/sci/ - Science & Math

no, SCIENCE.

>>4288622
MATH AND SCIENCE

why the parentheses in (a_n(x,y))^2?

If x^2 + y^2 > 2|x|, then each iteration increases the value; ((x^2 + y^2)/2)^2+y^2 > (x^2 + y^2)^2, and so on. If x^2 + y^2 < 2|x|, then each iteration decreases the value. If it's increasing, though, by the nature of squaring, the absolute rate of change increases, and if it's decreasing, the absolute rate of change decreases.

>>4288661
Rage, this isn't right at all.
The acceleration bit is crap.

wtf? is this shit even bounded?

I can't be expected to do math and watch MLP at the same time. Maybe ponymath can.

A quick question: to find the answer to this question is it necessary to have knowledge of math than only one person cursing mathematics would have or an engineer student is also supposed to know who to solve this?

>>4288682
No, you should be able to do this with just calc, if your calc was any good.

>>4288682

Same doubt here.

>>4288695

Yeah, it really wasn't... Thanks for the answer, though.

it converges if x < 1 and y < 1, yes?

How are they squaring an ordered pair? I think I understand the adding and multiplying, if ordered pairs are, in this case, in any way similar to vectors; but I've never squared a vector and never even imagined squaring an ordered pair.

tl;dr notation is confusing

>>4288682
Okay, so if it's increasing, it will always be increasing, and if it's decreasing, it will always be decreasing. If x < y, then the series is bounded above by y, y^2, (y^4 + y^2)/2, (y^8 + 2 y^6 + y^4 + 4 y^2)/8, etc. converges. Since, for y > 1, this is strictly increasing, yikes. The difference between each term and the last is (a^2 + y^2)/2 - a = ((a-1)^2 + y^2 - 1)/2 which, if y > 1, is positive and increasing as a increases. So if y > 1, x > y, this is a divergent sequence.

If y > 1, then by this same logic, the sequence approaches 1 because it's always increasing with at least y^2 - 1 each iteration. Once a > 1, the growth increases every step. If y > 1, the sequence is always divergent.

If y < 1, but the sequence is increasing, then (a-1)^2 > 1 - y^2. We have to figure out whether it's possible for a to jump from smaller than 1 to greater than 1.

Otherwise, for each one we have to prove they converge.

>>4288780
It's a function on two variables, not an ordered pair. <span class="math">a_n(x,y)[/spoiler] is either a function from x and y to a single value, or a function from an ordered pair to a single value, and the function operation takes precedence over squaring.

>>4288789
thanks
>why the fuck am i so retarded

>>4288783
Is it asking too much to say that each sequential term has to be strictly less than its preceding term? i.e.:
2|x| > x^2 + y^2 as in your post >>4288661

If this *is* sufficient, then the area should be a simple double integral in polar coordinates.

Area=8

>>4288849
2pi, in fact

>>4288849
No, because the series 1 - 2^(-n) is increasing, but convergent.

>>4288857
yeah, for sure the minimum area. but isn't it too strict?

I'm too lazy to read the case by case approach and see which one still has to be done. I'm wondering if the following would work:
Consider <span class="math">f_y:x\mapsto \displaystyle f(x)=\frac{x^2+y^2}{2}[/spoiler]. A necessary condition for <span class="math">a_n[/spoiler] to converge is that <span class="math">f[/spoiler] has a fixpoint. Then <span class="math">x[/spoiler] needs to be "close enough" to that fixpoint, but we'll see that later.

<span class="math">f_y[/spoiler] has a fixpoint if <span class="math">g_y:x\mapsto g_y(x) = f_y(x)-x= \frac{1}{2}x^2-x+\frac{1}{2}y^2[/spoiler] has a zero.
<span class="math">\Delta=1-4\frac{1}{2}\frac{1}{2}y^2=1-y^2[/spoiler], so a necessary condition for the convergence of <span class="math">a_n[/spoiler] is that <span class="math">|y|\leq 1[/spoiler].

Oh well, that didn't get us very far... But it was based on a fixpoint analysis, and fixpoints are cool. Also no calc is cool.

>>4288940
TN5, good to see you.

Can we prove that the sequence can't hop the fixpoint and increase unbounded?

If x>2 then the sequence is bounded below by x^(2n)/2^(2n-1). So for any y, we can find an a_n>y where n satisfies 2*(2*y)^(1/n)<x^2.

>>4288971

Absolute value of x>2.

>>4288971
If x>2, then it's divergent, so it's not in the set.

>>4288982

Exactly, so x<2, y<2-x. Area = 8.

>>4288989
Also, y <= 1. We haven't completely determined the set, so we have to do that first.

>>4288992

y>1 if x<2-y

You niggers solve things too quickly

>>4288963
I guess that if the fixpoint is lower than x and the sequence is increasing, or the fixpoint is higher than x and the sequence is decreasing, then you don't reach it. The problem is probably the same as what you already noted about jumping over "1": can we jump over the fixpoint?

>>4288994

wait that's wrong. but |x|=<2,|y|=<min(|2-x|,1) seems to work.

>>4288992
Do we have a lower bound for y? I'd say |y| <= 1 is worth trying to show since if you assume that the sequence converges for an arbitrarily large n then you could be motivated to say:
a_n+1 ~= an, so a_n ~= (a_n^2 + y^2)/2 -> 0 = a_n^2 -2a_n +y^2.

From here you could say:
a_n = 1 +/- sqrt(1-y^2) ; the realness of a_n gives the requirement y^2 < 1 from which a_n e [0,2] follows. No rigor to be seen, but it could be a place to start.

>>4289054
rather, y^2 <= 1 and the convergent value of a_n is between [0,2].

>>4289054
Yeah TN5 showed the y>=-1 part here >>4288940 (and I think mathfag worked on the absolute value everywhere, anyway, since the problem is symmetrical).

>>4289033
You're always increasing from 0, so you can't decrease away from a fixpoint.

dimW = imT iff vector space V is surjective: why is this?

Area=4+pi

The region will be symmetrical about <span class="math">x=0[/spoiler], so assume <span class="math">x > 0[/spoiler] and therefore all <span class="math">a_i > 0[/spoiler].

If <span class="math">a_n\rightarrow L[/spoiler], then <span class="math">L = \frac{L^2+y^2}{2}[/spoiler], so <span class="math">L = 1 \pm \sqrt{1-y^2}[/spoiler].

Let <span class="math">f(t) = (t^2+y^2)/2[/spoiler]. Then <span class="math">f'(t) = t > 0[/spoiler] for <span class="math">t>0[/spoiler]. Plotting <span class="math">f(t)[/spoiler] against <span class="math">g(t) = t[/spoiler] we see <span class="math">f(t) = t[/spoiler] for <span class="math">t = 1 \pm \sqrt{1-y^2}[/spoiler] and therefore

1) <span class="math">0 < t < f(t) < 1-\sqrt{1-y^2}[/spoiler] for <span class="math">t \in (0, 1-\sqrt{1-y^2})[/spoiler]

2) <span class="math">1-\sqrt{1-y^2} < t < f(t)< 1+\sqrt{1-y^2}[/spoiler] for <span class="math">t \in (1-\sqrt{1-y^2}, 1+\sqrt{1-y^2})[/spoiler]

3) <span class="math">1+\sqrt{1-y^2} < t < f(t) < \infty[/spoiler] for <span class="math">t \in (1+\sqrt{1-y^2}, \infty)[/spoiler]

In all three cases <span class="math">f(t)[/spoiler] is mapping the specified region into itself

1) If <span class="math">a_0 = x \in (0, 1-\sqrt{1-y^2})[/spoiler] the sequence <span class="math">a_n[/spoiler] will be increasing and bounded above by <span class="math">1-\sqrt{1-y^2}[/spoiler], and so convergent.

2) If <span class="math">x \in (1-\sqrt{1-y^2}, 1+\sqrt{1-y^2})[/spoiler], it's increasing and bounded above by <span class="math">1+\sqrt{1-y^2}[/spoiler], so convergent.

3) If <span class="math">x \in (1+\sqrt{1-y^2}, \infty)[/spoiler], it's increasing and the first term is already higher than either possible limit, so it diverges.

The region <span class="math">0< x < 1-\sqrt{1-y^2}[/spoiler] is half a square of side 2 and half a circle of radius 1, so the total region has area <span class="math">2^2+\pi \cdot 1^2 = 4+\pi[/spoiler]

>>4289288
Oops, case 2 is backward. In that case <span class="math">f(t) < t[/spoiler] and it decreases toward <span class="math">1-\sqrt{1-y^2}[/spoiler].

(I proofread and everything (._. ) )

>>4289288
That's the same answer the book got. Good job!

>>4289177

I promise I'll shut up after this bump, I just feel like this is a definition more than a proof but I've been asked to prove it.

>>4289349
Well then I guess the book was right.

>>4289357
>>4289177
Vector spaces aren't surjective, morphisms are.

if f: V -> W is onto, then by definition im f = W, so dim im f = dim W.

OTOH if im f != W then dim im f < dim W since you can't have a proper subspace with the same dimension as the whole space.

>>4289401
>since you can't have a proper subspace with the same dimension as the whole space.
You can...

>>4289288
...and the last line should say <span class="math">0< x < 1+\sqrt{1-y^2}[/spoiler] instead of minus

<span class="math">
\int_{0-}^{infty} e^{-st}. f(t). dt | s \in C
[\math][/spoiler]

<span class="math">
\int_{0-}^{infty} e^{-st}. f(t). dt | s \in C
\<span class="math">[/spoiler][/spoiler]

>>4290528

Story of my life.

<span class="math">
\int_{0-}^{infity} e^{-st}. f(t). dt \; | \; s \in C
[\math][/spoiler]

<span class="math">
\int_{0-}^{\infty} e^{-st}. f(t). dt \; | \; s \in C
[\math][/spoiler]

<span class="math">
\int_{0-}^{\infty} e^{-st}. f(t). dt | s \in C
[\math][/spoiler]

<span class="math">
{\infty}
[\math][/spoiler]

<span class="math">
{\infty}
\<span class="math">[/spoiler][/spoiler]

<span class="math">
{\infty}
\<span class="math">



\int_{0-}^{\infty} e^{-st}. f(t). dt \; | \; s \in C
\[/spoiler][/spoiler]

<span class="math">
{\infty}
\<span class="math">



\bigint_{0-}^{\infty} e^{-st}. f(t). dt \; | \; s \in C
\[/spoiler][/spoiler]

wtf is this nerd shit

>>4290995
>on 4chan
>calling someone a nerd

>>4291057
> on 4chan
> recapping what has happened in a thread

>>4291568
>on 4chan
>being a hypocrite