/sci/ - Science & Math

>>10067855
What is mc theorem?

>t. Brainlet self teaching herself analysis

>>10067855
Monotonía just prevents cusps, IVT does th rest.

>>10067910
Wtf are u even saying

Simple, you can recursively apply the extreme value theorem and then iterate up to n times

>>10067855

>>10067855
I'm not a mathlet so I have no idea what that even means. Isn't IQ meant to be independent of specialized knowledge?

>>10068036
(if f is monotonically decreasing apply the same argument with -f instead)

>>10067855
I too have taken real analysis

>>10068054
>real analysis
this was calc 2 stuff

>>10067855
For IVT to be true, f must be in C0.

wew

>>10068064
kek

>>10068040
its not an iq test they're just being cheeky and trying to gauge how many people here are brainlets who haven't taken the requisite maths to solve it. Its sort of an iq test as mathematical attainment is a good measure of intelligence but there are 110 iq mathematicians so its not really that useful.

>>10068062
Man that was years ago, I do remember chasing things of this nature down rabbit holes in real analysis.

>>10068064
>what is Darboux's theorem

God, you're ignorant. A function can have the Intermediate Value Property even though it is not continuous

>>10067855
Let x(n) be a sequence in [a, b] which converges to x(0) (x(0) being an element of [a, b]). We must prove that f(x(n)) converges to f(x(0)).

Given ε > 0, consider the inequality |f(x(n)) - f(x(0))| < ε. This means f(x(0)) - ε < f(x(n)) < f(x(0)) + ε.

For ε small enough, we can assume that both f(x(0)) - ε and f(x(0)) + ε belong to [a, b]. Since f has the Intermediate Value Property, there exist two real numbers v and w in [a, b] such that:

f(v) = f(x(0)) - ε
f(w) = f(x(0)) + ε.

We know that f is monotonic; let's assume it is increasing (the argument is analogous for decreasing functions). Then, since f(v) < f(x(0)) < f(w), it follows that v < x(0) < w. Since x(n) converges to x(0), for large enough n we always have v < x(n) < w; and from this it follows that for large enough n we always have f(v) < f(x(n)) < f(w); in other words we have f(x(0)) - ε < f(x(n)) < f(x(0)) + ε (because of the definition of f(v) and f(w)). This proves that f(x(n)) converges to f(x(0)) and therefore that f is a continous function.

>>10068138
>we can assume that both f(x(0)) - ε and f(x(0)) + ε belong to [a, b]

They should belong to [f(a), f(b)]. (Assuming that f is increasing).

>>10067855
This retard has been posting babby's first analysis questions and complaining about nu-/sci/ all week. They're obviously asking for homework help.

>>10067855
Here, I'll do your homework, anon. You're my bitch now.

[eqn]
\text{Let } f \text{ (without loss of generality) be a non-decreasing function, such that it satisfies the Intermediate Value Property.}\\

\text{Assume the contrary, that } f \text{ is not continuous. }
\text{Then } \exists r \in \mathbb{R} \text{ and } \epsilon_0>0 \text{ such that } \forall \delta > 0 : f(r) - f(r - \delta) \geq \epsilon_0 \text{. (A)} \\ \text{ Since } f \text{ is non-decreasing: }
\forall x \in [a, b]: (x > r \Rightarrow f(x) > f(r)) \land (x < r-\delta \Rightarrow f(x) < f(r-\delta)).
\\
\text{Take some fixed } \delta>0 \text{, and let } t =f(r)-\frac{\epsilon_0}{2}.
\text{By Intermediate Value Property, there exists } c \in [a, b] \text{ such that } f(c) = t = f(r) - \frac{\epsilon_0}{2} \Leftrightarrow \\
\Leftrightarrow f(r) - f(c) = \frac{\epsilon_0}{2} < \epsilon_0, \text{Which is a contradiction to (A). Thus, } f \text{ is continuous.}\\

\blacksquare
[/eqn]

>>10067855

Can I use the MC Hammer Theorem?

>>10068766
What a garbage "proof". Your statement (A) is completely wrong. Take the Dirichlet function. It's discontinuous at 0 but D(0)-D(h) = 0 for every rational h.

y=f/x

>Worthless math makes you the top 5% of /sci/
I bet you can get any fart huffing job you want after you can solve this, autismo.

>>10069042
t. biofag

>>10068970
That’s not monotonic though.

Suppose f has a discontinuity at x. Since it's monotonic f(x-) and f(x+) exist but f is Darboux, therefore at least f(x-) or (x+) doesn't exist. QED.
Also, what the fuck is the MC theorem?

The fuck y’all talmbout

>>10067874
I'M A GIIIIIIIIIIIIIRRRRRRLLLLL

>>10067855

>>10067855
Inverse image of a basic open set is an open interval. What's the hard part supposed to be?

>>10067855
Very likely a homework problem; I'll give you the idea and you must fill the gaps

Take x in (a,b) and e>0. Then there exist c,d so that a<=c<x<d<=b. By IVP, there exist c*, d* so that c<=c*<x<d*<=d and f(c*) and f(d*) are within e of f(x). By monotonicity, you are done.
Some gaps to fill:
- Elaboration on using IVP (explicit use)
- Why monotonicity above guarantees continuity at x
- Adapting the procedure for x=a or x=b

Low effort bait post too, everyone on this board can do such a trivial problem

>>10068066
There is literally no prerequisite for this, it follows directly from the definitions of continuity, monotonicity and IVP

>>10067855
F=R=x incorrect
Try this for size.
AB/CA /— T2+ACv.
4R-E=(F)
F=G/E
A=E
G=r+3/=34-S

>>10067855
do not know English :( and the automatic translator does not respond wellu
I'm just crazy and she's the genius

I haven't used college math in seven years. This was the power series? Probably not.

>>10068066
> the requisite maths to solve it
>literally calculus 1

>>10068040
If you have no idea what this means then that means you're a mathlet
Being a mathlet means you don't understand math, not that you do. The "-let" suffix implies lack of something.

>>10070799
>open set
>open interval
Is [] actually used as open set notation somewhere?

>>10068064
Let f(x)=2x if x<0.5 , and f(x)=x-0.5 if x>=0.5

>>10072321
No. A power series expansion of a function requires derivatives of all orders to exist (where, roughly speaking, we approximate the function using linear combinations of its derivatives, and the approximation becomes exact when we use derivatives of all orders). Having a derivative at all already implies the function is continuous, so you would end up assuming what you have to prove.

>>10067855
IT CAN ONLY SATISFY THE INTERMEDIATE LAGRANDE TEOREM IF IT IS CONTINOUS RETARD

>>10072352
OP meant that if f(a)=c and f(b)=d, for any f(a)<y<f(b) there's e such that f(e)=y.
He just goofed the text.

>>10072357
Actually he didn't. https://en.wikipedia.org/wiki/Intermediate_value_theorem

>>10067874
>femanon here

>>10072330
I've seen ]a,b[ to denote open intervals but never [a,b].

>>10072361
Scroll down, you retarded physicist.

>>10072350
Thanks anon. I looked up the definition of IVT. I thought maybe it had something to do with the radius of convergence or w/e, but like you said since it has a derivative it's continuous by MVT. I never had to take real analysis so the problem looked unfamiliar. Perhaps it was covered in calc but like I said it was seven years ago.

>>10067874
tits or gtfo

>>10072368
>I've seen ]a,b[ to denote open intervals
god I hate this notation so much
it looks so fucking stupid

*sip*
Yep
Buncha undergrads

>>10067855
Quick proof.

Let $y\in [a,b]$. It's clear that the supremum $L^-$ of $f(x)$ such that $x\in [a,y)$ exists and that it's equal to $\lim_{x\to y^-}f(x)$. Similarly, the infimum $L^+$ of $f(x)$ such that $x\in (y,b]$ exists and is equal to $\lim_{x\to y^+}$. We have $L^-\leq f(y)\leq L^+$. Any of these inequalities being strict contradicts the intermediate value property.

Undergraduate shit.

>>10072669

Let [math]y\in [a,b][/math]. It's clear that the supremum [math]L^-[/math] of [math]f(x)[/math] such that [math]x\in [a,y)[/math] exists and that it's equal to [math]\lim_{x\to y^-}f(x)[/math]. Similarly, the infimum [math]L^+[/math] of [math]f(x)[/math] such that [math]x\in (y,b][/math] exists and is equal to [math]\lim_{x\to y^+}[/math]. We have [math]L^-\leq f(y)\leq L^+[/math]. Any of these inequalities being strict contradicts the intermediate value property.

Sorry for the formatting issues... I don't browse /sci/ and came here just for a minute for shits and giggles.

>>10072655
Yeah, me too, but it's useful when you're dealing with functions from R^2 to R because you'll have a lot of open intervals and ordered pairs.

>>10072650
This

>>10072657
*Freshmen

>>10068564
i told him to try the kurisu image and he actually did it the absolute madman

what a faggot

>>10068970
>Dirichlet
>monotonic
just withdraw from the class jamal.

>>10067855
Let e>0.
Let x be in [a,b]
Use IVP to find l in [a,x) such that f(l)>=f(x)-e/2
Use IVP to find r in (x,b] such that f(r)<=f(x)+e/2
(If x is a or b, only one of these is needed)
f(l)<=f(u)<=f(x) for any u in [l,x].
-e/2<=f(l)-f(x)<=f(u)-f(x)<=0. e/2>=|f(u)-f(x)|.
f(x)<=f(u)<=f(r) for any u in [x,r].
0<=f(u)-f(x)<=f(r)-f(x)<=e/2. e/2>=|f(u)-f(x)|
(l,r) contains x and |f(x)-f(u)|<=e/2<e for any u in (l,r).

>>10067874
If you want a real answer: monotone convergence theorem

>>10067874
Fuck off back to the pit you came from roastie

>>10067855
Suppose [math]f[/math] were not continuous at [math]a[/math]. Then for some [math]\epsilon>0[/math], there is no [math]\delta>0[/math] for which [math]|f(a+\delta)-f(a)|<\epsilon[/math]. If such an [math]\epsilon[/math] existed, then there would be some value [math]b[/math] not in the range between [math]f(a)[/math] and [math]f(a)+\epsilon[/math] (or [math]-\epsilon[/math] wlog). But [math]f[/math] is monotone and satisfies the IVT, which means such a value must exist and must be between [math]f(a)[/math] and [math]f(a)+\epsilon[/math]. Thus [math]f[/math] must actually be continuous.

I could do that in last year of high school.
Don't remember shit now.

Let x in a,b and e>0.
If f(x)=f(a), we define a':=a, otherwise let m:=max (f(x)-e, f(a)). By the intermediate value, m is in the range of f and we pick a' such that f(a')=m. Since m<f(x) we have a'<x (f(x) cannot be equal to max (f(a),f(x)-e) which are both strictly infierior to it).
We do a similar thing for b: if f(x)=b we define b':=b, otherwise we pick b' such that (intermediate value) f(b')= min (f(b), f(x)+e). Again in this case, x<b'.

Then for every t in [a',b'], f(t) is in [f(x)-e,f(x)+e]. And in addition, [a',b'] is a neighborhood of x. Indeed it is clear that x is in [a',b']. But we saw that if x=a' then a'=a (resp. if x=b', then b'=b).
If a'=b', then x=a'=b'=a=b and the map is defined on a singleton and is trivially continuous. Otherwise, x is in ]a',b'[, or x=a and x in [a,b'[=[a',b'[,or x=b and x is in ]a',b']=]a',b]. In any case, [a',b'] is a neighborhood of x and the result follows.

Physicist here, it is obvious.

Proof: Think!

>>10067855
The preimage of any open interval in R is an open interval in [a,b].

>>10068062
>Having a Calc 2 class instead of studying real analysis alread

>>10067874
Instant de-railment.
AKA Cleo is not here anymore.