Math thread! Come in and have some pi...

[ctownballer04]

mathematician, your pm is full.

I tried to pm you this link.
http://forum.bodybuilding.com/showth...hp?t=168424483

plz respond

[ckuyook]

If I were MiscMathematician, I would want more textbooks

[K3Fallout]

How the hell does one develop number sense? Been studying for the GRE and all the word problems kick my ass. It's one thing to translate a paragraph into variables and numbers but I can never figure out how to put it all together sensibly.

[SlyOdysseus]

dont have enough bookshelf room. I plan to move half my chit to my new office tho. I've also gone digital



ayyyy

Should try evernote brah ... it sick

Also lol at how many maths book can be fond for free ... honestly astounding.

Are you planning to read that amount though? It takes me roughly one or two months to cover a book thoroughly - am I being a little bitch or what lol?

[K3Fallout]

It's something you need to get used to. Setting up models is one of the emphases of most mathematics courses. Pattern following in mathematics often leads to deficiency in this area, and it just so happens most of american students go about studying this way. In short, do a bunch every day. Write down all the given information, draw pictures and take a little time to make sense of them. Physics really helped me in this.

Thanks. What I've noticed a lot is I'll see the solution to a problem and go "oh ok I get it now" but as soon as I try the next problem that's similar it all leaves my head again and I can't apply general principles. You get my measlies

[SlyOdysseus]

It's better to find values a and b such that a and b are in the interval [-1,1] and it is the case a =/= b, with the property 0 lies between f(a) and f(b).

[turkey_server]

I'm trying to prove there are at least two values of x that satisfy the equation x^2=cox(x). I can assume x^2 and cos(x) are continuous on R.

Is my proof method okay?
Define h(x)=x^2 and g(x)=-cos(x) which are both continuous on R
Thus, f(x)=h(x)+g(x)=x^2-cos(x) is continuous on R.

So clearly f is continuous on R (and in particular, on [-10,10])
Now I'm not entirely sure how to word it but I want to use the IVT twice.

Choose big numbers so I don't even have to think about it. On [-10,0], f(-10) > 0 and f(0) = -1 < 0. IVT is satisfied (b/c you already showed continuity), so you're guaranteed a zero in (-10,0). Similarly for [0,10].

[ctownballer04]

It's better to find values a and b such that a and b are in the interval [-1,1] and it is the case a =/= b, with the property 0 lies between f(a) and f(b).

yes, I screwed all types of things up in my OP. Reposting now

Choose big numbers so I don't even have to think about it. On [-10,0], f(-10) > 0 and f(0) = -1 < 0. IVT is satisfied (b/c you already showed continuity), so you're guaranteed a zero in (-10,0). Similarly for [0,10].

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!

[ctownballer04]

Large deltas do not matter. It is not necessary to show delta cannot be larger than 1, because if there was any d' at all then there are infinitely many smaller than it all the way down as close to 0 as you want. You are contradicting that any such delta cannot exist, so it suffices to contradict the existence of a small one. I chose one that was convenient (the minimum of any delta d' that works and 1). If d' < 1 then d=d'<1. If d'>=1 then d=1. This is so I could use a convenient bound on the distance x might be from 2 to obtain a contradiction.

Here is a different approach which might be more useful in general practice.

Assume there is a d>0 such that for all x in (2-d, 2+d) we have 2 < f(x) < 4.

Let a_n = 2-10^{-n} (just a simple sequence that gets close to 2 from below). There is an N>0 such that x=2-10^{-n} belongs to (2-d, 2+d) for all n>N. This is intuitively clear. {a_n} = 1.9, 1.99, 1.999, 1.9999, 1.99999, .... For rigor, use the unboundedness of the naturals: Pick n such that 10^n > 1/d . Then 10^(-n) < d implies 2-10^(-n) > 2-d. Moreover 2-10^(-n) < 2.

So we know 2-10^(-n) belongs to (2-d, 2+d). Notice also that as n gets larger, a_n gets larger, so f(a_n) gets smaller. This is also evident from the graph, since f(x) is decreasing on (1,2). We know then that f(a_n) < f(a_m) if m<n, so we'll use that f(a_n) < f(a_1) for all n.

If x=2-10^{-n) then f(x) = 3-(2-10^(-n))^2 < 3-(2-10^(-1))^2 = -0.61




dont have enough bookshelf room. I plan to move half my chit to my new office tho. I've also gone digital



ayyyy

I can safely say that I follow now.

Thank you!

[iDontExistHere]

I'm lost right now. What do I do next? Can someone link a video on this topic ?

[iDontExistHere]

But im looking for cotangent?

edit: so sin(a)/cos(a) = tan(a) ..so cot(a) = cos(a)/sin(a)

[iDontExistHere]

and cot(2a)=cos(2a)/sin(2a)

Thanks for all your help, unfortunately I'm just not understanding it. So i moved to another problem.(possible to link a youtube video going in depth?)
Am I on the correct path for this problem?

[iDontExistHere]

so far you are good

Final product(finding it much easier treating these expression as regular a&b instead of sin/cos/ect:

[JimFromRaleigh1]

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?

[JimFromRaleigh1]

Your work is fine, but be careful with grouping symbols. See red above.

You have done all the work on the right-hand side of the equation. But remember the left-hand side.

y = (x+1)^2(x+2)^3
... => ln(y) = 2ln(x+1)+3ln(x+2)
... => y'*(1/y) = 2(1/(x+1))+3(1/(x+2))

So my next step would be to multiply both sides by (y/1)?

[JimFromRaleigh1]

Yes (just y, you have my permission to graduate from writing reciprocals like y/1!)

Then replace y by what it was originally equal to and simplify

d-u-n dun

[iDontExistHere]

Am i doing something wrong? It feels like it.

[casherr]

I just wanna do a shout out to MiscMathematician, thanks to him, i graduated highschool despite not knowing anything about algebra at first.
Thank you man.

[iDontExistHere]

I just wanna do a shout out to MiscMathematician, thanks to him, i graduated highschool despite not knowing anything about algebra at first.
Thank you man.

His is really great person. I also appreciate his effort to help me . Someone give this man a raise.

[Slacker23]

Am i doing something wrong? It feels like it.
[img]http://i.imgur.com/hdjhB7i.jpg[img]

You are.

You know that sin(x)=-2/3 with x in the 3rd quadrant.

This means that cos(x)<=0 which implies cos(x)=-√[1-(2/3)^2], not that cos(x)=cos(-√[1-(2/3)^2]).

Since you know cos(x), try to use it to find cos(x/2).

[ctownballer04]

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.

[Slacker23]

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

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.

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.

[ctownballer04]

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.

[ctownballer04]

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.

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)."

[ctownballer04]

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.

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.

[SlyOdysseus]

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.

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

[ctownballer04]

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: equality is obtain at both the end points.

(*) More generally: https://en.wikipedia.org/wiki/Brouwe...-point_theorem

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.

[SlyOdysseus]

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.

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.

[ctownballer04]

Consider f(x)=x^2. Does there exist a c such that f(c)=c in (0,1)?

If c was allowed to exist in the closed interval, the statement is true. And I suggest you try to prove it.

For the latter, geometrically what is happening is that if you draw any continuous function in the unit square, it must cross the line g(x)=x. But this may happen at the endpoints as well.

Yes! this helped w/ the intuition. Thanks. Before SlyOdysseus posted I was trying to draw geometric counter examples without thinking about f(x)=x lmao. Clearly there can't be a counterexample if the domain is defined at 0 and 1.

[ctownballer04]

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

Oops, I misunderstood your correction earlier. I posted c in (0,1) where in actuality the the statement is only true for c in [0,1]. Got it now.