/sci/ - Science & Math

2 + 2 = 5

Is that enough of a trick for you?

the word bed looks like a bed

>>1189806
radiohead?

I'm disappointed, /sci/.

For ax^2+bx+c,

x=(-b+-(b^2+4ac)^(1/2))/2a FTW

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a=b ! +a
2a = a+b !-2b
2a -2b = a - b ! simplification
2(a-b) = a - b ! /(a-b)
2 = 1

>>1189785

>>1189836
so so so much fail.
you cant just add a equals sign.
you ment the ax^2+bx+c=0
=> x=(-b+-(b^2+4ac)^(1/2))/2a

>>1189836
That's not a trick. That's entry level shit bro.

>>1189785
anyone know how this works

>>1191058
they're not similar, they have different slopes.

>>1191058
Bent hypotenuse

>>1191058
The slopes of the red and blue triangle are slightly different.

>>1189881
Division by zero

>>1189881
if a=b then a-b=0
when you divide by a-b you are dividing by zero, both sides of the final equation should be infinity, which would be true.

>>1191064
hate to break this to you...there is no blue triangle.
and anyways the hypotenuses are same slope and shit.

(taking this out of a book)

Take any four-digit number except an integral multiple of 1111 (i.e. any of the nine numbers with four identical digits). Rearrange the digits of your number to form the largest and smallest strings possible (allowing initial zeros as digits) and subtract the two. Apply this process again to the new obtained number. Within the total of seven steps, you will always reach 6174. At that point, proceeding any further is pointless since you can only continue to get 6174.

EX: choosing 8028
largest permutation: 8820
smallest: 0288
8820 - 0288 = 8532
largest: 8532
smallest: 2358
8532 - 2358 = 6174
7614 - 1467 = 6174

>>1191146
cool stuff

>>1191146
1000 - 0001 = 999
the game

>>1191146
But, why?

>>1191181
1000 - 0001 = 0999
9990 - 0999 = 8991
9981 - 1899 = 8082
8820 - 0288 = 8532
8532 - 2358 = 6174

Checkmate

>>1191181
9990 - 0999 = 8991
9981 - 1899 = 8082
....8532
....6174

>>1191203

Oh shit!

Say you have a tree, of 50cm diameter, and wrap a ribbon tight around it. If you cut that ribbon, added a meter to it and sewed it back together, then pulled it tight into a circle again, how far would it come up from the surface of the tree?
It turns out to be about 16cm from the surface of the tree, which sounds about right.

Now lets imagine a ribbon pulled tight around the entire world(assuming the earth is a perfect sphere). If we cut the ribbon and added 1 meter to it, then got a bunch of dudes to pull it tight all around the earth, how far up would they pull it from the surface, all around the world?

16cm, niggers.

this is because:
circumference (C) = 2*pi*r
so C + 1m = 2*pi*r'
where r' is the new radius
sub in C = 2*pi*r
2*pi*r + 1m = 2*pi*r'
r' = r + 1/(2*pi)
which is the old radius plus ~16cm.

Doesnt matter if its a hydrogen atom or the entire universe - if you add 1 meter to the circumference you increase radius by 16cm

bamp

>>1191318
That's SO counterintuitive.

>>1191318

So... when I increase the size of my shape in one dimension, a measurement of the shape of the same dimension increases in direct proportion.

Who would have thought.

Surely that's like saying no matter how long your ribbon is, if you add one metre to it then it will always turn out one metre longer?

>>1191318
>>1191318
bullshit, I have seen this in a thread before. its something like 128 meters.

>>1191687

You might be thinking of the problem where the ribbon is pulled tight to make a teardrop shape, and then the distance from the surface to the point of that shape is surprising.

What this poster did is just give expressions for the diameters of two circles, which is a considerably less complex problem.

one of my favorites.

try and prove it!

sage because OP's picture annoys me

>>1191687
o rry?
Well, if theres a problem with the maths then feel free to point it out, otherwise its proven.

>>1191769
On it

>>1191769
who can solve this?

>>1191828
everyone can solve this
(n(n+1)/2)^2

>>1191834
yeah except there are cubes over those

>>1191855
except you are fucking retarded

>>1191769
let Claim(k) by that the sum of the cubes of integers from 1 to k is equal to their sum squared.
Then Claim(k+1) is that (1+2+3+...+k)^2 + (k+1)^3 = (1+2+3+...+k+(k+1))^2
=> ...I got nothing

>>1191769
The answer is 4.

>>1191769
I've got it, but i cant figure out the math fonts. Basically all you have to do is compare formula of the sum of n^3 with the formula of the sum of n squared. You find that they are indeed equal.

>>1191660

Maybe less so if you phrase it differently. Suppose you take two similar N-dimensional objects. We know that each of their n-dimensional features are in direct proportion. In the original the two features in question were the one dimensional radius and one dimensional circumference, but it doesn't really matter. For example, if you took two similar square based pyramids, the ratio of areas of the bottom square to one of the sides would remain constant over all similar pyramids.

Since the ratio is constant (<span class="math">\frac{F1}{F2} = c[/spoiler]) we know that they change linearly with respect to one another (<span class="math">\delta F1 = c \delta F2[/spoiler]). Note that the initial values don't show up anywhere. The change in the first feature only depends on the change of the second and even then depends in a very trivial way (it's multiplied by a constant). In the original, the poster's constant was simply <span class="math">\frac{1}{2 \pi}[/spoiler] and since he was always considering the same change in circumference, the corresponding change in radius will also be the same.

So going back to our pyramid, imagine that the pyramid has a square base and each of its sides are equilateral triangles. If we increase the area of the base by 1 square meeter and keep the object similar to the original, we will always increase the area of the side by <span class="math">\frac{\sqrt{3}}{4}[/spoiler] no matter what the initial size of the pyramid was.

>>1192068

<span class="math">\delta F1 = c \delta F2[/spoiler] should read <span class="math">\Delta F1 = c \Delta F2[/spoiler]

>>1192068
True, it makes perfect sense when you look at it mathematically, as is shown in the original post. The 'trick' is in the fact that it seems like small change in circumference creates a big difference in the radius, when really 16cm is nothing compared to the radius of earth, so a small relative change in circumference is really producing a small relative change in radius. It just seems big because a small change relative to earths radius is a big change relative to the size of a person

>>1191769
thanks you for that, took me longer that it should have

>>1192119

I guess my point was that the general statement:

the ratio of changes of two n-dimensional feature of an N-dimension object is constant

is far less counterintuitive (to me... in fact I'd go so far as to say it's completely intuitive) than the statement:

If I change the circumference of any circle by a meeter the radius will change by about 16 centimeters.

But of course the second is just a special case of the first.

Actually the cognitive dissonance between the intuitiveness of the former and the latter is neat because it illustrates a phenomenon that comes up a lot in math. Often when you remove the specifics of a problem and only consider the necessary information, the solution becomes far easier. Once we stop imagining circles and ribbons and simply think of the features as being like dimensional, the conclusion is much easier to get our heads around.

Take the square of side length 4 that is centered at the origin. In other words its vertices are at (2,2), (2, -2), (-2, 2), and (-2, -2). Withing this square inscribe four unit circles in each of the four corners so that each is completely contained within the square. The centers of the circles are, of course, at (1, 1), (-1, 1), (1, -1), (-1, -1). Now, construct the largest circle you can which is centered at the origin but whose interior is disjoint from the interior of each of the four circles. Clearly, that circle will be tangent to all four of the other circles. It shouldn't be too hard to figure out that the radius of the center circle plus the radius of a corner circle is the distance from the origin to the center of one of the corner circles. That distance is easily calculated to be <span class="math">\sqrt{2}[/spoiler] and so the radius of the center circle is <span class="math">\sqrt{2} - 1[/spoiler].

Now, repeat the exercise with a three dimensional square (cube) of side length 4. This time there will be 8 corner circles (spheres). The radius of the center sphere will be <span class="math">\sqrt{3} - 1[/spoiler].

It shouldn't be terrible hard to convince yourself that in n-dimensional space we can construct an n-dimensional hypercube of side length 4, the <span class="math">2^n[/spoiler] hyperspheres of radii 1 in the corners and a center hypersphere of radius <span class="math">\sqrt{n} - 1[/spoiler] which is tangent to each of the corner hypersphere.

If you're me, you imagine a small center circle being surrounded by larger corner spheres which are in turn surrounded by a big square. For <span class="math">n \geq 9[/spoiler], however, that intuition fails. Say, for example, n is 25. Then the radius of the center sphere is <span class="math">\sqrt{25} - 1 = 4[/spoiler]. This means that along each axis, the center hypersphere goes twice as far as the hypercube. The cube contains the corner spheres, the corner spheres contain (in some sense) the center sphere, but the cube does not contain the center sphere.

>>1189785
<span class="math">Faggots really need to learn to use JSMath[/spoiler]

>>1192338
<span class="math">So~do~you.[/spoiler]

>>1191769
>>1191769

Assuming that means
test for N=1
1^2 = 1^3
1=1
True
Now assume <span class="math">S_k \to N=K[/spoiler] is valid.
Test for N=K+1 being valid
<span class="math">1^3+2^3+3^3 +...+ n^3 + (n+1)^3 = (1+2+3+...+N+N+1)^2

.....

Shit, I forgot how to do mathematical induction with a series on both sides![/spoiler]

N is a triangular number if and only if the square root of 8N + 1 is natural number.

>>1192061

What?

2*0=0 and 1*0=0

Realy this only proves that 0 = 0

congenital syphyllis?

>>1189784

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Take any n-digit number. Multiply it by 3 and then by a number of the folowing form:
3 repeats n times, then 6 repeats n-1 times, and then 7.
You will recieve your original number, in repetition.
Slightly more generalized form:
Take any n-digit number. Multiply it by a number of the folowing form:
1 and then n-1 zeros, repeated an arbitary m amount of times, and then 1. You will recieve the original number, repeated m times. When m = 2, the two methods are the same.

>>1191769


I DID IT!!!!!!

>>1189784
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