transgalactic Posted February 3, 2009 Share Posted February 3, 2009 how many times can we differentiate this splitted function on point x=0 ?? [math] f(x) = \{ x^{2n} \sin (\frac{1}{x}) ,x \ne 0 [/math] on {0,x=0} Link to comment Share on other sites More sharing options...
the tree Posted February 3, 2009 Share Posted February 3, 2009 I don't see why you shouldn't be able to differentiate it indefinitely. Crudely, [math]f^{(n)}(x) = \begin{cases} \frac{d^n}{dx^n} x^{2}\sin(x^{-1}), & x \neq 0 \\ \lim\limits_{x \to 0}\left ( \frac{d^n}{dx^n} x^{2}\sin(x^{-1}) \right ) , & x=0 \end{cases}[/math] Why would there be a problem there? Link to comment Share on other sites More sharing options...
timo Posted February 4, 2009 Share Posted February 4, 2009 I don't see why you shouldn't be able to differentiate it indefinitely. I do! Crudely,[math]f^{(m)}(x) = \begin{cases} \frac{d^m}{dx^m} x^{2n}\sin(x^{-1}), & x \neq 0 \\ \lim\limits_{x \to 0}\left ( \frac{d^m}{dx^m} x^{2n}\sin(x^{-1}) \right ) , & x=0 \end{cases}[/math] Why would there be a problem there? Because you have to show that the limit exists. Hint: Explicitely try n=0, n=1 and n=2 for the 1st derivative, first. You should already see a pattern from there. Link to comment Share on other sites More sharing options...
the tree Posted February 4, 2009 Share Posted February 4, 2009 Oh damn. I disregarded the n out of laziness, I wasn't expecting it to make a difference for some reason. Well, if you're willing to throw continuity to the wind then you could still go for: [math] f^{(m)}(x) = \begin{cases} \frac{d^m}{dx^m} x^{2n}\sin(x^{-1}) & x \neq 0 \\ 0 & x=0 \end{cases} [/math] Link to comment Share on other sites More sharing options...
D H Posted February 5, 2009 Share Posted February 5, 2009 Oh damn. I disregarded the n out of laziness, I wasn't expecting it to make a difference for some reason. Well, if you're willing to throw continuity to the wind ... That's the point of this exercise. You cannot throw continuity to the wind. A pre-condition for a function to be differentiable at some point is that the function must have both a limit and a value at that point, and these two must be equal to one another. Link to comment Share on other sites More sharing options...
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