MBTI Isotopes
by Mathandski, May 28, 2025, 4:27 AM
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This post has been edited 2 times. Last edited by Mathandski, May 29, 2025, 6:01 AM
The problem that angers me
by Mathandski, May 22, 2025, 5:04 PM
Quote:
(24TSTST7): An infinite sequence 
, 
, 
, 
of real numbers satisfies
![\[
a_{2n-1} + a_{2n} > a_{2n+1} + a_{2n+2} \qquad \mbox{and} \qquad a_{2n} + a_{2n+1} < a_{2n+2} + a_{2n+3}
\]](//latex.artofproblemsolving.com/4/d/a/4da3b88a20c42c1798141b8db086de341cfb9a67.png)
for every positive integer 
. Prove that there exists a real number 
such that 
for every positive integer .
, 
, 
, 
of real numbers satisfies![\[
a_{2n-1} + a_{2n} > a_{2n+1} + a_{2n+2} \qquad \mbox{and} \qquad a_{2n} + a_{2n+1} < a_{2n+2} + a_{2n+3}
\]](http://latex.artofproblemsolving.com/4/d/a/4da3b88a20c42c1798141b8db086de341cfb9a67.png)
for every positive integer 
. Prove that there exists a real number 
such that 
for every positive integer .With a 720 757 so far, I opened the third and final day of the TSTST hoping to get two solves. The AGC subject distribution was the worst-case scenario, but if I can solve P7 early, I might be able to pull it off.
Sunk cost
15 minutes into looking into the problem, I thought I had it. We let,
![\[b_1 = a_1, b_2 = a_2, b_3 = a_3, b_n = b_{n-3} + b_{n-2} - b_{n-1}\]](http://latex.artofproblemsolving.com/7/5/0/7505dc7d3c67e6188f0b89139ebb74f9644e45fa.png)
and compare 
with 
. Even better, using generating functions, has a really neat expression!![\[b_n = (an + b) (-1)^n + c\]](http://latex.artofproblemsolving.com/6/9/1/691bbf20e6f98b14c69ce8b284133b9d5cce2d14.png)
For an increasing sequence 
of positive reals,![\[\Longrightarrow a_n = (an + b + d_i) (-1)^n + c\]](http://latex.artofproblemsolving.com/5/c/1/5c15151c6d7db121d0a719bd3a5472614493a130.png)
15 minutes in and I was already close. I plowed deeper to find the critical hit; it seemed that I was getting closer and closer with every new discovery. Despite this, with every step forward, a new "minor flaw in my idea" would arise. I was always so close... I always needed a bit more time.Before I could realize, 2 hours had passed without a working solution. I have not even looked at the other problems.
Fatal stack overflow
Though I eventually successfully solved the problem, I had little remaining time to tackle the difficult P8/P9 on the test. 24TSTST7 was a frustrating experience that highlighted the flaws in my test-taking strategy.
Problem solving is navigating a graph trying to find a destination - what the problem writer wants us to prove. Each observation is a step that usually takes us closer to the result. Through experience, we build intuition on which steps take us closer to the solution. In this case, intuition misled me and I trusted it ignorantly. The first step in my solution took me into a dead end, and I sunk further and further by the lure of progress. I took dozens of small steps to asymptotically approach the solution without ever reaching it.
From a graph theory standpoint, the strategy I took was a depth-first search: trying the most promising path to the end until we get stuck. This strategy is usually highly efficient since intuition usually guides us onto the right path and most dead ends finish quickly when intuition fails. However, this problem was an instructive warning to me on what happens if things may go wrong.
In the future, perhaps I will experiment with BFS problem solving: enumerate plausible approaches and explore them concurrently or perhaps I will continue using DFS with a lower recursion depth limit. Even if there is no strategy that is definitively superior, I think this will bring me increased flexibility in competitions.
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First-Order Exact Differential Equation Shortcut
by Mathandski, May 2, 2025, 9:28 PM
Summary
In this blog post, I will prove an interesting algorithm which serves as a shortcut to solve exact differential equations. This was discovered and introduced to me by my math teacher.
Proposed Shortcut
Given an exact differential equation of the form
![\[M(t, y) + N(t, y) \frac{dy}{dt} = 0\]](//latex.artofproblemsolving.com/5/a/b/5ab7d1bbf763fbaf19a5acfb75f5a1fc7494d4e6.png)
An algorithm to solve for 
is as follows:
1. Let
and 
where we omit any products of integration.
2. If a certain term appears as an addend in both
and 
, we remove that term from
3. We write
and solve for where may be any constant.
Example Problem
![\[2ty - 9t^2 + (2y + t^2 + 1) \frac{dy}{dt} = 0\]](//latex.artofproblemsolving.com/2/1/d/21d871fb94785dda4ab537e61041dd495e0be58c.png)
Solve for in terms of 
.
Proof
Recall that and after step 
of our algorithm. Our main claim is as follows. We split the terms of into terms exclusively in terms of , terms exclusively in terms of , and other terms dependent on both and . We do the same with .
Remark: For example, for,
![\[F(t, y) = t^2y + 1434 \cos(y^2) + \ln(t + y) + \arctan(y) + t^{2t}\]](//latex.artofproblemsolving.com/8/e/4/8e42229df636b080d176ff934c8ab871eb13d96a.png)
We group 
together, 
together, and 
together.
More formally, for some functions
such that \textbf{each term of 
is dependent on both and }, we may write,
![\[H_1 (t, y) = h_1(t, y) + f_1(t) + g_1(y)\]](//latex.artofproblemsolving.com/7/6/5/765cd275a1991307d99a2c8cf0b68e428ccf640c.png)
![\[H_2 (t, y) = h_2(t, y) + f_2(t) + g_2(y)\]](//latex.artofproblemsolving.com/a/0/0/a00fb96f37e71764b37c81990fed9db21b94d11a.png)
Claim:
![\[h_1(t, y) = h_2(t, y)\]](//latex.artofproblemsolving.com/8/c/d/8cdb892db6fbf474d06537ff633a885b8f25a818.png)
Note that,
![\[\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}\]](//latex.artofproblemsolving.com/9/5/6/956e5969c4c7d6d551b0c8bcd8d35c79f1213c1e.png)
Plugging in 
and 
,
![\[\Longrightarrow \frac{\partial}{\partial y} \left( \frac{\partial H_1}{\partial t} \right) = \frac{\partial}{\partial t} \left( \frac{\partial H_2}{\partial y} \right) \quad (*)\]](//latex.artofproblemsolving.com/3/d/9/3d9e92c6fc5671f27faa1d192eba5d08bdea5d18.png)
However, notice that,
![\[\frac{\partial H_1}{\partial t} = \frac{\partial}{\partial t} (H_1(t, y)) = \frac{\partial}{\partial t} (h_1(t, y) + g(y)) = \frac{\partial h_1}{\partial t}\]](//latex.artofproblemsolving.com/0/0/c/00c83c8ca9d6875aa6196b03402b37a9b0bb6f24.png)
Analogously,
![\[\frac{\partial H_2}{\partial t} = \frac{\partial h_2}{\partial t}\]](//latex.artofproblemsolving.com/7/6/d/76daefd19b2456d1e0f2db3f5802037a96d92053.png)
Plugging this into 
,
![\[ \frac{\partial}{\partial y} \left( \frac{\partial h_1}{\partial t} \right) = \frac{\partial}{\partial t} \left( \frac{\partial h_2}{\partial y} \right)\]](//latex.artofproblemsolving.com/7/f/2/7f2a7c7635b6c7f58fa003da2a28358e5f595f8f.png)
We use Clairaut's Theorem, which states that if
and 
are continuous at 
,
![\[\frac{\partial }{\partial x} \left( \frac{\partial }{\partial y} \left( f(x, y) \right) \right) = \frac{\partial }{\partial y} \left( \frac{\partial }{\partial x} \left( f(x, y) \right) \right)\]](//latex.artofproblemsolving.com/2/9/9/2995196fc0d2629c58257e4a9e4e843c2b5a2c0d.png)
We apply Clairaut's Theorem to the left side of the prior equation,
![\[\iff \frac{\partial}{\partial t} \left( \frac{\partial h_1}{\partial y} \right) = \frac{\partial}{\partial t} \left( \frac{\partial h_2}{\partial y} \right) \]](//latex.artofproblemsolving.com/f/0/7/f07eba5c82e22ac9c24bef635f7dd645df1ab081.png)
We now do partial integration on both sides. Note that the product of integration becomes an arbitrary function in terms of since we are working with partial derivatives,
![\[\Longrightarrow \frac{\partial h_1}{\partial y} = \frac{\partial h_2}{\partial y} + f(y)\]](//latex.artofproblemsolving.com/1/1/8/1189ce72e64b38a0c2d49a5d3bb6f4806d1edff5.png)
We now integrate with respect to ,
![\[\Longrightarrow h_1(t, y) = h_2(t, y) + \int f(y) \, dy + g(t)\]](//latex.artofproblemsolving.com/c/d/6/cd6c30eaeff568dbbbfd4285a81e1b7b20eb0221.png)
Therefore, 
and 
differ by 
and 
, which are entirely functions in terms of and respectively. However, neither and have any terms which are independent in or from our definition of and . Therefore, the constant functions of integration are both zero and 
as desired. The claim is proved.
Now recall that
and 
are both zero since, under are algorithm, we clarified that we are omitting any constants of integration when writing and . Therefore, we may in fact write,
![\[H_1 (t, y) = h_1(t, y) + f_1(t)\]](//latex.artofproblemsolving.com/a/7/a/a7ae5abc07fbfaaec649fb8a4ea2700a9314005f.png)
![\[H_2 (t, y) = h_2(t, y) + g_2(y)\]](//latex.artofproblemsolving.com/a/0/2/a0289f949b0f273885d1ea07c612c8a3d036475a.png)
We now apply step
of our algorithm, which removes any duplicate terms appearing in both and from .
Claim:
![\[H(t, y) := h_1(t, y) + f_1(t) + g_2(y)\]](//latex.artofproblemsolving.com/b/6/a/b6a7f4395043de9cd158cbf0ea156256dd1bd3d0.png)
Is 
after we apply step of our algorithm.
Indeed, when we added,
Under step, we removed everything in that appeared in . Since 
, the entirety of 
was removed. However, since 
is a function in but there were no terms solely in terms of in , nothing from was removed. As a result, after step ,
![\[H_2(t, y) := g_2(y)\]](//latex.artofproblemsolving.com/3/7/0/37096a723786db65b1b0cc0a2b36317698ed9556.png)
Adding it to
, we get the desired result. The claim is proved.
It now suffices to show that this
has all the solutions to the differential equation.
Claim:
![\[H(t, y) = h_1(t, y) + f_1(t) + g_2(y)\]](//latex.artofproblemsolving.com/6/f/a/6fa6577437e0d43483badedfae65e94342d9d27d.png)
Satisfies 
and 
We may plug it in to prove it. Indeed,
's partial derivative with respect to is 
and adding on a does not change the partial derivative with respect to . The analogous conclusion can be made for 
. The claim is proved.
Therefore, the given differential equation,
![\[\iff \frac{\partial H}{\partial t} + \frac{\partial H}{\partial y} \frac{dy}{dt} = 0\]](//latex.artofproblemsolving.com/5/a/3/5a3ecc34c1168f3d030d833c577ce03c31d9edf1.png)
Applying the reverse chain rule,
![\[\iff \frac{d}{dt} (H(t, y)) = 0\]](//latex.artofproblemsolving.com/a/b/f/abf0f685c4f34eff460cb3f4971b02680e2f57af.png)
As desired, since is a function of ,
![\[\iff H(t, y) = C\]](//latex.artofproblemsolving.com/6/0/9/60920799b48c87befaf214aa2a3c41c90c0260ec.png)
In this blog post, I will prove an interesting algorithm which serves as a shortcut to solve exact differential equations. This was discovered and introduced to me by my math teacher.
refers to a partial derivative.Proposed Shortcut
Given an exact differential equation of the form
![\[M(t, y) + N(t, y) \frac{dy}{dt} = 0\]](http://latex.artofproblemsolving.com/5/a/b/5ab7d1bbf763fbaf19a5acfb75f5a1fc7494d4e6.png)
An algorithm to solve for 
is as follows:1. Let
and 
where we omit any products of integration.2. If a certain term appears as an addend in both
and 
, we remove that term from 3. We write
and solve for where may be any constant.Example Problem
![\[2ty - 9t^2 + (2y + t^2 + 1) \frac{dy}{dt} = 0\]](http://latex.artofproblemsolving.com/2/1/d/21d871fb94785dda4ab537e61041dd495e0be58c.png)
Solve for in terms of 
.![\[\frac{\partial}{\partial t} (2y + t^2 + 1) = 2t = \frac{\partial}{\partial y} (2ty + 9t^2)\]](http://latex.artofproblemsolving.com/8/e/5/8e53a6df127d43062f9123d902a2247f892f9089.png)
Therefore, we may apply the shortcut. We see that, omitting any products of integration,![\[H_1(t, y) = \int 2ty - 9t^2 \, dt = t^2y - 3t^3\]](http://latex.artofproblemsolving.com/3/0/f/30f7a54c2b6c7a3efdc44140a596e1fa6f4a7887.png)
![\[H_2 (t, y) = \int 2y + t^2 + 1 \, dy = y^2 + t^2y + y\]](http://latex.artofproblemsolving.com/a/f/6/af62b6174c443312956e05df52c95f2061d18132.png)
Note that 
appears in both ![\[H_2(t, y) := y^2 + y\]](http://latex.artofproblemsolving.com/f/5/8/f58570302ad0d705e2862d9b383d9e6359bec665.png)
We now write,![\[H_1 (t, y) + H_2(t, y) = C\]](http://latex.artofproblemsolving.com/e/8/e/e8e1b866b4f9ac92669a01fee471b7bcc001c548.png)
![\[\iff y^2 + y + t^2y - 3t^3 = C\]](http://latex.artofproblemsolving.com/a/c/b/acb84a4c2713d5393aea7abe9f5b79b57186c41e.png)
We may now simplify and solve.Proof
Recall that
and after step 
of our algorithm. Our main claim is as follows. We split the terms of into terms exclusively in terms of , terms exclusively in terms of , and other terms dependent on both and . We do the same with .Remark: For example, for,
![\[F(t, y) = t^2y + 1434 \cos(y^2) + \ln(t + y) + \arctan(y) + t^{2t}\]](http://latex.artofproblemsolving.com/8/e/4/8e42229df636b080d176ff934c8ab871eb13d96a.png)
We group 
together, 
together, and 
together.More formally, for some functions
such that \textbf{each term of 
is dependent on both and }, we may write,![\[H_1 (t, y) = h_1(t, y) + f_1(t) + g_1(y)\]](http://latex.artofproblemsolving.com/7/6/5/765cd275a1991307d99a2c8cf0b68e428ccf640c.png)
![\[H_2 (t, y) = h_2(t, y) + f_2(t) + g_2(y)\]](http://latex.artofproblemsolving.com/a/0/0/a00fb96f37e71764b37c81990fed9db21b94d11a.png)
Claim:
![\[h_1(t, y) = h_2(t, y)\]](http://latex.artofproblemsolving.com/8/c/d/8cdb892db6fbf474d06537ff633a885b8f25a818.png)
Note that,
![\[\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}\]](http://latex.artofproblemsolving.com/9/5/6/956e5969c4c7d6d551b0c8bcd8d35c79f1213c1e.png)
Plugging in 
and 
,![\[\Longrightarrow \frac{\partial}{\partial y} \left( \frac{\partial H_1}{\partial t} \right) = \frac{\partial}{\partial t} \left( \frac{\partial H_2}{\partial y} \right) \quad (*)\]](http://latex.artofproblemsolving.com/3/d/9/3d9e92c6fc5671f27faa1d192eba5d08bdea5d18.png)
However, notice that,![\[\frac{\partial H_1}{\partial t} = \frac{\partial}{\partial t} (H_1(t, y)) = \frac{\partial}{\partial t} (h_1(t, y) + g(y)) = \frac{\partial h_1}{\partial t}\]](http://latex.artofproblemsolving.com/0/0/c/00c83c8ca9d6875aa6196b03402b37a9b0bb6f24.png)
Analogously,![\[\frac{\partial H_2}{\partial t} = \frac{\partial h_2}{\partial t}\]](http://latex.artofproblemsolving.com/7/6/d/76daefd19b2456d1e0f2db3f5802037a96d92053.png)
Plugging this into 
,![\[ \frac{\partial}{\partial y} \left( \frac{\partial h_1}{\partial t} \right) = \frac{\partial}{\partial t} \left( \frac{\partial h_2}{\partial y} \right)\]](http://latex.artofproblemsolving.com/7/f/2/7f2a7c7635b6c7f58fa003da2a28358e5f595f8f.png)
We use Clairaut's Theorem, which states that if
and 
are continuous at 
,![\[\frac{\partial }{\partial x} \left( \frac{\partial }{\partial y} \left( f(x, y) \right) \right) = \frac{\partial }{\partial y} \left( \frac{\partial }{\partial x} \left( f(x, y) \right) \right)\]](http://latex.artofproblemsolving.com/2/9/9/2995196fc0d2629c58257e4a9e4e843c2b5a2c0d.png)
We apply Clairaut's Theorem to the left side of the prior equation,
![\[\iff \frac{\partial}{\partial t} \left( \frac{\partial h_1}{\partial y} \right) = \frac{\partial}{\partial t} \left( \frac{\partial h_2}{\partial y} \right) \]](http://latex.artofproblemsolving.com/f/0/7/f07eba5c82e22ac9c24bef635f7dd645df1ab081.png)
We now do partial integration on both sides. Note that the product of integration becomes an arbitrary function in terms of since we are working with partial derivatives,![\[\Longrightarrow \frac{\partial h_1}{\partial y} = \frac{\partial h_2}{\partial y} + f(y)\]](http://latex.artofproblemsolving.com/1/1/8/1189ce72e64b38a0c2d49a5d3bb6f4806d1edff5.png)
We now integrate with respect to ,![\[\Longrightarrow h_1(t, y) = h_2(t, y) + \int f(y) \, dy + g(t)\]](http://latex.artofproblemsolving.com/c/d/6/cd6c30eaeff568dbbbfd4285a81e1b7b20eb0221.png)
Therefore, 
and 
differ by 
and 
, which are entirely functions in terms of and respectively. However, neither and have any terms which are independent in or from our definition of and . Therefore, the constant functions of integration are both zero and 
as desired. The claim is proved.Now recall that
and 
are both zero since, under are algorithm, we clarified that we are omitting any constants of integration when writing and . Therefore, we may in fact write,![\[H_1 (t, y) = h_1(t, y) + f_1(t)\]](http://latex.artofproblemsolving.com/a/7/a/a7ae5abc07fbfaaec649fb8a4ea2700a9314005f.png)
![\[H_2 (t, y) = h_2(t, y) + g_2(y)\]](http://latex.artofproblemsolving.com/a/0/2/a0289f949b0f273885d1ea07c612c8a3d036475a.png)
We now apply step
of our algorithm, which removes any duplicate terms appearing in both and from .Claim:
![\[H(t, y) := h_1(t, y) + f_1(t) + g_2(y)\]](http://latex.artofproblemsolving.com/b/6/a/b6a7f4395043de9cd158cbf0ea156256dd1bd3d0.png)
Is 
after we apply step of our algorithm.Indeed, when we added,
Under step
, we removed everything in that appeared in . Since 
, the entirety of 
was removed. However, since 
is a function in but there were no terms solely in terms of in , nothing from was removed. As a result, after step ,![\[H_2(t, y) := g_2(y)\]](http://latex.artofproblemsolving.com/3/7/0/37096a723786db65b1b0cc0a2b36317698ed9556.png)
Adding it to
, we get the desired result. The claim is proved.It now suffices to show that this
has all the solutions to the differential equation.Claim:
![\[H(t, y) = h_1(t, y) + f_1(t) + g_2(y)\]](http://latex.artofproblemsolving.com/6/f/a/6fa6577437e0d43483badedfae65e94342d9d27d.png)
Satisfies 
and 
We may plug it in to prove it. Indeed,
's partial derivative with respect to is 
and adding on a does not change the partial derivative with respect to . The analogous conclusion can be made for 
. The claim is proved.Therefore, the given differential equation,
![\[\iff \frac{\partial H}{\partial t} + \frac{\partial H}{\partial y} \frac{dy}{dt} = 0\]](http://latex.artofproblemsolving.com/5/a/3/5a3ecc34c1168f3d030d833c577ce03c31d9edf1.png)
Applying the reverse chain rule,![\[\iff \frac{d}{dt} (H(t, y)) = 0\]](http://latex.artofproblemsolving.com/a/b/f/abf0f685c4f34eff460cb3f4971b02680e2f57af.png)
As desired, since is a function of ,![\[\iff H(t, y) = C\]](http://latex.artofproblemsolving.com/6/0/9/60920799b48c87befaf214aa2a3c41c90c0260ec.png)
We're so back
by Mathandski, Apr 25, 2025, 8:59 PM
Attachments:
This post has been edited 1 time. Last edited by Mathandski, May 2, 2025, 5:43 PM
Keeping my eye on Gmail
by Mathandski, Apr 21, 2025, 5:28 PM
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This post has been edited 3 times. Last edited by Mathandski, Apr 23, 2025, 6:28 AM
FAQs for Subjective Ratings (MOHs)
by Mathandski, Apr 3, 2025, 9:24 PM
This post has been edited 1 time. Last edited by Mathandski, May 21, 2025, 4:31 PM
Sprinting upstairs during USAMO
by Mathandski, Mar 21, 2025, 7:39 PM
This post has been edited 1 time. Last edited by Mathandski, Mar 21, 2025, 7:39 PM
Reason: aops won't allow "k*ll"
Reason: aops won't allow "k*ll"
Skiandmath
by Mathandski, Jan 27, 2025, 10:03 PM
This post has been edited 1 time. Last edited by Mathandski, Jan 27, 2025, 10:04 PM
Cash Bounty for Solution Errors
by Mathandski, Jan 13, 2025, 6:10 PM
This program is now closed (3/18) has passed!
If I made a critical error causing a point deduction of -5 or more, you will be awarded $15 (via PayPal or amazon gift cards). This includes misreading the problem, an error that causes my solution to solve a far easier problem, or an error causing my solution to solve an entirely different problem.
For minor errors causing a point deduction of -2 to -4, you will be awarded $10. For example, this includes failing to consider edge cases that require special attention, forgetting the point-wise trap, or explaining a non-trivial step with insufficient detail.
Similar to crypto, bounties will decrease exponentially over time (due to limited budget) but the reward you get
is guaranteed to be the amount it says at the time of your email being sent.
If you are interested, please read:
If I made a critical error causing a point deduction of -5 or more, you will be awarded $15 (via PayPal or amazon gift cards). This includes misreading the problem, an error that causes my solution to solve a far easier problem, or an error causing my solution to solve an entirely different problem.
For minor errors causing a point deduction of -2 to -4, you will be awarded $10. For example, this includes failing to consider edge cases that require special attention, forgetting the point-wise trap, or explaining a non-trivial step with insufficient detail.
Similar to crypto, bounties will decrease exponentially over time (due to limited budget) but the reward you get
is guaranteed to be the amount it says at the time of your email being sent.
If you are interested, please read:
instead of 
or 
instead of 
)This post has been edited 14 times. Last edited by Mathandski, Mar 20, 2025, 2:34 PM
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