# Differential Lambda Calculus

How do we convert Celsius to Farenheit? For the sake of argument, suppose we know to multiply by a constant p then add another constant q:

▶ Toggle boilerplate

type R = Double
convert:: R -> R -> R -> R
convert p q x = p * x + q


…but we have forgotten the constants. So we guess their values and hope for the best:

stab :: R -> R
stab = convert (1.23) (4.56)


Furthermore, suppose we remember water boils at 100 degrees Celsius or 212 degrees Farenheit. We run stab 100. Instead of 212 we see 127.56. We guessed wrong.

We can measure for how far off we are for a given input x and expected output y.

oops :: R -> R -> R -> R -> R
oops p q x y = (convert p q x - y)^2


In our case, the error is:

oops 1.23 4.56 100 212 == 7130.1136

How can we fix our guesses for p, q so that for all test cases x, y, the error oops p q x y is zero, or at least close to zero?

Let’s focus on p. Define:

focusp p = oops p 4.56 100 212

We approximate focusp around p = 1.23 with a straight line, using a method we later describe:

focusp (1.23 + dp) ~= focusp 1.23 - 16888 * dp

This minus sign tells us something. Nudging p upwards from 1.23 reduces the error focusp p quite a bit, so we decide to increase p by a small amount. If instead the coefficient of dp were positive, we would decrease p.

We switch focus to q and find a linear approximation in terms of a variable dq at q = 4.56, and adjust q similarly, at which point we repeat the whole process using our improved guesses for p and q. With luck, we’ll home in on the correct constants after iterating many times. This technique is known as gradient descent.

(We could be cleverer and modify our guesses so the approximate error is exactly zero, which is called Newton’s method. Does it work well for machine learning? Good question!)

Recall at our particular choices of p, q, x, y, the coefficient of dp is -16888. It turns out in general, the coefficient of dp is:

oopsp p q x y = 2*x*(convert p q x - y)


and that of dq is:

oopsq p q x y = 2*(convert p q x - y)


Next, we remember another piece of trivia: minus 40 is the same temperature in Celsius and Farenheit. This gives us a second test case, that is, convert p q (-40) == -40 when p, q are correct.

We repeatedly nudge our guesses according to our two test cases:

rate = 0.0001
step x y (p, q) = (p - oopsp p q x y * rate, q - oopsq p q x y * rate)
learn = iterate (step 100 212 . step (-40) (-40)) (1.23, 4.56)


where we’ve defined the size of a nudge to be 0.0001 times the slope of the linear approximation. This is because the steeper the slope, the greater the error, so we are correspondingly bolder about changing our guess.

We find:

learn!!50000 == (1.8000000000067133,31.99999999798702)

which is close to the true constants (9/5, 32).

Why choose 0.0001? This fussy parameter is known as the learning rate. The smaller it is, the longer it takes to reach a good answer. But if it is too large, those gentle nudges become violent shoves and our guesses are forever all over the map. Researchers have explored many ideas to tune the learning rate, but in our case, we picked 0.0001 because it happened to work well for our toy example!

The convert function is simple enough that there are direct ways of figuring out the right answer. However, we can imagine problems where the function involves thousands of guessed parameters. As long as we can compute linear approximations for the error with respect to each parameter, we can iteratively improve our guesses using the above process.

It remains to explain how we obtained linear approximations. In short, we used differential calculus.

## With no undue respect

We say "differential calculus" and not "derivative calculus", yet we tend to think in terms of derivatives and not differentials. Faced with an expression, our first instinct is to single out a variable and take a derivative with respect to this variable. Why?

The root cause is a myth perpetuated by generation after generation of teachers and textbooks, that claim the Leibniz notation $$dy/dx$$ is a mere mnemonic device. Woe betide those who dare to reason algebraically with it!

This is a lie. Below, we define the differential $$d$$ as a function from lambda terms to lambda terms, and $$dy/dx$$ is an algebraic expression like any other. As we’d expect, it means $$d$$ applied to $$y$$ divided by $$d$$ applied to $$x$$.

Knowing the truth lets us forget derivatives and study differentials such as:

$d(3 x^2 + 2 y) = 6 x dx + 2 dy$

A differential describes the effects of small changes in the inputs. More precisely, it is the best linear approximation to a given function at a given point. Unlike derivatives, no variable gets special treatment. Differentials respect all variables equally.

Above, we see changing $$x$$ by $$dx$$ leads to a change of approximately $$6x dx$$ in the output value, and similarly changing $$y$$ by $$dy$$ leads to an change of approximately $$2 dy$$. The plane through $$(x, y)$$ with slope $$6x$$ in the $$x$$-direction and $$2$$ in the $$y$$-direction lies tangent to the surface represented by this function.

If we discover $$y$$ depends on $$x$$, say $$y = x^3$$, then we can compute the differential $$dy = 3x^2 dx$$ and substitute into the above to get $$(6x + 6x^2) dx$$ using plain algebra.

After years of heavy use of derivatives, taking differentials may seem alien. Fortunately, a few lines saves us from figuring them out ourselves.

data V = S String | Dee V deriving Eq
data Expr = Con Int | Var V | Lam V Expr | Expr :@ Expr
| Inv | Sin | Cos | Exp | Log | Expr :+ Expr | Expr :* Expr | Expr :^ Expr

d :: Expr -> Expr
d expr = case expr of
Con _ -> Con 0
Var v -> Var $Dee v x :+ y -> d x :+ d y x :* y -> (x :* d y) :+ (d x :* y) x :^ y -> (y :* (x :^ (y :+ Con (-1))):* d x) :+ ((Log :@ x) :* (x :^ y) :* d y) Lam v x -> Lam (Dee v)$ d x
f :@ x | Lam (Dee v) y <- d f -> sub (Dee v) (d x) $sub v x y Inv -> lzdz$ Con (-1) :* (Inv :@ (z :* z))
Log -> lzdz $Inv :@ z Exp -> lzdz$ Exp :@ z
Sin -> lzdz $Cos :@ z Cos -> lzdz$ Con (-1) :* (Sin :@ z)
where
z = Var $S "z" lzdz x = Lam (Dee$ S "z") $x :* Var (Dee$ S "z")

sub :: V -> Expr -> Expr -> Expr
sub v x y = case y of
Var w | v == w -> x
Lam w b | v /= w -> Lam w $rec b a :+ b -> rec a :+ rec b a :* b -> rec a :* rec b a :^ b -> rec a :^ rec b a :@ b -> rec a :@ rec b _ -> y where rec = sub v x  Our variables differ from those of standard lambda calculus. In our world, a variable can be a run-of-the-mill variable like x, but it can also be a differential of a variable, such as d x. We may take differentials recursively, so if x is a variable, then so is d x, d d x, d d d x, ... (perhaps we should call these Peano variables). Instead of $$dddx$$, for example, mathematicians customarily write $$d^3 x$$. We define functions to pretty-print our expressions, or at least make them less ugly than the default. instance Show V where show (S s) = s show (Dee v) = "d " ++ show v instance Show Expr where show expr = case expr of Con c -> show c Var v -> show v x :+ y -> concat ["(", show x, " + ", show y, ")"] x :* y -> concat ["(", show x, "*", show y, ")"] x :^ y -> concat ["(", show x, "^", show y, ")"] x :@ y -> concat ["(", show x, " ", show y, ")"] Inv -> "(1/)" Sin -> "sin" Cos -> "cos" Exp -> "exp" Log -> "log" Lam v y -> concat ["\\", show v, " -> ", show y]  We supply a parser so it’s less painful to play with our functions. line :: Parsec String () Expr line = expr <* eof where expr = pwr chainl1 ((spch '+' *> pure (:+)) <|> (spch '-' *> pure (\x y -> x :+ (Con (-1) :* y)))) pwr = term chainr1 (spch '^' *> pure (:^)) term = apps chainl1 ((spch '*' >> pure (:*)) <|> (spch '/' *> pure (\x y -> x :* (Inv :@ y)))) apps = dOrApply id <$> many1 atm
dOrApply acc [Just one]     = acc one
dOrApply acc (Nothing:rest) = acc (d $dOrApply id rest) dOrApply acc (Just f:rest) = dOrApply (acc . (f :@)) rest atm = Just <$> (lam <|> num <|> between (spch '(') (spch ')') expr)
<|> dWord <$> many1 letter <* spaces dWord s = if s == "d" then Nothing else Just$ word s
word "sin" = Sin
word "cos" = Cos
word "exp" = Exp
word "log" = Log
word s     = Var $S s lam = spch '\\' *> do Var v <- apps string "->" *> spaces *> (Lam v <$> expr)
num  = Con . fromDecimal <$> (many1 digit <* spaces) spch :: Char -> Parsec String () Char spch c = char c <* spaces fromDecimal = foldl (\n d -> 10*n + fromEnum d - fromEnum '0') 0  Our user interface is peculiar. Function application is left-associative (to match the conventions of lambda calculus and combinatory logic), while d is right-associative (so that d d d x means $$dddx = d^3 x$$). Unlike Haskell, lambdas bind exactly one variable, so that we can more easily parse \d x as the lambda binding the differentiable variable $$dx$$. There is no unary minus, so we write negative integers as, for example, 0 - 42. Lastly, we add some code for the interactive demo at the top of this page: go :: String -> String go s = case parse line "" s of Left err -> "parse error: " ++ show err Right expr -> show expr #ifdef __HASTE__ main :: IO () main = withElems ["in", "out", "go"]$ $iEl, oEl, goB] -> do let setup button text = do Just b <- elemById button let act = do setProp iEl "value" text setProp oEl "value" "" void  b onEvent Click  const act when (button == "example") act setup "example" "d((\\z -> z*z)(p*100 + q - 212))" setup "implicit" "d(\\x -> \\y -> sin(x + y) - cos(x*y) + 1)" setup "lambda" "d ((\\z -> z^3) (x*x + y^2) - (\\z -> z*z) (x^2 - y*y))" setup "second" "d (d y / d x) / d x" let diff = setProp oEl "value" . go =<< getProp iEl "value" void  goB onEvent Click  const  diff void  iEl onEvent KeyDown  \key -> when (key == mkKeyData 13)  diff >> preventDefault #endif  ## A second opinion We first test our code by computing the second derivative of some variable $$y$$ with respect to a variable $$x$$: go "d (d y / d x) / d x" We get: (((d y*((-1*((1/) (d x*d x)))*d d x)) + (d d y*((1/) d x)))*((1/) d x)) Our program isn’t quite ready to take over our calculus homework because it fails to collect like terms and fold constants and such. However, we can manually simplify to get: \[\frac{d(\frac{dy}{dx})}{dx} = \frac{ddy}{dx^2} - \frac{dy}{dx} \frac{ddx}{dx^2}$

This is the correct way to write the second derivative.

The incorrect but widespread $$d^2 y / dx^2$$ notation for the second derivative with respect to $$x$$ is another pernicious consequence of failing to properly appreciate Leibniz notation.

Some might complain that the truth is too verbose. This is no justification for propagating falsehoods. If brevity is paramount, we can always write the unexpanded $$d(dy/dx)/dx$$ or use Arbogast’s notation $$D^2_x y$$.

It makes no sense to half-heartedly write pseudo-Leibniz terms. It is useless as a mnemonic because it is wrong, and because it is wrong it reinforces the myth that differentials cannot be algebraically reasoned with.

## To a certain degree

Our code shows that:

$d ((p * 100 + q - 212)^2)$

evaluates to (after simplification):

\begin{align} & & 200 * p * (p * 100 + q - 212) * dp \\ &+& 2 * p * (p * 100 + q - 212) * dq \end{align}

confirming our claims about the coefficients of dp and dq.

Our code lacks support for symbolic constants, but it’s clear how 100 and 212 can be replaced by x and y. Alternatively, we can compute d ((p * x + q - y)^2) and ignore the dx and dy terms.

## Think differential

In the dark ages, generalizations of our self-taught temperature conversion program were called neural networks. More recently, we’ve been saying deep learning. Even more recently, this area seems to have been rebranded as differentiable programming.

This reflects a shift in thinking. Although the underlying mathematics remains the same, the emphasis is no longer on crudely modeling the human brain, nor stitching together layers of matrix multiplications. The latest fashion is to program with differentiable functions and improve them with gradient descent, or perhaps other methods. See Conal Elliot, A Functional Reboot for Deep Learning.

In a 2019 paper, Bartlett and Khurshudyan overturn centuries of abuse of Leibniz notation, and bring differentials back to the realm of algebra. They show how the correct notation for the second derivative fits perfectly with the chain rule for the second derivative.

Our live demo shows off the second derivative. Other examples were taken from Wikipedia’s entries on implicit curves and the quadrifolium.

See also another paper by Bartlett, which discusses partial derivatives, and gives a shoutout to the rule:

$d(u^v) = v u^{v-1} du + \log(u) u^v dv$

which our code supports.

I suspect "perturbation confusion" is another casualty of the undeserved status of derivatives over differentials. An example in the paper goes straight for the derivative, asking for $$\frac{d}{dx} (x + y)$$. Similarly, they define a "derivative-taking operator" that we’ll write as D.

If we think with differentials instead, we have $$d(x + y) = dx + dy$$, and dividing by $$dx$$ gives $$1 + dy/dx$$, which doesn’t seem to be what they want. I believe they really meant $$\partial_x (x + y) = 1$$. Similarly, I believe they really meant D to be a "partial-derivative-taking operator", that is D (\x -> e) means \x -> partial x (d e), where partial x sets all differential variables to zero except for d x. With these definitions, there are no surprises; the example evaluates to what we expect.

A more modern paper agrees with my interpretation (see Section 4.3).

## Two weird tricks

It may seem we could beef up our code to get something like TensorFlow, that is, a system that automatically performs gradient descent on a given differentiable function containing unknown parameters. However, our approach for computing differentials turns out to scale poorly. Better is automatic differentiation, a vague-sounding term that encompasses two tricks:

1. Memoization. For every sub-expression $$f$$, we keep around $$(f, df)$$, so we avoid recomputing the same sub-expressions over and over again.

2. Using values instead of symbols. Rather than compute a formula for a differential that we later apply to particular values, we just compute with values all the time.

An example of the first trick: suppose we wish to compute $$d(f g h)$$. The naive method requires us to compute:

$(df) g h + f (dg) h + f g (dh)$

With memoization, we first compute and remember:

$(g h, d(g h)) = (g h, g (dh) + (dg) h)$

which we later use to compute:

$d(f g h) = (df) g h + f d(g h)$

The more functions in our product, the more memoization saves.

An example of the second trick: if we know $$(f, df) = (1, 2 dx)$$ and $$(g, dg) = (3, 4 dx)$$ then the product rule gives $$(f g, d(f g)) = (3, 10 dx)$$. Algebra is great, but sticking with values sure is easier than manipulating symbols.

Automatic differentiation has a forward mode and a reverse mode. The latter is also called backpropagation in some contexts. These modes relate to the chain rule, which in our implementation is hidden in lambdas and applications. Working through the details, we find the chain rule leads to expressions like:

$3 \times 4 \times 5 \times (dp + 2 dq + 4 dr)$

My understanding is that in forward mode, the multiplications associate to the right:

\begin{align} & & 3 \times 4 \times 5 \times (dp + 2 dq + 4 dr) \\ &=& 3 \times 4 \times (5 dp + 10 dq + 20 dr) \\ &=& 3 \times (20 dp + 40 dq + 80 dr) \\ &=& 60 dp + 120 dq + 240 dr \end{align}

while in reverse mode the multiplications associate to the left, which is more efficient:

\begin{align} & & 3 \times 4 \times 5 \times (dp + 2 dq + 4 dr) \\ &=& 12 \times 5 \times (dp + 2 dq + 4 dr) \\ &=& 60 \times (dp + 2 dq + 4 dr) \\ &=& 60 dp + 120 dq + 240 dr \end{align}

We can view the right-most factor as a list [1,2,4], in which case reverse mode is the result of applying a fusion law to forward mode:

map (3*) . map (4*) . map (5*) = map (60*)

By the way, the second trick shows up elsewhere. In certain areas of cryptography, we need to compute a function related to the Weil pairing. The numbers involved are so large that it’s infeasible to write the function in terms of two input points given symbolically, say $$(x_1, y_1)$$ and $$(x_2, y_2)$$.

Thus in practice, to compute this function on two given points, we roughly act as if were trying to derive a formula, but always use values instead of symbols. This is known as Miller’s algorithm.

Automated theorem proving profits from turning this trick on its head. Early theorem provers exhaustively tried every possible value for every variable. Later provers improved on this by by computing with variables instead.

Ben Lynn blynn@cs.stanford.edu 💡