Ben Grant
@bengrantmath.bsky.social
PhD candidate in math at UConn. Interested in cluster algebras, representation theory, algebraic combinatorics, dimer models, knot theory, and logic. he/him/his
https://sites.google.com/view/benjamingrant
https://sites.google.com/view/benjamingrant
For the longest time, I tried to convince myself that I didn’t find combinatorics all that interesting, and that if I found anything interesting about it, it was probably just because it was similar to something I liked about abstract algebra. Turns out it was the other way around this whole time
October 18, 2025 at 12:47 AM
For the longest time, I tried to convince myself that I didn’t find combinatorics all that interesting, and that if I found anything interesting about it, it was probably just because it was similar to something I liked about abstract algebra. Turns out it was the other way around this whole time
let alone one of the form we are interested in, x_n=f^n(x). So f cannot exist.
Thanks for reading!
/end/
Thanks for reading!
/end/
October 16, 2025 at 1:52 AM
let alone one of the form we are interested in, x_n=f^n(x). So f cannot exist.
Thanks for reading!
/end/
Thanks for reading!
/end/
contradiction with the assumption that (x_n) is an integer sequence, since L=1+1/k requires our (integer!!) differences between consecutive terms to eventually be arbitrarily close to 1+1/k, which no integer is. So there doesn’t even exist an integer sequence (x_n) with x_{n+k}-x_n=k+1 for all n,
October 16, 2025 at 1:52 AM
contradiction with the assumption that (x_n) is an integer sequence, since L=1+1/k requires our (integer!!) differences between consecutive terms to eventually be arbitrarily close to 1+1/k, which no integer is. So there doesn’t even exist an integer sequence (x_n) with x_{n+k}-x_n=k+1 for all n,
We can rewrite the LHS as a telescoping sum with k terms:
x_{n+k}-x_{n+k-1}+…+x_{n+1}-x_n=k+1.
Taking the limit on both sides as n—>infty (and letting L be the same limit as in the previous problem), we see that kL=k+1.
But then L=1+1/k, which is not an integer (since k is at least 2). This is a
x_{n+k}-x_{n+k-1}+…+x_{n+1}-x_n=k+1.
Taking the limit on both sides as n—>infty (and letting L be the same limit as in the previous problem), we see that kL=k+1.
But then L=1+1/k, which is not an integer (since k is at least 2). This is a
October 16, 2025 at 1:52 AM
We can rewrite the LHS as a telescoping sum with k terms:
x_{n+k}-x_{n+k-1}+…+x_{n+1}-x_n=k+1.
Taking the limit on both sides as n—>infty (and letting L be the same limit as in the previous problem), we see that kL=k+1.
But then L=1+1/k, which is not an integer (since k is at least 2). This is a
x_{n+k}-x_{n+k-1}+…+x_{n+1}-x_n=k+1.
Taking the limit on both sides as n—>infty (and letting L be the same limit as in the previous problem), we see that kL=k+1.
But then L=1+1/k, which is not an integer (since k is at least 2). This is a
use this technique for: suppose we are told to show that for any fixed natural number k at least 2, there does not exist a function f:Z—>Z such that f^k(x)=x+(k+1) for all x in Z. Run the same “replace f^i(x) with x_{n+i}” bit we did before and rearrange to get x_{n+k}-x_n=k+1 for all n \geq 0.
October 16, 2025 at 1:52 AM
use this technique for: suppose we are told to show that for any fixed natural number k at least 2, there does not exist a function f:Z—>Z such that f^k(x)=x+(k+1) for all x in Z. Run the same “replace f^i(x) with x_{n+i}” bit we did before and rearrange to get x_{n+k}-x_n=k+1 for all n \geq 0.
f(x)=x-4 does indeed satisfy this functional equation, but this is easy to verify).
I think this is a pretty neat method. Like I said, I wasn’t familiar with it until yesterday, so forgive me if this is a common trick you have seen before, since I certainly hadn’t.
Another quick application we can
I think this is a pretty neat method. Like I said, I wasn’t familiar with it until yesterday, so forgive me if this is a common trick you have seen before, since I certainly hadn’t.
Another quick application we can
October 16, 2025 at 1:52 AM
f(x)=x-4 does indeed satisfy this functional equation, but this is easy to verify).
I think this is a pretty neat method. Like I said, I wasn’t familiar with it until yesterday, so forgive me if this is a common trick you have seen before, since I certainly hadn’t.
Another quick application we can
I think this is a pretty neat method. Like I said, I wasn’t familiar with it until yesterday, so forgive me if this is a common trick you have seen before, since I certainly hadn’t.
Another quick application we can
In this case, x_0=x is some fixed x in Z and x_1=f(x_0)=f(x). We now know that x_1=x_0-4=x-4 by the above, so f(x)=x-4. But this holds for arbitrary x in Z, so we have our answer: the only function f:Z—>Z such that f(f(x))-4f(x)+3x=8 for all x in Z is f(x)=x-4 (one should technically also check that
October 16, 2025 at 1:52 AM
In this case, x_0=x is some fixed x in Z and x_1=f(x_0)=f(x). We now know that x_1=x_0-4=x-4 by the above, so f(x)=x-4. But this holds for arbitrary x in Z, so we have our answer: the only function f:Z—>Z such that f(f(x))-4f(x)+3x=8 for all x in Z is f(x)=x-4 (one should technically also check that
n+1 on, then it also is from term n on. This allows us to conclude that (x_n) is genuinely an arithmetic progression.
Therefore, all of (x_n) just depends on x_0: we see that x_n=x_0-4n for any n \geq 0. In particular, x_1=x_0-4.
Let’s go back to the case we were interested in to begin with.
Therefore, all of (x_n) just depends on x_0: we see that x_n=x_0-4n for any n \geq 0. In particular, x_1=x_0-4.
Let’s go back to the case we were interested in to begin with.
October 16, 2025 at 1:52 AM
n+1 on, then it also is from term n on. This allows us to conclude that (x_n) is genuinely an arithmetic progression.
Therefore, all of (x_n) just depends on x_0: we see that x_n=x_0-4n for any n \geq 0. In particular, x_1=x_0-4.
Let’s go back to the case we were interested in to begin with.
Therefore, all of (x_n) just depends on x_0: we see that x_n=x_0-4n for any n \geq 0. In particular, x_1=x_0-4.
Let’s go back to the case we were interested in to begin with.
arithmetic progression, but that is an arithmetic progression in its entirety. Indeed, if n is such that x_{n+2}-x_{n+1}=-4, then by massaging our two-step recurrence a tiny bit, we can see that
x_{n+1}-x_n=-1/3*(8-(x_{n+2}-x_{n+1}))
=-1/3*(8-(-4))
=-1/3*12
=-4.
So if (x_n) is arithmetic from term
x_{n+1}-x_n=-1/3*(8-(x_{n+2}-x_{n+1}))
=-1/3*(8-(-4))
=-1/3*12
=-4.
So if (x_n) is arithmetic from term
October 16, 2025 at 1:52 AM
arithmetic progression, but that is an arithmetic progression in its entirety. Indeed, if n is such that x_{n+2}-x_{n+1}=-4, then by massaging our two-step recurrence a tiny bit, we can see that
x_{n+1}-x_n=-1/3*(8-(x_{n+2}-x_{n+1}))
=-1/3*(8-(-4))
=-1/3*12
=-4.
So if (x_n) is arithmetic from term
x_{n+1}-x_n=-1/3*(8-(x_{n+2}-x_{n+1}))
=-1/3*(8-(-4))
=-1/3*12
=-4.
So if (x_n) is arithmetic from term
means that if the differences between consecutive terms *approach* -4, then at some point they actually *are* -4 and they stay that way (since these differences are integers!). Thus, (x_n) is eventually arithmetic with common difference -4.
Next, we can show that (x_n) is not just eventually an
Next, we can show that (x_n) is not just eventually an
October 16, 2025 at 1:52 AM
means that if the differences between consecutive terms *approach* -4, then at some point they actually *are* -4 and they stay that way (since these differences are integers!). Thus, (x_n) is eventually arithmetic with common difference -4.
Next, we can show that (x_n) is not just eventually an
Next, we can show that (x_n) is not just eventually an
L=lim_{n—>infty} (x_{n+1}-x_n). Then (after some mild regrouping on the LHS), we may take limits on either side of our recurrence to get L-3L=8, which gives L=-4. Therefore, the difference between consecutive terms in our sequence approaches -4. But recall that (x_n) is an INTEGER sequence. This
October 16, 2025 at 1:52 AM
L=lim_{n—>infty} (x_{n+1}-x_n). Then (after some mild regrouping on the LHS), we may take limits on either side of our recurrence to get L-3L=8, which gives L=-4. Therefore, the difference between consecutive terms in our sequence approaches -4. But recall that (x_n) is an INTEGER sequence. This
when applied to x_n for any n. The idea is that if we can determine properties of *any* integer sequence (x_n) satisfying this two-step linear recurrence, then we can do it for our specific sequence x_n=f^n(x).
We can first show that (x_n) is *eventually* arithmetic with common difference -4. Let
We can first show that (x_n) is *eventually* arithmetic with common difference -4. Let
October 16, 2025 at 1:52 AM
when applied to x_n for any n. The idea is that if we can determine properties of *any* integer sequence (x_n) satisfying this two-step linear recurrence, then we can do it for our specific sequence x_n=f^n(x).
We can first show that (x_n) is *eventually* arithmetic with common difference -4. Let
We can first show that (x_n) is *eventually* arithmetic with common difference -4. Let
about integer sequences (x_n : n \geq 0). Fix x, and let x_n=f^n(x) be the nth composite power of f applied to x (i.e., we’re taking f^0=id, f^1=f, f^2=f°f, etc.). Then the functional equation above becomes x_2-4x_1+3x_0=8 when applied to our specific x, or more generally, x_{n+2}-4x_{n+1}+3x_n=8
October 16, 2025 at 1:52 AM
about integer sequences (x_n : n \geq 0). Fix x, and let x_n=f^n(x) be the nth composite power of f applied to x (i.e., we’re taking f^0=id, f^1=f, f^2=f°f, etc.). Then the functional equation above becomes x_2-4x_1+3x_0=8 when applied to our specific x, or more generally, x_{n+2}-4x_{n+1}+3x_n=8
Reposted by Ben Grant
from now on I'm just going to truncate the taylor series for sin at the first term. sin(x) = 0 for all x. this approximation
- is efficient to compute
- has bounded error
- has excellent analytic properties
- has very high accuracy for small values, which occur frequently in applications
- is efficient to compute
- has bounded error
- has excellent analytic properties
- has very high accuracy for small values, which occur frequently in applications
October 8, 2025 at 5:21 PM
from now on I'm just going to truncate the taylor series for sin at the first term. sin(x) = 0 for all x. this approximation
- is efficient to compute
- has bounded error
- has excellent analytic properties
- has very high accuracy for small values, which occur frequently in applications
- is efficient to compute
- has bounded error
- has excellent analytic properties
- has very high accuracy for small values, which occur frequently in applications
the award or an HM, but because it feels like this reviewer didn’t take my application seriously at all, and I worry that I’m not the only person this happened to. End of rant, thanks for reading.
October 4, 2025 at 12:30 PM
the award or an HM, but because it feels like this reviewer didn’t take my application seriously at all, and I worry that I’m not the only person this happened to. End of rant, thanks for reading.
algebraic combinatorics— this reviewer was leaving comments about my desire to do research in “set theory,” about “supercompactness,” and about “the combinatorics of the first singular cardinal.” I did not mention any of these whatsoever.
I am mostly just disappointed, not because I didn’t receive
I am mostly just disappointed, not because I didn’t receive
October 4, 2025 at 12:30 PM
algebraic combinatorics— this reviewer was leaving comments about my desire to do research in “set theory,” about “supercompactness,” and about “the combinatorics of the first singular cardinal.” I did not mention any of these whatsoever.
I am mostly just disappointed, not because I didn’t receive
I am mostly just disappointed, not because I didn’t receive
were almost directly copied but with additional typos that I didn’t make (???), but by far the most annoying part to me is that other bullet points were about things that I just *did not talk about at all* in my application. For context, the focus of my proposal was in representation theory and
October 4, 2025 at 12:30 PM
were almost directly copied but with additional typos that I didn’t make (???), but by far the most annoying part to me is that other bullet points were about things that I just *did not talk about at all* in my application. For context, the focus of my proposal was in representation theory and
detailed reviews that demonstrated they, at a minimum, read and thought about my statement and proposal. The third reviewer, on the other hand, left as their comments a relatively incoherent mix of bullet-point-style remarks. Some of these points were directly copied pieces of my statements, some
October 4, 2025 at 12:30 PM
detailed reviews that demonstrated they, at a minimum, read and thought about my statement and proposal. The third reviewer, on the other hand, left as their comments a relatively incoherent mix of bullet-point-style remarks. Some of these points were directly copied pieces of my statements, some
question, but it seems (to me) like a very natural point of view nonetheless.
September 12, 2025 at 11:13 PM
question, but it seems (to me) like a very natural point of view nonetheless.
the “regular functions” on A^n into an object R_n (i.e., taking R_n=Hom(A^n,A)) should give you the “algebraic” dual of the “geometric” A^n. Then duals of mappings f:A^n—>A^m are just their images f^*:R_m—>R_n under Hom(-,A).
I am not sure if this gives a reasonable answer to the original duality
I am not sure if this gives a reasonable answer to the original duality
September 12, 2025 at 11:13 PM
the “regular functions” on A^n into an object R_n (i.e., taking R_n=Hom(A^n,A)) should give you the “algebraic” dual of the “geometric” A^n. Then duals of mappings f:A^n—>A^m are just their images f^*:R_m—>R_n under Hom(-,A).
I am not sure if this gives a reasonable answer to the original duality
I am not sure if this gives a reasonable answer to the original duality
This is perhaps naive and not-so-well-thought-out, but it seems to me like the question here might be interesting to look at from an algebraic geometry perspective. Morphisms A^n—>A appearing in an algebraic structure might be understood as “regular functions” on A^n as an “affine space.” Collecting
September 12, 2025 at 11:13 PM
This is perhaps naive and not-so-well-thought-out, but it seems to me like the question here might be interesting to look at from an algebraic geometry perspective. Morphisms A^n—>A appearing in an algebraic structure might be understood as “regular functions” on A^n as an “affine space.” Collecting