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[cedar kiln]

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[sturdy helm]

We've started learning about differential equations I was wondering if anyone had a good website or links to math videos for introduction to differential equations.

[past wave]

Khan academy

[sturdy helm]

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[astral mulch]

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[crimson sedge]

I’m trying to do this Lagrange problem, but it seems to complicated is there something I’m missing?

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[vapid kayak]

Need help with proving
"Claim: For any integers a,b,c,d,p∈Z, if a≡b(modp) and c≡d(modp), then ac≡bd(modp)."

[vapid kayak]

if someone knows how you could manipulate this into something

[mental trail]

refer back to the definition of x ≡y(modp)

(for example p|(x-y))

then your job is writing ac - bd as a multiple of p

(hint: -bc +bc)

[vapid kayak]

I don't quite understand.

[mental trail]

you wanna show ac - bd = (...)*p

my hint is writing ac-bd = ac - bc + bc - bd

[vapid kayak]

how does p relate to that

[mental trail]

?

[vapid kayak]

how would we define p that we could figure out to put it back into this equation

[mental trail]

a - b and c-d are multiples of p

use that info

[vapid kayak]

right and I want to prove that ac-bd = p*(some number)

so then I can set it equal to itself, how do I use - bc + bc

I mean what relation could they have

c is some remainder and b is some value that p divides

ah wait do we divide by c and b to get isolated terms we can define?

like

(ac-bc) + (bc - bd)

and then

(a-b)/c + (c - d)/b

then we'd have isolated terms that are multiples of p, so we could replace them with

like p*(n), for n in Z

so

(p(n))/c + (p(n))/b

I don't know how that helps me

<@&286206848099549185>

[mental trail]

what?

it's not dividing

it's factoring

[vapid kayak]

sure sorry so then factoring out c and b

oh

yeah yeah I see it now

wow that's dumb thanks

pnc + pnb

so ac-bd = pnc + pxb

[mental trail]

yeah, be careful it's not necessarily the same "n"

[vapid kayak]

I see, that's true

[mental trail]

$pn_1c + pn_2b$

[wraith dagger]

rafilou is not not born in 2003

[vapid kayak]

factor p?

$p(n_1c + n_2b)$

[wraith dagger]

Ayu beem

[vapid kayak]

I see, and since the term we have multiplying p is an integer because the terms in it are integers, we have that ac-bd = p(...) some integer

it doesn't matter than c and b are variables we already had or anthing like that

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[long chasm]

Here is my working out

I believe the y is right

But everything after I suspect somewhere a mistake occurred

I'd like help spotting the mistake

It's reduction order question**

<@&286206848099549185>

[cedar kiln]

@long chasm Has your question been resolved?

[long chasm]

No

<@&286206848099549185>

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[digital pawn]

I am currently in theorems and proofs in geometry. I am having a hard time trying to figure out how to start out.

[digital pawn]

Its just I don't get it, it does not say if it is getting triseced, or anything

Trisected*

[digital pawn]

Oh, thank you

.close

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[odd scroll]

hey

[odd scroll]

i need help

[devout urchin]

linear functions are functions that essentially only have one degree

y = 2x + 4

it represents constant change it's the same every time

rise/run

[fading parcel]

number 3 is linear because it is a straight line

[devout urchin]

i'd recommend learning your parent functions

[fading parcel]

all the others are non linear because they curve

[odd scroll]

o

thank you

[fading parcel]

ofc

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[maiden fractal]

my initial idea to solve this was to write the polynomial as:

x^100 = q(x) (x+1) + R

[maiden fractal]

all i know is that x^100 / (x+1) = (-1)^100 = 1

remainder theorem

i dont know how to proceed though

[crimson sedge]

you have your R as 1 from here

reminder theorem

[crimson sedge]

substituting it into the equation
we get
x^100 = q(x)(x+1) + 1

now do you see how we can find q(-1)

?

[maiden fractal]

uhmm

q(x) is now (x^100 -1) / (x+1)?

[crimson sedge]

yep!

brackets

[maiden fractal]

would that not be 0/0

[crimson sedge]

:O

[maiden fractal]

i dunno 😓

[crimson sedge]

we are supposed to simpligy (x^100 -1) / (x+1) so that we get a polynomial

how to factor x^100 -1

hmmm

[crimson sedge]

basically

excuse the bad hand writing

but the reason we cant solve this because we have (x+1) and x is -1

so what we need to do is find a way to remove the x+1 from the denominator

[crimson sedge]

can you think of something ?

[maiden fractal]

uhmm

i was thinking since x^100-1 is a difference of two squares it could be written as (x^50+1)(x^25+1)(x^25-1)

but idk if thats useful

[crimson sedge]

yes!

why did you stop

since we have to eliminate x+1 from the denominator, we need it in the numerator

[crimson sedge]

simply keep repeating this until you get x+1

[vivid tundra]

couldn't you just do polynomial long division from the start?
then just ignore the remainder

[crimson sedge]

as in

[crimson sedge]

at this step?

[maiden fractal]

oh but x^25 isnt a perfect square is it

[crimson sedge]

uh

[vivid tundra]

at x^100/(x+1) on the original question

a pattern will emerge really quickly

[crimson sedge]

can you show what you mean, i am dumb today

[maiden fractal]

i mean like

you cant simplify the numerator further i think?

not sure

[crimson sedge]

(x^5-1)(x^5+1)

[maiden fractal]

isnt that equal to (x^10 - 1)

[maiden fractal]

uhh

x^99 - x^98 + x^97.. right?

[vivid tundra]

yea

[crimson sedge]

oh yeah, alright time to sleep then

my bad

[maiden fractal]

all good!! dw

[maiden fractal]

okay

thank you!! got it

thank you everyone :3

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[vivid tundra]

np!

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[dire thicket]

hi guys, so I have a physics screening test today and this is their past paper

im stuck on question 2, I’ve thought about using pyth theorem + 1.1km but it seems wrong

ive also tried searching online but no videos seem to have a map similar to this, they usually have a guard

does anyone know how to solve this question? thank youu

[void sand]

you can use the Pythagorean theorem

just not in the way that you initially did it

[dire thicket]

specifically how

[void sand]

sorry for the wobbly lines

[void sand]

the "legs" of the triangle are going to be the blue and green lines

you can calculate blue by adding together 800m and the vertical component of the 1.1km

and the green by adding together 1.5km and the horizontal component of the 1.1km

[dire thicket]

do i get that by using sin cos tan?

[void sand]

you'd have to use two of the trig ratios in order to get the vertical/horizontal component of the 1.1km vector, yes

[dire thicket]

OHH i see thank you so much 😭😭

[void sand]

I saw that question

the answer is no

[dire thicket]

should be this right

[void sand]

you can calculate the angle

[void sand]

well, the angle you drew is unnecessary

[dire thicket]

oops

[void sand]

it's extraneous information, even if it is correct

your goal is to get red

and you get it via getting green and blue, and then using Pythagoras

green = all horizontal vectors on the path going from A to B added up
blue = all vertical vectors on the path going from A to B added up

[dire thicket]

yup ok

[void sand]

and you can use this to get the angle b/w A and B too

[dire thicket]

using Pythagoras?

[void sand]

well, that would get you the hypotenuse

to get the angle, you can reexamine the green and blue lines

let me try to draw

[dire thicket]

okok thank youu 😭😭

[void sand]

maybe a little complicated, but I can try to explain

inside the orange circle is what we need to find: theta

that angle is the angle between A and B

but you can clearly see that the small dotted triangle inside the orange circle looks exactly like the bigger one with red hypotenuse and dotted green and blue edges

so to calculate theta, we can just use the bigger triangle

note that tan(theta) = opp/adj = blue/green

so theta = arctan(blue/green)

but you already calculated blue and green, from part a)!

so it turns out there's very little work you need to do to get your theta

does that make any sense?

[dire thicket]

yes thank you so much!!

let me try calculating

[void sand]

[void sand]

[dire thicket]

oops

[void sand]

you mixed the two up

[dire thicket]

one sec

[reef halo]

you can find the vector from a to the 30 degree angle and add that vector to the 1.1km vectro

[void sand]

the angles don't match, so you can't

it will not be a straight line

[reef halo]

no

like adding the vectors from head to tail

[void sand]

that won't be the minimum distance

[reef halo]

not just adding the 2 lines

[void sand]

which is what we need

[reef halo]

this is what i mean

sorry in blue i meant to write u+v

[dire thicket]

wait is it this one

[reef halo]

yes

[void sand]

sure, but to do this in practice requires you to decompose both vectors into their components

and at that point, you may as well decompose all 3 of them right from the get go

[dire thicket]

should r be 2800 or 2024

[void sand]

unfortunately, I have to go to bed now

[void sand]

so I'm not going to be able to help you finish this one

but I have given you all the pieces you need already

[dire thicket]

thank you!

[dire thicket]

good night 😭

[void sand]

good luck with the problem, and I wish you a wonderful day or night!

[dire thicket]

.close

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[neon idol]

i have no idea

[loud gyro]

partial fractions

rewrite y as a generalised summation, and use partial fractions on it

[neon idol]

hmmm

i see

[loud gyro]

1+n^2+n^4 has a very nice factorization

[neon idol]

so like 1/(x^2+x+1)(x^2-x+1)?

[loud gyro]

thats just factorizing the denominator

[neon idol]

yea

[loud gyro]

partial fractions are 1/factor - 1/factor

[neon idol]

i need to find the numerator'

kind of confused on that one

[loud gyro]

can you write this, and simplify the addition?

youd see the numerator appear in some form or the other

oops typo

[neon idol]

numerator goes like uhh a(factor) + b(factor) = 1?

or no

[loud gyro]

= 1 is wrong

just a(factor2) + b(factor1)

[neon idol]

of

oh

i forgo

[loud gyro]

just write a/factor1 + b/factor2 and go at it

[neon idol]

do we need to find like the a and b furthermore?

[loud gyro]

yes

find suitable a,b to match the problem

[neon idol]

alright ill try to continue it, i gtg now

thanks for the idea and hint

i get it now

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[lethal canopy]

im not sure how to do (b)

[cedar kiln]

@lethal canopy Has your question been resolved?

[lethal canopy]

<@&286206848099549185> 🙏

[rotund yacht]

Not probability theory 😭

[lethal canopy]

yeah im so lost lol

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[wind forge]

$\begin{aligned} y^2+25&=x \\
10y&=x \end{aligned}$

[wraith dagger]

Urara4ya

[wind forge]

If
(x,y) is a solution to the system of equations shown, which of the following is an
x-coordinate of the solution?

[grave karma]

oh

[grave karma]

doesnt the first part refers to the property of the residuals in a simple linear regression model??

[lethal canopy]

yeah i did that part already

[wind forge]

umm

[grave karma]

oh then which part do u want help with?

specify it

[crimson sedge]

make a channel

[wind forge]

.close

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[grave karma]

[grave karma]

bruh

[wind forge]

what

[grave karma]

why did u close it?

ahm

[wind forge]

i think you guys were discussing previous question so just not to intervene

[grave karma]

i was helping them in this question

[grave karma]

in the second part u plug X into the estimated regression line equation

like

[lethal canopy]

ik but i have to show the proof

[grave karma]

substitute X(-), the mean of X value into the equation duh

[lethal canopy]

not really how that works

[grave karma]

:?

The resulting Y> will be the mean of the Y values

[grave karma]

let me do it one instance

[grave karma]

does this show work :d @lethal canopy

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[lethal canopy]

i guess so

[lethal canopy]

thx for the help

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[grave karma]

np

🫡

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[whole inlet]

why is a not a lower bound of {d, f}, I am guessing I am reading the hasse diagram wrong, but i know that it is read from bottom to top but in question it is given horizontally

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@whole inlet Has your question been resolved?

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@whole inlet Has your question been resolved?

[whole inlet]

<@&286206848099549185>

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[oblique lynx]

hi is this projection onto subspace correct?

[cedar kiln]

@oblique lynx Has your question been resolved?

[dawn junco]

it feels sus

proj W(x) = proj W1 (x) + ... + proj Wn (x), how do you know that is true here ?

@oblique lynx

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[cosmic marten]

Why is it that $E(E((X|Y)^2)= E(E(X^2|Y)) = \sum_{k,j} j^2\frac{p_{x,y}{j,k}}{k}$ , where X and Y are random variables?

[wraith dagger]

Pen

[cosmic marten]

I don't understand why the first equality holds, why is it not E(E(X^2|Y^2))?

[tough hemlock]

i need help , i don't remember how to find x and y

[cedar kiln]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

@cosmic marten Has your question been resolved?

[cosmic marten]

<@&286206848099549185>

The only reason that I can think of is a notation issue

Since, per definition

$E((X|Y)^2) $ should be $E(X^2|Y)$ otherwise you'd end up with $\sum_{j} j^2p_{X|Y}(j|k)^2$

[wraith dagger]

Pen

[cosmic marten]

bump

[cedar kiln]

@cosmic marten Has your question been resolved?

[cedar kiln]

@cosmic marten Has your question been resolved?

[cosmic marten]

anyone?

[cedar kiln]

@cosmic marten Has your question been resolved?

[cedar kiln]

@cosmic marten Has your question been resolved?

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[vernal sentinel]

Guys, how am i supposed to do this if i don’t know the total number of households in the whole Arka?

[slate lintel]

i don't think it matters, it's a big number

[cerulean sail]

(this is a binomial question - how many houses there are in total doesn't matter)

[vernal sentinel]

Ooooohh

[vernal sentinel]

Okayyy thank you so much

[vernal sentinel]

Thank you so muchh!

.close

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[sleek gale]

i found found the derivative of f(x) which i got as -8cos(2x) and then set that equal to zero and got pi/2 and 3pi/2 and im not sure why it’s wrong! please help if u can

[opaque root]

Show your equations

Cause you don’t get pi/2 and 3pi/2 solving that equation

@sleek gale

[sleek gale]

NVM I GOT IT I WAS SETTING COS X= 0 when it was COS 2X=0 so 2X is equal to pi/2 and 3pi/2

THANK YOU THO

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[opaque root]

There are 3 solutions between pi/2 and 2pi

And pi/4 is not one of them

[sleek gale]

i got 3pi/4,5pi/4, and 7pi/4! does that sound right?

[opaque root]

Yeah

[sleek gale]

cuz this is my last attempt

OK TYY

[opaque root]

That is correct

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[vagrant surge]

Please help to me find out the correct limits, i just can't figure out how this limit work

[crimson sedge]

wich one ?

[versed siren]

is this a test or graded assignment?

[rapid pumice]

We have to check which one is correct

[vagrant surge]

It's a quiz with 22 questions and this is one of them

[versed siren]

is it graded though?

[vagrant surge]

yep

It requires us to finish in 50 minutes and other 21 questions just beyond my knowledge

[versed siren]

Isn't this cheating then..?

[vagrant surge]

i submitted it yesterday but i still dont know how to do this question

[versed siren]

ok

I thought

[vagrant surge]

i wish i wasn't asian

[versed siren]

I'm asian too

So in the first limit

sinx/x will go to 0

can you see why

[vagrant surge]

yes

[rapid pumice]

Substitute 0 in that equation

Sinx/x i mean

It's 0/0 approaching

[versed siren]

why zero

its approaching inf

?

[rapid pumice]

He asked for zero, right?

[versed siren]

no?

[rapid pumice]

My bad

[versed siren]

the question is

which of these limits is right

[versed siren]

np

[rapid pumice]

Okay I got it

[rapid pumice]

In 1 st one sinx/x will 1, right?

I am solving the limit after 1 yr lol, but it was easy, not that difficult

[versed siren]

no, its 1 when x approaches ZERO but here its approaching INFINITY

,w limit sinx/x as x approaches infinity

[wraith dagger]

Results provided by WolframAlpha

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Requested by moonguy257

[vagrant surge]

i think that -1/x< sinx/x < 1/x and when x approach inf, lim of -1/x and 1/x will be 0 and so sinx/x will approach 0 as well

[versed siren]

You know L'Hopital?

[versed siren]

yea

another way you can think is that

-1< sinx < 1

[vagrant surge]

yep

[versed siren]

but x is going to infinity

so finite/infinite approaches 0

[rapid pumice]

Ohh ok

[rapid pumice]

Yeah there was formulas

I forgot

[versed siren]

yea so you can use that on the second term in the first question

$$ \frac{ a + 6^x}{3- 9^x} $$

[wraith dagger]

TheBlueMoonVampire257

[versed siren]

in this I mean

[vagrant surge]

oh so when x approach inf of like (a)^x with 0<a<1 then it will approach 0 as well

[versed siren]

but we don't have a^x here

we have 6^x

and 9^x

which are >1

[vagrant surge]

so it's (6/9)^x and 6/9 <1

[versed siren]

so they blow up to infinity

[versed siren]

ohh you mean like tht

I see

yea you could do that

since a<<<<6^x

and same for the denominator

(so neglect the finite terms)

so the prob becomes (6/9)^x as xgoes to inf

and we get 0

[vagrant surge]

ok and how about the 2nd one

[versed siren]

to evaluate we write it as e^( log ( lim ..... ) )

then

e^( lim( log(.....) ) )

since log is continuous

??

bro you here?

[vagrant surge]

yeah but i dont get it

[versed siren]

do you know that $$ x = e^{lnx} $$

[wraith dagger]

TheBlueMoonVampire257

[versed siren]

?

[vagrant surge]

yep

oh i get it

after that we use L'HOSPITAL and the function will approach -1 so it will approach e^-1 right ?

[versed siren]

yea

Idk if it will be -1

you should evaluate tht

[versed siren]

the next two problem also use this

sorry I can't walk you through it, I'm working on another problem

You can ping me though

[vagrant surge]

ok thank you for this helpful method, i have already sovled other questions with it <333

[versed siren]

Have you done all of them?

[vagrant surge]

yea

[versed siren]

!done

[cedar kiln]

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

[versed siren]

you can open another one after if you get more doubt

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[open tartan]

Circle the letters of the statements that are true for the implicit form of curve representation.
A. In many cases, the conversion to the explicit form is not possible.
B. Conversion to the explicit form is always possible.
C. Describe by so-called coordinate functions. ´
D. we make the function in y = f(x) form.
E. A closed shape can be created

[open tartan]

i got
T
F
T
F
T

Is this correct, or i should think about one of the questions

[pastel vault]

yep, that looks all correct

I'm not sure what they mean by "coordinate functions"

but yeah, implicit equations are in the form $f(x, y) = k$ for some $k \in \mathbb R$

or $f(r, \theta) = k$, you get the idea

[wraith dagger]

south's secret twin brother

[open tartan]

uhum

i think about coordinante functions

they meant this

that we have (x, y) given

"given"

anyway

thanks for the help

[pastel vault]

yeah I'd say it's true

implicit functions can be described in terms of x and y

they are not always described in x and y though, for instance, r and theta

the wording of option C has a slight issue

[open tartan]

ye

this ones sneaky

but the point is the same, right?

[pastel vault]

so it's ambiguous
if you justify your reasoning well enough a good teacher should give you credit

[pastel vault]

I'm hoping me and your teacher are on the same brainwave

lol

I mean actually r and theta can be also coordinate functions, in polar coordinates

I don't see the issue then

[open tartan]

can this mean anything else, that would make this false?

[pastel vault]

it's just maths terminology
coordinates can be any system, not just Cartesian
if we say the coordinate axes though, we mean Cartesian coordinates

[open tartan]

implicit curves can be described by coordinate-functions

[pastel vault]

yep that's correct

or are described is more accurate

[open tartan]

ye

so

we got the answer

thanks for the help

have a nice day

.close

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[next moss]

This is transform methods

[next moss]

how do i determine the outpute signal? Like how should i think?

[crimson sedge]

think about the values that u know

for the output

[next moss]

I dont understand exactly but i think we know that a0 = 1 and rest of a = 0 right, considering the input signal?

[crimson sedge]

yes

U can look fisrt A0

then A0+A1

then A0+A1+A2

etc

and u notice smthg

[next moss]

Are u talking about the sequence in the picture?

[crimson sedge]

yes

[next moss]

so it should become (1,1,0,1,0...) ? right?

the output

[crimson sedge]

why do u have 0 in third position ?

[next moss]

isnt a1 = 0?

[crimson sedge]

yes

but in third u have a0 + a1 + a2

so 1 + 0 + 0

[next moss]

ok but then shouldnt 5th also be 1 then?

[crimson sedge]

yes

actually they are all 1

but important

we agree that (an) = (a0, a1, a2, ...) right ?

[next moss]

mm

[crimson sedge]

mm= yes ?

[next moss]

ye

but

how does solutions say they get "impulse response" (1, 1, 1, 0, 0, ..)?

[crimson sedge]

oh my bad

i didnt read well

i thought they added the terms of An till A0

so it is because, when A0 doesnt appear anymore, the value will always be 0 since A0 was the only terms wich was not equal to 0

[next moss]

I understand why 1th, 2th and 4th are 1 because a0 is there but i dont understand 3th where there is only a1?

[crimson sedge]

uh the third is A2+A1+A0

so 0+0+1

[next moss]

why though? shouldnt then for example a2 equalt 1 (the 5th index)?

[crimson sedge]

if u agree with this, then A2 = 0

u see ?

[next moss]

I meant this

u said a1 = 1 because A2+A1+A0

but shouldnt a2 = 1 also

[crimson sedge]

isnt this the output when the entry is (an) ?

[crimson sedge]

here i talked about the 3rd term in the output

and sorry i have to leave

ill not be back soon too

[next moss]

https://tenor.com/view/emoji-gif-24664106

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[next moss]

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[regal hollow]

why arent my lambdas cancelling out??

[regal hollow]

green arrows

[past wave]

lambdas *?

[regal hollow]

yeah my fsult

[past wave]

fault* ?

lmao

[crimson sedge]

🤓

[past wave]

determinants?

[regal hollow]

yeah

[past wave]

There should be a +

Cos it's + then - then +

So - * - = +

[regal hollow]

which should be a +?

(-12 - 2l) - (-12), (-2l)(-4) = 8l

thats my thought process

[past wave]

This should be +4 instead of -4

[past wave]

.

[past wave]

That would give you a -8x

[regal hollow]

this is the original question

it has to be -4

[past wave]

I am trying to say that

To open the determinants

We take a11 * smaller det - a12 * smaller det + a13 * smaller det

So the -(-4) = +4

[regal hollow]

ah i didnt know that

thats just a rule?

[past wave]

thats why you are doing
(5-x)(-6-x) - (5)(-6)

[regal hollow]

is everything after that - (5)(-6) also -

ill rewrite it out

in one line

and see

[past wave]

rest is all good

only multiply -1 to -4

[regal hollow]

at the bottom i still dont understand where the 4 becomes positive

[regal hollow]

im reading this

[past wave]

see they mul -1 to b1

[regal hollow]

yeah now it all makes sense

i didnt know that was a thing at all

do they just take that into account when they do the sum?

e.g

horrible quality

in this example, different question

[past wave]

yupp

you should still multiply -1

[regal hollow]

so they just dont show that?

[past wave]

its how its defined

https://www.cuemath.com/algebra/determinants/

have a read here maybe

[regal hollow]

yeah i get it now

these examples confused me since they never mention that at all

when i write it out, should i do it like the example i sent or factor in the fact im doing -a2

[regal hollow]

right here shouldnt the second red arrow be a +?

[cedar kiln]

@regal hollow Has your question been resolved?

[cedar kiln]

@regal hollow Has your question been resolved?

[past wave]

Yeah it should lol

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[winged hemlock]

Here, there is something wrong, and I don’t know what it is. If I replace -4 at the beginning of the inequality...

[winged hemlock]

Here are the conditions that the root inequality must meet to avoid being negative (working in the real numbers)

[dire geode]

There are no real solutions to the first inequality

[winged hemlock]

I can’t see how to prove that conclusion

[dire geode]

Break your first line into two cases, x>0 and x<0.

You already kinda arrived at a contradiction with the x>0 case.

[winged hemlock]

this one?

[dire geode]

If x>0, this work arrives at x<0 at the last line, a contradiction

If x<0, you'll arrive at a contradiction sooner with much fewer steps

[winged hemlock]

thank you 😇

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[obtuse orbit]

not too sure how u do this

[past wave]

sister sage?

how is s5 going?

do you know about tree diagrams?

as it is randomly selected im gonna assume its 50-50 that its first-class or second-class

[obtuse orbit]

what is s5

[obtuse orbit]

yes i do

[past wave]

refernce to a series lol

[past wave]

use that maybe

[obtuse orbit]

i can try

wait how would i start the tree diagram

[past wave]

^

[obtuse orbit]

how will that help to draw it tho

[past wave]

you can start with 2 branches

each having 1/2 chance

sorry for late reply

@obtuse orbit

[obtuse orbit]

i dont think the question is right

if u work backwards you realise its not 50-50

its 60-40, but the only way of knowing that is from the answer

[past wave]

oh

lets say that its given in the ques that its 60-40

can you do it now?

[cedar kiln]

@obtuse orbit Has your question been resolved?

[obtuse orbit]

im not sure how to start the tree diagram

[past wave]

^

But this time one has 6/10 whole other is 4/10

[cedar kiln]

@obtuse orbit Has your question been resolved?

[obtuse orbit]

ohh i see

yea i should be able to do it now

ty

[cedar kiln]

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[twin mica]

Good afternoon, I'm writing a paper for IEEE and I'm having trouble for writing a sum notation. I have v_d and h={-5, +7, -11, +13} and I want to write v_d as the sumation of v_h in such interval, hopefully it can be understood

[dire geode]

Not really seeing what you're asking

What is v_d

[proper mortar]

do you want (-5) + (+7) + (-11) + (+13) but in sum notation?

[twin mica]

v_d=sumation of v_h in the interval h={-5, +7, -11, +13}

[twin mica]

i want v_-5 + v_+7 + v_-11 + v_+13 in sum notation

[proper mortar]

oh I see

$\sum_{i \in h} v_i$ this should be the simplest way

[wraith dagger]

rbit

[twin mica]

okay, seems good

[proper mortar]

so instead of providing a range for i, you specify it as all the elements of a set instead

[twin mica]

thanks!

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[reef slate]

Hi ! I need help with this exercise. I’ve spent time on it but I still don’t see how to start.
I tried, unsuccessfully, to find an upper bound for (a_n), and I also tried using the following relation:
[
a_{n+1} = a_n \cdot \left(\frac{u_{n + 1}}{u_n(1 + u_{n+1})}\right)
]

[reef slate]

Let ((u_n)) be a sequence of positive terms. For ( n \in \mathbb{N} ), we set:

[
a_n = \frac{u_n}{(1 + u_0) \cdot (1 + u_1) \cdot \ldots \cdot (1 + u_n)}
]

Show that the series ( \sum a_n ) converges and that:

[
\sum a_n \leq 1
]

[wraith dagger]

Cornelius

Cornelius

[cedar kiln]

@reef slate Has your question been resolved?

[reef slate]

<@&286206848099549185>

[static jackal]

Yea

[reef slate]

👋 Hi !

[static jackal]

Tell me the problem

[reef slate]

Yes, of course

I need to solve this problem

[static jackal]

Can we talk in dms

[reef slate]

ok 👍

.close

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[green scarab]

Can you have a basis of nullA [0 0 0] ?

[cedar kiln]

Please don't occupy multiple help channels.

[green scarab]

Or is [0 0 0] dependant, therefore not in a basis so its basically emptyset

[mental trail]

an empty basis doesn't mean emptyset

empty basis = basis of {0}

[cedar kiln]

@green scarab Has your question been resolved?

[south tundra]

The zero vector never forms a basis

[green scarab]

Are you guys not contradicting eachother rn? @mental trail @south tundra

[south tundra]

Not sure

Anyway, {0} is not linearly independent since 1*0 = 0 and 1 =/= 0, so you can't have that as a basis

[green scarab]

Okay then let say im finding the basis of a nullspace

I get the trivial solution with no free variabeles, x1 = x2 = x3 = 0

Will my basis be emptyset

[south tundra]

Yeah

[green scarab]

Appreciated thanks

.close

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[green scarab]

.reopen

[cedar kiln]

✅

[green scarab]

@south tundra wait so what will be my basis

The zero vector or the emptyset sign

[south tundra]

The empty set

[green scarab]

What is this then

Does this say empty set

[south tundra]

I'd assume they meant the empty set by empty basis

Which is weird since they stated the opposite before

[green scarab]

Do i need to write the o with a line between it

[south tundra]

Anyway the basis of {0} is {}

[south tundra]

$\emptyset$ is the basis, yes

[wraith dagger]

A Loner Bean

[green scarab]

Oh okay

And I have to put that sign in those weird brackets right

{}

[south tundra]

No

$\emptyset = { }$ already

[wraith dagger]

A Loner Bean

[green scarab]

Ohh okay

Thank you

.close

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[hexed vortex]

I am confused by these steps

[hexed vortex]

Were trying to write A as LU

so for the second one

they did R_3-2R_2

that means for the L matrix we'd do the inverse

we'd add 2 times row 2 to row 3

but why is entry (3,1) = 1?

shouldnt that be 5

2*2 + 1 = 5

how come they didn't apply the operation to that column?

got this but it’s wrong

[cedar kiln]

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@hexed vortex Has your question been resolved?

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@hexed vortex Has your question been resolved?

[cedar kiln]

@hexed vortex Has your question been resolved?

[cedar kiln]

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[stiff furnace]

How do I find the derivative of xabs(x)?

[bleak viper]

hint: $|x| = \sqrt{x^2}$

[wraith dagger]

Astar777

[stiff furnace]

So, f(x) is now equal to x^2?

Should I now apply the power rule?

[bleak viper]

how

[stiff furnace]

sqrt(x^2) is the same thing as x^(2/2) equalling x

x(x) = x^2

[bleak viper]

what, no

[stiff furnace]

ekaborate pls

[bleak viper]

$\sqrt{x^2} \ne x$

[wraith dagger]

Astar777

[bleak viper]

thats why

its |x|

[stiff furnace]

oh yes, because its the principal root meaning it must be 0 or greater

[bleak viper]

what you have now is $f(x) = x\cdot\sqrt{x^2}$

[stiff furnace]

I'm assuming from here apply product rule?

[wraith dagger]

Astar777

[bleak viper]

differentiate this (product rule)

[stiff furnace]

x/2sqrt(x) + sqrt(x)?

[bleak viper]

no

[stiff furnace]

oops i made a mistake

i did xsqrt(x) instead of xsqrt(x^2)

x+sqrt(x^2)

[bleak viper]

how x

[stiff furnace]

because the derivative of sqrt(x^2) is 1

[bleak viper]

said who

[stiff furnace]

what is it then

[bleak viper]

$f'(x) = \sqrt{x^2} + x \cdot \frac{1}{2\sqrt{x^2}} \cdot 2x$

[stiff furnace]

Wait, does this derivative of sqrt(x^2) require the chain rule?

[wraith dagger]

Astar777

[bleak viper]

yes

[stiff furnace]

I havent learned that yet lol

[bleak viper]

x^2 inside

so 2x

[bleak viper]

you havent learned chain rule?

[stiff furnace]

yes

i can learn it now

anyways, thanks

.close

[cedar kiln]

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[cedar kiln]

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[obtuse coral]

How do I do this?

[fresh geyser]

plugging in 1 for x works, so start by factoring out (x-1)

[dire geode]

Factor

[obtuse coral]

Why would I just randomly plug in 1, tho?

Isn’t there a different way of doing it?

[obtuse coral]

How?

[fresh geyser]

rational root theorem

[glacial field]

Figure out what two numbers add to -3 and multiply to 2

[fresh geyser]

with a cubic there's either 1 real root and 2 imaginaries or 3 real roots

[obtuse coral]

Isn’t there also a way to do this using Pascal’s triangle?

Or am I thinking of smth else?

[dire geode]

This is the easiest

[glacial field]

I believe that’s for expanding (a+b)^n

Using pisano leonardo

[obtuse coral]

Okay, got it

Thank u guys

[dire geode]

TIL Fibonacci's first and middle name

[bleak viper]

is his name even fibonacci?

iirc its a made up name (short for filius bonacci)

[cedar kiln]

@obtuse coral Has your question been resolved?

[cedar kiln]

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[wild pine]

Hello I need some help with sketching rational functions

My main problem is the sign chart

And the type of asymptote.

But it would be really helpful if someone could confirm my answers

For the sign chart I know im supposed to factor the rational functions but I'm really bad at it.

Factor it to get the numbers for the bottom of the sign chart

[pastel vault]

y = 2 is a horizontal asymptote

so the points to test are the horizontal asymptote, x = -2, and the root at x = 2

you want to test one point from each region:

1. x < -2
2. -2 < x < 2
3. 2 < x

so testing the signs of x = -3, 0, 3 for example will work

also your asymptote form is wrong

$y = \frac{2x + 4 - 8}{x + 2} = 2 - \frac{8}{x + 2}$

[wraith dagger]

south's secret twin brother

[wild pine]

Did i do the synthetic division wrong? Where do I get that?

[pastel vault]

so yep, bring the 2 down
then 2 * -2 = -4

add that to the -4 on the top row and you get -8

[pastel vault]

so the quotient is 2 and the remainder is -8, hence it's this

[wild pine]

I'm confused on the sign chart. The sign chart my professor wants us to use is usually a line containing multiple signs (+ or -) showing if they flip at certain points. I can give an example of what one looks like

[pastel vault]

I know what it looks like

[pastel vault]

read this again: where are you stuck?

[wild pine]

I'm stuck because I don't know where you are getting the -2 and 2s from and inequalities. I never had to use them when splving

[pastel vault]

those are what the question asks when they say "test point(s)"

the sign of the function changes at the roots, by definition of what a root is

the sign can also change before and after a vertical asymptote

[wild pine]

But what about the sign chart itself? Like what numbers to I write there and the sign changes?

They usually look like this

Just to confirm we have the same idea

My professors way (which i have to do) i have to factor the original linear function and use the numbers to put on the bottom of the sign chart. I'm confused on how to factor that problem and how to create the sign chart. Idk if there's supposed to be numbers on the top too

[pastel vault]

well, so you're talking about the roots then

to find the roots of a polynomial, you have to factor it

the issue here is that this is a rational function and not a polynomial

so the process of finding the roots is to set the numerator equal to 0
2x - 4 = 0, x = 2 like you did
sometimes the numerator can be a polynomial, like a quadratic or even cubic, so yes you would have to factor the numerator in that case

here is a sign diagram for another rational function

this shows how you need to consider the asymptote as well

in addition to all the roots

[pastel vault]

I don't know why what I said isn't entering your brain

roots: x = 2
asymptotes: x = -2

white circles on the sign chart: -2 and 2

[wild pine]

Sorry I'm a little slow. What about the signs themselves? Do they change from positive to negative or what do they start as or end?

[pastel vault]

that's why you have to test values in each region to know

you want to test one point from each region:

1. x < -2
2. -2 < x < 2
3. 2 < x
   so testing the signs of x = -3, 0, 3 for example will work

[wild pine]

What do you mean by test them?

Do it substitute those values for x?

I still don't understand the inequality signs because those aren't used on the sign chart

[pastel vault]

yes

[pastel vault]

so what I mean by 'regions' are the parts between the white dots, and also like away from them as well

imagine that -infinity exists

so between -infinity and -2 for example

so in the region x < -2, they have tested x = -4

[wild pine]

Ohhh so I test a point between each

[pastel vault]

yessss

[wild pine]

So i tested -10 and got 3 which is positive so it starts positive im assuming

[pastel vault]

yes that's correct

[wild pine]

So is this what it's supposed to look like?

And how do I know if open or closed circles?

[pastel vault]

open circles just mean the function doesn't exist at x = -2 and x = 2

[wild pine]

Ah ok

[pastel vault]

yeah, your sign diagram is now correct!

you could have saved yourself some trouble actually

[wild pine]

Tysm!

[pastel vault]

cause you don't need to test the region -2 < x < 2 twice

[pastel vault]

yeah so your graph is correct then

it's easy to see the shape for (ax + b)/(cx + d)

when you have anything harder then you really need the sign diagram

[wild pine]

Mm ok

[pastel vault]

np!

[wild pine]

Uhh idk how to close this thing to make it open again

The channel

.close

[cedar kiln]

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[dire radish]

Am i doing problem number 45 correctly? Something seems off

[dire radish]

just to be clear for that third line is product rule

but i had to move it doen since i didnt have enough space

am sorry

.close

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[crimson sedge]

@dire radish seems fine to me

or maybe you can just break the ln[(x^2+2)^2(x^4+4)^4] using

$\ln(ab)=\ln(a)+\ln(b)$

[wraith dagger]

yajat

[cedar kiln]

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[fossil coral]

Is my answers correct?

[fossil coral]

A little confused on 1 due to 4

<@&286206848099549185>

-_-

[pastel vault]

the domain is just $(-\infty, -9/2) \cup (-9/2, \infty)$

[wraith dagger]

south's secret twin brother

[pastel vault]

that's a fancy way of saying $x \ne -9/2$

[wraith dagger]

south's secret twin brother

[fossil coral]

ah. K.

thanks

[pastel vault]

np!

[fossil coral]

:)

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[tiny vault]

for c, d, and f conditions in the image. I don't know how these are graphable? if there are two f(f(x)) as x approaches -1 how does it approach from both sides to -inifnity and infinity. For the f what should i even do to address this?

[cedar kiln]

@tiny vault Has your question been resolved?

[cedar kiln]

@tiny vault Has your question been resolved?

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@tiny vault Has your question been resolved?

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[compact gulch]

what to do next

[cedar kiln]

@compact gulch Has your question been resolved?

[compact gulch]

<@&286206848099549185>

[crimson sedge]

Next, you need to solve for $k$ by factoring or completing the square for the equation $k²+k+2$.

[wraith dagger]

JustYourAverageDude

[crimson sedge]

@compact gulch

[compact gulch]

Okay but then how does that help us?

,w k^2 + k+2

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by ahbleugh

[crimson sedge]

If $k^2 + k + 2 \neq 0$, the matrix has a non-zero determinant, meaning it's non-singular (invertible).

[wraith dagger]

JustYourAverageDude

[compact gulch]

true but like how do I show that its not = 0

using discriminant

[crimson sedge]

Use the discriminant: $\Delta = b^2 - 4ac$. For $k^2 + k + 2$, $\Delta = 1^2 - 4(1)(2) = 1 - 8 = -7$. Since the discriminant is negative, there are no real solutions, so $k^2 + k + 2 \neq 0$ for all real $k$. Thus, the matrix is non-singular.

[wraith dagger]

JustYourAverageDude

[compact gulch]

That makes sense!

Thanks @crimson sedge

[crimson sedge]

np

[compact gulch]

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[gritty drift]

How do I factorize this and solve for the value of x? With the extra variable x on the farthest right, I don't know how to solve this.