help-19

[tulip thicket]

rules here

[mystic saffron]

do you know the definition of xor?

[worldly rune]

Ig you can write AoB as (A+B)bar(AB)

[tulip thicket]

net

[mystic saffron]

net?

[tulip thicket]

nah

[worldly rune]

Nvm you can't

[tulip thicket]

i idk what this stuff is

[mystic saffron]

do you have any book about it?

AoB = A'B + AB'

[tulip thicket]

okk

[mystic saffron]

You can verify using truth table

[mystic saffron]

yeah.. i was trying to get him to find that himself

[tulip thicket]

wait what are the commas

[worldly rune]

bar ig

[mystic saffron]

bar

[tulip thicket]

oh

[worldly rune]

Damn I wanted to figure this out myself

Thx alot icebal

[mystic saffron]

Wait this was your question?

[tulip thicket]

what rule is it?

[tulip thicket]

it’s mine

[mystic saffron]

Definition of Xor

[tulip thicket]

i write xor under justification?

[mystic saffron]

you probably should ask you professor about there not being any rule about xor

[tulip thicket]

i’m gr11

idk any of this ngl

[mystic saffron]

you can certainly see that there is no rule that uses xor

[tulip thicket]

like this?

[worldly rune]

Nvm you can write it like this

Yey

[mystic saffron]

You basically want 2-input xor to return 0 when there's even parity of 1's

you can intuitively see A'B and AB' are zero when A=B=0 or 1

Doing their OR will return 1 when 1's have odd parity

[tulip thicket]

he wants us to use justifications from the table

[mystic saffron]

That's weird xor is usually covered with its definition and truth table in the start

[tulip thicket]

wait lemme check

ngl i’m cooked

[odd edge]

@tulip thicket Has your question been resolved?

[tulip thicket]

@mystic saffron @mystic saffron is this right

[mystic saffron]

you should ask the prof about step 2

[tulip thicket]

he said it’s right

it’s just defn xor

he just wanted us to remember that

not on sheet

is the rest good

[mystic saffron]

No.

$\overline{A \oplus B} = \overline{A'B + AB'}$

$= (A'B)' \cdot (AB')' = ((A')' + B') \cdot (A' + (B')') = (A+B')\cdot(A'+B)$

$= AA' + AB + A'B' + BB' = AB + A'B'$

$\$
$\implies \overline{A \oplus B} + \overline{A + B} = AB + A'B' + \overline{A + B}$

$= AB + A'B' + A'B' = AB + A'B' = \overline{A \oplus B}$

[clever fjord]

icebal²

[worldly rune]

Counter example A=B=0

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[vapid aspen]

what in the world are R S T and point rule

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[vapid aspen]

<@&286206848099549185>

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[clever violet]

am i allowed to simplify h like this?

[elfin zodiac]

You can deduce that by yourself with another example

Can you simplify (6-2)/2 to 6?

[clever violet]

no

oh ok

so i can only simplify x's if its just multiplication and theres no addition or subtraction?

[elfin zodiac]

There is an explanation for that

If you have AB/B you can rewrite it as A(B/B)

And B/B is 1

[clever violet]

oh

[elfin zodiac]

So A * 1 = A

[jaunty tinsel]

You cant simply cancel out the x in the denominator because it is not a common factor in the numerator

[elfin zodiac]

When you multiply the numerator and the denominator by the same number

[clever violet]

ohhhhh

[elfin zodiac]

Is like multiplying by 1

But that doesnt happen with sum or subtraction

[clever violet]

oh ok

[elfin zodiac]

And it is importantant that such number multiplies and divides the whole thing top and bottom

[clever violet]

guys do yk what it means

how do i know which 3 sides are 12m?

this is what someone did but it said three im confused

[amber schooner]

what does the rest of the problem ask

[amber schooner]

so you don’t know what the other side is

that’s why they put x

on either side

of the 12

the picture should make sense

[clever violet]

I still dont understand why it says three sides of a trapezoid are 12

and they made 4 sides 12

[amber schooner]

that’s just a given

[amber schooner]

no they didn’t

[clever violet]

ohhhh

[amber schooner]

they made it 12+2x

[clever violet]

but the sides could've been different

idk

i get it

[amber schooner]

result would be the same

if you rotate it

doesn’t matter

[amber schooner]

yea

is this what you drew

[clever violet]

yes

[amber schooner]

and how did that work for you

[clever violet]

no i just drew it

[amber schooner]

oh

[clever violet]

Life is not working out

I have a question

[amber schooner]

welcome to the club

[amber schooner]

ask

[clever violet]

howd it go

from there to there

[amber schooner]

because the bases are the same you can combine them to just bh

and the height is the same as well

they’re the same triangle

so 1/2 bh +1/2 bh=bh

b=x

[clever violet]

ohhhhh

[amber schooner]

h is found from pythagoras

x^2+h^2=12^2

[clever violet]

howd they find domain

[amber schooner]

well how could the base of the triangle be greater than 12

[clever violet]

oh

would an algebraic way of doing it

setting h to 0

[amber schooner]

yes

if the long side or the bottom of the trapezoid is too long the sides won’t connect

or there will be zero height

so yea

[clever violet]

oh ok

[amber schooner]

a right triangle

cant have the hypotenuse being as long as one of the legs

then the other leg would be zero

and it won’t be a triangle

[clever violet]

oh ok

.ckidr

.clor

.close

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[clever violet]

.reopen

[odd edge]

✅

[clever violet]

would I do the product rule here or

[sly juniper]

i mean you can factor this

and then yes

but youd need to use chain rule for derivaitve of (144-x^2)^1/2

[slender rampart]

Yeah, you can factor our the (144-x²)½

[clever violet]

ooh

[sly juniper]

you could also write this as sqrt((x+12)^3(12-x))

but it probably still need a chain rule

so not sure if its easier way

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[vapid pawn]

In the polar equation d^2 = (r1)^2 + (r2)^2 - 2(r1)(r2)cos(theta - alpha), where r1 and alpha are the coordinates for the center of a circle, what is d^2?

[odd edge]

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[vapid pawn]

<@&286206848099549185>

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[proven dock]

.close

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[icy kindle]

is there a question or confusion?

[worn ledge]

ah sorry i was typing on my Amazon doodoo max

i was wondering how i can check my work?
would it be like proof by contradiction (i seen this in a separate book but the methods were a bit advanced for me)

[icy kindle]

you can either do some partial sums, or plug it into a series calc to check that it diverges.
but it looks like you did the divergence test and showed it fails so the series is divergent.

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[worn ledge]

Ty

.close

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[static mulch]

i have $\int_{0}^{4}\int_{0}^{\sqrt{2}}z\left(\sqrt{1+\frac{1}{4z^{2}}}\right)dzdx$

[clever fjord]

Jaxx

[static mulch]

correct one is $\int_{0}^{4}\int_{0}^{\sqrt{2}}z\left(\sqrt{1+4z^{2}}\right)dzdx$

[clever fjord]

Jaxx

[static mulch]

i used $dS = \sqrt{1+f_{x}^2+f_{y}^2}$

[clever fjord]

Jaxx

[static mulch]

with $f=z=\sqrt{y}$

[clever fjord]

Jaxx

[static mulch]

$f_{x}=0 and f_{y}=\frac{1}{2\sqrt{y}}$

[clever fjord]

Jaxx

[static mulch]

$dS=\sqrt{1+\frac{1}{4y}=1 +\frac{1}{4z^2}}$

[clever fjord]

Jaxx

[static mulch]

why is it 4z^2 and not 1/4z^2 is my concern

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@static mulch Has your question been resolved?

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@static mulch Has your question been resolved?

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@static mulch Has your question been resolved?

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[halcyon parrot]

can someone given an example of an invertible function

[merry finch]

the identity is an invertible function

f:ℝ -> ℝ

f(x) = x

[halcyon parrot]

ok mb

[halcyon parrot]

an invertible function which is neither symetric or is identity

[merry finch]

What does it mean to be symmetric

Oh

Ok

[merry finch]

Are functions of 1 variable always symmetric

[halcyon parrot]

if in a relation both a,b and b,a belong to the relation, the relation is symmetric

[merry finch]

And here f is the relation?

What about f(z) = z* where * is the complex conjugate

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[mystic saffron]

I need some help with a combinatorics problem:

For a student to pass an exam, they have to receive a 6, 7, 8, 9 or 10. How many ways can the student be graded for 12 passed exams, if for 3 subjects they can receive a grade bigger than 6, and for 5 subjects a grade smaller than 8?

[mystic saffron]

The exact numbers are not important to me more or so how I should start

I assume variations with repetitions is used

[slow light]

might be easier to subtract invalid combinations no?

[mystic saffron]

idk if i understand what an invalid combination means

i had a feeling subtracting was involved

[mystic saffron]

this is the best answer i could give

however i feel it may be wrong

where V with a line on top represents variation with repetition

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@mystic saffron Has your question been resolved?

[mystic saffron]

whatever

.close

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[loud echo]

help

[stiff wind]

you again dude

LOL

[loud echo]

im asking for help this time am i not

[zinc glacier]

show your work

[stiff wind]

yeah you are indeed

[loud echo]

look ik you don't care but i suck at math and i don't care for it so just give me some steps

dickheads

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[hard cedar]

not sure what to do with csc and cos since the example given to me was in cos and sin

[green elm]

can you just convert everything into sine and cosine and proceed from there?

[hard cedar]

ohh

ok hold on

[green elm]

what's up with that tiny little answer box

it's not like this is gonna reduce to a number

[hard cedar]

it expands the more u put into it

[green elm]

oh ok

[hard cedar]

i got that but

[tacit parcel]

isnt this just equal to 1?

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[prisma lily]

$\lim_{x \to 0^+} \frac{1}{4 - 2^{\frac{1}{x}}}$

[clever fjord]

pezet pdw

[prisma lily]

is this correct?

[mystic forge]

seems fine

wait

yeah its fine

[prisma lily]

alright thanks a lot have a nice day

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[arctic steeple]

heelp

[arctic steeple]

how do i calculate this

i know how to calculate a right triangle but idk abt this one

[slender rampart]

Take tangent of 40° to find the base in terms of x

Then do the same for the 20° one, you'll find the base in terms of x, now equate that to what u got from the 40° triangle + 400

[arctic steeple]

but how can i calculate the tangent of 40 if i dont have any numbers on that triangle

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@arctic steeple Has your question been resolved?

[tawdry cave]

Perhaps, you are just supposed to use a calculator for that.

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[mystic saffron]

what would i use to multiply this fraction

[cold swift]

i would say use lhopital’s

not sure if you can multiply by the conjugate to cancel things out

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[mystic saffron]

I cant

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[mystic saffron]

.close

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[pallid lynx]

how can i set this proportion up?

[mystic saffron]

trivial by jensens

[pallid lynx]

.close

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[mystic saffron]

a rectangular paddock is half as wide as it is long. it is completely enclosed by 3km of wire. its area in square km must be?

[mystic saffron]

a rectangular paddock is half as wide as it is long. it is completely enclosed by 3km of wire. its area in square km must be?

[meager hound]

Hi, how can I help you? Is your problem understand/draw the question?

@mystic saffron

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[silver spire]

I’ve been given two vectors and I need to write the position of those vectors when subtracted from one another

[silver spire]

You’re in the wrong channel

<@&286206848099549185>

[warped glacier]

Think of u - v as u + (-v)

So in other words you start at the origin, move in the direction of u

And then starting from when you left off (tip-to-tail)

Move in the opposite direction of v

So yeah, -v has the same length as v, but the opposite direction

Ah right sorry you've done 4 already

[silver spire]

Is 4 correct or nah?

[warped glacier]

No

[silver spire]

Damn

[warped glacier]

Should be (6 - 3, 2 - (-4))

So yeah (3, 6)

[warped glacier]

Also for question 5, the thing about vectors is that it doesn't change the vector if you start somewhere else

So you can add the vector (4, -2) to (-4, 2) to get (0, 0) as the starting point

So what must you do to the ending point to keep the vector the same?

[warped glacier]

Yeah, you're not being consistent here

[silver spire]

Damn

[warped glacier]

It's y value minus y value, x value minus x value

Ah shit sorry I got the vector wrong

So (3, -3) - (-4, 2)

= (3 - (-4), -3 - 2), sorry yes that's correct

So yeah there's the way where you subtract the two coordinates

[warped glacier]

This way gives you (3, -3) + (4, -2) = (7, -5)

You're doing the same thing, adding -v is the same as subtracting v

I just find it's easier to add than to subtract cause subtracting might make you do a sign error

[silver spire]

Gotcha gotcha

That makes sense

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[queen rune]

A small object is placed at the center of a large evacuated hollow spherical container. Assume that the
container is maintained at 0 K. At time t = 0, the temperature of the object is 200 K. The temperature of the
object becomes 100 K at t = t1 and 50 K at t = t2. Assume the object and the container to be ideal black
bodies. The heat capacity of the object does not depend on temperature. The ratio (t2/t1) is ____,

[ancient patrol]

is that the only question?

no additional information?

[queen rune]

Yep

[ancient patrol]

is t1 and t2 just any variables?

[queen rune]

Hol up lemme give you the original question

[ancient patrol]

idk which way but u can look at it 2 ways

okie

[ancient patrol]

thats weird

u can do it 2 ways ig

1 way is that the heat released is constant

2nd way is that the heat released is consistently decreasing depending on the temperature of the object

so like 1st is constant minus, 2nd is constant divide

idk what that question wants

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[sharp bay]

How can I find the set of limit points of the sequence $cos\frac{2npi}{3}$?

[sharp bay]

Oops

[clever fjord]

Kazutoraaaa

[sharp bay]

<@&286206848099549185>

[amber oyster]

$cos(\frac{2n\pi}{3})$

@sharp bay ?

[sharp bay]

Yes

[clever fjord]

Tittom_123

[sharp bay]

This is the correct version, yeah

But idk how to solve it

[amber oyster]

Find a limit for n?

[sharp bay]

It doesn't exist if it tends to infinite

But it's 1 if n = 0

[amber oyster]

What do you need to find?

[sharp bay]

The set of limit points

Values of n for which the limit exists?

[odd edge]

@sharp bay Has your question been resolved?

[odd edge]

@sharp bay Has your question been resolved?

[sharp bay]

<@&286206848099549185>

[amber oyster]

Maybe do some tests. Draw a graph of the function

[odd edge]

@sharp bay Has your question been resolved?

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[solid minnow]

Hi, I need help in LA 2 to prove that for all A, and for every inner product, <Av, u> = <v, A*u>

[solid minnow]

I have a direction but I got stuck. I thought about trying to prove this statement:
$$\giventhat{\forall A\exists B,orthogonal,base,,T:V\to V} | {\left[T\right]_{B}}=A\implies\forall A:,\left\langle Av,,u\right\rangle =\left\langle v,,A^{\ast}u\right\rangle $$

[clever fjord]

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

[solid minnow]

Btw, I am not even sure this statement is true

But I would like to know if it is / if so, for which inner products

[odd edge]

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[wanton sierra]

i just started power series representations and this is the most confusing thing ive done in math so far honestly, somehow my brain does not understand it like everything else.

how do i do this?

[forest sky]

think about if this series looks like any power series you've seen before

[wanton sierra]

i know the idea behind power series, you basically just want it to look like 1/1-r

i just have no idea how

assuming my algebra is right i can get it into this form:

[forest sky]

you have the right idea but i'm not sure what you did with the denominator

[wanton sierra]

first step i did was this:

or woops typo in there

should be 81/x^2

then i did this

yeah x^2f(x) and 1/(1- -81/x^4)) are the same line

[odd edge]

@wanton sierra Has your question been resolved?

[wanton sierra]

@forest sky still have no idea what im looking at with this problem

[wanton sierra]

.close

[odd edge]

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[desert cove]

Can someone explain how I can solve these kinds of problems?

[desert cove]

let me get a picture

tell me if translation is needed

[fresh ruin]

yea can you translate the question?

[odd edge]

@desert cove Has your question been resolved?

[desert cove]

solve the right angled triangle ABC (<C=90 degrees) off the given:

and then it's just the given stuff below

It's solved in the area below the question

i just need you to explain how to do it in general

[fresh ruin]

there's like 6 variables for this right triangle height setup and it gives you the equations to move between them

so when you're given two v ariables, you look for the equation with those two to get the third

and if that's not the variable you want you just use the third variable with something you have and keep going

[desert cove]

yeah i get that, but how about the actual solving part

i get the theory but im dogshit at solving

and i also don't know how to square root stuff

because we aren't allowed calculators

[fresh ruin]

probably just keep the numbers inside the square root, maybe simplifying with sqrt(a^2*b)=a*sqrt(b), like you can pull squares inside out as normal numbers

it's like a crafting system or whatever where you can combine two tools to make a new tool

[desert cove]

can u like solve the question

and break it down

?

[fresh ruin]

like this one? sure

[desert cove]

yeah

[fresh ruin]

is the question just find every single variable?

[desert cove]

yeah

just find the ones that arent given

[fresh ruin]

so I look at these equations and see the one with a and a1 is a^2=a1*c, so I can get c there

sqrt(10)^2=1*c so c=10

then I can look for equations with c and a1 and it's the easy c=a1+b1 so b1=10-1=9

[desert cove]

is there an easy way of finding the square root of things?

[fresh ruin]

if it's not the numbers 1,4,9,16,25,36,49,64,81,100,... you just leave them in

until you square it later

sqrt(10) is just like its own thing

[desert cove]

what are the formulas for this again?

c2=a2+b2

besides that

is there any other formulas

like with the height or maybe something else

?

[fresh ruin]

you have to really know all these ones

technically you can get them all by doing pythagorean theorem on all the three right triangles there yea

and whenever you have two sides of any right triangle, use the formula to get the third

[desert cove]

so all you need is to just reuse pythagoras

inbetween all the answers you get

?

[fresh ruin]

yea just keep checking if there are triangles to use it on

and eventually you get all the lengths

[desert cove]

wait so what are the formulas with orthogonal projections?

what do i do with them?

i forgot since ive been on spring break

and my math teacher just dumped this on me

[fresh ruin]

you're meant to memorize the 6 equations I think

[desert cove]

what are they?

[fresh ruin]

solving three a^2+b^2+c^2's simultaneously is like the bonus extra way of doing it

[fresh ruin]

these are everything you need for these problems

the 6 equations up top

[desert cove]

oh i see, got it

so i just need to memorise those

and i should get some clarity?

ill be sure to do that, can u break down some of the actual solving ill need to be doing with them?

like with the square roots for example

[fresh ruin]

this was the example, and now with a,a1,c, and b1, you can use the b^2 and h^2 equations to get the final variables

like for b^2=b1c you get b^2=90, so b=sqrt(90)

the way you deal with that square root is ask what 90 is made out of, in particular if there's any squares in its factorization

like is it divisible by 4,9,16,25,36,49,64,81,100,..

there's a 9 in there so you actually have b=sqrt(3^2 * 10), and you can pull squares outside like b=3sqrt(10)

[desert cove]

What about the process to the ones where

U find triangles inside

[fresh ruin]

this one is the other page just titled and ignore D

the others are just a^2+b^2=c^2 I think

[desert cove]

How would you solve that one?

[fresh ruin]

in the picture looks like they solve it with similar triangles instead of the 6 equations before, but either way works

[desert cove]

Can you show me how would be the easiest way?

[fresh ruin]

quickest way is to get AC it's pythag 6^2+8^2=10^2

for BQ you look at the area of the triangle of 1/2*base*height in two different ways

it's 1/2*6*8 because those are a base and height

and if you tilt it BQ is a height and AC=10 is the base, so 6*8=10*BQ

like anything involving height can be gotten through looking at triangle areas

[odd edge]

@desert cove Has your question been resolved?

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[alpine blade]

What’d I do wrong here? I know I forgot to square the other side in my work but my answer would have been wrong anyway, I believe.

[odd edge]

@alpine blade Has your question been resolved?

[alpine blade]

<@&286206848099549185>

[cyan moon]

Huh?

[alpine blade]

?

[odd edge]

@alpine blade Has your question been resolved?

[uneven horizon]

you did not square both sides of the equation correctly

wait nvm just saw what you said :(

here is my work I got a function in the same family as the answer i think

[alpine blade]

I got it now

It was w/ diff numbers but the same problem

[odd edge]

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[plucky sail]

Hi

[plucky sail]

On a hurry, to get this submission in

What did my friend do

To get this arc length,

ugh

[dawn quest]

they squared and then multiplied out

?

they just wrote y' squared as y' * y'

then multiplied it out

(substituting y' ofc)

[odd edge]

@plucky sail Has your question been resolved?

[sharp creek]

arc length is

integral root 1+(dy/dx)^2

right

[plucky sail]

.close

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[gleaming dew]

Calculate the sum of the digits of the integer 1234...199819992000

[fluid tundra]

first ask yourself how many times each digit appears

[gleaming dew]

I used that method, I just want to ask for the results for comparison

[fluid tundra]

well that would have been nice to specify...

[fluid tundra]

what number do you get?

[gleaming dew]

2001000

i think it's wrong

[fluid tundra]

that seems extremely big

much larger than a possible answer so yes i'd also say that

[fluid tundra]

as a hint, you may want to consider the digit 0... even when it doesn't appear

[gleaming dew]

no i removed it

haha

it's just a fun problem

[fluid tundra]

wdym "it's just a fun problem"

i'm confused what you're trying to say here

do you want to solve it or not?

[fluid tundra]

@gleaming dew does this make any sense

[gleaming dew]

I want to solve it, but if you're having trouble there's no need to try everything to solve it

[fluid tundra]

i did it in a minute

hence why i am asking you for your answer / thought process / etc

[gleaming dew]

oh

[fluid tundra]

so does it

that is indeed the answer though i have no idea why you felt the need to say it

i interpret that as "4. I got an answer and would like my work checked."

well in any case it's just vague wording

[gleaming dew]

yeah

[fluid tundra]

either way he hasn't been responding to the questions i've been asking him so

whatever

[gleaming dew]

I apologize for not making it clear

[fluid tundra]

he means "yes"

he just wanted the answer but for some reason never said that

[gleaming dew]

I'm not fluent in English so I have to translate what you want to tell me

[fluid tundra]

is this the right channel

[gleaming dew]

so when you suggested it, I didn't fully understand it

[fluid tundra]

"pythagorean theorem"

i'm not really sure how that is supposed to work

[fluid tundra]

i c

right triangle using what lengths

?

[gleaming dew]

Can you say more about this

I didn't think about the geometric method

[fluid tundra]

i am going to assume you are trolling on grounds of how you are intentionally not answering the questions i'm asking

[fluid tundra]

the method is nonsense

simple algebra is 100% the way you do this problem

[odd edge]

@gleaming dew Has your question been resolved?

[odd edge]

@gleaming dew Has your question been resolved?

[odd edge]

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[echo mango]

i need help with understanding how to do geometric series test

[cold swift]

given a series in the form $\sum_{n=0}^{\infty}a(r^n)$

[clever fjord]

y0shi

[long tinsel]

whats a geoemtric series "test"

[cold swift]

if $\abs{r}<1$, the series converges

[clever fjord]

y0shi

[cold swift]

i assume that they meant what a geometric series is

[echo mango]

its a series test

[long tinsel]

oh like that

[echo mango]

OHHHH is it like the 1/n^p

wait no

so like how would i solve this

where would i start

[long tinsel]

that's not a geometric series though

[echo mango]

i have no idea what test this is

thats the problem

[cold swift]

alternating series

[long tinsel]

just alternating series

[echo mango]

idk where to start

how do you know?

[long tinsel]

do you know the conditions for an alternating series to converge?

[long tinsel]

because each term changes sign

[cold swift]

(-1)^n shows that the series alternates

[echo mango]

OHHHHHH ok

[cold swift]

actually wait we shouldn’t do the alternating series test

it fails

[echo mango]

so whenever there is a (-1) its an alternating series

[long tinsel]

that's a little misleading because stuff like $\sum_{k = 0}^\infty (-1)^k\sin(k)$ would not be alternating

[clever fjord]

uoǝu

[cold swift]

that’s fair

[long tinsel]

you need to ensure that each term has alternating +/-

[cold swift]

yeah so the best way is prob listing a few terms out

[long tinsel]

basically whatever is multiplied with -1^n needs to always be postive or negative

[echo mango]

ohhh

[long tinsel]

it does?

[cold swift]

yeah

it fails the limit condition

[long tinsel]

yeah right

[echo mango]

so you know it fails because the terms arent all alternating?

[long tinsel]

no they are

[cold swift]

well you need to check the conditions for the alternating series test

[long tinsel]

$\sum_{k = 0}^{\infty} (-1)^ka_k$ where $a_k > 0$ converges iff $\lim_{k \to \infty} a_k = 0$ and $|a_{k + 1}| < |a_k|$

[echo mango]

so 2^1/n isnt 0 so it fails

[clever fjord]

uoǝu

[echo mango]

thats why?

[cold swift]

yep

[echo mango]

okay okay i get it

[cold swift]

so the alternating series test would be inconclusive here

we need to use a different test

[echo mango]

what is like ur thought process on which test to do next?

[cold swift]

well for most series i would usually start with the divergence test

since it’s relative quick

[echo mango]

okayyy

so i would do the limit n->infinity of (-1)^n 2^1/n

[cold swift]

yep

if the answer doesn’t equal 0 then it diverges

[echo mango]

um

i kinda forgot how to do the limit part 😭

[cold swift]

so if we just look at the $2^{\frac{1}{n}}$ bit

[clever fjord]

y0shi

[cold swift]

when n approaches infinity

what does this approach

[echo mango]

0?

and 2 to the 0 is 1

[cold swift]

yep

and (-1)^n when n approaches infinity?

[echo mango]

well

i guess so since infinity is a really large number

so ya?

[cold swift]

well i meant to ask what it approaches when n goes to infinity

[echo mango]

infinity i think

[cold swift]

well not exactly

(-1)^n is -1 if n is an odd number

and 1 is n is an even number

regardless of how big n is

so when it goes to infinity

the limit doesn’t exist

[echo mango]

ya like it keeps alternating right?

[cold swift]

yep

so $\lim_{n\to \infty} (-1)^n 2^{\frac{1}{n}}$ does not exist

[clever fjord]

y0shi

[cold swift]

which is not equal to 0

so it diverges by the divergence test

[echo mango]

but y?

[cold swift]

because (-1)^n alternates back and forth when n approaches infinity

[long tinsel]

because you don't know what it is

[echo mango]

ok

[cold swift]

however if it was let’s say (-1)^n 2^{-n}

that limit exists when n goes to infinity

since the 2^-n bit approaches 0

so the alternating part doesn’t really matter

[echo mango]

mhm cuz it gets multiplies by zero which would equal 0

[cold swift]

yep

[echo mango]

i get that part

so how would i do something with trig?

like that

so i did the divergence test and the tan of (1/infinity) is the same as tan (0)and thats 0

so where would i move to next?

[cold swift]

hmm

i can’t really think of a test that would work either

unless we can somehow integrate it

[echo mango]

so integral test?

but idk how to do that one...

[cold swift]

yeah the thing is i’m not sure if we can integrate tan(1/x) either

,w integral of tan(1/x)

[clever fjord]

Results provided by WolframAlpha

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

Requested by doctormysterio9

[cold swift]

yeah so no

[echo mango]

hm

[cold swift]

are you allowed to use a calculator for this or no

[echo mango]

yep

[cold swift]

cuz if you can, we can use the integral test actually

do you remember the conditions for the integral test?

[echo mango]

if it equals like a number then ist converges or if its infinity or dne it diverges?

[cold swift]

well yeah

but there are conditions that the series needs to meet before we can apply the test

[tidal idol]

can we not do comparison test?

[echo mango]

continuous positive and decreasing?

[cold swift]

hmm what can we compare to tho

[tidal idol]

1/n

note that 1/n <= 1 for all n

[cold swift]

can’t think of anything off the top of my head

true

ooh yeah that does work too

[tidal idol]

@echo mango is that enough of a hint?

[echo mango]

um

im kinda lost

is that like the geometric test?

[cold swift]

i would say use what @tidal idol recommended here

essentially we are using the direct comparison test

[echo mango]

so thats the direct comparison test

?

[tidal idol]

yes you need to use comparison test

[echo mango]

Like

Idk what the a sub n and b sub n mean

[tidal idol]

it's a sequence

a_n is a sequence of numbers a_0, a_1, ...
same with b_n

[cold swift]

a_n is just the part without the summation sign really

and we just compare it to another series

[echo mango]

Oh like a potential original series

Ohhhh like lgkoo was saying to compare it to 1/n?

[cold swift]

yep

so does the series of 1/n converge or diverge

[echo mango]

Im a little stuck on the work and how to work it out

So 1/n would be 1/infinity

?

Which would be approaching 0 and by divergence test it would be convergent?

So it converges?

[cold swift]

have you heard of the harmonic series before?

or the p-series test

[echo mango]

I heard of p series

Not harmonic

[cold swift]

alright so what are the conditions for the p series test

since 1/n is a p-series

[cold swift]

the divergence test is inconclusive if the limit equals 0

[echo mango]

Oh p has to be greater than 1 to converge and less than or equal to 1 to diverge?

How is it a p series tho if there’s no exponent

Or is the exponent just 1

[cold swift]

yep

[echo mango]

Ohhh so it can’t determine that it’s convergent only that it’s divergent

[cold swift]

the exponent is just 1

[cold swift]

yeah

or else it would be too easy

cuz then you can just use the divergence test for everything

[echo mango]

Right lol

[cold swift]

alright so the series of 1/n would diverge right

[echo mango]

So it’s divergent since p is 1 and it’s equal to 1

[cold swift]

yep

so now the hardest part

[echo mango]

So I would write it’s divergent based on the direct comparison test

[cold swift]

is tan(1/n) bigger or smaller than 1/n

[cold swift]

we have to meet the other condition first

[echo mango]

Oh ya

[cold swift]

if a_n is less than b_n and a_n diverges, b_n also diverges

[echo mango]

Tan would be

Um

[cold swift]

yep tan is smaller

[echo mango]

Wait would tan in this case be the b_n?

[cold swift]

yeah

i kinda wrote it the other way so

[echo mango]

So if the second series is less than the original one(the one we are comparing to) then it is divergent as well?

That’s kinda confusing

[cold swift]

if the series we are comparing to is less than the original and that series is divergent, the original is also divergent

it makes sense conceptually if you think about it

if a series is smaller than another series and it diverges, the bigger series must diverge since it’s bigger

[echo mango]

Hm

Makes a little bit of sense

I want to find like a problem that’s doing root test cuz I haven’t even practiced that one

And also the harmonic that you mentioned before

What is that?

[cold swift]

the harmonic series is basically what we compared our series to

$\sum \frac{1}{n}$

[clever fjord]

y0shi

[cold swift]

it’s a famous series and there are nice proofs online proving why it diverges

so basically if you see a series in this form, you can just say that it’s harmonic and it diverges

[echo mango]

Ok cool trick

[cold swift]

i would say root test isn’t really that common

but generally when you have something to the n power

you can apply it

and it follows the same conditions as the ratio test

if the limit to infinity is less than 1 then it converges and so on

[echo mango]

And what would I typically do in a root test

is this a root test?

cuz its to the nth power

[cold swift]

basically we have to do this:
$$
\lim_{n \to \infty}\left(a_n\right)^{\frac{1}{n}}
$$

[clever fjord]

y0shi

[cold swift]

yeah that works

[echo mango]

So I would do

(2^1/n-1)^1/n?

[cold swift]

well $a_n=\left(2^\frac{1}{n} -1 \right)^n$

[clever fjord]

y0shi

[echo mango]

right

[cold swift]

so our limit would be $\lim_{n\to \infty} \left( \left(2^{\frac{1}{n}}-1\right)^n\right)^\frac{1}{n}$

[clever fjord]

y0shi

[cold swift]

and you can see here that the n and the 1/n would cancel out

[echo mango]

Right

So we would be left with 2^1/n -1

[cold swift]

yep

and we just take that limit to infinity

[echo mango]

So it would be 0

[cold swift]

yep

[echo mango]

Which is less than 1

So it would be

Convergent

[cold swift]

exactly

[echo mango]

By the root test

[cold swift]

yep

[echo mango]

COOL

I’m starting to like this

[cold swift]

yeah series are kinda fun

[echo mango]

Ya now that I get it, it seems cool

This tutor I go to at my college hates series

[cold swift]

the hardest part is prob remembering all the conditions and the tests

[echo mango]

And every time I ask him he gets mad😂

[cold swift]

haha yeah its a lot of memorization

and also takes a good amount of thinking to choose which test to apply

[echo mango]

Ya i think i get how to do the tests now but i still can’t memorize the conditions

Is there a condition for telescoping?

[cold swift]

not exactly

the best way to do them is really just try to find the partial sum formula

and take the limit of that to infinity

well i mean for a series to be telescoping

it has to cancel out some terms that are previously there

so thats kinda a condition

[echo mango]

So like we would be left with the first term?

[cold swift]

honestly it depends on the series

if we have $\sum_{n=1}^{\infty} \frac{1}{n}-\frac{1}{n+1}$

[clever fjord]

y0shi

[cold swift]

we can list out a few terms and see that it looks like the terms are all canceling out

$(1-\frac12)+(\frac12-\frac13)+(\frac13-\frac14)$ and so on

[clever fjord]

y0shi

[echo mango]

I get that

So this one would be

1

[cold swift]

yeah

[echo mango]

So it would converge

[cold swift]

since the partial sum formula would be $S_n=1-\frac1{n+1}$

yeah

[clever fjord]

y0shi

[echo mango]

So what would I answer there

Like

I would know it’s 1/n+1 because it was in the starting series?

How come you changed the 1/n to 1

[cold swift]

so since the nth partial sum is that formula, we essentially just do this:

$\lim_{n\to \infty} S_n$

[clever fjord]

y0shi

[cold swift]

the partial sum formula there is created by just observing the series

like the sum of the first two terms would just be 1-1/3

and the first three terms would just be 1-1/4

and you can kinda see why the nth partial sum would be that

[echo mango]

OHHHH

cuz it would cancel out the -1/2+1/2 on the third series

[cold swift]

yep

[echo mango]

Which would be 1-1/3

[cold swift]

exactly

[echo mango]

Okay I see where you got that from

so is there like a ocndition for geometric series test

[clever fjord]

y0shi

y0shi

[cold swift]

you just have to identify r

and just see if its absolute value is less than 1

or not

[echo mango]

so it could be like

what is r? in this case

like

(-2)(-5^n-1)?

and -5 would be r?

and |-5| is 5 so it would be

5 >1 so it would diverge?

[clever fjord]

y0shi

[echo mango]

yes

[cold swift]

yeah

it would diverge

[echo mango]

okay i think i get it

would something like this be a geometric test?

[cold swift]

yeah

[clever fjord]

y0shi

[echo mango]

so i would do like

-3/2^2n+1?

or is that completely off

[clever fjord]

y0shi

y0shi

[echo mango]

how come

like where did the n go for the first (-3)

[cold swift]

so (-3)^(n+1) is just (-3)^n * (-3)^1

by the exponent rules

[echo mango]

right

but i dont get how u made that into the second picture

like what u converted

[clever fjord]

y0shi

[cold swift]

another exponent rule

[echo mango]

oh my lord im forgetting everything

okay so what would i do next?

[cold swift]

well so what would our r be?

[echo mango]

-3/8

so it would be convergent

since (3/8) is less than 1

what is the condition for a p series test?

[cold swift]

yep

[clever fjord]

y0shi

y0shi

[echo mango]

and p would be the exponent of 1/n^p

[cold swift]

yep

it’s kinda similar to the geometric series conditions

but flipped

[echo mango]

thats nice

easy to remember

and no condition for nth term test right

also, for the integral test and the both comparison tests is there condtions

[cold swift]

divergence test?

[cold swift]

yeah

[echo mango]

for the

[cold swift]

integral is positive decreasing and continuous

[echo mango]

integral test, it just has to be continuous and pos and decreasing?

for the direct comp test it has to be

sorry

for the integral test

how would i know if its pos decr and cont.?

like without a graph?

like if i have

but we replace the integral with the summation sign

and like

how did they know to use integral test here and how did they know its decreasing

[cold swift]

you can either graph the function

or look at its first derivative

you don’t really need to show much work to prove it either

the integral test can usually be avoided by comparison tests

for this specific series, we could’ve used the limit comparison test

but for some, you can only use the integral test

[echo mango]

noted

is there condition for the limit comparison test?

i have this but i dont really get it..

[cold swift]

yeah so basically that limit has to evaluate to a positive and finite number

for it to work

and a_n and b_n can be any positive sequence

[echo mango]

so like what we would do for that problem

[echo mango]

this

so we could do lim n-> infinity (all of that)

[cold swift]

yeah

a_n/b_n or b_n/a_n

it doesn’t really matter

as long as the limit approaches a finite positive number

[echo mango]

like i dont really undertsand what that means😭

[cold swift]

the result holds true

basically let’s say a_n is 1/n and b_n is 1/(2n+1)

[clever fjord]

y0shi

y0shi

[cold swift]

and we just have to evaluate this limit

[cold swift]

as long as it approaches a finite positive number, the results stated above would be true

[echo mango]

so in this case

it would be like lim n->infinity (3x^2/2x^3 +2)

would be

[cold swift]

well that was our original series

we need to compare it to something first

[echo mango]

right

so we would compare it to

oh like 1/n+1?

[cold swift]

well i would usually just take the highest degree terms

[clever fjord]

y0shi

[echo mango]

so we would divide by x^3

both top and bottom?

[cold swift]

now we need to find whether or not this series converges or diverges

[cold swift]

hmm? why

this would be our b_n

[echo mango]

wait no im thinking of something else

[cold swift]

and our a_n would just be the original

all good

[echo mango]

so infinity over infinity would be

1

[cold swift]

well infinity over infinity is an indeterminate form

so first of all

[echo mango]

oh

[cold swift]

does this series converge

[echo mango]

um

no? cause its not a finite number?

im not sure

[cold swift]

well we aren’t doing the limit comparison test yet

we need to see if our series that we are comparing our original to is convergent or divergent

so we essentially just choose a test to test it out

[echo mango]

oh ok

[clever fjord]

y0shi

[echo mango]

ya

[cold swift]

is this convergent or divergent

see how that resembles a harmonic series

we don’t even need the p series test here

[clever fjord]

y0shi

[echo mango]

so it diverges