Obnoxious Posted October 18, 2005 Posted October 18, 2005 Is the set of functions: {[math]f® = R |\frac{df}{dx} + 2f = 1[/math]} a vector space? I said no because it doesn't seem to have a zero vector, but I'm doubtful of my answer. Can someone help me prove its vector space validity (or lack thereof)?
Dave Posted October 18, 2005 Posted October 18, 2005 I'm not quite sure on my interpretation of your question. Are you asking for all functions f that have a fixed point at R, or the fixed points of the function f, or what?
Obnoxious Posted October 20, 2005 Author Posted October 20, 2005 No, I mean f of a real number is another real number. But I have no idea how to make the funky looking R with Latex
Tom Mattson Posted October 20, 2005 Posted October 20, 2005 Are you trying to say the following? [math]\left\{f:\mathbb{R}\rightarrow\mathbb{R}\mid\frac{df}{dx}+2f=1\right\}[/math] If so then you are correct. [imath]f\equiv0[/imath] doesn't satisfy the condition specified in the set definition. No, I mean f of a real number is another real number. But I have no idea how to make the funky looking R with Latex Type the following, without the spaces: [ math ]\mathbb{R}[ \math ]
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