help-19

[versed depot]

find X value

sub it back into x^2y

find the min product

[split mica]

Second derivative test?

Oh I forgot about that

nvm

[versed depot]

Yea

[versed depot]

I have another question tho

do you know the shadow problem

this is the question

i understand part A but i dont understand like what B is asking

[split mica]

So the tip of his shadow is going to be the point where the light blocked by the top of his head would have hit the floor

Do you need help with the math or how the question is asked?

[odd edge]

@versed depot Has your question been resolved?

[odd edge]

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[versed depot]

2hd part

[split mica]

In part b, the setup is the same as part a, but you are finding the speed of the bottom right vertex instead of the change in length of the shadow

[versed depot]

so dx/dt + dy/dt

so the tip of the shadow is x+y

?

[split mica]

Yes

[versed depot]

k

but like

the tip

that confuses me

like i would assume the tip is just that

ok yk what imma j memprize it

[split mica]

The tip is the one spot, but you're finding the speed of that one spot, which is the sum of the two changes in length, if that helps with the confusion

[versed depot]

ohhh

ye

that does make sense

can u help me with this

but i think i know how to do it

its just quotient rule

i got D but the answer key says E

[split mica]

could you show your work?

[versed depot]

(0)(sinx+cosx)-(3)(cosx-sinx)/(sinx+cosx)^2

[split mica]

That far is correct

I did it by taking the bottom to be ^-1 and doing the chain rule but it gets to the same spot

[versed depot]

Yea

and just simplyfying it out

[split mica]

so the 2 things you need to do and what make the two answers different is:

1. the - being distributed to turn -(cosx-sinx) into sinx-cosx
2. expanding the bottom

[versed depot]

but i got - in front of D

theres no negative in front

wait wdym

how does a negative being distrubted

how did u get the negative out

this is what i got so far

[split mica]

The negative that is in front of the fraction gets treated like a -1 and is distributed to the (cosx-sinx)

[versed depot]

so wont it just be -3cosx+3sinx

in the numerator

lets just ingore the denominator

we are just distributing the negative right

just flip signs lol right

[split mica]

yeah, but you also distributed the 3 which is not done in the answer

[versed depot]

ok

so

-3(cosx-sinx)

[split mica]

but you then distribute the negative and flip the signs

[versed depot]

can you show the work

so

3(-cosx+sinx)

[split mica]

Yes, that's it

[versed depot]

ok

-cosx+sinx is the same thing as sinx-cosx

[split mica]

Yes

[versed depot]

i dont get the denominator thing

[split mica]

The denominator you just have to expand and simplify

so (sinx+cosx)^2 turns into sin^2x+cos^2x+2sinxcosx

[versed depot]

ok

[split mica]

Then, sin^2x+cos^2x is a Pythagorean identity

[versed depot]

oh

yea

its E

mb

i have no idea to do this one

[split mica]

Do you know what an inflection point is?

[versed depot]

chaneg in concavity

[split mica]

Right

so, assuming you aren't going to go based off of the visual, how do you calculate concavity?

[versed depot]

second derative

check where it changes from the left and the right of critical points of second deriative

[split mica]

An important thing to note, an inflection point is specifically where the concavity changes sign

[versed depot]

Yes

uh

isnt that wat i said or na

did i say it wrong

[split mica]

If you think of concavity as concave up vs concave down then yes its the same thing

[versed depot]

Ye

lol

[split mica]

I don't think you'll run across anything where the distinction matters any time soon

[versed depot]

so the second deriative of that function is (n-1)nx^(n-2)

thats what i got

[split mica]

right

so for f(x) to have an inflection point, what must be true about the second derivative?

[versed depot]

uh

it changes from cu to cd or cd to cu

[split mica]

so it must at some point be cd and some other point be cu

but numerically, how do you quantify cu vs cd?

[versed depot]

cd = negative

cu = psotive

if the second deriaitve is negative on a certain point/interval it is concave down on that certain point/interval vice versa for cu

[split mica]

So for what integer values of n can there be both a positive and negative second derivative?

[versed depot]

huh?

the function is alwayspositive

no matter what n value the function is always positive

[split mica]

Keep in mind that x can be negative

[versed depot]

oh

bro

honeslty

i dont know

can u tell me

[split mica]

nothing around the x matters for this except the power, so, for these purposes the second derivative is x^(n-2)

if x^(2k), the result is always positive, but if x^(2k+1), the result is whatever sign x is

so there is no inflection point when x is to the power of an even number, but there is when it is to the power of an odd number

So, you need to check the parity of n-2

subtracting by 2 does not change if it is even or odd, so you just have to check the parity of n

so, when n is (even/odd), f(x)=x^n has an inflection point

[split mica]

This is more of something to know than a specific step

[versed depot]

but what if -x^2k

thats negative

[split mica]

because that means -(x^2k), where the - is applied after the power

[versed depot]

thats still negative

no bro

-x

its the same thing

[split mica]

when you substitute values for x, it would be (-x)^2k, which is still positive

[versed depot]

this makes no sense to me

[versed depot]

??

[split mica]

k is a positive integer

any positive integer

[versed depot]

if k was 3/2 and x was 1

[split mica]

2k is an algebraic version of an even number

[versed depot]

thats negative

fraction not an integer

[split mica]

I did not specify before, but k is a positive integer in this case

[versed depot]

ok

yea

it will always be positive

[versed depot]

this will always be postive

[split mica]

right

but x^(2k+1)'s sign depends on the sign of x

[versed depot]

wdym

sign of x

[split mica]

if x is negative the result is negative and vice versa

[versed depot]

yea ok

[split mica]

So, you need to know the parity (even or odd-ness) of (n-2)

[versed depot]

ok

[split mica]

if n-2 is even, then it is always positive, if odd, then it can be either positive or negative and therefore have an inflection point

[versed depot]

???

wdym if its odd it can be even

oh ok

[split mica]

My bad, mistype

[versed depot]

oh hok

thx that makes sen

but bro

we have 50 mcq

how can i look at the qurestion

and solve it quick

[split mica]

So the answer is just that if n is odd there is an inflection point

[versed depot]

yea

[split mica]

the easy more graphical way imo to just look at this one in particular and solve it quick is to know what the base shapes look like

[versed depot]

ye but we dont have tat

[split mica]

wdym?

[versed depot]

like

we arent given

graphing calculators

to do this

[split mica]

right, but you can solve this one pretty much instantly if you remember what the graph looks like

The specific shape is not important, the important part is that with the even ones it makes a u but with odd ones it makes an s-curve

[versed depot]

cant we just

quickly find second derativie

then look at the power

and just think of that

[versed depot]

then use logic from ehre

[split mica]

That works too, but you asked for something faster and remembering the shape is faster

[versed depot]

k

[split mica]

It's probably fine to just remember the answer to this question on a test because i doubt you'd get a question that is similar enough that would be made easier from a process that would not be solved by just remembering that answer

[versed depot]

ok

[versed depot]

can u help wth this

[split mica]

What have you done so far?

[versed depot]

i know a and b arent right

[split mica]

right

so what do you not understand to do the rest?

[versed depot]

i got C

how is it D

so find f(1)

which is 0

then find g(0)

which is DNE

[split mica]

right

[versed depot]

its not continious at x=1

so its true

[merry axle]

all are true

[versed depot]

no C isnt

g(0)

thats DNE

f(DNE) is no where

so thats false

wait no

cs theres a -1 at the end

so the dne squared would just be 0 or smth then u -1

so it would j be -1

so C is false

[split mica]

No, dne is a discontinuity

[versed depot]

wait bro

[split mica]

I think you need to read the answers more carefully

[versed depot]

ok

[split mica]

the answer is not that a discontinuity is at a point, it is that that is the only discontinuous point

[merry axle]

f(g(x))=(1/(x^2))-1

[versed depot]

?????

[versed depot]

what bro

what are u gys saying

[split mica]

D is the answer because 1 is not the only point where it is discontinuous

[versed depot]

explain why C isnt

[merry axle]

c is true

wtf

the composition is continuous except at 0

[split mica]

C is true and therefore not the correct answer, as it is asking for which is false

[versed depot]

how to do

f(DNE)

cs g(0) is DNE

what if f(DNE)

[split mica]

DNE

if DNE shows up anywhere the answer is DNE

[versed depot]

ok

moving on to D

how is D false

what ther x values

[split mica]

So, the function is discontinuous when the input to g(x) is 0, right?

the input is f(x) in that question, so you need to find when f(x) is 0 to find the discontinuities

[versed depot]

ye'

[versed depot]

yes

[split mica]

when is x^2-1=0?

[versed depot]

oh

its also at -1

[split mica]

Right

so 1 is not the only discontinuity

[versed depot]

wait

C is true bro

oh yea

bro

im stupid

do i just find deriative to do this question @split mica

[split mica]

not just, it is a necessary step but there is a second step

the derivative is the slope of the tangent, the normal is perpendicular (at 90 degrees) to it

[merry axle]

hey that have concluded?

d is false?

[versed depot]

wdym

[split mica]

The derivative is the first step, and putting in the x value gives you the slope of the tangent

but it asks for the line perpendicular/normal to it

[versed depot]

k got it

[split mica]

do you know how to find a perpendicular slope?

[versed depot]

u just solve for dy=dx

i got 1

[split mica]

I don't know what you mean by that, and it's not what I'm thinking of

[versed depot]

the answer key says -1

i plugged in the x value

and i got 1

[split mica]

So you found the slope of the tangent?

and got 1?

[versed depot]

yes

[split mica]

So that's why it's important that it asked for the perpendicular line

[versed depot]

so thats the opposite slope of the tangent

got it

[split mica]

but opposite is not just negative

it's the negative reciprocal

in this case it's just negative because the reciprocal of 1 is 1

but generally the perpendicular slope =-1/m, where m is the starting slope

[versed depot]

oh

ok

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[viscid crane]

is "measures indicated" x?

[viscid crane]

x and y?

well obviously

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[viscid crane]

.reopen

[odd edge]

✅

[viscid crane]

but is it possible tho

hmm

2y + 180 - 2y = 180

180 = 180 😭

[steep mantle]

what is it asking

[viscid crane]

if its possible for the measures to be indicated

o wait

nvm

x + (180 - x) = 180?

but thatll just make 180 = 180

i completed something similar before but that was with angles and not variables

im stuck stepbro 🥺

how do i figure out y and x

cuz x + 180 - x = 180
is just
180 = 180
which is pretty useless

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[slow sandal]

Two capillary tubes AB and BC are joined end to end at B, AB is 16cm long and of diameter 4mm whereas BC is 4cm long and of diameter 2mm. The composite tube is held horizontally with A connected to a vessel of water giving a constant head of 3cm and C is open to the air. Calculate the pressure difference between B and C. (In centimeters of a column of water)

I tried to solve this by solving for the pressure difference between A and B, which is rho * g* h which is 1 * 980 * 3 = 2940 Ba
Then I recalled the equation of continuity
a1V1 = a2V2 a1 = pi * 0.2 ^2 a2 = pi * 0.1^2 V1 = pi/8 * 2940 * 0.2^4/(η * 16) V2 = pi/8 * p * 0.1^4/(η*4)
We want to find p/(rho * g) to find the column of water

However when solving for p ( i got 47040) which corresponds to 48 cm of water, but the answer is 2.4 cm How did I go wrong by a factor of 20?
Here is the solution online (which I didn't understand)
https://youtu.be/K7SCruah6ds?si=cMvtRLFyX8eOxQeC

the image has different numbers but the situation is the same

[slow sandal]

also i tried the same with the example in the iamge

and im also off by a factor of 20

i dont know why

<@&286206848099549185>

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[glacial tapir]

What should I use to determine if this converges or diverges?
the arcsin throws me off

[pastel dew]

use its talylor expansion

[glacial tapir]

Aight thanks

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[mystic saffron]

Its like basic geometry but i forgot how to do it

[pale jay]

u did it right!

just continue the cross multiplication

the two triangles are similar triangles right?

[mystic saffron]

i think so

[mystic saffron]

with a equal sign in between?

[pale jay]

yes

then solve for x and ure done

[mystic saffron]

alright thank you

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[jagged lantern]

how do i antidifferentiate x(4x-1)^2 ?

[jagged lantern]

.close

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[mystic saffron]

Find the complex solution in:

[mystic saffron]

<@&286206848099549185>

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[sage helm]

Do these 2 symbols mean the same thing?

[nocturne brook]

without more context, probably not

I see the left as congruency, usually as in $a\equiv b\pmod n$

[clever fjord]

Flip

[nocturne brook]

right seems self-explanatory, defining something via an equation

[sage helm]

Context: If I wanted to replace the def symbol with the triple equal, I wondered if it would be correct use of the symbol:

[nocturne brook]

I've also seen the left to mean equivalence between two statements

[nocturne brook]

this is defining Tr[A], and I would be confused to see the equiv sign in its place

are you looking for an alternative symbol?

[sage helm]

I was just curious.

[nocturne brook]

I've seen $\coloneq$ more frequently than $\overset{{\rm def}}{=}$

[clever fjord]

Flip

[nocturne brook]

that's just me though

[sage helm]

Good tip.

Thank you for the info.

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[steady cargo]

The question says "State the value of the variable for which each question is undefined, then solve it." I'm mainly just confused on how I should start solving this, should I try to find an LCD?

[steady cargo]

I know how to solve it, just confused about the part that says state the value of the variable for which each question is undefined.

[mystic saffron]

if u want the find the value for which the eqn is undefined

it is undefined when x = -1 and x = 1

as denominator becomes zero

[steady cargo]

Oh I see, so how would that impact me solving the equation?

[mystic saffron]

idk abt that but the question said to find the value for which the eqn is undefined

[merry axle]

basically you solve and if you find a solution that makes the original equation undefined, then it is not a solution.

[steady cargo]

Okay thanks.

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[mystic saffron]

solution of the equation if u need it

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[mystic saffron]

Best way to revise maths ?

[vale vortex]

probably do maths

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[lofty summit]

i am doing a math midterm and this tripped me out

[umbral pasture]

hmm

did you notice the little ticks they have in the angles?

it means they are the same

[stone vector]

Please do not ask for help on exams. That is not allowed in the server

[spark lintel]

do i solve it or no

[ember oak]

!nosols

[odd edge]

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.

[ember oak]

Also, looks like OP got nuked

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[mystic saffron]

I know the answer is 2*pi

[mystic saffron]

But I need to prove that somehow

Where should I even begin?

[brittle beacon]

Noticing that x is 1/(1/x) might help

[ember oak]

(for non-zero x )

[brittle beacon]

[mystic saffron]

Yeah Im kinda lost

Idk if It's something obvious

or an identity of some kind that I'm forgetting entirely

[brittle beacon]

Do you know or remember any "standard" limits?

[mystic saffron]

if u mean basic ones like

lim(x->inf) [1/x] = 0 then sure

[brittle beacon]

Any ones involving trig functions?

[mystic saffron]

Nope we rarely mix trig functions with limits but i really felt compelled to try and get to the answer for this one

or understand where the answer comes frmo

[brittle beacon]

So you haven't seen the limit of sin(x)/x as x approaches 0?

[mystic saffron]

oh i have

once

i see

[brittle beacon]

If you're smart about it, then...

[brittle beacon]

...you can turn the limit you have into something that's kinda like that limit of sin(u)/u as u approaches 0

[mystic saffron]

Yea i get it now

thanks

I think this is what you do Either way, thanks again (answer for anyone curious)

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[ember oak]

you did good

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[winter spruce]

I'm beyond confused

[amber schooner]

is this not your work?

how could you be confused

[winter spruce]

yes but I think the identity should be verified, at least that's what Im understanding from what the exercise tells me to do

[amber schooner]

what

you’ve just proved it

from the induction hypothesis P(n) was true and you’ve just shown that it holds for P(n+1)

[winter spruce]

but shouldn't the outcome be (n+1)^2 again?

no cuz n is n+1

damn

[amber schooner]

((n+1)+1)

[winter spruce]

ight thanks lol

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[ember oak]

This is where people always end up getting confused

The notation is very particular

[tropic fulcrum]

can someone help me?

[odd edge]

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[mystic saffron]

Find tan ,if AB = 8√2 and AM = 11

[green glen]

what have you tried

[mystic saffron]

Putting 90 - theta

in thr other side

completing the polygon

[pseudo dove]

hmm
welp

if we add a height for the triangle ABM from point B

and name its intersection with AM, for example, D

we get a right triangle ABD with two 45-deg angles

pythagoras tells that AD = BD = 8

then DM = AM - AD = 3

now looking at triangle DBM

[mystic saffron]

√73

ah

it's 3/8

right

[pseudo dove]

i think so

for the angle BMD

[mystic saffron]

wow, thanks

yeah

they are similar triangkes

AAA

[pseudo dove]

oh yeah it's 3/8 in the end

tan BMD = 8/3

[mystic saffron]

3/8 yes

[pseudo dove]

gg

[mystic saffron]

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[vagrant sluice]

I an doing the ratio test and getting -3>x>3

[vagrant sluice]

but the

center is 5

how?

do i find R?

im tryna find radius of convergence

[cedar geyser]

Center is 5 because x-a

20-4x -> solve for x

[cedar geyser]

This would be ur radius of convergence

3

[vagrant sluice]

but its not

the answer key says 2

[cedar geyser]

Ok wait lemme try doing it

Yeah I agree it’s 2

I applied root test

[cedar geyser]

Is ioc 3<x<7?

I might be wrong

[violet pier]

I agree with you, Alex

[vagrant sluice]

so im doin it wron?

where

[violet pier]

-8 < 20-4x < 8
-28 < -4x < -12
7 > x > 3

[vagrant sluice]

why are we allowed to brnig out the 8 from the abs value

but not 20

or 4

[violet pier]

Because multiplying is nicer to signs than adding / subtracting

[vagrant sluice]

i see

so if its multiplyin or dividing i can bring it out

but if i cant multiply or divid without sub/adding first

use the def of a abs value

which would be, in thid example -8<20-4x<8

[violet pier]

Yup.

[vagrant sluice]

Ok thanks!

appreciate u both

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[hushed tide]

whats the next step after u make the line? howd they get those fractions man

[grand mural]

Use trigonometry

[grand mural]

Find this angle

[hushed tide]

45

[grand mural]

Now use trigonometry to find the coordinates

[hushed tide]

uuuuuhm

where did sqrt 2 come from

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[spare bramble]

Working on working with Percentages. Like, finding either the whole, part, or percentage itself in a question. Ran across a question and got 14.3 as the result, only for me to be told the answer was 14.2. My math was correct, but apparently the 3 transmogs into a 2.
There's also cases where I feel they're just making stuff up so I get it wrong intentionally. Like dividing a number against another number. It was 45 ÷ (I forgot). I got all the way down to the remainder of 36. Apparently, I'm supposed to keep dividing. Am I supposed to keep adding 0s until the remainder becomes 0 atp? I've seen cases where you don't don't do that however.
This is extremely confusing with none of it really being explained at all so I'm just lost as all hell

(All within the subject of working with Percentage Problems)

[steep mantle]

by remainder of 3, it means 14+3/15=14+0.2=14.2

[spare bramble]

14+3÷15=14+0.2?

[spare bramble]

Ah for reference the question was
15% of 213
So
213 ÷ 15
Got 14.3 as a result. Was looking up how I got it wrong. Apparently my math was right, but the 3 becomes a 2. How that is I don't know, nor is it explained. The remainer just becomes .2 instead of .3

So far this subject is the first and only one to genuinely give me hiccups

There's other stuff but I didn't take SSs or pictures. Only this instance

[tepid pelican]

15% of 213 isnt 213 / 15

it's 213 * (15/100)

e.g. 50% of 213 is (50 / 100) * 213 = 213 * (1/2)
100% is (100 / 100) * 213 = 213 * 1
25% is (25 / 100) * 213 = 213 * (1 / 4)

it makes sense if you think about it

[spare bramble]

I'll apply this logic tomorrow. This isn't how it was explained at all.
I was taught
Whole = Part ÷ Percentage
Part = Percentage × Whole

None of those other steps (the 15/100) was ever brought into existence

For me I'm doing the math right (for the most part) and there's some curve ball thrown into the equation that makes it far more complicated than it actually is, so I'm more confused than anything. So I'm left either thinking I don't understand it, I'm being taught poorly, or both.

[odd edge]

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[heady plover]

I am mainly confused about the ln^1/3 x part

,w plot 1/(ln^(1/3)x)

[clever fjord]

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[heady plover]

Shouldn't the whole function be convergent then?

[sharp oak]

Yes that's divergent

[heady plover]

Why

[sharp oak]

Increases without bound

[heady plover]

What about 1/lnx then?

[sharp oak]

I think it's a pretty chill function

[heady plover]

So..?😭

[sharp oak]

,w graph 1/ln(x)

[heady plover]

I have my answer, just want a confirmation

[clever fjord]

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[sharp oak]

That approaches 0 from the right.

Ooh wait, the other one isn't even defined near 0

But wait, why not? Doesn't seem like the ³√ should be doing that

,w graph 1/(cube root of ln(x))

[clever fjord]

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Requested by kaynex

[heady plover]

Omg why is it different from the graph I draw above

[odd edge]

@heady plover Has your question been resolved?

[feral scaffold]

it seems like your ^1/3 one was taking the principal root
and when lnx is negative, it's a complex number

and cube root on Kaynex's graph is taking the real valued root

[heady plover]

I still don't quite understand but, so in the end, it's convergent or divergent?😭

[feral scaffold]

from 0+ right?

[heady plover]

Yes

[feral scaffold]

lemme try it, haven't looked at the convergence

I think it's divergent, because the denominator goes to 0

[heady plover]

Ok thx

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[jaunty current]

I'm lost here, where did 5x^2 came from?

[strange aspen]

-6x^2+x^2=-5x^2

[jaunty current]

Wait, I don't understand

[strange aspen]

Yk how x-3x=-2x
It’s the same thing here

[thin kelp]

when u expanded the square one of the terms is -6x^2 which when added to x^2 becomes -5x^2

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[serene axle]

cos2theta formula

cos2x = 2cos^2x -1

[low locust]

() would improve the readability quite a lot

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[sinful prism]

i’m doing grade 11 trigonometric identities how do you know which side to start with when proving ls and right side are equal

[sinful prism]

also how do you know what to sub in and out

[nimble blaze]

many of those types of questions involve the application of one of the pythagorean trig identities
start with whichever side feels more complex

[sinful prism]

okay also i know to always sub out your tangeant but how do you know what else to sub out

[serene axle]

There is no definite "method", for trigonometry, you can only practice and develop your intuition

[sinful prism]

okay

can the answer be different sometimes as things can be = to eachother

[serene axle]

Yeah, the same answer can be written in several forms

[sinful prism]

alright good to know

thanks

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[sage terrace]

I have this problem on my test that I was able to sove, but I can't explain the process. Here's a picture of it

[odd edge]

@sage terrace Has your question been resolved?

[boreal apex]

you can do this using the concepts of differentiation

since 2x + y = 300
and we need to maximise Area
which can be described as a function of x
fx = x.(300 - 2x)
and then differentiate it

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[sage terrace]

.reopen

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✅

[sage terrace]

So the top line 2x+(300-2x)-300 = 0 does thar make sense? If so, how do I prove it?

[boreal apex]

300 = 300
so that makes sense

you don't need to prove it

you are naming one side x and another y

you have been given the length of fences

so just equate to it

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[steady tide]

i thought when heated up, the volume of the gas expands

why does V_i = V_f here?

oh nevermind, any change in the volume is negligible

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[long tinsel]

it is very dangerous

[steady tide]

it is indeed very dangerous

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[empty acorn]

Continuing based on this

[empty acorn]

Maybe one day I’ll just open up a forum

Only n = 6k is possible if not divisible by 2 or 3

@warped grove

[tepid pelican]

when n = 6k, n is divisible by both 2 and 3, or maybe im just misunderstanding the question?

If n is not divisible by 2 or 3 then n^3 + 5n is not divisible by 6

[tepid pelican]

So this is what you're trying to prove now

there are 6 cases:

1. n = 6k
2. n = 6k + 1
3. n = 6k + 2
4. n = 6k + 3
5. n = 6k + 4
6. n = 6k + 5

Now out of these 6 n's, which of them are not divisible by 2 or 3?

e.g. in the first case, n = 6k, 6k is divisible by both 2 and 3 and so n is divisible by both 2 and 3 as well

so it's not true that it's not divisible by 2 or 3

[odd edge]

@empty acorn Has your question been resolved?

[crude prairie]

I own you.

[mystic swan]

whatt?!

[empty acorn]

Only 2 and 6 I think are not divisible

Meaning like

Numbers 2 and 6 in the list you made

[odd edge]

@empty acorn Has your question been resolved?

[tepid pelican]

Yep, perfect!

sorry for late response

somehow i completely missed your message

If n is not divisible by 2 or 3 then n^3 + 5n is not divisible by 6

anyway, now this needs to be proven

only n which satisfy the condition of not being divisible by 2 or 3 are n = 6k+1 and n = 6k+5

now try substituting those n's into n^3 + 5n and see if the result is divisible by 6

[empty acorn]

So I could do a proof like

Okay so if I’m going for a contrapositive proof

[tepid pelican]

Let n be an integer. Prove that if n^3 + 5n is divisible by 6, then n is divisible by 2 or 3 or both

Wait I just realized that the theorem is false

take n = 1

[empty acorn]

If n is not divisible by 2 or 3 then n^3+5n is not divisible by 6

Sorry I just wanted

[tepid pelican]

then n^3 + 5n = 6 is divisible by 6

[empty acorn]

To restate here

[tepid pelican]

the problem is that the theorem is false

so we cant prove it

[empty acorn]

Hm

I need to come up with a true one to experiment with

We still never learn proof by cases in my course tho

It’s all either proof by induction

Biconditional

Direct

Contrapositive

Contradiction

Idk why we don’t learn proof by cases

[tepid pelican]

cases technically fall under direct proof

[empty acorn]

I think if I try it in an exam my professor will accept it

But I didn’t learn it as a part of the curriculum

Is there another one we can try

Hold on let me see if I can find one in the book

I said this in the other channel before it timed out

[tepid pelican]

n^3 + 4n should work I think

[empty acorn]

But I’m looking for a structured proof

Like

Sorry I’m still a bit groggy

I’m looking for a structured way to solve proofs

I don’t know why but I always fall short somewhere when solving proofs

[tepid pelican]

there is no universal way unfortunately

[empty acorn]

This is going to make me insane I think

Idk why I’m so shit at this!

Let’s try the n^3 + 4n one

I’m not even sure how to set rhis one up

I’m so frustrated

[stone vector]

proof by cases isn't really a thing. you can use cases in any of the more formal proof types

[tepid pelican]

it's exactly the same setup

[tepid pelican]

the only difference is that instead of substituing it in n^3 + 5n we substitute it into n^3 + 4n

[odd edge]

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[empty acorn]

.reopen

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✅

[empty acorn]

Fuck this bot

[empty acorn]

Is there a general strategy you apply for proofs, or not really?

Someone said there isn’t one

But surely there must be a structured way to think about these things

A more structured way at least

[warped glacier]

proofs is an incredibly broad category

or do you just want an answer for divisibility proofs

they've forced you to use the contrapositive so the proof becomes a bit clunky

otherwise just do n^3 + 4n = n(n^2 + 4)

if n is divisible by 2 or 3 or both 2 and 3, we are done

the only other possibility is that n^2 + 4 must be divisible by 6, which means n must be even (n^2 = 6k + 2, where n, k are integers)

wait

okay corrected

otherwise consider modular arithmetic

[warped glacier]

basically modular arithmetic is this but saves you from having to expand the parenthesese for n^2 or n^3

[empty acorn]

yes but when do you know what proof method to use

[warped glacier]

practice

[empty acorn]

and what's the strategy for making connections

so there's actually no solid like

guideline

[warped glacier]

yeah

[empty acorn]

🥹

[warped glacier]

you learn from experience after all

[empty acorn]

that's awesome

[warped glacier]

practice and use your brain is a lot of mathematics, that's just the truth

[empty acorn]

i don't know why but i have trouble finishing a lot of proofs

even if i use the definitions

here's another proof i'm trying from the book

here's the proof set i'm trying from the book

i'll post my answers here

[empty acorn]

These are a bit clunky

So advice is appreciated

[odd edge]

@empty acorn Has your question been resolved?

[errant gust]

the first one is wrong

$0.5 \notin Z$ but $\frac12 \in Q$

[clever fjord]

ashy!

[errant gust]

small detail for 2a, a prime is only divisible by 1 and itself, every number is divisible by 1 and itself

look at negative numbers for 2b

for 2c, there are 4 roots to the polynomial $x^4-1=0$

[clever fjord]

ashy!

[odd edge]

@empty acorn Has your question been resolved?

[empty acorn]

Damn that’s fair

Fuck

[errant gust]

$0.5 \notin \mathbb{Z}$ but $\frac12 \in \mathbb
{Q}$
terrible typo but you get the point

[clever fjord]

ashy!

[empty acorn]

fuck!

why didnt i spot this

@errant gust do you have any advice for getting better at these kinds of proofs

[odd edge]

@empty acorn Has your question been resolved?

[north badger]

what's the question?

[errant gust]

make sure your conclusions are valid

and dont make any assumptions

[empty acorn]

But aren’t you allowed to make assumptions if they’re a part of a definition

[errant gust]

if its a definition, then you arent assuming anything

i meant assumptions like $\frac{a}{b}\notin \mathbb{Z} \rightarrow \frac{a}{b} \notin \mathbb{Q}$

[clever fjord]

ashy!

[errant gust]

which is quite clearly wrong if you think about it

try thinking about the formal definitions of what the terms given in the question are, and how you can show they are equivalent

what is a rational number?

[solemn wadi]

@empty acorn yo u need help?

[empty acorn]

A number that is two integers being divided

[empty acorn]

Yeah

[errant gust]

so how can you show a number is not rarional

[solemn wadi]

can you show me the problem

[empty acorn]

If it can’t be represented by this

[errant gust]

try that

[empty acorn]

Wdym

[empty acorn]

Its up at the top I’ll link it

[empty acorn]

.

[solemn wadi]

all of them?!

[solemn wadi]

should i start with 1?

[errant gust]

okay these type of proving irrationals is the standard of proof by contradiction

assume sqrt 5 is rational

and find a contradiction

[solemn wadi]

im person who is interested in maths can you give me name of book you're using?

[solemn wadi]

$\documentclass{article}
\begin{document}
Let's assume that $\sqrt{5}$ is rational and can be written as a fraction $\frac{a}{b}$, where $a$ and $b$ are co-prime integers.\
If $\sqrt{5}$ is a rational fraction $\frac{a}{b}$, it must be in its simplest form, so $a$ and $b$ are co-primes.\
If you square both sides, you get $5 = \frac{a^2}{b^2}$.\
Multiplying both sides by $b^2$ gives $5b^2 = a^2$. This means that $a^2$ is divisible by 5, so $a$ must also be divisible by 5. Therefore, we can write $a = 5k$ for some integer $k$.\
$5b^2 = (5k)^2$\
$5b^2 = 25k^2$\
$b^2 = 5k^2$\ This also means b can be divisible by 5 contraction to a and b being co-primes and $\frac{a^2}{b^2}$ being in the simplest terms.
\end{document}$

[clever fjord]

whatihavedone

$\documentclass{article}
\begin{document}
Let's assume that $\sqrt{5}$ is rational and can be written as a fraction $\frac{a}{b}$, where $a$ and $b$ are co-prime integers.\\
If $\sqrt{5}$ is a rational fraction $\frac{a}{b}$, it must be in its simplest form, so $a$ and $b$ are co-primes.\\
If you square both sides, you get $5 = \frac{a^2}{b^2}$.\\
Multiplying both sides by $b^2$ gives $5b^2 = a^2$. This means that $a^2$ is divisible by 5, so $a$ must also be divisible by 5. Therefore, we can write $a = 5k$ for some integer $k$.\\
$5b^2 = (5k)^2$\\
$5b^2 = 25k^2$\\
$b^2 = 5k^2$\\ This also means b can be divisible by 5 contraction to a and b being co-primes and $\frac{a^2}{b^2}$ being in the simplest terms.
\end{document}$
```Compilation error:```! LaTeX Error: Can be used only in preamble.

See the LaTeX manual or LaTeX Companion for explanation.
Type  H <return>  for immediate help.
 ...                                              
                                                  
l.49 $\documentclass
                    {article}
Your command was ignored.
Type  I <command> <return>  to replace it with another command,
or  <return>  to continue without it.```

[solemn wadi]

if a and b are co-primes then its only common factor is 1

and ofc a and b are integers

thats only prove i know (learned from youtube)

also you can replace 5 for n and add some extras to be proof for sqrt(5)

if you have any other questions let me know

[solemn wadi]

@empty acorn

[errant gust]

!nosols

[odd edge]

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.

[solemn wadi]

sorry for that but cant u just read untill u get and do the rest by urself?

[errant gust]

seems like you just spoonfed them the entire proof

[solemn wadi]

yeah kind of

[empty acorn]

Yeah but I need reviews for the rest of the ones I wrote

I want to see what I’m lacking in my proof writing

[errant gust]

2 b and c are just wrong

counter examples do exist

[solemn wadi]

yeah i think book wants them to find counter examples

[odd edge]

@empty acorn Has your question been resolved?

[empty acorn]

Like which

I can’t think of a value of a that would be greater than b but has a square smaller than b

That seems quite nearly impossible

[velvet heart]

are a and b any integer ?

if so (-1) > (-2) is true but (-1)² > (-2)² is false

and x⁴ = 1 has 4 roots (1 real and 3 complex) so it's false to assume x must be 1 if you allow complex numbers

,w solve x⁴ = 1

[clever fjord]

Results provided by WolframAlpha

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Requested by wkaue

[velvet heart]

oh yeah theres also -1 as a solution

[errant gust]

2 real and 2 complex

[empty acorn]

For some reason I guess I thought it said natural numbers and not integers

[steep mantle]

what is even the question here?

[faint knot]

@empty acorn you here?

[empty acorn]

Yes

[empty acorn]

It is the first thing in the channel. I’m asking for clarification on what I did wrong in my proofs.

[faint knot]

Let’s do 2 first

[empty acorn]

Alright

[faint knot]

For 2a, the proof can be slightly shortened

For one, you can just have both primes be 2

also, when mentioning that a number is not prime, you could just say “58 = 2 * 29” instead of saying “58 = 2k for some k and k is 29”

Also, the beginning can be condensed by writing multiple = signs together

n = 7^2 + 3^2 = 49 + 9 = 58 = 2 * 29

It’s important to see how a proof shortens itself to be easier to read

Did you get all of that?

[empty acorn]

I guess this makes sense

But does this make my proof incorrect

[errant gust]

not really

[faint knot]

These are just steps to make your proof easier to read, none of these make your proof incorrect
(I would have told you that)

[empty acorn]

I’m not sure about the others I wrote down though

[faint knot]

For 2b, have you found a counterexample?

[faint knot]

@empty acorn you here?

[empty acorn]

For 2b

-2 and -3

That’s a counter example

[faint knot]

yep, what about a counterexample for 2c?

[empty acorn]

Hm

Oh

X = -1

Idk why that escaped me

Stupid

[faint knot]

With this, you can write out 2b and 2c just like you did for 2a

(FYI you can also choose 0 and -1 for 2b)

Go send it here when you’re done

[odd edge]

@empty acorn Has your question been resolved?

[empty acorn]

Will do 🫡

[odd edge]

@empty acorn Has your question been resolved?

[odd edge]

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[odd edge]

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[mystic saffron]

Hey guys

[mystic saffron]

I want to learn calculus one and probability and statistics in the next 3 months 2 months or so, I have some background in algebra and calculus but it has some gaps in it

What do you guys think I should start with first?

What would help me understand the other more?

[sharp oak]

Probablility does use calculus

[grave roost]

You can learn basic probability before calculus, but only up to counting methods

[mystic saffron]

Oh so it would be better to start with calculus right?

Because it depends on calculus

Also do you guys ever use workbooks to learn math? Why would you guys use workbooks when there are many online videos and courses?

Is it like the quantity and the quality of the problems there?

[sharp oak]

Having a problem set, and a chapter containing everything necessary do to that problem set

[mystic saffron]

That sounds awesome I see what you mean

[sharp oak]

But I agree that videos are great and relying on them is not a bad thing

[mystic saffron]

A mixture of both sounds really nice

Thank you guys I appreciate it

Take care!

.close

[odd edge]

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[timid sky]

Need an elaboration on this: is a matrix being unitary a sufficient or a necessary condition for the square-norm sum of any of its rows or columns being equal to 1?

[timid sky]

In mathematical terms:

\bigskip
A 3x3 matrix $S$ being unitary $\overset ?\Implies \sum_{i=1}^3 \abs{s_{ji}} = 1$

\medskip
For a matrix $S$, $\sum_{i=1}^3 \abs{s_{ji}} = 1\overset ?\Implies S$ is unitary

[clever fjord]

Aero

[outer wadi]

I would say nesscary?

I think lol

because if A is unitary then A = A^*

or if A is a real matrix A = A^T

meaning A columns are orthogonal

hmm wait

[grave roost]

Don't you mean A^-1 = A^T

[outer wadi]

oh damn did I get it mixed up with hermitian

yeah I did

it should be A^-1 = A^T

[grave roost]

At least, starting with the unitary property, $|AA^T|_1=|I|_1=3$

[outer wadi]

from the theorems in my book it would be nesscary I think

[clever fjord]

Cris

[outer wadi]

so if A coulmns fail to be a orthonormal set then A can't be unitrary

[grave roost]

Nvm, this is nonsense

[outer wadi]

they dont prove this one unforunately

[grave roost]

Confused matrix and vector norms

[outer wadi]

but I'd imagine you could prove it using b

[grave roost]

When you say square norm, don't you mean to square the entries

[outer wadi]

yeah

[timid sky]

it needs to be $\norm{a_{ij}}^2$ i believe

[clever fjord]

Aero

[grave roost]

I think in math speak it should be:
There exists some $i$ such that $||Ae_i||_2=1$ or $||A^Te_i||_2=1$ where $e_i$ is the $i$-th standard basis vector

[clever fjord]

Cris

[grave roost]

Given A would be unitary, $||Ae_i||_2 = ||e_i||_2=1$

[clever fjord]

Cris

[outer wadi]

Heres a proof that they must be 1

[green elm]

both the rows and the columns will have length 1 if the matrix is unitary

the converse is not necessarily true

[grave roost]

I think so

[green elm]

why

[grave roost]

You can build a matrix that satisfies the sum but is not even invertible

Wait

🤔

I think I spoke too quickly

[green elm]

for a simple counterexample, let A be the 2x2 matrix whose entries are all 1/sqrt(2)

[grave roost]

Do we have a counterexample for 3x3

The one I wanted to say is wrong

[green elm]

3x3 matrix whose entries are all 1/sqrt(3)

or in general, the n x n matrix whose entries are all 1/sqrt(n)

columns and rows all have length 1, obviously not unitary

[grave roost]

Perfect

[odd edge]

@timid sky Has your question been resolved?

[odd edge]

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[odd edge]

Please don't occupy multiple help channels.

[green elm]

duplicate

.close

[odd edge]

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[tepid pelican]

Think about the discriminant

what would it be if the graph was strictly above the x-axis?

[loud palm]

not geting

[tepid pelican]

do you know what the discriminant is?

do you know quadratic equations?

[loud palm]

D < 0

[tepid pelican]

What happens if D < 0?

[loud palm]

live above x

?

[tepid pelican]

yeah

[loud palm]

line

[tepid pelican]

if D < 0, then it has no roots

and therefore the graph must lie completely above the x-axis

[loud palm]

yayayyayayayayayaa

ah ha

[tepid pelican]

okay, so now we need to know when is the discriminant negative

for that we need the discriminant itself

can you find it?

[loud palm]

yusss

yussssss

a length 13

S length 20

13/20 yus

thnx

[tepid pelican]

Yeah, a must be between -15 and -2, so the total length of that interval is 13

[loud palm]

u r smart

.close

[odd edge]

Channel closed

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[loud palm]

.reopen

[odd edge]

✅

[loud palm]

is there any short cut rather than manually solving for each cases?

[tepid pelican]

I'd start by placing the white balls like this

_W_W_W_W_W_W_W_

[wild hull]

gap method ig

[loud palm]

ok

[tepid pelican]

now we can place at most 1 black ball in each gap

How many ways are there to do that?

[loud palm]

donno ... i'll learnn

[loud palm]

3

[tepid pelican]

Really? You could place the black balls in e.g.
gap 1, gap 2, gap 3
or in gap 4, gap 6, gap 7
or in gap 3, gap 5, gap 8

there are plenty of ways to do it

[loud palm]

i thaught i asked how many black balls r there

sry

i misread

[tepid pelican]

ah

im asking in how many ways can you place those 3 balls in the gaps

[loud palm]

so many TT

[tepid pelican]

if there can be at most one black ball in each gap

you dont need to count them manually

[loud palm]

8! / 3!?

[tepid pelican]

Why?

[loud palm]

i just guessed sry

[tepid pelican]

you essentially need to choose 3 gaps to place the balls in

[loud palm]

8c3

[tepid pelican]

8c3, yes

there are 8 gaps for us to choose from

and we need to choose 3

[loud palm]

ok

[tepid pelican]

,calc 8C3

[clever fjord]

The following error occured while calculating:
Error: Undefined symbol C3

[tepid pelican]

,w 8 choose 3

[clever fjord]

Results provided by WolframAlpha

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Requested by methisalwaysright

[loud palm]

ok

then?

[tepid pelican]

What else do you need

Currently you have the number of ways to arrange those balls if black balls cant be adjacent