4.3 Notation

Mathematicians use several notations for expressing derivatives. Myron supports two: Leibnitz notation and LaPlace notation.

Leibnitz notation combines a derivand and a derivator into what is essentially a binary operator: “derive this with respect to that”. Leibnitz notation is often used to produce standalone expressions; everything that is required to define the derivative is specified in the notation. Thus ⅆx^2ⅆx can be simplified to 2⋅x.

Leibnitz notation is sometimes treated as if the derivand and the derivator are separate terms. Thus the equation ⅆx:1÷ⅆy:1=1 can be transformed to ⅆx:1=ⅆy:1. To support this usage, Myron provides decoupled derivatives, in contrast to the combined, or coupled derivative in the previous paragraph. Decoupled derivatives that appear as operands of a divide operator act like the coupled derivative they resemble.

Sometimes the derivand is not an expression and is instead a function reference. ⅆf(x)ⅆx does not contain enough information to be transformed using derivation, but it does contain enough information to be transformed to LaPlace notation. That is, .{ⅆf(x)ⅆx.} and special Simplify produce f’(x). These two expressions are related; special Simplify on the latter reproduces the former.

Laplace notation extends to antiderivatives. Replacing the previous example with integration, special simplification of .{∫f(x) ⅆx.} produces f'{-1}(x)+ĉ.

LaPlace notation for higher-order derivatives and anti-derivatives uses a higher tick count. Nested derivatives like f’’’(x) transform (after three special simplifications) to ⅆⅆⅆf(x)ⅆxⅆxⅆx. This in turn can be transformed to the Leibnitz form ⅆf(x)ⅆx:3. Similarly, the antiderivative of f'{-3}(x) transforms to ∫∫∫f(x) ⅆx ⅆx ⅆx.

LaPlace notation is useful when the elaboration of a function is not relevant. For example, Newton's method for finding a root can be expressed in part using the recursive relation c_(n+1)=c_n-f(c_n)÷f’(c_n). To use this function in a concrete way, f and f’ must be bound to global definitions, say, f(x)→sin x and f’(x)→cos x. (The second function can be produced symbolically by converting a copy of the first to a differential function and simplifying the right side.) By choosing this f(x), Newton's method will find the root of sin x, i.e., the point at which the sin curve crosses the x-axis. Then the recursive relation must be expressed as a function.

Here, x is the number of iterations, descending recursively to 0, where iteration 0 returns the approximation 3. N(2) evaluates to 3.141592653300477 and N(3) evaluates to 3.141592653589793. N(3) differs from ℼ in the tenth decimal place.