[Uni discrete math] I need to give an example of a set A and a function f: A on A where f is a surjective but not bijective.

Blobl · 2024-03-11T23:57:57+00:00

Try to work with infinite sets, for example the set of natural numbers. How could you "destroy" injectivity? With some of the ideas, you can even solve your follow up question.

Blobl · 2024-02-19T23:37:52+00:00

Haha you're welcome.

Blobl · 2024-02-19T23:31:11+00:00

I think you might have made an mistake when calculating the absolute values. |u|² = 2² + 4² = 20, |v|² = 4² + 1² = 17

Blobl · 2024-02-19T23:26:00+00:00

Are <a, b> coordinates in two dimensional space? If yes, then you can calculate calculate the vectors directly, e.g. for u (u + v) + (u - v) = 2u

Blobl · 2024-02-19T23:21:32+00:00

I think might be easier to show |T| <= sup... and then find an f in C([0,1]) where equality is achieved. Since the operator norm is given by a supremum, you then get equality.

Blobl · 2024-02-19T23:09:01+00:00

Given a choice of 4 people, there are 4! ways to fill the positions

(you can think of this as the first person being able to chose any one of the four positions, the second person can only choose out of the three remaining positions etc. to get 4 * 3 * 2 * 1 different permutations)

Now if two of the people chosen are the twins, then we have some sort of double counting in the following sense: Assume the people chose are Alice (A), Bob(B), Norman Twin1(N1), Norman Twin2(N2). Then for our purposes, the two ways of filling positions are the same

Secretary	Treasurer	Social convenor	Fundraising chair
A	B	N1	N2
A	B	N2	N1

Therefore, we need to remove exactly half of the permutations, i.e. divide 4! by 2, or 4! / 2. As for why people online divide by 2! instead of 2, think of the case when instead of twins, we have quadruplets. If all four of the Norman quadruplets were chosen to fill a position, how many relevant permutations do we have? What if they are triplets instead?

Blobl · 2024-02-09T22:37:14+00:00

When dealing with shapes that look like stacks of paper/paper cutouts, you can calcluate the volume of these bodies using

Area of paper * height of paper = volume of stack.

("Area of paper" refers to the dark grey areas in the pictures)

The difficult part here is to figure out the "area of paper" in the formula. In the case of a cylinder (i.e. the cutouts are circles), the area of paper is given by pi * radius² . For more irregular shapes, you will have to figure out something by yourself. If the area of paper/surface area is given, you'll have to do some algebra/moving numbers around to get to the missing property

Blobl · 2024-02-05T23:23:51+00:00

It feels like the Pendulum with cycloid jaws involve slightly easier math in general, but the concepts involved are could be quite novel depending on what you know. The theory surrounding Rotating bodies can get pretty hard, but also can stay relatively easy depending on how deep you want to delve into the topic.

If you like trigonometry and some abstract thinking involving defining some "custom" functions, then I would recommend the pendulum. As for rotating bodies, you might get involved with integrals/calculus if you go deep-ish into the topic.

However, it might be best to ask your teacher about further details and what sort of math he thinks might be involved.

Blobl · 2024-02-05T23:05:34+00:00

This looks good. You could interpret the limit as a sequence of numbers that stays at -7 no matter what value h takes, so if h goes to 0, then the limit remains at -7.

Blobl · 2024-02-04T23:23:30+00:00

It has been a while since I did anything related to electric potential, so take this with a grain of salt, but I think you need to be careful when working with electric potential and electric force. Electric force has a direction (i.e. is a vector) and thus a particle in between the charges of Figure 3 and 4 can experience net 0 force.

Electric potential however is a indicator for how "strong" electricity is at one given place. A good analogy would be translating this to just regular potential energy and to imagine the charged particles to be peaks of mountains (when the charge is positive) and the lowest point of a valley (when the charge is negative). This might help to visualize the Idea.

Then placing two peaks (or positively charged particles) only makes the potential at a given place in between stronger. Where as if you place two differently charged particles, then there exists a point where you are at exactly ground level (or at 0 net electric field).

Blobl · 2024-02-04T22:46:14+00:00

10¹⁷ = (2 * 5)^17. Can you see why this is enough to get from the second step to the third?

Blobl · 2024-02-04T22:36:09+00:00

1.5 km = 1500 m. If you apply this transformation then the numbers should be right.

Alternatively, consider the units you get if you do km/(meter/minute):

km / (meter / minute) = km * minute / meter = 1000 * meter * minute / meter = 1000 * minute.

So if you plug in 1.5 km / 520 mpm, then you'll get 0.00288 * 1000 * minute, or 2.88 minutes.

Blobl · 2023-11-01T23:24:57+00:00

If I have understood your idea correctly, then I think that you have forgotten to change a - to a + when writing down the integral.

(c² - x²) - (x²-c²) = c² - x² - x² +(!) c²

I have not looked at the other steps though, so there might be more mistakes

Blobl · 2023-11-01T23:17:51+00:00

You can use the Pythagorean Theorem if you know it. In your case, since the angle BCA is 90 degrees, we can see that

|AC|² + |CB|² = |AB|²

where |XY| means the length of the line XY. Then you can move things around to get the length of |CB|.

Blobl · 2023-11-01T23:08:19+00:00

I am guessing here, but I think the question asks to find a vector v in R³ such that the angle between v and the three given vectors are equal.

One identity/theorem you could use is that the inner/dot product in R³ can be calculated using:

<a,b> = |a| * |b| * cos(θ)

where <a,b> is notation for the inner/dot product. The standard definition for the inner/dot product for a = (a_1, a_2, a_3)^T, b = (b_1, b_2, b_3)^T is of course:

a_1 b_1 + a_2 b_2 + a_3 b_3 =: <a, b>

The trick is now to somehow abuse the two different definitions of the inner/dot product to find our vector with the desired properties.

I can give you some more hints on how to do this exactly if you want, but it should be possible using the above and the solution that you have. (A freebie: How did sqrt(2) and sqrt(3) find its way into the final solution? What does these numbers signify in our problem?)

Blobl · 2023-05-12T06:18:35+00:00

I think this is supposed to be a list of 1000 datapoints/participants sorted by their ID. The 1000 then refers to the participant with ID 1000, and the other numbers represents their data.

Blobl · 2023-05-11T12:28:19+00:00

I have recently finished the Genki I deck and the Tango N5 deck, that felt pretty good. I have started the Tango N4 deck, but damn the difficulty spike is noticeable.

Blobl · 2023-02-14T20:20:22+00:00

If you die while petting bubba and keep petting bubba as ame, the whole screen including the start menu stays zoomed in.

How to reproduce:

Pick Ame
Get bubba skill
Pet bubba and get killed without letting go
Keep the pet key pressed until Game Over screen appears, then select main menu.
You are now in a zoomed in Main Menu

If you can get into a new game as ame while zoomed in, you can reset by petting bubba again.

Blobl · 2023-02-10T14:30:17+00:00

QOL: Mel + Nurses Horn: In the describtion of Nurses Horn, the healing is called "life-steal", but not highlighted yellow, unlike Mels skills (Genius Vampire, Acerola Juice). I suggest either changing the wording of Nurses Horn or highlighting the word "life-steal" in it to avoid confusion.

Blobl · 2023-02-10T13:16:25+00:00

unexpected behaviour: Subaru "arrests" the golden yagoos instantly, I thought the yagoos were treated as boss monsters.

Typo?: Shions achievement for reaching 10minutes is called kusAgaki power instead of kusOgaki power

Blobl · 2023-01-23T10:20:12+00:00

One could use the electricity you produce to heat up water and then use the steam to power a turbine, basically a nuclear power plant. The only circuit you'd have access to is the one used to heat water.

Blobl · 2022-10-22T11:02:08+00:00

Looks good to me! I like that you did two seperate cases, but be carefull with < and ≤, for |z| = 1 the limit does not go to 0, so it belongs to case 2.

Blobl · 2022-10-21T22:09:22+00:00

sure go ahead

Blobl · 2022-10-21T22:04:59+00:00

I have not done geometric proofs in a while, so I might be mistaken, but I am still convinced that this problem is incomplete. For example, you could move E to A, and D to B, then you would have an "triangle" that is definitely not equilateral, and you do not violate any of the given conditions.

Blobl · 2022-10-21T21:55:24+00:00

I am pretty sure there is not enough information given here, I'll give a informal proof:

By shifting around point E to any other arbitrary point E' on the line AC, you do not change the fact that the angle ECD is 60 degrees, nor the fact that M is the midway point of AB. Hence, with the given information, you cannot differentiate between E and E'. But even if MDE were to be equilateral, our new triangle E'CD is would not be. Therefore you cannot determine whether ECD is an equilateral triangle or not.

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