help-42

[old epoch]

like intuitevly yes since "gamma" is connected

but how can i show this formally?

[glad plover]

?

it's by definition of the function i think

xsin(1/x) always has a negative trough between any two positive peaks

[old epoch]

but we dont knwo what kind of function gamma is

[old epoch]

i mean gamma

like its saying that gamma is connected thus in gamma([a, a + delta]) only 1 peak at most

[glad plover]

well the point is that gamma is equal to xsin(1/x) essentially?

hmm

[old epoch]

ah wait how so tho? ig i dont really see that

[glad plover]

oh

i'm stupid

it's not xsin(1/x)

it's just sin(1/x)

i misread the first line lol

but anyway yeah so

[old epoch]

oh ok

[old epoch]

how is gamma still sin(1/x) tho?

[glad plover]

basically gamma covers all of A and B

but A is just the image of sin(1/x)

so gamma takes all of the values of sin(1/x)

[old epoch]

can it also take values frmo B tho

[glad plover]

yes

but either way if you look at the diagram you can see that you can't possibly go from one peak to the next without at least hitting the x-axis

i mean B is just a vertical line right

[old epoch]

yeah the diagram makes sense but i dont really see the connection with gam,a

[old epoch]

yes

[glad plover]

so if you go down B then you also hit the x-axis

[old epoch]

yeah but thats the set A, not gamma i think right

[glad plover]

look

the point is that gamma only takes values in A or B

so if it wants to get from one peak to the next

then it must go along the lines of A or B

and those lines force it to cross the x-axis

like, gamma is from [0, 1] to A U B by definition

the codomain is A U B

[old epoch]

yeah, but i mean cant it go like map two points in the interval to the peaks, and then everythign else to B?

[glad plover]

no because that wouldn't be path connected?

like what two points are you mapping here

[old epoch]

yeah so ig, am just trying to shwo that if it is connected and it maps two poitns in itneral to peak, then it must have zero crossing

[old epoch]

like any two points in [a, a + delta]

[glad plover]

ok

call them x and y

[old epoch]

yes

[glad plover]

then consider the midpoint

where is the midpoint

it is way the fuck over at B

way too far away

[old epoch]

yes

[glad plover]

so it couldn't be path connected

or no

that shows that it's not uniformly continuous, i mean

[old epoch]

but it could still be connected

oh ok hmm

like it makes kinda sense visually

[glad plover]

ok let's go like this

[old epoch]

but i just cant put this into mathematical words

[glad plover]

let's ignore B for now and just focus on A

A is just the image of sin(1/x)

[old epoch]

yes, then i agree it must have a zero crossing

[glad plover]

and it is simply a fact that sin(1/x) crosses the x-axis between any two peaks

so

[old epoch]

since its image is A yes

[glad plover]

now you just need to show that B can't help

which is also pretty easy

since if it tries to go to B, that violates uniform continuity

it's going really far away in whatever tiny interval

[old epoch]

hmm

[glad plover]

probably

i mean it's like

also

how can B even help

[old epoch]

idk maybe some weird contribution

[glad plover]

it is literally just one vertical line that crosses the x-axis anyway

[old epoch]

im thinking maybe suppose there is no zero crossing

then show a contradiction...

[calm coral]

@old epoch Has your question been resolved?

[calm coral]

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[tribal iron]

need help with deciphering this basic proof T_T im so sorry in advance

[tribal iron]

so basically, i think my instructor mentioned that the idea came from a + (-a) = 0 or the existence of the additive inverse and i get (i think) how it follows from that idea but it always bugs me since we're trying to prove -1*a = -a can we use it to be substituted???

because that's what happened i think in the left hand side from a+ (-a) = 0 but basically we have

a + (-1a)
then weuse multiplicative identity to expand it to
1a + (-1*a)

but when do i know it's ok to substitute what im trying to prove? i think im confused on something i dont really know, all i just know is im confused im sorry T_T

[elder pawn]

additive inverse is unique, thats what the lemma says

[tribal iron]

yes, i get that part but i sort of have this mindset that if you're trying to prove it you can't use it to prove something so how can we use -1⋅ a =-a in the lines we need to prove but im think im missing some nuance here help

[elder pawn]

we're not using -1a=-a

maybe it helps to reword the lemma

[tribal iron]

i mean in the first line

-a + (-a )
-a + (-1⋅a)

[elder pawn]

a+(-a)=0 is the additive inverse property

what the lemma says is that if b also satisfies the additive invese property, ie a+b=0, then b equals the additive inverse, ie b=-a

so all we need to do is show -1a satisfies the additive inverse property, ie a+(-1a)=0, then use the lemma to conclude -1a=-a

yes

[tribal iron]

oh! thanks, sorry i deleted my comment to reply to u lol

thank u so much! i might come back for more T_T

i love u

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[elder pawn]

ly2

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[cedar ether]

Theorem. Let $A,B$ and $C$ be matrices such that $A(BC)$ is defined. Then $(AB)C$ is also defined and $A(BC)=(AB)C$; that is, matrix multiplication is associative.\

How do I show $(AB)C$ is also defined given the hypothesis of the theorem? I'm afraid I don't really know what I'm supposed to show.

[potent lotus]

psie

[glass heart]

if A(BC) and BC are defined, what are the possible sizes for A,B,C?

[calm coral]

@cedar ether Has your question been resolved?

[cedar ether]

alright, so I guess saying A(BC) is defined implicitly assumes that BC is defined If yes, A needs to have the same columns, say n, as the number of rows in BC, and A can have any number of rows, say m. Then B needs to have the same columns, say p, as the number of rows in C, and C can have any number of columns, say q. So A is m x n, B is n x p and C is p x q. So given these sizes, we see that AB is defined as m x p and (AB)C is also defined as m x q.

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[river pike]

good morning

[river pike]

This is a bit wierd

[void umbra]

Think you'd find the critical points of the function first

[river pike]

dont we find the derivative first then the critical points

[void umbra]

Yea

In this case f'(x)=6-6x^-2

[river pike]

-2?

[void umbra]

So the critical pts are?

[void umbra]

What about it

It's power rule

[river pike]

I think you are going too fast for me

I dont follow

[void umbra]

Ok mb

So uh do you know how to find the derivative

Of this function

[river pike]

it follow the 3rd one

so prime for the f and prime for the g

[void umbra]

Yea we'd use the first rule and third rule

[river pike]

oh i mixed up the cx

[void umbra]

I mean a lot of these rules are just things that follow from other rules

Like the x^-1 derivative

But anyway using the rules

What's the derivative

[river pike]

so my final like derivative would be 6-6x^-2

[void umbra]

Yeps

[river pike]

and then i compare it to zero

[void umbra]

Equate to 0

Yea

Then solve for the x values

[river pike]

ok il try to do that brb

how do i remove the ^-2

[void umbra]

Multiply the whole equation by x^2 yielding a quadratic

[river pike]

can you show me visually

[void umbra]

6-6x^-2=0
6x^2-6=0

Ok now you cannot sell from here

[river pike]

oh ok you just switched them sides

both of them

[void umbra]

Huh

6-6x^-2=0
x^2(6-6x^-2)=0(x^2)
6x^2-6=0

Maybe more specific like this

[river pike]

theres a feature on this server that shows like images

usually

[void umbra]

Ok fine man

$$6-6x^{-2}=0$$
$$x^2(6-6x^{-2})=0(x^2)$$
$$6x^2-6=0$$

[river pike]

thank you i appreciate that

[potent lotus]

denzio321

[river pike]

ok i understand now

[void umbra]

Aight bet bet

So what are the solutions

[river pike]

so one of the critical pts is 1

[void umbra]

And the other

[river pike]

-1

but arent there supposed to be 3

[viscid rover]

so it says those points are either critical or not in the domain

you found -1 and 1 which are the critical points

now think about where that function is not defined

[river pike]

by undefined it means

[viscid rover]

meaning x cant have a certain value

[river pike]

0

[viscid rover]

exactly

and i suppose thats the C, or third point you are looking for

[river pike]

ok il try it

[viscid rover]

sure

[river pike]

ok il try to do the next part

@eternal shard i put it in the derived right

succes

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[rare inlet]

i have a doubt here

When trying to simplify this multiplication, we simplify root 3 with root 3 and we remain with 1. BUT do we also do 3-1? in the second term

what i am trying to say is after they reduce we remain with 1 right? and then what do we do with this one do we operate with it or pretent it does not exist i am confused

[desert nymph]

you cannot simplify across addition/multiplication

unless I'm missing something here

[rare inlet]

wait what?

[desert nymph]

er, cancel across

[rare inlet]

this is what i am trying to do

[desert nymph]

yeah you cannot cancel across subtraction. otherwise e.g. 2/4 = 2/(6-2) = 1/6

[rare inlet]

oh. so addition and substraction are considered like one term here so i can't simplify across with them

?

[desert nymph]

yep

[rare inlet]

I see

[desert nymph]

you can write 3 as (root3)^2 and then combine the fractions via normal multiplication
and I think it should simplify nicely

[rare inlet]

how can you write root 3^2 tho don't we multiply root 3-2 with 6 and 3 with 3-root 3?

[desert nymph]

probably don't even need to do that (roo3)^2 thing

yeah so your numerator will be $\sqrt 3 \times 2 \times 6$ and your denominator will be $ 3(3 - \sqrt 3)$

[potent lotus]

PrettyPrincessKitty FS

[rare inlet]

did I do something wrong because i think i am stuck

actually you don't have to answer i found it😭

here i can simplify root 3 -1 since they are the same term right?

[ripe python]

Yes they are cancelable

[calm coral]

@rare inlet Has your question been resolved?

[calm coral]

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[rare inlet]

I am doing something wrong here and idk where and what can someone please help me figure this problem out

[real ginkgo]

ur doing simplification right?

[rare inlet]

yea

[real ginkgo]

i think you simplified something wrong

[rare inlet]

I think so too but i did this 3 times and i still get 0.

i am not supposed to get 0

[ripe python]

You missed the rooted term

[real ginkgo]

it should be sqrt(6)

[rare inlet]

I don't get why tho lemme check

I rationalized tho

[ripe python]

Ohh my bad

Lemme check

[rare inlet]

nw

i am always doing small mistakes so somewhere most likely there is a issue

[ripe python]

Do you know the final answer?

[rare inlet]

-4root3

and that is 100% the right answer as it's from an exam.

tho 3 times i did this i always got 0 lol

[ripe python]

Ohh

I understood

Now

[ripe python]

Focus on this step

If you simply just this

It will be

[rare inlet]

ok lemme do it agian

again*

[ripe python]

(3√2 - 2√3) - (3√2 + 2√3)

or

3√2 - 2√3 - 3√2 - 2√3

= -4√3

[rare inlet]

wait what

i don't get this at all help

[ripe python]

Alright

[rare inlet]

I re-did that part and got the same

[ripe python]

Look at the right portion of the underlined substance

In this part when you multiply them

It becomes

6√18/6 - 6√12/6 - 3√6(√3/3 +√2/3)

So when you multiply the second part

And simplify all of this

It will become

√18 - √12 - √18 -√12

Did this part make sense to you?

[rare inlet]

?

[ripe python]

Now simplify the right side

see the 3s can be cancelled out

[rare inlet]

I think i see where the confusion was for me.

[ripe python]

Is it fine now?

[rare inlet]

a dobut that i have.

lemme show ya

[ripe python]

Yes sure

[rare inlet]

3√2 - 2√3 - 3√2 + 2√3 this is where i got it wrong first time

[ripe python]

Yes

[rare inlet]

it should have been -

[ripe python]

Yupp

[rare inlet]

but why.

[ripe python]

True

[rare inlet]

Cuz lize

like

that minus before the fraction does it not get multiplied with all terms

[ripe python]

It does

You're absolutely correct

You're stating that

[rare inlet]

so does this not become a +

[ripe python]

-a( x - y) = -ax + ay

But here you have to notice your previous step

https://media.discordapp.net/attachments/948770026914213909/1349394768580444240/1349390428272853093remix-1741791179424.png?ex=67d2f14f&is=67d19fcf&hm=05f5703c2358c92bdf3f4d45376f1e45795e24483af30b28df06e2c0b232584f&

Look at the right side

a negative is multiplying with two positives

[rare inlet]

this would give us both negative.

I see.

[ripe python]

and that's all

[rare inlet]

But that means I wrote it a little wrong. Sytanx speaking

[ripe python]

Yes

Syntax is important at times

[rare inlet]

imma rewrite this

[ripe python]

They are the primary error causers in these types of calculation

That'll be really good

[rare inlet]

that should do it

[ripe python]

Yupp

That looks perfect

Good job

[rare inlet]

i noticed. Literally just because of a small syntax error i wasted 30 mins haha

[ripe python]

It's hard to identify for sure

[rare inlet]

tysm

.close

[calm coral]

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[warm pewter]

Normally, what is the range of sin graph for questions like this?

[eternal shard]

criminal how they wrote that

sin(x) is usually bounded between -1 and 1

[warm pewter]

What about like the range of x?

[eternal shard]

What do you think?

[warm pewter]

Idk bruh it goes on forever

[eternal shard]

Are you allowed to input any number?

[warm pewter]

Yes except 0

Cos t>=0

[eternal shard]

t >= 0 includes t = 0

it's greater or equal

[warm pewter]

Ah yesss I forgot

Then

[eternal shard]

Also I am confused

you asked about x

and proceeded with t

[warm pewter]

Hahahah I guess I was just asking in general

Ok I will swap

[eternal shard]

in general, the input can be any real number

here it is restricted to non-negative numbers

[warm pewter]

But there can be different times bc may can make 0

Ohhh

But they would give the same outcome right

[eternal shard]

outcome of what

[warm pewter]

Laike if I put in sin(…)

[eternal shard]

laike skull

[warm pewter]

😭

😅😅

[eternal shard]

can you do a concrete example

or just forget what you were sayig

sayig skull

[warm pewter]

Ok

Yes

You know how to make sin(..)=0 there can be many values of t

[eternal shard]

yes

this is a consequence of sine being periodic for example

[warm pewter]

How do you know which value of t to use

[eternal shard]

that satisfies the equation sin(...) = 0?

[warm pewter]

For example, the maximum speed means a= sin(3pit) = 0 right

Yes

[eternal shard]

oh

now i see what you are asking

[warm pewter]

I’m bad at explaining my bad ahh

[eternal shard]

you could have just asked straight away, how do i find the maximum

well it does not matter

for every solution of t that makes sin(...) = 0 you get the same maximum speed

well almost

sin(...) = 0 could also give you solutions to the minimum speed

but yea you are right to point out, that there are of course infinite many solutions

but each will return the same maximum value, after all it's periodic

[warm pewter]

Ohhhh I seeeee

[eternal shard]

For example cos(0) = 1 so does cos(2π) = 1

[warm pewter]

So

That would apply to all cos, tan and sin

[eternal shard]

tan less

unless you restrict its domain, it doesnt have a min or max

also you for your question, you wouldn't really need calculus

i supposed you figured out the velocity function?

[warm pewter]

Ohhhhh

But for 0 and 2pi they would give the same right for tan

Yep here is my velocity equation

[eternal shard]

hmm

does v(0) = 1/3π

ok it does, the brackets confused me

[warm pewter]

My bad

Hahaahahahha

[eternal shard]

ok so you to find the max

[warm pewter]

Yupppp

[eternal shard]

2/3π aside

[eternal shard]

knowing some trig properties

-1/3π * cos(3πt) is bounded between -1/3π and 1/3π

so you want to see when -1/3π * cos(3πt) = 1/3π

[warm pewter]

Is that because

Max and min of cos is 1 and -1 so

They multiply with the number -1/3pi in front

[eternal shard]

you can look at it that way

if you remember acos(x) the factor a stretches the cosine function vertically

naturally you have 1cos(x) so it varies between -1 and 1

bute here -1/3π * cos(3πt) varies between -1/3π and 1/3π because a = 1/3π

[warm pewter]

Ohgggggg

I completely forgot

Ok I understand ur point

[eternal shard]

now yes -1/3π * cos(3πt) = 1/3π is equivalent to -cos(3πt) = 1 if you divide by 1/3π

cos(3πt) = -1

are you following so far

[warm pewter]

Ahhhhh

Yesssssssss

[eternal shard]

now you know that cos(x) = -1 if x = π or multiples

cos(3πt) = -1
you can basically think here, what must t be so that 3πt becomes only π

[warm pewter]

U mean odd multiples or nah

[eternal shard]

yes

[warm pewter]

t = 1/3

[eternal shard]

yes

of course again there are infinite solutions, but for this task knowing one solutions tells us the maximum already

[warm pewter]

T can be 1

Ohhhh

[eternal shard]

so now you would plug in v(1/3) and that should give you the max

[warm pewter]

Phhhhhhhhhhh

Bc if it is multiple of pi they would give the same so that does not matter

[eternal shard]

are you confused or why are you overreacting? 😓

[warm pewter]

Lmaoooo

Nah

[eternal shard]

yes, odd multiple as said

[warm pewter]

I get everything

Bc I was doing it wrong

[eternal shard]

so t = 1 would also work cause 3π is also odd multiple of π

[warm pewter]

Oh yesssss and t that makes 5pi and so on

[eternal shard]

yea

[warm pewter]

Omg thanks

[eternal shard]

but you dont need it

[warm pewter]

Yep

Ordrr

Can I just not find t

[eternal shard]

nah you need at least one

like a house with keys

[warm pewter]

Bc that time u said like we know what the max value is 1/3pi

[eternal shard]

you need at least one to enter

[warm pewter]

Ohhhh

[eternal shard]

unless you do something illegal skull

[warm pewter]

HAHAHHAAHHAHAHAA

I hope I can in maths

[eternal shard]

yes you can

then it's wrong

or you just re invented math

but the latter is less likely

[warm pewter]

Hahahahaha

I’m trying the last question rn

[eternal shard]

oh

it's basically similar to how you found v

s(0) = 1 keep in mind too

[warm pewter]

Ok! I will try rn

I got the answerrrr

Thanks sm for ur help

.close

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[gusty cliff]

I don’t understand the Va/2usin30=Vb/2u

How to they get that same direction equation

Cuz they both have different magnitudes🧍‍♂️

Is it like a ratio?

.close

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[remote mural]

Help

[remote mural]

$(3x-1)^2 - 9(x-1/3)^2$

[potent lotus]

James

[remote mural]

Factroise

[remote mural]

Idk where to start

[velvet osprey]

id start by looking at that second term

9 is 3^2, so you can rewrite it as [3(x - 1/3)]^2 ...

[remote mural]

Mm

AHH

[remote mural]

What happened to the negative that was attached to the 9

[velvet osprey]

i didn't touch it

i was having you focus specifically on the 9(x - 1/3)^2 bit

like if you wanna recontextualize it, sure: the expression as a whole becomes $$(3x-1)^2 - [3(x-\tfrac{1}{3})]^2$$

[potent lotus]

Ann

[remote mural]

Continue pls

[velvet osprey]

how can you simplify 3(x - 1/3)?

[remote mural]

Distributing 3 in it

[velvet osprey]

ok, do that. what happens then?

[remote mural]

$3x-1$

[potent lotus]

James

[velvet osprey]

right

so your expression is now

(3x-1)^2 - (3x-1)^2

[remote mural]

Ahhh

I see

Thanks

.close

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[keen flower]

<@&268886789983436800>

[calm coral]

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[remote mural]

a

[remote mural]

what is the differentiation of e^(1/4x^2)

wait

[clever monolith]

consider the chain rule

[remote mural]

e^((1/4)x^2)

[clever monolith]

could someone do this in latex

[remote mural]

😭

let me take ss

wait

i thought the derivative of e^ax+b is

ae^ax+b

how did we get -1/2

[clever monolith]

well its not that formula

as a is a constant

and x is raised ti the

2nd power

[remote mural]

what a

[clever monolith]

coefficient*

not constant

but if it helps, break it down into a series of smaller functions

[remote mural]

we do the uv thingy

but when we did dv/dx

i didnt get where -1/2 came from

[clever monolith]

u sub is for integration

just take the derivative and find the critical point

where the derivative evaluates to 0

[remote mural]

how did we get this

[clever monolith]

if you want a quick answer

[remote mural]

please

[clever monolith]

the square comes down so ylu get a coefficient of 2

which multiplies

the -1/4

[remote mural]

so 2*-(1/4)

[clever monolith]

long answer id have to write out

[remote mural]

ohhh alright

thanks

.close

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[nocturne heron]

I've seen that we can canonically plunge $R^2$ into $R^3$ with $(x,y) \mapsto (x,y,0)$, but is there a canonical plunge of $R^2$ into the sphere?

[potent lotus]

tm

[nocturne heron]

Well I found a formula for that, it’s horrible 😭 $(x, y) \mapsto \left( \frac{2x}{x^2 + y^2 + 1}, \frac{2y}{x^2 + y^2 + 1}, \frac{x^2 + y^2 - 1}{x^2 + y^2 + 1} \right)$

doesn’t even work

[marsh gazelle]

can someone help me with grade 8 fractions

[nocturne heron]

.close

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[limpid lagoon]

create a new channel

#❓how-to-get-help

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[normal spindle]

so i tried graphing this

$\text{fake time (hours)}=7+(\text{hours passed})-(\frac{(\text{hours passed})}{60})$

[potent lotus]

UCYT5040

[normal spindle]

and i got fake time=7.5 when hours passed = 0.5

0.5084746 to be a little more exact

the correct answer is 8:48 PM

which is not 0.5084746 hours after 7:00 AM

so what should i have done?

[limpid lagoon]

its stopped for 12 full hours

and the total stop time is the sum 1+2+3+...+12

the formula for the sum of the first n natural numbers is $\frac{n(n+1)}{2}$

[potent lotus]

Krish

[limpid lagoon]

substitute n = 12 hours into there

$\frac{12(12+1)}{2} = \frac{156}{2} = 78$

[potent lotus]

Krish

[limpid lagoon]

add 78 minutes to 7:30

and you get 8:48

[normal spindle]

How did you originally get 12 hours?

Or is that just 7am to 7pm?

[limpid lagoon]

because there are 12 full hours in between 7am and 7:30pm

it doesnt get stopped for the 13th hour because we check it before 8pm

[normal spindle]

Ahh see I thought like from 7:00am to 7:30am you subtract (1/2)

But the question does say at the end of the hour

Thank you for your help

.close

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[limpid lagoon]

yeah ofc

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[glacial elbow]

I'm really confused with axis for this, I thought that fx is cos and fy is sin , how do i know when each one is each and also when it is positive sign or negative sign

[strong halo]

just solve it your way.. you are the decider of axis. answer will be same (use |A+B| vector).. , however in the solution, he took the cos for the vector which is making the angle.. as we do and and sin with the x axis and then he inverted x in opposite direction to change its sign

[calm coral]

@glacial elbow Has your question been resolved?

[glacial elbow]

I might be wrong, but my way I would personally do cos(70) x 100
And sin 20 x 100

[glacial elbow]

Could I do this?

So finding the angle of the force from the x axis and angle of the force from the y axis

[strong halo]

absolutely

no no wait

it would be cos 70 and sin 70

[glacial elbow]

U wrote cos20 on that?

Ok and just one more question

Could I really figure it out how i want to figure out if the angle was something like this

So I could just do - cos and + sin

Because I saw this chart and it confused the hell out of me

[glacial elbow]

Sorry I mean - cos and - sin

[strong halo]

negative x is there as vector would be point toward the ... rod? and we require the force against as the axis assumed will be in opposite direction

just keep 1 thing in mind.. make two axes.. the one with which it is making theta.. write it as cos theta and other as sin theta

[glacial elbow]

What do you mean

[glacial elbow]

And with this; is it cos 70 and sin 70 or sin 70 cos 20?

[strong halo]

sin 70 and cos 20 both are equal

[strong halo]

wrote that to prove this..

[calm coral]

@glacial elbow Has your question been resolved?

[glacial elbow]

Ohh i see

Cos being from the x axis?

@strong halo

[strong halo]

your choice for choosing axis

[glacial elbow]

Last question

It's just for example this angle

It's - cos 60 and sin 60 right?

What if I was doing the angle from the y axis

So 30

[strong halo]

if u are assuming positive x to be right side then yep

[strong halo]

then cos 30 and - sin 30

[glacial elbow]

So I have the power to just do that

That makes it way simpler if so

[strong halo]

yessir

in questions.. they need answer

not the sign bullshit

[glacial elbow]

Just when you do this

Do you go cos from the x

And sin from the y typically

Like u personally

Or don't think about it much

[strong halo]

i just do the angle which is specified..

and i take the standard axes.. like you took

just keep 1 thing in mind.. lemme write that out

[glacial elbow]

Cuz like for P2 that could be to be 80 , or 10

[strong halo]

not 80 ,45 + 25 is 70

[glacial elbow]

Oh yea whoops

70 or 20

[strong halo]

see so while solving the magnitude of vector will be same.. only direction would change

[strong halo]

depends where you are taking its component

if you are talking along p3 then p3 = p2 cos 45

and along +x p2 cos 70

[calm coral]

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[mortal fern]

is xsinx an even function or an odd function?
I know that by definition odd function is f(-x)=-f(x) and even function is f(-x)=f(x).

Now I say that xsinx is an even function function since x is an odd function and sinx is an odd function and odd*odd is even function. is my reasoning correct here? will an odd*odd function always be even?

[swift laurel]

yes

you can prove that fact from the definition

[mortal fern]

ohh like this? -fx * -fx = fx = f(-x)

[swift laurel]

i would use multiple different function names but you get the idea

[mortal fern]

okay sweet. thanks @swift laurel

.close 🔐

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[pale prairie]

What’s the general process of proving something is discontinuous using epsilon delta for a piecewise function

[pale prairie]

First I guess you negate the definition

For all delta > 0 there exists epsilon > 0 such that if |x - x0| < delta, then |f(x) -f(x0)| >= epsilon

and then you fix delta > p

0*

After that do you just pick some random epsilon value and hope for the best ?

[potent smelt]

You can think of an epsilon value as a challenge. Your job is to pick a delta that depends on the epsilon as a response.

[pale prairie]

ok this is obviously incorrect but you usually try to do some manipulation with |f(x) - f(x0)|

[potent smelt]

So the idea is that there is some evil math genius who wants to prove you wrong

[pale prairie]

hmm

[potent smelt]

And he or she will provide you an epsilon

[pale prairie]

Wouldn’t delta be arbitrary

Ah

[potent smelt]

Your job is to either turn that into a delta, or show that there is no such delta

[pale prairie]

This is proving something is continuous

and ur goal is to uncover some epsilon

oh

[potent smelt]

If it's continuous then you can always find a delta

[pale prairie]

ohhh

[potent smelt]

If it is not continuous then there is some value of epsilon where you cannot find the delta that satisfies it

[elder pawn]

epsilon is before delta

forall and exists do not commute

[velvet osprey]

greek mfs upon hearing this:

[calm coral]

@pale prairie Has your question been resolved?

[elder pawn]

i wonder how the greeks reacted to weierstrass presenting the def for the first time

[velvet osprey]

idk

iirc the epsilon and delta used to stand for something

i think error and difference

[elder pawn]

error makes sense, i just thought we use delta bc its adjacent in the alphabet

oh nvm we probably used capital delta for difference way before weierstrass

[calm coral]

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[split sluice]

How do i get -5 for log^2 1/32 ? I dont understand how you get there, compared to how you get 2^5 =32

[split sluice]

I always get stuck with fractions i just dont get it

[kind axle]

What is 2^(-5)

[keen flare]

when you raise something to the negative power, you get 1/b^p

from b^-p

[split sluice]

But how do i get -5? I actually dont know how to answer it, i just got that answer online

[keen flare]

what's log base 2 of 32

[split sluice]

Thats 5

[keen flare]

then what about 1/32 following that b^-p = 1/b^p

[split sluice]

I dont know :((

[keen flare]

2^-5 = 1/2^5 = 1/32

$2^{-5} = \frac{1}{2^5} = \frac{1}{32}$

[potent lotus]

BuilderDolphin

[split sluice]

How did u get the 1/2^5

Like how did it go to a fraction

[keen flare]

when you raise something to the negative power, it's the reciprocal

$b^{-p} = \frac{1}{b^p}$

[potent lotus]

BuilderDolphin

[split sluice]

So because were given log^2 1/32 in our question. We already know the square root of 2^5 is 32 so we just put a negative in front of the answer always?

[keen flare]

well not the square root

2^5 is just 32

[split sluice]

Oh yea

[keen flare]

but yes if it's the reciprocal in this case 1/32, the answer is negative

for example log base 2 of 1/16 is -4

since it's the reciprocal of 2^4 or 16

[split sluice]

okay

So what about log^32 of 1/2

[keen flare]

log base 2?

or log base 32

[split sluice]

32 is the small number next to log

[keen flare]

oh base 32 then yes

same logic here

32^? = 2

[split sluice]

Like that

[keen flare]

yes

since it's < 1, the result will be negative

to make it simpler find log base 32 of 2 first

then you can just flip the sign

[split sluice]

So 2^5

And put it into a fraction?

[keen flare]

well 2^5 = 32

but you're trying to find 32^? = 2

base 32

[keen flare]

you can get the answer from here

[split sluice]

So half of 5?

[keen flare]

remember $(b^x)^y = b^{xy}$

[potent lotus]

BuilderDolphin

[keen flare]

what can you multiply 5 by to get 1

since you want 32^p = 2

[split sluice]

I dont know

[keen flare]

it would be 1/5

[split sluice]

Why

[keen flare]

$5 \cdot \frac{1}{5} = \frac{5}{5} = 1$

[potent lotus]

BuilderDolphin

[split sluice]

Oh right ive seen that before

Ok i think i need a break my brain isnt braining

I still dont get it

[calm coral]

@split sluice Has your question been resolved?

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[calm coral]

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[arctic thorn]

A city water department claims that the average daily water consumption per
household is 250 liters. A researcher suspects that the actual average is lower
and collects a random sample of 40 households, finding a sample mean of
240 liters with a standard deviation of 20 liters.
At a 0.10 significance level, test whether the average daily water consumption
is less than 250 liters.

[arctic thorn]

Is the null hypothesis u = 250 or u >= 250

I'm in denial here

[calm coral]

@arctic thorn Has your question been resolved?

[arctic thorn]

<@&286206848099549185>

[velvet osprey]

the null hypothesis is the water dept's claim

cause you're trying to see if you have enough evidence to reject that

[arctic thorn]

so they claim that the avg daily water consumption is 250L?

which is u = 250?

[velvet osprey]

mu not u, but yes

[arctic thorn]

could it also be mu >= 250?

[calm coral]

@arctic thorn Has your question been resolved?

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[warm pewter]

Why is 120 degrees the only angle that is allowed?

Here is the question

[wraith geode]

At what points does cos give -1/2

Nvm wait

[wraith geode]

Which part are you looking at ?

a or b

[warm pewter]

Part b)

[wraith geode]

And do you need help with the full solution or something specific

[warm pewter]

This part

I don’t know why it’s only 2pi/3

[wraith geode]

Are you sure arccos(-1/2) is 2pi/3 ?

[ancient grotto]

It is.

[warm pewter]

Yessss

[ancient grotto]

What is the other value, then?

And what happens if you use the other value?

[wraith geode]

It’s because cos is only negative at 120 degrees

Adding 180, that’s the 4th quadrant , it’s positive there

[warm pewter]

What about 240 degrees?

[wraith geode]

That’s the “coffee” 4th quadrant , cos is positive for any value there, so it would give +1/2 not -1/2

Sorry wait

240 you say

[warm pewter]

Yesss

[jovial pumice]

that is correct too

[warm pewter]

I think so tooo

But the mark scheme does not allow it??!!?!,!,!

Oh maybe

[ancient grotto]

@wraith geode cos(2pi/3)=-1/2. It’s negative in the second and third quadrants.

[warm pewter]

It’s not the same for sin

[jovial pumice]

wdym?

[wraith geode]

Yea ik

[jovial pumice]

there are infinite values of x in cos x = -1/2

but the principal solutions(in range of [0,2pi], are just going to be 2 which will be 2pi/3 and 4pi/3

[warm pewter]

Understood. Now which one do I use for time?

[jovial pumice]

both will give the same answer prolly, u can use both i think

[warm pewter]

The thing is sin(2pi/3) and sin (4pi/3) r not equal

[jovial pumice]

ohh yeaa, sry i dont know then

[warm pewter]

No worriesssss

Thanks anywayaa

[jovial pumice]

wlcm🥲

[winter elbow]

phenomenal stuff, gonna save it rq 😉

Don't forget to close the channel

[warm pewter]

Still can’t solve itttt

[warm pewter]

R u able to help?

[winter elbow]

oh, hasn't the question been solved already?

lemme check

[warm pewter]

Nopeee

[winter elbow]

I'm pretty bad at physic. I'm afraid that I can't provide any help on this question

[warm pewter]

It’s okay thank uuuu smmm

[warm pewter]

OHHHHH

So what does that mean?

[eternal shard]

well t >=0 and your first solutions are 2pi/3 or 4pi/3 so you'd take 2pi/3 according to the task

since it's the first value

[warm pewter]

Ohhhhhhh

Thank u smmmm

I finally know whyyy

[eternal shard]

np winnie

[warm pewter]

Saving my life as always

Okk see uuu

.close

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[calm coral]

@old plover Has your question been resolved?

[calm coral]

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[crystal salmon]

I was doing a certain problem and I faced an issue

[crystal salmon]

$\int_0^{\pi/2} \frac{x}{tan(x)}$ $dx$

[potent lotus]

bagelguy3

[crystal salmon]

First I applied U substitution

$u = tan(x)$

[potent lotus]

bagelguy3

[velvet osprey]

\tan

[crystal salmon]

$\int_0^{\infty} \frac{arctan(u)}{(u)(u^2 + 1)}$ $du$

[potent lotus]

bagelguy3

[crystal salmon]

I noticed that there are poles at

$u = i, -i$

[potent lotus]

bagelguy3

[crystal salmon]

So I tried contour integration

I figured I'd go for a counter clockwise semi-circle and the semicircle from theta = 0 to theta = pi would be 0

I did a bit of substitution and found out that

$\int_{-\infty}^0 \frac{arctan(u)}{(u)(u^2+1)} du$

[potent lotus]

bagelguy3

[crystal salmon]

would equal I

$2\pi{i}Res(f(z)$ at z = i$ $=$ 2I$

[potent lotus]

bagelguy3

[crystal salmon]

therefore

$\pi{i}Res(f(i))$ = $I$

[potent lotus]

bagelguy3

[crystal salmon]

Except the residue at i = infinity

<@&286206848099549185>

$f(z)$ = $\frac{arctan(z)}{(z)(z^2+1)}$

[potent lotus]

bagelguy3

[crystal salmon]

@velvet osprey @

[velvet osprey]

what

[crystal salmon]

help

[velvet osprey]

no

[calm coral]

@crystal salmon Has your question been resolved?

[crystal salmon]

.close

[calm coral]

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[unkempt nimbus]

guys, i have questions, how to quickly memorize arithmetic progression formula? what is the basic concept? how did we get Sn formula?

[amber bolt]

e.g. 7 10 13 16

you reverse the order and add
7 10 13 16
+16 13 10 7
23 23 23 23

[unkempt nimbus]

yeaa

[ivory lion]

i have an even simpler way

[unkempt nimbus]

the result is all 23

[ivory lion]

are you familiar with the idea that you can "move" the data around without changing the average?

for example, the average of:
5 7 9
move 2 from the 9 over to the 5
7 7 7

[amber bolt]

how i think of it is we can find the average number

by averaging ends

but it's not more compicated if you get all 23

[velvet osprey]

do you mean the n'th term or the sum of the first n terms

oh you did say S_n formula

my bad

[unkempt nimbus]

the Sn formula

[velvet osprey]

do you remember the story of how Gauss found a way to sum 1+2+...+100 quickly

when he was a kid

[unkempt nimbus]

this means that after we get the result, we just have to divide the 2 terms into n and then multiply it by 23.

[unkempt nimbus]

i don't know about that

[velvet osprey]

the story went that his teacher tried to keep the class busy for a while and so had them do this sum

but gauss realized (so it goes) that you could write a copy of the same sum backwards underneath it

so like

S =  1  +  2  +  3  + ... +  99 + 100
S = 100 +  99 +  98 + ... +  2  +  1

[unkempt nimbus]

its all 101

[velvet osprey]

and then adding those vertically you get pairs of terms adding to 101 yes

and there are 100 of them

and then you divide by 2 to get back from 2S to S

the spirit's the same for general arithmetic progressions.

[unique halo]

100 pairs? i think theres 50 pairs

[ivory lion]

not if you add 2 copies of ALL the terms

[unkempt nimbus]

yea, i think half of it

[velvet osprey]

no, the total when you add vertically is 2S and there are 100 "columns" in the setup

[unkempt nimbus]

e.g. given the arithmetic sequence –2, 1, 4, 7, …, 40. determine the number of terms in the sequence.

[ivory lion]

yeah you can compute that without too much difficulty

i have two methods

[unkempt nimbus]

the result is all 38

but we don't know how many numbers there are

[ivory lion]

want me to go over the two methods?

[unkempt nimbus]

yes please

[ivory lion]

so this first one is less prone to error i think

write out the formula for the nth term of the sequence

start with a_0 = -2

each term after that adds 3 so it would be

a_n = a_0 + 3n

you can check for yourself that this works

[unkempt nimbus]

yeaa

[ivory lion]

so the first term of the sequence is a_0

whats the n if a_n = 40?

[unkempt nimbus]

its 14

[ivory lion]

good

so how many terms are there from a_0 to a_14?

(be careful)

[unkempt nimbus]

266 i guess(?)

[ivory lion]

imagine i have this sequence:

a_0, a_1, a_2

can you see that there are 3 terms

you dont add the indexes, they simply tell you the position of the term, not how many terms they are

we said a_0 is the first term, not that a_0 has no terms

a_14 is equal to the last term, not that the last term counts as 14 terms

[unkempt nimbus]

i got it

it turns out quite simple

then how abt the Sn of geometric progression formula

[ivory lion]

i was gonna make a small detailed point as a warning

cause there is a common mistake here

[unkempt nimbus]

what kind of mistake?

[ivory lion]

so here we started the sequence on a_0

instead of a_1

so between a_0 and a_14 there are 15 terms, not 14

this might catch some students occasionally

you could go back and do it starting from a_1 instead, and it will still work, BUT

while the final count is slightly easier

the formula is different, and slightly more annoying

a_n = a_1 + 3(n-1)

[ivory lion]

compare to this

so up to you which one you want to use, just make sure you're paying attention to what the first term is defined as

[unkempt nimbus]

its longer than that

[ivory lion]

yeah

alternatively, to find the number of terms you can just

take the difference of the last and first term

and divide by the gap

so (40-(-2))/3

which gives you 14, the number of gaps

so the number of terms is 15

this is cleaner and quicker but way more prone to error because:

• the final +1 is less likely to be checked for
• this method only works for arithmetic sequences

[unkempt nimbus]

but the number of gaps is 14

[ivory lion]

yes

[unkempt nimbus]

and we use the number of the gaps

[ivory lion]

no you want the number of terms

if i have the sequence 3, 5, 7, there are 2 gaps, but 3 terms

a gap between 3 and 5, and a gap between 5 and 7

this is a dangerously common mistake

(7-3)/2 = 2

which is not the number we want

|--|--|

3 fenceposts but only 2 fences

[unkempt nimbus]

e.g. 30, 27, 24, 21,...., 12 can we use a_n=a_0+n.(difference betwen numbers)??

[unkempt nimbus]

is it mean that we cant use that formula every time?

[ivory lion]

yes, this can also work!

try it on simple examples and see for yourself

[ivory lion]

you can, as long as you have an arithmetic sequence and you are careful about the off by one fencepost counting

[unkempt nimbus]

idk but i cant find it with those method

[ivory lion]

show your work?

[ivory lion]

uhh from

12 = 30 - 3n

this step is correct, but the step after this is wrong

i think skipping steps caused you to mess up

[unkempt nimbus]

how it should be?

[ivory lion]

do it step by step and see if you can find your mistake

[unkempt nimbus]

yepp

i found my mistake

[ivory lion]

eyyyy!

ok now how that you found n, how many terms are there?

thats a different question

[unkempt nimbus]

the gaps is 6

there are 7 terms there

[ivory lion]

bingo

and if you count them manually

your answer should match the sequence

[unkempt nimbus]

I GOT IT!!!

but damnn, there is stil geometric progression formula

is there a simple way of geometric progression formula @ivory lion ?

[ivory lion]

there is

there is a slightly more tedious method that doesnt require too much creativity, but makes it a pain to derive and remember

[unkempt nimbus]

why??

[ivory lion]

but it is the standard derivation and it works

lemme pull it up hold on

[unkempt nimbus]

okaayy

[ivory lion]

https://en.m.wikipedia.org/wiki/Geometric_series

so you want to scroll down to where it says

"The partial sum of the first n+1 terms..."

and thats where the finite geometric series is derived

the page mentions some stuff on convergence but if youre just doing finite geometric series you can ignore that completely

this is the standard derivation

i have a simpler intuition for it, but it has a prerequisite

[unkempt nimbus]

??

[ivory lion]

do you have any questions?