help-39

[jolly parrot]

Edmund Cloudsley

[midnight haven]

so that the equation becomes

$$a^2 + 2ab + b^2 - b^2$$

[jolly parrot]

Edmund Cloudsley

[midnight haven]

now what since addition is associative i.e. (a + b) + c = (a + c) + b

we can say that

$$ = (a^2 + 2ab + b^2) - b^2$$

[jolly parrot]

Edmund Cloudsley

[midnight haven]

now I am sure this reminds of the familiar identity of (a + b)^2

therefore we can write this equation as

$$ = (a + b)^2 - b^2$$

[jolly parrot]

Edmund Cloudsley

[midnight haven]

this is known as the "completing the square" method

and it is quite commonly used for factoring equations

[wild temple]

Let me think twice.

Just 2 minute.

[midnight haven]

Ofcourse! Take as long as you require

This can be quite daunting to grasp

[wild temple]

Where did the 2ab gone?

@midnight haven

[midnight haven]

the 2ab is part of the expansion of (a + b)^2 now

we know that

$$(a + b)^2 = a^2 + b^2 + 2ab$$

[jolly parrot]

Edmund Cloudsley

[midnight haven]

this is just the opposite of this identity

of you'd like to know why

$$(a + b)^2 = (a + b)(a + b) = a^2 + ab + ab + b^2 = a^2 + b^2 + 2ab$$

[jolly parrot]

Edmund Cloudsley

[midnight haven]

or you can also use binomial theorem to expand this one

but that's bit of an overkill

$$(a + b)^2 = {}^2 C_0 a^2 b^0 + {}^2 C_1 a^1 b^1 + {}^2 C_2 a^0 b^2$$

[jolly parrot]

Edmund Cloudsley

[wild temple]

Alright. 1 minute.

[midnight haven]

take ya time

[wild temple]

After completing the square, what does the fourth part mean?

[midnight haven]

the 4th and 3rd part are both completing the square

[midnight haven]

in the third part, we are setting up our equation to apply completing the square method

in the fourth part, we are actually applying the completing the square method

in your case, $a = 2x^2$ and $b = 3m^2$

[jolly parrot]

Edmund Cloudsley

[midnight haven]

what's remix?

I have no clue what that is

[wild temple]

Oh the drawing method. I want to know how the third part looks like based in (a+b)²=a²+b²+2ab

Something like that.

[midnight haven]

oh I understand what you are trying to say

You want me to illustrate how the third part fits in the idea of completing the square method

[wild temple]

Yes

[midnight haven]

ofcourse

so esentially we have

[wild temple]

I made a draft in here.

[midnight haven]

$$2(4x^4 - 13x^2m^2 + 9m^4)$$

let us write $2x^2$ as $a$ amd $3m^2$ as $b$
$$$$
Then this equation would be

$$2[(2x^2)^2 - 13x^2m^2 + (3m^2)^2]$$
$$2[a^2 - 13x^2 m^2 + b^2]$$

let us subtract and add $+2ab$ and $-2ab$ so that we can set this up for completing the square method

$$2[a^2 -13x^2m^2 +b^2 + 2ab - 2ab]$$
$$2[(a^2 + 2ab + b^2) - 13x^2m^2 - 2ab]$$
$$2[(a+ b)^2 - 13x^2m^2 - 2ab]$$
now we can resubstitute $a$ and $b$
$$2[(2x^2 + 3m^2)^2 - 13x^2m^2 - 2(2x^2)(3m^2)]$$
$$2[(2x^2 + 3m^2)^2 - 25x^2m^2]$$

[jolly parrot]

Edmund Cloudsley

[wild temple]

Alright,give me few minutes.

[midnight haven]

take your time

I would have to take my leave now unfortunately

I shall let someone else continue

[wild temple]

Okay, when I have a problem I will ping you by then.

[midnight haven]

Sure youc an ping me

I would take some time to respond

[pearl pond]

@wild temple Has your question been resolved?

[wild temple]

Not yet, I'm analyzing on my copy.

[wild temple]

@midnight haven I'm bad at explaining. I am stuck again.

Does factorization have cancellation?

[midnight haven]

Yeah

You can canel stuff when you are factorising

[pearl pond]

@wild temple Has your question been resolved?

[wild temple]

@midnight haven can you explain me on note book I want to understand visually and orally at the same time ; it's like taking a lecture in class.

[wild temple]

My explanation,I feel like I explained everything wrong.

[midnight haven]

Your explanation is correct!

I just had a look at the video

[wild temple]

Can you make some corrections. I was explaining my best but my understanding is less and not complete.

[midnight haven]

It is complete ma man

I think you did it quite well

You took some time but you got there

[wild temple]

Hold on.

[midnight haven]

Unfortunately I would be unable to do that since that would involve me revealing my voice on this server.

Something I am not comfortable with

[wild temple]

I explained till there.

[midnight haven]

alrighty

[wild temple]

The last 3 steps can you tell me how?

[midnight haven]

for the parts after that, we have to use a different identity

in particular

this one

$$a^2 - b^2 = (a + b) (a - b)$$

[jolly parrot]

Edmund Cloudsley

[midnight haven]

in your case if you take $a$ as $2x^2 + 3m^2$ and take $b$ as $5xm$
you should be able to solve it

[jolly parrot]

Edmund Cloudsley

[midnight haven]

its all about recognising the corrrect identities

You may wish to review those before you proceed

https://byjus.com/maths/algebraic-identities/

You may use this link to brush up on those if you'd like

[wild temple]

5² = 25 so with x² and m² it will be (mx)²?

[midnight haven]

Yes

and together they would be

$$25x^2m^2 = 5^2 \times (xm)^2 = (5xm)^2$$

[jolly parrot]

Edmund Cloudsley

[wild temple]

The last step though, how to understand it?

[midnight haven]

It’s an application of this identity

[wild temple]

I don't understand because the last step has 4 operators (two "+") and (one "-" one "+")

[wild temple]

This = to one plus and one minus.

@midnight haven

Can we take this through dms.

[midnight haven]

I don’t quite understand ur question

[pearl pond]

@wild temple Has your question been resolved?

[pearl pond]

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[simple gyro]

im doing a practice paper right now for studying and using a textbook to help me, but i can't figure this out, the equation above is x^3 + 4x^2 - 13 = 0

[ancient smelt]

take the 13 and x^3 to the rhs

$$4x^2 = 13-x^3$$

[jolly parrot]

AnshumanNeon

[ancient smelt]

divide by four

$$x^2 = \frac{13-x^3}{4}$$

[jolly parrot]

AnshumanNeon

[ancient smelt]

then square root both sides

[vagrant sierra]

if it helps u understand u can put bracket around 13-x^3/4

helped me when i started these types of questions

[ancient smelt]

$$x = \sqrt{\frac{13-x^3}{4}}$$

[jolly parrot]

AnshumanNeon

[ancient smelt]

and there you go

[simple gyro]

thank you

[ancient smelt]

yeah it helps

to make it clear, i used latex

i don't use latex, but it helps me practice it

[vagrant sierra]

🙂 glad it does, im studying for my finals rnnnn 😭

[vagrant sierra]

never used that sounds nice

[ancient smelt]

haha

[ancient smelt]

this is latex

[ancient smelt]

but it converts to this then

[ancient smelt]

good luck

[vagrant sierra]

tysmmm u too

[jolly parrot]

convergence

[simple gyro]

another thing im confused about is iterations? there's a follow up question asking about it but i literally can't figure it out even using my textbook

[ancient smelt]

ah yes!

[ancient smelt]

what's the question. i didn't get what you meant by iterations

[simple gyro]

im nearing the end until my finals but ive never seen this question in my life

[summer gorge]

ah recursive functions

,w iterate

[jolly parrot]

Results provided by WolframAlpha

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Requested by convergence0091

[summer gorge]

iteration formula basically means to repeat the formula again and again @simple gyro

[simple gyro]

okay thanks

[summer gorge]

my suggestion would be to find x_3 in this case

[ancient smelt]

i haven't seen a question like that tbh

[summer gorge]

it is very commonly used in convergent sequences

[ancient smelt]

i haven't seen much so ig its just that i live under a rock

[pearl pond]

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[chilly jay]

what would be the first step to do this question?

[gilded loom]

Just write tan(πx/2) as 1/cot(πx/2)

Then 0/0 form

Can u do that from there?

[verbal whale]

Personally, I would make a substitution, u = x - 1

[chilly jay]

is the answer 2/pi?

[gilded loom]

Idk

[heavy lagoon]

rewrite it as tan(pi x /2)/(1/(1-x)) then use l'hopital

[gilded loom]

I didn't solve

[heavy lagoon]

shouldn't it be one?

[chilly jay]

this makes sense and makes things a little easier but, is there a specific rule as to where to apply the substitution?

[chilly jay]

yeah I got it

[gilded loom]

Yep right

[chilly jay]

okay, thank you

[gilded loom]

2/π

Mhm

[gilded loom]

Anytime!

[pearl pond]

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[midnight haven]

last quastion

[midnight haven]

Calcus 1
a 2 / b 2 / c -3 / d 2 , my answers

uh so for a. u get that lim x->2- f(x) = 2 ?

no I got -2

missclick

sorry but last one isI didn't find the solution

and for b

2

how did u get 2 ?

It’s a bit hard to explain since English is not my first language, but when I went above 4, I reached the two closest functions

what is ur language ?

There is no function when going down, and I went up and reached the result of 2

turksh

ah i dont speak turkish

"Could you explain how you found your solution if mine is incorrect?

Oh, right, my mistake. I thought it was the beginning of the blue line

mb

I thing C is correct

no

u found -3 ?

yes because goes to -3

when x= 0 - ?

-3 is not even on the graph

Can I do my solutions through print and share them again with you as long as the chat stays open

yes

i didnt rly understood what you meant but ok

Its looks like this

I did go left from the 0 and then I saw the blue line and its going to -3

x is on the horizontal line

u go at x=0-

and you watch above or under the point where you are

under

not really

it is exactly wher u are

I go litelbit left because its -

and then going to down side

[midnight haven]

for example, here, x was blocked at 4 since u look for the limit when x=4

so u watched above and under you until you meet the graph

[midnight haven]

and in this case, you meet the graph at y=4

yes I did and then up side I didnt find something

I went down side

[midnight haven]

here ?

yes sir

correct anser is 4 ?

[midnight haven]

u dont have to move here because it is 0

[midnight haven]

for the question before yes

when x tend to 4

I think I got it so I followed this solve style and then I looking for quastion D

answer is it 5?

or my old answer -3

none

can u draw how you get those answers please ?

sure

i can't understand how you think

for D

yes it is 2

how could u find 5 ?

i dont know my answer was 2

for 5 I try to go last part

but It was bad idea

okey I got it 2 quastion can we look for last one?

for e ?

yes

can you share the soliton with me?

I have no idea

it is undefined since the limits when x>0 and when x<0 are not the same

so what should write to answer gap

D.N.E ?

undefined is good i think

ok thank you for answering my all quastion sir

where do you live btw?

u said maybe I can try to speak ur own language so thats mean u are not from UK

lol

im french

ohh cool I am tryin to develop my english sorry for that

sorry for what ?

I passed my depertman with C1 certifation but still not enogh

[midnight haven]

my english bad

You needed to put in more effort

sorry about that

its ok bc i was doing the same mistakes before, so i can still unnderstand

[midnight haven]

uh me ?

yes

wait I write wrong I guess

I mean u tryed to hard because my english was bad

ah ok i just understood

thanx again

good luck

[midnight haven]

understandable enough

thx

[pearl pond]

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[proven kite]

Hey can someone just tell me why this is wrong?

[proven kite]

I'm guessing third line went wrong, but if that 1st term had 2 terms it would be fine I'm guessing. I think it's due to lack of conceptual knowledge of how factorization gives you roots

[dense jasper]

just b/c two things multiply to 36 doesn't mean that one of the factors is 36

ex. 4 and 9, 6 and 6, 3 and 12, 8 - 2 sqrt 7 and 8 + 2 sqrt 7, etc.

[proven kite]

If it was like (a-1) instead of a in the first term, would it be fine?

[dense jasper]

just b/c two things multiply to 36 doesn't mean that one of the factors is 36

[proven kite]

Why does this work then?

[dense jasper]

zero is special

It's the only number c for which you can say ab = c -> a = c or b = c without anything else

As a proof, suppose that a and b aren't zero

then ab clearly isn't zero, a contradiction

hence at least one of a and b is zero

[proven kite]

Ohh I see now

Thanks ❤

[dense jasper]

[proven kite]

.close

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[wild hamlet]

How we get logb(x)=ln(x)/ln(b)

[dapper spear]

https://www.chilimath.com/lessons/advanced-algebra/proofs-of-logarithm-properties/

see the steps for number four

[pearl pond]

@wild hamlet Has your question been resolved?

[pearl pond]

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[sage cobalt]

When doing subspace iteration using QR decomposition, my algorithm is convering too fast due to a bad stop condition. Although this stopping condition is mandatory for this assignment:

while(1)

    P = matrix * Q;
    [mQ, mR]= SN_gs(P); % QR decomp

    if(
      norm((eye(rows, rows) - (mQ*mQ')) * Q) 
      <= tol
      )
        mQ = Q;
        disp(runs);
        break;
    end

    Q = mQ;

    runs = runs + 1;
end

[sage cobalt]

Stopping condition is

norm((eye(rows, rows) - (mQ*mQ')) * Q) 

As defined in my assignment:

Tolerance level is 10^-6...

I see that mQ * mQ' gives almost a unity matrix... so maybe rounding errors

[pearl pond]

@sage cobalt Has your question been resolved?

[humble lintel]

Is your answer bad? There is nothing wrong with it converging too fast

[pearl pond]

@sage cobalt Has your question been resolved?

[sage cobalt]

My eigenvalues are just off

but my algorithm is just not working properly... sometimes it does not converge and keeps between 0.7 and 0.2... Im so fcked

[humble lintel]

You haven't really provided enough information to say anything, it sounds like this has nothing to do with your stopping condition because your assignment is unlikely intended for you to turn into a long-winded exercise in arithmetic precision and regularization techniques outside the scope of your class

[sage cobalt]

Some matrices work, some don't. I have no clue what is happening...
This is my algorithm:

while(1)

    P = matrix * Q;
    [mQ, mR]= SN_gs(P);

    disp(norm(Q - mQ));

    if(norm(Q - mQ) < tol)
        mQ = Q;
        break;
    end

    Q = mQ;

    runs = runs + 1;
end

It is subspace iteration using QR decomposition

[sage cobalt]

Well, it seems like it is... The implementation of subspace iteration is supposedly the exercise, but debugging it to work with the random matrix generator is the most difficult part ):

I have this matrix for example: (condition = 10)

This will not converge under any stopping condition I use.

This matrix works fine: (condition > 10^4)

[humble lintel]

Is it supposed to converge for your first example?

[sage cobalt]

Yes it should, but what it does is run only once. The stopping condition is already < 10^-10. But this algorithm is supposed to find eigenvectors, and it is concerning that it just stops without making any progress...

[humble lintel]

I don't know this algorithm, and I'm not looking at your work very hard, but it looks like the magnitude of your eigenvalues are the same there and I don't know how your algorithm is expected to converge on a local solution there

What happens if you take your first example and add a random diagonal matrix with very small values like between +-0.02

[sage cobalt]

Woow

The algorithm works with that little addition...

[humble lintel]

okay well that's a regularization technique, and the first thing I said was that I didnt expect your class to need you to go off on some regularization tangent

so I don't know what your are expected to know

[sage cobalt]

Interesting, the matrices that we use are generated using a provided matlab function...

[humble lintel]

the idea is that if your eigenvalues are poorly behaved, by the greshgorin circle theorem, if you perturb your diagonal by a very small amount, the probability of the new matrix being poorly behaved is on a set of measure zero and it can fix your problem by introducing a small error

[sage cobalt]

For smaller matrices this technique works, but once the size of the matrix rises, the iterations fall to 1 again

[humble lintel]

in your case, since it is randomly generated, the probability of a 2x2 matrix having complex eigenvalues and thus problematic is very high

so I guess in some sense you also got lucky

Like I said, I don't really know what you are doing. This looks like some sort of power method, but this isn't really something I use

[sage cobalt]

So it is possible that my other matrices also have complex eigenvalues which is causing problems, correct?

Or not?

[humble lintel]

My assumption is that if your matrix just happens to have two largest eigenvalues in magnitude

then your algorithm probably can't converge

so by perturbing your matrix a little bit in that case, you will move your eigenvalues around a little bit and hopefully you won't have your largest eigenvalues be the same in magnitude anymore

The problem is that in the 2x2 case, if you have a-bi and a+bi as eigenvalues

then |a-bi| = |a+bi|

so there is a chance that what is happening is that your algorithm is going towards the eigenvector corresponding to one eigenvalue, but then gets stuck trying to travel to the other eigenvector and gets caught in a loop

[sage cobalt]

This is very complex mathematics

[humble lintel]

You should worry about making sure your algorithm works when there is 1 largest eigenvalue

and just ask your instructor about the edgecases

There is a reason why nobody actually implements their own numerical linear algebra routines, they are finnicky and filled with edge cases

[sage cobalt]

You have a lot of knowledge

Thanks a lot for taking your time to share, I really appreciate it. I will read again and use your insight

Thank you

[humble lintel]

good luck

[sage cobalt]

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[oak patrol]

Where can I see the finding of all solutions x^2-y^3=17?

[oak patrol]

(x,y integers)

[mint wave]

I mean you could always graph it

[oak patrol]

It's endless, and as far as I know there are only 8 solutions here

<@&286206848099549185>

[paper egret]

set y to 0 and x to 0 and solve for both intercepts?

[oak patrol]

?

[pure pier]

x and y are different numbers

[oak patrol]

I think that’s not all solutions

[paper egret]

to find the x intercepts
x^2-0=17
x^2=17
x intercepts at plus minus sqrt(17)
for y intercept
0-y^3=17
y=cube root(-17)
y intercept at negative cube root of 17

[paper egret]

wait

only real numbers?

[oak patrol]

Yep

[paper egret]

yeah

that should be your answer then

[oak patrol]

.

[pure pier]

This question has actually got me thinking, what grade maths is this (or uni)

[paper egret]

interception for x values at
(-sqrt(17),0) (sqrt(17),0)
interception for y value at
(0, -cuberoot(17))

[oak patrol]

(282; 43) solution

[paper egret]

wdym

[oak patrol]

U can check))

[paper egret]

we can see your graph extends to positive and negative infinity which means the domain of X accepts all real numbers

[paper egret]

whered u get that solution

[oak patrol]

I saw on the graph

[pure pier]

I know these are the answers but no idea how they found them

[paper egret]

where'd u get that

[oak patrol]

Wolfram or smth

[pure pier]

Yea wolfram

[oak patrol]

All solutions

[pure pier]

Look at integer solutions

[acoustic path]

with calculator

this is, ofc, not easy to prove

[oak patrol]

Do u know how?

[oak patrol]

I understand it

[acoustic path]

http://magma.maths.usyd.edu.au/calc/

Enter: E:=EllipticCurve([0,0,0,0,17]); IntegralPoints(E);

this specific problem is mentioned in various literature:

it is talked about here: https://mathoverflow.net/questions/52979/integer-points-on-the-elliptic-curve-y2-x317?noredirect=1&lq=1

[oak patrol]

Do you have a solution without computer search?

[pearl pond]

@oak patrol Has your question been resolved?

[pearl pond]

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[sonic kraken]

jst need someone to explain this step to me

[sonic kraken]

should be a very quick question

[unborn abyss]

$x^2 - 6x + 10 = \blue{x^2 - 6x + 9} + 1 = \blue{(x-3)^2} + 1$

[jolly parrot]

Лайка

[sonic kraken]

oh wow

alr thx

.close

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[somber heron]

#help-39 message i thought having solving couple of this question i finally understand how to do this kinda question. But apparently not...i used the same method but somehow i got it wrong again and idk where i made mistake...

[somber heron]

(excuse of messy handwritting)

also here is the answer

could anyone be kind enough to work me through how to get that answer or point out what i made mistake in my work

[gleaming moss]

Check blue again

More specifically, what the bearing is relative to ( west of south )

[pearl pond]

@somber heron Has your question been resolved?

[somber heron]

ok, i checked again. But i still dont understand what i did wrong here...like it says 10 degrees west of south so doesnt that mean its like 10 degnvm as i was typing i noticed i was using the wrong side of the angle

i did the calc again with the correct bearing, it did match up the answer.

[somber heron]

but wait why is the direction W of N

the question asks what direction should they be walking to

[gleaming moss]

What do you mean

[somber heron]

.

Im assuming origin(0,0) is their home, after the last vector(green) arent they supposed to walk toward E of S direction and go back to origin

[gleaming moss]

?

[somber heron]

i do get the path that toward to their home(purple), in terms of vector, it should be pointing N of W, but the question is confusing to me.

[somber heron]

^

[gleaming moss]

I just re-read the question and yeah you are right. The answer should be 3deg S of E or 87deg E of S I believe

@somber heron

Ah no nevermind, this confused me because of your diagram

The tip of the green vector should be in quadrant 4. Which means the direction back to the origin will in fact be up and left

[somber heron]

oh hmm maybe cuz my drawing its not the actual scaling

let me try to put on desmo

[gleaming moss]

Yeah, the angles are too steep and the lengths all look equal

[gleaming moss]

Should look a bit more like the yellow path

[somber heron]

yeah

looks something like that

it now makes sense

Thank you so much for answering my question!

.close

[pearl pond]

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[pearl pond]

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[river lodge]

Five cards are drawn from a standard deck of cards. What is the probability of being dealt one pair?

Can someone explain why this is the answer:

(52x48x44x40x30)/52P5

where does 30 come from?

[pearl pond]

@river lodge Has your question been resolved?

[pearl pond]

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[simple elbow]

Need help

Need help

[pearl pond]

@simple elbow Has your question been resolved?

[pearl pond]

@simple elbow Has your question been resolved?

[pearl pond]

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[wraith badger]

if anybody here is familiar with the hp prime GDC, can somebody explain why i cant put this equation into the functions app to plot it?

[wraith badger]

this is the equation im trying to solve and i was trying to do it graphically by putting the line equations separately and finding where they intersect

[pearl pond]

@wraith badger Has your question been resolved?

[wraith badger]

what the freakity doodles

[pearl pond]

@wraith badger Has your question been resolved?

[pearl pond]

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[native atlas]

can't figure this one out

[native atlas]

just need some clarification

when it becomes ln(10x)^3

is that valid

im tired asf and i'm forgetting my log rules

.close

[pearl pond]

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[exotic scaffold]

yes

[pearl pond]

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[rich jolt]

i know the formula but not sure how the formula works

[exotic scaffold]

what is the formula

what have you tried

[rich jolt]

the formula would be 15-4+1c4

[pearl pond]

@rich jolt Has your question been resolved?

[vestal tapir]

(15−4+1) is how many gaps there is around 11 days

like if you need to choose 2 days out of 6
.D.D.D.D.
you put 2 days anywhere in the 5 gaps around 4
(6−2+1)c2

you can also do it differently, with stars and bars

[pearl pond]

@rich jolt Has your question been resolved?

[rich jolt]

but why does choose it causes gaps

[vestal tapir]

that's the condition

they can't be consecutive

meaning they can't touch

one chosen day fits in each gap, 0 is fine, two is not allowed

[pearl pond]

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[slate hawk]

Can someone check it for me

[dusk horizon]

This needs a lot of words.

Even if the calculations are right, I'd grade this at like 75% at most

[midnight haven]

fuck allat

[slate hawk]

What should i write on it

[midnight haven]

what's ur fav dinosaur

[slate hawk]

What am i missing.

[rough forge]

Hast du Induktion vesucht?

[dusk horizon]

Why are you writing this? How do you know it's true?

[slate hawk]

Ja.

Its given on the top

[dusk horizon]

Isn't that what you're trying to prove?

[rough forge]

base case of induction, written poorly that it works for some k in N

[slate hawk]

I just read some article on internet and we assume its true when when its n

[dusk horizon]

I know why khishi is writing it, but it needs to be written that you're assuming that part for the sake of induction

[rough forge]

well as said, written poorly

[slate hawk]

And use this to prove n+1

Yeahh im in subway and forgot my pen. So didnt want to write more stuffs.

[rough forge]

der will sagen, du solltest noch hinschreiben, dass die Aussage wahr ist für mindestens ein k aus N

und dass du das benutzen wirst im Induktionsschritt

[slate hawk]

Ach so.

Aber ist es wirklich so wichtig.

[rough forge]

entweder in worten oder mit logik

[rough forge]

ja eig schon

[slate hawk]

Was sollte ich schreiben.

Könntest du mir ein Beispiel geben.

[rough forge]

Es existiert ein n aus N so, dass f(n) wahr ist

[slate hawk]

Was warum k?

Nicht n?

f(n)?

[rough forge]

achso ja kannst n benutzen

[slate hawk]

Ach so.

Und ist das alles, das ich schreiben musste?

[rough forge]

Formal gesehen, würde man am Ende noch einen Schlussatz schreiben

Nach dem Prinzip der v.I. ist f(n) für alle n aus N wahr.

[slate hawk]

Was für einen Schlussatz?

Ach so.

[rough forge]

Es ist halt ein Beweis den du da machst

es ist nicht nur bloß umrechnen und umformen

[slate hawk]

Verstanden.

[rough forge]

und dagehören halt langweilige, repetitive aber wichtige Formalitäten

[slate hawk]

Okay. Danke.

[rough forge]

no problem

[slate hawk]

Ich habe andere Probleme

[rough forge]

mentale?

[slate hawk]

Mathematische und mentale beide

[rough forge]

du ich war zwar gut in stochastik, aber ich hab keine ahnung

[slate hawk]

Ich habe am Dienstag noch eine Prüfung.

Oh. Schade

[rough forge]

kanns es aber sein, dass da was fehlt

[slate hawk]

Einen Moment

I dont understand the sentence. Define suitable probablility space…

Muss ich alles schreiben die in wahrscheinlichkeitsraum sind?

[rough forge]

Also ein WK-Raum besteht aus einer Grundmenge Q, einer sigma Algebra und einem WK-Maß P

Also P ist sowas wie eine Abbildung

Sie ordnet Elementen (Ereignissen) aus der Sigma Algebra (eine Menge aus Teilmengen von Q, da sind halt verschiedene Ereignisse drinnen) und da ordnet die Abbildung P einem Ereignis eine Wahrscheinlichkeit zu

[slate hawk]

Nichts verstanden.

[rough forge]

Danke.

[slate hawk]

Was macht sigma algebra?

[rough forge]

Ok ein Bsp

Q = {1,2,3} dan wäre eine mögliche sigma Algebra Sigma = { {}, {1}, {2}, {1,2}, ... }

Man sieht ja, es ist eine Menge von Teilmengen von Q

{1} ist eine Teilmenge von Q

{2} und {1,2} auch

[slate hawk]

Hier muss auch Q selbst sein oder?

[rough forge]

Ja

[slate hawk]

P(omega) ist die großte Sigma algebra.

[rough forge]

Ja, das war die Potenzmenge wenn ich mich entsinne

[slate hawk]

Ja es ist

Was ist mit P?

Letzte P

[rough forge]

Jetzt bin ich nur bisschen verloren, also in der letzten Aufgabe, da war X ja die Differenz der Zahlen, die man gewürfelt hat mit Würfel 1 und 2

[rough forge]

Sie ordnet allen Element aus P(A) eine Wahrscheinlichkeit zu

also allen Elementen aus der sigma Algrebra

Also eine Abbildung P : Sigma -> [0,1]

[slate hawk]

Lass uns zuerst 2.a machen. Danke.

[rough forge]

Nein. Bitte.

Also ähm

[slate hawk]

Ist X -5 bis 5?

[rough forge]

Man würfelt die Würfel gleichzeitig nehme ich an

[rough forge]

ja

würde ich auch sagen

[slate hawk]

Aber es ist geschreiben, dass sie nacheinander gewürfelt sind

[rough forge]

ah ok

[slate hawk]

Jetzt zurück zu 3

[rough forge]

M wäre hier doch die Sigma Algebra

Weil man aus Q = {1,2,3,4,5,6} alle möglichen Ereignissen bildet denke ich

[slate hawk]

Ach so

Ist M nicht auch -5 bis 5?

[rough forge]

wobei ich denke M = Q x Q das kartesische Produkt

man hat ja Zahlenpaare z.B. (6,1) wo man dann eben die Differenz bildet 6-1

usw

[rough forge]

X variiert zwischen -5 und 5

[slate hawk]

Ok

[rough forge]

(3,2) und (2,1) beiden resultieren 3-2=2-1=1

[slate hawk]

Okay

[rough forge]

wobei lass mich nochmal nachdenken

[slate hawk]

Okay.

[rough forge]

Ich denke es würde mehr Sinn machen wenn Q = [1,6] x [1,6]

Q ist ja die Ergebnismenge, also die Menge aller möglichen Ergebnisse

[slate hawk]

Zum beispiel?

Ich habe gedacht, M ist die Ergebnismenge

[rough forge]

ok ich bin lost, ja

[slate hawk]

Ich bin auch

[rough forge]

M = [1,6] x [1,6]

es gibt kein Q in 2)

das wäre dann die a)

b) wäre jetzt einfach Teilmengen von M bilden, die alle das Resultat liefern

2i) wären alle geordneten Paare aus M, die eine negative Differenz liefern, z.B. (1,2) weil 1-2 = -1

[slate hawk]

Jetzt zurück zu 3 bitte

[rough forge]

Ah ok jetzt verstehe ich, M, A, ... , D sind diese Teilmengen

[slate hawk]

Ist omega=M ?

[rough forge]

was genau meinst du mit Sigma

[slate hawk]

Nicht sigma meine ich omega

[jolly parrot]

𝔸dωn𝓲²s

[rough forge]

Ok ja hätte ich auch gesagt

das meinte ich vorher mit Q lol meinter aber Omega

Omega = M

ja würde ich sagen

[slate hawk]

Ach so. Jetzt verstehe ich

Wie kann man jetzt die Potenzmenge definieren?

[rough forge]

Das wäre denke ich jetzt die Menge aller Teilmengen von Omega

P(Omega) = { {(1,1)}, ... {(6,6)}, {(1,1), (1,1)}, ... , {(6,6), (6,6)}, ...}

auch wenn {(1,1), (1,1)}, ... , {(6,6), (6,6)} eig keinen Sinn machen, weil das nie passieren kann

[slate hawk]

Ach so.

[rough forge]

man könnte den ja die Wahrscheinlickeit 0 zu ordnen

[slate hawk]

Was warum?

[rough forge]

weil diese Ergebnisse nie auftreten können

Man hat ja ein Paare

man würfelt einmal zwei Würfel

[slate hawk]

Ach so

[rough forge]

{(6,6), (6,6)} wäre z.B. eine Möglichkeit von 6, dass man zweimal hintereinander X = 0 bekommt

aber das ist ja nirgends hier gegeben, was das wirklich bedeutet

wobei man könnte auch simpel sagen P(Omega) = {A,B,C,D,E,M, ...}

[slate hawk]

Ach so.

[rough forge]

ich hab noch ... gemacht, weil es heißt ja alle möglichen Teilmengen deswegen könnte es ja noch weitere kombinationen geben

Jetzt suchen wir alle Wahrscheinlichkeiten
P({X in A}) = ?
...
P({X in M}) = ?

[slate hawk]

Wie hat du P definiert

Ist es wahrscheinlichkeitsmaß?

[rough forge]

ja

ich denke das ist ja was wir jetzt bestimmen sollen

[slate hawk]

Oder wahrscheinlichkeitsverteilung?

Was ist Unterschied?

[rough forge]

So wie es steht, ist wird das Bildmaß eines Wahrscheinlichkeitsmaß, als Wahrscheinlichkeitsverteilung bezeichnet, aber ich hab wenig Ahnung, ich hatte kein Maßtheorie wirklich

https://de.wikipedia.org/wiki/Wahrscheinlichkeitsmaß#Elementares_Beispiel

Also anscheindend ist das WK-Maß, die Abbildung, die den Ereignissen die WK zuordnet

und die WK-verteilung beschreibt eher wie die Wahrscheinlichkeiten zugeordnet werden

🍒 :

[slate hawk]

Tur mir leid. Ich war ein bisschen beschäftigt.

[rough forge]

Mentale Pausen sind okay.

[slate hawk]

Danke, dass du mir Erlaubnis gegeben hast

[rough forge]

hahahah

[slate hawk]

Okay jetzt wie definiert man es

[rough forge]

Das WK-Maß zu finden, ist denke ich einfach
P({X in A}) = ?
...
P({X in M}) = ?
zu bestimmen

Wie würde ich es machen...

Ich würde sagen

[slate hawk]

Ach so. Lass uns M finden

Ist es 1

[rough forge]

P({X in A}) = |A|/|M|

usw.

[slate hawk]

Ach so.

Was ist P(X in M)

Ist es 1

[rough forge]

M = [1,6]² = {(1,1), ... , (6,6)} haben wir gesagt

Ja

denke ich

[slate hawk]

Ist es 0 ?

[rough forge]

wieso

ist |A| leer?

[slate hawk]

Nein.

Aber A ohne M bedeutet nichts oder?

Ich bin verwirrt

[rough forge]

Ich auch, was meinst du A ohne M

[slate hawk]

Hier

Du hast geschrieben.

Ist es nicht A ohne M?

[rough forge]

|A| ist keine Menge :)

das ist die Mächtigkeit/Kardinalität von A

also die Anzahl an Elementen

demnach ist / nicht ohne sonder bedeutet geteilt

[slate hawk]

Ach sooo

[rough forge]

|A| geteilt |M|

[slate hawk]

chuldigung

Jetzt verstehe ich.

[rough forge]

vergeben und vergessen

[slate hawk]

Danke

Ich glaube es ist alles was ich fragen wollte.

Genug für heute.

[rough forge]

Danke

Genug für heute, sagte kirschi

[slate hawk]

.solved

[pearl pond]

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[pearl pond]

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[merry stirrup]

Find the number of diagonals in the convex polygon of n sides

[merry stirrup]

I know how to solve this problem but I want to work on a specific approach

Can I use Combinations to solve this problem

[oak quiver]

Yes you can

[merry stirrup]

How can I approach this problem by using combinations?

[midnight haven]

I did this myself without P%C

Once

[oak quiver]

Let's think of a 8 sided polygon

[merry stirrup]

Ok

[oak quiver]

Can you draw and send it?

[merry stirrup]

Yeah

[merry stirrup]

A diagonal can be created with two vertices and Those vertices can't be adjacent

[oak quiver]

Let's select the A point

[merry stirrup]

Yeah

[oak quiver]

How many points are there which I can choose now to form a diagnol?

[frigid horizon]

Write that algebraically

[oak quiver]

?

[merry stirrup]

Diagonals from A (everything joining A and Other points except B and H )
Diagonals from B
(everything joining A and Other points except C and A )
And so on

8

Points

[oak quiver]

Number of points ?

Nope

[merry stirrup]

Wait

5

[oak quiver]

Yup

For A there are 5

For B 5 too

[merry stirrup]

Yep

Yess

[oak quiver]

Therefore total lines 8*5

[merry stirrup]

n(n-3(

[oak quiver]

Is this correct?

[merry stirrup]

yes

[oak quiver]

Nope

There will be repeation

Can you figure out how?

[merry stirrup]

n(n-3) for the general one and because of the principle of double counting it is n(n-3)/2

[oak quiver]

Correct

Wait

n(n-3)/2

[merry stirrup]

Yep

[oak quiver]

Not n-1

[merry stirrup]

Oh sorry that is a typo

But can I use nCr to find this

Like there are n number of points and 3 points shouldn't be considered if I'm taking a points

[oak quiver]

You can

what are the total number of lines that can be formed from n nonlinear points?

[merry stirrup]

nC2

[oak quiver]

Now subtract n from it

Do you see why?

[merry stirrup]

Why?

[merry stirrup]

No actually

[oak quiver]

Because there are n sides in a n-gon

Total - unfavorable type thing

[merry stirrup]

I still don't understand

[oak quiver]

Gimme a min

[merry stirrup]

I know this thing

But how is it related

Maybe I don't know the meaning of n-gon

[oak quiver]

In this we will have total 8c2 lines

[merry stirrup]

Yes

[oak quiver]

Subtracting 8 we get

20

,w 8*7/2 - 8

[jolly parrot]

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[merry stirrup]

Yes but why do I even get this , can you explain geometrically

[oak quiver]

Now what we did here was

[merry stirrup]

Yep

[oak quiver]

So on we will get 56 lines

Divide by 2 for double counting

28

[merry stirrup]

Yes

[oak quiver]

Now you see we also included AB, BC, CA ...

So - 8

[merry stirrup]

Oh you're subtracting the Line made by the sides , and the line made by the adjacent sides

Ohhh

[oak quiver]

I guess

[merry stirrup]

Line made by sides are =8

[oak quiver]

Yup

[merry stirrup]

Oh yeah

[oak quiver]

Nuce

Nice

!done

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[merry stirrup]

in n sided polygon, n lines would be there which will be the sides of the polygon, they can never contribute to the diagonal that's why you have substracted it

Thanks

.close

[pearl pond]

Channel closed

Closed by @merry stirrup

Use .reopen if this was a mistake.

[oak quiver]

Yes exactly

I wasn't able to put this into words lol

[merry stirrup]

No problem thanks anyways

[pearl pond]

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[thick trench]

Just need some help with intuition, i think there are some rigorous definitinos in topology but for non-pathological examples, is the boundary of a surface basically the edges?

[pearl pond]

@thick trench Has your question been resolved?

[thick trench]

.close

[pearl pond]

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[midnight haven]

question. if 6 people are playing russian roulette probability of anyone getting shot is 1/6?

[midnight haven]

or will we do (5/6)*(1/6) for 2nd person and so on

like multiplying probability of previous people not getting shot

[honest spear]

,w binomial distribution

[jolly parrot]

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Requested by nyxzore

[honest spear]

either is happens or it doesnt it depends on the number of events

in your case n=6
p=1/6
q=5/6 where 1 = 1-p

[midnight haven]

sorry i dont understand what number of events has to do with this

[honest spear]

if you had 4 people the chance of someone getting shot would be different

because no one might get shot

[midnight haven]

i mean there are 6 people and 6 positions for the bullet to be in then what would be the chance of each person losing

because then one of them must get shot

i thought it would be safest to go first but according to my calculations im getting 6th place has best chance of not getting shot

[honest spear]

oh so you want to see how the order affects the probabilty

[midnight haven]

yes

[honest spear]

well 6th would be the least safe

because lets say i am player number 1

theres a 1/6 chance

if i dont get shot player number two had a 1/5 chance

3 => 1/4

[midnight haven]

nono i mean

[honest spear]

6 => 1

[midnight haven]

by that logic probability of 6th person getting shot becomes 1

but

thats not my question

i mean lets say 6 people are playing

what would be the best order out of 6 to go

like would the 1st person have most chance of getting shot or 6th

[honest spear]

I'd be the same I'm certain

because what is the inherent different between person place number 1 and place number 6

[midnight haven]

yes that was what i asked initially. like would it just be same or do we multiply it with probability of the previous people not getting shot

and if not then why

[honest spear]

well lets say for person 2 the probabilty that they get shot is the p(player 1 survives) and p(bullet in the right chamber)

$p(p 1 lives) = 5/6$ ; $p(p 2 dies) = 1/5$

[jolly parrot]

Nyxzore

[honest spear]

p(event) which is player 2 dying is p(a) *p(b)

that is 5/6 times 1/5 which is --> 1/6

try that same logic for the third person

[midnight haven]

ohh okay

thank you

[midnight haven]

i get it now

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[pearl pond]

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[pearl pond]

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[magic olive]

How I find M^-1

[dusty flame]

but i assume here

[magic olive]

Right I used that

[dusty flame]

that one of the matrix is a rotation

[magic olive]

But my matrix made out of 2 matrices

So how do I apply this

[dusty flame]

make it into one first

[magic olive]

But like in the mark scheme it’s separated

[dusty flame]

i see

[magic olive]

This is the answer

[dusty flame]

use this

then

[random ermine]

yes

[magic olive]

Ohhh

I did do that but then right

[random ermine]

simpilfy (AB)(B^-1A^-1)

[magic olive]

Why u multiply by AB again?

[dusty flame]

what ab

no

[magic olive]

I’m talking about ultrafilter

[dusty flame]

):

[magic olive]

AB is just M

[random ermine]

just do it and u'll see why

[magic olive]

ohh ok daddy

So assertive

Mb

This shit has me tweaking

[random ermine]

yes M = AB

[magic olive]

Lemme try

Like that gang?

But then if I multiply by M I’ll just get an Identity what?

@random ermine

[gleaming moss]

That means it’s correct

[magic olive]

Ohh so it’s just to verify the answer

Hi again lol

[gleaming moss]

Also, for rotation matrices, the inverse is just a rotation by -theta

[magic olive]

That makes a lot of sense

So like a clockwise rotation?

[gleaming moss]

Yes. You rotate by theta ccw so to go back you rotate by theta cw

[magic olive]

Oh and I got stuck because 1/det I got 4/3 instead of 1 because I forgot to add back the 1/4

Sorry guys💀

[magic olive]

Shiii

[random ermine]

i meant simplify that for general A and B

[magic olive]

That’s really logical right there

[random ermine]

what i'm trying to saying (AB)(B^-1A^-1) = A(BB^-1)A^-1 = AIA^-1 = AA^-1 = I so (AB)^-1 = B^-1A^-1

[magic olive]

Ohh-

That’s pretty but like what it mean?

[random ermine]

wdym

(AB)^-1 = B^-1A^-1 is the punchline

[magic olive]

My matrices are not good enough to understand this gang😭

[random ermine]

so if M = AB then M^-1 = B^-1A^-1

[random ermine]

hence M^-1 is the product of 2 matrices

[magic olive]

Right I got this but then the whole A(BB^-1)A^-1 stuff

Ion get that

[random ermine]

which step

[magic olive]

How u move the I to the RHS

[gleaming moss]

A matrix multiplied by its inverse will result in the Identity matrix. That is a fact in the same way 5* 1/5 = 1

[random ermine]

where

[gleaming moss]

So BB^-1 = Identity

[magic olive]

Right and then it does AIA^-1= AA^-1=I

[gleaming moss]

And Identity * A^-1 = A^-1, so you have AA^-1 which is again the Identity matrix

Yes

[magic olive]

But this

[random ermine]

because AI = I

I is like 1 for matrices

[magic olive]

Lemme process for a sec gang

Ohhhh so u just like remove the I then AA^-1 becomes identity once more?

Init🤓☝️

Waittt I get it

So smart

Which course y’all take chat?

[random ermine]

💀

[magic olive]

What that skull for

Did I petrify u❤️

Sorry chat😞

Yo I’m joking gang u don’t gotta do me like that💀💀💀

[gleaming moss]

Already graduated

[magic olive]

Shiittt so like in ur 30s?

[gleaming moss]

Only recently graduated relax

[magic olive]

Oh mb💀

What course u took b4?

[gleaming moss]

Pure maths

[magic olive]

Damnn

That’s tough

@random ermine didi been quiet

[random ermine]

stfu

[magic olive]

Oh

Again😳

[random ermine]

dawg

[magic olive]

Jkjk

Ight thanks for the help guys🙏

see y’all

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[pearl pond]

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[chilly ermine]

why is the horizontal asymptote not -6,6 here

[light helm]

perhaps they want you to enter actual equations

[toxic fractal]

because -6, 6 is a point.
you'd want y=-6, and y=6

[chilly ermine]

it doesnt allow that

[toxic fractal]

maybe you need to separate them with ; then?

[light helm]

do they give instructions / a description of how they want the values

[edgy wren]

y=6 ?

[chilly ermine]

oh youre right

thats weird

[toxic fractal]

"," is the decimal separator

[chilly ermine]

first time theyve done this

[toxic fractal]

and this is why tests should either tell you the notation they want or be done HANDWRITTEN

[surreal relic]

Agree

This is so frustrating

[chilly ermine]

it just says comma seperated

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[pearl pond]

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[toxic fractal]

.then they need to learn what a comma is

[pearl pond]

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[midnight haven]

yes

[worldly crown]

M

[unborn abyss]

ai spam

[worldly crown]

huh

[unborn abyss]

go reread our rules and come back when you have

fwiw the approach is right

oh. yeah I think that's implicit

[pearl pond]

@buoyant hazel Has your question been resolved?

[pearl pond]

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[solar gyro]

Could someone plz help me w/ the circled number?

[cursive wraith]

can you show rest of the graphs?

I think you first gotta find f(pi)

the graph of which I'm guessing is the one on the top left I can't see fully

[pearl pond]

@solar gyro Has your question been resolved?

[solar gyro]

No, I got it

Thx, tho

[pearl pond]

@solar gyro Has your question been resolved?

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[dapper ferry]

Should c) not be 19.7<k<26.9 without the local maxima and local minima to be included as there would only be 3 real roots only and not all as it is explicitly stated?

[dapper ferry]

?

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[spring crystal]

Bro what’s the point doing derivative if ur not even using it

[spring crystal]

Representing function as power series by differentiation

Question is express 1/(1-x)^2 as a power series

[midnight haven]

hm looks like u used differenciation

[spring crystal]

Yeah I don’t get it at all

Like isn’t that power series is representing the 1/(1-x) not even the original question??

[midnight haven]

from 1/(1-x) to 1/(1-x^2)

[spring crystal]

This is question

[midnight haven]

and ur answer is correct