help-26

[jade thunder]

we just showed the right side

if i stick the right side one in here

i get this

so there's 2 maxes going on

[opal vault]

$\forall x\in \bR^n, |x| < 1$, $|\partial^\alpha\varphi(x)| \leq \frac{2^{d/2}\sup_{|x'|<1}|\partial^\alpha\varphi(x')|}{(1+|x|^2)^{d/2}} \leq \frac{C_{d,\alpha}}{(1+|x|^2)^{d/2}}$

[thorny flame]

rafilou2003

[jade thunder]

what happened to the 2^d/2 for the C_d,alpha

[opal vault]

definition of C d alpha

it's bigger than the term on the right

[jade thunder]

oh it's inside

[jade thunder]

ok before i ask where the middle term came in here

we have C_d,alpha = max(p*, 2^d/2 thing)

and p* = max(thing1, thing2)

so we have 2 maxes going on

[opal vault]

C and Cdalpha are different, remember that

$C = \sup_{|x'|<1}|\partial^\alpha\varphi(x')|$

[thorny flame]

rafilou2003

[opal vault]

it's this C here

[jade thunder]

what's the '

[opal vault]

so that we don't get confused with x, already being used

so we use a different dummy variable

[opal vault]

mainly the x used here

[jade thunder]

ah it's just used to descibred the furthest (\partial^\beta \varphi(x)) can take us

[thorny flame]

Frosst

[jade thunder]

thats just this

[opal vault]

yes

the furthest it can take us inside the unit ball

[jade thunder]

right it's not really a varaible since we take hte sup of that

so really there's only "1" value that gives us the sup

or maybe a bunch of them that gives the same sup either way

it's not a variable

[opal vault]

yes

[jade thunder]

well i suppose it depends on varphi and beta

but that's not the point here

[jade thunder]

ok still trying to comprehend this here

[opal vault]

So for $|x|<1$, $|\partial^\alpha\varphi(x)| \leq C$

[thorny flame]

rafilou2003

[opal vault]

by definition of sup, right?

[jade thunder]

true

[opal vault]

but we showed this

[jade thunder]

ah

that's also true

so you bound that C by the thing on the right

with another C

that's indeed true again

[jade thunder]

and that's the middle term

[opal vault]

yes

And finally, by definition of Cdalpha

[jade thunder]

and that in turn is yet smaller than the C_d, alpha

[opal vault]

the numerator is smaller than Cdalpha

by its definition

[jade thunder]

ah

because it is at leas thte same as the middle term

because if that was the biggest we'd have pikcedi t

if p* was bigger then of course it's bigger

[opal vault]

$b\leq \max(a,b)$

[thorny flame]

rafilou2003

[jade thunder]

indeed

[opal vault]

so giving us the final inequality

$\forall x\in \bR^n, |x| < 1$, $|\partial^\alpha\varphi(x)| \leq \frac{C_{d,\alpha}}{(1+|x|^2)^{d/2}}$

[jade thunder]

but that's only for |x| < 1

[opal vault]

Yes, now we need to prove it for |x| >=1

[jade thunder]

oh dear

[thorny flame]

rafilou2003

[jade thunder]

that's a lot of steps they've hidden in 1 line

[opal vault]

For $|x| \geq 1$, we know by definition that $|x|^d|\partial^\alpha\varphi(x)|\leq p^*_{d,\alpha}(\varphi)$ right?

[thorny flame]

rafilou2003

[opal vault]

definition of supremum

[jade thunder]

from here?

[opal vault]

well yes

[jade thunder]

oh that's the wrong one

this one

[opal vault]

yes

[opal vault]

so now we divide this inequality by |x|^d

[jade thunder]

if the left one is bigger

what happens

then |x| >=1 is bounded by the same bound as for |x| < 1

ok that makes sense

that would be like the e^-x^2 example

the overall upper bound is in |x| < 1 and for |x| > 1 it's still bounded by the upper bound of |x| < 1

ok now if the right one is bigger

[jade thunder]

then we indeed have this inequality sure

either we took the left one and p* is bigger

or we take the right one and we are at equality

ok good

[opal vault]

all good?

[jade thunder]

yes

good

[jade thunder]

now we can do this good thing

so now we have $|\partial^\alpha\varphi(x)|\leq \frac{p_{d,\alpha}^*(\varphi)}{|x|^d}$

[thorny flame]

Frosst

[opal vault]

so now we just need to find a constant to bring us back to the denominator we want

[jade thunder]

hmm

i see what you mean

[opal vault]

ah I think we need 2^d/2 again maybe

no doesn't work

Ah I got it

[jade thunder]

but we wanna make the left side still big right

[jade thunder]

we want to show that the right side is smaller than the right side here

[opal vault]

Now $1\leq |x|$ right?

[thorny flame]

rafilou2003

[jade thunder]

that is true

[opal vault]

So $(1+|x|^2)^{d/2} \leq 2^{d/2}|x|^d$

[thorny flame]

rafilou2003

[jade thunder]

is that from before

[opal vault]

yeah but we need to change it because |x| >= 1

[jade thunder]

so we had this

[opal vault]

$1+|x|^2 \leq 2|x|^2$, is that alright?

[thorny flame]

rafilou2003

[jade thunder]

now you multiply it but something thats > 1 on the right

true

that inequality will still hold

[opal vault]

so now we raise this inequality to the power d/2

[jade thunder]

wait does that work

[opal vault]

gives this

[jade thunder]

no it doesnt

hold up

[opal vault]

we're on |x|>=1 now

[jade thunder]

this comes from the fact that |x| >= 1 right

[opal vault]

yes

[jade thunder]

ok sure

we added |x|^2 to both sides

now we raise both sides to d/2

ah

that is indeed true

[opal vault]

ok good

[jade thunder]

we do get this

[opal vault]

So uh I didn't anticipate this but I'll have to change Cdalpha a bit

[jade thunder]

will it change the stuff from before?

[opal vault]

$C_{d,\alpha} = \max(2^{d/2}p^*{d,\alpha}(\varphi),2^{d/2}\sup{|x|<1}|\partial^\alpha\varphi(x)|)$

[thorny flame]

rafilou2003

[opal vault]

no because we only change the left term

[jade thunder]

ah

[opal vault]

and we used the right term on what we did before

[jade thunder]

here we divide both sides by 2^d/2

[opal vault]

So from this, $\frac{1}{|x|^d}\leq ?$

[thorny flame]

rafilou2003

[jade thunder]

then we sub that in for the denominator

uh huh

very clever

$|\partial^\alpha\varphi(x)|\leq \frac{p_{d,\alpha}^(\varphi)}{|x|^d} \leq \frac{p_{d,\alpha}^(\varphi)}{\frac{(1+|x|^2)^{d/2}}{2^{d/2}}}$

[thorny flame]

Frosst

[jade thunder]

so we have this now

$|\partial^\alpha\varphi(x)|\leq \frac{p{d,\alpha}^(\varphi)}{|x|^d} \leq \frac{2^{d/2}p_{d,\alpha}^(\varphi)}{(1+|x|^2)^{d/2}}$

[thorny flame]

Frosst

[jade thunder]

aha

[opal vault]

yep that's good enough

[jade thunder]

that's the other part of our C_d,alpha

[opal vault]

So $\forall x\in \bR^n, |x| \geq 1$, $|\partial^\alpha\varphi(x)| \leq \frac{C_{d,\alpha}}{(1+|x|^2)^{d/2}}$

[thorny flame]

rafilou2003

[jade thunder]

and also for |x|< 1

so then it's just for all x

[opal vault]

done

[jade thunder]

that is a lot of stuff

well, the other guy did hope i could get to page 3

alright thank you so much for your patience

i will look through this again in the morning

[opal vault]

good luck

[jade thunder]

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[ruby torrent]

how do i

[ruby torrent]

answer

[fast yarrow]

A looks like simple multiplication since 3000 every year, so total money is 3000×7. 7yr for 19 to 25th birthday

B is Geometric progression with
a = 400
r = 2
Sum of Geometric B= a(r^n-1)/(r-1) i think

And tabular form smth like this probably

[topaz sinew]

@ruby torrent Has your question been resolved?

[ruby torrent]

is this for the option A and B?

I did this for option A

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[proud smelt]

I need help in finding the domain of this function

[proud smelt]

y=(x^2-x-2)^-1

Idk what to do in order to find the domain

[acoustic tangle]

You just need to make sure that x^2 - x - 2 is not equal to 0 as 0^-1 is undefined (Assuming you are asked to find the maximal domain within reals)

[proud smelt]

oh so all I need to do is solve x^2-x-2≠0 ?

[acoustic tangle]

Yeah

[proud smelt]

but I don't understand why it can't be equal to 0

[acoustic tangle]

Because 0^-1 is undefined

[proud smelt]

oh

I just realized

thanks man

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[potent terrace]

|x-3| < 2|x|
how do i solve this linear inequality? do i first divide by x to get it to the other side?

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[slender smelt]

Hello there, I'd like to know what is the procedure to getting the inverse function of f(x)= 1/x+2, which is (1/x)-2

[long stirrup]

it's not the correct reverse

[acoustic tangle]

You mean $\frac1{x+2}$?

[slender smelt]

for real?

[thorny flame]

A Lonely Bean

[slender smelt]

yes

[acoustic tangle]

You should have used parenthesis there then lol

1/(x + 2)

[long stirrup]

ah ok

[slender smelt]

my bad

is it possible to move 2 out of the division by changing its sign to negative?

sadly, no

english isn't my native language so I'm not very familiar with terminology in english

[slender smelt]

from what I understand, you get the inverse by swapping x with y

you mean to say that the inverse of x+2 is y-2?

I'm starting to think I wasn't taught how to get inverse functions properly, or that there is something I don't know about reciprocal functions

As i was taught, you start by expressing the function as y = 1/(x+2)

and when you get the inverse, you swap y with x

I'm more curious about how 2 is out of the division in the inverse, and I don't see how swapping x with y explains how the 2 is changed

ok, I understand that, what I don't understand is how 2 is adding to x in the division on f(x) but on the inverse it's outside of the division

the problem is more related to algebra here I think

[thorny flame]

Idris

Idris

Idris

Idris

Idris

[slender smelt]

each day I realize how blind I can be from seeing things

I was brainstorming on this for hours yesterday

not really

for like an hour and a half prob

but damn do I suck at algebra

Yea, don't know how to thank you

sorry for any annoyance

.close

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[placid valley]

Guys, I have a confusion. in a complex number of the form: a+bi. a is the real part and b is the imaginary part, however b is not an imaginary number(when squared will produce a negative) b is a real number what makes it imaginary is i. So why do we refer to be as the imaginary part? Why do we not consider the imaginary part as bi because b itself is not an imaginary number. Or am I wrong?

[placid valley]

I think you meant to write in a help channel rather than an already existing channel for me

[hallow depot]

i don 't see the point ....the immaginary part is the real in front of i

[placid valley]

but why is it imaginary? imaginary part=imaginary number?

[hallow depot]

no

no idris he talked about imaginary number not complex

[placid valley]

yeah but b is a real number, then why do we call it the imaginary part?

[odd pagoda]

we just call it imaginary part because it is in front of the i

its not supposed to be imaginary itself

in fact its nice that it is real because then we can treat it like we can treat any real number

no

just b

[placid valley]

I'm saying that only i is imaginary, b is real so why do we refer to is specifically "imaginary part"

[odd pagoda]

because names

what do you want to hear

[hallow depot]

its the definition !

[odd pagoda]

I can also call it orange part but that would be an even worse name

[hallow depot]

gabe if you want b immaginary then immaginary part of -1=i*i would be i ...that doesn t make sense

[placid valley]

yeah that's fair enough, I was just confused that whether b is an imaginary number itself

[odd pagoda]

no. b is the imaginary part

thats the whole point of why gabe is asking

[placid valley]

exactly

[odd pagoda]

because it is attached to the i

[placid valley]

hmm yeah that make sense, i was only confused by the fact that whether b is an imaginary number, but as it turns out it is a real number but considered as the "imaginary part" of the complex number only because it is the coefficient of an imaginary number.

[hallow depot]

chatgpt best math teacher ,,,

[odd pagoda]

!nogpt

[topaz sinew]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[placid valley]

am I right above?

[odd pagoda]

well specifically it is the coefficient of i

like if you had a+b*(2i) then the imaginary part would be 2b

[placid valley]

yeah but itself is a real number, only its called imaginary part because it is the coefficient of i

I was just stupidly thought that imaginary part=imaginary number, which made this whole confusion but now it's clear

[placid valley]

well I wouldnt say so, it can make you VERY confused

[hallow depot]

i was ironic

[placid valley]

ohh, I thought you were making a statement, my bad 🙂

[hallow depot]

an ironic statement

[placid valley]

yeah

"dog"?

wow, I never would have guessed that you use an animal to describe that something is bad.

Anyway thanks guys for your help, I appreciate it

[hallow depot]

its like "you are an animal!"

it s an insult

[odd pagoda]

which is definitely totally related to the animal

[hallow depot]

greatest of all time?

[placid valley]

I've never heard anyone saying: "you are dog at maths" or smth

Its def not Irish 😄 I'd guess its american

ohh 😄

well speaking for Ireland I've never heard anyone saying it

ohh so thats why I never heard it then 😄

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@placid valley Has your question been resolved?

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[topaz panther]

I need help

[topaz panther]

This is where Im stuck, cant seem to match the left hand side with right hand side

[hallow depot]

induction?

[topaz panther]

Yeah

[hallow depot]

wow beautiful sigma symbol

[topaz panther]

Haha thanks

[hallow depot]

ok got it

[topaz panther]

Im stressed ok

[drifting swift]

yeah it's a bit too wide imo

it looks like you traced it with a ruler

anyway lets see whats up

[topaz panther]

I dont know if Ive done the steps incorrectly or if IM just too dumb to see the answer

[drifting swift]

some typographical alignment issues here

[hallow depot]

$\sum_{k=0}^{n-1}(k+2)2^{k+1}=n2^{n-1}+\sum_{k=0}^{n-2}(k+2)2^{k+1}$ right?

[drifting swift]

ok so uh

[thorny flame]

everg

[drifting swift]

lets just rewrite this into a more manageable form

$\sum_{k=0}^n (k+2) \cdot 2^{k+1} = (n+1) \cdot 2^{n+2}$ by the looks of it

[thorny flame]

AnnGhost

[topaz panther]

So that part is p-2 right hand side + p-1 left hand side

should equal to p-1 right hand side

[drifting swift]

we want to show $(p-2) \cdot 2^{p-1} - (p-3) \cdot 2^{p-2} = (p+1) \cdot 2^p$ by the looks of it

[thorny flame]

AnnGhost

[drifting swift]

unless i fucked up

which i could've

[topaz panther]

That doesnt look right

[drifting swift]

really evil of them to run the sum to n-2, ngl.

such a minefield for fuckups.

[topaz panther]

Yeah I know, first time Ive stumbled upon one

But I assume that if the sum equals to p-2 for an example during the assuming part, all N = to p-2

Is that correct?

[topaz sinew]

@topaz panther Has your question been resolved?

[topaz panther]

Nein

[obsidian moth]

Have you tried induction?

If that's what you were doing in the paper you sent please forgive. You can use your induction hypothesis to substitute the left-hand-side for the right-hand-sight when you do the inductive step, and then simplify from there

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@topaz panther Has your question been resolved?

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[median zenith]

yo

[median zenith]

how do you go about calculating the edge of a graph that has an edge like where the equation has a root in it

for example here

and this being the formula

[quartz flax]

find the point where the value under the root becomes negative

[median zenith]

basically what are the coords

ohhh right

domain

soo here that would be

-2x^2 = -12

x^2 = 6

[quartz flax]

yes

[median zenith]

right

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[vast juniper]

Can i get help on this question:

[topaz sinew]

Please don't occupy multiple help channels.

[proper lily]

oo this is fun

[vast juniper]

can i carry on here

I'm getting that its 1/n^p * 1 * 2 * 3 ... * n-1 * n

[proper lily]

let's look at an example term

[vast juniper]

but as n tends to infinity, 1 / n^p tends to zero but n! tends to infinity

[proper lily]

no clue what youre writing... id need latex visual

but lets look at the term n = 5

[vast juniper]

mb

i know it diverges

but i dont know how to show it

[proper lily]

ah

[vast juniper]

because the numerator tends to infinity at a faster pace than the denominator tends to infinity

so it goes to infinity but that is a very short explanation

[proper lily]

indeed, I'm thinking of a good way to show it

[north spruce]

The idea is you can write out the two equations, n(n-1)(n-2)...(3)(2)(1) and n*n ...p times... n

the factorial term can have a growing number of factors

[proper lily]

id say showing that it monotonically increases is sufficient

[north spruce]

while the n^p term has only ever p factors

so, at a point in the sequence when n! has more than p factors which are all ~n in size, then the numerator will grow faster

as it will have a higher degree in n

[vast juniper]

monotonically meaning it is always increasing right but i wouldnt know how to do that for all n (would i have to differentiate a factorial?)

[vast juniper]

oh ok

[vast juniper]

im trying to understand sorry if im taking some time

[proper lily]

yeah thats my problem with that method lol

possible but a lot of work

[north spruce]

in particular, if look at a factorial n(n-1)(n-2)...(3)(2)(1), we can always truncate to get a strictly smaller number

[vast juniper]

ahhh

[north spruce]

so, as long as n is larger than 6

[hallow depot]

diverges

[vast juniper]

factorial is basically unlimited whereas n^p is limited right

[vast juniper]

why 6 tho?

[north spruce]

exactly!

[north spruce]

we then can write out the first terms we can write as (n)(n-1)(n-2)(n-3)(n-4)(n-5) of our sequence and it will be smaller than n!

so in particular, n! is larger than a 6th degree polynomial

[vast juniper]

yes i figured that out by intuition just need to explain it

[north spruce]

thats just with 6 as an example

[hallow depot]

$\frac{n!}{n^p}=\frac{n}{n}\frac{n-1}{n}...\frac{n-p+1}{n}\frac{(n-p)!}{1}$

[thorny flame]

everg

[north spruce]

you could instead choose larger than p

[hallow depot]

the first p terms goes to 1

the last one diverges

[vast juniper]

oh i didnt know that

[north spruce]

to then get something that is degree p+1 in the numerator and p in the denominator

so the limit would go to infinity

It's pretty cool! it shows that n! grows faster than ANY polynomial in n

at least asymptotically

[vast juniper]

expect for n^n right because it tends to zero if p = n

[north spruce]

yeah!, (n^n) is larger than degree n though since its degree becomes unbounded as n gets larger

[north spruce]

In general we have a heirarchy of polynomial< exponential like (2^n or 3^n) < (n!) < (n^n)

[vast juniper]

i dont understand how you have formed the terms above the fraction, is n! = n * n-1 * ... * (n-p+1) * (n-p)! ?

[hallow depot]

n! is by definition n(n-1)(n-2)...(2)1

[vast juniper]

yh that makes sense, i learnt this in big O notation

[hallow depot]

now I expanded only the first p terms

[north spruce]

its exactly the same thing! the big O dertermines how it behaves in limits at infinity

[hallow depot]

now its more like small o

[vast juniper]

Thank you all for helping, i believe Ive got a good explanation now. I appreciate all the help!

what does that f emoji mean lol?

nvm now ive got to get back to work. Special thanks to @north spruce for explaining the concepts thoroughly!

.close

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[hallow depot]

what a pen!

[lyric mountain]

I'm trying to isolate y. Is there a natural log rule I'm forgetting?

[acoustic pecan]

$e^{ln(x)}=x$

[thorny flame]

AℤØ

[hallow depot]

so y(x)=kx

[acoustic pecan]

huh

oh

yeah

[lyric mountain]

wait

oh okay

So, help me recall this rule. If I multipy both sides by e, would I be able to rewrite it as e^ln(y) = e^(ln(x) ) + e^C

[vast juniper]

e^ln(y) = e^(ln(x) ) * e^C (times e^C)

then y = x * e^C but since e^c is just a constant you can write y = x * A => y = Ax

[lyric mountain]

wait I'm not sure I follow this part

e^c is a constant and I follow that part

[acoustic pecan]

youre not multiplying by e

[lyric mountain]

we usually replace it with k

no?

oh wait

[acoustic pecan]

no, youre making each side an exponent of e

[lyric mountain]

yeah

so it's;
y = x + e^c?

right?

[acoustic pecan]

$$e^{ln(y)}=e^{ln(x)+c}$$

[lyric mountain]

o

o

[thorny flame]

AℤØ

[acoustic pecan]

you can split the right side by the fact x^a * x^b= x^(a+b)

[lyric mountain]

I can rewrite this as
y = e^ln(x) * e^c tho

right?

[acoustic pecan]

no

how would that work

[lyric mountain]

why not

[vast juniper]

times e^c man

[acoustic pecan]

thats like saying if i have 5=3+2 then 2^5=2^3+2^2, 32=8+4

[lyric mountain]

is it correct now then?

[acoustic pecan]

this is fine

just simplify

and replace the constant with some simpler notation if you want

[lyric mountain]

k!

I will use K!

Stewarts book uses k

so it's y = xk!

ughhhhh math is so much nicer when it's not for hw

ty ty both

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[gritty matrix]

I'm confused on this

[gritty matrix]

the teacher explained us and I understood most, but the things underlined in cyan

is what's confusing me

like how did the teacher get those variables

[acoustic pecan]

the shortest distance between a point and a line, is the length of line (perpendicular to the main line) that joins them

the slope of the main line is 1

so the slope of our joining line is -1 since theyre perpendicular

(main line is y=x, joining line is CP)

[gritty matrix]

it's always gonna be -1

or only for this question?

[acoustic pecan]

not necessarily no

but the product of the slopes will always be -1

like if the main line was y=1/2 x, the slope of the joining line would be -2

[gritty matrix]

yes you filp it

how did they know its 1?

I know after it got flipped which = -1

[acoustic pecan]

y=x, y=1*x

m=1

[autumn mica]

y = -x - 2 the teacher wanted a line that's at a right angle perpendicular to y=x so they flipped the slope to -1 but when they plugged in the point (5.5,1.5) to get the equation they made a mistake It should be y = -x + 7 not y = -x - 2

'b' is where the line crosses the y-axis and the teacher was trying to find it but got it wrong for the point (5.5 1.5) b should be 7 not 6

[acoustic pecan]

its (-3,5), not 5.5
y-1.5=-(x--3.5)
y-1.5=-x-3.5
y=-x-2

[gritty matrix]

ohh, and did it mention the slope in the question?

[gritty matrix]

where

[acoustic pecan]

im just saying the teachers work is fine

anyway

[gritty matrix]

so the y= -x + 7?

[acoustic pecan]

the slope was from the fact the main line is y=x, so the slope is 1

[acoustic pecan]

no

[gritty matrix]

or was the teachers work correct

[acoustic pecan]

it is y=-x-2

[gritty matrix]

but they said

[acoustic pecan]

they were mistaken

[gritty matrix]

ohh

ok

[acoustic pecan]

they read the point C wrong

[gritty matrix]

and this

the underlined cyan

where did they get that from

[acoustic pecan]

y=x is given in the question

[gritty matrix]

ohhh

[acoustic pecan]

its the yellow line of this diagram

[gritty matrix]

so could I also have used elimination?

[acoustic pecan]

no?

oh

[gritty matrix]

or only substitution

[acoustic pecan]

you mean to find the intersection?

[gritty matrix]

like the teacher here used substitution

could I have used elimination instead

[acoustic pecan]

if you wanted to use elimination, feel free to

[gritty matrix]

okay thanks

[autumn mica]

yes

[gritty matrix]

ty

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[neon iron]

hi

part d

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[neon iron]

seems like greedy on highest urgency?

[fluid belfry]

?

[neon iron]

question is above

[fluid belfry]

oh mb

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[slow tusk]

what did they do in this step? why does 8y+2x become 4y+x

i know it’s divided by 2, but why?

[fallow chasm]

They multiplied both sides of the equality by 8y+2x. So it cancels out on the LHS and becomes 1/2(8y+2x) = 4y+x on the RHS.

[slow tusk]

Im stoopid

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[pearl pulsar]

I'm trying to show that the function f:(-1,1)-> R f=x/(1-x^2) is bijective. For surjectivity, I got the quadratic equation: 1+-sqrt(1+4a^2)/2a but I'm not sure how to show that one of these is in (-1,1)

[pearl pulsar]

I'm trying to show that the function $f:(-1,1)\to\mathbb{R}, f=\frac{x}{(1-x^2)} {}$ is bijective. For surjectivity, I got the quadratic equation: ${} x=\frac{1\pm\sqrt{ (1+4a^2) }}{2a} {}$ but I'm not sure how to show that one of these is in (-1,1)

[thorny flame]

Jesses

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[pearl pulsar]

<@&286206848099549185>

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[gritty mirage]

Here is a few tips:
You want to show that the absolute value of your expression is less than 1.
Since |a+b>|a-b| for all positive a and b, you want to consider the version with a neqative sign

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[atomic anchor]

Describe the sequence of rigid transformations that shows ABC is congruent to A’B’C’ ?
A (5,7), A' (-2,7)
B (0,2) , B' (3, 2)
C' (7,4), C' (1,9)

[atomic anchor]

I tried to do this but it was too difficult for me I need somebody to help me out

What I did was

I was already kinda out of ideas so I just reflected onto the y axis

But then I can’t translate point C correctly

So that idea was wrong

And now I’m stuck

Uh

<@&286206848099549185> sorry, could you please help me out here?

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[molten marlin]

can someone help me with this

i just started integrals

i dont understand how they got that'

[silver trench]

we know that the maximum value of the interval is 2

[molten marlin]

yes

[silver trench]

we dont know what the true area of under the graph of that region is

but we know that i cant exceed 6

[molten marlin]

why not?

cuz of the interval and maxvalue?

max value

[silver trench]

yes

[molten marlin]

2*3

[silver trench]

let me draw you a graph

[molten marlin]

?

alr

[silver trench]

this would be the only possible way for the area to be 6

[molten marlin]

yea

[silver trench]

is if the value of the function between 0-3 is constant (2)

[molten marlin]

oh ok makes sense

also

it cant be A or B cuz we dont know what the graph looks like and dont know if its increasing or decreasing right

or no?

[silver trench]

yup

[molten marlin]

what about C

[molten marlin]

@silver trench

[silver trench]

it could be c, since it has a max value of 2

so we cant rule it out

[molten marlin]

oh alr

ty for the help

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[ornate magnet]

What have you tried

[quaint plover]

i am clueless right now

[ornate magnet]

ζ= rF sinθ

rsinθ= r perpendicular

[quaint plover]

is the perpendicular 0.45m?

[ornate magnet]

Do you know this?

[ornate magnet]

Yes

[quaint plover]

what is the torque then?

nvm i have the calculate it

is the answer 45?

[ornate magnet]

Yes

[quaint plover]

thank you

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[limpid mica]

trying to solve this ode with these initial conditions but im lost

[limpid mica]

this is what i got so far but idk if it is right

i dont think it makes sense to me because wouldnt i need to find e^pi?

[sweet shard]

e^pi is just a number

,calc e^pi

[thorny flame]

Result:

23.140692632779

[limpid mica]

hm

it still does not fit the answer (i have the key for it) im just lost on how to get to it

as c1 is 1, c2 os -1

i got c1 as 1

i got this

i am so tupid i did the derivative wrong

i got derivative e^x in then product rule as multiplying another e^x, for some reason

ty

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[neon iron]

I don't know where to start

i don't want the answer

i just don't know where to start

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@neon iron Has your question been resolved?

[neon iron]

No

<@&286206848099549185>

[magic ingot]

you should start by writing the numers that use only th edigits 1,2,3,5 and 7

then look at the first 31 numbers which are divisible by 5

add those numbers up

and ur done

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[mortal spear]

I think I got it, just need someone to point up my mistake

[mortal spear]

we re-writte the denominator

and use trig-subs

where

and

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[native moth]

.solved

[cosmic jackal]

.reopen\

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[sharp dew]

is there any part of your question which makes you confused?

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@barren garnet Has your question been resolved?

[barren garnet]

I tried to solve it with formula but it is so lengthy is there any short method?

[sharp dew]

ofc

[barren garnet]

Please guide

[sharp dew]

the bottom one is quite commonly used to solve questions revolve standard deviation.

[barren garnet]

Yes. I used this formula

But it became complicated

In last term

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[barren garnet]

.close

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[eternal spoke]

,w plot |x|

[thorny flame]

Results provided by WolframAlpha

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

Requested by masteroogway0599

[eternal spoke]

Why does the graph have negative values

[junior acorn]

it doesnt tho

[eternal spoke]

Oh yeah im dumb

Thanks

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[neon iron]

\def \P{\mathrm P}
Does [
\map \P{\overline A \cap \overline B} = 1 - \map \P{A\cup B}
]

[neon iron]

just a quick question

[carmine marten]

yes (by de morgan)

[neon iron]

ty

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[neon iron]

If each coded item in a catalog begins with 2 distinct letters followed by 3 distinct nonzero digits, find the probability of randomly selecting one of these coded items with the first letter a vowel and the last digit even.

[neon iron]

i have [
\map \P A = \f{25\cdot 8 \cdot 4 \cdot 5 \cdot 4}{26 \cdot 25 \cdot 9 \cdot 8 \cdot 7}
]

would this be true perhaps

hmm

[lean meteor]

5 x 25 x 4 x 8 x7

why u get 4 there

[neon iron]

yeah 7

typo

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[lament halo]

.close

[neon iron]

Q6

[drifting swift]

to cut away all the extra distracting questions:

$x^2 - 2x - \log_2|1-x| = 3$

[thorny flame]

AnnGhost

[neon iron]

Number of roots of the equation x2 – 2x – log2 | 1 – x | = 3 is

[drifting swift]

this is your equation and you want to find how many roots it has, yes?

[neon iron]

yes

[drifting swift]

ok

progress?

[neon iron]

umm yes

[drifting swift]

!show

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

[neon iron]

i dont have a phone rn

i can tell

got 3 in rhs

lhs*

then completed the square

then simplified a bit

[drifting swift]

and what did you end up with

[neon iron]

2^(x+1)(x-3) + x = 1 ,,,, for x <1

2^(x+1)(x-3) = x -1 ,,,,, for x >1

[drifting swift]

this is suspicious

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[neon iron]

?

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[neon iron]

can anyone help me with this i was absent from school when we learnt it and dont understand it

[weak zealot]

A force of 200N acts over the area of 40cm^2 calculate the pressure

can any one help me

[craggy haven]

this isn't your channel, find an unclaimed one up top

[craggy haven]

look up law of cosines

[neon iron]

YE CANT UNDERSTAND IT

[weak zealot]

oh sorry

[weak zealot]

i am new in this server

[craggy haven]

okay, please stop shouting. perhaps you might find a video lecture useful? there are plenty available

[neon iron]

i tried calculating 4x but my answers were all wrong

[lone nest]

This looks more like a law of sines problem.

[neon iron]

can someone help?

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@neon iron Has your question been resolved?

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@neon iron Has your question been resolved?

[neon iron]

no

[wicked hemlock]

sin rule states that sinA/a = sinB/b = sinC/c, a being the length opposite to angle A. essentially it's just plugging in numbers... that would give you a basic angle (which, in this case doesn't happen to be obtuse). use the ASTC quadrants to find an obtuse value of C. the quadrants part might be a bit more difficult to understand, i'd recommend consolidating with your teacher or watching a video explanation for that

this is the quadrant thing i'm referring to. apologies if i'm not using the correct terminology

[wicked hemlock]

man.. im sorry

[candid pilot]

where is the question

[twin linden]

Hello?

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@neon iron Has your question been resolved?

[candid pilot]

@neon iron Show us the question

[neon iron]

its up there

[candid pilot]

Have you learned sine rule?

$$\frac{\sin(A)}{A} = \frac{\sin(B)}{B}$$

[thorny flame]

Ratatatat

[candid pilot]

Use this to solve the problem. Remember the angle is obtuse

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[toxic stirrup]

i got to lim (cos(8.5/x^8)*8.5/x^8)/(-9/x^16)

but what do i do next

[eternal wing]

I'd say start from scratch again since that's not the correct approach

Have you learned about this limit?

$\lim_{x \to 0} \frac{\sin x}{x}$

[thorny flame]

RedstonePlayz09

[eternal wing]

@toxic stirrup

[toxic stirrup]

yeah

its 1

[eternal wing]

Good

Try to transform your limit into something similar

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[toxic stirrup]

I’m not sure how I could to that

[eternal wing]

You want sin(something)/something

Just keep the value of the expression the same, but move it and manipulate it in such a way, so that you have sin(?)/?

If whatever is inside the sin and in the denominator approaches 0, the limit of that part will be 1

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[violet current]

Can some please help me in this? Is it correct? If not please tell me what is the correct answer and explain it please

[violet current]

<@&286206848099549185>

[topaz sinew]

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[fluid belfry]

dude like atleast give them some time to even read your question

[topaz sinew]

@violet current Has your question been resolved?

[violet current]

Okay sure why not (read the English part)

Sorry

[spark grove]

it is correct

[violet current]

Sure? Then which are the f’’ (second derivative)?

Which represents

[spark grove]

f'' on the first picture is the red graph

f'' on the second picture is the blue one

whenever a graph is decreasing, the derivative is negative

[violet current]

So basically, the lower the graph, the more the derivative is increasing, like going from the first to second derivative, right?

[spark grove]

could you clarify what you mean by "the lower graph"

sorry my english isnt perfect

[violet current]

Like in the first picture, I can see that the red line is going more straighter/lower, so it means that it is the second derivative

Like when the graph is being more straight does that mean it going through more the derivatives?

f’ straight, and f” more straight

[violet current]

It’s totally fine

So am I right and got the idea?

[spark grove]

i look at these problems by trying find f(x) first. The (simple) definition of a derivative is a rate of change, so when a function is increaing, that means its rate of change is positive, which means the derivative becomes positive there. If you look at the first picture, the blue curve begins increasing at x=-1. you can notice that the black curve becomes positive at x=-1, so therefore, the black curve is the derivative of the blue curve. notice how when the black curve begins decreasing at x=1.5, the red curve becomes negative, so therefore, on the first picture, the red curve is the derivative of the black curve.

[spark grove]

when the graph of a function becomes straight (horizontal), its rate of change (slope) is 0, so the derivative of the function is at y=0

[spark grove]

this is a way to think about it, but its not necessarily correct. I think that you are thinking that the derivative function of a polynomial is one degree lower (for example, the derivative of x^2 is 2x) which does make it look straighter in most cases. I wouldn't use that for this type of question though, because you only get to see the graph of the functions, not the equations of the functions itself

[violet current]

Hmmmmm, that was an excellent explanation tbh. I got the idea mate

Tysm for helping

🤝

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[rare jewel]

Confused about this, only question I really dont know lol

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[rare jewel]

any help

: (

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[verbal crater]

the population proportion is just X/n, or the amount of people that think they dont need a television divided by the total amount of people surveyed

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[graceful leaf]

hello

when i type the -3.5 for X in my calculator is error sadly

is this undefined or something like that

[verbal crater]

how did you type it

[graceful leaf]

i take pic 1 sec

idk i did the same format for the other ones i guess this one is different

[verbal crater]

huh weird

[whole sleet]

Return ur calculator

[graceful leaf]

oh

have to write in those paranthese ok

[obtuse tundra]

No

U need to use the minus at the bottom

(-)

[graceful leaf]

oh

ok

i try 1 sec

[whole sleet]

uh

I think your calculator broken man

[verbal crater]

never used one of those calculators lol, always used a casio

[graceful leaf]

I tried that and still error

[graceful leaf]

maybe

i just got it tho

is this something possible without calculator

or its too much

[sand ivy]

@graceful leaf u can simplify first using exponent laws:

(1/5) ^ (-2.5)
= ((1/5) ^ (-1)) ^ (2.5)
= 5^2.5

the error probabably has to do with it having trouble doing negative exponents so if u do that by hand u should be able to type it in now

or try using negative sign (left of enter) instead of subtraction sign in ur exponent

[topaz sinew]

@graceful leaf Has your question been resolved?

[graceful leaf]

wow thnxs very much

let me try ok

i forgot to put

whole thing

tahts why the error

Lol

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[rigid vortex]

i'm struggling to understand what the logic from 'inspect these summations further...' and how this changes the indices on the summations so you change the sum to n into a sum to infinity

[rigid vortex]

basically, how does n>=k warrant the step that comes after?

[odd pagoda]

it might help to write out the first few terms of all the sum notations

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[south ferry]

Hello, I have the following logistic regression definition

[south ferry]

it's clear to me, that we can basically treat it as a probability and use it for classification problems.

yet I'm confused by $P(Y=y | x;\theta) = \sigma(y\mathbf{x}\cdot\mathbf{\theta})$.

resp. confused by how the loss is defined.

if we e.g. look up the logistic unit we see

so why do we have the y in the argument of our sigmoid?

[thorny flame]

Pascal

[south ferry]

hmm I guess mathematically it kinda makes sense, just somewhat weirdly notated (at least for me) but I still don't understand what's going on exactly.

[strange whale]

re pascal

to go back to hyperplanes

we're trying to do binary classification here, i.e. we're trying to see where the points x with y=1 are located in space, and same for the x's with y=-1

the way we're doing that here is with a hyperplane

the hyperplane splits the x-space in 2

on one of the half-spaces you'll consider that y=1

on the other one you'll consider that y=-1

that's just basic binary classification

now logistic regression tries to add some probability info

the closer you are to the hyperplane, the less sure you are about the result (1 or -1)

and conversely the farther you are, the more sure you are about the result

just trying to give a very high level view of what's going on here

[south ferry]

yeah I get that mostly but I fail to properly interpret the math here.

[strange whale]

you want to say something else or no ?

[south ferry]

We have a bunch of data points with labels and we draw a line that separates them (in a 2D case) i.e. we tune theta. Right? Now how do we tune theta, we do that by defining a loss function and then solveing the resulting optimization problem. In our case, the loss function is the cross entropy i.e.

H(p;q) = - sum_i( p_i* log(q_i)) (iirc)

Now p is basically the "distrbution of our labels" i.e. "y". We only have i=1 so we get -y*log(q_i)

hmm i remember that yesterday, when I looked at that I didn't fully get the connection which is probably why I struggle now.

I guess I'd rather bother a TA with that.

I really wanna see it being derived properly

[strange whale]

hoping the TAs are willing to

I gtg pretty soon also

unfortunately

[south ferry]

they have to 😄

otherwise math stackoverflow 😛

[strange whale]

I'll force them to, whatever the cost

[strange whale]

the ultimate answer

just wondering, what's that book you got ?

[south ferry]

ah just lecture notes from a deep learning course

[strange whale]

ah okey

[south ferry]

it's not public but I think the course is based on

https://www.deeplearningbook.org/

[strange whale]

alright alright

[south ferry]

maybe

Logistic regression is basically just taking a linear assumption e.g. f(x) = x * theta and then defining a loss function lik e.g.

• Perceptron
• least squares
• log loss

and then optimizing theta, right?

[strange whale]

well ultimately yes that's the goal

[south ferry]

funny how I probably did all that stuff a shit ton of times but never really tried to understnad the math behind if properly.

[strange whale]

to seperate the -1 and +1 classes in the nicest way

nicest being defined by the loss here

[south ferry]

yes

thanks you too

anyway I have to leave. thanks again.

have a nice day you two

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[wintry oar]

Just a quick log question. Log10(3)=m, log10(5)=n Q:find log15(30) in m and n

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[wintry oar]

<@&286206848099549185>

[twin linden]

hi

[wintry oar]

Hi

[wintry oar]

Meter?

I think it can be rewritten as log10(30)\log10(15)

[wintry oar]

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[drifting stream]

How do i solve (-2/cot^2(3t)), i need to derivate.

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@drifting stream Has your question been resolved?

[steel violet]

u need derivative of that right

[drifting stream]

Yed

Yes*

[steel violet]

first write 1/cot^2(3t) as tann^23t

then use chain rule

[drifting stream]

-2 tan^2(3t)?

[steel violet]

yes now use chain rule

[drifting stream]

I dont remember how to do that

I derivate 3t?

[steel violet]

in chain rule we follow the rule power function angle

first we do derivative of power of tan^2(3t)

so it will be 2tan(3t) right

[drifting stream]

Oh yes yes

[steel violet]

then we do deriivative of function that is tan3t as sec^2(3t)

then we do derivative of 3t

[drifting stream]

-2*2 sec^2(3t)?

[drifting stream]

-4 sec^2(3)?

[steel violet]

no each step has to be multiplied

it will be -2* 2tan3t* sec^2(3t)*3

[drifting stream]

Thats the final answer?

[steel violet]

yes

[drifting stream]

Oh so u dont delete tan

U just add sec

But where did that las *3 came from?

[steel violet]

derivative of 3t

[drifting stream]

Ohhhh

Tytyty

Helped me a lot ❤️

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[fluid belfry]

how did they get h= sqrt 3 L/2

[drifting swift]

either trig or pythagoras

you can split the equilateral triangle in half with an altitude

it'll have hypotenuse L and legs L/2 and h

[fluid belfry]

yeaa im getting it by pythagoras then

Thankyou

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[late crypt]

Why is arg(z^25)=arg z=(i-1)

[rigid ivy]

what is z

[late crypt]

X+yi