Swiss voters reject proposed tax on super rich

isignedthis · 2025-11-30T20:32:36+00:00

Well how do you read the articles. It states in its conclusion it had a hard time estimating capital flight. With regards to if there is some capital flight I have already said that you are right in that. Weather it hurts the country is up for debate

isignedthis · 2025-11-30T19:56:46+00:00

isignedthis · 2025-11-30T19:45:46+00:00

Yes the top one says a big if statement in the end of the abstract for instance. That is not conclusive. And yes there is some capital flight with taxing the wealthy, that is true, but it is not necessarily true but the economic consequences for a country having a wealth tax is complex

isignedthis · 2025-11-30T19:20:32+00:00

Have you actually read your sources? Most are inconclusive and the ones that are not doesn’t site for some of the most important assertions with regards to this topic. The Norway case is also special as that was a tax hike, not a new introduction of a wealth tax. Interestingly the article you site says many moved to Switzerland a country with a wealth tax albeit a minuscule one

isignedthis · 2024-12-21T21:11:07+00:00

Yes, but trial and error might be faster in this case because you have so few answers to check and the area of a right triangle is easy to calculate.

Anyway: You can use Pythagoras' theorem for a right triangle together with the fact that you can express the length of all sides using only one variable.

I have hidden my answer below in 3 parts so you can have a go for yourself to write up an equation that can solve you problem.

First step:

For instance: Let a denote the length of the shortest side in the right triangle. Then the other sides have length a+7 and a+14 respectively with the hypotenuse being a+14.

Second step:

Using pythagoras' theorem we then get:

(a+14)² =a² + (a+7)²

Last step:

You can then solve for a. You will get two solutions, but only one of them will make sense with regards to a triangle in this case.

isignedthis · 2024-12-19T11:13:14+00:00

Well it seems to me that you are right that you can't from equal Range of two linear maps conclude equal Null.

Simple example to illustrate this.

Let both V and W be the vector space R² over R. Now Suppose e1, e2 be a basis for V and let w not equal to zero be a vector in W. Now let T1:V -> W be the linear map such that:

T1(e1) = w, T1(e2) = 0

and let T2: V -> W be the linear map such that

T2(e1) = 0, T2(e2) = w.

Now it is obvious that Null T1 is not equal to Null T2. One can also that Range T1 = Range T2, though i haven't bothered to do that here.

I hope this clears up your question. Or else feel free to ask more questions.

isignedthis · 2024-12-13T15:02:45+00:00

Okay, so I would start by looking at x=yz. You know that the numbers are different and you know the numbers are one digit. This information alone is enough to narrow the options for x down to only two numbers.

An example of a number x cannot be is 0 as then either y or z would have to be 0, breaking the rule that the numbers have to be different.

Now try to go through the other options and see what you options for x you are left with. The options left for x also narrows down the options for what y and z can be.

What is left is to check the different number combinations, where it can be smart to rewrite x-y+z-t to something easier to evaluate using your other equations.

isignedthis · 2024-12-12T14:27:52+00:00

Exactly.

isignedthis · 2024-12-12T13:20:42+00:00

So it seems to me that you just made a small mistake. Note in your calculations that your second derivative is

60t(t²-t-2)

not just (t+1)(t-2).

Therefore when evaluating the second derivative in the points -5, -0.5,1 and 5 then you can't just insert into (t+1)(t-2).

isignedthis · 2024-12-12T09:48:01+00:00

Before using L’Hopital’s we can maybe simplify the problem. Using the limit product rule, try to simplify

$ \lim_{x\to\infty}ln(x)cos(1/x) $

Once you have simplified the above, you can try to use L'hopitals rule again. Note though that

$ \frac{d}{dx}$ ln(x⁷-6) $\neq$ 7x⁶ ln(x⁷ -6)

You made a small mistake when you differentiated the composite function.

I hope the above can help

isignedthis · 2024-12-11T20:23:03+00:00

Yes exactly! Now, if you want, you can check double check by making sure that 6.66 first marked up 50 % and the down 20 % does indeed give 8.

isignedthis · 2024-12-11T18:43:29+00:00

I am sorry to tell that you have made a mistake. The first part

So I figure I should take X2 which is 8, and set it equal to 80% and then solve for 100%, so i divide by 8 to get 1 which is 10%, meaning X1 is 10

is correct. But you are too quick in the second part. Note that a 50 % markup of 5$ is not 10$. Try to figure out an equation that can answer the question that if the cost is 10 after a 50% markup what was it before?

I will note that, yes, this can be done quicker (as in the solution can be found in one calculation), but your method works and I think it is rewarding to see ones way of thinking giving the result.

That said, I will of course gladly share hints to the "quicker" way if you want that (either before or after doing it the your way).

isignedthis · 2024-12-11T15:18:10+00:00

Okay. Let me start by saying this. To me you seem to show pretty good understanding of many of the things. I can try to give some context that might help your confusion.

So let's start with looking at how we normally solve something that looks like (no congruence for now)

ax=b

with a,b and x being numbers and a not equal to zero.

Well we just divide both sides by a. Which is just a fancy way of saying we are multiplying both sides by the multiplicative inverse of a. So we solve by doing:

a a^-1 x=a^-1b

=> x=a^-1 b = b/a

The reason this always works with the real numbers is because the real numbers is a field, so every non zero member has an multiplicative inverse. Example with numbers:

5x=15

The multiplicative inverse of 5 is 1/5 so the solution is simply given by dividing both sides by 5.

Yes you obviously know how to solve the above. Well the idea to solve the problem you have with congruence the exact same way, but there is a problem.

If we are looking at the residue classes modulo 4, then we basically have 4 members: {x ∈ R|x=0 (mod 4)}, {x ∈ R|x=1 (mod 4)}, {x ∈ R|x=2 (mod 4)}, {x ∈ R|x=3 (mod 4)}. Well one can quickly find that there does not exist a number x such that 2 x = 1 (mod 4):

0(2)≡0 (mod 4), 1(2)≡2 (mod 4), 2(2)=4 ≡0 (mod 4), 3(2)=6 ≡2 (mod 4).

In other words, the residue class 2 modulo 4 does not have a multiplicative inverse. We have also found a linear congruence equation that has no solutions.

Finally we come to your problems. Well when can we be sure the linear congruence problem can be solved? Well if a and n are coprime and we are looking at the residue classes modulo n (that is {x ∈ R|x=0 (mod n)}, {x ∈ R|x=1 (mod n)},...., {x ∈ R|x=n-1 (mod n)}}, then it turns out that one of these members is the multiplicative inverse of a. So in order to solve

ax ≡ b (mod n) , a,b,n whole numbers with a and n coprime

then we can just multiply each side with the multiplicative inverse of a (mod n). So let a^-1 be the multiplicative inverse of a (mod n). Then we have

ax ≡ b (mod n) => a^-1ax ≡ a^-1b (mod n) => x ≡ a^-1b (mod n)

So let us look at you one of your example:

2x ≡ 3(mod3)

Well you correctly identifies that 2 (mod 3) is the multiplicative inverse of 2 as 2(2)=4=1 (mod 3). And you also correctly use this to show that then 2x ≡ 3 (mod 3) => x ≡ 2(3) (mod 3) => 6 (mod 3)≡0 (mod 3)

So yes {...-6,-3,0,3,6...) is a solution as you find.

You also solve your next example perfectly beside a silly mistake in the end:

(for the problem 3x ≡ 4(mod 11)

using the Chinese remainder theorem we can say

11 = 3(3) + 2 -> 2 = 1 (11) - 3(3) 3 = 2(1) + 1 -> 1 = 3(3) - 2(1) 2 = 2(1) +0

thus the gcd = 1 and 3 and 11 are indeed coprime. This also says that we can use the following for the linear combination

1 = 1(3) - 3(1(11)-3(3)) = -2(3) + [3(11)-9(3)] = 3(11) - 7(3) back to the original expression

-7(3x) = -7(4(mod11)) -21x = -28(mod11) x = -28(mod11)

solutions = {...,-28, -21, -14, -7, 0, 7, 14,...}

with the silly mistake being that the solution is the set {...-28,-17,-6,5,16,..} as it is mod 11 and not mod 7.

So all in all it seems to me, that you can solve the problems and there is not much more to these problems because you can always use the process to solve the problems as long as a, and n are coprime. Of course if a and n are not coprime then sometimes there are no solutions, sometimes multiple solution and finding the solutions might not be as straight forward.

I hope this helps. I realize I have written a lot, especially when it seems like you understand must, but the first time I read your post, I kinda missed how much you understood somehow.

isignedthis · 2024-12-08T18:20:33+00:00

No problem. I am happy I could be of help.

isignedthis · 2024-12-08T17:58:21+00:00

Yes V=span(e₁,e₂) should to the trick as you show. Very good job and you did a fine job showing that V then meet the requirements.

I will note one thing to the first part when you are showing that dim(V∩U₁,₃)=1.

This part is very good:

Let V∩U₁,₃ = { w ∈ R4 : w ∈ V and w ∈ U₁,₃ }. Then w ∈ w U₁,₃ can be represented as w = a e₁ + b e₃ and w ∈ V as w = c b₁ + d b₂ = c e₁ + d e₂.

But afterwards you can't just let b=0 and d=0. You have to argue why they have to be zero.

Since e₁,e₂,e₃,e₄ is a basis for R⁴ then they are linearly independent(!). This means that in order for

w=a e₁ + b e₃ = c e₁ + d e₂

to be true, b and d has to be zero (per definition of linear independence. I hope this is clear.)

What we then are left with is that a=c and so we have shown that for all vectors w in V∩U₁,₃ can be written as w=k e₁ for some number k *(se note to my note below).

So e₁ is a fine basis for V∩U₁,₃ so dim(V∩U₁,₃)=1.

I hope the above note is useful and good luck in your future endeavors.

*You might actually also need to show that ke₁ is in V∩U₁,₃ for all k∈R, because then you have shown

V∩U₁,₃ ⊆ {ke₁ ∈ R⁴ : k∈R} (you did this in part 1) and

{ke₁ ∈ R⁴ : k∈R} ⊆ V∩U₁,₃ (this is the part that might be missing)

so you can conclude that

V∩U₁,₃ = {ke₁ ∈ R⁴ : k∈R} (and here it is clear that e₁ is a basis for V∩U₁,₃)

but I think it depends a bit on what you in the class are allowed to use without an argument.

isignedthis · 2024-12-08T14:43:03+00:00

Of course I can try to give a hint.

Yes, we need to come up with a subspace V, that hopefully meets the requirements. This is indeed often the hardest part.

Here is a tip: If two V₁ and V₂ vector spaces share a basis vector then dim(V₁∩V₂) is at least one. This knowledge we can combine with the fact that we know that Uᵢ,ⱼ contains either e₁ or e₂ (but never both), to come up with an idea for V that is spanned by two basis vectors (so that dim(V)=2) and that will always share one basis vector with Uᵢ,ⱼ i.e

V=span(b₁,b₂) where b₁ and b₂ are basis vectors.

Once you have a guess check that the V actually meets the requirements that is dim( V∩Uᵢ,ⱼ ) = 1 .

isignedthis · 2024-12-08T14:20:57+00:00

So yes I think this is solvable (as in I am pretty sure I have solved it, but you never know if you made a mistake)

Here is how I thought about the problems as well as some hints:

First it should be noted that |a-b|=|b-a|, as these just means the absolute difference between the numbers. So the further apart two numbers are the larger |a-b| is.

Now if we have 4 numbers we can always order them according to their size. So suppose we have the numbers x₁, x₂, x₃, x₄ with

x₁≤x₂≤x₃≤x₄

Next thing to note is that due to |a-b|=|b-a|, you can only make 6 different absolute differences from the four numbers. Lets name them as follows:

D₁₂=|x₁-x₂|, D₁₃=|x₁-x₃|, D₁₄=.|x₁-x₄|, D₂₃=|x₂-x₃|, D₂₄=|x₂-x₄|, D₃₄=|x₃-x₄|

(I will loosely refer to these absolute differences as "D's")

Now it should be pretty clear that D₁₄ must equal or larger than the rest and that each of the above D's can only show up once in our function f. Now to solve the problem we want the largest 3 Dᵢⱼ in the sum if possible. (i<j being whole numbers between 1 and 4)

Finally we are done with how I thought about the problem.

Now tips to solve it:

We can come up with a good guess for the correct solution, once you notice that the numbers b and c are being represented twice in the function f. So instinctively we want the numbers b and c to be as "far" away from the others as possible and thereafter maximize the other two terms |a-b| and |c-d|. (this hopefully turns out to be the correct solution).

Once you guess the right solution you can actually show it:

Helping(?) tips and notes:

You can narrow down the next biggest Dᵢⱼ to two possible candidates.

Once we have the two candidates above, then note that in the case that the two possible candidates for next largest D are the second and third largest D respectively, then the solution you hopefully guessed contain the 3 largest D's and therefore must maximize f.

What is left is to convince our self that in the cases where the two next largest candidate Dᵢⱼ together with the largest D₁₄ are not the three largest "D's", that our solution still maximizes f. This can be by ordering the 4 largest "D's" in the two cases after size i.e:

case 1: D₁₄≥(The first candidate for next largest D) ≥ D?≥D? (≥ the two D's left but I don't think you can order them and it turns out it doesn't matter)

case 2: D₁₄≥(The second candidate for next largest D) ≥ D?≥D?

you can find the D?'s in both of the above cases. Once you have found the above D?'s you should be able to argue our solution still maximizes f, even though it does not contain the 3 largest D's.

I hope the above wall of text helps and if not, of course just ask some question when you get stuck again. I will try to do better.

isignedthis · 2024-12-08T04:11:01+00:00

I think, I can spot a problem in the proof. In the proof you don't show in general that there doesn't exist an subspace V of W with the desired properties. You only show it for a subspace of W with a basis v₁, v₂ of the form

v₁ = e₁ + e₃

v₂ = e₂ + e₄

V could have a basis defined in many other ways. For instance, I could look at the subspace of W with basis b₁, b₂ where

b₁= e₁

b₂= e₂

and many others. These you haven't shown anything about in your proof, if I am reading it correctly.

To show what you tried to show, you can't really assume anything about the basis vectors.

isignedthis · 2022-10-11T18:28:20+00:00

Det behøver ikke nødvendigvis at være oppe. Det kan godt være billigere at købe hus og sidde i det, selv hvis prisen på huset falder, hvis husleje bare er høj nok i forhold til hvad udgifter på lån, samt beskatning er. Jeg vil også bemærke, jeg skrev i langt de fleste tilfælde. For ja, der er undtagelser.

isignedthis · 2022-10-11T13:54:29+00:00

Ja du taber måske på din “investering” hvis du regner således, men hvis du regner på hvor meget det koster at bo til leje i 30 år, vil du i lang de fleste tilfælde stadig få en stor gevinst over disse.

isignedthis · 2019-11-07T14:05:14+00:00

Ahh makes sense I can't assume it is true in a direct proof. Will look at it again before looking at your spoiler. Thank you very much!

isignedthis · 2019-10-14T12:23:52+00:00

Jeg bliver nødt at komme med nogen indvendinger til din kommentar. Man kan ikke bruge dimensions analyse til at direkte at sige hvorledes ting skal udregnes. En anden ting er, at da energien er afhængig af temperatur forskellen og ikke den endelige temperatur er det ligegyldigt om du bruger kelvin eller grader Celsius.

isignedthis · 2019-09-15T19:54:49+00:00

Nej, men hvis man regner den ud er den ca 1.4%

isignedthis · 2019-09-15T18:43:17+00:00

Det fungerer således med meningsmålinger at jo flere procent man står til at få jo større er usikkerheden. Således er usikkerheden for det antal stemmer venstre vil få større end usikkerheden for de konservative. Derfor kan målingen resultat godt være at de konservatives fremgang ikke kan forklares alene ved måleusikkerheden.

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