2.2.9 Advanced Topic: Working with OperatorsYou can think of an expression like f[x] as being formed by applying an operator f to the expression x. You can think of an expression like f[g[x]] as the result of composing the operators f and g, and applying the result to x.
| Composition[f, g, ... ] | the composition of functions f, g, ... | | InverseFunction[f] | the inverse of a function f | | Identity | the identity function |
Some functional operations. | This represents the composition of the functions f, g and h. | |
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Composition[f, g, h]
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| You can manipulate compositions of functions symbolically. | |
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InverseFunction[Composition[%, q]]
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| The composition is evaluated explicitly when you supply a specific argument. | |
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You can get the sum of two expressions in Mathematica just by typing x + y. Sometimes it is also worthwhile to consider performing operations like addition on operators. | You can think of this as containing a sum of two operators f and g. | |
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| Using Through, you can convert the expression to a more explicit form. | |
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This corresponds to the mathematical operator  . | |
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Identity + (D[#, x]&)
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| Mathematica does not automatically apply the separate pieces of the operator to an expression. | |
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| You can use Through to apply the operator. | |
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| Identity[expr] | the identity function | Through[p[ ,  ][x], q] | give p[ [x],  [x]] if p is the same as q | | Operate[p, f[x]] | give p[f][x] | | Operate[p, f[x], n] | apply p at level n in f | | MapAll[p, expr, Heads->True] | apply p to all parts of expr, including heads |
Operations for working with operators. | This has a complicated expression as a head. | |
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t = ((1 + a)(1 + b))[x]
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| Functions like Expand do not automatically go inside heads of expressions. | |
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| With the Heads option set to True, MapAll goes inside heads. | |
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MapAll[Expand, t, Heads->True]
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| The replacement operator /. does go inside heads of expressions. | |
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| You can use Operate to apply a function specifically to the head of an expression. | |
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