Maths geeks

[vizIIsto]

So I used to be very good at maths. I feel like I should challenge myself to some problems, but there are certain concepts in maths that I haven't been taught at school.
I'll give a few problems here which I'd encourage you maths geeks out there to try out as well. They're all video's with the answer explained at the end of the video.

For example, this:
Maths problem 1
Was easy to solve. It takes a few seconds to get to the right way of attacking this problem, but I managed to figure it out quite quickly.
For those that want to solve it, but want me to prove that I got the right answer,

the answer is 20 metres: Y runs 0.95:1 to X, and Z runs 0.931:1, so Z runs 0.931:0.95, which is also 0.98:1, so Z runs 980 meters in the time Y runs 1 km, so the answer is 20.

However, problems like in the following video:
Maths problem 2
Are way beyond my range. I don't even know what the symbol to the left means in the first problem mentioned.

I like being challenged to complicated maths problems, as long as I know that I know enough to solve the problem. When I need to make use of certain symbols and formulas that I have not been taught, it's too hard for me.
Basically, all simple calculations (not like maths, just adding, subtracting, multiplying, square roots, etc) are good for me.

Here's another puzzle that I struggled with because radiuses are out of my range:
Maths problem 3

I'm looking for maths geeks to gather here and present each other with maths problems.

Whether they are too complicated for someone like me or too easy for someone else, anything goes here.
Here's another puzzle which looks impossible, but for which you need to think out of the box:
Maths problem 4

[Ircher]

An easy problem that the OP should be able to solve is this:
1) Prove that (a + b + c)³ = a³ + b³ + c³ + 3(a + b)(b + c)(c + a).

Another fun problem that was on my final exam for one of my classes last semester was this:
2) Recall that Fn(A, B) denotes the set of functions with domain A and codomain B. Prove that Fn(N, {0, 1}) has the same cardinality as Fn(N, {0, 1, 2, 3}). That is, show that there exists a bijection between the two sets.

This problem requires zero (!) computation, but it may be difficult for the OP if they are not familiar with elementary set theory and cardinality. Note that N in Fn(N, {...}) denotes the set of natural numbers. That is, N = {1, 2, 3, ...}.

[DrDolittle]

Spoiler: sol to ircher

Fn(N,{0,1}) is isomorphic to all decimals in base 2 (i.e., F(n) = in <=> 0.i1 i2 i3 i4...)
Fn(N,{0,1,2,3}) is isomorphic to all demicals in base 4

the sets' cardinalities are the same since they are both decimals

Here's a stupid one:

I like juice a lot, but I only have a liter of it in a one liter jar. For the 12 days of Christmas: the first day I drank 1/12 of the jar, and refilled the jar with water. The second day I drank 2/12 of the jar, and refilled the jar with water etc. For the 12th day Christmas day, I rewarded myself and drank 12/12, or all the liquid in the jar.

How much water did I drink in total?

[DeathRowKitty]

Spoiler: sol to ircher

Fn(N,{0,1}) is isomorphic to all decimals in base 2 (i.e., F(n) = in <=> 0.i1 i2 i3 i4...)
Fn(N,{0,1,2,3}) is isomorphic to all demicals in base 4

the sets' cardinalities are the same since they are both decimals

To nitpick a little bit, you've shown a bijection between F_n(N,{0,1}) and decimal

representations

in base 2 and between F_n(N,{0,1,2,3}) and decimal

representations

in base 4. There's only a countable set separating the set of all decimals in those bases from the set of all decimal representations in those bases of course, but, given the choice of sets in the question, I suspect the intended resolution to this is to also note that, since 2 and 4 have the same distinct prime divisors, the set of numbers with two base 2 representations is the same as the set of numbers with two base 4 representations, allowing you to use the obvious bijection with no clever trickery required for these specific sets.

[DrDolittle]

DRK point well taken, but why can't I map
f \in F_n(N,{0,1}) to r \in R+ to g \in F_n(N,{0,1,2,3,4}) directly by using the decimal representation map?
It doesn't seem like there is a countable set separating.

[DeathRowKitty]

Those wouldn't be bijections. Consider for example the functions corresponding to the binary strings .01111111... and .1000000... Both map to 1/2 in R.

[vizIIsto]

An easy problem that the OP should be able to solve is this:
1) Prove that (a + b + c)³ = a³ + b³ + c³ + 3(a + b)(b + c)(c + a).

I know this isn't a complete answer, but because (a + b + c)³ can be written as (a + b + c)(a + b + c)(a + b + c), you'd assume you can simplify by taking away the a*a*a, b*b*b and c*c*c from the formula. I know that doesn't necessarily equal removing those letters from each pair of brackets, but then you'd get: (b + c)(a + c)(a + b) + a*a*a + b*b*b + c*c*c
I don't know do I have to go in more depth about the other 24 permutations of (a + b + c)³?

[Ircher]

I think you just have to expand it all out and then simplify. It's a lot of algebra.

[vizIIsto]

So this problem faced me today:

This question appeared on an exam. It went as follows:

"The product of the three integers x, y and z is 192.
z = 4, and p is equal to the average of x and y.
What is the minimum value of p?"

The four answers were: a) 6 b) 7 c) 8 d) 9,5

But... No one gave the right answer.

What is the right answer?

[Xtoxm]

uhh -49/2 if we're being pedantic?

[vizIIsto]

For those that want hints, here's three hints to my problem:

Spoiler: Hint 1

Nobody had the right answer. Doesn't that seem a little suspicious to you?

Spoiler: Hint 2

You're probably this far that you've solved xyz=192 and z = 4 means x*y=48. Now, try to consider all possible combinations of two integers which multiply to 48.

Spoiler: Hint 3

The reason that no one gave the right answer is because there's something wrong with the question. Or the answers. Or both!

Tbh the fact that I give three hints here kind of makes this feel like a Layton puzzle, of which I'm a big fan of.

[DrDolittle]

viz what grade are you in

[tictac]

Is this a place where I can rant about math?
I'm gonna do it anyways.
Why does np.fft.rfft give me freaking fractional-hertz frequency-bins?
how is it even calculated if sample-length isn't diviseble by wavelenght, so the sines don't zero average??
Why even call it "frequency domain" if it's more about sculpting individual waves, and info stored in phase like multi-headed fricking gramophone.
I'm feeling super salty at fast fourier rn.

[tictac]

^twas waves/sample, not waves/fft_size like I thought. way more useful.

next thing:
Amplitude of A*sin (ax)+B*sin (bx) seems to be abs((A+B)cos((b-a)*x\2)) when A==B
What about when A!=B?

[tictac]

^oh doh

Spoiler: answer

sqrt((A*sin(ax)+B*sin(bx))^2+(A*cos(ax)+B*cos(bx))^2)

[tictac]

lol

...well, for some value of "answer"

[GeneralWu]

The first forum mafia game I ever played was on a math website.

Here's a cool math question:
Find the area of a circle with side lengths of 3, 4, and 5.