help-13

[floral arrow]

I meant less, somehow I fell into the trap

[latent citrus]

ahhh ok thank you

that approach makes sense

[floral arrow]

You can try visualizing the "truthness" on the various intervals

For A, P needs to be true for x^2 on [0, 1) and maybe for other values

For B, P needs to be true for x^2 on [0, 1) and only there

For C, P needs to be true for x^2 on [0, 2) and maybe for other values, and for D, only on [0, 2)

[latent citrus]

so becuase C has a wider range it needs to be true for it is more restrictive

[floral arrow]

Exactly

[latent citrus]

and so anytime C is satisfied, A is also satisfied

but they can't both be true by the question

thank you for your help 🫶

[floral arrow]

You're welcome

[drowsy raptor]

can we close this now

[cedar kiln]

@latent citrus Has your question been resolved?

Channel closed

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[cedar kiln]

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[young flume]

how do i solve this using the theorem above

[dire geode]

sec(x) = 1 / cos(x)

[young flume]

this where im at

[dire geode]

that doesn't equal what you started with

$\frac{ab}{c} = \frac{a}{c} \cdot b$

[wraith dagger]

riemann

[young flume]

o

how does that work

is the b not also being divided

[dire geode]

this is correct

,tex .limit rules

[wraith dagger]

riemann

[dire geode]

you can use the quotient one now

or product depending how you want to look at it.

[young flume]

but it says evaluate using this theorem

[dire geode]

yes you used that theorem incorrectly

in this step

use this

[young flume]

is that correct perchance

[dire geode]

all correct there

[young flume]

alrighty

thnx

wait whyd i need to do the division law

lmfao

feels like an extra step

[hollow trail]

shouldn't the bottom limit just be or cos x

also you're mixing x and theta

[young flume]

whoopsie

.close

[cedar kiln]

Channel closed

Closed by @young flume

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[cedar kiln]

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[visual thicket]

Hi, can someone find me to find the complex root of this polynomials equation z^4+2z^3+3z^2-6z=0

[visual thicket]

I have done this

[buoyant latch]

Guess

[buoyant latch]

Okay that gets you z = 0

[flint plinth]

well one root is hopefully obvious

[buoyant latch]

Guess another one

[chrome elk]

They want to find the complex root

[visual thicket]

Yep, the problem is the other 3

[buoyant latch]

Ya just guess

Okay just tell them the answer

Good one

[chrome elk]

Sorry

Just woke up, wasn't thinking

[visual thicket]

I use WolframAlpha but I don't how I suppose to say to my teacher I just guess the answer

[buoyant latch]

You look at it

And give it your best guess

There really isn’t a good method at this point

You can always if you really need to, use the cubic formula but that’s like hitting a nail with a hydraulic press

[royal finch]

You know there is a real root from the cubic and the rational root theorem tells you that if it is rational, it must be one of +-1,2,3,6. If none of those are a solution, then the problem is likely going to require a calculator.

This kind of exploits the fact that problems made for school can't be too mean, so the rational root theorem helps you sniff out a solution. This doesn't work in "real life" for whatever little that means, but you are working with polynomials a computer algebra system can instantly solve for you.

[visual thicket]

But how you find the 2 root complex?

[royal finch]

the quadratic formula

[visual thicket]

But how do I remove the 2 first root? Because we didn't see in class the one for third degree and for fourth degree?

[royal finch]

You would use polynomial long division or undetermined coefficients to find the quadratic

You have z^3 +3z^2 + 2z-6. Find the root real root, call it r. Then do polynomial long division with z-r, or write (z-r)(Az^2+Bz + C) = z^3 +3z^2 + 2z - 6 and solve for A,B,C.

[visual thicket]

Thanks

[cedar kiln]

@visual thicket Has your question been resolved?

[visual thicket]

Not yet, I'm doing the exercises and see if it work

[fossil dawn]

react to the bot please

if not, the bot will close it anyway

[visual thicket]

Thanks

[fossil dawn]

nps. just a quick reminder that the bot can't really see messages the way you think it can

[visual thicket]

That true

I do not use that server as often

[cedar kiln]

@visual thicket Has your question been resolved?

[visual thicket]

It work

[cedar kiln]

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[quick thunder]

Can someone help me, o have no idea where to go from here

[cedar kiln]

Please don't occupy multiple help channels.

[wicked mantle]

.solved

[cedar kiln]

Channel closed

Closed by @wicked mantle

Use .reopen if this was a mistake.

[wicked mantle]

Please stick to your original channel

[quick thunder]

Wait where was my original channel

[wicked mantle]

#help-4

Someone is already typing

[quick thunder]

My bad

.close

[wicked mantle]

No worries

[quick thunder]

How do I close again

[wicked mantle]

I’ve already closed it for you

[quick thunder]

O

[wicked mantle]

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

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[vapid grotto]

I’m so confused what’s going on how did we move the X+1

[hollow trail]

divided both sides by x + 1

[vapid grotto]

Oh … whoops

[indigo lagoon]

Anything else?

[vapid grotto]

Also I don’t understand why the domain is y does not equal 1 like I understand after looking at the answer key why she did it but I don’t know how I would know to do that

[slender atlas]

the denominator of a fraction cannot equal 0.

[hollow trail]

the only input that could make the inverse function undefined is the one that makes it a division by 0

[vapid grotto]

But -1 is outside fo the fraction

For the problem before solving

[slender atlas]

you're talking about the domain of the inverse function, isn't it?

[distant path]

Basically,
$$-1=\frac{5}{x+3}-1$$
has no solutions.

[wraith dagger]

Will

[vapid grotto]

J(X)=5/x+3 -1

[slender atlas]

in your domain, you gave y.

[distant path]

Because this would require
$$\frac{5}{x+3}$$
to be zero, which isn't possible for any finite x.

[wraith dagger]

Will

[buoyant latch]

Do you know why we swap the x and y positions?

[slender atlas]

the original function's domain would involve x, not y. so quoting y as the domain implies you're quoting the domain for the inverse function, not the original.

[vapid grotto]

Because they are inverse?

[buoyant latch]

That’s not really an answer

[reef quartz]

you're describing the process frost wants to know why you do that

[vapid grotto]

Then I have no idea

[slender atlas]

may I know if you know the definition of an inverse function?

[vapid grotto]

Not off the top of my head

[buoyant latch]

It’s very enlightening to know why you actually swap the letters

I didn’t know this until I got to uni either

[slender atlas]

I'll leave this to frosst then, sorry for intruding.

[crimson sedge]

math is hard work

[buoyant latch]

We want to find f^-1(y) = … with some y on the right side

That’s the inverse function

[vapid grotto]

Yes

[distant path]

Swapping the x and y has a very nice geometric intuition too.

[buoyant latch]

We only know what happens if we put x into f, ie. f(x) = 5/(x+3) -1

[vapid grotto]

Waiti was mixing up two things

So I did know X and y are swapped

[buoyant latch]

So what if we put f^-1(y) into f itself?

We get f(f^-1(y)) = 5/(f^-1(y) + 3) - 1

But the left side by virtue of being the inverse is just y

So we have y = 5/(f^-1(y) + 3) - 1

[buoyant latch]

If we now solve for f^-1(y) we get this

This process is the same as if we just swapped x and y around

Another thing to consider is that f^-1(y) has domain that is the image of f

If for any number b, there aren’t any x such that f(x) = b, then you couldnt put b into the inverse, so f^-1(b) doesn’t make sense

Cos you’re asking, what input gives b when I put it into f? And the answer is no inputs give b when you put it into f

[vapid grotto]

I mean I understand how to get domain mostly I don’t understand how to get range

[buoyant latch]

Range and image is the same thing, ill use the word range now

[vapid grotto]

Wdym by image

[buoyant latch]

You need to look at 5/(x + 3) and see what values this can be

Image is another word for range

[vapid grotto]

How can I tell what values it can be

[buoyant latch]

Well it’s a stretched and shifted hyperbola

That means there’s a horizontal asymptote so the function can’t attain that value

[vapid grotto]

What’s a hyperbola

And a asymptote

Could I use a graphing calculator and find this out I kinda left mine in my class on Monday so thags why I’m not using it…

[cerulean sail]

As an alternative way, are you happy that the only way you can have a ("legal") fraction equal to zero is if the numerator was equal to zero?

[vapid grotto]

Wait is it just the y intercept ?

[cerulean sail]

Not quite here, no

[cerulean sail]

(and if you're allowed to, then fair enough )

[vapid grotto]

For a line aka mx+b would the range be infinity both ways

[vapid grotto]

Would it be obvious to see the range on the pixely screen

[cerulean sail]

It might be, depending on how much you can see of the plot, but graphs like 1/x have a particular shape to them (and the one you have is basically a "transformed" version of that one)

,w plot y = 1/x

[wraith dagger]

Results provided by WolframAlpha

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

Requested by chartbit

[vapid grotto]

Before I go on with more of my homework are these two correct

[cerulean sail]

You can see that for this one, the only value you're not allowed to put into y = 1/x is x = 0 (because dividing by zero is illegal ), and that you can basically get any value you want out of 1/x, apart from 0, right?

,rccw

[cerulean sail]

Careful here, you added 6 to both sides, left hand side should have been x + 6

[vapid grotto]

Oh whoops

[cerulean sail]

Happy with the second one if you wanted, you could rewrite both of them into the "y = mx + b" format to make it clear that you get a line, but then that may not be required anyway

[vapid grotto]

I have no idea what to do next and I don’t understand how to find the range I think it’s infinity that’s thags me guessing

[cerulean sail]

We shall come back to the range later for now

[vapid grotto]

Yea sorry

[cerulean sail]

Remember that we wanna make $y$ the subject here, and right now, we're at
[
x(y + 1) +3 = 2y
]
What would be nice is to get everything with $y$ on the same side, and everything without $y$ on the other side, if you get me?

[wraith dagger]

@cerulean sail

[vapid grotto]

Im also stuck in the same position for the next one

[vapid grotto]

I don’t know if I can do this but can I devide y+1

[cerulean sail]

You shouldn't, because that would undo what we just did by multiplying everything by (y + 1) the step before, and we would be right back where we started

[vapid grotto]

Do I distribute the X

[cerulean sail]

You just read my mind

Yep, we should

[vapid grotto]

Also for square root for the range would it be whatever the domain is to zero. For example the square root of X+3 the range would be [3,infinity ) or maybe [0,infinity )

[cerulean sail]

The range for sqrt{x + 3} would be [0, inf), you can get e.g. 0 out of the square root when x = -3, and 1 out of the square root when x = -2

[vapid grotto]

Im not sure what to do next because I can subtract 4 bevause I’ll still have 2y on that side

And the same thing happens with this problem

[cerulean sail]

Be careful, you should have the left hand side as xy + x + 3

You're multiplying both the y and the 1 by x

[vapid grotto]

Why are we multiplying?

[cerulean sail]

Because we're expanding out x(y + 1), which is xy + 1x

[vapid grotto]

Ohh

Even after that I’m still stuck

Do I divide by t

Y

[cerulean sail]

Don't divide by y (because that will mean that the x and the 1 will still have y's in them )

[cerulean sail]

There is another way for us to get everything with y's on one side of the = sign, and the things that don't have it on the other side though

[vapid grotto]

I have no idea

[cerulean sail]

You don't? Well how could you not, I'm gonna take things away from you

(do you like my hints )

[vapid grotto]

How would I subtract?

If that’s what the hint means

Because X and y are together

You would have subtract them both

[cerulean sail]

And why would that be a problem?

I say we try it anyway

[vapid grotto]

Wouldn’t I have to decide by X now but that would make everything X

[cerulean sail]

Not yet, there is another thing we can do instead, now we have that right hand side

Notice how both terms have a y in them, right?

[vapid grotto]

Yea

[cerulean sail]

Can you think of something we can do to that RHS to make it look a bit "different"?

[vapid grotto]

Not ready

Really

[cerulean sail]

Not really? but how will I- alright, I don't think I have a good way to hide my hint here

We could factor that side, do you agree?

[vapid grotto]

I’m not sure how I would do that

[cerulean sail]

Well, both terms have a common factor of y, right?

[vapid grotto]

Yea I have no idea how to find it tho

[cerulean sail]

One second

So, right now, we have
[
{\color{red} 2} {\color{green} y} - {\color{yellow} x} {\color{green} y}
]
I'm sure you're also happy that factoring is the "opposite" of expanding, and that last time, we tried to expand
[
{\color{green} x } ({\color{red} y} + {\color{yellow} 1})
]
which we turned into
[
{\color{red} y} {\color{green} x } + {\color{yellow} 1} {\color{green} x }
]

[wraith dagger]

@cerulean sail

[cerulean sail]

Does that give any ideas as to how we could factor that 2y - xy?

[vapid grotto]

No…

Could we move on from these for now

I’ll try to ask my teacher

I got two hours of psych jrp testing first and second block in the morning so I really need to go to bed 😭

Should of not procrastinated

[prisma pike]

Always good to get more sleep

Gn frfr

[cerulean sail]

That's fair enough

[cerulean sail]

Awwwww best of luck with those, hope they go well

[vapid grotto]

I still got to do math tho lol

Can I solve anymore or is this it

[cerulean sail]

I will say for now though, that you can factor that as y(2 - x) (as in general, ab + ac can factor into a(b + c), that's just applying it here )

[cerulean sail]

And you can simplify that down much more

[vapid grotto]

How would I start?

[cerulean sail]

Well, if I asked you the Totally Unrelated™ question of "how would you simplify something like $3\qty(\frac{p}3))$, would you know how to do it?

[wraith dagger]

@cerulean sail

[vapid grotto]

Multiply by 3?

I’m bad at simplifying

[cerulean sail]

If you did, do you know what we'd get?

[vapid grotto]

I think I’m getting confused but how would i multiply it from only 1 side

[cerulean sail]

How do you mean, only 1 side?

[vapid grotto]

I think I’m getting confused since we worked for two sides with the other problems

I don’t know how I would multiply by 3

Also just realized I don’t need to find the domain for this rigjt

[cerulean sail]

Yea the difference here is that everything is over 3, so like we could treat the numerator as one "object"

[vapid grotto]

Sorry I’m having like a brain fart I still have no idea

[cerulean sail]

Alright, a more simpler silly question

What's 3/3?

[vapid grotto]

1

[cerulean sail]

Yep

[cerulean sail]

You're happy that the 3's here would cancel each other out for that reason, right?

[vapid grotto]

Ohhh

[cerulean sail]

It's one of those things that are quite simple, but it's hard to hint at without either outright saying it or being too vague

(btw please ping replies, I may have switched to another chat so may not notice if you reply! )

[vapid grotto]

Would anybody be able to check these. Ignore the highlighted ones those ones I’ve decided I’ll ask my teacher

[cerulean sail]

Watch out for this one

18 is good

For this one, everything should be under the cube root, so make sure that you show the root extends over everything, including the -5 (or otherwise add to make that clear!)

19 and 21 I'm happy with

Careful of the second equals of the top line, it's better to just remove it imo as those aren't equal

This part, you may want to try writing it out again

[cerulean sail]

hiii

[cedar kiln]

@vapid grotto Has your question been resolved?

[cedar kiln]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
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[queen plover]

how would i even begin to do this question?

[prisma pike]

Possibly helpful diagram

[cedar kiln]

@queen plover Has your question been resolved?

[queen plover]

what would i do now?

[indigo lagoon]

You should include w into your calculation as well

@queen plover Please ping me when you're back

[queen plover]

@indigo lagoon

[indigo lagoon]

Alright

I'm back

[queen plover]

me too

this is how mine currently looks like

would you solve for the bottom

?

that would yield 0 so idt thats it

[indigo lagoon]

no, don't mark w there

it makes things confusing

[indigo lagoon]

yep

Do you see a triangle in the graph?

[queen plover]

yea

the one with R

[indigo lagoon]

Alright, can you represent the sides of the triangle with the given variables?

My Internet sucks rn, so I can't draw anything on it to demonstrate

[queen plover]

uh idk how

[indigo lagoon]

Have you found this segment?

This is the triangle I was referring to

[queen plover]

i havent but idk where i would even start with that

[indigo lagoon]

okay, so this segment is w/2, do you know why?

[queen plover]

because its the midpoint of the hole

[indigo lagoon]

yes

so this segment is h-r, because the segment above it is the radius

[queen plover]

Oh i see

because the ball is lodged into the ground?

[indigo lagoon]

ya understand?

[queen plover]

the blue segment is w-r?

and that top variable is L?

[indigo lagoon]

mb, it's w/2

lmao

I'm fumbling

[queen plover]

its okay

i get what u mean

[indigo lagoon]

that's r

[queen plover]

oh ok

[indigo lagoon]

it's radius

[queen plover]

that segment h-r of the rectangle is because the height of the triangle is greater than the radius right?

so to solve for the difference you subtract the radius from the max height.?

[indigo lagoon]

the height of the triangle is greater than the radius
No, it has nothing to do with the height of the triangle, you're basically just remove the extra part of the segment to get the height of the triangle

[queen plover]

oh okay

[indigo lagoon]

and for the rest, you know what to do

[queen plover]

yea i believe so

thank you!

[indigo lagoon]

no worries

Anything else?

@queen plover

[queen plover]

just this

[indigo lagoon]

If you're done with the question, type .close to close the channel

[queen plover]

its another question if you dont mind 😭

[indigo lagoon]

oh, sure

[indigo lagoon]

This one is much tricker

[queen plover]

Idk where to even start with this one

[indigo lagoon]

ah, I see how to break this down

[queen plover]

how

this is what i have rn

[indigo lagoon]

dawg, I'm laggy badly

q

hello

[queen plover]

hi

its alright take your time

[indigo lagoon]

Alright my Internet's back

phew

This is what I drew, but seems like you get the idea already

[queen plover]

yea idk the values

tho

for AC i believe itll be a-b at the very least

[indigo lagoon]

yeah, you don't know about the value but you get smth interesting

Let's set radius as r in the first place

[queen plover]

this is what i have atm

[indigo lagoon]

and try to represent AC with the variables in hand

[queen plover]

a-b** in the pic sorry

[queen plover]

yea

this is what i have

[wicked mantle]

No, that’s not a-b

Sorry, my hotspot crashed for some reason, so I gotta reply on my main account which is on my phone

[queen plover]

No problem i appreciate your help a lot

[wicked mantle]

We never know if the centre is at the same level of the higher ground, yk what I mean?

[queen plover]

yea

[wicked mantle]

So AC would never be simply “a-b”

In fact, if you’re looking for AC. You’ll have to add up all the segments and deduct the redundant part

[queen plover]

oh okay

[wicked mantle]

What you should be looking for is orange part - r - b

[queen plover]

a+c-r-b?

[wicked mantle]

A is a point

Not a segment

yes

And that’s segment AC

[queen plover]

i see

[wicked mantle]

Now we’ve done with everything needed

Do you agree that angle ACB is theta/2?

[queen plover]

so sin(40.35)* (a+c-b-r)

=r

[queen plover]

Yes

[wicked mantle]

Fantastic work

Now you know what to do

[queen plover]

Yup im gonna solve it really quick brb!

thank you for your help

[wicked mantle]

No worries, ping me when you’re ready for the second question

Ping this account not the other one. I’m not going to take out my pc again LOL

[queen plover]

got it, thanks again!

@wicked mantle the second question would just be sin(theta/2)= 7.9/(a+c-b-7.9)?

[wicked mantle]

yeah second one is just doing it reversely lol

[queen plover]

alright thank you! i have no more questions

[wicked mantle]

!done

[cedar kiln]

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

[wicked mantle]

Have a good one

.solved I’ll close it for you :)))

[cedar kiln]

Channel closed

Closed by @wicked mantle

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[slender atlas]

,rccw

[wraith dagger]

Couldn't find an attached image in the last 10 messages.

Sep 24, 2025, 6:33 AM

[cedar kiln]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[wispy storm]

,rccw

[slender atlas]

this channel is closed unfortunately.

please claim another one.

[wispy storm]

Okay

.close

[cedar kiln]

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[elder bobcat]

i dont understand part B

[elder bobcat]

what does it mean lol

the a can be 0?

[hidden mural]

but you understand part A?

[hidden mural]

is 0 > 0?

[elder bobcat]

no

i saw that it says a>0

[hidden mural]

so can alpha be 0?

[elder bobcat]

no... i think?

[hidden mural]

if alpha = 0 , is alpha > 0 ?

[ancient lodge]

Ignore what I said earlier (I assumed alpha had to be an integer and misread \geq 0)

[prisma pike]

Bro understands man it's just because of what civil service pigeon wrote 😭

[elder bobcat]

no... rigt?

[hidden mural]

why ru unsure

[prisma pike]

Because you keep questioning bro about it bro

And because of what civil service pigeon wrote

[hidden mural]

let him speak

[elder bobcat]

im confused xD it says a>0 so a cant be 0....

[hidden mural]

and why is that confusing?

[prisma pike]

Sir😭

[elder bobcat]

i saw the [0, 0], [-1, -1]...

and uhm yeah i dont get part A either

its just part B looks harder lol

[hidden mural]

do you know how to take the union of 2 sets

let's start with A

[elder bobcat]

what 2 sets?

[hidden mural]

any

[elder bobcat]

A = {1, 2, 3}
B= {3, 5, 6}

can i use this

[hidden mural]

go on

[elder bobcat]

union is {1, 2, 3, 5, 6}

[hidden mural]

what is [1,1] U [2,2]

[elder bobcat]

{1, 2}

[hidden mural]

is the number 0.5 in the set [1,1]?

[elder bobcat]

but like this question looks so weird comapred to this

no

[hidden mural]

what is the defintion of [1,1]

[elder bobcat]

its a set?

[hidden mural]

what set

[elder bobcat]

any set...?

[hidden mural]

[1,1] can be any set?

[elder bobcat]

i dont know.... i dont get it.. lol

i dont understand what this is saying either

[prisma pike]

[-a,a] represents the subset of real numbers that fulfills the inequality shown

[hidden mural]

i'm done here ✌️

[prisma pike]

Is basically what the set notation says in words

[prisma pike]

Thanks

[hidden mural]

why thanks

[prisma pike]

For being helpful

[hidden mural]

np, feel free to take over this channel

[elder bobcat]

so uhm... it says if x is a member of all real numbers.... then...
if i let x=1

need that pic again lol

okay if i let x=1... now what... ;-;

a can be anything positive?

i mean

like 1000

[prisma pike]

Maybe we can break this into parts

Firstly the interval notation

[elder bobcat]

[-a, a]

[prisma pike]

Yes so that represents the subset of real numbers which is greater than or equal to -a

And less than or equal to a

Is that ok with you

[elder bobcat]

i think so..

[prisma pike]

Alright

So let's start with a.)

You see that big union operation to the left of the interval

[elder bobcat]

yeah i do

[prisma pike]

What does that notation mean

[elder bobcat]

it wants elements that exists at least one time in a bunch of sets

or its looking for*

[prisma pike]

I mean yes I guess that's partially true

But like lowk

This is a new type of operator

[elder bobcat]

what do you mean a new type of perator?

operator*

[prisma pike]

You know your summation operator

[elder bobcat]

yeah

with the lower and upper bounds?

[prisma pike]

Yea this is kinda like that

But with unions

[elder bobcat]

is the a>0 the lower bound...?

[prisma pike]

I'll draw this out hold on

[elder bobcat]

okokoo

[prisma pike]

If S is the set of real numbers greater than 0

Then the expression in question 5a.)

Is just equal to the union of the intervals as shown above

[elder bobcat]

it's equal to [-s1, s1]U[-s2,s2]...?

[prisma pike]

Where S1 is the first element in the set

S2 is the second

Etc

Actually nvm it might not be that notation lowk

Have you not seen this in like your textbook

[elder bobcat]

I looked at the picture it made sense but then when I looked back to the problem it looks different lol

He said we don't use a textbook just his notes and lecture

Wait let me find his note

Okay I don't think he ever posted the notes about that notation

The closest thing to it is indexed sets

And someone asked him about 5a) the other day and he said its all real numbers for the answer because a just stretches onto infinity or something I don't remember his exact words

[prisma pike]

Then I think yea it's this then

[dawn junco]

well you can see it in that way yes, it's just a simplified notation for an indexed union they're using

let $I$ = positive real numbers,
$U_a = [-a, a]$
then you're looking at $\bigcup_{a\in I} U_a$

[wraith dagger]

aPlatypus

[elder bobcat]

oh it looks like this now

[dawn junco]

yea

[elder bobcat]

the middle part

can i do like

case scenario

because i still dont get why he said the answer is all real

[dawn junco]

why not

what cases are you thinking of?

[elder bobcat]

need the pic again lol

like plugging in number

[dawn junco]

well what numbers

you can't exactly list all real numbers and check each of one is in that union

if you don't wanna be stuck for your whole life and more

[elder bobcat]

so like it says x is a memer of all real number

member

[dawn junco]

yea

[elder bobcat]

like

if x=2

how do i check that

i want an example... i think im lost even on that

[dawn junco]

well is there an interval among all these [-a, a] which has 2 in it ?

that's how you check something in is such kind of union

[elder bobcat]

i think there is...?

[dawn junco]

give one then

[elder bobcat]

[-2,2]... i think?

so like

[-2,2] is one of the sets

and its a union

[dawn junco]

in math speak, $x \in \bigcup_{a\in I} U_a \iff \exists a \in I, x \in U_a$

[wraith dagger]

aPlatypus

[elder bobcat]

so just that set is enough

[dawn junco]

well [-2, 2] is part of that union

and you just need to be in one set to be in the union

so yes 2 is in the union

[elder bobcat]

does that mean anything bigger than 0 works..?

like

a cant be 0 so

[dawn junco]

yes the exact same argument works

[dawn junco]

well 0 is in all these intervals really

[elder bobcat]

so is the answer the natural numbers?

[dawn junco]

why natural

[elder bobcat]

WAIT

all real numbers but not 0 and not negative

wait

ohhh yeah 0 is in all of them

[dawn junco]

for the moment you've shown that all positive real numbers are in the union yes

you ain't finished tho

[dawn junco]

yea

well you just need to check it's in one of them if you're lazy

like 0 is in [-1,1] or whatever

that's pretty uncontroversial

[elder bobcat]

yeah but i do not know how to check all the numbers xD

it says x is all real

wait

[dawn junco]

well what about negative now

[elder bobcat]

it says [-a,a]

so like

if x is -1000 then it is in [-1000,1000]

right?

so it works

[dawn junco]

yes indeed

[elder bobcat]

so isnt it all real numbers...?

[dawn junco]

it is

[elder bobcat]

omg yay

okay uhm now i need to think about the uh intersection...

uhmmm so if x = 2

wait

thats impossible right

i think..?

[-1,1], [-2,2]...

but

[-1,1] doesnt have 2 in it right

[dawn junco]

right

what about 0 tho

[elder bobcat]

uhmm [-1,1], [-2,2]...

[dawn junco]

it's not so obvious 0 ain't in the intersection

[elder bobcat]

x=0

wait

[dawn junco]

yeah but a=0 isn't allowed, positive a only

[elder bobcat]

0 is in all of them

oh wait

but it says x...?

wait it says x can be all real so x can be 0..?

but like

it says [-a,a] so i can use a=1

and its

[-1,1] and 0 is in the inverval

i mean

the x=0 is in the interval of the a's

and if i do the interval [-2,2], x=0 is also in it

[dawn junco]

sure

what if you shrink the interval a ton tho ?

that's the main issue here

when a is very very close to 0

[elder bobcat]

uhmmm if a=0.1 then [-0.1, 0.1], x=0 is still in it xD

[dawn junco]

sure

what if a = 0.01, 0.001, 0.0001, 0.00001, ... etc

[elder bobcat]

0 is still in it...

right..?

[dawn junco]

well at least you're convinced 0 is in it, good

how do you really prove it is in the intersection tho, we can't just plug in a's for all eternity

[dawn junco]

you prolly should have seen some criterion like this but for intersections

[elder bobcat]

there exists an "a" in I, such that x is a member of Ua

[dawn junco]

ok that's for unions

[elder bobcat]

oh

[dawn junco]

what about intersections

[elder bobcat]

switch the there exists to for every?

[dawn junco]

indeed

[elder bobcat]

for every "a" in I, x is a member of Ua

is logically equv to the intersection of Ua

wai

how do i apply that xD

to this

[dawn junco]

well you've already shown that everything not 0 is not in the intersection

[dawn junco]

so what you're left to do is prove this for x=0

[elder bobcat]

for every a in I, 0 is a member of Ua

[dawn junco]

yeah so if you actually use the meaning of I and Ua, you're left with
for every a>0, 0 in [-a, a]

or for every a>0, -a <= 0 <= a

[elder bobcat]

is that the proof for x=0?

its just... for every a>0, 0 is in all [-a,a]

[dawn junco]

yes

I mean you're just saying it right now, no proof

but the proof ain't hard sure

[elder bobcat]

how do i like show that all a contains 0 xD

like you said i cant uhm keep going forever xD

keep going down forever..

[dawn junco]

doesn't this sound a bit easier to prove ?

it's just another restatement of what you gotta do

[elder bobcat]

thank you xD i finally got a question done for once

.close

[cedar kiln]

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[ornate echo]

Hey guys

[ornate echo]

Does anyone know a proof of this:??

I've translated it here:

My textbook doesn't provide a proof

And it doesn't seem obvious at all for me

s_{x} is the standard deviation of the samle

And $\bar{x}$ is the sample mean

[wraith dagger]

stoicindiehacker369

[ornate echo]

Would appreciate any insights on this

In the textbook they just used an arbitrary sample and an arbitrary k value

And basically it says: "yo, 0,10 is smaller than 0,25 so this holds true for this one arbitrary scenario"

[cedar kiln]

@ornate echo Has your question been resolved?

[cedar kiln]

@ornate echo Has your question been resolved?

[cedar kiln]

@ornate echo Has your question been resolved?

[cedar kiln]

Channel closed

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[loud nexus]

One can prove rather straightforwardly, by Mellin transforms, that
$$I=\int\limits_{0}^{\infty}\frac{J_{0}^{2}(t)J_{1}(t)}{t}\mathrm{d}t=\frac{1}{2\sqrt{\pi}}G^{1,2}{3,3}\left(\left.\begin{matrix}\frac{1}{2},\frac{1}{2},\frac{3}{2}\0,0,0\end{matrix}\right|4\right)\approx 0.524866$$\
\
By employing Slater's theorem, we can obtain
$$\frac{1}{2\sqrt{\pi}}G^{1,2}{3,3}\left(\left.\begin{matrix}\frac{1}{2},\frac{1}{2},\frac{3}{2}\0,0,0\end{matrix}\right|4\right)= {}{3}F{2}\left(\left.\begin{matrix}\frac{1}{2},\frac{1}{2},-\frac{1}{2}\1,1\end{matrix}\right|4\right)$$\
\
Obviously, the left-hand side is a real number.\
But since here $|z|=4>1$, the right-hand side must be evaluated by an analytic continuation of ${3}F{2}$ and does not output a real number. \textbf{The above equality is false.}\
\
In fact, the correct version is
$$\frac{1}{2\sqrt{\pi}}G^{1,2}{3,3}\left(\left.\begin{matrix}\frac{1}{2},\frac{1}{2},\frac{3}{2}\0,0,0\end{matrix}\right|4\right)=\Re\left({}{3}F_{2}\left(\left.\begin{matrix}\frac{1}{2},\frac{1}{2},-\frac{1}{2}\1,1\end{matrix}\right|4\right)\right)$$\\
\textbf{I wonder what transformations I can do to the Meijer G function to obtain a reduction in terms of hypergeometric functions without using this problematic analytic continuation.}

[wraith dagger]

如月あやみ　Kisaragi Ayami

[cedar kiln]

@loud nexus Has your question been resolved?

[ornate echo]

Hey is that a new question?

Or an answer (sorry if I seem naive haha. It's because I am)

[loud nexus]

It's a new question. Your question was timed out.

Please read the bot's messages.

[loud nexus]

<@&286206848099549185>

[last apex]

You should post this in the advanced channels

probably #real-complex-analysis, #advanced-number-theory or #advanced-analysis

[ornate echo]

yeah i noticed that haha

[loud nexus]

Maybe the first channel there is the most suitable

Thanks

I'll close this help channel for now

.close

[cedar kiln]

Channel closed

Closed by @loud nexus

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[cedar kiln]

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[gleaming mantle]

Question: How do I read this? I'm not really familiar with the sum notation

[dawn junco]

you're talking about the i+j=k thing right ?

[gleaming mantle]

Yes, the thing in the middle with the sum notation

[dawn junco]

my question is, are you confused with sums completely (like is the sum k=0 to m+n part a problem to you also), or is it just the second sum (with i+j=k) you got a problem with ?

[gleaming mantle]

The first thing also confuses me, specifically why is it m+n and not just m or n? And the second has me totally lost. I understand the conditional but don't understand what the a and b are

[dawn junco]

well what do you expect the degree of the product of two polynomials to be ?

[gleaming mantle]

Oh yeah damn of course

[dawn junco]

if you multiply a degree m and a degree n, you expect to have a degree m+n term there

[gleaming mantle]

Yes I didn't think of that

[dawn junco]

so you gotta go up to m+n

now for the inside thing, let's take a small example

deg 2 * deg 2

(a2 x^2 + a1 x + a0) * (b2 x^2 + b1 x + b0)

if you expand this out what do you get ?

[gleaming mantle]

a2b2x^4 + a2b1x^3 + a2b0x^2 + a1b2x^3 + a1b1x^2 + a1b0x + a0b2x^2 + a0b1x + a0b0

[dawn junco]

right if you collect all the x^4's, x^3's, x^2's, ... together now ?

[gleaming mantle]

a2b2x^4 + (a2b1 + a1b2)x^3 + (a2b0 + a1b1 + a0b2) x^2 + (a1b0 + a0b1)x + a0b0

[dawn junco]

right look at the coefficient for x^2 for example

a2 b0 + a1 b1 + a0 b2

it's a sum of things, that are products of one coeff in a, one coeff in b

[gleaming mantle]

Yes the indexes are 2

[dawn junco]

and the sum of the two indices is 2, cause it's the term in x^2 in the product

for x^3, you want sum of indices to be 3, etc...

[gleaming mantle]

Yes

[dawn junco]

that's what this sum i+j=k means

for the coeff of degree k in the product, you take all combos of coeffs 1 in a 1 in b where the indices sum to k, and you sum all of these

[gleaming mantle]

Okay that makes sense

Thznk you, qI will wait a little to close this helpchannelnbecause I might have a related question

[dawn junco]

it's obvious from context but yeah i and j have to be non-negative integers and that's not apparent in the sum

you can't look at coeffs that don't exist

[dawn junco]

sure

[gleaming mantle]

So I have this, it is related to he last thing. I understand what it says now, but I'm wondering how I could think of that, is there like an order of operations for summation symbols or how do people figure that out?

Specifically from step 1 to step 2, where everything is a summation

[slender atlas]

inside out, I believe.

like parentheses.

[gleaming mantle]

I figured it out. Thank you all for the help

.close

[cedar kiln]

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[hasty path]

how i do this?

[hasty path]

why is 1 / sin(pi-x) smh 1 / sin(x)?

oh wait cast diagrams

nvm

.close

[cedar kiln]

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[crimson sedge]

How to start this?

[smoky cliff]

Is that j a quaternion?

[crimson sedge]

not sure what that means

im doing complex numbers qs

[smoky cliff]

What is j?

[crimson sedge]

rt-1

j^2 = -1

[opal hinge]

Wut

Isn't that i

[smoky cliff]

Thats what we called it in my days

[crimson sedge]

im doing engineering maths so it might be diff in our case

cause of electrical eng

[smoky cliff]

Its not

[crimson sedge]

i is current

is it not?

[smoky cliff]

Oh

[crimson sedge]

well amps

[smoky cliff]

Maybe

I mean

[crimson sedge]

its in my book anyways

[smoky cliff]

Im pretty sure i always saw i in circuits

But nvm

Also weird that its j8

And not 8j

But

You solve them

[crimson sedge]

in my book

[crimson sedge]

yea idk why

[smoky cliff]

Exactly the same way

[crimson sedge]

gets me confused too

[smoky cliff]

As with real

Isolate variable

And substisute to second

[crimson sedge]

so z = 6 + j8 - jw

[smoky cliff]

Yes

[crimson sedge]

hm

i get wrong val for 2

w

[smoky cliff]

You divided wrong

This

[crimson sedge]

what abt it?

[smoky cliff]

Not true

[crimson sedge]

why?

im doing divison of complex numbers

[smoky cliff]

And you do it incorrectly

[crimson sedge]

ok how do I do it?

or where did i go wrong?

[smoky cliff]

Well

How did you come to that second thing?

[smoky cliff]

That here

[crimson sedge]

using this where z1 = x1 + jy1

[smoky cliff]

Ah wait

Its actually good

I misread

[crimson sedge]

but i went wrong somewhere right

[smoky cliff]

And you know the answer is wrong?

[crimson sedge]

yes

so?

[cedar kiln]

@crimson sedge Has your question been resolved?

[ancient lodge]

,w 4z+3w=23, z + i*w=6+8i

[wraith dagger]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[ancient lodge]

4 times 8 isn't 36

[crimson sedge]

Bruhh im dumb

Lol

.close

[cedar kiln]

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[crimson sedge]

Help

[tropic oxide]

image coming?

[crimson sedge]

Is it correct

[tropic oxide]

the spelling of isosceles is broken but the proof looks OK to me

[crimson sedge]

Before sas rule

Just before it

@tropic oxide are you here

@tropic oxide

[tropic oxide]

...

dont spam ping me, it is upsetting.

[crimson sedge]

I am not

[tropic oxide]

you pinged me twice, one minute apart.

[crimson sedge]

But listen one thing

[tropic oxide]

do you want to ping me 17 more times? it is really upsetting that you ping me so often. one ping was enough.

anyway let me reread the proof

[crimson sedge]

I write that a b is equal to AC because since sides and angles are equal so hypotenuse will also equal

[tropic oxide]

i dont see even a single mention that angle A is shared between the two triangles

[past wave]

That seems incorrect

[tropic oxide]

did you purposefully stay silent about that

[past wave]

You should use RHS here

[crimson sedge]

My appolocheese

But there is nothing for it

[tropic oxide]

i need to go, sorry

[crimson sedge]

Because in my book RHS rule is in the next exercise

[past wave]

Oh

Can you send a pic of the question?

[past wave]

You can use asa here too

By angle sum in a triangle, acf and abe are equal

@crimson sedge You there?

[crimson sedge]

Ok

So I think it has been solved

Thank you

! close

[past wave]

Welcome bro

I'll close it for you

.close

[cedar kiln]

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[tropic oxide]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[wild vessel]

@sinful abyss do you know the first step ?

Because after the first step it will be normal solving

[worldly oasis]

do you know what is rationalise?

[tropic oxide]

y'all let OP speak please.

[sinful abyss]

nvm guys i got it

[wild vessel]

Rationalise means you have to remove the irrational part

[wild vessel]

👍

[worldly oasis]

irrational part from denominator

[wild vessel]

Nice

[wild vessel]

Yupe

[tropic oxide]

if you're done you can .close this channel

[gray hamlet]

!done

[cedar kiln]

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

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

@sinful abyss Has your question been resolved?

[cedar kiln]

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[foggy berry]

I have to factor completely

I'm just looking for the method to find the numbers

[nimble mountain]

what are the factors of 3

what are the factors of 20

[foggy berry]

2,10

4,5

1,3

10,-7

[nimble mountain]

trial and error with the different combinations and see which gives u that

[foggy berry]

That's it?

Ik it's -10, and 6 but I don't want to have to guess

[nimble mountain]

u can use the quadratic formula

[foggy berry]

How do you do that again

[nimble mountain]

that gives u the roots directly.

from the value of x here, u can find the factors.

[foggy berry]

So how would you use the quadratic formula for this equation

[nimble mountain]

a=3, b=-4, c=-20

taking y=1 for simplicity.

[foggy berry]

Ok I see

.close

[cedar kiln]

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[cedar kiln]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[marsh mesa]

Why $\int_0^0 \cos(x) dx$ its not 1?

[wraith dagger]

Reginald Puddingface

[marsh mesa]

Wouldn't the underlying area be 1?

[tropic oxide]

no

[celest flume]

there is no area in that integral

[tropic oxide]

area 1 means something like a 1×1 square

what you've got is a zero-thickness vertical line of length 1

height 1 perhaps

but its width is 0

[marsh mesa]

Clear

[tropic oxide]

$\int_a^a f(x)\dd{x}=0$ always, regardless of what $f$ is.

[wraith dagger]

Ann

[gray hamlet]

yea

[marsh mesa]

Thanks

.close

[cedar kiln]

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[final moth]

why did they multiply the denominator by 1/h just to multiply the denominator by h again afterwards this thing sucks so they dont explain where it came from lol

[iron wren]

its just to clean it up and get h in the upper denominator

[final moth]

umm Oh wait I think i get it the 1/h got rid of the h on the bottom originally so the fractions on the top became the main fraction and bc they multiplied 1/h on top as well the h is on the bottom it all clicked

[hexed ravine]

The value of the function at a chosen point is the chosen point divided by one more than the chosen point.

• The value of the function at a nearby point is the sum of the chosen point and the change, divided by one more than that sum.
• The difference quotient is found by subtracting the first value from the second value, then dividing the result by the change.
• After simplifying, the difference quotient becomes one divided by the product of one more than the chosen point and one more than the sum of the chosen point and the change.

[final moth]

i understand it now

thank you

[hexed ravine]

Do you want it in more simple terms?

[final moth]

no its okay your explanation is good

[hexed ravine]

Any more questions?

[final moth]

nah thats it thank u

[cerulean sail]

You didn't get this from GPT, did you?

[hexed ravine]

Chat gpt?

[cerulean sail]

Yea, or anything like it

[hexed ravine]

Nope!

Or do I just act like a bot :\

[cerulean sail]

It seemed quite susly worded and laid out to me

[hexed ravine]

Ok I’ll stop acting “sus”

[cerulean sail]

As long as you know you can't answer questions with it, that's all fine

[hexed ravine]

[cedar kiln]

@final moth Has your question been resolved?

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lapis kindle]

Hey so I have a pretty basic high school level question but im struggling , I don’t know what Im doing wrong

[lapis kindle]

We are doing algorithmic average or however it is called in English

Anyways

The question is :” there are 25 students in a class with the average height of 168 , when John gets taken out of the equation , the resulting average is 167.5 , how tall is John ?”

[lapis kindle]

I tried 2 different ways but I keep getting the answer wrong

[cerulean sail]

How did you try, if you don't mind sharing please?

[lapis kindle]

Yes sure

First thing I wrote is “168/25-J ( for John ) = 167.5/24

I got like 155

So that’s wrong

Second thing I wrote is 167.5/24 +J= 168/25

But now that I’m thinking about it that’s probs wrong too

😭

uhm

[cerulean sail]

Both are equivalently saying the same thing may I asked why you divided by 25 and 24?

[lapis kindle]

I know 😭

Because 25 and 24 are the amount of people in the class

Mmm

Idk

I keep trying to solve it like an equation

And I’m starting to think it’s not

[cerulean sail]

Sure, maybe let me ask a slightly different question, how would you find the average height of a set of people, if I gave you some of them?

[lapis kindle]

I would add up all of the heights and divide it by the number of people

[lapis kindle]

But if you only gave some , like 3 out of 7 , I would add up the 3 heights +4x /7 = average

[cerulean sail]

Cool, yep but you know that 168 is the average height of all the 25 students in the class, so when you add up everyone's height, including John, and divide that by 25, you get 168, right?

[lapis kindle]

Yup

And John’s height is above 168

Because him leaving negatively impacted the average

[cerulean sail]

Sure, so what could we do to get the sum of everyone (including John's) heights, from that 168?

[lapis kindle]

Maybe

Put the 168 in the middle

And there is 25 people

So go 168 - 1 , 168 -2 and so on 12 times and 168 + 1 , 168 + 2 12 times

/25

Which is essentialy 168/25

And

Idk

😭

[cerulean sail]

Hmmmm, it may be worth being a bit careful, let's walk through basically "what we did" to get that 168

[lapis kindle]

Ok sure

So the 168 was a result of all of the classmates heights being added up and divided by 25

[cerulean sail]

Yep, so if we wanted to find out what all the classmates' heights added up was, what could we do, do you think?

[lapis kindle]

168 Times 25?

[cerulean sail]

Yep, that's it well done