help-23

[fair hound]

sum everything by coefficient, you've not simplified yet

the 3 will appear

[proven smelt]

wait what

[fair hound]

also, 3k+3 = 3(k+1)

k^3 + k^3 + k^3 = 3k^3

etc

[proven smelt]

didnt i do that?

[split kayak]

Theres an easy way to show that 3k + 3 divides the other expression considering that this is a polynomial

[fair hound]

oh mb didn't see it

[split kayak]

time to recall your calc classes ig

[fair hound]

so uh yeah, you can already factorize by 3, 3k^3+9k²+15k+9 = 3(k^3+3k²+5k+3)

your problem is about factorizing k^3+3k²+5k+3 by k+1

[vague phoenix]

long division

[split kayak]

Noooo

[vague phoenix]

[proven smelt]

pleaase

[fair hound]

you can be way smarter than using long division, for example, (k+1)^3 = k^3+3k²+3k+1, add 2(k+1) on both sides and done

but there are at least 3 or 4 smart ways to do it

[split kayak]

You could just test for the quite obvious root

[vague phoenix]

I automatically think of long division

[proven smelt]

k = 1?

[split kayak]

3k + 3 = 3(k+1)

What value k would be the root in that expression?

[proven smelt]

-1 i mean sorry

[split kayak]

As i was trying to hint that

If the degree 3 polynomial we have here is divisable by 3(k+1), then what does that mean?

[proven smelt]

I have no idea how calculus can show thi

[split kayak]

Id assume youre not too far into calculus then, right?

[proven smelt]

I mean depends, i may know

[split kayak]

There are multiple ways to express any given polynomial

One of them is the obvious one

[proven smelt]

uhh

oH

if u sub in k = -1, u get

0

?

[split kayak]

yeaa

[proven smelt]

oh my so u can litreally just do that

[split kayak]

If a polynomial P(x) has a root "x0", then that means that it can be written as P(x) = Q(x) * (x-x0)

Where Q is another polynomial of 1 degree less

[split kayak]

And even so, if it so happened that k = -1 was not a root of P(k)

Then the original statement was false

[proven smelt]

oh wait i think i saw this tech somewhere

[split kayak]

so you have a <==> (if and only if) relation

[proven smelt]

ok wait i got it divisible. so i know it is divisible by k = -1, but this shows (k + 1) is a factor, not 3(k + 1) right?

[split kayak]

3 is a constant

[proven smelt]

so do i nede to factorise 3 from 3k^3 + 9k^2 + 15k + 9

and then show k = -1 for factorised bracket = 0?

[split kayak]

You could go to try with both

You will notice that theres only a 1/3 difference in factors.

[proven smelt]

i dont understand sorry?

[split kayak]

gimme a sec

The threes are only constants

And for polynomial division, factors dont alter the condition for divisibility

You could divide any combination of the 4 versions for these polynomials between eachother and they will always work

Since "factors" for us, are not scalar (real) numbers, but our polynomial factors of the form (x - a)

[proven smelt]

so 4(k+1) is a factor too?

[split kayak]

consider that in the expression 3(k+1) the three is not part of the factor itself

its only the k + 1 part-

[proven smelt]

okay. but if its divisible by (3k+3), then wouldnt 3 be like part of the division too?

[split kayak]

Its divisible by (k+1), it turns out that (3k+3) is just a way we write 3(k+1)

The polynomial would be divisable for (4k+4), (7k+7), etc....

For our case, we care about 3k+3 because its what we got from the a+b+c expression

[proven smelt]

oh, so 4(k+1) is a factor too

I see

[atomic gorge]

It will be easier if u chose n-1 n n+1

[proven smelt]

yeah i saw that in the solns. is this just intuition?

[atomic gorge]

Ye u just need to realise that calculating (n+2)^3 is harder than (n-1)^3

The lazier u are the more you’re convergingto the right answer

[safe radish]

@proven smelt Has your question been resolved?

[proven smelt]

thanks guys

[safe radish]

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[cerulean dune]

whats BN?

[vague phoenix]

What have you done?

[cerulean dune]

nothing idk what is BN

[warm warren]

feel question has typo

[vague phoenix]

Missing spaces?

[warm warren]

no i feel the point b/w B and C is N but idk.

[vague phoenix]

N here is something else idk

[cerulean dune]

so the question is wrong?

cuz this is from an old test if theres something wrong they wouldve meantionned it

[near sky]

feel like N is on BC

[safe radish]

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[solar inlet]

can any1 send the working for this

[safe radish]

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.

@solar inlet Has your question been resolved?

[safe radish]

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[upper forge]

i dont even know where to start

[agile scaffold]

use rule 2!

[agile scaffold]

(this is incomplete r*log(a) = log(a^r), and the change of base, vice versa)

[upper forge]

Oh

Oh

I am so dumb

I just did rlog(a)

[agile scaffold]

no worries, you might just be confused, without a calculator this is not too trivial either

[upper forge]

lemme try now

[agile scaffold]

the question can be solved using rule 1 and rule 2 only tho, keep that in mind!

[upper forge]

yeah i realized that

i think i can do it now

[dense furnace]

why not rule 3 and then rule 1?

[agile scaffold]

rule 3 then rule 1 would make it really ugly when it doesnt need to be

im assuming youre raising 16/15 to the power of 7?

[agile scaffold]

you have a pretty interesting website btw, some of them were nice reads

type .close when youre done

[upper forge]

it reduces to log2

[upper forge]

oh thank you

i just started my 11th grade. i will continue to explore my artistic side 😭

.close

[safe radish]

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[agile scaffold]

mm thats cool!

[agile scaffold]

oh this works too, yes, i did not see that

youre right @dense furnace

[dense furnace]

it’s ok , you learned something new today 🫶

[agile scaffold]

hahaha no i got tunnel visioned, splitting it into a bunch of log 2s and log 3s and log 5s and subtracting and adding is also pretty satisfying

[safe radish]

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[daring sonnet]

How do I do part b

[safe radish]

@daring sonnet Has your question been resolved?

[tardy mango]

The conjugate root theorem gives you $-1-\sqrt{7}i$ is also a root

[flat frigate]

Civil Service Pigeon

[tardy mango]

and by vieta's the sum of the roots is -3/2

[safe radish]

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[daring sonnet]

But the last root is x = 1/2

[ivory moat]

how do i ask

[tardy mango]

Mhm. Your point is?

[safe radish]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[tardy mango]

,w -3/2 - (-1-i sqrt 7)-(-1+i sqrt 7)

[flat frigate]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[tardy mango]

Hello?

[safe radish]

@daring sonnet Has your question been resolved?

[safe radish]

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[fringe dock]

sadly i had some internet issues so im gonna reask the question as i couldnt reply to the messages

[fringe dock]

regarding this working out, i am unsure how it goes from line 1 to line 2

because shouldnt it be done like this?

[round egret]

you can factor it out

[fringe dock]

but what is wrong with my approach

[round egret]

huh what is?

[fringe dock]

this

[round egret]

your approach looks fine

[fringe dock]

but it does not lead to the same answer

i think what i showed isnt clear sorry

[fringe dock]

this is the worked out solution

[fringe dock]

the black crossed outline is me

[round egret]

are you cancelling the denominator out fully?

[fringe dock]

yes

[round egret]

ah

[fringe dock]

because there are 2 terms on the top

[round egret]

yeah you can't do that

you are cancelling the squared part

[fringe dock]

but ive seen something really similiar

maybe it didnt have the subtraction sign in the middle...

[round egret]

$\frac{x}{x^2} = \frac{1}{x}$

[fringe dock]

hmm

[flat frigate]

MxRgD

[fringe dock]

yes

[round egret]

as an example

[fringe dock]

i know that

[round egret]

the x doesn't get fully canceled

[fringe dock]

yeah

[round egret]

it's what happens here

the denomaintor doesn't get canceled fully

[fringe dock]

but there is 2 terms on the numerator not one

[round egret]

right but you can factor it out

[fringe dock]

yeah

thats true

i think im mixing it up if they were all being multiplied

because lets say that subtraction sign was a multiplication sign

[fringe dock]

would this be correct?

yeah it would

i think

[round egret]

yes as long as you don't fully cancel out the denominator

[fringe dock]

wait u would

e

wait

okay

now it should be fine

2 terms on top, 2 terms on bottom

[round egret]

huh

was the minus meant to be multiplication?

[fringe dock]

no this is a new case

suppose it wasnt a subtraction

[round egret]

ah okay

[fringe dock]

and it was multiplication instead

i think that is what im getting confused by

[round egret]

then yeah you can cancel fully

[fringe dock]

so the denominator becomes 1, right?

[round egret]

yep

[fringe dock]

okay perfect

i think that was the issue

thank you

[round egret]

np

[fringe dock]

.solved

[safe radish]

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[safe radish]

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[daring orchid]

hlo

[daring orchid]

mhm

[safe radish]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[daring orchid]

proof that if n was not a multiple of 5 then the rest of n^4/5 is 1

[verbal cloud]

do you know modular arithmetic?

[daring orchid]

I would greatly appreciate the help

[daring orchid]

yes

[tardy mango]

(In the future, please make the first message you send the actual question since that's what the bot pins. This saves us the effort of having to change the pin manually.)

[verbal cloud]

then just do each case

[daring orchid]

u mean 5k+1 5k+2....

[verbal cloud]

pretty much

you could do n^2 first

then just reapply

[daring orchid]

there is the way of table

like this

@verbal cloud

[spark stump]

Yes use this table for x²

Then x⁴

There are another way
You want to prove 5|x⁴-1

Means 5|(x-1)(x+1)(x²+1)

for x=1(mod5) or x=-1=4(mod5) it's done

Now you can just prove for x=0(mod5), x=2(mod5) and x=3(mod5)

[safe radish]

@daring orchid Has your question been resolved?

[daring orchid]

oh

[daring orchid]

mention me

[stoic torrent]

-# mark the bot else it will close the channel

[daring orchid]

wait

im reading

[daring orchid]

proof that if n was not a multiple of 5 then the rest of n^4/5 is 1

this is my question

[tawny vessel]

what do you mean by "the rest of n^4/5"?

[daring orchid]

The rest of the division

or remainder

[tawny vessel]

oh

[spark stump]

Yeah for x is a multiple of 5
Means x=0(mod5)
In this case x⁴-1=4(mod5)
So you can do check it for x=2,3(mod5)

[tawny vessel]

remainder

[spark stump]

Or you can just prove it with Fermat's little theorem

[daring orchid]

x is not a multiple

of 5

[spark stump]

So you want to prove that if x not multiple of 5
Then x⁴=1(mod5)

[daring orchid]

i want to proof that if x is not a multi then the remainder of n^4 divasible by 5 is 1@spark stump

[spark stump]

Do you know modular arithmetic?

[tawny vessel]

first you'd represent x as 5k + m, where k is an integer and m is any integer in [1, 4]

[daring orchid]

yes at least

[spark stump]

انتي عربية؟

[daring orchid]

عربي

😔

[spark stump]

تمام نتحدث بالعربية

[daring orchid]

انت مش عربي؟

[spark stump]

لا أنا عربي

[daring orchid]

اوك حسبتك صيني متعلم عربية

انا عارف انجليزيتي ضعيفة 😔

[spark stump]

من أي دولة

[daring orchid]

الجزائر

[spark stump]

عظيم يمكننا التحدث بالفرنسية

أنا من المغرب

[daring orchid]

لا احكي عربي

[daring orchid]

والنعم

[spark stump]

تمام
انتي تريدين إثبات أن باقي قسمة x⁴ على 5 هو 1
اذا كان x ليس مضاعف ل5

[daring orchid]

واه

بلطابلو بغيت نديرها

[spark stump]

اوكي
نعملها بالطابلو

[stoic torrent]

-# .

[spark stump]

اذا كان x=1 (mod5)

كم ستكون x² mod 5

[daring orchid]

I'm not finished

[daring orchid]

1

[stoic torrent]

Yeah ik just reminding you to rewct with the cross so that bot doesnt close

[spark stump]

هو يقصد اضغطي على زر لا

قبل ما يحذفه البوت

[spark stump]

اذا كم ستكون x⁴ mod 5

[daring orchid]

1

[spark stump]

هذا في حالة 1

الآن لننتقل لحالة 2

x² mod 5 ماذا ستكون

[daring orchid]

4

[spark stump]

و x⁴ mod 5

[daring orchid]

1

[spark stump]

لقد أثبتناها في حالة 2

الآن في حالة 3

x² mod 5

كم ستكون

[daring orchid]

4

ومبعد 1

[spark stump]

نعم

الآن تبقت حالة 4

كم ستكون x²
و x⁴ في هذه الحالة

[daring orchid]

1و1

نظن

[spark stump]

نعم

هكذا تم حل السؤال

[daring orchid]

بصح علاش تجبس عند 5

ح

[spark stump]

لأن x=5(mod5)
هي نفسها x=0(mod5)

لأن
5=0(mod5)

وفي هذه الحالة يعني أن x مضاعف ل5

وأنت لا تريدين إثباتها في حالة x مضاعف ل5

[daring orchid]

علاش منديرش 6 مالا

ولا 7

علاش تحبس عند 4

[spark stump]

لأن x=6(mod5)
هي نفسها x=1(mod5)

لأن 6=1(mod5)

و 7 تصير 2

و8 تصير 3

وهكذا

[daring orchid]

بصح معلبالكش لازم تجرب كل الاعداد

مثال 124342534

فالرياضيات يخي متقدرش تدير حاجة بلا برهان

ولا تخدم ب n

هو علبالي بلي هدرتي غالطة بصح حاب نفهم علاه

[spark stump]

لكن الmodular arithmetic
جاءت لتسهيل الحسابات

في هذه الحالة هناك شيء يسمى
Classe d'équivalence

والعمل عليها فقط يعتبر بمثابة العمل على جميع الحالات

[daring orchid]

وشنوا هيا هذي

[spark stump]

بالنسبة ل5 نأخذ الأعداد الصحيحة الطبيعية الأصغر من 5 ونعمل عليها

Classe d'équivalence الخاصة ب5
هي 0,1,2,3,4

[daring orchid]

نعم

لكن الاعداد بعد هذه

ماذا نفعل لها

[spark stump]

الأعداد بعد 4 تصير في نفس الدائرة

مثلا 5
تصير 0

لأن
5=0(mod5)

[daring orchid]

سما ديما كاين هذا الدوران

[spark stump]

هناك قاعدة
a=b(modx)
Et b=c(modx)
تعطي
a=c(modx)

6 تصير 1

و7 تصير 2

وهكذا

[daring orchid]

لكن لازم تثبت بلي 6=1 modx

سع قبل ما تخدم

ولا لا

[spark stump]

نعم في هذه الحالة x كم تساوي

[spark stump]

نحن لدينا x=5 في حالتك

Delete this

[daring orchid]

استغفر الله

ثاني واحد نتلاقاه هك هنا

[daring orchid]

نعم

[spark stump]

<@&268886789983436800>

[spark stump]

و كم باقي قسمة 6 على 5

[daring orchid]

1

ماذا عن 1086074569806

مثال

لازم تجرب

لا تستطيع التكهن مسبقا

[spark stump]

هذا لديك
1086074569806=108607456980×10+6
=108607456980×2×5+6

هذا العدد باقي قسمته على 5 هو 6

وباقي قسمة 6 على 5 هو 1

اذا باقي قسمة هذا العدد على 5 هو 1

الأمر يعتمد بشكل كبير على كيفية تعاملك مع الرقم

[daring orchid]

اوك

got it

ثانكيو ويليام جيمس مورياريتي

[spark stump]

العفو

اكتبي . ثم solved

[daring orchid]

@tawny vesselthank u toooo

.solved

[safe radish]

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[safe radish]

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[round egret]

<@&268886789983436800>

[safe radish]

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[round egret]

why have there been so many of them

[lilac crane]

Easter special

[safe radish]

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[versed terrace]

hints wanted: volume of an n-ball

using various integral calculus methods i've found the area of a circle (in terms of its circumference), the volume of a sphere, and the surface area of a sphere.

i've also found that for an n-ball, the relationship of its volume to its surface area is given by S=nV. i found this by integrating over (n-1)-ball shells:

int_0^1 S r^(n-1) dr = V
S = V / int_0^1 r^(n-1) dr
S = V / (1^n / n)
S = nV

my next goal is to find the volume of an n-ball, but i'd like to do this as independently as possible, with only very minor hints. however i've been stuck for a few days, so i'm resorting to asking here. what methods of attack should i try?

[versed terrace]

i've learned all of the content in ap calc ab/bc and understand the concept behind (but am bad at) trig substitution

for a unit circle found its volume both by integrating over triangle slices and integrating over circular shells. for a unit sphere i found its volume by revolving the function sqrt(1 - x^2) over the z-axis (integrating over circles about the z-axis.) for the surface area of a unit sphere i divided the sphere into circular shells and integrated over their angle from the origin

[safe radish]

@versed terrace Has your question been resolved?

[blazing swallow]

$S = nV$ is not necessarily true - for a 2-ball (the sphere in 3 dimensions), $S = 4\pi r^2$ and $V = \frac 43 \pi r^3$, but $2V = \frac 83 \pi r^3 \neq 4 \pi r^2$. however, you are close: think about the specific case of the 2-sphere and try to find an operator that takes you from the volume to the surface area

[flat frigate]

haseeb ♥

[blazing swallow]

2 ideas come to mind on how to proceed with your problem. option 1 is to find a function of (n-1) variables that you can revolve to get the volume of an n-sphere. option 2 is to look into spherical co-ordinates in n-variables: if you know spherical coordinates in 2 and 3 dimensions, try to generalize to n dimensions, then integrate the parametrized n-sphere. both of these approaches require multivariate integration, which im not sure you're familiar with

then, use the "operation" from before to obtain the surface area of the n-ball

note, it is not a clean answer, but there is a closed form for said volume

[safe radish]

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[versed terrace]

.reopen

[versed terrace]

thank you! i should have mentioned that this is specifically for a unit n-ball, and that i incorrectly assumed that the n in an n-ball referred to its dimensions

[versed terrace]

i assumed a sphere was a 3-ball, for example. (i'm trying not to look too much into this to not get spoiled lol)

so for a unit sphere, to obtain the surface area you'd multiply by 3

similarly for a unit circle you can multiply by 2. i believe the integration i did proves this for all dimensions

[versed terrace]

^

[blazing swallow]

made the same mistake when i started studying integration 🙃

i also didnt realize you were talking about unit n-balls, so that does work.
but if you're curious, there is an operation for n-balls of any size that translates between surface area and volume, and it ties in really nicely to some fundamental theorems in vector calculus. but i won't spoil

[versed terrace]

ooh interesting; i will look into this!

okay so we have an n-ball of radius r (in dimension n+1) with volume V and surface area S. we have that the unit n-ball has a volume U and surface area Z. we know that Z=(n+1)U

V = U * r^(n+1), meaning U = V / r^(n+1)
and
S = Z * r^(n)

therefore S = (n+1) U * r^(n) = (n+1) (V / r^(n+1)) * r^(n) = (n+1) (V / r)

does that look right?

[safe radish]

@versed terrace Has your question been resolved?

[blazing swallow]

you seemed to have substituted U = (n+1)*Z, when in fact we have Z = (n+1)*U. so the (n+1) should be in the denom

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[desert pasture]

Consider $(x_1,x_2)$. Let the same be generated by $p(x)$. Then there exists a polynomial $q(x)$ such that $p(x)q(x)=x_1$. Then WLOG the degree of $P$ is atmost $1$ and is a multiple of $x_1$. But then $x_2$ cannot be generated by the same. We thus have a contradiction.

[flat frigate]

Wai

[opaque fern]

I think all of this is fine but could be made a bit more rigorous

Like you should handle zero divisors for example

[desert pasture]

mhm

how do you mean

*mean

[opaque fern]

A principal idea domain must by definition be an integral domain

So lile

You must establish that either R has zero divisord

Or R is an integral domajn

[desert pasture]

ah

got it

thanks

I can fix that so 'll close this

.close

[safe radish]

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[twin prawn]

for no 2, is this correct?

[frigid locust]

didnt write the question correctly

supposed to be a negation for P(x)

[twin prawn]

oh

wait

there

instead of using negation laws at the end for P(x) its indempotent

[frigid locust]

not sure about the 2nd statement tbh

actually nvm

couldve done this for ease

[twin prawn]

i see

but its correct overall

[frigid locust]

unsure about the 2nd to last step

im thinking about it

[twin prawn]

the third to last step all have disjunctions between

i used commutative

to move them

for the second last step, i then used associative

then for P(x) i used idempotent, and for Q(x) negation

[frigid locust]

hmm ok

well Q(x) v ~Q(x) = t (tautology)

its not equal to Q(x) like u wrote in last line

negation law

[twin prawn]

oh woops

so its just Ax(P(x))

[frigid locust]

Ax, (P(x) v t)

[twin prawn]

why did we not do anything with P(x) V T?

[frigid locust]

u should do the law no. 8

i was just saying that Ax(P(x)) was incorrect

[twin prawn]

i see

so truth value is T

[frigid locust]

ig it fully simplifies to true for all x

damn , pretty clean answer

[twin prawn]

hell yea

.close

thanks

[safe radish]

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[twin prawn]

.close

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[daring orchid]

halo

[daring orchid]

ummmmmmmmmmm

[vague phoenix]

Helo

[daring orchid]

i wan to proof that 851^3n +851^2n +851^n +2 divisible by 7

[daring orchid]

hiiiiii hru

[vague phoenix]

I am good

[daring orchid]

thank god

[vague phoenix]

Try to factorise out 7?

[worldly lantern]

heard about ?modular arithmetic

[daring orchid]

yes

[daring orchid]

how

[worldly lantern]

apply mod 7 and just use properties of modulo

[vague phoenix]

Wait forget what i said

[daring orchid]

Before this, I studied the remainders of 2^n/7

[daring orchid]

it looks to long

[worldly lantern]

hint : 851 = 4 (mod 7)

try to simplify from here

[daring orchid]

yes than i got 2^n +4^n +3

and

[worldly lantern]

where did 3n term go

u straight away

set it 1

cuz its 8^n ?

[daring orchid]

wait

[daring orchid]

no

[daring orchid]

its this

yo

[worldly lantern]

so

after this

infact it might be kinda wronf

it should be
4^3n + 4^2n + 4^n + 2

simplify after this

[daring orchid]

yes u can simplify it

[daring orchid]

2^n +4^n +3

[worldly lantern]

i got
3 + 2^4n + 2^2n

ah

okay

[daring orchid]

than

wat

wata sih

[worldly lantern]

write it as 2^n + 2^2n + 3
take three cases for n in mod 3

n is either 0 , 1 or 2 mod 3

and then use Fermat's Little theorem

[daring orchid]

i dont get this

[worldly lantern]

after u get this

take 3 cases

since every number can be written either as
3k
3k+1
3k+2

[daring orchid]

why 3

[worldly lantern]

2^3k is 1 mod 7

itll reduce stuff

[daring orchid]

ye

than

[worldly lantern]

in all the three cases after solving youll get 0

hence proved

[daring orchid]

You mean after compensation

[worldly lantern]

compensation of what

[daring orchid]

3k and n

like that

replacement

[worldly lantern]

no no

[daring orchid]

wat

[worldly lantern]

wait we did something wrong

[daring orchid]

😭

wat

.solve

[worldly lantern]

i dont get it

[daring orchid]

i solve it

[worldly lantern]

how

[daring orchid]

oh no i solve another one

[daring orchid]

get back

[daring orchid]

What do you think about studying the remainders of 4^n and 2^n?

[worldly lantern]

sure

[worldly lantern]

,rccw

[worldly lantern]

this is not divisible by 7 for n= 0,3,6....

[daring orchid]

how did u get this

[worldly lantern]

factored out

[daring orchid]

from 4^n

[worldly lantern]

wrote
4^n as 2^2n

[daring orchid]

oh

[worldly lantern]

infact this is never divisible by 7

!original

[safe radish]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[daring orchid]

cycel length in those 2 is 3

[worldly lantern]

could you provide the original question please

[daring orchid]

this is it

no more no less

[daring orchid]

.

@worldly lanternit is normal to mention the helpers

?

[worldly lantern]

sure you can for now

but in this case

we cant prove the question since its false

[daring orchid]

<@&286206848099549185> 🙏

[gilded stag]

<@&268886789983436800>

don’t use ai

@worldly lantern

[worldly lantern]

oh

sure

i didnt actually

[gilded stag]

All good ❤️

[worldly lantern]

i mean
i solved it myself first but
mb

[daring orchid]

u can help us also

[gilded stag]

Was going to but hard to get that removed 😅

[junior smelt]

(if you did use it to answer helpee questions, we can remove you from the server, just so you're aware )

[civic musk]

this?

[daring orchid]

yes pls

and thanks

[worldly lantern]

lmao i dont use AI
i was just checking if the question is even valid or not
turns out it isnt valid

ill keep that in mind tho

and i apologize

[civic musk]

find reminder pf 851 when by 7

[daring orchid]

He was asking questions, but he wasn't answering me.

[gilded stag]

Isn’t this a sequence question

[daring orchid]

dk

[junior smelt]

I'm not saying that's the case, I'm just giving a general "letting y'all know"

[daring orchid]

ok thank u

[verbal lily]

You don't go and use 3k directly
The general method is to investigate powers of 2 modulo 7 and then notice that they do have a period

[gilded stag]

851^3n +851^2n+3n of 851 forms a sequence

[worldly lantern]

guys

the question cant be proved

i just tested it out

[verbal lily]

Give a counter example

[daring orchid]

can anyone do this

[gilded stag]

Give counter examples bro

[gilded stag]

we are trying to

Pls wait

[worldly lantern]

lmao what
take n=anything and it will never be 0 mod 7

[daring orchid]

sry

[worldly lantern]

literally anything

[gilded stag]

No what

[verbal lily]

Ah yes n=1

n=1 is a counter example

[obsidian scaffold]

Ping?

[worldly lantern]

every n belonging to natural numbers is a counterexample

[daring orchid]

I think the question requires extracting patterns for n

[worldly lantern]

,rccw

[daring orchid]

😭

[gilded stag]

patterns = sequence

[daring orchid]

oh

my bad

[verbal lily]

This is another statement
Anyways he can follow the method of finding a period to prove that the expression modulo 7 is never 0 by exhaustion

Or he

[gilded stag]

No your good haha

im abit confused with what others have to say tho

[verbal lily]

Nta dziri mateleme?

[daring orchid]

hih 😔

[daring orchid]

anyone have smth to say?

[worldly lantern]

why do yall love patterns and sequences

[gilded stag]

Patterns are the same as sequence

[worldly lantern]

lmao

[daring orchid]

i dont

[verbal lily]

Because this is the general method
Tricks come with practice

[worldly lantern]

no they arent 😇

[gilded stag]

son

[worldly lantern]

modular arithmetic is the "method"

[daring orchid]

im crinne

[gilded stag]

patterns are not the same as sequence

forget what I said

Sorry

[worldly lantern]

sequence is any array of number
pattern has a "pattern"

[gilded stag]

yep just realised I made a mistake here

[daring orchid]

ur good

[worldly lantern]

you cant learn integration by estimating area with rectangles
although the rectangle method is "general"
using integration is better
similarly
patterns are helpful for basic questions

[daring orchid]

i think the pattern is smth like 2k 2k+1

???

i mean here

[worldly lantern]

are you sure the question that you asked says
prove that this is divisible by 7
or does it say
prove that this is never divisible by 7

[daring orchid]

the first one

ooooooooh

wait

Deduce the remainder of the division.

😭

[worldly lantern]

the remainder isnt constant

there are 3 different remainders

[daring orchid]

remainders

[verbal lily]

I am sure he gave you another thing
Can you send the whole exercise you found this?

[worldly lantern]

thats easy then

smh

[verbal lily]

He said deduce 🗿

[spring bison]

Wait why not use that 851 congruence 4 modulo 7?

[daring orchid]

ya5o jbthm mlkitab

[worldly lantern]

yeah
it can be done easily by modulo

[worldly lantern]

?

[worldly lantern]

we were doing that earlier

[spring bison]

it did not work out?

[verbal lily]

Eb3at

Fayen l9ytih

[worldly lantern]

is this a language

[verbal lily]

Yes he said he found it in a textbook

[worldly lantern]

woah

[verbal lily]

Translation
Find the remainders of 4^n over 7
Deduce the reminders of (the expression) according to the values of n

So yes this is very easy

[daring orchid]

😭

no its not

[verbal lily]

Rwa7 prv hna ytelfouhalek

[daring orchid]

okayyy

[daring orchid]

kain hada tanik y7ws y3rf al7l

jitk prv

[worldly lantern]

you solved it yet ?

[safe radish]

@daring orchid Has your question been resolved?

[verbal lily]

Yes it been solved

[safe radish]

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[vapid cypress]

This one

[safe radish]

Please don't occupy multiple help channels.

[vapid cypress]

@peak estuary

[dusky totem]

Sorry what’s the actual question

What are we supposed to do

[vapid cypress]

the characteristic eqn of AB is given as in the pic

we have to find characteristic eqn of AB

i mean

BA

mb

[dusky totem]

I see

[vague phoenix]

.pin

[vapid cypress]

traces are equal

that is provable easily

[flat frigate]

robin.dabanc_

[dusky totem]

Where lambda_i are the eigenvalues

Since determinants and traces are equal trivially, you just have to show that “X”’s are the same in your picture

[vapid cypress]

is there a way to do it without eignenvalues

[safe radish]

@vapid cypress Has your question been resolved?

[vapid cypress]

<@&286206848099549185>

[dusky totem]

This is the cleanest way

You need very very basic understanding

@vapid cypress do you know what eigenvalues are

[vapid cypress]

just that Av = lamba(v)

thats abt it

[dusky totem]

Yeah that’s enough pretty much

[vapid cypress]

and |A-lambda* I | = 0 is characteristic eqn

[dusky totem]

Yup

In fact, the characteristic is written as a polynomial in lambda. Then by Cayley Hamilton theorem it’s shown that a matrix satisfies its own characteristic equation

That’s why you’re allowed to replace lambda by your matrix (AB in this case) in the polynomial

From here try comparing coefficients of the two polynomials using Vieta’s (the roots of the poly in lambda are called eigenvalues); and the trace, determinant and “X” will pop out

[vapid cypress]

is the characteristic eqn interms of lambda just replacing A with lambda in whatever i sent?

oh k

[dusky totem]

Yea

[vapid cypress]

how exactly do u prove it?

[dusky totem]

It’s a non elementary proof

[vapid cypress]

actually??

[dusky totem]

You can look up cayley Hamilton

[vapid cypress]

ok

ok what now

[dusky totem]

In the sense it’s not under pre uni

[dusky totem]

This

Can you write tr(AB) in terms of the roots of the polynomial in lambda

[vapid cypress]

wait the order of the matrix changes when u go from ab to bba

so determinant ba is not equal to determinant ab

first of all

[dusky totem]

You never told me A and B are not square tho

That’s what I assumed

[vapid cypress]

ahh sht

its there a bit above

[dusky totem]

Mb

[vapid cypress]

A is 3X2, B is 2X3

[dusky totem]

Do you know about ranks?

[vapid cypress]

no

[dusky totem]

https://tenor.com/view/wallahi-say-wallahi-say-wallahi-bro-ishowspeed-tiktok-gif-14576999865344060530

Idk a method without ranks, if I find a way I’ll get back to you

[safe radish]

@vapid cypress Has your question been resolved?

[vapid cypress]

ok thanks

<@&286206848099549185>

[empty anvil]

how do i remove helper role

[vapid cypress]

lol

idk

go to roles and channels at the top?

[safe radish]

@vapid cypress Has your question been resolved?

[vapid cypress]

<@&286206848099549185>

[zealous tundra]

can u repost the question

[vapid cypress]

.close

[safe radish]

Channel closed

Closed by @vapid cypress

Use .reopen if this was a mistake.

[lone grail]

.open

.reopen

[last heath]

for future, check pinned messages

[safe radish]

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[muted sapphire]

Is there a way to solve this using complex analysis, and if not, why not?

[plucky elk]

Standard residue theorem probably

[muted sapphire]

The answer is pi * e * erfc(1)

and the only solution I found online is using feynmans technique of differentiating under the integral sign. But I wanted to know if it's possible using residue theorem

I can't land on pi * e * erfc(1) using complex analysis personally

[plucky elk]

What contours did you try

[muted sapphire]

What would you do first? I have no clue anymore :(

just a circle contour around the poles at i and -i? That would be my first instinct

that would give 2pi i (e/2i - e/2i) which is 0

but heres the thing, only having z = i as the pole would give pi * e

which looks alot like pi * e * erfc(1)

And my question is pretty much, why doesn't this approach work and, if it's possible, how would it work.

<@&286206848099549185>

[fleet willow]

partial fractions and cauchy's integral formula probably

and then probably something stupid with ML

[muted sapphire]

Partial fractions doesn't work for non-polynomials

[fleet willow]

x^2 + 1 = (x+i)(x-i)

[muted sapphire]

How would you continue?

[fleet willow]

hmm actually maybe that doesn't work

i'm still a beginner at this

[muted sapphire]

Me too, that's why I asked here :D

[fleet willow]

lol

yeah maybe just ignore me

[muted sapphire]

<@&286206848099549185>

[whole hill]

Only way I can think to solve is is using the complimentary error function. Have you learnt what that is yet?

[muted sapphire]

Not used really, just the definition. But I wanted to know if the integral is possible using complex analysis/ contour integration

[whole hill]

It should be possible

[muted sapphire]

and if it's not possible, why not? I know there are a lot of ways of solving this.

then what would be your course of action

[whole hill]

I’d use Feynman’s Trick, and differentiate under the integral sign

[muted sapphire]

.

[whole hill]

For this particular integral, you technically could. But you wouldn’t be able to do it using a regular semicircular contour

[muted sapphire]

Why is that exactly?

whats holding the semicircular contour back

[whole hill]

The fact that e^-z^2 will end up growing like e^y^2 in the upper and lower half planes. Because of this, the arc integral does not vanish

So the idea of using the residue theorem is valid. You’d just have to rewrite e^-x^2 as a Fourier Transform

[muted sapphire]

why would e^y^2 growing mean that the arc integral doesn't vanish exactly?

Im sorry but it's a bit a -> b therefore c -> d for me

[whole hill]

That’s ok bro, let’s take it a bit at a time. Do you agree that in order for the semi circle method to work, the integral over CR must go to 0 as R goes to ♾️?

[muted sapphire]

Yes

[whole hill]

So, for our (e^-z^2)/(z^2 + 1), at the very top of the arc, where z = iR, we’d have z^2 = -R^2. You following up to this point?

[muted sapphire]

Yes

yeah e^R^2 /(1 - R^2)

[whole hill]

So, yeah. It’s not that the integral is undefined. It’s just that it’s just that the limit not 0

More over, it’s grows without bound

And since the semicircle method requires the arc length component to essentially disappear (go to 0), that method wouldn’t work on this particular integral

It’s not a condition of the Cauchy Residue Theorem. It’s a condition for the semicircular contour method to work. We work around it by USING the Cauchy Residue Theorem.

[muted sapphire]

I'll look into it more, thank you so much for your help!

[whole hill]

I’m glad you’re curious enough to keep looking more into stuff. Have a good one man, and no worries 😊

This is why first year Uni math is boring to me 😭

[muted sapphire]

Hahahahahaha

[whole hill]

Because there is way cooler maths out there

[muted sapphire]

So true man

alright good day

.close

[safe radish]

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[tender meteor]

@muted sapphire still need help ?

We can use 2 known Fourier transform + Parseval

And thats all

[muted sapphire]

I would love to!

hear how you would solve it

[tender meteor]

Do you know fourier transform?

[muted sapphire]

A little bit

but perseval I've never heard of

[tender meteor]

Ok in this case we can use 2 FT that are known

[safe radish]

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[tender meteor]

The idea is that $I=\int_{\mathbb R}\frac{e^{-x^2}}{x^2+1}dx$ is the integral of the product of two functions whose Fourier transform we know well

[flat frigate]

alee

[tender meteor]

Let's take the convention $\widehat f(\xi)=\int_{\mathbb R} f(x)e^{-ix\xi}dx$

[flat frigate]

alee

[tender meteor]

And $f(x) = \frac1{2\pi}\int_{\mathbb R}\widehat f(\xi)e^{ix\xi}d\xi$

[flat frigate]

alee

[tender meteor]

And the known formulas are:$\widehat{\frac1{1+x^2}}(\xi)=\pi e^{-|\xi|}$ and $\quad \widehat{e^{-x^2}}(\xi)=\sqrt{\pi}e^{-\xi^2/4}$

[flat frigate]

alee

[tender meteor]

using Parseval in the form $\int_{\mathbb R} f(x)g(x)dx = \frac1{2\pi}\int_{\mathbb R}\widehat f(\xi)\widehat g(-\xi)d\xi$

[flat frigate]

alee

[muted sapphire]

Yes

[tender meteor]

You get $I = \frac1{2\pi}\int_{\mathbb R}
\bigl(\sqrt{\pi}e^{-\xi^2/4}\bigr)\bigl(\pi e^{-|\xi|}\bigr)d\xi$

[flat frigate]

alee

[tender meteor]

And since all Is even

You get $I =\sqrt{\pi}\int_0^\infty e^{-\xi^2/4-\xi}d\xi$

[flat frigate]

alee

[tender meteor]

Can you proceed ?

[muted sapphire]

I think so.

Thank you so much!

I'll write it all down in my notebook

[tender meteor]

If you want i can finish

[muted sapphire]

I think I got it from here :)

[tender meteor]

Oh ive got I = πe erfc(1)

Is It correct?

[muted sapphire]

Yes 👍

[tender meteor]

👍

[muted sapphire]

incredible

[tender meteor]

Let me know if you still need help

[muted sapphire]

I will thanks!

[safe radish]

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[safe radish]

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[safe radish]

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[lavish skiff]

university calculus I, related rates. How do i even find an equation to relate the variables?

[queen parcel]

i always start these by drawing a picture

idk about you, but when i learned this stuff, i couldn't create an equation with my mind, so a picture always helped me

[lavish skiff]

i did

i just cant think of a formula id use for this

[safe radish]

@lavish skiff Has your question been resolved?

[plucky elk]

https://m.youtube.com/watch?v=W3qoQIlfG0E

[safe radish]

@lavish skiff Has your question been resolved?

[lavish skiff]

ty riemann 🙏

i didnt know this was an actual form of a problem

.close

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[worthy shoal]

can someone verify wether this is correct

wrong labels for axis , but other than that

[safe radish]

@worthy shoal Has your question been resolved?

[tardy mango]

The origin should not be shaded

[worthy shoal]

other than that?

[tardy mango]

pretty sure the you should have two disjoint regions anyway so

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[vale onyx]

wat

[vale onyx]

whenever i do it I get either cos inverse -2 or sin inverse 2 which are both not real

[midnight steeple]

What’s the expression for the derivative you got?

[vale onyx]

-4 + 2 sin theta

[warm warren]

thats.......wrong?

[midnight steeple]

That’s wrong

[vale onyx]

what is it?

[warm warren]

ok, do it step-by-step.

[vale onyx]

it can be this or -4 -2 cos theta

[warm warren]

differentiate term by term

so $y = 6 - 4 \sin \theta -\cos^2 \theta$. What's the derivative of the first term,i.e. 6? It's 0. What about the second term, i.e. $-4\sin\theta$?

[flat frigate]

Annie Maqionde

[vale onyx]

it would be -4 right?

[warm warren]

why?

derivative of $\sin \theta$ w.r.t $\theta$ is $\cos \theta$.

[flat frigate]

Annie Maqionde

[vale onyx]

aa i was thinking u have to treat sin and cos as x

is derivative of cos theta sin theta

[warm warren]

the $x-axis$ is clearly given in terms of $\theta$, and as a rule, for such kind of questions, its differentation w.r.t. the variable of the x-axis, i.e. $\theta$

[flat frigate]

Annie Maqionde

[warm warren]

-sin

$-\sin\theta$

[flat frigate]

Annie Maqionde

[vale onyx]

ok

so -4cos theta + 2sin^2 theta?

[warm warren]

What is the derivative of $\cos^2\theta$(hint chain rule)

[flat frigate]

Annie Maqionde

[warm warren]

hint for the hint: chain rule = $\frac{dz}{dx} = \frac{dz}{dy}\cdot \frac{dy}{dx}$

[flat frigate]

Annie Maqionde

[vale onyx]

ok i was googling the chain rule just now

cos theta (-sin theta)?

u only differentiate one of them

[warm warren]

$2\cos\theta \cdot (-\sin \theta)$

[flat frigate]

Annie Maqionde

[vale onyx]

ok

-4 cos theta - 2 cos theta (-sin theta)

[warm warren]

correct.

now simplify it

[safe radish]

@vale onyx Has your question been resolved?

[safe radish]

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[granite iris]

Find integers x, y, z such that
56x + 98y + 140z = gcd(56, 98, 140)

I don't know where to begin.

[agile scaffold]

use the identity that gcd(a, b, c) = gcd(gcd(a,b), c)

this is an extended bezout identity

[granite iris]

Oh I did not know that existed, thanks.

[agile scaffold]

npnp

[brisk pike]

Fuck

[granite iris]

Wait how do I solve for the integers x y and z? I can inspect that it is obviously 2x56 - 1x98 = 14 but this isn't all solutions?

[spring bison]

Treat it as diophantine equations

[agile scaffold]

think ab - ba - 0c = 0

you can add n amounts of 0 and it does not change the original equation

you might have to solve for bc - cb and ca - ac too but i think this can work

So 3 cases? (i think should work)

[granite iris]

.close

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[soft lava]

Can somebody help me with 40 I have no idea how to do it

[quiet plume]

Can you write down what a half time life means?

[soft lava]

Half life?

[quiet plume]

Yes

[soft lava]

I mean how long smth break downs I don’t really know tbh

[quiet plume]

It's how long it takes for half of that thing to decay

Like if I told you some substace starts with a given quantity A and has a half life of 2 years, would you be able to write down an equation for how much is left after a given time passed?

[soft lava]

Would I put it equal to half the thing? But what number would I use the 10 or the long decimal?

[quiet plume]

Let's just start with a half life of one year.

After one year how much is left if you started with a quantity A with half life 1 year?

[soft lava]

.987?

[quiet plume]

Half life of 1 year mean every year half of what you have is gone.

[soft lava]

Yea I put it in the calculator

Cus u divide by 1 by both sides with equals the same thing then out it in I thought

[quiet plume]

What I mean is if you start with some number A and I tell you every year we divide what you have by 2, what's left after 1 year?

[soft lava]

I’m not sure

Do I have to put it in log I forgot to do that

[quiet plume]

If you have 100$ and every year you burn half of it, after the first year how much do you have left?

[soft lava]

I’m really bad at word problems

[quiet plume]

If you lose half of 100$ how much do you have left?

[soft lava]

50

[quiet plume]

Ok, so after 1 year, you are left with 50$.

And then after the second, year, you'll have 25$

Then 12.50$ after the third year

And so on

In other words we could say that the amount of money you'll have left after $t$ years is $100 \left(\frac{1}{2}\right)^t$ right?

[flat frigate]

Azyrashacorki

[quiet plume]

This is what a half life of 1 represents : every year, half of what you have decays

If you had a half life of 300 years, then every 300 years, half of what you have decays

This is what a half life represents. So you have your initial equation $20e^{-0.0001216t}$ and you want to write it in the form $20\left( \frac{1}{2}\right)^{\frac{t}{h}}$ where $h$ will be your half life.