After College
by djmathman, May 23, 2016, 2:07 AM
home life seems so boring in comparison >.<
I've been thinking a bit about the whole college application process as of late. Don't get me wrong, it's not that I don't like CMU - I'm extremely happy where I am right now, and I definitely feel that I made the right choice. But sometimes, it's interesting to reflect on how my life might be different had my thought process been different.
I remember once during my AMA when I was asked about why I chose CMU one of the reasons I gave was the following:
But now that I think about it, I'm not entirely sure that I would have gone to MIT even if I had gotten in. After all, there is a sort comfort that comes from being at a place where the amount of sheer brainpower floating all over the place isn't super overwhelming. At CMU, there's enough interest in mathematics that I feel comfortable and sociable with others but not so much that I'm constantly facing imposter syndrome, and that sort of balance feels really nice. (Of course, had I gotten in, it probably would have been because I would have made USAMO the previous year, and so maybe I'd be slightly better at math than I am now. Oh well.)
Also darn 2018 AMC/AIME submissions are due by the end of the month, and I don't feel like I have enough good AIME proposals to make up for all the times I pestered the MAA to let me write problems for them D:
I've been thinking a bit about the whole college application process as of late. Don't get me wrong, it's not that I don't like CMU - I'm extremely happy where I am right now, and I definitely feel that I made the right choice. But sometimes, it's interesting to reflect on how my life might be different had my thought process been different.
I remember once during my AMA when I was asked about why I chose CMU one of the reasons I gave was the following:
djmathman wrote:
I didn't get into MIT. (This wasn't as much of an issue as I grew older, but even in 11th grade I was dead set on getting into MIT and going there, despite what anybody else said about it.)
IMC 2010.1 wrote:
Let 
. Prove that ![\[\int_a^b (x^2+1)e^{-x^2} dx \geq e^{-a^2} - e^{-b^2}.\]](//latex.artofproblemsolving.com/4/b/8/4b81207bed4a768652c4373d94ae1f1aa4be748c.png)
. Prove that ![\[\int_a^b (x^2+1)e^{-x^2} dx \geq e^{-a^2} - e^{-b^2}.\]](http://latex.artofproblemsolving.com/4/b/8/4b81207bed4a768652c4373d94ae1f1aa4be748c.png)
as desired.IMC 2010.2 wrote:
Compute the sum of the series ![\[\sum_{k=0}^{\infty} \frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)} = \frac{1}{1\cdot2\cdot3\cdot4} + \frac{1}{5\cdot6\cdot7\cdot8} + \cdots.\]](//latex.artofproblemsolving.com/5/6/0/5602abbee969b9d697fbf42a691ad47882539927.png)
![\[\sum_{k=0}^{\infty} \frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)} = \frac{1}{1\cdot2\cdot3\cdot4} + \frac{1}{5\cdot6\cdot7\cdot8} + \cdots.\]](http://latex.artofproblemsolving.com/5/6/0/5602abbee969b9d697fbf42a691ad47882539927.png)
We can use a similar reduction process a few more times to obtain the full simplification ![\[\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)} = \dfrac16\left(\dfrac{1}{4k+1}-\dfrac{3}{4k+2}+\dfrac{3}{4k+3} - \dfrac{1}{4k+4}\right).\]](http://latex.artofproblemsolving.com/8/c/5/8c50d507c953b8c1a6631eb4e5b9b38c99409a7e.png)
It thus remains to compute the sum ![\[\sum_{k=0}^\infty \left(\dfrac{1}{4k+1}-\dfrac{3}{4k+2}+\dfrac{3}{4k+3} - \dfrac{1}{4k+4}\right)\]](http://latex.artofproblemsolving.com/7/d/3/7d363779773b1ab21e13c5c5809390e35497d521.png)
and then divide the result by 
to obtain our final result.
for which the function is defined, let ![\[F(x) = \sum_{k=0}^\infty\left(\dfrac{x^{4k+1}}{4k+1}-\dfrac{3x^{4k+2}}{4k+2}+\dfrac{3x^{4k+3}}{4k+3} - \dfrac{x^{4k+4}}{4k+4}\right).\]](http://latex.artofproblemsolving.com/b/f/9/bf9d49a99d02e3aaa40ed87b4645c60debe69afa.png)
Observe that 
In an attempt to integrate this, note that for all 
,![\[\dfrac{(1-x)^3}{1-x^4} = \dfrac{(1-x)^2}{(x+1)(x^2+1)} = \dfrac{2}{x+1} - \dfrac{x}{x^2+1} - \dfrac{1}{x^2+1}.\]](http://latex.artofproblemsolving.com/8/4/3/843931d565c9b65c1f42258c7e46613e0031f322.png)
Each of these terms is easily integrable, and we obtain ![\[F(x) = 2\ln(x+1) - \dfrac12\ln(x^2+1) - \arctan x.\]](http://latex.artofproblemsolving.com/d/3/c/d3c495bfa735f4f28f947b008de764c025fac069.png)
Note that although 
is not defined for 
,
exists, and so by continuity this expression for 
is also valid for ![\[F(1) = 2\ln 2 - \frac12\ln 2 - \arctan 1 = \dfrac32\ln 2 - \dfrac{\pi}4\]](http://latex.artofproblemsolving.com/6/7/4/674c667cc0ebe28e267e3923fd1d9671048f4076.png)
and so the desired sum is 
IMC 2010.3 wrote:
Define the sequence 
inductively by 
and 
for each 
. Compute ![\[\lim_{n \to \infty} \frac{x_1 \cdot x_2 \cdot x_3 \cdot ... \cdot x_n}{x_{n+1}}.\]](//latex.artofproblemsolving.com/4/d/0/4d03bf968b6c215e086bc9f985844ea7299968a2.png)
inductively by 
and 
for each 
. Compute ![\[\lim_{n \to \infty} \frac{x_1 \cdot x_2 \cdot x_3 \cdot ... \cdot x_n}{x_{n+1}}.\]](http://latex.artofproblemsolving.com/4/d/0/4d03bf968b6c215e086bc9f985844ea7299968a2.png)
,![\[\dfrac{x_1x_2\cdots x_n}{x_{n+1}} = \sqrt{\dfrac{x_{n+1}^2-4}{x_{n+1}^2}}.\]](http://latex.artofproblemsolving.com/6/3/7/637dd58fb3c2ba3de7a74070f329c1f79a4eeff8.png)
Proof. We proceed by induction. The base case of 
is trivial. For the inductive step, assume the result holds true for some 
.
as desired. 
![\[\lim_{n \to \infty} \frac{x_1 \cdot x_2 \cdot x_3 \cdot ... \cdot x_n}{x_{n+1}} = \lim_{n\to\infty}\sqrt{\frac{x_{n+1}^2-4}{x_{n+1}^2}} = \sqrt{\lim_{n\to\infty}\left(1 - \dfrac{4}{x_{n+1}^2}\right)} = \boxed{1}.\]](http://latex.artofproblemsolving.com/9/b/e/9bea6d9bc3d37e581630005dd33b66e224e91d20.png)
Also darn 2018 AMC/AIME submissions are due by the end of the month, and I don't feel like I have enough good AIME proposals to make up for all the times I pestered the MAA to let me write problems for them D:
This post has been edited 3 times. Last edited by djmathman, May 23, 2016, 2:19 AM
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