dsb Report post Posted October 11, 2005 Prove the function f: R-->R given by f(x) = x^2 is continuous. Share this post Link to post Share on other sites
rasberrybeard Report post Posted October 11, 2005 >> Why. dear god WHY? *gut wrenching sobs from corner* Share this post Link to post Share on other sites
Tanyia Report post Posted October 11, 2005 (edited) Easy if you've taken analysis Take a point p in R. Let delta>0 and take a ball of radius delta around p, such that d_R(x,p)<delta. We know |f(x)-f(p)|=|x^2-p^2|=|(x-p)(x+p)|(less than or equal to)|x-p|*|x+p|<delta*|x+p| Thus we can choose delta small enough such that for any epsilon>0, delta*|x+p|<epsilon. QED. Edit: I LaTeX'ed it for easier viewing: http://www.geocities.com/tiggerchickn/proof.pdf Edit: I had some interesting = and <'s...probably a function of me writing this proof in about 60 seconds And, after some discussion with my husband, I should have said what delta was and gone from there Edited November 4, 2005 by Tanyia Share this post Link to post Share on other sites
Quinticus Report post Posted October 11, 2005 .. .... ...... ........ ...... .... .. There went my last brain cell! Thanks dsb...you just ruined my century. Share this post Link to post Share on other sites
Kendai Report post Posted October 11, 2005 Ah, I remember this... but I took my last math class in my senior year in high school and I ain't never going back! Share this post Link to post Share on other sites
Mad_Raven Report post Posted October 12, 2005 Well....im doing this right now in school but no word about continious thingies..... errrmmmm......i think its all a conspiracy! ... .. . . .. ... by the gnomes! *gasp* Share this post Link to post Share on other sites
tuga_elf Report post Posted October 12, 2005 translate plz Share this post Link to post Share on other sites
mauriciom Report post Posted November 2, 2005 Easy if you've taken analysis Take a point p in R. Let delta>0 and take a ball of radius delta around p, such that d_R(x,p)<delta. We know |f(x)-f(p)|=|x^2-p^2|<|(x-p)(x+p)|(less than or equal to)|x-p|*|x+p|=delta*|x+p| Thus we can choose delta small enough such that for any epsilon>0, delta*|x+p|<epsilon. QED. Edit: I LaTeX'ed it for easier viewing: http://www.geocities.com/tiggerchickn/proof.pdf 208367[/snapback] u proved that x^2 has limit for all p in R but u didnt pruve that is conti... a easy way is to diff(x^2)=2x=0 so the only critical point is x=0... so lim-->x+=0 of x^2 must be = to lim-->x-=0 of x^2 doing that u have 0=0 and f(x=0)=o so u have that x^2 is conti... Share this post Link to post Share on other sites
ssjgohan Report post Posted November 3, 2005 *pukes* Share this post Link to post Share on other sites
RolandR1 Report post Posted November 3, 2005 :drools wow....... Share this post Link to post Share on other sites
Arnieman Report post Posted November 3, 2005 I thought I knew math... at one point, I did. But this... I haven't the slightest idea what you are talking about. Share this post Link to post Share on other sites
ttlanhil Report post Posted November 3, 2005 (edited) okay, a simpler one for those of us who don't do a lot of maths... (x^2 means x squared, for those not familiar with the symbology) given: a = ba^2 = ab a^2 - b^2 = ab - b^2 (a + (a - = b(a - (a + = b a + a = a 2a = a 2 = 1 explain how/why 8^P (those who can do the one above, don't do this one... and yes, I figured this one out shortly after I saw it) be careful trying to figure this out though, or else... Edited November 3, 2005 by ttlanhil Share this post Link to post Share on other sites
Dorplein Report post Posted November 3, 2005 (edited) a=0 Edited November 3, 2005 by Dorplein Share this post Link to post Share on other sites
ttlanhil Report post Posted November 3, 2005 a=0 216430[/snapback] a and b can be any number... you're closish though Share this post Link to post Share on other sites
Tanyia Report post Posted November 3, 2005 u proved that x^2 has limit for all p in R but u didnt pruve that is conti... a easy way is to diff(x^2)=2x=0 so the only critical point is x=0... so lim-->x+=0 of x^2 must be = to lim-->x-=0 of x^2 doing that u have 0=0 and f(x=0)=o so u have that x^2 is conti... 216323[/snapback] I proved it fit the definition of continuous (I spose I should have said that at the end), without assuming we knew anything about differentiation etc. Share this post Link to post Share on other sites
Dorplein Report post Posted November 3, 2005 (edited) given: a = b a^2 = ab a^2 - b^2 = ab - b^2 <------all numbers become 0 here (a + B)(a - B) = b(a - B) (a + B) = b a + a = a 2a = a 2 = 1 Edited November 3, 2005 by Dorplein Share this post Link to post Share on other sites
Placid Report post Posted November 3, 2005 Easy if you've taken analysis Take a point p in R. Let delta>0 and take a ball of radius delta around p, such that d_R(x,p)<delta. We know |f(x)-f(p)|=|x^2-p^2|<|(x-p)(x+p)|(less than or equal to)|x-p|*|x+p|=delta*|x+p| Thus we can choose delta small enough such that for any epsilon>0, delta*|x+p|<epsilon. QED. Edit: I LaTeX'ed it for easier viewing: http://www.geocities.com/tiggerchickn/proof.pdf 208367[/snapback] YEAH! Share this post Link to post Share on other sites
Tanyia Report post Posted November 3, 2005 \o/? Somehow I thought you'd appreciate the nerdiness of that :lol: Share this post Link to post Share on other sites
Arnieman Report post Posted November 3, 2005 given: a = b a^2 = ab <--Multiply both sides by a a^2 - b^2 = ab - b^2 <--Subtract b^2 from both sides (a + b)(a - b) = b(a - b) <--Factoring (a + b) = b <--Divide both sides by (a - b) a + a = a <--Substituting the given 2a = a <--Grouping (1a + 1a = 2a) 2 = 1 <-- divide both sides by a Now, to assume a random number for a, we'll try 5. 5 = 5 5^2 = 5 * 5 <--25 = 25, True 5^2 - 5^2 = 5 * 5 - 5^2 <--0 = 0, True (5 + 5)(5 - 5) = 5(5 - 5) <--0 = 0, True (5 + 5) = 5 <--10 <> 5, False (Also an error for dividing both sides by 0) 5 + 5 = 5 <--10 <> 5 2 * 5 = 5 <--10 <> 5 2 = 1 <--2 <> 1 Share this post Link to post Share on other sites
Tanyia Report post Posted November 4, 2005 After mauricio poked me and I spoke with my husband, I realized I should have said what delta was and gone from there So, here's the updated version: Take a point p in R. We WTS that for any epsilon>0, there exists a delta>0 : d_R(x,p)<delta=>|f(x)-f(p)|<epsilon. Let delta>0 and take a ball of radius delta around p, such that d_R(x,p)<delta. We know |f(x)-f(p)|=|x^2-p^2|=|(x-p)(x+p)|(less than or equal to)|x-p|*|x+p| Let delta=sqrt(epsilon). Then d_R(x,p)<sqrt(epsilon)=>|x+p|<sqrt(epsilon) and |x-p|<sqrt(epsilon). Thus we have |f(x)-f(p)|<|x-p|*|x+p|<delta*delta<sqrt(epsilon)*sqrt(epsilon)=epsilon. \blacksquare Share this post Link to post Share on other sites
Banshen Report post Posted November 8, 2005 (edited) okay, a simpler one for those of us who don't do a lot of maths... (x^2 means x squared, for those not familiar with the symbology) given: a = ba^2 = ab a^2 - b^2 = ab - b^2 (a + (a - = b(a - (a + = b a + a = a 2a = a 2 = 1 explain how/why 8^P (those who can do the one above, don't do this one... and yes, I figured this one out shortly after I saw it) be careful trying to figure this out though, or else... the thrid line is what stuffs it up coz ( a-b ) will be 0 if a=b, so to remove ( a-b ) from both sides you would get 0/0 which is undefined. however looka t what your doing in the second line also. you are seeting it up so that the LHS =0 = RHS. but the third line is where the divide by 0 happens. Edited November 8, 2005 by Banshen Share this post Link to post Share on other sites
Arnieman Report post Posted November 8, 2005 be careful trying to figure this out though, or else... the thrid line is what stuffs it up coz ( a-b ) will be 0 if a=b, so to remove ( a-b ) from both sides you would get 0/0 which is undefined. however looka t what your doing in the second line also. you are seeting it up so that the LHS =0 = RHS. but the third line is where the divide by 0 happens. Look at my post on the end of the last page... good work, tho. And I had to leave that cartoon, it's funny. Share this post Link to post Share on other sites
Banshen Report post Posted November 8, 2005 ah yes i see that now lol. i didnt see your proof Share this post Link to post Share on other sites
mauriciom Report post Posted November 12, 2005 After mauricio poked me and I spoke with my husband, I realized I should have said what delta was and gone from there So, here's the updated version: Take a point p in R. We WTS that for any epsilon>0, there exists a delta>0 : d_R(x,p)<delta=>|f(x)-f(p)|<epsilon. Let delta>0 and take a ball of radius delta around p, such that d_R(x,p)<delta. We know |f(x)-f(p)|=|x^2-p^2|=|(x-p)(x+p)|(less than or equal to)|x-p|*|x+p| Let delta=sqrt(epsilon). Then d_R(x,p)<sqrt(epsilon)=>|x+p|<sqrt(epsilon) and |x-p|<sqrt(epsilon). Thus we have |f(x)-f(p)|<|x-p|*|x+p|<delta*delta<sqrt(epsilon)*sqrt(epsilon)=epsilon. \blacksquare Thats better =) Share this post Link to post Share on other sites