help-42

[remote mural]

Tbh

[young zephyr]

i meant sequence*

[remote mural]

I have trouble distinguishing

I think so

[young zephyr]

im no good at math so dont take what i say too seriously

[golden frost]

sequence is just a list of numbers

series is a list of added numbers

[remote mural]

So the context of that problem it’s a sequence

I think

I just don’t ljke how it says it doubles every four hours

Cuz now getting the ratio is hard

[golden frost]

is n in days or hours?

[remote mural]

That’s what I’m wondering

I think I have to convert it to days

Because an is in the days measurement

But idk

[golden frost]

if it's in days then do 6(n-1) for the exponent

cuz it doubles every 4 hours

so it doubles 6 times a day

[remote mural]

You mean 6^(n-1)?

I’m tryna get the R value

Not the n value

[golden frost]

oh r is 2

cuz it doubles

[remote mural]

So it’s 3 million(2)^6(n-1)?

[golden frost]

yeah

looks right

[remote mural]

Your correct

What what’s the n value?

[golden frost]

the day which it is

so in this case day 2

[remote mural]

Ohhhthats what I was doing wrong

I assumed n would be 1

Thanks

.close

[calm coral]

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[verbal otter]

how do you find the equation of a cubic graph

[verbal otter]

is there a video

beacuse for me the equation is k(x - a)(x - b)(x - c)

but i see this function and i get confused f(x) = ax³ + bx² + cx + d

[calm coral]

@verbal otter Has your question been resolved?

[thick sinew]

do you have a sample?

[verbal otter]

Lemme get the question rq

[thick sinew]

if you're given a graph and want to determine the function you can use k(x - a)(x - b)(x - c) as you mentioned

now you just need to multiply out the brackets

which gives you the function in form ex³+fx²+gx+h

[verbal otter]

sorry for the trash quality 💀

[thick sinew]

ok so we can see the x-intersections

-2 1 3 I presume

so your function would be k(x+2)(x-1)(x-3)

and you still need to find k

@verbal otter so far clear?

[verbal otter]

Yeah thanks for the help

[thick sinew]

can you find k?

[verbal otter]

not really sure how to do that

[thick sinew]

so far we used all intersections with the x axis

meaning where the function is equal to 0

so let's find another point on the graph whose value we have

the y-axis intersection

is at 6

meaning f(0) = 6

and we know f(x) = k(x+2)(x-1)(x-3)

so let's set these two equal to find k

f(0) = k(0+2)(0-1)(0-3) = k*6

and since f(0) = 6 ====> k = 1

[verbal otter]

Ill try to send a better photo

[thick sinew]

did I not read it right?

now we'd have that the function is g(x) = 6(x+2)(x-1)(x-3)

[verbal otter]

No its just hard for me to understand lol

[thick sinew]

yea so x-intersections are at -2 1 3 right

of g(x)

@verbal otter ?

[verbal otter]

ye

[thick sinew]

meaning we know g(x) = k(x+2)(x-1)(x-3)

since we have the roots

that part clear?

[verbal otter]

wait so why do we put it in that order instead

i thought it was (x-2)

[thick sinew]

nop since we want -2 to be a root

meaning that if our function has the factor (x+2)

then if we insert x = -2

that factor becomes 0

:)

[verbal otter]

ohhhh

[thick sinew]

that's why you flip the signs

k so we have g(x) = k(x+2)(x-1)(x-3)

now we need k

the graph g(x) intesects the y-axis at 6

so we know g(0) = 6

.-.

[verbal otter]

wait so is 6=k or just g(0)

[thick sinew]

obv we need to find that out first

g(0) = k(x+2)(x-1)(x-3) = k*6

and since we know g(0) = 6

==> k = 1

[verbal otter]

thanks for the help bro

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[bleak socket]

I don't know where to start with a problem like this- I was always really bad at the 3D conceptual stuff.

[marsh valley]

This triangle they're talking about lies inside a plane. Now with three points, you should be able to get two vectors in that plane. Do you have a tool that, if given two vectors, will spit out a vector orthogonal to both?

[bleak socket]

Cross product?

[marsh valley]

Indeed.

So try and find two vectors using the coordinates you're given. Then use the cross product to find a vector orthogonal to the plane the triangle is in (and hence orthogonal to the triangle itself).

You'll have to normalize it, since they are asking for a unit vector.

[bleak socket]

This correct so far?

[marsh valley]

Careful, you have to take the cross product of vectors not points.

So for instance, a first one could be the vector given by the side with endpoints (4,2,2) and (1,0,-4), which you can get the vector for by subtracting those two, giving (3, 2, 6)

[bleak socket]

Oh, right, I forgot about that part!

[bleak socket]

So then I could get another possible vector by doing (4,2,2) and (0,2,1) right?

[marsh valley]

Yes, that would work.

[bleak socket]

Which would be 4, 0, 1?

Alright, let me try again with this in mind!

[bleak socket]

This look good so far?

[marsh valley]

Yes, that should be right!

[marsh valley]

.

[bleak socket]

Oomf, forgot about that part too.

Admittedly I don’t understand normalizing it all that much, what does that mean again?

To make it orthogonal right?

I thought we had already done it in the form of the cross product here?

[marsh valley]

Now you have the vector that is orthogonal to the triangle.

YOu want it normalized, so you have to divide it by its own norm.

[bleak socket]

I’m sorry, I don’t believe I quite understand what you mean.

[marsh valley]

RIght now, the vector you have has some length that isn't 1.

You want it to have the same direction, but length 1.

If you divide the vector by the length it has, then it keeps the same direction, and the resulting vector has length 1.

[bleak socket]

So divide it by itself? I think I’m recalling unit vector definition now from what you’ve said, although it’s a bit fuzzy since I still don’t have the greatest handle on it.

[marsh valley]

Yes somewhat. You divide by the length of the vector.

For instance, the vector (1,1) has length sqrt(1^2 + 1^2) = sqrt(2). This isn't 1. So if we take the vector

1/sqrt(2) * (1, 1) = (1/sqrt(2), 1/sqrt(2)), this new vector has length 1 but it still has the same direction.

[bleak socket]

Oh, I remember that length calculation- so it’d be the coordinates I have currently divided by the length?

Which would be the square root of each coordinate squared and added together?

[marsh valley]

Yes I think you got it

Try and do it and show me what you get.

[bleak socket]

So that’d be 2, 21, and -8 divided by the square root of 509?

[marsh valley]

That should be it.

[bleak socket]

That’s very messy…

[bleak socket]

How does this look?

[marsh valley]

I think that's good if they want 2 digits of precision.

[bleak socket]

Got it correct, thank you!

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[bleak socket]

Not sure where i went wrong here?

[bleak socket]

https://www.youtube.com/watch?v=CD0MXPhRkvQ

I followed this video's instructions.

[quaint sphinx]

do u have work

[bleak socket]

That last bit next to the k says (36-40)

Nevermind, I just solved it- it was a calculation issue. Thank you!

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[remote mural]

What did I do wrong?

I did every step correctly

[crystal forge]

what was your reasoning for setting up this equation? 15 = 3n + 2

[harsh siren]

15 is your n

not a_n

[remote mural]

15 is an

Oh

What

What’s an then

[crystal forge]

these are your a_n values, you listed them before, what you mena to do is find the last one i assume? that would be plugging in 15 for n to find the a_15

[harsh siren]

you dont need to know it

you have a, d and n

then just use the sum of arithmetic progression formula

[remote mural]

Why is n 15

And not an

[harsh siren]

a_n is the score available on the nth question

not the total number of questions

15 is your NUMBER of questions

[crystal forge]

a_n represents how many points the nth question is worth doesn't it? the test has 15 questions, 15 is not a particular amount of points a question is worth

[remote mural]

Ohh

Did I get an and n correctly this time?

[harsh siren]

two weeks is 14 days no?

[remote mural]

I’m not thinking straight today

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[desert falcon]

How many ways can you arrange the letters in the word ARRANGEMENT such that no two As are adjacent and all Rs are adjacent

[thick sinew]

neat question

[desert falcon]

Maybe if it was just one condition say, no two As together, I could have gone about it by doing 9! × 2C2. ah no there's reptitions. it's confusing

[thick sinew]

can you find the number of arrangements if no restrictions are set?

[desert falcon]

that's 11!/(2!×2!×2!×2!) right?

[thick sinew]

exactly!

now, let's look at the second restriction

two R's are adjacent

hm, we could see them as one element in that case

since they're always next to each other

so we only have 10! / (2! * 2! * 2!)

so far clear right

[desert falcon]

yes got it

[desert falcon]

yup makes sense

[thick sinew]

ok, now what we can do is:

we look at the negated first restriction

so we'll look at all cases where A's are together and R's are together.

and we subtract these cases from the cases where R's are together

because then we get the cases where R's are together and A's aren't together

[desert falcon]

alright yes

[thick sinew]

to express it in set theory: |notX and Y| = |Y| - |X and Y|

where X, Y are your restrictions here

[desert falcon]

okayy. i can't visualize this but I get that in set theory it's this

okay I can

[desert falcon]

Will this be 9!/2!×2!?

[thick sinew]

yup!

since we have 9 elements

and 2 duplicates twice

[desert falcon]

right!

[thick sinew]

therefore:

10! / (2! * 2! * 2!) - 9! / (2! * 2!)

[R's adjacent] - [R's & A's adjacent] = [R's adjacent and A's not adjacent]

and the result is 362880 🦇

[desert falcon]

yess

[desert falcon]

oh damn okay

[thick sinew]

as a general advice, it's often easier to consider the opposite case

don't get too hung up on specifically finding the combinations for the desired case

bc subtracing the opposite is sometimes easier

[desert falcon]

say just for this one restriction where we have to find out the number of arrangements with no two As adjacent.
And I know one approach now is to consider the opposite case and subtract it from 11/2!×2!×2!×2!

Although if I were to do it directly, how could I do it?

[thick sinew]

yup!

[thick sinew]

hm, would have to think a bit, makes the cases quite annoying

probably by tree splitting

but maybe there's a faster way

[desert falcon]

by directly I mean, I saw this one video where he did it by thinking of this as gaps between the letters.

so it was no vowels together in a word.
so he arranged all the consonants and now there's a number of spaces between them for the vowels

[tawdry lodge]

lunatic mind helping me out when u finish this?

[desert falcon]

sorry if I'm taking some timee

[thick sinew]

you can occupy your own channel here and passengers will eventually come by and guide :)

[tawdry lodge]

i did and ive been waiting for 30 minutes lol

[thick sinew]

nono you occupied the channel, you can keep it as long as you wish

the channels don't run out

others can simply open a new one

[desert falcon]

oh okay

[thick sinew]

oh k, will look in a sec

[tawdry lodge]

ty and good luck harry

[thick sinew]

hm I think tree splitting would be the best I guess?

but it's still somewhat horrible in contrast to doing the opposite case

[obtuse pebble]

@tawdry lodge sir can u help me

[thick sinew]

.-.

[desert falcon]

It was this

[desert falcon]

yes I understand

[tawdry lodge]

im in the 10th grade lol dont know logarithms

[thick sinew]

yea in this case it's easier to do this separate aspect

first placing down the consonants and then the vowels

would work for the initial as well true

[desert falcon]

why was it 6P3?

[thick sinew]

what do you use APB as notation for?

[desert falcon]

permutations

[thick sinew]

because you place three vowels onto 6 possible spots

[obtuse pebble]

ok

[thick sinew]

-D-G-H-T-R-

each - is a possible spot

for the vowels

and you distribute 3 of them onto the 6 spots

[desert falcon]

in our case, or in any case where say all the vowels were same we'd do 6C3 tho right?

[thick sinew]

that's by 6P3

[thick sinew]

yea since order is irrelevant

[desert falcon]

Okay okay I've got it

thank you!

[thick sinew]

np 🐛

[desert falcon]

there's a long queue for you ig so I'll close ty. .close

.close

[calm coral]

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[thick sinew]

I just do as I like (as everybody does)

essentially voluntary guidance for strangers :)

[desert falcon]

true. It's voluntary

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[lost ice]

i forgot how to change my bounds....

[lost ice]

when R=0, and R=1, what is x?

omfg

nvm

so tired

.close

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[modern peak]

let's call the point at which the rectangle touches the circle "P"

we know that OP = CD, correct?

[final zephyr]

Yes

[modern peak]

well, we know OP is the radius

which is 6

done

and also

!done

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[final zephyr]

So its 6

[modern peak]

yes

[final zephyr]

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[tawdry lodge]

@final zephyr if ur just looking for answers in this server you wont learn anything im guessing this is a test for your math class or something

[modern peak]

nah it's probably practice

[final zephyr]

It was practise test online

That is failed a week ago

That is

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[mossy wigeon]

for integration by parts here

would it make sense to consider u to be ln(2x+1)

[exotic falcon]

perchance

[mossy wigeon]

im using the tabular method and its pretty unending

[dreamy hull]

It helps if you know the indefinite integral of log

[mossy wigeon]

any tips on where to stop?

[quaint sphinx]

look for the line u know how to integrate

[mossy wigeon]

so i integrate until idk how to integrate anymore?

[quaint sphinx]

no

[mossy wigeon]

im not too sure if i nudertsand

ah

[exotic falcon]

don't use tabular method for this if you're getting confused

[quaint sphinx]

it helps to understand what's actually going on when applying ibp

[exotic falcon]

just do basic integration by parts

look at the integral you get

seee if you can solve that

[quaint sphinx]

👀

[mossy wigeon]

yeah essentially im just getting a bucnh of u v

[exotic falcon]

tabular method is like a speed up trick for IBP, but you have to really know what is going on like Mqnic said

[mossy wigeon]

but usually it stops when i can lets say simplify it

but when it comes to ln idk wheres the line i should stop at

[exotic falcon]

Try it without using the tabular method

Show what the result is

(if you get stuck)

[mossy wigeon]

here

[exotic falcon]

at a brief glance it looks like you got it

are you still stuck?

[mossy wigeon]

mhm not on this problem in particular but

i wanna ask how i can translate

where to stop here on the tabular

[exotic falcon]

If you don't know where to stop using the tabular method don't use it

Normal integration by parts is all you need

I've never used that tabular method in any of my classes

[mossy wigeon]

thing is my teahcer knda requires it

[exotic falcon]

I doubt that

[mossy wigeon]

at least he said idk if hes gonnna grade it differently

[exotic falcon]

That would be very dumb, as the tabular method is just integration by parts

written less cleanly as to make you go faster

[mossy wigeon]

true

alos im redoing it cuz i did it worng apparently

[mossy wigeon]

wait

just curious

cuz i think i figured it out

on tabular do we stop when we can find a product on the row that can be integrated

anyway ty!

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[thick sinew]

did I utilise the properties of the binomial series correctly?

[calm coral]

@thick sinew Has your question been resolved?

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@thick sinew Has your question been resolved?

[thick sinew]

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[hidden ice]

https://i.imgur.com/6CjyWas.png

[hidden ice]

I dont know what I did wrong

u = 3x+9

(u-9)/3 = x

So sub it in to get

((u-9)/3)(u)^9

[gloomy badge]

Remember, dx = du/3

[hidden ice]

Im not following sry

Which step was wrong

[gloomy badge]

No wrong steps so far, just reminding you that dx has to be changed into du/3

[hidden ice]

I still dont follow sry

[gloomy badge]

OK, when you substitute, you have to substitute for the dx part as well

[hidden ice]

Which dx part

Could you show me please im not following

[gloomy badge]

The dx part in the integral

[hidden ice]

Why does it become du/3

[gloomy badge]

Well, you're saying u = 3x+9, right

(back in 2m)

[hidden ice]

Ahhhh yes

I think I remember this rule now

That was what I did wrong 😇

Thank you! I think I can finish it now

❤️

.close

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[burnt saffron]

.reopen

[leaden bluff]

i need help to solve it

[remote mural]

try to convert that (i^2)/(4^i)

[leaden bluff]

that is what i have now

teacher explained how to do it, but havent finished

i also know that sum of i/4^i = 4/9

and sum of 1/4^i = 4/3

but it's from i =0 to oo

[leaden bluff]

here we have k = 1 not 0 in the last line and i don't know why

someone please help

gpt saing that the answear is 5\9

but when i am splitting the last equation i have:

sum of (2k-1) / 4^k

sum of 2k / 4^k - sum of 1 / 4^k

second sum is 4/3 from what i know

so we have

sum of 2k / 4^k + 4/3

sum of 2k / 4^k is my question i have no idea what to do with it. I tried the next:

sum of 2k/4^k is two sums of k/4^k which is from what i know 4/9. so:

2 * 4/9 which is 8/9

then we have

8/9 - 4/3 = - 4/9

which is not correct as i suspect

and i don't know where is my mistake

here is my messi notes if it will help

I have explained my problem as detailed as possible for me. Please help me to solve it and if you need something else - ask

[leaden bluff]

what do you mean

convert to what

[remote mural]

sry for my bad english 😢

[leaden bluff]

no worries

[remote mural]

i mean this and stuck 😢

[leaden bluff]

not really helpful honestly

[calm coral]

@leaden bluff Has your question been resolved?

[leaden bluff]

.close

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[exotic gate]

Questions: (1) what is an algebraic subvariety? it does not seem to match the definition on wikipedia where it is the set of zeros while here it is a polynomial system, (2) what is the set sigma in the example supposed to be? a set of polynomials? (3) how is sigma calculated in the example?

[exotic gate]

R is the vector space representing systems of n polynomials in n variables with complex coefficients

[calm coral]

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[noble notch]

Working on a lab, could someone confirm that I'm doing my math when substituting values of k into the chi-square distribution equation?

[noble notch]

here’s k=2, k=4 will follow shortly

above is k=4

[calm coral]

@noble notch Has your question been resolved?

[noble notch]

<@&286206848099549185>

Please help

Oh wait I just realized sumn

The gamma function is in terms of k/2, not x, silly me

new k=4

new k=2

oops- factorial should be on the outside of the parentheses

I finally got it I think, lemme draw them graphs

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[drowsy tide]

after I put 3x on the same side with 8 do I divide by -2?

[dull wagon]

depends on what the question is asking for

[drowsy tide]

The slope and y-intercept

[dull wagon]

show your attempt

[drowsy tide]

If I divide by -2 I get: Y=3/2x+4

[dull wagon]

you didn't divide properly

[drowsy tide]

i meant 4

[dull wagon]

wrong sign

[drowsy tide]

I realized that right after I typed it

negative and a negative

makes a positive

[dull wagon]

looks ok now

[drowsy tide]

Thank you

.close

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[drowsy tide]

.reopen

[calm coral]

✅

[drowsy tide]

I dont know how to simplify

[karmic sonnet]

are you aware of this identity? :

[potent lotus]

Reaper

[drowsy tide]

Yes

[karmic sonnet]

do you see something similar in the question?

[drowsy tide]

Yes

(x+7)(x-7)

[karmic sonnet]

correct

now does something cancel out?

[drowsy tide]

the numerator

[karmic sonnet]

correct

what are you left with now?

[drowsy tide]

x^2-49

or (x+7)(x-7)

[karmic sonnet]

you just said the numerator got cancelled

what did you cancel it with?

[drowsy tide]

Idk

[karmic sonnet]

isn't it x-7/(x-7)(x+7) ?

do you see the (x-7) gets cancelled?

[drowsy tide]

Yea because the first and second are the same

[karmic sonnet]

the numerator and one term in denominator is same

[drowsy tide]

so it would just be x+7

[karmic sonnet]

1/(x+7)

and that is your answer

[drowsy tide]

Oh

ok

.close

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[coral moss]

i have this

i need help on setting the integrals up lol

functions are
2x=1-2y-y^2
y=-x
y=-3

[calm coral]

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[karmic sonnet]

do you know the form of a line when it passes through a particular point (x1,y1) and has slope m?

$(y-y1)=m(x-x1)$

[potent lotus]

Reaper

[hushed folio]

so do i find the gradient

[karmic sonnet]

yes and you can do that by using the 2 points given to you

[karmic sonnet]

and use this specific equation ^^

[hushed folio]

is it the same as y-b=m(x-a)

[karmic sonnet]

yes

[hushed folio]

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[calm coral]

@onyx quiver Has your question been resolved?

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[hallow mountain]

how would someone solve this using grade 12 advanced function, i dont quite know how to check if something is a possible solution

[hallow mountain]

<@&286206848099549185>

[calm coral]

@hallow mountain Has your question been resolved?

[untold drum]

you need to know something about csc. and you need to know some spacial values for such functions. can you express csc in terms of cos or sin?

@hallow mountain

[hallow mountain]

csc is just 1/sin i think

right

i would like to know how do i plug in 5pi/3 to get possible solutions

[untold drum]

just a moment.
write 1/sin instead of csc. what do you get?

[hallow mountain]

is it 1/sinx or 1/sin * x

or is it the same

[untold drum]

1/(sin(x))

[hallow mountain]

1/sinx + 2/root(3) = 0

do i put 5pi/3 in x

[untold drum]

now modify this equation to get sin(x)= ...

[hallow mountain]

i dont see how i can do that without multiplying a sinx to the 2/root(3) than i dont know how i would remove it from the 2/root(3)

[untold drum]

is 1/sin x = -2/root(3) a possible step?

[hallow mountain]

yes but how i would remove the 1/ from sinx

[glacial nacelle]

Just plug in 5pi/3 into the equation

That should be easier than solving the equation

[hallow mountain]

since it doesnt equal 0 doesnt it mean its not a solution

but the question says it is

[glacial nacelle]

,w 1/sin(5pi/3) + 2/sqrt3

[potent lotus]

Results provided by WolframAlpha

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

Requested by dork9399

[hallow mountain]

?

[glacial nacelle]

Your calculator is set to degrees my friend

[hallow mountain]

yep

[glacial nacelle]

yea

[hallow mountain]

k thx i was super confused

[glacial nacelle]

lol

happens all the time

[calm coral]

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[tiny stump]

[tan(x)csc(x)]/cos(x)

[tiny stump]

I cant tell if its 1 or sec^2

[ivory pilot]

You want to take its derivative or just want to evaluate it

[tiny stump]

I got 1/cos/cos

Is it one or sec^2

1/cos^2 or it is 1/1

[ivory pilot]

$\frac{\tan(x)\csc(x)}{\cos(x)} =
\frac{\tan(x)}{\sin(x)\cos(x)} =\frac{\sin(x)}{\sin(x)\cos^2(x)}$

[potent lotus]

Sherif Player

[ivory pilot]

Then you can see what value you should get

[tiny stump]

So its sec^2 (x)

[ivory pilot]

Yeah

[tiny stump]

Well ok thanks

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[latent zinc]

Not sure where to begin

I have never dealt with a function where they specify "on the entire real axis" or one thats a piece-wise or whatever its called

[civic dirge]

"Differentiable on the entire real axis" = "differentiable everywhere"

[latent zinc]

right, okay

[civic dirge]

It should be pretty easy to imply that most points are differentiable. Any giving you any trouble?

[latent zinc]

Well its just the general setup

Im not sure how to word or write it

In fact, I don't even know what happens when we square a two-piece like this

[civic dirge]

It's not really important that it's a piecewise function

What definition of ReLU are you using here?

[latent zinc]

sec

lemme open the book

[civic dirge]

So f(x) = x² for any x > 0, clearly differentiable

f(x) = 0 for any x < 0, clearly differentiable

[latent zinc]

you mean x >= 0

?

[civic dirge]

But the point x = 0 is weird

As that's a join between y = 0, and y = x²

We can refer to the definition of the derivative, to see how this should be differentiated

[latent zinc]

why is this point weird?

[civic dirge]

To the left, we've got y = 0.

To the right, we've got y = x².

Is this even differentiable?

[latent zinc]

mhm

[latent zinc]

Let me get the definition

But

from the definition we see that it's x^2 for x >= 0

and not just for x > 0

@civic dirge

[civic dirge]

True! I misspoke

Now the issue we've got is whether or not this limit exists

[latent zinc]

mhm

[civic dirge]

Like you're saying, the right-hand limit obviously exists, because that's all just x²

[latent zinc]

What do you mean when you say right hand and left hand derivative?

Im so confused

[nimble badge]

https://en.wikipedia.org/wiki/One-sided_limit limit not derivative

[latent zinc]

oh, okay

[latent zinc]

Rigth, we have a right-handed limit

Which is x^2

Are you here?:)

<@&286206848099549185>

[civic dirge]

No I meant left-handed derivative

That is, the limit for the definition of the derivative, but only taken as the left handed limit

We can "better expose" this, by subbing -h in for h:
lim [f(x - h) - f(x)] / -h

Does that exist for x = 0?

[calm coral]

@latent zinc Has your question been resolved?

[cloud sun]

kheryhvargen zunichan therthurkiler

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[foggy plume]

Does anyone know how they came up with this? I couldn't find any other resources online aboout it

[foggy plume]

It kind of looks like the inverse hyperbolic tangent

like

[calm coral]

@foggy plume Has your question been resolved?

[calm coral]

@foggy plume Has your question been resolved?

[foggy plume]

<@&286206848099549185>

[clever ruin]

Perhaps something related to this?

Probably better to ask in #real-complex-analysis

Also $\tan(ix)=i\tanh(x)$

[potent lotus]

Max

[foggy plume]

It's not really complex analysis though...

that's why I was scared to ask there...

[clever ruin]

$\frac{1+\frac{b}{a}i}{1-\frac{b}{a}i}=\frac{a+bi}{a-bi}$

[potent lotus]

Max

[clever ruin]

They might have just got the identity wrong?

[clever ruin]

Someone is more likely to recognize it there

[foggy plume]

I verified it though, it works

[clever ruin]

Engineers induction?

[foggy plume]

how would you go about starting that then?

[clever ruin]

Oh that was a joke lol

Engineers induction as in subbing in values

$a+bi=re^{i\theta}\Rightarrow \frac{a+bi}{((a+bi)(a-bi))^{1/2}}=e^{i\theta}$

[potent lotus]

Max

[clever ruin]

I think this is how they did it

So I've divided both sides by the magnitude of $a+bi$ (which is $\sqrt{(a+bi)(a-bi)}$)

[potent lotus]

Max

[foggy plume]

but it's not....

[clever ruin]

It should be

[foggy plume]

the magnitude is $\sqrt{a^2-b^2}$

[potent lotus]

OmegaAplha

[foggy plume]

because this is the split complex numbers

I'm sorry if I confused you with that

[clever ruin]

Split complex numbers?

[foggy plume]

yeah where j^2=1

j not equal to 1 or -1

[clever ruin]

Huh okay lmao

[foggy plume]

https://en.wikipedia.org/wiki/Split-complex_number

[foggy plume]

I like some of your ideas though, some of these formulas could be good starting points for finding the solution

[clever ruin]

I got no idea tho, I guess alot follows from CA tho like how Trig identities follow to Hyperbolic

Good luck

[foggy plume]

thank you 🙏

[calm coral]

@foggy plume Has your question been resolved?

[foggy plume]

I think I get the gist of how it works now.

.close()

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.close

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[wraith gale]

Can someone help me understand tensor products?
for v.s. U,V : U⊗V has elements z=x⊗y ; with x and y from U and V respectively
The problem is that absolutely every source refuses to define x⊗y in any concrete manner.
All I know is that there is some kind of bilinear property between x and y
My textbooks says "x⊗y should define a bilinear (vector-valued) function of x and y."
The dot product is a bilinear function of x and y, right? x•y
So is the dot product in U⊗V?????
On the other hand, it seems that these elements z exist for each pair of x and y as opposed to being
And I have had physics friends define x⊗y as being a "tensor" that could take the form of
[x1y1, x1y2, x2y1, x2y2],
or
[x1y1, x1y2,
x2y1, x2y2]
depending on who you ask.
The more I look into it, the less consistent and more confusing it gets.

[azure rose]

You might get some more help in #old-network, there is a physics specific discord there.

[wraith gale]

I'm not in physics at all. I am taking the second course in a linear algebra series. It is more abstract and mathematical.

[unkempt quail]

tensor product is a very general object

in the same way that a "product" is a general object

it is a famously tricky concept when you first encounter it

@orchid tendon find your own help channel, read #❓how-to-get-help

[orchid tendon]

Someone else said 5

[orchid tendon]

I don’t see a available channel

Oh okay thx

[unkempt quail]

#help-24

[wraith gale]

I really don't get the 'generality' of it. Is there any kind of explanation?

[unkempt quail]

well the vector/matrix example from your physics friends is correct

in fact the tensor product of two vectors is both of these things

[wraith gale]

ok, so how is that object related to or representative of a bilinear function as my textbook states?

[unkempt quail]

neither is the "correct" one

@peak lava don't GPT post, come back in a couple hours

here's a good article

i would stick about to explain further but i gotta run

[wraith gale]

aight, ill look into the article. thanks

.close

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[grave bane]

how do i solve a depressed cubic say x^3+x+1=0

[grave bane]

real solutions only btw

[neon sage]

are you aware of cardano's formula

[grave bane]

no

[neon sage]

well it's a direct result for solving cubics of that form

you can avoid having to memorize it by following the same substitutions used in its derivation

[grave bane]

okay ill check that out

[calm coral]

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[patent tundra]

So for this question you should apply rule of product axiom and rule of addition axiom

[prime topaz]

is this for part i and ii respectively?

<@&286206848099549185>

<@&286206848099549185>

[sick zealot]

for part i), the only way that the player would have no chance of winning is if both dice show the same number, so just show the probability of rolling 2 of the same numbers

[calm coral]

@prime topaz Has your question been resolved?

[prime topaz]

oh i see so is it 1/6

wait so how do u do part ii)

[sick zealot]

you just work out the probability that the player rolls a winning roll on the third die. its a bit more involved than the first part but its not too much more complicated

[prime topaz]

but how..

[sick zealot]

on the third die, a 1 and a 6 can never be winning rolls, if they roll a 2, with a 1/6 probability, the first two dice rolls must be 1&3 or 1&4 or 1&5 or 1&6, and you work out those probabilities. there might be a more elegant way but if i was doing the question in an exam i would just brute force it.

and you do that for each roll of the third die

[prime topaz]

ohhh got it

thanks 🙂

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[remote mural]

What to do next? I can't seem to do anything with (x^2+2) to remove (x-1)

[clear delta]

what happens if you substitute x = 1 in there?

[granite jay]

牛顿皇帝……

6

能拍清楚点吗，看不清

Are you there?

@remote mural

[winter elbow]

Can you make it more clear? It’s blurry

[granite jay]

Yes

[remote mural]

wait

[remote mural]

the top becomes 0

[granite jay]

OK

[remote mural]

which is why i need to get rid of (x-1)

[granite jay]

Can you make it more clear? It’s blurry

[clear delta]

what's wrong with that

[granite jay]

I cant see it clearly

[remote mural]

Isn't that against the rules in limit

Wait

......

[winter elbow]

No, why?

[remote mural]

oh NVM

BRUH

[clear delta]

would you have a problem with $\lim_{x\to2}\f{x-2}{7}$?

[remote mural]

0/3 would be 0

[potent lotus]

hayley!

[remote mural]

its only 0/0 that would be a problem

my mistake

.close

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[remote mural]

sorry for the hassle

[granite jay]

LOL

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[alpine cedar]

Discrete mathematics problem
Especially the b one

[clear moat]

what did you try

[alpine cedar]

ahh
nothing

couldn;t understand anything

I need your help

[clear moat]

you already know how to represent a even number E(x)

for any could be said as \forall

[alpine cedar]

yeah
wait I know about these symbols and logical thing

[clear moat]

cool

[alpine cedar]

how can I represent it into symbol

what will be the right way

[alpine cedar]

yeah
a guy who knows maths is a cool guy
🙂

hey there

[clear moat]

do you mean how to write out in TeX?

[alpine cedar]

what is tex?
can you explain to me please

I mean I have to represent it into symbols and formulla

hey

[clear moat]

whats O(x)

@alpine cedar

$$(E(x) \forall x \in \mathbb{Z}) \rightarrow (E(x) + 1 = O(x))$$

[potent lotus]

ℕaive

[clear moat]

I assume O(x) is any odd.

[calm coral]

@alpine cedar Has your question been resolved?

[thick sinew]

@alpine cedar did you solve it? :)

[calm coral]

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[next vigil]

can someone help with this?

[calm coral]

@next vigil Has your question been resolved?

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@next vigil Has your question been resolved?

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[leaden tinsel]

i need to define DF using a and b, i know that ∆AFD is similar to ∆CEF

[leaden tinsel]

Please help, also why are vectors so hard.

[calm coral]

@leaden tinsel Has your question been resolved?

[calm coral]

@leaden tinsel Has your question been resolved?

[leaden mulch]

is the a vector AB or BA?

[leaden tinsel]

its BA and AD as shown by the arrows.

sorry if i'm unclear.

[leaden tinsel]

ok nvm i found the solution its DF=(5/7)DE=5/7(-b-a+(3/5)b)

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[viscid spruce]

help

[opaque fulcrum]

@viscid spruce

please post the question

[viscid spruce]

ok

wait

nvm i got it

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[tired gyro]

I have a true/false question. This is the statement:
Let V be a set that contains the 0-vector, and such that whenever u and v are in then u+v is in V. Then V is a vector space.

[tired gyro]

I know that I need to check the axioms to verify that it is a vector space, but I'm not sure what to say about the scalar ones.

[calm coral]

@tired gyro Has your question been resolved?

[tired gyro]

<@&286206848099549185>

[uneven tangle]

True?

[tired gyro]

That is what I assumed, but I'm not sure if the addition was defined to be standard addition.

The only thing I can really say for sure is that it fulfills additive closure and additive identity.

[potent lotus]

BotToes

[tired gyro]

I see, so since the statement jumps to the assumption only based on the additive axioms and not the scalar axioms, I can say it's false since it doesn't mention V being closed under scalar multiplication.

Ok, I understand now. Thanks!

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[orchid palm]

This doesn't seem right

I'm not sure how I can proceed with the -4sec(theta)^3 * sec(theta) * dtheta

Ok I figured out it's right

I just need help with one thing

Since its x^2-16, does that mean x is -4sec(theta), or is it +4sec(theta)

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[paper grail]

hello!

[paper grail]

i think i'm misunderstanding this question on self-dual connectives in predicate logic

phi is just some sentence and f its truth function

[solar spruce]

try asking in #foundations

[paper grail]

so my idea was by induction

okay

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[random stratus]

i have to show that it exists 2 constants A ans B such as there's this equality
P is a polynom of degree 1

I have no idea how to do it, developing P doesn't seem that much helpful at first glance

[mortal orbit]

well P = ax+b

so start from there to evaluate the integral

[calm coral]

@random stratus Has your question been resolved?

[random stratus]

i found the integral is equal to a + b

so a + b = A(a*alpha + b) + B(a*beta + b)

Idk if this could be true

<@&286206848099549185>

[tall mason]

yes that is correct

assuming that it is required for any or all P(x) = ax + b

as in a and b are arbitrary

[random stratus]

well A and B doesn't depend on a and b so that's fine

[tall mason]

gl :)

[random stratus]

i solve the system and i found B = 1/alpha - beta and A = -B
i think that's correct

thx for the help

[tall mason]

A + B = 1

A = 1 - B

[random stratus]

ye i'm dum

[random stratus]

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[vale sorrel]

A lot of the time, people use f'_n uniformly convergence to f' to prove f_n uniformly convergence to f. Does the reverse implies? Does f_n -> f uniformly implies f'_n -> f' uniformly?

[mortal orbit]

if f_n -> f uniformly, sometimes f' doesn't even exist even though f_n' does

[vale sorrel]

hmmm, what about if f', and f'_n exist?

[mortal orbit]

still you can find counterexamples

take fn(x) = x^2sin(1/(x+(1/n))) on [0,1]

this converges uniformly to f(x) = x^2sin(1/x)) when x non 0 and f(0) = 0

fn and f are both derivable

yet fn' cannot converge uniformly to f' because f' is not continuous whereas fn' is

[mortal orbit]

small update, i had to change to [0,1] to have uniform convergence

[vale sorrel]

thanks!! I'll mull over this counterexample!

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[mortal orbit]

i changed the function again sorry

now it should be good

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[zenith wind]

Pete’s Sno-Cone Stand sells three different size sno-cones. Pete will earn $6 for each large sno-cone,
$3 for each medium, and $2 for each small that he sells. Suppose Pete sells twice as many small snocones as he sells large sno-cones. If Pete sells 55 total sno-cones and earns $175, how many of each size
sno-cone did Pete sell can you help me put it in augmented matrix

[zenith wind]

hello

.close

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[teal hull]

how can I approach this

[calm coral]

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[old falcon]

Having trouble again with this indexing

My question is only about the sentence beginning "Argue that there is a strictly increasing..." and before.

I don't really understand how what is stated is equivalent to x being cauchy here

dont we need to put some restriction on k?

I understand that we could equivalently state whatever like

[mortal orbit]

we never said it's equivalent

we just said cauchy => this

[old falcon]

so it could be more restrictive

i still don't believe it

i guess i dont mean equivalent in that way

i mean, how do we know this will be true given (x_n) is cauchy

[mortal orbit]

use cauchy definition with a well chosen epsilon

also actually it is equivalent to being cauchy I just noticed

[old falcon]

i may need help

ive been playing with it

i get that you could just let m,n > N in whatever cauchy defn you have

then just take n_k to start at the m such that m>N

and fix n>m>N to fix that problem

then we may as well just select epsilon = 1/k^2 for all it matters

but this doesnt resolve the issues does it

[mortal orbit]

that's precisely it

[old falcon]

oh

[mortal orbit]

select epsilon = 1/k^2

[old falcon]

right and there doesnt need to be any care put into that

[mortal orbit]

this gives us N such that for m,n > N, ...

[old falcon]

okay

hmm maybe i am getting tired

thank you again 🙂

[mortal orbit]

take n_k = N+1

and voilà

[old falcon]

hmmm

is that necessary

i guess yea you want to describe n_k

or, its possible

anyways

i think i can struggle forward

thanks

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[pine schooner]

can someone help me find a counterexample to show this statement is false? (or tell me why its true but i think its false)

I was thinking like what if you just made a set out of the terms of the infinite series $\sum_{n=1}^{\infty} \frac{\sqrt{2}}{n^2}$ ?

[potent lotus]

goobybalooby

[pine schooner]

or does that not do it? are the irrationals closed under addition?

[graceful dust]

that might be a little difficult to work with

what intuition leads you to that?

[pine schooner]

well im thinking $\sum_{n=1}^{\infty} \frac{1}{n^2}$ is bounded because it converges to $\frac{\pi^2}{6}$, but all the terms would be rationals

[potent lotus]

goobybalooby

[pine schooner]

so then just multiply everything by an irrational number

[graceful dust]

o sure if you can use that then you're good

[pine schooner]

oh fosho?

[graceful dust]

yea

[pine schooner]

i mean i think i can use it lol we havent proved it but its come up in other hw problems so i hope i can use that

and the irrationals are closed under addition? is that true?

[graceful dust]

you also need that sqrt(2)*pi^2/6 is irrational

[graceful dust]

no

e.g. sqrt(2) + -sqrt(2) is rational

[pine schooner]

i think we can take this as a given

[graceful dust]

and it isn't even know if pi + e is irrational

for example

[pine schooner]

hmmmm okay so maybe its not a good one to use then

[graceful dust]

well it's not hard to prove all the partial sums of this series are irrational?

[graceful dust]

this one rather

[pine schooner]

how can you do it if the irrationals arent closed under addition?

[graceful dust]

for nonzero rational p and irrational q, p*q is irrational

[pine schooner]

just factor our sqrt(2) for every term?

[graceful dust]

yea

[pine schooner]

right

okay that makes sense. thank you!

[graceful dust]

[pine schooner]

.close