<title>By: Geoff</title>
		<link>http://xorshammer.com/2008/08/11/smooth-infinitesimal-analysis/#comment-67</link>
		<dc:creator>Geoff</dc:creator>
		<pubDate>Sat, 13 Sep 2008 14:57:28 +0000</pubDate>
		<guid isPermaLink="false">http://xorshammer.wordpress.com/?p=3#comment-67</guid>
		<description>Ah, of course you&#039;re right.  I think I was confused by the notation.  Thanks for the quick reply.

I guess my next question would be: how does one define the &quot;area under a curve&quot; function?  I assume we would like to do it without limits.  

Now that I look over the above again, I notice that there&#039;s something similar with the arc length function: you mention that it&#039;s not formally defined, so instead work with what reasonable properties it should have.   But there has to be some nice way to formally define these notions; perhaps I just need to get Bell&#039;s book.

As a side note, I&#039;m amazed by how much more elegant some of the proofs with smooth infinitesimal analysis are than with regular calculus.  In particular, the proofs of the product and chain rules are so much nicer than their usual counterparts.</description>
		<content:encoded><![CDATA[<p>Ah, of course you&#8217;re right.  I think I was confused by the notation.  Thanks for the quick reply.</p>
<p>I guess my next question would be: how does one define the &#8220;area under a curve&#8221; function?  I assume we would like to do it without limits.  </p>
<p>Now that I look over the above again, I notice that there&#8217;s something similar with the arc length function: you mention that it&#8217;s not formally defined, so instead work with what reasonable properties it should have.   But there has to be some nice way to formally define these notions; perhaps I just need to get Bell&#8217;s book.</p>
<p>As a side note, I&#8217;m amazed by how much more elegant some of the proofs with smooth infinitesimal analysis are than with regular calculus.  In particular, the proofs of the product and chain rules are so much nicer than their usual counterparts.</p>
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		<title>By: mkoconnor</title>
		<link>http://xorshammer.com/2008/08/11/smooth-infinitesimal-analysis/#comment-66</link>
		<dc:creator>mkoconnor</dc:creator>
		<pubDate>Sat, 13 Sep 2008 01:27:26 +0000</pubDate>
		<guid isPermaLink="false">http://xorshammer.wordpress.com/?p=3#comment-66</guid>
		<description>Hi Geoff,
I&#039;m glad you liked the article and find the subject interesting!

Axiom 5 doesn&#039;t really axiomatize the fundamental theorem of calculus, since all Axiom 5 says is that antiderivatives of functions exist.  The fundamental theorem of calculus says something more: it says that not only do antiderivatives of (suitable) functions $latex f(x)$ exist, but that furthermore the function $latex F(x)$ giving the (signed) area under the graph of $latex f$ from $latex {0}$ to $latex x$ is a specific example of an antiderivative.  (Also, that all other antiderivatives are equal to $latex F$ plus some constant.)

The notation is confusing though, because in Axiom 5, I said that antiderivatives of functions exist, and furthermore I denoted the antiderivative of $latex f$  which takes the value $latex {0}$ on input $latex {0}$ by $latex \int_0^x f(t)\,dt$.  This could be confusing, because normally $latex \int_0^x f(t)\,dt$ is defined as a Riemann sum, which is the classical formalization of the ``area&#039;&#039; concept, and so with that notation, the statement that the derivative of $latex \int_0^x f(t)\,dt$ is $latex f(x)$ is indeed the fundamental theorem of calculus.  But with the notation I used in this article, it&#039;s just a definition.

As an aside, you can prove a version of the fundamental theorem of calculus in smooth infinitesimal analysis, in much the same way that you can find the area of a cone.

If you want to not take the existence of antiderivatives as an axiom, you might want to take a look at Bell&#039;s book ``A Primer on Smooth Infinitesimal Analysis,&#039;&#039; which does a lot of stuff without that axiom, but in its place he assumes that if $latex f&#039; = g&#039;$ and $latex f(0) = g(0)$ then $latex f = g$.</description>
		<content:encoded><![CDATA[<p>Hi Geoff,<br />
I&#8217;m glad you liked the article and find the subject interesting!</p>
<p>Axiom 5 doesn&#8217;t really axiomatize the fundamental theorem of calculus, since all Axiom 5 says is that antiderivatives of functions exist.  The fundamental theorem of calculus says something more: it says that not only do antiderivatives of (suitable) functions <img src='http://l.wordpress.com/latex.php?latex=f%28x%29&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='f(x)' title='f(x)' class='latex' /> exist, but that furthermore the function <img src='http://l.wordpress.com/latex.php?latex=F%28x%29&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='F(x)' title='F(x)' class='latex' /> giving the (signed) area under the graph of <img src='http://l.wordpress.com/latex.php?latex=f&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='f' title='f' class='latex' /> from <img src='http://l.wordpress.com/latex.php?latex=%7B0%7D&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='{0}' title='{0}' class='latex' /> to <img src='http://l.wordpress.com/latex.php?latex=x&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='x' title='x' class='latex' /> is a specific example of an antiderivative.  (Also, that all other antiderivatives are equal to <img src='http://l.wordpress.com/latex.php?latex=F&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='F' title='F' class='latex' /> plus some constant.)</p>
<p>The notation is confusing though, because in Axiom 5, I said that antiderivatives of functions exist, and furthermore I denoted the antiderivative of <img src='http://l.wordpress.com/latex.php?latex=f&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='f' title='f' class='latex' />  which takes the value <img src='http://l.wordpress.com/latex.php?latex=%7B0%7D&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='{0}' title='{0}' class='latex' /> on input <img src='http://l.wordpress.com/latex.php?latex=%7B0%7D&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='{0}' title='{0}' class='latex' /> by <img src='http://l.wordpress.com/latex.php?latex=%5Cint_0%5Ex+f%28t%29%5C%2Cdt&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='\int_0^x f(t)\,dt' title='\int_0^x f(t)\,dt' class='latex' />.  This could be confusing, because normally <img src='http://l.wordpress.com/latex.php?latex=%5Cint_0%5Ex+f%28t%29%5C%2Cdt&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='\int_0^x f(t)\,dt' title='\int_0^x f(t)\,dt' class='latex' /> is defined as a Riemann sum, which is the classical formalization of the &#8220;area&#8221; concept, and so with that notation, the statement that the derivative of <img src='http://l.wordpress.com/latex.php?latex=%5Cint_0%5Ex+f%28t%29%5C%2Cdt&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='\int_0^x f(t)\,dt' title='\int_0^x f(t)\,dt' class='latex' /> is <img src='http://l.wordpress.com/latex.php?latex=f%28x%29&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='f(x)' title='f(x)' class='latex' /> is indeed the fundamental theorem of calculus.  But with the notation I used in this article, it&#8217;s just a definition.</p>
<p>As an aside, you can prove a version of the fundamental theorem of calculus in smooth infinitesimal analysis, in much the same way that you can find the area of a cone.</p>
<p>If you want to not take the existence of antiderivatives as an axiom, you might want to take a look at Bell&#8217;s book &#8220;A Primer on Smooth Infinitesimal Analysis,&#8221; which does a lot of stuff without that axiom, but in its place he assumes that if <img src='http://l.wordpress.com/latex.php?latex=f%27+%3D+g%27&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='f&#039; = g&#039;' title='f&#039; = g&#039;' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=f%280%29+%3D+g%280%29&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='f(0) = g(0)' title='f(0) = g(0)' class='latex' /> then <img src='http://l.wordpress.com/latex.php?latex=f+%3D+g&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='f = g' title='f = g' class='latex' />.</p>
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		<title>By: Geoff</title>
		<link>http://xorshammer.com/2008/08/11/smooth-infinitesimal-analysis/#comment-65</link>
		<dc:creator>Geoff</dc:creator>
		<pubDate>Fri, 12 Sep 2008 15:20:14 +0000</pubDate>
		<guid isPermaLink="false">http://xorshammer.wordpress.com/?p=3#comment-65</guid>
		<description>I recently became very interested in smooth infinitesimal analysis, and found your article a useful guide when trying to understand some of the earlier works (Kock, Reyes-Moerdijk, etc.).  

However, there is something that bothers me about this setup, and I wonder if you have any ideas on it.  Specifically, the problem is Axiom 5.  This axiomatizes the fundamental theorem of calculus!  Is there no way to define an area function using infinitesimals, then show that its derivative is the original function?

I think the idea of taking as an axiom &quot;every curve is made up of infinitesimally small line segments&quot; a beautiful concept, but it seems to me to be giving up too much to also have to axiomatize the existence of an anti-derivative.     

Do you know of any way to not have to take this as an axiom?</description>
		<content:encoded><![CDATA[<p>I recently became very interested in smooth infinitesimal analysis, and found your article a useful guide when trying to understand some of the earlier works (Kock, Reyes-Moerdijk, etc.).  </p>
<p>However, there is something that bothers me about this setup, and I wonder if you have any ideas on it.  Specifically, the problem is Axiom 5.  This axiomatizes the fundamental theorem of calculus!  Is there no way to define an area function using infinitesimals, then show that its derivative is the original function?</p>
<p>I think the idea of taking as an axiom &#8220;every curve is made up of infinitesimally small line segments&#8221; a beautiful concept, but it seems to me to be giving up too much to also have to axiomatize the existence of an anti-derivative.     </p>
<p>Do you know of any way to not have to take this as an axiom?</p>
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		<title>By: Mommo</title>
		<link>http://xorshammer.com/2008/08/11/smooth-infinitesimal-analysis/#comment-2</link>
		<dc:creator>Mommo</dc:creator>
		<pubDate>Mon, 11 Aug 2008 06:19:19 +0000</pubDate>
		<guid isPermaLink="false">http://xorshammer.wordpress.com/?p=3#comment-2</guid>
		<description>I don&#039;t understand a word/character/symbol but it looks great!</description>
		<content:encoded><![CDATA[<p>I don&#8217;t understand a word/character/symbol but it looks great!</p>
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