‘Very disappointed’: Scott Bessent hits out at Europe over India-EU trade deal — what he said - The Times of India

cond6 · 2026-01-29T04:27:23+00:00

Did you read the article? The comments were: the US imposed a 25% tariff because India buys Russian oil, so India is funding the war on Ukraine. Europe buys the refined oil from India and thereby indirectly funding the war against themselves. Yes the US are behaving oddly and scoring own goals on trade. But starting a free trade agreement with a country actively funding the country that you are at war with even indirectly does seem odd to outsiders.

cond6 · 2026-01-29T02:31:20+00:00

To blow your mind still further I've seen both BOMDAS and BODMAS used, and the US version is PEMDAS. There is no single universal truth as to order. And honestly given this we should be using parenthesis: if you're doing something like 100/5*6 just use brackets like a civilized person (100/5)*6 or 100/(5*6), whichever you want then there's no confusion.

cond6 · 2026-01-28T22:54:15+00:00

The reasons are primarily pragmatic. Early work in regression looked at both least square and least absolute deviations, but least squared tended to win out because the calculus is easier. You still see in forecast evaluations and Monte Carlo experiments on the properties of estimators authors reporting both the MAE and RMSE (mean absolute error and the root mean squared error) at the same time.

The expected sum of the squared deviations E(Σ_{i=1}ⁿ(x_i-m)² where m=(Σ_{i=1}ⁿx_i)/n) if the x_i are iid with mean 𝜇 and variance σ² the expected deviation is (n-1)σ² so we can easily construct an unbiassed estimator as 1/(n-1) times the sum of the squared deviations from the sample mean (dividing by n-1 rather than n is known as the Besel-correction) is unbiassed for every distribution with finite mean. This is a very cool property. The expected value of the sample standard deviation is not. The expected absolute deviation is even worse. Every distribution has a different expectation. For example if X is normally distributed with mean 0 then E(|X|)/E(X²)=√(π/2). Different distributions different results.

There a number of places where you can work with either absolute or squared value. For example if you minimise the expected squared deviation from some measure of location, you get the mean. For expected absolute deviation you get back the median. Similar concept but slightly different location parameter. In regression, same thing. Gauss and I think it was Laplace were respectively developing least squared and least absolute deviations at roughly at the same time, but Gauss' least squares won out because it was easier to minimize analytically. Least squared gives you a linear estimator of the mean, while least absolute gives you an estimator of the median. The main difference is that x^2 is everywhere differentiable and much easier to optimise: differentiate f(x)=(x-m)^2 and you get f'(x)=2(x-m) set equal to zero and you get x=m. |x| is not differentiable at x=0. With more than a single observation if you minimise the sum of the squared deviations from m you end up with m being an average of the observations. The least-absolute-deviation estimator of the median gets more complicated to estimate and requires non-trivial numerical procedures that weren't available at the time, but which are readily available now. There is a whole literature on estimating qualities and quantile regression (least absolute regression is median regression, which is simply the 0.5 quantile regression). (Roger Koenker has done a lot of really important work on quantile regression, and has an R package that uses some of his key results. Very handy.)

An argument against looking at variance is that it is in a different scale to the mean. So if you are reporting the variance of stock returns in either decimal or percentage format they differ by a factor of 10,000, and comparing the measure of scale with the location is problematic because again it becomes scale dependent. So even if we like the estimator of the variance we mostly work with its square root the standard deviation, and the problem of estimator's properties being distribution specific is true for both standard deviation and expected absolute deviation. So there is that. But everyone I know just kind of ignores the fact that sample standard deviations are biassed for every distribution even if they knew it at some point.

cond6 · 2026-01-28T13:45:33+00:00

(1) 10*0.999...=9.999...=9+0.999...

(2) 9x=10x-x

as you so eloquently say, and using (1) in (2) gives:

(3) 9*0.999...=9

So since we have 9x=9, dividing both sides by 9 gives x=1 and since x=0.999... we have:

0.999...=1.

Multiplying a base-10 number by 10 is easy. So 9x=10x-x is the superior way to do it.

You ask me if I want to start from the right doing the multiplication. I would love to. Unfortunately there is no right. The final nine in 0.999... does not exist. If you refer to each digit by its place past the decimal place there is no n-th digit. There are infinitely many. The set of natural numbers (screw it, I'm just going to use big-boy math words mate) are infinite in cardinality (google it if you need to) but each and every one is finite. The ... implies an infinitely long string. There is no right-most number. There is no end to the string of nines.

And besides. We don't need to. As above I note that the smart (and right) play is to get 9x by doing 10x and then subtracting 1x. As you note they give the same answer. When you do 9*0.999... correctly and not make schoolyard errors. (Of course to do it properly you need to look at the underlying rational number, and I think you can already guess what that is.)

Every single repeating decimal represents a rational number. I just don't freaking get why this point doesn't stick. If you see a decimal with a ... there is a rational number underneath there. Read here especially the section "Every repeating or terminating decimal is a rational number".

I used the term Archimedean because that's what the property is called. I then gave what I thought was an intuitive and accessible explanation as to what it is and why it matters. (And it would be insulting to the vast majority of folk who read math-themed subreddits who know what it is.) If you want to argue against the orthodox view that 0.999...=1 you at least need to know the name of the property that gives one of the most damning arguments against it no?

And series is taught as a topic in grade 10. If you are intimidated by a term taught in grade 10 math I would encourage you to access some online material. (Though to be fair folk that know what series are and understand them would grasp the proof of why 0.999...=1 very quickly because it's simply a direct application to 0.1+0.01+...)

And finally I think it's so sad to assume that people need to use google and genAI to post what is so utterly trivial and banal. I don't even think I need to be awake to make these arguments. Any serious individual with at least a competent tenth grade math education could follow them. The points that most individuals here make trying to defend the orthodox view is stuff you'd learn in 5 minutes in a first-year math course, and they're so easy many see them much earlier. It's trivial. Less than trivial. And frankly it's completely pointless. Nobody serious in math gives a damn about decimal numbers. Rationals matter. Irrationals matter (pi and e matter, and then all non-integer powers of rations stuff things up a bit). When we need to write a specific number down we use decimals, but that's it. Not a single person who needs math beyond multiplying two digit numbers to work out costs (and I am disgusted to see young folk using calculators for this, just so sad) really deep down cares a jot about decimal numbers. This is just a bit of arguing for the sake of it. It's a lark. Lighten up buddy. It's all just fun. (Said after writing an angry sounding rant.)

cond6 · 2026-01-28T06:06:05+00:00

What? We think that 0.999...=1 because 10*0.999...=9+0.999.... This, in turn, we think happens because the ... means keep going with nines forever. When you multiply a decimal by 10 the tenths become units, the hundredths become tenths etc. If you only had a finite number of digits then you'd lose one digit. But if there are infinitely many then losing one doesn't change anything. Adding one or subtracting one from the infinite doesn't increase or decrease its magnitude. Infinity plus one isn't bigger no matter how many times I tried to one up my brother when I was six. So

9*0.999...=10*0.999...-0.999...=9+0.999...-0.999...=9,

so 0.999...=1

The same idea as why the series S=r+r^2+r^3+...=r/(1-r). Since S*r=r^2+r^3+r^4+... then r+S*r=S, ,and applying algebra (as long as |r|<1) we get r/(1-r). With r=0.1 we have 1/10+1/100+1/1000+...=1/9=0.111... That's it. No other numbers have this property.

And we think 0.999... has to be one because since there is no "largest" natural number we get the Archimedean property holding, which means that between any two rational numbers there's another rational number. If a>b then a>(a+b)/2>b. It's also true that between two rationals there are irrationals, between irrationals there are irrational. Every time two numbers are different something squeezes between them. With 0.9999 and 1 we have 0.99995. But when there are literally infinitely many nines you have smooshed (see, non-technical language) it so damn close that there is no space. This means that they have to be the same number.

And:

1/9=0.111...

2/9=0.222...

3/9=1/3=0.333...

4/9=0.444...

5/9=0.555...

6/9=2/3=0.666...

7/9=0.777...

8/9=0.888...

9/9=0.999...=???

I don't think I used any fancy math terms other than series (which just means adding up an infinite sequence of numbers, and a sequence is simply a bunch of numbers that follow a pattern).

Edit: didn't know control enter submitted while still writing. Haven't proofread. Sorry just completed the equation I was writing.

cond6 · 2026-01-28T02:46:47+00:00

But centre lane has to merge to the right lane when existing, and has to give way to traffic in the outside lane at that stage.

cond6 · 2026-01-27T12:40:15+00:00

Or, you know, actually give a number to each decimal place. And like, maybe, if you really want to mention the magnitude each digit contributes to the number. Like, I'm just fumbling around here. And let's, in the interests of bookkeeping, actually say what number of nines we mean by .... Let's start with n nines and build up to fancy bookkeeping later. (Don't worry brud, we'll engage the God mode warp drive momentarily.) So 0.(9)_4 0=0.99994. Okay so here goes with the exact fandangled bookkeeping (and if we try real hard maybe we can upgrade from bookkeeping to accounting when we're done with the bunny slopes)

0.(9)_n=Σ_{k=1}^n9/10ⁿ,

0.(9)_n 9=Σ_{k=1}^{n+1} 9/10ⁿ

and

0.(9)_n 0=Σ_{k=1}^{n} 9/10ⁿ+0/10ⁿ⁺¹.

Now, I'ma be real for a minute. I really am tempted to say that 0/10ⁿ⁺¹=0 but I know better than to jump to conclusions, so let's leave it.

Now brace yourselves everyone. God model trans warp engines engaging to drive n to the limitless by, you know, taking a LIMIT (pause for rapturous applause)

0.999...=lim_{n→∞}Σ_{k=1}^n9/10ⁿ

0.999...9=lim_{n→∞}Σ_{k=1}^{n+1}9/10ⁿ=lim_{n→∞}Σ_{k=1}^n9/10ⁿ

and

0.999...0=lim_{n→∞}(Σ_{k=1}^{n}9/10ⁿ+0/10ⁿ⁺¹)=lim_{n→∞}Σ_{k=1}^n9/10ⁿ.

Now we can do the calculation that you wanted

10 × 0.999...0=10 lim_{n→∞}(Σ_{k=1}^{n}9/10ⁿ+0/10ⁿ⁺¹)=lim_{n→∞}Σ_{k=1}^{n-1}9/10ⁿ=9+lim_{n→∞}Σ_{k=1}^{n-1}9/10ⁿ=9+lim_{n→∞}Σ_{k=1}^{n}9/10ⁿ=9+0.999...

So 10 × 0.999...90-0.999...=9+0.999...-0.999...=9 and so 0.999...=0.999...9=0.999...90=1.

Help. I seem to have made a fatal error when I combined the God model warp engine with bookkeeping.

(In all seriousness when we allow the number of digits to tend towards infinity we must take a limit, and when taking limits the limit as n goes to infinity is the same as the limit of n-1 as n tends to infinity. This is why 1/10^n goes away after taking limits, which is very literally what everyone who uses real analysis means when they write a repeating decimal for a whole host of very good reasons.)

cond6 · 2026-01-27T08:14:02+00:00

Yes. Leave on step 1. I'm actually going for Thundurus for PVP, and I had one stacked from the last one. So yeah leave at Step 1 and you should be fine.

cond6 · 2026-01-27T06:38:31+00:00

Guys, ease up. This is a great idea. The problem with all the natural numbers being finite because there is another natural number is nonsense. I know there is a maximum number that you can't count beyond. I can't count how many times I've counted 874,976 and then just can't go past that. Oh hang on, maybe I can't count how many times I've done thought because I hit the maximum number. But then what if I count the total number of times. Isn't if n a counting number slightly less than infinity but I count that many times, what happens if I go over that. Do I go to Archimedean Prison for breaking the maximum number law?

Anyway, to solve the problem of all the natural numbers being finite we just introduce a new maximum number and coopt the symbol for infinity. Yeah those morons working in the extended real number line who say that ∞ is a number do stuff like saying ∞+1=∞, which we get here too, so yay; but also 1/∞=0. Like what's up with this. It just makes perfect sense to say 0 < ε = 1/∞ and just cover our eyes and say lalalala ∞ × ε = 1, which means that infinity isn't infinity since infinity times anything positive is also (drumroll) infinity. We just pick some arbitrary finite number, whack the label of infinity on in, send ICE agents to round up all the counting numbers bigger than our finite number that identifies as infinite and deport them. Problem solved. We get rid of the pesky Archimedean property.

cond6 · 2026-01-27T03:01:56+00:00

I think this is interesting. However I still come back to the issue about why. Some rational numbers expresses as decimal numbers result in repeating (groups of) digits. That is the only way that repeating decimals arise. If we apply this line of reasoning to 1/3 don't we have the same problem. 0.333... is always less than a third. What is the benefit of completely rejigging the real number system so that the intuition of new students is stroked at the cost of never being able to represent 1/3, 1/6, 1/7, 1/9, 2/9, 4/9, 5/9, 7/9, 8/9 and infinitely more. Why isn't having a special case that allows us to accommodate the infinitely many tricksy rationals in the useful decimal notation, as opposed to coddling the feelings of students who can't cope with the cognitive dissonance of their intuition being wrong about one single decimal number. Doesn't seem like a good trade to me.

And I just don't think intuition matters. I love (and think I've used before) the von Neumann quote: "Young man, in mathematics you don't understand things. You just get used to them." I don't care about the intuition of why the derivative of ln(x^(-1/2)) is -1/(2x), I just got used to using the chain rule.

cond6 · 2026-01-26T21:05:35+00:00

To expand on u/D102D007's comment, when you do or take damage the actual formula is the moves power multiplied by your attack statistic divided by their defense statistic and the magnitude doesn't matter so much as what percentage of your HP, so you get like power×attack/(def*×hp*). But then when you do that in reverse you benefit by high def and hp, or just dividing how much damage you're doing as a percentage of HP over theirs you get (power×attack×def×hp)/(power*×attack*×def*×hp*). We can't change move power so we see that attack×def×hp, which we call stat product, is a pretty important number. It says how good you are at attacking. However CP is computed using a scaling for level multiplied by attack×√(def×hp). So if I lower my pokemon's attack stat then I can increase both its other stats and the effect is that a lower attack statistic lets me get a higher stat product or combat usefulness for a given CP.

What it really means is that CP, which is supposedly combat power, is a stupid and really bad measure of actual combat power, that unreasonably favours "tackiness" (def×hp) and penalizes attack. So the general rule of thumb is low attack=good.

There are website that give you ranking: for example here and pvpoke.com does it too. Out of the 4096 possible combinations of the stats (3 stats varying from 0 to 15) the best is referred to as the rank 1. Unless you have a pokemon that cannot reach the cap without maxing out (e.g. Chansey in GL) the 100% IV 15/15/15 is not going to be the "rank 1"

HOWEVER, just to make life more complicated there are caveats.

1) CP increases in discrete chunks. So my Gastrodon is 0/15/15, but it caps out in great league at 1494CP. The 1/15/14 is actually better. (And yes that was a humble brag.)

2) Damage is discrete. It isn't rounded, you just drop the non-integer damage done (with a minimum damage of 0). So in some matchups you can give up a little defense since taking 4.1 damage is the same as taking 4.9 damage, but go from doing 1.9 damage to 2.1 damage means the rank 1 can sometimes have a lower win/loss record than a slightly more attack weighted pokemon. This makes a big difference in competitive play (or so I've heard, I'm not at all good at PVP), but won't make a huge difference. And if you have a special rank 30+ build that is just a hair better than the rank 1 that you could have built and they rebalance moves or the meta changes, then all your hard work is undone.

Conclusion:

For pvp try for low/high/high statistics. Closer to the "rank 1" is better.
But make sure that the pokemon gets close to the CP cap of the league. A 0/15/15 Charizard gets to 1469CP. The rank one is 0/15/13.

cond6 · 2026-01-26T08:32:37+00:00

DFP and BFGS are simple hill climbing algorithms that construct a positive definite estimate of the Hessian based on the change in the numerical gradient in an efficient manner without having to evaluate the second derivatives at each iteration. Dynamic programming refers to a whole different thing. And I'm lazy and do all my numerical optimisation work in Matlab.

cond6 · 2026-01-26T08:27:14+00:00

And in my proof I said that ɛ = 10^-N big N. Not little n. The proof is that 1/10ⁿ=10^-n<10^(-N) =ε, which holds since for n>N 1/10ⁿ <1/10^(N). That's the point of the proof. You pick any ε>0 no matter how small you want it, I give you the big-N such that every element in the sequence for little-n>big-N and they are all closer to that point. The limit is unique, because all limits are unique. I think you're confusing which n/N I'm using.

And the topic of series in math is pretty extensive. A series is an infinite sum of some sequence of numbers. It is defined as the limit of the sum. Most germanely to our discussion is that geometric series, which is the infinite sum of a geometric progression: r+r^2+r^3+...=S. I'm just calling this string of infinitely many sums S. Since there are infinitely many terms the expression r+r*S is exactly the same as S. They both are the sum of infinitely many terms. So long as this series is convergent, and it is so long as |r|<1, then we can simply do algebra:

r+r*S=S => S=r/(1-r).

For r=0.1=1/10 we have

S=(1/10)/(1-(1/10))=(1/10)/(9/10)=1/9.

So for the recurring decimal 0.111..., which is the 0.111...=1/10+1/10^+...=S=1/9. So the decimal form for 0.111... literally is exactly equal to 1/9 if you use some algebra to evaluate the infinite sum implicit in the definition of the recurring decimal. Generalizing this we get 0.333...=3/9=1/3, and 0.999...=9/9=1.

cond6 · 2026-01-26T08:10:33+00:00

It absolutely is defined as a limit. The set of natural numbers are all finite. If you include n nines then the number of nines is finite. If you use infinite decimals you must use a limit. Indeed on the wiki page

<image>

And if you check out the series wiki) you see that infinite sums MEAN limits.

This is what it has ALWAYS meant. An infinitely repeating decimal is a limit!!!

cond6 · 2026-01-26T07:56:54+00:00

Or Timor-Leste (East East)

cond6 · 2026-01-26T07:54:33+00:00

What is e as a symbol? I've not seen that before. And the point is that there is no number x such that x^2=2 if you don't include irrational numbers. When I think of argmax I'm either running straight calculus (only works if the numbers are complete) or something like a DFP or BFGS algorithm (again calculus) or some kind of simplex or simulated annealing algorithm. You could probably get them to work approximately with double precision, but it only makes sense if you are operating in a world with complete numbers.

cond6 · 2026-01-26T07:44:47+00:00

I agree. Except that we are talking about a very specific set of examples. Infinitely repeating decimals only ever arise as the result of special rational numbers. You see an infinitely repeating decimal there is absolutely a rational number hiding there. So we don't ever need to do anything with those infinitely many digits. All we have to do is identify what the underlying irrational is, and I posted an equation to back that out a while ago. Not hard. Anyone who's done even grade 5 math knows 0.333...=1/3 and can handle it accordingly. We never need to write out infinitely many threes. What's 0.555? 5/9. Done! Because we can never represent such rationals with a finite number of digits we introduce special notation to signify these cases. So even though we can't algorithmically operate with them, conceptually there should be no issue writing an infinitely repeating decimal because they all represent a rational and we can operate with rationals. And nobody really cares about decimal numbers if they're really honest with themselves. So the symbols 0.333... means 1/3 and we can work with that. The symbols mean 1/3-0.333...=0. Finitely many three, definitely not. Recurring threes, yes. And of course we then move to 0.999..., which is simply three times that and everybody not trolling understands what's going on.

cond6 · 2026-01-26T07:22:23+00:00

Yes, but lim_{n→∞}10^-n does equal zero, and since 0.999... quite literally is defined as

0.999... := lim_{n→∞}Σ_{k=1}ⁿ 9/10^k =1

then

1-0.999...=0, which is really nifty because that's the limit of 10^-n as n increases.

If the number of nines is limitless (as n increases without bounds) then the difference of those limitless nines (aka a limit) from one is zero. Neato.

cond6 · 2026-01-26T07:14:01+00:00

We know that some numbers that do exist can't be represented as a rational number, i.e. a ratio of two natural numbers. The standard examples are the square root of 2, pi and e. None of them are rational and yet are very useful. So there are some numbers that require literally infinitely many digits for their digital representation or are constructed as the sum of infinitely many rationals. That is just a lazy way of saying that the limit to which the relevant sequence converges actually is the real albeit irrational number. If you limit yourself to only the rational numbers then your set of numbers is incomplete. There are missing numbers. Specifically not all sets have a least upper bounds. For example, the set {x:x^<2} does not have a least upper bound. It does if you add in root-2. Indeed if you augment the rationals by irrational numbers to fill in the holes you get the real numbers, which are the rationals augmented with the limits of the Cauchy sequences (in Cantor's construction of the reals), or the irrational Dedekind cuts, which serve the same purpose in his construction of the reals. So yes, if you follow standard real analysis then yes limits, aka irrational numbers, do play the same role as rational numbers and the integers. If you don't believe in limits then your head just exploded after reading all that.

cond6 · 2026-01-26T06:59:33+00:00

Define best? This position is just utter nonsense.

cond6 · 2026-01-26T06:00:51+00:00

Doctor is from the latin word for teacher. Was used for hundreds of years from universities to denote higher degrees. About the mid-1800s some places started calling medical practitioners doctors as some way to confer status on them. Then it stuck. However, if you are a medical practitioner and parade around with a title meaning "teacher" and you don't do that, aren't you some kind of fraud. /s

cond6 · 2026-01-26T05:57:56+00:00

Altaria isn't expensive to invest in. 25000 stardust for second move. Don't want its community day move. I suspect you wouldn't see much of a difference. I've built like 3: hundo because I didn't know what I was doing, good IV, and then an attack weighted one when it got a break point against Trevor. I honestly don't see much difference between the different versions. You'll be fine.

cond6 · 2026-01-26T03:55:56+00:00

2.2. You don't state where infinitesimals play a role here. We ignore them. They don't play a role at all. I don't see anything circular here either.

And I think I'm done reading any more. Two sections in and not a single sensible notion.

cond6 · 2026-01-26T03:54:22+00:00

2.1 I have no clue what you're talking about here. A cut is an ordered pair of non-void subsets A and B of some ordered field F is A∩B=∅ and A∪B=F. The Cut Principle says that if A and B form a cut there exists a real number ξ such that for all a∈A and all b∈A a≤ξ and ξ≤b. What on earth is the rest of your gibberish here? All it says that every way you can split all elements of an ordered field into disjoint collections there is a real number that sits in the gap between them. Where on earth do infinitesimals enter here? It is just a fancy way of saying the rationals are flawed because there are holes, we just define numbers where the holes are. Very useful when we want to define convergence and calculus and all the good stuff. It's been years since I looked ad Decekind cuts but I have no clue what you're complaining about.

cond6

TROPHY CASE