help-17

[marble steeple]

right

[wind geyser]

You have to also consider cases where $\cos\theta=0$, which gives two principal values

[twin meteor]

@wind geyser

[wind geyser]

which are the missing solutions

[marble steeple]

right im lost

[wind geyser]

lets do a simpler example
How do you solve $x^2=x$?

[marble steeple]

sqrt it

[twin meteor]

@wind geyser

[wind geyser]

typo

[marble steeple]

take it to one side

factorise

x(x-1)

[wind geyser]

what are the solutions, 0 and 1 right

[marble steeple]

yup

[wind geyser]

what you could also do is divide both sides by $x$, you get $x=1$

[twin meteor]

@wind geyser

[marble steeple]

but we are missing the 0 now i see

[wind geyser]

yes

similarly, cancelling $\cos\theta$ implies you chose to make it non-zero

[twin meteor]

@wind geyser

[wind geyser]

but the question doesnt ask us to consider it non-zero

[marble steeple]

righttt

[wind geyser]

therefore we consider a separate case where it is 0

[marble steeple]

and well ofcourse cos is 0 at 90 and 270

[wind geyser]

yes

[marble steeple]

so the rest of the equation doesn't matter?

[wind geyser]

it does

[marble steeple]

no but in the case that cos is 0

[wind geyser]

yes?

[marble steeple]

oh

wait then im confused

[wind geyser]

lol

When we cancelled the x-es in $x^2=x$, we lost a solution

[twin meteor]

@wind geyser

[marble steeple]

okay so what should i do instead

[wind geyser]

you could do some crazy factoring with that trig equation, or better: whenever you cancel something, you could set that to 0 separately

obviously it doesnt apply to constants and stuff that cannot be ever 0, say exponentials

[marble steeple]

yeah okay

and then how does that apply to this question when i apply setting it equal to 0

what does the equation then become

[wind geyser]

$2\cos\theta=\sqrt{3}\cot\theta$

$\implies 2\cancel{\cos\theta}=\sqrt{3}\frac{\cancel{\cos\theta}}{\sin\theta}$

$\implies 2=\sqrt{3}\csc\theta\textbf{ OR }\cos\theta=0$

[twin meteor]

@wind geyser

[marble steeple]

oh right

thank you i understand very well now

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[river minnow]

!status

[vocal sleet]

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

[vast shale]

have you tried anything ?

[next quartz]

i got it , nvm

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[viscid bridge]

find the total number of different arrangements of the 8 letters in the word TOMORROW that have an R at the beginning and an R at the end, and in which the three Os are not all together.

[viscid bridge]

guys i really don't understand how to do this

[uneven flicker]

Do you know how to start?

[viscid bridge]

i'm thinking of finding the arrangement that have R at the beginning and an R at the end

[kind light]

by "three O's are not all together" do you mean there arent any O's that are adjacent or rhere arent any "OOO"

[uneven flicker]

|| first arrange the Rs then TMW and then use gap method for OOO||

[viscid bridge]

i think the Os aren't side by side

i'm not sure too, it is a question from cambridge

[viscid bridge]

ohhhh okok i'll try

is it like that?

[uneven flicker]

Can you explain what you have done?

Sorry had to go @viscid bridge

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[tribal lynx]

Can you help me 1+1

[oak magnet]

<@&268886789983436800>

[uneven flicker]

Can you give some context?

Really depends

Can vary from 0 to 2

[full hatch]

Maybe they know the answer 😂

[oak magnet]

Thats why we should always ask, thats personnaly out of my hand xd

[vocal sleet]

@tribal lynx Has your question been resolved?

[fervent wasp]

1 = cos^2(theta) + sin^2(theta)
1 + 1 = cos^2(theta) + sin^2(theta) + cos^2(theta) + sin^2(theta)
= 2 (cos^2(theta) + sin^2(theta))
Because cos^2(theta) + sin^2(theta) = 1
Therefore 2( cos^2(theta) + sin^2(theta) ) = 2 * 1 = 2

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[uneven lynx]

could someone help me figure out the divergence or convergence of this series using rabbe's test

[sly sierra]

a simple comparison test should suffice

the summand is greater than 1/sqrt(n+1) if n >= e

[uneven lynx]

I understand that, but I wish to know the procedure if I were to try and do it via rabbe's test

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[halcyon valley]

I'm trying

Nop it's getting big

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[stiff adder]

I have tried the variable seperation method in both the cases where y > 0 and y <= 0

but i dont understand how to proceed

i do have the two distinct solutions, one is y = x^2 / 4 and other is y = -x^2/4

[bitter pilot]

You could rewrite |y| = sqrt(y²)

[stiff adder]

hmm i dont get it how does rewriting it help

[bitter pilot]

How did you get the first solution?

[stiff adder]

variable seperaiton

i assumed y > 0

for y < 0 i got -x^2/4

[bitter pilot]

can you show your work

[stiff adder]

initial condition at x = 0 got me C = 0

for both cases

[bitter pilot]

ok i see it looks good

[stiff adder]

how do i justify the picard thing

nvm i got it

close

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[bitter pilot]

mind explaining?

[stiff adder]

the derivative of the function f(y) = sqrt(|y|) isnt continouos as required by picards existence and uniqueness theorem

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[bitter pilot]

but isn't that a violation

[stiff adder]

nono it doesnt violate the theorem, the theorem says if its the derivative isnt continuous then u can have multiple solutions

which is what we see

[bitter pilot]

ohhh

ok i misunderstood the theorem, i thought it says that there is a unique solution so i was confused

thank you very much

.solved

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[torn timber]

you got helpful back

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[summer ore]

Hi, can anyone tell me is my method for finding eigenvectors for Eigenvalues that involved complex number correct or wrong?

Or i need to solve like this?

[brittle wasp]

method looks fine

[summer ore]

Hmm, so both methods also correct?

[summer ore]

I remember I learnt eigenvectors will different if we use different method, for me I think the second method is also correct too. Initially I thought eigenvectors just negative or positive value might be differ

[brittle wasp]

You can get any scalar multiple when doing different methods

[summer ore]

Oh, I see. Thanks

That solve my confusion

Really appreciate

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[summer ore]

.reopen

[vocal sleet]

✅

[summer ore]

I think I spot the mistakes

I forget the x_1

[brittle wasp]

I didn't look at the calculations. Too much for me. 😵‍💫

[summer ore]

I will make that as 0 🤣

x_2 will also become 0 if I don’t use the second method

[summer ore]

Anyway, thanks for that

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[marble steeple]

i've done this before but now i do not understand it

[vocal sleet]

@marble steeple Has your question been resolved?

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[viral shell]

Do you know how to arrange them if T and M weren't fixed?

Yes

[kind light]

yes

[uneven flicker]

T and M are in the same place?

Next to each other maybe?

[viral shell]

You take the factorial of the total number of letters but then divide by the factorial for the number of repeated letters

If they're in the same place you ignore T and M and arrange the rest

[uneven flicker]

Yup

Also divide by something

As O is repeated 3 times

And R too

Lmao

by 6

[viral shell]

You divide by the factorial for repetitions

O is repeated 3 times so 3!, R is repeated twice so 2!

Yes

Yes

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[mild trench]

am i correct

coz for some reason AQ is

im very confused

wait lemme rework out Q

[bronze osprey]

yeah do that

[mild trench]

omg...

is it

[bronze osprey]

while you're at it, pls send the full q

[mild trench]

(10,-3)

[mild trench]

i did

o wait

[bronze osprey]

there should be information above

ok

[mild trench]

i did my Q(x,y) wrong ill do it again

[bronze osprey]

,calc 10^2 - 8 * 10 + 9 + 10 * -3 + 1

[twin meteor]

Result:

0

[bronze osprey]

this is now correct

[mild trench]

now i need to work out

AP correct?

coz i got AQ

[bronze osprey]

what did you get for AP and AQ?

[mild trench]

for AQ i got

6
2

lemme work out AP

[bronze osprey]

yeah that's correct

[mild trench]

im stuck

idk wat to do to get AP

[bronze osprey]

well, AP and AQ are perpendicular

so the slope of AP is?

[mild trench]

it will be

well i know slope of AQ is 1/3

so ap will be -3

[bronze osprey]

this should be 1/3

great, so do you see (2, -6) also has a slope of -3?

or on second thoughts, (-2, 6)?

[mild trench]

yes

[bronze osprey]

yeah so given that A = (4, -5)

that's now enough info to find P

[mild trench]

how did u find out 2,-6 btw

[bronze osprey]

you want AP = AQ of course, radii so lengths are equal

by Pythagoras this is true for (-2, 6) and (6, 2)

[mild trench]

ohhh

will

AP be

-2
6

[bronze osprey]

no

AP = (-2, 6) is the direction vector

[mild trench]

2,1

[bronze osprey]

A = (4, -5) is a position vector

yes

[mild trench]

i see now

i heard somebody use the discriminant, is this possible for this part of the question?

[bronze osprey]

cool then just spam point-slope form or y = mx + c to get the equations of l1 and l2

[mild trench]

got it

[bronze osprey]

technically it is possible but then it would be an awful lot of work

that method would involve setting y = -x/3 + k

where you don't know k

then yes, set the discriminant equal to 0 when you sub that y into the equation of the circle

one solution = one intersection, so tangent

[mild trench]

ah i see

[bronze osprey]

this vector rotation strategy is super useful

if you have a vector (x, y)

then (y, -x) and (-y, x) will be perpendicular and the same length

[mild trench]

do u mean (-y,x)

[bronze osprey]

oh right yeah

[mild trench]

i get it now

[bronze osprey]

typed the same thing twice haha

[mild trench]

wait u see everything youve told me and all the steps and stuff (other then the discriminant) do u mind writing it like a recipe list

like a step by step

[bronze osprey]

starting from which part?

[mild trench]

start from this info

itll mean a lot

[bronze osprey]

1. find vector AQ (find Q by solving the quadratic where you sub in x = 10 in the circle equation)
   now if vector AQ = (x, y)

2. possible perpendicular vectors with the same length are (-x, y) and (y, -x)
   you want this direction vector to be negative from the diagram, so choose (-x, y)

3. (position vector of A) + (direction vector AP) = position vector of P

4. the slope of l1 is the same as y/x from AQ, and hence the slope of l2 is -x/y

5. point-slope form with slope = y/x and coordinates of P for l1
   slope = -x/y and coordinates of Q for l2

[mild trench]

tysm, W explanation

that is all

you helped a lot

[bronze osprey]

no worries!

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@mild trench Has your question been resolved?

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[vast shale]

i want to know why does the trigonometric function change in
the identities:
sin (90-x) = cos x
but not in
sin (180-x) = sin x
i memorized it as 90 and 270 change and 180 and 360 keep it the same but whats the reason behind it ?

[uneven flicker]

Sin(A+B) = ?

do you know this identity?

[oak magnet]

A-B so

[vast shale]

yeah im slightly familiar with it

[uneven flicker]

Can you tell what it is?

[vast shale]

sin (A-B) = sin(A) x cos(B) - cos(A) x sin(B)

same goes for A+B but instead just put a plus instead of the minus up here ^

@uneven flicker

[uneven flicker]

Yup correct

Now do for 90-x and 180+x

[oak magnet]

180-x so

[oak magnet]

You can also construct it with the unit circle

But some algebraic manipulation is good too

[vast shale]

sin (90-x) = sin (90) x cos (x) - cos(90) x sin (x) =
1 x cos (x) - 0 x sin (x)
sin (90-x) = cos (x) ?

oh it actually did change

sin (360-x) = sin ( 360) x cos (x) - cos (360) x sin (x)
sin ( 360-x) = 0 x cos (x) - 1 x sin (x)
sin ( 360-x) = - sin (x)

i thinkk i get where its from now thanks alottt

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[vast shale]

.reopen

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✅

[vast shale]

oh wait i just realized something, this identity applies to sin, but what about other trigonometric functions like cos, tan, cot csc?

i know i asked regarding sin, but i was wondering if there are other identities for other functions

[uneven flicker]

There is one for cos too

Cos(A+B) = cosA cosB - sinA sinB

Similarly there are for tan and all too

You can google it

[vast shale]

what should i search? cause i don’t really know their names

Trig sum identities

alright thanks alot for the help guyss

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[golden rapids]

I need some solid explanation for how this question works, I cant seem to get this in my head

[jade gyro]

anyone can help me with this?

3. The Lagrange polynomial that interpolates the Points (1,3),(3,5)and(4,2)

a.
P(x)=12(x−3)(x−4)−52(x−1)(x−4)+23(x−1)(x−3)

b.
P(x)=52(x−3)(x−4)−12(x−1)(x−4)+43(x−1)(x−3)

c.
P(x)=52(x−3)(x−4)−32(x−1)(x−4)+13(x−1)(x−3)

d.
P(x)=12(x−1)(x−4)−12(x−3)(x−4)+23(x−1)(x−3)

[golden rapids]

Bud, you gotta make your own channel😭

[jade gyro]

sorry .. i didnt know that!
im just looknig for help im new here
how to make my own channel

[golden rapids]

Its ight. Make your own channel by going to the channel I circled and post your question

[jade gyro]

thx .. im sorry again

[vocal sleet]

@golden rapids Has your question been resolved?

[golden rapids]

Nuh uh

[golden rapids]

.

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@golden rapids Has your question been resolved?

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[ripe karma]

Find all integer solutions to the equation. x3+y3+z3 = 33

[ripe karma]

its basically like x cubic + y cubic + z cubic = 33

I think its called a Diophantine equation

[queen root]

like

i js looked up this equation

and js found such monstrous solutions

apparently its a famous historical diophantine equation

[ripe karma]

yea i know

was just doing math research for fun akhi when i came upon these

[hard atlas]

so you knew that the answers are huge and just wanted to troll here?

[queen root]

then i cant answer u :]

i have no idea how to answer

[ripe karma]

troll?

i need help to understand it

[hard atlas]

what exactly is there to understand

there is a huge solution

even that took ages to find

[ripe karma]

if u cant answer stop whining then its alr

[hard atlas]

what kind of answer do you want to hear

no one knows all solutions

no one knows whether there is even more than one

and you wont find a world expert in the topic in a random discord chat

[ripe karma]

i did this a few days ago with friends and we were close to doing it js needed extra help

chatgpt ..... i dont rely on it so much it lied to me sometimes

nvm found it!!

https://blog.computationalcomplexity.org/2019/04/x-3-y-3-z-3-33-has-solution-in-z-and.html

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[nimble tendon]

i need help w this lesson, i missed it and i literly dont understand anything at all

[halcyon valley]

What's the name of the lesson u missed

[buoyant saddle]

how do i make my basics stronger

idk shit in math and i have roughly 2 weeks

[halcyon valley]

?

[halcyon valley]

#❓how-to-get-help

[nimble tendon]

at the top

[halcyon valley]

Mhm i c

What do I have to explain?

[nimble tendon]

i just dont know how to get started and stuff

[halcyon valley]

Okk so what do u know about quadratic equations?

[nimble tendon]

uhhh, from the lesson before i learned that like how to figure out if its quadratic or linear

and how to graph

[halcyon valley]

Mhm okk

So do u know what x and y intercepts are?

@nimble tendon

[nimble tendon]

oh shoot

yeah i do

[halcyon valley]

Then u should be able to solve a and b

[nimble tendon]

okay yeah

[halcyon valley]

Ok so like

What topics do i explain

Tell me the things and I'll explain

@nimble tendon

[nimble tendon]

i js dont get this partt

like

i see what it says but idk how to do itt

[halcyon valley]

Okk I c

Ok first let's see what a vertex is

Vertex of quadratic equations is either the minimum or maximum of the function

Do u get this part?

[nimble tendon]

yeah i think i doo

[halcyon valley]

Ok lemme draw

U get it now?

[nimble tendon]

OOHHH YEAH

yeah i knoww that

[halcyon valley]

Mhm

So the vertex of quadratic equations lies at -b/2a for the equation ax²+bx+c=0

[nimble tendon]

wait wdym

[halcyon valley]

Do u know calculus? So I can explain the derivation

[nimble tendon]

im in grade 10

[halcyon valley]

Ahh okk

Then the longer one

But fine

Do u know "completing the square method" ??

[nimble tendon]

oh yah i do

[halcyon valley]

Yes so that for ax²+bx+c=0

[nimble tendon]

its the one with the a h and k right

[halcyon valley]

Uh idk how u solve
Different people use diff methods

[nimble tendon]

this is what my teacher wrote

[halcyon valley]

Mhm another way I see

Ok so let's do this then

What is axis of symmetry?

U there? @nimble tendon

[nimble tendon]

oops yah sorry

[nimble tendon]

its like the line that goes through the vertex

[halcyon valley]

Mhm

Basically dividing it into 2 equal halves right

[nimble tendon]

yeah

[halcyon valley]

Ok so

U said u know how to find intercepts

[nimble tendon]

i doo

[halcyon valley]

So axis of symmetry lies in the middle of the intercepts

[nimble tendon]

yeah cause theyre symetrical

[halcyon valley]

Yep

So find x intercepts first

[nimble tendon]

okay uhh

[halcyon valley]

Just take this itself

[halcyon valley]

.

[nimble tendon]

wait so the x intercepts are 0,0 and 6,0 for this equation

[halcyon valley]

Yep

[nimble tendon]

wait i thought i knew how to find x-int but i dontt i only know y-int

[halcyon valley]

U just found x-int???

[nimble tendon]

nahh those r from the lesson answers i looked and then realised i had no clue what to do

[halcyon valley]

Ok soo

Consider this itself

x²-6x=0

Take x common

x(x-6)=0

So either x is zero or x-6 is zero

So u get x=0,6

[nimble tendon]

ohhhh

[halcyon valley]

Yepp

[nimble tendon]

oh wait this is so easy

wait so

u take the x intercepts and just use the average equation?

to find the aos

[halcyon valley]

Yes

[nimble tendon]

wait then the coordinate of the vertex

[halcyon valley]

Is the x coordinate of that aos

And yea I gtg sry man

I'll reply later if u have questions
U can dm me

[nimble tendon]

its okayy ty for all your helpp

.close

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[willow lava]

how can i show that this is not continuous at (0,0), ive tried approaching on quite a few lines

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[native schooner]

how do you go from the original question to 3*cos(x) <= g(x)/x <= 3? (squeeze theorem)

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[wooden sapphire]

hey

[wooden sapphire]

can u help me understand graphing

<@&286206848099549185>

[marble sluice]

for Jim, his one time setup fee is $10, its the constant term in the equation y = mx + b

so b = 10

it then says each card will cost 7 dollars per box

thats the rate/slope in which the graph increases

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[marble sluice]

so for jim his equation will be y= 7x + 10

[coarse escarp]

can you help me

[wooden sapphire]

blud

its occupied by me

thesres 4 available ones

[marble sluice]

for kimberly, her shipping fee was 20$

one time fixed cost

so her b value would be b = 20

5 dollars per box, so each box X would cost 5 dollars each, your equation would be 5x + 20

[wooden sapphire]

the answer is 25 total right

[marble sluice]

set the equations equal

and solve

[wooden sapphire]

its graphing it thts my proble

[marble sluice]

so for jim

his y intercept is

10

as when x is 0, the constant term remains

so thats where you start the graph

each box costs him 7 dollars

, if you want to know the cost at 1 box,

you simply plug in 1 into your x

and you get a y value

just plot that point on your graph

same for kimberly

[wooden sapphire]

ok ty

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[vast shale]

if cos theta equals 4/5 and theta is 270<theta<360

then what is sin(180-theta) + tan(360-theta) + 2 sin (270-theta) ? i solved it as -29/20 but just want someone to check over my steps

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

<@&286206848099549185>

[fervent sphinx]

hi

what did you get for theta

oh

[fervent sphinx]

Yes, your solution is right! Good work!

[vast shale]

alright thats great, thank you alot ^^

.close

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[fervent sphinx]

yw bye

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[gusty crow]

3. Two toys are started at the same time each with a different battery. The first battery has a lifetime that is exponentially distributed with mean 100 minutes; the second battery has a lifetime that is exponentially distributed with mean 75 minutes. Let X represent the lifetime of the first toy, and let Y represent the lifetime of the second toy. Find:
   a. The joint pdf of X and Y. (Hint: are X and Y independent?)
   b. The probability that the first toy outlasts the second.

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@gusty crow Has your question been resolved?

[gusty crow]

<@&286206848099549185>

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@gusty crow Has your question been resolved?

[ornate seal]

Do you know what is joint pdf

And what pdf of exp distribution

[gusty crow]

Yea

[ornate seal]

Which part are you stuck?

[gusty crow]

im not syre how to make th ejoint pdf

we need to find beta right

?

wait nvm beta is the mean its given

so for f(x) = 1/100 * e^(-x/100)

and f(y) = 1/75 * e^(-y/75)

@ornate seal

do i just multuplythese two?

ok so fXY(x,y) = 1/7500 * e^(-x/100-y/75)

i got part a

i dont understand part b

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[hidden sage]

help me pls

[tacit axle]

what about this is incorrect? i simply did 0.5 into each v so 0.5 * (v1+v2+v3+v4+v5)

oh

.close

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[tacit axle]

i included the final point

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[sweet crystal]

how do you know if while using euler's method to solve for an approximate answer that you have an under approx or over approx

[slate ember]

Is this the help channel

If it is

Can someone help me with this

Im just an intern

[hushed pewter]

Second derivative

[sweet crystal]

what do I do with it after I get that

[hushed pewter]

!occupied

[vocal sleet]

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).

[sweet crystal]

plug in my X and Y for a particular point

see if it's negative or positive?

[slate ember]

Can someone simplify?

[sweet crystal]

go to help channel 1

[hushed pewter]

You have a function f(x) and you want to approximate it at at some x=a, correct?

[slate ember]

Okay

[robust spruce]

you should read #❓how-to-get-help

[sweet crystal]

yea

well I already did the approximation part

I just need to know it's an over or under approx

[hushed pewter]

Oh. Correction, you know what f(a) is, and you want to approximate at some x=b near a

[sweet crystal]

well it gives you f prime and also that f passes through a certain point (x ,y)

im approximating with eulers

lmk if you know how to know if it's an over or under approximation

[hushed pewter]

You use the first derivative approximation to basically determine if f is increasing or decreasing near x=a. So consider the second derivative. If f’’(a)>0, then the derivative will be increasing. Which means that you are under approximating

[sweet crystal]

ah ok tysm

.close

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[hushed pewter]

Btw you can always cheat and solve the differential equation

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[granite patrol]

This is a factoring question and I got the factor by grouping part but I’m not sure what to do next

[left bough]

The goal is to have a factor that are the same for both terms

do you happen to be missing a z?

[sly sierra]

,rotate

[granite patrol]

OH HELP

apparently i cant read a question from a textbook correctly

My bad 😭😭😭🙏

[left bough]

why is it that the #1 problem why people get questions wrong is that they can’t read

[granite patrol]

Idk 😭🙏 its like late not really and i havent slept in 2 days 💀

[left bough]

i’m sorry

don’t overwork yourself

[granite patrol]

HELP WHY ARE YOU APOLOGISING 😭🙏

[granite patrol]

i dont let myself sleep until ive perfected everything 💀 i cant let myself fail otherwise i will explode 🗣️🔥

[left bough]

no one is perfect, that shouldn’t be a worry

[left bough]

it’s called sympathy

[granite patrol]

HEJDHSJSHS 😭😭😭

[granite patrol]

nah i just like to perfect things completely otherwise i feel like i failed and that it wasnt worth it

if im going to do something 😭 im going to give it my all

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[granite sail]

P(x) polynomial with degree 3
P(x)/x^2 + 3x + 4, the remainder is 7x+25
P(x)/x^2-3x-4, the remainder is 23x+25

Is there exist polynomial P(x) such that both condition satisfied

[vocal sleet]

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@granite sail Has your question been resolved?

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@granite sail Has your question been resolved?

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[leaden scroll]

hi how do i graph thiss

[vocal sleet]

@leaden scroll Has your question been resolved?

[leaden scroll]

<@&286206848099549185>

[sharp rain]

Hello @leaden scroll

Again.

Lol

!status

[vocal sleet]

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

[leaden scroll]

2

this is what i have so far

hi lol

[sharp rain]

So you done with graphing bit?

[leaden scroll]

yes

[sharp rain]

The graph should decrease from x= 1 to 2 and then increase after

Also for the approximation of area

For 2(a)

Make rectangle in interval
1-2, 2-3 , 3-4, 4-5

Then find areas which would be value * interval size

And then add em up

You got it?

@leaden scroll

[leaden scroll]

yes i think so

i just plug in the x values right?

but doesnt it end up being weird because of the ln

[sharp rain]

Keep it in terms of ln

We will take care of it in the end

When we add up all the areas

[leaden scroll]

so f(2) = 2 - ln(4)

[sharp rain]

Start with f(1) no

?

[leaden scroll]

oh yeah

[sharp rain]

f(1)* 1+f(2)* 1+f(3)* 1+f(4)* 1

[leaden scroll]

is 1 - 2 ln(1) =1 ?

[sharp rain]

Why the +1

?

Oh you fixed it

[leaden scroll]

yuh

[sharp rain]

Yeah it's correct

[sharp rain]

Yup

Add em up

Eh eh

Why f(5)

4 rectangles

@leaden scroll

[leaden scroll]

huh

do we count that 5

[sharp rain]

Wait

[leaden scroll]

hm

[sharp rain]

Oh I messed up

I thought the rectangle were below the graph

Ok

Then

f(1) will not come

My fault

[leaden scroll]

its okayy

wouldn't all the numbers just stay like that otherwise they would be decimals or no?

[sharp rain]

I don't think I understand your point

[leaden scroll]

oh wait nvm

14 - ln(4) - ln(9) - ln(16) - ln(25)

[sharp rain]

Yup

[leaden scroll]

when you add them up

[sharp rain]

14 - 2 ln 120

[leaden scroll]

ohh okay

then for c we do the exact same but for the midpoints right?

[sharp rain]

Yup

And in 3

Make rectangles at intervals of 1/2

[leaden scroll]

and you add those with the right endpoints

[sharp rain]

Great!

[leaden scroll]

so do i add 14 - 2ln(120) to that

[sharp rain]

To do 3

?

[leaden scroll]

?

uh

[sharp rain]

Add this to what

[leaden scroll]

the 12 - ln(2.25) - ln(6.25) - ln(12.25) - ln(20.25)

14 - 2ln(120) + 12 - ln(2.25) - ln(6.25) - ln(12.25) - ln(20.25)

[sharp rain]

To get what?

In 3 you gotta fit 8 rectangles between 1 and 5

If you add these areas would overlap

And you get an answer bigger than intended

[leaden scroll]

ohh so this is just a more exact answer?

[sharp rain]

For 3
Do 1/2 ( f(1.5)+f(2)+f(2.5)+f(3)+f(3.5)+f(4)+f(4.5)+f(5))

Make sense?

So yeah

Kinda just add 6(i) and 6(ii)

And half it

Cuz the intervals has become 1/2

@leaden scroll

[leaden scroll]

ohh okay

ty

im going to bed now but ty for the helppp

[sharp rain]

Ur welcome

Gn

Bye bye

[leaden scroll]

byeee

.close

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[turbid scarab]

Hi

[turbid scarab]

I got the answers of a and b just idk what does c means

<@&286206848099549185>

E

Oh nvm the answer is just about the constants in p(x)

It’s solved

@vocal sleet

[tawny fjord]

!1c

[vocal sleet]

Please stick to your channel.

[tawny fjord]

This is the second channl you've opened thus far

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[violet saffron]

How would you solve this?

[violet saffron]

I get up till the point where n = k + 1

[lyric hamlet]

what working out do u have so far?

[lyric hamlet]

u wanna prove n=k+1, start from lhs, via algebra u want to eventually get to the point were ur able to factor 21 out of an equ thus proving its divisible by 21

[violet saffron]

Yeah but im not able to do that

[lyric hamlet]

show what u have so far

[lyric hamlet]

so u have ur n=k+1, u can start off by splitting the 4^K+1+2 into 4^K+2 times 4^1

then u use ur assume part, to sub into the 4^K+2 part

[lyric hamlet]

btw ur assume is suppose to say smt like 4^k+2 + 5^2k+1 = 21M, for some integer of M

[violet saffron]

Oh and you can sub it n = k into that

same with 5?

[lyric hamlet]

so essentially once u have ur n= k+1 u want to to eventually get it to equal 21x (something) to prove its divisible by 21

and to do that, once u have an equ for k+1, u want to use ur assume part at some point in it and sub it in

here i made it possible to sub in 4^k+2 by separating it a bit and then sub the assume part and the rest is algebra soup

[lyric hamlet]

what 5

[violet saffron]

Nevermind

I understand what you've done thank you

Cheers

[lyric hamlet]

try anfsee if ths makes sense

i hope it does

lmk if it doesnt

[lyric hamlet]

ohh okay nwws

[lyric hamlet]

u bascially approach these same steps for any divisbility induction proof qns

and for prove inductions qns every k+1 part u want to sub in ur assume somewhere in it

[violet saffron]

You assume n = k to be divisible and you sub it in

Yup

[lyric hamlet]

yuh

[violet saffron]

Oh my god that makes so much sense

[lyric hamlet]

yayyy

[violet saffron]

Ur a genius

Thank you 🙏

[lyric hamlet]

im not but thanks 😭

no worries!

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[jagged cargo]

my first thought is that the lnx doesnt even matter

you can just remove it

so what you would have is

[somber portal]

you cant just remove stuff

[jagged cargo]

$\frac{(x-1)^3}4+\frac{2(x-1)}{x+1}\ge 0$

[jagged cargo]

can i?

its given that x >= 1, so lnx will always be positive

[arctic mantle]

uh

fungus you actually just

solved it

maybe

hide the solution

????

;-;

[jagged cargo]

thus you can just pop the lnx out and still conserve the inequality

[arctic mantle]

*sighs*

[jagged cargo]

no?

[arctic mantle]

yes-

never mind. i'm dumb

physics has corrupted me (the subject)

[jagged cargo]

@untold arrow you following me?

[somber portal]

what does lnx being positive have to do with the inequality being proven?

[jagged cargo]

i am simply removing the lnx

yknow, to make your life easier by not having log?

[somber portal]

you... cant?

you cant just remove things "to make life easier"

[jagged cargo]

take this to hlounge please

[torn timber]

another victim.....

[vocal sleet]

@untold arrow Has your question been resolved?

[jagged cargo]

consider some inequality a < b

note that a is on the smaller side, so really i can subtract whatever i want to a and the inequality would retain

thus a < b implies a - c < b for positive c

they made it clear here that x >= 1, thus lnx >= 0. it is clear that you can simply pop it out and the inequality would still be retained

it sounds rather hacky, but i have personally used this plenty of time during epsilon-delta proofs

in other words, consider some kind of scale ⚖️

lets say the left side is higher than the right side, so the right side must be heavier right?

then i can simply remove stuff on the left side, the scales still lean towards the right side

is that clear?

[cobalt crypt]

-4 <= 5 so 20 - 4 <= 5 uh huh

very factual

[jagged cargo]

wut

[inner osprey]

real and factual and true and based

[jagged cargo]

thats not what i said

[cobalt crypt]

its exactly what you're suggesting

[somber portal]

no

[cobalt crypt]

that $-\f {2(x - 1)} {x + 1} \le \f {(x - 1)^3} 4$ is sufficient for $\ln(x) -\f {2(x - 1)} {x + 1} \le \f {(x - 1)^3} 4$

[somber portal]

-4 <= 5 so -4 -20 <=5

[cobalt crypt]

ln(x) is positive bro

wot are you on

[somber portal]

He's saying that $\ln(x) -\frac{2(x - 1)}{x + 1}\le \frac{(x - 1)^3} 4$ implies $-\frac{2(x - 1)}{x + 1} \le \frac{(x - 1)^3} 4$

[cobalt crypt]

thats not the goal of the question...

[twin meteor]

LordFelix

[inner osprey]

[cobalt crypt]

lets just change up the question to something else more convenient

then everything will be solved!

[torn timber]

proof by convenience 🔥

[jagged cargo]

:fire

:

[wet solstice]

I'd actually take the idea of simplification from the others.

Not quite 'take out the lnx'

But the (x>=1) is quite useful

If you prove a few useful 'subfacts' you can prove the inequality

[inner osprey]

kill the lnx 🫵

[cobalt crypt]

simpliest way out is probably just differentiate 3 times

[inner osprey]

that is quite an effective way of killing the lnx

[cobalt crypt]

🫵

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[vast shale]

question one, its stupid asf but it wants the interval of the real values of k, i forgot how do u properly solve stuff like that

[vast shale]

so k^2 -36 < 0 ?

ohhh

yeahh i think i can manage

i’ll prob struggle w sth else later but lets see

thankss

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[night wave]

how xe^(-1/2) dx , gives -2e^(-x/2) ?

[night wave]

can anyone please help me to understand this, I was very confused about this and the ai chat didn't made any sense.

[vocal sleet]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[night wave]

I was preparing for my exam and learning from a youtube video for a continuous variable solving problem. In a certain step the instructor performed integration on the xe^(-1/2) and came up with -2e^(-x/2) which didn't made sense to me.

[unborn beacon]

$\int xe^{-\frac{1}{2}} , dx$

[twin meteor]

Roman_Garland

[drifting jackal]

That says e^(-x/2) and not xe^(-1/2) dx

[night wave]

I am sorry, I was mistaken

[unborn beacon]

This?

[drifting jackal]

e^(-x/2)

[night wave]

yes

i was confused how it resulted as -2e^(-x/2)

sorry if it was a basic question, I am new to all this and preparing solely based on online so i can pass my exam

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[tall trellis]

Is it true that if I is an ideal of a noetherian ring R, and M is a finitely generated R-module then ((0):I)_M=0 implies there is an M-regular element in I?

[dull bear]

(you may wanna also try #groups-rings-fields too for this question btw )

[vocal sleet]

@tall trellis Has your question been resolved?

[tall trellis]

I'm taking this question to #advanced-algebra

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[feral sage]

How do i change the bounds like this here?

[feral sage]

the function X is periodic every 2 pi in omega

if we substitute z = omega + 2 pi we get the same integral but from pi to 3 pi

how do we get from 0 to 2p

i

[flat whale]

w = w + pi

[feral sage]

X isnt periodic every pi

cant do that

[flat whale]

w is the independent variable, not X

https://tutorial.math.lamar.edu/classes/calcI/SubstitutionRuleIndefinite.aspx

you can do $u = \hat{\omega} + \pi$ and then $\hat{\omega} = u$

[twin meteor]

riemann

[feral sage]

again,

$X(e^{j(\hat{\omega}+\pi)}) !=X(e^{j(\hat{\omega})})$

[twin meteor]

amnesia

[flat whale]

gonna have to show 7.2 then

[feral sage]

ah

ok

one sec

sorry

@flat whale where x[n] is a discrete time signal

[flat whale]

$e^{i \pi} = -1$

[twin meteor]

riemann

[flat whale]

and

,tex .exp rules

[twin meteor]

riemann

[feral sage]

yes

ah wait i got it

the negatives cancel each other out

X(e^(jw)) spits out a negative

arigato guzaymasu sir

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[hidden turret]

How do you find where two functions intersect algebraically
f(x)=cos(x)
g(x)=ln(x)

[peak axle]

If you can set them equal to each other and solve for x

[hidden turret]

ok now what?

[peak axle]

If you can solve for an x, they intersect at x,f(x)

[hidden turret]

yes but i dont know how to isolate x

[peak axle]

You can generally 'undo' a function by apply its inverse to it

[hidden turret]

ok

so do i use arccosine

[peak axle]

I.e. if $f(x) = x^2$ and $f^{-1}(x)=\sqrt{x}$, $f^{-1}(f(x)) = x$

[twin meteor]

m. frost

[peak axle]

You can, or you can use the inverse of ln(x) which may be easier

If oyu know it

[heavy yoke]

for two functions which are such different classes i would say you are unlikely to find an algebraic solution

[hidden turret]

conflicting ideas

[peak axle]

Or at least unlikely to find a solution that isn't in terms of two nested functions lol

[heavy yoke]

you can of course find it numerically to any precision

[hidden turret]

numerically??

[heavy yoke]

like x = (some decimal number) and you can find as many decimals as desired

[hidden turret]

i think im trying to do that

how do i do it without using like wolfram or something

[heavy yoke]

well if you have the intersection of two functions, you can always rearrange the equation to be about finding the roots of one function

[hidden turret]

so thats when it equals 0 right

[heavy yoke]

yes

[urban edge]

surely plugging in "cos(x)=ln(x)" into wolfram will give you solutions

[hidden turret]

yeah i know but thats highkey cheating

[urban edge]

not really

[peak axle]

They don't appear to want to have to use wolfram

[urban edge]

oh without

my bad

[peak axle]

If they did, they would have just put x=arccos(ln(x)) into it

Or what you said

[heavy yoke]

then you can use one of the many algorithms for finding roots of functions numerically

[hidden turret]

algorithms is this like the youtube algorithm

[urban edge]

its a set of steps

[heavy yoke]

an algorithm is basically a set of instructions, like a recipe

[urban edge]

which is what the youtube algorithm does

[hidden turret]

so what algorithm can find the point

[urban edge]

it follows its instructions

[heavy yoke]

an example of an algorithm which can find the roots of functions is the newton-rhapson algorithm

[hidden turret]

newton raphson that sounds cool how do you use it

[heavy yoke]

do you know derivatives yet?

[hidden turret]

like simple ones yeah

the power rule and stuff

[heavy yoke]

would you be able to find the derivatives of the given functions?

[hidden turret]

wait a sec i have this formula table brb

ok i found it

derivative of ln x is 1/x

and derivative of cos x is -sin x

what do you do now

[heavy yoke]

ok so we start with the equation
cos x = ln x
we need to rearrange it so that one side is 0

[hidden turret]

oh do you just move the ln x to the other side?

and set it equal to 0

[heavy yoke]

that works

so what do you end up with?

[hidden turret]

cosx - lnx =0

which makes sense when they are equal

[heavy yoke]

yes. so now we are going to call the left side of that an equation some new function. so come up with a function name (letter)

[hidden turret]

i guess just f cause thats normal

[heavy yoke]

ok. so we are interested in where f is 0

we first have to come up with a starting guess (some value of x close to where f(x) = 0)

[hidden turret]

hmm maybe try like 1 cause its pretty close

[heavy yoke]

sure

so we will call our starting guess $x_0$ and our first value will be given by the equation [ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} ]

[twin meteor]

cloud

[hidden turret]

uhh a bit confusing so do you just put 1 in for x0

[heavy yoke]

yes

but the equation is the same no matter what we chose for x0, which is why i wrote it like that

[hidden turret]

ok so i got $1- \frac{cos(1)-ln(1)}{-sin(1)-\frac{1}{1}}$

[twin meteor]

Banana Steeler

[heavy yoke]

so we can put that into a calculator and get a decimal number out

[hidden turret]

ok let me check

ok nice i got it its 1.293407993

wow thats a cool method

is this the answer?

[heavy yoke]

that's the first step. the next step is to call $x_2$ our next answer and do the same thing again
[ x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} ]

[twin meteor]

cloud

[heavy yoke]

and basically keep going until you are satisfied that you're close enough to the answer

[hidden turret]

oh so i just put the number i got into the thing again

[heavy yoke]

yes

[hidden turret]

okay

okay for some reason i did it 3 times and it isnt changing anymore

i think that means i got the answer

im gonna check

oh wow it worked

i got 0 when i subtracted them

Thanks

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[tawny nimbus]

What is 12x3

[vocal sleet]

Please don't occupy multiple help channels.

[drifting jackal]

Please don't open multiple channels for this

Use a calculator

[sly sierra]

duplicate of #help-32

.close

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[indigo wave]

Hello, can someone help me solve this integral.

[candid magnet]

factorize denominator