Math Nerds Answer Math Questions

Okay: Try to find the point of intersection of a parabola and a linear function:

The system in question: -x²+8x-7=x+3. I've solved things like this in the past but for some reason I forget how.

The first thing you should try is to move all terms to one side and set the other side equal to 0, and solve for x=0.

Oh, okay.

thanks.

Yes, it's actually very easy.

-x2+8x-7 = x+3

x2-7x+10 = 0

(x-2)*(x-5) = 0

x = 2 or x = 5

edited 4th Jul '15 3:37:56 AM by Aetol

One particular thing I never really understood in high school math is the application of integrals. Specifically, how to calculate the area between lines and curves. For example, how should one work this out:

Determine the area bounded by y = x², x = 3, and the x-axis.

edited 4th Jul '15 5:32:02 AM by DarkDestruction

That one's simple. The boundaries of the area are :

• Left : x=0 (it's not really a boundary, but the point where y=x² joins the x-axis)
• Right : x=3
• Top : y=x²
• Bottom : y=0 (the x-axis).

So you integrate f(x) = x²-0 between 0 and 3.

Function : f(x) = x2

Primitive : F(x) = x3/3

Solution : int(f(x) ; 0..3) = F(3) - F(0) = 9 - 0 = 9

The problem would be more complex (but still solvable) if more boundaries were defined by functions : for example, the right boundary might be defined by x=5-y/2.

edited 4th Jul '15 6:56:03 AM by Aetol

Nevermind, it's been explained quite clearly. Thanks, Aetol!

edited 4th Jul '15 6:11:56 AM by DarkDestruction

Hey, I'm a bit stuck on a topic. Given a matrix of N^m representing tuples of length m, enumerating the natural numbers (so a N^2 matrix would have (0,0), (0,1), (1,0) and so on), how would one find the x-th member?

The Cantor pairing function might be helpful. It can give you an unique id for any n-tuple. It's shown to be invertible for N², but not for arbitrary dimension. So hopefully with a little jiggery-pokery you can find the n-tuple for any given id.

Thanks! That looks to be just about what I need.

How do I prove that F(3n) (That's the function for the Fibonacci sequence) is even for all integers greater than or equal than 1? I know I need to use strong induction, but I'm stuck. Do I need to go farther than F(3(n+1)) ? I can't even prove that.

I don't remember exactly how induction works, but I think I just managed to prove that thingy.

Okay, you check to see that the base case works. F(3) = 1 + 1 = 2, so we know that's true.

Now we check to see whether the induction is true. That is, assuming that F(3n) is true, then F(3(n+1)) must also be true.

We begin by assuming that F(3n) is even by default (I'm not sure whether this is a valid step, but it's a step I took).

F(3n) = F(3n-1) + F(3n-2) -> even.

Now we check for the inductive case.

F(3(n+1)) = F(3n + 2) + F(3n + 1) = 2F(3n + 1) + F(3n)

Already, we know that F(3n) must be even because that is given by our base assumption.

2F(3n + 1) / 2 = F(3n + 1)

While this might be a bit of a stretch, we know that all the numbers in the Fibonacci sequence must be integers (I'm not sure how you'd go about showing this, but since since the first few numbers in the sequence are integers, and since every other number is formed by adding the previous numbers, then all the numbers in the sequence must be integers), so 2F(3n +1) must be even. Therefore, if F(3n) is even, then F(3(n+1)) must also be even. Since we've already shown this is true for the base case, this must be true by extension for all other cases in the sequence we need to prove. You could go on and elaborate further on the numbers in the sequence, but that would require more work than you need to do. You might also have to define what "even" means, based on how its defined in your textbook or whatnot.

I don't know if this proof is rigorous enough for someone doing a math major, but this is the gist of the proof that I came up with.

edited 7th Dec '15 10:06:25 AM by CathariSarad

The way I was thinking it, you have to consider F(3(n+1)), not F(3n+1). So, in other words:

• We know F(3n) is even, because it's your hypothesis. This leaves us with two possibilities:
  • F(3n-1) and F(3n-2) are both even. Then by keeping on iterating the Fibonacci sequence, it's trivial that F(3n+3) is even.
  • F(3n-1) and F(3n-2) are both odd. Therefore, since F(3n) is even, F(3n+1) will be a sum of odd (F(3n-1)) and even (F(3n)) and therefore odd. F(3n+2) will likewise be odd. And since F(3n+1) and F(3n+2) are both odd, then F(3n+3) is even.
    
    That should do it.
    
    I think there might be an error there — you're jumping from 2F(n+1) being even to F(n+1) being even. That doesn't follow.

edited 7th Dec '15 10:21:26 AM by Ninety

I never said F(3n+1) was even. I said 2F(3n+1) was even because F(3n+1) was an integer; i.e. 2F(3n+1) is divisible by 2.

Because F(3(n+1)) is the addition of two even quantities, F(3n) and 2F(3n+1), it is thus even assuming F(3n) is even for n = 1, 2, 3, ...

edited 7th Dec '15 10:43:17 AM by CathariSarad

Is that even necessary to demonstrate? As far as I'm aware, the Fibonacci sequence is defined for integers.

Maybe? If it's not explicitly mentioned as such in the problem statement it would probably be a safe bet to go about proving it.

Well it's pretty obvious. Though, when you think about it, you need strong induction to prove it.

That edited-in one is way more succinct than mine. Though if they require being really formal with induction, maybe it can serve.

What's the notation for the Fibonacci sequence? I mean, I'm not sure I understand the f(X)s you're writing down here because I'm not used to the notation. For me F(3n) looks like F(3*n), if you know what I mean.

F(3n) here means "the (3*n)-th member of the Fibonacci sequence for a given n".

Ahh, that makes more sense. I see it now, I think.

Just out of curiosity, what else did you think F(3n) could mean ?

Here's a different way to do it (just noticed you wanted strong induction, but here it is anyway).

First, assume there exists some n for which F(n) is even. Then, F(n+1) has to be odd. If it wasn't, then F(n-1) would be even since F(n+1) = F(n) + F(n-1). The same argument would apply to F(n-1) and force F(n-2) to be even, and all the way back to F(1). But, F(1) = 1, which is odd.

Thus, F(n+1) is odd if F(n) is even.

Now, F(n+2) has to be odd since F(n+2) = F(n+1)[odd] + F(n)[even].

Then, F(n+3) has to be even since F(n+3) = F(n+2)[odd] + F(n+1)[odd].

Therefore, if F(n) is even, then so is F(n+3).

From here, we note that F(3) = 2 is even, and thus F(3n) must be even for all integers n.

You're making no less than three inductions in that proof, the first is described informally and the third is glossed over (because it is trivial). It's a bit of an overkill.

I prefer CathariSarad's solution. It's simple, elegant, and a weak induction.

edited 8th Dec '15 6:51:00 AM by Aetol

I didn't recognize n as the n-th number of the Fibonacci sequence. I thought F(3n) was a function I didn't know somehow related to it. I mean, F(n) = F(n-1) + F(n-2) is how you'd define it, but that didn't come to my mind as I've never studied Fibonacci.