help-27

[worthy magnet]

not that I know of

simplify the fraction first

corss out tan x on the bottomand on top

left with (1-x cos x)/x

[wanton scarab]

ok so tanx * cosx = sinx

that's an identity

[worthy magnet]

oh shit yes

I fucking forgot

[wanton scarab]

so tanx(1-cosx)

should resolve to tanx-sinx i think

if im not brain dead

[worthy magnet]

so u read tan x(x) as tan x?

[wanton scarab]

im still tryna figure out what happens to the x

lol

[worthy magnet]

u know what let's reorganize

[wanton scarab]

but i think

tanx(x) * cosx

is just

x sin(x)

[worthy magnet]

yes

and ( x sin x )/(x tan x)

should be cos x

[wanton scarab]

so wouldnt the top be

tanx(1-xcos(x))

which should resolve to

tan(x) - x sin(x)

[worthy magnet]

yes

top is that

but consider the bottom

[wanton scarab]

yea so now

since sin(x) is tanx*cosx

since on the bottom we are dividing by tanx

i can just turn it back to cosx

[worthy magnet]

yep

[wanton scarab]

xcosx

but the x is on the bottom

so remove the x as well

[worthy magnet]

no

[wanton scarab]

tanx also divides into tanx by 1

so wouldn't it be

1/x - cosx

[worthy magnet]

yes

[wanton scarab]

ok so i've factored it but isn't the limit still undefined

since 1/x is undefined

approaching 0

[worthy magnet]

l'hopital?

[wanton scarab]

no idea what that is

2nd week of calc

[worthy magnet]

hmmm basically there is a misprint

I suspect u need L'hopital's rule

ask ur teacher

[devout snow]

@wanton scarab Has your question been resolved?

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[wanton scarab]

yea im pretty sure he misprinted

lmao

[devout snow]

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[quartz gale]

hi how to solve y = x^(2x)

[quartz gale]

im not sure if power rule is applied or what

[long kettle]

Wdym "solve"

[quartz gale]

i mean taking the derivative with respect to x

or solving for dy/dx

[long kettle]

Generally, when you have something like f(x)^g(x), it might be a good idea to take the natural log of both sides

Since there's a nice property that ln(a^b) = blna

[quartz gale]

is there a list of properties like those

i dont remmeber them

if i have a list of these i can refer to them

like same with lnx = 1/x

and logb x = 1/xlnb

like is there a list for these

i forget them

[long kettle]

lnx is not 1/x, unless you mean the derivative of lnx

[quartz gale]

i meant dertivate sorry

[long kettle]

I'm sure you could look up "log properties" and "famous derivatives" and such

[quartz gale]

so for the y = x^(2x) dont we also use implicit differentiation or nah? to solve for dy/dx

[long kettle]

After taking the natural log of both sides, you will

[quartz gale]

so lny = 2xlnx

[long kettle]

Yep

Then you can take the derivative of that with product rule

[quartz gale]

or i can also use u(x)^v(x) = u(x) ^v(x) (ln(u(x)v(x)'

[long kettle]

I'm not familiar with the formula for the derivative of u(x)^v(x)

I never bother to remember it, I just take the ln of both sides

[quartz gale]

the answer is

but im getting different

2 +2lnx

[long kettle]

Show your work

[quartz gale]

2x*1/x + 2lnx

[long kettle]

That's the derivative of 2xlnx

But don't forget that's not y

That's ln(y)

Gotta take the derivative of that as well

[quartz gale]

how do i take for that

[plush knot]

implicit differentiation

do the chain rule

[quartz gale]

im not sure how to do for ln

[plush knot]

d/dx [ln(_)] (← with inside y) · d/dx [y]

Remember that you are considering y as a function of x

try writing it for a minute like ln(f(x)) and try finding the derivative of that

and then replace f(x) with y and f'(x) with dy/dx

[quartz gale]

is it dy/dx * 1/x ?

[plush knot]

where did x come from?

[quartz gale]

derivate of ln?

no wait

[plush knot]

it's the derivative of ln(f(x)), so you should get|| 1/(f(x)) · f'(x)||, and f(x) is y

[quartz gale]

its 1/y * dy/dx

[plush knot]

Yes much better

That's it

[quartz gale]

so 1/y * dy/dx = 2+2lnx

but answer is

not sure were x^2x is from

[plush knot]

Yes the one we found is the implicit derivative

We have
1/y · dy/dx = 2 + 2lnx

[quartz gale]

how to get this then

[plush knot]

Now, bear with me

[quartz gale]

ok

[plush knot]

let's solve for dy/dx, multiply on both sides by y

dy/dx = (2 + 2lnx) · y

Yes?

[quartz gale]

correct

[plush knot]

Now, what's the value of y

[quartz gale]

ohhhhhhhhhhh

[plush knot]

[quartz gale]

thank you

[plush knot]

[quartz gale]

i love this so much better than paying money to a tutor

only thing is that it takes a bit more time but its fine

[plush knot]

I think it's not the same thing

if I pay a tutor i want them to actually teach me stuff

Here usually people just answer questions or link to material

Which is still pretty good

[quartz gale]

well usually i require answers to my doubts/questions

thank you @plush knot @long kettle

not sure how to close this chat or if it does automatically

[long kettle]

".close"

[quartz gale]

.close

[devout snow]

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[solar whale]

.close

[devout snow]

Channel closed

Closed by @solar whale

Use .reopen if this was a mistake.

[restive river]

glad we could help

[devout snow]

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[orchid pike]

I need help

[orchid pike]

I have the following data points

x1= -8 and x2= 2 and Y=(-infinity;3}

using that how do I get y= ax2 + bx + c

[regal roost]

I dont believe its possible to get a quadratic from two points?

[orchid pike]

what else would I need?

[regal roost]

Wait so just to clarify you only have 2 positions in the form (x,y) right?

[long kettle]

You have more than two points, assuming Y is representing the range

[orchid pike]

The highest point is (3,5)

[long kettle]

You sure?

Is Y representing the range?

[orchid pike]

actually I meant (5,3)

and yes

[long kettle]

x1 and x2 are roots, right?

[orchid pike]

yes

[long kettle]

So what do you know about the equation if x1 and x2 are roots?

Maybe in terms of some sort of factorization?

[orchid pike]

Are you able to solve it regardless of knowing anything else?

[long kettle]

Yes, there's enough info

[orchid pike]

could you give me the answer?

[long kettle]

No

I'm not going to just give you the answer

[orchid pike]

Could you explain it to me from the start?

So the quadratic equation is y= ax^2 + bx + c

[long kettle]

.

[long kettle]

.

[orchid pike]

So am actually trying to help a friend. I don't know the math, haven't taken algebra in some time (last year). This is their thought process, are you able to help out?

@long kettle

[long kettle]

Looks right

Wait

Hold up

Let me think rq

[orchid pike]

Ok

[long kettle]

Here's my problem

The parabola points down

So the a should be negative

Problem is, I can't find the mistake in the working out

OH

The midpoint isn't 5

She did (8 + 2)/2

But she should've done (-8 + 2)/2

[orchid pike]

Ok, lemme see if that helps her

[devout snow]

@orchid pike Has your question been resolved?

[orchid pike]

@long kettle is it -3/25

[long kettle]

Yes

[devout snow]

@orchid pike Has your question been resolved?

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[kindred agate]

hello mr "why are you even helper role" "useless fuck"

[long kettle]

I'm sensing a history here

[kindred agate]

yeah

this dude pinged helper twice

within 7 minutes

and then said i was a useless fuck

[cunning mulch]

well you didnt help did you

with the helper role?

[kindred agate]

you can become a helper today too

[long kettle]

[kindred agate]

and get 500 pings

everyday

and get annoyed by people

like you

[lone ravine]

You know none of us is getting paid?

[long kettle]

Yeah idk why anyones a helper given the endless pings

[cunning mulch]

the problem isnt getting paid

its the fact he has the role and chooses to answer like a dick

[kindred agate]

we're fucking volunteers

[cunning mulch]

?

[long kettle]

I doubt they answered like a dick. I think they responded to you clearly breaking the rules

[meager blaze]

what's ur doubt

[kindred agate]

just check my message history

in the other channel

[long kettle]

"You dumbass" seems a little harsh ngl

[kindred agate]

kind of

but that was at the very end

[meager blaze]

i doubt even mods get paid

[kindred agate]

this is a discord server ofcourse nobodies getting paid

[meager blaze]

yea

[kindred agate]

if you want to be a mod it's because you're interested in math

[meager blaze]

ig yea, everyone is here to learn

[lone ravine]

Of course not. But my point is that you can't demand a volunteer to answer a question

[kindred agate]

yeah

[cunning mulch]

i didnt demand him to answer im waiting for any helper to help with my question

if he cant help he doesnt need to respond

[kindred agate]

don't break any rules and i won't respond

take like

a 5 second look

at the rules

then come back

[cunning mulch]

omg i pinged after 7 minutes instead of 15

lock me up

[kindred agate]

you pinged twice

[meager blaze]

can u show ur working, where are u having doubt in

[kindred agate]

that's the issue

not the 7 minutes

[cunning mulch]

alright mb

[kindred agate]

anyway ima let lvn help u

[long kettle]

The problem isn't that you pinged early---people do that constantly---the real problem is you proceeded to argue

[kindred agate]

also did you block me?

[cunning mulch]

no i can see ur chat

[kindred agate]

ok

yeah that's weird

ah

ok

anyway

lvn seems like they know the answer to this

so ima let them help u

[cunning mulch]

i dont know how to find the points a and B given the magnitude

also does OA just mean that it goes from the origin to point A?

[meager blaze]

do uk how can u calculate length of segements by using coordinates?

yea

[meager blaze]

O is considered origin

[cunning mulch]

distance formula?

[meager blaze]

yup

@cunning mulch do u want help? or can u proceed on ur own

[cunning mulch]

im working through the distance formula rn but im not sure what to do after

[meager blaze]

a sec

this is the distance between 2 points

you'll get a+b=sqrt2, a^2+2b^2=3
solving them would give u a and b

[long kettle]

||OA + OB|| ≠ ||OA|| + ||OB||

[meager blaze]

so what would it be? length of OA is a and OB is b.

this isnt vector addition?

[long kettle]

Wait, OA and OB aren't vectors

Yeah, ignore me

[cunning mulch]

thank you

.close

[devout snow]

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[agile spire]

I'm having trouble starting this problem off - are my bounds 0-8? I'm not sure how to get A(x) with this given info

is A(x) just x^3 - 2?

[devout snow]

@agile spire Has your question been resolved?

[agile spire]

.close

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[signal reef]

can someone please help me with (2)

I’m not quite sure how QP + QB >= PB

I would understand if they’re equal but greater than?

[devout snow]

@signal reef Has your question been resolved?

[devout snow]

@signal reef Has your question been resolved?

[signal reef]

<@&286206848099549185>

[deep cove]

B, P, Q may not lie on the same line, hence they can form a triangle, and the Triangle Inequality applies; QP, QB, PB are all side lengths of such triangle, so we can claim sum of any two is larger than the remaining one.

[devout snow]

@signal reef Has your question been resolved?

[devout snow]

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[hushed pendant]

I have a question, how would you plot a y axis that has a zero? Would the x and z only be plotted and make a triangle of somewhat? And how would we find the final point if it's not a square or rectangle?

[devout snow]

@hushed pendant Has your question been resolved?

[hushed pendant]

<@&286206848099549185>

[uncut crow]

it's hard to mark individual points like that on a 3d plot but it might help for visualization if you pretend the y axis isn't there

[hushed pendant]

So just ignore the ya axis

Y

[uncut crow]

yea you can do that in this case

[hushed pendant]

Then how would we connect it?

[uncut crow]

wait what does the question ask?

[hushed pendant]

Let me show you the example my teacher did. It's shows what she asks for.

Like this

[uncut crow]

oh weird ok

[hushed pendant]

It's like she made a rectangle and then plotted a point. I honestly nvr grasped this

[uncut crow]

then yea, triangle sounds ok

or uhhh

more like a rectangle

[hushed pendant]

So we would just plot the x and z?

How?

Like this is what I did

[uncut crow]

something like this maybe?

[hushed pendant]

Ahhhhh okay soooo how did you get the other dot like the one floating in space?

Like how did end up in -2 z

[uncut crow]

it's kinda hard to see

hmm how to explain

[hushed pendant]

I know 😭 it's just confusing but yet so simple I think

[uncut crow]

if you go down 3 on the z axis and then out 3 in the negative x direction

you'll end up at the red dot

even though it kinda looks the z position is -2 now

it's not really

[hushed pendant]

Like an optical illusion kinda?

[uncut crow]

not sure about that

[hushed pendant]

Okay I'm starting to see how yiu got that point ...

[uncut crow]

yay

[hushed pendant]

But I don't feel confident doing these problems

Can we do another and this time can I tell you how I would work it out and then you tell me if I do it right?

[uncut crow]

sure

[hushed pendant]

Ok thank you smch

[uncut crow]

here, -1 in the x direction is in red, 4 in the y direction is in blue, and 3 in the z direction is in green

[hushed pendant]

Seeing it in colors makes it much easier. It like the numbers work as directions to make the point

[uncut crow]

the numbers are more like how much to go

x y z is for the directions

[hushed pendant]

So for this one we would ignore the x because it has a zero and go left on the y axis and 3 up on the z

[uncut crow]

yep

[hushed pendant]

So the dot would be here?

Like a normal graph kinda because the x is not there anymore?

[uncut crow]

yea that looks good enough

[hushed pendant]

Ok I think I'm getting better now

😀

[uncut crow]

woooo

[hushed pendant]

Thank you for your help. Your really nice and patient. Sorry if I bother to much in math.

[uncut crow]

awww thank you so much

you don't bother me at all

[hushed pendant]

Well that's nice to hear. Hope you have a nice day. Bye 👋:>

[uncut crow]

you too ^-^

[hushed pendant]

.close

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[restive river]

Really simple question but I cant seem to find an answer on google. What is the inverse function of a -1 exponent?
I know to solve something like (x^-1) = 1/2 you would flip it, because a negative exponent is kinda just a fraction indicator. But what is the exact thing that's used to inverse it?

[pseudo basin]

the function f(x) = x^-1 is its own inverse actually

[devout snow]

@restive river Has your question been resolved?

[restive river]

oh

thanks!

[devout snow]

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[devout snow]

@frail spruce Has your question been resolved?

[frail spruce]

.ckise

.close

[devout snow]

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[frail spruce]

solved it

ty

tho

[devout snow]

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[deep fable]

if x was -4-2√2/2 do i simplify by dividing -4 and 2? or do i just keep it as the same

[pseudo basin]

do you mean $-4 - \frac{2\sqrt{2}}{2}$

[woven radish]

Ann

[deep fable]

yes

[pastel belfry]

take lcm

then take 2/2 as common

[pseudo basin]

...overcomplicated

to simplify this just cancel the 2s in the fraction

and get -4 - sqrt(2)

[deep fable]

oh ok

thnks

[pastel belfry]

turns out to be

[pastel belfry]

ye i was about to tell the same thing lol

[deep fable]

.close

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[tiny rain]

would this be right for part a?

i kinda feel that the way im reading the diagram is wrong

[arctic field]

that theres an equal number of equations and variables doesnt mean its enough to fully determine the system

and how did you get those equations?

@tiny rain

[tiny rain]

hi

oh um

so the first row was obtained by taking the road from 610 to 640

so x4 + x1 + x2 = 640 - 610

[tiny rain]

can i ask why?

[arctic field]

what about the 450 and 310 coming in

[tiny rain]

oh right

[arctic field]

we discussed this in the question you asked the other day

[tiny rain]

sorry whoops

[arctic field]

we explicitly constructed counterexamples to this

[tiny rain]

uhhh

this?

[arctic field]

yes

we constructed a system for (d)

$\begin{bmatrix} 1 & 1 \ 1 & 1 \ 1 & 1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}$

[tiny rain]

ah wat

[woven radish]

snow

[arctic field]

this is one example where you have 3 equations and 2 unknowns

but the 2 unknowns are not uniquely determined

[tiny rain]

um wait

can we express that in the normal syntax form

lol

[arctic field]

i wouldve thought this is more normal but um

ill see if i can find out how to put bars in matrices

cant remember

[tiny rain]

$\left(\begin{array}{cc|c}
1 & 1 & 0\
2 & 2 & 0\
3 & 3 & 0
\end{array}\right)$

[woven radish]

HellO

[tiny rain]

i found this

[arctic field]

yeah that

thatll work

[tiny rain]

a ha

[arctic field]

[
\brk {
\begin{array}{@{} c c | c @{}}
1 & 1 & 0 \
1 & 1 & 0 \
1 & 1 & 0
\end{array}
}
]

[woven radish]

snow

[tiny rain]

right

so this wouldnt give us an answe

okay that makes sense

[arctic field]

definitely not

[tiny rain]

ahh i see

okay so

how would i read that picture lol

[arctic field]

for example

just taking this intersection

you would have

$x_3 + x_4 = 520 + 480$

[woven radish]

snow

[tiny rain]

AH

that makes

SO MUCH more sense

okay

imma get to it

[tiny rain]

harro does this make more sense

then to find out whether we would have enough information, i will convert it to reduced row-echelon form?

[arctic field]

that 1 shouldnt be there

but yes

[tiny rain]

gimme 1 min

[arctic field]

do row reduction and see if its invertible

[tiny rain]

ok im back

[tiny rain]

whats invertible?

[arctic field]

okay dw about it if you dont know

[tiny rain]

is ti this?

[arctic field]

just means if theres a unique solution

its only unique if rref is identity

[tiny rain]

oh so it is

[arctic field]

is it

i dont believe it is

[tiny rain]

oh

[arctic field]

,w rref{{0, 0, 1, 1, 1000}, {0, 1, -1, 0, -270}, {1, -1, 0, 0, 330}, {1, 0, 0, 1, 1060}}

[tiny rain]

are u referring to this

[woven radish]

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[arctic field]

yes

[tiny rain]

oh so theres infinite solutions

[arctic field]

yep

cuz rref isnt identity

[tiny rain]

so we cant solve it

[arctic field]

nope

we cant

[tiny rain]

ohhh

so we can only solve it iff theres only 1 solution?

[arctic field]

well

we can solve it

but the solution isnt unique

[tiny rain]

umm wym by unique

like every variable must be unique?

[arctic field]

as in

[arctic field]

here

the solution looks like

\begin{align*}
x_1 + x_4 & = 1060, \
x_2 + x_4 & = 730, \
x_3 + x_4 & = 1000
\end{align*}

[woven radish]

snow

[arctic field]

the last row is all 0s so it tells us nothing

and we can see that the solution depends on the value of x_4

we could set x_4 to anything

[tiny rain]

right

so its unsolvable

sort of

[arctic field]

its solvable

but only in terms of the "free variable" x_4

[tiny rain]

oh wait

ohhh

this question is asking us for the exact values of all 4

so "solvable" yes

but to find exact values of all variables, no

[arctic field]

yes

[tiny rain]

ahHhh

[arctic field]

the question wants us to determine x_1 x_2 x_3 x_4 uniquely

but we cannot

so the answer is no we cant do it

[tiny rain]

rightt

ok lemme work it out

dont go pls XD

@arctic field ummm

i got this

[arctic field]

that

is not what wolfram got

awks

[tiny rain]

sad

[arctic field]

did you correct this

the 1 should be a 0

[tiny rain]

yes

[arctic field]

well

i got nothin

[tiny rain]

sad moments

[arctic field]

oh gosh

[tiny rain]

did i make a mistake

[arctic field]

i really hate reading rref working

its just voodoo

[tiny rain]

sorry

[arctic field]

no its fine

that was just a general complaint towards how linalg is taught to students

[tiny rain]

oh wot

how would u recommend

learning rref

[arctic field]

theres a formulation of row reduction where you can use pretty specific matrices to do the algorithm in a deterministic way

where the equals signs arent completely wrong

dw about it

[tiny rain]

o.o

lemme check as well

[arctic field]

this is a subtraction

so youd get 1060-330

[tiny rain]

omg

lets just check the REF form first

before rref

[arctic field]

except that also looks wrong

because even in REF you'd get a 0 row

[tiny rain]

sad

[tiny rain]

ah

ye

[arctic field]

the 1120 seems sus because you shouldve subtracted

[tiny rain]

yes

[arctic field]

and the last row also seems sus

because that shouldve been an addition

[tiny rain]

lemme just redo from the third step

AH

YES

,w rref{{0, 0, 1, 1, 1000}, {0, 1, -1, 0, -270}, {1, -1, 0, 0, 330}, {1, 0, 0, 1, 1060}}

[woven radish]

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[tiny rain]

nvm

man

[arctic field]

this looks almost right

just turn it into rref form

[tiny rain]

how did i totally forget about rref

im stupid

umm

YES

ill just get rid of the -

ok nice

yay

[arctic field]

amazing

[tiny rain]

this is the same as being high on drugs

anyway

so now for part ii

its just substituitutititign

[arctic field]

yes

[tiny rain]

v nice

gimme a sec

[arctic field]

except a couple extra tuis seem to have crept into your spelling 🤔

[tiny rain]

yes

drug effect

[arctic field]

ah

ofc

understood

[tiny rain]

lemme do dis

one secs

[arctic field]

wait

your first row is missing a 1

um

[tiny rain]

oh wait

theres a 1

nvm

careless

[arctic field]

youve found it?

good

[arctic field]

LOL

[tiny rain]

huh

why is it like this

sadge

[arctic field]

LOL

[tiny rain]

supposed to be like this

[arctic field]

youll have to wrap it in ```

[tiny rain]

                                .---..---.   '   _    \   
   .              __.....__     |   ||   | /   /` '.   \  
 .'|          .-''         '.   |   ||   |.   |     \  '  
<  |         /     .-''"'-.  `. |   ||   ||   '      |  ' 
 | |        /     /________\   \|   ||   |\    \     / /  
 | | .'''-. |                  ||   ||   | `.   ` ..' /   
 | |/.'''. \\    .-------------'|   ||   |    '-...-'`    
 |  /    | | \    '-.____...---.|   ||   |                
 | |     | |  `.             .' |   ||   |                
 | |     | |    `''-...... -'   '---''---'                
 | '.    | '.                                             
 '---'   '---'                                            ```

a ha

nvm

ok anyway we did it

reeeeeeeeeeeee

tysm boss

[arctic field]

great

[tiny rain]

i will go now

thanks alot

.close

[devout snow]

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[restive river]

I was trying to calculate the limit of this function, when χ goes to 1, and the function was in an indetermined form.

The function that's with the square root.

[restive river]

The result: ^

[hallow hazel]

@restive river use l hopital

[restive river]

I did, and I got rid of the indetermined form, but it's one over zero now, and the interval is: ]1, positive infinity [

So the limit has to be from the right side of the interval, right?

Higher values than one, right?

@hallow hazel

Because it usually comes with that case, when χ is lower than 1.

[hallow hazel]

that's given yes

[restive river]

But it could not be lower than one, because the interval.

So, the limit will be positive infinity?

@hallow hazel

[hallow hazel]

what is a

limit x tends to a?

[restive river]

Yes, now I'm doing it x to 1.

When x tends to 1, lim f(x) = positive infinity?

After getting rid of the indetermined form.

@hallow hazel

[hallow hazel]

it's not defined

[restive river]

Yes, but isn't that 1/0 = infinity?

[hallow hazel]

yea

[restive river]

Then what's the limit?

Of f(x)

[hallow hazel]

1 min

[restive river]

No problem.

[hallow hazel]

at 1 it's infinity I think

[restive river]

Yes.

Got it.

[hallow hazel]

or not defined

[restive river]

Final results?

Sorry the pic is flipped

[hallow hazel]

,rotate

[hallow hazel]

yes

[restive river]

Correct?

[hallow hazel]

think so, I will verify now

@restive river correct

sorry I was little busy

[restive river]

No worries.

And thank you a lot

[hallow hazel]

and yes it's correct

at x--> infinity it's 0

[devout snow]

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[jovial blaze]

How to find sec(x)= - square root of 2

[jovial blaze]

I was told there is two answers but I'm confused since the domain and range of any inverse is restricted to Quadrant 1 and 2

[long sundial]

so first of all we dont like working with the reciprocal functions when it reaches this point ie. trig function = constant

so first change this equation to something equivalent that involves a sin, cos or tan instead of secx

[jovial blaze]

cos(x) = - square root of 2

[long sundial]

close

-1 <= cosx <=1

and $-\sqrt{2} < -1$

[woven radish]

Max..

[long sundial]

so clearly youve made a small mistake somewhere

what is secx = ? in terms of cosx

[jovial blaze]

It says to give the answer on an interval of bracket 0, 2pi bracket

[0,2pi]

@long sundial

<@&286206848099549185>

[devout snow]

@jovial blaze Has your question been resolved?

[long sundial]

yeah

[long sundial]

first of all

what is the definition of secx?

[devout snow]

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[reef pebble]

.

[reef pebble]

so this is the problem i am doing and I have what i believe is the answer but there is one step i apparently messed up at

so what i've done is convert tangent into sin/cos and then cancelled it giving me
int of cos^6xsin^3x

I understand the next step is to do substitution of sin and get only cos where u = cosx and du = sinxdx

but once i get to cos^6x sin^2x sinxdx is where i get confused

i know you need to convert the sin^2x into cos^2x using sin^2x + cos^2x = 1 so when i subtract cos^2x on both sides i get sin^2x = 1 - cos^2x

but the solution has it as cos^2x - 1

so I wanna know why cos^2 -1 is the way it is

[meager blaze]

the solution also took a negative outside so its -(1-cos^2x)

[reef pebble]

ah i didnt notice that

any reason of why they would do that?

[proven minnow]

du = -sinxdx

[proven minnow]

so that its evident that -sinxdx will be du

[reef pebble]

ah word thank you so much

that makes sense now

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[loud dirge]

For a polynomial p(x), the value of p(3) is −2.
Which of the following must be true about p(x) ?
A) x − 5 is a factor of p(x).
B) x − 2 is a factor of p(x).
C) x + 2 is a factor of p(x).
D) The remainder when p(x) is divided
by x − 3 is −2

[loud dirge]

my SAT test is a few weeks ahead and i still don't know how to solve this 😦

[winter patrol]

read up on factor theorem and remainder theorem

[loud dirge]

i don't get your point yet

[winter patrol]

did you read up on what i suggeted?

[loud dirge]

give me a minute

[loud dirge]

i got your idea man

D is the correct answer right ?

[winter patrol]

yes

[loud dirge]

okay

thanks for your help

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[inland hinge]

Hey, I've been wondering about this problem for a while and I don't have any clear solution in mind (note: it's not a problem from any olympiad, i was just wondering if there is some bound).

You are given a set of n: a_1, a_2, ..., a_n different non negative integers. Each integer is not greater than x. What is the number of different sums when you sum each integer with every other. More formally, if you'd take a set {a_x+a_y | x, y <= n} and remove duplicates what would be the size of it?
It doesn't have to be exact bound, I just need it in big-O notation. (obviously, it's O(n^2) but I'm wondering if there is any better bound)

[stone stump]

what do we know about n vs x?

I assume in general all of these could be distinct assuming x is large enough

[inland hinge]

we know nothing about them (if it makes it easier you can e.g. assume that x > n or sth like that)

[inland hinge]

yes, that's true if x was infinitely large then we could take 2^0, 2^1, 2^2, ..., 2^(n-1) and all pair sums would be different

[inland hinge]

actually, if you want you can assume that x < 10^8

maybe that'll be easier

[stone stump]

well ok that's still pretty large

unless n is also around that

[inland hinge]

n is < 10^5

[stone stump]

which is essentially nothing compared to 10^8

[inland hinge]

yeah

can you maybe think of a worst case scenario for the greatest number of different pair sums?

[stone stump]

worst case is probably if we have the numbers 1, 2, 3, ..., n. then the sums are between 3 and 2n-1

[inland hinge]

hmmm

so it's kind O(n)-like

[devout snow]

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[neon aspen]

Which part do you want help with

[safe trench]

part b

im not sure how to formally prove altho i can somehow visualise it?

[neon aspen]

You show it's a subspace like you normally would

Show that it's a subset and is closed under vector addition and scalar multiplication

[safe trench]

yepp

proof b is similar to a right

[neon aspen]

Yes

[safe trench]

can we just say that the elements of the sets exist in V hence they are closed under vector add n scalar multp of V

[neon aspen]

Well yea but you need to show that it's closed within W1 n W2

Let a,b be vectors in W1 n W2

Because that's a subset of V, a and b are in V and because V is a vector space (given) a+b is in V

But you want to show that a+b is in W1nW2

[safe trench]

shucks i was j gna use an element starting from the set of W1nW2

[safe trench]

im not too sure on how to go abt doing this

[neon aspen]

To show that it's in W1nW2 you need to show it's both in W1 and W2

Because its an intersection

[safe trench]

Yessir

Do we use the fact that W1 and W2 are subspaces of V...

[neon aspen]

Yea you use the fact that they're vector spaces

So they're closed under va and sm

[devout snow]

@safe trench Has your question been resolved?

[devout snow]

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[heavy terrace]

"Find the lowest value for the function y and the value x for which y=2x^2+x"

[heavy terrace]

so the answer is y=-1/8 for x=-1/4 but I wonder

how do I get to the answer?

[main gull]

The minimum of the function?

Just find the vertex

[plush knot]

Find the lowest value for the function y
What function?

[main gull]

@plush knot

[heavy terrace]

y=2x^2+x

[heavy terrace]

that's it?

[main gull]

Yes

That's it

The vertex is going to be the min/max of the parabola

[heavy terrace]

finding it is with -b/2a

right?

[main gull]

That's for the x coordinate

[heavy terrace]

yeah yeah

[plush knot]

oh mb, I read something wrong, yes, dldh is right

[heavy terrace]

then I substitute

[main gull]

Yes

[heavy terrace]

ty a lot @main gull

forgot it, it was in my textbook as well

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[dire slate]

P: For every even integer n, there is an odd integer m such that n + m is odd.
Q: There is an odd integer m such that for every even integer n, n + m is odd
a. Do these statements mean the same thing? If not, explain the difference.

[dire slate]

b. Write the negations of P and Q

Thats a question related to foundation of mathematics

for a. i put i. No, the order of the n and the m have been switched changing the meaning of the statement.

for b. i. There exists an odd integer n, for every even integer m such that n+m is even.
ii. For all even integers m such that there exists an odd integer n, n+m is even

I was wondering if this was correct

?

[devout snow]

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[devout snow]

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[tough perch]

Hello do you know how to represant sin(x+y+z)=1 in R3 ?

[tough perch]

I didn't know at all how to do with sinus

It's just the cartesian equation plane : x+y+z= pi/2 +2.k.pi ?

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[digital egret]

Im having trouble one this

Im trying to use a matrix but i ended up with 0 0 0 0 on the bottom

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[rapid harness]

Hey, I don't understand what the question is asking, I am not sure how to relate binary strings to the power set of n integers

[rapid harness]

Could someone show me an example as to relating a binary string to a power set

[topaz axle]

ugh

{1,2,3}
0 0 0 {}
0 0 1 {3}
0 1 0 {2}
0 1 1 {2,3}
1 0 0 {1}
1 0 1 {1,3}
1 1 0 {1,2}
1 1 1 {1,2,3}

i would present it as proof, what do they want me to say

(that's why i said ugh)

[rapid harness]

ah

can I do this proof and get back to you with the solution, can u lmk if it looks alright

[topaz axle]

okay

[rapid harness]

thanks for the example, i was confused to how they related

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[devout snow]

Please don't occupy multiple help channels.

[restive river]

For this question

to be honest I don't know how to approach it

i don't understand the relationship of these regions to lines

can you identify which line is which in the graph from the equations

oh right I had done that, forgot to mention it my bad

so a line divides the plane in 2 regions right ?

[restive river]

right

this makes sense

ahh

yes that is correct

so 3 lines makes different regions

alright

[restive river]

wel it depends on how they intersect

oh right

must take that into account

so I'm guessing it's a region created by 3 lines intersecting

so it must either be F, D, B, C

for the line y=mx+c
y> mx+c represents one of the region and y<mx+c another

[restive river]

i don't quite get this, if y = mx + x, how can there be a region where one is smaller than the other if they're both equal to each other

so which region does y< x(1/2) +1 represent the above or below ?

[restive river]

well i'm guessing below

So it would be the lowest one

F

pure guesswork though

[restive river]

just substitute a point if it satisfies the inequality then the region represented by the inequality contains that point if it does not satisfy the inequation then the other region is represented by the inequality

Ah I see

Nevermind I wasn't being very intelligent I get it now

with respect to first line, where does the origin lie above or below ? does it satisfy the inequation ? what can you conclude ?

I see so this is the thought process

Using the origin

yes (0,0) is easy to substitute

but you can use any known point

aight so if i understand correctly, when mx + c > y it means that the point is further from the y axis to the left/right then it is from the x axis to up/down

[restive river]

uh what point exactly, every point of that region ?

not entirely sure

yeah

about that

well isn't it intrinsic for example for a theoretical point 4, 3) it must be that mx + c > y because it is 4 points away from the y axis whereas only 3 points away from the x axis

disregarding the fact that 4 > 3, if we only look at how far they are from the axises

well what if m or c is negative

you could have said this if the equation was like y>x

oh yeah didn't think about that lol

thank you for your help dude

now I understand it

better prepared for the Exam

have a good day

welcome

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[devout snow]

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[lost wedge]

how do i prove this

basically i need to show that P has q linearly independent rows right

how do i do that

[devout snow]

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[devout snow]

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[wraith aurora]

90% sure I’m right but just wanna make sure I have the correct limits

[wraith aurora]

For when x is approaching infinity it’s 2 and when it’s approaching negative infinity it’s 0?

Just from looking at the graph as the problem is not showcased in the photo

<@&286206848099549185>

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[devout snow]

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[gentle ridge]

I think i put my problem in a occupied lobby sorry but I am need help if anyone wants to help me.

[gentle ridge]

thank you for reading and the problem is a integration by parts problem.

[violet wind]

Ah

So for integration by parts

u gets differentiated to become du

ah that's what that does huh

and dv gets integrated to become v

[gentle ridge]

I dont get how many times I am suppose to integrate.

[violet wind]

for that reason you usually want to choose u as something that simplifies when it gets differentiated

[gentle ridge]

oh so dont use sin?

[violet wind]

yeah

you should use the polynomial part as u

[gentle ridge]

wow that sounds easier actually Let me try that instead lol did not even think to swicth them

[violet wind]

you'll definitely still need to do it a few times

[gentle ridge]

ok good that sounds much better lol.

[violet wind]

it's a tedious problem for sure

[gentle ridge]

yes indeed thank you.

[violet wind]

no problem

if you finish you can type .close to close the channel

[gentle ridge]

oh yeah I forgot thank you type.close

type.close

[violet wind]

.close

I think only you can do it

[gentle ridge]

lol

.close

[devout snow]

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[violet wind]

lol

[devout snow]

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[formal pecan]

Course: Complex Analysis

What type of question is this called so I can look up instructions?

[devout snow]

@formal pecan Has your question been resolved?

[formal pecan]

-close

.close

[devout snow]

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[opaque parrot]

.Ask

.Ask Probability

How do I do this?

[opaque parrot]

This was a question in my exercise book but I don't know how to do it

[devout snow]

@opaque parrot Has your question been resolved?

[violet wind]

Ah

So there are n^2 total possible outcomes, right

That are equally likely

[devout snow]

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[devout snow]

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[restive river]

Consider the function: f(x) = 4x - 9 / x + 3

a. determine the coordinates of the x- and y- intercepts

How do i find those intercepts?

[ionic belfry]

just set the function equal to 0 to find x int
like f(x)=0 and find those values of x
for y-int set the value of x = 0, in this case its undefined so no y-int

[restive river]

to find x int i need to set f(0)

but for y intercept?

[ionic belfry]

yes

[restive river]

y intercept is the same?

[ionic belfry]

no

[ionic belfry]

this is y-int

[restive river]

y int f(0)

for x?

oh -y int?

[ionic belfry]

for x-int you make the entire thing equal 0

f(x) = 0

=> (4x-9)/(x+3) = 0

[restive river]

whatt

i dont understand sorry

so i consider the function to be 0?

[ionic belfry]

solve this equation and find the values of x

[ionic belfry]

yes

[restive river]

(4x - 9) / (x+3)

(4(0) - 9) / (0+3)

for x intercept?

[ionic belfry]

x-int literally means where the graph touches the x-axis and x-axis means y=0

[ionic belfry]

nope

thats y-int

[restive river]

what about x int?

[ionic belfry]

@restive river

[ionic belfry]

@restive river

for x-int you make the entire thing equal 0
f(x) = 0
=> (4x-9)/(x+3) = 0
=> (4x-9) = 0
=> x = 9/4

@restive river is this clear?

[restive river]

wait

[restive river]

so thats for x intercept?

[ionic belfry]

yes

and we have no y-int

are you done?

[restive river]

alright got it