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07:24 AM on 11/11/09 
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John JD Dorian
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If anyone has an interesting math problem to talk about, or needs help with something they're working on, bring it in here!
05:40 PM on 11/11/09 
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John JD Dorian
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no one, huh...
07:13 PM on 11/11/09 
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Fullblast
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Do you know anything about Boolean Logic? Stuff makes my head hurt....
07:59 PM on 11/11/09 
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John JD Dorian
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Do you know anything about Boolean Logic? Stuff makes my head hurt....

i very much do. ask away.
08:15 PM on 11/11/09 
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Truman122
Gandhi On Ice
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Do mathematical physics problems count...?
09:05 PM on 11/11/09 
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John JD Dorian
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Do mathematical physics problems count...?

no reason not to give it a try.
05:20 PM on 11/12/09 
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John JD Dorian
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someone ask something!!!
05:20 PM on 11/12/09 
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John JD Dorian
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or i will...
11:32 AM on 11/15/09 
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tonyC4L
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This question was on my analysis midterm last week:

Let f: R -> R be defined by f(x) = 0 if x is rational, x if x is irrational. Show that lim x->0 f(x) = 0 but lim x->x0 f(x) does not exist for x0 != 0.

Epsilon-delta proofs are welcome along with equivalent arguments using sequences.
03:45 PM on 11/15/09 
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John JD Dorian
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St Louis, MO
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This question was on my analysis midterm last week:

Let f: R -> R be defined by f(x) = 0 if x is rational, x if x is irrational. Show that lim x->0 f(x) = 0 but lim x->x0 f(x) does not exist for x0 != 0.

Epsilon-delta proofs are welcome along with equivalent arguments using sequences.

what you are essentially proving is that f is continuous at 0 but nowhere else. alright, here goes:

Proof:
(i) At x=0, f is continuous. For if we let e > 0, we can choose d = e, and then if |x - 0| = |x| < d, then |f(x) - f(0)| = |f(x)| is either 0 for x in Q or |x| for x not in Q. In the former case, we have |f(x)| = 0 < e, and in the latter we have |f(x)| = |x| < d = e, so we are done.

(ii) Let x ≠ 0. Then, f is not continuous at x. To show this, we choose e = |x| > 0, and let d > 0.

If x is in Q and x > 0, then we are sure that there is some y in the interval (x, x+d) such that y is not in Q, so that f(y) = y. If x < 0, we can similarly choose a y in the interval (x-d, x). Either way, for this y we have |x-y| < d, yet |f(x) - f(y)| = |0 - f(y)| = |f(y)| = |y| > |x| = e, so f is not continuous at x.

If x is not in Q, then we are sure that there is some y in the interval (x-d, x+d) such that y is in Q, so that f(y) = 0. But for this y, we have |x-y| < d, yet |f(x) - f(y)| = |f(x)| = |x| = e, so f is not continuous at x.

In either case, f is not continuous at x.

////
06:06 PM on 11/15/09 
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BPerone201
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Hackensack, NJ
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I'm probably going to be popping up in this forum a lot.
I've been failing Algebra 2 since last year and repeating.
07:42 PM on 11/15/09 
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John JD Dorian
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St Louis, MO
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I'm probably going to be popping up in this forum a lot.
I've been failing Algebra 2 since last year and repeating.

please, bring it on!
10:46 PM on 11/15/09 
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tonyC4L
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Lake Forest, CA
Male - 25 Years Old
what you are essentially proving is that f is continuous at 0 but nowhere else. alright, here goes:

Proof:
(i) At x=0, f is continuous. For if we let e > 0, we can choose d = e, and then if |x - 0| = |x| < d, then |f(x) - f(0)| = |f(x)| is either 0 for x in Q or |x| for x not in Q. In the former case, we have |f(x)| = 0 < e, and in the latter we have |f(x)| = |x| < d = e, so we are done.

(ii) Let x ≠ 0. Then, f is not continuous at x. To show this, we choose e = |x| > 0, and let d > 0.

If x is in Q and x > 0, then we are sure that there is some y in the interval (x, x+d) such that y is not in Q, so that f(y) = y. If x < 0, we can similarly choose a y in the interval (x-d, x). Either way, for this y we have |x-y| < d, yet |f(x) - f(y)| = |0 - f(y)| = |f(y)| = |y| > |x| = e, so f is not continuous at x.

If x is not in Q, then we are sure that there is some y in the interval (x-d, x+d) such that y is in Q, so that f(y) = 0. But for this y, we have |x-y| < d, yet |f(x) - f(y)| = |f(x)| = |x| = e, so f is not continuous at x.

In either case, f is not continuous at x.

////
Very nice, thank you. We actually have not learned the definition of continuity yet in my class, so I didn't make that connection when taking the test, but after reading the definition and then your proof it all makes sense.

I'm probably going to be popping up in this forum a lot.
I've been failing Algebra 2 since last year and repeating.
well, the thing about us math people is that we like doing math. so please, ask questions!
10:57 PM on 11/15/09 
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tonyC4L
tech-savvy at-risk youth
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Lake Forest, CA
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And for the record I'm mostly interested in problems in differential equations (ODEs and PDEs). I'm also taking complex analysis right now and I like that too.
11:32 PM on 11/15/09 
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concernedparent
I'm very concerned.
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San Jose, CA
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Could someone explain how to find the second order taylor polynomial for a multivariable function? For example:

f(x, y) = 2/(x^2 + y^2 + 2), at a = (0,0)



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