Thank you so much! I'm pretty sure I follow everything now. Going to try and write it up on my own w/o looking again loll.
edit: Actually let me clarify again lol.
So you chose d=min{d',1} because if d=1, we'd have 1<x<3 => 2<f(x)<4 which is obviously not true because when x<2, more specifically, when x is in the interval (1,2), 2<f(x)<4 is a false inequality. So this shows that no delta greater than 1 could work, and then we chose delta to equal the minimum of 1 and some arbitrary d' which will take care of all the very small cases where d is less than 1. Then you showed if delta is<1, we are at 3/4 which is certainly less than 2..
So we are done.
Is my logic all okay here or am I misstating something?
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07-26-2015, 09:45 AM #5521Trading/Investing Thread Crew
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07-26-2015, 10:23 AM #5522
mathematician, your pm is full.
I tried to pm you this link.
http://forum.bodybuilding.com/showth...hp?t=168424483
plz respond
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07-26-2015, 12:40 PM #5523
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07-26-2015, 02:24 PM #5524
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07-26-2015, 02:30 PM #5525
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07-26-2015, 02:47 PM #5526
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07-26-2015, 03:08 PM #5527
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07-26-2015, 03:19 PM #5528
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07-26-2015, 03:19 PM #5529
yes, I screwed all types of things up in my OP. Reposting now
Ayyyy. ty sir. Long time no see.
I got confused as I have two separate theorems in my text for the IVT where they're really the same thing. I'm just confusing myself. Pretty certain I can do this in like 2 seconds now. brb writing my thoughts lol. Will implement your idea of a larger interval so no calculations need to be made to check.
edit: just realized I actually have nothing to write up. lol. It's as simple as your post and can be complete in 2 lines after showing continuity.
thanks!Last edited by ctownballer04; 07-26-2015 at 03:24 PM.
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07-26-2015, 04:04 PM #5530
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07-26-2015, 05:05 PM #5531
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07-26-2015, 05:45 PM #5532
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07-26-2015, 06:57 PM #5533
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07-26-2015, 07:14 PM #5534
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07-27-2015, 07:40 AM #5535
Where am I going potato here?
Use logarithmic differentiation to find derivative of the function. I am assuming that means to add a log into the mix.
y=(x+1)^2(x+2)^3
=ln(x+1)^2(x+2)^3
=2ln(x+1)+3ln(x+2)
2(1/x+1)+3(1/x+2)
=5x+7/(x+1)(x+2)
However the book is showing (x+1)(5x+7)(x+2)^2
Where did I go wrong?Last edited by JimFromRaleigh1; 07-27-2015 at 07:47 AM.
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07-27-2015, 11:55 AM #5536
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07-27-2015, 12:07 PM #5537
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07-27-2015, 01:42 PM #5538
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07-27-2015, 01:46 PM #5539
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07-27-2015, 02:08 PM #5540
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07-27-2015, 02:17 PM #5541
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07-27-2015, 02:19 PM #5542
I'm working on a hw problem very similar to a problem I posted last week but this time applied to continuity instead of limits.
It's again the function f: R-->R
f(x)=x for rational values of x
f(x)=x^2 for non-rational values of x
However, this time instead of proving limits exist or limits don't exist, I'm supposed to prove it's continuous at every continuous point and then prove it's not continuous at all other points.
So based on my previous homework assignment, I can use the fact that I proved limits exist at 0 and 1 and the limits at 0 and 1 were equal to f(x) evaluated at 0 and 1, then I can apply the theorem that says this happens iff the function is continuous at these points to prove it's continuous at 0 and 1.
Now I'm curious what the best approach is to prove nothing is else continuous? So far with these problems, I've only proven individual points are discontinuous. Am I supposed to do some type of proof by contradiction for three cases w/ x_0 being an interval in all three cases, namely (-infinity,0), (0,1), and (1,infinity). I hope there is an easier way cause I don't know if I can pull that off lol.Trading/Investing Thread Crew
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07-27-2015, 02:26 PM #5543
The way I'd approach it is this:
By the density of the rationals and irrationals in the reals you know that, for any real x and positive real d there will exist a rational x_r and an irrational x_i such that |x-x_i|<d and |x-x_r|<d.
Pick an x that is neither 0 nor 1 and let r=|x-x^2|>0. Show that, for any positive real d, there will exist some real number x_d such that |x-x_d|<d and |f(x)-f(x_d)|>r/2.
The intuition is that there's a sequence of rationals {x_n} and a sequence of irrationals {y_n} such that x_n,y_n -> x but f(x_n) -> x and f(y_n) -> x^2, but since this isn't the definition you saw in class you can't use it directly and must "translate" it into a metric argument.Ex-Ex-Fatass crew
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07-27-2015, 02:35 PM #5544
Also, is the following just poor notation or am I misunderstanding:
"Supposed f:[0,1]-->[0,1] is continuous. Then, there exists a c in [0,1] such that f(c)=c"
My last hw problem is similar to this and I need to understand this statement I think better to move forward w/ it.
My mind is all full of fuk. If c is a fixed number I don't get how f(c)=c every time, that seems like it'd only happen if we were very lucky w/ a nice function. Are they using c as an arbitrary value where the c in f(c) could be a different value than c itself? If so I feel like they should just use different variables.Last edited by ctownballer04; 07-27-2015 at 03:30 PM.
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07-27-2015, 02:48 PM #5545
I need some time to break down this post and think through what you're saying. I made need to clarify somethings here in a minute though. I can make that argument you're talking about, we have that as a theorem, just not our definition.
Our definition is :"f is said to be continuous at a if for any epsilon>0, there exists a delta>0 such that if x is in D and |x-a|<delta, then |f(x)-f(a)|<epsilon"
However, one of our theorems pertaining to properties of continuity is really just another definition and I think the same as what you're saying. "Let f: D->R and let a be in D.Then f is continuous at a if and only if for any sequence x_k in D that converges to a, the sequence {f(x_k)} converges to f(a)."Trading/Investing Thread Crew
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07-27-2015, 03:03 PM #5546
I understand the first line.
I understand the intuition behind the third line since we have that theorem.
I don't really know follow the second line. What is r here? I don't entirely see where that equation is coming from. I follow the second half of this line as it appears to be just the negation of our epsilon/delta contuinity definition but you chose epsilon r/2. If that's correct, I follow everything but the intuition behind the r equation.
Thanks you for the help.Trading/Investing Thread Crew
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07-27-2015, 03:06 PM #5547
More explicitly the theorem is this, where we let A = [0,1] (*):
Suppose f: A -> A and f is continuous (recall we have fixed A to be the closed interval [0,1]). Then there is an element (i.e. there exists at least one element) in A - call it a - such that f(a) = a.
We can prove this, in this specific instance where A is a (closed) interval in R by the IVT theorem.
Note your statement of the theorem as of present is false:
Let f: A -> A be the function mapping x to x^2. Then for every y in (0,1) we certainly do not have f(y) = y. However, equality is obtained at both the end points, which stresses we cannot exclude the endpoints in the statement of our theorem (nor can we set A to be an open interval).
(*) More generally: https://en.wikipedia.org/wiki/Brouwe...-point_theorem
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07-27-2015, 03:12 PM #5548
Sorry my point was I didn't think this would hold for every continuous function f, not every point. That was bad wording on my part.
I just didn't realize this theorem was actually true and I thought I wrote something down wrong or it was poor notation used. This concept never clicked w/ me in this class apparently. I follow much better now, I'm retarded for not knowing this considering we studied Kakutani's fixed-point theorem for like months straight in Game Theory. Those game theory classes were so far beyond me at points that I didn't learn as much as I would have if I had already taken analysis and topology. Now I struggle to relate stuff back from those classes to more "normal" math classes because I just never had the clearest picture.
Will rep when off RC. Appreciate it man.Trading/Investing Thread Crew
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07-27-2015, 03:13 PM #5549
The idea is this:
f is cts at x iff every sequence in the domain converging to x converges to f(x).
Now, if y is not 0 or 1, then y^2 will obtain a different value than y. In particular, as the rationals (and therefore the irrationals) are both dense subsets of the reals, we can find a sequence of rationals and irrationals converging to y, but then the corresponding sequence which is the image of the rationals (or irrationals) under f then converge to y (and y^2 respectively) which concludes our analysis.
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07-27-2015, 03:21 PM #5550
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