/sci/ - Science & Math

[Anonymous No.5287809]

Lets do it /sci/!

[Anonymous No.5287823]

69

[Anonymous No.5287880]

infinite discontinuality

[Anonymous No.5287980]

>>5287809
><span class="math">\underset{0\leq x\leq 1}{\mathrm{max}}|f'(x)|[/spoiler]
><span class="math">0\leq x\leq 1[/spoiler]
><span class="math">\leq[/spoiler]
>implying the derivative exists on the boundary
>ISHYGDDT

[Anonymous No.5287992]

This question defies the first fundamental theorem of calculus. If the function f(x) is really smooth and continuous as you claim, F (1) - F (0) should be equal to the integral of f(x) with respect to x on the domain-interval (0,1).

[Anonymous No.5288022]

>>5287980
>implying you cannot extend <span class="math"> C^1[/spoiler] function by continuity of their derivative.

fuck off

[Anonymous No.5288035]

f is continuous and alpha < 1, thus:
<span class="math">\left | \int_{0}^{a}{f(x)dx}\right | \leq \int_{0}^{a}{\left | f(x)dx}\right | \leq \int_{0}^{1}{\left | f(x)dx}\right |[/spoiler]
We also know:
<span class="math">0 = \left | \int_{0}^{1}{f(x)dx} \right | \leq \int_{0}^{1}{\left |f(x)dx}|[/spoiler]
and:
<span class="math">\int_{0}^{1}{\left |f(x)dx}| \geq 0[/spoiler]
So:
<span class="math">\int_{0}^{1}{\left |f(x)dx}| = 0[/spoiler]
Which means that:
<span class="math">\left | \int_{0}^{a}{f(x)dx}\right | = 0[/spoiler]
And that your problem is stupid.

[Anonymous No.5288037]

>>5288035
It's cool that the only equation you managed to properly type is enough to know you are wrong.

Good try though.

[Anonymous No.5288065]

use taylor lagrange with F; F being a well chosen antiderivative of f

[Anonymous No.5288069]

>>5288065
int f(x) dx over [0,1]=0 so that means there exists c in }0,1[ such as f(c)=0, right? (Rolle theorem)
let F(x)= int f(x)dx over [c,x].
then using taylor lagrange, F(x)=F(c)+(x-c)f(c) + R(x)
with F(c)=f(c)=0
etc

[Anonymous No.5288071]

Let's say there's alpha such that this isn't true and let k be the upper bound of |f'|.

<span class="math">\left| \int_0^{\alpha} f(x)\mathrm{d}x \right| >\frac{k}{8}[/spoiler]
Let's assume wlog that the integral is >= 0:
<span class="math">\int_0^{ \alpha} f(x) \mathrm{d}x >\frac{k}{8}[/spoiler]
Thus <span class="math">\int_{ \alpha}^1 f(x) \mathrm{d}x < -\frac{k}{8}[/spoiler].

So we're basically solving the same problem on both sides.

The intuition now is that the "best" alpha is 0.5 (others are harder because one side will be <0.5 and that side will still have to compensate for the larger other side and integrate to more than k/8 in absolute value), and the best way to shape a function to have a high integral in (0;0.5) while having a 0 integral over (0;1) is to have a constant slope (e.g. f(x)=1-2x). That gives an integral between 0 and 0.5 which is the area of a triangle of sides 0.5 and 1 (so 1/4), while the max of |f'| is 2: 2/8=1/4, we're there.

Now to prove that, just use the same reasoning and replace every "this is the worst case" and "this is the best case" with inequalities instead.

[Anonymous No.5288161]

>>5288069
>F(c)=f(c)=0
that's the part, that I have problems with. why can you do that?
What I got is
<span class="math">\int f(t)dt = F(x)-F(0) \leq F(c)-F(0)+ f(c)(x-c)+max(f'(x))\frac{(x-c)^2}{2} [/spoiler]
you can choose the c now, that either the f(c) term or the F(c)-F(0) term disappears

[Anonymous No.5288178]

F(c)=0 because it is the integral from c to c of something.
f(c)=0 because that's how I defined c!

[Anonymous No.5288192]

>>5288178
so <span class="math">\int_{a}^{b}f(x)dx=0 [/spoiler] for every function f that has an antiderivative F, because F(b)-F(a)=0-0=0.
Are you serious right now?

[Anonymous No.5288199]

>>5288178
what the fuck can you even read before insulting other people?
I'm saying:
<span class="math">F(x)=\int_{c}^{x} f(t)dt[/spoiler]
so <span class="math">F(c)=\int_c^c f(t)dt = 0[/spoiler]
gee, what an asshole.

[Anonymous No.5288214]

>>5288199
sorry man, my fault.
but I still don't really get, what you do with that afterwards
now you have
<span class="math">
\int_{c}^{x}f(t)dt=\frac{(x-c)^2}{2}f'(\xi)
[/spoiler]
with <span class="math"> c< \xi < x [/spoiler]

[Anonymous No.5288631]

This is rustling my jimmies.

[Anonymous No.5288660]

>max{infinite set}

[Anonymous No.5288679]

What is this sorcery, /sci/? Am I looking at differential calculus, set theory or some mysterious combination thereof?

Would someone please define this subdiscipline of whatever it is?

[Anonymous No.5288723]

>>5288679
Analysis.

[Anonymous No.5288760]

I looked up the solution. Now I'm feeling stupid. It's so easy, once you got the right approach.

[Anonymous No.5288766]

>>5288760
Where'd you get it? I'm curious?

[Anonymous No.5288773]

>>5288723

I mean what is the discipline or subdiscipline? "Analysis" being part of a standard college calculus curriculum? Or Linear Algebra? What class would I find such concepts in?

[Anonymous No.5288776]

>>5288773
Analysis is probably part of the standard curriculum if you're studying mathematics.

[Anonymous No.5288778]

>>5288766
Look at OP's filename and use google to find something like http://amc.maa.org/a-activities/a7-problems/putnamindex.shtml


>>5288773
The class is called "analysis" and is part of every undergrad math curriculum.

[Anonymous No.5288800]

>>5288778
>is part of every undergrad math curriculum.

I understand that FFS I just wanted to know if it was present in the first three semesters of elementary calculus OR if it pops up later in some advanced course such as linear algebra or advanced calculus.

I'm assuming it comes in the last portions of elementary, mandatory calc but not sure...

[Anonymous No.5288812]

>>5288800
Real Analysis is what you take in the very first semester.

[Anonymous No.5288813]

>>5288800
>advanced course such as linear algebra or advanced calculus
oh you americans and your shit tier schools, you have to pay a fortune for

[Anonymous No.5288815]

>>5288800
What the fuck are you asking? Analysis is a course on its own. Calculus is what you do in high school: Intuitive explanations of why derivates are legit and some symbolic manipulation. In analysis you learn WHY this shit works.

[Anonymous No.5288819]

>>5288800
>>5288812
It differs per school. I got a whole year of linear algebra the first year. Got real analysis only 4th term. So it depends on what your university/college thinks is best.

[Anonymous No.5288820]

>>5288800
>linear algebra
>advanced

[Anonymous No.5288840]

>>5288820
People who can't into math tend to think first year calc and linear alg are the end of math, and that everything falls into one of those 2 categories after that.
Atleast that's how they are in america.
>Be in 9th grade
>start learning group theory on my own
>next year turn my friend onto it
>by 12th grade I have taught myself a large part of complex analysis among other topics
>tfw you have saved not only yourself, but your jewish friend from the scourge that is the american school system
feels good man

[Anonymous No.5288841]

>>5288819
I had Linear Algebra I + II, Analysis I + II, a programming class, a discrete math class, group theory and Experimental Physics I + II in my first year.

[Anonymous No.5288845]

>>5288841
That sounds like a standard curriculum for the first year.

[Anonymous No.5288855]

>>5288812
would you consider an spivak based course to be real analysis?

[Anonymous No.5288961]

>>5288815

I've taken set theory, geometry, algebra, trig and first year calc in HS. Nowhere in the curriculum do I recall anything called "analysis". I've had to prove theorems in plane geometry as well as in set theory and part of that proof appeared to look like this "analysis" you speak of.

I've never been asked to prove differential equations like how you prove corollories of theorems but rather been asked to apply them to problems where you plug numbers in.

Capiche?

>>5288820


It comes AFTER calc and requires calc as a prerequisite. Unless the opposite is true in your school? Or they just let you waltz into it after Algebra?

[Anonymous No.5288973]

>>5288840

>>tfw you have saved not only yourself, but your jewish friend
>your jewish friend

Off topic somewhat but what is your opinion on Ashkenazi prowess in the math/science niche? Stormfags assert jews have very high verbal IQs and somewhat low "spatial" intelligence whatever that translates to in academia but I thought it meant they sucked at math. However, to me the opposite seems true but just wondering what the general consensus is about the matter?

[Anonymous No.5288981]

>>5288961
>It comes AFTER calc and requires calc as a prerequisite.
Yes, calc is taught in high school, linear algebra is the first math course in university.

>Or they just let you waltz into it after Algebra?
Algebra is usually taught after linear algebra. Linear algebra only requires a rough understand of what a group and a field are, while algebra covers these in depth.

[Anonymous No.5289002]

>>5288981
>Algebra is usually taught after linear algebra. Linear algebra only requires a rough understand of what a group and a field are, while algebra covers these in depth.

wtf! Here's how it is in my school:

1. Learn Rithmatic
2. Learn Algebra
3. Leanr Geometry, same time as Algebra if you want
4. Learn Trigonometry AFTER you're done with Geometry.
5. Now you can move to Calculus, pleb.
.
.
.
College/College level stuff where you repeat anything you didn't already cover in HS and/or pass with AP tests. Afaik Linear algebra is listed with a course number HIGHER than that of basic algebra, geometry and trig. I think the course numbers are even higher than calc.

[Anonymous No.5289009]

>>5288973
He was pretty good with it, although I helped him alot.
I had an extreme niche for abstract algebra especially, he was better with actual calculations and such

[Anonymous No.5289041]

>>5289002
Are you seriously telling me that you learned about sylow theorems, galois theory and noetherian rings in fucking high school before having heard of calculus for the first time? Not that you need calculus for those, but that's fucking weird. Where did you go to school?

[Anonymous No.5289093]

>>5289041
Are you retarded?
>http://en.wikipedia.org/wiki/Elementary_algebra

[Anonymous No.5289096]

>>5289093
So you think because you know how to solve a linear equation, you're entitled to talk about "higher" math? You're truly a cretin.

[Anonymous No.5289118]

>>5289096
Who the fuck are you talking to? That was my only post. The OP clearly doesn't know what any of that stuff is. So stop being an elitist prick when you're probably just a stupid undergrad shit yourself.

[Anonymous No.5289120]

>>5289096
Guy who taught himself abstract algebra and some other mid-tier math during highschool here.
I had no problem learning anything during this time armed with the skills from algebra I
Learn calc and review some geometry, and youll be fine

[Anonymous No.5289122]

>>5289118
Oops, not OP. But the guy who started that conversation.

[Anonymous No.5289131]

>>5289118
> clearly doesn't know what any of that stuff is
That's what we're trying to tell. He doesn't want to get it though.

>>5289120
I know the math and of course it's easy to learn it on your own. I'm not the retard poster who doesn't know what he's talking about.

[Anonymous No.5289148]

>>5289131
I have no idea who is who anymore.
But if you want to learn something, learn it.
If you dont know where to start, start anywhere.
Mid-tier math is pretty wide open, find something you like, and back track until you find something you understand, then go forward again

[Anonymous No.5289153]

>>5289131
>That's what we're trying to tell. He doesn't want to get it though.
That's your own faults. This thread is nothing but idiots projecting their own education system onto him, and using that "misunderstanding" to act elitist. Make no mistake, the guy is saying stupid stuff, but you all came across as idiots as well.

For example, he clearly thinking of high level math as the highest level of mandatory math for the average non-math major. If someone had just pointed out that Analysis was not in general part of that curriculum, instead of saying stupid things that depend on your specific school and program, then it wouldn't have taken over the thread.

[Anonymous No.5289165]

>>5289153
Knowing undergrad math is not elitism. He asked what math is required to understand OP's problem. We told him it's a course called analysis. He was too retarded to understand such a simple statement and went on asking analysis was part of calculus or linear algebra. There's not much we can do for him, if he understands neither math nor simple english sentences. And you're wrong btw. Analysis is a mandatory undergrad course for every math student.

[Anonymous No.5289236]

>>5289165
>Analysis is a mandatory undergrad course for every math student.

At a semi-decent school. Some will actually try to mask it as "Advanced Calculus."

[Anonymous No.5289241]

>>5288660
Anyone want to reply to this? Half of the proofs I've seen in real analysis/topology have relied on the fact that you can only take the max of a finite set.

[Anonymous No.5289256]

>>5289241
What the fuck are you talking about? On a compact set a continuous real valued function has a minimum and a maximum.

[Anonymous No.5289264]

>>5289256
I don't remember specific proofs, but I remember having to appeal to the "finiteness" of compact sets at some point to ensure we could find a maximum epsilon or something.

[Anonymous No.5289369]

>>5289264
Perhaps because finite sets are guaranteed to have a maximal element?

(At least subsets of R)

[Anonymous No.5289410]

>>5288679
please be troll

[Anonymous No.5289670]

>>5289410

What's so trolly about it? I've just never seen a proof like that in my calc book that I recall although I haven't done ALL of calc, just some in HS. The "belongs to" sign and the "such that" sign just reminded me of set theory. Maybe in this "analytics" or whatever it means something else. *shrug*

http://en.wikibooks.org/wiki/Discrete_Mathematics/Set_theory

[Anonymous No.5289964]

Since the function
<div class="math">g(y) = \int_0^y f(x) dx</div>is continuous, it attains its maximum at a point c in [0, 1]. If c is at the boundary, we have g(c) = 0 immediately. Otherwise the continuous derivative f(c) = 0. So for all <span class="math">u \in [0,1][/spoiler]:
<div class="math">|f(u)| \le \int_c^u |f'(x)| dx \le |u - c| \max_{0 \le x \le 1} |f'(x)|</div><div class="math">|g(u) - g(c)| \le \int_c^u |f(x)| dx \le \frac12 |u - c|^2 \max_{0 \le x \le 1} |f'(x)|</div>Take either u = 0 or u = 1, whichever has <span class="math">|u - c| \le 1/2[/spoiler]. We obtain
<div class="math">g(c) \le \frac18 \max_{0 \le x \le 1} |f'(x)|.</div>Similarly, the minimum value of g satisfies
<div class="math">g(c') \ge -\frac18 \max_{0 \le x \le 1} |f'(x)|.</div>Thus
<div class="math">\int_0^\alpha f(x) dx \le \frac18 \max_{0 \le x \le 1} |f'(x)|.</div>for all <span class="math">\alpha \in [0,1][/spoiler].

[Anonymous No.5290134]

>>5289410
nah, he's just another retarded CS major that think CS belong in /sci/

[Anonymous No.5290150]

>>5289964
Nice.

[Anonymous No.5290218]

>>5289670
No, this IS calculus.

What you've been told is calculus is really just what they tell the engineers. This is calculus for mathematicians, aka analysis.

[Anonymous No.5290223]

>>5290218
>replying to an obvious troll post

Can't you see that poster wasn't serious at all?

[Anonymous No.5290259]

>>5290223
>replying to that obvious troll post

[Anonymous No.5290265]

>>5290259
Is it true that when you reply to trolls they steal some of your karma?

[Anonymous No.5290310]

>>5290218

Don't listen to that deluded troll-hunting wannabe whiteknight retard. I'm just comparing the HS calc I took along with some set theory I also took to the diagram in the OP like I said.

I don't understand the difference, if any, between the calc for engineers, CS majors and mathematicians. Is regular HS calc a tiny subset of this vast mysterious landscape that comes in college?

[Anonymous No.5290321]

>>5290310
>Is regular HS calc a tiny subset of this vast mysterious landscape that comes in college?

Yes. Now go read Rudin.

[Anonymous No.5290371]

bump

[Anonymous No.5290928]

>>5289964
|g(u) - g(c)| \le \int_c^u |f(x)| dx \le \frac12 |u - c|^2 \max_{0 \le x \le 1} |f'(x)|

could somebody explain this part with more details? I understood the rest, but could not figure out how this transformation works

[Anonymous No.5290941]

>>5290928
<span class="math">|g(u) - g(c)| \le \int_c^u |f(x)| dx[/spoiler] or <span class="math">\int_c^u |f(x)| dx \le \frac12 |u - c|^2 \max_{0 \le x \le 1} |f'(x)|[/spoiler]?