help-10

[austere yew]

Understand?

@stoic elm

?????

[stoic elm]

yep

thank youu!

.close

[obtuse pebble]

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[obtuse pebble]

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[mental vector]

Trying to understand this solution. This is a problem from survival analysis but my issue is with the calculus

I'm struggling to understand how we go from the first red line to the second red line (using integration by parts)

I've tried writing this out in the usual \int uv' = uv - \int u'v notation but I think I'm misunderstanding how to handle the v' (i.e. dS(t))

[obtuse pebble]

@mental vector Has your question been resolved?

[obtuse pebble]

@mental vector Has your question been resolved?

[obtuse pebble]

@mental vector Has your question been resolved?

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[outer siren]

allo calc question on the jacobian

[outer siren]

change of variables from cartesian -> spherical

heres the math stack exchange post

i had the exact same question

https://math.stackexchange.com/questions/4472765/computing-the-jacobian-for-the-change-of-variables-from-cartesian-into-spherical

can someone further explain this person's answer?

why can i just take the absolute value of a jacobian?

[timid silo]

The sign of the Jacobian is telling you whether or not the change of variables preserves (if the sign is positive) or reverses (if the sign is negative) the orientation of space

like

its the difference between saying

[outer siren]

ITS U AGAIN

ITS ALWAYS U

THE MAGICAL HUMAN WITH THE MAGICAL ANSWERS WTH

[timid silo]

oh frick i forgor

[outer siren]

ohh

so if i want to preserve the orientation of space

and my jacobian is negative

i can just take abs() value of it

does it mean that other than the sign, the determinant of a jacobian matrix always has the same value?

even if i move the columns/rows around?

yes

LOL

[timid silo]

sorry i experienced brain damage remembering this

[outer siren]

its okie

take ur time

[wild swallow]

[timid silo]

,, \mat[v]{a &b \ c &d} \tss{and} \mat[v]{b &a \d&c}

[timid silo]

there u go oh my god

anyways

[outer siren]

yes i see u have swapped the columns!

[timid silo]

yes

pretty much

[wild swallow]

okay but people use | | to denote det, and also absolute value

so how many do you want there

[timid silo]

[outer siren]

tis ok iget the msg

[wild swallow]

,, \mat[V]{a &b \ c &d} \textss{and} \mat[V]{b &a \d&c}

[tardy epoch]

6

big, Big, BIG, large, Large, LARGE

[timid silo]

if u r swapping columns/rows in a det u r just changing the sign in the det

[outer siren]

okay umm

then how would i know

which order the columns should be in

[timid silo]

i mean it doesnt matter if it gives a negative just abs it

[outer siren]

can i do that on an exam

LOL

[wild swallow]

yeah just abs it

[timid silo]

i mean yeah you can justify why interchanging columns in a determinant makes it negative

[outer siren]

ah

is this always true

or is this just a particular case

[timid silo]

yes its always true

[outer siren]

does this apply to rows too?

i think it does

[timid silo]

yeah

i think it was like

[outer siren]

interesting

so if i took that matrix

swapped two columns

[timid silo]

Column flipping is a multiplication by permutation matrix with determinant -1

[outer siren]

id still get the same positive determinant

yes

okay thank u

[timid silo]

[outer siren]

magical human always answering my q's

whether 2am at night

or 10 am in the morning

wtf

[timid silo]

i only answered your probability question the other day

[outer siren]

no ive seen u before more

i username changed so u prob dont remember

but remember u

[timid silo]

really?

[outer siren]

LOL

yeh

[timid silo]

what

[outer siren]

like twice more

[timid silo]

you remmeber

[outer siren]

4 times in total now

i REMEMBER U

[timid silo]

my old account

[wild swallow]

so did he

[timid silo]

[outer siren]

:ooooo

u a magical old man

[wild swallow]

hes 83 years old

[outer siren]

i will go back to studying byebye and ty!!

[wild swallow]

this is his retirement plan

[outer siren]

:oooo

老人家

[wild swallow]

good luck

yeah

[outer siren]

.close

[obtuse pebble]

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[wild swallow]

very 老

[outer siren]

HAHhwha

[timid silo]

[wild swallow]

and very wise

[outer siren]

yes

[obtuse pebble]

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[orchid warren]

A cone has a radius of 4cm and a slant height of 10cm. What is the perpendicular height?

[pseudo lance]

Hello lrmsyy

@orchid warren

[swift stream]

Use the Pythagoras theorem

[obtuse pebble]

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[graceful marten]

I started Modulus Functions and I still cant seem to get the hang of it

I dont get why its wrong

Oh i think I do actually

No i dont

I would understand if you cant have negative Y values

as its reflected off X axis the | asdsd |

?

@stuck olive Channel is occupied

[gentle tangle]

Create a help channel and post ur problem

Don't post ur problem in an occupied help channel

And dont forget the 15 min rule

[obtuse pebble]

@graceful marten Has your question been resolved?

[obtuse pebble]

@graceful marten Has your question been resolved?

[obtuse pebble]

@graceful marten Has your question been resolved?

[glass depot]

if u gonna draw graph, u will see y= mod( 3x + 4) and y= x will never meet, so there is no solution of this, but if u want to know whats wrong in this solution, then I will say, over here question is wrong, probably it would have to mod( 3x + 4) = mod(x) that make sense, since if u take any value of x then it is impossible to get a positive and negative value of that same x , and this question will also make sense if it was mod( 3x +4) = y , since u gonna get , two straight lines

so in short question is wrong over here

[obtuse pebble]

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[lament atlas]

i know that when 2 radiuses are different the radius of the sum of the 2 series is equal to the smallest of the 2 radiuses

but is this rule still valid in the case when we take a linear combination of the 2 series such as in this case?

please tag me if you answer

[brisk grove]

Well what's the radius of convergence of 3a_n?

and -5b_n?

[lament atlas]

im not sure

[brisk grove]

What happens if you multiply each term by a constant?

Either in the series of a_n or b_n

Does the convergence radius change?

[lament atlas]

i think no

but im nor really sure

to why that happens

[brisk grove]

It doesn't

If a_nx^n converges for some x, then multiplying each term by a constant would still converge

[lament atlas]

is there a reason for that?

[brisk grove]

I mean, it's just multiplying by a constant...

Not sure how to explain it

If the series converged to L, it would now converge to cL if you multiplied each term by c

The partial sums are c times bigger.

[lament atlas]

yeah you are right

x stays the same

thank you

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[low spade]

5(x+y)=xy help pls

[twilit meteor]

what do u have to do with that

[leaden lion]

x,y are integers?

[low spade]

yes

[twilit meteor]

divide by xy on both sides

u get 1/x + 1/y = 1/5

if u simplify

[leaden lion]

xy-5x-5y+25=+25

Now can you factorise lhs

[twilit meteor]

u'll get x-5Xy-5 = 25

[low spade]

asks for the answer range of x and y in response

[leaden lion]

@low spade

[leaden lion]

.

[low spade]

Okey thanks

.close

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[queen solar]

Does anyone know how to solve this integral. Its from a length of a cardioid curve i applied the formula but im stuck on this integral

If it had a 4 instead of 5 it would be easy but i dont know how to solve it like this

[lilac flame]

Over the given interval, sin theta > 0

So you can use sin theta = sqrt 1- cos^theta

Try u = sqrt 5 + 4 cos theta

Or just u=5+4cos

Although it may still be impossible with special functions

[last pilot]

,w antiderivative of sqrt (5 + 4cos theta)

[warm shale]

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[last pilot]

yep

doesnt seem to have elementary antiderivatives

there ya go

[queen solar]

Oh

[lilac flame]

F

[last pilot]

XD

[queen solar]

And this was on my analysis 1 final 😂

[last pilot]

LMAO

[lilac flame]

The integral is definite tho

[last pilot]

for sure

but no antiderivatives

[lilac flame]

True

But definite integrals may have nice values

Without standard anti derivatives

[last pilot]

agreed

[queen solar]

I think the prof made a mistake tho

[last pilot]

,w integrate sqrt (5 + 4 cos theta) from 0 to pi

[warm shale]

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[last pilot]

does that look familiar @lilac flame ? i got no clue

[lilac flame]

Nothing off the top of my head

[timid silo]

try taking 4 common from inside the root

[queen solar]

What is Ε(θ/2|8/9)

[lilac flame]

Elliptic something

[timid silo]

you would get 5/4 ( an angle in a right angled triangle)
write it in terms of cos

then

[lilac flame]

Literally all fractions can represent ratio of sides of right angled triangles

[timid silo]

use cosc + cosd = 2cos(c+d/2)cos(c-d/2)

might help

[timid silo]

yeah so ?

[lilac flame]

But 5/4 isn't cosine of any real value

[timid silo]

yeah it isnt

[lilac flame]

....

[timid silo]

ik

[lilac flame]

???

[timid silo]

you would get a single term in root tho

instead of two

might help

[lilac flame]

...how?

[timid silo]

this one is a single term

[lilac flame]

Do you want to go to C (complex)

[lilac flame]

...

Whats d?

[timid silo]

?

bruh XD

theta

[lilac flame]

So whats c

[timid silo]

nah c is that value

5/4

cosine

[lilac flame]

C is

[timid silo]

not complex lmao

[lilac flame]

Cos ^-1(5/4)?

[timid silo]

yeah

[lilac flame]

You do realize

Cos^-1 5/4

Isn't real??

[timid silo]

mhm

[queen solar]

Anyway i think the professor made a mistake there is no way anyone could have solved this

In analysis 1 at least

[lilac flame]

You sure you set up the integral right

[queen solar]

Yup

Everyone was stuck on this

[timid silo]

uh wait XD

its 5/4 lmao i thought 4/5

mb guys

yeah some mistake ig

[queen solar]

It was question 2

Sory its in greek

But the function is r=1+2cosθ

So i just took the formula r^2 + (dr/dθ)^2

Sqrt

Of that i mean

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[obtuse pebble]

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[brisk grove]

@timid silo Have you tried what I told you?

[timid silo]

take 2 common from the denominator

[brisk grove]

Show your work

[timid silo]

multiply divide denominator by x^2

[brisk grove]

@timid silo Stop

[timid silo]

and numerator by x

1

[brisk grove]

Don't give out solutions

[timid silo]

sorry bud

[brisk grove]

I can see where you are going with this, but let's allow him to try it for himself

All good

[timid silo]

;-; okay

[brisk grove]

Good

Keep simplifying

Ok good. Now try to manipulate this, so that you can use the limit of sin(x)/x.

Why is there an x in the denominator at the first line?

And where did the x^2 go to in the second line?

Make sure you didn't forget anything, but yes good progress

Same thing for sin^2(x/2)

That's sin(x/2) * sin(x/2)

[queen solar]

He cant use l'hopital right?

Cause that would be easier

Okok

[brisk grove]

I'd say that's overcomplicating it

The trick to solve this limit of adding terms we need is very useful for other limits

A good thing to learn

[queen solar]

No i get that

Just asking

[lilac flame]

Just read it

Mb

[queen solar]

Try to make limits you know

You know x cant be 0

[lilac flame]

Intuitively you may notice that the denominator has a higher degree than the numerator

[lilac flame]

So you expect it to diverge

So you can't get rid of all the terms in the denominator

So try to get rid of the numerator

Tell you are left with something like 1/x or 1/sin x etc

[queen solar]

Also u sub could help

[lilac flame]

You are going right here

What's sinx /x

[queen solar]

You need to solve this right?

Yes

[lilac flame]

So you can get rid of it

So do that

[queen solar]

And focus on the other term

Try to make known limits

You know x cant be 0

[lilac flame]

Try to see if you can get rid of the x in the numerator too

You know you can destroy x's using sines

And there are 2 sines in the denominator

So that's nice

[queen solar]

You can simplify the x/2 if you want

With u sub

How can u make the limit of sinu/u here

It will be more complicated that way

There is an easier way

Also

U didnt square the 2

Its (2u)²

[lilac flame]

(Ignore this, I misread)

The constants don't really matter

Just thik of it as u^2/sin^2 u

Now

If it was just u/ sin u, could you do it?

Yes

[queen solar]

W8 u need to prove that right?

Only sinx/x is known

Okay

Try diving with something so you can make those limits

Yup

Think of something simpler

If you have (sinx/x)²

Its the same thing right

Its just 1²

And you have the sin²x

Look at the denomination

sin²u same thing

Its the same thing

You just have (x/sinx)²

Just dont look at the ^2

You have x/sinx

And you know x can not be equal to 0 because it is approaching 0

So you dont need to worry diving by 0

If x cant be 0

You can divide by x

And not worry abou it being 0

Try doing it

You have x/sinx

If you divide both top and bottom by x

What is the result

Yes

Can you solve it now?

Yup

[obtuse pebble]

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[queen solar]

Npnp

[obtuse pebble]

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[warm shale]

kisnar

[tardy epoch]

Why did you flip the inequality sign

log is monotone

[gilded needle]

try it with x=0 to see that it's not equivalent

[wild swallow]

monotone increasing or decreasing?

[tardy epoch]

Pain

[warm shale]

kisnar

[tardy epoch]

log(x) is monotone increasing in x

Yes

[viral blade]

Why are you just turning the e^ into a log

Can you explain the step you're performing here

I saw the equation that you're trying to solve

[warm shale]

kisnar

kisnar

kisnar

kisnar

[viral blade]

right

the step after that

how did you get to this?

[viral blade]

I swear I replied

discord app gaslighting me

hmm

so I think the way to think about it here is to think about taking the ln of both sides

rather than just applying the definition of ln

then you use the fact that ln is an increasing function

So if x>y, then ln(x)>ln(y)

And in that sense, the inequality really isn't flipping

Very understandable source of confusion though

[warm shale]

kisnar

kisnar

[viral blade]

Yeah that's correct

[warm shale]

kisnar

[viral blade]

Nice

Wolfram and some other places use "log" as ln

And then specify log_10 when taking base 10

That's because ln comes up a lot more in advanced math then log_10

Yeah it is

I've made that mistake before too

Np!

[obtuse pebble]

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[cursive fox]

by applying l,hospital in these

[cursive fox]

in the part where you get

"2x+1/1"

how did it turn to 3

woudlnt it be 2 since both derivatives of those ones are zero since they are both constants aswell

[fierce vale]

why not factor ?

[gilded needle]

what do you mean how did it turn to 3, you are taking the limit as x->1

(2x+1)/1 is 3 when x=1

[cursive fox]

im having a hard time

wait

first is x^2+x-2/x-1

then 2x+1/1

[cursive fox]

so you mean ill only apply the derivative thingy on 2x?

[cursive fox]

,

[gilded needle]

no

that's not what i mean

you got to (2x+1)/1 by applying l'hopital, yes?

ohh

[cursive fox]

yes

[gilded needle]

i bet i know the issue

you're trying to apply it again, aren't you

[cursive fox]

yes

i did

[gilded needle]

yea you can't do that

[cursive fox]

oh why

[gilded needle]

you can only apply it when the limit in the num and denom are both 0 or both infinite

in this case that's not true, the denom is 1, you can't apply l'h anymore

[cursive fox]

it does lead to 0/0

i dont get

[gilded needle]

no i mean (2x+1)/1 is no longer 0/0

it's 3/1

which is the limit

you have to stop applying l'h when you get something that's not 0/0

[cursive fox]

ah

icic

2(1)+1

oki thx

.close

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[crude jasper]

hello idk the English term of this subject but i need to know how to solve these 2 tasks

[crude jasper]

it just says calculate the route SB'

and calculate the route SB

[obtuse pebble]

@crude jasper Has your question been resolved?

[atomic umbra]

What you refer to as route is known as a line segment in English.

,tex $$\triangle SA \prime B\prime \sim \triangle SBA$$

[warm shale]

@atomic umbra

[atomic umbra]

(This is for part A)

SB’ corresponds to SB the same way SA corresponds to SA’

SA/SA’ = 4/6

Therefore, SB/SB’ 4/6

SB’ = 6 * SB / 4

SB’ = 6 * 5 / 4

SB’ = 7.5

The length of SB’ is 7 cm

[obtuse pebble]

@crude jasper Has your question been resolved?

[obtuse pebble]

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[static relic]

I did it wrong but how am i supposed to do it

[static beacon]

It wanted you to graph both lines and see where they intercept

Put them into the form y=mx + c

[static relic]

do both lines meet at 3,-4?

the line i drew is y=4/-3x +0

and the second one is y=5/-3x +1?

[static beacon]

Nearly

[static beacon]

+1?

[static relic]

-1?

since its 3/-3

so they meet at -3,4?

would this be the correct graph?

would -3,4 be the solution?

would it be (-3,4) or [-3,4]?

ok

thank you so i think my mistake was mostly a silly mistake and i only did one of the lines

.close

[obtuse pebble]

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[hallow hatch]

i got the value for F'(x) right, but im not sure what to do for the other parts

[velvet hound]

so what can you tell about first derivative

[hallow hatch]

what do you mean?

[velvet hound]

like what does F'(x) tell you

or you know what

because first derivative is slope

and its increasing

the slope has to be positive

so F'(x) has to be > 0

you just solve for x

[hallow hatch]

right

so the inequality 1/ln(3x)>0?

[velvet hound]

yeah

oh wait

i see what you mean

oh wait

so if you graph it

you will notice there is a VA

so what you will do is try to figure out where that is, which is pretty easy, and then find the interval from there

[hallow hatch]

vertical asymptote?

[velvet hound]

yeah

[hallow hatch]

isnt the VA at 1/3

[velvet hound]

yeah

and the function is positive after 1/3

so interval would be 1/3 to infinity

[hallow hatch]

oh ok

[hallow hatch]

says its wrong

also [1/3,inf)

[velvet hound]

thats the answer?

[velvet hound]

oh oop

aren't you supposed to type infinity instead of inf or is that acceptable

oh im dumb

its greater than 6

[hallow hatch]

oh

[velvet hound]

so it would be 6 to infinity

[hallow hatch]

so [6, inf)

[velvet hound]

yeah

[hallow hatch]

what about the last one though

is it none

oh yeah

it is

thank you foryour help!

.close

[obtuse pebble]

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[timber osprey]

Is there a way to factor $$ x^2 - 17x + 72 $$?

[warm shale]

JusticeIroquois

[wooden cipher]

yes

check the factors of 72

[timber osprey]

8 and 9?

[wooden cipher]

yes

[timber osprey]

what do I do with the -17?

ah, I see it now

I knew I was having a big block here. Thanks

$$can it be either (x-8)(x+9) or (x+8)(x-9)$$

[warm shale]

JusticeIroquois

[timber osprey]

no wait

(x-8)(x-9)

[obtuse pebble]

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[agile delta]

Can someone help me with this?

[pearl blaze]

where is the question?

[agile delta]

Message is sending

There we go

[golden night]

do you know how to move a graph on the x axis?

[agile delta]

Kinda

[golden night]

well

you cant solve it unless you know that

[agile delta]

Okay learned it

<@&286206848099549185>

[velvet hound]

alr what did you get

ok actually how would you translate the graph to the right

[obtuse pebble]

@agile delta Has your question been resolved?

[obtuse pebble]

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[grizzled perch]

Help with part ii Im stuck.

[obtuse pebble]

@grizzled perch Has your question been resolved?

[grizzled perch]

<@&286206848099549185>

[glossy basalt]

are you using power series for part (ii)?

[glossy basalt]

i mean here

[grizzled perch]

Yes

[glossy basalt]

that's a good approach

[glossy basalt]

if you continue with the remaining 2 fractions with power Series up to x² term, you should get the answer

[grizzled perch]

I want to ask about the 4(2-x)^-1 and 8(x-2)^-2

How do i bring out the 2 properly

[glossy basalt]

i see i see

4(2-x)^(-1) need some specific alegebra, while 8(x-2)^(-2) youll need differentiation
and sorry for the slow reply, I'm on the streets

[grizzled perch]

its alright

[glossy basalt]

I can't read a thing on the screen with the bright sun lol

[grizzled perch]

is it alright if you show me how?

i just realised (x-2)^-1 i cant rotate the sign because its not commutative

oops

[glossy basalt]

yes, finally some shade for me lol

4(2-x)^(-1)
2(1-x/2)^(-1)

and then
8(x-2)^(-2)
= 8(2-x)^(-2)
=2(1-x/2)^(-2)

and do you know the power series for (1-x)^(-2)

[grizzled perch]

wait we call it binomial expansion not power series i never learnt power series before

[glossy basalt]

i see

[grizzled perch]

so 2^-1 is 1/2 and i multiply 4 by 1/2 and it becomes 2?

[glossy basalt]

but i bet you have learnt that expasion for (1-x)^(-2)?

[glossy basalt]

correct

[grizzled perch]

yes im fine with the expansions just confused with the algebra

i got it now thank you

[glossy basalt]

Cheers!

if you are done with this question, you can .close this channel 🙂

[grizzled perch]

.close

[obtuse pebble]

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[kindred marten]

not sure how to do this

[timid silo]

(y2-y1)/(x2-x1)

Then use algebra

[kindred marten]

ohh ok ty

lemme

solve that rq

(1/b-1/a)/(b-a)

i dont know how to algebra

[polar fossil]

$\frac{\orange{\frac1b-\frac1a}}{b-a}$

[warm shale]

hayley!

[polar fossil]

simplify the orange part into one fraction first

[kindred marten]

i think i may be the opposite of a math prodigy

bc i cant figure out how to do that

:D!!

[polar fossil]

the common denominator would be ab

[kindred marten]

.

i keep thinking

they gotta multiply by the same number for some reason

idk why

a/ab-b/ab

/

b-a

[polar fossil]

okay so if we combine those two orange ones together?

[kindred marten]

a-b/ab

[polar fossil]

$\frac{\orange{\frac{a-b}{ab}}}{b-a}$

[warm shale]

hayley!

[polar fossil]

we can write this as $\orange{\frac{a-b}{ab}}\cdot\frac1{b-a}$

[warm shale]

hayley!

[kindred marten]

yes

[polar fossil]

and then a-b and b-a have a very special relationship with each other

[kindred marten]

omg

they

cancel each other

or smth

bc b-b and a-a

[polar fossil]

hmm

how about you try some examples

pick random values for a and b

and calculate a-b and b-a

[kindred marten]

10 and 5

10-5 * 5-10

idk

[polar fossil]

no, not multiplyint

[kindred marten]

oops

[polar fossil]

just calculate a-b and calculate b-a

so 10-5 and 5-10

what do you get for those?

[kindred marten]

-5

5

[polar fossil]

yep ok try another example

how about 4 and 7

[kindred marten]

3

-3

[polar fossil]

do you have a theory as to the relationship between a-b and b-a?

[kindred marten]

fwb???

theyre opposites

idk

[polar fossil]

opposites sure, we might say in math that one is the negative of the other

which we can write like
a-b = -(b-a)

[polar fossil]

can you use this to finish simplifying this expression?

[kindred marten]

1/ab

maybe

[polar fossil]

maybe?

[kindred marten]

i do not trust myself with math atp

if that isnt it then no idea what it is :DDDD

[polar fossil]

make the substitution a-b = -(b-a) to this

[kindred marten]

-1/ab

i hope

[polar fossil]

yes, that seems to be correct

[kindred marten]

yay

ty for being patient w me

[polar fossil]

this comes up a lot btw

[kindred marten]

.close

[obtuse pebble]

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[grizzled lagoon]

How to proof this? Is it true

[kind hawk]

well in mod 7 you have 3=-4 and 5=-2

and then you are done

[grizzled lagoon]

Ok, I am trying to find the therom to proof it

Ok i got it

Thanks a lot

How can i use modular arithmetic to get the remainder 37*16! - 1753^99 is divided by 17

[glossy basalt]

this is fun, what have you tired?

[obtuse pebble]

@grizzled lagoon Has your question been resolved?

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[edgy needle]

excuse me

[edgy needle]

can someone plz explain what my teacher did for this proof

he used a conditional statement

[torpid cave]

Which part is unclear?

[edgy needle]

can u walk me through it?

to be honest all of it looks confusing

why did he do x+y - 2sqrt(xy)?

[torpid cave]

\begin{align*}
\frac{a + b}{2} & \geq \sqrt{ab} \
a + b & \geq 2\sqrt{ab} \
a - 2\sqrt{ab} + b & \geq 0\
\sqrt{a}^2 - 2 \sqrt{ab} + \sqrt{b}^2 & \geq 0 \
(\sqrt{a} - \sqrt{b})^2 & \geq 0
\end{align*}

[warm shale]

casework

[graceful marten]

When you copy make sure u copy it correct

[edgy needle]

is it not correct

he used a conditional statement and took away the inequality

im confused

[torpid cave]

Because AM - GM is equivelent to square inequality

At least for the basr case for n = 2. Then later you do induction

[edgy needle]

what is induction sorry?

[edgy needle]

is this different to his method?

[torpid cave]

Irrelevant right now

[edgy needle]

he kept telling me to think backwards

but idk what he means by that

[torpid cave]

Well you know tha classic inequality $x^2 \geq 0$

[warm shale]

casework

[edgy needle]

hmm yes

[torpid cave]

Well

$\frac{a + b}{2} \geq \sqrt{ab} \iff (\sqrt{a} - \sqrt{b})^2 \geq 0$

[edgy needle]

for the symbol there does it mean they both rely on each other

[torpid cave]

It means that they are basically equivelant

[edgy needle]

ah ok

[warm shale]

casework

[torpid cave]

This is probably better way to write it

[torpid cave]

One implies the other

[edgy needle]

ahh

[torpid cave]

Like lets say

$\text{water is falling from the clouds} \iff \text{its raining}$

[warm shale]

casework

[edgy needle]

how about the grass is wet, it's raining?

[torpid cave]

If its raining water is falling from the clouds and if water is falling from the clouds its raining (i assume so)

[edgy needle]

but it can also be mildew

?

or like a hose

[torpid cave]

In that case

$\text{its raining} \implies \text{grass is wet}$

[warm shale]

casework

[edgy needle]

so its raining is what we hypothesise

and grass is wet is like

the result

[torpid cave]

Not really if its raining grass is wet

If grass is wet it could just be dew

[edgy needle]

yea :/

[torpid cave]

So just because its wet we cant say that its raining

[edgy needle]

so we cqnt do that?

the conditional statement?

[torpid cave]

Yes we cant say $\iff$

[warm shale]

casework

[edgy needle]

but we can only do the one arrow facing to the right?

[graceful marten]

Is this what you want to teach him

[torpid cave]

Yea basically

But im just giving examples rn

[edgy needle]

first day back at school and i am mind boggled

[graceful marten]

ah ok

[edgy needle]

im australian

year 11

[torpid cave]

What is that?

[torpid cave]

Oh ok so basically you do proofs

[edgy needle]

idk

[torpid cave]

I wanted to say you dont really need logic if you wont be doing proofs

I mean its good to know logic but a typical person doesnt need it in life

[edgy needle]

yes u do hahaha

[torpid cave]

By logic i mean "logic"

[edgy needle]

like math logic?

[torpid cave]

https://en.m.wikipedia.org/wiki/Logic

[edgy needle]

...

[torpid cave]

Yes

[edgy needle]

thats what im doing hey

:D

[torpid cave]

So... any questions?

[edgy needle]

hmm yeah

[edgy needle]

how does it turn into (sqrt(a) - sqrt(b))^2?

[torpid cave]

You know the identity
$(x - y)^2 = x^2 - 2xy + y^2$

[warm shale]

casework

[edgy needle]

because when u expand (x-y) (x+y)?

u get dat?

[torpid cave]

$(x - y)^2 = (x - y)(x - y)$

[warm shale]

casework

[edgy needle]

oh

[torpid cave]

Thats what the square means

[edgy needle]

ohhhh

ok let me try the question again :P

btw did u do it different to my teacher?

he did this here highlighted,

[torpid cave]

Its same thing

[edgy needle]

o ok i see

thx

[edgy needle]

is this wrong

[torpid cave]

Its ok

[edgy needle]

my classmate said it is faulty

[torpid cave]

Oh i see i mean its not xyxy but xy + xy

[edgy needle]

yeasss hahahah

oops

i was like smt looks wrong here 😭😭

[torpid cave]

But the idea is ok

[edgy needle]

he said

it is true that (x-y)^2 greater than or equal 0 it is faulty because we cannot prove that the result is true by assuming the result is true

which sounds like non sense

[edgy needle]

is he right?

[torpid cave]

Ask him where did you assume the result is true

[edgy needle]

he says

u started with the original equation

and u are trying to provei t

so u dont know if its true or not

[torpid cave]

Classic

In proving inequalities it is very common to start with original inequalitie. You just need to make sure all your steps are reversible

[edgy needle]

huh?

so is he right

[torpid cave]

Not really but he has a point basically you just put $\iff$ after every line

[warm shale]

casework

[torpid cave]

Like this

\begin{align*}
\frac{a + b}{2} & \geq \sqrt{ab} \
a + b & \geq 2\sqrt{ab} \
a - 2\sqrt{ab} + b & \geq 0\
\sqrt{a}^2 - 2 \sqrt{ab} + \sqrt{b}^2 & \geq 0 \
(\sqrt{a} - \sqrt{b})^2 & \geq 0
\end{align*}

[warm shale]

casework

[torpid cave]

Someone could argue this is wrong

But just make sure you put this in proof

\begin{align*}
& \ \ \ \frac{a + b}{2} \geq \sqrt{ab} \
& \iff a + b \geq 2\sqrt{ab} \
& \iff a - 2\sqrt{ab} + b \geq 0\
& \iff \sqrt{a}^2 - 2 \sqrt{ab} + \sqrt{b}^2 \geq 0 \
& \iff (\sqrt{a} - \sqrt{b})^2 \geq 0
\end{align*}

[edgy needle]

oh so now this is correct?

[torpid cave]

Yeah basically put that symbol like that to write that every step is reversible

[warm shale]

casework

[torpid cave]

I cant get the spacing quite right

Its important to mention not everything is reversible

This is 1 example

Note that the reason you can square it is because you are working with numbers in R^+

In that case $a \geq b \iff a^2 \geq b^2$

[warm shale]

casework

[torpid cave]

Generally squaring inequalities in R is a big no

But you have extra info

[edgy needle]

why not?

[torpid cave]

$-3 > -5 \implies 9 > 25$

[warm shale]

casework

[torpid cave]

You see squaring negative values flips the sign

$1 > -2 \implies 1 > 4$

[warm shale]

casework

[torpid cave]

Squaring positive on one side and negative on other is well it depends on the absolute value

I mean this is supposed to be a contradiction i dont know how to do not implies in latex

[obtuse pebble]

@edgy needle Has your question been resolved?

[obtuse pebble]

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[steady patrol]

w=1/z +i and z=1+i. find w in polar form

[steady patrol]

please

[cyan sinew]

what'd you try?

[steady patrol]

idek how to do it

listen can you hop on a call please

[cyan sinew]

not really 😭

[obtuse pebble]

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[main wasp]

how I can work out if there are any similar numbers in two sequences (2^n, 5+ 2^n)

[covert arch]

its never gonna be similar

its up by 5

[pine sail]

I think it's more like 2^a vs 5 + 2^b

[high lily]

wdym by "similar"

[main wasp]

Any same numbers

[pine sail]

n is a natural number right?
You can just see one of them is always even while the other is always odd (when n is not 0)

[main wasp]

yeah

so it’d be 2, 4, 8, 16… 7, 9, 15

[pine sail]

11?

[main wasp]

am i mathing right now

basically 2N and N would have 2, 4 in common

[pine sail]

yeah

[main wasp]

hmm i guess that could work

any other ways? if not i’ll close

.close

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[wanton kraken]

If A is a set with 71 elements and < is a partial order in A, then which of these statements are correct:

1. If A has 70 minimal elements, then in A there's 70 maximal elements.
2. If A has 71 minimal elements, then in A there's 71 maximal elements.
3. In A there's at least 2 maximal elements.
4. In A there's a minimal element.

[wanton kraken]

Need help concluding the answers

so, right off the bat, Ik that there's a minimal element

so 4)

nvm

confused minimal with least biggest

[cunning burrow]

minimal would be least element. If all elements are the same they would all be minimal/maximal

[wanton kraken]

but we don't know if they are the same

[cunning burrow]

If A has 71 minimal elements, then in A there's 71 maximal elements.

The keyword is if

[wanton kraken]

so that's true for sure

but can't claim 1), because it's about 71

[cunning burrow]

Yes that is correct

[wanton kraken]

can't say that there's at least 2 maximal elements

so we exclude 3) as well

since it's about <

whereas there would be 1 maximal

Ig?

[cunning burrow]

correct

[wanton kraken]

so 2) for sure, but I kind of am unsure about 4)

[cunning burrow]

There will be least 1 maximal

[cunning burrow]

Why are you unsure about 4

[wanton kraken]

the formal definition is

for a minimal element

[cunning burrow]

Image if we had a set of number. {1, 2, 7, 3, 9}. No matter what set we construct there is always going to be a least element.

[wanton kraken]

yes, but aren't least and minimal different?

[cunning burrow]

No to the best of my knowledge. How would you define them?

[wanton kraken]

I just remember that in lectures they told us to know that there's a difference

oh I have the formal definition

this is for least

minimal

[cunning burrow]

I am afraid I can't help you

[wanton kraken]

oh alright ty though

.close

[obtuse pebble]

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[vivid mirage]

I believe I've got all parts sorted except for e. I'm not sure what half would be

[obtuse pebble]

@vivid mirage Has your question been resolved?

[obtuse pebble]

@vivid mirage Has your question been resolved?

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[pine sail]

!status

[obtuse pebble]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[wet flume]

1 @pine sail

[pine sail]

...

I saw that.

[wet flume]

ok... i thought you didn't see so i pinged you

[pine sail]

Continuity would imply the limit of f as x approaches 0 = f(0)

Can you solve for the limit?

[wet flume]

no... i am a complete new to this

i wanted to solve it with steps but i am not sure where to start

[pine sail]

A function 'f' is continuous at x = a, iff\
$\lim_{x \to a} f(x) = f(a)$

In your case, a is 0.

You want to find

[wet flume]

ok

[pine sail]

$\lim_{x \to 0} \f{\sin(4x)}{9x}$

[warm shale]

! What the hell am I doing here?

[wet flume]

ok

[pine sail]

Now do it.

[wet flume]

should i use the f(x)+f(x+h) formula?

[pine sail]

Which formula?

[wet flume]

one sec i am solving it...]

[wet flume]

i solved this and got 4/9

[pine sail]

That's correct.

[wet flume]

now what should we do?

oh wait that's it?

what about the k^2?

[pine sail]

well 4/9 is limit of f as x -> 0

And k^2 is f(0)

[warm shale]

! What the hell am I doing here?

[wet flume]

so this is the answer?

[pine sail]

And you know this.

[pine sail]

It's not.

[wet flume]

oh

[wet flume]

yeah

[pine sail]

So what do you think, k is?

After all that?

[wet flume]

0?

[pine sail]

...

How?

[wet flume]

there is no x in k so..0?

[pine sail]

Wait what?

[wet flume]

sorry!! the thing is i only know how to solve it if it is in this form

[pine sail]

yeah

But try to understand what it means for a function to be continuous.

Like I said previously.

Now you know the limit.

And it is known that f(0) is k^2

[wet flume]

how ?

[pine sail]

It's given.

"f(x) = k^2, x = 0"

[wet flume]

ok

[pine sail]

My bad, I've edited that.

[pine sail]

Which means, the limit is equal to f(0)

You can solve for k from there.

[wet flume]

ok

but i still don't get it.. what should i do after solving the first one

[pine sail]

What's the limit?

Of f as x approaches zero.

[wet flume]

0?

[pine sail]

How?