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[sterile nimbus]

How do I go about proving this?

[sterile nimbus]

So far, all I have is: We have set X = {1, 2, ..., n}, and set Y = {1, 10, 11, 100, 110, 111, ..., 11111... (n times)}. So, I need to prove that there is a function f: X -> Y

Although I have no idea how to use the pigeonhole principle there

<@&286206848099549185>

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@sterile nimbus Has your question been resolved?

[mossy lion]

use the fact that if you have n+1 integers, there is at least one pair which is divisible by n.

now choose a set of n+1 integers such that the difference of any two of them only has digits of 0 or 1.

[mossy lion]

idk if you can just use this or you have to prove it as a lemma for your proof. but to prove it you need the PHP

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[atomic stream]

Anyone know how solve this

[plucky lynx]

Try rewriting the second equation in terms of y = etc, think about why we are doing this. Then ask yourself, we are trying to find an x satisfy the the two equation. Would do you think we should do?

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[ebon sluice]

7x-(6-2x)=3

[ebon sluice]

then plug that into the first equation

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[elder lark]

How would I convert boolean algebra to algebra?
Like assuming all my variables: A B C D are 1's or 0's, what would this expression look like which would give the same output?:

[brazen patrol]

?

given a variable x

sigh

[elder lark]

hmm?

[brazen patrol]

so ANDs are like multiplying and PLUS are like adding

[elder lark]

Yeah, what are nots

[brazen patrol]

the opposite

[elder lark]

what algebraic expression can I use for a not

[brazen patrol]

you don't need one

[elder lark]

But if I wanted one

I'm pretty much trying to convert the boolean algebra into normal algebra so my calculator can use the solve function

if that makes sense

[brazen patrol]

eh, i mean you could always try to invert it i guess

I don't think that works though

not gives you the "inverse" in the boolean algebra world

im not sure if you can convert that into "normal algebra"

[elder lark]

hmm

bizzare

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[lone heart]

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[raw geyser]

How do I solve this using the Alternate Series test?

[raw geyser]

I know it converges

[forest harness]

you want to find the value it converges to?

[raw geyser]

nvm

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[alpine sable]

Let $z_{0}=a+bi$, where $i=\sqrt{-1}$. Let $z_{n}=z_{n-1}^{2}$. If $a^{b^{c}}=a^{bc}$, then $z_{n}=\left(a+bi\right)^{2^{n}}$. Is that correct?

[ocean seal]

3317

[fiery adder]

well what are you trying to show

[alpine sable]

I'm trying to find whether there are some complex numbers that when squared an infinite number of times "converge" to some complex number.

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

Maybe I could use $(a+bi)^{n}=\sum_{k=0}^{n}\frac{k!a^{n-k}\left(bi\right)^{k}}{k!\left(n-k\right)!}$ for that purpose.

[ocean seal]

3317

[alpine sable]

Well screw it all.

.close

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[buoyant kayak]

is this a quiz

can you screenshot the title

what have you tried

what have you tried

for the questions

you asked

about

what have you tried

refer to your notes

attempt something

then go to bed and take a zero on the assignment

because you're not going to attempt it

you thought wrong, perhaps read #rules #❓how-to-get-help #info

<@&268886789983436800>

thanks you too

[tame falcon]

Thanks

.close

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[lone heart]

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[steel turret]

hello, i asked help for this question yesterday but im a little confused abt something

[steel turret]

im clear on the operation for c

but i have no idea how to get the actaul money mrs blake won

(trouble on b)

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@steel turret Has your question been resolved?

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[thorny sinew]

I’m assuming it has something to do with the squeeze theorem

And the fact that the limit as n approached infinity of something in the form (a^n / n!) = 0

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[magic needle]

hey can someone explain this bc

when ilike

simplify it

isplit up the exponent into 1/2 * n

then i get

3^1/2(3^n)

which makes me think

the commonr atio is 3

but

its root three

which is the outside term which is usually 1

[magic needle]

at least thats what i found off the quizlet verified answers for the book

[slender gull]

Do you know geometric progression?

[magic needle]

i have not heard of it so id guess not

im in precalc

we're j doing sums and sequences

[slender gull]

Yes that's a sequence.

[magic needle]

oh wait no i have yea

[slender gull]

Yeah.

[magic needle]

i j looked it up

[slender gull]

Right.

[magic needle]

yea our teacher didnt call it that LOL

[slender gull]

That's fine.

So yeah, do you know how to calculate the sum of terms like

3^1+3^2+3^3...3^n

[magic needle]

yep

[slender gull]

Then what's the problem here?

[magic needle]

when i try to find the ratio myself

i get 3

which i explained in the text belowthe pic

but

apparetnly its root 3

so idk what i did wrong

[slender gull]

That is actually incorrect.

[magic needle]

can i not split itup into 1/2 * n?

[slender gull]

$$(3^{\frac{1}{2}})(3^n) \neq 3^{\frac{n}{2}}$$ if that's what you did.

[ocean seal]

Sakata Yaksha

[magic needle]

that is

why is that like wrong

[slender gull]

Laws of exponents.

$a^m × a^n = a^{m+n}$

[ocean seal]

Sakata Yaksha

[magic needle]

oh wow im dumb

i forgot ab that

[slender gull]

Right, that happens.

[magic needle]

oo okay okay tysm

[slender gull]

You're welcome.

[magic needle]

have a good one

.close

[lone heart]

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[bleak harbor]

Don't even know where to begin with this one

wait...oh ok so I think I got x...?

because x = cos theta + 1/2 (1-cos theta) because that's how the point P is defined right...?

so simplifying that wouldn't it be x = 0.5(1-cos theta)?

[lone heart]

@bleak harbor Has your question been resolved?

[bleak harbor]

apaprently its supposed to be 1+cos theta
but idk why

oh wait did i make an arithmetic error

wow that fixes everything

nvm

.close

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[pastel jasper]

Maybe a little bit out of context

But, is the r^ the usual vector hat?

because it doesn't really make sense to me since pj = m(omega)*r and not the hatted r

The mentioned vectors are probably 3d

Which wouldn't make sense with the original hat

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[pastel jasper]

<@&286206848099549185>

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@pastel jasper Has your question been resolved?

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@pastel jasper Has your question been resolved?

[rustic jungle]

So what's your question

If you're just asking what is r^, it's just the normal vector of r which means a multiple of vector r that has length 1

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@pastel jasper Has your question been resolved?

[pastel jasper]

@rustic jungleNot sure if i understood that, and yes that's waht im asking

[rustic jungle]

So r is a vector with some lenth l

And r^ is r divided by l

So this new vector r/l has length 1

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[brittle sun]

Hey I'm taking a couple of math courses in university, and the material isn't always very clear. I discovered 3blue1brown for Linear Algebra and it's been helping me a ton with conceptualizing things.
I've been wondering if there's a similar channel that also talks about discrete math? I can't find much relating to it on 3blue's channel.

[brittle sun]

Right now I'm struggling with 'Equivalence Relation'

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@brittle sun Has your question been resolved?

[tacit arch]

book recommendation questions go in #book-recommendations

.close

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[brittle sun]

.reopen

[lone heart]

✅

[brittle sun]

I was talking about video channels though, as in visual aid...?

I can elaborate if needs be.

Books aren't very useful to me(Although they can definitely fill a similar role in most cases) as I'm not a native English speaker and 'math talk' is almost completely alien to me. I've found that with videos, I can figure out the subject at hand by visual context

Hence the question

[tacit arch]

help channels aren't meant for video recommendations either. go to #discussion or #chill or just google some

.close

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[alpine sable]

hi how to factor? is there a way to not forget it or something

[next spade]

wdym?

[prime badge]

what

[alpine sable]

ye

like (x+1)(x+1) or sth

[plucky lynx]

It varies depending what the expression is.

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[lone heart]

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[rotund compass]

can somebody help me with this. i have found the equations of area and perimeter for both square and triangle

[rotund compass]

what’s the next step

[royal meadow]

add the areas and find when they're minimised

[rotund compass]

how. i did and got nowhere

given that the side total perimeters of each can’t surpass 10

<@&286206848099549185>

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[rotund compass]

.reopen

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✅

[royal meadow]

what's the total area

@rotund compass

[rotund compass]

That's what I'm trying to find out. I am optimizing the two shapes so that the area is the smallest with a combined perimeters of 10

[royal meadow]

you have the area of the square and the area of the triangle in terms of s, right

what are they

[rotund compass]

Yes

[royal meadow]

what are they

[rotund compass]

Sqrt(3)/4 s

And s^2

[royal meadow]

nope

s should be s^2 in the first one

that's all

well no

[rotund compass]

?

[royal meadow]

ok

so

the side length of the triangle is s

but the side length of the square doesn't have to be s

but it does depend on s

because the perimeters add up to 10

[rotund compass]

Ok makes sense

Yeah

[royal meadow]

so the perimeter of the square is 10 - the perimeter of the triangle = 10 - 3s

so what's the side length of the square now in terms of s

so what's the area of the square from that

[rotund compass]

Perimeter of both is 10

I might just not be following

[royal meadow]

nope

the total perimeter is 10

[rotund compass]

Yes

10 = 3s + 4x

Is the equation on the perimeters

[royal meadow]

sure

so what's x in terms of s

[rotund compass]

(10-3s)/4

[royal meadow]

yes

so what's the area of the square

if that's the side length

[rotund compass]

^2

[royal meadow]

yes

so what's the total area? what's the area of the triangle plus the area of the square?

[rotund compass]

mb s instead of y

[royal meadow]

yeah

,rotate

[royal meadow]

ok so basically now you want to expand that out

and simplify it

and then you can find the minimum however you like

completing the square or calculus or w/e

[rotund compass]

i’ve come to this now do i find the derivative

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@rotund compass Has your question been resolved?

[rotund compass]

<@&286206848099549185>

[rotund compass]

bro what

.close

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[vapid steppe]

is it possible to find the area of this kite

[karmic rapids]

yes

[marsh rapids]

yes

[karmic rapids]

you'll need this: https://en.wikipedia.org/wiki/Geometric_mean_theorem

[vapid steppe]

how

[vapid steppe]

ok how to use that here

[karmic rapids]

you have a two right triangles there

both are equal

so you just gotta find the area of one and multiply it by 2

you first gotta apply the geometric mean theorem

and then the pythagoreans theorem twice

[vapid steppe]

what is volume again

[rotund compass]

inside area

[drowsy lagoon]

area instead of volume?

[karmic rapids]

same thing

[vapid steppe]

ok how to do it

is 13 the altitude?

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[covert nimbus]

help pls

[limpid spade]

I don't see

a quadriluetql

quadrilateral

[covert nimbus]

what

there has to be

[limpid spade]

where is it

[covert nimbus]

idk, thats the question given to me

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[pale arch]

hiii why is it telling me the variable is wrong Lol

[pale arch]

is a NOT what is supposed to be used here???

NEVERMIND I GOT IT

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[covert nimbus]

.reopen

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[alpine sable]

im stuck on how to convert the below to a series form

i suspect I need to adjust the denominator

[hidden heart]

maybe think about the -1^n

when n = 1 d/dx(ln(x)) = 1/x not -1/x

[alpine sable]

oh......

[sudden hinge]

@alpine sable
$$f^{(1)}(x) = \frac{1}{x}, f^{(2)}(x) = -\frac{1}{x^2}, f^{(3)}(x) = \frac{2}{x^3}, f^{(4)}(x) = -\frac{6}{x^4}, \ldots$$

[ocean seal]

Navix

[sudden hinge]

$$\implies f^{(n)} = (-1)^{n+1} \frac{(n-1)!}{x^n}$$

[ocean seal]

Navix

[alpine sable]

Hmm yeah

Thanks!

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[frozen flume]

I'm a little new to this, but I need to find a set A that:
$$Is not\N , every x\in A also means that 2x\in A and finally 1\in A$$
Any help?

[ocean seal]

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

[frozen flume]

Would $$\z$$ work?

[ocean seal]

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

[sudden hinge]

\mathbb{Z}?

[frozen flume]

yes that one

[sudden hinge]

Z should work

[frozen flume]

just making sure

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[lone heart]

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[mint linden]

hey 👋 i need some math help

<@&268886789983436800> <@&286206848099549185>

[sly mantle]

pls dont ping mods for math help

[mint linden]

ok sorry

can you help me tho?

[sly mantle]

is this a test?

[lone heart]

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[candid torrent]

i don't need help finding the first two derivatives

i just don't know what a maclaurin series is and i don't understand it when i google

[placid zinc]

Too bad let's go through them

[candid torrent]

what

[placid zinc]

Jk. So there exists a best possible quadratic that approximates e^(sin(x)) around x = 0

1 sec I'll demonstrate it

[candid torrent]

alright

[glass lichen]

It's the Taylor series @ x=0

[candid torrent]

i don't really know what a taylor series is

is it an approximation of a function?

[placid zinc]

,w graph e^sin(x) and (1/2)x^2 + x + 1

[ocean seal]

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[placid zinc]

There we have it. We can see the blue is a complicated function, and the red is just a quadratic. The red sticks to the blue really well around x = 0

[candid torrent]

I see

[placid zinc]

There's a reason why this works: The first three derivatives of both functions are the same at x = 0

[candid torrent]

wait

i found this

[placid zinc]

Well, I mean f(0), f'(0), f"(0) all match

Exactly. Using this notion, you can create that series right there

[candid torrent]

so i do f(0) = e^sin(0)

[placid zinc]

I calculated f(0), f'(0), f"(0) and plugged em in.

[candid torrent]

oh i see

so can you like

tell me what this is doing

it just approximates the function at x = 0?

is that all?

[placid zinc]

Yes. It's a quadratic approximation.

[candid torrent]

thank you mr kaynex. you have helped me

[placid zinc]

Much like how a single derivative is a linear approximation

[candid torrent]

I see.

i will now close the channel

unless you have more to say

[placid zinc]

Well, there's another important thing. For some functions, including more terms will make the approximation better.

You can calculate sin real easy using this, ect.

[candid torrent]

i see

.close

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[ashen blade]

hi

[ashen blade]

I'm doing a geometric series convergence question and was wondering

so I have some Σ (to infinity, n = 1) of (3^(n+1)) * (4^(-n))
which I simplified to
3 * Σ(3/4)^n

my question is in regards to the values of "a" and "r", with a denoting the first term and r denoting the part within the sum

for the geometric series formula Σar^(n-1)

[raven rover]

a: first term of the sequence
r: common ratio

Yup

[ashen blade]

so my question is, why would a = 3/4 here?

according to the textbook, that's the answer

but isn't this set to be in terms of (3/4)^n?

oh wait, I guess it wouldn't matter?

I was confused because I thought a = the first term of the entire sequence altogether, which was something like 3 * (3/4)^1 = 9/4 but that's wrong

[raven rover]

You start at n = 1, so your first term is (3/4)^1 = 3/4

[ashen blade]

so you don't make use of any constants and only find the first term within the series?

[candid torrent]

this is unrelated but

[ashen blade]

wait and one more thing ...
would 3 * Σ(3/4)^n technically be a worse way of expressing geo series compared to something like (9/4) * Σ(3/4)^n-1?

[candid torrent]

i like your komi san pfp

[ashen blade]

thanks! komi is waifu 👀

[placid zinc]

So indeed, first term a = 9/4

[candid torrent]

ew

komi is like too young man

[placid zinc]

Shen might have missed the 3 multiplier

[candid torrent]

but the anime very wholesome

ok bye

[ashen blade]

later

[placid zinc]

Then r = 3/4

[ashen blade]

so my textbook is wrong then?

https://i.imgur.com/2hYqenw.png

or would you only take the value within the Σ? which is always going to be 3/4

[placid zinc]

,w sum from n = 1 to infinity of 3^(n+1)4^(-n)

[ocean seal]

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[placid zinc]

OH I see what they did

They ignored the 3, then multiplied it back in later

You could include the 3 to make a = 9/4, and that's including it from the start

Same answer.

[ashen blade]

oh .......................

you're right, thank you

my mistake for not making note of that

but why did they have to complicate it ... jeez

that's annoying lol

[placid zinc]

That's a weird way to do it imo

But I can see why it works haha

[ashen blade]

official answers smh

yeah

[placid zinc]

Whatever is your preference

[ashen blade]

thank you for your time!!!

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[near hinge]

word problem

an experiment begins with 200 bacteria that grow at such a rate that the population doubles every 5 hours
write an equation that describes the situation

[near hinge]

im blanking on how to model this

is this a sort of compounding-type problem?

the formula is p(1+(r/n))^nt where t is time, n is times compounded over a period of time, and r is the rate

what would n be?

<@&286206848099549185>

ive got y=200(1+1/n)^5

[still vigil]

This seems more like a differential equation to me rather than an compounding type

[near hinge]

it doubles so i think the rate is just 1 right

[still vigil]

because it can just be modeled with $$ \dv{N}{t} = N $$

[ocean seal]

Toru

[near hinge]

sorry i havent used that equation before

[still vigil]

AH

ok

[near hinge]

formula i have used for these sorts of problems is A = P(1+r/n)^nt

[still vigil]

i think if it were to be compounded then it should be
y=200(1+1/100)^x ?

[near hinge]

P is the principal amount, r is the rate (in percent increase), t is time and n is amount of times compounded in that time

and it doubles every 5 hours

so how would i write that though

if i make my T variable 5 hours, can i just make N = 1 ??

[still vigil]

yeah

that could work

[near hinge]

so when X = 1 on the graph, thats 5 hours?

[still vigil]

yes

got to check that x=0 means y=200, x=1 means y =400

[near hinge]

yes and yes

alright lets see if this works

i have to find when the bacteria is 10,000

so its 5.6438562

so times 5 thats 28.2 hours?\

ok that is right

thanks so much dude you saved me from going to bed at midnight the night before a test

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[cunning trout]

I don’t understand what this means and why it’s the way it is for example a+b=-b/a.
This is what’s avoiding me from doing questions based on this exercise.

[lone heart]

@cunning trout Has your question been resolved?

[cunning trout]

Ok I see

So how would I do question 2 the first part (2 i .a)

[cunning trout]

<@&286206848099549185>

[lone heart]

@cunning trout Has your question been resolved?

[lone heart]

@cunning trout Has your question been resolved?

[ionic jewel]

cant remember the name of the theorem, but its probably related to wherever you got those problems

hmm

its -b/a though

for ax^2 + bx^2 + cx + d

@cunning trout

https://brilliant.org/wiki/cubic-equations/

some random link with info

[cunning trout]

Ok

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[simple mauve]

I have a word "TALLAHASSEE"
L repeated 2 times
S repeated 2 times
E repeated 2 times
A repeated 3 times

how to find the number of ways in which 2A's are adjascent?

[wind hawk]

You mean for any combination of those letters?

[simple mauve]

yeah

[wind hawk]

It might be easier to find all the ways they are adjacent

And then subtract that from all possible combos

[simple mauve]

but then it includes combinations where 2A's are not adjascent

[wind hawk]

Yeah isn’t that what you want

[simple mauve]

oh sorry i want where 2A's are adjascent

[wind hawk]

Ah ok

Well gut instinct is to split it up into when all 3 are next to each other first

And the other letters shouldn’t matter

There’s 8 ways of lining them up as 3 next to each other

Then, if they’re not all 3 next to each other, it’s as if we only have 2 A’s

Repeat the same process, we get 10 ways

18 total

(?)

Like this:
AAAxxxxxxxx,
xAAAxxxxxxx,
…
xxxxxxxAAAx,
xxxxxxxxAAA

[simple mauve]

but if they are not three next to each other, it can be like none of the A's are next to each other + 2A's are next to each other

[wind hawk]

Well if they’re not 3, we can ignore the one on it’s own

[wind hawk]

And then we just do this with 2

Like this:
AAxxxxxxxxx,
xAAcxxxxxxx,
…
xxxxxxxxAAx,
xxxxxxxxxAA

And then for each way

There’s 8/9 letters left

Which can be arranged in their own way

[simple mauve]

and substract where 3 A's are next to each other from this?

[wind hawk]

So multiply that as well

Yes

[wind hawk]

So to calculate this fully:
There’s 10 combinations of AA with some # of combos of the rest of the letters

Multiply the number of combos of the rest of the letter by 10

[wind hawk]

Then do the same with the remaining letters here

And subtract

Does that make sense?

[simple mauve]

hmm let's consider if i take atleast 2 A's are together

so if i take all combinations where two A's are together and multiply the rest as permutations ?

[wind hawk]

Yeah I think so

So how many ways can you arrange:

L repeated 2 times
S repeated 2 times
E repeated 2 times
A repeated 1 time

Then for each position:
AAxxxxxxxxx

[wind hawk]

The xs can be arranged that many ways

[simple mauve]

(10!/2!* 2!* 2!) times arranged? taking AA as one

11 letters total so taking AA as one

[wind hawk]

Hm

I thought of it as theres 9 letters left

And so there’s all the ways those 9 letters can be arranged

[wind hawk]

Times the 10 positions

[wind hawk]

But that might work too(?) I’m not very familiar with combinatorics tbh

[simple mauve]

hm ok

[lone heart]

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[slate quiver]

am trying to find the LCM of two numbers
was wanting to know why there is only 1 5 and not two as it appears in both the prime factor for both numbers, would it not be 5^2 instead of 5^1?

[slate quiver]

I have highlighted the numbers I am addressing in red

[buoyant kayak]

"for each prime factor, find where it occurs most often as a factor and write it that many times in a new list"

[slate quiver]

so if it only occurs once in both data sets it only will be used once in the superset

[buoyant kayak]

yep

[slate quiver]

so the one that has more of that number overides the smaller set

[buoyant kayak]

👍

[slate quiver]

cheers

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[crisp rock]

hi

[crisp rock]

A number is formed by starting with a single digit, doubling it and writing down the unit digit of the product as the next digit. For example, starting with 2 you get 248624...

When starting with 3 what will be the sum of the first 2015 digits?

i calculated the pattern was 39713971...

2015/4 = 503 r 3

[trim plinth]

good

[crisp rock]

then

[trim plinth]

that'll be (3+9+7+1)*503

[crisp rock]

yes

[trim plinth]

+3+9+7

[crisp rock]

• 19

but

that was 10079

the answer said 10071

[trim plinth]

exactly

wait

[crisp rock]

im confused

[trim plinth]

the answer is wrong?

no that's not 3971

362486248624…

[crisp rock]

???!

[trim plinth]

it goes 3624862486…

[limpid spade]

372736274727r7

[trim plinth]

3*2=6

[crisp rock]

isnt it

powers of 3

[trim plinth]

6*2=12 => 2

it says double the digit

[crisp rock]

oh... i guess i didnt understand the problem

[trim plinth]

and write it down

[crisp rock]

thanks

[trim plinth]

is this ur homework

[crisp rock]

ye

thanks

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[trim plinth]

2015-2=2013
2013/4=503 r 1

so…

I guess you'll know

[crisp rock]

ye

[trim plinth]

so it's 3+6+(2+4+8+6)*503+2

which equals 10071

glad to help.

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[crystal tiger]

Show that for a sequence $X_n$ of random variables, $X_n \to X$ in probability if and only if p(X_n, X) $\to$ 0. We have $p(X_n, X)$ defined as follows:$\$ $$p(X,Y) := E{\frac{|Y-X|}{1+|Y-X|}}$$

[ocean seal]

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

[crystal tiger]

and 2nd question:$\$ Let $B_n$ be a countable set of Borel subsets of [0,1]. Suppose that there is $\delta >0$ such that the Lebesgue measure of each $B_n$ is greater than $\delta$. Show that there is at least one point which belongs to infinitely many $B_n$.

[ocean seal]

Neko

[crystal tiger]

please ping me if I don't respond. If I don't come, I will be asleep, but will greatly appreciate a hint in the morning

[alpine sable]

second one is just trivial application of the pigeohole principle

and just use teh definition on the first one

[crystal tiger]

Could you explain 💀

the second one

I think I can do the first one

The point is in the subset

Is this supposed to be a hint?

What is that supposed to mean

Lol

Not a philosopher sorry

and 2nd question:$\$ Let $B_n$ be a countable set of Borel subsets of [0,1]. Suppose that there is $\delta >0$ such that the Lebesgue measure of each $B_n$ is greater than $\delta$. Show that there is at least one point which belongs to infinitely many $B_n$.

[ocean seal]

Neko

[alpine sable]

you can just argue by contradiction here

suppose that there is no such point

what can you infer from that

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[trim plinth]

Prove, for the given set L=[0,1,....,n] for any given number n which is an integer that lies in the interval [0,∞), that, if all values of 2^L[i] for all integral values of i=0,1,…,n, where L[i] denotes the ith term of the list L, was to be made into a list M, then there will always be a unique expression of the sum of some terms in M that will equal to any given integer between the interval [1,2^(n+1)-1].

[trim plinth]

someone help

pls

[lone heart]

@trim plinth Has your question been resolved?

[trim plinth]

AND SO IF THEREFORE SOMEONE CAN HELP HERE?

[limpid spade]

say it louder

[oak perch]

I am not sure I got the meaning Are you asking why any integer between 1 and 2^(n+1)-1 can be expressed uniquely as sum of some elements of {2^0,2^1,2^2,…,2^n}?

This is just 2-adic

First we show uniqueness, if Σa_k2^k=Σb_k2^k, where those a_k,b_k are 0 or 1, then Σc_k2^k=0 where c_k=a_k-b_k.now c_0 must be 0, because c_0=-c_1 2^1-c_2 2^2-…is divisible by 2, but |c_0|<=1. So c_0=0, now c_1=-c_2 2^1-c_3 2^2-…. So similarly c_1=0…. Continue this process all c_k=0 so a_k=b_k.
Next there are 2^(n+1) those different expressions, because Σa_k 2^k you have two choice of a_k, this gives you 2^(n+1) different consecutive numbers from 0 to 2^(n+1)-1 so it covers all of them, all integers from 1 to 2^(n+1)-1

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[alpine sable]

Question 29

[alpine sable]

Please help

I have been trying to figure out this inequality and trying to find it's solution sets

I can unable to do so. Because it's coming in complex roots. And I can't really plot them on the real number line

Adversely. Taking x²+x>-1
And then telling x²+x>=0 is going to be invalid as there are infinite numbers between -1 and 0

So if anyone could help me out a bit

[silver adder]

all x satisfy x^2+x>-1

[alpine sable]

Could just explain it to me

[dense sleet]

In theory yes the lowest point of x^2 + x is -0.25

[alpine sable]

Oh I get it

Thanks

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[lone heart]

@stoic monolith Has your question been resolved?

[dense sleet]

Well for 5x^2 + 8 to be odd 5x^2 must be odd

[jagged imp]

you also posted this in #discrete-math !please don't multipost <3

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[stoic monolith]

.reopen

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✅

[stoic monolith]

so is this correct?

i think this is incomplete

[jagged imp]

It is correct. which part do you think is incomplete

[stoic monolith]

the x^2 part?

[jagged imp]

the fact that x^2 is odd implies x is odd? It could be expanded upon but it seems like the proof writer assumed the reader was already familiar with that fact

[stoic monolith]

can u help me familiarized with it?

like i need adetailed solution please . it will help me understand the topic more

[jagged imp]

I'll prove it by contradiction. Suppose that x^2 is odd and x is even(this is the negation of [x^2 is odd implies x is odd]). since x is even, we can write x=2k for some integer k. Then, x^2=(2k)^2=4k^2=2(2k^2) is even, which contradicts our assumption that x is even

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[alpine sable]

Can someone show me how to get the square root of 294.75

[alpine sable]

I've got till here and now I'm stuck

[wanton hare]

use a calculator

[alpine sable]

We can't use one for our exams so

[wanton hare]

oh

well

then

pray

[alpine sable]

Yeah it's a drag

[wanton hare]

thats a gg

[alpine sable]

But I'm looking for actual answers

[wanton hare]

i normally just

use the long but reliable method

trial and error

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[alpine sable]

Here why we need to do u substitution?

Can't we simply say:

- cos^4(x) * sin(x) + c

[glass lichen]

No

You can differentiate that and see its wrong

[alpine sable]

I see. Let me try

Yes, it will be wrong.

[alpine sable]

But may I know because of the which rule it will be wrong?

[glass lichen]

Cause you didn't integrate properly

As in, did not follow an actual rule for doing integrals

[alpine sable]

I see,

Thank you 🙂

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[viscid jewel]

Hello so Basically if i say
sin^2x
i can lift the square and put it on the entire equation so it becomes

[viscid jewel]

so the question is

If its sinx^2 instead

can it still become this:

[gray isle]

no

the notation is suboptimal but $\sin x^2$ is interpretted as $\sin(x^2)$

[ocean seal]

ℝamonov

[viscid jewel]

thank you so much @gray isle, life savor

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[tranquil kestrel]

may i know if imy solution is right?

[oak perch]

Do it by definition…

When you are done I can check whether your answer is correct

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[alpine sable]

the answer is 4, because (n-1)^2. however I just watched short youtube video[less than 3 minutes] saying the degree of freedom is n-1. so my question is why sometimes its n-1 and sometimes it could be (n-1) * (n-1) ?.

[alpine sable]

I understand why n-1 is right, and why (n-1)^2 is right, but I found this a bit confusing

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@alpine sable Has your question been resolved?

[alpine sable]

<@&286206848099549185>

[tranquil kestrel]

Yes my answer is A, is it right?

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@alpine sable Has your question been resolved?

[oak perch]

Yes

[solemn grove]

degrees of freedom is like the sample dimensions right? so for linear things, you know it’s n-1, that’s the number of spaces, between the data and here it’s (n-1)^2 cuz two dimensions or something. Idk, I only know the /(n-1)

[raw fractal]

I think it depends on the test. A test for independence (with the contigency table) has (n - 1)² degrees of freedom. A test for goodness of fit would have (n - 1) degrees of freedom.

[alpine sable]

Rylo, PhenomPlasma. Thanks for helping me

!close

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[craggy valve]

Hello
So if I have a 4 boxes and want to pick two balls from it. What is the combination of 2 balls with different number?

Box 1 (1,2,3)
Box2 (1,2)
Box3 (1,2,3)
Box4 (1,2,3,4)

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@craggy valve Has your question been resolved?

[fluid eagle]

wdym by 'what is the combination'

[craggy valve]

@fluid eagle
What is all of the possible outcome that two balls are different?

[craggy valve]

<@&286206848099549185>

Help _:(´ཀ`」 ∠):

[remote heron]

you have to list them?

@craggy valve

or count

[craggy valve]

Just count is okay

[remote heron]

i would say just do exhaustive

if it were me and youre really not sure

[craggy valve]

Okay I will look it up

[remote heron]

like box 1, we draw a 1

then we can get:

well, can you drow again from the same box?

is that allowed?

[craggy valve]

No

So it is combination without repetition?

I just studied about this for two days. Forgive me if I am wrong 😂

[remote heron]

its not 100% clear

but yea, i think so

like draw one ball, hold on to it

then pick another box and draw from it?

[craggy valve]

Yes!

[remote heron]

so yea, i think you could just count em out

like box 1, ball 1

we can draw a 2 from any other box

or a 3 from box 3 and 4

or a 4 from box 4

so 6

then check 2 in the same way

then 3 the same way

now the catch is that when you are done checking box 1 in this way, you move on to box 2

but you cannot check aganist box 1 at all now

all the combinations with box 2 and box 1 are accounted for

so you just check against box 3 and 4

then with box 3, only against box 4

[craggy valve]

Thanks a lot!

I get the image now!

[remote heron]

[craggy valve]

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[cedar vault]

alr so

i integrated dy/dx to get y = ((kx - 1)^6)/3k + c

from there i put the x and y values to get

1/3k + c = 1

or c = 1 - 1/3k = (3k - 1)/3k

and then when x = 1

((k - 1)^6)/3k + c = 8

=> ((k - 1)^6)/3k + (3k - 1)/3k = 8

=> (k - 1)^6 + 3k - 1 = 24k

and im lost here

how do i even solve a 6th degree polynomial

we would need to solve this algebraically btw

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@cedar vault Has your question been resolved?

[hardy gulch]

maybe the minimum point of the curve will tell you something

[tacit arch]

i guess you're supposed to be able to look at this and see that 3 is a solution for k

,calc 2^6 - 21 * 3 -1

[ocean seal]

Result:

0

[tacit arch]

and also k=0 is another solution

once you divide through by k, you still have a fifth order polynomial, so it's still hard to find why k=3 is a solution other than by inspection

might be easier to see if you let $t = k-1$ so that $t^6 - 21 (t+1) - 1 = t^6 - 21t - 22 = 0$ ?

[ocean seal]

riemann

[tacit arch]

,calc 2^6 - 21 * 2 - 22

[ocean seal]

Result:

0

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[alpine sable]

is there a simple way to explain what the first symbol means?
someone told me to calculate an are under a function

not sure i get it

It's basically that it's just the symbol for integration

[buoyant kayak]

what a nice function

[drowsy lagoon]

the symbol is integration symbol

[alpine sable]

2 numbers next to it are threshold and limit of the range of the area which the symbol combined with function represent?

[buoyant kayak]

that is a very confusing sentence to read

[glass lichen]

Refer to the definition of a definite integral

[alpine sable]

okay i'm gonna check on yt

.close

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[calm wave]

How to explain

[alpine sable]

Solve for the turning point and y intercept ig

[calm wave]

.close

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[fervent cloak]

Why cant I cancel this

[buoyant kayak]

test?

[tight locust]

Division distributes over sums

Because division is the reciprocal multiplication

[buoyant kayak]

bro

[tight locust]

This is not emphasized enough in algebra classes

Too many people I work with fail to see the connection between x/3 and 1/3 x

Just as an example

Another thing I see them struggle with is the idea that 8 = 8/1. They don't like 1 in the denominator

And they also struggle with the idea that the sum or difference of multiples of a number is also divisible by that number. I gotta put together a manifesto on this shit haha

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[humble cairn]

Does this type of table have a specific name?

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@humble cairn Has your question been resolved?

[mild mulch]

I think it's just called a "matrix table"
https://www.ibm.com/docs/en/tnpm/1.4.3?topic=layout-matrix-table-format

[fluid basin]

i'd call it a data table personaly

[mild mulch]

I just call it a table as that's the only sort of table I work with

[fluid basin]

although i'm confused at what this table represents

[mild mulch]

It might be what the temperature feels like? (very mathematical, not subjective)

[fluid basin]

could be meterological theta

theta is a measure of heat that combines both temperature and the heat of vaporization of water vapor present as humidity, but it's almost always expressed as an equivalent temperature in kelvin and those are definitely not in kelvin

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[jovial mesa]

So far: Let g_1,g_2 be in N_G(H) then g_1Hg_1^(-1) = H and same with g_2

[remote heron]

someone will help but fyi #groups-rings-fields is active

[jovial mesa]

I wanted to set them equal to each other but i got stuck, unless I am supposed to do something else

[placid zinc]

You can go ahead here, you've got my interest at least lol

[jovial mesa]

Oh okay cool

thank you

[jovial mesa]

this is all i have so far

[placid zinc]

So you are proving closure here? You'll definitely need it, but there are easier group axioms to prove

[jovial mesa]

Proving closure yes

[placid zinc]

Well okay. So you've got the conditions on g1 and g2 correct.

What does this condition do to g1g2?

Remember that, for ' as inverse:
(g1g2)' = g2'g1'

[jovial mesa]

it's also in N_G(H)?

[placid zinc]

Well it had better be, lol

But why is it?

[jovial mesa]

wait, i was starting closure under operation

so i think there is where i want to show that g1g2 is in N_G(H)?

[placid zinc]

Yus. If that's true, what condition is imposed on g1g2?

Is this condition true, simply because of the condition imposed on g1 and g2?

[jovial mesa]

I'm lost

Um, so you're asking if g1,g2 are in the subgroup, then g1g2 is also in there?

[placid zinc]

Okay. So if g1g2 is in N_G(H), then:
g1g2H(g1g2)' = H

[jovial mesa]

Isn't that what I have to show?

[placid zinc]

Right. You want to show this is true

[jovial mesa]

cant i set g1Hg1' = g2Hg2'?

[placid zinc]

So that's like saying all H = H

Remember that H here is a placeholder for "any element of H"

[jovial mesa]

Oh okay, so I am not able to do that?

[placid zinc]

So g1Hg1' and g2Hg2' are not the same element, but do both belong to set subgroup H

So no, since they are not equal, you can't set them equal

Or, wait, I guess the H being in both sides makes this fine

Yeah actually that can work

Sick let's go with this then!
g1Hg1' = g2Hg2'

[jovial mesa]

okay bet

[placid zinc]

Can you carry out some algebra to turn this into
g1g2H(g1g2)' = H?

[jovial mesa]

could i multiply the right side by g2

g1Hg1'g2 = g2H

g2'g1Hg1'g2 = H but idk if that works since we don't know if it's abelian

[placid zinc]

Oh, huh, weird that gives the wrong order lol

[jovial mesa]

yeah

wait!

omg

H = gHg'

could we use that?

but i think that's not useful in this case

[placid zinc]

The problem is that you really do want to write the form exactly:
g1g2H(g1g2)' = H

So I propose starting with this instead:
g1Hg1' = g2'Hg2

[jovial mesa]

oh right

[placid zinc]

Eh. I don't love that either

I guess you could argue that inverses exist first, and then this works better

[jovial mesa]

g1Hg1'g2' = g2'H
g2g1Hg1'g2' = H

wait

uh

so

if they both equal H

[naive valley]

you are trying to show that if g1 and g2 normalize H, then so does g1g2?

and ' means inverse here?

[jovial mesa]

g1Hg1' = g1(g2Hg2')g1'

since H = g2Hg2'

🤯

[placid zinc]

That's much more elegant, yes

[jovial mesa]

it works yes?

[placid zinc]

Finally realizing that g1Hg1' = H finishes it

g1(g2Hg2')g1' = H
g1g2H(g1g2)' = H
g1g2 is in the normalizer when g1 and g2 are

[jovial mesa]

well if i could get to (g1g2)H(g1g2)'

then that would equal H

therefore, g1g2 is in N_G(H)?

[naive valley]

that's the whole point, to show that (g1g2)H(g1g2)' = H

which you just did

[jovial mesa]

yes

[naive valley]

that shows that N_G(H) is closed under multiplication

two more steps to show it's a subgroup

then you have to show it's a normal subgroup

[jovial mesa]

one more step yeah? closure under inverses

[naive valley]

one more step beyond that, you have to show it's nonempty

that's easy, though - show it contains the identity

[jovial mesa]

I don't think i gotta do it since we never really did that in our homework. My professor just said closure under operation and inverses

[naive valley]

your prof is wrong then 😆

[jovial mesa]

most likely

[naive valley]

the empty set is closed under multiplication and inverses, but it's not a subgroup

[jovial mesa]

oh nvm

its cuz

he said

if we show a (lets say H in group G) H is closed under operation, using inverses we can get the identity

[naive valley]

only if you assume H contains at least one element

[jovial mesa]

I guess that's why we never did that step

[naive valley]

then yeah, closure under inverses gives you the identity

[placid zinc]

No you need to define an identity to then define what an inverse is

"Why is 1/2 the inverse of 2"
"Well because 1/2 × 2 = 1"
"Wait what's 1 got to do with it?"
"I don't know"

[naive valley]

well the 1 is already defined in G

[placid zinc]

As the identity?

[jovial mesa]

if you wanna see an example we did

[naive valley]

yes, G is a group, it has an identity

[jovial mesa]

from our textbook

[placid zinc]

That's not what I said lol

[naive valley]

H will also have the identity if it contains an element and is closed under inverses

[naive valley]

paste the definition of subgroup

[jovial mesa]

Okay

and this

[naive valley]

"for S is nonempty" (in the second screenshot) - they seem to be tacitly assuming this even though they didn't state it in the first part

[jovial mesa]

so if you have an element and the inverse, then you have an identity

[naive valley]

the reason this matters is that it's easy to (advertently or inadvertently) define some condition that no elements of G actually satisfy, yet is closed under addition and inverses

people have actually made similar mistakes in group theory, showing that "if G is a group satisfying such and such property, then here are a bunch of consequences" when it turns out that there is no group satisfying that property 😆

[jovial mesa]

yeah lol

I guess to save us time, it's to show closure under operation and inv

because i know some do identity, operation and then inverses