/sci/ - Science & Math

[spoiler]Hey mod, are you still mad?
<div class="math"> \def \d#1{{{#1}\atop{#1}}}\def \e#1{\d{\d{\d{\d{\d{#1}}}}}}\def \f#1{\e{\e{#1}}}\smash{\f{{-------------------------------------------------------------------------
----------------------------------------------------------------------------------------------------
-------------------------------------------------------------------------------------}}} </div></spoiler>

inb4 someone tries to just take the derivatives of n^2/(n^2+1)

f(x) = f(1 / (1/x)) = (1/x^2) / (1 / x^2 + x^2/x^2) = (1/x^2) / ((x^2 + 1) / x^2) = 1 / (x^2 + 1).

The taylor series of this is 1 - x^2 + x^4 - x^6 + ...
So for odd derivatives we'll get 0. And for even, we'll get -n! for n = 2, 6, 10, ... and n! for n = 4, 8, 12 ...

I think I still have to show that either this is the only such function, or that all functions of this type will have the same derivative at 0.

Apply the following inductively to <span class="math">F(x) = f(x) - 1/(x^2+1), F(x) = f'(x) - [1/(x^2+1)]', etc.[/spoiler]

Lemma: Let <span class="math">F(x)[/spoiler] be infinitely differentiable and suppose there is a strictly decreasing sequence <span class="math">x_1 > x_2 > \cdots > 0[/spoiler] with limit zero such that <span class="math">F(x_i) = 0[/spoiler] for all <span class="math">i[/spoiler], and also <span class="math">F(0) = 0[/spoiler]. Then <span class="math">F'(0) = 0[/spoiler] and there exists another strictly decreasing sequence <span class="math">y_1 > y_2 > \cdots > 0[/spoiler] with limit zero such that <span class="math">F'(y_i) = 0[/spoiler].

Proof: The sequence of difference quotients <span class="math">(F(x_i) - F(0))/x_i[/spoiler] is identically zero, and since <span class="math">x_i \rightarrow 0[/spoiler], <span class="math">F'(0) = \lim_{i \rightarrow \infty} (F(x_i) - F(0))/x_i = \lim_{i \rightarrow \infty} 0 = 0[/spoiler] by continuity of <span class="math">F'[/spoiler]. The sequence <span class="math">y_i[/spoiler] exists by Rolle's theorem, choosing a <span class="math">y_i[/spoiler] between each pair of consecutive <span class="math">x_i[/spoiler]'s.

Therefore <span class="math">f^{(n)}(x) - \frac{d^n}{dx^n} 1/(1+x^2)[/spoiler] is zero at <span class="math">x=0[/spoiler] for all <span class="math">n[/spoiler] and the answers are what >>4266592 said.


Now pony tiem

[spoiler]<div class="math"> \def \d#1{{{#1}\atop{#1}}}\def \e#1{\d{\d{\d{\d{\d{#1}}}}}}\def \f#1{\e{\e{#1}}}\smash{\f{{WHY HELLO , /SCI/}}} </div></spoiler>

>>4266619
Very niec.

[spoiler]<div class="math"> \def \d#1{{{#1}\atop{#1}}}\def \e#1{\d{\d{\d{\d{\d{#1}}}}}}\def \f#1{\e{\e{#1}}}\smash{\f{{█████████████████████████

█████████████████████████████████
██████████████████████████}}} </div></spoiler>

>>4266619
>Therefore f(n)(x)−dndxn1(1+x2) is zero at x=0 for all n

It looks like you went to a lot of effort to prove something completely trivial.

>>4266870

>>4266560
Fuck, thanks for posting it in LaTeX.

It's impossible to read that image when you're using a dark theme.

>>4267113
protip: press ctrl+a

[spoiler]>>4267350
thats the retard way of circumventing sigouneys faggotry, dumass. it gives all pics a retarded blue tinge</spoiler>

>>4267388
thanks for offering an alternative, instead of just insulting people.

[spoiler]>>4267473
>insulting people
you didnt really get offended by 'dumass' did you? you fucking overly sensitive weak-assed pussy!

i DO have an alternative actually, one i could give you in less than 5 seconds, but seeing as you're a fucking whiny bitch i think i'll leave you to rot.

eat shit, hun.</spoiler>

bump so I dont see the face of the whore from EK's avatar in the front page

[spoiler]>>4267558</spoiler>

>>4267567

You think this is a motherfucking game?

[spoiler]>>4267572</spoiler>

>>4267473
http://userscripts.org/scripts/show/123446

>>4266870
>It looks like you went to a lot of effort to prove something completely trivial.

It looks like you don't even understand the question.

Poasting so I don't have to see this pony every time I refresh the page.

>>4268074
u mad?

>>4266870
>prove something completely trivial.
When you took analysis, is that what you wrote on your exams?

I can imagine:

>implying you don't know my name
>implying you need to know the date

>implying i will do your homework
>implying this isn't trivial
>implying a 5th grader couldn't do this proof
>implying this isn't babby calculus
>this involves harmonic conjugate functions? laughing_undergrads.tiff
>implying i'd waste my time

alone with a bunch of poorly drawn sketched images of george costanza in the margins.

>>4267350
No, I didn't mean that.

Even without this blackout thing, it's really hard to read those with a dark theme.

There's a difference between the blackout and a dark theme. I really appreciate what mods did.

PROTIP to >>4268308 : Just open the .gifs in a new tab.

PROTIP to OP: Resave the pictures as .jpgs before posting them so they don't show up as white-on-black in the thumbnail.

poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni

poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni poni

Is it weird that ponymath gives me a semi?

>>4268898
pics or it didn't happen

>>4266870
And that's how you prove something.

It was posted a while back. But does anyone have the archive of the problems of this website?

>>4266619
0 points.

You never proved that such a decreasing sequence exists. Also, the function is n^2/(n^2 + 1), not 1/(n^2+1)

>>4269352
The initial sequence is 1/2, 1/3, 1/4, ... and f(x) = (1/x)^2/(1+(1/x)^2) = 1/(x^2+1)

>>4268316
If you open the gifs in a new tab and you're using some dark themes, it will still open with a dark background.

For fuck's sake, stop pretending you understand my problem

>>4268964
>>4268656
>>4268632
>>4268605
>>4268170

>>4268632

>>4269445
Make a greasemonkey or userscript that handles clicks on .gif images on 4chan and instead of opening them in a new tab, opens a local tab like http://localhost/scigif/1326874867078.gif.html (which does not exist). Then make another greasemonkey that handles pages http://localhost/scigif/*.gif.html, and make it something like (modulo a mistake or two that I've likely made in something that I haven't tested):
var gif_location = [some regexp to get the URL from the window.location];
document.body.innerHTML = "<div style='bgcolor=white'><img url='" + gif_location + "' /></div>";

That'll open your gifs on a white background. I could make you the scripts if you can't do it but would really like to be able to read stupidly coded gifs.

>>4269727
> confirmed for full retard

>>4269792
Well that would work. Maybe it's not optimal, maybe it has a few annoying problems like you can't use ctrl+s to save the picture in its new tab because it would try saving the html instead, so you have to right-click+save it. But that would work anyway.

If you have a far better solution that has the same "low risk" has unprivileged user scripts, that is as simple to code, and that also allows directly viewing "black on transparent" gifs on a dark background theme, please suggest it. If not, don't pretend that you can judge my solution.

>>4269840
How about.....click the image. It appears as black text on white background in a new tab.

>>4269844
Have you read >>4269445 ?
If you have a dark theme, at least in some browsers, it looks like opening the image in a new tab opens in on the background color of the theme, which is NOT white.

So make a custom style sheet in the browser so images appear on white backgrounds.

>>4269857
I don't know if you can make a custom stylesheet for images that are opened in a new tab, therefore not in an html document, therefore not subject to CSS. Or well, maybe that's possible but I don't know how to do it.

Oh. Well, just change the background to white then.

That awkward moment when your knowledge of math doesn't extend outside of Pre-Calculus, and therefore can't use theorems and shit to prove this shit. Feels bad man.

>>4270133

>theorems and shit to prove this shit

Get the fuck out of here. This is the wrong place for your flippant way of acting like total nigger.

>>4270245
>saging a sticky

>>4270245
Neither is this the place for racism. Ergo GTFO applies to you too.

>>4270256
>Thinking Nigger on 4chan is racism
You must be new here.

>>4270133

I feel kinda bad for your having done all that work when it's all wrong...

Now I know how only really heartless people can be teachers and not go insane...

The first post explicitly told you you can't do it that way.

YUDOTHIS??????????

So, 1/n - 1/(n+1) goes to zero for large n, so taking n to be an integer (large) the derivative at zero looks like

lim n->infinity
(f(1/n) - f(1/(n+1)))/((1/n)-(1/(n+1)))

which is defined by the problem

I'm too lazy to plug in the numbers, but I think it's zero. I don't know where to go from there though

>>4266592
What are you doing? The function isn't defined that way for x not in natural numbers.

>>4266619
Your proof is erroneous. You completely disregarded the denominator which tends to zero as i goes to infinity. This results in an indeterminate form. I'm not even sure how you are "using the continuity of F' ".

>>4271267

Doesn't matter that the denominator goes to zero because the numerator IS zero. So you have 0/x_i = 0. The limit of zero is zero.

What you are thinking of is when both numerator and denominator are non-zero but approach zero. Then you have an indeterminate form.

>>4266560
$\frac{n}{x}$

>>4271267
That's a function that satisfies the requirements.
The post below it shows that any function f has the same derivative at 0.

It's not an indeterminate form. lim x->0 of 0/x = 0. He's using continuity to show that the limit exists, so that the derivative must be equal to the lim i -> inf of f(xi) - f(0) / (xi) = 0

>>4266592

I believe this is the correct answer, but for the wrong reasons (the author even admits it's not enough). You should use this answer to conjecture a formula and use the limit definition of the derivative, which becomes:

f'(0) = lim_{n->\infty} n * ( f(1/n) - f(0) )

You then continue inductively for higher derivatives. It's much easier if you know the general form of the nth derivative (it looks like a binomial coefficient).

>>4271267
Oops, should be continuity of the difference quotient, not F'. If h is cts then <span class="math">\lim_{i\rightarrow \infty} h(x_i) = \lim_{x\rightarrow 0} h(x)[/spoiler] if <span class="math">\lim x_i = 0[/spoiler]

I FAIL

I can't even get the Taylor series right.

>>4272129

op is a fag

>>4266619
this is why this boards makes me feel dumb every fucking time
i have no god damn clue what the hell you are talking about

>>4272894
Yeah because not studying maths at university level makes you dumb? It's actually not hard to understand what ponymath writes (harder to come up with it, which is the point of putnam), but obviously that requires the math knowledge. And lack of knowledge isn't dumbness. Thinking that is is, however, is a bit dumb.

>>4272894
*hugs*

Don't worry about it. The math is surprisingly simple. It's only the symbols that might confuse you.
The work like

a = something really easy like a meter
a+b+c+d+e = something a bit more complex, but once you understand, it becomes easy
and we can call that "f"
and f is basically the equation above, and once you understand the equation above, "f" becomes easy to use in other equations like
f+g+h+i= ?
And each of those letters is also something more complex, but with study, they become easily understood. I mean, you can't make a really good cake from scratch, can you? But a professional chef can because he practices and understands the ingredients.
Same with this. You are not dumb, you just need to study and understand.

For some basic stuff, I recommend "Understanding Physics" by Issac Asimov. It's good for beginners, and can lead you on the path to understanding the more complex stuff.
Just work hard and STUDY.

>>4272979
You sir, are a bro.

>>4272991
Really?
I made this thread >>4272960
But the people there seem to dislike me.
I suppose people truly are shallow and dislike any sort of divergence from the norm.

why is this problem still up? it's time for a new problem.

Wut are limit of polynomial interpolation of the points (1, f(1)), (2, f(2)), (3, f(3), ..., (n, f(n)) as n goes to infinity? I can has the derivative of thems?

>>4273788
oops i ment (1, f(1)), (1/2, f(1/2)), (1/3, f(1/3)), ... (1/m, f(1/m)) as m goes to infinity

try this...
instead of E=mc2
put E=t

Okay, stand back bitches, math is about to happen.

Maybe now somebody can finally update this old stickey.

f(1/n) = (1 + (1/n)^2)^-1
Taylor expand about 1/n = 0 (use the binomial theorem):

f(1/n) = (-1)^i * (1/n)^2i

f^n(0) is therefore 0 is n is odd and (-1)^(n/2) * (1/n)^n if n is even

>>4272979
im really fucking late to answer, but anyway
thanks for cheering me up
sometimes i seem to forget that 4cha­n is not compleatly made of assholes (like /b/)

>>4269840
Well that's pretty good, but I'd rather leave my 4chan theme dark.

Still, thanks a lot.


.... mods, why is this thread still alive? It's been like three days, did Putnam website suddenly die?