help-39

[carmine juniper]

But R c R

[ancient slate]

Because R is a set

[carmine juniper]

So we can also mark it as a set

[ancient slate]

{R} would be a set with the letter R in it

[carmine juniper]

So it wouldn’t mean real numbers

[ancient slate]

Its like you dont say if A is a set , you dont say {A}

[ancient slate]

yes

[carmine juniper]

Oh I get it

.close

[pearl pond]

Channel closed

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Use .reopen if this was a mistake.

[carmine juniper]

Why?

Okay

[ancient slate]

Its the same

It would be a set with A in it not the set A

[proper nova]

imagine a set would be a bag right?

so if you say {A}, it would be a bag containing a bag

[carmine juniper]

But if A is alone then it’s okay?

[ancient slate]

Yep

[proper nova]

if A is alone then it's just a bag

[ancient slate]

Yea

Its just a bag labeled A

[carmine juniper]

It’s like saying I’m eating unhealthy food and saying I’m eating a hamburger

[ancient slate]

Same for R

[proper nova]

erm

[ancient slate]

I dont understand that concept but ig

[compact ridge]

yep, that's the issue

you'd be saying all unhealthy foods are hamburgers

R, the collection of all real numbers, is all the unhealthy foods

{R}, the single element, is one of the unhealthy foods

[carmine juniper]

So conclusion

R can be written as {R}

Cause R is a set

Still

[civic drum]

🤨

1 is not an element of {{1}}

because the only element in that set happens to be {1}, which is not the same as 1

but 1 is certainly an element of {1}

make of that what you will

[proper nova]

no

[carmine juniper]

Why gal

But the set is a set of 1

[proper nova]

use the bag thing i told you

{{1}} is a bag containing a bag which contains 1

we define $a \in R$ if the bag R contains an object a

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

now we apply this to the set {{1}}

{{1}} has an object {1}

not 1

[civic drum]

well for starters

1 is a natural number and {1} is a set

[carmine juniper]

Only that

[proper nova]

yep

[carmine juniper]

Only {1}

[proper nova]

yep

[carmine juniper]

So it’s the same

[proper nova]

?

[civic drum]

"a bag containing the apple isn't the same as the apple"

[carmine juniper]

But

1 c {{N}}

[civic drum]

no

[carmine juniper]

Yes

[nocturne grail]

@proper nova

[proper nova]

are they talking about the problem now?

[proper nova]

🙏

[cinder flower]

@civic drum hi

[civic drum]

Yikes

[proper nova]

use $\LaTeX$

[jolly parrot]

1 divided by 0 equals Infinity

[civic drum]

[pastel umbra]

!noai

[pearl pond]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[civic drum]

heyy

[pastel umbra]

This exists precisely for cases like this

[cinder flower]

you’re so pretty

[proper nova]

!redir

[pearl pond]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[proper nova]

you can complement your friend in there

[carmine juniper]

She could’ve been more tanned

[pastel umbra]

If R is a box, {R} is the box that contains the R-box

These are emphatically different objects

[iron basin]

is this ragebait

[pastel umbra]

you implying they know better?

[civic drum]

💅 you're prettier!

[cinder flower]

[pastel umbra]

Again, Layla and Kizzy, not here

[pearl pond]

Available help channel!

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[humble lintel]

<@&268886789983436800>

[spare lark]

<@&268886789983436800>

[pearl pond]

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[pearl pond]

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[hollow mirage]

how would i go about this?

[latent quail]

Ye, don’t advertise here

[cursive wraith]

No ads.

[brave sluice]

you can integrate along the x-axis

to do that, you need to find the limits for the integral, and find the integrand

actually, a lot of the options have logarithms

so you should integrate along the y-axis instead

oh yeah, and it's also because there's dy rather than dx

[pearl pond]

@hollow mirage Has your question been resolved?

[pearl pond]

Channel closed

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[hollow mirage]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[hollow mirage]

i tried doing that and keep getting the answer #4, which is apparently incorrect

[brave sluice]

4 seems right

[ivory basin]

Might be easier to do it for x and then do a substitution ig?

[hollow mirage]

alright thank you guys, ill try it out again

[brave sluice]

i integrated w.r.t x and got the same answer

[ivory basin]

Yeah 4 seems to be correct

[hollow mirage]

ahh, can i see how you did it to see if it's the same as mine?

[plush bramble]

<@&268886789983436800> troll

[lilac jackal]

@rain burrow do not troll

theyre ded

[slow oak]

kil stab murder die

[tepid flame]

hello

anybody here studying for the december sat

and could help me with certain math problems?

[lilac jackal]

this channel is occupied, pls see #❓how-to-get-help

[plush bramble]

#discussion or #math-discussion

[tepid flame]

👍

[pearl pond]

@hollow mirage Has your question been resolved?

Channel closed

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[pearl pond]

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• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
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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.

[stable mango]

hi

[final shore]

ahh yes nice legible drawing

[stable mango]

abc and def are equilateral triangle

[stable mango]

chromebook finger drawing 😓

yes

[final shore]

2 and 3

sidel engths

[stable mango]

question: what is the ratio of the area of efcb vs dec

[final shore]

great

think about how area scales

[stable mango]

uh

[final shore]

compared to length

[stable mango]

its square

[final shore]

2 dimensions vs one dimension

yes square

so take the length ratio

and square it

[stable mango]

so the length is 2:3

right?

so i square it

[final shore]

length ratio right

yes

[stable mango]

4:9?

[final shore]

correct!

not so bad right

[stable mango]

thats the answer?

[final shore]

yes

[stable mango]

that was simple

[final shore]

wait

[stable mango]

?

[final shore]

is ur question

what is the ratio

between the triangle areas

[stable mango]

nope

[final shore]

def and abc

[stable mango]

ratio between trapazoid efcb vs triangle dec

[final shore]

bruh what

dec

[stable mango]

yes bro

its dumb

my dad gave me the question

[final shore]

yeah 0 idea

good luck with that bro

your lucky to have a dad like that I will say

[stable mango]

ok

[final shore]

is the smaller equilateral triangle

exactly in the middle

of the larger

[stable mango]

both triangles are equalateral

[final shore]

yeah but

where is the second one

in relation to the larger

is it in the direct middle or not

[stable mango]

oh

[final shore]

this seems like a very hard question imo

[stable mango]

i think my dad meant def

i think

[final shore]

bruh

then it's way easier

but igtg to bed sorry

[stable mango]

ok gn

[snow owl]

Close the channel if yr doubt is cleared

[stable mango]

i need help

[snow owl]

What is the question?

[stable mango]

here

[snow owl]

Can't understand the question

[stable mango]

same

[snow owl]

Is the inner tri touching the outer tri?

[stable mango]

ill come back later

if i need help

[stable mango]

idk

[snow owl]

Ok

[stable mango]

i think the question wrote wrong

you can close

[snow owl]

U write .close

To close

[stable mango]

.close

[pearl pond]

Channel closed

Closed by @stable mango

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[pearl pond]

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[cinder spoke]

hey i wanna ask about something simple i really struggle with factoring my mind comprhend anything but that can anyone give me simple way to factor equation easily

[snow owl]

Which factorisation?

Like single variable?

[cinder spoke]

yeah for example smthn simple like

x^2-4x+4

[unkempt yacht]

find a pair of numbers that multiplies to +4 and sum to -4

[snow owl]

The best foolproof method: find roots of the eqn

Make use of quadratic formula

This is the most straightforward method

[plush bramble]

https://www.hancockcollege.edu/mathcenter/documents/Diamond Method of Factoring.pdf

[cinder spoke]

okay thank you guysss

[plush bramble]

That's the simplest

[cinder spoke]

ill check it out

[unkempt yacht]

not fool proof by the way, consider x^4 - 27x^2 + 1

[cinder spoke]

i have final exam next monday, i wish i had yall mind for a day

.close

[pearl pond]

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[pearl pond]

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

Which one?

@soft otter

[soft otter]

id like to start with the second one

i dont understand how to approach a trig sub problem

like how to figure out what to sub in

i can sort of identify that this would be trig sub

but idk how to do it

[prisma kernel]

Okay, multiply and divide it by 8 and then add and subtract 4

Separate out the 8x + 4 and -4

For the 8x + 4 fraction, u-sub denominator

[soft otter]

wait

[prisma kernel]

Sure

[soft otter]

im not sure how/why you did this

[prisma kernel]

Okay, so let me ask you something

If the question didn't have an x at the numerator, then would you be able to do this?

[soft otter]

itd still be trig sub right?

i wouldnt be able to

[prisma kernel]

Okay do you know the formula for 1/sqrt(x² + a²)?

[soft otter]

or wait

[soft otter]

no

is that maybe arctan or something like that

[prisma kernel]

No, that'd be just 1/(x² + a²)

[soft otter]

oh

some what close

maybe tan

[prisma kernel]

How about we take x = atany

[soft otter]

so its tan?

[prisma kernel]

Okay see, $\frac{dx}{sqrt(x^2 + a^2)}$

[jolly parrot]

doctorstrangejr

[prisma kernel]

We're trying to integrate this

[soft otter]

right

im given this

[prisma kernel]

Take x = atany, it becomes $\frac{a(secy)^2dy}{asecy}$

[soft otter]

but id like to know it off the top of my head

[jolly parrot]

doctorstrangejr

[prisma kernel]

Oooooo, well the formula I'll give you is in log, not hyperbolic function though

[soft otter]

oh wow how do you end up in log form

its probably not necessary for me to know im curious but i should save it for another day i guess

[prisma kernel]

Well we can just take it to be inverse hyperbolic sine then

I'm worried about the 'a' though, it won't be 1 here

Okay so basically, we can solve 1/sqrt(x² + a²)

So $\frac {xdx}{sqrt(4x^2 + 4x + 5)}$

[soft otter]

sry i was lagging

brain

[jolly parrot]

doctorstrangejr

[soft otter]

i understand so you used a trig identity to make it asecy? in the bottom?

[prisma kernel]

That was for 1/sqrt(x² + a²)

Here we got a x in the numerator

So we gotta deal with that first

[soft otter]

brb

[soft otter]

right

[soft otter]

so how do you try and do this?

u said multiply by top and bottom by 8 and then add/subtract 4

8xdx/8sqrt(4x^2+4x+5+4)-4?

[prisma kernel]

Yes

So basically, what would be the differential of the function of the denominator in the root?

[soft otter]

like a long ish expression with chainrule

4( 8x+4/sqrt4x^2+4x+9)?

[soft otter]

why do we take the derivative/

[prisma kernel]

okay so what we're trying to do here, is get a constant in the denominator

[soft otter]

is there a way to interpret this without memorizing the whole table?

[prisma kernel]

whenever you've got a $\frac {(Ax + B)dx}{sqrt(Cx^2 + Dx + E)}$ type function

[jolly parrot]

doctorstrangejr

[prisma kernel]

You manipulate the numerator in such a way that you get two functions, where the linear function is basically the derivative of the quadratic in the denominaor's root, and the constant function can easily be solved using the formula of 1/sqrt(x^2 + a^2)

[prisma kernel]

more intuition after one point if you do it a sufficient amount of times

So, taking your question

$\frac {xdx}{sqrt(4x^2 + 4x + 5)}$

[jolly parrot]

doctorstrangejr

[prisma kernel]

The derivative of the denominator is 8x + 4

So first we multiply/divide by 8

$\frac {8xdx}{8sqrt(4x^2 + 4x + 5)}$

[jolly parrot]

doctorstrangejr

[prisma kernel]

Then we add and subtract 4

$\frac {(8x + 4 - 4)dx}{8sqrt(4x^2 + 4x + 5)}$

[jolly parrot]

doctorstrangejr

[prisma kernel]

Then we separate them

$\frac {(8x+4)dx}{8sqrt(4x^2 + 4x + 5)}$ - $\frac {4dx}{8sqrt(4x^2 + 4x + 5)}$

[jolly parrot]

doctorstrangejr

[soft otter]

i see

[prisma kernel]

You basically u-sub the denominator in the first one and it's just $\frac {log(4x^2 + 4x + 5)}{8}$

[jolly parrot]

doctorstrangejr

[soft otter]

not sure how you get log there

[prisma kernel]

Well just take $4x^2 + 4x + 5 = u$

[jolly parrot]

doctorstrangejr

[prisma kernel]

Post differentiation --> $(8x + 4)dx = du$

[jolly parrot]

doctorstrangejr

[prisma kernel]

oh wait

yeah you're right, I made an error

It'd be $\frac{du}{8sqrt(u)}$ not $\frac{du}{8u}$

[jolly parrot]

doctorstrangejr

[soft otter]

even if it was du/u itd be ln instead of log right

[prisma kernel]

Eh I meant that only, but you're right, that's an important nuance

So now we can integrate this one

[soft otter]

i see so this version of the problem doesnt require trig sub?

[prisma kernel]

Not yet (also it completely doesn't if you remember some formulae, like 1/sqrt(x^2 + a^2)

Now here - $\frac {4dx}{8sqrt(4x^2 + 4x + 5)}$ (let's ignore the -1/2 for now)

[jolly parrot]

doctorstrangejr

[prisma kernel]

So simply $\frac {dx}{sqrt(4x^2 + 4x + 5)}$

[jolly parrot]

doctorstrangejr

[prisma kernel]

We break the quadratic in such a way that it's a sum/difference of squares, where one is a function of x and the other is a constant

Now $4x^2 + 4x + 5 = (2x+1)^2 + sqrt(3)^2$

[jolly parrot]

doctorstrangejr

[prisma kernel]

So we write it like that

$\frac{dx}{sqrt((2x+1)^2 + sqrt(3)^2)}$

[jolly parrot]

doctorstrangejr

[prisma kernel]

Now firstly we'll take (2x + 1) as t to remove any constant multiplied to x

so $(2x + 1) = t$

[jolly parrot]

doctorstrangejr

[prisma kernel]

Alternatively, instead of doing this headache, we can just take 4 common here and bring it out of the root as 2

So $\frac {dx}{2sqrt(x^2 + x + 5/4)}$

[jolly parrot]

doctorstrangejr

[prisma kernel]

(Again keep 2 for later)

$\frac {dx}{sqrt(x^2 + x + 5/4)}$

[jolly parrot]

doctorstrangejr

[prisma kernel]

You following me so far?

[soft otter]

yes

[prisma kernel]

Okay

[soft otter]

but one thing

not sure how u did the factoring

[prisma kernel]

Factoring where?

[soft otter]

here

[prisma kernel]

Okay so a trick is

Like I had 4x^2 yes

And there's only one x function after factoring (ax + b)² ± c²

So 'a' I can simply find by the root of the x² function (which is 2 here)

For b, once I have a, I just look at the x term I had, because that term would be 2ab

And that term I know, so 2ab = [term of x]

Here we can find b

Now whole square it, and by whichever value the constant is off, the root of that is c

For eg, $(9x^2 + 10x + 60)$

[jolly parrot]

doctorstrangejr

[prisma kernel]

Holy it's getting on my nerve

Okay so here we find it in terms of (ax ± b)² ± c²

a²x² is just 9x², so a is 3

(3x + b)² is 9x² + b² + 6xb

6xb = 10x

b = 5/3

Now b² is 25/9

And we need 60

So sqrt(60 - 25/9) is c

Seems like a lot but when you do it, it makes a lot more sense

[soft otter]

it does seem like a lot

[copper finch]

Calculate this (m+)

[soft otter]

it does make sense

[prisma kernel]

It's that kind of a thing, you can do it quickly but explaining/understanding it for the first time takes a lot of time

[prisma kernel]

?

Okay so our integral was $\frac{dx}{sqrt(x^2 + x + 5/4)}$

[jolly parrot]

doctorstrangejr

[soft otter]

yep

[prisma kernel]

x² + x + 5/4 = (x + 1/2)² + 1

(See how I do it so fast, it's really simple once you get it)

That makes it $\frac{dx}{sqrt((x + 1/2)^2 + 1^2)}$

[jolly parrot]

doctorstrangejr

[soft otter]

do you go through all the steps in your head?

[prisma kernel]

In a way, although I do need to see it written, otherwise it becomes a mess in the head

[soft otter]

yeah id imagine

ok so that simplifies

[prisma kernel]

Now the integral of 1/sqrt(x² + a²) is log(sqrt(x² + a²) + x)

So it simply becomes log(sqrt(x² + x + 5/4) + x)

Now add the constants, and that's your integral

[soft otter]

im not entirely sure if thats equivalent though

shouldnt the original expression on the denominator (sqrt4x^2 +4x+5) be in the final answer somewhere

oh but maybe once the constant distributes

[prisma kernel]

Yeah it is

I just took the 4 out at the start to make it easier

[prisma kernel]

But we did have sqrt(x² + x + 5/4)

[soft otter]

i see

this problem was very hard to wrap my head around

[prisma kernel]

It's more about familiarity

I saw the integral and instantly knew what to do, after some practise you would too

[soft otter]

i need to study all night i have a lot of sets of problems to sort of get down

[prisma kernel]

Well practise is a virtue man, the more you practise the better you get

[soft otter]

are volumes and areas easier, we only have to create the integral not solve it

[prisma kernel]

Area under curve?

And volume as in rotating it around an axis?

[soft otter]

like areas between curves and volume mostly cross sections and rotated about some axis

yep

[prisma kernel]

Ahhh yeah that could be a headache too

However it depends on the level of questions you have

[prisma kernel]

Yeahhhh looks like a headache

Although I'm not much experienced in these

[soft otter]

what makes these hard?

[prisma kernel]

I mean just the visualisation

And you gotta know the concepts well enough in case you're ever confused somewhere

[soft otter]

not sure what makes these hard to visualize i mean i think x^3 and 4x is pretty simple to look at

[prisma kernel]

Well good for you if you can

Then it shouldn't be much of an issue

[pearl pond]

@soft otter Has your question been resolved?

[soft otter]

are u still up?

i have this

[prisma kernel]

Yes

[soft otter]

and i made it into (u^2 -1) root 4 u du

u=e^x +1

oh wait

i made a fatal algebra mistake

[prisma kernel]

(u - 1)²?

(u - 1)² (u)¼ du

[soft otter]

oh ok does that work

[prisma kernel]

Well, open the whole square

Then multiply each term with u¼

Then integrate all 3 of them individually

[soft otter]

ahh

ok got it

[prisma kernel]

Nice

[glacial mulch]

id do

u^4 = e^x + 1

it simplifies to something quite nice

just the power rule

[soft otter]

oh wow

yeah that was kinda weird to workout

[glacial mulch]

ooooh

believe it or not

just set the denominator equal to u

because youll get constant + sec^2x

which is very similar to the numerator

[soft otter]

yeah i think i got it

[compact ridge]

if you differentiate the denominator, then multiply by some constant, you get exactly the numerator

[soft otter]

i just multiplied by 2

[prisma kernel]

Or just take a 2 out

After that it's simply the differential of the denominator

[soft otter]

how do u take 2 out?

[prisma kernel]

Like take it common

2(2 + sec²x), and bring the multiplying 2 out of the integral

[soft otter]

ah

[snow owl]

Is yr doubt solved?

[soft otter]

this is inverse

[humble root]

Hi! The easiest way to do this question is to show by example that g(x) has no inverse function.

Under what circumstance does a function have an inverse function?

[soft otter]

one to one

[humble root]

Good! So,

Try and show that this function is not one-to-one. That is, can you think of two values of x that have the same output?

[soft otter]

do i just guess?

0 and 1 seem to be

[humble root]

Perfect! That's it.

When x = 0 and x = 1, the function values output to the same thing. So, the function does not have an inverse on R.

[soft otter]

if i had to show f(x) was an inverse function how would i do that

the e term is confusing

[humble root]

Hint: Consider the derivative of f(x).

What can you say about the derivative? (Specifically, what can you say about its sign?)

[soft otter]

Positive?

[humble root]

Good! (Well, you need to show me that in your working :) )

[humble root]

Now, if f'(x) is always positive, what does that imply?

[soft otter]

since its always increasing it will be one to one

but how would i find the inverse function if i could

[humble root]

Yep!

[humble root]

That's a more interesting question

Don't think it is possible here algebraically

Best thing you can do is swap x and y.

[soft otter]

i see

[soft otter]

these questions seem a bit more abstract for antiderivatives idk im not sure

[pearl pond]

@soft otter Has your question been resolved?

[pearl pond]

@soft otter Has your question been resolved?

[reef kestrel]

How far have you come?

In these types of problems, a good starting point is trying to find a counter example for the statements

[pearl pond]

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[pastel mango]

Can anyone help me with this shii?

[pastel mango]

Is there anyone here?

[uneven cypress]

@pastel mango just calculate A, A square and A cube

[brittle tinsel]

I think factoring like that makes it easier personally

[pastel mango]

Yeah i did that

(A-2)(A^2+1)

[uneven cypress]

Afterwards write 2 as the unit matrix times 2 and then add componentwise

[pastel mango]

The problem is, if I subtract 2 from the Matrix A, what would the result be 😅

[uneven cypress]

@pastel mango .

[pastel mango]

Alr but, a pic would help me a lot 🥲

[pastel mango]

Ouuu

[uneven cypress]

This case we use $I_3$. Sorry I do not know how to typeset matrices in laTex

[jolly parrot]

Adhi

[blissful cloak]

,, X=X\cdot I_n=\begin{bmatrix}x&0&\cdots&0\0&x&\cdots&0\\vdots&\vdots&\ddots&\vdots\0&0&\cdots&x\end{bmatrix}

[jolly parrot]

ΠαϳαμαΜαμαΛλαμα

[uneven cypress]

thanks man

[pearl pond]

@pastel mango Has your question been resolved?

[pearl pond]

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[lofty bobcat]

Lesson 8: Find m so that the inequality mx – 3m + 2 > 0 has x > 0 as its solution

Lesson 9: Find m so that the inequality x + m ≥ 1 has x ≥ -2 as its solution

Lesson 10: Find m so that the inequality 2x – m < 3(x – 1) has x > 4 as its solution
can someone help me pls, i'm new to inequality
I need the answer, I'm new to inequality

[severe quarry]

Have you done something similar in class? There are basically two ways of doing this; purely algebraically or geometrically visualizing it

[toxic lichen]

I need the answer, I'm new to inequality
jsyk:

!noans

[pearl pond]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

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[pearl pond]

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[hearty fiber]

On what basis is the highlighted step done ?

[hearty fiber]

Upon taking -ve value of the modulus we get m=29/2

Why is that incorrect ?

[dense jasper]

https://www.desmos.com/calculator/ooptmpntfp

the question specifies that this angle must be acute

[pearl pond]

@hearty fiber Has your question been resolved?

[pearl pond]

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[sick lantern]

hi, i have this law of composition and i need to prove its an abelian group

i got trough all axioms but i dont understand commutativity

i get to x^{\ln y} = y^{\ln x}

WTF

i know if i take ln of both sides they are equal but is that the correct way to do it?

[plush bramble]

yes

[sick lantern]

or i could write x as e^ln(x)

and i get the same result

[humble root]

Wait, just to be rigorous to the formatting, you should start with the expression
x * y = x^ln(y)
Then taking the natural log of both sides.

The point is, you want to show that ||ln(x * y) = ln(y * x)||, then use this to show that the group is abelian.

[sick lantern]

yup, i got it

thanks guys

.close

[pearl pond]

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[humble lintel]

I personally think the more important point is that ln(x) is one-to-one.

[pearl pond]

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[burnt skiff]

I don’t understand for b

[toxic lichen]

!occupied

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[toxic lichen]

have you made a diagram?

please claim your own channel

#❓how-to-get-help for instructions

@burnt skiff you there?

[burnt skiff]

Ye

Sorry

@toxic lichen

[toxic lichen]

right ok so you've done part a right

so you know that vector GG' = GB + GC

do you also know the vector sum GA+GB+GC = 0?

[burnt skiff]

Yes

But I can’t think of a way to use this

[pearl pond]

@burnt skiff Has your question been resolved?

[toxic lichen]

consider that G'A = G'G + GA and likewise for B and C

and G'G = -GG'

[burnt skiff]

Hmm

A sec

And how could I tie this to G’B as well

@toxic lichen

[toxic lichen]

my idea actually was that you could first just add all three together

and get G'A + G'B + G'C = 3 * G'G

[burnt skiff]

Don t you have a full demonstration I’m quite sure my small brain won’t put piece to piece bits of information

[toxic lichen]

i'd rather not give away the full solution, no.

[burnt skiff]

Then maybe elaborate ?

A line

So I understand

I don’t know what bumble jumble you are trying to prove

I found another solution

Thank you ☺️

.close

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[pearl pond]

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[brazen bluff]

can someone help me here

[brazen bluff]

im not understanding what hes doing here

this is a weird notation, i dont understand how he calculats with it

[inland ivy]

seems like a kind of ratio test to me

[west vault]

What do you not understand? The algebra? Or why he did what he did?

[brazen bluff]

the algebra

only the algebra

the generall idea i completly understand

but i never seen it writen like this 13....*(2n+1) and i never calculated with it

[west vault]

It's the product of the first n + 1 odd numbers

It'd probably be better if you rewrote in terms of factorials and exponents: $1 \cdot 3 \cdot 5 \cdots (2n + 1) = \frac{(2n+1)!}{n!2^n}$.

[jolly parrot]

jewels!

[west vault]

(That should make the ratio test easier to apply now)

[brazen bluff]

is this the whole term

[west vault]

No just the denominator inside the square

[brazen bluff]

or only 1 * 3....*(2n+1)

lel

is this some common knowledge?

[west vault]

Uh

It's not hard to derive

[jolly parrot]

jewels!

[brazen bluff]

not getting this

[rustic atlas]

Try on small examples

[autumn fossil]

$2\cdot4\cdot6\cdot8\cdot10=\left(2\cdot1\right)\cdot\left(2\cdot2\right)\cdot\left(2\cdot3\right)\cdot\left(2\cdot4\right)\cdot\left(2\cdot5\right)=\left(2\cdot2\cdot2\cdot2\cdot2\right)\cdot\left(1\cdot2\cdot3\cdot4\cdot5\right)=2^{5}\cdot\left(1\cdot2\cdot3\cdot4\cdot5\right)$

[jolly parrot]

MathIsAlwaysRight

[west vault]

Factor all the 2s out

[autumn fossil]

not all of them, just one 2 from each factor

[toxic lichen]

in the product 2 * 4 * 6 * ... * 2n there are n factors and each one can have a 2 extracted from it

when you do that to all of them and put all those extracted 2's on the left you get 2^n * (1 * 2 * 3 * ... * n)

does that make sense to you

[pearl pond]

@brazen bluff Has your question been resolved?

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[minor tapir]

hi I was just wondering if I can find a series representation by manipulating this function

[minor tapir]

But I think something is not right lol

It’s not matching the answer, and im not really sure with my integration tbh

[cinder flower]

what’s going on in the first 3 lines?

[lofty chasm]

You sure that you are differentiating correctly?

[cinder flower]

equal signs don’t make sense there

[west vault]

It seems like they integrated instead

[lofty chasm]

Kinda but even then

[west vault]

I think you're better off starting with $\frac{1}{1 - x} = \sum_{n \geq 0} x^n$, and manipulating both sides to get what you want

[jolly parrot]

jewels!

[pearl pond]

@minor tapir Has your question been resolved?

[pearl pond]

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[calm pike]

why it is hard to find sources which write clearly about second linear differential equation. Can someone suggest me please?

[pearl pond]

@calm pike Has your question been resolved?

[pearl pond]

@calm pike Has your question been resolved?

[inland ivy]

Khan academy is always a good go to

[verbal whale]

Also "Paul's online math notes", I believe

Even though honestly I don't remember if they cover diff eq as well

[calm pike]

yah but in this section Paul not attach practice...

[plush bramble]

This isn't a math question. Just go Google resources and books

[pearl pond]

@calm pike Has your question been resolved?

[pearl pond]

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[pearl pond]

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[unique viper]

This doesn't appear to be orthonormal. <x,x> = 0 for rho2, rho3, rho4, rho5
My calculations may be wrong.

[unique viper]

<p2,p2> = 1 + 7z + 7z^2 + 3 + 3 = 7(1 + z + z^2) = 7(0) = 0
<p4,p4> = 9 + 3(-3) = 0

Should be (21 without normalisation) 1 for both of these.

Thanks

Fair - I thought I was just being blinded by the fact it's 4am

Ahah

chatgpt is often wrong

Don't think that's really in the spirit of this channel 😆

Fair though. Bedtime

.close

[pearl pond]

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[unique viper]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[pearl pond]

@unique viper Has your question been resolved?

[west bluff]

What's your question? And also, which representative have you picked for that semidirect product?

[unique viper]

The character table aligns with one I was given for a question
The rows do not appear to be orthonormal (see my working) wrt the standard character inner product.

My question is: is my working wrong?

For reference,
21<p1,p1> = 1 + 3*1 + 3*1 + 7*1 + 7*1 = 21
Which is what we want for <p2,p2> etc.

We also want <pi,pj> = 0 for i =/= j

[west bluff]

Which course is this problem in? Because, while I'm familiar with that semidirect product, I don't recall it being endowed with an inner product or a notion of normality

At least, not when I saw it

[unique viper]

Representation theory

[west bluff]

Hm, might not be able to help you then, not too well versed in that discipline, apologies

[unique viper]

An example of a table which works is here - this is the character table of S3

<V,V> = 4 + 2*1 = 6

<V,1> = 2 - 2*(-1) = 0

etc.

Not only that, but the columns should be orthonormal wrt the standard dot product

[unique viper]

works here

[unique viper]

not here

[unique viper]

No worries - could you point me towards a channel which could help? Thinking of #advanced-algebra but that's not quite rep. theory

[west bluff]

If I remember correctly, that's the channel used for representation theory, yeah

If not, you can also try in #groups-rings-fields , maybe you'll have more luck there

[unique viper]

Thank you

[west bluff]

Np, hope someone can help you

[unique viper]

I'll ask there and go to bed

[unique viper]

Thanks^^

.close

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[pearl pond]

@void grail Has your question been resolved?

[pearl pond]

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[frosty grail]

hello

[frosty grail]

i’m trying to determine whether this function under Z -> Z is injective, surjective or bijective

so the correct solution i was given is that it’s not injective but it is surjective

[toxic lichen]

have you considered whether it is in fact a function at all

[frosty grail]

hm

[toxic lichen]

also "Z -> Z is injective" is bad form

Z -> Z isnt the name of the function; f is.

[frosty grail]

what would be better to write?

[toxic lichen]

"f is injective"

[frosty grail]

i see

[toxic lichen]

can you send the original problem btw

[frosty grail]

ofc

[toxic lichen]

i wanna make sure there's no misread bullshit

[frosty grail]

wait that’s a really weird function

isn’t that just a vertical line?

[toxic lichen]

oh well you went and obliterated the floor function brackets.

[frosty grail]

that does NOT pass the horizontal line test

[toxic lichen]

what vertical line

do you know what the notation $\floor{x}$ refers to?

[jolly parrot]

Ann

[frosty grail]

OH

that was a floor function? i didn’t even realize that 💀

we are rounding to the smallest integer number

[toxic lichen]

RTFQ strikes again

we're rounding DOWN to the NEAREST integer.

[frosty grail]

ok

with the floor function i’m not quite sure how i’d go on about doing this

[pearl pond]

@frosty grail Has your question been resolved?

[toxic lichen]

i mean you kinda have to understand how it works on a deeper level than pushing symbols around

like floor(n/2) should clue you in that maybe it behaves in some significant way on the odd numbers vs the even numbers

[pearl pond]

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[pearl pond]

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[feral urchin]

.close

[elfin geyser]

I am unsure of the circled bit

[pearl pond]

@elfin geyser Has your question been resolved?

[elfin geyser]

<@&286206848099549185>

[pearl pond]

@elfin geyser Has your question been resolved?

[elfin geyser]

<@&286206848099549185>

[stark tulip]

i don't really know what i'm looking at but you should have the equation
[ s^2 X(s) - sx(0) - x(0) + 4X(s) = \frac{e^{-2s}}{s} ]
where i think the part you need help with is the RHS?

[jolly parrot]

THERMOONUCLEAR BOMB

[stark tulip]

or needed

with the initial values we have $$(s^2 + 4)X(s) - 2 = \frac{e^{-2s}}{s}$$ so $$X(s) = \frac{2}{s^2 + 4} + \frac{e^{-2s}}{s^3 + 4s}$$

[jolly parrot]

THERMOONUCLEAR BOMB

[stark tulip]

the first part of the rhs on the second eq is clear

i think you have it from here?

@elfin geyser

[elfin geyser]

I am confused because according to the second shift theorem, this is the transform of the heaviside function, but is f(t-a) just 1?

[stark tulip]

yah, exactly

[elfin geyser]

oh, ok, thanks!

.close

[pearl pond]

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[carmine marten]

close enough

[pearl pond]

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[indigo stone]

On the BC side of triangle ABC, point D is taken so that BD/AB = DC/AC. Prove that AD is the bisector of triangle ABС

[proper nova]

hold up

[earnest warren]

What can we use? are trigs allowed?

[proper nova]

congruency alone won't solve this

[indigo stone]

Only basic ones. Sin, cos, tg

[proper nova]

angle measures

[orchid geode]

.

[indigo stone]

How did u find this

[proper nova]

i drew my own diagram on geogebra

your measures might be different

[indigo stone]

well there arent any measures

triangle is arbitrary

[still hamlet]

won't it?

since the third side is common

[proper nova]

AD you mean?

[still hamlet]

yeah

[proper nova]

how does a triangle which has full acute angle

congruent to a triangle which has 1 obtuse angle

explain

[still hamlet]

AB/BD=AC/DC=AD/AD=1

[still hamlet]

no angles have been given

[proper nova]

which it isn't

[proper nova]

then draw a diagram, and look using your eyes

[earnest warren]

extend AB, E is in AB so that AE=AC

This's a big big hint

Reason behind it is because we are given BD/AB=DC/AC

It looks like SAS case in similar triangles

[proper nova]

no ratios have been given 1 which is according to your logic

[earnest warren]

$\frac{BD}{AB}=\frac{DC}{AC}=\frac{BD+DC}{AB+AC}$

[jolly parrot]

Alexis_Fx

[still hamlet]

oh shit i forgot the two ratios are not equal to AD/AD my bad

i just assumed they were, without realising there's no reason for that

[earnest warren]

bruh wait I meant SAS

[pearl pond]

@indigo stone Has your question been resolved?

[pearl pond]

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[sharp smelt]

Let $v$ be an eignvector of $T$. Then $T(v)= \lambda v$. $\phi(T(v))= \lambda \phi(v)$]

[jolly parrot]

wai

[sharp smelt]

this doesn't help much I suppose

<@&268886789983436800>

[brisk scarab]

(xristina spam)

[sharp smelt]

I suppose what I did was wrong?

[brisk scarab]

I mean it didn't get you anywhere useful that I can see

[sharp smelt]

yea

I need to start with an eigenbasis and assume F=C

idt this is doable in F=R

ykw I'll do this tomorrow

I'm too sick ( atually) to do this anymore

.close

[pearl pond]

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[sharp smelt]

Tysm

and so sorry

[sharp smelt]

wait I just use the isomorphism b/w V and V'?

[pearl pond]

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[icy patio]

i have 2 equations: x + y = 50 and xy = 87, i need to solve for x and y in simplest form(not fraction) what i'vev done is made new equation, y = 50 - x and then substituted for y, x(50 - x) and then simplified, 50x - x^2 = 87, last i made it a quadratic equation, 50x - x^2 - 87 = 0 but i dont know how to finish with the formula, i do know it is: -b +or- sqrt(b^2 + 41c)/2a but cant figure out what a b c is

[slim forge]

Hello

I am here to help @icy patio

[icy patio]

so what is a b c i can do the formula by meselfbut cant figure what a b c is

[toxic lichen]

rearrange it so the terms are in descending degree order

-x^2 + 50x - 87 = 0

[plush bramble]

You can just help yenno

[toxic lichen]

now can you see what a, b and c are?

[slim forge]

Sorry I was helping another person

[toxic lichen]

i do know it is: -b +or- sqrt(b^2 + 41c)/2a
this is badly written

also actually incorrect

the stuff under the root is b^2 - 4ac and not b^2 + 41c

[icy patio]

i meant to do 4ac

[plush bramble]

Then stop spamming everywhere and just help

[toxic lichen]

and then also brackets

[slim forge]

what do you think I am doing

[plush bramble]

.

[slim forge]

well yk I have 12 other tickets too

so im replying to this guy then this

[plush bramble]

Spamming "I'm here to help" in 10+ channels is not useful.

[slim forge]

Well I am letting them know

So they do not go afk

[plush bramble]

You're just flooding their channel instead of helping. Stop

[toxic lichen]

apparently i blocked this guy but i have no recollection of what for

[verbal whale]

You're not the only helper here 😅 there are at least 20 online I believe

[verbal whale]

Ohh that explains it

[slim forge]

who is she anyway

you can keep me blocked

[icy patio]

ok actually what is the value of a b c

[slim forge]

ok let me help

[toxic lichen]

rearrange it so the terms are in descending degree order
-x^2 + 50x - 87 = 0

[slim forge]

let me solve the equation

[icy patio]

ok

so is a = -x^2, b = 50x and c = -87

[slim forge]

Ok I’ve solved it

[toxic lichen]

no

a, b and c never involve x themselves

b is just 50, not 50x

a is -1, not -x^2

[slim forge]

A is -1 and b is 50 c, -87

[icy patio]

why is a = -1 from -x^2

[slim forge]

Oh Mb it’d -87

[toxic lichen]

-x^2 = (-1)x^2

[verbal whale]

$$ax² \quad \quad -x²$$

[jolly parrot]

Alberto Z.

[icy patio]

oke

[verbal whale]

Can you compare these?

a is the number (coefficient) multiplied by x²

[icy patio]

oh

so a = -1, b = 50 and c = -87

got it

so finally sub each value to get ansewr: $-50 + $sqrt($50^2 - 4*$-1*$87) &/ &2*-2

[jolly parrot]

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

[icy patio]

or *-1 at end

to get answer

.close

[pearl pond]

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[pearl pond]

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[midnight haven]

Hi! I'm trying to find out how many digits 2^n has, I've kinda played around and got the ceiling of log_10 2^k, would this be the expected answer?

[toxic lichen]

digit count of N generally equals floor(log_10(N))+1, but for all n that aren't perfect powers of ten this does match up with ceil(log_10(n))

so powers of two are not very special aside from never coinciding with powers of 10

[midnight haven]

Ok thanks !

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[exotic gale]

i need help with number theory especialyl modular airthmetic i dont understnad a thing

[exotic gale]

what is multiplicative inverse

[final shore]

we say b is the multiplicative inverse of a if a x b = 1

[autumn fossil]

So for example in mod 5, multiplicative inverse of 3 is 2. Because 3 * 2 = 6 = 1 (mod 5)

[final shore]

where 1 is the multiplicative identity

[exotic gale]

is mod just like base. so mod 5 is base 5

[final shore]

absolutely not

[autumn fossil]

Not at all

[exotic gale]

no like the digits they use no?

[final shore]

I see what you are thinking here

before I try to explain this

what level of math are you at

[exotic gale]

mathcounts middle school competition

[final shore]

oh lol

okay then

as a simplified explanation

just think of a mod b as the remainder of a when divided by b

if we take the number 10 and divide it by 7, we get 1 and a remainder of 3

so 10 mod 7 = 3

[exotic gale]

so 12 mod 5 for instance equals 2 mod 5

ok i understand it now thanks

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[pearl pond]

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[pearl pond]

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[pearl pond]

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

@coral jackal Has your question been resolved?

[pearl pond]

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[fair comet]

This is what I have rn

but epsilon1 is supposed to be greater than 0

which rn my two solutions are not

[cinder flower]

is this supposed to be your proof, or are you just doing work to figure out how to write it?

[pine jay]

You can avoid this by assuming additionally that delta is less than 1

[fair comet]

trying to write the proof

altho I've found myself stuck

[fair comet]

not sure what thatd do tho? can't seem to get any part into |x-4|

[pine jay]

Oh sorry I misread

Okay so you can do |f(x) - 7| < 1 whenever 0 < |x-4| < some delta’

So with ur delta_1 from
before, choose delta = min(delta_1, delta’)

Obviously you should probably use some other notation if you’d like for this delta’

But notice what this lets you do

@fair comet lmk if that’s unclear or not

[pine jay]

Sorry saying to choose delta this way is obviously wrong since you want to find it, but hopefully it’s clear how you can find it now

[fair comet]

I don't think this is what you meant but I think it works?

[pine jay]

Hm, not clear what happens here for me

Could you try and point out what youre doing

And why

[fair comet]

if I choose epsilon1 to be the min of 1 and epsilon/16 then
for epsilon1=1 then 6<f(x)<8
plugging in upper bound of 8 for f(x) and epsilon/16 reduces to epsilon which proves the limit?
I'm not entirely sure if I can choose epsilon to be multiple values the way you can with delta tbh

[pine jay]

I see that looks good

[pine jay]

You should preferable instead do this from the choice of delta, i.e you know e.g. that there’s delta_1 such that |f(x) -7| < 1, and there’s delta_2 such that |f(x) - 7| < e/16 whenever 0 < |x-4| < delta_1 and delta_2 , i.e whenever 0 < |x - 4| < min(delta_1, delta_2)

[fair comet]

right that makes sense, but the issue is I don't know how to use delta in my proof as I can't get any part to be in terms of |x-4|
I might be missing something really obvious I'm not sure

[pine jay]

That’s fine

You don’t have to

You know how to bound |f(x) - 7| and that’s enough

Try and write it all up formally

And notice you’ve done exactly what was needed for the definition of a limit

I’d argue a bit differently for how to bound |f(x) + 8| since you already set that up before nicely imo

[fair comet]

ohhh ok thanks :)

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[pearl pond]

@coral jackal Has your question been resolved?

[dense jasper]

did you see my message from last time

[coral jackal]

Yes

But I don’t rlly see how that helps in solving this

I get that it’s derivative is this graph given here

[dense jasper]

mhm

What can you say about $g$ being increasing/decreasing on each of $(a,c)$ and $(c,b)$?

[jolly parrot]

Civil Service Pigeon

[pearl pond]

@coral jackal Has your question been resolved?

[pearl pond]

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[hollow sonnet]

$r={ (x,y) \in \mathbb{R}^2 \mid y=tx^2+8x+10-t, t \in \mathbb{R}}, {A = {t,$ where t makes R_r \subset \mathbb{R}+}$. B=(-2,4). Find A' \cap B.

[hollow sonnet]

R_r is defined as the range of function r

[naive zinc]

What is A prime?

[jolly parrot]

d1 shitposter
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[hollow sonnet]

defined as the complement of A

all R not A

computed t to be the real numbers ranged (2,8)

[naive zinc]

$r={ (x,y) \in \mathbb{R}^{2} | y=tx^{2}+8x+10-t, t \in \mathbb{R} }, A = {t ,$ where $R_{r} \subset \mathbb{R} }$. $B=(-2,4)$. Find $A^{c} \cap B$.

[hollow sonnet]

so A' would become R-(2,8)

[jolly parrot]

Cogwheels of the mind

[hollow sonnet]

complement of A union B should be uhhh

(2,4) if im not wrong

[naive zinc]

$y=t((x+\frac{4}{t})^{2}+\frac{10-t}{t}-\frac{16}{t^{2}})$

[jolly parrot]

Cogwheels of the mind

[naive zinc]

R^+ is positive reals or non-negative reals?

[hollow sonnet]

pretty sure it's positive reals

includes zero

?

[naive zinc]

?

Does it include 0 or not

And what is your answer?

[hollow sonnet]

not defined within course syllabus

so it should be assumed to include zero

[naive zinc]

Anyway, I got A=[2,8] / (2,8) depending on R^{+} include or not include 0

[hollow sonnet]

ic

answers match

nice

[naive zinc]

Okay

[hollow sonnet]

.solved

[pearl pond]

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[pearl pond]

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[fallen lotus]

What do we mean by this?

[fallen lotus]

isnt y=cx-c^2 a straight line?

y' = c

im so confused

[bronze heath]

which one?

[feral pasture]

yea

ur right that for each fixed c that funciton is a straight line

but

[fallen lotus]

16

[feral pasture]

what i assume ur confused on is the point of the exercise is to check its a solution

[fallen lotus]

no i am kinda confused that this differential equation has a straight line as one of its solution, which feels kinda counter intuitive?

and the graph

[feral pasture]

theres nothing that says a differential equation cant have straight lines as solutiuons

[fallen lotus]

the straight lines are just the slopes of the parabola y=x^2/4

[fallen lotus]

which is also a solution of this differential equation

[fallen lotus]

i mean yeah thats true

[fallen lotus]

i think ill focus on this, is there a connection/relation/anything why the slopes of the solution are also the solution to this equation?

[feral pasture]

yea if u solve the DE for y you get y = xy' - (y')^2 so it literally says the solution curve must lie on one of the lines y = mx - m^2 which is why those straight lines and their envelope y = x^2/4 all satisfy it