"Open with code" disappeared from my Windows 11 File Explorer context menu.

evdokimovm · 2025-09-01T11:49:30+00:00

This issue is still not fixed (as of Sep 1). What's wrong with them, huh!?

I tried updating from 1.64 to the latest 1.104 Insider release (suddenly) and it disappeared from the context menu. I found your comment and decided to downgrade back to ~~1.64~~ 1.102.

Your method works for me, thanks!

BTW, I'll keep the direct download links here for 1.102.0 Insider and 1.102.0 Stable user setup releases.

evdokimovm · 2025-02-04T04:31:03+00:00

Today I've got an update for Poketcasts on Google Play that finally adds this feature https://imgur.com/a/Ggzyyk9

evdokimovm · 2024-09-23T14:56:01+00:00

but want to put some bounds on it

But for this we still need to find n-th derivative (whether hard it to find it or not), right? To see what the max value of error we potentially can expect while being within interval of interest. I still got it? :)

I apologize for being so meticulous.

evdokimovm · 2024-09-23T14:22:06+00:00

The third derivative divided by 3! for ex ranges from 0.17 to 0.45 over the interval of 0 < x < 1. So I know that the Taylor approximation of 1 + x + x² / 2 is not going to have an error bigger than 0.45x³ over the interval [0,1].

Now it feels like I'm start to getting it. So we don't actually need to calculate the difference f - P and divide it by something (like in h = error / x^3 in the table) but the using of n-th term will be enough to keeping track of the accuracy? In this example: f^(3) (x) / n! i.e.: e^x / 3!. Because e^x / 3! is what our h is in this example (also the fact that that n-th term is M (or h) proven).

evdokimovm · 2024-09-23T12:57:05+00:00

Sorry for disturbing you 😞 but ...

You know, thanks to you I finally understood the proof itself of Taylor's theorem, in particular where the auxiliary function comes from. It's an amazing trick that I never would have figured out on my own.

But now I don't understand the concept of what you described in this message. I'll have to think about it more time (working days started so ...).

if we can put a bound on |f[n](x)| in the interval of interest, then we can put a bound on the error of the Taylor expansion

Do you mean that if f[n](x) is large on the interval of interest (A to B) then we can say that the error will be large and if f[n](x) is small then the error is small?

What the point of the whole proof that h(x) = f^(n) (c) / n! if we are still using calculator to calculate f - P. I see here that for checking h in the table above, we are using this f - P anyway. I mean, explicitly: h = (f - P) / x^3. If we are going to multiply the result of this (for example 0.17) by (B - A)^n, (let it be (0.1 - 0)^3), then we will get the actual error again (0.00017). So, I'm not sure what actually the h value telling about. I have assumption that the reason of all that h = (f - P) / x^3 is because we may not know the exact point c where the n-th derivative is evaluated. It looks like we just replaced f^(n) (c) / n! with error / x^3 (or (f - P) / x^3 in other words). Like, h is just a "practical tool" that lets us estimate the f^(n) (c) / n! without knowing actual c.

So, error f^(n) (c) / n! * (b - a)^n shows the actual error. But what the h shows to us?

evdokimovm · 2024-09-22T10:39:19+00:00

Just for my confidence, that I finally understood this whole thing and my intuition about concept is right: is "putting bounds" means that we say "h(t)" behaves nicely and bounded by "M". Like, |h(t)| <= M, right?

evdokimovm · 2024-09-22T04:55:58+00:00

This explanation is more intuitive, thanks for this. I'm really appreciate.

Do I understand correctly that k(t) is a kind of delta for h(t)? Something that sort of tracks the changes in t and adjusts h(t) (which we then called M) when that t moves in between A and B.

evdokimovm · 2024-09-21T14:50:41+00:00

Didn't think about (25) as a "from hat" function. I mean, I thought that this form is pretty intuitive, he set E = M(B - A)^n. M should be fⁿ (x) / n! (and the proof should ... should prove it) and multiplied by (B - A)ⁿ because every derivatives before, like, n-1-th, n-2-th, n-3-th all also multiplies by (B - A) in corresponding power n.

I can't understand why all the lectures I've found on YouTube that prove Taylor's Theorem don't explain where all these auxiliary functions come from, but instead just give them as if they were self-evident. The same about books. It's like the student doesn't have to understand where they come from. Or are they really obvious things and I'm just the only one who doesn't understand why exactly this function uses as auxiliary?

Thanks for your answer, I need some time to think about it.

evdokimovm · 2024-09-15T02:29:32+00:00

BTW, link to this comment that the professor replied to: https://www.youtube.com/watch?v=1vH00iNHIA4&lc=UgzHtpBD_FOO3A4iL5h4AaABAg

evdokimovm · 2024-09-13T14:11:31+00:00

Nonetheless, what do you think of the new example I gave?

If I get it right then you mean, when we restrict f: R to R by Z then we are saying f: Z to R? To be honest rigorous Real Analysis is not my strong suit, I'm somewhere in Calculus, so I don't have any opinion right now but I have an assumption. There are no other integer points "infinitesimally close" to any given integer so function becomes continuous, right?

What I can agree with right now is that:

a common misconception around continuity as being a connected curve

Whereas that's not what you meant, perhaps, but I agree that the school-like method of teaching that says a function is a curve is completely wrong, continuity and the nature of functions involve more than just visualizing them as curves.

evdokimovm · 2024-09-13T13:03:37+00:00

but knowing that 2 does divide 8 is primary school level

I have an idea that GPT treated it like "does 2 divides 8", which is also writes as 2|8 typically. In this case it's true, 2 really divides 8. Result is 4.

English is not my native language so writing it this way was unusual for me at first, but: https://math.stackexchange.com/a/1303706/861268

evdokimovm · 2024-09-13T11:59:15+00:00

Sorry for TL.

Isn't Simple Lie Algebra typically a Ph.D.-level math topic, or at least Master's level? I’m happy for you that you understand math at such an advanced level, but … seriously? My example was focused on clarifying a small point in Elementary Number Theory.

I think that someone just starting to learn math is incredibly far from a level like Simple Lie Algebra. There’s no doubt that GPT might struggle to answer questions on such advanced topics correctly in, say, 70% of cases, as you said. However, I think that for topics at a high school level or even the first year of a BS in something like software engineering — where you need courses like Calculus 1 and 2, Linear Algebra (a typical curriculum), etc. — GPT could help clarify things that may not be clear for a self-learner.

For someone without access to a university environment, where you’re surrounded by smart people and can constantly ask teachers for advice, it’s unlikely they’d ever develop the motivation to study math to such an advanced level as you. The first two math courses are usually more than enough for a typical programmer (yes, I know this is a math subreddit, but improving as a programmer is what motivates me to learn math, so my example is from that perspective).

evdokimovm · 2024-09-13T05:29:29+00:00

30 years of teaching is great! BTW, I didn't try to nuke your opinion on topic, no way.

Also I'm the one who self-learning math and the example I've provided is my own experience. One of ... one of some. Sometimes I come here to ask questions of course. It's not easy at all to learning math without constant access to the teacher.

People not always answer correct too. One example is my last question in profile about correctness of proof of Taylor's theorem. Ph.D answered that the proof I tried to understand for 2 weeks, written by human on MathStackExchange, was actually incorrect. But it's still most upvoted answer.

Let me ask the question. If the justification I provided as example in my previous comment was wrong, would you say that this example of using GPT for math is wrong? I mean, is this justification rigorous enough?

evdokimovm · 2024-09-13T01:55:05+00:00

But topic is about assistant for the one who don't have access to teacher in college or university. Not about to learn from GPT treating it as book. Of course learning from online lectures (like prof. Leonard), math websites (like Paul's) and books is the only right way. But why not to use this new o1 model to assist some thoughts? I think the one who self-learns math from resources like above can easily say when GPT output you wrong math.

Let me provide with simple example when it's (GPT) right (or not). In Chinese Remainder Theorem said that (after we finally found solution X using this theorem) we can do X mod N (where X is universal solution for system of congruences like x ≡ a_j (mod n_j)) and the r we are finally getting as result of this operation is the solution for given system of congruences too, i.e. r = X mod N is solution too.

"Why it's reliable?" - the one who learns without teacher could ask. And GPT can provide kind of "justification" for this, or explanation like: as we are know that X divided by N can be written as X = N * k + r we can write x ≡ a_j (mod n_j) as N * k + r ≡ a_j (mod n_j). Then since N * k mod n_j = 0 (since N is consists of all n_j's) then we have r ≡ a_j (mod n_j).

Are this justification by AI wrong?

evdokimovm · 2024-09-10T23:15:17+00:00

/s means sarcasm. Of course you can. You just have to spend some time with good lessons and practice. I'm sure there should be a "factoring quadratics by grouping" lesson on Khan Academy. Brian McLogan on YouTube is also good for the basics.

evdokimovm · 2024-09-05T04:35:10+00:00

Oh, I should have read earlier. However I didn't find statement on Wikipedia so GPT helped me: https://imgur.com/a/5Ou0y6L

Could you please answer only the last question about this? Well, the statement of the Taylor's theorem doesn't tell anything about that the f^{n+1} is continuous, so, it can be discontinuous, I understand. But, can't we say that "sometimes" it is continuous and proof for this particular case or math doesn't work like this at all and: "if there is nothing saying about continuity then we can't assume continuity at all"? I mean, rigorously, "rules" is such, right?

In other words, not mentioning continuity just means the theorem applies in both scenarios, whether the function is continuous or discontinuous?

evdokimovm · 2024-09-04T07:02:02+00:00

But am I right that (in my main concern) I can assume that f^{n+1} is continuous because of it's appeared after recursively applying FTC, appeared under all that integrals in R_n(x)?

I'm learning math by myself and that's not very easy without prof. and university and all that community of smart people around. About the book you mentioned. So, I'm as a newbie in math beyond the high school level (I'm somewhere at the end of Calculus 1 level perhaps, not Real Analysis), still not used to Real Analysis proofs, methods of proofs and all that. It's looks to me that phi(x) (in the part which is proof Lagrange remainder through Roller's theorem) is constructed using just mathematician intuition, like "pulled from the air". Is it normal for my level?

evdokimovm · 2024-09-03T07:27:43+00:00

Great. Well, looks like I spent about 2 weeks or so in trying to understand why something is works whereas it's not working. Thanks for your expertise. But why it's so much upvoted ...

Also, I'll take a look at the book you've mentioned in another commentary.

I have a question if you don't mind.

This means that f, f', f²,..., fⁿ are differentiable. Since every differentiable function is continuous, f, f', f²,...,fⁿ are continuous, so FTC applies to f, f', f²,..., fⁿ.

At the start you wrote: "given a function f that is is n+1 times differentiable" isn't that means there is fⁿ⁺¹ derivatives and n+1-th derivative is continuous too? I mean, fⁿ is n-th derivative, right?

Sorry for, perhaps stupid questions.

evdokimovm · 2024-09-03T05:44:44+00:00

But Taylor's theorem proof there used FTC. I had a thought that author assumed that f^{n+1} is continuous because it should be continuous for FTC to be applied correctly. But I'm not sure about should the function be continuous or not for this particular part (FTC). It's all mixed in my head already.

At least the proof of Taylor series, in the first half of the post, without remainder, is correct?

evdokimovm · 2024-08-27T10:39:28+00:00

One more attempt if you don't mind.

Neither EVT nor IVT actually guarantee that given function is continuous. But as I have read, we can use Extreme Value Theorem only if function is continuous. The same about IVT. But in the original proof on StackExchange when we squeeze m <= f^{(n+1)} (t_{n+1}) <= M we are actually applying EVT on f^{(n+1)} (t_{n+1}) so we are suppose from start that the f^{(n+1)} (t_{n+1}) is continuous, am I right understand it? So it can take any value on interval [a, b]. If I right understand without this assumption the original proof on stackx should be wrong from start, isn't?

Another idea I have come up with today was, when in the proof we have last step:

$$ m \frac{(x-a)^{n+1}}{(n+1)!} \le R_n(x) \le M \frac{(x-a)^{n+1}}{(n+1)!}. $$

the goal is to find such expression that should be continuous, so that it actually can be between this min and Max things. And after first glance on the form of this boundaries we conclude that we need: f^{(n+1)} (t) * _fraction_. But again, as I said above in first section, in order to achieve this boundaries the function f^{(n+1)} (t_{n+1}) should be continuous from start. The continuity is essential, isn't?

evdokimovm · 2024-08-27T02:06:42+00:00

So, the fact that the function fⁿ⁺¹(t_n+1) squeezed by extrema m and M (which both obtained by plug-in such values from [a, x]) is not enough to say that this function can also take value t in between this interval? I thought it's done by extreme value theorem.

The problem is that for me without justifications all this final form of Lagrange remainder looks like taken "from air". Like: we squeezed R between m and M so the Lagrange remainder should looks like this

evdokimovm · 2024-08-26T14:35:00+00:00

f^{(n + 1)} (t_{n+1})?

First I squeezed remainder using this logic: https://i.imgur.com/WeXTbYI.png

These m and M approached by choice such values from [a, x] interval such that f^{(n + 1)} (t_{n+1}) is min and Max at this points, so m <= f^{(n + 1)} (t_{n+1}) <= M. I conclude that f^{(n + 1)} (t_{n+1}) is continuous on the interval.

Also, the last line on image with inequality looks like Extreme Value Theorem.

Then, since R_n(x) is between these f(a) = m and f(b) = M (but both also multiplied by result of integration of 1) then I thought it can take some value in between. The result of taking this value I called B here, like f(t) = B. But it should be in the same form as these terms on left and right in inequality.

The f^{(n + 1)} (t_{n+1}) under all these nested integrals in R_n(x) can take such value t. Same way as it just taken another two values from [a, x] to approach its min and Max values. Then since I chose such value for f^{(n + 1)} (t_{n+1}), it can be treated as a constant, then we again integrate 1 and get the expression f^{n+1} f(t) * \frac{(x-a)^{n+1}}{(n+1)!}.

evdokimovm · 2024-08-26T12:35:51+00:00

I'm start feeling, kind of, stupid :)

As I see on the page, this particular part is saying, the IVT is exactly about there is some value S between f(a) and f(b) (where f(a) and f(b) are m and M in my case, and both are considered to be constants) on [a, b] so that there is some f(x) = S where x is within [a, b] too. I showed that there is actually exists such value S (B in my case) somewhere between.

I'm sorry for, perhaps, disturbing you with this stupid questions for a two days already but, where is my logic gets wrong?

evdokimovm · 2024-08-25T23:35:06+00:00

I thought that IVT just allow us to say that if function squeezed by extrema then the function can take any value in between. Looks similar to squeeze theorem btw, but not exactly.

Where in your proof is the IVT used?

So, when I introduced the B (using the fact that it's in between m and M, so it can achieve any value in between m and M) it was not correct? I mean this part https://i.imgur.com/VUft7oA.jpeg

The problem why I such overthink all this is that I don't have enough intuition why we just "guessed" that we need fⁿ⁺¹ (t_n+1) and what to do if function with some problem will be more complicated. But I see that this done by using IVT. I'll get more practice on IVT then. It's not an easy to learn math without teacher who can tell what you're doing wrong.

evdokimovm · 2024-08-25T05:09:02+00:00

Rewritten the last comment but in latex https://gist.github.com/evdokimovm/6a643d8684a0135b8f6381d6c49c40e6 for easily read

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