help-13

[cedar kiln]

@livid swan Has your question been resolved?

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[gloomy bough]

Ye, not of human ilk, aid in the unraveling of this inquiry.

[violet flume]

what have u tried

[violet flume]

.

[gloomy bough]

circle area

[violet flume]

what does that mean

[gloomy bough]

I have endeavored to calculate the area of the circle.

[violet flume]

but why?

what does that have to do with the problem

[gloomy bough]

What dost thou signify, wherefore, brother?

[gloomy bough]

We must ascertain the extent of the square, or do we not?

[warped drum]

why dont you do it like in your drawing?

blue and red

[cedar kiln]

@gloomy bough Has your question been resolved?

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[little junco]

Please help me with part b.

[little junco]

I am stuck in that part.

[gritty galleon]

Where?

Have u done any work?

[cedar kiln]

@little junco Has your question been resolved?

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[indigo blaze]

Find the sum 1+2i+3i^2+4i^3+...ni^(n-1)

[vagrant elbow]

take the derivative of the geometric series

and plug in i

[indigo blaze]

how is this a geometric series

[vagrant elbow]

it's the derivative of the geometric series

$$f(x) = a + ax + ax^2 \cdots ax^n$$
$$f'(x) = a + 2ax + 3ax^2 + 4ax^3 \cdots anx^{n - 1}$$

[wraith dagger]

jan Nejon

[long swan]

a = 1, x = i

[vagrant elbow]

yes thank you

[long swan]

you're welcome!

[vagrant elbow]

@indigo blaze do you know how to compute derivatives?

[indigo blaze]

I do

[vagrant elbow]

and the geometric series formula

well then this shouldnt be too hard for you then

[indigo blaze]

yes

[languid bison]

Another way is
S = 1 + 2i + 3i^2 + ... = (1 + i + i^2 + ...) + S i.

Well but yea, this is just discrete differentiation

[long swan]

you will also need the closed form

take the derivative of the finite sequence formula

wrt r

[indigo blaze]

So i should take the formula of the geometric series and apply it on 1+i+...i^n

And then take the derivative

right?

[vagrant elbow]

apply it on x

differentiate

and then substitute x = i

[languid bison]

Ig you can deform i as well

[indigo blaze]

thank you all for your help

.close

[cedar kiln]

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[crimson sedge]

Guys how do I maximize this trig expression

I will send the original que wait

[vast pike]

[crimson sedge]

The third one

[vast pike]

,rotate

[crimson sedge]

@vast pike U rude my writing is best!!

[crimson sedge]

The third one

[vast pike]

the book writing is better

[crimson sedge]

Nope

[glass sky]

megacosm for the win

[crimson sedge]

Any idea how we would maximize it?

[cloud inlet]

find function

take derivative

put derivative = 0

[crimson sedge]

.

[cloud inlet]

why is there a matrix ?

[vast pike]

so you have to maximize this one?

[crimson sedge]

Oh no it's modulus not matrix

[crimson sedge]

yeah

[cloud inlet]

are u using determinants?

[crimson sedge]

Yes

To calculate the area of that triangle

[cloud inlet]

ok diff determinants one row or column at a time and add them

[crimson sedge]

yeah

Ok I think I know what to do lol

Nvm

No I don't

[crimson sedge]

Yes

The angles add up to 0 symmetrically lol

Somehow

[cloud inlet]

use this rule for differentiating determinants

[crimson sedge]

Lengthy

[cloud inlet]

idk bruh but this stuff works

[crimson sedge]

Also I don't have any intuition for this so I am not gonna do it lol

[cloud inlet]

ok ur wish

[crimson sedge]

If all angles become equal that would mean the area is 0

Which is basically minimum area

lol

But we need maximum

[cloud inlet]

check the double derivative test

[crimson sedge]

Is weed legal in your country

[glass sky]

?

is bro ok

[crimson sedge]

by the time you guys were suggesting ideas out of my league i solved the 4th one

you guys should try it

its a nice question

[glass sky]

its a elllipse property

[crimson sedge]

yes

but prove it

lol

[glass sky]

already have

smh

[crimson sedge]

idk any property

how did you ?

[glass sky]

homogenisation

homogenised the line with ellipse

[crimson sedge]

you can also theory of equation

[glass sky]

i think

then a+b = 0

[crimson sedge]

and product of slopes = -1

thats basically extended version of homo

[glass sky]

which module is that btw?

istg this looks exactly like mine

but i dont think i have 3 excerises in ellipse

[crimson sedge]

youre old

youre 12th class

student

[glass sky]

no?

[crimson sedge]

my module is new

[glass sky]

im in 11th dude.

institute?

[crimson sedge]

then your class people are scamming you

[glass sky]

^

[crimson sedge]

whats yours

[glass sky]

ftj

[crimson sedge]

same

3 exercises

ok

[glass sky]

intresting , is that ellipse module?

[crimson sedge]

yes

[glass sky]

lemme check i dont remember tbh

[crimson sedge]

a

.close

[cedar kiln]

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[glass sky]

oh btw your question

i think you could not use sin (a-b) and maximise the product using am gm

[cedar kiln]

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[dry cargo]

Hey there can someone please take a look am i went something mistake?

The ans givem min x is 12

[sonic robin]

x is maxima

you need maximum

They asked area not the value of

x

substitute the value of x in A(x)

[dry cargo]

yea i understand but look at my working steps my x is 8 i cant figure it out where's the mistake...

[sonic robin]

the

congruency

part

redo the congruency part

[dry cargo]

huh

[sonic robin]

yes

[dry cargo]

how

[sonic robin]

it should be um

TP/TS

=

PQ/Su

now solve

[dry cargo]

is same ans as i did bro

y = 96 / x

[sonic robin]

oh yeah

its correct

then

[sonic robin]

answer isnt 192?

i see no mistake

then

yup youre right

there's no mistake

the value of x is also correct

[sonic robin]

dude

you did get the answe

[dry cargo]

But how the x is 12 ?

[final meteor]

you switched the places of x and y

[sonic robin]

who said its 12?

yeah

[final meteor]

in your figure

[sonic robin]

.....

you said you did

but you didnt

[cedar kiln]

@dry cargo Has your question been resolved?

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[cedar kiln]

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[dire turtle]

AB is a diameter AC = CD

AP = PC = CQ = DQ

how can i prove that triangle APO and BDQ are Congruence triangles

thats what i have dont so far

[opal basin]

Do you see why angle CDB = angle CAB?

[dire turtle]

yeah

and also AP = QD

so now how do i prove that AO = BD

[opal basin]

It seems like BQD and APQ are right angles, is that somehow given?

[dire turtle]

nope

you mean APO?

[opal basin]

Yes, misread

[dire turtle]

yeah so it wasnt given

[opal basin]

Maybe send the original question

[dire turtle]

tho ADB and ACB are 90

[opal basin]

yes, by Thales

[dire turtle]

AB is a diamater in circle O.
AC and CD are equal
P is the middle of AC
Q is the middle of CD

we can draw out OD as a radius if that helps

as i think we need to prove that AO = BD

[opal basin]

It might not be the quickest way, but you can prove that APO is 90, by proving APO and CPO are congruent

something similar can be done for Q as well

[dire turtle]

well i dont even need to prove that they are congurent

[opal basin]

I mean, it's obvious, but that means that angle CPO is angle APO and they sum to 180, so both must be 90

[dire turtle]

yeah

because a line from the center that divides the chors into two is also forming that 90 degree

how can that be helpful tho

[opal basin]

Same applies to angle BQD and BQC, right?

[dire turtle]

oh

yeah i see it

so that is why they are congurent

now i gotta prove that OQ = QB

[opal basin]

is that the next exercise?

[dire turtle]

yeah its the second section

i mean

we already have 90 degrees

so maybe we can prove its an isosceles triangle

talking about ODB

[opal basin]

QB = OP, by the previous exercise

[dire turtle]

ok

[opal basin]

you meant equilateral probably

[dire turtle]

yeah

because of the radiuses

[opal basin]

yes

but you can also prove both things with one congruency

[opal basin]

because of this, proving OP = OQ is enough

[dire turtle]

but i need to prove that OQ=BQ

[opal basin]

yes, but BQ = OP, right?

[dire turtle]

why is OP = OQ?

oh

i see

because the chords are equal

so that means the distance from the center is equal

[opal basin]

yes, to be more exact maybe, you can prove that APO and DQO are congruent

unless that's a theorem you are allowed to use

[dire turtle]

yes there is

[opal basin]

oh, then it's fine

[dire turtle]

i have this sentence which i can use

thanks!

.close

[cedar kiln]

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[dire turtle]

cricles N and M are equal while their radiuses are 8 cm, they are tangent to each other at point A. BC is a tangent to both circles at point B and C. and there is another small circle which is tangent to BC and the other circlces.

i need to calculate the radius of the small circle

what i did was this

the radiuses are all parallel and BCMN is Parallelogram

i cant see what to do next

[split pike]

use pythagoras

[dire turtle]

in what triangle

[split pike]

think

this is easy

[dire turtle]

?

in what triangle

[dapper raven]

half the equilateral triangle

[dire turtle]

what equilateral triangle

[dapper raven]

how many equilateral triangles are there in the image?

[dire turtle]

zero?

[dapper raven]

my apologies

isosceles triangle

not equilateral

[dire turtle]

i see now

if i half the isoscele triangle

i will get two right triangles

[dapper raven]

sorry for the misunderstanding

wrong word

[dire turtle]

all good

that is because MA = AN right?

so OA form a 90 degree

[dapper raven]

yes

[dire turtle]

O is the middle of the small circle

lets say the raidus is r

so what would that make AO?

8 - r?

[dapper raven]

you can find all sides of those right angle triangles by using some clever ideas

yes

[dire turtle]

and NO = 8 + r

AN = 8

[dapper raven]

yes

there you go

[dire turtle]

yeah now i see

[dapper raven]

now you can find the size of O

[dire turtle]

my guess its two

[dapper raven]

that's what i got

[dire turtle]

why is DA 8 again

D being the point where

the small cricle

is tangent to BC

oh ok

its a square

i see

thanks

.close

[cedar kiln]

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[gloomy bough]

What I did just added 200 for upcoming number 500+700+900+1100+1300+1500. Is there any other ways to solve this?

[tropic oxide]

if you know how to find the sum of an arithmetic progression in general, you can view this one as a progression with 6 terms, first term 500 and common difference 200

[gloomy bough]

I have heard of that but not sure

[cloud inlet]

ok lets analys it

500 + 200 + 200 + 200 + .....

6th birthday?

500 + (200 * 5) = 1500?

[gloomy bough]

Wait what

[cloud inlet]

oh sh it says 200 more

[gloomy bough]

yeah

[cloud inlet]

im bad in english sorry let me reanalyse it

[gloomy bough]

I know how to solve it in simple way

but how to use sum of nth term

[cloud inlet]

ok got it

Sn = n/2 (2a+(n-1)d)

here a = 500
d = 200
n = 6

[gloomy bough]

oh

wait

whats n

[cloud inlet]

Sn is the answer

[gloomy bough]

and d

like what does n and d stands for

[cloud inlet]

n is the parameter used to define this relationship meaningfully
1st,2nd,3rd,....6th
but for 1st birthday its not in AP , AP starts from 2nd birthday

[gloomy bough]

oh that makes sense

[cloud inlet]

n is number of iterations, and d is the difference

[gloomy bough]

ok

[cloud inlet]

Sn = 4500)

4500 + 500 = 500

[gloomy bough]

oh okay

[cloud inlet]

wait wat but theres no answer like that

[gloomy bough]

its K

[cloud inlet]

yeah im also bad in basic arithematic lel

so yeah can u solve it?

[gloomy bough]

6/2 x (2 * 500 + (6-1) x 200)

there

equals to 6000

[cloud inlet]

wait how come its 6/2 and not 5/2??

[gloomy bough]

you said its the nth term

[cloud inlet]

oh yeah sorry

[gloomy bough]

and its asking for 6th term

[cloud inlet]

500 + 500 + 200 + 500 + 200 + 200 right?

nice k

[gloomy bough]

thank you so much bro

.close

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[mighty shuttle]

How would I compare $\frac{dx}{log(x)}_e^{e^2}$$ and $\int$\frac{e^x dx}{x}_1^2$ without Li(x)?

[mighty shuttle]

.close

[cedar kiln]

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[mighty shuttle]

.close

[cedar kiln]

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[meager pike]

does this apply to closed subsets as well?

[crystal raptor]

It does, and you shouldn't have too much trouble.proving it using this proposition

(Hope the fireworks were fun)

[meager pike]

haha yes they were

OH yeah now i got you here 😛

[meager pike]

I think i got the idea for that other one down

its pretty simple I think. A closure is literly a closed set. it has all its limit points

[crystal raptor]

that was a result from a lemma right

[meager pike]

its a mixture of lemmas i think

because its taking into account what an interior point is

what a boundry point is

and that a closed set is defined as its limit points, which are related to an arbitary sequence

[crystal raptor]

its this proposition

[meager pike]

yes

[crystal raptor]

A bar is closed so all limit points are inside it and so you have one half of the set inclusion

${x \mid \exists (a_n) \subset A s.t. a_n \to x} \subseteq \bar{A}$

[wraith dagger]

ΣΑC

[crystal raptor]

so that proposition gives you that set inclusion

you just need the other direction

[meager pike]

Let F be closed, and x_n a convergent sequence of elements of F. Since F is closed, it is equal to its own closure, e.g. it has its interior points and its boundary points. The limit point is either an interior point or a boundary point, and F contains both.

[crystal raptor]

sure thats true just seems a bit redundant when you could just cite this proposition^^

or are you just restating the prop

bc i see ur using F and not Abar

[meager pike]

i just used the prop to make this as my proof for answering the question

since i thought this prop was the solution essentially

[crystal raptor]

i mean its one half of the "if and only if" yes

you are trying to show that these two sets are equal

${x \mid \exists (a_n)_n\ \subset A\ s.t.\ a_n \to x} = \bar{A}$

[wraith dagger]

ΣΑC

[crystal raptor]

that proposition tells you that the left hand side set is contained within the right hand side set

[meager pike]

the LHS represents the limit points from the set, and the right hand side is saying that the closure is all points plus the boundry right?

[crystal raptor]

sure yeah

your definition of closure is interior union boundary

youre trying to show its equivalently all limit points

[meager pike]

man i gotta say, this stuff here makes me question my hobbies

[crystal raptor]

in a good way or a bad way lol

[meager pike]

haha, its fine. its like a crossword puzzle

tbh i dont undestand. I get what the closure is and I understand the idea of limit points being sequences

i understand their eqivilent statements

if you have a closure then it has all its limit points

[crystal raptor]

just for context (chartbit lurking :p )

[crystal raptor]

okay but this is only one half of the "if and only if"

your proposition has allowed you to show "all limit points lie inside the closure", you now need to show that "the closure is contained inside the set of all limit points"

then those two sets will necessarily be equal

@cerulean sail do you wanna jump in? i gotta start getting ready

[meager pike]

no no its okay

its enough

[crystal raptor]

okay maybe go and think about proving this second quotation

[meager pike]

haha its going through my brain dont worry 😛

have a good night and happy new year 🙂

[crystal raptor]

you've spoke about the ideas before, open balls and stuff. try splitting it into two cases, interior and boundary

okay happy new year!!

[meager pike]

okay an interior point is less than r, and an boundry point is r with the balls and such then

so its gonna be the union of points that are less than r and points that are less than or equal to r

haha something like that maybe

.close

[cedar kiln]

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[crimson sedge]

Find a basis of P2 in which the vector x+3 has all coordinates equal to 1

[crimson sedge]

i am completely lost with this problem

can someone explain to me how to do it?

[crimson delta]

so you want to find a basis (b1,b2,b3) such that x+3=1*b1+1*b2+1*b3

[crimson sedge]

ahhhh

and we know b1 b2 b3 will be in the form of $a_0 + a_1x + a_2x^2$

[wraith dagger]

Derivative

[crimson sedge]

is b1 b2 b3 in the form (1,x,x^2) ...

since it says p2

[raw gulch]

i just looked at your problem and i guessed it at once, look:

$\left{ \overrightarrow{b_{1}},\overrightarrow{b_{2}},\overrightarrow{b_{3}} \right}=\left{ 3,x-x^{2},x^{2} \right}\\rank\begin{pmatrix}
3 &0 &0 \
0& 1 &0 \
0& -1&1
\end{pmatrix}=3$

[wraith dagger]

Joanna Angel

[crimson sedge]

yes but how do i get that

[raw gulch]

you can try such method:

$\begin{pmatrix}
1 \
1\1

\end{pmatrix}=P^{-1}\begin{pmatrix}
3 \
1\0$

[wraith dagger]

Joanna Angel
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[raw gulch]

$\begin{pmatrix}
1 \
1\1

\end{pmatrix}=P^{-1}\begin{pmatrix}
3 \
1\0
\end{pmatrix}$

[wraith dagger]

Joanna Angel

[raw gulch]

P is the transition matrix from the old basis to the new basis, note that your problem has many solutions, so you just need to guess the system and then show that it is a basis, as I show the rank of the matrix

[crimson sedge]

i didnt learn rank

i just learnt coordinates, dimensionm basis, span and linear combination

[raw gulch]

In every linear space, all bases are equinumerous, this is a very well-known theorem and from this, I can conclude that every system of independent vectors in a given space, which is equinumerous to any basis of this space, is itself a basis of this space.,

The basis of the space of polynomials of at most degree 2 has three vectors, so every system of three independent vectors in this space is also a basis

[raw gulch]

So all you really need to do is guess three vectors that are independent, and this will be your solution.

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[dire geode]

#latex-testing

[wraith dagger]

Lucas
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[flat forge]

Sorry

now it's good

[cedar kiln]

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[flat forge]

Hello,
I'm doing an exercise on the analaytical way of solving the quantum harmonic oscillator.
I first start with schrodinger's equation

$-\frac{\hbar^2}{2m} \frac{d^2\phi(x)}{dx^2} + \frac{1}{2}m\omega^2x^2\phi(x) = E\phi(x)$

Then I do a substitution to nondimensionalize x $x =\tilde x / \beta$ with $\beta = \sqrt{\frac{m\omega}{\hbar}}$

By doing so I find :

$\frac{d^2\psi(\tilde{x})}{d\tilde{x}^2} - (\tilde{x}^2 - 2\epsilon)\psi(\tilde{x}) = 0$ with $\epsilon=\frac{2E}{\hbar\omega}$

Then it is suggested to use a solution of the form :

$\psi(\tilde{x}) = e^{-\frac{\tilde{x}^2}{2}}h(\tilde{x})$

So I plug it in and derive the following differential equation for $h(\tilde{x})$:

(From now on i'll use Lagrange's prime notation for derivative with respect to $\tilde{x}$)

$h''(\tilde{x}) - 2\tilde{x}h'(\tilde{x}) + h(\tilde{x})(2\epsilon - 1) = 0$

It is then said to use a solution of the form :

$h(\tilde{x}) = \sum_{m=0}^{\infty} a_{2m}\tilde{x}^{2m + p}, \quad a_0 \neq 0$

And now I have to show that there exist two independent solutions that corresponds to p=0 and p=1.

So first I derived h two times:

$h'(\tilde{x}) = \sum_{m=0}^{\infty} (2m + p) a_{2m}\tilde{x}^{2m + p - 1} $

$h''(\tilde{x}) = \sum_{m=0}^{\infty} (2m + p)(2m + p - 1) a_{2m}\tilde{x}^{2m + p - 2}$

and plugged it into the equation and I get :

$\sum_{m=0}^{\infty} a_{2m} \tilde{x}^{2m + p} [(2m + p)(2m + p - 1)\tilde{x}^{-2}-(4m+2p-2\epsilon+1)]=0$

So my guess is that the sum cannot equal zero since It would mean the wave function is always zero.
So now I think I should try to solve the following but I have no ideas.
$[(2m + p)(2m + p - 1)\tilde{x}^{-2}-(4m+2p-2\epsilon+1)]=0$

Thanks in advance !

[wraith dagger]

Lucas

[cedar kiln]

@flat forge Has your question been resolved?

[dire geode]

A series is zero if all the coefficients are zero

Oh you say that

It's unclear what your question is

Show the original question

[flat forge]

Sure, wait 5 min while I'll translate it

[wraith dagger]

Lucas

[flat forge]

(There are still some questions after in case you were wondering)

[dire geode]

Yea I think you just apply this and make observations about what p has to be

You just need to group the coefficients for each power of x by either shifting the index for one of the sums, or shift the power of x

[flat forge]

I'm not sure to understand, isn't what i've done (at the end of my initial post) what you've just said?

Because I get $(2m+p)(2m+p-1)\tilde{x}^{-2}=4m+2p-2\epsilon+1$

[wraith dagger]

Lucas

[flat forge]

and it seems strange

[royal finch]

I havent tried this problem because this amount of notation exceeds my own laziness, but I suggest starting by simplifying the problem by setting p = 0 and trying to find the coefficient corresponding to x^0 in your power series representation. What system of equations do you get when you do this?

[dire geode]

This isn't shifted

Your coefficients in a power series shouldn't depend on x

[royal finch]

If you are having trouble finding this by just observing the equation and being careful about your choice of m. Note that you can extract the nth coefficient of any power series by taking the nth derivative, dividing by n! and evaluating at x = 0.

But you probably shouldn't do it that way because this should be easy enough by solving it via "inspection"

[dire geode]

e.g., find the coefficients of the polynomial

(1+2x+3x^2) +(4x+5x^2+6x^3)

[flat forge]

Yeah I got it i think

So since I have the following equation

$h''(\tilde{x}) - 2\tilde{x}h'(\tilde{x}) + h(\tilde{x})(2\epsilon - 1) = 0$

[wraith dagger]

Lucas

[flat forge]

I see that the h' and the h will get factorized nicelly

But I'll have trouble with h''

So I should do something so that the sum depends on x^(2m+p) and not x^(2m+p-2)

To factorize it

[royal finch]

I can't decipher what you are saying, but I get the impression you didn't try either of the approaches because I think your response would be less vague if you did.

[flat forge]

I'm sorry I'm not really familiar with power series, I'm trying to "verify" if I understood what you were saying

[flat forge]

But from this message I had the impression i was on the right track with what i've said

[royal finch]

Just tell me, in the case that p = 0, what value you need to set m to for you to get information about the coefficient of x^0

[flat forge]

Well, I would say m=0 ?

[royal finch]

Is that the only m?

[flat forge]

for x^2m=x^0 I think so

[royal finch]

What happens when m = 1?

[flat forge]

I get x^2?

[royal finch]

Take your whole equation here

[flat forge]

sure

[royal finch]

and plug in p = 0 and m = 0, then plug in p = 0 and m = 1

The point is that when m = 1, you will have an x^(2(1) + 0) coefficient multiplied by a x^-2 which gives you x^0

So to extract the coefficient of x^0, you need to look at multiple values of m because of that x^-2 term. This is what Riemann meant by needing to shift the indices

Because you need multiple m's, this is going to give you multiple equations for your unknowns and you can reduce it to a linear algebra problem with a nontrivial solution.

[flat forge]

Ok, but then how should I shift the indices ?

[dire geode]

https://math.libretexts.org/Bookshelves/Differential_Equations/Elementary_Differential_Equations_with_Boundary_Value_Problems_(Trench)/07%3A_Series_Solutions_of_Linear_Second_Order_Equations/7.02%3A_Review_of_Power_Series

Do example 7.2.3

And the one after

Or review your own calculus book

[flat forge]

Oh I think I did something wrong right from the start then

Because I did not shifted the indices when deriving the sums

Wait no

The term a_0 =/ 0 in my case

So it's still correct

But then for h'' I can rewrite it as :

$h''(\tilde{x}) = \sum_{m=-1}^{\infty} (2m + p + 2)(2m + p + 1) a_{2m+2} \tilde{x}^{2m + p}$

[wraith dagger]

Lucas

[flat forge]

Oh

I have two solutions

for h''=0 at m=-1

p=0 and p=1

Ok

It' still somewhat strange but It looks better

$\sum_{m=0}^{\infty} \tilde{x}^{2m + p} \left[ a_{2m+2} (2m+p+2)(2m+p+1) + a_{2m} (2\epsilon-1-4m-2p) \right] \+ a_0 p(p-1)\tilde{x}^{p-2}=0$

[wraith dagger]

Lucas

[flat forge]

I find this equation

[cedar kiln]

@flat forge Has your question been resolved?

[cedar kiln]

@flat forge Has your question been resolved?

[flat forge]

Ok I got it

from this equation I can deduce two things

For the term in x^2m+p i can find the following recurrence relation

$\frac{a^{2m+p}}{a^{2m}}= \frac{4m+2p-1-2\epsilon}{(2m+p+2)(2m+p+1)}$

[wraith dagger]

Lucas

[flat forge]

This is the condition for the first term to equal 0. Now for the second one it is oviously p=0 and p=1
therefore I have two "couples" of conditions:

p = 0 and the recurrence relation, and the same for p=1

And if somehow a_0 wasn't non-zero, I would only have the recurrence relation

thanks for your help and happy new year !

.close

[cedar kiln]

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[gloomy crater]

The base of a solid is a circular region of radius 3. For each cross-section perpendicular to the x-axis is an equilateral triangle. What is the volume of this solid?

[hot isle]

what do you think the solid looks like

[gloomy crater]

i can picture it but idk how to describe it

[hot isle]

if you draw a circle and then draw small triangles from the edge to big triangles in the middle

it kinda looks like a cone

which it should be

then using equilateral triangle ratios, you can find the height of the cone

solve for the volume

[gloomy crater]

how do you get the function

[hot isle]

do you need a function?

wait i might be trippign

[hot isle]

i found a solution online to a similar problem, since i think im wrong i can send you the link instead

[gloomy crater]

the equation of the circle is x²+y²=9

so y=±√(9-x²)

[hot isle]

yeah

[gloomy crater]

so top function minus bottom function

√(9-x²)-(-√(9-x²))

2√(9-x²)

ohhh

f(x)=2√(9-x²)

and the volume is $\frac{\sqrt{3}}{4}\int_{-3}^3 (f(x))^2 \dd{x}$

[wraith dagger]

mXd

[hot isle]

i mean you dont need to use integrals for this

[gloomy crater]

i’m in ap calc so that’s what we have to do xd

[hot isle]

😭

ngl idk how to do this with regular integrals

im thinking of using polar integratoin but like idt that's what ur going for

oh wait i got it

[hot isle]

yeah this makes sense

[cedar kiln]

@gloomy crater Has your question been resolved?

[cedar kiln]

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[summer lintel]

is this correct

[earnest socket]

What’s the question

[summer lintel]

just let last part

i want to express the solution of the matrix [3,-1,0,0]

[dire geode]

Solution to what

[cedar kiln]

@summer lintel Has your question been resolved?

[cedar kiln]

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[viscid scaffold]

so I am analyzing the probabilities of 5 card hands where the player is dealt 2 cards, then 3 more cards are dealt onto the table.

if the player recieves say 2 jacks, what's the probability that when the dealer deals the house cards, the five card hand will contain 3 of a kind?

i'm not quite sure how i should approach the 2 cards + 3 cards, but I do know how to calculate the total probability of the hand even occurring

[solid juniper]

the conditions under which the five card hand will contain 3 of a kind are: the house contains exactly 1 jack, or the house has 3 of a kind

so you just want the probability of drawing exactly 1 jack or 3 of a kind in 3 draws from the remaining 50 cards

[cedar kiln]

@viscid scaffold Has your question been resolved?

[viscid scaffold]

ok so the probability of making it a triple should be something like
$P(\text{make triple w/ jack})+P(\text{draw 3 of a kind (not jack)})$

[wraith dagger]

notnick

[lyric latch]

Try composing the event into being dealt 3 cards one-by-one (you will need to decompose it further, but this is a start)

[viscid scaffold]

well can't i calculate the probabilities with n(E)/n(S)?

so i should just be able to find the total selections of just jack and 3 of a kind dealt out in those 3 cards

and divide by 50c3

[lyric latch]

ah yeah you can

i think

yeah you can

how would you decompose your (3 of a kind) event?

[viscid scaffold]

i would say ${12\choose1}{4\choose3}/{50\choose3}$

[wraith dagger]

notnick

[viscid scaffold]

choose a card out of the remaining 12 * 3 out of 4 suits

[lyric latch]

I think this is right

[viscid scaffold]

so if i just add the two events (since its OR) i should be at my answer then

so like

[lyric latch]

yeah

cos the events where you draw jack and where you form a triple are already exclusive

[viscid scaffold]

$\frac{{2\choose1}{12\choose2}{4\choose1}^2}{{50\choose3}}+\frac{{12\choose1}{4\choose3}}{{50\choose3}}$

[wraith dagger]

notnick

[lyric latch]

I'm still kind of uncomfortable with the 4C3 / 50C3

[lyric latch]

what about say (48/50) * 3/49 * 2/48 ?

[cedar kiln]

@viscid scaffold Has your question been resolved?

[cedar kiln]

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[viscid scaffold]

why so? shouldn't that be the way to choose 3 suits

[viscid scaffold]

it's all the ways to select 3 suits for 4 cards

multiplied by the 12 different cards no?

since we know none of those 12 have been dealt

[cedar kiln]

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[cedar kiln]

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[sonic robin]

I've tried using properties

wasnt able to simplify

<@&286206848099549185>

[cedar kiln]

@sonic robin Has your question been resolved?

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[sonic robin]

.reopen

[cedar kiln]

✅

[sonic robin]

<@&286206848099549185>

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[sonic robin]

1

bymistake

i tried

i tried using properties

i might be doing it wrong

i mean

wrong approach

[final oak]

What have you tried?

[sonic robin]

i used properties

[final oak]

can you be more specific?

[cloud inlet]

he meant odd even function right?

[sonic robin]

yes

f(a+b-x)

[cloud inlet]

unfortunately it needs symmetric intervals

oh kings rule

[sonic robin]

Ive tried queens rule

as well

[final oak]

to me it seems like you just have to do u-sub 2 times

[sonic robin]

the limit half

[cloud inlet]

unfortunately queens rule doesnt work

it needs 2 times the limit

[final oak]

5x^4 in numerator gives easy u-sub for 1+x^5

[cloud inlet]

tried
1 + x^5 = u?

[sonic robin]

then i got stuck

the power

[final oak]

write x^5 as u-1

[cloud inlet]

yeah

[final oak]

separate the terms

for the new one there is a new u-sub

try p=1/u

or any other variable name you prefer

[sonic robin]

okay i'll try

[cloud inlet]

u goes from 1 to 2

[final oak]

yeah keep changing the limits and signs according to your substitution/whatever other properties you might use

[sonic robin]

ahhh its getting complicated

[final oak]

what did you get?

[sonic robin]

werid thingy

ill do it again

[final oak]

numerator was 1+(u-1)^2022 and denominator u^2024 right?

[cloud inlet]

kasper should we try binomial splitting?

[sonic robin]

thats the issue

[sonic robin]

what is thta

[cloud inlet]

u can seperate out 1 and u-1^2022

[final oak]

with 2022 power that would be very impractical

[cloud inlet]

left side one is easy

yeah

[final oak]

yeah for the right side one

you have (u-1)^2022/u^2024

write it as (1-1/u)^2022*1/u^2

and then do the substitution for 1/u

[cloud inlet]

hey yea

if u keep 1/u = t

[sonic robin]

um

[cloud inlet]

-u^-2 du = dt

u^-2 is in bottom

multiply and divide by -1

[sonic robin]

im confused

[cloud inlet]

new integral is

bro

which step

[sonic robin]

why the 1/u

[cloud inlet]

cus u see another subs for 1/u

that is u^2 in denoms

[sonic robin]

okay

[sonic robin]

okkk

[cloud inlet]

new integral limits are from 1 to 1/2

[sonic robin]

now?

[cloud inlet]

now sub 1-t as k

k goes from 1/2 to 0

there will be a minus sign

[sonic robin]

its so looooong

[cloud inlet]

use it reverse the limits

or whatever

it doesnt matter

final result after integration i got is

k^2023/2023

from 1/2 to 0

but whoever made this integral is just insane

.close?

[sonic robin]

um

this question was supposed to be solved

in under 3min

[cloud inlet]

yeah we did

it barely took us

simple question with 3 u susbtitituns

[sonic robin]

I think they did use some property

they used only 1 subs

in the first step

[cloud inlet]

ahh integrals are like sea

u can use whatever that clicks

theres is nothing wrong or right

[sonic robin]

yeah

thanks though

have a nice day

.close

[cedar kiln]

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[upper garnet]

Happy new year!

[upper garnet]

My wishes are that you all pass your math exams!

And become Gausses, or Eulers!

[hollow minnow]

Happy new year!

Do you have a question?

[upper garnet]

No

[tropic oxide]

discussy

or chill

[hollow minnow]

Okay, I'll close.

.close

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[crimson sedge]

Me again

I came back

I cannot find any relation between the fn

To have a closed form

[opal basin]

I think you have to prove that f_n(1) = e^n

[crimson sedge]

It's not always the case

[opal basin]

yeah, nvm

[crimson sedge]

Im trying the derivatives

And each time i get a polynomial multiplied by exp x

Helloooo? 🥺

[sleek condor]

what did you get for the first few? n up to 3 or 4 for example?

[crimson sedge]

f_0(x) = e^x
f_1(x) = xe^x
f_2(x) = xe^x + x^2e^x
f_3(x) = xe^x +3 x^2e^x + x^3e^x

hello @sleek condor

oh no

f4 is wrong

no nvm

[sleek condor]

sorry i gotta go

[crimson sedge]

<@&286206848099549185>

<@&286206848099549185>

.close

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[mighty tree]

. A uniform rod 𝐴𝐵 of mass 10𝑘𝑔 and length 6𝑚 rests in equilibrium with 𝐴 on rough
horizontal ground. The rod is resting on a smooth peg at 𝐶, where 𝐴𝐶 = 4𝑚. The angle
between 𝐴𝐵 and the ground is 𝜃,
Where tan 𝜃 = 0.4
Given that the rod is on the point of slipping,
Find the coefficient of friction between the rod and the ground giving your answer to 2
significant figures.

[mighty tree]

Not sure how to get to the correct answer

[cedar kiln]

@mighty tree Has your question been resolved?

[cedar kiln]

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[toxic coral]

this question doesnt make sense to me

these were the only options

but shouldnt it be 0,0 and then 5,8??

SOMEONE HELP

IM GOING CRAZY

[limber marsh]

wait like yeah what

[toxic coral]

DONT JUST SAY THAT

HELP

[tropic oxide]

calm down

[limber marsh]

lol chill out i have like 4 phrases

[tropic oxide]

allcaps helps nobody

are we sure those are the only 3 answer options

[toxic coral]

YES

[tropic oxide]

yeah the ones you've shown are all wrong

is it possible at all to skip this question

[arctic condor]

It sounds very urgent for some reason
Being calm and focused will help to think better
💕🐾
🐾💕

[tropic oxide]

or is this an "if i don't do this i will literally get expelled from school and my life will be ruined" situation

[toxic coral]

ok.

[toxic coral]

i was actually helping a friend but then i came here because i thought i was going crazy

[low plover]

Since 5,8 is not in option take 8,5

[tropic oxide]

and are the stakes high at all for the friend?

[toxic coral]

that isnt there either

[toxic coral]

hmmm

im not sure

[low plover]

option third has 8,5

[toxic coral]

they are sure there are no more options.

[toxic coral]

ok wait

[limber marsh]

oh wait oh my god yeah

[low plover]

wait

[tropic oxide]

smart boy i am not sure it's a good idea to try to look for the least wrong answer of 3 obviously wrong ones

[low plover]

the firsr option is (b)

so there will be an option (a)

[toxic coral]

WHAT

[tropic oxide]

it is ??

[limber marsh]

oh wait

[toxic coral]

STOP CONFUSIN

[tropic oxide]

this is option 3

it is wrong

[low plover]

see first one is marked (b) there will be an (a)

[toxic coral]

what da hel

are u saying

[tropic oxide]

hes saying maybe your friend missed a 4th answer option

which is actually the first

but who knows

maybe the question itself is fucked

[toxic coral]

👍

[cedar kiln]

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[toxic coral]

@tropic oxide @low plover @limber marsh there was a 4th option that my friend missed...

[cedar kiln]

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[mighty tree]

. A uniform rod 𝐴𝐵 of mass 10𝑘𝑔 and length 6𝑚 rests in equilibrium with 𝐴 on rough
horizontal ground. The rod is resting on a smooth peg at 𝐶, where 𝐴𝐶 = 4𝑚. The angle
between 𝐴𝐵 and the ground is 𝜃,
Where tan 𝜃 = 0.4
Given that the rod is on the point of slipping,
Find the coefficient of friction between the rod and the ground giving your answer to 2
significant figures.

[mighty tree]

Not sure how to solve

[low plover]

you assumtion that Rc is horizontal is wrong

rc will be applied perpendicular to the rod

[mighty tree]

Ah

Is that all that's wrong?

[cedar kiln]

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[hollow storm]

can somebody please help?

[hollow storm]

THEOREM The set N is infinite.
Proof. By contradiction: suppose that there exists an application f: N --> In bijective; f|In is injective and has, as immediately occurs, an image that is a proper subset of In. Thus In is placed in one-to-one correspondence with one of its proper subsets, which is absurd.

[soft zealot]

What part do you want help on

[hollow storm]

The proof, I don't get it

Visually I don't understand

[soft zealot]

What is ln or In or wtv

[hollow storm]

In = {0,1,2,...,n-1} if the set has cardinality n

[soft zealot]

Alright, so obviously those are finite

You can try proving any injection In -> In is actually a bijection

Then that might make things more obvious

[hollow storm]

What is f|In ?

[soft zealot]

the function f but only taking elements of In as inputs?

[hollow storm]

But isn't the domain N?

How are the inputs elements of In?

[kindred gale]

its a new function called f| ln : ln --> ln and f| ln( x )=f(x)

called f restricted in ln

the arguments says we have found a way to put ln (finite set) into something smaller (still finite) than it self by a bijective map

which is absurd

and thus such application f can not exist, leading the conclusion that f must be not finite (infinite)

[cedar kiln]

@hollow storm Has your question been resolved?

[hollow storm]

Okay I get it now thank you!!

.close

[cedar kiln]

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[bronze pivot]

Guys explain this to me i cant understand orthonormal only orthogonal

[kindred gale]

orthogonal == perpendicular
orto+normal == perpendicular(ortho) + length 1 (normal)

[bronze pivot]

What is this dirac delta looking thing

Sorry im on ipad rn i cant latex

[kindred gale]

so orthogonal only asks delta(i,j) = 0 when i is not j

[hollow minnow]

that is just the dirac delta

[bronze pivot]

Ohhh i see

inner product with others in the set is 0

innerproduct with itself is 1

[hollow minnow]

Yes

[bronze pivot]

Whats the point of number 3

[dire geode]

Have you learned what a basis is

[bronze pivot]

Yeah

[dire geode]

Apply the definition of basis to S and T for the given v

[cedar kiln]

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[sinful field]

so basically i have to find the positive and negative coterminal angles

normally i would resort to add 360 and subtract 360

but i saw somewhere that -920+360(2)
and -920+360(3) also work

how does that work ?

[cedar kiln]

@sinful field Has your question been resolved?

[sinful field]

.close

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[kindred hornet]

can i get some help on what to do?

[native vale]

i mean if ur allowed to compute u can pretty much just use cas ig

[kindred hornet]

This is what I drew

[kindred hornet]

I don’t think I’m allowed to use a computer

[native vale]

i guess they dont mean computer when they say "compute"?

[native vale]

also yeah this looks right

[native vale]

u know where to go from here?

[kindred hornet]

Not really

[native vale]

i would try finding the side length of the small square

[kindred hornet]

How would I do that?

Do I have to use a theorem

[native vale]

umm not really
i figured it out u just need to see that this is equal to the radius of the circle

[kindred hornet]

The black line is the radius right

[native vale]

yes

[kindred hornet]

How does that help?

[native vale]

use pythagoras on this triangle lol

in terms of side length of small square and radius

and the big square ofc

[kindred hornet]

Well

The bottom leg of the triangle is half the side length of the small square

And the hypotenuse is the radius

Not sure about the other leg

@native vale I have to introduce variables?

[native vale]

yes

[native vale]

call the side length of small square s or something like that

[native vale]

the other leg would be half the side length of big square plus side length of small square

u can find half the side length of big square pretty easily

[kindred hornet]

Oh yeah

But what about the radius of the circle

That is a variable too

[native vale]

no u can find that out pretty easily by using pythagoras on this

[kindred hornet]

Oh

Should be sqrt(2)/2

Thank you for your help!

[cedar kiln]

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[cedar kiln]

@kindred hornet Has your question been resolved?

[weak ice]

If my little sister is 10 years old and im 5 years older than her. How much older am i?

[kindred hornet]

Huh

[south abyss]

15

uh?

[kindred hornet]

.close

[cedar kiln]

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[crimson sedge]

im a bit confused as to how E(X0 + X1 ...) = the answer

isnt part b asking for the expected number of stickers the collector must buy to fill his album

wouldnt summing those individual geometrically distirbuted events just tell me the expected amount of stickers i would need to collect to obtain a new unique sticker ?

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

I am trying to prove the fact that $x + x = x$ and $x \cd x = x$ in a Boolean algebra using the postulates. I will just prove the former and conclude the latter by the duality of a Boolean Algebra. 

\vs{5 mm}
\env{alignat*}{{2}
x + x & = (x+x) \cd 1 \q &\c b{x \cd 1 &= x} \\
&= (x+x)\cd (x+x') &\c b{x +x' &= 1} \\
&= \bs{x\cd(x+x')} + \bs{x' \cd(x+x')} &\c b{x \cd (y + z) &= (x\cd y) + (x \cd z)} \\
&= \bs{(x\cd x) + (x\cd x')} + \bs{(x'x) + (x'x')} &\\
&=\bs[\big]{(x\cd x) + \underbrace{(x\cd x')}_{=0}} + \bs[\big]{\underbrace{(x'\cd x)}_{=0} + (x'\cd x')} &\c b{x \cd x' &= 0} \\
&=(x\cd x) + (x'\cd x') 
}

[crimson sedge]

either something went wrong or im going no where with this

like what do i do now

okay fuck something definitely went wrong lol

the last line is saying [
(x\cd x) + (x' \cd x') = x + (x'\cd x') = (x+x') \cd (x+x') = 1 \cd 1 = 1
]

[crimson sedge]

so what went wrong

oh its this i somehow summoned an extra x'

[nimble mountain]

it feels so weird

[crimson delta]

what postulates are you working with? I never worked with boolean algebras but I always thought idempotence was part of the axioms

[crimson sedge]

the Huntington postulates

i will post them

[nimble mountain]

in your third line what substitutions are you using?

[crimson sedge]

i like

foiled it basically by the distribution of the multiplication operator over addition

[crimson delta]

ok interesting. the 4b and the fifth one must be quite powerful. cause the other ones are just standard ring stuff

[nimble mountain]

like whys the second one a x' multiplied by stuff

shouldnt it be x multiplied by stuff

[crimson sedge]

.

i made a mistake yeah

[nimble mountain]

ok um

ive no idea what u are even doing so ill get out nwo

[crimson delta]

do you have 0x=0?

[crimson sedge]

no thats a theorem i will also be proving

but like

the issue is now

im still going in circles i think

i mean is it fine if i take some of the other theorems as fact and then use them to prove the implication

[crimson delta]

wait is there not even associativity?

[crimson sedge]

no

i dont have associativity

i mean associativity is valid

but its not a postulate

the proof for that is very convoluted i believe

[crimson delta]

it follows from the other ones? very interesting

[crimson sedge]

yes

[crimson delta]

hmm ok. sry but I cant help you. but interesting stuff

[crimson sedge]

no worries

i will skip proving associativity and De morgan because those are pretty long i believe