help-49

[quick creek]

yeah

[formal rune]

perfecto

[quick creek]

!done

[midnight plank]

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

[formal rune]

.close

[midnight plank]

Channel closed

Closed by @formal rune

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[small jasper]

.close

[midnight plank]

Channel closed

Closed by @small jasper

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[twilit field]

problem 11

[twilit field]

So I've proven LH for one iteration, but idt that's enough here, is it

[nova yoke]

try working straight from the definition of derivative?

[twilit field]

the limit is only iterated once, which is my problem

$

[nova yoke]

can you apply LH once and then work with the definition of derivative?

[twilit field]

Well, not the way I proved LH, but I'll think

Like The way I prived LH was

$\lim_{h \to 0} \frac {f(h)/h}{g(h)/h}= f'(h)/g'(h) by defn

[tawdry kraken]

Interesting

Yeah the limit being written only once is annoying

[nova yoke]

^ this seems to work if my scratchwork is correct

mostly just need to argue why you can apply LH

[twilit field]

I think the bigger question is arguing why using LH doesn't kill the limit if it's still of the form 0/0

[nova yoke]

wdym by kill

[twilit field]

the operator

[nova yoke]

well what's the statement of LH?

the limit is still there in the conclusion...

[twilit field]

Well I was going off question 7

[nova yoke]

how did you prove question 7, did you use LH?

[twilit field]

well, f(x)=g(x)=0

[nova yoke]

(i don't think it necessarily applies, does it?)

[twilit field]

so I divided the num and denom by t

and then used the quotient rule along with the defn of derivative

[nova yoke]

ok fair enough

maybe this can be applied to your problem, but just using LH seems more straightforward

(if you're allowed to)

[twilit field]

I'll have to see, this is for a RA class 😭

[nova yoke]

what's theorem 5.13, mentioned in the question?

[twilit field]

LH

l'hopita;'s lol

[nova yoke]

yea there you go then

[twilit field]

fair enough, we didn't prove it in class though, so lemme prove it

thanks!

[nova yoke]

yw

[twilit field]

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[scenic wharf]

My proof for (2)

The theroem 3.2.2 states that for two convergent sequences (xn) and (yn) if (xn) has limit x and (yn) has limit y , then (xn+yn) has limit x+y

[midnight plank]

@scenic wharf Has your question been resolved?

[ionic crane]

looks good

[scenic wharf]

Okay thanks

.close

[midnight plank]

Channel closed

Closed by @scenic wharf

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vivid yoke]

Consider triangle ABC, where the lengths of the sides are AB = 3, AC = 4, and BC = 5, respectively. Construct circles using AB, AC, and BC as diameters. Let (C) denote the circle that is internally tangent to these circles, ensuring that each of them is completely contained within (C). Compute the radius of circle (C)

[flat spire]

adventure time

[vivid yoke]

I even have adventure time tag next to my name , I already watched it

[flat spire]

ohhh

[midnight plank]

@vivid yoke Has your question been resolved?

[midnight plank]

@vivid yoke Has your question been resolved?

[vivid yoke]

Average Geo help ticket

[midnight plank]

@vivid yoke Has your question been resolved?

[raw vector]

What have you tried?

[vivid yoke]

eh, nothing

[raw vector]

💀

Try using coordinates

[surreal charm]

eh is it like the tangent point and centre of C, centre of circle is collinear

[raw vector]

I see an easy solution with coordinates rn

[vivid yoke]

I'm lwk trying to finish this problem as soon as possible, the chance of a 2D geo appear on my test is less than getting a jackpot so I don't even care if I understand the solution or not

[raw vector]

Let A be the origin, find the coordinates of all the points in the pic

[vivid yoke]

k

I can do that

what's next

[raw vector]

Let O(x,y) be the center of (C)

You have OD=R-R_(C)

Similarly you get a system of equations

Done

[vivid yoke]

how? OD is unknown, so is R

Similarly OE and OF

That would be 3 equations for 4 variables

[raw vector]

You can calculate OD in terms of x,y

It's 3 variables only

[vivid yoke]

huh?

don't get it

hey can you briefly show me how you make the system?

I might have to go to sleep, I don't feel good

I'll leave this ticket open

[quick creek]

so let A be the origin and the center of the big circle R(x,y)

the three small circles have a radius of 1.5, 2, 2.5 each

use the distance formula and solve it by combining

[raw vector]

Do you know how to calculate the distance between two points?

[vivid yoke]

Ah

Okay

.close

[midnight plank]

Channel closed

Closed by @vivid yoke

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[scenic wharf]

I had to prove ,
The sequence (-1)^n is divergent.

And my idea is to prove it by contradiction , let x be the limit of sequence (xn):= (-1)^n

And considering
ε= |1-x| /2

[scenic wharf]

Then 1 does not belong to ε neighbourhood of x

[lyric charm]

what if x was 1

[noble forge]

Exactly. Why don't you just use the Cauchy criterion?

[scenic wharf]

Haven't came across it yet

[noble forge]

Then use a fixed value for epsilon

[scenic wharf]

ε>0

[noble forge]

Not dependent on x

[scenic wharf]

Why does dependency on x cause problem?

[noble forge]

The definition of a limit requires epsilon > 0

By picking a constant epsilon you ensure your proof holds up for any x

[lyric charm]

doesn't answer my question

when x = 1, your epsilon is zero as defined here

[scenic wharf]

Understood

I was trying to form a smaller interval

[lyric charm]

i would suggest you actually write out in quantifier logic what it means to say $\lim x_n \neq x$

[grand pond]

Ann

[lyric charm]

with full quantifiers properly

no halfbakes, no silent shuffling of order, proper

[noble forge]

You can pick an arbitrarily small epsilon. I don't quite get why you'd want to define it like that

[scenic wharf]

∀ ε>0 ∃ k(ε) such that ∀n≥k , the terms xn satisfy|xn-x|<ε

[lyric charm]

incorrect

do you know how to negate statements with quantifiers?

[scenic wharf]

I wanted to define a ε neighbourhood such that 1 is outside of ε nbd of limit x

[scenic wharf]

I don't know

[lyric charm]

when negating a statement with a quantifier, you swap the type of each quantifier (every ∀ becomes ∃ and vice versa) and then negate the statement buried inside them all.

$\lim x_n = x$ means: $$(\forall \ep>0)(\exists k)(\forall n \geq k)[|x_n - x| < \ep]$$
$\lim x_n \neq x$ means: $$(\exists \ep>0)(\forall k)(\exists n \geq k)[|x_n-x| \geq \ep]$$

[grand pond]

Ann

[scenic wharf]

That's new for me

[lyric charm]

this is propositional logic

or wait no

not propositional, first-order

[noble forge]

Do you intuite what it says?

[scenic wharf]

Yes ,

I have to find some ε

Above every k, some n doesnt satisfy that limit criteria

Thank u

.close

[midnight plank]

Channel closed

Closed by @scenic wharf

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[hushed glen]

Need some help with LaTeX, trying to create pictures like this. I got no clue what I'm doing, any help?

[runic hamlet]

we have #latex-help but it would probably be easier for you to make the images in some other software and then include the png/pdf/...

[cerulean oyster]

As Denascite mentioned, you can import images to your projects.
Or, if you want to learn the actual method to do it in LaTeX, check on TikZ

[supple sigil]

Go ask in #latex-help

Also this is not a bad figure to use tikz for because it's mostly programmatically placed simple geometric shapes

[midnight plank]

@hushed glen Has your question been resolved?

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[small warren]

Can I please get help with this geometry problem?

[small warren]

I've drawn a little diagram so far to try and understand the problem better.

If I draw an altitude from the tip of the big triangle to the midpoint of its base, then that line segment must equal 1.75 meters, correct?

[supple crypt]

yes till the red line

[topaz sage]

yes

[small warren]

Yea

[supple crypt]

r u allowed to use calculator

[small warren]

Yep

[nimble sequoia]

I‘d use similar triangles

[small warren]

Thanks, in the video I was watching that was given as a hint

So that's likely what I should do, I don't know how exactly though

I suppose that the altitude of the smaller triangle to the bigger triangle is 1:1.75

[supple crypt]

2-0.25

[small warren]

And that 2x:big-base has to be the same ratio

[small warren]

1.75

[supple crypt]

did u find the ratio

[small warren]

Nope

I suppose the ratio is 1:1.75

Right?

[supple crypt]

yes

[nimble sequoia]

Actually I‘d recommend for you to male the ligne wirh 1m standing above it till the left Corner and then give the Points of the triangles names

[small warren]

so 2x:y is 1:1.75?

if y is the big base?

[nimble sequoia]

If I‘m not mistaken then you could also just use Thales‘

[small warren]

I haven't learned Thales' theorem, what's that?

[small warren]

Standing above what?

[nimble sequoia]

Pretty much just a ratio for similar triangles in specific situations (more like to see better that they‘re similar)

[small warren]

ah okay

[nimble sequoia]

I mean the black ligne x lengths over the base

[small warren]

I still don't understand what you mean, sorry. Can you draw it maybe?

[nimble sequoia]

But if you really haven‘t learnt it, maybe with ratio like you wanted to do would be easier

[small warren]

1.75:1 is really nice

[nimble sequoia]

I meant that one

[small warren]

That's 2 meters long

Will that help?

[supple crypt]

@small warren label the diagrom first

[small warren]

Okay

[nimble sequoia]

Yesh maybe that‘s a better way, normally there are many solutions

[nimble sequoia]

Yeah wanted to work with these ratios with help of Thales‘ but the way with ratio is maybe better, I just don‘t see a good way for it

[small warren]

Is this good?

[supple crypt]

yea and mark the point between GC as H

[small warren]

Okay

[supple crypt]

sorry mb by between i meant on the red line

[nimble sequoia]

I‘d do BF/BG = FD/AG <=> 1/1.75 =x/(0.75-x)

[supple crypt]

dont say

[small warren]

Like this?

[supple crypt]

no on the same line as E but in the line GC

[small warren]

So construct line GE and make point H its midpoint?

I don't understand

sorry

[supple crypt]

np

i mean GH=x

H lies on GC

[small warren]

It does equal x yeah

Oh

is it 0.25 meters away from G?

So just rotate H about G by 90 degrees?

And make H' be H?

[sour solstice]

I have an idea to solve this problem.

Maybe we can use coord geometry

[small warren]

I haven't learned coord geometry yet, this should be able to be solved without that

Sorry

[small warren]

Ah ok!

[sour solstice]

Ah ok

[small warren]

How do I know what HC's measurement is?

[supple crypt]

so do u know similarity

[small warren]

Is it known?

[small warren]

Yes

[supple crypt]

u need X right

[small warren]

For what?

[supple crypt]

DF is also X

[small warren]

Yes

[supple crypt]

to answer the question

[small warren]

Yea

Absolutely

[supple crypt]

then use similarity to find ratio in the triangle AGB and DFB

[small warren]

Is the ratio 1.75:1?

[supple crypt]

yes

[small warren]

i see

[supple crypt]

but also find DF/ AG

[small warren]

DF:AG = 1:1.75?

[supple crypt]

1 tip for similarity if u want to find the larger side make it so the ratio is more than one but if u want to find the smaller side make it less than 1/(eg. X is smaller side so the ratio should be 1/1.75 to make it easy.

[supple crypt]

yea but find DF and AG

[small warren]

I see! Ty

[supple crypt]

DF= X

AG = ...

find AG

[small warren]

x:AG=1:1.75

AG=1.75x

right?

[supple crypt]

yea but u can find AG in terms of X by doing AH-GH

[small warren]

So 1.75-x

[supple crypt]

this wouldnt help to find X although it is technically correct

[small warren]

i see

[supple crypt]

0.75-x

AH is 1-0.25 right

[small warren]

yeah 0.75 my bad

Because the height of the rectangle is 1 meter

[supple crypt]

yes

thenDF/AG=BF/BG

[small warren]

sorry i mean width

[supple crypt]

height is 1m doesnt matter tho

we know BF/BG u told it

[small warren]

Yep

[small warren]

Yeah

[supple crypt]

tell?

[small warren]

1:1.75

[supple crypt]

yes, we also found AG

[small warren]

AG is AC-GC

[supple crypt]

yes but we dont know GC he know GH

we just used AG-GH

[small warren]

GH is x

[supple crypt]

and AG?

[small warren]

1.75x

<@&268886789983436800>

[sour solstice]

My solution to this problem is seeing the ratio of the vertical distance traveled to the total distance. as we can see after quick calc, we see the total vertical distance being 3/4 m and the horizontal distance being 11/4 m. Now, with similar triangles, we see the ratio of the lower, smaller triangle to the upper one is 1:1.75, as said by @supple crypt , now, the merged ratio is 2.75. Now you guys are super duper close, so should I help out a bit to finish the problem?

[supple crypt]

yes but that doesnt help to find X, we used 0.75 remember

[small warren]

You can solve it for yourself if you'd like, but I think I'm close to solving it with the help of @supple crypt , and thank you all

[sour solstice]

got it

I'll put solution in spoilers

[small warren]

Sorry I don't understand

I kind of like half understand

I dont know

[supple crypt]

we foung AG. AH-GH

AH is 0.75

gh is X

So AG is 0.75-X

[small warren]

That makes sense yeah

Perfect so far

!noans

[midnight plank]

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

[supple crypt]

soX/0.75-X=1/1.75

[sour solstice]

I meant to put it in spoilers to hide it...

[small warren]

i dont think thats an appropriate factoid, i used it because i dont think you should be saying that here

[sour solstice]

sorry

[small warren]

GH/AH=BF/FG?

you're good @sour solstice

[supple crypt]

DF/AG=BF/BG

[sour solstice]

tysm.

[small warren]

Ah i see

[sour solstice]

also @supple crypt ur rly close

[small warren]

i'm the one needing help here

[sour solstice]

oops

[small warren]

np

[sour solstice]

but the equation he put up there...

[small warren]

No that was me

[supple crypt]

tbh he could help better

[sour solstice]

I can't bro, I get too inclined and give the answer😭

[small warren]

$1.75x=0.75-x?$

[grand pond]

Aurora

[supple crypt]

just tell if i am going correct even i dont trust myself

[sour solstice]

you are

[sour solstice]

this is good

[small warren]

2.75x=0.75

x=0.75/2.75

[sour solstice]

now just simplify!

[small warren]

3/4 /

how do you convert 2.75 to a fraction?

[sour solstice]

alright 2.75 = 11* 0.25 -->

fuck off <@&268886789983436800>

[small warren]

@supple crypt Thank you very much for all the help! You were really nice and made sure I understood everything

<@&268886789983436800>

[supple crypt]

<@&268886789983436800>

[small warren]

Yep

so 11/4?

[sour solstice]

yes

[small warren]

or 44/4?

okay

[sour solstice]

so now (3/4)/(11/4)

[small warren]

Which is (3/4)*(4/11)

[sour solstice]

yep

[small warren]

So 12/44

[supple crypt]

11X/4

[small warren]

Divide both sides by 2, 6/22

[sour solstice]

and then

[small warren]

What's that

[sour solstice]

6/22 --> ?

[small warren]

3/11

And we're done!

x=3/11

[sour solstice]

and that's correct

[small warren]

if everything follows logically

which i don't see why it wouldn't

[sour solstice]

using coord geometry got me 3/11 too

[supple crypt]

can u check @small warren

[small warren]

From the video?

[supple crypt]

pls xplain i also want to learn

[small warren]

Also, there's a coordinate geometry unit in Khan Academy

[sour solstice]

coord geometry is a pretty good thing

[supple crypt]

the answer

[sour solstice]

oh ok

u want my full explanation?

[small warren]

I fast forwarded, the answer is in fact 3/11!

[supple crypt]

what curriculum dou follow

[small warren]

Thank you all so very much

[small warren]

Khan Academy

[sour solstice]

epik

[supple crypt]

no like in school

[small warren]

This was a problem on their website, he said to pause the video and try to figure it out, but I was stuck

[small warren]

Algebra 1, Geometry, Algebra 2, pre-calculus

[supple crypt]

u didnt learn coordinate geometry?

[small warren]

Maybe I'm not sure

[supple crypt]

lol

[small warren]

XD

[supple crypt]

btw what is pre calculus

[sour solstice]

coord geometry is very simple, should I copy paste notes (one paragraph) of coord geometry?

[supple crypt]

and what is algebra 1 and 2

[small warren]

it's just like more algebra 2 and trigonometry in preparation for calculus, sometimes it also introduces basic calculus concepts like derivatives and stuff like that

[supple crypt]

wow

[sour solstice]

its not the worst

[small warren]

algebra 1 usually has quadratic functions and stuff like that, while algebra 2 has more advanced stuff like dividing polynomials and logarithms and imaginary numbers

[sour solstice]

Its simple enough

[small warren]

i'm going to close this channel now if thats ok

[sour solstice]

got it

[supple crypt]

u have imaginary numbers and logs

k

bye

[small warren]

Ok bye everyone tysm

[small warren]

Yep!

.close

[midnight plank]

Channel closed

Closed by @small warren

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[zealous schooner]

How do I find the radius of convergence of the power series?

Without complex methods

[fallow scarab]

did you try ratio test

[zealous schooner]

What do I do with the Bernoulli numbers bit

Like I don't even know where the original series converges

[fallow scarab]

either find their exact form or find a few common ratios for small n and develop a hypothesis then prove it with induction

[zealous schooner]

I know there's simple poles at +- 2pi so the original series should converge for |z| < 2pi

but like

I can't use that

[zealous schooner]

hmm

I don't follow

[fallow scarab]

$B_4 / B_2 = ?, B_6 / B_4 = ?$

[grand pond]

riemann

[zealous schooner]

-1/5, -5/7, -7/5, -25/11, ...

doesn't seem to be much pattern there

[fallow scarab]

maybe it's just sufficient to use alternating series test then

do you know an asymptotic formula for B_n

[zealous schooner]

Well I know of one, but the method I know of to derive it uses complex methods

[fallow scarab]

try for a softer bound then like B_{2n} <= n^a for some a

alternatively, i suppose depending on interpretation of the problem, you just use that if the power series converges to the function, then the power series must converge where ever the function is defined and you just use the branch that includes x=0

[midnight plank]

@zealous schooner Has your question been resolved?

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[rain wasp]

how did i do here?

[midnight plank]

@rain wasp Has your question been resolved?

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fickle sierra]

Hello, could someone check if this proof looks good please?

\begin{Theorem}
Suppose $A$ and $B$ are sets. Then, $A \subseteq B$ if and only if $A \cap B = A$.
\end{Theorem}

\begin{proof}
~
\paragraph{$\Rightarrow$}
Let $x \in A$.
Because $A \subseteq B$, we have $x \in B$.
By the definition of the intersection, $A \cap B = \{x : x \in A \text{ and } x \in B\}$.
Because every $x \in A$ is also in $B$, this means that $x \in A \text{ and } x \in B$ is true.
Thus, $A \cap B$ contains every element of $A$.
Therefore, $A \cap B = A$.

\paragraph{$\Leftarrow$}
Let $A \cap B = A$ and $x \in A \cap B$.
Recall that $A \cap B = \{x : x \in A \text{ and } x \in B\}$.
So, $x \in A$ and $x \in B$.
$x \in A \cap B$ implies $x \in B$, therefore $A \cap B = A \subseteq B$.
\end{proof}

[grand pond]

Mor Bras

[scenic wyvern]

(forward direction) when showing equality of sets, you need to argue both A \subseteq B and B \subseteq A. here you asserted that A \cap B \subseteq A, but you have not shown that A \subseteq (A \cap B).

(reverse direction) it might be better to let your arbitrary element start in A rather than A \cap B. as written you're using x in (A \cap B) to show that x is in B, which is always true by defn and thus redundant.

[fickle sierra]

I see, let me rewrite this

[bold peak]

Cuz this is slightly unclear, you wanna show that x ∈ A → x ∈ B, you have started with x ∈ A ∩ B → x ∈ B, which is true by definition as Hana put it

The changes necessary are quite minimal though, since you're starting with A = A ∩ B

[midnight plank]

@fickle sierra Has your question been resolved?

[fickle sierra]

Yes, I'm trying to rewrite the only if part so that x in A implies x in B

Here's the proof:

\begin{proof}
~
\paragraph{$\Rightarrow$}
Let $x \in A \cap B$.
Then, by the definition of the intersection, $x \in A$ and $x \in B$.
$x \in A \cap B$ implies $x \in A$, hence $A \cap B \subseteq A$.

Now, let $x \in A$.
Because $A \subseteq B$, we have $x \in B$.
By the definition of the intersection, $A \cap B = \{x : x \in A \text{ and } x \in B\}$.
Then $x \in A \cap B$.
$x \in A$ implies $x \in A \cap B$, hence $A \subseteq A \cap B$.

We have that $A \cap B \subseteq A$ and $A \subseteq A \cap B$, therefore $A \cap B = A$.

\paragraph{$\Leftarrow$}
Let $A \cap B = A$ and $x \in A$.
Since $A \cap B = A$, we have $x \in A \cap B$.
Recall that $A \cap B = \{x : x \in A \text{ and } x \in B\}$.
So, $x \in A$ and $x \in B$.
$x \in A$ implies $x \in B$, therefore $A \subseteq B$.
\end{proof}

[grand pond]

Mor Bras

[bold peak]

Yup that looks good

[fickle sierra]

Thank you very much, and thanks Hanako too!

.close

[midnight plank]

Channel closed

Closed by @fickle sierra

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[scenic wharf]

To prove (g) , my idea :-

I took :
lim(p(t))=lim(ak t^k + ak-1 t^(k-1) +... )

Then i distributed limit as:
lim(ak t^k )+ lim (ak-1 t^(k-1))+....

Then i took the constant out .
ak lim(t^k) + ak-1 lim(t^(k-1)) + ...

And finally used :
lim(a^k)= (lim(a))^k

[trail sphinx]

polynomials over the real numbers are (________) functions?

[scenic wharf]

Continuous

[trail sphinx]

yess

recall the sequential charaterization of continuity.

[scenic wharf]

Haven't studied continuity in analysis yer

[trail sphinx]

Take a any seqeuence that converges to x in the domain

[trail sphinx]

the sequence of images converges to the image of the limit.

[scenic wharf]

I have only studied limit theorems

[trail sphinx]

This looks fine, but you haven’t argued any step

[scenic wharf]

It was just an idea

An approach

[scenic wharf]

Could you suggest, some arguments

[trail sphinx]

The last step is what one you should really argue

you pushed the limits inside the argument

which is only possible for continous functions

[scenic wharf]

I know only these

And without the notion of continuity

[trail sphinx]

Have you already proven all the properties listed?

[scenic wharf]

Yes using induction

[trail sphinx]

You don’t need continuity, you just have to prove the properties

[trail sphinx]

suggestions

1. Write the complete polynomial, don’t put …
2. write where each letters belongs to. For example, a_k in R, k in N. (Otherwise things might not work)

3.first step is basically addition law of limits (but see this through induction)

4. a similar law
5. One might need to argue seperately, if not already done.

ah right last one, you might use the product rule repeatedly k times

[scenic wharf]

What do we mean by arguing

A claim ?

[trail sphinx]

Your proof looks fine, write where each letters belong to, and write the limit laws

[scenic wharf]

Ok thank u

[trail sphinx]

The whole point of this example is just see induction in action

[scenic wharf]

.close

[midnight plank]

Channel closed

Closed by @scenic wharf

Use .reopen if this was a mistake.

[trail sphinx]

Looks correct

Maybe one thing I like to add is,

the way you should look at algebraic limit theorem is not that it only gives some nice properties.

But it also gaurantees the existence of the limits. For examples an-> a , bn -> a.

We might say lim an + bn = lim an + lim bn

(The theorem also guarantees those limit exists (a fixed real number) each time you apply these propertie

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[untold sentinel]

$$\sum_{n=0}^{k}\frac{\operatorname{nCr}\left(k+1,n\right)\left(k-n+1\right)^{2}}{\left(k-n+1\right)^{2}+n^{2}}=2^{k}$$

[grand pond]

zeta theta beta eta

[untold sentinel]

1

and $k \in \mathbb{Z}^{+}$

[grand pond]

zeta theta beta eta

[midnight plank]

@untold sentinel Has your question been resolved?

[silent dock]

glow Write $m = k+1$ and define:
[
T_n = binom mn fr{(m-n)^2}{(m-n)^2+n^2}.
]
Consider $T_n+T_{m-n}$.

[grand pond]

Oléagineux Distilliànus VIVII

[untold sentinel]

whats the motivation behind those two steps

[silent dock]

the first step is just to simplify the expression

the second step is motivated by binom{m}{n}=binom{m}{m-n}

[flat spire]

rainbowlud

[untold sentinel]

ok

so its just like kings rule

\begin{align*}
I &= \int_{0}^{\infty}e^{\cos\left(x\right)+e^{-x}-x}\cos\left(\sin\left(x\right)\right)dx \
&= \mathfrak{Re}\left(\int_{0}^{\infty}e^{\cos\left(x\right)+e^{-x}-x}e^{i\sin\left(x\right)}dx\right) \
&= \mathfrak{Re}\left(\int_{0}^{\infty}e^{e^{ix}+e^{-x}}e^{-x}dx\right) \
&= \mathfrak{Re}\left(\int_{0}^{\infty}\sum_{n=0}^{\infty}\frac{e^{ixn}}{n!}\sum_{m=0}^{\infty}\frac{e^{-mx}}{m!}e^{-x}dx\right) \
&= \mathfrak{Re}\left(\int_{0}^{\infty}\sum_{n=0}^{\infty}\frac{e^{ixn}}{n!}\sum_{m=0}^{\infty}\frac{e^{-x\left(m+1\right)}}{m!}dx\right) \
&= \mathfrak{Re}\left(\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{1}{n!m!}\int_{0}^{\infty}e^{-x\left(m+1-in\right)}dx\right) \
&= \mathfrak{Re}\left(\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{1}{n!m!\left(m+1-in\right)}\right) \
&0\le n\le k\le\infty,\quad k=n+m,\quad m=k-n \
I &= \mathfrak{Re}\left(\sum_{k=0}^{\infty}\sum_{n=0}^{k}\frac{1}{n!\left(k-n\right)!\left(k-n+1-in\right)}\right) \
&= \mathfrak{Re}\left(\sum_{k=0}^{\infty}\sum_{n=0}^{k}\frac{\operatorname{nCr}\left(k,n\right)}{\left(k-n+1-in\right)k!}\right) \
&= \left(\sum_{k=0}^{\infty}\frac{1}{k!}\sum_{n=0}^{k}\frac{\operatorname{nCr}\left(k,n\right)\left(\left(k-n+1\right)+in\right)}{\left(\left(k-n+1\right)-in\right)\left(\left(k-n+1\right)+in\right)}\right) \
&= \mathfrak{Re}\left(\sum_{k=0}^{\infty}\frac{1}{k!}\sum_{n=0}^{k}\frac{\operatorname{nCr}\left(k,n\right)\left(\left(k-n+1\right)+in\right)}{\left(\left(k-n+1\right)^{2}+n^{2}\right)}\right) \
&= \sum_{k=0}^{\infty}\sum_{n=0}^{k}\frac{\operatorname{nCr}\left(k,n\right)\left(k-n+1\right)}{\left(\left(k-n+1\right)^{2}+n^{2}\right)k!} \
&\operatorname{nCr}\left(k,n\right)\left(k+1\right)=\frac{k!\left(k+1\right)}{n!\left(k-n\right)!} \
&= \frac{\left(k+1\right)!\left(k-n+1\right)}{n!\left(k-n+1\right)!} \
&= \operatorname{nCr}\left(k+1,n\right)\left(k-n+1\right) \
\therefore ; &\operatorname{nCr}\left(k,n\right)\left(k-n+1\right)=\frac{\operatorname{nCr}\left(k+1,n\right)\left(k-n+1\right)^{2}}{\left(k+1\right)} \
I &= \sum_{k=0}^{\infty}\frac{1}{k!}\sum_{n=0}^{k}\frac{\operatorname{nCr}\left(k+1,n\right)\left(k-n+1\right)^{2}}{\left(k+1\right)\left(\left(k-n+1\right)^{2}+n^{2}\right)} \
&= \sum_{k=0}^{\infty}\frac{1}{\left(k+1\right)!}\sum_{n=0}^{k}\frac{\operatorname{nCr}\left(k+1,n\right)\left(k-n+1\right)^{2}}{\left(k-n+1\right)^{2}+n^{2}} \
&= \sum_{k=0}^{\infty}\frac{1}{\left(k+1\right)!}\sum_{n=0}^{k+1}\frac{\operatorname{nCr}\left(k+1,n\right)\left(k-n+1\right)^{2}}{\left(k-n+1\right)^{2}+n^{2}} \
&= \sum_{k=0}^{\infty}\frac{1}{\left(k+1\right)!}\sum_{n=0}^{k+1}\frac{\operatorname{nCr}\left(k+1,n\right)n^{2}}{\left(k-n+1\right)^{2}+n^{2}} \
&\underset{\text{Averaging}}{=} \frac{1}{2}\sum_{k=0}^{\infty}\frac{2^{k+1}}{\left(k+1\right)!} \
&= \frac{1}{2}\sum_{k=1}^{\infty}\frac{2^{k}}{k!} \
&= \frac{1}{2}\left(\sum_{k=0}^{\infty}\frac{2^{k}}{k!}-1\right) \
&= \frac{e^{2}-1}{2}
\end{align*}

[grand pond]

zeta theta beta eta

[untold sentinel]

does this look right

[untold sentinel]

<@&286206848099549185>

[midnight plank]

@untold sentinel Has your question been resolved?

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[untold sentinel]

does this look right

[next kayak]

Yes?

[untold sentinel]

read

does my working look right/can i improve on efficiency

[midnight plank]

@untold sentinel Has your question been resolved?

[bold peak]

Imma be real with you mate, no one is reading all that

[midnight plank]

@untold sentinel Has your question been resolved?

[midnight plank]

@untold sentinel Has your question been resolved?

[midnight plank]

@untold sentinel Has your question been resolved?

[ashen sage]

🤓

[small jasper]

The preferred notation for the binomial coefficient is $\binom{n}{r}$. Besides this, other than justifying why you can swap the sum and integral (which might be deemed trivial tbf), this is fine.

[grand pond]

Civil Service Pigeon

[midnight plank]

Channel closed

Closed by @untold sentinel

Use .reopen if this was a mistake.

[untold sentinel]

thx

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[novel flax]

if both a1x+b1y=c1 and a2x+b2y=c2 pass through point P, then the line a1x+b1y+ k(a2x+b2y)=c1+kc2 will also pass through the point P but will have a different slope

[novel flax]

can someone explain why this is true?

[fallow scarab]

did you try calculating the slope of the new line

and different slope than what?

[novel flax]

the slope would be -(a1+ka2)/(b1+kb2)?

[novel flax]

uhh

[fallow scarab]

?

feel free to think more about your question

[novel flax]

oh can i think of it like P is the sol for both the lines

and P is also the soln for the system of these 2 eqns

so if i plug in P, i basically get LHS=RHS or smtg like that?

[surreal charm]

This is equation for family of concurrent lines

L1 + kL2 =0 represents lines that pass through the intersection of line L1 and L2

[novel flax]

ohh

right right

got it thanks

.close

[midnight plank]

Channel closed

Closed by @novel flax

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[coral belfry]

Hi. This is an example of my own, to see if I understand a certain part of combinations right. So assume we have 6 different types of spheres, and we wanna pick 3 spheres at random. The order we pick them in does not matter, and we can pick a sphere of the same type up to 3 times if we want. The formula to calculate how many different ways there are to pick them should be C(6-1+3 , 3) , right?

[noble forge]

Yes

[fallow scarab]

you need more information

[coral belfry]

oh?

[fallow scarab]

how many spheres of each 6 different types do you have

[lusty python]

i think it's just $6^3$

[grand pond]

1 divided by 0 equals Infinity

[coral belfry]

unspecified amount

[lusty python]

no need for $\binom{n}{k}$

[grand pond]

1 divided by 0 equals Infinity

[fallow scarab]

then your "pick 3 spheres at random" isn't defined

[supple sigil]

they said there are at least 3 of each type

[coral belfry]

but we pick them in any order, and we dont place them in discernible places

[lusty python]

-# (with each type consisting only 1 sphere)

[supple sigil]

which is enough for this question

[surreal charm]

We have few spheres of 6 different types, we choose 3 at random such that max 3 of same type?

[lusty python]

i didn't see he said that

[coral belfry]

everyone has a different opinion lol

[coral belfry]

yeah

[junior flower]

[supple sigil]

we can pick a sphere of the same type up to 3 times if we want.

[lusty python]

they only said you can only pick up the same type 3 times

[coral belfry]

mhm mhm

[lusty python]

that does not implies there are at least 3 spheres

[coral belfry]

wha

up to 3 of the same type

[supple sigil]

?

[lusty python]

you could put the sphere back and pick it up the same type

[coral belfry]

i basically assumed that there are at least 3 spheres of each type

[noble forge]

They said you can pick any type up to 3 times, so it shouldn't matter

[coral belfry]

it doesnt matter if there are more, since we arent doing probabilities

[supple sigil]

OK but that is an irrelevant difference

[surreal charm]

So what we want is non negative integral sols for x1+x2+..x6 =3 such that xi <=3(doesn't matter)

[coral belfry]

HAHAH WHAT

it should be a simple combinatorial exercise

[junior flower]

no

bluds...

[surreal charm]

Eh idk

[coral belfry]

blud gathering

[hard umbra]

truly a blathering

[junior flower]

anyway your answer is correct, though your question is a little poorly worded

[coral belfry]

what would be a better phrasing?

[junior flower]

how many 3-combinations are there of the multiset {1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6}

[coral belfry]

hmm i see

thats a way to avoid ambiguity for sure

[supple sigil]

your original phrasing is fine IMO

[coral belfry]

thanks Plant

[junior flower]

why are you guys ing me

[surreal charm]

How is that different from what i said

[coral belfry]

im guessing bc the exercises are meant to resemble real life scenarios

[coral belfry]

whatd you say lemme see

[junior flower]

you are considering order too much

[coral belfry]

oh scoobs the integral guy

[surreal charm]

Uh i just did stars and bars

[noble forge]

These multisets can be infinite for all it's worth

So I think their wording is sufficient

[junior flower]

i didn't want to explain what \infty \cdot 1 meant

[coral belfry]

yeah i shouldve prolly specified that theres at least 3 spheres in each type even tho its implied

no objections?

[surreal charm]

-# slayla tell me my mistake after thhis done

[junior flower]

oh i misread what you wrote. ok sure i agree with that

[coral belfry]

alright thank you everyone

.close

[midnight plank]

Channel closed

Closed by @coral belfry

Use .reopen if this was a mistake.

[junior flower]

i also thought you were the guy who said 6^3 so i was not giving you much benefit of the doubt

my bad

[silent dock]

you're welcome fijo

[junior flower]

thank you pure

[hard umbra]

oure

[silent dock]

always happy to help

[silent dock]

you're seeing tings

[junior flower]

snow go away

[hard umbra]

tings

[junior flower]

this is a cop free zone

[hard umbra]

cree zone

[silent dock]

crone

[lusty python]

CFZ

(C-F)Z

[lusty python]

in literature, we respect order

in math, uhm

-# we don't for multiplication and addition

[lusty python]

i can name it ZFC if i wanted to

[junior flower]

[supple sigil]

blud son is here

[silent dock]

[supple sigil]

what is causing you agony my son

[silent dock]

so i just saw lotr the first one

[junior flower]

no bf is causing me agony

[silent dock]

and i left the movies thinking i can't wait to see the second and the third films on big screen

then i realised it’s only the first film’s 25th anniversary, so they’re only showing the first film

[lusty python]

trick shot

[supple sigil]

I know you're a patient blud

palud

[junior flower]

[silent dock]

[lusty python]

what is this creepy ahhhh 😭

[silent dock]

you cannot say that to plantmojis

[junior flower]

repent now

[silent dock]

repow

[lusty python]

-# i repent that i didn't say that earlier

[junior flower]

you're gonna regret this

[coral belfry]

wtf is happening

[lusty python]

-# i regretted didn't say the emoji was creepy

[silent dock]

is it tuff to quote bludself for the 15 year olds

[lusty python]

😭

[junior flower]

when the channel closes and the pure clique is here, anything can happen

[silent dock]

having a good and civilised conversation

[lusty python]

-# i don't regret talking to a nitro booster

[coral belfry]

thanks for the help cooly youre so helpful

[silent dock]

truly all capes not heroes wear

[hard umbra]

the plique

[junior flower]

i hope one day i can find a combinatorics bf

[hard umbra]

benerating function

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[silent dock]

.combriend

[junior flower]

.waiting for some blud to accidentally open this again

[coral belfry]

Hi. Suppose we have 6 different types of spheres. Name each type as 1,2,3,4,5,6. Assuming we have 2 spheres of type 1, 3 spheres of type 2, and the rest types have just 1 sphere each, how can we evaluate how many ways there are to choose 3 spheres at random and at once, without putting them in any particular order after picking them up?

[lyric charm]

are you doing this for probability purposes

[coral belfry]

no just combinatorics

[lyric charm]

ok so just counting how many multisubsets of size 3 the multiset {1,1,2,2,2,3,4,5,6} has

[coral belfry]

honestly idk how to do that

using a formula i mean

[lyric charm]

p sure you can't avoid finicky casework.
3 twos
2 twos + 1 something else
1 two + 2 ones
1 two + 1 one + 1 something else
etc.

[coral belfry]

oh i see

can we not use the choose function and divide by something?

that was my suspicion but i couldnt think of what to divide it by

[lyric charm]

not that i can think of

the overcount factor wouldn't be consistent

[coral belfry]

alright thank you

.close

[midnight plank]

Channel closed

Closed by @coral belfry

Use .reopen if this was a mistake.

[shell wigeon]

Tbf there are only four cases here

[coral belfry]

i was looking for a general formula

[lyric charm]

think you're SOL on that

[shell wigeon]

Yea there is none

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[midnight plank]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[kind ledge]

if log(a) 3 = 0,47, log(a) 5 = 0,69 and log(a)20 = 1,28 determine the value of each expression

[kind ledge]

log(a) 15

[slender walrus]

what have you tried?

[kind ledge]

well i tried first to do 15/10 because my calculator does log10

[slender walrus]

this is a non calculator question

[surreal moon]

(sorta)

[slender walrus]

also you're not given that a = 10

[kind ledge]

idk i didnt try anything

[slender walrus]

do you know your log laws?

[kind ledge]

yes

[slender walrus]

specifically product to sum

[kind ledge]

like log(c) (m x n) = log(c) m + log(c) n

[slender walrus]

yes

[kind ledge]

what about that law

[slender walrus]

apply that

[kind ledge]

log(a) 15

where are the 2 products

[slender walrus]

15 is a product of which two positive integers?

[kind ledge]

7.5 x 2

[slender walrus]

technically 15= 7.5 * 2 and you could apply laws based on that, but that doesn't help you much here
consider what information you're given

reminder: you're given info about

log(a) 3
log(a) 5

[kind ledge]

oh we got to find a

with that

hmm

[slender walrus]

don't think about finding a

think of applying

log(c) (m x n) = log(c) m + log(c) n
to log(a) 15
that involves
log(a) 3
log(a) 5

[kind ledge]

idk my teacher hasnt really taught this to us specifically

[slender walrus]

try not to overthink

you know the required law

all you need to do is apply it

[kind ledge]

oh so like 3 x 5

[slender walrus]

yes, 15 = 3 * 5

now apply the law accordingly

[kind ledge]

log(a) 15 = log(a) 3 + log(a) 5

[slender walrus]

yes

[kind ledge]

wait but if i have log(a) 20

what do i do with that

its for bigger numbrs?

[slender walrus]

nothing for this part

[kind ledge]

what if it as log(a) 75

instead of 15

[slender walrus]

similar idea

[kind ledge]

its a new law tho

right

[slender walrus]

no

same law

(don't need the 20 here yet either)

[kind ledge]

alright

well i cant do 75 = 5x3

[slender walrus]

try a product of more than 2 numbers

[kind ledge]

5x3x5

[slender walrus]

idea is to multiply and/or divide based on the values to you're given to get the desired number
that works

[kind ledge]

wait how does it work cause we dont have 2 log(a) 5

its only 1

.close

[midnight plank]

Channel closed

Closed by @kind ledge

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[midnight plank]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[buoyant linden]

just want to double check something

[buoyant linden]

is it true taht

$\neg (\neg (P \implies Q)) = \neg P \lor Q$

[grand pond]

Flatus

[buoyant linden]

also what symbol should I use for imaginary? I've seen

[pine thicket]

P implies Q is equivalent to not P or Q, so yes

[buoyant linden]

$i\mbb{R}, \mbb{I}, \mbb{M}$

[pine thicket]

and you have a double negative, so those disappear

[grand pond]

Flatus

[buoyant linden]

wdym

[buoyant linden]

like in this case, do the negations cancel each other out

or do i just

apply it in the order

[pine thicket]

not(not(P)=P, thats a double negative

[buoyant linden]

to get not P or Q

[buoyant linden]

but

[pine thicket]

i shouldve used a different variable

[buoyant linden]

$\neg (P\implies Q) = P \land \neg Q$

[grand pond]

Flatus

[pine thicket]

sure you can use de morgans

[buoyant linden]

wait so is this true?

[buoyant linden]

oh

i see

i didn't realise P implies Q is equivalent to not P or Q 😭

i see

.close

[midnight plank]

Channel closed

Closed by @buoyant linden

Use .reopen if this was a mistake.

[simple cove]

um

[pine thicket]

you can check with a truth table

[simple cove]

where can i get math help

chatgpt isnt working for this one problem

[buoyant linden]

just go to another math channel

to start it

like #help-7｜zen1thxyz

that one's unoccupied

[simple cove]

like what?

um

im confused

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fickle sierra]

Hello, could someone check if this proof looks good please?

\begin{Lemma}
Suppose $A$ and $B$ are sets. Then, $A \subseteq B$ if and only if $A \cap B = A$.
\end{Lemma}

\begin{Theorem}
Suppose $C$ is a set. Then, there exists a unique $A \in \powerset{C}$ such that, 
for every $B \in \powerset{C}$, $A \cap B = B$.
\end{Theorem}

\begin{proof}
Let $A \in \powerset{C}$.
Either $A = C$ or not.
Suppose $A \neq C$.
If $B = C$, then there is a $x \in B$ that is not in $A$.
So, $A \cap B \neq B$.
Now suppose $A = C$.
For any $B \in \powerset{C}$, we have that $B \subseteq A$.
By lemma 1, this means that $A \cap B = B$.
Therefore, $A = C$ is the element of $\powerset{C}$ such that for every $B \in \powerset{C}$, $A \cup B = B$
\end{proof}

[lyric charm]

did you type the thm correctly

you seem to be mixing up \cup vs. \cap

[grand pond]

Mor Bras

[midnight plank]

@fickle sierra Has your question been resolved?

[fickle sierra]

Yes, it was a typo, now it's fixed

[fickle sierra]

.close

[midnight plank]

Channel closed

Closed by @fickle sierra

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[surreal charm]

Arrangement of numbers from 1 to n such that no two numbers are consecutive

[dreamy lichen]

https://oeis.org/A002464

that oeis sequence is the solution, there doesnt seem to be any particularly nice expression

there is a recursive one though

[midnight plank]

@surreal charm Has your question been resolved?

[surreal charm]

Ty

[dreamy lichen]

Wait you want no two consecutive numbers to be consecutive?

so like 5 6 isnt allowed but 6 5 is?

[surreal charm]

does that make a difference

[dreamy lichen]

well it does

[dreamy lichen]

this is where no 2 consecutive numbers are adjacent, which is a different thing. Both 5 6 and 6 5 would be disallowed

[surreal charm]

Ok so

How do we do

If it's the other case

[dreamy lichen]

Maybe we could try inclusion exclusion

[raven zinc]

https://tenor.com/view/pokemon-card-gif-5994659789676629325

https://tenor.com/view/heres-my-card-gif-2339690228405306031

[dreamy lichen]

Stay on-topic here, for off-topic, go to #chill

[dreamy lichen]

||Starting with all arrangements, we get n!.
Now we need to count arrangements where 1 pair is wrong, there are (n-1 choose 1) ways to pick the first element of the consecutive pair, then we treat the consecutive pair like a single unit and together with the remaining n-2 numbers, there will be n-1 units to permute. So the total number of such arrangements will be (n-1 choose 1) * (n-1)!...

For 2 pairs wrong, it should be (n-1 choose 2) * (n-2)! etc..||

[surreal charm]

oh

an expression

[dreamy lichen]

i doubt it will have a closed form tho

[surreal charm]

hm..

right

[dreamy lichen]

hmm... this thing aint working for some reason (nvm it works, i just counted 2,3,1 as valid for some reason)

fails at n=3 already

[surreal charm]

given n integers from 1 to n, how many ways choose k integers that none of those chosen integers are consecutive.

i miread the question

[dreamy lichen]

oh, now this is different

i think this was done via stars & bars or sth like that? But i cant recall the exact method

[surreal charm]

hm,,

[vivid yoke]

Let a_n be the numbers of way
Say the last number n can't be right next on the right of number n-1, take a_{n-1} we have the total n gaps to put the number n-1 ( the only invalid case is next to n-1 on the right ) , but n-2, n, n-1 case is still valid and hasn't been included in the a_{n-1} case, we can count this case by take a_{n-2}, then we let n and n-1 be on the right of n-2
Hence a_n= (n-1)a_{n-1} + a_{n-2}

Ehhhh

[dreamy lichen]

is this for the current problem or for the previous one?

[vivid yoke]

Ig this's the way, tho it isn't linear

Oh the one he said 56 disallow but 65 isn't

[dreamy lichen]

apparently the question was just misread

so now we just choose k integers s.t. none of them are consecutive

[surreal charm]

hm..

[surreal charm]

i take that for prev

[vivid yoke]

oh that's less interesting

[surreal charm]

nah it was confusion

[dreamy lichen]

if u just look it up, you should be able to find the sol. But I dont wanna spoil it for myself

[surreal charm]

uh if i'll post it

[dreamy lichen]

"number of ways to choose k non-consecutive intgers out of n integers"

[surreal charm]

this counts for consecutive numbers adjacent to each other?

https://math.stackexchange.com/a/1347532

this is what you are referring

[dreamy lichen]

yeah, u should prolly be able to find some answers on MSE

[dreamy lichen]

somehow our results differ at n = 4

[surreal charm]

ok i understand that

hm.. let's get back to the fucked one

[dreamy lichen]

1 3 2 4
1 4 3 2
2 1 4 3
2 4 1 3
2 4 3 1
3 1 4 2
3 2 1 4
3 2 4 1
4 1 3 2
4 2 1 3
4 3 2 1

This suggests 11, which is what I got via inclusion-exclusion

[dreamy lichen]

this gives me 10

seems like it slightly undercounts? Or my list above is wrong

[surreal charm]

idk what you mean tbf

an represents what

[dreamy lichen]

python confirms this

$\sum_{k=0}^{n-1}\left(-1\right)^{k}\operatorname{nCr}\left(n-1,k\right)\cdot\left(n-k\right)!$

[grand pond]

MathIsAlwaysRight

[dreamy lichen]

this formula seems to work for low n

it should be an = (n-1)a_(n-1) * (n-2) a_(n-2)

[surreal charm]

this is for cases n(n-1) is allwed but not (n-1)n?

[vivid yoke]

Maybe?

Imma check my working later, I'm having dinner atm

[dreamy lichen]

maybe u just missed this (n-2) factor

if its added, it works

[surreal charm]

would you mind explaining later coz i dont understand

[vivid yoke]

Sure

[dreamy lichen]

Btw by some rearrangement, we get

$\sum_{k=0}^{n-1}\left(-1\right)^{k}\operatorname{nCr}\left(n-1,k\right)\cdot\left(n-k\right)!=\sum_{k=0}^{n-1}\left(-1\right)^{k}\cdot\frac{\left(n-1\right)!}{k!\left(n-k-1\right)!}\left(n-k\right)!=\sum_{k=0}^{n-1}\left(-1\right)^{k}\cdot\frac{\left(n-1\right)!\left(n-k\right)}{k!}=\left(n-1\right)!\cdot\sum_{k=0}^{n-1}\left(-1\right)^{k}\cdot\frac{\left(n-k\right)}{k!}$

[grand pond]

MathIsAlwaysRight

[dreamy lichen]

$\left(n-1\right)!\cdot n\cdot\sum_{k=0}^{n-1}\left(-1\right)^{k}\cdot\frac{\left(n-k\right)}{n\cdot k!}=n!\cdot\sum_{k=0}^{n-1}\left(-1\right)^{k}\cdot\left(\frac{1}{k!}-\frac{k}{n\cdot k!}\right)=n!\cdot\left(\sum_{k=0}^{n-1}\left(-1\right)^{k}\cdot\frac{1}{k!}\right)-\left(\sum_{k=0}^{n-1}\left(-1\right)^{k}\cdot\frac{1}{n\cdot\left(k-1\right)!}\right)$

[grand pond]

MathIsAlwaysRight

[dreamy lichen]

now the first summation is expansion of e^-1

the second summation is just an error term which tends to 0 as n gets large

so asymptotically, it's gonna be n! * e^-1

to be a bit more exact, the error term approaches e^-1 / n

so $n!\cdot\left(e^{-1}+\frac{e^{-1}}{n}\right)$ should be a decent estimate

[grand pond]

MathIsAlwaysRight

[dreamy lichen]

dang its way more decent than i expected

at n=1, the error is .25 at n=2, the error is .1 at n=3, the error is .05

[dreamy lichen]

rounding this gives the exact answer

(n! + (n-1)!) / e

[vivid yoke]

You're right

I know what went wrong

[vivid yoke]

My worst nightmare, complex expression

So, I'll adjust my argument a bit

[surreal charm]

fr

[vivid yoke]

i,n,i+1 wouldn't be included in the a_{n-1} case

So we group i,i+1 as one

Hence the numbers of way to arrange like what the problem ask for but in this case there is exactly 1 invalid pair i,i+1 for n-1 numbers is a_{n-2}

There're in total, n-2 pairs ig?

That makes it (n-2)a_{n-2}

is that right

[surreal charm]

now

xplain

pwease

[vivid yoke]

Okay so

[dreamy lichen]

ah, i think i get what u mean now, seems right

[vivid yoke]

Let a_n be the number of ways to arrange

[surreal charm]

to arrange n numbers?

[vivid yoke]

Yes

1,2,...,n

So, take a_{n-1} if we put n in an arrange, there're in total n gaps

But one of them is invalif

Namely the gap next to n-1

But that hasn't covered all the cases

i,n,i+1 would also be valid, but we can not get this case because there's no two consecutive numbers in a_{n-1} for us to put n in between them

[surreal charm]

ok

[vivid yoke]

To count these cases, as I said, we group i,i+1 as one number

Now instead of n numbers we have n-2 numbers ( 1 of them is a pair of numbers )

So there're a_{n-2} ways, and there're (n-2) consecutive pairs

[surreal charm]

oh

[vivid yoke]

Total (n-2)a_{n-2}

Add to the previous (n-1)a_{n-1}

That's it really

[surreal charm]

oh

ty

[surreal charm]

so your recurrence matches inclusion exclusion

[vivid yoke]

np

[surreal charm]

that was smooth

.close

[midnight plank]

Channel closed

Closed by @surreal charm

Use .reopen if this was a mistake.

[vivid yoke]

I don't like excl-incl

Cuzz the algebra step

[dreamy lichen]

i like it because I can do asymptotic analysis on it :]

there was no algebra really

[dreamy lichen]

u could end up here and itd be fine

[vivid yoke]

Big expression scares me

[surreal charm]

i haven't seen both until today

🥀

[dreamy lichen]

it scares me when someone else writes big expressions

[vivid yoke]

Yeah, but I don't think JEE has that much counting

So it doesn't matter much

[surreal charm]

yh it doesn't

[vivid yoke]

My exam have a lots counting

[surreal charm]

i justed stumbled across this

[vivid yoke]

That sucks for preparing

[surreal charm]

oh

vn entrance is tuff

[vivid yoke]

not as tuff as JEE

Or gaokao

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fringe crag]

Hi guys, I'm writing on this channel because I don't really know where to write here. I'm a physics student and I'm struggling with math formal proofs. Where Can I ask for help?

[fringe crag]

.close

[midnight plank]

Channel closed

Closed by @fringe crag

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vague wigeon]

i got p as -1

cuz the LHD of |x - 1| will be -x +1

then i tried to diff g(x) cuz its 0/0 form

but im kinda stuck

i am not able to get n and m values

i tried to diff it and also try to apply expansions

[surreal charm]

!show

[midnight plank]

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

[vague wigeon]

its rough but it didnt work

[surreal charm]

Yh but what did you get

[vague wigeon]

by doing that i got -n/m(2x-2)

where x tends to 0

is the value of p right?

[lethal owl]

Hmm what did you get after trying l'hopital

[vague wigeon]

n(x-1)^ n-1

in the numerator

[lethal owl]

Ok

[vague wigeon]

and in deno -mtan(x-1)

[lethal owl]

Use the sin(x-1) over cos(x-1) better

[vague wigeon]

then i did sinx/x when x tends to 0 is one

so i divide n multioly my x-1

[lethal owl]

So you should get at the end
$$\frac{n}{m} \frac{x-1}{\sin(x-1)} (x-1)^{n-2} \cos(x-1)$$

[grand pond]

Sherif Player

[lethal owl]

What do you think of the value of n if the limit is non zero

[vague wigeon]

shouldnt there be a -ve

[lethal owl]

Yeah

[vague wigeon]

wait

[vague wigeon]

im not sure

[lethal owl]

$$-\frac{n}{m} \frac{x-1}{\sin(x-1)} (x-1)^{n-2} \cos(x-1)$$

[grand pond]

Sherif Player

[vague wigeon]

oh

[lethal owl]

n should be 2 to cancel the x-1 term

[vague wigeon]

n/m should be able to cancel the entire term

so that only 1 remains

is that it?

[vague wigeon]

what

oh

i see

[lethal owl]

Limit as x goes to +1 for (x-1)/sin(x-1) is 1
And for cos(x-1) is also 1

Then the only thing remaining is for (x-1)^{n-2} to be defined and non zero so we know that n = 2

You can see now what is the value of m

[vague wigeon]

yea

both are 2

[lethal owl]

Yes

[vague wigeon]

can we use

nvm this way better

tysm 🙏

[lethal owl]

You are welcome

[vague wigeon]

.close

[midnight plank]

Channel closed

Closed by @vague wigeon

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[cloud bronze]

Are there restrictions every single time u do polynomial divison?

[wet torrent]

x+2 cannot be equal to zero because then youre dividing something by 0

[runic hamlet]

if you divide by e.g. x^2+1 then you dont have to worry about dividing by zero

[cloud bronze]

Ok