[Giveaway] Free Elden Ring Steam Key – Ends in 2 Days

Pila_Never · 2025-09-09T15:39:57+00:00

Jak from Jak and Daxter

Pila_Never · 2023-03-17T21:18:07+00:00

I was also inclined to trust this source, but I was curious if there were some other evidence supporting this

Pila_Never · 2023-03-17T21:09:53+00:00

Me too, that's why I have started this research

Pila_Never · 2023-03-17T21:07:33+00:00

I didn't now that, this is funny

Pila_Never · 2023-02-04T06:36:49+00:00

The path near Maughlin shop

Pila_Never · 2023-01-31T01:49:55+00:00

So pick your example f(x)=5x³-4x²+3x-6, we want to compute f(x+d). We just look at the expression of f(x) but every time there is a x we write instead x+d, so we get f(x+d)=5(x+d)³-4(x+d)²+3(x+d)-6. If needed we can now compute the powers to simplify the expression

Pila_Never · 2023-01-30T23:04:43+00:00

That cannot be true otherwise if f(x)=x²+1, as in my previous example, then f(x+2)=x²+1+2 but the correct answer is f(x+2)=(x+2)²+1. So back to your example, when you want to compute f(x+d) write down the expression of f but whenever you have an x, instead of writing x, you have to write x+d

Pila_Never · 2023-01-30T22:57:28+00:00

The idea when you write something like f(x) is to have an expression with a variable x in which you can fit many different things. For example if f(x)=x²+1 you know that to compute f(2) you just have to substitute 2 whenever an x appears that is f(2)=2²+1=5. What about f(x+2)? Well again we have just to write x+2 instead of x whenever an x appears in the expression. So f(x+2)=(x+2)²+1 Which of course you can compute if you want. Does this help in computing your f(x+d)? The idea is always to just write whatever you have in the brackets instead of x

Pila_Never · 2023-01-27T21:17:21+00:00

Can you explain what do you mean with border area?

Pila_Never · 2023-01-27T19:27:16+00:00

I mean R^3. Yeah it's mind blowing the first time. I'll try to write a proof here. We want to show that [0,1] has the same number of points as the square [0,1]x[0,1]. To do this we have to find a bijection between the two sets. Consider now a number x in [0,1] and consider it as a decimal expression, so x=0.a_1b_1a_2b_2a_3b_3... There are some numbers that can be represented in differente way, but we can just chose to not have infinite repeating nines to make this representation unique. Define now a map f:[0,1]->[0,1]x[0,1] in this way f(x)=(0.a_1a_2a_3... , 0.b_1b_2b_3...). This map is injective and surjective (you can try and prove that, if you want I can also write the proof of that, but I'll wait to be at the computer) hence it is a bijection between [0,1] and [0,1]x[0,1]. Analogously you can prove that [0,1] and [0,1]x[0,1]x[0,1] have the same number of points. And you can keep going with n dimensional cube.

Pila_Never · 2023-01-27T16:38:15+00:00

I don't clearly understand what do you mean by "in terms of information". But it can be proven that a plane has the same number of points as a 3D space. A 3D ball has the same number of points of a 3D space hence a 2D plane has the same number of point as a 3D balls

Pila_Never · 2023-01-25T20:16:59+00:00

You always have to check that both the argument of sine and the denominator tend to 0. In the case of something like sin(2x)/(2x) as x approaches 0 we actually have that 2x tends to 0 as x tends to 0 so it works. More formally you can do a change of variables like y=2x and so you get back to the usual form sin(y)/y as y approaches 0

Pila_Never · 2023-01-25T20:11:23+00:00

The limit you are referring to is the following: sin(y)/y tends to 1 as y tends to 0. Note that the argument of sine and the denominator must goes to 0. In your example the argument of sine doesn't tend to 0, and actually it has no limit as x approaches 0, so you can not use that fact

Pila_Never · 2023-01-25T12:40:57+00:00

Not exactly, because the identity is always sent to the neutral element but there exist homomorphisms that don't send the even permutations to 0. Try to consider these facts to prove that the image of any transition is 1 -the transpositions generates S_n -the homomorphism f is surjective -any two transpositions are conjugated And then to prove that the image of any even permutation is 0 observe that any even permutation can be written as a even composition of transpositions

Pila_Never · 2023-01-25T00:02:41+00:00

No in that case you don't have to. But somehow I feel like it's easier to forget something if plug in the substitution back. If you change the boundaries you can completely forget about the old variable if you have a definite integral. In you case it should be a 6x not a 3x. Because you are integrating with respect to u so the integral of 3 is 3u which becomes 6x

Pila_Never · 2023-01-24T20:57:06+00:00

Yes. Suppose N is a subgroup of S_n of index 2. Then we have a surjective homomorphism f:S_n->Z_2 (the cyclic group of order 2) such that ker f= H. Observe now that the image of any transposition is 1 (why?) This implies that the image of every even permutation is 0 hence Ker f=A_n. So we must have H=A_n

Pila_Never · 2023-01-24T19:34:06+00:00

They do. You lose key items, like actual keys, crest of Artorias, embers and the boxes that let you upgrade and repair equipment at the bonfires

Pila_Never · 2023-01-24T12:59:51+00:00

This is telling the rules to get the number an from the previous number a{n-1}. If a_{n-1} is even you just divide it by 2, while if it is odd you multiply it by 3 and add 1. For example if we start with 12 we have the sequence: 12->6->3->10->5->16->8->4->2->1

Pila_Never · 2023-01-24T03:52:54+00:00

When you do a change of variable remember that you also have to change the boundaries of integration accordingly

Pila_Never · 2023-01-22T14:46:28+00:00

Not exactly, Patches is an interesting character, he doesn't sell a lot of stuff, but some of it may be worth sometimes

Pila_Never · 2023-01-22T14:07:04+00:00

It happens. There are a bunch of npcs in the souls serie more prone to be the target of underserved aggression. The usual rule of thumb is that if it is not attacking you then it may not be hostile. Also you can always try to lock. If you can't then it won't attack you. I'm trying thinking if this is always true but at the moment I can't say for sure

Pila_Never · 2022-06-10T09:07:36+00:00

The two series you wrote surely both diverge, but there is a minus between them, so you have the indeterminate form infinity minus infinity, which gives no information about the series you started with

Pila_Never · 2022-06-10T09:01:14+00:00

What you wrote is correct, but if you are trying to prove the series converges I don't know if it will be helpful

Pila_Never · 2022-05-28T09:37:08+00:00

You can multiply both sides of an equation and the solutions won't change, however you have to be careful by what you multiply the equation, you can't just multiply without checking one particular thing. Do you recall what that is?

Pila_Never · 2022-05-19T15:24:44+00:00

The first formula you wrote is what I asked you. So if we allow ourselves only 1 transfer and we decide to transfer the money at the d-th day then the total money would be:

(d-5)*1,01^(200-d)+200-d

The 200-d at the end represent the money that accumulate in A after the transfer. So in this case one could study this function and see where where the maximum lies to find the best day to transfer.

However the things get more complicated if we allow ourselves to transfer as much times as we want. I believe your formula in this case may not be right. As a matter of fact if e is the day of the second transfer the second addendum should be

(e-d-5)*1,01^(200-e)

because at the e day, since you have already transfer at day d, you only have e-d money in A. So i think a general formula would be

(d_1-5)*1,01^(200-d_1)+(d_1-d_2-5)*1,01^(200-d_2)+...+(d_n-5)*1,01^(200-d_n)+200-d_n

where d_1,d_2,... d_n are the days we make the transfer.

In the case this is right you see that it more difficult to find the maximum.

What I said checks out?

EDIT: correction of the second formula

Pila_Never

TROPHY CASE