# What did Alan Turing mean when he said he didn't fully understand dy/dx?

Alan Turing's notebook has recently been sold at an auction house in London. In it he says this:

Written out:

The Leibniz notation $\frac{\mathrm{d}y}{\mathrm{d}x}$ I find extremely difficult to understand in spite of it having been the one I understood best once! It certainly implies that some relation between $x$ and $y$ has been laid down e.g. $$y = x^2 + 3x$$

I am trying to get an idea of what he meant by this. I imagine he was a dab hand at differentiation from first principles so he is obviously hinting at something more subtle but I can't access it.

• What does intuitive differentiation notation require?
• Does this notation bring out the full subtlety of differentiation?
• What did he mean?

See here for more pages of the notebook.

• Is there access to the rest of the page, for example, to what a) was? – Alamos Apr 30 '15 at 10:25
• They also quote "What is the way out? The notation (d/dx f(x, y))x=y,y=x hardly seems to help in this difficult case." which maybe is on the same page or close to it. He might have been thinking, among other things, about the notation pragmatically. It can be a mess, as in that example. – Alamos Apr 30 '15 at 10:34
• I guess no one could know the answer without further inside information. – Jonas Meyer Apr 30 '15 at 13:02
• To understand Turing's difficulty, try computing derivatives on Turing machine! – Robert Lewis Apr 30 '15 at 20:03
• When I was a maths undergraduate, I heard a (possibly apocryphal) quote attributed to Lamb (1849-1934), who was taught mathematics by Maxwell and Stokes, and who produced a revised edition of his textbook on calculus at the age of 70: "When I was 16, I understood calculus perfectly. Now I am 66, I am not sure I understand anything at all about it". Maybe Turing was following the same path! – alephzero Apr 30 '15 at 21:41

I'm guessing too, but my guess is that it has something to do with the fact that beyond his initial introduction to calculus Turing (in common with many of us) thinks of a function as the entity of which a derivative is taken, either universally or at a particular point. In Leibnitz's notation, $y$ isn't explicitly a function. It's something that has been previously related to $x$, but it actually is a variable, or an axis of a graph, or the output of the function, not the function or the relation per se.

Defining $y$ as being related to $x$ by $y = x^2 + 3x$, and then writing $\frac{\mathrm{d}y}{\mathrm{d}x}$ to be "the derivative of $y$ with respect to $x$", quite reasonably might seem unintuitive and confusing to Turing once he's habitually thinking about functions as complete entities. That's not to say he can't figure out what the notation refers to, of course he can, but he's remarking that he finds it difficult to properly grasp.

I don't know what notation Turing preferred, but Lagrange's notation was to define a function $f$ by $f(x) = x^2 + 3x$ and then write $f'$ for the derivative of $f$. This then is implicitly with respect to $f$'s single argument. We have no $y$ that we need to understand, nor do we need to deal with any urge to understand what $\mathrm{d}y$ might be in terms of a rigorous theory of infinitesimals. The mystery is gone. But it's hard to deal with the partial derivatives of multivariate functions in that notation, so you pay your money and take your choice.

• I always used Lagrange notation. Fortunately, I wasn't ever docked points for this on a test. And for partial derivatives of multivariate functions you just put the variable you are taking the partial derivative of as a subscript. I never liked dy/dx notation because it suggests more algebraic manipulations than you're entitled to in some cases. – Atsby May 1 '15 at 2:44
• Atsby - Funnily enough, I always use dy/dx notation because it suggests algebraic manipulations that may be possible - compare the chain rule in Liebniz vs Newton/Lagrange notation. – Peter Webb May 1 '15 at 13:31
• It means "I, too, am guessing." The implication is that other people are guessing and so is he. In other words: "Other people are guessing. I am also guessing." – Michael Laszlo May 2 '15 at 6:09
• An easy generalization is to write $f_x$, $f_y$ and so on for partial derivatives. – cfh May 2 '15 at 8:50
• I very much like this answer, but don't agree with the conclusion about the partial derivative notation. It would be quite easy to introduce a consistent notation like say $f_{,1},f_{,2}$ to denote partial derivatives of a function $f:\mathbb{R}^2\to\mathbb{R}$ with respect to the first and last argument, and allow $\partial z/\partial x$ only when $z$ is a variable depending on $x,y$. – Michael Bächtold Oct 24 '16 at 8:01

I'll try my hand and give one possible rationale behind Turing's confusion over the notation $\tfrac{\mathrm{d}y}{\mathrm{d}x}$. The short answer is that he appears to take issue with the notation on the grounds that differentiation is a mapping between two function spaces but $y$ looks like a variable.

To answer the first part of your question regarding depth and his meaning, I base my answer on the discussion in the linked webpage. Based on the dating of the notes and the discussion in the pictures, I doubt that his contention is regarding actual interpretation in terms of differentials $\mathrm{d}x, \mathrm{d}y$ and is instead more pedantic in nature. Earlier in the discussion, he talks about indeterminates and the difference between an indeterminate and a variable. Later, he writes "What is the way out? The notation $\tfrac{\mathrm{d}}{\mathrm{d}x} f(x, y)_{x=y,y=x}$ hardly seems to help in this difficult case". From this I gander that it's primarily the fact that he doesn't like the use of $y$ as something being differentiated. He states that $y = x^2 + 3x$ as if to say that you could alternatively just rearrange the equation in terms of $x$; however, in taking the derivative, you get another function - that is, you have a function $f(x)$ that becomes $g(x)$ as a result of differentiation $D:f(x) \rightarrow g(x)$. Thus, $y$ can't be a variable but rather a function if we want differentiation to be well-defined. Think of it as something akin to abuse of notation.

Regarding intuition and subtlety, Leibnitz's notation does indeed provide both although in truth $$\dfrac{\mathrm{d}}{\mathrm{d}x} f(x) = g(x)$$ is clearer than using $y$. From a more intuitive point of view, you can think of the derivative as a variation in $f$ with respect to $x$ and indeed it's this notion that is more important than the one regarding secants and tangents. The subtlety of the notation is readily apparent when people are using the chain rule, even though you're not really cross-multiplying (for instance it doesn't work with second order derivatives). The subtlety becomes more apparent when you abstract even further to vector calculus or even further to differentiable manifolds.

• Looks like this is closest to a realistic answer. – nbubis Apr 30 '15 at 15:09
• But you have to take into account the historical origin of Leibniz' symbols : plot a curve in the $x-y$ plane and consider a point $A(x,y)$ of the curve. Leibniz coined the term differential for the infinitely small difference $dx$ and $dy$. Then $dy/dx$ represents the slope of the tangent line of the curve at point $A$. For Leibniz there is no "general" function $f$ here, but a curve ... – Mauro ALLEGRANZA Apr 30 '15 at 19:45
• See The Early Mathematical Manuscripts of Leibniz (1920, Dover reprint), page 137. – Mauro ALLEGRANZA Apr 30 '15 at 20:21
• Thanks Mauro! I can't say I'm completely surprised in the same way that classical differential geometry is more concerned with curves compared to modern expositions that deal with functions (granted I'm no expert). When I mention differentials, what I meant was the definition as linear maps, namely pushforwards, and not as infinitesimals. It is wholly possible he was indeed craving a rigorous definition of the latter as pointed out by others who refer to Robinson. – BaNaL May 1 '15 at 10:57

I'm no expert on Alan Turing at all, and the following will not directly answer your question, but it might give some context. Following the link provided, I also found the following page, which might shed some more light on things:

It says the following (my apologies for any errors in transcribing, I'm grateful for any corrections):

A formal expression $$f(x) = \sum_{i=1}^n \alpha_i x^i$$ involving the indeterminate' (or variable) $x$, whose coefficients $\alpha_i$ are numbers in a field $K$, is called a ($K$-)polynomial of formal degree $n$.

The idea of an indeterminate' is distinctly subtle, I would almost say too subtle. It is not (at any rate as van der Waerden [link added by me] sees it) the same as variable. Polynomials in an indeterminate $x$, $f_1(x)$ and $f_2(x)$, would not be considered identical if $f_1(x)=f_2(x)$ [for] all $x$ in $K$, but the coefficients differed. They are in effect the array of coefficients, with rules for multiplication and addition suggested by their form

I am inclined to the view that this is too subtle and makes an inconvenient definition. I prefer the indeterminate $x$ [?] be just the variable.

I think one thing to keep in mind here is that at this time anything relating to `computability' was not as clear as today. After all, Turing (an Church & Co.) were just discovering the essential notions.

In particular questions of intensionality vs. extensionality could have been an issue. It might be that Turing was pondering on the difference between functions (and also operations on functions) from a purely mathematical point of view (i.e., functions as extensional objects) vs. a computational point of view (i.e., functions as some form of formal description of a calculation process, which a priori can not be looked at in an extensional way).

All of this can be still seen in the context of the foundational crisis of mathematics (or at least strong echoes thereof). Related to this, are of course, questions of rigour, formalism, and denotation. This, in turn, is where your quote comes in. As others have outlined, Turing might have asked the question, what $\frac{\mathrm{d}y}{\mathrm{d}x}$ is (from a formal point of view), but also what is denoted by it, and (to his frustration) found that the answer to his question was not as clear as he wanted it to be.

• The word you're not sure of appears to be 'array' to me. – postmortes May 1 '15 at 12:05
• @postmortes Yes, thanks, this looks right to me. – godfatherofpolka May 1 '15 at 12:06
• It's funny to me that Turing thinks it is too subtle to treat polynomials with coefficients in $K$ as formal expressions and prefers to regard them as functions $K \rightarrow K$, considering the importance of finite fields in computer science. They are exactly the fields where this change in point of view really wrecks the algebraic structure of polynomials. – KCd May 2 '15 at 0:58
• @KCd True, but I guess if you do not yet have things like programming languages (or more formal versions thereof) to talk about, polynomials are not the worst thing to talk about the fact that two things (when treated as a black box w.r.t. their output) can look the same and yet be quite different. Nowadays we've got it easy, we can just think of different sorting algorithms, for example. It's actually quite interesting to think about the fact that these notions, now so clear to us, were very far from clear less than one hundred years ago. – godfatherofpolka May 2 '15 at 5:40

This phrase from the description catches my eye, and prompts me to provide a different perspective from the other answers.

This, Turing's wartime notebook on logic, is the first time a manuscript by him has ever come to public market.

The concept of a derivative is a rather natural notion, and it is easy for young students of mathematics to intuitively understand. What Turing is writing about is the Leibniz notation for infinitesimals in the context of mathematical logic.

In the 60s, Robinson published his non standard analysis, which successfully defines Leibniz's infinitesimals in a manner that is consistent with the properties of the reals. Note the date: this was published the decade after Turing died. I believe this rigor is what Turing was searching for, and in his notes he is commenting about the difficulty of the task.

To expand upon zoli's reply a bit, the Leibniz notation intuitively makes sense in that the derivative of an explicit formula is algebraically defined as $\frac{\Delta y}{\Delta x}$ for any $(x, y)$ pair around which the function is defined (i.e. 'the relation is laid down').

Of course it's all speculation, but perhaps Turing meant to say that the notation was harder to understand (i.e. unclear, not necessarily difficult) later in his career, when for example considering higher order derivatives or trying to think about derivatives without the constant mental reminder of the explicit relation imparted by the notation.

We, who learn differentiation based on calculating the limit of quotients of differences, consider the Lebniz notation to be only a notation. We do not try to grab the intuitive meaning of $dx$ and $dy$. We don’t even care.

I guess that Turing was trying to understand how Lebniz could reach his results based on this vague notation. I guess that he refers to two things:

1. For Leibniz $\frac{dx}{dy}$ must have had a special intuitive meaning if the relationship between $y$ and $x$ was given. So for Leibniz it was not only a notation then.

2. Turing perhaps had gone through that leibnizian vague understanding before learning the standard approach. Turing must have been thinking about his own losing this once existing transcendental understanding that got destroyed in his mind by the mathematical discipline developed in the 19th century.

• By standard approach do you mean from first principles? Is there a way of maintaining this intuitive link or has the notation irrevocably lost it? Should we even attempt to achieve this with notation? Sorry if I'm being slow or vague here; both your response and Fasermaler's were very helpful. – Matta Apr 30 '15 at 10:36
• You cannot be slower than I am. (Vaguer -- perhaps : ) By standard approach, I mean the $\epsilon, \delta$ discipline. As far as the the term "irrevocable": the non-standard analysis of -- p.ex. Rosbinson's gave a firm basis for direct handling infinitesimals. However, I am not an expert on that. But, I have my personal experience with "understanding something in a way before" then completely rejecting that way. (I am not a Turing though ... unfortunately) – zoli Apr 30 '15 at 10:46
• I'm not an expert in non-standard analysis either, but from what I know, it is probably not worth learning, since a lot of it can be done easily in the standard reals using formal manipulation of asymptotic expansions in conjunction with raw ε-δ definitions. It seems that in every advantage of non-standard reals over raw ε-δ approaches, asymptotic methods do far better. In fact, asymptotic methods are so versatile yet highly deterministic that modern computer algebra systems use them. – user21820 Apr 30 '15 at 13:41
• @Matta: Anyway there is another interpretation of those things using differentials, which you may want to look up as well. I believe those get quite close to the original idea of looking at the relationship between the small changes in the related quantities. – user21820 Apr 30 '15 at 13:42