Obnoxious Posted October 18, 2005 Share 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)? Link to comment Share on other sites More sharing options...
Dave Posted October 18, 2005 Share 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? Link to comment Share on other sites More sharing options...
Obnoxious Posted October 20, 2005 Author Share 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 Link to comment Share on other sites More sharing options...
Tom Mattson Posted October 20, 2005 Share 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 ] Link to comment Share on other sites More sharing options...
Obnoxious Posted October 20, 2005 Author Share Posted October 20, 2005 Yeah, that's what I meant, thanks! Link to comment Share on other sites More sharing options...
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