help-27

[kind ridge]

done

[faint zinc]

And then use change of base

[kind ridge]

i think i have not studied that

just 3 properties i have studied

[runic grove]

log base c a / log base b a = log base b c

[kind ridge]

oh

[faint zinc]

[
\frac{\log_a(b)}{\log_a(c)} = \log_c(b)
]

[woven radish]

OmnipotentEntity

[kind ridge]

how do to detrive this

[runic grove]

Hey my thing is also correct, right

Or did I fuck up

[faint zinc]

Probably

Didn't notice before I sent

Phone

[runic grove]

Hm whatever

[kind ridge]

so i am getting

[runic grove]

This is correct afaik so use this

[kind ridge]

12 raised to power log base 12(2)

[kind ridge]

yeas

[runic grove]

Which is 2

[faint zinc]

It is correct, sorry for trampling

[runic grove]

No worries

I didn't mind

[kind ridge]

what?

[faint zinc]

By the definition of a log

[runic grove]

a^log b c = c^log b a

[kind ridge]

its 12 raised to log_12(2)

[runic grove]

Yeah, that's 2

Let log_12(2) be x

[kind ridge]

oh

[runic grove]

So 2 = 12^x

Sub x back in

2 = 12^log_12(2)

[kind ridge]

where did 2 came from

[runic grove]

Re read

[kind ridge]

12 raised to power x

how is that 2

like how did u knew that its 2

[faint zinc]

$b^{\log_b(a)} = a$ just in general, it's the definition of a log

[woven radish]

OmnipotentEntity

[runic grove]

It's the definition

[kind ridge]

oh boi

[faint zinc]

Recall that the log is the solution to $b^x = a$

[woven radish]

OmnipotentEntity

[faint zinc]

We rewrite $b^x = a$ as $\log_b(a) = x$

[woven radish]

OmnipotentEntity

[kind ridge]

yes

[kind ridge]

yes

[faint zinc]

Then replace x

[kind ridge]

but

this

so log_12(2)

is

2 power x=12?

[faint zinc]

It is the x that satisfies that equation, yeah

Sorry no

My bad

You have it backwards

12^x = 2

Not 2^x = 12

[runic grove]

,w 12^log_12(2)

[woven radish]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by _nofluxgiven

[runic grove]

yes finally

[kind ridge]

oh

[kind ridge]

so 12 raised to power 12 raise to power x

=2

dem

bruh but how to solve

this

[faint zinc]

Hmmm? That's been explained by both myself and flux

Do you have a more specific question?

[kind ridge]

uh no

[faint zinc]

Ok one more try

[kind ridge]

so 2 is the answer?

oh

mb

i thought we had to find the value of x

mb

[faint zinc]

The definition of the log.

Given the equation:

$b^x = a$

We say

$\log_b(a) = x$

[woven radish]

OmnipotentEntity

[faint zinc]

Do you follow up to this point?

[kind ridge]

yes

[remote sun]

Where did you get this Qs from?
I saw exact same in server last week

[faint zinc]

Excellent!

[kind ridge]

idk

some random guy gave me

[kind ridge]

i asked this question previously

but then i didnt knew log

now i know basics

so wanted to know how to do this with log

[faint zinc]

Now in this problem we have $12^{\log_{12}(2)}$.

Consider for just right now the expression $\log_{12}(2)$

[woven radish]

OmnipotentEntity

[faint zinc]

We can use the definition of the log to write $\log_{12}(2) = x$ means $12^x = 2$.

[woven radish]

OmnipotentEntity

[faint zinc]

Do you follow up until this point?

@kind ridge ^

[kind ridge]

yes

[faint zinc]

Awesome!

Now note that the x in $12^x = 2$ and the x in $x = \log_{12}(2)$ are the exact same value.

[woven radish]

OmnipotentEntity

[faint zinc]

Knowing this, we can perform a substitution of the variable

[kind ridge]

ydx

yes

[faint zinc]

$12^x = 12^{\log_{12}(2)}$

[woven radish]

OmnipotentEntity

[kind ridge]

true

base is same

so

[faint zinc]

So on the right we have the value we want to find.

[kind ridge]

yes

[faint zinc]

And on the left, we have an expression that we know is equal to some other value

$12^x = 2$

[woven radish]

OmnipotentEntity

[faint zinc]

So we have $2 = 12^x = 12^{\log_{12}(2)}$

[woven radish]

OmnipotentEntity

[faint zinc]

If we consider only the far left and the far right, we have established the identity we wanted to.

[kind ridge]

yes

damn

[faint zinc]

@kind ridge I'm sorry if this was a little plodding, I was trying to be as explicit as possible.

[kind ridge]

no no i understood

[faint zinc]

In general, $b^{\log_b(a)}= a$

[woven radish]

OmnipotentEntity

[faint zinc]

It's what the log "does." It "undoes" an exponent

@kind ridge anyway, I'm going to go afk, it's a bit late here. Best of luck!

[runic grove]

!done

[devout snow]

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[kind ridge]

yess

[kind ridge]

best of luck to you too

i am closing

.close

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[devout snow]

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[humble tundra]

|| 12^((1-a-b)/2(1-b))
= 12^((1-b)/2(1-b) - a/2(1-b))
= 12^(½ - a/2(1-b))
= 12^½ / 12^(a/2(1-b))
= 12^½ / (12^(a/(1-b)))^½
60^a = 3 -> a = ⁶⁰log 3
60^b = 5 -> b = ⁶⁰log 5
1-b = 1-⁶⁰log 5
1-b = ⁶⁰log 60 - ⁶⁰log 5
1-b = ⁶⁰log(60/5)
1-b = ⁶⁰log 12
a/(1-b) = ⁶⁰log 3 / ⁶⁰log 12
a/(1-b) = ¹²log 3
12^(a/(1-b)) = 12 ^ ¹²log 3
12^(a/(1-b)) = 3
12^½ / (12^(a/(1-b)))^½
= 12^½ / 3^½
= (12/3)^½
= 4^½
= (2²)^½
= 2 (c) ||

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[faint zinc]

!nosols

[pseudo basin]

!nosols

[devout snow]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

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[bleak arrow]

If I have the inequality y < -1 + x, and I set y = 2, i get x > 3 in return, right? But if you were to graph x > 3, only a portion of it would be in the region satisfied by y < -1 + x

[thick ledge]

Well yeah

You've inserted a value of y

[bleak arrow]

shouldn't the solution be in the region?

[thick ledge]

You'll get a subset of the solution

Not the entire one

[thick ledge]

It has to be

[acoustic leaf]

when you set y = 2 that means you go from considering the entire plane to just the line y = 2, so the region satisfying both the equation and the inequality is given by the blue line

[thick ledge]

you put y = 2

So all x >3 and y = 2 satisfy the inequality

[bleak arrow]

i see

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[dreamy atlas]

genuine asking

[dreamy atlas]

how to make a graph with a constant value,

i've been asked for making a graph of this function v(t) = 5m + C*e^(-2t/m)

My goal is to graph v(t) and y(t) for 0≤t≤3 s.

m and c, both are constant values

i've tried it on desmos, but is it already correct?

[mystic scarab]

Use x instead of t

Because that's the name of the x-axis in Desmos (honestly I don't know if you can change its name)

So you'll have v(x) = 5m + C•e^(-2x/m), {1 ≤ x ≤ 3}

[devout snow]

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[grim ginkgo]

where did the 4 and 3 come from?

[acoustic leaf]

reading the graph

[grim ginkgo]

yes thats what i need help with

[acoustic leaf]

counting grid squares

[grim ginkgo]

i dont see where the 3 and 4

come into play

[grim ginkgo]

oh

still dont get this

why is one x-1

anmd the other x + 1

[acoustic leaf]

distributing out the multiplication

[grim ginkgo]

ah alright

thanks

.close

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[bleak arrow]

How to find the intersection of:

y = -0.5+x
y <= -0.5+x

(without graphing)

[olive snow]

well one is included in the other

[bleak arrow]

is there a way to use algebra? for example plugging in the first equation into the second?

but it didn't work

i got something trivial

[olive snow]

but

yes it is

the first one describe a line that is included into the second one

so the intersection is that line

[polar chasm]

just graph it

[bleak arrow]

Ok.

[smoky gyro]

yaku

[olive snow]

ouais ?

[smoky gyro]

salut'

[olive snow]

salut le s

[bleak arrow]

.close

[devout snow]

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[devout snow]

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[fallow spear]

Illustrate the representation of irrational numbers on number line

Can someone show me how to do this

[devout snow]

Please don't occupy multiple help channels.

[pseudo basin]

duplicate #help-17

.close

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[arctic pendant]

Help

[arctic pendant]

Guys is my solution for this correct? And how to make it more formal or is it sufficiently rigorous?

I am really clueless about formalization since I am somewhat bad at math formalization 😔

[pseudo basin]

seems fine to me

[arctic pendant]

Can it be made more formal like actual get a epsilon (this is easy just define N_epsilon) but the setup should eventually lead to an arbitrary m so the sum goes boom

How to actually naturally derive such m so I can make it look like there is no black box or anything

[pseudo basin]

what epsilon

you want to prove N is countable

you assume towards a contradiction that N is not countable

like do you already know the sum of an uncountable number of positive terms diverges to +∞?

[runic prawn]

this may not be measurable so i'd say the way you wrote it is not rigorous

[runic prawn]

this is the main idea tho

[pseudo basin]

actually ig like

yeah to rigorize it

you can divvy up all the points into countably many bins in a logarithmic sorta fashion

[arctic pendant]

The trick is that using the fact the uncountable many points can’t be assigned to one sequence so it eventually gets blown up but I find it hard to formalize it…

[runic prawn]

i think this is correct

but u should probably prove it

and u can assume |F| is countable instead of finite

[arctic pendant]

Because one sequence can’t represent uncountable many points so you won’t get an arbitrary epsilon

[pseudo basin]

like for each $n \in \bZ$ make $N_n := { x : \mu({x}) \in (2^n, 2^{n+1}] }$

[woven radish]

Ann

[arctic pendant]

But you won’t get one sequence

Since N is not enumerable so you can’t represent it in one sequence

[jaunty mantle]

Wait why not

[pseudo basin]

this is a seq of sets

[jaunty mantle]

Is that union not just omega

[runic prawn]

it's an uncountable union

[pseudo basin]

which are disjoint, countable in number, and whose union gives N

[arctic pendant]

Instead you admits one epsilon but the supremum is outside I can choose arbitrary m>0 let it goes to infinity

[runic prawn]

you mean \mathcal{A}

no

oh wait

[arctic pendant]

This part is okay though I can use monotone convergence

[pseudo basin]

i don't understand what issue you're seeing here

[runic prawn]

yeah it's Omega

wait no it isn't

it's N

[pseudo basin]

N_0 consists of all points whose weights lie in (1,2]

[arctic pendant]

Like having a decaying sequence doesn’t matter

[jaunty mantle]

It’s omega without all the null sets

[pseudo basin]

N_3 consists of all points whose weights lie in (8,16]

[pseudo basin]

????

[runic prawn]

null singletons

[jaunty mantle]

Clearly the null set is measurable otherwise you wouldn’t be able to tell its null

[pseudo basin]

N_-1 consists of all points whose weights lie in (1/2, 1] etc.

[arctic pendant]

Because the assumption is that the point isn’t countable your decay won’t be fast enough

[runic prawn]

which null set

[pseudo basin]

i wasn't even done lmfao

[jaunty mantle]

Wouldn’t it follow immediately that it’s measurable

[pseudo basin]

i was just presenting a starting pt

break up N into a countable union of sets

whence you note that at least one of said sets must be uncountable itself

[runic prawn]

this = N

[jaunty mantle]

Well omega \ ∪ x in N

[runic prawn]

so Omega\N

how do u know it's null?

if x is in Omega\N then {x} is null yes

[jaunty mantle]

Ah I see, you don’t know the entire thing is null unfortunately

[runic prawn]

yep

[arctic pendant]

But I ain’t really thinking it that way I feel it’s more explainable using monotone convergence theorem though

I can essentially express F as a sequence and make it a partial sum correct?

And then I pull a limit

S_n : Sum_F_n mu(x)

And since f_n is increasing we have continuity from below (here should be MCT I don’t know but maybe continuity also works)

S_n upwards to sup_m S_m

Hence I can essentially write lim S_n = sup_|F| sum mu{x}

It’s easy to think it that way instead of measurability and we are talking about sets inside a sigma algebra

So it’s really meaningless but the point is how to formalize this

This is actually a common technique in combinatorics of how to sum up an uncountable set, but I don’t have detailed proof for it however I have a slice of lecture note

Here

[drifting sinew]

,rccw

[arctic pendant]

Yes but still even this is true I need to formalize this so I actually learn

[devout snow]

@arctic pendant Has your question been resolved?

[arctic pendant]

How’s this trick, it’s well defined right

I am too dumb to actually think how to justify the blown up of the sun using epsilon but maybe I can get away from it with outer measure using monotonicity property

[long jetty]

Hello everyone

I need help for studies

[devout snow]

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[teal kettle]

i need help with solving these somthing keeps going wrong becouse the answer in the book dosent match up

[lilac crescent]

What doesn't match exactly?

Also send the coordinates of the points

[teal kettle]

the answer

[lilac crescent]

Show the coordinates of the points

A,B and C

[teal kettle]

give me a moment)

im doing that rn

you need to click on the photo to see all the numbers)

discord squishes long photos.

the black one is abc the blue one is A1B1C1

[lilac crescent]

both BC and AB are wrong

,,(a+b)^2 \ne (a^2+b^2)

[woven radish]

<rajel />

[lilac crescent]

You did $(2-1)^2=2^2-1^2$

[woven radish]

<rajel />

[lilac crescent]

Which isn't correct

[teal kettle]

why

[lilac crescent]

Because this isn't the formula

,,(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2

[woven radish]

<rajel />

[lilac crescent]

Same for a-b

@teal kettle look here:
(2-1)^2=1^2

Because 2-1 is 1

And it's different than 4-1=3

[teal kettle]

so before i do the 2 thing i must do whatever is in the parethesis?

[lilac crescent]

Yep , but do you understand why

[teal kettle]

becouse the parethesis means that you gotta do whatever is in the parenthesis first?

like (2x2)-2

4-2

smth like that?

[lilac crescent]

sure

You can also do (2-1)(2-1)

Which is 1×1

Just know that $(a-b)^2 \ne a^2-b^2$

[woven radish]

<rajel />

[teal kettle]

btw i dont really understand one other bit hold on

[lilac crescent]

Sure

[teal kettle]

okay so this one was shown to me by my teacher but i dont know what he did at the end with root of 20

i tried to imitate what i think he did

but i think i didnt get it right

bc the numbers came out wrong

for A1B1C1

[lilac crescent]

,,\sqrt{a \cdot b}=\sqrt{a}\cdot \sqrt{b}

[woven radish]

<rajel />

[teal kettle]

oh nevermind i think i got it right then

but they came out wrong bc i didnt follow the parethesis rule

ty anyways tough

[lilac crescent]

!done

[devout snow]

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

[teal kettle]

can i try to solve it by myself before we rap this up

[lilac crescent]

Sure

[teal kettle]

to make sure

ok

this is correct right?

[teal kettle]

did anything go wrong or is this fine?

[lilac crescent]

BC seems fine

but the perimetre isnt correct

[teal kettle]

yeah i know il correct that later i just didnt solve all of the sides yet

[lilac crescent]

alr your fine

[teal kettle]

yay

ok thank you

!close

.close

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[devout snow]

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[cosmic pelican]

I need someone to do this question "Logically" Please
How many bit strings of length eight contain either three consecutive 0s or four
consecutive 1s? Show your work details.

[polar chasm]

can it have both?

what have you tried btw?

[pseudo basin]

can it have a longer run of 0s or 1s

[cosmic pelican]

yea it can have both

[pseudo basin]

like would 11111100 qualify

[cosmic pelican]

yea

i just want a final answer to make sure im right

i got 110

[pseudo basin]

plausible but show your work

[cosmic pelican]

im on my laptop idk how to take pic of my paper XD

i ahve an idea one sec

[lost laurel]

type your work if possible

[cosmic pelican]

Question 2 Answer (Minimal Explanation):

Total number of 8-bit strings = 256

We need to count how many strings have either three consecutive 0s or four consecutive 1s.

Let:

A = number of strings with at least three consecutive 0s

B = number of strings with at least four consecutive 1s

Some strings may have both, so we use the inclusion-exclusion principle:

Number of strings that satisfy the condition = A + B - (A and B)

From known pattern enumeration:

A ≈ 56

B ≈ 76

A and B ≈ 22

Final answer = 56 + 76 - 22 = 110

So, 110 bit strings of length 8 satisfy the condition.

by plus i mean U and by and i mean upside down U

aka union and intersection

[devout snow]

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[worn flax]

Why is it that we can choose epsilon here

[worn flax]

(I was under the impression that the definition goes for all eps...)

[sonic smelt]

The limit definition states that for every positive epsilon exists N s.t. …; In particular, you can choose a suitable value for epsilon and the existence of N will still hold

Since the epsilon is arbitrary

[sonic smelt]

Yeah, meaning you can replace it with any positive real number

[sullen island]

i.e. we're not trying to show a limit statement is true, we know it is true already and we're exploiting that fact to get to that conclusion about sum a_n

big epsilon is at your doorstep

[worn flax]

ah ok. thanks both. I have seen the "we choose eps" to be a certain value, eg 1/2 before, but not "choose eps = L". But I guess it is the same thing here since we are treating L as a constant

[lilac crescent]

whats the book ?

[worn flax]

just course notes my prof wrote

[lilac crescent]

aha

[worn flax]

.close

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[grim ginkgo]

could this be an answer?

[north roost]

What’s the question?

[grim ginkgo]

my bad thought i put it in the first image

[north roost]

Ye

[grim ginkgo]

why is that tho

i mean theres multiple

but what type of working out

do i or can i do

[north roost]

All would work

Integral fo the absolute value of the function

Is just the sum of absolute function

Meaning it makes all negative positive

That sounds confusing…

Hmm

Do u agree that a definite integral from a to b is the sum of the function (height) multiplied by some infinitesimal width dx from a to b?

[grim ginkgo]

yes

i suppose

[north roost]

Ok

So generally

If the function at a certain x is negative

Then the product between f(x) and dx is negative, right?

So if when using definite integral to find the total area under a graph, u can get value that is less than the actual value

Because of these discrepancies

So to fix it

I will make all values of the function positive

Hence the |f(x)|

[grim ginkgo]

i slighty understand

so i have to put

the

[ ]?

[north roost]

Yes the absolute

[grim ginkgo]

and i cant just do f(x) dx

[north roost]

To make all output of f(x) positive

[grim ginkgo]

ah alright

[north roost]

Ye

So area = definite of |f(x)|

Hence the expression

[grim ginkgo]

ok thanks

[north roost]

The remaining is just application of definition of absolute

Since f(x) <= 0 over that interval |f(x)| = -f(x) by definition of the absolute function

The last one is one property of integral

[grim ginkgo]

makes sense

[north roost]

So

Is there any part that seems unclear?

[grim ginkgo]

i dont really have questions i understand it now

[north roost]

Ah ok

[grim ginkgo]

your a good explainer

u a math teacher?

[north roost]

Nah

Just a guy who’s trying to learn math ;-;

[placid minnow]

what a nerd smh

[grim ginkgo]

alright imma close ts

thanks for everything

.close

[devout snow]

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[devout snow]

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[gleaming socket]

I am confused how Stokes theorem applies such that the curve integral of the C is equal to surface integral of S. I do understand how curve integral of C is equal to surface integral of R as C is literally the border curve of R. But don't get S

[acoustic leaf]

it's the border curve of both R and S

[gleaming socket]

Interesting. I've seen other examples where divergence theorem is used for flux of the whole shape and then the plane flux separated

[acoustic leaf]

in general the same closed curve can be the boundary of many different surfaces

[gleaming socket]

Thanks for the clarification

[wary ruin]

Is this the same @acoustic leaf

[acoustic leaf]

sort of, but in this case in order to match the normal vector directions it has an opposite orientation

[devout snow]

@gleaming socket Has your question been resolved?

[devout snow]

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[spring oasis]

guys lets say I have a polynomial P € R^n[X] and a poly Q € R^m[X]
how to show that deg(P+Q) = max{deg(P), deg(Q)}

[sturdy estuary]

Idk

[spring oasis]

please ping when replying

[sturdy estuary]

Can u help me

I’m in help 9

it’s easy maths probably for u if u doing this stuff

[spring oasis]

seems hard, idk

[faint gorge]

Maybe do a case distinction, where deg(P) >= deg(Q) and deg(P) < deg(Q). Then also write down how P, Q look like

[spring oasis]

i forgot to add n,m < infty

P € <1, X, . . ., X^n>

Q € <1, X, . . ., X^m>

[sturdy estuary]

that looks hard

What topics

[spring oasis]

linear algebra / polynomials

[sturdy estuary]

rlly

is this uni maths

[spring oasis]

deg(P) >= deg(Q) when n > m or when n = m

[spring oasis]

either uni or pre university level depending on the country

idk

[sturdy estuary]

pre uni???

Gawd damn

[faint gorge]

yes

[spring oasis]

deg(P) < deg(Q) when n < m

🤔

[olive snow]

Yes

But what if Q = -P ?

[spring oasis]

what will happen with Q = -P?

[olive snow]

You tell me

P+ Q =?

[spring oasis]

P - P = 0

[olive snow]

Ok and what is the deg of 0 ?

[spring oasis]

1?

[olive snow]

Is it the max between deg p or deg q ?

[spring oasis]

no

[olive snow]

Thats why your statement is wrong

You have that deg(P+Q) <= max(...)

Not equality

[spring oasis]

I didnt know

is the deg of 0 0 or is deg 0 1

?

fair enough

[olive snow]

Cuz its x^0

So its 0

Nvm

[spring oasis]

?

[olive snow]

deg(constant polynomial) is 0

[faint gorge]

If $P = \sum_{i=0}^n a_ix^i$ and $Q = \sum_{j=0}^m b_jx^j$ for example if $n > m$ then
[ P+Q = (a_0+b_0)+(a_1+b_1)x + \dots + (a_m+b_m)x^m + a_{m+1}x^{m+1} + \dots + a_nx^n ]
where you have $n+1$ independent basis vectors.

[olive snow]

n+1 mein spezi no ?

Anyway, from what adonis wrote you can prove the inequality @spring oasis for every cases

[spring oasis]

how ?

[olive snow]

See that deg(P+Q) = deg(P) here

And if you switch the role of P and Q

You will have deg(P+Q) = deg(Q)

And when its equal it will be under the <= case

[faint gorge]

and the case work where if degP >= degQ then you have degP and else degQ is precisely what max does for you

[olive snow]

So its fine

Cuz if its equal but with a terrible unlucky moment, the leading coefficient are cancelling each others, well you don't have equality holding

But its fine since the formula is <=

[spring oasis]

I dont understand

[olive snow]

Yes

[spring oasis]

why x^m+1

[stark oxide]

its in P

[spring oasis]

if j goes up to m

[olive snow]

Its like

[faint gorge]

to show that b_m+1 doesnt exist anymore

[spring oasis]

ahh

[olive snow]

[0,m] include in [0,n]

[spring oasis]

yeah hahah

[olive snow]

Its just to explicitely show the end of Q

[spring oasis]

what is the definition of deg

[spring oasis]

i mean in here deg(P+Q) = n+1

[stark oxide]

highest exponent in a polynomial

[spring oasis]

yeah

[stark oxide]

with non-zero coefficient

[spring oasis]

ok

[spring oasis]

i mean in here deg(P+Q)=n+1 but idk how to show

I need to show that the basis has cardinality n + 1?

[faint gorge]

you could use the fact from before where R^n[x] has dimension n+1

[spring oasis]

noo because the highest exponent in the polynomial is n

[stark oxide]

ye you right

[faint gorge]

this is in R^n[x] so it must have dimension n+1

[stark oxide]

maybe deg is something else

[spring oasis]

for getting a cardinality of a basis being n + 1 I need to show that there are n +1 linearly independet bvecvtors

[spring oasis]

yeah I think this already shows they are liunearly independent but we would need to describe abbasis

{a0 + b0, (a1 + b1)x, ... , (am + bm)x^m, am+1 x^m+1m, ..., anx^n}

then this clearly is dim n + 1

but this is when P is not -Q

i guess when P = -Q will be important for the n = m case

,, P = \sum_{i = 0}^{n} a_i x^i \quad Q = -\sum_{i=0}^{n} b_i x^i \quad \text{when n = m} \ P + Q = 0?

[faint gorge]

why would you consider P = -Q?

[woven radish]

renato

[spring oasis]

because when deg(P+Q) <= max{deg(P), deg(Q)}

I thought it was deg(P+Q) = max{deg(P), deg(Q)} for a sec

[faint gorge]

oh is ee

[spring oasis]

no but its irrelevant forget it

[faint gorge]

i thought too

[spring oasis]

,, P = \sum_{i = 0}^{n} a_i x^i \quad Q = \sum_{j=0}^{m} b_j x^j \quad \text{when n = m} \ P + Q = (a_0 + b_0) + (a_1 + b_1)x + \dots + (a_n + b_n)x^n

{(ai + bj), (ai +bj)x, . . ., (ai + bj)x^n}

that is clearly dim n +1 or dim m + 1 since n = m

[faint gorge]

yes

but you need to correct the coefficients

[spring oasis]

what?

[faint gorge]

you wrote a_i+b_j

everywhere

in the end it's a_n+b_n

[woven radish]

renato

[spring oasis]

yeah good catch 😓

{(a0 + b0), (a1 +b1)x, . . ., (an + bn)x^n}

is a basis of cardinality n + 1

[faint gorge]

this clearly shows that the basis vectors are linearly independent because it implies a_i+b_j=0 for all 0<=i,j<=n (but a_i = -b_j is not an issue if we assume that P is not -Q in the first place)

[spring oasis]

yeah

I think its a good argument, what about the case when n < m

[faint gorge]

it's the same argument

[faint gorge]

you bring it into this form and draw a conclusion about the coefficients

[spring oasis]

,, P = \sum_{i = 0}^{n} ai x^i \quad Q = \sum{j=0}^{m} b_j x^j \quad \text{when n < m} \ P + Q = (a_0 + b_0) + (a_1 + b_1)x + \dots a_{n+1}x^{n+1} \dots + a_mx^m

[woven radish]

renato

[spring oasis]

I think I get it now

[faint gorge]

i mean there is no way they are not independent because each basis vector has a different power

[spring oasis]

yeah

exactly

and we are assuming the coefficients are non zero i think

but is also possible that a1 = -b1

and a0 = -b0

and so on

but that is covered because

deg(P+Q) <= max{deg(P), deg(Q)}

like is an inequality not necessarily equal

[spring oasis]

I think I kinda get it but not necessarily the coefficients of the polynomial all need to be zero

for example

deg(1 + 0X + X^2) = 3

no?

no, my bad is 2

deg(1 + 0x + x^2) = 2

but is tricky because the basis {1, X, X^2} has dim 3

[faint gorge]

is it = or <=?

[spring oasis]

<=

[faint gorge]

ahhhh

[faint gorge]

i mean yea if basis vectors vanish because of a_i = -b_j then it only decreases the dimension

[spring oasis]

how is the dimension of the basis related to the degree

because

deg(1+x+x^2) = 2

but dim(<1,x,x^2>) = 3

[faint gorge]

yes it isn't actually

i keep talking about dimensions lmao my baad

but i mean that if for example the dominant powers cancel out because of a_i = -b_j then it would reduce the degree anyway

[spring oasis]

yes

its <= so its fine

[faint gorge]

yea

[spring oasis]

is fine, I had plenty of fun, before my pizza arrives

I will continue it later, thx

.close

[devout snow]

Channel closed

Closed by @spring oasis

Use .reopen if this was a mistake.

[faint gorge]

Enjoy your pizza!

[devout snow]

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[steady creek]

im confused by this question

isn't the given 200 the magnitude?

[wind mason]

No

[misty crest]

at 20 degrees brother

[wind mason]

The magnitude of the force vector is different than that of the force

[pseudo basin]

eh???

[misty crest]

🤔

[wind mason]

Idk how to word it better

[pseudo basin]

what is the diff between "the force vector" and "the force"

[faint gorge]

i wish i had directions

[wind mason]

Like the force of 20 degrees

[heavy current]

they seem to have messed up the calculation for |F|?

[acoustic leaf]

yeah it should be

[heavy current]

that expression evaluates to 200 N indeed

[pseudo basin]

how does one mess up the calculations this bad

a force of 200N has magnitude 200N

[heavy current]

the magnitude of the force is 200 N for sure

I wonder if they meant the x component of the force?

no, it couldn't be

the magnitude is just 200 N

idk what they were going for in this problem

[acoustic leaf]

they show the correct calculation for the magnitude, the final number is just wrong

not that it's particularly necessary since the magnitude was given

[heavy current]

even without the given magnitude, the calculation is clearly wrong because sin^2(x) + cos^2(x) = 1

[devout snow]

@steady creek Has your question been resolved?

[devout snow]

Channel closed

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

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[cold kraken]

is this true

[heavy current]

try some examples :p

[cold kraken]

would you mind helping with this question

[dire mountain]

yeah!

you know that a, b, c, d > 0, right?

[cold kraken]

yes

[cold kraken]

are we tlaking about this question?

[dire mountain]

alrighty. and you have a >= b and c>= d, that's your hypothesis

[dire mountain]

the one from before

or did you already get that one figured out

[cold kraken]

ohh okok tysm

[cold kraken]

yeah kind of

so i mainly need help with the one after

thank you so much

<@&286206848099549185>

<@&286206848099549185>

[thorny isle]

hello guys

[cold kraken]

hellooo

[thorny isle]

if you dont mind can i share a problem of jee advanced maths

exam

complex numbers?

[cold kraken]

sorry but im afraid my question hasnt been resolved yet 😭

[thorny isle]

which one

am and gm one?

[cold kraken]

this one

yes

do u know how to do it?

[pseudo basin]

!occupied

[devout snow]

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).

[thorny isle]

basic concept is

[cold kraken]

yeah no i understand the basic concept

[thorny isle]

a+b/2> or equal to root ab

but let me try once

bro this is tough

good question ngl

[cold kraken]

ikr...

and this is yr 10..

[thorny isle]

which grade are u

[cold kraken]

did u manage to crack the question

[cold kraken]

10

[thorny isle]

damn ur good

im at 12 grade

[cold kraken]

nah but ive got no clue how to solve it

[cold kraken]

ahhhh damn

[thorny isle]

i think first step

a+b+c+d = z

if u notice

a(y-1)+z lly b(y-1)+z lly c(y-1)+z lly d(y-1)+z

if u again notice its going

product i mean big pie (cyclic)

[cold kraken]

wait sorry whats y

[thorny isle]

big pie( its a operator like u see sigma as a summation lly for product its a fucktion

sorry

its x

i cosidered y as x

considered

[cold kraken]

ohhh okok

[cold kraken]

wait so what does it do

[thorny isle]

it keeps products the functions

∏
​

its this

one

[cold kraken]

oh

but what can i do with it

[thorny isle]

like u know this function

∑( u know this sign)?

u might have learned in statistics

[cold kraken]

thats sigma right

[thorny isle]

yep

it sums all the no

simmilarly

[thorny isle]

product we denote this

[cold kraken]

ohhhh right

i think i kind get it

[thorny isle]

yep

now use the equation

∏^4 Ai greter or equal to (a1+a2+a3+a4)^4/4
i=1

solve u might get x > or equal to 2

[cold kraken]

ohhhhhh icic

this helped a lot

thank you os much!!!

[thorny isle]

im noob at maths bro

i took help from a engineer rn

he explained me

im a failure dont thank me

but im glad

dont let ur talent go waste

[cold kraken]

HELPPP TYSM

UR ACUTALLY RLLY GOOD

[cold kraken]

my god thats such a flatter

but thanksssss

.close

[devout snow]

Channel closed

Closed by @cold kraken

Use .reopen if this was a mistake.

[thorny isle]

which country u from

[cold kraken]

australia

wbu

[thorny isle]

india

u have knowledge of complex no?

[cold kraken]

ahhh icic

[cold kraken]

no...

[thorny isle]

limits?

[cold kraken]

yes

[thorny isle]

beautiful

[cold kraken]

like functions wise?

[thorny isle]

ye

take a jee advanced paper of mathematis

[cold kraken]

i do know abit abt it not much tho...

[cold kraken]

whats that

[thorny isle]

worlds toughest entrance exam

[cold kraken]

oh hell nah

no way

if i can barely do amgm shi

no way..

[thorny isle]

bro ur a 10th grader

first practice from basics

u will be

a champion

i trust u bro

[cold kraken]

HAHAHA yes guys im gonna keep dreaming

[thorny isle]

im feeling the next euler in u

[cold kraken]

oh hell nah bro

appreciate the compliment but

nah thats over the top

[thorny isle]

i think u should work hard

good luck pal

[cold kraken]

thanks man u too!!

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[restive river]

hi, can someone help me with pemdas?

[polar chasm]

any specific question

[restive river]

5x - 10 = 2x + 5x

i wanna be able to learn how to solve this

[polar chasm]

okay, thats a linear equation

[runic grove]

subtract 5x from both sides

divide by 2 both sides

[polar chasm]

firstly, get everything with x on one side and the rest on the other side

[runic grove]

your goal is to isolate x

[restive river]

5x - 2x + 5x like this?

[polar chasm]

You cant simply move stuff around

to change the equation, you must do either of those:

1. add something to both sides
2. subtract something from both sides
3. multiply both sides by some number

4. divide both sides by some number

so if you e.g. want to move the 5x from the left side to the right side, you must subtract it

5x - 10 = 2x + 5x, if we subtract 5x from both sides, we get:
-10 = 2x + 5x - 5x

is this step clear?

[restive river]

yes

[polar chasm]

Okay, now try simplifying 2x + 5x - 5x

2 apples + 5 apples - 5 apples = ?

[restive river]

uh

2 apples + 5 apples - 5 apples = 2

[polar chasm]

2 apples to be exact

and in the same way
2x + 5x - 5x = 2x

[restive river]

yes

[polar chasm]

so we can simplify it to just 2x

making the equation
-10 = 2x

if 2x = -10, what does one x equal?

[restive river]

number

[polar chasm]

which number though?

[restive river]

any but it needs to match the = -10

[polar chasm]

yeah, but there are not many numbers which match that

in fact there is only one

x is some number

and 2x must be -10

2x is 2*x

what number gives -10 when it's multiplied by 2?

[restive river]

uh

is the x 5?

[polar chasm]

almost

but 2*5 is 10, not -10

you gotta make it -10

2 * ? = -10

[restive river]

hm

[polar chasm]

negative numbers exist too, x can be negative

[restive river]

ok

is it 2*5

[polar chasm]

2*5 is 10

we need to make it -10

[restive river]

or 2*-5

[polar chasm]

2*-5, yes

if thats what you were typing

[restive river]

yess

[polar chasm]

x is -5

we can take the equation 2x = -10, divide both sides by 2 and get
x = -10/2 = -5

so x is -5

3x = 12
try this now

[restive river]

this becomes so understandable

[restive river]

the x is 4 so its 3x4=12

[polar chasm]

perfect

5x = -20

[restive river]

hmm

[polar chasm]

It's again negative

[restive river]

5(-4) = -20

[polar chasm]

so x = -4

you can also take -20 and divide it by 5

5x = -20
x = -20/5 = -4

what about this one:
3x = 0

[restive river]

x is 0 then?

[polar chasm]

Correct, x is 0 / 3 which is just 0

how about -2x = 10

[restive river]

hm

-2(5) = 10

am i wrong?

[polar chasm]

-2*(5) isnt 10

-2 * 5 is -10

[restive river]

oh yeah

so its

[polar chasm]

you can again take 10 and divide it by -2

x = 10 / (-2)

[restive river]

-2*(-5) = -10

this correct now?

[polar chasm]

x = -5, correct

so to recap, when we have something like
5x = 40, we can divide both sides by 5 and get
x = 40 / 5, which is just 8. so x = 8

lets try something a bit more difficult now

3x - 2x = 10

[restive river]

3(5) - 2x = 10 ?

[polar chasm]

3x - 2x is similar to 3 apples - 2 apples

[polar chasm]

x is there twice

you have to start by adding those x's together

until you get something nice such as
2x = 10
or
3x = 15
or something like we did earlier

start by simplifying 3x - 2x

[restive river]

i have to guess 2 numbers of those x?

[polar chasm]

x is the same number

you dont need to guess it, you can be systematic about it

3x - 2x can be simplified

3 apples - 2 apples is what?

[restive river]

1

[polar chasm]

1 apple to be exact

in the same way, 3x - 2x is just 1x

or just x

[restive river]

3x - 2x = 10

[polar chasm]

so this immidiately simplifies to x = 10

because 3x - 2x is just 1x, which is x

how about
2x + 3x = 10

[restive river]

hm

it could be any number?

[polar chasm]

Nope

3x + 2x is like 3 apples + 2 apples

[restive river]

2x + 3x = 10
2(2) + 3(5)=10

10=10

[polar chasm]

Why 2(2) and 3(5)?

[restive river]

2*5

=10

wait im wrong

[polar chasm]

Yeah, its not simple as that though

We have to start by adding 2x and 3x

2 apples + 3 apples is 5 apples

So whats 2x + 3x?

[restive river]

5

[polar chasm]

5x, not just 5

[polar chasm]

2 apples + 3 apples is also 5 apples and not just 5

So the equation simplifies to 5x = 10

Because 2x + 3x is just 5x

[restive river]

oh

[polar chasm]

Now can you solve this?

[restive river]

5x = 10
5(2)=10
10=10

[polar chasm]

X is 2

[restive river]

yes

[polar chasm]

And we can verify that

3x + 2x = 10
3(2) + 2(2) = 10
6 + 4 = 10
10 = 10

x + 8x + 3x = -24 try this now

[restive river]

makes sense

[restive river]

hm

so x is negative then?

[polar chasm]

You dont have to do it in one step

Algebra is all about doing things step by step

The first step here will be simplifying x + 8x + 3x

it's like apple + 8 apples + 3 apples

[restive river]

x + 8x + 3x = -24
(-5) + 8(2) + 3(3)=-24

[polar chasm]

no, not this

dont try to guess x

be systematic

x + 8x + 3x is what?

[restive river]

x + 8x + 3x

[polar chasm]

Yeah, but can you simplify it?

[restive river]

11

[polar chasm]

try recalculating it

and youre forgeting x

apple + 8 apples + 3 apples is?

[restive river]

11

[polar chasm]

try again

you are off by 1

[restive river]

12?

[polar chasm]

Yes, 12 apples

never forget that part

in the same way, x + 8x + 3x is 12x

note the "x", it's not just 12, it's 12x

x + 8x + 3x = -24
okay, so if x + 8x + 3x is 12x, what does the equation simplify to

[restive river]

x + 8x + 3x -24
-12 - 12 = -24

[polar chasm]

uh not quite

also it was
x + 8x + 3x = -24, not -24

x + 8x + 3x is the same thing as 12x

[restive river]

you put -24

[polar chasm]

oh right, then it's = -24

[polar chasm]

anyway, because its the same thing as 12x, it just simplifies to
12x = -24

[restive river]

but how is it 8x + 3x = 12?

[polar chasm]

x + 8x + 3x is 12x

not 8x + 3x = 12

x + 8x + 3x is 12x

in the same way as
apple + 8 apples + 3 apples is 12 apples

[restive river]

isnt it 11? where do you get other 1 from

[polar chasm]

from the x

1 apple + 8 apples + 3 apples is 12 apples

1 + 8 + 3 is 12

x is the same thing as 1x

the 1 is just ommited

I'll have to go soon btw

https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-equations-and-inequalities/cc-6th-intro-equations/v/variables-expressions-and-equations

[polar chasm]

try watching this and doing some exercises

if you watch and try the whole unit, you should be able to understand linear equations

https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-equations-and-inequalities

here is the whole unit

[restive river]

could you show me the whole equasion? i wanna know how would it looked like

[polar chasm]

x + 8x + 3x = -24, now notice that x + 8x + 3x is same as 12x
12x = -24, now we divide both sides by 12
x = -24 / 12 = -2

2x + 3 = 5x - 9
Firstly, we will try to move everything with x on one side, and everything without x on the other side.
So subtract 2x
3 = 5x - 9 - 2x
Now add 9 to both sides
3 + 9 = 5x - 2x
Now notice that 5x - 2x is same as 3x and 3 + 9 is 12
12 = 3x
Now divide both sides by 3
4 = x

x = 4

ill have to go now, the above are few examples

[polar chasm]

to learn more, check this

[restive river]

that all looks complicated

is there a shorter way to explain?

[devout snow]

@restive river Has your question been resolved?

[hidden sparrow]

x + 8x + 3x = -24

Step-by-step:

Add all the x's:
x + 8x + 3x = 12x
So it becomes:
12x = -24

so, to get the value of x, 12 will go to the other side, as u know that if a number is multiplied by a variable, it will be divided by the value on the other side. Therefore:
x = -24 ÷ 12 = -2 or x = -24/12

✅ Answer: x = -2

[restive river]

makes sense

@hidden sparrowgive me a equasion

[hidden sparrow]

x + 10 = 5

[restive river]

huh

thats impossible to solve

oh

x + 10 = 5
5 - 10 = 5
5=5

[hidden sparrow]

nope

[restive river]

now?

[hidden sparrow]

x + 10 = 5
x = 5 - 10
x = -5