little help please... :)

[Sarahisme]

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

[Sarahisme]

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?

[Sarahisme]

...although i dunno how part 2 is going to work...

[Sarahisme]

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

[Sarahisme]

so what do peoples think?

[Sarahisme]

lol, anyone at all?

[ydoaPs]

Q.E.D

how was all of that related to Quantum ElectroDynamics?

[Sarahisme]

how was all of that related to Quantum ElectroDynamics?

 

 

what are you talking about ? this is the maths problems section!

[ydoaPs]

QED=Quantum ElectroDynamics, right?

[Sarahisme]

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 )??

[Sarahisme]

anyone? ... please :S ????

[Sarahisme]

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]

[DQW]

QED is an abbreviation of Quod Erat Demonstrandum, which is Latin for "...which was to be demonstrated."

 

Shall look into the problem later today.

[Sarahisme]

lol see i knew it meant something to do with proving!

 

thanks DQW

[DQW]

How do you get

 

[math]\frac{|x - 4|}{|\sqrt(x) +2|} ] <\delta\frac{1}{\sqrt(5) + 2}[/math] ?

[Sarahisme]

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