help-38

[tawny escarp]

yea but that 100 can only give 1 99, not 2 😭

[trim jolt]

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[trim lichen]

.reopen

[trim jolt]

✅

[trim lichen]

the other 99 was already there

[tawny escarp]

first line has 1 99, the other line has 3 99

in this this exact problem comes on my exam, i will just know to add 99 3 times, so its not a biggie, i can remember that

but im just baffled

[trim lichen]

💩 ×100 = 💩 ×1 + 💩 ×99

does this make sense to you

[tawny escarp]

guys guys

seriously am i being trolled

loooook

there is one more 99

100 poops cannot give 199 poops

[trim lichen]

nobody is saying 99 poops

the 99 that you circled in black is PART OF the poop

along with the 10^2k

[tawny escarp]

but what kind of math is this? what is the point of just adding numbers? how did i get 99x10^2k + 99x10^2k OUT OF only one instance of 99x10^2k

its like 1 poop gives 2? how

[trim jolt]

@near leaf Has your question been resolved?

[trim lichen]

it is called the distributive law

[tawny escarp]

i am aware of that law... could you please pinpoint exactly where it was used, there are no brackets here

[trim jolt]

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[tawny escarp]

its oke, let it stay closed, help someone else who is in more need than me

thank you for helpppp, apprecaite itt

appreciate

[trim jolt]

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[granite raptor]

Can someone help me understand this pls

[granite raptor]

I thought to find the least term I just need to find a value of m which is less than 1/5000

which I could do by substituting m = 4 into 1/(n)(10^n) ; given m = n

<@&286206848099549185>

id rly rly appreciate it if someone helps bc my exam is tmrw :(((((

<@&286206848099549185>

[trim jolt]

@granite raptor Has your question been resolved?

[granite raptor]

no

<@&286206848099549185>

[trim jolt]

@granite raptor Has your question been resolved?

[granite raptor]

.close

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[stone sluice]

Is there's a difference between finding factors of any number and prime factorization of any number.? And we write factors in which form?? Like 1,2,3...or 1 times 2times3 ??

[arctic silo]

prime factors are factors which are prime numbers

and normal factors include both composite and prime

and we usually write factors as ,ultiple of the prime factors and the powers of prime factors

ig

[stone sluice]

I came across this example.can you make me understand what is happening?

@arctic silo

[arctic silo]

aight here they just did hit and trial

in future u will learn about how to represent factors in terms of the number's prime factors

basically in this example they used symmetricity and hit and trial

[stone sluice]

What are the factors of 63? What would be answer of this??
1,3,7,9,11,21,63 or 3^2 times 7

@arctic silo

[arctic silo]

the first one

dont think about it too much rn

for now just use the firstanswer

u will learn about what i am saying in the future

[stone sluice]

.close

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[bronze crest]

Hey guys if anyone is fit enough to riddle this then feel free to express it

[bronze crest]

I have this signal

I need to recreate the mathematical curve

and to me it looks like exponential saturation functions

rising with a positive saturation, falling with a saturation of 0

so how can I do this?

I could attempt to recreate one half of the signal

using e^(-x)

[trim lichen]

this looks periodic enough to just make a fourier series out of it 💀

[bronze crest]

is there much to do to achieve this?

would you be able to do it if you had to

Ann

[trim lichen]

ughghhhhh dunno

youd have to do it numerically i presume

i dont have the energy for it rn tho

[bronze crest]

its ok thanks

yk what fuck it chatgpt did it

i just wanted a white background

but I would have to cut out the curve and put it into a white background thats why I wanted to recreate it as a function instead

.close

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[knotty token]

Hello All

[knotty token]

I am familiar with the idea of completing the square

and factoring out the a part of the quadratic trinomial

to actually do it

Yet, this is literally the most brain boggling and confusing thing

someone please just elegantly explain it

[spiral ocean]

ok

so first lets rewrite this in descending order for clairty

${-qx^2 + 12x + q}$

[solid kiln]

k

[spiral ocean]

after that, in completing the square what u do next is factoring out the coefficient in front of x^2

so

${-q(x^2 - \frac{12}{q}x) + q}$

[solid kiln]

k

[knotty token]

-q (x^2 + 12/q x) + q

[spiral ocean]

yes

[knotty token]

you should know

that when completing the square

we use this formula which I dont see anywhere else

[spiral ocean]

what formula?

[knotty token]

(x+b/2)^2 - (b/2)^2 + c

its always just a faster version of the formula i dont see why people never use it

[spiral ocean]

well

too lazy to remember

[knotty token]

also its the only type I know

[spiral ocean]

bruh

[knotty token]

is c just 0

(12/q)/2

6

1q

(x+6/q)^2-(6/q)62

should we expand or what...

[spiral ocean]

...

[spiral ocean]

this formula doesnt assume a is 1 i think

[spiral ocean]

mhm..

[knotty token]

Help me bro

[spiral ocean]

i am aware of the answer

[knotty token]

Im confused the ... is just (6/q)^2

[spiral ocean]

but lets go the traditional method, ok?

[knotty token]

36/q^2

[spiral ocean]

ur formula doesnt account for a ≠ 1

[knotty token]

bro

fine

normal methdo

[spiral ocean]

ok

${-q(x^2 - \frac{12}{q}x) + q}$

[solid kiln]

k

[spiral ocean]

from here

we can see that 12/q can be rewritten as this

${-q(x^2 - 2 \times\frac{6}{q} \times x) + q}$

[solid kiln]

k

[spiral ocean]

so to make it fit the formula for complete square

we add and subtract (6/q)^2

right?

[knotty token]

bathroom break

sorry

let me catch up

ok sure

yeah its equivalrent

[spiral ocean]

ok

${-q(x^2 - 2 \times\frac{6}{q} \times x + \frac{36}{q^2} - \frac{36}{q^2}) + q}$

[solid kiln]

k

[spiral ocean]

${-q(x^2 - 2 \times\frac{6}{q} \times x + \frac{36}{q^2})- q\left(\frac{36}{q^2}\right) + q }$

[solid kiln]

k

[knotty token]

Yo K i think ti would be helpful

if i learnt what completing the square is

beyond that formula

ill come back again cause i also really need to have lunch

[spiral ocean]

ok

[knotty token]

Thanks anyways

.close

[trim jolt]

Channel closed

Closed by @knotty token

Use .reopen if this was a mistake.

[warm valley]

you have to factor out a before u can use this

[knotty token]

I know

didnt i do that w/ q

[warm valley]

theres one with a but its goofy

[knotty token]

broooooooooo

[warm valley]

because instead of (2a)^2 is like 4a

so its too annoying to learn

[knotty token]

@warm valley @spiral ocean

.reopen

[trim jolt]

✅

[knotty token]

its the same thing

I understand

36/q

but isnt is supposed to be squared

(6/q)^2 = 36.q^2

so its solved

broo im gonna crashout bro

[split chasm]

a factor of q cancels

also there's a slight error in the tex above

[knotty token]

ok so

yes i understood the error

wait noo

[split chasm]

(missing - sign)

[knotty token]

ok wait

you need to add them

to make it into the form

one second

[split chasm]

${-q(x^2 - 2 \times\frac{6}{q} \times x + \frac{36}{q^2})- q\left(\red{-} \frac{36}{q^2}\right) + q }$

[solid kiln]

ℝαμOmeganato5

[knotty token]

bro

what

[warm valley]

i just realised

[knotty token]

am i looking at

[warm valley]

you dont even need to do all of this

you can just solve by equating coeff

by expanding

[knotty token]

guys

[warm valley]

the other side

[knotty token]

if i was as smart as you guys

i wouldnt be the one asking

please simplify this

can you go back to

[knotty token]

this

${-q(x^2 - \frac{12}{q}x) + q}$

[solid kiln]

GabeNewelGlazer

[split chasm]

to complete the square you want to introduce
+(b/2)^2 - (b/2)^2
right?

[knotty token]

sure

like x + (b/2 ....

[warm valley]

-q(x^2 - (12/q)x) + q
-q(x^2 - (12/q)x + 36/q^2- 36/q^2) + q ((b/2) = 12/2q = 6/q)
-q((x - 6/q)^2 - 36/q^2) + q

-q(x - 6/q)^2 -q(-36/q^2) + q

-q(x - 6/q)^2 + 36/q + q

[knotty token]

explain the second step

in my igcse video

it doesnt mention that

[warm valley]

ok so we can remove evrything

and loom atthis

x^2 - (12/q)x

whats b here

[knotty token]

12/q

[split chasm]

what does your igcse vid say to do?

[warm valley]

ok

so what would we do next

[knotty token]

(x+6/q) - 36/q^2

[warm valley]

no

it would be minus

[split chasm]

ah, so they skipped the step where the square is actually completed
and jumped straight to the end

[warm valley]

yeah u can skip what i did

i just wanted to show u

[knotty token]

oh

ok

so i have

[split chasm]

can, but you shouldn't when starting out

[knotty token]

bro all completing square questions are the same

this one is weird cause theres two variables

and usually its a coefficient in a

ok

so

let me attempt this

-q(x^2 - (12/q)x) + q

(x^2 - (12/q)x)

(x + b/2)^2 - (b/2)^2

(x + 6/q)^2 - (6/q)^2

brooo

[split chasm]

wrong sign

[knotty token]

whta do you mean

oh

x -

ok

then

[warm valley]

youre forgetting the minus

where x is

[knotty token]

-q((x - 6/q)^2 - (6/q)^2 ) + q

guys

this is incorrect

because of that

[split chasm]

its ok so far

[knotty token]

fr?

[split chasm]

yes, now redistribute that -q on the outside

[knotty token]

what do you mean so like

-q((x-6)^2)) - (36/q^3) + q

[split chasm]

no

g(h - j) = ?

[knotty token]

gh - jg

[split chasm]

yes, apply that here

[knotty token]

wait

bruh

the q cancels out

the q^2

correct?

[split chasm]

yes

[split chasm]

the idea of completing the square isn't necessarily restricted to messing with the constant term.
there was a question i saw recently here where completing the square can be used to nicely factorise
x^4 + x^2 + 1

[knotty token]

-q((x-6)^2)) - 36/q + q

[warm valley]

hidden quadratics ❤️

[knotty token]

a − b (x − c) 2

we

- (36+q^2)/q

[split chasm]

put a \ before the -

[knotty token]

- (36+q^2)/q-q((x-6/q)^2))

[split chasm]

why did your 6/q turn into 6

[knotty token]

is this the answer bro

i guess

so its solved

[warm valley]

btw

[split chasm]

you messed up signs again

[knotty token]

that question wasnt even hard it wasw just confusing

[warm valley]

u didnt have to do any of that

[split chasm]

didn't catch it earlier

[warm valley]

u coulve expanded the a c b

and equated coeff

and solved like taht

wouldve been faster

[split chasm]

-q * -36/q^2 is 36/q
but you had -36/q

[knotty token]

can you use latex

OH

ok

bro

its fine

I got it now

thanks

i think this was all unnecesasry

the skipping the square

well

[warm valley]

i mean i wouldve done this without factorising

[knotty token]

if i knew this was the methodi would have just solved it

bro

i loved you all

thanks for being there for me

🐺

[warm valley]

these type of questions are too time consuming to actually solve

[knotty token]

🖤

[warm valley]

aura

[knotty token]

.close

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Channel closed

Closed by @knotty token

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[split chasm]

Please don't make such comments in this server.

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[dapper moon]

I need help understand how to find the common ratio of a Geometric Series.

The problem:

Summation n=1, (1 + 2^n) / (3^n+1)

[trim lichen]

$\sum_{n=1}^{\infty} \frac{1 + 2^n}{3^n + 1}$

[solid kiln]

Ann

[trim lichen]

this?

[dapper moon]

Yes.

[trim lichen]

this isnt a geometric series as such

[dapper moon]

Oh wait-

Not correct. 3 is raised to the power of n+1

[trim lichen]

$\sum_{n=1}^{\infty} \frac{1 + 2^n}{3^{n+1}}$

[hollow cypress]

3^(n+1) for future reference

[solid kiln]

Ann

[dapper moon]

That is a better way to state it. Thank you.

[trim lichen]

this still technically is not a geometric series as such.

but it is equal to the sum of two different geometric series each with its own common ratio.

[dapper moon]

Ok. So these are two different geometric series stitched together? Or is that too oversimplified?

[trim lichen]

added together

not stitched

added in the literal sense like +

this is $\sum_{n=1}^{\infty} \frac{1}{3^{n+1}} + \sum_{n=1}^{\infty} \frac{2^n}{3^{n+1}}$

[solid kiln]

Ann

[trim lichen]

for both of these, the common ratio is rather straightforward to identify

[dapper moon]

Lefthand side: 1/3 * (1/3)^n

Right-hand side:
1/3 * (2/3)^n

But the main goal for this problem is to get the sum of this series.

I've never taken two ratios for one geometric series

[hollow cypress]

im inclined to call xy

!xy

[trim jolt]

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

[hollow cypress]

that's what xy is

[dapper moon]

Got it.

Then the original problem.

Find the sum of the series

$\sum_{n=1}^{\infty} \frac{1 + 2^n}{3^{n+1}}$

[solid kiln]

Doc

[dapper moon]

What I understood:

You could usually get the sum of a Geometric Series, so that was where I started, and why I asked about it.

Geometric Series' Sum = a / (1 - r)

But now, I do not know if I am progressing correctly with the question.

[hollow cypress]

did this help

[trim lichen]

this is not one geometric series

this is NOT one geometric series

as i have already stated before

which i believe i might have to repeat a few more times

[trim lichen]

these are not LHS and RHS, they're not separated by an equals sign

Geometric Series' Sum = a / (1 - r)
apply that, to each of the two geometric series independently, and then add the results

and please try not to overthink yourself into inability to do fraction artihmetic

[dapper moon]

Ok, here's what I've got.

Geometric Series 1

(1/3) / (1 - 1/3) =
(1/3) / (2/3) =
(1/3) * (3/2) = 1/2 or 3/6

Geometric Series 2

(1/3) / (1 - 2/3) =
(1/3) / (1/3) =
(1/3) * (3/1) = 1

1 + (1/2) = 3/2

[trim lichen]

the first term in the 2nd geometric series is 2^1/3^2 = 4/9 and not 1/3 as you wrote.

[dapper moon]

Got it. Give me a few minutes.

[trim jolt]

@dapper moon Has your question been resolved?

[dapper moon]

From the top. I even forgot a crucial step.

Geometric Series 1

(1/3) [ (1/3) / (1 - 1/3) ] =
(1/3) * (1/3) / (2/3) =
(1/9) * (3/2) = 3 / 18 -> 1/6

Geometric Series 2

(1/3) [ (2/3) / (1 - 2/3) ] =
(1/3) * (2/3) / (1/3) =
(2/9) * (3/1) = 6 / 9 -> 2/3

(1/6) + [ (2/3) * 2 ] = 1/6 + 4/6 = 5/6

The exterior 1/3 came from the 3^[n+1] = 3^n * 3

[hollow cypress]

spot on

[dapper moon]

Oh thank lordy.

@trim lichen Thank you for showing me that Geo-Series can be split apart.

And also thank you @hollow cypress for clarifying on xy. I'll keep that in mind going forward for this server.

[hollow cypress]

np

!done

[trim jolt]

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[dapper moon]

.close

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[boreal shuttle]

$N,a,b \in \mathbb{N}$

[solid kiln]

<rajel />

[boreal shuttle]

prove that if N is a multiple of 4 then $a,b$ are even

[solid kiln]

<rajel />

[boreal shuttle]

$N=a^2 + b^2$

[solid kiln]

<rajel />

[boreal shuttle]

i can assume N=4K

if a^2 + b^2 adds to 4k

i believe they can be odd ?

idk how would that mean they are even

[opal badge]

Try proof by contradiction

or if you think it's actually wrong then come up with a counterexample

[boreal shuttle]

so we assume they arent even

[opal badge]

yeah, you assume at least one of a or b isn't even

[boreal shuttle]

no its not , it just me trying to think

[opal badge]

i see

But yeak, suppose (for example) a = 2d-1 for d >0 and see what contradiction might pop up

[boreal shuttle]

then a^2=4(d^2-1) + 1

which isnt a multiple of 4

but what about b

i can do the same for it

[opal badge]

You have
$$
N = 4k = 4(d^2-1) + 1 + b^2
$$
You are assuming this holds, so whaqt condition would $b^2$ have to satisfy for you to be able to factor a 4?

[solid kiln]

Nanigov

[boreal shuttle]

b^2 = 4k'-1

so the +1 goes with -1 and then we factor with 4

[opal badge]

so b^2 is one below a multiple of 4, what does that tell you about the parity of b?

[boreal shuttle]

i cant find any value of b that satistifies this

every odd number tends to be a multiple of 4 + 1

anyways its an odd number

even if i cant find any example of it

4k' is even and taking one from it makes it odd

[opal badge]

well with that you have b^2 is odd, why does that guarantee b is also odd?

(i assume you meant 4k')

[boreal shuttle]

ofc

[boreal shuttle]

yep it should guarantee's it

[opal badge]

mhm, so now you have that the sum of the squares of two odd numbers is a multiple of 4, can you see the contradiction?

[boreal shuttle]

odd x odd = odd

[boreal shuttle]

how does it contradict ?

two odds can add up to an even number

[opal badge]

yeah, but you aren't dealing with two random odd numbers, it's specifically two odd squares

write a = 2d-1, b = 2c-1 and see what issue might pop up

[boreal shuttle]

i see , we get a multiple of 4 + 2

which isnt a multiple of 4

[opal badge]

yep

[boreal shuttle]

thanks

.close

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[spare lance]

could anyone help explain how we come to this conclusion?

[solid kiln]

Nanigov

[spare lance]

Yes

[opal badge]

aight

[spare lance]

i understand the concept of vertical asymptotes, that the limit as x approaches the point of interest should be +/- infinity

just unsure of this case in particular

[jaunty scarab]

not 0, it has to be +/- inf

[spare lance]

yep mb

[jaunty scarab]

can we see the rest of the context?

[spare lance]

its a 14 mark scaffolded question that ultimately results in sketching the functions graph, the first and second derivatives of the function are also provided

not really any more context

[opal badge]

Do you understand why x = 1 is a candidate for a vertical asymptote, i.e. why you'd think to check the limit x->1?

[spare lance]

yes, its e raised to a rational function and you can set the rational function's denominator equal to zero to figure out a potential vertical asymptote

it ends up being an asymptote if the limit of f(x) approaching that value is infinity

[opal badge]

mhm, and do you know why there aren't any other candidates for a vertical asymptote?

[jaunty scarab]

we kind of need all the rest of the context tho

!original

[trim jolt]

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

[spare lance]

to be honest, not entirely

we were just taught what the definition of the vertical asymptote is

all the examples in the textbook are just ones of obvious rational functions

so if you gave me a regular function with a vertical asymptote that isnt clear from the start (because the function is rational) i wouldnt really know where to start

[opal badge]

That is a common issue

[spare lance]

this is question (e)

[opal badge]

I see

[jaunty scarab]

okay, so you want to obtain all values that make the function tend to infinity

so limf=+-inf

[spare lance]

yep as per the definition of VAs

but how do i find those values?

[jaunty scarab]

since you got the exponential function, and that's annoying to work with, you can take the limit of the logarithm of f

[opal badge]

So for $f$ to have a vertical asymptote at $x = a$ you need
$$
\lim_{x\to a} f(x) = \pm \infty
$$
in particular this means that \textbf{if} $x = a$ is a vertical asymptote, \textbf{then} the limit of $f(x)$ as $x \to a$ doesn't exist.

The idea then is that the candidates for vertical asymptotes are all points where the limit might not exist, in this case $e^{\frac{1}{x-1}}$ is continuous everywhere except at $x = 1$ (where it is not defined), so everywhere $x \neq 1$ there cannot be an asymptote (because the limits exist)

[solid kiln]

Nanigov

[jaunty scarab]

that's not correct tho. The limit when x->a can exist if there's a vertical asymptote

[spare lance]

could you name an example of that?

[jaunty scarab]

1/x^2

limit +inf when x tends to 0

[opal badge]

+inf is still a type of divergence

[jaunty scarab]

doenst mean the limit does not exist

[spare lance]

im pretty sure a limit that equals infinity doesnt exist

[opal badge]

im pretty sure the usual definition for a limit to exists is that it has a real value

at least at the calc 1-3 level

[spare lance]

i was going to say, at least for this level

im not studying towards a maths degree, currently in engineering so the maths isnt going to get anywhere near a pure maths level

okay, so i understand why we look at x=1

we want to look at discontinuities basically

[jaunty scarab]

okay, if both sides being equal limit but not finite is not considered as the limit existing then sure. We usually consider them here

[spare lance]

but could you explain how it evaluates to infinity as x approaches 1 from the left and zero as x approaches 1 from the right?

[jaunty scarab]

usually, you get a limit of +/-inf when you're dividing by close to zero

[opal badge]

do you know about lateral limits?

[spare lance]

nope

[solid kiln]

Nanigov

[spare lance]

yes

[opal badge]

Using that fact, what can you tell me about
$$
\lim_{x\to 1^+} \frac{1}{x-1} ;;\text{ and };; \lim_{x\to 1^-} \frac{1}{x-1}
$$

[solid kiln]

Nanigov

[spare lance]

this is where i get a bit tripped up

okay, the first one is positive infinity

and the second one is negative infinity

the first one i figured out fine intuitively but the second felt a bit harder for some reason

obviously 1/(x-1) is just 1/x shifted a unit to the right so that also makes it easier

[opal badge]

mhm

You're one step away, now you want to evaluate (i'll save space by writing $\pm$)
$$
\lim_{x\to 1^{\pm}} e^{\frac{1}{1-x}}
$$
what would you do from here? (Mind there that the denominator is $1-x$ rather than $x-1$)

[solid kiln]

Nanigov

[spare lance]

lmao

[opal badge]

I could be 80 years old and ill still forget some or other dollar sign when typesetting something

[spare lance]

well, id evaluate the limit of 1/(1-x) as x approaches 1 from either side, because e is independent of x

[opal badge]

more correctly you can do that because e^x is continuous so lim e^x = e^(lim x)

but yeah

[spare lance]

$$\lim_{x\to 1^+} \frac{1}{1-x} $$

just want to see if i got that latex right

how do i trigger the bot?

[opal badge]

huh, idk why it didn't trigger

[spare lance]

weird

anyway

[opal badge]

you might be missing a } or smtn

[solid kiln]

la garce

[spare lance]

well, that was fun

i guess i would just let x be 1.1, 1 - 1.1 is -0.1

1/-0.1 is -10

then let x be 1.01, 1 - 1.01 is -0.01, so 1/-0.01 is -100

so itll approach negative infinity?

[opal badge]

correct

and that's more or less the intuition for it

though formally you'd factor out a -1 from the denominator and then factor it out of the limit

[opal badge]

and then use this observation here

[spare lance]

alright

so then i would evaluate e^-infinity

ohhhh

its 1/e^infinity

[opal badge]

ding ding ding!

[spare lance]

e^infinity is evaluated as infinity

so 1/infinity is zero

bruh

[opal badge]

yup

[spare lance]

oh my goodness

okay, well it makes sense

i think to make evaluating VAs easier i'll just write out physically x being 0.1, 0.01, etc to visualise it better

[opal badge]

and it's all the same logic the case where x-1^-, but the signs are flipped so you get e^inf = inf

[spare lance]

yeah

[opal badge]

that's as good a way as any

[spare lance]

alright, thanks for your help

[opal badge]

though it's also recommended you get familiar with other notable asymptotes, for example ln(x) as x ->0^+ etc

np

[spare lance]

honestly i enjoy all the calculus and am pretty good at differentiation (and in high school integration)

but all the other stuff like sketching graphs and extrema i really dislike

[opal badge]

same tbqh

it's cool that that's something you can do by just looking at the derivatives but It does get a bit boring once you do it a couple times

[spare lance]

yeah i suppose

thats why im looking foward to uni integration

i did most of the 1st year techniques in high school but am looking forward to a bit more rigourous of an approach next semester

anyway, thanks for your help

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[tall pebble]

How does the Exponent rules apply here? I dont quite get it

[patent crow]

you can write 7/9 as (9/7)^-1

and (a^b)^c = a^(bc)

[tall pebble]

thanks, thats useful to know

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[tender kettle]

I'm not sure what to do next. I've calculated everythng once but chatgpt and my uni friends have all different answers.

[tender kettle]

Is it okay to do the Gaussian elimination like this or do i have to do it in the stair system?

[clear cloud]

Its original but it works

I mean there is technically nothing wrong but it may leads to some errors, depending on what you are doing with it

[tender kettle]

okay perfect. So i have to use a variable for x1 because i didn't use it before or do i have to use x3?

[clear cloud]

I don't get it

[tender kettle]

okay the last line is a "zero line" so i can put anything for one variable which i didn't use right?

[clear cloud]

It means that for any (x1,x2,x3), the last line will be true

What are you asked to do ?

Find alpha values such as ...?

[tender kettle]

i need a solution for the equation.

wait

Using the Gaussian algorithm, determine all solutions (x1, x2, x3)^T ∈ R^3 depending on the parameter α ∈ R:

The last line says that it could be everything so i have to but a variable for x3 and calculate the anserws (x1,x2) in proportion to the variable

[clear cloud]

It means that there is solutions iff a = -2

[tender kettle]

yeah but now i have to calculate the system with a = -2

[clear cloud]

Which make x1 as a free variable

[tender kettle]

okay so x1 not x3

[clear cloud]

Yeah, with you way of writing the free variable is x1

Doesn't mean that your friend who writes x3 as free variable are in the wrong btw

Ah and yeah avoid gpt, it seems good at math but its not really great to do the exercices

[tender kettle]

yeah i know they used the standart methode

okay so i have x1 as b so x3 equals -3b right?

[clear cloud]

Yep

And x2 ?

[tender kettle]

i think 1 - b/2

[clear cloud]

Don't think and be sure

[tender kettle]

yeah i'm sure haha

[clear cloud]

Great

As a bonus for geometric interpretation

The set of solutions is actually a line of R³ when a = -2 and is the empty set for a != -2

[tender kettle]

but is this right or wrong then? we have diffrent answers

[clear cloud]

Which makes sense cuz when a = -2 this is the intersection of 2 planes in R³ that are not parralel and so intersect is a line

[clear cloud]

I can't tell for their solution but it seems to be another parametrization wrt another variable than yours

Its probably equivalent

[tender kettle]

okay did I understand this right that it looks like this?

[clear cloud]

But since they let x2 as free and not you thats why the difference

[clear cloud]

Exactly

[tender kettle]

oh wow thank you. Thats a great explanation

[tender kettle]

Thank you for your help. It really helped me grasp the concept more especially beacause of this

[clear cloud]

You're welcome

[clear cloud]

They are bunch of interesting things with this kind of triple planes system

Using barycenter and mass-center

For example

You can show that in a tetraedric, the medians planes of the base (orthogonal to the sides + cutting it in the middle) are meeting in a unique point, which is the center of the circomscribed sphere

Anyway

[tender kettle]

Yeah i hope i will never have to do this hahaha

[clear cloud]

It sounds scary but its not harder than what you did

[tender kettle]

oh okay, i'm not really that into math but i have to pass the exams for mechanical engineering

i think i have to pass math 1-4

[clear cloud]

Viel Glück

[tender kettle]

Danke

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[bleak ridge]

like idk

[left oriole]

the slope should always be negative when y < 0 and x < 5, look at the formula for dy/dx

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[cyan copper]

for this, shouldnt sin^2θ cannot equal 1 also be a restriction

did the book just forget to mention that?

[cyan copper]

this is the answer for 5b from textbook

[hidden dew]

well sin(90) and sin(270) is 1 so I don't see they missed something

[cyan copper]

oh

[cyan copper]

so can i not just say cosθ cannot equal 0 and sin^2θ cannot equal 1 instead

well ig i gotta ask my prof that

[hidden dew]

yea

[cyan copper]

alr cuz i think its easier

ill also double check w my prof

alright thank you !

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[jagged python]

hi i just started learning vectors and im kinda confused

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[jagged python]

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[worn roost]

Why’s the answer here D? I keep getting B

[supple copper]

Show your work

[hidden dew]

is there a mistake in the question, cuz what is a 0th term ?

[worn roost]

Nvm, I got it

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[real herald]

Im so confused on whats going on here, are the alphas acting like scalers

[real herald]

im so confused on how the elements from different fiields ended up together; here is the proof so far:

How are they putting elements of E into F[x]?

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@real herald Has your question been resolved?

[real herald]

<@&286206848099549185>

[real herald]

the goat

oh nvm

THE GOAT

lmfao

[whole coral]

Remember that E is F[x]/(p), tl;dr “F[x] but we consider that p(x) is also zero too”

(Back again for a little bit )

[real herald]

lmaoo nw

uhhh

hmm

[lusty delta]

E extends F

[real herald]

yes

Not F[x] tho

[lusty delta]

p is a polynomial with coefficients in F

this is enough

this is similar to how you can view an element like i in C as a root of x^2 + 1 even though x^2 + 1 is a polynomial in Q[x], as C extends Q

[real herald]

oh hmm

so we have it that

E extends F

so for some p(x) in F[x]

alpha is in E

so we are putting that into F[x]?

and why is that zero?

ok how about

like a more explicit example

[lusty delta]

the idea is setting alpha = x + (p) and considering p(alpha)

[lusty delta]

p(alpha) is the image of p in E, which is zero since p is taken to be ideal of the quotient

[real herald]

oh

im still a little lost as to how they just put it in but that makes sense

Ok so what ur saying is

[lusty delta]

maybe it feels like a strange construction at first, it works though since E extends F

[real herald]

@lusty delta ok let me draw this out

We have an homomorphism from

F to E =$\rightarrow F[x] / \langle p(x) \rangle$

[solid kiln]

BOSS

[real herald]

not surjective but it is injective

so we know the kernel of this

is p(x)

right

the ideal gereated by p(x) persay

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[frail heron]

English?

[solid kiln]

938c2cc0dcc05f2b68c4287040cfcf71

[urban copper]

,, \begin{amatrix}{3} 1 & -1 & 1 & 0 \ -1 & k - 1 & -4 & -3 \ 0 & 0 & k^2 - 1 & 1 + k \end{amatrix} \ R_1 + R_2 \to R_2 \ \begin{amatrix}{3} 1 & -1 & 1 & 0 \ 0 & k - 2 & -3 & -3 \ 0 & 0 & k^2 - 1 & 1 + k \end{amatrix}

[solid kiln]

938c2cc0dcc05f2b68c4287040cfcf71

[urban copper]

,, \rightarrow \begin{cases} k - 2 &= 0 \ k^2 - 1 &= -3 \ 1 + k &= -3 \end{cases} \rightarrow \begin{cases} k &= 2 \ k^2 &= -2 \ k &= -4 \end{cases}

[solid kiln]

938c2cc0dcc05f2b68c4287040cfcf71

[trim lichen]

hmmmmm that looks sus

can you verbalize what you're doing

like your goals

[urban copper]

rank

[urban copper]

only if we got full rank we have a unique solution

max rank for a 3x4 matrix is min{rows, cols} = min{3,4} = 3

we need strictly that the rank is < 3

because if rank = 3 then we have unique solution I guess

[trim lichen]

mmm i mean im not sure where you are being k^2-1=-3 from for example

[urban copper]

rank = dim(R) = dim(Col)

[trim lichen]

i would think that at this point you would want to first stop and consider what happens with k=2 (solve the system for this k), then proceed under the solution that k ≠ 2

[trim lichen]

yes and? why should k^2-1 be -3?

[urban copper]

what about k = -1

[harsh siren]

how you got negative sign

-3?

[trim lichen]

that will come later

first look at k=2 since that zeroes the pivot in the 2nd row

only LATER worry about the 3rd row

one thing at a time

do not try to do everything all at once

[urban copper]

I was trying to think if for example for R2 = R3 then we have one row full of zeros

[trim lichen]

k^2-1=-3 does not make R2 and R3 equal, this is the wrong direction to think in

[urban copper]

it was three equations that I proposed, but it didnt worked out

[urban copper]

.

[trim lichen]

should i repeat my suggestion/instruction

[urban copper]

where did k = 2 came from

[urban copper]

ohh, you are looking it in terms of columns I think

[trim lichen]

i would not phrase it that way

[urban copper]

the dimension of the colspace drops down by 1 when k = 2, no?

colspace = span of the columns

[trim lichen]

eh sure. i would think the row space, but their dimensions are equal anyway

so ultimately you are right in that k=2 makes the rank drop

and that's why we are looking at it

[urban copper]

,w {{1,-1,1,0},{0,k-2,-3,-3},{0,0,k^2 - 1, 1+k}} where k = 2

[solid kiln]

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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[urban copper]

ah I see

the rank is 2 here

[harsh siren]

yes

second and third row are same

[urban copper]

yes this is perfect

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[vagrant imp]

how do I solve this Integral?

[vagrant imp]

like I tried some stuff but I keep getting stuck

[empty prairie]

!showyourwork

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

[vagrant imp]

I've tried by parts but I get another integral which isnt direct, should I apply by parts again?

[trim lichen]

no good

you can make a substitution here

i can confirm to you that it will cook

[vagrant imp]

um secx+tanx=t?
secx dx=dt/t
but there's another secx, so should I use by parts on secx * secx/(secx+tanx)^9/2?

or is this too complicated and there's a simpler way?

I was also thinking tanx=u but then i would have to write secx as sqrt(1+u^2) which doesnt seem to be any good

[vagrant imp]

can you give me a hint, I can't figure it out

[vagrant imp]

<@&286206848099549185>

[final fulcrum]

Integration is in general very difficult

[native shuttle]

PEKKA

This question looks familiar

[vagrant imp]

I thought I've done it before too, but idk, maybe I've done a similar looking Qs

[native shuttle]

secx+tanx=t

also

1+tan^2x=sec^2x

so sec^2x-tan^2x=1

[vagrant imp]

this? but I have sec^2x in numerator..

[native shuttle]

secx-tanx=1/t

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[floral lodge]

When you use u substitution on sec the bottom disappears into du, and then the the secx^2 becomes u^2

[trim lichen]

yes that's the route

consider sec(x)-tan(x) = 1/t also

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[vagrant imp]

ahh i got it thank you very much!

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[frank mango]

idk how to prove

[round imp]

Connect RT

And use tangent secant theorem

That angle RUT is angle RTP

@frank mango

[frank mango]

this kind of questions can draw on the diagram or use working ?

[round imp]

idk

They asked to prove

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[lunar stirrup]

can somone help me in greatest integer inequalities?
for example [x]=(-infinity,-2]U(3,infinity)

[lunar stirrup]

how would i solve for x?

[vivid glade]

what do the square brackets around the x denote?

[lunar stirrup]

greatest integer function i think in other countries they call it floor function

[lunar stirrup]

not really an inequality but im trying to find solution set for x from this

[vivid glade]

it sounds more like it'd take a set as an argument

$[A]=\max(A\cap\mathbb Z)$

[solid kiln]

Flip

[vivid glade]

our floor function only takes a real number

[lunar stirrup]

sorry didnt really understand

[thorn bay]

thats standard, but, the student wrote floor(x)=set of numbers, which is weird

its like saying x²=[0,1)

[vivid glade]

agreed, so I'm confused

[lunar stirrup]

could u explain for that particular example

ok let me send the original q for more clarity

[vivid glade]

that's way clearer, the original question unfortunately couldn't have been derived from this one

[lunar stirrup]

why not

first case mod opens positive

then take [x]=t and solve

[lunar stirrup]

u get something LIKE this

[thorn bay]

numbers aren't intervals

[thorn bay]

like this, it doesn't make sense

[lunar stirrup]

why not?

[thorn bay]

because x^2 is a nunber and [0,1) is a set of numbers

[vivid glade]

numbers are elements of intervals, not elements themselves

[thorn bay]

and they arent the same object

= means "the same object"

[lunar stirrup]

isnt this saying x^2 belongs from 0 to 1 with 0 included

[thorn bay]

no, its saying x^2 IS all numbers between

[lunar stirrup]

yes

[thorn bay]

= means equal to, in all respects, identical in every way

[vivid glade]

to clarify, we are imagining x is a single number. it has a single value, we just haven't written it

[thorn bay]

Clark Kent = Superman

[lunar stirrup]

oh so u guys are saying it should be a belongs sign instead of an equal to sign?

[vivid glade]

what interval equals 3? can't make sense in this setting

unless if we're doing dedekind cuts but incorrectly

or whatever. inventing a new number system as collections of sets from an existing one

[lunar stirrup]

could u help me solve the question i think that would clarify my doubt

[vivid glade]

that's read better

well, what have you done on your own at this point?

[lunar stirrup]

can u give me 10 mins? ill finish dinner and brb

[vivid glade]

I have to leave by then, got an appointment in 40mins

[lunar stirrup]

oh man

ill ask someone else to help then

thanks anyway

[vivid glade]

all is well

[gentle crag]

Hello,
still need help?

[lunar stirrup]

yes pls

[gentle crag]

what have you done so far?

[lunar stirrup]

[x]-2/([x]-3)>=0

(when mod opens positive)

[x] not equal to 3

[gentle crag]

what does “mod opens positive” mean?

[lunar stirrup]

like when [x]>=0

[gentle crag]

okay, so in general what solution do you get?

for [x] that is

[lunar stirrup]

(-infinity,2]U[3,infinity)

[gentle crag]

what is there, x?

[lunar stirrup]

huh?

[gentle crag]

I don't understand what that set is supposed to represent

(-∞,2]U[3,∞)

[lunar stirrup]

set for [x]

[gentle crag]

how did you reach that conclusion?

[lunar stirrup]

wavy curve method

[lunar stirrup]

this is only when [x]>=0 btw

[gentle crag]

if [x] ≥ 0 then it cannot obtain all values in the set (-∞,2] or it could be negative, don't you agree?

[lunar stirrup]

oh yes correct

[gentle crag]

[x] would be more specifically in ( [0,2] U (3,∞] ) ∩ ℤ

[lunar stirrup]

oh ok yeah correct

[gentle crag]

also 3 is excluded, you said [x] ≠ 3 is the solution

[lunar stirrup]

yeah

[gentle crag]

what if [x] < 0 ?

[lunar stirrup]

let me quickly tell for negative one second

[lunar stirrup]

[x] = [-3,-2]

[gentle crag]

first of all, [x] = -3 is not allowed

[lunar stirrup]

oh right

[x]=(-3,-2]?

[gentle crag]

secondly, are you sure that's the correct interval?
if you try manually [x] = -1 do you get a solution or not?

[lunar stirrup]

no sorry i missed something

let me try again

i think its (-infinity,-3]U[-2,infinity)
(havent removed -3 yet)

[gentle crag]

you should prolly stop at 0 since [x] < 0 and not go up to ∞

[lunar stirrup]

yeah (-infinity,-3)U[-2,0]

[gentle crag]

Also it's worth noticing that you could've avoided calculating it manually again (if you did)

if you have to solve
( |[x]|-2 ) / ( |[x]|-3 ) ≥ 0
you can call y = |[x]|, so that y is a nonnegative integer
then the inequality (y-2)/(y-3) ≥ 0 has solution y ≤ 2 ∨ y > 3 which for our y becomes 0 ≤ y ≠ 3
then |[x]| = y is just [x] = ±y
so the domain is just [x] ≠ ±3

but anyway
now the question is: “what real x have floor different from ±3?”

[lunar stirrup]

one second i undersstood till y<=2 and y>3

after that could u tell me what u did

[gentle crag]

for an integer to be less than or equal to 2 and strictly greater than 3, it just means that it's not 3

[lunar stirrup]

oh ok yeah

[gentle crag]

so if y is not 3 and [x] = ±y, then [x] is neither 3 nor -3

[lunar stirrup]

ohh wait so 3 is the only value of y not allowed?

[gentle crag]

yeah (among the positive values, I assumed y = |[x]|, so y ≥ 0)

[lunar stirrup]

oh ok got it

[gentle crag]

so really we have to describe the real numbers x such that [x] ≠ ±3

[lunar stirrup]

so x=(-infinity,-3)U[-2,3)U[4,infinity)

[gentle crag]

you shouldn't write x = SET

x ∈ Set

you could write “belongs/is in” or something like that

or use the LaTeX support

[lunar stirrup]

oh ok lol that symbol isnt there in my keyboard

so i thought people will understand if i write =

but ok point taken

[gentle crag]

neither in mine, using special combinations
but you just have to find a workaround

[lunar stirrup]

👍

thanks a lot

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[marsh forum]

Let $a$ and $b$ be positive integers,not both zero, and let $d$ be their greatest common divisor , the positive integer that generates the subgroup $S= \Z a + \Z b$. So $\Z d = \Z a+ \Z b$. Then \begin{enumerate} \item d divides a and b \item if an integer $e$ divides both $a$ and $b$ , it also divides $d$. \item There are integers $r$ and $s$ such that $d= ra +sb$ \end{enumerate}
\
Proof:
\~\
\begin{enumerate}
\item Follows as $d$ is the greatest common divisor of $a$ and $b$.
\item As $d \mid a$ and $d \mid b$. It follows $d \in \Z a+ \Z b$. Furthermore as $e \mid a \land e \mid b$, it follows that $e \in \Z d$. Thus $e \mid d$
\item Follows as $d$ is an element of $S$
\end{enumerate}

[solid kiln]

What a wonderful world !
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[trim lichen]

looks like you are missing the enumitem package. add \usepackage{enumitem} to your preamble

[marsh forum]

Will do

thanks

Is that why I'm still getting this error

[marble wharf]

the question is strange. is it taken for granted already that the number d with dZ=aZ+bZ is the same as the gcd? cause if so, there is barely anything to do...

[marsh forum]

I can send a picture of my book

[marble wharf]

and that is the interesting part of all of this

[marsh forum]

Yea, this confused me too, Let me quote the book

[marble wharf]

ok so you arent supposed to know what a gcd is

[marsh forum]

I should probably prove d Z= a Z + b Z

no?

[marble wharf]

thats given

from that equality you are supposed to prove those properties

[marsh forum]

I meant I have to prove this

[marble wharf]

forget that they used the words gcd

it is given that dZ=aZ+bZ

now prove the properties

(a)-(c)

[marsh forum]

What do we mean by the addition of two groups, the same as that of the addition of two vector spaces?

[marble wharf]

middle of the page

(2.3.4)

[trim lichen]

yes

as in, set of all sums of (thing from one)+(thing from the other)

this is a general notation used whenever you talk about subsets of some, in general, additively-written group

[arctic silo]

yo dood which book is dis?

[marsh forum]

Artin's Algebra

[arctic silo]

aight thx 😄

[marsh forum]

I think I'll think about this a bit and then send a proof here ( I'll try to send it today, but tomorrow is likely)

[arctic silo]

btw which exam u preparing for?

[marsh forum]

no exam

[arctic silo]

dayuum respecc

[marsh forum]

I have this in my course next sem, so I;m studying it

[arctic silo]

😮

[marsh forum]

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[rancid mountain]

DE Question:
$x y^{\prime \prime} - 4x^2y^\prime + (4x^3-x)y = -3xe^{x^2} \sin{(2x)}$

Tried to reduce to normal form and got:
$v^{\prime \prime} + v(4x^4 - 8x^3 + 4x^2 - 1) = -3 \sin{(2x)}$

Can't solve this second DE. Any ideas of methods I can use?
I'm supposed to stick to one of those: Euler-Cauchy, D-operator, Variation of Parameters, or Reduction of Order.

[solid kiln]

Youssef Samy

[rancid mountain]

Oh I made a pretty stupid mistake that just cost me a buncha marks this morning 😭

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[devout arch]

Hi

[devout arch]

Does anyone know how they got the area of the rhombus

Hello

Hi

Hi

Hi

[fierce lake]

Yeah

The language is a bit weird

Basically what they mean by altitude is the length of the perpendicular drawn from point of intersection of diagonals to any one side of thr rhombus

[empty prairie]

,rccw

[fierce lake]

The area of the rhombus is basically 4 times area of the triangle formed by drawing diagonals

To the rhombus

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[indigo tusk]

hellpppp me i am in danger my maths teacher cant solve it i have two min to solve

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[lean kraken]

How can I start 76

[charred trail]

Because there is a modulus, find out where ln(x)/x becomes negative (the x coordinate I mean), because you will have to add this negative area

Once you split it up, you can remove the modulus, use u substitution to solve the integral

[lean kraken]

It is negative when we have 0 to x=1

[charred trail]

Correct

So now, split the integral between 1/e and 1, and then 1 and e^2

[lean kraken]

Then I apply ilate?

@charred trail

Opps no

Substitution

Logx=t

[charred trail]

Yes

[lean kraken]

5/2

I got

,w integration 1/e to e^2 |logx/x|

[solid kiln]

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Requested by andy03738

[lean kraken]

1/e is less than 1

We need modulus

[charred trail]

Sorry yeah oops

Mixed up the numbers

[lean kraken]

India se h?