Sarahisme Posted August 15, 2005 Share Posted August 15, 2005 hey, its been awhile since i could do these, could i have a little bit of help, if there is someone willing out there? Cheers Sarah Link to comment Share on other sites More sharing options...
Sarahisme Posted August 15, 2005 Author Share Posted August 15, 2005 ok so far this is what i have gotten: [math] |\sqrt(x)-\sqrt(4)|=|\sqrt(x)-\sqrt(4)|\times\frac{|\sqrt(x)+\sqrt(4)|}{|\sqrt(x)+\sqrt(4)|}=\frac{x-4}{|\sqrt(x)+\sqrt(4)|} [/math] Let: [math] \frac{3}{2}<x<\frac{5}{2} [/math] Then [math] \frac{x-4}{|\sqrt(x)+\sqrt(4)|}\leq|x-4|\times\frac{1}{|\sqrt(\frac{3}{2})+2|}=|x-4|\times\frac{1}{\sqrt(\frac{3}{2})+2} [/math] So [math] |\sqrt(x)-\sqrt(4)|\leq|x-4|\times\frac{1}{\sqrt(\frac{3}{2})+2} [/math] Let [math] |x-4|\times\frac{1}{\sqrt(\frac{3}{2})+2}<\epsilon [/math] [math] \therefore |x-4| < (\sqrt(\frac{3}{2})+2)\epsilon [/math] So choose [math] \delta = (\sqrt(\frac{3}{2})+2)\epsilon [/math] [math] \therefore |x-4| < \delta \implies |h(x) - h(4)| < \epsilon [/math] well, yep, hows that? Link to comment Share on other sites More sharing options...
Sarahisme Posted August 15, 2005 Author Share Posted August 15, 2005 ...although i dunno how part 2 is going to work... Link to comment Share on other sites More sharing options...
Sarahisme Posted August 16, 2005 Author Share Posted August 16, 2005 ok i've got another method....here we go We want to find a [math] \delta > 0 [/math] (depending on [math] \epsilon [/math]) so [math] |x-4| < \delta \implies |h(x)-h(4)| < \epsilon [/math] [math] i.e. |\sqrt(x) - 2| < \epsilon [/math] Note [math] |\sqrt(x) - 2| = \frac{|x - 4|}{|\sqrt(x) + 2|} [/math] Choose [math] \delta [/math] so that [math] \delta\frac{1}{|\sqrt(x) +2|} < \epsilon [/math] Note [math] |x-4| < \delta \implies |x| < 4 + \delta [/math] Choose [math]\delta[/math] to be < 1 then [math] |x| < 5 [/math] Then choose [math] \delta = min(1,\epsilon(\sqrt(5) + 2)) [/math] Then [math] |x-4| < \delta \implies |x|<=5, \ Since \ \delta<=1 [/math] Then [math] |\sqrt(x) - 2| = \frac{|x - 4|}{|\sqrt(x) +2|} [/math] [math] <\delta\frac{1}{\sqrt(5) + 2} <= \epsilon, \ since \ \delta <=\epsilon(\sqrt(5) +2) [/math] Q.E.D Link to comment Share on other sites More sharing options...
Sarahisme Posted August 16, 2005 Author Share Posted August 16, 2005 so what do peoples think? Link to comment Share on other sites More sharing options...
Sarahisme Posted August 17, 2005 Author Share Posted August 17, 2005 lol, anyone at all? Link to comment Share on other sites More sharing options...
ydoaPs Posted August 17, 2005 Share Posted August 17, 2005 Q.E.D how was all of that related to Quantum ElectroDynamics? Link to comment Share on other sites More sharing options...
Sarahisme Posted August 17, 2005 Author Share Posted August 17, 2005 how was all of that related to Quantum ElectroDynamics? what are you talking about ? this is the maths problems section! Link to comment Share on other sites More sharing options...
ydoaPs Posted August 17, 2005 Share Posted August 17, 2005 QED=Quantum ElectroDynamics, right? Link to comment Share on other sites More sharing options...
Sarahisme Posted August 17, 2005 Author Share Posted August 17, 2005 oh ok , no i can't remember what is means, its a mathematical thing you put at the end of proofs, (like the little square boxes that are sometime used) anyway, how was my proof (ignoring the Q.E.D bit )?? Link to comment Share on other sites More sharing options...
Sarahisme Posted August 17, 2005 Author Share Posted August 17, 2005 anyone? ... please :S ???? Link to comment Share on other sites More sharing options...
Sarahisme Posted August 17, 2005 Author Share Posted August 17, 2005 or how about this: (i think this is the one): [math] Want: \ |x - 4| < \delta \implies |\sqrt(x) - 2| < \epsilon [/math] [math] x - 4 = (\sqrt(x) - 2)(\sqrt(x) + 2) [/math] [math] |(\sqrt(x) - 2)(\sqrt(x) + 2)| < \delta [/math] [math] x>=0 [/math] [math] \therefore \sqrt(x) + 2 >= 2 [/math] [math] \sqrt(x) - 2 < |(\sqrt(x) - 2)(\sqrt(x) + 2)|<\delta [/math] So Choose [math]\epsilon = \delta[/math] Link to comment Share on other sites More sharing options...
DQW Posted August 17, 2005 Share Posted August 17, 2005 QED is an abbreviation of Quod Erat Demonstrandum, which is Latin for "...which was to be demonstrated." Shall look into the problem later today. Link to comment Share on other sites More sharing options...
Sarahisme Posted August 17, 2005 Author Share Posted August 17, 2005 lol see i knew it meant something to do with proving! thanks DQW Link to comment Share on other sites More sharing options...
DQW Posted August 18, 2005 Share Posted August 18, 2005 How do you get [math]\frac{|x - 4|}{|\sqrt(x) +2|} ] <\delta\frac{1}{\sqrt(5) + 2}[/math] ? Link to comment Share on other sites More sharing options...
Sarahisme Posted August 18, 2005 Author Share Posted August 18, 2005 don't worry about it, i never really understood, i'll just have to work it out some other time thanks for the help though Link to comment Share on other sites More sharing options...
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