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I've prepared one Calculus and one Number Theory problems for now. If they are both solved I'll post others.
Starting from tomorrow I think I'll try and do one hard problem a day with the field of study depending on the weekday like some user suggested. Other problems are very much welcome. It would be cool if during the weekend we could have one roundup thread where no one posted any more problems but we attempted all unsolved questions from that week and someone archived the whole thing.
DAILY MATHS CHALLENGE #2
Justin Ross
Other urls found in this thread:
en.wikipedia.org
mathworld.wolfram.com
twitter.com
Christian Green
And the NT problem.
Connor Sanchez
For the calc one:
For each n >= 1, [math]x_{n+1}^3 = x_n^3 + 1 + \frac{1}{3x_n^3} + \frac{1}{27x_n^6}[/math]. Hence, we can write [math]x_n^3 \ge n + \sum_{j=1}^{n-1}\frac{1}{3x_j^3}[/math].
To complete the proof, we need only prove that [math]\sum_{j=1}^{n-1}\frac{1}{x_j^3}[/math] is unbounded. But we have [math]x_{n+1}^3 \le x_n^3 + 3[/math], hence [math]\forall n \ge 1, x_n^3 \le 3n[/math].
Finally, we get [math]\sum_{j=1}^{n-1}\frac{1}{x_j^3} \ge \sum_{j=1}^{n-1} \frac{1}{3j}[/math], which concludes.
Zachary Ortiz
Good initiative. I'll gladly ponder both problems. Too few people in this board value problem-solving over memes
Leo King
different insight (which is overkill but meh)
The sequence is trivially increasing and unbounded (boundedness would yield convergence to 0, which is a contradiction),
hence [math]\frac{1}{x_n} [/math] converges to 0.
Note that [eqn] \begin{align} x_{n+1}^3 &= \left(x_n + \frac{1}{3 x_n^2} \right)^3\\
&=x_n^3\left( 1+\frac{1}{3x_n^3}\right)^3\\
&=x_n^3 \left(1 + \frac{1}{x_n^3}+o\left(\frac{1}{x_n^3}\right) \right)\\
&=x_n^3+1+o(1)
\end{align}[/eqn]
Hence [math] x_{n+1}^3-x_n^3[/math] converges to 1 and Cesaro mean theorem yields [math]x_n^3 = n+o(n)[/math].
Since [math] x_{n+1}^3 = x_n^3 + 1 + \frac{1}{3x_n^3} + \frac{1}{27x_n^6} [/math], [math]x_{n+1}^3 = n+1 + \frac 13 \sum_{k=1}^n \left( \frac 1{x_k^3} + \frac 1{3x_k^6}\right) [/math]
and it suffices to prove that [math] \sum_{k=1}^n \left( \frac 1{x_k^3} + \frac 1{3x_k^6}\right)[/math] is unbounded which is clear since [math]\frac 1{x_n^3} = \frac{1}{n} + o\left( \frac{1}{n} \right) [/math]
David Foster
Its only true for all even Numbers M. We can write a(k) for a generall M as a(0) for another M'=M^(2k), so we only have to look at the first iteration. Wich only gets integer if M is even
Luis Scott
wrong, for example with [math]M=9[/math], [math]a_4 = 2789204756584545 [/math]
Alexander Kelly
Ok, its not completely correct, but the general idea should be right, ill write the Tex when i get home
Jace Butler
This is true for M equal to all numbers in the union of the following sets:
S[1] = {2, 4, 6, 8, ...}
S[2] = {3, 7, 11, 15, ...}
S[3] = {5, 13, 21, 29, ...}
S[4] = {9, 25, 41, 57, ...}
...
S[n] = 2S[n-1] - 1
...
Which might be the entire set of natural numbers excluding 1, but I don't know for sure.
Carter Johnson
I've proved there's an integer term for any M which is not [math] 1 \pmod 4[/math], but I'm stumped dealing with the case [math] 1 \pmod 4[/math] ...