[QUOTE=Bradyns;40204106]Drawing pictures can be helpful.[/QUOTE]
Tru dat
Although I'm 99% sure it's a typo and should be 6cm... 7x13 seems a lot more plausible than something with the square root of 90091 :v:
EDIT: Or maybe not
[QUOTE=Joey90;40206790]Tru dat
Although I'm 99% sure it's a typo and should be 6cm... 7x13 seems a lot more plausible than something with the square root of 90091 :v:
EDIT: Or maybe not[/QUOTE]
I thought the same, 7*13 seems much more likely an answer given the level of the question
Just did a 17 questions homework to prove that lim(1+1/4+1/9+1/16+...+1/n²) is irrational.
I find it astonishing how this proof is done using such diverse methods as complexs, integration, sequences, etc... The way it is done seems so quirky, yet it yields the desired result. Thinking about it, the matematicians who conceived that proof must have spent quite some time on it. :v:
[QUOTE=_Axel;40209464]Just did a 17 questions homework to prove that lim(1+1/4+1/9+1/16+...+1/n²) is irrational.
I find it astonishing how this proof is done using such diverse methods as complexs, integration, sequences, etc... The way it is done seems so quirky, yet it yields the desired result. Thinking about it, the matematicians who conceived that proof must have spent quite some time on it. :v:[/QUOTE]
I remember the first time I saw the 'proof' of the limit by factorising sin(x)/x at school and it totally blew my mind :v:
The problem of the limit was really big at the time - [url]http://en.wikipedia.org/wiki/Basel_problem[/url] - Liebniz couldn't do it and neither could any of the Bernoullis... Took Euler to really get close to the right answer.
[QUOTE=Joey90;40209702]I remember the first time I saw the 'proof' of the limit by factorising sin(x)/x at school and it totally blew my mind :v:
The problem of the limit was really big at the time - [url]http://en.wikipedia.org/wiki/Basel_problem[/url] - Liebniz couldn't do it and neither could any of the Bernoullis... Took Euler to really get close to the right answer.[/QUOTE]
Wow, that's a really elegant and straightforward way to solve it. Much better than the proof in my homework, but I guess that's because the latter aims to use integrals and infinite sums.
Is this technique of expanding and factorising functions useful in more widespread cases ? I assume said functions would need to have an infinite number of easily definable roots to be subject to such techniques, though ?
-snip, okay; sorry.-
[QUOTE=mastoner20;40210334] The premise behind portal is you're changing multi-dimensional planes from one multiverse to another. Therefore by going through one end of the portal, you are coming out of the other through a separate multiverse, and thus re-instituting the frame of reference in reference to you.[/QUOTE]
Ehhh, what? Also, the physics discussion is probably a better place for this.
[QUOTE=_Axel;40210202]Wow, that's a really elegant and straightforward way to solve it. Much better than the proof in my homework, but I guess that's because the latter aims to use integrals and infinite sums.
Is this technique of expanding and factorising functions useful in more widespread cases ? I assume said functions would need to have an infinite number of easily definable roots to be subject to such techniques, though ?[/QUOTE]
The problem with it is that while it is elegant and straightforward to demonstrate, making it rigorous is a little trickier (you have to prove that the series expansion does actually converge to the right thing - the [url=http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem]Weierstrass Factorisation Theorem[/url] does this, but obviously you'd then want to prove that too...)
You can use a similar technique on, for example, the Riemann Zeta function although I don't really know enough about it to say how useful it is...
Alright I think I almost have a better argument against that function.
Let f(x) be a function which returns to x after q compositions of f when x = p/q in lowest terms. Then f(x) is not continuous. To show this, assume otherwise and consider four points on the graph, an orbit which includes a rational with denominator 4 in lowest terms. I don't know what an orbit is but I think it means every point you reach after iterating the function enough times.
Label the x coordinates of these points a, b, c, d from least to greatest. I'll say for this case f(a) = d, f(d) = b, f(b) = c, and, as required, f(c) = a. By the intermediate value theorem, there is a point e between a and b for which f(b) < f(e) < f(a). That is to say, (e, f(e)) is in the rectangle determined by [a,d] and [b,c]. f(x) is continuous at e unless the limit as x approaches e isn't equal to f(b).
Choose ε small enough so that f(b) ≤ f(e)-ε. Let δ be positive. Since f(x) is continuous at (c,a) there is an interval J inside the interval [b,c] for which all f(x) lie in [e-δ,e+δ]. In that interval, periodic points will have the last point in their orbit have an x-coordinate between a and b, and their last y-coordinate will be the first x-coordinate.
Define g as the point between c and d such that f(g) = e, again guaranteed by the intermediate value theorem. f(x) is continuous at g too, so set ε[sub]g[/sub] = δ. That is, the epsilon corresponding to g is the largest delta corresponding to e. If necessary, shrink ε so that ε = δ[sub]g[/sub], where δ[sub]g[/sub] is the largest delta corresponding to g.
Then, finally, I consider the behaviour of periodic points in J. For brevity let's call a number with even denominator there h. f(h), being in J, lies in [e-δ,e+δ]. If it's period 2, f(f(h)) = h. Since f(h) is between a and b and h is between b and c, f(f(h)) is less than c and hence delta must be smaller. Assume, then, that its period is greater than 2 and f(f(h)) lies within ε of e. Since ε = δ[sub]g[/sub], further iteration yields more points lying within some rectangles whose coordinates I don't really feel like typing out.
The final point in h's orbit, denoted f[sup]-1[/sup](h) with no assumptions of invertibility, lies in J. Its x coordinate, thanks to the restrictions on ε[sub]g[/sub] lie within δ of e but its y coordinate isn't within ε for the previously stated reasons. Normally, you could just shrink δ a little more. Since this applies to every point in J with period 2k, the points who are in [e-δ,e+δ] but outside [e-ε,e+ε] form a dense subset of R. By definition e is either a member or limit point of this set, but since it is in [e-ε,e+ε] it must be a limit point. Continuity isn't allowed.
There are still some errors and ambiguities and this could all be clearer, but I want to make sure the central argument doesn't have some huge hole in it before I press on.
Anyway, this is for a very specific case of four points with period four and specific inequalities relating them all. This is just one case of 16 (only 14 are admissible and by inspection most are highly similar). However, I think this argument follows from the inability to form a monotonic sequence with four terms which also forms an orbit. Justifying this as well as the argument above would disallow all continuous functions with that property.
I just took a unit exam on integral applications after having been up for 24 hours. Scored 11/14 on the multiple choice part, here's hoping for similar results on the teacher scored portion.
I really should stop taking exams without proper sleep/food.
[QUOTE=Krinkels;40214222]Alright I think I almost have a better argument against that function.
Let f(x) be a function which returns to x after q compositions of f when x = p/q in lowest terms. Then f(x) is not continuous. To show this, assume otherwise and consider four points on the graph, an orbit which includes a rational with denominator 4 in lowest terms. I don't know what an orbit is but I think it means every point you reach after iterating the function enough times.
Label the x coordinates of these points a, b, c, d from least to greatest. I'll say for this case f(a) = d, f(d) = b, f(b) = c, and, as required, f(c) = a. By the intermediate value theorem, there is a point e between a and b for which f(b) < f(e) < f(a). That is to say, (e, f(e)) is in the rectangle determined by [a,d] and [b,c]. f(x) is continuous at e unless the limit as x approaches e isn't equal to f(b).
Choose ε small enough so that f(b) ≤ f(e)-ε. Let δ be positive. Since f(x) is continuous at (c,a) there is an interval J inside the interval [b,c] for which all f(x) lie in [e-δ,e+δ]. In that interval, periodic points will have the last point in their orbit have an x-coordinate between a and b, and their last y-coordinate will be the first x-coordinate.
Define g as the point between c and d such that f(g) = e, again guaranteed by the intermediate value theorem. f(x) is continuous at g too, so set ε[sub]g[/sub] = δ. That is, the epsilon corresponding to g is the largest delta corresponding to e. If necessary, shrink ε so that ε = δ[sub]g[/sub], where δ[sub]g[/sub] is the largest delta corresponding to g.
Then, finally, I consider the behaviour of periodic points in J. For brevity let's call a number with even denominator there h. f(h), being in J, lies in [e-δ,e+δ]. If it's period 2, f(f(h)) = h. Since f(h) is between a and b and h is between b and c, f(f(h)) is less than c and hence delta must be smaller. Assume, then, that its period is greater than 2 and f(f(h)) lies within ε of e. Since ε = δ[sub]g[/sub], further iteration yields more points lying within some rectangles whose coordinates I don't really feel like typing out.
The final point in h's orbit, denoted f[sup]-1[/sup](h) with no assumptions of invertibility, lies in J. Its x coordinate, thanks to the restrictions on ε[sub]g[/sub] lie within δ of e but its y coordinate isn't within ε for the previously stated reasons. Normally, you could just shrink δ a little more. Since this applies to every point in J with period 2k, the points who are in [e-δ,e+δ] but outside [e-ε,e+ε] form a dense subset of R. By definition e is either a member or limit point of this set, but since it is in [e-ε,e+ε] it must be a limit point. Continuity isn't allowed.
There are still some errors and ambiguities and this could all be clearer, but I want to make sure the central argument doesn't have some huge hole in it before I press on.
Anyway, this is for a very specific case of four points with period four and specific inequalities relating them all. This is just one case of 16 (only 14 are admissible and by inspection most are highly similar). However, I think this argument follows from the inability to form a monotonic sequence with four terms which also forms an orbit. Justifying this as well as the argument above would disallow all continuous functions with that property.[/QUOTE]
I couldn't quite follow the argument all the way through I'm afraid... I suspect you might be making some unfounded assumptions but I'm not really sure. However your last couple of sentences prompted me to do this:
It's quite straightforward to see that such an f cannot be monotonic (and hence cannot be bijective):
To start with, since f(n)=n for all integers, we know if at all it must be monotone increasing. Then just think about any 2 element orbit {a,b} with a<b. Then f(b)=a < b=f(a) which contradicts it being monotone increasing.
The problem is that being bijective when restricted to the rationals does [i]not[/i] in general imply that it's bijective on R.
If we assume that no [i]irrational[/i] maps to a rational (i.e. every rational has a [i]unique[/i] preimage) then it's easy, but that doesn't seem obvious either.
If you want to try and tidy up your argument, my advice would be to look at a 2 element orbit rather than 4 (I don't think you get any useful extra information with 4) and to draw a clear diagram as to what's going on.
Could a kind soul please explain the difference between asymptotic equality and asymptotic behaviour to me? I fail to see the difference between the two:
[img]http://gyazo.com/24432b7bc7dd2ba242b6c3c4f50f846d.png[/img]
[img]http://gyazo.com/8f731b7ec04a2cfcd3298e5ecbb47c1c.png[/img]
Well it seems like asymptotic equality has to do with their absolute difference going to zero, while identical asymptotic behavior has to do with their ratio going to 1.
For equality:
f(x) - g(x) = e(x), e(x) goes to 0
For behavior:
f(x) / g(x) = 1 + e(x), e(x) goes to 0
I think you can see this a lot with rational functions. For example, the functions f(x) = x[sup]2[/sup] - 4 and g(x) = x[sup]2[/sup] + 3 have identical asymptotic behavior, because f(x)/g(x) approaches 1 as |x| gets large, but not asymptotically equal because f(x) - g(x) = -7 for all x. I hope this is of some help, I haven't studied this before so hopefully someone can provide a more formal explanation.
[QUOTE=agentalexandre;40224185]Could a kind soul please explain the difference between asymptotic equality and asymptotic behaviour to me? I fail to see the difference between the two:
[img]http://gyazo.com/24432b7bc7dd2ba242b6c3c4f50f846d.png[/img]
[img]http://gyazo.com/8f731b7ec04a2cfcd3298e5ecbb47c1c.png[/img][/QUOTE]
Pretty much exactly what account says
Both require that f(x)/g(x) -> 1 but for Asymptotic [i]equality[/i] we also require that f(x) - g(x) -> 0
[QUOTE=Joey90;40220113]I couldn't quite follow the argument all the way through I'm afraid... I suspect you might be making some unfounded assumptions but I'm not really sure. However your last couple of sentences prompted me to do this:
It's quite straightforward to see that such an f cannot be monotonic (and hence cannot be bijective):
To start with, since f(n)=n for all integers, we know if at all it must be monotone increasing. Then just think about any 2 element orbit {a,b} with a<b. Then f(b)=a < b=f(a) which contradicts it being monotone increasing.
The problem is that being bijective when restricted to the rationals does [i]not[/i] in general imply that it's bijective on R.
If we assume that no [i]irrational[/i] maps to a rational (i.e. every rational has a [i]unique[/i] preimage) then it's easy, but that doesn't seem obvious either.
If you want to try and tidy up your argument, my advice would be to look at a 2 element orbit rather than 4 (I don't think you get any useful extra information with 4) and to draw a clear diagram as to what's going on.[/QUOTE]
I can't follow it either haha. I should sleep more before posting.
[IMG]http://imageshack.us/a/img507/2637/pic1dn.png[/IMG]
So, you have a, b, c, d, and e as above.
[IMG]http://imageshack.us/a/img818/1136/pic2w.png[/IMG]
Epsilon is more than zero and delta is such that (e-delta, e+delta) is the shaded region there. All x in the shaded region have y values within epsilon of e.
[IMG]http://imageshack.us/a/img690/8750/pic3hi.png[/IMG]
There is a point in the middle rectangle whose y value is e. Draw a strip of height 2 delta around that point. There is an interval where the x coordinate is less than c but the function is in the strip. That interval is J.
[IMG]http://imageshack.us/a/img607/5089/pic4xo.png[/IMG]
So, start in J. This goes to a point in the strip, which is within delta of e. I thought, when I started writing the first post, that I could just continue iterating and then end up with the point below the left rectangle but with an x coordinate within delta of e. It was at this point I noticed two more iterations of the function doesn't necessarily give a point within delta of e. It was also at this point that that post became total nonsense.
So, at around eleven-thirty I tried to think of a way to ensure a bunch of points still end up within delta of e. I didn't do a good job of that.
Anyway, at first I thought having infinitely many points outside the three boxes might somehow make it discontinuous. I don't know if it's true or false.
Krinkels, did you use a program to make those diagrams? If so, what program? [Looks like paint]
[QUOTE=Krinkels;40225126]So, start in J. This goes to a point in the strip, which is within delta of e. I thought, when I started writing the first post, that I could just continue iterating and then end up with the point below the left rectangle but with an x coordinate within delta of e. It was at this point I noticed two more iterations of the function doesn't necessarily give a point within delta of e. It was also at this point that that post became total nonsense.
So, at around eleven-thirty I tried to think of a way to ensure a bunch of points still end up within delta of e. I didn't do a good job of that.
Anyway, at first I thought having infinitely many points outside the three boxes might somehow make it discontinuous. I don't know if it's true or false.[/QUOTE]
Yeah, that's pretty much the problem... You can't really say much about where the neighbourhood of f(e) goes... it could essentially go anywhere! (Doesn't even have to be on the graph shown)
The problem I'm having is that there's not really a problem with it being periodic on a [i]finite[/i] number of denominators (even when enforcing the bijectivity when restricted to rationals). The trouble is it seems 'implausible' that you can do it at [i]all[/i] denominators.
If it [i]was[/i] possible, I'd probably construct it by defining inductively on all rationals (since they're countable) and then since it's a dense set, it's enough to define the whole function (so long as it is then continuous!)
[QUOTE=Bradyns;40235925]Krinkels, did you use a program to make those diagrams? If so, what program? [Looks like paint][/QUOTE]
Yeah it's paint.
-snip- found it:
[URL]http://en.wikipedia.org/wiki/Infinite_monkey_theorem[/URL]
You will be bored by this newbie question but how can I prove that P(A∩B)=P(A)∩P(B)? P is a Power set and A and B are sets and c=subset €=element.
I thought about inserting (A∩B) in the definiton of a Power set. P(A∩B):={X|Xc(A∩B)}=>XcA and XcB and x€X. I am new to this so I don't really know how to structure proofs. Does my approach even make sense?
[QUOTE=Desuh;40308849]You will be bored by this newbie question but how can I prove that P(A∩B)=P(A)∩P(B)? P is a Power set and A and B are sets and c=subset €=element.
I thought about inserting (A∩B) in the definiton of a Power set. P(A∩B):={X|Xc(A∩B)}=>XcA and XcB and x€X. I am new to this so I don't really know how to structure proofs. Does my approach even make sense?[/QUOTE]
That's exactly the right approach:
P(A∩B) = {X | Xc(A∩B)} = {X | XcA and XcB} = {X | XcA}∩{X | XcB} = P(A)∩P(B)
Can't get over this:
[url]http://thatsmathematics.com/mathgen/[/url]
Hey guys, quick question for Linear Alg. We're doing Subspaces/basis, and doing proofs there of. I'm getting hung up on this one particular one, and have no idea why.
"Prove the given sets of vectors are subspaces of R^m: The set of all scalar multiples of the vector <1,2,3>(of R³)."
How should I get this sucker started? I'm assuming I start with letting V be a subset of R³ where v=(1,2,3) and v€V, then showing c€R³ and c*v€R³ for all Real-numbers, c.
Yes? (I'm letting the Euro key be "an element of", because I don't really feel like digging through character map for it, sorry.)
[QUOTE=mastoner20;40319648]How should I get this sucker started? I'm assuming I start with letting V be a subset of R³ where v=(1,2,3) and v€V, then showing c€R³ and c*v€R³ for all Real-numbers, c.
Yes? (I'm letting the Euro key be "an element of", because I don't really feel like digging through character map for it, sorry.)[/QUOTE]
Do you mean c is in R, rather than R[sup]3[/sup]?
I'm not sure what you mean by "start with letting V be a subset of R³ where v=(1,2,3) and v€V."
The way I'd do it is to start by saying, "Let V be the set of all scalar multiples of <1,2,3>." Then show that 0 is in the set, (trivial) show that any scalar multiple of a vector in V is in V, and show that the sum of any two vectors in V is in V.
[QUOTE=mastoner20;40319648]Hey guys, quick question for Linear Alg. We're doing Subspaces/basis, and doing proofs there of. I'm getting hung up on this one particular one, and have no idea why.
"Prove the given sets of vectors are subspaces of R^m: The set of all scalar multiples of the vector <1,2,3>(of R³)."
How should I get this sucker started? I'm assuming I start with letting V be a subset of R³ where v=(1,2,3) and v€V, then showing c€R³ and c*v€R³ for all Real-numbers, c.
Yes? (I'm letting the Euro key be "an element of", because I don't really feel like digging through character map for it, sorry.)[/QUOTE]
So the set you're interested in is
[img]http://latex.codecogs.com/gif.latex?V%20=%20\{(\lambda,2\lambda,3\lambda)\in\mathbb{R}^3%20|%20\lambda%20\in%20\mathbb{R}\}[/img]
Now you look at your definition of a subspace - it must be closed under addition and scalar multiplication. (Technically it also needs to be nonempty, but that's obvious)
So take any two vectors v and w in V, they can be written
[img]http://latex.codecogs.com/gif.latex?v=(\lambda,2\lambda,3\lambda)\quad%20w=(\mu,2\mu,3\mu)[/img]
now we need to check
[img]http://latex.codecogs.com/gif.latex?v+w\in%20V[/img]
and
[img]http://latex.codecogs.com/gif.latex?c.%20v\in%20V\quad\forall%20c\in\mathbb{R}[/img]
but this is easy if you just substitute in the definitions of v and w.
Thanks guys, I just figured out what I was doing wrong in my proof as I hit refresh. I was begging the question and assuming what was to be proven, which is where part of the logical issue for me was.
Also, the of c€R³ was a typo. I did mean c€R (I meant to say it was a scalar multiple.)
I need help. I've got some doubling time questions to do and I've completely forgotten how to do that stuff. I need something to go off of.
My question states that a bacterial culture doubles every 20 minutes. After an hour, there are 32000 bacteria. What is the initial size of the culture? How many bacteria are in the culture after 6 hours?
Well, the equation would look like
I*2[sup]t/20[/sup]
so plugging in t = 60 (1 hour) we get I*2[sup]3[/sup] = 32000. I think it's pretty easy to solve from there.
I've figured out that much already, it's unfortunately the steps afterwards that I've forgotten.
EDIT:
No wait I've got it. The initial value is 4000.
From there, the amount of bacteria after 6 hours is:
P=400(2)^18
P=1,048,576,000
Man, I'm finding certain proofs legitimately fun. I just did one for my topology midterm that involved appealing to the fact that finite products of connected spaces and compact spaces are connected and compact respectively, and then appealing to the generalized version of the extreme value theorem, and then the generalized intermediate value theorem. Slick.
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