DAILY MATH CHALLENGE THREAD #4

Welcome all to the 4th DMCT! Work on math problems, post solutions, and feel free to add your own problems!

Today's problem is pic related. Note that F_n indicates the nth Fibonacci number.

Other urls found in this thread:

en.wikipedia.org/wiki/Sophomore's_dream
twitter.com/SFWRedditVideos

This is easily done using the closed form for the fibonacci sequence.

Let's try for a solution that doesn't use that.

1/3?

Just to clarify:
F_1 = 1
F_2 = 1
F_3 = 2
?

It is? By closed form you mean (Phi^n - phi^n) / sqrt(5)?

Nope

Yeah, 1, 1, 2, 3, 5, etc

>Nope
whoops, 4/15?

Yes, given that closed form you just break the given sum into two geometric series. I mean, it's a bit messy but conceptually easy.

[eqn]\sum_{n=1}^\infty \frac{F_{n}}{5^n} = \sum_{n=1}^\infty \frac{\phi^n - \psi ^n}{\sqrt{5} *5^n} = \frac{1}{\sqrt{5}} \left( \sum_{n = 1}^\infty \left(\frac{\phi}{5} \right)^n - \sum_{n=1}^{\infty} \left( \frac{\psi}{5} \right)^{n} \right)[/eqn]

I hope it compiles well, it did in the preview. At the last step, you just calculate the geometric series and whatever it gives is the result.

Sorry if someone has already solved it.

If you want a nice topology problem, proof that in the real kolmogorov line, a set A is compact iff:

1.- You can put the set inside a set of the form [a, b)
2.- A is closed
3.- There's no strictly increasing sequence inside of A

Another version asks to prove the same but instead of conditions 1 and 2 you have that any strictly decreasing sequence inside of A eventually converges in A.

reee im bad at algebra, 5/19

>Yeah, 1, 1, 2, 3, 5, etc

In that case, we have
[eqn]S = \sum_{n \ge 1} \frac{F_n}{5^n} = \frac15 +\frac{1}{25} + \sum_{n \ge 3} \frac{F_{n-2} + F_{n-1}}{5^n} [/eqn]
and most likely (Im too lazy) you just write the last expression in terms of S again and then you can solve for S. (this is valid as long as we assume the series actually converges)

Oh I see, there's a much nicer solution though.

Correct!

Yeah that's the idea

[eqn]S = \sum_{n = 1}^{\infty} \frac{F_n}{5^n} = \frac{1}{5} + \frac{1}{25} + \frac{2}{125} + \frac{3}{625} + ...[/eqn]
[eqn]\frac{S}{5} = \frac{1}{25} + \frac{1}{125} + \frac{2}{625} + \frac{3}{3125} ...[/eqn]
[eqn]\frac{4S}{5} = \frac{1}{5} + \frac{1}{125} + \frac{1}{625} + \frac{2}{3125} + ...[/eqn]
[eqn]\frac{4S}{5} = \frac{1}{5} + \frac{S}{25}[/eqn]
[eqn]\frac{19S}{25} = \frac{1}{5}[/eqn]
[eqn]S = \frac{5}{19}[/eqn]

Here's my work

I guess its fairly obvious the series converges. For example, by induction we have that F_n is at most say 2^n. So this method of "solving for S" is valid since S is a finite real number

b) Evaluate [eqn]\sum_{n=1}^\infty \frac{F_n}{N^n}[/eqn]

Is that supposed to be a lowercase "n" on the bottom? If so I doubt this one has any kind of nice answer

define the predecessor function in the untyped lambda calculus

Is there even a closed form for [eqn]\sum_{n=1} ^{\infty} \frac{1}{n^{n}}[/eqn]

sort of
en.wikipedia.org/wiki/Sophomore's_dream

Here's a kinda stupid one, but hopefully one of you finds it amusing. Let X be a non-empty set. Show that X admits a poset structure with a greatest element without using AC

for you, my children

crap forgot photo

> A combinatorial problem

I wonder why they're so popular in these threads.