help-23

[safe radish]

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[nimble vine]

claim

[nimble vine]

[
\Delta(x)=
\begin{vmatrix}
x^2 + x & 2x - 1 & x + 3 \
3x + 1 & x^2 + 2 & x^3 - 3 \
x - 3 & x^2 + 4 & 2x
\end{vmatrix}
= a_0 x^7 + a_1 x^6 + a_2 x^5 + a_3 x^4 + a_4 x^3 + a_5 x^2 + a_6 x + a_7,
]
[
\text{find } \sum_{k=0}^{7} a_k.
]

[flat frigate]

oppenheimer

[gray mulch]

The sum you're looking for is just the determinant at x=1

[peak estuary]

dont know how to give a hint without spoiling it ||plug in x=1||

oh well got spoiled anyway

[nimble vine]

oh

right

ty

.close

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[glass edge]

Got stuck

.close

[safe radish]

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[faint hornet]

do you still need help

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[thin lion]

,rotate

[thin lion]

For first one, I have -n

Second one I have 1/n + 1

But for the last one which I’m unsure on

I have made cases, xn = 0 if n is even and 1 if n is odd

It took a while to decipher what exactly the last one is saying, is it like, since it says there exists, as long as I have ht case where it’s 0 then it works?

[brave wolf]

Any sequence which contains 0 and either diverges or converges to sth else is valid counterexample

or you could also have sth like
1 for odd n, 1/n for even n

[thin lion]

Oh alright, as long as 0 is in there somehow

[brave wolf]

yeah

[thin lion]

Ty ❤️

.close

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[brave wolf]

you can also have sequence without 0 which is still a valid counterexample though

so its not iff

[thin lion]

.reopen

[safe radish]

✅ Original question: #help-23 message

[faint hornet]

yes i was about to say this

[brave wolf]

this doesnt have terms arbitrarily close to 0

[quasi bison]

that's just 2

unless you meant n/(n+1)

[desert pasture]

yes

[faint hornet]

if you have a sequence that doesn't converge to 0 but has a subsequence doing so is what he was getting at ^

[brave wolf]

you need to have a sequence that converges to 0 and then combine it with some other sequence which doesnt converge to 0

[thin lion]

Ah alright

Yea so I mean the case that I said

Has a subsequence which is essentially x_(2k)= 0

[faint hornet]

yeah

[thin lion]

And in Mathsisalwaysrights case, had 1/n which isn’t 0 but tends to 0

Alright

That works

.close

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[safe radish]

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[thin lion]

,rotate

[viral geyser]

hello

[thin lion]

Idk if I used the max part correctly

,rotate

[faint hornet]

this is good but i think you shouldn't reuse the same variable twice

[thin lion]

Can I just go from definition of b converging to then just combining the inequality?

[thin lion]

u mean the N?

I had a N1 and N, idk if it’s visible 😭

[faint hornet]

oh it is visible, i am just blind

[faint hornet]

but yes

[thin lion]

Ah alright so this would be fine

.close

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[faint hornet]

yh but make sure you explicitly say for n > N_1, depending on how pedantic your lecturer is, you could lose some credit

at the end i mean

[thin lion]

Ah alright 👍

[obsidian oracle]

By the way, as an alternative thought
Have you shown the squeeze theorem yet?
As this theorem shows you |a_n - l| -> 0 as n to infinity
so if you know the equivalence between |a_n-l| -> 0 and a_n -> l, the proof is finished

[thin lion]

Oh so 0<= |a_n -l| <= |b_n| —>0

Oh that’s pretty neat

[safe radish]

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[forest crow]

Hi, i'm doing polynomials and factoring, and i'm struggling to see how this becomes (x-1)(x^2+x+2). I'm used to dividing and factoring normal polynomials but i'm kind of at a loss when this doesn't well look like a normal polynomial.

Basically i think it would help if someone broke down how they do polynomial division for this equation. Thanks!

[glacial cairn]

Are you asking about polynomial division specifically or just about why x^3+x-2 = (x-1)(x^2+x+2)

[forest crow]

I guess the second half. I have done other divisions no problem but this task specifically i'm a bit stuck at since it doesn't follow the normal ax^3+bx^2+cx+d routine

[glacial cairn]

You're giving me conflicting answers here...

Are you asking (1) how to divide x^3+x-2 by x-1 using a polynomial division algorithm or (2) how to factorize x^3+x-2 in any way

[forest crow]

Apologies, i'm not that good with speaking in math, and with english as 2nd language. I think 1 is what i'm after, a breakdown of how that would be done

[glacial cairn]

Ok so we have the dividend x^3 + x - 2, which we'll rewrite x^3 + 0x^2 + x - 2, and the divisor is x - 1

1a) divide x^3 by x (first terms of dividend and divisor) to get x^2

1b) multiply x^2 by the divisor x-1 to get x^3 - x^2

[red basalt]

hahhhah noobs

[glacial cairn]

1c) subtract that from the dividend to get x^2 + x - 2

[red basalt]

i am o;ympiad toppre

[glacial cairn]

So far so good?

[red basalt]

rair 2

AIR 2

[glacial cairn]

<@&268886789983436800>

[nimble vine]

Don't troll

[red basalt]

U BOTS

[lean otter]

Was Abt to do so but thought I should ask first thanks I will do the same in the future

[glacial cairn]

When it's obviously bothering you really don't need to hesitate

[junior smelt]

You've been silenced, do this again when your timeout expires and you'll be forcibly removed from our presence

[forest crow]

Yes, actually rewriting the equation adding 0x^2 was what i forgot i could do, must have has a brainfart. I think i can manage to solve it myself from here on, thanks a lot!

[glacial cairn]

Oh alright

Good luck then

[forest crow]

Thanks!

.close

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[frigid spruce]

it doesn't even sound threatening when you say stuff like that 😭

[junior smelt]

Why be threatening when you have all the power

[nimble vine]

Bit the wise

[frigid spruce]

[safe radish]

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[serene gazelle]

So for 10 I got 95C5, but for 11 I'm unsure how the solution manual is getting there answer

[quasi bison]

do you want the basic idea or do you want specifically the solution manual's stuff explained

[serene gazelle]

Like they took off the 4 and 2 from 100 to get 94 then did bar and star and got 97C3

No I want to know why it works

I wanna know what's happening

[quasi bison]

subtract 6 (a.k.a. 4+2) off both sides to get

(w-4) + (x-2) + y + z = 94

(w-4) and (x-2) can be treated as variables with a >=0 constraint

then it's the same as 10 with different numbers

[serene gazelle]

Hold on

Oh i see

Oh

Yeah but why does that work

[quasi bison]

well give the variables new names like w' and x'

there's a bijection between the solution set in problem 11 and the solution set of {w'+x'+y+z=94, all variables >=0}

[serene gazelle]

So for 12 it would be a similar process

[quasi bison]

almost the exact same process, really

[solar hazel]

this is easiest to see through generating functions

[safe radish]

@serene gazelle Has your question been resolved?

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[solar hazel]

the answer to 11 is the coefficient of $x^{100}$ in $$(x^4 + x^5 + \cdots)(x^2 + x^3 + \cdots)(1 + x + x^2 + \cdots)^2,$$ which is equal to
$$x^6(1 + x + x^2 + \cdots)^4.$$ so now we can see the $x^{100}$ coefficient we were looking for is the coefficient of $x^{94}$ in $(1 + x + x^2 + \cdots)^4$. and that's equal to the number of solutions to $w+x+y+z = 94$ with $w,x,y,z\geq 0$

[flat frigate]

sleighla

[solar hazel]

you don't have to mind me just wanted to explain this for fun

[safe radish]

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[sharp raft]

x/8 = 4

why do we multiply by 8?

[sharp raft]

like how will it turn into just x?

[plucky elk]

https://www.khanacademy.org/math/arithmetic-home/multiply-divide/division-intro/v/the-idea-of-division

[normal moss]

You've got one eighth of x

[peak estuary]

are you asking why x/8*8=x or why we are multiplying the equation by 8

[normal moss]

Say you've got a massive pizza. And you split it into slices among 8 friends. We know, because we are the ones who split it, that each friend got 4 slices.
How many slices of pizza are there?

[warm edge]

The idea is we multiply the left side of the equation by 8, so that 8 * x/8 = 1 * x = x, so you get only x. But if we multiply the left side, we must also do the same for the right side, hence 8 * 4 = 32.

[normal moss]

It's fascinating when someone asks a question and then they are instantly captured by aliens

[faint hornet]

or perhaps they just became enlightened and figured it out on their own

[safe radish]

@sharp raft Has your question been resolved?

[sharp raft]

but isnt it just

8x/8?

,w 8x/8

[flat frigate]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by fuzzials

[normal moss]

[sharp raft]

oh thats how it works

.close

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[normal moss]

fascinating-er

[safe radish]

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[stoic ravine]

It says that I can select any points in order to get the answer but it doesn't look like I can

[austere goblet]

why not? what have you tried?

[stoic ravine]

I tried counting how much it goes down versus horizontally

Like in the image

I picked the points based off the numbers on the chart

[austere goblet]

and what is the slope you got?

[stoic ravine]

When I connected the points together, it looked like the y axis was going down by one

[austere goblet]

first of all, which two points are you using?

[stoic ravine]

3 and -1

sorry

[austere goblet]

each point should have two coordinates, as a reminder

I assume you meant (-2, 3) and (-1, 0)

as from the diagram

[tardy mango]

hi

[austere goblet]

in that case, do double check that the y-axis is going down by just 1 unit, becuase it doesn't seem that way to me.

[austere goblet]

random pigeon hi

[tardy mango]

I just wrote this much for one question send help ok I leave ygs alone

[stoic ravine]

this is what I actually meant

[austere goblet]

uh...? what does this do?

ok actually before we proceed, may I ask how you've been taught to find the slope of a line given two points?

[trail portal]

,,\m

[flat frigate]

calvin

[stoic ravine]

To place the points on any y axis and x axis

[trail portal]

use two pairs of integer points and assign values

[austere goblet]

well, I'll let calvin handle this then

[trail portal]

,,(x_1,y_1), (x_2, y_2)

[flat frigate]

calvin

[trail portal]

i believe your understanding of a slope is wrong

the slope is a rate of change

in linear (line) functions

the slope (m) is always the same

thats why its a line

its a ratio between the change in y and change in x

hence the formula

[flat frigate]

calvin

[trail portal]

does this make any sense?

[stoic ravine]

Just confused on how M is relevant

[trail portal]

m is the slope

its the thing you’re looking for

thats what the question asks for right?

[stoic ravine]

Yes, just dont know how I would apply it

[austere goblet]

I'll just come back in to remind you of this.

back to Calvin

[trail portal]

you don’t know how to apply it or find it?

[stoic ravine]

y2-y1

[flat frigate]

calvin

[trail portal]

this is the slope formula

,,(x_1,y_1), (x_2, y_2)

[flat frigate]

calvin

[trail portal]

use these points to find the slope

By substituting them in

[(x_1, y_1), (x_2, y_2)] [(-2, 3), (-1, 0)]

[flat frigate]

calvin

[trail portal]

if you don’t understand lmk

@stoic ravine

[trail portal]

basically

find easy points that lie on the line

tell me your misconceptions

[trail portal]

use whole numbers

lemme annotate a diagram up rq

tell me the coordinates of the blue green and purple points

[stoic ravine]

sure

-12/4

[trail portal]

not quite

[stoic ravine]

-4/1?

[trail portal]

do you know how to label points

[stoic ravine]

yeah I think that's my issue

[trail portal]

points are in the format (x,y)

eg. (1,0)

[stoic ravine]

I used to think that points were in the format of y over x

[trail portal]

the black point is (1,0) on the plane

[trail portal]

yeah thats a big mistake

its always (x,y)

now try again

[stoic ravine]

1/-4? wouldn't the numbers be reversed

[trail portal]

use a comma and brackets

and tell me which point youre doing

[trail portal]

still wrong

ill find a video for you

https://youtu.be/a1DXlvwnqUw?si=ip8wYX_QcKhCo0Xn i highly advise you to watch this

close the ticket and reopen later

ping me i have to go rj

rn

[stoic ravine]

ok

[safe radish]

@stoic ravine Has your question been resolved?

[elfin comet]

U still stuck?

[safe radish]

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[grim scarab]

.rotate

,rotate

[grim scarab]

I got :
0, 4π/3, 2π/3, -2π/3, -4π/3

[plucky elk]

!show

[safe radish]

Show your work, and if possible, explain where you are stuck.

[quasi bison]

$\frac{4\pi}{3} \notin (-\pi, \pi)$

[flat frigate]

Ann

[grim scarab]

🤦‍♂️

Omg why do I always forget that

[nimble vine]

Mb

[grim scarab]

,rotate

[quasi bison]

,rotate 180

[quasi bison]

fuck

[warm ocean]

[nimble vine]

,rcw

[grim scarab]

Indeed

[quasi bison]

-sin(x)[2cos(x) + 1] = 0.

[grim scarab]

I still put cosx = -1/2

[quasi bison]

even number of mistakes cancelling out

[grim scarab]

So my work is right

Just no 4π/3

And -4π/3

[plucky elk]

No

[grim scarab]

🤨

[plucky elk]

You can still get points marked off if your work has mistakes

[grim scarab]

Nono I just meant the answer

[plucky elk]

???????

[grim scarab]

I meant answer

😅

Cause i had to translate my work

[plucky elk]

Even your words are wrong

[verbal kite]

lol

[nimble vine]

[solar hazel]

[grim scarab]

I just woke up😔

Got finals

So to confirm the final answer is correct if I remove 4pi/3 and -4pi/3

.close

[safe radish]

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[safe radish]

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[jagged bane]

someone help

[verbal kite]

What are the options?

[quasi bison]

try writing this in sigma notation

may make things clearer

[jagged bane]

yeah i tried it isnt doing anything

[quasi bison]

can you show your sigma notation anyway

also come to think of it

[verbal kite]

The expression is equal to itself. Just write it in sigma notation and you're done.

[quasi bison]

i just realized this looks like the m'th order forward difference operator on x^m

[jagged bane]

wat

the options were m! (m+1)! m+1Cm and mPm-2

[quasi bison]

two options only?

[jagged bane]

those are the 4

[quasi bison]

... dude, put some commas

or newlines

anyway ok do you know what "forward difference" is

[jagged bane]

this is a eleventh grade question

just so u know

[quasi bison]

do not call me "brother".

or any variation thereof.

[jagged bane]

ok ok can u plz tell me what it is

[quasi bison]

"it"?

[jagged bane]

the question

[quasi bison]

i'd like you to edit out the word "brother" first.

[jagged bane]

now can u plz solve it

[quasi bison]

ok let's try this

for a sequence, are you familiar with the concept of first differences, second differences, third differences etc.

yes or no

[jagged bane]

yes

[quasi bison]

ok

``forward'' just means you subtract $a_{n+1}-a_n$ and not $a_n-a_{n-1}$ (which would be called a backward difference)

[flat frigate]

Ann

[quasi bison]

you can do the same difference thing to a function

and you can also speak of second and third etc differences

$\Delta f(x) = f(x+1)-f(x) \ \Delta^2 f(x) = f(x+2) - 2 f(x+1) + f(x) \ \Delta^3 f(x) = f(x+3) - 3 f(x+2) + 3 f(x+1) - f(x)$ \ and so on and so forth

[flat frigate]

Ann

[quasi bison]

the coefficients here are binomial coefficients with sign alternation

[jagged bane]

ok so

plz continue

[quasi bison]

what you've got there is the m'th forward difference of the function x^m

do you know what happens if you take the m'th difference of a sequence whose rule is an m'th degree polynomial

[jagged bane]

no

what do u mean by mth difference

[quasi bison]

ok lets make it more concrete

the 2nd differences of a quadratic sequence are constant. if a_n = n^2, what's the value of the second difference

[jagged bane]

i dont understand what the 2nd differnce means

[jagged bane]

ok so i just got my hands on the solution and i dont know wat u are talking about

[quasi bison]

i dont think we can continue down this path

ok can you share the solution

what does it say & what does it do

[jagged bane]

yeah wait a sec

[feral linden]

I think not just look like, exactly this

[timber karma]

is the answer ||m!||?

[jagged bane]

yep

[quasi bison]

ok what kind of ass crack did they pull the FIRST line out of

[feral linden]

Yes

[quasi bison]

e^(xy) (e^y - 1)^m ??

[jagged bane]

yeah i dont know

[quasi bison]

also old factorial notation L

[jagged bane]

what am i supposed to dream about this

can anyone atleast explain me the solution

[quasi bison]

they expand (e^y - 1)^m via binomial theorem and then e^y itself via its taylor series

or so one would think

[feral linden]

The author multiplied by y^m /m! of your expression and added up I think

[kind seal]

It does also seem similar to an expression one would obtain using the generalised PIE. I wonder if there could be a more combinatorial solution. m! is also a simple solution which could arise through some combinatorial bijection.

Edit: Doesnt seem like it. So ignore this.

[jagged bane]

this one also plz

[feral linden]

Might be actually, if you prove E_m(x) is irrelevant with x, E_m(0) by inclusion-exclusion is exactly number of surjections (thus bijections) from {1,2,..,m} to itself

[kind seal]

I thought of E_m(0) but proving it is independent of x put me off

[kind seal]

What have you tried for this one

[jagged bane]

.

[feral linden]

Just remove one layer, first find order of 10=3 mod 7, say r, then calculate power of 10 mod r

[jagged bane]

yeah can you solve it using binomial plz

plz

[feral linden]

Can you tell me order of 3 mod 7?

Minimal k such that 3^k=1 mod 7

[jagged bane]

6

[feral linden]

Okay now you calculate 10^k mod 6, which will be periodic too

(4^k mod 6)

It will be a very nice sequence

[jagged bane]

but i havent been taught any of this and this question is given to me while i am taught binomial

so there must be a way to do it using that right

[kind seal]

Congruences work on the basis of the Binomial Theoem so ig its fine

You could just show it using the Binomial theorem in the same way we prove the basic properties of congruences

[jagged bane]

ok thanks guys

[kind seal]

You got it?

[jagged bane]

yeah

[safe radish]

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

[jagged bane]

.close

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[fresh vine]

given equation is x^2-2(m+1)x+m^2-2m+5=0. Find m so equation has 2 solution x1 , x2 that satify:

[fresh vine]

I got to this step

idk what to do next

[glacial crow]

can you pls translate it

[verbal kite]

x1 * x2 = c/a and x1 + x2 = -b/a

[fresh vine]

I did

[verbal kite]

There will be three equations in three variables.

No way. The expressions inside the square roots are perfect squares. Lmao.

[fresh vine]

so i did the absolute value

[snow birch]

Aren"t there 2 variables in the question :x1 and x2 ?

[verbal kite]

Oh.

[fresh vine]

yeah

it is 2 solution for the equation

they are*

[split kayak]

for my own sake and whoever else comes later;

[split kayak]

$x_{1,2} = (m+1) \pm \sqrt{(m+1)^2 - (m^2-2m+5)}$

[split kayak]

which simplifies to $m+1 \pm 2\sqrt{m-1}$

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[fresh vine]

calculate the delta we got m>1

[fresh vine]

x1 + x2 =2(m+1)>0
x1x2=x^2 -2m+5 = (x-1)^2 +4 >0

so x1 and x2 must be positive

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[frosty leaf]

can someone help me to translatew this ito an equation

[quasi bison]

let x represent the price difference in dollars from the 'default' $40

in other words, suppose the price for widgets is set at (40+x) dollars

can you write down how many sales are made this way?

[frosty leaf]

i know rev = 700*40$

now these changes should represented as(700 - 20)*(40+2)

[quasi bison]

you don't know that they specifically increase the price by 2 dollars

[frosty leaf]

this will give a quadratic so 700 - 20)*(40+2x)

[quasi bison]

700*40 is only the revenue when the price is 40

also i said the price should be 40+x and not 40+2x. why this 2 coefficient?

[frosty leaf]

but its +2x why +x ?

otherwise it would have said increase of 1

[quasi bison]

it says

for each 1-dollar increase ...

[frosty leaf]

oh

sorry

[quasi bison]

price = 40 + x
sales = ?

don't try to put revenue yet

do it one step at a time

[frosty leaf]

700-10x ?

[quasi bison]

right

[frosty leaf]

hold on

[quasi bison]

now you can write down the revenue function

[frosty leaf]

i was biased because i knew it needed a quadratic

why is it x ?

in quantity

[quasi bison]

why is what x?

what does "i quantity" mean?

[frosty leaf]

700 - 10(smth))

why is smth = x ?

[quasi bison]

yes...

it says: for every dollar in price increase, quantity drops by 10

[frosty leaf]

we are talking about quantity and price right?

[quasi bison]

the relationship between price and quantity is given explicitly by the question here

[frosty leaf]

yeah, quantity drops by 10 but why do i multiply this by x and the prixe + x ?

what is x here

i fail to recognize

[quasi bison]

.

the very 1st msg i sent

[frosty leaf]

oh

so x is the price?

[quasi bison]

no.

x is the difference between the price and $40.

in other words, the price is $(40+x).

[quasi bison]

which i also repeated here

so we have now
price = 40 + x
qty = 700 - 10x

are you happy with this now or is it still causing doubt

[frosty leaf]

im still confused with the notation.

to me i understand nenw revenue is determined by (700 - 10 * smth) *(40 + smth)

i dont uderstand how to pin the x

why is it not 2 different variables

[quasi bison]

this "smth" is what i am calling "x".

[frosty leaf]

why x for both

[quasi bison]

it is not two different variables because price and quantity are directly related

as the question tells you

[frosty leaf]

ok

[quasi bison]

when price goes up by $1, then qty goes down by 10 units

it is a linear relationship

[frosty leaf]

ok makes sense

[quasi bison]

that is why the variable is the same here

[frosty leaf]

ohhh

so +1 = x ,

[quasi bison]

"+1 = x" is not a good way to write it

[frosty leaf]

sorry 1 = x

[quasi bison]

still no

[frosty leaf]

x = (1,2,3..

[quasi bison]

still no

x is x

1 = x means specifically that we set the price at $41. we don't want that yet.

[frosty leaf]

ok

[quasi bison]

R = (700 - 10x)(40 + x)

we want the maximum value of this

we now forget about revenue and price and shit for a moment

and treat it as just another quadratic

do you know how to find the coordinates of the max/min point of a quadratic function yes or no

[frosty leaf]

yes, if a>0 min will be -b/2a

if a<0 max will be -b/2a

[quasi bison]

ok can you go and do it

[frosty leaf]

thank you Ann

[quasi bison]

yw

!done

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[hybrid heath]

could you walk me through this process?
i am struggling to understand this

[hybrid heath]

It is about moments in statistics

i only got to know that moments are used to check the shape and size of the data without looking at the original frequency distribution.

moments have mean,standard deviation, skewness and kurtosis

[quasi bison]

which point do you want to start with? 2.15.1?

[quasi bison]

the idea of moments as a whole is rather abstract but yes that is the gist of their 'practical' usage

[hybrid heath]

i don't get why they are writing like this

i want to know which one is a proof and which one is going to be used as a formula

[hybrid heath]

from 2.15

[viral geyser]

hi

[hybrid heath]

ok..i will do it later

[viral geyser]

is this solved

[hybrid heath]

.close

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[frosty leaf]

is there a faster way ?

[quasi bison]

apply $a_n = c(a_{n-1}+1)$ twice to get $a_4$ directly in terms of $a_2$

[flat frigate]

Ann

[quasi bison]

namely $a_4 = c(a_3 + 1) = c[c(a_2+1)+1] = c^2 a_2 + c^2 + c$

[flat frigate]

Ann

[quasi bison]

you cannot assume a3 = 21 and a1 = 1 though

that's unjustified

[frosty leaf]

a4 = (c^2)*6+c^2+c

[quasi bison]

those brackets are a bit misplaced...

[frosty leaf]

and plug it back in for a3?

[quasi bison]

no. a_4 is known to be 66.

so you have 7c^2 + c = 66. solve for c.

[frosty leaf]

oh nice

how would you solve this without using (-b+-sqrt(D))/2a

viettas formulas wont give an answer , just the relationship between the solutions

[quasi bison]

-# brackets ...

honestly if i was banned from using the QF i'd probably suffer through completing the square or something

why do you want to avoid the QF though?

[frosty leaf]

its for the gmat

[quasi bison]

gmat?

[frosty leaf]

i am constrained with time pressure

2 mins per q

look it up

its sat for grads

[quasi bison]

hm...

$a_4 = c^2 a_2 + (c^2 + c)$ and $a_6 = c^2 a_4 + (c^2 + c)$

[flat frigate]

Ann

[quasi bison]

maybe we could try to like, avoid finding c itself somehow

a_6 = 67c^2 + c

but this i don't know how to do in 2 minutes

maybe you could try applying the rational root theorem on 7c^2 + c = 66 in hopes of getting a rational solution

,w 7c^2 + c = 66

[flat frigate]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by ann_dec

[quasi bison]

c=3, by the looks of it...

just cross your fingers hoping that c is an integer and guess&check

thats the only thing i can think of when there's time pressure

[frosty leaf]

yeah i figured, thanks again !

[kind seal]

Or you could use factorisation ig by writing c=22c-21c. But again, this is not guaranteed to work all the time (here, its pretty decent)

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[fervent pulsar]

got the correct solution

[fervent pulsar]

jst was wondering

ok

arent the areas symmetrical

and if so

that means each small one =3

and then if u integrate mx-x^3

from root m to 0

ull get

m^2/2 = 3

m^2=6

m= root 6

which is incorrect

but why is that

[feral linden]

m^2/4=3

[timber karma]

this is correct

[feral linden]

Not /2

[timber karma]

because both functions are odd

[fervent pulsar]

oh shoot.

wow.

thats embarrasing

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[keen tulip]

write x = n + t where n = floor(x) and t = {x} and split the integral over intervals [n,n+1)

[solar hazel]

yet another yuri server tag

[keen tulip]

[safe radish]

@wraith scroll Has your question been resolved?

[quasi bison]

op gone

[keen tulip]

unfortunate

[wraith scroll]

I got distracted

[quasi bison]

do you understand this hint

or instruction, rather

[wraith scroll]

should I be able to 😭

can't really understand the last part

[quasi bison]

do you know that integrals can be split up into pieces

[wraith scroll]

yea

[quasi bison]

ok right

[wraith scroll]

n and n+1 just don't make sense for me for bounds

[quasi bison]

what @keen tulip is suggesting is that you should split your integral into the integral from 0 to 1, and then from 1 to 2, and then from 2 to 3, and then from 3 to 4, and so on and so forth.

[wraith scroll]

and idk why the right parentheses is open

oh

[quasi bison]

$\int_0^{\infty} f(x) \dd{x} = \sum_{n=0}^{\infty} \int_n^{n+1} f(x) \dd{x}$

[flat frigate]

Ann

[wraith scroll]

I was trying infinite series but I was manually computing each step for a pattern 💀

[worthy citrus]

I'm getting ln2

[queen ingot]

|| yeah it turned into the alternating harmonic series (mercator series with x = 1) ||

[wraith scroll]

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[sudden lotus]

I am trying to prove that a group of order 30 always has a normal subgroup of order 5. This is my work until now but i am a little stuck here. If someone needs some translations tell me!

[sudden lotus]

From the place where i wrote question marks is the place where i am not sure about my work at all

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[sudden lotus]

<@&286206848099549185>

[feral linden]

I don’t understand what you are doing. I would simply number of sylow 5 groups m_5=1+5k | 6 and exclude 6

And the handwriting is a little hard to read

Following your ideas in the beginning, if m_5=6 I ended up m_2=3 and m_3=1

You somehow excluded m_2=3, it should already cause a contradiction

I see, m_2|5, so contradiction. Your method makes sense. And the way I obtained m_2=3: notice that there are 24 elements of order 5, one identity. The remaining 5 elements, since one sylow-3 subgroup exists, two of them are of order 3 (m_3=1,10, so m_3 has to be 1, no room for 20 order 3 elements ). Thus three order 2 elements left, thus m_2=3. Combine with m_2|5, contradiction

@sudden lotus

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@sudden lotus Has your question been resolved?

[sudden lotus]

Im so sorry. I am using Sylow and then finding options for n_3 and n_5. Now i didnt include n_2 because i didnt think it would be of any importance in this situation since i was looking for a normal subgroup of order 5. Now the option n_5=6 and n_3 = n_2=1 would be the only non trivial situation that has to be researched so thats why i was focussing on it. i was looking for then a method for which to find a contradiction to n_5=6 so that i can prove that the only possibilty would be n_5=1 and thus i would have proven that you have a normal subgroup of order 5

[feral linden]

I see. Anyway, it worked in the end

[sudden lotus]

yes the actual question wouild be if my final method is good enoigh

well to say it better if it is mathematically correct

[feral linden]

It is correct. You found out it can only be (m2, m3, m5)=(1,1,6), I found out it can only be (3,1,6). Thus contradiction

[sudden lotus]

Perfect thank you !

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[steel plume]

can anyone help me please with part a

[pine mountain]

I would simply plug in x=3 and y=0 and solve for k

[mossy lotus]

if the circle passes thru some point, you can sub in the coordinates in the equation and it satisfies the equality of RHS = LHS

[steel plume]

ohh its that simple thanks

[quasi bison]

it is indeed

[versed wave]

it's just that shrimple

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[steel plume]

.close

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[quasi bison]

https://tenor.com/view/shrimp-as-that-clash-royale-hee-hee-hee-haw-gif-25054781

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[gritty reef]

the variance of 20 observations is 5. if each observation is multiplied by 2, find the new variance of the resulting observations.

[gritty reef]

i know that i have to do k^2 * variance (here k=2)

but how do i prove it

[plucky elk]

Use the variance formula

Var(X) = ... And Var(2X) = ...

[gritty reef]

then?

[plucky elk]

That would prove

[quasi bison]

what's your defn of variance

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[gritty reef]

i got my answer thanks

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[quasi frost]

Can someone help me with a contour integral, its of 1/z²+1 dx

[quasi bison]

dx?

also what contour?

also is the function z^-2 + 1 or is it 1/(z^2 + 1)?

[quasi frost]

Shit mb dz

its 1/(z²+1)

[quasi frost]

closed

[quasi frost]

idk what information you want from me 😭😭😭

[quasi bison]

show me a picture of the original problem

[quasi bison]

that was it?

[quasi frost]

yeah the dz bit cut off

[quasi bison]

theres like 3 different contours you could integrate that around, with different answers each time.

uncrop the entire img please

if you're gonna keep refusing i'll say that the contour is actually the circle |z|=0.001 and the integral will be zero

[quasi frost]

my mate was lost and asked me for help and I was so lost

[quasi bison]

so this is not even your question?

[quasi frost]

That was it

no but I was curious

[quasi bison]

if you're gonna keep refusing i'll say that the contour is actually the circle |z|=0.001 and the integral will be zero

[quasi frost]

cuz it didn't make sense

[honest perch]

the answer is an integer

[mossy lotus]

well, this is not analytic everywhere, so based on whether/how many of the poles are inside the contour would change the answer

[honest perch]

regardless of the contour

[quasi frost]

where can i learn abt this in depth bc none of my books have it and I find it easier to learn like with an actual book

[honest perch]

oh wait nvm

[mossy lotus]

any book on complex analysis would do

[quasi frost]

okay thanks

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[south sparrow]

∀ (G : Grp), ∃ (T₁ T₂ : Grp → Grp), T₁ G = G ∧ T₂ G = G ∧ T₁ ≠ T₂

[quasi bison]

symbol soup both unnecessary and probably not correct anyway

[astral glacier]

Unnecessary yes, but seems correct. Still completely unnecessary

[delicate shore]

The two quantifiers should probably be switched

[astral glacier]

True actually

[quasi bison]

thats the part that made me issue this comment

[astral glacier]

Fair, I'm bad at ordering quantifiers myself

[south sparrow]

∃ (G : Grp), ∃ (T₁ T₂ : Grp → Grp), T₁ G = G ∧ T₂ G = G ∧ T₁ ≠ T₂

T₁ = identity on objects and arrows:
T₁: A ↦ A, B ↦ B
AB ↦ AB, BA ↦ BA

T₂ = identity on objects,
maps arrows differently:

T₂: A ↦ A, B ↦ B
AB ↦ BA, BA ↦ AB

Thankyou

[quasi bison]

what does "AB" mean here

[south sparrow]

def T1 (G : Grp) : Functor Grp Grp := { obj := λ X => X, map := λ X Y f => f, ... }

def T2 (G : Grp) : Functor Grp Grp := { obj := λ X => X, map := λ X Y f => swap f, ... }

[delicate shore]

Is this Lean

[south sparrow]

⬦◻

[delicate shore]

What

[south sparrow]

maybe qed

[delicate shore]

How is this related to your question

[hard crest]

hey friendo if you're gonna ignore all of the helpers you should just post on livejournal or something

[keen tulip]

[south sparrow]

yes, related

find functor, I found functor maybe

[quasi bison]

what's "swap f"

[keen tulip]

you can’t define a functor by swapping two “arrows” unless you have a systematic rule that sends every homomorphism to another homomorphism functorially

[south sparrow]

I am reading and appreciating the helpers contributions

[hard crest]

oh i do see the thumbs up now

[south sparrow]

Nice

[mortal sandal]

have you found one such functor yet?

[south sparrow]

Not yet, one sec

[feral linden]

I suppose you can imitate what we do when we localize a category

Nvm, didn’t work

Oh, each group choose an automorphism of it should work

(Mapping any homomorphism f: G->H with a_H ^-1 f a_G I meant)

[south sparrow]

like so?

f₁ : ℤ → ℤ
f₁(n) = 2n
T₁(f₁) = f₁

0 ↦ 0
1 ↦ 2
2 ↦ 4
3 ↦ 6

T₂(f₁) = -f₁

0 ↦ 0
1 ↦ -2
2 ↦ -4
3 ↦ -6

[feral linden]

No. For any G, you select a automorphism a_G of it. Make sure at least one of them is not identity.

T_2(f: G->H)=a_H ^-1 f a_G

[south sparrow]

T₁(f) := f
T₂(f) := f⁻¹

[feral linden]

I suggest you read

(Actually though my T_2 works, now I prefer T_2(f)=a_H f (a_G)^-1, anyway, both work)

[south sparrow]

Nice, thank you @feral linden !

[feral linden]

Np

[mortal sandal]

or you could just map each morphism to the trivial homomorphism

[feral linden]

This doesn’t work

T(1)=1 so T(f)T(f^-1)=1

[mortal sandal]

oh right identities

[south sparrow]

Thanks all!

.close

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[neon pier]

okay im kinda confused on how to convert these 2 into logical statements. Lets start with the 2nd one bc that's simpler, mainly an issue of what notation is allowed, and what I came up with is:
P(x) = x can do it
∀x(P(x)) -> P(Jones)
I'm not sure if P(Jones) is legal though because it just feels wrong to write even though I'm pretty sure it should be correct

As for the first one, I'm not sure whether I need a quantifier for "someone who got a D". What I think the answer should be is:
P(x) = x failed the test
Q(x) = x got an A
R(x) = x will tutor someone who got a D
∀x(¬P(x)) -> ∀x(Q(x) -> R(x))
The someone part confuses me because I feel like that should be an existential quantifier but I can't think of a way to put that into the statement in a way that makes sense

[brave wolf]

I'm not sure if P(Jones) is legal though because it just feels wrong to write even though I'm pretty sure it should be correct
Might be better to introduce a constant J for Jones, then u can say P(J)

How did you interpret the sentence "If anyone can do it, Jones can"?

1. "If everyone can do it, Jones can"
2. "If someone can do it, Jones can"

[neon pier]

i interpreted it as if everyone can do it, jones can, since that's generally what you'd mean when you say anyone can do it

[neon pier]

alright yeah that's probably better than having a word

[brave wolf]

R(x) = x will tutor someone who got a D
Might be better to instead have T(x, y) as "x will tutor y"

then you can do the someone thing via quantifier

And btw it's often good to name the relations s.t. its "clear" what they mean, e.g.
F(x) - x failed
A(x) - x got A
D(x) - x got D
T(x, y) - x will tutor y
then when you or someone else reads the FOL sentence, they dont have to look back at the definitions that often

[neon pier]

Okay so if we let R(y) = y got a D then will something like this be fine?
∀x(¬P(x)) -> ∀x ∃y((Q(x) and R(y)) -> T(x, y))

[neon pier]

okay yes that makes sense, probably better to name them something more relevant

all my years of coding have not translated to math it seems

[brave wolf]

yeah, seems right

[neon pier]

alright thanks mate

.closr

.close

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[brave wolf]

oh wait, actually, on second look im not sure

[neon pier]

.reopen

[safe radish]

✅ Original question: #help-23 message

[brave wolf]

∀x ∃y((Q(x) and R(y)) -> T(x, y))

this part looks kinda off

It's true as long as there is someone who didn't get D

which means its actually always true

[spice laurel]

Guys what’s 1 plus 1 I forgot

[brave wolf]

you could just pick the y which didn't get D and R(y) would be false, making the -> trivially true

you should deal with Q(x) adding the Ey quantifier

that way you want make this mistake

first deal with "everybody who got an A" and only then do "will tutor someone who got a D"

[brave wolf]

Don't troll in the help channels please, go to #chill

[neon pier]

oh okay so move the ∃y just before the R(y) basically?

[spice laurel]

Please I need help with my maths assignment

[brave wolf]

you'll have to fix the connectives as well, -> should be somewhere else

[spice laurel]

Solve for all real values of x:

\log_2!\big(\log_3(x-1)\big) + \log_2!\big(\log_3(9-x)\big) = 1

[neon pier]

fish if i remember from that one minecraft mod hope it helps

[brave wolf]

go to your own channel e.g. #help-27 is currently free

[neon pier]

okay ill look over it once properly just a min

[spice laurel]

Gng I thought you can solve this

[neon pier]

i feel like theres an assumption in the statement that there has to be at least one person that got an A and at least one that got a D

just based off how it sounds in english

[brave wolf]

Yeah

it's just that what you wrote would be true even if the english statement wasnt true

ill give a specific example

∀x(¬P(x)) -> ∀x ∃y((Q(x) and R(y)) -> T(x, y))
Consider a class with Adam who got A and Daniel who got D and and Bob who got B. In this class, Adam tutors Bob.

∀x(¬P(x)) this is true as nobody failed
∀x ∃y((Q(x) and R(y)) -> T(x, y))
Pick y to be Bob. Then R(Bob) is false, Bob got B. So the -> simplifies to False -> True which is True. Hence the whole statement is also gonna be true for all x

so the whole-whole statement is True -> True which is true

but the english statement isnt true in this class

[neon pier]

ah i see

[brave wolf]

It'll help to deal with Q(x) before introducing the exists quantifier for someone

∀x(¬P(x)) -> ∀x (Q(x) -> x tutors someone who got D)

∀x(¬P(x)) -> ∀x(Q(x) -> R(x))
In fact, you can use ur original thing, just replace R with what it means using quantifiers

[neon pier]

okok lemme try to do that

∀x(¬P(x)) -> ∀x(Q(x) -> ∃y(R(y) ∧ T(x, y))

does that make sense?

second part is only true when they both have a D and get tutored

@brave wolf

[brave wolf]

yep, this is right

[neon pier]

alright lovely

thanks then ill go back to trying to inflict more pain upon myself with more of this

.close

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[real tendon]

Hi, i dont know how to solve these tasks

[quasi bison]

,rcw

[real tendon]

I used photomath to solve this one but i didnt quite understand it

[quasi bison]

https://www.youtube.com/watch?v=gfnVHwhEe6U

[quasi bison]

this is a great way to confuse yourself and be stuck in samsara and never achieve enlightenment.

[real tendon]

How to solve without fraction

[quasi bison]

what do you mean "without fraction"

you definitely didn't watch the video yet

but if you wish: https://www.youtube.com/watch?v=Fd5ys4PQ-aM

[real tendon]

Yes thats it thank you

Close.

[quasi bison]

the command is .close

[real tendon]

close.

.close

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[tropic canopy]

Um i have no idea on what to do here i need to find the limit of the sequence but how do i do it when both are roots as i cant divide by the biggest X

[tropic canopy]

atleast how do i start

[quasi bison]

you can still think in terms of fractional powers of n

the top grows as n^(3/2) and the bottom grows as n^(5/3)

[tropic canopy]

so i devide by n^(5/3)? but how would that even look in terms of the other numbers like 3n

[upper moat]

Rationalize it can you

[vagrant ice]

the idea is to rewrite this as $\frac{n^{3/2} \sqrt{1 + 2/n^3}}{2^{1/3} n^{5/3} \sqrt[3] {1 + 3/n^4 - 1/n^5}}$

[flat frigate]

south

[vagrant ice]

and then 2/n^3, 3/n^4, -1/n^5 all go to 0 as n -> infinity

[tropic canopy]

ye but where do the 2^(1/3) come from

[vagrant ice]

oh, $(2n^5)^{1/3} = 2^{1/3} (n^5)^{1/3}$

[flat frigate]

south

[tropic canopy]

ah ok

i think i got it

.close

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[tropic canopy]

tnx btw

[vagrant ice]

npnp

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[tender meteor]

Can i rewrite It as

$\int^{\pi}_0 e^{\sin x + \cos x} \cos(2x) \mathrm dx$

[flat frigate]

alee

[tender meteor]

?

[charred summit]

Yes, that is right

[tender meteor]

Maybe it's better that I don't have that cos(2x)

Maybe its better

(cos x - sin x)(cos x + sin x)?

[charred summit]

Yeah do a u sub and use that identity

[tender meteor]

$\int^{\pi}_0 e^{\sin x + \cos x} (\cos x + \sin x)(\cos x - \sin x)\mathrm dx$

[flat frigate]

alee

[tender meteor]

so u = sinx + cosx?

[charred summit]

Mhm

[tender meteor]

u = e^(sin x + cos x)?

[charred summit]

Which one do you think will get you the cleanest integrand?

You can try both

[tender meteor]

du = e^(sin x + cos x) (cos x - sin x) dx

1/[u (cos x - sin x)] du = dx

This One i think

[charred summit]

I think you should try the first one real quick before you move on with the other 😉

[tender meteor]

Sorry but

u = sin x + cos x it is not globally invertible

So i have to split the integral?

@charred summit

[charred summit]

Well for definite integrals, invertibility is not required according to the definition. For indefinite integrals, it's often convenient because you eventually need to write it back in terms of x

[tender meteor]

On 0,π is not monotone

[charred summit]

But yeah you can split the integral in 0,pi/4 and pi/4, pi

If you really want to follow that rule

But it's not necessary

[tender meteor]

But

$\int^1_{-1} e^{x^2} \mathrm dx$

[flat frigate]

alee

[tender meteor]

u = x²

Then u(-1) = u(1) = 1

$\int^1_{1} \dots \mathrm du = 0$

nonsense (the integral is positive)

[buoyant dragon]

i dont think that its correct

[charred summit]

Yeah, it is not

[tender meteor]

I say the thing above

[charred summit]

dx = du/2x

[tender meteor]

I mesn this

[tender meteor]

.

Isn't what I wrote a counterexample?

[obsidian oracle]

The conditions for variable change are, according to the definition, that the new variable u = f(x), with f C^1 and bijective (injective) on the interval of integration

[charred summit]

Your substitution in your counter example isn't correct

[obsidian oracle]

f(x) = x^2 is not bijective from [-1,1] to [0,1]

[flat frigate]

alee

[tender meteor]

Yes, I know, I meant if you don't break the integral

That is, the substitution should be done correctly by breaking the integral

So in my case I have to do this too

[tender meteor]

I mean this

I have to do it like this

[obsidian oracle]

if you want to do the variable change u = cos(x) + sin(x), yes you need to split the integral in 2

[obsidian oracle]

so that on each split interval, the variable change follows these conditions

[charred summit]

$This is the substitution theorem for definite integrals. It does not require (g) to be invertible. If the integrand is of the form [f(g(x)),g'(x),] then we have[\int_a^b f(g(x)),g'(x),dx= \int_{g(a)}^{g(b)} f(u),du,] so in this case there is no need to split the integral$

[flat frigate]

bartdestinkerd34
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[charred summit]

Splitting will work of course, but it's just extra work in this case

[obsidian oracle]

that requires you to specifically have an integral of the form f(g(x))g'(x)dx

mmh

[charred summit]

Which we have in this case if we do the u sub

[obsidian oracle]

wait

[charred summit]

$[
u = \sin x + \cos x, \qquad
u' = \cos x - \sin x, \qquad
f(u) = u e^{u}.
]$

[flat frigate]

bartdestinkerd34
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[charred summit]

Sorry I'm using AI to translate to latex cause I don't know how to use latex hahaha

[tender meteor]

But maybe I understood

You mean I do u = sin x + cos x

But I don't change the integration extremes?

So I don't have to split the integral

In this case I use substitution only to find the primitive so I don't need u to be invertible etc.

@charred summit

[charred summit]

Yes, exactly

In your counterexample, this substition theorem does not apply

Which is why you have to split there

But not in the exercise you're doing

[tender meteor]

👍

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@tender meteor Has your question been resolved?

[safe radish]

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[loud cedar]

Math - Rounded Park
Front Page

• Make sure to have enough space for the strip of colored paper and make sure that the strip is cut cleanly (be generous with the strip but not too much)
  Second Page
• The actual Park
• Overlay the grid to the park and place some landmarks on the x- and y-axis
• Maximize the space and fill it of possible
• Limit the landmarks (5-7 only)
  Third Page and onwards
• Isolate each landmark, make sure to zoom in/out to maximize the space
• EACH page is reserved for ONLY one landmark
• The circular landmark will be placed on the left side of the page while the right side is reserved for the FINAL (this means no solutions) equation in general and standard form
  2nd to the last page
• This is where your table is located (only one table)
• Area/Landmarks, Center, Radius, Equation in Standard form, and Equation in general form
• Maximize space
  Last Page
• This is your solution page (can be 1-2 pages)
• The ONLY solution needed is for the general form (no need for standard)

[loud cedar]

This is my design and I need help with all every math parts ty

[still charm]

This is AI

[slim pulsar]

hi

[loud cedar]

WHAT

These are examples from my classmates

[slim pulsar]

ok so remember that we set fountain's center at (0, 0)

and the lawn's center is also (0, 0)

the radius of the lawn is 20

and the radius of the fountain is 5

so then we can start with the landmark at quadrant I

you can set it as a pond or whatever

[loud cedar]

Which is lawn center there

I only see the fountain

[slim pulsar]

like the biggest circle

the entire thing

is the lawn

the main lawn

[loud cedar]

This is mine btw

Where did rlly the radius 5 came

How did u get that

[slim pulsar]

well we supposed it

because that's the only we can't prove to have the biggest area

it's optional really

so then for the pond in quadrant I

the easiest way, which is basically the right way, is to make its center at (10, 10) and radius 3~5

you can choose which radius you want

do the same thing with quadrant II, quadrant III, and quadrant IV

which means that your coffee shop in quadrant III should have its center at (-10, -10)

the parking lot's center at (10, -10)

and the playground thingy in quadrant II's center is at (-10, 10)

welps do you understand so far

uhh

[daring furnace]

Pls

Help

[slim pulsar]

bro this is not open

[daring furnace]

Ok lemme send it for u in your dm

[slim pulsar]

nonono

don't

i don't understand this either

go to a open help

like help-25

[daring furnace]

Oh sorry

Ok ok

[slim pulsar]

it's ok

[safe radish]

@loud cedar Has your question been resolved?

[safe radish]

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[lean geyser]

well why does f(x+1) make the graph go back instead of up by 1 tell me where im making the mistake in my working out:
so f(x) = x+1
and so if we input 1 we get a return of 2 and making it to a graph we get 1,2
with f(x+1)
we input 1
it becomes 1+1+1 which is three
so the output is 3
and its 1,3

[hard crest]

yeah so for a linear graph going vertical looks very similar to going horizontal

try with f(x) = x^2

[lean geyser]

okay

WHOOPS I DID IT WRONG

f(x) = x^2
use 1 as input to get output of 1 (1,1)
f(x+1) our input it again 1 so thats our x coordinates and then 1+1 =2 for our input and then we square our input to get 4 therefore 1,4

why do we go back when we do that? is the fact that f(x+1) manipulates the x number we put in? so instead the value becomes 2, 4?

but that still doesnt make it go back

[hard crest]

hmm

we "move the graph to the left" because when we plot the point at x = 0, we're plotting the original x = 1 point

so like using f(x) = x^2

and let's say g(x) = f(x+1)

if we look at g(6) that's f(7)