help-0

[nimble fern]

i think it's good for (10) too

so, they are both good

[lone heart]

@fresh nebula Has your question been resolved?

[fresh nebula]

they are both 100% correct?

because im really confused on how to treat |-x| < 1

[nimble fern]

I can't guarantee, but after checking, they looks good to me

,w infinite sum 1/sqrt(n)

[ocean seal]

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[nimble fern]

,w infinite sum (-1)^n/sqrt(n)

[ocean seal]

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[nimble fern]

what do you mean by treat |-x|<1

[fresh nebula]

do i just say x < 1

or do -1 < -x < 1

then do 1 < x < -1 but that wouldnt make sense

so do i do -1 < x < 1

[nimble fern]

well, |x|=|-x|

[fresh nebula]

ah okay, i guess I was confused because the videos I watched on this said: to get rid of absolute value symbol, u also put -1 < on the other side

okay thank you

.close

[lone heart]

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[median oar]

Ok so I think I might be understanding the LRT better

[median oar]

If we want to test θ = θ₀ vs θ < θ₀ or θ > θ₀ then these conditions satisfies the Karlin-Rubin theorem, and thus a UMP test can be constructed.

If we want to test θ = θ₀ vs θ = θ₁ then these conditions satisfies the Neyman-Pearson lemma, and thus a UMP test can be constructed.

Any other test (that doesn’t satisfy either lemmas) has no UMP. We can instead construct a likelihood ratio test. The -2log likelihood ratio of the MLE over H₀ to the MLE over all space is used. This statistic is used because it has desirable asymptotic behaviour.

Is this correct

[lone heart]

@median oar Has your question been resolved?

[lone heart]

@median oar Has your question been resolved?

[lone heart]

@median oar Has your question been resolved?

[stone pond]

Yeah this sounds good to me

Note also that the Newman Pearson lemma guarantees more than just the existence of a UMP test

Specifically it demonstrates that the MLE is the UMP

For simple hypotheses

Likewise the KR theorem gives that the threshold test is the UMP

Actually scratch that last part

This is only the case if the likelihood ratio is strictly increasing iirc

[median oar]

Yeah there’s that part too

I’ve shortened a bit of the first 2 points in their conditions for the theorems

Ok fantastic, I’m not going crazy

When we have some wack ass hypotheses, there’s no guarantee we can find a UMP test

Thus we do the LRT which is defined to use the MLE over all space for one of the things

Then everything is good in the limit

Ok thank you very much

.close

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[regal lotus]

18,11,4 the common difference between these numbers is -7 right?

[fair harbor]

I need help with a alegbra problem

[chilly idol]

yes -7 is the common difference

[echo socket]

This channel is taken, open your own one

[chilly idol]

#❓how-to-get-help

[regal lotus]

so it's alright if I use -7 when finding the arithmetic sequence?

[chilly idol]

yup

[echo socket]

What exactly do you mean by finding the arithmetic sequence?

[chilly idol]

you can use the -7 as your d variable

[chilly idol]

I'm assuming that he/she is using the arthmetic sequence formula?

[regal lotus]

yeah

I see, thanks 🙏

.close

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[small junco]

Is this DNE?

[echo socket]

No, why do you think the limit does not exist?

[small junco]

cuz i thought step functions are all DNE

[vale wigeon]

this is a one sided limit tho

x goes to -2 from below.

[small junco]

gotcha, do i just plug in?

[little blade]

Imagine that x=1.999..., which is 2-

Then [x]=1

Therefore 2[1]-2 = 0

[tardy stag]

it gives you a graph, you might use it

[echo socket]

Btw it approaches -2 from below rather than 2

[small junco]

so -1.999 and -2.001

[little blade]

Only -2.0001

Because given $-2^-$

[ocean seal]

Unique

[small junco]

hmm so from the left would be -2.0001

so step function -2.0001 is -3

-3*2 - - 2.0001?

[little blade]

Yes you are right

It will be -3

[tardy stag]

-4?

[small junco]

wouldnt it be -4?

yea

[little blade]

See, you are amazing 👏

[small junco]

bet thank you so much

.close

[lone heart]

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[small junco]

-2, 1 correct answers?

[echo socket]

Nope, there's just one removable discontinuity

[small junco]

so would it just be -2?

[echo socket]

There is a discontinuity at x = -2, sure, but it is not removable

[small junco]

oh its 2 right

[echo socket]

Yeah

Could you explain why as well?

[small junco]

its 2 bcuz at a single point can fill a hole

[echo socket]

Informally speaking, yeah, but removable discontinuity occurs when the limit of f(x) as x -> a exists, but is not equal to f(a)

And that was the case here at a = 2

[small junco]

gotcha that makes sense

[lone heart]

@small junco Has your question been resolved?

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[small junco]

we are asked for B, i put -2/3 is that correct?

[left isle]

yeah that's what i got

did you find a?

[small junco]

we dont need to

[left isle]

im confused

it asks you to find the values of a and b so that it's continuous

you are only asked to find b by your instructor?

[lone heart]

@small junco Has your question been resolved?

[young finch]

dont you need a to find b

[left isle]

you can find b first to find a

[coarse void]

you get simultaneous equations for a and b that you can solve to get both

[left isle]

one may be easier to find than the other, but the order in which you find them doesn't particularly matter

[lone heart]

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[tender ruin]

<@&286206848099549185>

[lone heart]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[tender ruin]

Q no. 1

[young finch]

do you know the definition of a function

[tender ruin]

Yeah

[young finch]

so do you see an issue with g being a function

maybe looking at where the two parts of the piecewise combine

[tender ruin]

Hm

I am so confused 🤗

[young finch]

alright

the issue is its saying at x=2, g(x) equals 4 and 6

[tender ruin]

😃

[young finch]

you said you know the definition of a function right

this should obviously contradict the statement that g(x) is a function

[tardy tapir]

ncert

[lone heart]

@tender ruin Has your question been resolved?

[lone heart]

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[nocturne trench]

the answer is D but i’m not sure why (for number 8)

[nimble fern]

I'm thinking its (E), would it be a typo for the answer?

[nocturne trench]

isn’t both limits 2 for E

[nimble fern]

yea

[chrome flower]

isnt fx exitst at x=4

[nocturne trench]

yeah i also thought that

[nocturne trench]

mb i forgot we’re choosing the correct statements 😭😭

[chrome flower]

wth

[nimble fern]

wtmh

[nocturne trench]

yeah idk i’m really confused

i’m staring at it and istg i see smth at 4

[lone heart]

@nocturne trench Has your question been resolved?

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[languid bolt]

!status

[lone heart]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[alpine sable]

substitute x with 2a+5

[languid bolt]

[ancient cargo]

its that simple

[languid bolt]

yes

[alpine sable]

yup

[ancient cargo]

wt abt this

[alpine sable]

same logic applies here

[ancient cargo]

so i substiude x with z-1

[alpine sable]

yes

[ancient cargo]

does z-1= zero matter?

[alpine sable]

z-1 is not equal to 0

[ancient cargo]

f(z-1)

[alpine sable]

yes cause after you sub x with z-1 you can equal that to 0

[ancient cargo]

so 2x^2 when subsituded= 3x^2?

[alpine sable]

you'll get a simple second power equation

[ancient cargo]

or 3x^3

[alpine sable]

nope

its $2(z-1)^2$

[ocean seal]

male-blu

[ancient cargo]

so (2z-2)^2

[alpine sable]

$2(z^2-2z+1)$

[ocean seal]

male-blu

[alpine sable]

powers before multiplication

[ancient cargo]

yep

thx

[alpine sable]

send me the entire thing when you re done

[ancient cargo]

so 2z^2+7x-14?

[alpine sable]

you have to sub x at 7x as well $7(z-1)$

[ocean seal]

male-blu

[alpine sable]

and you have to solve this equation $2(z^2-2z+1)+7(z-1) -11 = 0$

[ocean seal]

male-blu

[ancient cargo]

so then 2z^2+3x-23=0

how to i solve for z

[alpine sable]

nope

its $2z^2 +3z -16 = 0$

[ocean seal]

male-blu

[ancient cargo]

i mustve gotten somthing wrong while simplyfing

[alpine sable]

use the determinant formula (i dont know if thats the word in english)

[ancient cargo]

its all g

thx

.close

[lone heart]

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[odd fulcrum]

hi can i get some help answering this question

[odd fulcrum]

ill show wat i have done so far

[native cloud]

Please ping me once you got the answer

I wanna know the answer also

[odd fulcrum]

is this corect

so far

[alpine sable]

i dont think its correct

[odd fulcrum]

o

wat did i do wrong

[alpine sable]

well , the -3 out of the absolute value should be 3, right?

[odd fulcrum]

oh true

then is everything else correct

[alpine sable]

but you havent proved it yet

[odd fulcrum]

is that with the |x=2| < S < 1

[alpine sable]

let me solve it, give me a second

[odd fulcrum]

okay

[alpine sable]

according to my calculation $\abs{x-2}<\frac{\epsilon}{18}$

[ocean seal]

male-blu

[odd fulcrum]

can i see how u did it

[alpine sable]

writing with this shit sucks btw, but since i started yesterday im getting better. you gotta give me some time , wait

[odd fulcrum]

okay i can wait

[alpine sable]

ok , im back , sorry my laptop died

[odd fulcrum]

lol ur fine

[alpine sable]

what do we know about $\abs{x-2}$

[ocean seal]

male-blu

[odd fulcrum]

it equals epsilon

righrt

[alpine sable]

no, its less than delta

and delta , according to our assumption is less or equal to 1

wich means we know at least |x-2|<1

[odd fulcrum]

ohh so

|x-2| < S < 1

[alpine sable]

i dont know what the S is

[odd fulcrum]

oops

delta

[alpine sable]

but wait

if |x-2|<1 it means that x+2 < 5

now at the |x+2||x-2|<epsilon/3 substitute |x+2| with 5

you end up with |x-2|<epsilon/15

now do the last part of solution

[odd fulcrum]

which is the

-1 < x-2 <1 right

[alpine sable]

no , we already implied it when i arrived at x+2 <5

[odd fulcrum]

o

now im confused then lol

wats the last sstep

[alpine sable]

|x-2|<1 means -1<x-2<1 means 3<x+2<5 means |x+2|<5

in the second step i just added 4 to each side

[odd fulcrum]

delta = e/-15 right

[alpine sable]

now what you do is $\abs{-3x^2+7+5}=\ 3\abs{x+2}\abs{x-2} <3\dot5\dot\abs{x-2}$

[ocean seal]

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

[alpine sable]

we know |x-2|< delta

the expression $ 3\abs{x+2}\abs{x-2} < 15\delta$

sub delta with $\frac{\epsilon}{15}$

[ocean seal]

male-blu

[odd fulcrum]

i couldnt see this one

[alpine sable]

$3\abs{x+2}\abs{x-2} < 15\delta$

[ocean seal]

male-blu

[alpine sable]

and when you sub delta , you'll conclude that the entire expression is less than epsilon, wich is what we wanted to prove

and you're done

btw thank you for asking me , if i could do this in latex , is helping me learn something new

[odd fulcrum]

and im done?

no ty

u r helping me way more

[alpine sable]

bro no wait

delta is epsiolon/15 , and at the end th expression would be just less than epsiolon

epsilon*

also send me the complete proof

[alpine sable]

last step is wrong, also its epsilon/15 instead of epsilon/-15

[odd fulcrum]

o

[alpine sable]

$3\abs{x+2}\abs{x-2} < 15\delta\3\abs{x+2}\abs{x-2} <15(\frac{\epsilon}{15}\3\abs{x+2}\abs{x-2} <\epsilon$

[ocean seal]

male-blu

[odd fulcrum]

and im done

[alpine sable]

yup

[odd fulcrum]

TY

[alpine sable]

anytime, keep up your good work.

also dont forget to .close

[odd fulcrum]

o true

ill be back soon blu

.close

[lone heart]

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[raven girder]

$a_n + \frac{4}{a_n} - 4 \geq 0$ with $a_n > 2$

[ocean seal]

madmike

[raven girder]

How do I simplify the left equation

so it gets obvious that it's true

I am lost

[zealous lichen]

why use a_n

if they are all the same

just use x?

[raven girder]

it's a recursive sequence

[alpine sable]

$(a_n)^2 -4a_n + 4>=0$

[raven girder]

I am trying to prove that it's monotonous decreasing

[ocean seal]

male-blu

[tardy tapir]

AM GM

[alpine sable]

you can multiple each side with an knowing an is positive

[median oar]

Is that am gm

[tardy tapir]

it is

[raven girder]

what is am gm

[median oar]

This looks better

[raven girder]

ohh right

[median oar]

Am gm is what people say when you want to use the property of arithmetic and geometric means

And their relationship

$\frac{a+b}{2}\geq \sqrt{ab}$

[ocean seal]

Frosst

[median oar]

But I don’t think that’s the best method here

[tardy tapir]

it is the fastest*

[wild trail]

This looks like squaring

[raven girder]

yeah I have no clue what that is

[wild trail]

I'm writing a instead of an to make it quick

[raven girder]

ok so it seems I can use pq formula now

[alpine sable]

you said you have to study the monotony?

[raven girder]

yes

I think I can do this

not sure what it's called in English

[ruby current]

isn’t LHS just a perfect square

[wild trail]

$a + \frac{4}{a} - 4 = \sqrt{a}^2 - 2 \sqrt{a} \frac{2}{\sqrt{a}} + \left(\frac{2}{\sqrt{a}}\right)^2$

[ocean seal]

numbily

[wild trail]

exactly

[raven girder]

what is a perfect square

[wild trail]

something which has a proper square root

for example 9, 16, 100 are perfect squares

[raven girder]

ah okay

[alpine sable]

what are y'all doing

[wild trail]

but 2, 6, 23 doesn't

[ruby current]

same thing you did but before multiplying

[alpine sable]

bro , just now at the euation that i gave you, just find the first derivative, study the sign and determine the monotony

[wild trail]

Does OP know all these concepts?

[alpine sable]

he does, cause he is asked to find the monotony, right @raven girder ?

[raven girder]

umm

[wild trail]

nope

[raven girder]

yes

but I am clueless

I thought I can just bring it into a simple form

where it says x > 2

qed

[alpine sable]

the first derivative is $2a_n -4$

[ocean seal]

male-blu

[wild trail]

mike do you know what monotonous decreasing mean?

if not this is pointless

[raven girder]

yes I know

[alpine sable]

it is negative for an<2 and positive otherwise. Which means in its domain the function is monotonously increasing

[raven girder]

a_n+1 <= a_n

[wild trail]

also @raven girder what is an again?

like what was the recurrence relation

[wild trail]

alr

bruh, just use induction

[raven girder]

I did

result is a_n > 2 for all n in N

[alpine sable]

yes, induction helps in this case lol

[ruby current]

What you did was fine

Just write the LHS as a perfect square

Then you’re done

[wild trail]

ie a{n+1} - an <= 0

well then as tushar said

write it as perfect square and you're done

[raven girder]

um wait I got a solution but idk if it's legit

[ruby current]

if you have trouble writing the LHS as a square, multiply both sides by a_n² and use that

[ruby current]

you’re done at your third-last line

because the square of any real number is >= 0

[raven girder]

oh I see

ok thank you everyone

and @alpine sable for the tip

[ruby current]

last two lines are nonsense

[wild trail]

yeah mike was done at the 3rd last step itself

you might wanna use <=> (if and only if) instead of equality @raven girder

anyway

use .close to close the help channel

[raven girder]

I did didn't I?

ok

.close

[lone heart]

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[lone heart]

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[lone heart]

@short compass Has your question been resolved?

[granite badger]

for the first one for example: x + 5y^2 - y + 7 = 0
based off what you wrote B^2 - 4AC
0^2 - 4(0)(5) = 0

your quadratic term is any term with a degree of 2

in this case 5y^2 is your quadratic term

type of conic can be determined by the rules you wrote:
A = C which is 0 = 5 (reject)
A = 0 or C = 0 which is 0 = 0 or 5 = 0 (accept)
A != C which is 0 != 5 (accept)
A != C which is 0 != 5 (reject)

your notes are somewhat confusing tbh

i would just complete the square..

[lone heart]

@short compass Has your question been resolved?

[lone heart]

@short compass Has your question been resolved?

[lone heart]

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[carmine fossil]

Question number l (L)

[somber eagle]

do you know the circle theorems?

[carmine fossil]

yes

I think all I need for this question is the cyclic quad ones?

[somber eagle]

i think

[carmine fossil]

yes I do know them

[somber eagle]

ok well look at the corners that have the x and y angles?

what can you tell about them

[carmine fossil]

Their sum is 180

[somber eagle]

yeah

do you know what one of those quad angles are?

[carmine fossil]

x is 85

[somber eagle]

yeah

[carmine fossil]

but it's 85 + y + angle MKJ

[somber eagle]

tbh im still trying to find out how to get y

geometry isnt my strong suit

[carmine fossil]

I don't understand just this one question

[somber eagle]

all you need is angle MKJ, so maybe there is some way to find that

[carmine fossil]

yeah but I just can't find it

[somber eagle]

neither, i should have probably found both before trying to help

[carmine fossil]

The answer is supposed to be
X = 85
Y = 35

but idk how to get there

sorry but I have to eat I'll brb in around 10 mins

[tardy tapir]

use this for MJ to find angle MKJ

[somber eagle]

i lied, my bad

[lone heart]

@carmine fossil Has your question been resolved?

[carmine fossil]

ty

[carmine fossil]

it's alr

.close

[lone heart]

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[deft stag]

Find the sum of roots x^=1

[deft stag]

X^2=1*

Than answer is zero right

[karmic grove]

Yes

[somber eagle]

yeah

[deft stag]

Alr thx

.close

[lone heart]

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[lone heart]

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[brisk ruin]

i need help with a geometry review

[terse turtle]

Ask away

[brisk ruin]

well im supposed to be finding thr 90 degree angle

[terse turtle]

Thr?

[brisk ruin]

thats not an answer it looks like i have see which ones equal 90 degrees i think

[brisk ruin]

has to be any of the selections

[wispy hazel]

do you want the answer straight up?

[wispy hazel]

the means 'the' lol

[brisk ruin]

that would help very much

[wispy hazel]

just try to compare the options with the grid

[brisk ruin]

i dont rlly know how to do thar

[wispy hazel]

ok

take XWY

[brisk ruin]

does that equal 90?

[wispy hazel]

What angle does XW make with the positive x axis?

then find the angle YW makes with positive axis (go counter clockwise)

the angle XWY would be the difference between the two right

think a bit about this

[brisk ruin]

can yu go in depth with that

do i add the x wich is 120 and w 60?

[wispy hazel]

the angle XW makes with x axis is 120

[brisk ruin]

now what do i do

[wispy hazel]

you see those markings right
0, 30, 60, 90

[brisk ruin]

yes

[wispy hazel]

the tell you the angle it makes with the x axis

now what angle XW makes with x axis looks like this

[brisk ruin]

whats next

[brisk ruin]

120 angle

[wispy hazel]

WY makes 210 degrees angle with the positive x axis

so if you subtract both of these

[brisk ruin]

i got 90

[wispy hazel]

you would get the angle XWY

that is correct

[brisk ruin]

is that only one or is there more

[wispy hazel]

ok now do you understand the concept of an angle between two lines?

[wispy hazel]

check all of them

[brisk ruin]

kinda do i jus take the number on and and the number on w and subtract them

[wispy hazel]

uh

do you understand what angles are

[brisk ruin]

yes

[wispy hazel]

ok then you can do the subtraction thing for this question

[wispy hazel]

not always W

it can change right

here it is always W

[brisk ruin]

for bwh i got 90 i subtracted b which is 180 and h which is 90 and i got 90

am i right

[wispy hazel]

but if you're doing another problem
you have to identify the middle point as some other letter

[wispy hazel]

yes

angle BWH = 90 degrees

[brisk ruin]

and bwd i also got 90

[wispy hazel]

correct

[brisk ruin]

and kwr which is 90 also

[wispy hazel]

yes

[brisk ruin]

ok i think i understand it ow

now it's asking thx is created by rotating @wispy hazel

some letters

[vale wigeon]

!onechannel @brisk ruin

[lone heart]

Please stick to your channel.

[brisk ruin]

oh srry

need help with this

[lone heart]

@brisk ruin Has your question been resolved?

[quartz mica]

reflection

honestly don't understand the question

[lone heart]

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[tall talon]

Just asking out of curiosity:
Is it possible to find at least one set of the exact values of a, b, c in a quadratic equation ( in its basic form ) given the value of delta and the quadratic formula for this equation?
(I have no idea on how to solve this, but this question seems cool)

[lone heart]

@tall talon Has your question been resolved?

[tall talon]

.close

[lone heart]

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[alpine sable]

integral x/cos^2(x/2)

[frigid mirage]

!status

[lone heart]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[alpine sable]

1

[frigid mirage]

first let's make it look a bit cleaner.
$I = \int x\sec^2(x/2) dx$

[tardy tapir]

IBP

[frigid mirage]

this^

[ocean seal]

nebula40

[frigid mirage]

that's integration by parts if you aren't familiar with the abbreviation

[alpine sable]

bruh

f

i am so dumb

thanks btw

.close

[lone heart]

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[lone heart]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[nimble fern]

do you know what does it mean by f(X) for set X?

[tawny radish]

No, not really. I know some set identity rules but I dont know how to use them here.

Also on this problem there are functions included

[nimble fern]

Recall:
f(A) for set A subset X (domain of f)
f(A)={y in Y| y=f(x) for all x in A}
where Y is the range of f

that is e.g.

f(x) = x², A={1,2,3}
then f(A)={1,4,9}

all good till here?

[tawny radish]

still processing in my brain

but you can continue

[nimble fern]

okay, now using the definition we have
f(A n B) = {y in Y | y=f(x) for all x in (A n B)}

[tawny radish]

ok

[nimble fern]

now we need to proof that all y in f(A n B) is in f(A) n f(B)

thats the goal

[tawny radish]

I thought that we would use some identity rules to prove it

[nimble fern]

I'm not sure what kind of identity rules you have.

[tawny radish]

neither do I thats why I asked here, but thank you

this is different perspective for me

[nimble fern]

okay, so, all is set, we can start now

For all y in f(A n B), there exists an x in (A n B) such that y=f(x).

x in A and B
then
x in A and x in B

so, for x in A, y=f(x) in f(A)
and for x in B, y=f(x) in f(B)

since x in A and x in B,
y=f(x) in f(A) and y=f(x) in f(B)

i.e. y in f(A) and y in f(B)
which means
y in f(A) and f(B)
which means
y in f(A) n f(B)

it follows that
f(A n B) is a subset of f(A) n f(B)

done

(side note: the reverse statment is not necessarily true, so be careful)

[tawny radish]

@nimble fern thank you very much and appreciate for the help

[nimble fern]

any questions so far?

[tawny radish]

not for now

.close

[lone heart]

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[proud wyvern]

How do I find this infinite sum? Not a homework problem

[summer bolt]

i need help with this

[alpine sable]

,rotate

[lone heart]

Please read #❓how-to-get-help

[summer bolt]

someone help?

[wary stream]

Read #❓how-to-get-help this channel is taken, you need to open your own

[proud wyvern]

Heres a better picture

[wary stream]

What's the goal?

[proud wyvern]

Find the sum

On wolfram alpha it says its -ln(1-x) / x which I want to learn how to do

[dawn zephyr]

set that equal to S then multiply both sides by x

xS =∑ xn/n

[proud wyvern]

Like this?

[dawn zephyr]

yes

[proud wyvern]

How does this help

[dawn zephyr]

can you proceed on your own now?

[dawn zephyr]

oh

can you differentiate?

[proud wyvern]

Yes

[dawn zephyr]

wait, there should be an easier way

[dawn zephyr]

differentiate and send your progress

it should come out as a geometric series

[proud wyvern]

This is the context

[dawn zephyr]

which you get the sum of, then integrate and divide by x to get the infinite sum of the original series

[proud wyvern]

Do I assume S is a constant

[dawn zephyr]

you can leavee that as is

[dawn zephyr]

here,
set the sum from 0 to infinity instead of 1 to infinity

you can then replace the n with (n+1)

that should give a neater series once you differemtiate

[proud wyvern]

Like this?

[dawn zephyr]

you can remove that negative 1 if you set n=1 under the summation symbol

let me write it out

[proud wyvern]

My a_n term becomes x^n / n+1

[proud wyvern]

Also, i got this from watching a video from michael penn and turning your 0th integral to a first

[proud wyvern]

@dawn zephyr is this correct

[dawn zephyr]

it is but that's not getting us anywhere

sorry for the ass handwriting..

[proud wyvern]

I didnt know that swapping index trick so thanks for teaching me that

[dawn zephyr]

x^n-1

that's a geometric series that you should be able to sum

[dawn zephyr]

does this make sense and can you proceed on your own afterwards?

[proud wyvern]

This is supposed to be the end result

[dawn zephyr]

yea

you integrate the infinite sum of x^n-1

then divide by x to get that

1 + x^2 + x^3 + x^4...
that's the series

[proud wyvern]

Does the sum still start at 1

[dawn zephyr]

yes, I didn't do the shifting bit

so, 1/1-x

integrate that

[proud wyvern]

Based Off the screenshot you posted is this what you are trying to say

[dawn zephyr]

uhhh

no...

[dawn zephyr]

first I multiplied both sides by x

does that part make sense?

[proud wyvern]

Oh the d/dx is still there in left hand side

At the end

[dawn zephyr]

x^n-1 multiplied by x is x^n

then I differentiated

derivative of x^n / n is x^n-1

[proud wyvern]

So you meant this?

[dawn zephyr]

I didn't finish it there,
you find the inifinite sum of the derivative, then integrate and divide by x to get the result

[dawn zephyr]

yes

[proud wyvern]

Well based of this the new sum looks like a geometric series

[dawn zephyr]

yea that's what you should've been trying to do

you know how to sum a geometric series, use the formula, then integrate that

something like this

[proud wyvern]

I get it now

[dawn zephyr]

good job

[proud wyvern]

The step at the beginning was something I wouldnt have thought

Just some exponent manipulation

[dawn zephyr]

me neither, I was trying random bullshit to try and get a geometric series as the derivative

[acoustic beacon]

Isn't 1/1-x the answer to the infinite sum though or am I wrong.

[dawn zephyr]

this is the original question

it's not a geometric series

[proud wyvern]

If I want to make a new question do I have to close this one. The new question is related to this one

[acoustic beacon]

Yeah I get that.
d/dx(Sx)=S
S=infinite sum of x^n-1 from n=1

[dawn zephyr]

d/dx(Sx) does not equal to S...

[proud wyvern]

Thats why the integral cancels it

[dawn zephyr]

uhhh, I hope you're not thinking the d's and x's cancel out...

[dawn zephyr]

yes, we integrate to get rid of differentiation notation

[acoustic beacon]

I was assuming S was supposed to be a constant.

[proud wyvern]

Thanks to your help I know this now

[proud wyvern]

Same

[proud wyvern]

@dawn zephyr is this allowed

[dawn zephyr]

idk, not a regular here

[acoustic beacon]

It makes sense why it's not though if you think about it for a minute.

[dawn zephyr]

risky

[proud wyvern]

Can I ask a different question

[dawn zephyr]

ask away

[acoustic beacon]

You can. This channel is all yours. You can use it until you close it.
Though it will close after a while if left unattended because of inactivity and such.

[dawn zephyr]

is that n-2?

[acoustic beacon]

Well, it's a geometric series again but the first term is x^-1 instead so I think x^-1/1-x would work.

[dawn zephyr]

it's the same series then..

[proud wyvern]

Maybe this?

[dawn zephyr]

don't over complicate things

the sum upto infinity of a geometric series is

a/1-r where a is the 1st term and r is the common ratio

absolute value of r must be less than 1 for there to exist an infinite sum

[proud wyvern]

I get that part

[dawn zephyr]

plug in the 1st few terms

if n = 1, x^n-2 = 1/x
if n = 2, x^n-2 = 1
if n = 3, x^n-2 = x

so the series is

1/x + 1 + x + x^2 + x^3 ....

so the infinite sum will be,
1/x / (1 - x)
= 1 / x(1-x)

[acoustic beacon]

Yup. Formula for reference.

[proud wyvern]

If Im being honest the technique you showed me I understand it better the mutliply both sides by X

[dawn zephyr]

whatever's more intuitive to you

that's not incorrect by any means

but it's still built upon the foundations laid by the original formula

[proud wyvern]

Will I be seen as greedy if i ask another question

[dawn zephyr]

no, ask

not sure if I'll be around long enough to answer though, someone else should help you out

[proud wyvern]

How would I go about doing this. Im thinking of a u sub but it might not be that easy

I dont know if its non elementary

This is my idea

But one of the bound would be ln(0) which is undefined

Actually no maybe this

<@&286206848099549185> its been 15 min ish

.close

[lone heart]

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[alpine sable]

anyone can help me

[alpine sable]

You know the rules of exponent?

• (ab)² = a²b²

And that

• (a²)³ = a⁶

Use these to get the answer

[tulip bear]

You put all terms in the inside of the parenthesis to the power of 3

[alpine sable]

Exponents of all terms*

[alpine sable]

@alpine sable ??

nah

i skipped

but this one

@alpine sable

is

a) 1 , 4, 7

Hmm??

b) -1, -4, -7

[alpine sable]

is a) and b) correct for this one

Do you not try for n = 0 as the first term??

idk

You do

First term n=0
Second n=1
Third n=2

I believe it's Arithmetic Progression

Tell me the answer now, cuz what you told are the second, third and fourth terms

@alpine sable

[alpine sable]

Yeah?

[alpine sable]

Yeah I mean , we here start from n=0 , it depends upon the teacher

You can start from 1 too, then it's correct

in algebra u stsrt it with one

so

Yeah no worries

a) 1, 4, 7

b) -1, -4, -7

Correct

dang my friend is smart cuz idk what algrbra even is

ty for the confirmation

ig

m

Yeah welcome

i think i might be joking

Though you shouldn't have had skipped that exponents one

but i stuggle on a + a

its a(squared)

[alpine sable]

I do too, used to, a lot

anyways cya

Yeah

@alpine sable close the channel

If you're done

no

is this true

the trick is whenever you encounter a question such as write the rule for this sequence: 3, 5, 7, 9...
you first wanna always write a(n) = a(n-1) then you count by how much the sequence is going up or down by subtracting the 1st term from the 2nd term so 5-3 = 2 so you would write down a(n) = a(n-1) + 2 because its adding 2 to the term before it to make the next term, then to get full credit you need to write down the first term a(1) so in this case the first time is 3 so you would put a(1) = 3

Yes, it is true

so can u help me

Yes, I can

7 AND 8

Hmm 7 first

What's the first term?

10, 8, 6, 4....

What's the difference between each consecutive terms?

I'm guessing ur tryna say whats the difference between each number?

Yeah

m it looks like to me its -2

Good

So can you write any AP as
(First term) - (Difference of terms)(nth term)

So here

what

?

It will be
10 -2n

the answer?

Wait a sec

Yeah

after. i write this down lemme try 2

But I'll have to see if your teacher wants n to start from 0 or 1

ok so

b)

30, 25, 20, 15

and difference is -5

Yep

so

Yep

relating back to this

Yep

So can you write any AP as
(First term) - (Difference of terms)(nth term)

that

so

first tetm

30

Yep

30-5n

?

Good

That's correct

cuz. difference is

Now the third one?

-5

OMGGG

i love thiss

ok thord

ait lemme write it down

Yep

third

2, 1, 0, -1

so

difference is -1

Yep

soo

its gonna be

So can you write any AP as
(First term) - (Difference of terms)(nth term)

therefore

2-1n

?

Yep

That's correct too

EZ PEZ LEMON SEUEZZ

can u help me with

[alpine sable]

Now Q8

oh yh mb 8

(8

))8

Hmm lemme see

Use
So can you write any AP as
(First term) - (Difference of terms)(nth term)
Again

wait this is e,

3+4n

First of all , tell me what would be the correct AP??

ap

[alpine sable]

Yess

whats that again

m

whats ap again

apple

?

[alpine sable]

Hmm leave it,

kay

Arithmetic progression, that's what these series are called

When you add or subtract the same number over and over again

rightt

3, 3+2 , 3+2+2, 3+2+2+2...

ok

i call that a pattern

.

No worries

[alpine sable]

So you got 3 + 4n

ye

i. got it

is thst the asnwer

of why he is wrong

how is that the answers

answer

Yeah you got the correct one

Now compare it with what Brett did

And find his error

m

[alpine sable]

Btw

dont

really

The formula you made , to find the uhhh nth term put in the value of n as n-1

get

it

Same

Like Brett is wrong in so many ways

But let's keep that on stay

k

[alpine sable]

You understood this?

no

Okay do you see

When 3 +4n

i got it

3-4n

?

Noo

always minus?

oh oops

Nooo

kkkkk

i didnt understand it btw

@alpine sable

@alpine sable @alpine sable

What you did is correct

I'm just telling you that
Whenyou put n= 0 you get the 1st term
N=1 you get 2nd term
N=2 you get 3rd term

[alpine sable]

Sorry was afk

[alpine sable]

Do you get this @alpine sable

no

Hmm

oh i got it

now

so like

Try putting values of n=0,1,2 in your formula

if i put n=0

You will get the 1st,2nd,3rd.. terms

then like n=6 will be my 7th term

?

Yesss

w

Phew

W

Now

but that wont stay in my head

im telling u rn

.

What value of n would be your 10th term?

wel

is n=0

?

if yes then 11

if n=1 then 10

correct?

[alpine sable]

Write it down somewhere, and read it today evening, tomorrow evening, a week after, a month after
It'll always be there for your lifetime

[alpine sable]

Ummmmmmmmmmmmmmmmmmm

No

i write on dis paper

bruh

[small gyro]

hi im new

[alpine sable]

At n=9 you get the 10th term

thats literally what i said

.

Idk anyway

and like at n=4 u get 5th

So

Yeah

Correct

Now find the 10th term for me in 3+4n

ok so why is brett wrong

uhh

is it like

[alpine sable]

please

3

(and send the question again, please)

[alpine sable]

No...?

wait im not done

Yeah

Waiting

3+11n

.

N=4 gave you 5th term
N= what will give you 10th term?

ohh 9

9

9

Ep

Yep

3+9n

3+9n

No

i brain ded for a sec-

bruh

n = 9 you don't change the formula

3+4n where n= 9 to get the 10th term.

so the answer is..

3+4(9) = 3+36 = 39.

Try it yourself..

I give you like 40 seconds

Go

@alpine sable @alpine sable

Yeah

im going afk for like 20mins

I'll go to sleep in 5

do not close channel

bruhh

I don't have the permission to keep it open or close

kay

i got 20 more questions

.

[alpine sable]

There will be other helpers ig

tmrw?

the channel is kinda ded tho

Yeah or you can dm them

k

Or you could ask it in some other channels

Whatever you wish

ur timezone??

+5:30 hrs GMT

whats the time for u rn

Lmao it's 2:15 am here

[tardy tapir]

@alpine sableso you're lurking around here

[alpine sable]

You? Maybe 10 ish

[alpine sable]

I am I need to sleep 😭

[tardy tapir]

leaving me there to do physics

[alpine sable]

I just did 50 questions of phy I was so damned exhausted I came here to solve sums

(:

whatt

For me life is like wake up, study, sleep, study, sleep, new day

its 2:15am ur tine

[alpine sable]

Yeah old frens nvm

[tardy tapir]

not only for you

[alpine sable]

Yep

for me its 3pm

.

jk jk jk j lj kj kj l jo j k jkjkhj

jk

I said jk

its

9:49pm

10pn?

Yeah

UK

sure 10pm

yea

[ocean seal]

The current time for oldieidlog is 02:19 AM (IST) on Fri, 25/08/2023.