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/sci/ - Science & Math

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>> No.1330806 [View]

Don't worry about notation. You'll get used to it with experience, if something is unclear in the lectures just ask the lecturer that's the best (and most natural way) to learn.

>> No.1330312 [View]

Science is for understanding life, through understanding we can solve problems in our everyday life, think oil crisis for instance. Even all the sciences that seem to contribute nothing to everyday life (mathematics) form a part of the net that is science.

>> No.1325888 [View]

>>1325875
The problem with that is that you'll have to proof it with almost no knowledge of sin x, apart from it being continuous, you can't even use that it's C^1. I don't know if it's possible, don't think I ever saw a proof.

>> No.1325775 [View]
File: 11 KB, 681x508, pic.png [View same] [iqdb] [saucenao] [google]
1325775

>>1325769
Aaaand that's the wrong picture, my bad.

>> No.1325769 [View]
File: 10 KB, 681x508, proof.png [View same] [iqdb] [saucenao] [google]
1325769

Found a nice proof in an elementary calculus book;
consider the picture on the left;
the small triangle 0,(1,0),(cos x, sin x) has area
(sin x)/2,

the big triangle has area
(sin x)/(2cos x).

The area in the middle (the diskpart) has area x/2.
So we now get for x close to 0;

sin x /2 < x / 2 < sin x / (2cos x) =>
1 < x / sin x < 1 / cos x =>
1 > sin x / x > cos x;

now use the squeeze(pinch) theorem to obtain desired result.

>> No.1325588 [View]

>>1325406
Pretty sure the rest of Europe is the same way, I can certainly vouch for the Netherlands.

>> No.1318434 [View]

For every product
of any compact spaces;
this too is compact.

>> No.1317905 [View]

Continuous f,
From ball to ball nothing more.
Yet has a fixed point.

>> No.1317878 [View]

Polynomials,
complex and of order n.
n roots shall it have.

>> No.1313337 [View]

Don't really get the context, but since these are all min/max, maybe just; (simplified notation)

We have min_s f(s,y) <= f(x,y) for all x. So;
max_t min_s f(s,t)<= max_t f(x,t) for all x so
max_t min_s f(s,t) <= min_s max_t f(s,t).

>> No.1313040 [View]

>>1312976
You can even integrate over functions from triangles into your space :D.

>> No.1312955 [View]

>>1312951
You can use this identity for integration, but be wary of using it for derivations of functions.

>> No.1312951 [View]

>>1312934
If y is a function of x alone, yes.

<div class="math"> dy = \frac{dy}{dx}dx </div>
in the notation Josef used before.

>> No.1312864 [View]

>>1312833
Well you can consider the 'local' property to be small, and the 'global' property to be big. As for the equivalence classes, they still kind of use regular derivatives of paths in your manifold. In that sense there still is the 'tiny' step approach.

>> No.1312800 [View]

>>1312715
Well officially nothing small, but the cotangent bundle adds a vectorspace to each of your points, I guess that can be consider the small part? Though of course you glue them all together smoothly so shouldn't really matter.

>> No.1305958 [View]

>>1305930
Because Erdös was fucking awesome, that's why.

>> No.1297100 [View]

>>1297088
>Einstein o.o.

Yeah there is an analytic solution, it's something like it converges for

e^-1 < x < e^(1/e). (something among those lines)

>> No.1296404 [View]

>>1296048
The only way I see I could link whatever your "heuristic" notion might be and inner products is to consider shortest distances. I.e.
in R^2; take a line and a point, the shortest path between the two is perpendicular to the line you started with.
in R^3; the same with a point and a plane.

This can be transfered back into notions for inner products and orthogonal decompositions.

>> No.1295090 [View]

>>1295084
Why not? It's definitely a Cauchy sequence, since
<span class="math">1^m - 1^n < \delta[/spoiler] for all n and m. And since the real numbers are complete every Cauchy sequence has a limit.

>> No.1295077 [View]

You could define it as the limit of the sequence
<span class="math">(1^n)_{n \in \mathbb{N}} [/spoiler].

>> No.1294471 [View]

Because calculating a real integral using only the definition of integral is terrible.

>> No.1244131 [View]

>>1244110
Yeah, I figured
>Check for n=1;
>2+6 =8.
Would suffice, apparently not.

>> No.1243956 [View]

>>1243948
yeah should be division by two, made an error. 4a +1

>> No.1243876 [View]

>>1243868
Woops at b=2;
8a+1 should be 4a+1

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