help-27

[gray oriole]

idk man

[vagrant skiff]

I'm just getting you to state a lower bound

well

[gray oriole]

infinity

[vagrant skiff]

you just said your sequence is always positive

[vagrant skiff]

a number, please

[gray oriole]

0

[vagrant skiff]

good

[gray oriole]

wait thatt was a fluke

[vagrant skiff]

do you see why that fits the definition of a lower bound

[gray oriole]

am i gonna have to guess these bounds

[vagrant skiff]

uhm

no

you can often analytically find them

in this case, since a_n > 0 (positive) for all n, 0 is a lower bound

does that make sense

[gray oriole]

you do 10 I'll do 11

[vagrant skiff]

what

[gray oriole]

well i can use ratio method

wait

[gray oriole]

this sequence is strictly decreasing

(10)

[rain summit]

!noans

[devout snow]

The purpose of this server is to help you learn; please don't ask for direct answers. Ask for guidance, explanations, or feedback instead.

[rain summit]

we can only guide you on how to do it

not provide solutions to you

[hollow aspen]

So if the sequence is strictly decreasing, and it's bonded below, what does the monotone convergence theorem tell us?

[gray oriole]

How do I determine that it is bounded below?

[opal herald]

10?

uh

[hollow aspen]

Oh, I'm sorry I thought that you already did that part. Is there a natural number n that yield anything negative? How about 0?

[opal herald]

sequence: an = 10^n/(2n!)

ya need to prove lim a(n+1)/an as n->infinity = 0

[gray oriole]

Well I tried putting n = 10 and n = 11 in the sequence, my calculator shows extremeley small values for it

[gray oriole]

it is

Isnt this test for series?

The question is about sequene

[opal herald]

wdym

[gray oriole]

this is the ratio test for convergence of series right

[opal herald]

ye

[hollow aspen]

For every natural number n, each term is a positive rational number

[gray oriole]

well can I apply it on questions about sequences?

[gray oriole]

yeah

I mean I know it is 0 but I have to prove it

[opal herald]

ye

unless ur lecturer didnt teach you this

[hollow aspen]

If you can show that 0 is the greatest lower bound (you've already shown that it's decreasing), then by the MCT the sequence converges to 0

[gray oriole]

yea how to show that 0 is the greatest lower bound

thats the question

[fossil locust]

every term in the sequence is positive

positive number divided by another positive number

[uneven coral]

eh, ghost ping

[gray oriole]

alright

so can i write that its gonna be zero cuz smallest non negative number is 0?

[fossil locust]

yeah, basically the logic is that 0 is smaller than every positive number, so it has to be a lower bound

so any negative number is not the greatest lower bound

[fossil locust]

(ping)

[gray oriole]

alright

ty!

.close

[devout snow]

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[hollow aspen]

It's obvious that 0 is a lower bound—this is enough to show that the sequence converges, but if you want to show that 0 is the greatest lower bound (you need this if you want to claim the sequence converges to 0), your professor might want to see some work there

[devout snow]

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[coral forum]

Hi! I need help with the task below.

The function $f$ is given by
$f(x)=x+a, 0\leq x \leq2$ and $a>0$
(a): Find $a$ such that the area under the graph of $f$ is 3.

Thanks in advance!

[woven radish]

dream

[coral forum]

can someone check what i have done wrong here

[twilit shell]

you didn't take the integral

[coral forum]

ohhh

[twilit shell]

the integral of x+a from 0 to 2 is not x+a

[coral forum]

silly meeee

omg

i forgot

thank youuuu

[coral forum]

so its $\int (x+a)=\frac{1}{2}x^2+ax$

[woven radish]

dream

[coral forum]

and now i can find the area

[twilit shell]

yes

[coral forum]

$a=\frac{1}{2}$

[woven radish]

dream

[tender wharf]

dream you doing math on new year eve

[coral forum]

mhm

i am

[tender wharf]

ye same

[coral forum]

bahaha

[tender wharf]

lol

[coral forum]

i just have one more task to finish

[tender wharf]

ic

[devout snow]

@coral forum Has your question been resolved?

[cold haven]

@coral forum

[devout snow]

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[iron chasm]

im not enterly sure on what i have missed.

Its 4 marks.

[glacial tide]

you just restated the provided information

and affirmed the conclusion

you need to prove why the conclusion is true

given the information they provided

[knotty salmon]

haywik I just learned this stuff I can help you 100%

[iron chasm]

so hwo do i do that?

[iron chasm]

ah pls.

[knotty salmon]

so u construct AC and BD first right?

[glacial tide]

contruct triangles

use sas congruence

[knotty salmon]

and then u gotta prove that triangles ADC and triangle DBA are congruent

[iron chasm]

how do i prove it?

[knotty salmon]

AD congruent to AD

the two angles are congruent

and DC and BA are congruent

so then......

it's what congruency

do u understand

[iron chasm]

and how do i write that?

i get its just the same thing in opposite directions.

[knotty salmon]

so basically u write construct AC and BD first

and then write because AB = CD, angle DAB = angle CDA, and AD = AD

[iron chasm]

afk sorry, ty for help.

[knotty salmon]

therefore triangel ADB is congruent to DAC

so then AC = BD because of CPCTC

congruent part of congruent triangles are congruent

yeah np

[devout snow]

@iron chasm Has your question been resolved?

[iron chasm]

k so here im at

[crude niche]

try making triangles

[iron chasm]

triangles?

[crude niche]

connecting to bd and ac

or like

bd is a line

and ac is a line

[twilit shell]

i think theyve already done that

[iron chasm]

i dont understand

i get the line part

[twilit shell]

you state ADB \cong DAC w/out proof, you should be a bit more explicit about why ADB \cong DAC

[crude niche]

mb

[iron chasm]

isnt that said when the angles are mentioned?

[crude niche]

maybe i shouldn’t be doing math when im walking 😂

[iron chasm]

dedication.

[twilit shell]

like

sure, but you know that they're congruent by SAS congruence

just to be clear you should state that

[iron chasm]

no??

confused

i prefer algebra, this imagray stuff is annoying

[twilit shell]

like

[crude niche]

try using coordinates?

[twilit shell]

ADB being congruent to DAC is a result of AD = AD, DC = BA, and <DAB = <CDA

so ADB is congruent to DAC by SAS congruence

you should just state its by SAS

[iron chasm]

yh but how do we prove AC = BD?

we are told DAB = CDA tho

[twilit shell]

these three facts are all things that the problem gives you

from this you can deduce that ADB is congruent to DAC by SAS Congruence

this is correct

i am just pointing out that you should be more explicit

and state that its by SAS Congruence

instead of just saying it outright

[iron chasm]

but how is that proof if your just reating the question?

[iron chasm]

so reword it basicly.

[twilit shell]

yeah

justify it

[iron chasm]

seems very mathey, i shall try it

[iron chasm]

so like this?

[twilit shell]

yes

and then from there you can try to prove AC = BD

[iron chasm]

so like

[twilit shell]

uhh

not quite

you know that ADB is congruent to DAC

which means that they're the same shape

and so their sides are all the same

[iron chasm]

right yeah

[twilit shell]

so

because ADB and DAC are congruent

we know their corresponding sides are the same length

so AC = DB follows

[iron chasm]

okay...

so add this

current answer:

[twilit shell]

yeah

if you want to you can say its by CPCTC idk

[iron chasm]

k, so this is 2/4 marks.

[twilit shell]

idk what else to do then

[devout snow]

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[deep abyss]

Let PQR be right angled isosceles triangle, right angled at P(2,1) If the equation of the line QR is 2x + y = 3 then the equation representing the pair of lines PQ and PR is

[deep abyss]

I got 3x²-3y²+8xy-20x-10y+25=0

But the answer provided by my teacher says 3x²-3y²-8xy-20x-10y+25=0 is right

[deep abyss]

I suppose thts wrong?

[feral agate]

How did you approach this

[deep abyss]

Found the eqn of two lines

[devout snow]

@deep abyss Has your question been resolved?

[deep abyss]

Cant show my work but the eqns were $$3y+x=5$$ $$y=3x-5$$

[woven radish]

ch3rry

[deep abyss]

<@&286206848099549185>

[cobalt epoch]

Yo

[rain summit]

hm

[lusty moth]

hm

[coarse flume]

I mean i got the same answer as you idk

[rain summit]

graphical check

seems like your answer is correct

[deep abyss]

.close

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[vivid estuary]

have a question about order, for the exercise we want to show A1 U A2 U A3 is no different from A2 U A1 U A3, where K = {1,2,3}. But what if K, the index set, has no order? Does this still make sense

[vivid estuary]

oh yeah it would

I think i’m confused about something else one second

ok nvm let me try doing this again

i’ll prob be back

[vital edge]

Yes

The logical definition of union might help

$a \in \bigcup_{i \in I} A_i \to \exists i \in I \text{ such that } a \in A_i$

[woven radish]

Xavier 🌺

[vivid estuary]

thanks that does help me understand it more

[devout snow]

@vivid estuary Has your question been resolved?

[devout snow]

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[modest shadow]

Hello. Im trying to prove what i wrote in the top-most part of the page. Is this a good approach? If what I wrote is right then i think im pretty close to arriving at a contradiction

[modest shadow]

that epsilon part seems faulty tbh

[harsh sierra]

its looks good

just choose an appropriate epsilon

[modest shadow]

can i not let it be arbitrary?

or do you mean for the next part of the proof?

[harsh sierra]

for the next part

to arrive at a contradiction to supC < supA + supB

[modest shadow]

alright alright thank you

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[foggy iron]

i dont understand what im supposed to do for part c

[pseudo basin]

let g(x) = f(x) + ln(k).
you want to find the mean value of g(x) over [0,3].

which is very easy to do if you know that g is just f shifted up by a constant

[foggy iron]

yess i agree

[foggy iron]

oh so am i just integrating lnk over [0,3] and then once ive found that, i just add it to part b?

[pseudo basin]

not even this.

no, you just add ln(k) to part b.

or you could do the redundant route of redundancy and integrate the constant ln(k) over [0,3] then divide that by 3 and add the result to part b

[foggy iron]

got it, thanks!

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[modest shadow]

,tex
Hi there. Suppose we have the sequence
$a_n = \frac{2^n}{\sqrt{n} \ n!} $ . I have proven that it is convergent, but not at which number. How would we go about proving the number it converges to? I missed the lecture on this kinda stuff oops

[woven radish]

fijokazż

[harsh sierra]

do you have a guess what the answer might be?

[modest shadow]

prolly 0

[harsh sierra]

right

[modest shadow]

i saw some people doing a(n+1)/a(n)

and taking a limit

but idk why

[lost laurel]

you could use the epsilon-delta defn, but that will be very painful

[harsh sierra]

no need epsilon delta

[modest shadow]

ye our prof told us we dont need to do that

[lost laurel]

oh, if the mod of that limit is less than 1 the limit is 0

you can infact prove that

[modest shadow]

the mod?

[lost laurel]

absolute value

[modest shadow]

ohh

hmm yeah thats what i saw ppl do

is that the only easy way?

[harsh sierra]

no

$2^n \leq n!$ for large n

and dividing by sqrt(n) reduced this to 0

[woven radish]

ExpertEsquieESQUIE

[modest shadow]

i dont see how that proves it

[lost laurel]

@harsh sierra in an introductory real analysis class wouldn't this be too much though ( assuming this is for an RA class)

[harsh sierra]

this is pretty easy

simple induction

[modest shadow]

ive proven 2^n/sqrt(n) <= n!

for large n

i just dont see why we just infer that the limit is 0

[stone stump]

we cant just from that

[harsh sierra]

this implies $\frac{2^n}{n!} \leq 1$ and dividing by $\sqrt{n}$ gives $$0 \leq \frac{2^n}{\sqrt{n} n!} \leq \frac{1}{\sqrt{n}}$$

[woven radish]

ExpertEsquieESQUIE

[modest shadow]

ohhhh

i was thinking abt that

yeah thats def better for me

thank you!!

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[modest shadow]

hello. supposing we have already proven that if b_2n and b_(2n-1) dont converge to the same number implies b_n doesnt converge, is this proof of b_n not converging fine?

[stone stump]

why does n!>3^n matter

[modest shadow]

so that they dont converge to 0

[stone stump]

write that

[modest shadow]

yeah true

here

[stone stump]

yes thats fine

[misty crest]

i would still reword this because of that first sentence you wrote

[modest shadow]

what would be best to write?

[misty crest]

you see what’s wrong with it right

[modest shadow]

ive proven that if they cant converge to the same number then our original sequence doesnt converge at all

unless you mean to switch the ending with the start

[misty crest]

you should first argue that for all but finitely many n, n!/3^n > 1 and -n!/3^n < -1 then say so they can’t converge to the same number. this would make more sense

[modest shadow]

yeah i see ur point

[misty crest]

the problem is it doesn’t follow from one being positive and the other being negative

take 1/n and -1/n for example

they both converge to 0

despite one being positive and the other being negative

[modest shadow]

yes thats what i noticed after i wrote half the sentence

thats why its worded kinda wrong

thank you

.solved

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[tropic vault]

Can someone help me with this graph

[olive snow]

For 2+ and 2- its the opposite way

-infinity is down

[tropic vault]

Is it like this

[pearl relic]

right so like yakubros said you have to change the direction (2- means left of 2 and 2+ means right of 2)

to get the limit at -2 you need a graph that looks something like this, where theres an open circle and then a point below it

[pearl relic]

so that isnt a function
those two lines overlap to the left of -2

[tropic vault]

Idk how to make it continuous tho

[pearl relic]

how to make what continuous

[pearl relic]

thats not continuous?

[tropic vault]

Cause it has to from both sides equal it tho

[pearl relic]

right

[tropic vault]

Not continuous that’s what I meant sorry

[pearl relic]

again, it'll look like this

[tropic vault]

Idk how to make it both sides equal

[pearl relic]

(replace a with -2)

[tropic vault]

It has to be filled in tho cause it’s one of the criteria

[pearl relic]

it has to be filled in at (-2,-3)

[tropic vault]

Yes. I know

[pearl relic]

but the (-2,4) must not be filled in

[tropic vault]

Doesn’t a line go there but I don’t understand how cause of this one

[pearl relic]

,w graph -1/x in interval [-3,0]

[tropic vault]

Oh from the top

[pearl relic]

or the bottom

either works

[tropic vault]

This

[woven radish]

Results provided by WolframAlpha

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

Requested by boredamogi

[tropic vault]

Is that right or no

[pearl relic]

yes that good

but see if you can make it go slower so that it reaches infinity at 2

if that makes sense

[pearl relic]

yeah exactly

so thats 4 of those

now all you need is

,,\lim_{x\to 2^+} F(x)=\infty

oh wait im sorry

no thats going the wrong way

that limit goes to positive infinity

we want negative infinity

you can flip the arrow upside down to get -infinity

[tropic vault]

How will it approach 0

[pearl relic]

this thing

[pearl relic]

so you only flip the right half

[tropic vault]

Doesn’t that make it have a cusp

[pearl relic]

not what a cusp is

but also, cusps are okay here

[pearl relic]

as long as the limit still works

[pearl relic]

okay perfect

and then all you need is that first thing

,,\lim_{x\to2^+}=\infty

[woven radish]

bored amogi

[pearl relic]

can you fill in the rest of the function to get that

[pearl relic]

yeah thats it

you got it

do you understand how i got all this?

[tropic vault]

I mean I understand it I’m not really good with drawing tho

[restive river]

Dw u got this ᕙ⁠(⁠ ⁠:⁠ ⁠˘⁠ ⁠∧⁠ ⁠˘⁠ ⁠:⁠ ⁠)⁠ᕗ

[tropic vault]

Thank you for helping me I appreciate it

.close

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[pastel rose]

Yk to take the limit of this, why am I getting different results when taking it separately vs when I just combine the fractions

[pastel rose]

Lim as n tends to inf

Combining it gives me 1

But I’m confused on how to work it out without combining

Since if I divide by the highest power, I’d be tending towards 0 in the denominator for the first fraction

[toxic flower]

is answer 3?

[ocean scroll]

I just did it and got 1

The first term is 2n-2+O(1/n)
The second term is -(2n-3+O(1/n))

[pastel rose]

Nah I checked on desmos and it’s 1

I get 1 asw by combining

[toxic flower]

oh yea i see

[pastel rose]

Doing it separately gives 3

[toxic flower]

my mistake

[ocean scroll]

If you take limits on each term you will get infinity-infinity

[pastel rose]

Yea is that from this?

[pastel rose]

What’s O?

Oh u divided by n^2

[ocean scroll]

O(1/n) means there are 1/n terms (or smaller terms) I don't care about as n->infinity e.g. 3/n+2/n^2

it should be 1/(n+1) my bad

[pastel rose]

Alright ty

.close

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[dire depot]

If effectiveness and C_r are known, how do I get the value of NTU

[pseudo basin]

doesn't look like you can get it out of there cleanly ngl

also this is tiny and hard to read

$\ep = 1 - \exp\sqb{\frac{1}{C_r} NTU^{0.22} \curly{\exp\sqb{-C_r NTU^{0.78}}-1}}$

[woven radish]

Ann

[pseudo basin]

is that what your formula says?

[dire depot]

Yeah

[pseudo basin]

yeah i would not expect a neat algebraic solution here sorry

what do you need it for

[dire depot]

So iteration?

[dire depot]

As in what the question is?

[pseudo basin]

yes

i'd like to know the context and the original question & goal

and like what tools you've got at your disposal here

[dire depot]

It's a heat exchanger question so I was abit tentative on sending it

I've got all the other theory understood but idk how to get NTU from this basically

I'll send it still

Here

[pseudo basin]

hmmm

well ok a cursory glance of the relevant wikipedia page confirms your formula really is that

but also i would have no idea how you're expected to solve it

except with like desmos or some shit

or a graphing calculator

[dire depot]

NTU is between 0 and 1

But I say iteration but I don't even remember how to do that

I'll research thanks tho

.close

[devout snow]

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Use .reopen if this was a mistake.

[acoustic leaf]

yeah if you're allowed a graphing calculator then a numerical solver would work

[dire depot]

I've got a classwiz fx90 I believe

I'm not that savvy on this stuff so idk if it's capable

[acoustic leaf]

from looking up the user's manual online, it seems that the "equation" app includes a numerical equation solver

[dire depot]

Yeah I just found out to boutta try it out

[devout snow]

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[turbid spoke]

can someone please give me a hint on what to do here?
apparently this is supposed to be done "global" but the only thing i can think of is it break it up and do it "local", as in the split the two sums into two smaller sums, but i don't think that's the correct approach

[dapper fable]

i did this problem with induction

let the statement be P(n,m). since P(n,n) is obviously true then we only need to show that P(n,m) implies P(n+1,m)

[woven radish]

Copter

[dapper fable]

and compare cases for floor(x/i) <= n etc

@turbid spoke

[devout snow]

@turbid spoke Has your question been resolved?

[turbid spoke]

okay thanks i'll try that

.close

[devout snow]

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[vivid estuary]

“An empty union makes sense and is empty, however an empty intersection does not make sense”

[vivid estuary]

not sure if i’m understanding this right but why wouldn’t the empty intersection be empty

[frozen holly]

eh itd be weird because the intersection is the set of all items that are in all the sets in the intersection

[vivid estuary]

by empty union are they referring to the union of empty sets

[frozen holly]

but that would be literally everything, because everything is in all those sets

(since theres none of them)

so you would have no upper bound on what youre actually considering

like when you remove items from a intersection (of N sets) you expect the resulting set to get (potentially) bigger, but never smaller

setting the empty intersection to empty is completely at odds with that.

and of course theres no set of all sets or other such nonsense, so

[vivid estuary]

what they say empty intersection are they referring to the intersection of empty sets

[frozen holly]

i do not think so

[vivid estuary]

oh

[frozen holly]

I think they are referring to an intersetion of N sets where N happens to be zero

like in

I should better not be the empty set

hence why they say non-empty family of sets.

[vivid estuary]

oh ok I see

[frozen holly]

[vivid estuary]

ok yeah i see what you mean now earlier thanks

.solved

[devout snow]

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[quartz bloom]

Can somebody give the correct statement of Archimedean property of R.

In my book they have mentioned
If x ∈R , then there exists n ∈N such that , x≤n

While in online platforms they have used strict inequality.

Which one is correct

[frozen holly]

both versions imply the other

if s is in R \ N, they mean exactly the same thing, if x is in N, it just means you have to choose n+1 instead of n if the inequality is strict

[quartz bloom]

In corollary 2.4.6

Here we have ≤ inequalities in both n-1 and n

Is this correct

Also they have proved that one of these is strict inequality

[upper schooner]

,rccw

[uncut crow]

there isn’t really a “the correct statement”

[quartz bloom]

Then. .

[uncut crow]

they are equivalent and like… not in a tricky way

it’s pretty much the same whether you use > or >=

[quartz bloom]

Then this is causing problems

[rain summit]

I can think of another proof without using the well ordering principle

[frozen holly]

I don't think that's the issue

[uncut crow]

hm the statement of the corollary is true, but they probably meant for the second inequality to be strict

[rain summit]

How come

[quartz bloom]

Yes

How can y possibly satisfy both inequalities

By the statement they have used for Archimedean property

[frozen holly]

Cuz the one they proved implies the one they stated

[rain summit]

$n_y$ you mean?

[woven radish]

1 divided by 0 equals Infinity

[quartz bloom]

I think the equality should be only with n y

[uncut crow]

x < y implies x <= y

[rain summit]

You mean the statement that says: $E_y$ has a least element, which we denote by $n_y$?

[woven radish]

1 divided by 0 equals Infinity

[quartz bloom]

But isn't there a possibility that y can be equal to x...

In second one

[rain summit]

You mean the other way?

[uncut crow]

yes oops

[quartz bloom]

No . .the Archimedean property statement in the book defines the set Ey , such that ny≤y

[uncut crow]

no actually, not oops

what i wrote is what i meant

[quartz bloom]

.

[uncut crow]

x <= y means x < y or x = y

and x < y implies x < y or x = y

[rain summit]

So x < y implies x <= y

[quartz bloom]

Ok so I can just prove anything for x<y

And say , that holds for x≤y

.close

[devout snow]

Channel closed

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[rain summit]

Another proof: Choose $n_y = \left \lfloor y \right \rfloor + 1$

[woven radish]

1 divided by 0 equals Infinity

[rain summit]

Then we can ensure that $n_y - 1 \leq y \leq n_y$

[woven radish]

1 divided by 0 equals Infinity

[uncut crow]

this is just hiding the well-ordering principle

[rain summit]

Ig so

[uncut crow]

floor and ceil are defined with the same ideas that are used in the given proof

[quartz bloom]

But you see if we have proved for x<y we can't say x≤y

[uncut crow]

we can tho

if we are trying to prove x <= y and we prove x < y, that’s fine

x <= y follows

[quartz bloom]

Whaaa

Ok if that's true , it clears my doubt

[devout snow]

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[sharp cloud]

here is the equations I did 2x+5+3y-35=180=>2x+3y=210(eq1) 3x+10+2y-10=180=> 3x+2y=180(eq2) 2x+3y=210=> 6x+9y=630(eq3) 3x+2y=180=>6x+4y=360(eq4) eq3-eq4= 5y=270 y=54 but there is no 54 in it

[dry oxide]

you solution seems correct; the options might be typed incorrectly

[sharp cloud]

so i am correct should I report it?

[dry oxide]

yes

[sharp cloud]

thanks

how to stop this

[dry oxide]

.close/.solved

[sharp cloud]

.close

[devout snow]

Channel closed

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[devout snow]

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[dapper token]

Hi

[dry oxide]

do you have a question

[short hare]

hi 🤩

[uncut crow]

yajat

[dry oxide]

slayla

[devout snow]

@dapper token Has your question been resolved?

[tender wharf]

sleigh

[devout snow]

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[inner anchor]

Let $G$ and $H$ be finite groups and $\theta : G \rightarrow H$ a random morphism. Show that for every $g \in G$, $o(g^{\theta})$ is a divisor for $o(g)$ as well as $|H|$. Also show that if $gcd(o(g),|H|)=1$, that $g^{\theta} = 1$.

[woven radish]

Nico

[inner anchor]

The hint says that I need to look at the restriction $\theta |_{\langle g \rangle }$

[woven radish]

Nico

[vagrant skiff]

what's the image of the restriction

[inner anchor]

It's something isomorphic to ${\langle g \rangle }$, right?

[woven radish]

Nico

[inner anchor]

(Yeah or everything is just equal to 1)

No it's isomorphic to ${\langle g \rangle } / Ker(\theta)$

[woven radish]

Nico

[inner anchor]

Oh right, but $o(g^{\theta}) = |g^{\theta}| = o(g)/Ker(\theta)$, no?

Wups

$o(g^{\theta}) = |g^{\theta}| = o(g)/|Ker(\theta)|$

[woven radish]

Nico

[vagrant skiff]

ker(theta) is a subgroup of all G

while ker(theta|_{<g>}) is a subgroup of <g>

[inner anchor]

Oh oops

Yeah I meant ker(theta|_{<g>})

And I'm guessing that o(g^theta) is a divisor for |H|, since im(theta|_{<g>}) is a subgroup of H?

[vagrant skiff]

yes

that follows immediately from Lagrange

[inner anchor]

Yea

[vagrant skiff]

the last part should be easy

[inner anchor]

Yeah

Bc I literally have a lemma that says that if gcd(|G|,|H|) = 1, that theta must be trivial

Just replace G by <g>, and if gcd(|G|,|H|) = 1 and o(g) is a divisor for |G|, then also gcd(o(g),|H|) = 1

[vagrant skiff]

sure

o(g^theta) | o(g) and o(g^theta) | |H| implies o(g^theta) | gcd(o(g), |H|) so if gcd(o(g),|H|) = 1 then o(g^theta) = 1 meaning g^theta = 1_H

[inner anchor]

Right

Thanks

[vagrant skiff]

[inner anchor]

.close

[devout snow]

Channel closed

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[devout snow]

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[deep abyss]

C is the circle x² + y² + 2gx + 2fy + c = 0. AB is a variable chord subtending a right-angle at the origin O. OP is perpendicular to AB. Prove that P traces a circle.

[deep abyss]

All i know is that we need to homogenize the eqn wrt line AB

but idk if i should go on to introduce 4new variables for coordinates of A and B

Given in ques.

Oh tht point is P btw

Where the math folks at

<@&286206848099549185>

[untold lance]

The image isn't exactly correct, maybe trying to draw it properly might help you.

O is the origin.

[deep abyss]

Thts the given image...

[untold lance]

What you have to show is that the locus of P also follows an equation of a circle

[deep abyss]

Understand tht.

K

[untold lance]

For the general equation of the circle, do you know what would be the centre of it in the R^2 plane?

[deep abyss]

How many times do i have to state the diagram is given in the ques

I didn't draw it

[deep abyss]

(-g,-f)

[untold lance]

Nice

Now as the chord subtends 90 degrees at O, we have some relation between the position of A and B

Not every chord will subtend a right angle

[deep abyss]

Origin isnt the center btw...

...wtf

I dont get it

[untold lance]

my bad

[deep abyss]

Welp could u clear the msgs then ?

[untold lance]

I've deleted mine

Okay, use the normal form of a line.

If the angle made by OP with positive x-axis is alpha, then the equation of AB would be xcos alpha + y sin alpha = p

where p is the length of OP

[deep abyss]

Sure then

[untold lance]

From that OP is perpendicular on AB drawn from O.

This gives an equation for OP.

[deep abyss]

Okay..

Then

[untold lance]

Solving for intersection gives you P, right?

This gives an expression for P.

[deep abyss]

Tbh idt this is a good idea

[untold lance]

I'll try the computation for myself, if anyone has any nice geometric solution they might help you. I'll get back after I complete the computation.

[deep abyss]

Okie

[tender wharf]

hmm..

[deep abyss]

Idk where we r going w these but its gon be long aff

[tender wharf]

finding equation of chord and then substitute in circle's equation

[untold lance]

The line also needs to be a chord of the circle, hence you must check for real solutions in the case of intersections. Make it a quandratic equation over a single variable and make sure the discriminant is positive. Not equal to zero, we're not solving for a tangent here.

[deep abyss]

..

[untold lance]

That is preciesly what I'm doing rn but yeah it's a bit of computation. If you have an alternative way please check\

[tender wharf]

let's see

[deep abyss]

Dang im so dumb

Can i js .. <@&286206848099549185> is tht allowed 😭

[fallow tartan]

Whot

[untold lance]

Once in 15-20 mins yes

afaik

!noping

[devout snow]

Please do not ping individual helpers unprompted.

[untold lance]

oh

[fallow tartan]

.close

[untold lance]

that's not the factoid I was looking for but yeah

[celest frigate]

Its called !15m

[untold lance]

my bad

!15m

[devout snow]

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

[deep abyss]

Been more than 15

[untold lance]

yeah yeah

[deep abyss]

Just asking if twice is allowed

[untold lance]

If someone has any idea please check

[deep abyss]

But anywho back to thw ques

[celest frigate]

Someone was helping you out 2 mins ago?

Dont spam here

[fallow tartan]

[deep abyss]

<@&268886789983436800>

Im already stressed🙏😭

[winter torrent]

get out of here and get your own channel if you need help

[celest frigate]

P is the midpoint or point of intersection? @deep abyss

[deep abyss]

Nopies

Thtd make the ques ezz lol

[celest frigate]

Whats P then?

[deep abyss]

...

[celest frigate]

It does not give what P is. I read it

[deep abyss]

Read the ques thts all the info

OP is perpendicular to AB

[celest frigate]

Yeah but OP is a line

And P could be any point

[deep abyss]

Only which does not make OP not perpendicular to AB

[celest frigate]

No then P could lie anywhere on the perpendicular

Its locus would span the entire plane then

[deep abyss]

P lies on AB

[celest frigate]

Yes thats what I was asking

[untold lance]

AB is a variable line which always subtends a right angle at the origin. In this case, AB must be a secant of the circle. Then a perpendicular line passing through origin meets AB at P. That line is OP.

[deep abyss]

Tht was p obvious tho?

[untold lance]

I mean they were asking this so I just clarified

[celest frigate]

It never read meets so I was clarifying

[deep abyss]

Ok anyways

[untold lance]

Makes sense, but there really isn't exactly anything else given so we can assume that without loss of generality.

[celest frigate]

You want a non analytical way right since you used a computational way

[untold lance]

Yeah, I'm currently solving for the equations and stuff so if There's anything else it'll be helpfukl

[deep abyss]

Wut

[celest frigate]

Do you know some cyclic quadrilaterals?

[deep abyss]

The diagram confirms it i think. We arent exactly assuming

[deep abyss]

As in?

[celest frigate]

Like properties of cyclic quadrilaterals

[deep abyss]

Yh

[celest frigate]

Ok and do you know that the perpendicular from the centre bisects a chord?

[deep abyss]

Yh

-# doubt there's use of tht here

[celest frigate]

And what about the length of the cords? They should be same, you agree

[deep abyss]

Lenghtbof what chords

[celest frigate]

Chords yes

AB

AB has a fixed length I mean

[deep abyss]

Its variable..

[celest frigate]

The angle is fixed though right?

[deep abyss]

Angle with?

[celest frigate]

At the centre

It subtends 90 degrees

[deep abyss]

No

[celest frigate]

Ooh mb

O is the origin and not the centre

[deep abyss]

:/

[tender wharf]

do you know how to do that?

[deep abyss]

Yh

[tender wharf]

then homogenize

[deep abyss]

Not when you call it center

[deep abyss]

Sure

[deep abyss]

..

[tender wharf]

and then apply perpendicular posl condition

sum coefficients of x^2 + y^2 should be = 0

[deep abyss]

How so

[tender wharf]

?

[deep abyss]

I mean how do i apply

POSL

And why would it help😭

[tender wharf]

basically a homogenous equation in second degree passes through origin

<@&268886789983436800>

[deep abyss]

Ik

[faint zinc]

@lusty moth #chill

[deep abyss]

Ima go cry now

[tender wharf]

the homogeneous equation would represent OA and OB

[deep abyss]

So i dont really get what you're trying to do. Can u idk show me?

[tender wharf]

mhm.. basically chord subtends 90 degree angle

P is lying on the chord

we homogenize the 2nd degree equation wrt to chord

now if posl are represented by Ax2+2Hxy+By2=0then for then tan(angle b/w them) = 2sqrt(H^2−AB)/(A+B)

theta is 90 so make the denominator zero

[deep abyss]

.... a+b=0...then...?

[tender wharf]

then simplify it

you would get an equation in the coordinates of P

[deep abyss]

..how

[tender wharf]

did you find eqn of chord?

assume foot of perpendicular (a,b) so slope of OP =b/a

[deep abyss]

Nuh uh

[tender wharf]

use this

[deep abyss]

Id rather (h,k)

That's smart

[tender wharf]

your wish

what would be slope of chord

[deep abyss]

-h/k

[tender wharf]

right

[deep abyss]

And we're done damn

[tender wharf]

lmao

[deep abyss]

No genuinely

Now homogenize

[tender wharf]

yes

[deep abyss]

a+b=0

Thts it

[tender wharf]

right

[deep abyss]

Thts gives the locus

[tender wharf]

right

[deep abyss]

I feel even more dumb now

[tender wharf]

basically homogenizing led you use coeffx^2+ y^2 = 0

hence it was efficient

[deep abyss]

Yh thts also in the soln

You're one smart fella xD

[untold lance]

Good

[deep abyss]

Ill keep this channel open tho for maybe another approach altho im sure this is the best one.

[untold lance]

I'm too old for these things

Yeah I think this is fine

[tender wharf]

going with this ig messy calculation but this would be more direct

[untold lance]

I am still stuck with like 5 unkowns so yeah I'm not going to contiunue

[tender wharf]

nah, you had mentioned it so i tried

[deep abyss]

Oh what did i mention..

[tender wharf]

not worth the effort

[deep abyss]

[tender wharf]

homogenization

[deep abyss]

Oh well

Thankyou for trying btw everyone

[untold lance]

Okay finally got it

P has equation x^2 + y^2 = k^2 for some k

so it's always a circle around the origin?

just for double checking

[deep abyss]

Have you proved tht?

[untold lance]

no I had some mistake there

but yeah after correction I got something liker

x^2 + y^2 + gx + fy + c/2 = 0

After that I tried the method you guys discussed just now

it gives me the exact same expression just way faster

I think my brain is fried but thank you

[deep abyss]

Same :>

[devout snow]

@deep abyss Has your question been resolved?

[wheat vector]

so have we figured it out

yet

.close

[deep abyss]

Yes

But im wondering if theres another method

[wheat vector]

oh their might be let me check

x^2 + y^2 + gx + fy + c/2 = 0 so i know you have that but anything that you dont understand?

[pseudo basin]

you cannot close other people's channels unless you have the Helpful role or higher

[wheat vector]

oh

hmm okay

[deep abyss]

Uhm...

[deep abyss]

..

[untold lance]

I think OP is asking if there is any other methods which can be implemented to solve this except the one people disacussed here

[wheat vector]

true

[untold lance]

Brute force works but I think you'll age like 10 years by the time when you arrive at the answer

Source: me, I tried

[deep abyss]

😭

[modern lance]

Play MC

There's a way do to this

And you don't have to use coordinate system at all

I'll make a digram hang on

[untold lance]

YES This is what i was looking for thank you smart geometer <3

[modern lance]

So, OP^2=AP*PB

[untold lance]

(Similarity conditions, yes)

[modern lance]

You're not gonna like it

vec(IP)=PB/BA * vec(OA) + PA/BA * vec(OB)

square both side

[untold lance]

Wait, why is this true?

[modern lance]

IP^2=PB^2/BA^2 * R^2 + PA^2/BA^2 * R^2 + 2cos(...)R^2*PB*PA/BA^2

[modern lance]

Ehh I can show why later

[untold lance]

yeah my bad, I'll let you do your stuff first

Just saying it isn't obvious

[modern lance]

Cosine law: IA^2+IB^2-AB^2=2cos(...)*IA*IB

plug that into the above equation

IP^2= PB^2/BA^2 * R^2 + PA^2/BA^2 * R^2 + PB^2/BA^2 * ( 2R^2-AB^2)/AB^2

uh wait

Then you compute IP^2+OP^2

which would give you constant

[deep abyss]

Go on ima bk later for explanation 😌

[modern lance]

According to this

1/2sqrt(2IP^2+2OP^2-IO^2)=MP

IO^2 is a constant ofc

So is 2IP^2+2OP^2

Thus MP is constant

Done

[tender wharf]

that is cool

[modern lance]

Gimme a min, I'm making a proof for this one

Here we go

Much stronger statement would be: if A,C,B are colinear by that order and O is a random point, OC= xOA+ yOB

then x+y=1

[modern lance]

the other way around is also true

[devout snow]

@deep abyss Has your question been resolved?

[modern lance]

-# What are you looking for? are two solutions not enough

[untold lance]

Got it, law of triangles

Nice proof

[deep abyss]

No sry i hvnt read them yet

[devout snow]

@deep abyss Has your question been resolved?

Channel closed

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Use .reopen if this was a mistake.

[devout snow]

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[boreal helm]

yo

[boreal helm]

sorry wrong channel

.close

[devout snow]

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Closed by @boreal helm

Use .reopen if this was a mistake.

[willow helm]

[devout snow]

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[ebon lily]

if I have a function f(x), that is 4.207267830296800x^{3}-24.082704661729231x^{2}+46.693121021503860x-24.496440720995176, is there anyway to create another function that would maintain a uniform thickness of 0.21 between the two, uniform as in it is perpendicular to the surface. something like a curve offset. the black function is f(x), i got the red function g(x) by doing a horizontal translation 0.21 to the right but is that mathematically correct, it is around what i'm looking for with uniform thickness (the background is just something I want to model with the functions). I also got this green function z(x) by doing f(x) -0.21 sqrt(1+ f'(x)^2)

[ebon lily]

is there a like a mathematically correct way to get that perfect uniform thickness to f(x) that is simpler than curve offsets, since I need to be able to explain this in my own terms as to how it works

[ebon lily]

ai is telling me both are wrong, and i should do something like "f(x) - 0.21/sqrt(1+(f'(x)^2))" but that is giving me this purple one

[proud perch]

Hint: perpendicular slope means negative reciprocal

[ebon lily]

i was on that track and i got this but then it ended up looking like the purple one, i think i did smth wrong

[ebon lily]

oops the integral isnt supposed to be there in the last image

g(x)2 is just the function itself without the integral

[devout snow]

@ebon lily Has your question been resolved?

[ebon lily]

<@&286206848099549185> sorry for the ping, i have no idea how to get this perpendicular offset as a function, if i were to just use the horizontal translation is that a valid approximation

[heady breach]

can we see the original question or problem you're trying to solve?

[devout snow]

@ebon lily Has your question been resolved?

[ebon lily]

it’s like an investigation, i want to find the volume of an object, and i’m using volume of a solid revolution to do that, for that one polynomial aspect, it has a thickness which i’m finding the volume for. I am assuming uniform volume of 0.21 across the whole segment, so i wanted to establish a parellel curve or offset curve type function that would have uniform thickness of 0.21 from the original f(x), and then i would use the volume of a solid revolution formula to get the volume,

does that make sense

sorry if it’s unclear

in a sense the overall question is just, establishing a function that creates a constant uniform thickness of 0.21 (below) f(x)

uniform thickness as in a parallell curve

[proud perch]

Yeah

Solved it

Can't write it as a function

In general

But you can write it parametrically

Let me know if you want the full derivation

@ebon lily

[ebon lily]

oh shoot tysm

[ebon lily]

yea thta would be helpful

[proud perch]

Do you know parametric relations? This can't be a function in general.

[ebon lily]

i've heard about it

it's like a set of rules for the x and y coordinates?

[proud perch]

Yeah I won't bother with the proof then. I'll just send you the desmos

[ebon lily]

ty!

[proud perch]

https://www.desmos.com/calculator/zbks5nju6v

Just change f

[devout snow]

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[toxic flower]

$$I = \int_{0}^{1} \frac{\ln(\ln(1/x))}{1+x^2} , dx$$

[woven radish]

Shikhar

[toxic flower]

any help?

no idea where to start

<@&286206848099549185>

[spark nymph]

Hi

Have u tried l'hospital? @toxic flower

[vital sedge]

bro

what is this integral

[spark nymph]

Wait fuck I am drunk

So

Wair

[vital sedge]

I'd say take it complex

where did you get this question

[spark nymph]

Yeah that might be the way

Or u could try doing some substitutes

[vital sedge]

trust me won't work here

[spark nymph]

Because u would get ln (-ln x)/(1+x^2)

[vestal prawn]

67

[spark nymph]

Well the function is well defined because between 0 and 1 ln x is better 0 and minus infinity

So I am unsure if

[vital sedge]

it gives an insane value

[spark nymph]

Complex numbers will hemp

Help

[vital sedge]

what the actual fuck

[spark nymph]

Maybe we could do an tan substitution,m

Or sum?

Or an x= e^t to get rid of one ln

?

[spark nymph]

Wait is that the gamma function

That's crazy

[vital sedge]

yes

[spark nymph]

Yeah good luck

[devout snow]

@toxic flower Has your question been resolved?

[toxic flower]

um.. in integral?

<@&286206848099549185>

[toxic flower]

assignment

[vital sedge]

first let u = 1/x

[toxic flower]

hm..

[toxic flower]

got this after simplification

$I = \int_{1}^{\infty} \frac{\ln(\ln(u))}{1+u^2} , du$

[woven radish]

Shikhar

[rotund umbra]

I can make it look nicer using x = e^{-t}

also with cosh(t) = (e^t + e^-t) / 2

[stoic lotus]

You can now use the substitution u = e^t

Hint: || it'll allow you to use hyperbolic functions and arrive to known results ||

[toxic flower]

um.. dont think so

$I = \int_{0}^{\infty} \frac{e^t \ln(t)}{1+e^{2t}} , dt$

[stoic lotus]

What'd you get

[woven radish]

Shikhar

[toxic flower]

any nice form?

[stoic lotus]

Yes. Divide the denominator and the numerator by e^t

[grand edge]

this is the malmsten integral

[toxic flower]

$\frac{1}{2} \int_0^\infty \frac{\ln(t)}{\cosh(t)} dt$

[woven radish]

Shikhar

[toxic flower]

dont think any known result would help

[grand edge]

I'll try and figure out a solution, I'm 99% sure you need to expand as geometric series

Oh you just get laplace transform of ln nice

$-\gamma\cdot\frac{\pi}{4}-\sum_{n=0}^{\infty}\frac{\ln\left(2n+1\right)}{2n+1}\left(-1\right)^{n}$

[woven radish]

Roy

[grand edge]

I wonder if differentiating the dirichlet beta function could help here

You'd have to appeal to already known values though since they're very nasty to derive

[toxic flower]

hm.. that'd help ig

lemme try

[grand edge]

yeah that'd be fairly easy

[toxic flower]

L-function it is?

dirichlet beta. right?

[grand edge]

https://en.wikipedia.org/wiki/Dirichlet_beta_function

[toxic flower]

yea yea ik about this

[grand edge]

ah I didn't know it was called L-function

\begin{align*}
I &= \int_{0}^{\infty}\frac{e^{-x}\ln\left(x\right)}{e^{-2x}+1}dx \
&= \sum_{n=0}^{\infty}\left(-1\right)^{n}\int_{0}^{\infty}e^{-\left(2n+1\right)x}\ln\left(x\right)dx \
&\mathcal{L}\left{\ln\left(x\right)\right}\left(s\right)=-\frac{\gamma+\ln\left(s\right)}{s} && \text{Laplace transform of $\ln(x)$}\
I &= \sum_{n=0}^{\infty}\left(-1\right)^{n}\mathcal{L}\left{\ln\left(x\right)\right}\left(2n+1\right) \
&= -\sum_{n=0}^{\infty}\frac{\gamma+\ln\left(2n+1\right)}{\left(2n+1\right)}\left(-1\right)^{n} \
&= -\gamma\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n+1\right)}-\sum_{n=0}^{\infty}\frac{\ln\left(2n+1\right)}{2n+1}\left(-1\right)^{n} \
&= -\gamma\cdot\frac{\pi}{4}-\sum_{n=0}^{\infty}\frac{\ln\left(2n+1\right)}{2n+1}\left(-1\right)^{n}
\end{align*}

[woven radish]

Roy

[toxic flower]

the second term appears to be $L'(1, \chi_4)$

[grand edge]

If you want to derive the laplace transform for ln(x) you can differentiate the laplace transform of t^n with respect to n

[woven radish]

Shikhar

[grand edge]

😵‍💫

Hmm

[toxic flower]

woa... some nice simplification

[grand edge]

derived here #calculus message

[toxic flower]

$$\frac{\pi}{4} (\gamma + \ln(2\pi)) - \pi \ln \Gamma\left(\frac{3}{4}\right)$$

[woven radish]

Shikhar

[toxic flower]

the log summation simplification

[grand edge]

$$\beta\left(s\right)=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n+1\right)^{s}}$$
$$\beta'\left(s\right)=-\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}\ln\left(2n+1\right)}{\left(2n+1\right)^{s}}$$
$$I = -\gamma\cdot\frac{\pi}{4}+\beta'\left(1\right)$$

[woven radish]

Roy

[toxic flower]

yea yea this thing it is

[grand edge]

Deriving B'(1) is nasty

Wikipedia has the result ${\displaystyle \beta '(1)={\tfrac {\pi }{4}}(\gamma +2\ln 2+3\ln \pi -4\ln \Gamma ({\tfrac {1}{4}}))={\tfrac {\pi }{4}}(\gamma -\ln 2+2\ln {\tfrac {\pi }{\varpi }})}$

[woven radish]

Roy

[grand edge]

Soo ☠️

[toxic flower]

Kummer's Fourier series relates to this

[grand edge]

Ah, I don't know what that is

[toxic flower]

nvm.. that gets us to this result

gamma cancels

[grand edge]

https://www.youtube.com/watch?v=59bTqwaoY-A

I just found this

[toxic flower]

$$-\frac{\pi}{4} \ln(2\pi) + \pi \ln \Gamma\left(\frac{3}{4}\right)$$

[woven radish]

Shikhar

[toxic flower]

hm..

[grand edge]

It may have a more original derivation of B'(1)

[toxic flower]

my prof gave results for some basic l functions

[grand edge]

So you're allowed to use the known values without proving them?

[toxic flower]

dirichlet functions.. that was last chapter

L function were a whole unit

so yes.. can use them

[grand edge]

very cool

saves you a lot of the trouble from deriving them

https://en.wikipedia.org/wiki/Carl_Johan_Malmsten

He evaluated all these integrals in 1842