Am I a brainlet?

I want to see if I am a brainlet.

Give me your hardest problems (that you know the answer to) and I will solve them using my brain powers.

No baby problems like pic related, we want extreme problems.

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when u look in a mirror why is text flipped horizontally but not vertically ?

i don't even know how one goes around solving continuing fractions

...

whoa never thought about this

It's not flipped horizontally. The mirror presents the object however the light reflects it (so straightforward), meaning that it is flipped perpendicular to the xy plane

Not the hardest problem I know, but one I doubt you can Google or Wolframalpha an answer for.

Consider a 2D scalar field [math]E[/math] obeying the Helmholtz equation:

[math]\nabla^2 E(\vec{r}) + k^2 E(\vec{r}) = j\omega \mu J(\vec{r})[/math]

where

[math]k^2 = \omega^2 \mu_0 \epsilon(\vec{r})[/math],

where [math]k[/math] is wavenumber, [math]\omega[/math] is frequency, [math]\mu_0[/math] is a constant, [math]\epsilon[/math] is a function space, and [math]J[/math] is a source term

Integrating this, we can reformulate the equation as

[math]E(\vec{r}) = E^{(i)}(\vec{r}) + k^2 \int_{A^\prime} E(\vec{r}^\prime) \chi(\vec{r}^\prime) G(\vec{r},\vec{r}^\prime) dA^\prime[/math]

or

[math]E(\vec{r}) = E^{(i)}(\vec{r}) + E^{(s)}(\vec{r})[/math]

Where [math]E[/math] is the total field, [math]E^{(i)}[/math] is an incident field, [math]\chi = \frac{\epsilon}{\epsilon_0} - 1[/math] is contrast, and [math]G(\vec{r},\vec{r}^\prime) = \frac{1}{j4}H_0^{(2)}(k|\vec{r} - \vec{r}^\prime|)[/math] is the 2D Green's function associated with the PDE.

Assume you have an object that determines an unknown [math]\epsilon[/math] that you probe with incident field [math]E^{(i)}[/math]. Given a finite number of point measurements of the scattered field [math]E^{(s)}[/math], solve for the [math]\epsilon[/math].

This is simple. You should be able to do this.

it's very intuitive
you should be able to see it

repeating ones are easy
[math]x=2+\frac{15}{2+\frac{15}{2+\frac{15}{2+...}}}\\
x=2+\frac{15}{x} \\
x^2=2x+15[/math]
and just solve the quadraic

Not him, but that's really intuitive. Thanks user.

>He put in the time to Tex something that he knows OP doesn't know the notation for and therefore can't solve even if he was a genius

What did he mean by this?

But you aren't proving that the thing converges. You can only apply this if you prove that it converges.

Don't follow his brainlet advice. He clearly hasn't even taken real analysis 1 because this shit is the first thing they tell you: You can only apply that if you prove it converges.

If you take his advice and then go on with other infinite fractions you will find that it is quite easy to "prove" by mistake that 1 = 2.

Why do my spaghetti-o's keep making lightning?

maybe i'm stalling because i can't figure out what the fuck i did wrong in my solution

I have a problem for you that I do know the answer to. Suppose that you have a function f that is defined on the natural numbers and outputs numbers (that can be whole, real, rational, complex, doesn't matter).

Consider the following sum:
[eqn] \sum_{k=1}^{n} f( gcd(n,k) )[/eqn]

Where gcd is the greatest common divisor. And suppose there exists some function g so that that sum equals this other sum:
[eqn] \sum_{d|n}^{} f(d)g \left(\frac{n}{d}\right)[/eqn] for each and every natural number n.

This, if you are not acquainted with the notation, means the sum of the divisors of n. In other words, you sum considering all the numbers d such that d properly divides n.

This is the question:
Version 1: If you don't know about number theory and arithmetic functions:

Assume that this function exists and is unique. Find a way to compute a value table for g and then write down at least the values for g(1), g(2), g(3), g(4), g(5).

Version 2: If you know about number theory and arithmetic functions or you don't know anything about these things but are feeling adventurous.

Prove that such a function exists, is unique, and tell me exactly what that function is.

Note: In case it isn't clear, the function g you are looking for must have this effect for all functions f. Not just one in particular. In other words, your g does not depend on what f you choose. It is a global g that does this for every function f.

Also, g is a function defined on the natural numbers. You must not think about or care what values g has in non-natural points.

If I can't prove that triangles are similar (I am a computer scientist, we have no triangles here), but I just guess it, get x = 40 and everything suits well with it - does it counts as a solution?

Oh no, I am retarded.

Why is there no Banach space with a countable basis?
(Hint: Baire)

It's still a decent way to test pattern recognition and problem solving probably, in someone who hasn't seen them before.

A separable Hilbert space is a Banach Space, so that isn't true.

Very very interesting.

I'm stepping out for a bit. Here's what I have so far:

g(1) = 1
g(2) = 1
g(3) = 2
g(4) = 2
g(5) = 4

Very nice. All of those are correct. I am more interested in how you got them though.

When I was given this problem I used some tricks I know from what is called Dirichlet multiplication to set up an expression of g using another special function but you probably don't know these tricks so what did you do?

Also, if you want to keep going I can tell you which function g represents (it is one of those famous functions) and then you can try to prove that indeed that famous function has this property.

It's some kind of inversion...

;^)

laughable problem
2ad^(2)-3ae+7b^(3)d-4b^(2)e

what's your solution then FAGGOT?

The same but first seeing if it converges.

Let [math] n [/math] be an odd integer [math] >2 [/math] and let [math] f(x)\in \mathbb{Q}[x] [/math] be an irreducible polynomial of degree [math] n [/math] such that the Galois group [math] Gal(f/\mathbb{Q}) [/math] is isomorphic to the dihedral group [math] D_n [/math] of order [math] 2n [/math]. Let [math] \alpha [/math] be a real root of [math] f(x) [/math]. Prove [math] \alpha [/math] can be expressed by real radicals if and only if every prime divisor of [math] n [/math] is a Fermat prime.

Prove [math] \operatorname{Ext} _\mathcal{A}^1\left( { - , - } \right)[/math] classifies short exact sequences in [math]\mathcal{A}[/math] and [math] \operatorname{Ext} _\mathcal{A}^2\left( { - , - } \right)[/math] classifies isomorphism classes objects in [math] {\operatorname{D} ^{\left[ { - 1,0} \right]}}\left( \mathcal{A} \right)[/math].

>It's another case of retarded faggot Texes something that has notation he knows OP does not understand. And also does not try to rephrase the question using only elementary terms or even give short descriptions of what his notation means

>It's another case of retarded faggot Texes something that has notation he knows OP does not understand. And also does not try to rephrase the question using only elementary terms or even give short descriptions of what his notation means

What is the sum of every multiple of 3 or 5 that's under 10,000?

basic galois theory
basic homological algebra

what are you a brainlet?

Why should I assume OP doesn't understand basic mathematical notation when he/she's already posted continued fractions?

does this assume all triangles add to 180? if it does than this is ez pz shit

Show your work Without using calculators brainlets

>he/she

Oh holy fuck just go back to the reddit or tumblr shithole you came from.

Also, because you shouldn't just assume that. That is why IQ tests don't make you do high level math even though high level math is the most accurate form to measure intelligence assuming the subject understands the notation.

>Oh holy fuck just go back to the reddit or tumblr shithole you came from.
i'm pretty sure i've been here longer than you

>Also, because you shouldn't just assume that.
why not?

>That is why IQ tests don't make you do high level math even though high level math is the most accurate form to measure intelligence assuming the subject understands the notation.
OP posted continued fractions so I assumed he understood basic notation that every undergraduate sees, so I don't know why you even mention IQ tests

You asking this question proves that you are a brainlet, sorry.

Choose a number.
If it's even divide by two, and if it's odd times it by three and add 1.
The goal is to find a cycle that never reaches 1.

If you can't solve this simple problem gtfo of this board

>assuming every number is even part of a cycle
brainlet.

Find length AB

The idea is to find a cycle.
I did not imply every number would induce a cycle.
>what is reading comprehension

AB = BA

NEXT

>what did you do

Well I noticed that g(a/b) is equal to the number of integers that have a gcd of b with "a". [which is really interesting] So for g(x/1) it's the number of integers that have a gcd = 1 with x.

Now that you mention famous function, it's eulers phi function!

>>>that you know the answer to
>>>that you know the answer to
>>>that you know the answer to

...It's not the answer I have on the card, so I can't accept it.

Not OP but gonna try it anyways.
I assume the "hard" part is that you have to sum unique terms so you can't repeat numbers that are multiples of 3 and of 5.

The sum of the multiples of 3 less than 10000 is
[math] \sum_{n=1}^{3333} 3n = 3\sum_{n=1}^{3333} n = 3\frac{3333*3334}{2} = 16668333[/math]
The sum of the multiples of 5 less than 10000 is
[math] \sum_{n=1}^{1999} 5n = 5\sum_{n=1}^{1999} n = 5\frac{1999*2000}{2} = 9995000[/math]
The sum of the multiples of 15 less than 1000 is
[math] \sum_{n=1}^{666} 15n = 15\sum_{n=1}^{666} n = 15\frac{666*667}{2} = 3331665[/math]
And the answer is [math]16668333 + 9995000 - 3331665 = 23331668 [/math]

Find all lines that are equidistant to 3 non collinear points (equidistant means the shortest distance from point to line is the same in all 3)

i found that OD = 2.sqrt(7).r
...don't wanna spend more time

I'll provide the answer on a similar problem to help you brainlets out.

Choose a number.
If it's even divide by two, and if it's odd times it by three and MINUS 1.
The goal is to find a cycle that never reaches 1.

Answer is 7
7 20 10 5 14 7. . .

>I'll provide the answer on a similar problem to help you brainlets out.
but that similar problem is trivial

if you knew the answer to it with +1 why not collect the prize money?

A natural number, N, that can be written as the sum and product of a given set of at least two natural numbers, {a1, a2, ... , ak} is called a product-sum number: N = a1 + a2 + ... + ak = a1 × a2 × ... × ak.

For example, 6 = 1 + 2 + 3 = 1 × 2 × 3.

For a given set of size, k, we shall call the smallest N with this property a minimal product-sum number. The minimal product-sum numbers for sets of size, k = 2, 3, 4, 5, and 6 are as follows.

k=2: 4 = 2 × 2 = 2 + 2
k=3: 6 = 1 × 2 × 3 = 1 + 2 + 3
k=4: 8 = 1 × 1 × 2 × 4 = 1 + 1 + 2 + 4
k=5: 8 = 1 × 1 × 2 × 2 × 2 = 1 + 1 + 2 + 2 + 2
k=6: 12 = 1 × 1 × 1 × 1 × 2 × 6 = 1 + 1 + 1 + 1 + 2 + 6

Hence for 2≤k≤6, the sum of all the minimal product-sum numbers is 4+6+8+12 = 30; note that 8 is only counted once in the sum.

In fact, as the complete set of minimal product-sum numbers for 2≤k≤12 is {4, 6, 8, 12, 15, 16}, the sum is 61.

What is the sum of all the minimal product-sum numbers for 2≤k≤12000?

The Collatz conjecture is baby's first unsolved problem, who do you think you're fooling?

If you think it's impossible, why don't you provide a proof for that?

>If you think it's impossible
i don't, can you into reading comprehension?

What's the smallest equilateral triangle possible that can fit 18 unit circles?
What about 30?

Unit circle = radius of one for brainlets

I can personally attest that if there is an answer, it's greater than 30 million

>Computer science
>No geometry
OUT

Let's remember computer science is a spectrum.
We're not all this retarded

why the fuck would someone draw that picture

I've learned the Internet takes more of a why not or who's going to stop me approach

X = 30

The answer is [math]AB=(\frac{8}{29}\sqrt{2+19\sqrt{7}})r[/math]

>times it
>babby can't into multiply
And God blessed them, saying,
Be fruitful, and times it, and fill
the waters in the seas, and let
fowl times it in the earth.
Gen.1:22

Let p(x,y) and q(x,y) be two-variable polynomials. Assume that
[math] p_xq_y-p_yq_x=c[/math]
where c is some non-zero constant and the subscripts denote partial derivatives.

Let F(x,y)=(p(x,y), q(x,y)) and let r,s be the components of F^{-1}. Thus

[math] F^{-1}(x,y)=(r(x,y), q(x,y))[/math]

Show that r and s are both polynomials.

ahh typo: q should s

also F^{-1} is the compositional inverse of F, not 1/F

I'm thinking it can be solved like this:
taking O as the origin we can deduce the equations that describe each shape. Then find A and B as the intersection of OD and the upper-middle circle
then simply put the coordinates into the distance formula

It can be done purely using trigonometric identities.

X=100

Jean is shorter than Brutus but taller than Imhotep. Imhotep is taller than Jean, but shorter than Lord Scotland. Lord Scotland is twice the height of Jean and Brutus combined but only one-tenth of the height of Millsy. Millsy is at a constant height of x-y. If Jean stands exactly one nautical mile away from Lord Scotland, how tall is Imhotep?

today i learned that 180-40-(130-x) = 170-x

Bumping so other can see this spaz faggot

x=80

I solved this a while ago

You have (190-x) + x + 30 = 180

desmos.com/calculator/q5r7uoofrz

I'm getting 20 degrees.

Not too difficult. There's two ways to solve this one. If you feel proud of yourself for figuring it out, remember that von Neumann solved it instantly (the hard way)

Two bicyclists start twenty miles apart and head toward each other, each going at a steady rate of 10 mph. At the same time a fly that travels at a steady 15 mph starts from the front wheel of the southbound bicycle and flies to the front wheel of the northbound one, then turns around and flies to the front wheel of the southbound one again, and continues in this manner till he is crushed between the two front wheels. What total distance did the fly cover?

how big is the fly?

>you should feel proud of yourself for solving something in the most tedious way possible

>What total distance did the fly cover?

15 miles

The total time the fly was alive was 1 hour (10 miles * 10mph). So the fly is 15mph * 1hr = 15 miles.

What the fuck is the hard way?

does this have infinite solutions?

No

desmos.com/calculator/q5r7uoofrz

No. X = 30

x=20 you mong

Geometric sequences

i am a mathematician and I understand these things and I understand that you are an insufferable faggot

Op said post math problems. So how is posting a standard algebra problem an issue?

autist

t.brainlet who doesn't actually understand the problem

...

The joke is that it wasn't tedius for von Neumann lol.

The triangle at the beginning of your problem should be the other way

X=20
it is a supplementary angle which adds up to 180 degrees
10+20=30
30+70=100
100+60=160
x+160=180
-160 -160
x=20

68.125%?

25% + 15% + 50% = 68.125%

>25% + 15% + 50% = 68.125%

I'm fairly certain that's incorrect

Yeah, that doesn't make sense

25% + 75%*15% + 75%*85%*50% = 68.125%

A rectangle, HOMF, has sides HO=11 and OM=5. A triangle, ABC, has H as the intersection of the altitudes, O the center of the circumscribed circle, M the midpoint of BC, and F the foot of the altitude from A. What is the length of BC?