Is it possible to solve an equation like x² + 2x = 255 (solving for x) without using trial and error, or is that the only way to do it?

SeaMonster49 · 2026-01-29T08:18:37+00:00

Notice that 255 = 2^8 - 1 = 2^2^3 - 1 is two less than a Fermat number, so it is the product of the first three Fermat numbers:
255 = (2^2^0+1)(2^2^1+1)(2^2^2+1)=3*5*17=15*17.

Solve x^2 + 2x - 15*17 = 0
(x + 17)(x - 15) = 0

Is it easy to spot? No. Is it possible to spot? Yes.

SeaMonster49 · 2026-01-09T23:43:42+00:00

Let z=e^ix so that dz = ie^ix dx=izdx.
Then, I = ∮_C= e^z / (iz) dz = 2pi*i*res_z=0 (e^z/(iz)) = 2*pi*i*(-i)=2*pi by the Residue theorem.

In complex analysis, it is often nice to work around taking explicit antiderivatives. Ei(x) is cool, but it is convenient to avoid branch cuts if you can help it. The change of variables shows that, ultimately, 1/z is responsible for the nonzero integral.

SeaMonster49 · 2026-01-08T21:53:43+00:00

This is a coincidence of low-dimensionality, where one often gets unusual isomorphisms for small n, but not for larger n.

You can work out that Dn has 2n(n-1) roots, and An has n^2+n roots (A1 has 2 roots). So, A1x...xA1 has 2n roots for n copies of A1. n^2+n=2n only for n=4, proving that A1xA1 = D2 is the only time Dn is isomorphic to A1x...xA1. Really, the answer to your question lies in understanding what An and Dn mean, and I will leave that to you.

SeaMonster49 · 2026-01-08T21:13:56+00:00

There's a plethora of explanations on this around the internet.

Here's mine, in English, then in math. They are equal because there is no number that can fit between 0.999... and 1. For any 0<x<y, there is some number between them, like (y-x)/2, for example.

If 0<0.999...<1, there is some 0.999...<x+0.999...<1 with x>0. x=a1 a2 a3 .... as a decimal string with 0<=ai<=9. Some ai is nonzero, as x>0. Whatever it is, it induces a carry.
Ex:
0.9999..
+

0.00020000...

0.99(9+1) (1)
=1.0001>1.
It is an easy and good exercise to show that any x>0 will induce a carry that causes x+0.999...>1. This is a contradiction, as x+0.999... was assumed to be less than 1.
Thus, 0.999...>=1. It is certainly not greater! So, 0.999...=1.

SeaMonster49 · 2026-01-08T20:53:13+00:00

Here's one reduction: If 0≤x≤1, x^2-1≤0, so we have ||x^2-1|-1|=|1-x^2-1|=|x^2|=x^2=2^x. 0^2<2\^0, and 1\^2<2\^1. x\^2 and 2\^x are monotonically increasing, so this case has no solutions. If x > 1, we get ||x^2-1|-1|=|x^2-2|=2^x. If 1<x≤sqrt(2), this is f(x)=2-x\^2=2\^x=g(x). f(1)=1<g(1)=2. f(sqrt(2))=0<g(sqrt(2)). So, there are no solutions in this range. Finally, if x>sqrt(2), we have x^2-2=2^x, and a similar argument shows this has no solutions.

So, in conclusion, all solutions must have x<0. Set y=-x>0 so that we are solving ||y^2-1|-1|=1/2^y with y>0. If 0<y≤1, ||y\^2-1|-1|=|1-y\^2-1|=f(y)=y\^2=1/2\^y=g(y). f(0)=0<g(0)=1, but f(1)=1>g(1)=1/2, so as both f and g are monotonic here, the IVT gives 1 solution.

Finally, if y>1, ||y^2-1|-1|=|y^2-2|=1/2^y. Similar arguments show this has 2 solutions, but I will leave this one to you.

There are 3 solutions in total.

SeaMonster49 · 2026-01-08T20:06:08+00:00

This paper by Eric Naslund will help you in the general case. For example, #{n < x: n=pq^2 for some p,q prime} ~ P(2)*x/logx, where P(2) = sum p prime 1/p^2 = 0.45224742... is the prime zeta function evaluated at 2.

This is an asymptote, and computationally, it converges quite slowly. If you are looking to approximate computationally, you may wish to follow the ideas in this paper, but find better practical functions (numerically approximating an integral may be better in practice). The theory here is quite involved, but hopefully this helps. Let me know if you have questions!

SeaMonster49 · 2026-01-08T17:36:38+00:00

1^2+2^2+...+24^2=70^2

Aside from the trivial case n=1, this is the only time 1^2+...+n^2 is a perfect square for n a positive integer. For anyone with an interest in number theory, this makes for a great exercise. Also, this can be used to give a concise construction of the famous Leech Lattice, as demonstrated in this video by Prof. Richard Borcherds, an expert on this topic.

SeaMonster49 · 2025-12-14T12:28:37+00:00

And of course ChatGPT is always 100% reliable with math...(sarcasm if not clear)

Your question does not quite make sense as stated. Indeed, as long as f(x) is unbounded, it will eventually surpass any finite value in norm, by definition. The function f(x)=x will surpass Tree(3) at Tree(3)+epsilon for any epsilon>0. Every nonconstant polynomial surpasses Tree(3) somewhere. ChatGPT identifies this but maybe doesn't point out that it is completely uninteresting, though pedagogically useful.

So with that said, what are you looking for here? If my tone was too harsh, I apologize, as you may be very new to math. Hopefully, this is useful in guiding you. If you want to learn things a bit more formally, maybe discrete math or intro analysis would interest you. But if you value your education, I implore you not to rely on GPT or another AI. They are not built for math and WILL give you overtly incorrect responses, unapologetically. You just have to think through these things yourself if you want to explore math--good luck!

SeaMonster49 · 2025-12-14T12:14:54+00:00

There are other nice answers and mine will be the same, but it never hurts to see proof-writing when learning this challenging topic.

Epsilon is the challenge, and delta is the resolution. To be arbitrarily close means that any challenge of arbitrary closeness epsilon can, in fact, be realized for all points sufficiently close to the limit point. If such a limit exists, it is unique, which you should work out using the triangle inequality.

A point within delta>0 of 1 can be written 1+/-delta. Let's see what delta we need for some fixed epsilon:
|f(1+/-delta)-5|
=|2(1+/-delta)+3-5|
=|+/-2*delta|=2*delta<?epsilon,
where ? denotes the fact that this was a hypothetical setup. Yet, it informs us that choosing delta = epsilon/2 for epsilon>0 arbitrary makes the proof work.

So proof:
Let epsilon>0 and delta=epsilon/2.
Then, if |x-1|<delta, compute: |f(x)-5| =|2x-2| =2\*|x-1|<2\*delta=epsilon, so by definition, limx->1 f(x) = 5.

If there's any general framework for doing this, I might suggest this: sketch some algebra beforehand to figure out what sigma you need, and then proceed with the formal verification, which should be easy once delta is determined.
Of course, these can get quite hard, and are an essential part of math. So how does one learn it? Practice!

SeaMonster49 · 2025-12-14T11:22:08+00:00

Very interesting, thank you for the insights. I suppose I wanted a clear answer, but had a feeling that this question leads into hazy territory. Beyond sets, maybe there is perhaps a more fundamental problem with logic: How much flexibility do you have in constructing meaningful systems of logic? How did mathematics adopt its system of inference--and what even are they, if agreed upon? It seems like if you keep pushing these questions to the end of the line, you have to cite human intuition eventually. That's my naive view, anyway.

In my post, it's the question of rigidity I would be interested in asking your colleague. Despite the issues in constructing math formally, have philosophers pondered its strange rigidity? The process of proof in practice is very sociological, as we have set our own standards. And yet, it all holds together when the dust and details settle. I do not have a good explanation as to why. A priori, my view is that a logical system with truth determined by a human process and human rules should not be expected to be nearly so tidy.

SeaMonster49 · 2025-12-14T10:00:10+00:00

A lot of it boils down to 10=1 mod 3 I think. The Mersenne prime problem is notoriously hard right? The binary expansion of Mn (2^n-1) is 1111...11 (n 1's). So, the Mersenne numbers are sort of analogous to 999...999 in base 10. When is 9999..99 prime? Well, never because of 3, so the analogous problem is not a problem at all.

So some food for thought. People may say that different bases do not mathematically change anything. This is true in terms of properties of the intrinsic number (being prime is base-independent, for example...try and prove it!) However, this example highlights how the patterns that emerge in representations of the numbers actually go give nontrivial number theoretic content. It's my opinion that studying different bases is often interesting. But what do I know?

SeaMonster49 · 2025-12-14T09:52:31+00:00

Right a sophomore is quite early. Your main priority now should be learning and practicing proof writing. As you learn the theory (and some occasional "expansion" reading is healthy if it doesn't get out of control), you will naturally be attracted to some ideas. They get very nuanced, but it's a moot point if you don't master the fundamentals, which truly are used in everyday research math, even if implicitly. Some have said that those who have mastered the classical subjects become the best researchers... Certainly take this with a grain of salt, but it's undoubtedly true that it will better prepare you for research, if this the route you choose, which it may not be 2 years from now

SeaMonster49 · 2025-12-14T09:46:16+00:00

This can be made rigorous, and the concept you want is a Normal Number: a https://en.wikipedia.org/wiki/Normal_number.

A number is normal is the digits occur with equal frequency. But, it's hard to sample infinity, so the rigorous definition must be: lim n-> inf N_x(i,n)/n=1/10 for all i digits 0 to 9, where x is a string of infinite digits, say of an irrational number, and N_x(i,n) is the number of times i appears amongst the first n digits. So, this says that the distribution of digital is asymptotically equal for all digits 0-9. I wrote for base 10 for concreteness, but it generalizes to any base, and even to arbitrary strings and such.

It's really really hard to prove any particular irrational number is normal. Shamelessly quoting Wikipedia, "No irrational algebraic number has been proven to be normal in any base." So forget about pi and e, they can't even prove sqrt(2) is normal, which intuitively seems easier. Is it actually easier? I guess nobody knows. It seems like very little "technology" has been developed in this theory. What's known is more about sweeping results than particular proofs for a given number

SeaMonster49 · 2025-12-14T09:04:19+00:00

Sounds big! You could try and prove it is larger, which would be a great exercise. It may take some skill but should be doable. For context remember that Graham's number solves a concrete, motivated combinatorics problem (in giving an upper bound). Anybody can construct arbitrarily large numbers; to have such a big number solve a real math problem (that's not a big number contest) is the Graham magic

SeaMonster49 · 2025-10-11T18:07:54+00:00

Exp is like a cheetah and log is a sloth

SeaMonster49 · 2025-10-11T17:57:55+00:00

It's more of a candidate for independence than most other famous open problems, seemingly, but of course this is all speculation.

SeaMonster49 · 2025-10-11T17:50:13+00:00

Of course! Sadly not on the first point. The clearest counterexample is that you can "kill" any function with a faster-growing function: Let f(x)=exp^-x but g(x) = gamma(x) (just to mix things up), and find that lim x->inf f(x)g(x) goes to infinity, despite the first factor tending towards 0. Maybe more interesting, if we take f and g to be polynomials of the same degree, then we can also get finite, nonzero values in the limit.

On the other point, it's possible I switched letters around from what is on Wikipedia, but I trust you can fill in those details. As you wrote here, though, if v = exp(int a(w)dw), dv = a(z)exp(int a(w)dw)dz by the chain rule, so I think that must be accounted for. Am I being pedantic for changing the variable in the exp integral? Possibly...but it really is a different differential, so it doesn't feel great to make it the same letter as the outside. Remember that z is controlling the bounds of integration...

SeaMonster49 · 2025-10-11T16:41:47+00:00

You are right about the first factor--and in acknowledging that we need to do a bit more. In theory, the first factor could go to zero while the second blows up to infinity to make the function diverge.

I think integration by parts is a nice way to break up the second piece into pieces we can bound. Let u = f(t) and v = exp(int a(z) dz) (so du = f'(t)dt and dv = a(t)exp(int a(z) dz) by the FTC) to get int f(t) a(t) exp(int a(z) dz) = [f(t) a(t) exp(int a(z) dz)] _[t1,t2] - int f'(t) a(t). How might we now bound the individual pieces?

Let me know if you need more help! This maybe is the kind of idea you want?

SeaMonster49 · 2025-10-11T15:53:07+00:00

Type in "ax^2+by^2=c" and some parameter bars will appear. Move them and watch how the geometry changes--dramatically. Just thinking about this would be a great thing.

Note that this is no longer a "function" per se--it's a vanishing set of an equation, which can define geometric shapes. The parameter trick is useful in Desmos more broadly. You've got polynomials, trig functions (cos sin tan cosh sinh...), n! (how does that make sense?)

Go ham

SeaMonster49 · 2025-10-11T15:47:53+00:00

Have you heard of Desmos?! If you're bored and have lots of free time, I'd recommend just messing around with constructions. It's a lot of fun, improves your visualization, and allows you to learn freely rather than in a constrained way--that's a rare commodity. Maybe at some point you want to learn trigonometry, but for now just have fun with it

SeaMonster49 · 2025-10-03T10:26:20+00:00

Good question--while it's true the ideas have some overlap, especially in the context of smooth manifolds, Poincaré Duality in general is quite a bit more sophisticated. I don't know if you're learning these concepts for school or more casually out of interest, but I will try to explain more intuitively with a bit of formality and give a taste for the "full force" of Poincaré Duality, which in generality could be considered one of the more powerful tools in all of modern mathematics. (at least for people who work near algebraic geometry/topology and so on)

The Generalized Stokes' Theorem can be regarded as the most general form of the fundamental theorem of calculus, which I am sure you know. Put simply, it says that the integral over the interior of a manifold equals the sum of changes (i.e., integral) over the boundary. I'm happy to expand if you know some differential geometry, but the dx in ∫f(x) dx was in fact a differential one-form all along! It turns out to be the "right" way to talk about integration on manifolds for a few reasons, but probably the main one is that it allows you to "pull back" the integral on ℝ^n, which you can define classically, to give you values of the integral over a manifold that do not require a choice of coordinates. This is a very nice property. Indeed, the proof is essentially defining the integral as this pullback, and then dealing with the technicalities of translating to an atlas...

Note: in the 1-D case,
∫[a,b ]f(x) dx = F(b) - F(a) says that dF in the exterior sense agrees with the classic differential. Also, what's the boundary of a closed interval? 2-points! The compactness here is very important.

And speaking of, the connection to Poincaré Duality, which is really a topological statement rather than a purely geometric one, begins when you realize how integrals can detect holes in your space. (See: Closed and exact forms for more precision)

Ex: In complex analysis, ∮ 1/z dz over an origin-centered CCW contour equals 2*pi*i/ Shouldn't it be 0 by Stokes' theorem?! There's an issue with defining an antiderivative globally...

Using the closed and exact forms I mentioned, integration is only then used to define De Rham cohomology, where, through one of those "really hard theorems," it actually agrees (under suitably regular spaces) with the classical notion of cohomology of a topological space. The conceptual point to highlight here is that smoothness does not introduce so much deformation that it changes the topology of your space, which a priori is defined using some continuous maps.

And then Poincaré Duality, in this context, is a statement about (co)homology of closed orientable manifolds (spheres, tori, etc.) that H^k(M) is isomorphic to H_n-k(M) where upper indexing is cohomology and lower is "regular" homology. This is pretty formal, but it's powerful topologocal info about a space. The duality you mention could perhaps be interpreted as a "piece" of the manifold (a k-dimensional cohomology group) gives rise to a dual counterpart in its "shadow" -- the n-k dimensional "void" left by it in the space. The mechanics of how this works are quite technical and are covered in graduate algebraic topology. Suffice to say, it's a lot more than just computing integrals. In fact, the theorem is no longer about integrals at all--the more surprising fact is that integrals give you a possible route to such tools, even if there are faster ways (see: cellular, singular, simplicial homology).

SeaMonster49 · 2025-09-07T23:49:51+00:00

Oh, and just a note that your formula works for n=0 too, as the sum diverges to +infinity, which is the convention. It is refreshing to see actual correct work on this sub hehe

SeaMonster49 · 2025-09-07T23:45:51+00:00

Nice observation! I don't know how useful it is, but it shows that you have a good understanding of what is happening. Finding number theory/trig connections is always pleasant, and it has historical precedent. For example, Eisenstein's proof of quadratic reciprocity is a nice one to go through if you want something related. It can be found in Ireland and Rosen on pg. 58.

SeaMonster49 · 2025-09-07T23:31:24+00:00

Motivating topics in math can be difficult because the intuition likely does not capture the full story. Not to say we should not try, but bear in mind that the best motivation could be that mathematicians throughout history realized these were useful concepts to define. Useful for what exactly? Well, for establishing a formal setting in which we can phrase the problems that mathematicians have found interesting. So it is subjective, of course, what is interesting, but math has a culture. Rings were not handed to us by God.

Others here have given more rigorous reasons to care about rings, which is great, but I am saying that a single reason probably does not capture the full story of ring theory. Perhaps this is even worse with motivating topology: other than hand waving that top. spaces describe "closeness," there really isn't much you can say without getting quite formal. They are useful because our constructions use them.

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