help-43

[exotic maple]

z1 with z2?

[wanton patrol]

yea

they're asking for $z_1 \cdot z_2$

[boreal girder]

mango

[wanton patrol]

you know both of them

[exotic maple]

how do i multiply these 2 now

like

3 + 3i = 6i?

[wanton patrol]

uhh nonono

that's like the biggest sin in math

[exotic maple]

OH

LOL

HAHAHHAHAHHAHAHAHA

[wanton patrol]

so you can't do stuff like

$3 + 3x = 6x$

[boreal girder]

mango

[wanton patrol]

or $3 + 2i = 5i$

[boreal girder]

mango

[wanton patrol]

it's the same reason why you can't add integers with square roots for example

well you can but like

not in the sense of $3 + \sqrt{2} = \sqrt{5}$

[boreal girder]

mango

[exotic maple]

so how do multiply them

[wanton patrol]

okay so

replace z_1 with this

and z_2 with this

[exotic maple]

so i just write these 2 down?

[wanton patrol]

yea

u just replace the letter with those numbers

[exotic maple]

okay finally

[wanton patrol]

what do you have

[exotic maple]

what do you mean by that

[wanton patrol]

like

replacing z_1 and z_2 with those numbers is enough, but it really depends if your teacher is kind or not

you might be required to simplify that too

and i think that you're required to simplify it, since you do get a nice answer

[exotic maple]

its okay for now, seriously thank you very much tho. I will aslo ask the teacher for some help in advance i have 1 week to study but i cam here early purely because i dont understand SHIT about maths. and a lot of people say its very easy and im upset because i cant solve this shit

[wanton patrol]

no problem, just make sure to study at least 1-2 hours a day

[exotic maple]

i still have 3 more others to learn in advance so i think im calling it quits for tonight i will start early tommorow and ask for help and maybe i have more energy

[wanton patrol]

you'll get it down easily

you can use wolframAlpha to check your answers, btw

[exotic maple]

yeah no i think i can learn it in here to be honest, i tried chatgpt but didnt work but

in here people really have the nerves to help you out

and i couldn't be more thankful

[wanton patrol]

yea chatgpt sucks

[exotic maple]

Thank you for your time, means a lot

[wanton patrol]

no problem, good luck on your test

[exotic maple]

see you later in advance thanks to everyone here who tried helping 😄

how do i close this

so people can take the channel

[wanton patrol]

do .close

[exotic maple]

.close

[compact pewter]

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[scenic sluice]

I am struggling on the definition of a product topology, it says that the product topology is defined by the basis B = {U x V | U and V are open sets respsectively in X and Y}

I feel like there should be more resriction on what U and V can be

If i take U and V just to include one element of X and Y wouldnt this be different to taking a different open set of X and Y which have more elements??

[calm trout]

what do you mean by taking U and V to only include 1 element?

you don't know that 1 element sets are open

[scenic sluice]

If i had (X,T)={∅,{a},{a,b},{a,b,c},X} and (Y,T)={∅,{1},{1,2},Y}

to be open in a topology just means that the set in the topology right so, wouldnt {a} be a valid open set and i could choose {1} to be an open subset of Y.

Then the product of these two would be quite smaller than if i picked a different set

[calm trout]

?

wdym picked a different set

{a} and {1} are open

so {(a, 1)} is open in X x Y

[scenic sluice]

Wait so is the product topology defined uniquely based on the U and V I pick?

That’s what I don’t get

[calm trout]

wdym pick?

[scenic sluice]

Like in the definition it makes me think that there is a U and V I pick from X and Y then that creates the basis for th product topology

Or is for Ui and Vi subsets of X and Y

[calm trout]

B = {U x V | U and V are open sets respsectively in X and Y}

B is the set of all U x V where U and V are open sets of X and Y respectively

[scenic sluice]

Okay I’m stupid

[calm trout]

where is Ui and Vi coming from

[scenic sluice]

I thought we were picking a unique set to generate it

Like Ui and Vi are all the subsets of X and Y

[calm trout]

why is it indexed

[scenic sluice]

Like Ui is the open sets of X for all i

Maybe I just forget the index

[calm trout]

but like where is the index coming from

[scenic sluice]

No where just typed it

[calm trout]

ok so it shouldn't be indexed then

[scenic sluice]

Alright

[calm trout]

you take U an open set of X, V and open set of Y, then you declare U x V to be open in X x Y

[scenic sluice]

Ok

[calm trout]

letting U and V vary, all these product sets U x V are open in X x Y

the topology is generated by them

if $W \subsete X \times Y$, then there exist $\UU \subsete T_X$ and $\VV \subsete T_Y$ such that [ W = \Union_{U \in \UU, V \in \VV} U \times V ]

[scenic sluice]

Okay thank you i think that makes sense

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[hard karma]

how to solve this chat?

[thick comet]

What have you tried?

[hard karma]

nothing i need to know the proper way to solve

bcs usually i just combine the x and add the exponents but theyre negative this time

so im wondering whats the right way to combine

[near bluff]

if the negative is fazing you, can you handle the 6?

if you can handle the 6, how would you do it?

[karmic stirrup]

its the exact same way! in-fact, if the negatives are bothering you, you can just cancel them out in this case!

[hard karma]

okay makes sense i was just trying to make sure i was doing it correct

[karmic stirrup]

you do realize that its -x^2 multiplied by -x^5 and thus its like multiplying -1 * -1 * x^2 * x^5

because multiplication is... what's the word

assicoacoateivive?

anyways yeah you can move the negative around terms that are multiplied

-2 * 3 = 2 * -3

[near bluff]

(associativity has to do with rebracketing. you're looking for commutativity.)

[hard karma]

so 6x^10?

[karmic stirrup]

wonderful! ty for correcting me

[hard karma]

okay thanks guys

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[hard karma]

is this logic correct?

[hard karma]

cuz they didnt show me how to do this type of equation so i wanna make sure its right before i do all the others

[merry latch]

No it doesn't work like that.

[hard karma]

okay how do i solve it

[icy sentinel]

no, since since a and 1 are not of the same base you cannot use that rule

[merry latch]

In order to use those rules you need to have the same base.

There's nothing to really "solve" honestly.

Dividing by 1 doesn't do anything.

So you just have $2a^{-3}$.

[boreal girder]

Ho-Oh's rainbow

[hard karma]

i also have this equation

[merry latch]

But then for any negative exponent, you are always allowed to make it a positive number by simply moving it to the denominator.

[hard karma]

so thats what i mean how do i solve this type

[merry latch]

So you get $\frac{2}{a^3}$.

[boreal girder]

Ho-Oh's rainbow

[hard karma]

okay i see

[merry latch]

Same idea, anytime you see a negative exponent, you can always flip it and change the sign.

[hard karma]

so like this

[merry latch]

Since the negative exponent is on the numerator, you can move the power to the denominator instead.

Yeah exactly!

[hard karma]

okay tysm!

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[hard karma]

back again lmao what does it mean by degree cuz it gives me the answers but half of them dont make sense to me

[hard karma]

wait i get it

no i dont mvm

[near bluff]

do you still need assistance?

[hard karma]

yes

[near bluff]

oh.

so I suppose you have not learnt what the degree of a polynomial is?

[hard karma]

no i havent

[near bluff]

the degree of the polynomial is the highest power of the variable present.

[hard karma]

okay so how is 7’s degree 4 is it bcs its one term and each exponent together equals 4?

[near bluff]

yes. that middle term has a degree of 4 because x contributes 3 and y contributes 1.
so the whole poly has a degree of 4.

[hard karma]

okay thank you i get it now

they didnt even teach me that sigh

[near bluff]

no worries. anything else?

[hard karma]

nope ty!

[near bluff]

all the best then!

[hard karma]

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[kind surge]

Hints!!

[dense mason]

which one

[small mason]

i dont know the name but u need to use that theorem about areas of triangles

that tell you about their ratio of areas

[kind surge]

there is one question only

[dense mason]

[abc] here is area triangle u mean?

[kind surge]

6 ratio by median?

[small mason]

If 2 triangles have the same "height"

then their areas will have the same ratio as their "bases"

[kind surge]

In parallel lines?

[small mason]

is this familiar?

[small mason]

uhhh no..

maybe someone else can explain / get the name of this theorem

[dense mason]

dont think that theorem have a name

but anyways, if A = 1/2bh then A1/A2 also cant be found if same height anyways

[small mason]

if you draw a random triangle XYZ
then on the line YZ you put W somewhere in the middle

Then [ XYZ ] : [ XYW ]
is equal to
YZ : YW

[dull moon]

Well, I'd first use Menelaus to find EF/FB and then use the equal-base-unequal-heights thing. If that's not in your syllabus, well then use similarity ig.

[dense mason]

That one doesnt need to be used

and irrelevant anyways

[kind surge]

In which triangle should I use menelaus?

[dense mason]

that is not thales

[kind surge]

?

[dull moon]

ABE

for CD

[small mason]

there arent that many triangles to try 🥺

[kind surge]

For ABE~

AC/EC.AD/DB.BF/FE=1

(AC.AD.BF)=EC.DB.FE

5.1.BF=4.3.FE

BF/FE=12/5

What next?

@dull moon

Base is BC

And we have BFC area

12~144 so 17×12

ABC area

@dense mason

Please check

[dull moon]

If you have BFC area, you can find CFE area

(same height, different bases)

[kind surge]

Same height how?

[dull moon]

*its the same height)

its just the length of the perpendicualr dropped from C onto EB, for both cases

[kind surge]

How can you say it is perpendicular?

Are you saying CF is same for BFC and CFE triangle and base is different

BF and FE base

CFE area would be then 5.12=60

@dull moon

[sleek meadow]

Yeah i didnt understand this part also

[dull moon]

no no im not saying its perpendicular

Ok let's construct a perpendicular from C onto EB. call this CX

then the area of $CFE = \frac{1}{2}EF\cdot CX$

[boreal girder]

Annie Maqionde

[dull moon]

and that of $CFB = \frac{1}{2}FB\cdot CX$

[boreal girder]

Annie Maqionde

[dull moon]

so $\frac{CFE}{CFB} = \frac{EF}{FC}$, is it not?

[boreal girder]

Annie Maqionde

[kind surge]

Got it nice

But how CFE area will help?@dull moon

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[sleek meadow]

<@&268886789983436800>

[sleek meadow]

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[rotund sphinx]

<@&268886789983436800>

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[royal furnace]

how do i do this q??

[royal furnace]

idk where to start

[vivid breach]

cut the belt up into circular pieces and straight line pieces

to do that, you first gotta figure out where on the wheels the curve becomes straight

find that by doing this:

[winged lion]

<@&268886789983436800> weird username

[thorny urchin]

Thx for notifying us. But it was deemed to be fine.

[vivid breach]

then:

to help is this new red line drawn, parallel to the outer blue one

figure out the length of that red line first

then its the same as the blue part

that will figure out the straight line piece

for the curved pieces, youll first need this green angle:

that way, you can find the purple angle for the arc

think and it will be true: this other arc is two green angles (use parallel lines to prove this)

with that, you just have to figure out the numbers in order then add the pieces together

go ask if youre stuck, this is just the general process

maybe it makes more sense when you put numbers to things

[compact pewter]

@royal furnace Has your question been resolved?

[royal furnace]

ohh

[royal furnace]

??

how is that even weird

[royal furnace]

oh so we use tangent in this case

ye i get it but how do u even think of that

[vivid breach]

click the ❌ to tell the bot youre still here

right I wasnt sure how to explain that

[royal furnace]

is this just lots of practice

[vivid breach]

no

[dense mason]

https://tenor.com/view/kryonax-skull-gif-26476587

[royal furnace]

oh

[vivid breach]

practice isnt going to tell you what that angle is

[royal furnace]

no i mean like making the rectangle

[vivid breach]

well first you still gotta trust that its a right angle in the first place

[royal furnace]

cuz i labeled it i got tangent and stuff but i didnt know what to do after that

[vivid breach]

is that intuitive for you

[royal furnace]

uh idk

[vivid breach]

alr lets consider that first

[royal furnace]

wdym

[vivid breach]

why is this a right angle

[royal furnace]

ye i know thats right angle cuz of tangent rule

[vivid breach]

what told you its a tangent then

[royal furnace]

because u made it a tangent?

[vivid breach]

you cant cite me as a source for the problem yk

cmon a real reason

I drew that because it wasnt obvious

[royal furnace]

oh its tangent cuz it touches the circle once

[royal furnace]

no but i drew that already as well

[vivid breach]

yep

well you still have to mention it first

[royal furnace]

like i knew that was tangent angle bu ti didnt know what to do from there which u did the rectangles

we need to do that

[vivid breach]

yea up next is that

[royal furnace]

but doesnt that go into circle proofs

[vivid breach]

that one I dont exactly know how to introduce

for this, heres something you can consider

we essentially have a shape thats like this:

[royal furnace]

ye

ohhhh

ive done that shape many times

[vivid breach]

yep

now here, its a bit harder to spot the shape

[royal furnace]

ye cuz i didnt recognise it

[vivid breach]

but you draw enough lines and you make the connection

you figure that out, you get the lines

[royal furnace]

hmm ic

[vivid breach]

as for the green angles, you know how to get those?

[royal furnace]

ye cosine or sine rul

e

[vivid breach]

I dont know what you just mentioned

wdym by rule

[royal furnace]

like sin a/ a = sinb/b

or cos c = a^2 + b^2 - 2ab cos c

[vivid breach]

no theres something more direct you can just do here

[royal furnace]

sine cos tan?

[vivid breach]

arcsin, arccos, or arctan

you know these turn ratios -> angles

[royal furnace]

ye

[vivid breach]

so you identify a trig ratio on the right triangle here

then you arcsin, arccos, or arctan it (choose the correct one based on the sides youre dividing for the ratio) to get the green angle

[royal furnace]

wait so after finding green we double it then do either 360- that or 2pi- that right

[vivid breach]

yep

[royal furnace]

then use l= theta r formula

alr

[vivid breach]

nice

[royal furnace]

yep

alr then i get it now

ty for ur help

[vivid breach]

the sine and cosine rule would just be more longwinded versions of this

dont forget that you can use them directly, just for this case

[royal furnace]

hm ye cuz this is right angle triangle

[royal furnace]

yep

[vivid breach]

in most geometry problems the symmetry in those rules are more convenient

alr good

[royal furnace]

got it

alr then ty

[vivid breach]

np

[royal furnace]

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[dense mason]

<@&268886789983436800>

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[pine cloak]

Can someone explain transformations of graphs, on how to solve questions like this? How do you know by what scale it's being transformed?

[rapid crescent]

do you know what happens to the graph if for example you do f(x)+1?

@pine cloak

[pine cloak]

yea

verticla shift up

[sleek meadow]

Thats right

[rapid crescent]

yes

and in the graph instead how it was moved

i mean in your exercise

[pine cloak]

it compressed leftside thats all i can notice

[rapid crescent]

and how do you move a graph to the left?

[pine cloak]

f(x+c) I think?

[rapid crescent]

yes

[pine cloak]

but how to know how much compressed? and how to move

[rapid crescent]

now here I think a square is displacement of 1?

[pine cloak]

yes

[rapid crescent]

so I assume it was moved left 1.5 from the figure?

right?

[sleek meadow]

Idts

look at othe parts of the graph asw

[rapid crescent]

that's what I think is compression

[sleek meadow]

Oh okay i thought you simply meant its just shifted sorry!

Carry on!!

[pine cloak]

how doyou know

[dire pine]

is it not move left 3

[sleek meadow]

nope its compressed

[dire pine]

it's move left 3 then compressed, you can't get the transformation simply by compressing

[inland kite]

I dont think it moves, initial left point stays at the same spot

[dire pine]

you compare the graphs, for example, for the linear part of f(x) on the left, it moves right by 3 units, but for the linear part of g(x) on the left, it moves right by 1.5 units

[dire pine]

scaling starts from x = 0 though?

[rapid crescent]

if you only think about the linear part from y=-x-2 it becomes after the transformation -2x+5

[dire pine]

what are you tryna say

[dire pine]

but when you compress a graph, it doesn't start from that initial left point, so there is definitely translation

[rapid crescent]

which then applying the same to the linear part on the right becomes different

[dense mason]

note that this is compressed horizontally, not shifting

perhaps you can find dilation factors

[dire pine]

how is it different..

[dense mason]

there are 2 types of dilation factors: from x-axis, and from y-axis

[dire pine]

g(x) = f(2x+3) or am I missing something

[dense mason]

wrong

g(x) doesnt shift

since the left-most point still remain the same, seems like it has been "squished"

[dire pine]

no but it literally does..

[dense mason]

then explain the point (-3,1)?

does it move?

and if you say g(x) is f(x) shifted, then all of the points must also shift with the same constant unit

[dire pine]

plug -3 into g(x) = f(2x+3)

[dense mason]

you dont really understand here right?

[dire pine]

all i understand is that a compression would look like this

which must then be shifted

so like wdym it's only compression 😭

compression starts from x = 0

so you can't compress from (-3, 1)

[dire pine]

or in my case i moved 3 first, then compressed it but either way works

[dire pine]

It doesn't move because it coincidentally overlaps

[dense mason]

I see

[drifting flax]

I mean it does move

Just ends up at the same place

[dense mason]

But it cant move that much for 3 units

[drifting flax]

1.5

[dense mason]

Since you compressed it, it can go for 3/3 only

Look at the lowest point

[drifting flax]

2x + 3, 2*(x + 1.5)

[compact pewter]

@pine cloak Has your question been resolved?

[dire pine]

@pine cloak what do you still not understand

[rapid crescent]

i got it now...

[pine cloak]

wahts the final answer

[dire pine]

i can't tell you that directly

[pine cloak]

alright

how doyou know how much is compressed

[rapid crescent]

seeing how much y changes

the red line is y=-x-2 right?

which in the new graph is translated horizontally to the right by +1.5

so f(x+1.5)

y=-(x+1.5)-2

ok so far?

@pine cloak

[pine cloak]

yea

[rapid crescent]

but now

look

and

in g(x) at x=-3 y is 1, while at our current point it is not

so we have y=-(x+1.5)-2 after the traslation

and therefore assuming that at x=-3 the function is equal to 1 then you have

so we need to find the new slope

that comparing the points from before and after (-3,1) and (-1.5,-2) m=(-2-1)/((-1.5)-(-3))=2

so the new line is y=-2(x+1.5)-2

so you acted on f like this

f(x+1.5)

and the f(2(x+1.5)) = f(2x+3) = g(x)

[compact pewter]

@pine cloak Has your question been resolved?

[compact pewter]

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[tulip rain]

where do I go from here

[dense mason]

Let x be Andy candies

Then we have ratio 1:6:14, so what is candy of luke and tina in term of x?

Using x would be easier to imagine of

[tulip rain]

x:6x:14x I think

[dense mason]

Correct

If Tina gives away 3/7 candies, then what is the number of candy being given away interm of x?

[tulip rain]

14x

[dense mason]

She only give away 3/7 of her candy, but she have 14x

Then?

[tulip rain]

wwould I do 3/7 X 14 which is 6

[dense mason]

So the candy give away is 3/7*14x, right?

Correct!

So 6x here go from Tina to Andy, but Andy have x candy, then number of candy Andy have is?

[tulip rain]

6?

[dense mason]

Dude

X+6x=?

[tulip rain]

oh shit 7x lmao

[dense mason]

So if tina give away 6x candy, what does she have left?

[tulip rain]

if she has 6x candy and given away is that not 0?

[dense mason]

Dude

Tina have 14x

Please read carefully

[tulip rain]

oh 8x then

[dense mason]

she gives away 12(1/2)% of the rest, what percentage this is?

[tulip rain]

12.5%

[dense mason]

The rest is 8x candy? Then she gives away how much candy?

[tulip rain]

im confused

i dont wht i need to do with the 8x

[sleek meadow]

Okay so Tina was left over with 8x candies after giving away some to Andy right?

[tulip rain]

yes

[sleek meadow]

Now the question says that she further gives 12.5% of her remaining candies to Luke

So what did shw give to luke?

[tulip rain]

ok so would I do 8x X 12.5 or do I divide it

[dense mason]

Multiply it by 12.5/100

Since it is percentage

[sleek meadow]

Multiply!

[tulip rain]

i got 1 as the answer

[dense mason]

Correct

This means 1x

She gave this to Luke, which originally have 6x

How much he have now? How much Tina have left?

[sleek meadow]

This 1x candy was given to luke by tina as minhh said

So tina has 1x less candy now right?

[tulip rain]

yes

So 7x is left over?

[dense mason]

Correct

Now check the candy that other 2 have

Check again

[tulip rain]

ye bc it has to add to 21x I think

[dense mason]

First time tina giveaway, how much candy she gave

Recall what i told you before

[tulip rain]

she gave away 6x

wwhich left her with 8x candy

[dense mason]

Who receive the 6x

[tulip rain]

andy

[dense mason]

What candy that andy have initially?

[tulip rain]

1x

[sleek meadow]

How much was given to Andy?

[tulip rain]

6x so he would have 7x in total

[sleek meadow]

Okay good!

[sleek meadow]

btw

[dense mason]

For the second time, how much tina give away

[tulip rain]

Tina had 14x gave 6x to Andy so he has 7x in total so that leaves tina with 8x

then she gives 1x to Luke

[sleek meadow]

Okay, thats after the first giveaway!

Yup!!

[sleek meadow]

Thats the second giveaway

[tulip rain]

so she has 7x left over

[sleek meadow]

Now you tell me, how much do all three have finally?

[tulip rain]

Andy has 7x Luke has 7x and tina has 7x as well

[sleek meadow]

Good!

So well is your question answered now?

[tulip rain]

im not sure if it answers the question ill check the mark scheme give me a sec

[sleek meadow]

Why not?
The question asked was if Tina was right at the end, aka all three having the same number of candies

[tulip rain]

yes thats the answer thank you

[sleek meadow]

And you did find that at the end, all three did have the same number of candies!!

[sleek meadow]

Your welcome !!

[dense mason]

It asked if the statement is correct

[tulip rain]

holy shit thank you @dense mason and @sleek meadow u 2 dealt with bs lmao

[dense mason]

Good luck for your exam buddy

[tulip rain]

thank you

[dense mason]

No problem buddy

Anyways if you are done please type .close :))

[tulip rain]

yes but im gonna need to keep this open to answer the question on a pdf since i need to show it to my tutor

[dense mason]

Np buddy

[tulip rain]

.close

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[drifting drum]

how do the normal formulas work for this

when the arc for example doesn't touch C

[sleek meadow]

What even is this significance of C in this Question?

[drifting drum]

bro wtf am i doing

.close

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[still tundra]

https://www.youtube.com/watch?v=q1U6AmIa_uQ

[still tundra]

In the surface area problem

its asking for the minimum

what if its asking for the maximum? what are the tweaks?

I still don't understand because the general process seems to be the same for me

the only thing you need to do is to pretty much create a general equation (or use equations depending on what is given), derive then equate to 0

the problem im having is what to do to specifically find maxima or minima

[thick comet]

If anything you could check the sign of the second derivative and this will tell you whether it's a maximum or a minimum.

In general though, depending on the constraints, you may not be guaranteed the existence of a maximum and a minimum

[still tundra]

yeah, but the thing is the general steps feel loosely the same

so i might feel like im doing something correct but the second deriv test says otherwise

[thick comet]

Yes, they are the same. You find critical values and then classify them.

[still tundra]

pretty much, cause optimization is basically in concept just finding local maxima/minima depending on whats gien

given

[thick comet]

The wording of the problem and the information you're given will usually ensure whatever (min or max) you're looking for exists

[still tundra]

i see

[thick comet]

And again, in general the problem may be posed in such a way that talking about a maximum in the same context doesn't make sense.

For the surface are for instance, there is not maximum in the problem given in the video

[still tundra]

but what if i have a problem and its asking me to find both the maxima and minima instead of one or the other?

like this one for example

[thick comet]

Then you may just find critical values and check whether they are maxima or minima using the second derivative.

[still tundra]

right right, they can contain multiple critical values

so if one of them can indicate a maxima, the other can indicate minima

i gotchu

[thick comet]

Yeah, if they ask for both, generally, you would expect them to be different critical values

Unless the function is constant or something

[still tundra]

right

thanks brother

.close

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[compact pewter]

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[echo thorn]

How do I factor $x^2-2yx-24y^2$?

[boreal girder]

Vortac

[echo thorn]

I get -6, 4

but I'm confused about the yx

[quaint gulch]

treat it as a 2nd degree polynomial in x

discriminant etc

[echo thorn]

What do you mean?

[quaint gulch]

when you have a polynomial ax²+bx+c, its discriminant D is b²-4ac
and its roots are (-b +- sqrt(D))/2a

here you have b = -2y and c = -24y²

so D is 4y²+96y² = 100y²

the roots are (-2y+-10y)/2

so -6y and 4y

[echo thorn]

isn't it yx though?

[quaint gulch]

well I just told you to treat it as a polynomial in x

so the coeff are 1, -2y, and -24y² in this order

[echo thorn]

but I thought coefficients just include the numbers?

[sleek meadow]

well isnt -2y also a number

coefficients need not be constants if that what you mean

[echo thorn]

this is what I meant

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[compact pewter]

Please don't occupy multiple help channels.

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[icy crystal]

struggling on where to start here

i just learned about the L operation, linear independent solutions / homogenous solutions, but i wasnt taught how to find the particular or homogenous solutions myself

[rigid perch]

you solve homogeneous second order constant coefficient differential equations with the characteristic polynomial, see here: https://tutorial.math.lamar.edu/Classes/DE/SecondOrderConcepts.aspx
and for more in-depth treatment of the cases you encounter:
https://tutorial.math.lamar.edu/Classes/DE/RealRoots.aspx
https://tutorial.math.lamar.edu/Classes/DE/ComplexRoots.aspx
https://tutorial.math.lamar.edu/Classes/DE/RepeatedRoots.aspx

[trail cave]

why do people just assume everyone knows/can digest more complex math notations

this is not as helpful to a new learner as you would think

[winged lion]

lmao what

[icy crystal]

i will review the provided materials and then depending on such a thing i will join in this argument for or against you

one moment

[trail cave]

someone new to math isn't going to get what you just conveyed

[winged lion]

doesnt mean it's harmful, what's your point?

[trail cave]

the whole skill in teaching people is letting them grasp concepts without using notation

[winged lion]

If OP has questions they can ask

[rigid perch]

i think maybe we should have this conversation in #「helpers-lounge」 and keep the channel focused on the help

[icy crystal]

thanks! very helpful, i will certainly be back to ask you abojt nonhomogenous equations as well

but for now that’s all

.close

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[modest egret]

another channel i dont have acess to

[compact pewter]

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[vale blade]

i don't quite understand this definition

[vale blade]

this is about projectors

[thick comet]

It’s a property. It means that the vectors of F are exactly those vectors that get mapped to themselves under the projection

Or like p(x)=x

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[twilit crane]

,rccw

[twilit crane]

is this a correct proof by contradiction

[digital delta]

In the first case why is x or y congruent to 0 mod 4?

[twilit crane]

the product is to be congruent to 0

to either one of them should be congruent to 0 to do that

we could also have both of them be equal to 2 mod 4
but we reject that cuz we assumed that x is not equal to y

[digital delta]

x = y = 2(mod 4) doesnt mean x =y

like take x = 6 and y = 10

[twilit crane]

ohhh right

[digital delta]

we could also have x = 1 mod 4

and the equation would be satisfied

[twilit crane]

mhmm

how do i prove this then

[digital delta]

I think contradiction might be the right way

but mod 4 is probably not enough

[twilit crane]

mod 7 ?

[digital delta]

especially considering that x and y are bounded by (p-1)/2

maybe work mod p

expand this and some terms will cancel mod p you get some congruences work with that

[twilit crane]

i just get
(xy)² = k² (modp)

[digital delta]

you can factor

now

[twilit crane]

so xy = +- k (modp)

[digital delta]

i guess you could write k = axy + mp where a is {-1,1}

and sub back

this is not easy

[twilit crane]

maybe we can try to bound k

idk if that helps

[digital delta]

idk

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[hexed copper]

Why is da and E parallel?

[near sierra]

electric field through vector area

[rigid perch]

dA points outward because it is always normal to the surface, E points outward for this case because of the spherical symmetry

[compact pewter]

@hexed copper Has your question been resolved?

[little gorge]

Why does that look like- uh nvm

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[crisp fossil]

I NEED HELP WITH CALCULUS III HOMEWORK BRO PLZ I DONT UNDERSTAMND TJOIS

[craggy sluice]

I might be able to help, can you like post a problem?

[crisp fossil]

i need help like starting these problems like i have no idea what to do i know i ahve to figure out if it converges or diverges but i lowkey forgot

[patent garnet]

i dont know how to be rigorous about it

but i can tell which converge and diverge

[crisp fossil]

yeah im barely starting and i already need help wait how can u tell thoooo

[patent garnet]

ok so

i know it sounds silly

but i think of a wall

and im standing a bit away from it

now someone tells you to take a step equal to half the distance to the wall

and so you do

they tell you to do it again, but now the diatance changed

sinceyou are closer to the wall

but if they told you to take twice the distance

then you would pass the wall -> diverge

if you take portions of the distance, smaller than it

-> converge

as long as the steps are smaller than 100% of the distance you will get closer to the wall

bur never reach it

but*

How is this relevant to your problems?

The idea sort of works for 89, 90, 91

[fading flicker]

let n+1000=k, if n=1, k=1001, so you can rewrite it as summationfrom k=1001 to infnity of 1/k(you can also use this for 88), you can then apply basic calc knowledge to know this is approximately lninf-ln1001 which just tells us divergenceif

[patent garnet]

worksfor 92 as well

[crisp fossil]

so 87 and 88 both diverge then right

OOOOO

okay

i can make n =1001 and then it turns into 1/n

and that diverges

and i can do the same for second one n=1+10^80

[fading flicker]

series in the form summation from k=0 to inf ar^k converges if |r|<1, because fractions get infinitesimally smaller,,, their task now is to express it as a series first
this works for 89 to 91, 92 is a bit more trivial but you can express it as the summation of (1-sqrt(pi/3)) * (find this one)^2n from n=0 to +inf,

[fading flicker]

yes

[patent garnet]

yeah sry like i said im not rigorous at all sry to poster if I just confused you

[fading flicker]

@crisp fossil express the series in 89-92 as summations in the form ar^k
where a is a constant(usually the starting term) and r is how much you multiply to it each time

I can't really say much abt this I only started studying about this recently

[patent garnet]

oshit yes

i get what you are saying now

yes this is what i meant but you managed to translate my nonsense

and you also make it clear that you can reach SOMETHING

[fading flicker]

be wary of no 91 btw

[patent garnet]

for 91 i thought like this

[fading flicker]

you explained it like gojo explained infinity

[patent garnet]

I dont know what gojo is but he sounds like a silly guy

[fading flicker]

(1/2+1/4+1/8+...)x=x, that is only if you reach infinity though

[crisp fossil]

for number 89 isnt it 1/10^n-1 how do i go from thjere like do i

[fading flicker]

well

r is less than 1

so what does that tell us

[crisp fossil]

um

converges

[fading flicker]

^

[fading flicker]

yes

[crisp fossil]

thats all

so 1/10 is smaller than 1 converges

90 the r is e/pie

[fading flicker]

yeah

[crisp fossil]

thats less than 1

[fading flicker]

which is?

[crisp fossil]

converges

okay

[fading flicker]

yes

what about 91

[crisp fossil]

let me see

[fading flicker]

be wary of it

[crisp fossil]

91 converges 92 diverges?

[fading flicker]

yes and are you sure about 92

oh wait yeah ur right

[patent garnet]

how does 91 converge

no way

Does it? If so I need to revise

i have been wrong before and will be wrong again 🙂

[compact pewter]

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[patent garnet]

oh nvm i didnt see the exponents lol

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[hazy obsidian]

Let $a>1$ be an integer, Prove that every element of the set [ {a^2+a-1,a^3+a^2-1,...,} ] are pairwise coprime

[boreal girder]

Copter

[hazy obsidian]

i dunno how to do this

i tried compare a^k+1 +a^k -1 and a^l+1 + a^l-1 but that didnt get me anywhere

[cursive harbor]

Hmmm let's call the terms A_n

$A_n = a^{n+1} + a^n -1$

[boreal girder]

Xavier 🌺

[cursive harbor]

Clearly $A_n \equiv (-1) \mod a$

[boreal girder]

Xavier 🌺

[cursive harbor]

Also $A_{n+1} - aA_n = a-1$

[boreal girder]

Xavier 🌺

[cursive harbor]

Unless I fucked up my calculations

From here you can use Euclid's algorithm to find gcd

@hazy obsidian you with me?

[hazy obsidian]

yea

[hazy obsidian]

wait, on what😭

sorry i kinda suck

[cursive harbor]

No worries lol

Well you can use the algorithm to show that $gcd(A_m, A_n) = gcd(A_m, A_{m+1})$

[boreal girder]

Xavier 🌺

[cursive harbor]

(do you see how)

And now it becomes easier cuz you have an explicit formulation for $A_m$ and $A_{m+1}$

[boreal girder]

Xavier 🌺

[hazy obsidian]

ohh cuz you get A_n from it

[cursive harbor]

Ye, you break down the problem from "two arbitrary elements" to "one arbitrary element and its successor"

[cursive harbor]

Id say try and show this first. Then we can move on

(or alternatively if you can continue yourself from here, feel free to)

[hazy obsidian]

= (A_n, a-1) and take An mod a-1 = 1

[cursive harbor]

Yup

[hazy obsidian]

okay, thanks!

.close

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[forest token]

I've done v = mu(a)+ lambda(b)

Then I used the condition given and got v.c = 1

How do I solve for both mu and lambda

[thorny walrus]

This question was wrong I think

JEE didn't give bonus for it but you can't find both

[forest token]

What was wrong in it? The same thing I said?

[thorny walrus]

Yes

You can't solve both mu and lambda

[forest token]

Alright thank you 👍

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[craggy shuttle]

I kinda know what to do but I need help putting it in a clean efficient proof

[thick comet]

For the first part it should be relatively straightforward to prove $[x]_n \subset {x + kn : k\in\Z}$ and vice-versa.

[boreal girder]

Azyrashacorki

[thorny walrus]

What's the described equivalence relation here?

[craggy shuttle]

$\mathbb{Z}/n\mathbb{Z}$

[boreal girder]

arachi

[craggy shuttle]

so you would prove two inclusions

[thick comet]

I feel it would be appropriate. You would be showing that the representatives of [x]n are exactly those integers of the form x+kn.

[craggy shuttle]

I did it like this

[thick comet]

Looks right.
You're essentially doing the same thing but both sides at the same time with iffs.
Then the second part arises since the remainder mod n is unique between 0 and n-1 by the division algorithm,

[craggy shuttle]

so this is good?

[thick comet]

At a quick glance, yes.

[compact pewter]

@craggy shuttle Has your question been resolved?

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[carmine garden]

is there any efficient way of doing this

[carmine garden]

because finding the double partial deriative and then fiding the variance of that ain't pretty

Okay, even once I find the double derivative, the expectation is a function of the sample no?

Like the double derivative wrt sigma^2 is

$\frac{-1}{\sigma^4} - \frac{1}{2} \sigma^6} \sum (x- \mu)^2$

[boreal girder]

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

[carmine garden]

then using the defn of fischer information we get $\frac{1}{\sigma^4} + \frac{ \sum (x_i- \mu)^2}{2 \sigma^6}$

[boreal girder]

Wai

[carmine garden]

*fisher

okay going to grab dinner now

byee

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[serene coral]

Can anyone refresh me on how to do these kinds of system of equations?

[serene coral]

I've some more kinda but let's do this first

[solemn matrix]

Here, i got something for you to try

This is a classic factorization problem

@serene coral

[solemn matrix]

So look at the first equation

I plugged in some random big value for $y$, like $100$ here and solve for $x$ (note that this is in a draft)

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

So when i solve it

[serene coral]

it gives -100

[solemn matrix]

Alright

[serene coral]

,w solve for x: x^2 + 2*(100)^2 + 300x = 3(x + 100)

[boreal girder]

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[solemn matrix]

Yea

So you got 2 results, $x = -100$, which is also $x = -y$

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

Which means you would have a $x + y$ in the factorization of the 1st equation

[boreal girder]

1 divided by 0 equals Infinity

[serene coral]

i can see (x+y) in the rhs

[solemn matrix]

In other words, your first equation could be written into $(x + y)(something) = 0$

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

That's a guide for you to factor

[serene coral]

got it into (x+y)(x+2y-3) = 0

[solemn matrix]

And looking at $x = -197$ tells me that $x = -2y + 3$, because $x$ is $-200 + 3$

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

Yep

And from that you can substitute each case into the 2nd eq and solve normally

[serene coral]

But how about when both equations have square roots in them?

[solemn matrix]

That's for another time

That's a different kind alr

[serene coral]

Well I'm revising the advanced part of systems of equations so...

[solemn matrix]

Here your first equation looks like that it can be factored out

So we can try the trick where we substitute y = something and guess the factors

[serene coral]

but like i said, what if it looks ugly?

[solemn matrix]

That's a different kind alr

You got any examples about the ugly roots in both equations?

Like most of the time, the solutions to these are ugly numbers, so you gotta pick some good ones from the internet

[serene coral]

it's in the book, which i can't snap a pic rn

[solemn matrix]

Aight

[solemn matrix]

Do you notice that when you swap out $x$ and $y$ in the first equation, you get the second one?

[boreal girder]

1 divided by 0 equals Infinity

[serene coral]

yeah i kinda noticed that but didn't know what to get out of it except both roots are swappable

[solemn matrix]

So if you subtract both sides of the equation

You'll get a $x - y$ factor guaranteed

[boreal girder]

1 divided by 0 equals Infinity

[serene coral]

How can you be sure abt that?

[solemn matrix]

Or probably $\sqrt x - \sqrt y$ if the equation is ugly

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

You will always have a $x = y$ solution when you solve it out

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

For instance, if $x$ solves out to be something

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

And if i flipped the 2nd equation to the 1st equation, i can also solve for $y$ the same way i did for $x$

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

I think this question, there's a $\sqrt x - \sqrt y$ factor

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

So here, try to expand both sides out

$(1 + x\sqrt{y})^2$ is $x^2y + 2x\sqrt{y} + 1$

WLOG, $(1 + y\sqrt{x})^2$ is $xy^2 + 2y\sqrt{x} + 1$

[boreal girder]

1 divided by 0 equals Infinity

[serene coral]

i got $2(\sqrt{x}-\sqrt{y})(x+\sqrt{xy}+y+11)=0$

[boreal girder]

Thomas

[solemn matrix]

The first case will always solve out to $x = y$

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

<@&268886789983436800>

[serene coral]

but what about the second?

[solemn matrix]

Usually you would have to prove that there's no pairs satisfy this

[solemn matrix]

So look at this, the systems must have the condition $x, y \geq 0$ right?

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

So the $LHS \geq 11$, but $RHS = 0$

[boreal girder]

1 divided by 0 equals Infinity

[serene coral]

it doesn't say x, y >= 0

in fact it doesn't give any restriction except x, y are real numbers

[solemn matrix]

But there's $\sqrt{x}$ and $\sqrt{y}$ in the original system

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

And if either $x$ or $y$ is negative, then the system is invalid

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

(assuming you're working in reals)

[serene coral]

so if the roots are interchangeable then some variation of x-y must be in the final factorization right?

[solemn matrix]

Yep

Something that leads to $x = y$ will be in there

[boreal girder]

1 divided by 0 equals Infinity

[serene coral]

,w solve 2*(1+x sqrt(x))^2 = 9x sqrt(x)

[boreal girder]

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Requested by thomas2406

[solemn matrix]

Looks neat enough

(im lucky my teacher taught me these tricks for me to help you)

[serene coral]

any other tips/tricks that you can pass on?

[fiery goblet]

just practice more you will get the feeling

you will have the observation skill

just look at this system of equation

you will see how symmetrical it is

lead us to the first hypotheses
x=y or x=-y

[serene coral]

2 months until the admission test 😭
and other parts in the exam too

[solemn matrix]

I remembered there's also one type of systems where you have to let $y = kx$

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

Can be applied if 2 equations share the same degree

[serene coral]

what if they're asymetrical?

[fiery goblet]

that would be too hard

unless

[solemn matrix]

,w asymetrical

[boreal girder]

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Requested by iaminfinityiq

[fiery goblet]

they imply a clue inside

[serene coral]

no like normal asym system

just x^2, y^2, x, y, xy, constant

[fiery goblet]

it depend because not all problems are the same

[solemn matrix]

If one of your equations are degree 1 the yk what to do right?

[serene coral]

x^2 + 6y = 6x, y^2 + 9 = 2xy

[solemn matrix]

💀

[serene coral]

but they never seem to be sadly

[solemn matrix]

Is that from your book or you made that up?

[serene coral]

book

,w x^2 + 6y = 6x; y^2 + 9 = 2xy

[fiery goblet]

there are several way you can try, for example try to subtract the two equations together

[solemn matrix]

Aight

I got smth

[boreal girder]

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[solemn matrix]

I got something that shouldn't be applied in here but somehow works

[serene coral]

do tell

[solemn matrix]

Check $x = 0$ first

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

If $x$ ain't 0 then let $y = tx$ and substitute

[boreal girder]

1 divided by 0 equals Infinity

[serene coral]

nah it doesn't work on second eq

[solemn matrix]

Dw i tried

First equation is factorizable

[fiery goblet]

you can try to add the equations together

[solemn matrix]

Or you can do that and it will get you something good

[serene coral]

(x-y)(x-y-6)+9=0

[solemn matrix]

Did not expect that

😭

[serene coral]

i can't get the constant in

[fiery goblet]

bro not just 🙏🏼
(x-y)^2 -6(x-y) + 9

you see the x-y ?

that's similar to x^2 -6x +9

[solemn matrix]

Blud ain't completing the square 😭😭😭

[serene coral]

i forgot to do that and just factorized 😭

so we get x-y-3=0

which can fit back into first eq

[solemn matrix]

Looks good

[serene coral]

x^2 = 18

gtg

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[drifting drum]

not sure

how to

[drifting drum]

do part b

[azure vault]

|t-a| <= b is the same as ? <= t <= ?

[drifting drum]

yeah but i cant put my finger on what it is

[azure vault]

write it down

|y| <= b is the same as max(y,-y) <= b (so y <= b and -y <= b)

so -b <= y <= b

[drifting drum]

well yeah

i meant like

idk how to put it in |t-a| <= b

[azure vault]

let y = t-a

so that means...

[drifting drum]

wait

[drifting drum]

okay this makes sense

but in the question we have 2 different limits

[azure vault]

do this first

we'll come back to what role 5 and 13 play afterwards

[drifting drum]

|t-a| <= b
so t-a <= b, a-t <= b

[azure vault]

directly use this