Difference between revisions of "Inverse"
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===Compositional inverseInverses and composition=== | ===Compositional inverseInverses and composition=== | ||
If | If <math>f</math> is an invertible function with domain <math>X</math> and codomain <math>Y</math>, then | ||
: <math> f^{-1}\left( \, f(x) \, \right) = x</math>, for every <math>x \in X</math>; and <math> f\left( \, f^{-1}(y) \, \right) = y</math>, for every <math>y \in Y. </math>.<ref name=":2" /> | : <math> f^{-1}\left( \, f(x) \, \right) = x</math>, for every <math>x \in X</math>; and <math> f\left( \, f^{-1}(y) \, \right) = y</math>, for every <math>y \in Y. </math>.<ref name=":2" /> | ||
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: <math> f^{-1} \circ f = \operatorname{id}_X</math> and <math>f \circ f^{-1} = \operatorname{id}_Y, </math> | : <math> f^{-1} \circ f = \operatorname{id}_X</math> and <math>f \circ f^{-1} = \operatorname{id}_Y, </math> | ||
where | where <math>id<sub>''X''</math> is the [[identity function]] on the set <math>X</math>; that is, the function that leaves its argument unchanged. In [[Category Theory]], this statement is used as the definition of an inverse [[morphism]]. | ||
Considering function composition helps to understand the notation {{math|''f''<sup> −1</sup>}}. Repeatedly composing a function with itself is called [[iterated function|iteration]]. If {{mvar|f}} is applied {{mvar|n}} times, starting with the value {{mvar|x}}, then this is written as {{math|''f''<sup> ''n''</sup>(''x'')}}; so {{math|''f''<sup> 2</sup>(''x'') {{=}} ''f'' (''f'' (''x''))}}, etc. Since {{math|''f''<sup> −1</sup>(''f'' (''x'')) {{=}} ''x''}}, composing {{math|''f''<sup> −1</sup>}} and {{math|''f''<sup> ''n''</sup>}} yields {{math|''f''<sup> ''n''−1</sup>}}, "undoing" the effect of one application of {{mvar|f}}. | Considering function composition helps to understand the notation {{math|''f''<sup> −1</sup>}}. Repeatedly composing a function with itself is called [[iterated function|iteration]]. If {{mvar|f}} is applied {{mvar|n}} times, starting with the value {{mvar|x}}, then this is written as {{math|''f''<sup> ''n''</sup>(''x'')}}; so {{math|''f''<sup> 2</sup>(''x'') {{=}} ''f'' (''f'' (''x''))}}, etc. Since {{math|''f''<sup> −1</sup>(''f'' (''x'')) {{=}} ''x''}}, composing {{math|''f''<sup> −1</sup>}} and {{math|''f''<sup> ''n''</sup>}} yields {{math|''f''<sup> ''n''−1</sup>}}, "undoing" the effect of one application of {{mvar|f}}. | ||
Revision as of 14:25, 6 August 2021
The notion of inverse in mathematical operation is simply the anti-operator.
Excerpt from Wikipedia
The following paragraph is copied from Wikipedia.
Compositional inverseInverses and composition
If is an invertible function with domain and codomain , then
- , for every ; and , for every .[1]
Using the composition of functions, we can rewrite this statement as follows:
- and
where is the identity function on the set ; that is, the function that leaves its argument unchanged. In Category Theory, this statement is used as the definition of an inverse morphism.
Considering function composition helps to understand the notation f −1. Repeatedly composing a function with itself is called iteration. If f is applied n times, starting with the value x, then this is written as f n(x); so f 2(x) Template:= f (f (x)), etc. Since f −1(f (x)) Template:= x, composing f −1 and f n yields f n−1, "undoing" the effect of one application of f.
- ↑ Cite error: Invalid
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