help-0

[rocky roost]

U have two aps linked together
One for odd terms one for even ul have to write the gen term separatly for n being odd and even

[marsh sedge]

like... an=a(n-1)d stuff

how do i express the alternating in a formula?

[rocky roost]

Yes

[sour mica]

split the single series into two

[rocky roost]

a2n = 6 + (n-1)9
a2n+1 = 1 + (n)9

[marsh sedge]

thanks for the first question :) ill just wait for the 2nd one now

[alpine sable]

the differnce is ap

[lone heart]

@marsh sedge Has your question been resolved?

[alpine sable]

so like write a bunch of an - an-1 n add them all up

[marsh sedge]

can you elaborate? thank you. was this for question c? because b is fine

[alpine sable]

a2 - a1
a3 - a2
...
an - an-1
add them n ur gonna get something

[marsh sedge]

would it be correct to write the formula for letter c as an=(an-1)-(n/3)?

[lone heart]

@marsh sedge Has your question been resolved?

[marsh sedge]

.close

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[brisk spindle]

help me bro

[brisk spindle]

how do u remove the y

the calculus part is easy but how do u do part (a)

[cedar kelp]

to eliminate y, you have to use the fact that total surface area of the brick is 600cm^2

[brisk spindle]

ok and

idk what

to do with that infomration

all i can make is an equation

that has y in it

[cedar kelp]

you know how total surface area is calculated?

[brisk spindle]

Ye

it should be 2(2x^2+yx+2xy)

[cedar kelp]

exactly

set it equal to 600

and you'll get another equation in terms of y and x

form here you can get the value of y
and substitute that in the expression for volume

[brisk spindle]

oh

Bruhh so much algebra

I think i know how to do it tho now

thanks bro

[cedar kelp]

yw

[brisk spindle]

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[left wharf]

okay this is kind of embarrassing but can someone help me rediscover why my argument works

[left wharf]

Back when I wrote this, it made perfect sense in my head

now I barely even remember what bezout's identity even is and most concerningly I don't understand why gcd(2, k) would belong to the ideal

it's possible that the proof is flawed to begin with but I vividly remember that it used to make sense to me

[lone heart]

@left wharf Has your question been resolved?

[mortal trellis]

bezouts identity says that there exist integers x,y such that gcd(a,b)=xa+by. so the gcd of the two numbers a,b can be written as an integer linear combination of them

as ideals are closed under integer lin combinations, it follows that if a and b are in the ideal, then the gcd also has to be

I want to point out that you can make the argument much quicker. the number n has to be odd, so gcd(2,n)=1 immediately, as 2 only has the factors 1 and 2

@left wharf

[left wharf]

I get that, by why is k in the ideal

[left wharf]

this makes a loooot more sense though XD

and that generalizes too, because if you add something to the ideal (p), then it cannot have p as a factor, and therefore gcd(p, n)=1, and bezout lemma says that you have a linear combination of p and n that gives 1, so therefore 1 is in the ideal

ohhh shit I'm dumb

yeah now I understand the original argument too

[mortal trellis]

oh I didnt even catch that. yeah that doesnt immediately have to be true

of course it will be true because the ideal turns out the be all of Z

[left wharf]

gcd(2, n) belongs to the ideal because of bezout and therefore gcd(2, k) also does and that's the argument

but yeah it's dumb I'll rewrite that according to your suggestion, thank you so much

thank you a second time this is very cool

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[strange fractal]

excuse me how do i write limh->0 inside word

[strange fractal]

like this

[alpine sable]

There's probably an option to use math symbols in it

[gray isle]

they have limit right there

[strange fractal]

ohhhhhh

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[warped topaz]

So to get the slope of the line l we get the derivative of 2^x which is (2^x)/ln(2)

Then sub in x = 3 so we have (2^3)/ln(2) = 8/ln(2)

[winter light]

Shouldn't ln(2) be multiplied instead of divided?

[warped topaz]

Oh

Yep sorry

so (2^x)ln(2), sub in x = 3 gives 8ln(2)

[winter light]

Yep

[warped topaz]

Then we do the same for 3^x, giving us the slope of line l 9ln(3)

So then to get the slope between them we use arctan

So its arctan(9ln(3) - 8ln(2)), right?

Which is arctan(ln(3^9) - ln(2^8))

[winter light]

There's a formula for the angle between two lines, but honestly I don't remember it at the moment

[warped topaz]

arctan(ln((3^9)/(2^8)))

[winter light]

Here it is

[warped topaz]

Oh wait what

I thought I just had to do arctan

[winter light]

No

But the concept behind is that of arctan indeed

[warped topaz]

Is there something else im getting mixed up with?

Cause I know I remember using arctan to get the angle between 2 lines or something

[winter light]

I can only suggest you to revise your lecture notes, probably the teacher has told you something about this kind of exercises with arctg

[warped topaz]

I will find it 1 moment

[warped topaz]

This one

First you take arctan of the slope of l, which gives pi/6, then you take arctan of k, which gives -pi/6

pi/6 - (-pi/6) = pi/3

Yep so arctan(9ln(3)) gives 84.22 degrees and arctan(8ln(2)) gives 79.78 degrees

Which is the answer they wanted

I knew I wasnt crazy 😂

Thanks!

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[frigid ore]

f(x)=(x^3 + 2x^2)/2x differentiate with respect to x

[frigid ore]

hello

[alpine sable]

Hi

What have you tried?

[frigid ore]

dy/dx= (1/2x)x^3 + (2/2x)x2

then what to do

or what to do before

[gray isle]

do you have a pic of your work,
its a bit hard to read that

[frigid ore]

no pic

phone dead

[gray isle]

missing ^2 at the end

and ideally you should also put () around the denominators to avoid ambiguity

and also

you haven't actually differentiated anything yet

so very inappropriate to put dy/dx =

all it seems you've done is split the fraction so what you currently have would still be f(x)

[frigid ore]

dy/dx= (3/2x)x^2 + (2/x)x

[gray isle]

no

how are you getting that

are you able to draw your work on paint
or display it on something capable to typing/displaying math like desmos

[frigid ore]

what is the answer?

[gray isle]

we don't just give out answers here

$f(x) = \frac{x^3 + 2x^2}{2x} \ \
f(x) = \frac{x^3}{2x} + \frac{2x^2}{2x}$ \ \
simplify each fraction

[ocean seal]

ℝαμΩℕωⅤ

[frigid ore]

ok

dy/dx = x + 1

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[round forge]

The author makes these claims in succession (besides the written claim of mine at the bottom), and the point is that I have zero clue as to how (a, a] is an element of C. I fully understand that the empty set is equal to a set (a, a], but by definition of c, we must have the left bound less than the right bound for any interval in C. This is clearly not satisfied by (a, a]

[rose sigil]

!original

[lone heart]

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.

[rose sigil]

this doesn't make any sense at all lol

post whatever the 'author' wrote pls

[round forge]

Ok

Sorry kinda big picture for continuities sake but the circled parts are what I specified

[rose sigil]

ok ic now

this looks like a misunderstanding

a semiring is a collection of sets

so C = (a,b] isn't right

[round forge]

I understand that C is the collection of all sets that can be written like that with a < b, yes, that was handwavy

But no set in C would be (a, a]

[rose sigil]

well (a,a] is in C

or i would even say, (0,0] is in C

C contains all intervals of form (a,b], where a and b are any real numbers

so that includes (0,0]

[round forge]

No it doesn’t, there are restrictions on a and b and one such restriction is that a < b. Surely that’s not the case and I’m wrong but it seems like that’s what every set in the collection has to satisfy

[rose sigil]

that's not a restriction on the sets in C

that's something inside the definition of the sets that are in C

a and b can be anything

(0,0] is empty because there are numbers x with 0 < x and x <= 0

$\mathcal{C} = {(a,b] : a,b\in\bR}$

[ocean seal]

:bending_skull:

[rose sigil]

right from the definition, (0,0] is in C

[round forge]

But a is less than b! That is also in the definition

0 not less than 0

[rose sigil]

$(0,0] = {x\in\bR: 0 < x \text{ and } x \leq 0} = \emptyset$, right?

[ocean seal]

:bending_skull:

[round forge]

Yes

I don’t have a problem with any (a , a] equaling the empty set

That’s fine

[rose sigil]

C is the collection of sets of form (a,b] where a and b are real numbers

setting a and b equal to whatever real numbers you want gives you an element of C

[round forge]

Just to clarify

[rose sigil]

due to this condition, (a,b] might be empty depending on what a and b are

but any set of that form is still in C

even if a = b

[round forge]

But then if you can contradict the definition, why not make b < a? What is (3, -7] supposed to mean?

[rose sigil]

that's also the empty set

[round forge]

Alright then if that’s the case it makes sense, thank you

[rose sigil]

it's not a 'contradiction'

it's just, the set contains elements that satisfy a condition but there aren't any that satisfy the condition

[round forge]

The way I read this is “you can write a set in C as long as it satisfies these conditions: (the conditions in the interval notation)”

But clearly that’s not the case

It does not need to satisfy a less than b even though the interval notation says that

So I see that if it doesn’t satisfy those conditions, it’s still a legit set in C, it just equals the empty set

Makes sense I thinkkkk

[rose sigil]

yep

[round forge]

Thank you

[rose sigil]

[round forge]

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[warped topaz]

Im really not sure what to do with this one

I see there is a quadratic in the denominator so I bet I need to set the discriminant of that > 0 at some point to find where there are 2 values of y or something

But yeah I have no idea what the first step is

[weak wharf]

You set g(x)=p

then rearrange and use the quadratic formula

[warped topaz]

Why do we set it equal to p? sry

[weak wharf]

you will get something like $x_{12}=A\pm \sqrt{B}$

[ocean seal]

NulledOutChicken

[weak wharf]

Because this is how you find intersections. f intersects g at a <=> f(a)=g(a)

[warped topaz]

Ah ok

I will try

i got it! Thanks

❤️

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[solar wolf]

help

[alpine sable]

ok

first example

choose an random number for x

[weak wharf]

where exactly are you unsure?

[alpine sable]

and see if you get a ratio of 1/2

[solar wolf]

the last two

[alpine sable]

oh

[weak wharf]

try to look up how to divide fractions by fractions

you should be able to solve that on your own

[modern sedge]

If you divide pizza into 3 pieces, then you take 1 of the pieces and divide it into 4 pieces and take 1, how large will it be (what fraction of the og pizza)

[solar wolf]

nvm guys i solved it

sorry i think i was just really tired yesterday

[alpine sable]

np

[modern sedge]

Alright, if that's all you can close this channel with .close

[solar wolf]

do i just type close or

[modern sedge]

.close

this

[lone heart]

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[modern sedge]

(it's already closed now)

[junior kraken]

How ?

Bro got superpowers

[modern sedge]

green privilege

[junior kraken]

Huh it didn't work earlier?

[modern sedge]

only green people can close other people's channel

and mods ofc

[junior kraken]

Nah i was a helper once , I mean then it wasn't available

[modern sedge]

Helper or helpful?

it's a different role

[junior kraken]

Helpful green tag ik bro 😭

First you get active-- very active --- helpful

[modern sedge]

It's quite an old feature, maybe you were helpful a really long time ago

[lone heart]

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[modern sedge]

.this isnt a rule btw, at least afaik

[compact osprey]

how do you solve this?

[alpine sable]

ok

first

calculate that number

[dusk minnow]

na

[alpine sable]

actually

yeah

you dont have to

just do some calculation you will end up with the prime factor decompose of it anyway

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@compact osprey Has your question been resolved?

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[sick lion]

Decide in natural numbers: x^3 - 3 = 2y^2

[sick lion]

I know that x=5 mod 6

[twin nimbus]

so how much do you know about elliptic curves?

[sick lion]

nothing

[twin nimbus]

what level of math are you studying?

[sick lion]

Olympiad

[orchid parrot]

Can it be helpfull if we write it ?
(x-1)(x^{2}+x +1) = 2(y-1)(y+1)

x-1 divides 2(y-1)(y+1)

[sick lion]

Why?

[orchid parrot]

From equation

[sick lion]

Sorry, got it

[livid sage]

rhs is 2(y^2 + 1) not 2(y - 1)(y + 1)

[orchid parrot]

Yeah

I made mistake

[sick lion]

Maybe we can use it?

[twin nimbus]

we can solve for x^3, x^3 = 2y^2 + 3, now, assuming x = 5 mod 6, and working under mod 6, the cube actually is transparent. That is a^3 = a mod 6. So we have 5 = 2y^2 + 3, or y^2 = 1 mod 6. The y^2 becomes 1 whenever y = 1 or 5 mod 6. So now you have a limitation mod 6 on y.

but I don't feel like this is significantly closer to a solution 😦

[sick lion]

That’s y^2 = 1 mod 6

But so far it doesn't give anything

[lone heart]

@sick lion Has your question been resolved?

[sick lion]

when substituting x=3k, you see that y is divisible by 3 and by substituting y=3l and dividing by 3 one gets mod 3 that 1=0

[orchid parrot]

Yea

[sick lion]

=> x=5 mod 6

And y = 1 mod 6

[orchid parrot]

We can view this in Gaussian numbers Z[i] but I think its overcomplicating

[sick lion]

I think there's a solution without them

[lone heart]

@sick lion Has your question been resolved?

[sick lion]

.close

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[heavy vale]

Not sure how to do this at all

I thought it was looking for points of intersection

But I don’t think it is after looking at answer

Answer is F

[pseudo ice]

Why do you not think it is [points of intersction]? And by "points of intersection", assumedly you mean where y = tan(x) intersects y = 100x?

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@heavy vale Has your question been resolved?

[near hollow]

Yeah plot 100x, plot tanx and see where they meet

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[rustic magnet]

https://discord.com/channels/268882317391429632/1275579342281506876

[lone heart]

Please don't occupy multiple help channels.

[rustic magnet]

I need help

[pure dust]

with what?

@rustic magnet

[rustic magnet]

https://discord.com/channels/268882317391429632/1275579342281506876

[pure dust]

alr lemme see

[limpid turret]

.close

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[limpid turret]

Stay in one channel please

[lone heart]

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[slow island]

How do I find the values?

[slow island]

One of them is correct, but I don't know which one it is

[sour verge]

You found a value for which h(x) = 4, not -4.

[forest marsh]

There is actually a unique answer to h(x) = -4

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@slow island Has your question been resolved?

[slow island]

I figured it out

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[fervent lichen]

I need to use complex numbers for this i think cuz it is on the complex number section

I know cis^n (theta) = cis (ntheta)

But idk how can i use that

[carmine reef]

do you know the complex definition of cosine

[fervent lichen]

Like if u have z = a + bi the cos is a over sqrt (a^2 + b^2)?

[carmine reef]

no

[fervent lichen]

So no

[carmine reef]

Do you know e^iθ

[fervent lichen]

Yeah

[carmine reef]

what's e^iθ + e^-iθ

[fervent lichen]

Uhm

Idk

[carmine reef]

e^i(-θ) ?

[fervent lichen]

Is it?

[carmine reef]

No

[fervent lichen]

Oh

[carmine reef]

Just the right term

[fervent lichen]

Yeah i still do not know 😦

[carmine reef]

what's e^iθ

[fervent lichen]

Cis (theta)

But idk if that is the answer you want

[carmine reef]

expand

[fervent lichen]

Cos + isin

Oh yeah i could do that

Hold on

2 cos (theta)

Ohhhhhhhhhhhhhhhhhhh

Interestinggggggggggggggggggggggg

Thanks

[carmine reef]

Np

[fervent lichen]

Wait can cis (a) = 0?

Hmmmmmm

If not, x = cis (a)

And then done cuz u can just do 0 = 0 for every n 🙂

Uhm i do not think cis (a) can be equal to 0

Cuz we would have cos a = i sin (-a) and idk how to solve that

.close

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[silent goblet]

does anyone know how we got from the second to last step to last step

[lone heart]

Please don't occupy multiple help channels.

[silent goblet]

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[fossil basalt]

How do you do this question

[fossil basalt]

It is from an sat book

[hushed locust]

start by finding how much money he earned in his second and third years, in terms of d

[fossil basalt]

How would you do that

[hushed locust]

use the information given at the top of the page

[fossil basalt]

I got 3x+300= d

[hushed locust]

what is x

[fossil basalt]

There is no total amount of money

X is the money he made in his first year

Mb

[hushed locust]

we are given that d is the amount of money he earned in the first year

[fossil basalt]

Isnt that 3d+V

[hushed locust]

3d + V is the average amount of money, d is the amount of money raised in the first year

[fossil basalt]

Oh mb I can’t read

So second year would be 6d+2v

And third year would be 18d+6v+300

Then I average and solve

[hushed locust]

in the second year he raised twice as much money as the first year. in the first year the money he raised is d

[fossil basalt]

V=100 then right

[fossil basalt]

Yeah that was the answer thank you

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[neat rune]

How is this incorrect? They share the same exact thing meaning more than one point

So how is this wrong lmao

A and B are the same thing

Is it because it says ( angles ) and not ( angle )

[arctic lintel]

it says a pair of angles

angle MRS is the same as angle SRM

same thing for B

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@neat rune Has your question been resolved?

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[brisk spindle]

6 SOLUTIONS?? ffs

how to get ur 6 solutions

[proven leaf]

cosine is a periodic function

if we shrink the period (which is what 3x does)

it oscillates faster

therefore we'll end up with more solutions within a certain domain

than if the period wasn't shrunk

[brisk spindle]

so its 360/3

so the period is 120

but shouldnt it be 60

because ur not really finidng the period, only the number of times it crosses the line at -1/2

right

so how does that yield 160 200

[proven leaf]

right within 0<x<=360

[brisk spindle]

ok lemme pullup desmos

Oh wait

sorry i assumed that the intercepts were equally distanced but i forgot we not talking about the x axis 💀

So ur supppoed to tkae the two solutions i originally found

and add 120 to them

?

[charred jewel]

your range is 0 <= x < 360, but you're solving for cos(3x), so you can extend your range to 0 <= 3x <= 1080

[brisk spindle]

so i shouldve left it all at 3x

and then found the solutions by adding the period

and then divide all the soltuions by 3

[charred jewel]

adding 360 yes

then divide all by 3

[brisk spindle]

so just adding 120 in the end

is the extension of the range nessecary for working points?

[charred jewel]

?

[brisk spindle]

like think about it

i already found X

not 3x, x

so then i can just add the period of this wave to my solutions

[brisk spindle]

no need for this funny buissness right

[charred jewel]

i suppose you can but i always do this

[brisk spindle]

ok

thanks

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[void sable]

number 12 plz

[drifting seal]

Is that we need to simplify it?

[alpine sable]

just multiply it dawg

[drifting seal]

Given: (a+bi)(c+di) = ac + bci + adi -bd

[lone heart]

@void sable Has your question been resolved?

[void sable]

i got 14-rad 70 i plus rad 140i plus 5 rad 2

i just dont know what to do next

@drifting seal

@alpine sable

[modest robin]

group the imaginary part and the real part

and that should be it

[drifting seal]

Ye

[void sable]

ok i got it right then

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[rustic minnow]

how do I find the domain and range, im confused with the gaps, please help thank you.

[rocky roost]

It takes the value where the closed dot is made

[sour haven]

how can i open a help channel?

[rocky roost]

And dosent take the value for the open dot

[rocky roost]

Go to #help-4

And enter ur question

[rocky roost]

Since it's a piecewise function

[rustic minnow]

for the range?

[rocky roost]

Yea for the range look for the values the function can take

[rustic minnow]

What would the domain be

I mean range

[rocky roost]

For the first one it will be -1 to 3

For the second it will be -infinity to infinity

For the third one you should make brakets since the function takes all values between 2 And 4

[rustic minnow]

so for number 2 the range and the domain will be the same answer?

[rocky roost]

No the domain is -3 to infinity
But range is -1 to 3

[rustic minnow]

ur confusing me, u said it would be -infinity to infinity

For the range

[rocky roost]

Ohh wait yes for the number two range and domain are same

Sorry i thought u were Reffering to number 1
Nvm I'm just high

[rustic minnow]

HELP

u helping me tho so I appreciate it 😭

the third one what is the domain would it be -5 to 5?

[steel yacht]

Can u help me

[rustic minnow]

bro I need help as well

[limpid turret]

!occupied

[lone heart]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

@rustic minnow Has your question been resolved?

[rocky roost]

Nope look closely the u can substitute x values from -5 to 1 but between 1-2 function is not defined so the domain should be [-5, 1] u (2, 5]

[rocky roost]

Ok
Not to be rude sorry but do u know what domain and range are?
Again I'm just asking not taunting you

If not I can help u understand

[rustic minnow]

kinda, like the domain are looking at the x values and the range the y values?

[rustic minnow]

thank u sm 🙏🏽🙏🏽

again sorry but for number 4 the domain would be -3 to 3? n the range would be -1, 1, and 3 or am I just lost

[rocky roost]

Yes values for which the function is defined

[rocky roost]

Yes

[rocky roost]

You're welcome

[rustic minnow]

OK LAST QUESTION THEN ILL STOP BOTHERING U

are these correct or no 😭😓😓

[rocky roost]

I've answered u in the other help channel

[rustic minnow]

IT TIMED OUT GRRRR

wait why is it not -20? i thought it was bc it has the lowest point

[rocky roost]

How come -20 is the lowest point

[rustic minnow]

i assumed bc it was a point 🥲

has

n it’s the lowest one

[rocky roost]

Huh? The lowest value y is taking here is less than 5

[rustic minnow]

oh!

[rocky roost]

Where what I can see is that graph is only defined for positive

[rustic minnow]

I thought since it didn’t have a point I didn’t think it was the answer

[rocky roost]

bruh
It doesn't matter wheather it's a point or not if the function is defined there it can be the maxima Or minima

[rustic minnow]

sorry bruh im hella slow 😭

so the range would be -4 to 39 o no…

[rocky roost]

Np
It's good to be slow

[rocky roost]

Why minus?
Is the value that y is taking -ve

[rustic minnow]

omg I got confused it would be 4

[rocky roost]

Yes perfect

[rustic minnow]

and the domain wouldn’t be 39, to infinity??

[rocky roost]

Can x = 1 here

Look is the function defined for all x other than for x = 0 , 16 , 61

So what all values can x exactly take

[rustic minnow]

so it would be 0, 16, 61?

and infinity o no

[rocky roost]

it would be all real no exept these ones

[rustic minnow]

how would I put it as an answer

[rocky roost]

(0, 16) u(16, 61) u(61, infinity)

sorry for the late response

[rustic minnow]

thank u so much for helping me along the way 🙏🏽🙏🏽

bless ur soul

saved a life right here

goodnight n thank u again!!

[rocky roost]

It's morning here but ok gn to u
Your welcome

[lone heart]

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[spice venture]

Hi I'm stuck here.
what I tried first is find the area using pi r^2 where I got 78.5 using that 5in as r then idk what to do next.

[livid sage]

to find the area of the shaded region, compute the area of the square minus the the area of the circular sectors

[spice venture]

is the area of the square the ones with r?

while the sectors are abcd?

cuz if so then the sectors are equal to each other?

[livid sage]

yes to your second two questions

(i don't understand your first question)

[spice venture]

oh wait brain dead sorry

I followed this site to solve but I can't understand it either

while for the sectors I used 78.5in where I'm not sure myself if its the one to use

or should I consider the radius to get the sector?

[lone heart]

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[spice venture]

.close

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[mild fractal]

None of this is in my course yet, but I was wondering what other ways there are to integrate expressions like 1/(1+x^5) and my first guess was to write it in the form of it’s Taylor series

[mild fractal]

What are the conditions that a function/expression must follow to have a Taylor expansion (in layman terms)

[hushed locust]

well a function that can be represented by a taylor series is called an "analytic function": https://en.wikipedia.org/wiki/Analytic_function

one thing that's worth noting is that taylor series will only converge to their function inside a particular interval. for some functions that interval is the entire real number line.

[mild fractal]

I see I see

[hushed locust]

but for 1/(1 + x^5) we would expect that to have a radius of convergence of 1

so if we expanded at x = 0, we would expect the series (and hence its integral) only to converge in (-1, 1)

[mild fractal]

ohhh

can we only expand it about specific points?

[hushed locust]

we can expand it about any point

but the radius of convergence is 1, so the interval of convergence would be (a-1, a+1)

[mild fractal]

hmmm

So if I wanted to write the original function in terms of its Taylor expansion entirely, so that it covers all its domain and range , would I have to make a piece wise function?

[mild fractal]

sorry for pinging >.<

[keen plinth]

if you integrate the taylor series and you're able to simplify the resulting series back into an expression which is defined everywhere then that will be the integral

[mild fractal]

yesyes that was my goal originally

[keen plinth]

that's probably not easy

you might have an easier time doing partial fractions over C

[hushed locust]

apologies, this was incorrect. the radius of convergence does change based on the point (i am up too late, sorry)

[keen plinth]

the radius of convergence is kind of irrelevant here though because whatever power series you get will define a unique analytic extension

[mild fractal]

I see I see

[mild fractal]

still out of interest, is there a meaningful way to attempt this

[keen plinth]

well you could try expanding out the geometric series

it looks like you might have to some kind of roots of unity filter though

[mild fractal]

but that would only apply for |x|<1

[keen plinth]

there's a unique analytic continuation, so it doesn't matter

[mild fractal]

I dont really know what an analytic continuation is

[keen plinth]

like if you have a power series that's defined on some ball, theres a unique analytic function that extends it on some maximal subset of C

so if you can work out the answer for |x|<1, then you have the answer everywhere

[mild fractal]

woaaaa this is genuinely so convenient

I wonder how the proof of it goes tho that’s beyond my scope

[keen plinth]

it's called the identity theorem

[mild fractal]

thanks a lot, I’ll check that out

one more thing

the geometric series I got for |x|<1 was 1 -x^5 +x^10 -x^15+…

how would that look like after its analytic continuation?

[keen plinth]

seems right

well you know it comes from 1/(1 + x^5)

[mild fractal]

yep

[keen plinth]

so that's the analytic continuation

even if the power series is only defined for |x|<1

[mild fractal]

ohhhh

So the original expression is the analytic continuation

I see

[keen plinth]

but if you had some random power series that didn't come from an expression that you expanded

you could try unexpand it

[mild fractal]

I see I see

makes sense

[keen plinth]

like 1 + 2x + 3x^2 + 4x^3 + ... is the derivative of 1 + x + x^2 + x^3 + ... for example

so you could "unexpand" it by simplifying it to 1/(1 - x)^2

that extends the power series analytically to all of C except at 1

[mild fractal]

oooh

in gp cases it’s like the reverse process of finding the sum

alright, got it, thanks a lot c:

[keen plinth]

[mild fractal]

Have a good day/night!

.close

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[autumn acorn]

I have to find values for rho, t and theta. This is a swap of coordinates I have to use to find a parametrization of the curve of intersection between an iperboloid and a cylinder that are $9y^2-4z^2=9$ and $9x^2+4z^2=36$

[ocean seal]

x_Shadow_x

[autumn acorn]

I've already found rho and t that are easy, but I'm stuck to find theta

it should have a range of [-pi, pi]

but how

[lone heart]

@autumn acorn Has your question been resolved?

[lone heart]

@autumn acorn Has your question been resolved?

[autumn acorn]

<@&286206848099549185>

[autumn acorn]

I think it goes well every theta since it always satisfy both equations. but is it a problem if it goes from 0 to 2pi instead of -pi to pi as solutions?

[lone heart]

@autumn acorn Has your question been resolved?

[lone heart]

@autumn acorn Has your question been resolved?

[lone heart]

@autumn acorn Has your question been resolved?

[terse gull]

@autumn acorn

still need help?

is this parametrics ?

[autumn acorn]

yup

then I tried to draw it on geogebra

so

and setting theta with a range of -pi to pi or 0 to 2pi nothing changes

so I don't think it was a problem to consider a range of [0, 2pi] instead of that one of solutions

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[lone heart]

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[signal shale]

the question is find the value of p and q

[wind cloak]

what does it mean mathematically when the golf ball is on the ground

[signal shale]

I think its saying like the ball was static 3 meters to the left from point A. Then it was hit and then traveled until 10m from the point it was static which was 3m away from point A

if you get what im saying

[signal shale]

quadratics for anyone wondering

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[eager iron]

Could someone just walk me through how we solve this by "extending" with x-1?

I got the right answer using L'Hopitals, but would like to know how the alternative approach works

[subtle light]

so we can factor out the denominator

that would be the first step

[eager iron]

well i'm confused about how they wrote the answer

but we're essentially doing sin(x-1)/x^3-1 * (x-1)/(x-1) yeah?

and then we can move then sin(x-1) over to the right one and that one to the left?

[subtle light]

i mean yeah that could work too

[eager iron]

and we get sin(x-1)/x-1 * x-1/x^3-1

[subtle light]

mhm

[eager iron]

and now we can suddenly split the limits?

[subtle light]

since both limits exist we can just do each limit individually

yep

[eager iron]

and that's okay when we have multiplication between the terms? or always?

[subtle light]

it’s okay as long as both limits exist

[eager iron]

interesting

[subtle light]

if they don’t then splitting up wouldn’t give you the correct answer

[eager iron]

well yeah then we get 1 * (x-1)/(x^3-1)

[subtle light]

mhm

and now just factor out the denominator for the second factor there

[eager iron]

and then how would you easily factor out x^3-1?

[subtle light]

using the difference of cubes

[eager iron]

don't think i've heard of that one before

[ocean seal]

y0shi

[eager iron]

o

why have we not been taught this

[subtle light]

haha

but yeah if you use that

the x-1 would cancel out

and it’s just substitution

[eager iron]

yeah then we get (x-1)(x^2 + x + 1)

1/x^2+x+1

[subtle light]

yep

[eager iron]

and then limit goes to 1 so in total we get 1 * 1/3

awesome noodles

[subtle light]

mhm exactly

[eager iron]

cheers mate, you're a legend

[subtle light]

yw!

[eager iron]

have a nice day ;D

.close

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[gilded marten]

help

[gilded marten]

im struggling

i just dont get it

on the more harder ones

[alpine sable]

plug x

in the value

and find y value

[alpine sable]

plug x

[gilded marten]

how

[junior kraken]

In place of x

[gilded marten]

6x6 is 36

[junior kraken]

Write -6

[gilded marten]

is now -36

[junior kraken]

-6×-6 = 36

[cinder compass]

if x=2 what's 3x

[junior kraken]

-1×-1 = 1

[gilded marten]

my head is spinning

i simplified it

and its now

36 + -48 + 7

[junior kraken]

,calc 36-48+7

[ocean seal]

Result:

-5

[gilded marten]

hmm

imma keep trying it

[alpine sable]

what part of it are you struggling with

The substitution?

[lone heart]

@gilded marten Has your question been resolved?

[gilded marten]

like when i simplify everything

and i think everything is correct it becomes wrongh

[lone heart]

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[crimson crag]

hello

[crimson crag]

I need help with this question:

investigate for which values of b does the equation lack any real answers

sqroot(2x + 1) = x + b

I plugged it into geogebra and slid the thing

but I'm wondering if this can be solved algebraically

[blissful rapids]

Is the answer 1 and -5/3?

[crimson crag]

let me check

[blissful rapids]

If so then you could square both sides and get a quadratic in x
Then the discriminant (a quadratic in b) would need to be negative for no real solutions to exist

[crimson crag]

b is bigger than 1

[blissful rapids]

I meant to write b ε R - {-5/3, 1}

But my answer is still incomplete

[crimson crag]

how would I use the quadratic formula

[blissful rapids]

Ok I corrected my solution

Wait

[blissful rapids]

We know that no real solutions of x exist

So the discriminate for this quadratic must be negative

You need to calculate the discriminant

[crimson crag]

yeah but how would the quadratic look like

I’m used to only one sole variable when I use it

[blissful rapids]

x² + 2x(b - 1) + (b² - 1) = 0

The coefficient of x is 2(b - 1)

And the "constant" term is b² - 1

[blissful rapids]

The correct name for that term would be the term independent of x but people are used to calling it the constant term

[crimson crag]

hold on

how did you get this

[ember heart]

then you can substitute x1 or x2 in x1x2. does this help?

[crimson crag]

sorry what is x1 and x2?

[ember heart]

roots of quadratic equation

[lone heart]

@crimson crag Has your question been resolved?

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[sturdy spear]

I can’t figure out what I’m doing wrong:

Suppose that a wind is blowing in the direction S45°E at a speed of 40 km/h. A pilot is steering a plane in the direction N60°E at an airspeed (speed in still air) of 300 km/h. The true course, or track, of the plane is the direction of the resultant of the velocity vectors of the plane and the wind. The ground speed of the plane is the magnitude of the resultant. Find the true course and the ground speed of the plane. (Round your answers to one decimal place.)

[sour mica]

I think your planes direction is wrong? N60E means 60 deg east of North, so the angle should be the other way

src https://www.mathsteacher.com.au/year10/ch15_trigonometry/11_directions/23dir.htm

Otherwise the calculations seem right

[sturdy spear]

if it was the other way, what would the angle be when calculating it? like 300cos(angle) and 300sin(angle)?

[sour mica]

I mean you have 60 deg wrt x axis in your figure

but the source I linked seems to say it should be 60 deg wrt y axis in 1st Quadrant

[sturdy spear]

yeah I get that

[sour mica]

or 30 deg wrt x axis

[sturdy spear]

if it was that way would the calculation change then?

I’ll try with 30 def

deg

[sour mica]

so your components would be flipped

<259,150>

[sturdy spear]

gotchu

thanks

I was going insane lol

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[stoic barn]

I’ve been trying to look everywhere but I haven’t found a solution…

I’ve been stuck but I’m trying to convert 53/22 into a mixed number percent ** not** a mixed number into a percent.

[mortal magnet]

convert the fraction to a percent, then convert the whole number next to it into a percent, then add them together

thanks for clarifying, "mixed number percent" is not a real math term

did it tell you the percent was incorrect?

[stoic barn]

It’s because it can’t figure out the fraction part. I have 240%

[mortal magnet]

have you tried 241%

[stoic barn]

But idk how to get the fraction

[stoic barn]

It should look like 241 x/y %

Or 240

[mortal magnet]

again thats not a real math term

you should try 241%

oh I see what you mean

[stoic barn]

Here an example

[stoic barn]

I did. The software won’t take it though

[sour mica]

you mean like $\frac{10}{7}$ becomes $1\frac{3}{7}$?

[stoic barn]

What?

[mortal magnet]

for this, you multiply 53/22 by 100, simplify it into a mixed fraction, then put a % sign after it

didnt realize that was called a mixed number percent

[ocean seal]

Bacter10Fr4g

[stoic barn]

Thank you!

[mortal magnet]

np

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[acoustic needle]

hi i need help in quadratic formula, sometimes when i’m solving for the root, the 4ac is always bigger than b² is it normal or i’m just dumb because i always ends up getting negative, what i do is i’ll get the value of b² first then i do -4(a)(c) then b² - 4ac

[cedar kelp]

always?

[sweet tundra]

It’s fine if you get that, just check if you’re missing any negatives for a or c

[azure needle]

Not always

[acoustic needle]

my substitution are correct, if it’s fine what would be the next step if the root is negative?

[sweet tundra]

the solution will be imaginary

[azure needle]

If 4ac >b² then the roots are imaginary

[sweet tundra]

the root(-1) becomes i

[azure needle]

Because in the quadratic formula you get something like ( -b+√(-n))/2a

Where n is a number

[acoustic needle]

how to find the value of x if the roots are imaginary? my teacher didn’t gave an example of it😭

[azure needle]

Do √(-n)=√(-1)$\cdot$√n

[ocean seal]

faiyrose

[azure needle]

Thanks 👍🏿

[acoustic needle]

lastly where do i get the number of the n?

[azure needle]

$\sqrt{-1}$=i

[acoustic needle]

i’m failing 9th grade 😭

[azure needle]

;-;

[ocean seal]

David

[azure needle]

No you won't

[acoustic needle]

ah i see, it’ll be always imaginary if root is negative

lastly where do i get the number of the n?

[azure needle]

👍🏿

[azure needle]

|b²-4ac|

[azure needle]

This

[acoustic needle]

W

[azure needle]

If b²-4ac is -ve then you have the case √-n

Where n is the absolute value of b²-4ac

[acoustic needle]

i’ll just turn it to positive if 4ac>b² because of the ||?

what do you call that thing again

[azure needle]

Yes

[azure needle]

Absolute value of

[acoustic needle]

Thanks for the help mate

really means a lot

[azure needle]

You're welcome 😁

Is that all then

[acoustic needle]

yes

that’s the only part where i’m struggling

thanks again

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[atomic steppe]

I am looking for exercise book (with solutions) recommendations for introductory real analysis please 🙂

[outer lark]

try searching 'real analysis' in #book-recommendations

[atomic steppe]

.close

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[alpine sable]

Hello

[alpine sable]

why is the integral here for d theta from 0 to pi

i figured the dr but

d theta

[quick mango]

Yes papa

[mortal magnet]

dtheta goes from -pi/2 to pi/2 instead of from 0 to pi

after you integrate dr, you have to integrate cos(theta)^3
to do that, split into cos(theta)^2 cos(theta) then rewrite as (1 - sin(theta)^2) cos(theta)
this lets you use u-sub u = sin(theta)

[alpine sable]

its alright i know how to continue

but i dont know why my professor choose 0 to theta

when it is from -pi/2 to pi/2 imo

[mortal magnet]

you should take a bigger screenshot

[alpine sable]

oh ok

this is the professor notes

[mortal magnet]

I have a feeling the professor just misremembered the bounds for r = cos(theta) to be from 0 to pi
it should just be a typo

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[vapid steppe]

why is the code not working

[mortal trellis]

is that the command for the dot product?

wait, did you run the other cell before? it says In[1]

[vapid steppe]

oh wait i forgot to run the top command

i ran it before and then quit and now i came back

thats why

[vapid steppe]

anyway do u know how to do the problem

.close

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[reef pier]

would it not be more straightforward to use impulse-momentum theorem?

F * delta(t) = delta(mv)

oh

yeah i just realized it's equivalent

F = ma = m dv/dt = d/dt (mv), so Fdt = d(mv)

both approaches give you 0.07

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[jaunty bloom]

How do i get an equation for this along with drawing the line of best fit

algebra 2^

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@jaunty bloom Has your question been resolved?

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@jaunty bloom Has your question been resolved?

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[hoary herald]

Can a primitive root exist under modulo m if m is not a prime?
What's the exact condition, and is it proved?

[rose sigil]

https://en.wikipedia.org/wiki/Primitive_root_modulo_n

it is stated there

https://oeis.org/A033948

[hoary herald]

It's very useful, ty for the quick resp. too.
So 9 has it even though it's not a prime because it's p^k with odd prime p and k > 0.

[rose sigil]

yep

[hoary herald]

And it's proved also.
I'll close this, thanks again

.close

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[thin swallow]

What does it mean to find the derivative of a function with respect to anything other than dx

[thin swallow]

Like $\frac{d\sin{x^2}}{dx^2}$

[ocean seal]

PianoDolphin

[thin swallow]

I always write derivatives with the little dash up top and haven't reallly used this notation

It means to write find the dervative of sinx and replace x by the inside function ik that

But what is it asking?

[pallid scarab]

you can write chain rule definition:
$$\frac{df(x)}{dg(x)} = \frac{df(x)}{dx} \frac{dx}{dg(x)}$$

[ocean seal]

rafilou2003

[thin swallow]

I know the chain rule

I write it using the dash defination

But like what does this even mean d(sinx)/dx^2

[pallid scarab]

it means "how much does sin(x) instantly change when x^2 instantly changes"

[waxen flame]

A dx^2 in the denominator usually implies a second derivative.

But it is missing a corresponding d^2 in the numerator.

[pallid scarab]

but here

didn't even notice

[thin swallow]

I think that happens in the numerator

[waxen flame]

Unless d(sin(x))^2 was implied.

[pallid scarab]

yeah it's not d(x^2)

it's dx^2

meaning second derivative

[pallid scarab]

did you mean $\frac{d^2\sin{x}}{dx^2}$?

[ocean seal]

rafilou2003

[thin swallow]

no no

So we are asking, how much times does this function change as we change the value of x^2, but the the question is how are we changing the value of x^2

?

by changing x right, like inputting 2.000001 instead of 2

[pallid scarab]

well you can link change in x^2 back to change in x

with the RATE OF CHANGE of x^2

dx^2/dx

[thin swallow]

yes but then my confusion is that how is this any different than just saying df(g(x))/dx

isn't this asking the same thing, how does f(g(x)) change instantly with a very small change in x

or rather how many times is that change bigger than the small change in x

[pallid scarab]

? you're not looking at how f(g(x)) changes with a change in x

first of all you're looking at f(x)

and then you're looking at how it changes with a change in g(x)

so if you wanna see it like this

[neat folio]

pardon me if I'm wrong but in the case of x^2 vs. x imagine taking x = 0.0000001 and inputting it through the function and in that case you can see that the increase in y in that case will be different for x^2 and x

[pallid scarab]

you could write f(x) as f(g^-1(g(x))

and so f(g^-1(g(x)) with a change in g(x) makes more sense

[thin swallow]

I said $\frac{df(g(x))}{dx}$

[ocean seal]

PianoDolphin

[pallid scarab]

I agree that's what you said

and I said that's not what we're looking at

we're not looking at change of f(g(x))

we're looking at change of f(x), with respect to g(x)

so more like

[thin swallow]

But how is this any different from df(g(x))/dg(x)

Here we are asking how much does f(g(x)) change when we change g(x)

[thin swallow]

here we are asking how much does f(g(x)) change when we change x

but the only way to change g(x) is by chaning x

oh wait

[pallid scarab]

but

[thin swallow]

I got it

[pallid scarab]

ok

when you CHANGE g(x)

it's like you did "+ 1" to g(x)

[thin swallow]

The value of x changes differently

right?

[pallid scarab]

when you change x

you do "+1" to x

yeah

[thin swallow]

ahhhhh

well that was a confusing couple moments ll

lol*

that is what I get for always using f` notation

for derivative

[pallid scarab]

yeah, to explain a bit more

[pallid scarab]

this is f'(g(x))

[thin swallow]

yeah that I know

[pallid scarab]

this is [f(g(x))]'

[thin swallow]

yes that I know too