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The Road to Reality Study Notes: Difference between revisions
→6.3 Higher derivatives; $$C^\infty$$-smooth functions
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In general, a function can be smooth for many derivatives and the mathematical terminology for general smoothness is to say that $$f(x)$$ is $$C^n$$-smooth. It can be seen that $$x|x|$$ is $$C^1$$-smooth but not $$C^2$$-smooth due to the discontinuity at the origin in the derivative. In general $$x^n|x|$$ is $$C^n$$-smooth but not $$C^{n+1}$$-smooth. In fact, a function can be $$C^\infty$$-smooth if it is smooth for every positive integer. Note that negative integers for $$x^n$$ immediately are not smooth for $$x^{-1}$$ (discontinuous at the origin). | In general, a function can be smooth for many derivatives and the mathematical terminology for general smoothness is to say that $$f(x)$$ is $$C^n$$-smooth. It can be seen that $$x|x|$$ is $$C^1$$-smooth but not $$C^2$$-smooth due to the discontinuity at the origin in the derivative. In general $$x^n|x|$$ is $$C^n$$-smooth but not $$C^{n+1}$$-smooth. In fact, a function can be $$C^\infty$$-smooth if it is smooth for every positive integer. Note that negative integers for $$x^n$$ immediately are not smooth for $$x^{-1}$$ (discontinuous at the origin). | ||
Penrose notes that Euler would have required $$C^\infty$$-smooth functions to be defined as functions, and then gives the function:<math>h(x)=/ | Penrose notes that Euler would have required $$C^\infty$$-smooth functions to be defined as functions, and then gives the function:<math>h(x)=</math> \overset{0 if x\le0}{e^{-\frac{1}{x}}} if x>0</math> as an example of a $$C^\infty$$-smooth function but one that Euler would still not be happy with since it is two functions stuck together. | ||
=== 6.4 The "Eulerian" notion of a function? === | === 6.4 The "Eulerian" notion of a function? === | ||