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This weeks math discussion

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Easy if you've taken analysis :inquisitive:

 

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 :P

 

 

And, after some discussion with my husband, I should have said what delta was and gone from there :)

Edited by Tanyia

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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*

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Easy if you've taken analysis :devlish:

 

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...

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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 = b

a^2 = ab

a^2 - b^2 = ab - b^2

(a + B)(a - B) = b(a - B)

(a + B) = 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...

cats.gif

Edited by ttlanhil

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a=0

216430[/snapback]

a and b can be any number... you're closish though

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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.

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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 by Dorplein

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Easy if you've taken analysis :P

 

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!

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       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

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After mauricio poked me and I spoke with my husband, I realized I should have said what delta was and gone from there :P 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

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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 = b

a^2 = ab

a^2 - b^2 = ab - b^2

(a + :)(a - :P = b(a - :P

(a + B) = 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...

cats.gif

 

 

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 by Banshen

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be careful trying to figure this out though, or else...

cats.gif

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. :)

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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 =)

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