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• CommentRowNumber1.
• CommentAuthorMike Shulman
• CommentTimeSep 13th 2013

In planning how to introduce derivatives to my calculus students this semester, I found myself going down the following interesting path. I wonder whether anyone has been down this path before? Is it a well-known path in disguise?

First recall that the category of fields has no initial object, and more generally its forgetful functor to sets has no left adjoint: free fields do not exist. If we restrict to the category of extensions of some fixed field $K$, such as $\mathbb{F}_p$ or $\mathbb{Q}$ or $\mathbb{R}$, then the field $K$ is the initial object, but free objects still do not exist.

However, the field of rational functions $K(x)$ seems to be “as close as you can come” to the free extension-of-$K$ on one generator. Specifically, for any extension $L$ of $K$ and any element $a\in L$, there is a unique largest partial morphism $\phi_a:K(x)\to L$ sending $x$ to $a$. To see this, note that there is a unique ring homomorphism $\phi_a:K[x] \to L$ sending $x$ to $a$. Let $P\subseteq K[x]$ be the kernel of $\phi_a$. Since $L$ is a field, and in particular an integral domain, $P$ is a prime ideal. Moreover, if $f\in K[x]\setminus P$, then $\phi_a(f)$ is invertible in $L$. Therefore, $\phi_a$ extends uniquely to the localization of $K[x]$ at $P$ (i.e. the result of formally inverting the complement of $P$ in $K[x]$), which is the subring of $K(x)$ consisting of all rational functions whose denominators do not vanish at $x=a$.

Now let $f\in K(x)$ be a rational function; we can compute its derivative in the following “purely algebraic” way. There is an obvious inclusion $\alpha : K(x) \hookrightarrow K(x,dx)$, and another inclusion $\beta: K(x) \to K(x,dx)$ specified by $\beta(x) = x+dx$. (These are also special cases of the construction above, where $L=K(x,dx)$ and $a$ is $x$ or $x+dx$ respectively; in both cases $P=0$ so the partial map is in fact total.)

Now consider the element $\frac{\beta(f) - \alpha(f)}{dx} \in K(x,dx)$, which we can regard as a formal way to compute the difference quotient $\frac{f(x+dx) - f(x)}{dx}$. Call this element $\Delta f$. Now we have our largest partial map $\phi_0 : K(x,dx) \to K(x)$ such that $\phi_0(x) = x$ and $\phi_0(dx) = 0$. If $\Delta f$ is in the domain of $\phi_0$, then $f$ is differentiable, and its derivative is $\phi_0(\Delta f) \in K(x)$.

Once I’d thought of this, it reminded me of the following definition of the derivative. Given $a$, consider the difference quotient $\frac{f(a+dx) - f(a)}{dx}$ as a function of $dx$. Obviously this is defined for $dx\neq 0$. Thus, there is at most one continuous function $h$ of $dx$ defined everywhere and such that $h(dx) = \frac{f(a+dx) - f(a)}{dx}$ for $dx\neq 0$. If such a function exists, then $f$ is differentiable at $a$, and its derivative is $h(0)$. This definition is equivalent to the usual one — it uses limits in the guise of continuity — but it feels very similar to the algebraic definition above. Somehow the algebraic definition has replaced the geometric notion of “continuity” with some kind of “algebraicity”. Which makes me wonder whether I am unknowingly doing algebraic geometry?

What I would really like to know is whether there is some way to extend the “algebraic” definition to compute derivatives of non-algebraic functions as well, like exponentials, logarithms, and trigonometric functions. My first guess was be to replace the theory of fields by some larger theory, like the theory of exponential fields, but I haven’t been able to progress from there.

• CommentRowNumber2.
• CommentAuthorZhen Lin
• CommentTimeSep 13th 2013

Interesting. A more common approach is to introduce the ring of dual numbers, $K [\epsilon] / (\epsilon^2)$, and observe that $f (x + y \epsilon) = f (x) + f' (y) \epsilon$ for all polynomials $f$. And, of course, there’s all that business with Kähler differentials.

For transcendental functions, John Baez gave a nice argument here for the fact that the module of (algebraic!) $\mathbb{R}$-derivations $C^{\infty} (\mathbb{R}) \to C^{\infty} (\mathbb{R})$ is precisely the space of vector fields on $\mathbb{R}$, and the corollary that the module of smooth 1-forms is the bidual of the module of Kähler differentials. It’s a nice way of getting around the fact that algebraic derivations can misbehave on general power series, but it does use a fact from real analysis, which might make it unsuitable for a first calculus course. Based on this, I get the feeling that it might be necessary to go all the way to $C^{\infty}$-algebras to get the right answer in general…

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeSep 13th 2013
• (edited Sep 13th 2013)

Back when I studied physics, this is how the experimentalist professors would differentiate in their lectures: let $\epsilon$ be a small change in the system, so small that $\epsilon^2 = 0$, then it follows that…

That experimental physicists like this shows that this is something that students may tend to like.

One could probably argue that experiemtal physicists invented this “synthetic differential geometry” long ago, at a vague level, and that it took mathematicians an aweful long time to agree that it makes good sense. It seems to be one instance of this Zen koan about the journey of understanding stuff:

When you do not understand, then a tree is just a treee.

When you are beginning to understand, a tree is no longer a tree.

After you have understood, a tree is again just a tree.

Well, this expresses only that one gets back to where one started. The Zen-piece about it is the implicit message: there is nontrivial monodromy along the way.

• CommentRowNumber4.
• CommentAuthorMike Shulman
• CommentTimeSep 13th 2013

Yes, I’m familiar with the approaches using infinitesimals: nonstandard analysis and synthetic differential geometry. This one seems to me to be different from both of them. In particular, nothing is ever said to be nilsquare, nor does an ordering ever enter the picture.

In case it wasn’t clear, by the way, I wasn’t planning to tell my calculus students about function fields and localizations. (-: I was just thinking about the simplest way to describe the procedure of taking a derivative: write down the difference quotient, simplify it as much as possible, then set $dx=0$. This doesn’t make sense if you insist on regarding $x$ and $dx$ as particular numbers, since we can’t have divided by $dx$ if it’s zero — hence why you have to introduce circumlocutions like limits or infinitesimals (of one flavor or another). But it struck me that it makes perfect sense regarded as a syntactic operation on the algebraic expressions that define a rational function. The above is the result of my thinking about how to express that in language that would be the most familiar to a mathematician. (-:

• CommentRowNumber5.
• CommentAuthorMike Shulman
• CommentTimeSep 13th 2013

Oh, another thought I had for dealing with transcendental functions is to work with the field of formal Laurent series. I expect this would probably work if I wrote it down carefully enough; the only problem is that it seems to require defining transcendental functions as power series, which is against the spirit of an intro calculus class (we want to derive the Taylor series of such functions by taking their derivatives).

• CommentRowNumber6.
• CommentAuthorZhen Lin
• CommentTimeSep 13th 2013

I agree that there is something syntactic about the difference quotient story. Nilsquares are just a particularly simple approach for dealing with derivatives of power series: in the numerator of the difference quotient, terms of order $(d x)^2$ or smaller can’t influence the value of the derivative, so we pre-emptively discard them.

Come to think of it, this partial evaluation technique is probably an example of what Weil-school algebraic geometers mean when they talk about ‘specialisation’. Nowadays, we regard the field $K (x, y)$ as being (the local ring at) the generic point of the affine plane $\mathbb{A}^2_K$. Taking an element $f$ of $K (x, y)$ and “setting $y = 0$” amounts to finding an open subspace $U \subseteq \mathbb{A}^2_K$ on which $f$ is “defined” (read, is in the image of the colimit homomorphism $\mathcal{O} (U) \to K (x, y)$) and then taking its germ at the generic point of the subscheme $\{ y = 0 \} \subseteq \mathbb{A}^2_K$ (if it is in $U$).

• CommentRowNumber7.
• CommentAuthorMike Shulman
• CommentTimeSep 13th 2013

Okay, there’s a bit of algebraic geometry; thanks.

Is there any object which contains all the formal power series rings $K[[x]]$, $K[[x-a]]$, etc. in the same way that the function field $K(x)$ contains all the local rings $K[x]_{(x)}$, $K[x]_{(x-a)}$, etc.? I want to do something like “complete at the generic point” but I can’t make sense of that.

• CommentRowNumber8.
• CommentAuthorTodd_Trimble
• CommentTimeSep 13th 2013
• (edited Sep 13th 2013)

Here is an idea which I think is very similar to Mike’s.

The basic idea is to assume that we have rings (with elements denoted $g$, $h$, etc., meant to suggest functions) such that we can somehow formalize a statement of the following sort: given $g$ in “one variable”, there exists a unique $h$ in two variables such that

$g(y) = g(x) + (y-x)h(x, y)$

and then define the formal derivative of $g$ by the equation $g'(x) = h(x, x)$.

So suppose we have a Lawvere theory $T$ containing the theory of commutative rings. It could be for example commutative rings itself, or it could be exponential rings, or it could be $C^\infty$ rings, etc. We can think of $T$ as given by free algebras, and there will be substitution maps $sub_i: T(n) \times T(m) \to T(n+m-1)$ and an “identity $x$” – formalize the notion of cartesian operad however you like.

Call $T$ differentiable if we have the axiom: for every $g$ of arity $n+1$, there exists a unique $h$ of arity $n+2$ such that

$g(y, x_1, \ldots, x_n) = g(x, x_1, \ldots, x_n) = g(x, x_1, \ldots, x_n) + (y-x)h(x, y, x_1, \ldots, x_n)$

interpreted in the obvious way. Then $h(x, x, x_1, \ldots, x_n)$ would be a formal partial derivative of $g$ in its first argument.

I imagine that every $T$ over commutative rings admits a “differentiable completion”, so that such completion is an idempotent monad. Of course, for $T = CRing$, the theory is already complete. So is $T = C^\infty Ring$, by Hadamard’s lemma. I don’t think $T$ the theory of exponential rings is itself differentiably complete, but in the completion we have the following formal calculation: define $h$ by $E(y) = E(x) + (y-x)h(x, y)$. Substituting $x = 0$, we have $E(y) = 1 + y h(0, y)$. Therefore $E(y-x) = 1 + (y-x)h(0, y-x)$. Now substitute this into the exponential equation:

$E(y) = E(x)E(y-x) = E(x)(1 + (y-x)h(0, y-x))$

to derive $h(x, y) = E(x)h(0, y-x)$. Thus we have $E'(x) = h(x, x) = E(x)h(0, 0) = E(x)E'(0)$, just as expected.

Maybe this idea is already somewhere in the literature?

• CommentRowNumber9.
• CommentAuthorMike Shulman
• CommentTimeSep 13th 2013

Brilliant, Todd! That’s exactly the sort of thing I was looking for. I don’t know whether that’s in the literature; the closest I can think of is that IIRC “Hadamard’s axiom” in SDG is equivalent to the more usual “Kock-Lawvere axiom”.

To connect this up with my first comment, if $T$ is a theory over $CRing$, we can define a “$T$-field” to be a $T$-algebra whose underlying ring is a field. Is there then a “$T$-field of fractions” construction taking any $T$-algebra (or perhaps $T$-integral-domain) to a $T$-field? And does the “differentiability” of $T$ extend to elements $g$ of the $T$-field of fractions of $T(n+1)$? If so, then we could write $h(x,y) = \frac{g(y)-g(x)}{y-x}$ in the $T$-field of fractions of $T(2)$, and then substitute $x$ for $y$, which would be basically the same as what I did (up to a change of variables $y = x + dx$) by writing $\Delta g = \frac{g(x+dx)-g(x)}{dx}$ and substituting $0$ for $dx$.

It seems almost as though the “$T$-field of fractions” obviates the need for differential completeness, since $\frac{g(y)-g(x)}{y-x}$ should be a perfectly good $(n+1)$-variable thing therein that we could call $h(x,y)$. I guess the point is whether it remains valid to substitute $y=x$ therein, which in my original description corresponded to whether $\Delta g$ was in the domain of $\phi_0$.

The argument for the derivative of the exponential function is nice, but of course it depends on assuming we are in a differentiably complete setting, which is roughly the same as assuming that the exponential function is differentiable. It’s a standard argument (at least, it’s in my calculus textbook) that assuming $E$ is differentiable at $0$, then we have $E'(x) = E(x) E'(0)$. What I’d really like (which is probably impossible, but I’ll mention it anyway) is some sort of “algebraic” argument for why $E'(0)$ exists, akin to how we can calculate $\frac{(x+dx)^2-x^2}{dx} = 2x + dx$ and then just “set $dx=0$”.

Finally, differentiable-completeness looks like an orthogonality property. Is that what was behind your guess that there is an idempotent differentiable completion?

• CommentRowNumber10.
• CommentAuthorTodd_Trimble
• CommentTimeSep 13th 2013

Just one quick reply for now: there is no algebraic argument that $E'(0)$ exists, because (for example) using Hamel bases of $\mathbb{R}$ as a $\mathbb{Q}$-vector space, there are lots of discontinuous homomorphisms $\mathbb{R} \to \mathbb{R}$, and then we just compose with $\exp: \mathbb{R} \to \mathbb{R}^\ast$ to get a discontinuous exponential function. But these are wild, and relatively modest assumptions (e.g., that $E$ is measurable) are enough to force $E$ to be of standard type.

• CommentRowNumber11.
• CommentAuthorMike Shulman
• CommentTimeSep 13th 2013

Okay, makes sense. )-: Is there a simple argument that (say) if $E$ is continuous then it is differentiable?

• CommentRowNumber12.
• CommentAuthorZhen Lin
• CommentTimeSep 14th 2013
• (edited Sep 14th 2013)

@Todd #8

That’s really elegant!

@Mike #7

It seems unlikely. How does one add/multiply two power series expanded about different points? We have a good answer for that when it comes to rational functions over different denominators, which is one reason we can have a field of fractions.

Perhaps the issue is that non-trivial Hausdorff spaces cannot have generic points. Consider the lattice of open subsets of an irreducible topological space $X$, not necessarily sober. Because every non-empty open subset is dense, the poset of non-empty open subsets has the rather unusual property of being cofiltered. In fact, it is a completely prime filter, which is why the soberification of $X$ has a genuine generic point. Of course, if $\mathcal{O}$ is a sheaf of rings on $X$, then the colimit of $\mathcal{O}(U)$ as $U$ varies over the non-empty subsets of $X$ is precisely the localisation of $\mathcal{O}$ at the generic point. We could naïvely apply this construction to the case where $X$ is $\mathbb{C}$ (in the usual metric space sense) and $\mathcal{O}$ is the sheaf of holomorphic functions, but I suspect the whole thing collapses in the colimit.

It seems that the right topology to consider in this case is the one where the non-trivial closed subsets are the discrete subsets (in the standard topology). (This is the topology associated with the Zariski spectrum of the ring of entire functions, which is an integral domain!) This has a generic point, and we can consider the localisation of the sheaf of holomorphic functions at the generic point. Concretely, it is the ring of holomorphic functions defined on the complement of a discrete set, modulo the obvious equivalence relation. It is a field because the zero locus of a holomorphic function is discrete, but it is bigger than the fraction field of the ring of entire functions because it contains functions like $\exp (1/z)$ that have essential singularities in $\mathbb{C}$. By construction, it contains all convergent power series (expanded about any point). This might be the best compromise that avoids exotica…

• CommentRowNumber13.
• CommentAuthorTodd_Trimble
• CommentTimeSep 14th 2013
• (edited Sep 14th 2013)

Re #11: I guess it depends what is meant by simple! For mathematicians, one could argue like this. We know $E(0) = 1$. Argue that

$\int_0^a E(t) d t$

must therefore be non-zero for some $a$; call this quantity $C$. Then for any $x$ we have

$C E(x) = E(x) \int_0^a E(t) d t = \int_0^a E(x)E(t) d t = \int_0^a E(x+t) d t = \int_x^{x+a} E(s) d s$

so that $E(x) = \frac1{C} \int_x^{x+a} E(s) d s$. Since the integrand on the right is continuous, the right-hand side is differentiable in $x$ by the fundamental theorem of calculus. But then the left side $E(x)$ is differentiable.

I don’t know if you’re hoping for something simple for freshman students, but it’s possible this argument could be manipulated or rearranged into something simple for them as well.

• CommentRowNumber14.
• CommentAuthorMike Shulman
• CommentTimeSep 14th 2013
• (edited Sep 14th 2013)

That was exactly my question. (-: I see what you’re saying, though, that rational functions with different poles have nontrivially intersecting domain, whereas power series about different points might have disjoint discs of convergence (and a formal power series, over an arbitrary ring, doesn’t necessarily have a meaningful radius of convergence at all).

holomorphic functions defined on the complement of a discrete set

Aren’t those what people call “meromorphic functions”?

@Todd: another nice argument! I might be able to tell that to freshmen once they’ve had integration and FTC, but since we do derivatives before integrals (reversing history), it wouldn’t really help.

Is the theory of $C^\omega$-rings (if that means anything) differentiably complete? It seems that it should be…

• CommentRowNumber15.
• CommentAuthorTobyBartels
• CommentTimeSep 14th 2013
• (edited Sep 14th 2013)

The rigorous development of calculus us definitely simpler if one does integration before differentiation, no question about that. But the traditional calculus curriculum is looking towards applications, and it's more important that students know derivatives.

It's tedious (at best) to even define (and prove the existence of) the standard exponential function, if you only have limits. How much nicer to see that $x \mapsto \int_0^x \frac{\mathrm{d}t}t$ is strictly increasing, hence invertible!

• CommentRowNumber16.
• CommentAuthorMike Shulman
• CommentTimeSep 14th 2013

I do also think that derivatives are easier to understand at first than integrals. Also, it’s hard to compute many integrals until you have FTC, and you need to know derivatives to even state that, so doing integrals first would be kind of frustrating.

• CommentRowNumber17.
• CommentAuthorZhen Lin
• CommentTimeSep 14th 2013

Meromorphic functions on $\mathbb{C}$ are (partial) functions that have Laurent series expansions at every point in $\mathbb{C}$ – which forces their singularities to be poles of finite order. In particular, $\exp (1/z)$ is not meromorphic on $\mathbb{C}$.

What I said earlier was not correct – the zero locus of a holomorphic function can have accumulation points, e.g. $\sin (1/z)$. So the ring I described is not a field, unfortunately.

• CommentRowNumber18.
• CommentAuthorUrs
• CommentTimeSep 14th 2013
• (edited Sep 14th 2013)

Todd, Mike, this is called a Fermat theory.

• CommentRowNumber19.
• CommentAuthorTodd_Trimble
• CommentTimeSep 14th 2013

Yeah, I think one pedagogical issue is how one even talks about exponentials and logarithms to freshmen. Certainly I’ve begun Calculus II courses by introducing $log t = \int_1^t \frac{d t}{t}$, and developing some properties, then defining $\exp$ to be the inverse. But for Calculus I, there is some tacit assumption (I guess!) that they have seen some exponential functions – but not necessarily defined rigorously – and then somehow appeal to some intuitions they may have developed along the way, such as the fact that any exponential function – to be any good – ought to be at least continuous. I think I’d be willing to go further and assume differentiability as well (and remark to the side that this is actually overkill, i.e., that continuity is enough, but proving that fact is best left to a second course in the calculus, after one has developed some integration theory).

But maybe I’ll think on this a little further, and see whether you can easily get around integrals (as in comments #13 and #14).

Is the theory of $C^\omega$-rings (if that means anything) differentiably complete? It seems that it should be…

What it means or conveys to me is the theory whose morphisms $n \to 1$ are globally defined real-analytic functions $f: \mathbb{R}^n \to \mathbb{R}$, since that is the standard meaning of $C^\omega$ in mathematics (whether that’s a really good notation is anyone’s guess!). So the question is whether there is a Hadamard lemma for that theory. The answer is ’yes’, I’m pretty confident.

Getting back to #9:

It seems almost as though the “$T$-field of fractions” obviates the need for differential completeness, since $\frac{g(y)-g(x)}{y-x}$ should be a perfectly good $(n+1)$-variable thing therein that we could call $h(x,y)$. I guess the point is whether it remains valid to substitute $y=x$ therein, which in my original description corresponded to whether $\Delta g$ was in the domain of $\phi_0$.

Yeah, I think we’re discussing basically the same thing but in slightly different ways. If the $T(n)$ are integral domains, it should be possible to define their fields of fractions as $T$-algebras, and then the Hadamard axiom is that $h(x, y) = \frac{g(y)-g(x)}{y-x}$ is represented by a global element $h \in T(n+1)$ (which is not always guaranteed of course). Really what I was doing was working safely within the confines of equational theories (where we can talk about free structures in the usual traditional way, without worrying about domains), and then concentrating the key property needed as a Hadamard lemma condition. But as I said at the outset, the idea is very similar to yours.

Finally, differentiable-completeness looks like an orthogonality property. Is that what was behind your guess that there is an idempotent differentiable completion?

Oh, nothing quite as nice-sounding as that! (Tell me more what you have in mind!) It was really more just the idea that $DiffTh$ is fully faithfully embedded inside $CRing\downarrow Th$, and that we get the reflection by closing up under Hadamard elements $h(x, y)$. Possibly it doesn’t work quite so smoothly as that – I’m vaguely uneasy around theories $T$ where the underlying commutative rings $T(n)$ are not integral domains.

• CommentRowNumber20.
• CommentAuthorTodd_Trimble
• CommentTimeSep 14th 2013

Urs: Ah, I thought this whole business sounded familiar! Thanks for that.

• CommentRowNumber21.
• CommentAuthorMike Shulman
• CommentTimeSep 14th 2013

@Zhen: Ah, okay, Wikipedia misled me. The first sentence of its page says “In complex analysis, a meromorphic function on an open subset $D$ of the complex plane is a function that is holomorphic on all $D$ except a set of isolated points, which are poles for the function.” It wasn’t clear to me from that phrasing that “which are poles for the function” was an additional condition being imposed rather than an observation. I am tempted to edit the page so that the sentence ends with “that are poles for the function”, but I’m worried that that might be wading into deeper waters of Wikipedia culture than I want to get into…

• CommentRowNumber22.
• CommentAuthorMike Shulman
• CommentTimeSep 14th 2013

@Todd: Defining the field of fractions as a $T$-algebra doesn’t seem quite straightforward, since if we just take the ordinary FoF of a $T$-algebra, it may no longer be a $T$-algebra; it seems that maybe some infinite iteration is required?

$DiffTh$ is fully faithfully embedded inside $CRing\downarrow Th$, and that we get the reflection by closing up under Hadamard elements $h(x, y)$.

That’s basically what “orthogonality property” says in fancier language. Let $G$ be the theory under $CRing$ freely generated by an operation $g$ of arity $n+1$, and $H$ the theory under $CRing$ generated by operations $g$ and $h$ of arities $n+1$ and $n+2$ respectively which satisfy the Hadamard equation. Then a theory under $CRing$ is differentiable (or I guess we should say, “Fermat” — thanks Urs!) iff it is orthogonal to the obvious map $G\to H$. The reflection is then constructed by the usual transfinite methods, which amounts to taking repeated pushouts along the maps $G\to H$ (to add Hadamard elements) and $H+_G H \to H$ (to enforce uniqueness of Hadamard elements).

• CommentRowNumber23.
• CommentAuthorMike Shulman
• CommentTimeSep 14th 2013

I think my Calculus I students definitely believe in the exponential function. I don’t know whether their pre-calculus courses even address the issue that it’s not obvious how to define $a^b$ when $b$ is irrational; they may just implicitly rely on the intuition that “anything you can do with some numbers you can do with all numbers”. I certainly don’t remember issues of that sort being raised in my own pre-calculus education.

I wonder whether the Hadamard-Fermat property would be a good way to teach differentiation. This semester I’m introducing differentiation informally/procedurally, using the approach I described in #4, and planning to discuss limits later, somehow. But if I phrased it in terms of the Hadamard-Fermat property, then a natural way to proceed would be to say that we want the Hadamard function to be “continuous” and then proceed to figure out what that means. I’ve always thought that it’s easier to motivate defining continuity than defining limits in general.

• CommentRowNumber24.
• CommentAuthorTodd_Trimble
• CommentTimeSep 15th 2013

I wonder whether the Hadamard-Fermat property would be a good way to teach differentiation.

It definitely sounds plausible as a good way, but I’ve never tried it myself. I can imagine that students who have had calculus before might resist being taught an alternative POV, but then again, it might be good for them. The method gibes well with how we should be teaching the finding of tangent lines $y = g(a) + h(a)(x-a)$ from $g(x) = g(a) + h(x)(x-a)$ (at given point $a$).

Regarding this business about continuous exponential functions, I wrote up something on my web here. The third proof uses no integration theory and is pleasantly geometric (let me know if you see any problem with it).

• CommentRowNumber25.
• CommentAuthorMike Shulman
• CommentTimeSep 16th 2013

In my experience, students who’ve had some calculus before are the bane of every calculus class. d-: I’ll have a look at what you wrote soon, I hope; today was an… interesting… day.

• CommentRowNumber26.
• CommentAuthorTobyBartels
• CommentTimeSep 18th 2013

@Mike #21: Sometimes people talk about poles of infinite order (although not on the English Wikipedia's page on poles!), so even using ‘that’ could be ambiguous. I edited it myself just now to explicitly refer to Laurent series.

• CommentRowNumber27.
• CommentAuthorTobyBartels
• CommentTimeSep 18th 2013

I teach precalculus (called ‘College Algebra’: precalculus without trigonometry, which is the subject of the following course) more often than any other, and this is how I introduce exponential functions: First, I make a table of values for $2^x$, using various rational values of $x$, each properly calculated as a whole-number root of a whole-number power of $2$ (or the reciprocal thereof), and graph these points. Then I draw the curve through them, and state without proof that every value lies on this curve. I then point out that this gives a meaning for $2^x$ when $x$ is irrational, stressing that this could not have been calculated before, using whole-number roots and powers. Then I repeat the process for $(1/2)^x$ and for $(-2)^x$, showing how the latter breaks down. (Since $(-2)^{1/3}$ is a negative real number in our books, and anyway the students don't know any definition of $(-2)^{1/4}$ until the end of the trig course, we can't even fix this by allowing imaginary values. There simply is no continuous extension of the $(-2)^x$ that these students use.) I think that the students find this a bit overkill, but I want to give them the idea of how mathematicians come up with their definitions, as well as why we only use positive bases.

But I wouldn't pretend to do this rigorously.

• CommentRowNumber28.
• CommentAuthorTobyBartels
• CommentTimeSep 18th 2013

@Todd, nice —convexity is a useful tool!

• CommentRowNumber29.
• CommentAuthorTobyBartels
• CommentTimeSep 18th 2013

This semester I’m introducing differentiation informally/procedurally, using the approach I described in #4, and planning to discuss limits later, somehow.

In my ‘Applied Calculus’ courses (for bio and econ majors, no trig), I start with differentiation and don't get to limits until almost halfway through the course. (I also have a heavy emphasis on differentials as a practical tool for calculation, so we don't actually see difference quotients until the application to tangent lines.)

I’ve always thought that it’s easier to motivate defining continuity than defining limits in general.

In my regular calculus courses (for math and science majors), I'm expected to do rigorous definitions (although the book doesn't actually have a correct definition of the Riemann integral!), although not always proofs, so I do limits before derivatives. But I've decided to do continuity before limits next time (term starts October 2), for reasons of motivation, as you say. I considered doing derivatives before limits, with the Hadamard definition, but decided against it; we want to calculate them, and limits of difference quotients are good for that. (I emphasize differentials in this course too, but I can't define them until after I've defined derivatives, and I don't want to defer calculations of practical examples too much.)

• CommentRowNumber30.
• CommentAuthorMike Shulman
• CommentTimeSep 18th 2013

@Toby: what do you mean by “differentials”? In your second paragraph you said you can’t define them until after derivatives, but in your first paragraph you implied you use differentials to define defivatives before introducing difference quotients?

(My approach allows using difference quotients for derivatives before limits.)

• CommentRowNumber31.
• CommentAuthorzskoda
• CommentTimeSep 18th 2013
• (edited Sep 18th 2013)

Sometimes people talk about poles of infinite order

This is completely colloquial and careless (I am not necessarily negative, this is a leisure mode of work) – one can and should just say in formal text essential singularity.

However, the article Laurent series is incorrect if one goes beyond the special case of Laurent series with complex coefficients for a meromorphic function around a finite point in the complex plane. Laurent series is a power series allowing negative powers. Period. Convergence etc. are additional considerations which are sometimes silently assumed by a convention in a specific context. (Remark. In general the series from $-\infty$ to $+\infty$ could be looked upon as a pair of two series. It is most apparently so when defining the convergence properties.) So this makes sense over other fields, and people in algebra talk so much about the spatial case of a Laurent series of formal variable (and the “ring of formal Laurent series”). Of course, in complex analysis, one restricts to Laurent series which define meromorphic functions in some neighborhood. Then, on a Riemann sphere, in usual coordinates, this means for finite points that only finitely many negative powers are nonzero, but for the point at infinity this means that only finitely many POSITIVE powers are nonzero. So it is not by definition, even in the special case of complex analysis on Riemann sphere, that finitely many negative powers are coming with nonzero coefficients. If you have a Laurent series with coefficients which are operators on a complex Hilbert space, then one can occasionally have good convergence properties and analyticity of the sum when the series is infinite on both sides. for example, say that we have normal operators and some block decomposition which is simultaneous for all coefficients. Then in some block we can have one finite $k$ limiting the principal part, while in some other block (an operator on another orthogonal subspace) we have larger $k$, but we have no uniform bound. So we have analytic operator valued function which has principal part which is infinite, but when restricting to any of the blocks it is finite! Also, we can have similar situations in representation theory (including vertex operator algebras) where one has infinite series which is finite for every relevant representation, so this may as well apply for the principal part at some $z\in\mathbb{C}$.

To add, the principal part of the Laurent series with complex coefficients of a meromorphic function on a Riemann sphere is not the part with negative terms, but again it depends if it is around finite point or around the infinity.

I looked at several standard and several nonstandard English references about this Laurent terminology when writing a Laurent series entry in certain Croatian database of scientific terms, so I am not making it up.

• CommentRowNumber32.
• CommentAuthorzskoda
• CommentTimeSep 18th 2013

I revised Laurent series.

• CommentRowNumber33.
• CommentAuthorTobyBartels
• CommentTimeSep 18th 2013

Zoran, how about Puiseux series, then? Is it wrong that these include all Laurent series, or is the definition of these wrong as well?

• CommentRowNumber34.
• CommentAuthorTodd_Trimble
• CommentTimeSep 18th 2013
• (edited Sep 18th 2013)

If you want to treat Laurent series over a general field $k$ as forming a ring, then they should have a well-ordered support; in particular, all but finitely many negative-power monomial terms should vanish. I’ll haven’t checked Zoran’s revision yet, but this point should not be obscured. (Edit: and looking at this now, I see that it is obscured. A further revision will be necessary.)

It is nevertheless true that sometimes one wants full freedom of powers in both directions, even in formal algebraic contexts where convergence issues do not come into play. One example is the Dirac distribution supported at $z=0$, which is sometimes introduced formally as $\delta(z) = \sum_{n \in \mathbb{Z}} z^n$.

• CommentRowNumber35.
• CommentAuthorTobyBartels
• CommentTimeSep 18th 2013

My approach allows using difference quotients for derivatives before limits.

It allows you to calculate derivatives through difference quotients before limits; but not, if I understand you, to actually define them. (There's also the Hadamard method of defining them using continuity rather than limits as such, but that's not what you're talking about, right?) Of course, that depends on how rigorous you want something to be before it counts as a definition; but you can't calculate the derivative of the sine function (even assuming its existence and basic properties, particularly the angle-sum identities), right? (In principle, one could do infinite series before derivatives, define sine and cosine using these, and calculate their derivatives, taking it for granted that absolutely convergent series behave well, but that wouldn't fit into the U.S. calculus curriculum very easily!)

what do you mean by “differentials”?

Not simply what it says in the book, where differentials are introduced as merely a tool for approximation! In the textbook that we use, differentials are suddenly revived in the chapter on integration as a technique to calculate integration by substitution, but in fact they are an excellent method to calculate pretty much everything in differential and integral calculus, including (perhaps especially) multivariable calculus, and I stress them throughout.

See my handouts for the latest terms that I taught of Applied Calculus, regular calculus, and multivariable calculus (which incorporates improvements to the regular calculus notes). You might also like to look at the notes on integration of differential forms from the multivariable calculus course, although I'm still not very happy with those; it's viewing worked examples that really teaches those ideas.

• CommentRowNumber36.
• CommentAuthorTobyBartels
• CommentTimeSep 18th 2013

I remember now that this dispute about Laurent series came up when we were writing the Wikipedia article in 2002; it's still on the Talk page. We came to the same conclusion as Zoran then.

• CommentRowNumber37.
• CommentAuthorzskoda
• CommentTimeSep 18th 2013
• (edited Sep 18th 2013)

Toby, 36. I wrote the first version of the entry Puiseux series (somebody essentially extended it later), indeed, assuming only Laurent series in the sense with finitely many negative terms. We should eventually have the $n$Lab self-consistent, which I made not so, but I am not yet sure how to resolve all of this in a reasonable manner, respecting the tradition, logic and generality in optimal way.

Todd: mentioning a ring, the definition of Laurent series has coefficient in a ring (it can even be noncommutative, but then the variable $t$ will commute with the coefficients, unless we have some weakly skew version); the fact that the Laurent series themselves form a ring is not treated yet, though used in the theorem about the Puiseux series ring. Now it will be interesting to have your remarks about the ordering of the ground field to have the Laurent series in this or that generality (with finitely or infinitely many negative power members allowed) forming a ring.

• CommentRowNumber38.
• CommentAuthorTodd_Trimble
• CommentTimeSep 18th 2013
• (edited Sep 18th 2013)

Well, I did some revising at Laurent series to try to clear up the relevant issues, adding some more material and references along the way.

I don’t think it’s necessary to argue about this – I am fine with the general point Zoran made – but it’s important to try to respect the basic correctness of the content (in its appropriate context) that was already there before Zoran’s revision.

Zoran, Toby, please take a look and let me know if you disagree with my revision. (Edit: I see there are still some residual inconsistencies – the Idea section should be improved, and there are perhaps slight inconsistencies in notation.)

• CommentRowNumber39.
• CommentAuthorzskoda
• CommentTimeSep 18th 2013
• (edited Sep 18th 2013)

Nice improvements and extensions, Todd !

• CommentRowNumber40.
• CommentAuthorMike Shulman
• CommentTimeSep 18th 2013

It allows you to calculate derivatives through difference quotients before limits; but not, if I understand you, to actually define them.

Well, I haven’t given a definition that a mathematician would consider rigorous, but at the beginning of this thread I sketched a rigorous definition along these lines which works for rational functions. Todd then explained that at least for polynomials, it’s the same as the Hadamard definition, but using “polynomiality” instead of “continuity”. I am going to give an actual definition of derivatives in terms of limits later on — although I’m probably not going to give an actual definition of limits — but since my students don’t start out even really knowing what a “definition” means in mathematics, I figured that a method for calculation was good enough to get started, and I wanted to let them dive into some usable calculus without mashing their head through limits first.

You’re right that it doesn’t work for trig functions; in fact that was the last question I asked back in #1, which Todd answered negatively (at least for the exponential function) in #10.

• CommentRowNumber41.
• CommentAuthorMike Shulman
• CommentTimeSep 18th 2013

This is getting a bit off-topic from the beginning of this thread, but has anyone tried categorifying Laurent series? If the coefficients were sets or groupoids instead of numbers, then the problematic coefficients $h_n = \sum_{k \in \mathbb{Z}} f_k g_{n-k}$ in the product of two doubly-infinite Laurent series would make perfect sense.

• CommentRowNumber42.
• CommentAuthorTodd_Trimble
• CommentTimeSep 18th 2013

Mike #40: your second paragraph reminds me how I’ve longed wished for a scenario where I could teach a course in the calculus, but taking advantage of complex numbers. Once you really explain the exponential function as a function of complex numbers, so many things make sense and fall into place, all those facts from trigonometry for all example are pretty much spin-offs from Euler’s great identity $e^{i x} = \cos(x) +i\sin(x)$ (which has held me in thrall ever since my eyes fell upon it by accident, while reading the Encyclopedia Britannica as a teenager). Using complex numbers would make for a tremendous unification for calculus students, if done right. (Would need a special sort of student, perhaps.)

It might be possible to adapt the argument mentioned in #24 to this scenario, but I haven’t tried. (By the way, thank you Toby #28!)

Mike #41: my mind immediately leaps to e.g. $\mathbb{Z}$-graded vector spaces and the like.

• CommentRowNumber43.
• CommentAuthorMike Shulman
• CommentTimeSep 18th 2013

Yes, in some ways, it would be lovely to be able to use complex numbers in an intro calculus course. On the other hand, we do a lot of drawing of graphs in calculus, and the two-dimensional graphs that we can draw on paper can be seriously misleading for functions of complex variables.

Re: categorified Laurent series, I guess maybe one question to start with is, what is it that distinguishes a (formal) Laurent series from a $\mathbb{Z}$-indexed family of numbers? Or rather, what does it mean when we choose to regard a $\mathbb{Z}$-indexed family of numbers as a formal Laurent series?

• CommentRowNumber44.
• CommentAuthorTodd_Trimble
• CommentTimeSep 18th 2013

what is it that distinguishes a (formal) Laurent series from a ℤ-indexed family of numbers?

As far as I’m concerned, nothing.

Or rather, what does it mean when we choose to regard a ℤ-indexed family of numbers as a formal Laurent series?

For me, it just has to do with how we define multiplication (it’s the convolution product wrt the monoid $\mathbb{Z}$). Here “multiplication” either has to do with a ring structure, or with a module structure (over some smaller structure like Laurent polynomials).

• CommentRowNumber45.
• CommentAuthorTobyBartels
• CommentTimeSep 18th 2013

Wikipedia says that a formal Laurent series must have only finitely many nonzero terms with negative exponents. However, there is no citation; it's possible that Axel Boldt and I made this up back in 2002, just to ensure that they form a ring.

• CommentRowNumber46.
• CommentAuthorTobyBartels
• CommentTimeSep 18th 2013

@Mike #40: What I mean is, you have (at first) no definition (however lacking in rigour) that defines the derivative (when it exists) of an arbitrary function (or even just of those that will appear in the course). I think that that's all right, by the way; but that's what I meant when I said that you couldn't define them before limits.

• CommentRowNumber47.
• CommentAuthorMike Shulman
• CommentTimeSep 18th 2013

I feel like we really need two different words: a $Laurent_1$-series is a $\mathbb{Z}$-indexed family of numbers written as $\sum_{n\in \mathbb{Z}} a_n x^n$, and a $Laurent_2$-series is a $Laurent_1$-series with only finitely many nonzero terms with negative index. It seems to me that both of them are equally “formal”.

I see what you mean about derivatives of arbitrary functions.

I guess the rig of categorified $Laurent_1$-series is just the Day convolution monoidal structure on $[\mathbb{Z},Whatever]$.

And finally, while we’re speculating about ways to teach calculus, another thing that would be interesting would be to try using power series early on.

• CommentRowNumber48.
• CommentAuthorTobyBartels
• CommentTimeSep 19th 2013

I suppose, Mike, that your method is to teach a (correct but maybe not fully rigorous) definition of the derivative of a rational function (and actually you should be able to handle other algebraic functions) so that the students can get in and do calculations, then bring in limits, then give a definition of the derivative of a more general function. The important theorems (possibly unproved but perhaps at least stated) are then that the two definitions are unambiguous and agree (so that any function has at most one derivative).

• CommentRowNumber49.
• CommentAuthorTobyBartels
• CommentTimeSep 19th 2013

I have a translation of a short Russian book, Infinite Series by A. I. Markushevich, which treats infinite series at a precalculus level. The definition of convergence of an infinite series (hence sequence) appears, but no other limit. One could then do calculus after this.

• CommentRowNumber50.
• CommentAuthorTodd_Trimble
• CommentTimeSep 19th 2013

I feel like we really need two different words: a $Laurent_1$-series is a $\mathbb{Z}$-indexed family of numbers written as $\sum_{n\in \mathbb{Z}} a_n x^n$, and a $Laurent_2$-series is a $Laurent_1$-series with only finitely many nonzero terms with negative index.

Maybe you’re right. I might propose (for internal nLab purposes) Laurent series vs. restricted Laurent series, respectively.

I guess the rig of categorified $Laurent_1$-series is just the Day convolution monoidal structure on $[\mathbb{Z},Whatever]$.

That’s really what I was driving at in #42.

And finally, while we’re speculating about ways to teach calculus, another thing that would be interesting would be to try using power series early on.

According to Vladimir Arnol’d (in Huygens, Barrow, Newton, and Hooke), that was really Newton’s emphasis as he was developing the calculus.

Finally,

On the other hand, we do a lot of drawing of graphs in calculus, and the two-dimensional graphs that we can draw on paper can be seriously misleading for functions of complex variables.

Well, this course I daydream about would probably have to be for a special sort of student, and probably would take two semesters. I admit that I haven’t thought about such a course in great detail, but the thought is tantalizing nonetheless because there is such a lot of insight one can gain by dipping in now and then into the complex numbers. Near the beginning of the course I would like to explain how $i$ or $-i$ is essentially a 90 degree turn. Then at some point, Euler’s identity, and why on an intuitive level it must be true, with accompanying discussions of logarithmic spirals and the like. Power series and analytic continuation would be another thing to touch on at some point (for instance, the fact that the MacLaurin series for $\frac1{1+x^2}$ has a radius of convergence $1$, even though it seems so tame as a function of a real variable, should be connected with the fact that it has a pole in the complex plane at distance $1$ from the origin). I don’t think I’d want to do a crypto course in complex variables (Cauchy’s theorem and all of its consequences), but toward the end of the course it might be very interesting to give a taste of some of those consequences. For example, I think one could explain without too much pain (and without Cauchy’s theorem) that complex polynomial functions are open mappings – which of course makes them hugely different from real polynomials – and this leads to an understanding of the fundamental theorem of algebra. Giving a taste of conformal mappings (e.g. Moebius transformations) would also be very cool.

Sorry – I know this is quite a digression from the thread!

• CommentRowNumber51.
• CommentAuthorMike Shulman
• CommentTimeSep 19th 2013

@Toby #48: Yes, that’s what I’m doing; sorry if I wasn’t sufficiently clear in explaining it. It’s a new experiment this semester, so I can’t say yet how well it works.

• CommentRowNumber52.
• CommentAuthorMike Shulman
• CommentTimeSep 19th 2013

Todd: Yes, I was also under the impression that many of the early mathematicians doing calculus used infinite series very much (and, from a modern perspective, perhaps insufficiently rigorously). Your idea for a course sounds really interesting! (This thread seems to be held together by digressions at this point anyway.)

• CommentRowNumber53.
• CommentAuthorTobyBartels
• CommentTimeSep 19th 2013

Mike, you were perfectly clear; I was just slow to reach that interpretation of it.

• CommentRowNumber54.
• CommentAuthorMike Shulman
• CommentTimeSep 20th 2013

@Todd #24: The convexity argument is very nice! The $\geq$ in the last sequence of inequalities is actually a $\gt$, right?

• CommentRowNumber55.
• CommentAuthorTodd_Trimble
• CommentTimeSep 20th 2013
• (edited Sep 20th 2013)

Thanks! Took a little while to come up with that one. :-)

Yes, that should in fact be a $\gt$.

• CommentRowNumber56.
• CommentAuthorMike Shulman
• CommentTimeSep 23rd 2013

There is a common “proof” that $\sin'=\cos$ which goes like this.

$\frac{\sin(x+dx)-\sin(x)}{dx} = \frac{\sin(x)\cos(dx) + \cos(x)\sin(dx)-\sin(x)}{dx} = \sin(x) \frac{\cos(dx) - 1}{dx} + \cos(x) \frac{\sin(dx)}{dx}.$

We also have

$\frac{\cos(dx) - 1}{dx} = \frac{(\cos(dx) - 1)(\cos(dx)+1)}{dx(\cos(dx)+1)} = \frac{\cos^2(dx) - 1}{dx(\cos(dx)+1)} = \frac{-\sin^2(dx)}{dx(\cos(dx)+1)} = \frac{\sin(dx)}{dx} \cdot \frac{-\sin(dx)}{\cos(dx)+1}$

Now assuming that $\sin$ and $\cos$ are continuous, we have $\lim_{dx\to 0}\frac{-\sin(dx)}{\cos(dx)+1} = 0$, so as long as $\frac{\sin(dx)}{dx}$ is bounded we have $\lim_{dx\to 0}\frac{\cos(dx) - 1}{dx} = 0$. Thus, for $\sin'=\cos$ it suffices to show $\lim_{x\to 0} \frac{\sin(x)}{x} = 1$.

Now consider the right triangle with vertices $(0,0)$, $(\cos(x),0)$, and $(\cos(x),\sin(x))$. Its area is $\frac{1}{2}\sin(x)\cos(x)$. This area is less than the area of the corresponding sector of the unit circle, which is $\frac{x}{2\pi} \cdot \pi = \frac{1}{2} x$. It is also greater than the area of the corresponding sector of the circle of radius $\cos(x)$, which is $\frac{x}{2\pi} \cdot \pi \cos^2(x) = \frac{1}{2} x \cos^2(x)$. Thus

$\frac{1}{2} x \cos^2(x) \lt \frac{1}{2}\sin(x)\cos(x) \lt \frac{1}{2} x$

and so

$\cos(x) \lt \frac{\sin(x)}{x} \lt \frac{1}{\cos(x)}.$

Since $\lim_{x\to 0} \cos(x) = 1$, the squeeze theorem applies.

In addition to continuity of $\sin$ and $\cos$ and the addition formula $\sin(x+y) = \sin(x)\cos(y)+\cos(x)\sin(y)$, this argument uses only the above inequalities, which it justifies geometrically in terms of areas (which are supposed to understand intuitively). Probably this could be made precise using integration. I wonder whether there is an analogue of Todd’s third argument using convexity which works for trigonometric functions? I want to “translate” across $e^{i\theta} = \cos(\theta)+i\sin(\theta)$, but it seems that neither argument works in the complex domain, since the inequalities stop making sense.

• CommentRowNumber57.
• CommentAuthorTodd_Trimble
• CommentTimeSep 23rd 2013

I think we can apply a convexity argument, and I will see about putting up such an argument on my web page and let you know when (if) I’m done, but first: what makes this whole business slightly awkward is that the argument $x$ of $\sin(x), \cos(x)$ is an arc length along the unit circle, and the very notion of arc length requires some integration (or at least some analysis) to make it rigorous.

One could instead try a development along the following lines. We’ve proved that if $E(-)$ is a continuous exponential function, then $E$ is in fact differentiable and in fact $E'(x) = E'(0)E(x)$; then we define the standard exponential function by an appropriate rescaling which sets $E'(0) = 1$. This in effect defines $e$ as $E(1)$. We could try a similar tack here: assuming we have continuous functions $S(x)$, $C(x)$ which satisfy the addition formulas

$S(x+y) = S(x)C(y) + C(x)S(y); \qquad C(x+y) = C(x)C(y) - S(x)S(y)$

(whence we derive a whole bunch of other stuff like $S(0) = 0$, $C(0) = 1$, $S(x)^2 + C(x)^2 = 1$, etc.), one sets out to show that the derivative exists at some point and thence every point, and derive

$S'(x) = S'(0)C(x); \qquad C'(x) = -S'(0)S(x)$

and then define $\sin(x), \cos(x)$ by $\sin(x) = S(x)$ and $\cos(x) = C(x)$, rescaling if need be so that $S'(0) = 1$. This is actually a pretty sane way of going about things; it’s closely connected with a definition of $2\pi$ from the exponential function, as the positive generator of the kernel of the map $\mathbb{R} \to S^1: t \mapsto e^{i t}$.

• CommentRowNumber58.
• CommentAuthorMike Shulman
• CommentTimeSep 24th 2013

Yes, that’s the sort of argument I was hoping to see!

• CommentRowNumber59.
• CommentAuthorTodd_Trimble
• CommentTimeSep 24th 2013

Okay, I have something here. It’s not the prettiest, perhaps, but maybe it can be polished a little more. At least it’s pretty honestly rigorous at the beginning calculus level (I think). :-)

• CommentRowNumber60.
• CommentAuthorTobyBartels
• CommentTimeSep 24th 2013

what makes this whole business slightly awkward is that the argument $x$ of $\sin(x), \cos(x)$ is an arc length along the unit circle, and the very notion of arc length requires some integration (or at least some analysis) to make it rigorous

Not that you don't still need integration (or other analysis), but this argument can also be viewed as an area rather than as an arclength. The bad thing is that the area subtended by the arclength $x$ is (on a circle of unit radius) $1/2\, x$ rather than $x$ itself; the good thing is that the analogous area when using a hyperbola in place of the circle is still (half of) the argument of the hyperbolic trigonometric functions, while the arclength is not.

The constructivist in me notices that area is approximated from both above and below (so is naturally a located real number, which in a sheaf topos is a continuous map), while arclength is approximated only from above (so is naturally only an upper real number, which in a sheaf topos is an upper semicontinuous map). I don't know if there's any significance to that; after all, the arclength that appears here happens to be located anyway (assuming that you stop at a point with located coordinates, which is only fair).

• CommentRowNumber61.
• CommentAuthorMike Shulman
• CommentTimeSep 24th 2013

@Todd: very nice, again!!

Has anyone studied a notion of “trigonometric field” analogous to that of “exponential field”?

• CommentRowNumber62.
• CommentAuthorTodd_Trimble
• CommentTimeSep 25th 2013

Thanks, Mike. It would be nice if the arguments were less ad hoc. But I can imagine some modified form of the arguments someday finding their way into a classroom; it’s interesting how much one can do from scratch, so to speak.

I’ve never heard of anyone developing a theory of trigonometric fields. [I do remember one vaguely related little snippet of information: the ring $\mathbb{R}[\sin(x), \cos(x)]$ is not a UFD because both sides of the equation $\sin^2(x) = (1 - \cos(x))(1 + \cos(x))$ are prime factorizations (due to Hale Trotter). Somehow this ring is begging to be “normalized” (having the singularity from its spectrum removed), which I suspect involves adjoining a square root of $-1$ so that one gets Laurent polynomials in a formal $\exp(i x)$.] But anyway, it’s an interesting suggestion!

• CommentRowNumber63.
• CommentAuthorMike Shulman
• CommentTimeSep 25th 2013

What exactly does “$\mathbb{R}[\sin(x), \cos(x)]$” mean? Are you adjoining two indeterminates which satisfy all the polynomial identities satisfied by sine and cosine?

• CommentRowNumber64.
• CommentAuthorTodd_Trimble
• CommentTimeSep 25th 2013

I just meant the ring of functions obtained by applying polynomials to the usual sine and cosine functions. I guess that means the same as what you just said.

• CommentRowNumber65.
• CommentAuthorMike Shulman
• CommentTimeSep 25th 2013

Okay. That is indeed an interesting observation!

• CommentRowNumber66.
• CommentAuthorjcmckeown
• CommentTimeSep 26th 2013

Getting back to #8,

The basic idea is to assume that we have rings (with elements denoted $g, h$, etc., meant to suggest functions) such that we can somehow formalize a statement of the following sort: given g in “one variable”, there exists a unique h in two variables such that

$g(y)=g(x)+(y-x)h(x,y)$

This is simply true in the ring of polynomials over a field, where it’s “the root theorem”, applied to the polynomial in $t$ :

$p_x ( t ) = g ( x + t ) - g ( x )$

That is, because $p_x (t) = 0$, so $t$ is a factor of $p_x(t)$ :

$p_x( t ) = t h_x ( t )$

for a unique polynomial $h_x \in k[t]$. One has to do some argument to resolve that $h_x$ is also a polynomial in $x$, but that’s not hard. Which is my round-about way of arguing (not original to me) that calculus is about polynomial approximations to real functions.

• CommentRowNumber67.
• CommentAuthorMike Shulman
• CommentTimeSep 26th 2013

Yes, Todd said that:

Of course, for $T = CRing$, the theory is already complete.

• CommentRowNumber68.
• CommentAuthorTobyBartels
• CommentTimeSep 26th 2013

The claim that calculus is about polynomial approximations brings us back to Taylor series and the possibility of building calculus upon them.

• CommentRowNumber69.
• CommentAuthorUlrik
• CommentTimeSep 28th 2013

I’m a little late to this game, but the idea of teaching undergraduate analysis with an early emphasis on series occurs in the 2008 book by Hairer and Wanner, Analysis by Its History, doi:10.1007/978-0-387-77036-9.

It starts with “prehistory”, Descartes and interpolation, leading to the binomial theorem, first for positive integral exponents, then the binomial series for rational exponents, then Leibniz’ construction of the exponential function, Euler’s theorem for the exponential function, etc. etc.

Derivatives and integrals are treated in only after a very long chapter 1. Just after the introduction of the derivative of the parabola $y=x^2$, there is a long historical discussion, including the statement that:

Lagrange (1797) rejects the infinitely small straightaway and tries to base analysis on power series (“One knows the difficulties created by the assumption of infinitely small quantities, upon which Leibniz constructs his Calculus.”) He introduces the name derivative and uses for $dy/dx$ the notation (see quotation).

The quotation referred to is:

We shall call the function $f x$ a primitive function of the functions $f' x$, $f'' x$, &c. which derive from it, and we shall call these latter the derived functions of the first one.

The reference is to:

J.L. de Lagrange, Théories des fonctions analytiques, contenant les principes du calcul différentiel, dégagés de toute considération d’infiniment petits, d’évanouissans, de limites ou de fluxions, et réduits à I’analyse algébrique des quantités finies, A Paris, de I’imprimerie de la republique, Prairial an V (1797), deuxième édition: Gauthier-Villars, 1813.

• CommentRowNumber70.
• CommentAuthorzskoda
• CommentTimeSep 30th 2013

Ignoring nonanalytic functions in physics amounts to ignoring nonperturbative effects.

• CommentRowNumber71.
• CommentAuthorTobyBartels
• CommentTimeOct 7th 2013

Pace my comment #29, I have now started this term's Calculus course, doing continuity before limits. You can see the handout on continuity for my motivation of both the idea and the definition. (There is not a lot of detail in this, as it's intended only as a supplement to the textbook. I refer them to the textbook for most examples and methods, but not for the definition.)

• CommentRowNumber72.
• CommentAuthorDavidRoberts
• CommentTimeOct 8th 2013

Not to advertise too much, but a colleague of mine has this book, which is a rigorous undergrad treatment of real analysis, starting from the complete ordered field axioms. Not for the students in the courses Toby and Mike are teaching, I would imagine, but perhaps some analogous students who are a little further along in their studies.

• CommentRowNumber73.
• CommentAuthorTobyBartels
• CommentTimeOct 22nd 2013

Here are definitions of limits from a day or two after the one about continuity. Next time, I'll combine these into a single document, which I'll probably expand. (I also have a handout on differentials, but that is getting a bit far afield.)

• CommentRowNumber74.
• CommentAuthorDavidRoberts
• CommentTimeOct 23rd 2013

A typo: the second $x \to \infty$ should be $x \to -\infty$

• CommentRowNumber75.
• CommentAuthorTobyBartels
• CommentTimeOct 23rd 2013

Thanks!


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Free calculator | Multiplication game | Expand | This calculator focuses on speed and ease of use and provides all basic operations with complex numbers. countdown solver | Limit calculator | Factorization | The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. countdown maths solver | Complex Number Calculator The calculator will simplify any complex expression, with steps shown. Instructions:: All Functions . Calculator | , the For the calculation of the complex modulus, with the calculator, simply enter the complex number in its algebraic form and apply the complex_modulus function. vector product calculator | Antidifferentiation | The scientific calculator supports three ways to enter a complex number: You can enter the number in Cartesian form: The cartesian form includes the real portion, and the imaginary portion of the complex number.The real portion is a real number, the imaginary portion is a real number multiplied by the imaginary unit i.For example, you can enter a number such as 3+2i. Fractions | For example, you can convert complex number from algebraic to trigonometric representation form or from exponential back to algebraic, ect. The complex number calculator is able to calculate complex numbers when they are in their algebraic form. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i2 = −1 or j2 = −1. Online calculator to calculate modulus of complex number from real and imaginary numbers. Its user interface is designed to be usable as a standard pocket calculator. Simplify fraction calculator | Square root calculator | ¯z z ¯ (or sometimes with a star z∗ z ∗) and is equal to ¯. function Graphics | arcsin calculator | the result 2+6*i is returned. arctan | Le Complex Number Calculator est un ordinateur créé en 1939 par George Stibitz et Samuel Williams. Dividing Complex Numbers Calculator is a free online tool that displays the division of two complex numbers. Expand a product, Fraction | Matrix Calculator | The calculator handles purely real math calculations effortlessly too. The complex number calculator calculates , to divide complex numbers 1+i et 4+2*i, enter online, the multiplication of complex numbers online applies to the algebraic form of complex numbers, Differentiate function online | scalar product calculator |, Graphing calculator | complex_number((1+i)/(4+2*i)) , after calculation, the result 3/10+i/10 is returned. cosh calculator | Scientific calculator online | Español; When a number has the form a + bi (a real number plus an imaginary number) it is called a complex number. conjugate ¯z =a−ib z ¯ = a − i b with a=R(z) a = ℜ ( z) the real part and b=I(z) b = ℑ ( z) the imaginary part. complex numbers 1+i et 4+2*i, enter To use the calculator, one need to choose representation form of complex number and input data to the calculator. This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. complex number. Extended Keyboard; Upload; Examples; Random; Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Complex number calculator. Taylor series calculator | Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. Find more Mathematics widgets in Wolfram|Alpha. real part Example: Determine the conjugate of z=1+i z = 1 + i is to calculate ¯. determinant calculator | Equation calculator | Online plotter | Calculus derivatives | Taylor polynomial calculator | Angle Between Vectors Calculator. arccos calculator | Complex number is the combination of real and imaginary number. Site map As imaginary unit use, (1+i) (3+5i) = 1*3+1*5i+i*3+i*5i = 3+5i+3i-5 = -2+8, pow(1+2i,1/3)*sqrt(4) = 2.439233+0.9434225, pow(-5i,1/8)*pow(8,1/3) = 2.3986959-0.4771303, (6-5i)^(-3+32i) = 2929449.0399425-9022199.5826224, equation with complex numbers: (z+i/2 )/(1-i) = 4z+5i, system of equations with imaginary numbers: x-y = 4+6i; 3ix+7y=x+iy, multiplication of three complex numbers: (1+3i)(3+4i)(â5+3i), Find the product of 3-4i and its conjugate. The complex number calculator allows to perform calculations with complex numbers (calculations with i). Dividing Complex Numbers Calculator is a free online tool that displays the division of two complex numbers. Complex Number Calculator. Derivative calculator | arctan calculator | tan calculator | countdown numbers solver | Contact | It can be written in the form a + bi. Factorize expression online | hyperbolic coth calculator | Entering real or complex numbers is simple. Online calculator. A complex number is a number that can be expressed in the form a + bi where 'a' and 'b' are real numbers and 'i' is the imaginary unit, which satisfies the equation i 2 = -1. combination calculator online | BYJUâS online dividing complex numbers calculator tool performs the calculation faster and it displays the division of two complex numbers in a fraction of seconds. acos | Calculator online | When typing the imaginary part of a complex number in the appropriate field of the calculator, make sure that the symbol “ i “, representing the imaginary unit, is adjacent to the numeric part without space. Simplifying expressions calculator | Fraction calculator | English. Simplified fraction calculator | Il est conçu pour faire des opérations sur des nombres complexes. Solve equations online, Factor | The complex number calculator allows to calculates the sum of Online graphics | As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = −1 or j 2 = −1.The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). Trigonometric Form of Complex Numbers Calculator. arccos | cos calculator | limit finder | Factorize | Subtracting Complex Number Calculator. Example 1: to simplify $(1+i)^8$ type (1+i)^8 . In the following description, $$z$$ stands for the complex number. Expand and reduce math | This calculator operates using post-fix or RPN notation. tangent hyperbolic calculator | The maximum number of decimal places can be chosen between 0 and 10. Solution To see more detailed work, try our algebra solver Calculate integral online | Antidifferentiate | Maclaurin series calculator, Calculus online | All you want to provide is the given inputs in the below box & tap on the calculate button to get the final result along with a detailed solution. Conic Sections Trigonometry. permutation calculator | The complex number online calculator, allows to perform many operations on complex numbers. Calculate antiderivative online | It was designed by an engineer from the start specifically to handle complex math. Solver | Simplify expression online | Similar Algebra Calculator Adding Complex Number Calculator. Draw functions | Example: The complex number z z written in Cartesian form z =1+i z = 1 + i has for modulus √(2) (2) and argument π/4 π / 4 so its complex exponential form is z=√(2)eiπ/4 z = (2) e i π / 4 dCode offers a complex modulus calculator and a complex argument calculator tools. The calculator will generate a … Not only numbers … Simplifying complex expressions The following calculator can be used to simplify ANY expression with complex numbers. Complex Number. A1. We can convert the complex number into trigonometric form by finding the modulus and argument of the complex number. Equation system | B1 ( a + bi) A2. ... Equations Inequalities System of Equations System of Inequalities Polynomials Rationales Coordinate Geometry Complex Numbers Polar/Cartesian Functions Arithmetic & Comp. Online calculator | Expand math | complex numbers online, to calculate the sum of complex numbers 1+i and So, a Complex Number has a real part and an imaginary part. Tangent equation, Online math games for kids : Worksheets on Complex Number. Complex Number Calculator Precision 45 1.0.1: Un outil maniable, rapide, fiable, précis si vous devez exécuter des calculs avec des fonctions complexes. 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# Non split extension of lie algebra?

Is there a Lie Algebra $\mathfrak g$ so that the extension

$0\xrightarrow{}\mathfrak h\xrightarrow{}\mathfrak g\xrightarrow{}\mathfrak q\xrightarrow{}0$

does not split, i.e. $\mathfrak g$ is not a semi-direct product of $\mathfrak h$ and $\mathfrak q$? As I understand it, $\mathfrak q$ should not be identifiable with a subalgebra of $\mathfrak g$, right? Since $\mathfrak h$ is already an ideal in $\mathfrak g$ as the kernel.

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## 1 Answer

I guess the Heisenberg algebra will do. This is the algebra $\mathfrak{g}$ with basis $a, b, c$, and products $[a, b] = c$, $[a, c] = [b, c] = 0$. Take $\mathfrak{h} = \langle c \rangle$. Then $\mathfrak{q}$ is a two-dimensional abelian algebra, and all such object in $\mathfrak{g}$ contain $c$.

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I think I follow, $\mathfrak g/\mathfrak h$ would be abelian because $\mathfrak h = [\mathfrak g, \mathfrak g]$, and two-dimensional because $\mathfrak h$ is one-dimensional, and it must contain c, otherwise it wouldn't be closed under the lie bracket. But what does that mean? That $\mathfrak q$ cannot be the quotient of the other two and therefore not a subalgebra of $\mathfrak g$? –  rabipelais Jan 20 at 23:56
Ah, I think I see it, now, how dumb. Because q contains c it is not a direct product to h. Thanks. –  rabipelais Jan 21 at 3:08
@rabipelais Precisely. Sorry, I could have added a couple of words to make this clear. –  Andreas Caranti Jan 21 at 6:50
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The pulley and string shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium…

Q: The pulley and string shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle θ should be

(a)0°

(b)30°

(c)45°

(d)60°

Ans: (c)
Sol: Let T = tension in the string

T = mg

For equilibrium of √2 mg

$\large 2T cos\theta = \sqrt{2}mg$

$\large cos\theta = \frac{1}{\sqrt{2}}$