help-39

[rancid steppe]

my answer sheet confirms it as x(a-b)(1+x) but it's the same tho right? ab=ba

[warm copper]

yea

[rancid steppe]

ty

.close

[pearl pond]

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[rancid steppe]

$x^2(a-1) -9a + 9$

[jolly parrot]

Simon James B

[rancid steppe]

to factor out 9 as a common factor do we consider the - ?

so it would be $x^2(a-1) -9(a -1)$ but i confuse when we use there signs and when we have left 1

[jolly parrot]

Simon James B

[warm copper]

wdym?

[rancid steppe]

how to we fector out from here

[warm copper]

and when we have left 1
I'm confused what you mean by this part

[rancid steppe]

like we get a-1

[warm copper]

Factor out what's in the brackets

[warm copper]

yeah

[pearl pond]

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[mighty meteor]

this is a tuff one. I need your help, math nerds!

[iron stream]

Bro

Have you tried calculating A²

I feel like you will get it Idempotent

[mighty meteor]

imma be honest with you bro, i didnt even try to solve it ye

[iron stream]

Do yk how to write matrix in wolfram ?

[mighty meteor]

i just have a friend who said its hard

[mighty meteor]

uhm...no

[iron stream]

Ok wait

[mighty meteor]

ah, so is the point to find A^n?

i think ive seen this type of exercises before

hold on

[fringe raft]

can you diagonalize the matrix?

[mighty meteor]

i have not heard of that in my life

what is that even supposed to mean

,,diagonalize,,

[jolly parrot]

Gustavo

[fringe raft]

find S, so that S^-1 A S is a diagonal matrix

[mighty meteor]

aaaah, its that formula

yea idk how to use it

[iron stream]

Finding A² and A³ would have led you to the conclusion

That the diagonal elements aren't changing

[mighty meteor]

wawawiwa

thank you a lot Mr. Xor

you always solve my problems

ight bye

.close

[pearl pond]

Channel closed

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Use .reopen if this was a mistake.

[mighty meteor]

.reopen

[pearl pond]

✅

[mighty meteor]

nvm

[mighty meteor]

the multiple choices are:
a) 30799
b)30789
c)30790
d)30800
e)30795
f)30788

[iron stream]

WTH wolfram gave us the wrong output

No worries

But

You will have to calculate A² now

[mighty meteor]

sure

[iron stream]

And A³ too, so we find a sequence

[mighty meteor]

sure ill do that

[iron stream]

Did it ?

[mighty meteor]

yea

one sec

[iron stream]

Ok

[mighty meteor]

im gonna send it

here

i wrote them again for u

[iron stream]

So 1st element doesn't change

[mighty meteor]

looks like it

[iron stream]

Second 1,10,28
Third 1,-8,-26

So you see the pattern

Yeahh

1,1+9,10+9*2

And 1,1-9,-8-9×2

[mighty meteor]

huh?

[iron stream]

Yeah for 2nd and third element of the diagonal

[mighty meteor]

its 1, 1+9, 10-9*2

[iron stream]

1,10,28

[iron stream]

It's 10+9*2

[mighty meteor]

ah, i see, you were talking about the term in the middle for all 3

[iron stream]

Yeah that's what we're concerned about

Now can you guess the next term ?

For second it is, 28 +9*3

And for third it is -26 - 9*3

[mighty meteor]

so the formula is 10+9*(n-1)(-1)^n

that should be it

or wait, maybe not

[iron stream]

Why so ?

[mighty meteor]

scratch that

its wrong

[iron stream]

Can you calculate A⁴ to verify the result

[mighty meteor]

oh god

sure

[iron stream]

Just calculate diagonal elements

[mighty meteor]

here

[iron stream]

,calc 28+27

[jolly parrot]

Result:

55

[iron stream]

So we're correct

[mighty meteor]

oh shiiiiii

ok but still

[iron stream]

The difference is always constant between second and third term

[mighty meteor]

how would you find the formula

[iron stream]

But now the question says to find the absolute addition value

[mighty meteor]

for A^59 i think

yes thats the matrice

[iron stream]

Yeahh

[mighty meteor]

so we gotta figure it out in terms of the exponent

[iron stream]

Now we will explicitly calculate our sequence

u is second diagonal element

u2 = 1+9

Hey say something

[mighty meteor]

im still here

[iron stream]

You know GP formula ?

(r^n -1)/(n-1)

[mighty meteor]

i forgot it

[mighty meteor]

right, thats it

[iron stream]

a = 1 here

[mighty meteor]

mhm

[iron stream]

And we know the third element

Is just 2 less than the second

No need to separately compute it

[mighty meteor]

hmm okie

[iron stream]

,calc (9^59 - 1)/8

[jolly parrot]

Result:

2.49584763877e+55

[mighty meteor]

wait, the third term is 55 right?

so that wouldnt be it

[iron stream]

That's u4

u1 = A¹

[mighty meteor]

ok so it should be 28 then

[iron stream]

,calc 1+9+9^2+9^3

[mighty meteor]

9^2 is 81

[jolly parrot]

Result:

820

[mighty meteor]

which is already bigger than 28

[iron stream]

Wait we had 3

Wth

We had 3

[mighty meteor]

3 what?

[iron stream]

Replace the 9s with 3 s wthhh

Lemme retype

Oh nah nah nah

It's arithmetico-geometric progression

[mighty meteor]

huh?

[iron stream]

Let me redo

I'm on 3 hrs of sleep, i must sleep after this

😭😭

[mighty meteor]

sure

[iron stream]

u1 = 1
u2 = 10 = 1 + 9
u3 = 28 = 1+ 9 + 2×9
u4 = 55 = 1+ 9 + 2×9 + 3*9

So

u_n = 1 + 9(1+2+...+(n-1))

so u_n = 1 + 9×(n-1)(n)/2

Now check this

[mighty meteor]

thats 10621

for n=59

which is def not in the multiple choice

[iron stream]

,calc 1+ 95859/2

[jolly parrot]

Result:

15400

[mighty meteor]

right, my bad, still not there

[iron stream]

,calc 15400*2 -2+1

[mighty meteor]

u4 is wrong btw

[jolly parrot]

Result:

30799

[iron stream]

Huh 💀

[mighty meteor]

oh nvm

i read it wrong

[iron stream]

,calc 1+943/2

[jolly parrot]

Result:

55

[iron stream]

I must apologise for that error !

[mighty meteor]

why did you do -2 +1?

[iron stream]

Because third element is 2 less than second

And our 1st element is 1

[mighty meteor]

so you calculated the third element when you did that?

[iron stream]

third element= second element -2

So i just doubled the second

And subtracted 2

second + second -2 + 1
= 2×(second) -1

[mighty meteor]

its -(second-2)

isnt it?

[iron stream]

I'm still confused why wolfram gave that weird output!

[iron stream]

No we need the addition of absolute values

[iron stream]

Read the question again

[mighty meteor]

so what does that mean

addition to absolute values

does it mean its in modulus?

[iron stream]

Modulus

Absolute values refers to the modulus value

[mighty meteor]

oh wow

thats really cool

[iron stream]

Else our answer would be

2 +1

[mighty meteor]

damn, thats right

ok so the answer is a)

[iron stream]

I was doing just that before a read your options, and then re read the question

I will be going to bed ! Have a good day ! And sorry again ! I messed real bad !

[mighty meteor]

fair

ok bro

goodnight

.close

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[mellow idol]

from the first recursion relation i got the 2nd result, is it correct?

[mellow idol]

i was able to veirfdy it for 3 cases but wanna know if its correct

i basically need to show that the first one diverges for all n, if the 2nd one holds true it trivialises that

<@&286206848099549185>

also is it fair to say that the harmonic series sum can excede 10^5 at some point

[pearl pond]

@mellow idol Has your question been resolved?

[mellow idol]

<@&286206848099549185>

if someone could gimme a yes or no that'd be great! ^^

[hallow ice]

yes the harmonic series diverges so exceeeds any upper bound

[hallow ice]

looks right to me based on a quick calculation

[mellow idol]

alright, thanks!\

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[heady hearth]

I gotta question on b, is there only one x intercept and if so is it 3,0?

[frosty creek]

Almost. $(x+3)^2=0$ is true when $x+3=0$ or $x=-3$

[jolly parrot]

Calculustache

[heady hearth]

Oh shoot i meant to put -3 thats my fault

So there is only one x intercept right?

[frosty creek]

Yeah. Because if you factor (x+3)^2 you have two (x+3) terms, meaning only one spot where it can equal zero

[heady hearth]

Ok gimme a sec

[bitter lodge]

Additional info in case you find it helpful: if the multiplicity of the root is a positive even number (in this case, 2 from (x + 3)^2), then the function comes to touch that root, then turns back around

[heady hearth]

So your saying when its a positive in the parenthesis?

Also what makes this top one have 2 different x intercepts?

[frosty creek]

Graphically, you can think of the -4 moving the parabola down from where it barely touched the x axis to intersecting it at two points.

[heady hearth]

So it move the whole graph down by -4 and where its at without the -4 is just at the point where moving it down causes it to go through the x intercepts twice?

[frosty creek]

Yeah

[heady hearth]

Ok and how would you be able to tell all of this without a graphing calculator

[frosty creek]

You start with $(x+3)^2-4 =0$
$$(x+3)^2=4$$
$$(x+3)^2=\pm\sqrt{4}$$
Which has two answers and therefore two roots.

[heady hearth]

Wait so this is a quadratic and so its already mostly solved but by completing the square?

[jolly parrot]

Calculustache

[frosty creek]

Yes

[heady hearth]

Ahhh ok this makes sense but the -4 is not moved yet

So you move it now and then square root it and that gives plus or minus

This is only because of this -4 but if you didnt have that you could do the thing with factoring?

[frosty creek]

Yeah. Factoring or the quadratic formula.

[heady hearth]

Ok but thats for the one without -4 but the one with negative four you have to, but why?

[frosty creek]

Have to do what?

[heady hearth]

Why not just do the thing with the equation that doesnt have -4 with the equation that does have -4?

[frosty creek]

It is basically the same thing: move all constants to the right side and take the square root.

[heady hearth]

Oh ok so both is the square root but for the one without the -4 its just the same making it have 1 x intercept

[frosty creek]

Yeah.

[heady hearth]

For this one first subtract the four?

Then square root it but since its - make it i?

[frosty creek]

Yes. It will be a complex number

[heady hearth]

Because its a negative square root?

[frosty creek]

Yeah

[heady hearth]

Plus or minus 2i=x+3

[frosty creek]

Yup

[heady hearth]

Subtract 3 so its 3 plus or minus 2i=x ? Im doubting this as the numbers are in opposite placement from what i like

[frosty creek]

Negative three plus or minus 2i

[heady hearth]

Thats right thx for catching my mistake

[frosty creek]

np

[heady hearth]

So you really cant determine the roots?

[frosty creek]

Not in the real numbers. That isn't to say they don't exist, but that you need a larger set of numbers to describe them.

[heady hearth]

Yea so you cant really graph it?

[frosty creek]

No.

[heady hearth]

Alr

This is confusing me its saying the sketch a graph of it and see if you can determine the roots but how do they expect us to graph it just by looking at it?

[frosty creek]

Any parabola in the form $(x+a)^2$ will have a single x intercept, because it touches the x axis at one point (namely x=-a). So when you shift it up 4 units, it is no longer touching the x axis and will have no real roots

[jolly parrot]

Calculustache

[heady hearth]

Whats the money sign stand for?

[frosty creek]

Just code for rendering this.

[heady hearth]

Ah ok

So if it has lets say +8 on the outside and a plus in the parenthesis it wont hit the x axis? Because its raised by 8 and the plus on the inside makes it go up?

[frosty creek]

Yeah

Assuming the parabola "points" upwards

[heady hearth]

Would you determine that by the plus or minus in the parenthesis?

[frosty creek]

You would determine that based on whether or not it has a minus outside the parentheses.

[heady hearth]

But that doesnt make sense what if its a negative on the our side but it foes down

[frosty creek]

Yeah. So if it has a minus sign, then adding a positive number doesn't stop it from intersecting the x axis.

[heady hearth]

Wait so what do you mean namely x=-a?

[frosty creek]

If $x=-a$, then $(x+a)^2$ is the same thing as $(-a+a)^2=0$. No other value can make $(x-a)^2$ zero because no other value can make $x-a$ zero.

[heady hearth]

This last part doesnt make sense i think its a typo it sats no ther

[frosty creek]

If x=-a, then (x+a)² is the same thing as (-a+a)²=0. No other value can make (x-a)² zero because no other value can make x-a zero.

[jolly parrot]

Calculustache

[frosty creek]

Alr I have to go.

[heady hearth]

Ah ok this last part doesnt make sense to me tho

[pearl pond]

@heady hearth Has your question been resolved?

[pearl pond]

@heady hearth Has your question been resolved?

[heady hearth]

@helper

<@&286206848099549185>

This is the question ignore all the work

This is the work ive done

[spiral pivot]

Interesting that the minimum is outside of the range of the two zeroes

This kind of implies to me that it's perhaps something like an exponential times a quadratic or something of the sort perhaps?

[heady hearth]

All i know is i must find a but i forgot how to and i dont wanna cheat and look at the answer

[spiral pivot]

There is no a that will satisfy this requirement.

If you're talking about ax^2 + bx + c

[heady hearth]

No im pretty sure ot something that multiplies to everything to make it true

[spiral pivot]

And I'm pretty sure there isn't

[heady hearth]

The minimum or maximum is the vertex right?

[spiral pivot]

Yes

And the vertex must always lie between the zeroes

But -8 isn't between 1 and -7

So this isn't a quadratic

[heady hearth]

So to get the vertex from the x intercepts you add the x intercepts and divode by two?

[spiral pivot]

Yes, which would place it at -3

Oh, is -8 the minimum y value?

[heady hearth]

Wait

Would i make it so something times the equation i have =-8?

[spiral pivot]

f(-3) = -8 yeah

[heady hearth]

Wait im confused

[spiral pivot]

Ok you have a quadratic right?

f(x) = a(x-r)(x-s)

[heady hearth]

Yes

[spiral pivot]

You know r and s are 1 and -7

[heady hearth]

The second one isnt minus tho

[spiral pivot]

So you have f(x) = a(x-1)(x+7)

[heady hearth]

Yes

[spiral pivot]

So we have f(-3) = -8

So f(-3) = -8 = a(-3-1)(-3+7)

Solve for a

[heady hearth]

Wait could we just assume the f() is just y?

[spiral pivot]

Yes, typically f(x) = y

[heady hearth]

Ok wait lemme process this

Wouldnt it be positive 3?

Wait nvm

I read it wrong

[heady hearth]

I dont understand this part

@spiral pivot

[spiral pivot]

I evaluated the function at -3

By replacing the x with -3

[heady hearth]

Hold on the x intercepts we have are 1 and 7 so we got (x-1) and (x+7)

[spiral pivot]

Yes

[heady hearth]

That vertex doesnt match the one it gives

[spiral pivot]

The one what gives?

b/2a ?

[heady hearth]

No the vertex it gives -8

We get -3 tho

[spiral pivot]

Yup, that's why I started out confused as well

[spiral pivot]

This

[heady hearth]

So we have to make the x-8 x+7 =8

[spiral pivot]

The vertex is at (x, -8) not (-8, y) like I thought

[heady hearth]

-8 mb

[spiral pivot]

The vertex is at (x, -8)

-8 is the y coordinate

[heady hearth]

Correct

[spiral pivot]

Not the x coordinate

The x coordinate is where the vertex must be, which is -3

[heady hearth]

So wouldnt it just be -8 is = what we multiply it by which will stand for a times the x-1 abd x+7?

@spiral pivot

[spiral pivot]

I explained, the vertex is at -3

[heady hearth]

The one we got

[spiral pivot]

This is independent of a

[heady hearth]

What do you mean independent of a?

[spiral pivot]

Doesn't matter what number a becomes, the vertex is decided by r and s

[heady hearth]

This doesnt make sense can you change the x intercepts? Cause i dont thibk you can

@spiral pivot

Bro i need help pls

[spiral pivot]

You can't change the x intercepts, which means that the x coordinate of the vertex is fully specified from the given information

And it's -3

A better question is do you know how to find the vertex?

It's simply the average of r and s, the two roots

[heady hearth]

Wait so we found the x but what it says about the vertex is the y?

[spiral pivot]

Yes

[heady hearth]

Wait i thibk i get it now lemme write it down and show you

[spiral pivot]

Exactly

[heady hearth]

Ok but i dont understand why would we weote it in this equation ?

[spiral pivot]

Trying to solve for a

[heady hearth]

Why though?

[spiral pivot]

Need to know a to write the equation

[heady hearth]

Why isnt x^2+6x-7

[spiral pivot]

Because it's a times that

[heady hearth]

Why though?

[spiral pivot]

Because a(x-r)(x-s) is the formula for the quadratic

But x^2 + 6x - 7 is just (x-r)(x-s)

Going afk. Best of luck!

[heady hearth]

Alr

[spiral pivot]

Sorry I have to jet

[heady hearth]

Ah ok

[heady hearth]

What os the differences between a quadratic function and equation

<@&286206848099549185>

.close

[pearl pond]

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[spiral pivot]

@heady hearth the difference between a function and an equation is the equation finds the roots of the function

0 = ax^2 + bx + c

Vs

f(x) = ax^2 + bx + c

Top is the equation, the bottom is the function

But that being said, not everyone who writes math problems is careful, for your question I would write the function.

[pearl pond]

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[plush sundial]

what did u try

[tranquil cobalt]

just angle chasing

[tranquil cobalt]

yes

and u know that triangle is isoceles

cuz all the sides r equal

those who know 💀 💀 💀

[stoic imp]

45

yipee

[tranquil cobalt]

yes i think

good boy

[stoic imp]

.solved

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[smoky mantle]

I do not know what to do at all 😭

[limpid lily]

The maximums (a kind of extreme point) will occur at the endpoints or when the first derivative is zero.

So, it might happen when x = -4 or x = 3.

It might happen when (x^2 + 3) cos(x - 5) is zero.

[smoky mantle]

do I just plug in for -4 and -3?

wait no

I'm sorry, I'm usually not this stupid I swear

I'm so tired from doing my other assignments that I can't even keep my eyes open long enough to try to understand the question

It's due in a couple of hours and I haven't slept yet

😭

It's a struggle out here frfr

I have like 4 more of these to do and another 5 questions for another section and atp I just want to get it over with 🙏

[limpid lily]

Have you done integrals or antiderivatives yet?

[smoky mantle]

I don't think I've done integrals yet no

[limpid lily]

Have you done antiderivatives?

[smoky mantle]

haven't done that either

I got this assignment out of nowhere and it's based on completion

but I can't complete it if I don't get it right wdoadja

[limpid lily]

OK, so let's look at the endpoints.

x = -4.

Let's say we have a graph of f(x).

And let's say that x = -4 is a maximum.

[smoky mantle]

okay

[limpid lily]

What would be happening with the derivative?

At x = -4?

[smoky mantle]

I'm not sure

oh

do I just plug in for f'(-4)?

[limpid lily]

No.

[smoky mantle]

oh

[limpid lily]

The derivative is the slope of f.

So, if f(-4) was a maximum, it would be going down as you go to the right.

Because it's higher at the maximum than the stuff around it.

Does that make sense?

[smoky mantle]

yes

[rose robin]

I don't think you can just plug in values
you need to take the integral then solve for the constant

[limpid lily]

So, the slope would be negative.

Is f'(-4) (note f', not f) negative (or zero)?

[smoky mantle]

would it be 0?

[limpid lily]

It could be.

But they give you the formula for f'(x), right?

[smoky mantle]

yes

[limpid lily]

Fill in -4 for x in that and see if it's negative or zero.

[smoky mantle]

(x^2+3)cos(x-5)

-13*cos(-9)

I got 11.844 pos

what

[limpid lily]

It can't be -13.

[smoky mantle]

hold on

I did something wrong

it's 13cos(-9)

I got a negative number

-11.844**

[plush sundial]

for fining maxima or minima u shud equate f'(x) = 0 and find the possible values

[smoky mantle]

so equate (x^2+3)cos(x-5) = 0?

[kindred stone]

https://tenor.com/view/heartthrob-sheldon-big-bang-theory-sheldon-cooper-jim-parsons-gif-5225040

[plush sundial]

yes

[smoky mantle]

okay I'll try doing that

[limpid lily]

It can also be the endpoints.

For example, y = x^2 in the interval [-5, 10] will have two maximums, both at the endpoints.

[smoky mantle]

welp thanks for trying to help but I honestly am not comprehending anything you're saying

like my brain is deep fried

[limpid lily]

You should sleep.

[smoky mantle]

I'll just accept the loss and go to sleep 🫡

good night 😭

[limpid lily]

You too.

[smoky mantle]

.close

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[pulsar berry]

Find the sum of the series

I have no idea how to approach this

[gleaming musk]

is there no context to this>

seems uncharacteristic of questions to just hae 4 variables undefined without any purpose

but for starters there are some rearranging u could do to start of

the denominator could be rewritten as something much cleaner, and same for the other term (u^n/2)/z^{n/3}

[pearl pond]

@pulsar berry Has your question been resolved?

[pulsar berry]

Yeah that’s what is not clear to me

It can be written like this too

But what is not clear to me is how to deal with x, y, z, and u

[gleaming musk]

nothing immediately jumps at me tbh

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[gleaming musk]

.reopen

[pearl pond]

✅

@pulsar berry Has your question been resolved?

[pulsar berry]

Same

It feels impossible

[pulsar berry]

<@&286206848099549185>

[stable dune]

Man this looks like some special functions type of thing

I dont know many of those off the top of my head tho

Wouldn't we want x-y

The sum on the bottom is raised to the n for each variable

[hot canyon]

Oh shiii

Then it's a issue (@_@;)

[stable dune]

Yeah uhh

This seems real tough

Ramanujan-like even

[hot canyon]

T-T where's the context for the question

[hearty dew]

I dont think you can get a specific sum of the series

[hot canyon]

Can you send a SS of the context? Or the complete question?

[stable dune]

You definitely can't as there's 4 variables in it

[hearty dew]

I was just thinking that there might be properties we can use that simplifies things out

but I can't think of any

[stable dune]

Honestly this is probably some sum for a special functions course

Like u gotta realize that one of these is a representation of the digamma or smth crazy

[hearty dew]

I kinda agree with you on that one

[pulsar berry]

Its just that nothing else

Is from analysis 2 class

[hearty dew]

Is this an advanced mathematics course?

[stable dune]

Yes

It's like super rigorous multivariable calculus

[hearty dew]

Oh my

Im going to pretend I didnt see this channel

[pearl pond]

@pulsar berry Has your question been resolved?

Channel closed

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[pearl pond]

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[hearty condor]

Let f be a function that its domain includes a neighborhood of x0, U.
The limit of f(x) as x approaches x0 exists.
Does a neighborhood of x0 exist, such that every t in that neighborhood, the limit of f(x) as x approaches t exists?

[hearty condor]

I don't see how that statement is wrong. But I need to prove it

[autumn trellis]

okay, so it is wrong

[hearty condor]

Really?

[autumn trellis]

i think you are restricting the images in your mind to continuous functions

[hearty condor]

I tried checking f(x)={x} or floot function

[autumn trellis]

if you remove this assumption you may come up with counterexamples

[hearty condor]

Floot

Floor

[autumn trellis]

eh i mean that is no where near as pathological as function can get

thats continuous almost everywhere

[hearty condor]

Well what's a function that's continous not almost everyehere

[autumn trellis]

well, think of something LIKE $f(x) = 1$ when $x$ is rational and $f(x) = 0$ when $x$ is irrational

[jolly parrot]

4573r01d|)d357r0y3r 45²

[hearty condor]

The D function?

[tropic saddle]

you can find a function thats continuous almost everywhere and for which the claim is still wrong

[autumn trellis]

modify this one a bit to get your CE

[tropic saddle]

real analysis has some quite cursed counterexamples

[stable dune]

Wouldn't dirac delta be a counterexample

[hearty condor]

Yeah because there are no limits in it

Yeah but how do we add a limit to it. 😭 if we make a neighborhood around a specific point then we can use that neighborhood as the example

[autumn trellis]

okay what if you made the f(x) = 1 part to f(x) = x...

[hearty condor]

Hmmm

Then 0 would have a limit

[hearty condor]

My prof actually talked about this function lol

[cinder thistle]

thats not rlly a function on reals

[autumn trellis]

more crucially, no other point does. so neither does any neighbourhood of 0...

[autumn trellis]

question doesnt specify it has to be over the reals

[hearty condor]

Yeah he also talked about that 😭

[cinder thistle]

also, i never understood how dirac delta function is defined, as in how is it a function

[hearty condor]

Well im in calc 1 so it does I just didnt mention that

[tropic saddle]

and if you are up for more fun, consider the function $$f(x) = \begin{cases} \frac1q & x=\frac p q, \gcd(p,q)=1 \ 0 & \text{ else}. \end{cases}$$ and think about where that function is continuous

[jolly parrot]

Denascite

[hearty condor]

I dont think thats fun 💔

[tropic saddle]

then ignore it

[autumn trellis]

||this is a classic, continuous at all irrationals/||

[hearty condor]

What is gcd again?

[tropic saddle]

greatest common divisor

aka, p/q is fully reduced

[hearty condor]

Ah

[autumn trellis]

well anyways i think you have a few counterexamples now

[hearty condor]

Yes, thanks a lot

.close

[pearl pond]

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[stiff mauve]

can someone help me with 1a

[gleaming musk]

what have u done so far

[stiff mauve]

this is supposed to be the answer

but

this is what i have

[gleaming musk]

ok before anything

im not saying it was the fualt here but just be careful since ur 4s do look like a

[stiff mauve]

true i get what u mean

[gleaming musk]

do u have more working out or did u just skip a bunch of stpes

it should reasoning ur 2nd column be all 0 since

the first entry is a multiple of 2 of the 2nd entry

[stiff mauve]

i mean i divided the 2nd row by two

so we have 0, a-2, a, a-2

and then

Row 3 - (new) Row 2 would become my new row 3

owww

[stiff mauve]

i did this operation wrong

forgot to do the 4th (b) column

im still stuck

[gleaming musk]

ur row 3 column 4 is still wrong

before u did ur subtraction

ur row 3 should be
0 a-2 a^2 | 2a-3 right?

and ur row 2: 0 a-2 a| a-2

[stiff mauve]

yes @gleaming musk

[gleaming musk]

u should have an a left in ur column 4

[gleaming musk]

when u do the subtraction

2a-3-(a-2)

a-1

[stiff mauve]

right

[gleaming musk]

ur column 4 is 1 isntead of a-1 for row 3

[stiff mauve]

ye i changed it now on my paper

how can i continue after this?

[gleaming musk]

i mean its echelon form already tbh

even tho u can simplify it

[stiff mauve]

shouldnt echelon be 1 1 1 on diagonals

[gleaming musk]

not always?

[stiff mauve]

its either 1

or 0

[gleaming musk]

no?

[stiff mauve]

right?

why not

[gleaming musk]

the quesiton doesn't specific reduced

it just says echelon

ok let say for the sake of quesiton u want to reduce this further

ur 2nd row should be pretty obvious

0 a-2 a | a-2

[stiff mauve]

divide evrtyhing by a-2 right

[gleaming musk]

nope

what is that row saying

[stiff mauve]

x2 = 1

x3 = 0

[gleaming musk]

oh wait sorry im looking at the wrong thing

row 3 not row 2

0 0 a^2-a | a-1

[stiff mauve]

x3 = 1/a

[gleaming musk]

=?

[stiff mauve]

1/a x (a^2-a) = a - 1

[gleaming musk]

wait

hold up

sorry let me backtrack

[gleaming musk]

what do u have here

im pretty sure it is in echelon form right here

idh paper on me so maybe i have it wrong in my mind

can u send a photo of what u have written down back here

[stiff mauve]

@gleaming musk

[gleaming musk]

ok

can we reduce the third row a little bit more

actually dont

[gleaming musk]

this is fine already

its in row echelon form

1a) is only asking for echelon form

i think u have the wrong definition of echelon

[stiff mauve]

why are they distingqushing cases for a = 2 then

[gleaming musk]

for a free variable

u end up with a column of 0s

[stiff mauve]

does echelon basically mean there is a pivot in each row/column

[gleaming musk]

and they are just showing that it is still in echelon form

[gleaming musk]

pivot in each row

[stiff mauve]

ahh and for a = 2

[gleaming musk]

dont confuse it with reduced

[stiff mauve]

there isnt a pivot in row 3?

[acoustic pagoda]

bro is on the flow of reduction lol

[gleaming musk]

wdym

[stiff mauve]

so for a = 2

2a - 4 is = 0

so the first pivot in the 2nd row is in column 3

[gleaming musk]

oh thats what u meant

yeah sure

its just specificy when the variable is free and u end up with that situation

[stiff mauve]

and so there is no pivot in the third row roght

[gleaming musk]

yes

and then u have to manipulate it into echelon form

also before

dont confuse reduced with just normal echelon form

[stiff mauve]

yes

ty

[pearl pond]

@stiff mauve Has your question been resolved?

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[pearl pond]

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

[ebon burrow]

1

[ebon burrow]

1. I don't know where to begin

[tropic saddle]

have you just tried filling in some numbers in a 3x3 grid and seeing what the dog and cat would write?

[ebon burrow]

Although I know of the 3 x 3 grid

[ebon burrow]

Yes

9 6 1
2 5 8
3 4 7

@tropic saddle

[cinder thistle]

the cat wont get 84

[tropic saddle]

what do the cat and dog write?

[ebon burrow]

The products of a row and columns as sets

Which must be equal (unordered sets)

@tropic saddle

[tropic saddle]

not sure why you are bringing sets into this

also all sets are unordered

what actual numbers do the cat and dog write for your grid

[ebon burrow]

What?

[tropic saddle]

the cat writes down three explicit numbers

which ones

[ebon burrow]

Cat: product of each row
Dog: product of each column

[cinder thistle]

oh

the the dog wont get 84

[ebon burrow]

Ok

[cinder thistle]

but the cat gets 84

[tropic saddle]

but what are those products

I am asking you to calculate them

[ebon burrow]

Oh

But depends on the grid

[tropic saddle]

well yes

but you gave a grid

which you claimed worked

[ebon burrow]

It doesnt

I was wrong

[tropic saddle]

ok then try a different grid

see if you can get one to work

this is expected to take some time experimenting

[ebon burrow]

Let's see

@tropic saddle

9 6 1
3 5 8
2 4 7

Dog = {54, 120, 56}

Cat = {54,120,56}

@cinder thistle

[cinder thistle]

yes this works

[ebon burrow]

@cinder thistle what to do next

[cinder thistle]

notice 5 and 7 are completely unshared

try showing 5 is impossible

[ebon burrow]

?

How?

[cinder thistle]

i mean

that a 5x5 grid is impossible

[ebon burrow]

But how and also why?

@cinder thistle

@tropic saddle

[tropic saddle]

experiment

[ebon burrow]

But 5! Many combinations are too much

120 (many of them are equivalent)

[tropic saddle]

there are a lot more

but thats not the point

just try out a few

see what fails

pay attention to primes

[ebon burrow]

Ok trying

Doesn't work for 5 x 5

[tropic saddle]

why

[ebon burrow]

Idk

I wrote a comp program to print all such grids

Works for 3

(which i write)

[tropic saddle]

sigh

[ebon burrow]

But doesn't work for 5x5

[tropic saddle]

do it by hand

[ebon burrow]

I told you, there's no possibility

I tell you man

[tropic saddle]

yes but you dont know why

[ebon burrow]

Also, it is kinda urgent

[tropic saddle]

if you had done it by hand you would know why

[ebon burrow]

I have been stumbling on it for like 5 days

Tomorrow is the submission

[tropic saddle]

ok unlucky for you

[ebon burrow]

Factor problem

[tropic saddle]

but this is an exercise which requires time spent by you

trying out stuff

[ebon burrow]

ikr

I could do 6/8 problems

But can't do this and one more

[ebon burrow]

Doesn't work for 5

I did IT!

[tropic saddle]

why not

[pearl pond]

@ebon burrow Has your question been resolved?

[pearl pond]

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[regal herald]

<@&268886789983436800>

[iron stream]

abcdefghijklmnopqrstuvwxyz

[pearl pond]

Channel closed

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as this action is irreversible, and this channel may abruptly lock.

[pearl pond]

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[exotic gale]

Can someone please help me with this question

[exotic gale]

At first, I tried to solve it using brute force

But I think there should be a smarter way to solve it

<@&286206848099549185>

<@&286206848099549185>

[cinder thistle]

Mod 10

[pearl pond]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[pearl pond]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[exotic gale]

sorry for spamming the command my pc was glitchy

[exotic gale]

I don't see how this correlates with modular arithmetic

Can you please elaborate

[wet swallow]

All I see is the unit digits repeat

[cinder thistle]

Unit digits

Are the remainders when divided by 10

[exotic gale]

oh okay

[iron stream]

3+5+7+9+1 = 25

25*2 = 50

So 10 terms of course!

[wet swallow]

@exotic gale

[exotic gale]

why'd you tell me the answer

but okay

thanks nevertheless

.close

[pearl pond]

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[pearl pond]

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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[stoic imp]

In this problem, $a$ and $b$ are positive integers.

When $a$ is written in base $9$, its last digit is $5$.

When $b$ is written in base $6$, its last two digits are $53$.

When $a-b$ is written in base $3$, what are its last two digits? Assume $a-b$ is positive.

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[autumn fossil]

this could probably be converted to a modulo problem

idk if there's a simpler way, but the modulo one is quite short

[autumn fossil]

do you know how modulo works?

[stoic imp]

sort off

the remainder after the division, no?

[autumn fossil]

so how could you rephrase the first statement about a in terms of modulo?

[stoic imp]

a = x*9^n + y *9^n-1 + ... + 9^0 * 5

idk

[autumn fossil]

there could be some coefficients as well

[autumn fossil]

but we could also avoid modulo and work with sth like this

would you be able to convert a to base 3?

how many digits of the base-3 expansion would we get to know?

[stoic imp]

a = x * 3^2n + y * 3^2n-2 + ... + 3^0 * 5

[stoic imp]

how do I even know that

at the very least one

[autumn fossil]

The digit can't ever be 5 in base 3 expansion

[stoic imp]

5 in base 9

,w 5 in base 9 to base 3

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[autumn fossil]

5 = 1 * 3^1 + 2 * 3^0

thats the logic behind the 12

[stoic imp]

5 = 9^0 * 5 = 5

[autumn fossil]

yeah, that's in base 9

[autumn fossil]

in base 3 we have to write the expansion with 3's

like that

[stoic imp]

a = x * 3^2n + y * 3^2n-2 + ... + 3^1 *1 + 3^0 * 2

[autumn fossil]

this is also not true

a = ....12 base 3

Do we need any other digits though?

We have 2 digits of a in base 3

and in the end, we want 2 digits of a - b also in base 3

so knowing 2 digits of a should suffice

so we can now move onto b

did you understand how a was done?

[autumn fossil]

(this is correct now btw, luckily for us, we wont need the full expansion though)

[stoic imp]

b = ... + 6^1 * 5 + 6^0 * 3

[autumn fossil]

try rewriting it in base 3 now

[stoic imp]

5 = 3^1 * 1 + 3^0 * 2

ahh is so confusing

3 = 1* 3^1

b = ... + (1+1)*3^1 + 2 * 3^0

[autumn fossil]

let me check...

[autumn fossil]

why do you need to write 5 like this?

[autumn fossil]

we have this expression

and all we need is to rewrite it in form ... + sth * 3^1 + sth * 3^0 while preserving equality

you could even just write it in your favorite familiar number system, base 10 and then convert it manually to base 3

so like 6 * 5 + 3 = 33

and then write that in base 3

33 = 27 + 6 = 3^3 + 6 = 3^3 + 2 * 3 = 3^3 + 2 * 3^1

oh wait

i really just wrote 6 = 3 + 2 *facepalm*

[pearl pond]

@stoic imp Has your question been resolved?

[stoic imp]

33 = (3^3) + 2(3^1)

[autumn fossil]

yep

and taking the last 2 digits of that, what do we get?

(in base 3)

[stoic imp]

12

[autumn fossil]

you sure?

[stoic imp]

,w 33 in base 3

[autumn fossil]

there are some missing 0's basically

namely 0 * 3^2, 0 * 3^0

[stoic imp]

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

[autumn fossil]

almost

[autumn fossil]

it was 2(3^1) here

1(3^3) + 0(3^2)+ 2(3^1) + 0(3^0)

so it would be this

now the last 2 digits of that are 20

in base 3

so now the last part is subtracting a and b in base 3

[stoic imp]

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

[stoic imp]

hmm

[autumn fossil]

. 12
- 20

and we need to do that subtraction base 3

[stoic imp]

a = ... + 1(3^1) + 2(3^0)
b = ... + 1(3^3) + 0(3^2)+ 2(3^1) + 0(3^0)

[autumn fossil]

okay, this'd work too

b = ... 2(3^1) + 0(3^0)

btw we could also write b like this, becasue we only need 2 digits

and now you just need to subtract them

[autumn fossil]

either you can do the subtraction directly in base 3, or if you don't wanna do that you can just do it through this and then convert it back to base 3

[stoic imp]

a = ... + 0(3^3) + 0(3^2) + 1(3^1) + 2(3^0)
b = ... + 1(3^3) + 0(3^2)+ 2(3^1) + 0(3^0)

[autumn fossil]

remember that we only need the last 2 digits

we only need the 3^1 and 3^0 terms

[stoic imp]

a = ... + 1(3^1) + 2(3^0)
b = ... + 2(3^1) + 0(3^0)

[autumn fossil]

and now subtract them

[stoic imp]

a-b = ... + (1-2)(3^1) + (2-0)(3^0)

but is negative number

[autumn fossil]

yeah

how do we solve this issue in decimal system?

[stoic imp]

-1x3 + 2

[autumn fossil]

If you were to subtract e.g.
100 + 10 + 4
40 + 3

so
114
-43

1 - 4 is also negative

how do we solve that?

[stoic imp]

?

[autumn fossil]

do you know the subtraction algorithm?

the one where you place the numbers under each other and then compute the difference digit by digit

it looks like this when on paper

do you know that?

[stoic imp]

I got lost

I understand tiling here

a-b = ... + (1-2)(3^1) + (2-0)(3^0)

[stoic imp]

are you asking me if I know subtracting

[autumn fossil]

Yeah, cool

and we ran into the issue that 1 - 2 is negative

what im trying to point out is that the same issue happens when we do subtraction in base 10

[autumn fossil]

using this algorithm

e.g. when we try to subtract
114
-43

we run into the issue of 1 - 4 being negative

when computing the tens digit

[stoic imp]

why 114 43

I got lost again tbh

[autumn fossil]

just a random example

e.g. means for example

[autumn fossil]

anyway, how do we solve the issue of 1 - 4 being negative when doing subtraction in base 10?

[stoic imp]

its 4 -3 = 1 in units place then 11-4 in tenths place

[autumn fossil]

Yeah

how did you get 11?

[stoic imp]

idk how to explain it is how math works

[autumn fossil]

you basically added 10 to the 1

and 10 is the base we're working in

and exactly the same method works for base 3

we can just add 3

so the 1 - 2 would turn into 4 - 2

a-b = ... + (1-2)(3^1) + (2-0)(3^0) = ... + (4-2)(3^1) + (2-0)(3^0)

we are basically adding 3*3^1

and what happens algebraically is that it becomes 3^2

so it moves behind the dots and it's no longer our issue

since we only care about the last 2 digits

[stoic imp]

okay i get it

so?

a-b = ... + 2x3+ 2

[autumn fossil]

yepp

so the last 2 digits are...?

[stoic imp]

2, 2

[autumn fossil]

yeah

22

you can also subtract the numbers directly using the same algorithm as in base 10, except replace 10 with 3

[autumn fossil]

so like
12
-20
the last digit is 2 - 0, we carry nothing. Then we have 1 - 2 so we add 3 to get 4 - 2 which is 2 and we carry 3, but that's no longer our issue since we only cared about last 2 digits

[autumn fossil]

Yeah, modulo was the other approach

a = 5 (mod 9) is equivalent to 5 being last digit of a in base 9

similarly, b = 33 (mod 36)

and so b = 6 (mod 9)

a - b = 5 - 6 = -1 = 8 (mod 9)

and 8 = 2 * 3 + 2 which is 22 base 3

[pearl pond]

@stoic imp Has your question been resolved?

[pearl pond]

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[orchid cobalt]

i got 20.29

and i did (8.2)/tan68

but the answer is 3.31!

[plush bramble]

,calc 8.2/tan(68 deg)

[jolly parrot]

Result:

3.3130150518483

[orchid cobalt]

oh

OH WAS I SUPPOSED TO USE THE INVERSE ON THE CALC?

[warm current]

what

[orchid cobalt]

ok wait that didnt work

mb

[orchid cobalt]

so how would i put this on gdc-?

[warm current]

what is tan(45) in your calc

[orchid cobalt]

1

[plush bramble]

What's tan(68) on your calc

[orchid cobalt]

guys im still getting 20.29

[orchid cobalt]

2.47

rounded

[plush bramble]

,calc tan(68 deg)

[jolly parrot]

Result:

2.4750868534163

[plush bramble]

Right

[orchid cobalt]

oh wait, i forgot to divide.

wait lemme try dividing

[plush bramble]

No idea what you're doing wrong then

[orchid cobalt]

omg i got it!!

[plush bramble]

,calc 8.2tan(68 deg)

[jolly parrot]

Result:

20.295712198014

[orchid cobalt]

wait what would the tan-1 be for then?

[plush bramble]

I see

That's inverse tangent function

[orchid cobalt]

yea, kinda silly of me! thank you so much though!

[plush bramble]

Also called arctan

[orchid cobalt]

you wouldnt use it in this case right?

[plush bramble]

No you would not

[orchid cobalt]

alright thank you so much! ❤️

[limpid lily]

The -1 doesn't mean reciprocal. It means undo this function. So, like if you have tan(50 degrees) and you run tan^-1 on it, you get back 50, loosely.

,calc tan(50 deg)

[jolly parrot]

Result:

1.1917535925942

[limpid lily]

,calc atan(1.1917535925942) * 180 / pi

[jolly parrot]

Result:

50

[orchid cobalt]

ohh wait, that makes sense!

so itls like f^-1!!

[limpid lily]

Right.

[orchid cobalt]

thank you so much!!

[limpid lily]

No problem.

[orchid cobalt]

https://tenor.com/view/cat-gif-4504947027691887877

.close

[pearl pond]

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[pearl pond]

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[wheat remnant]

Is this a valid approach and are these graphs correct?

[wheat remnant]

<@&286206848099549185>