help-0

[empty kindle]

then its fx = 0

[marsh rapids]

Yes

[empty kindle]

oh

f^2 is never negtaive

[marsh rapids]

So f^2 = 0. Then you should easily be able to prove f = 0 as well

[empty kindle]

yes

[marsh rapids]

Now that you have the idea that's pretty much the only step that takes some rigorous computation to show

[empty kindle]

but

i have a question

we know that for any polynomial

P

integral P*f = 0

but since continuous functions can be estimated with polynomials

so P can estimate any continuous function

so instead of P we can write any function?

[marsh rapids]

After the fact that's obvious because f is 0

[empty kindle]

yes but

by estimating

i mean

|P-g|<e

can we replace

P with g

[marsh rapids]

But yes, you could show that. That makes sense because the integral will be 0

[marsh rapids]

Oh you're trying to show the equality rigorously ?

[marsh rapids]

I thought not

[empty kindle]

i wasnt but now i am 😄

[marsh rapids]

The idea is to use Bn(f) as your polynomial that goes to f. Then you can show that the sequence of integrals goes to the integral of f^2 and prove that that implies the integral of f^2 is 0

[empty kindle]

:x

but integral Bnf f(x) +f-f =0

so integral f^2 = 0 ?

but it is less than

idk what im doing tbh 😄

[marsh rapids]

You have successfully proved something guaranteedly positive is greater than the absolute value of 0. There was an attempt

Try and show that the difference of the integrals goes to 0 instead. That's guaranteed to lead to the correct conclusion, you just need to figure out the workings

[empty kindle]

ok ill continue later

ty very much

.close

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[sage vine]

When calculating the area between two curves, you integrate f(x) - g(x), where f(x) > g(x) on that intervall
When calculating the volume between two surfaces, you integrate twice, once for both surfaces and then take the difference just like with the area between two curves, you take the volume of the surface that is above the other surfaces volume, no?
Example
Calculate the volume between the two paraboloids
z = x^2 + y^2
z = 2 - (x^2 + y^2)

[sage vine]

shouldn't you calculate the volume between the two as
V = V1 - V2
where V1 = volume under blue, V2 = volume above red

looks like my professor is doing the opposite which is confusing

question put in simpler terms, how do you find the volume between the blue and red parabloids using double integrals

specifically, what's the intuition, the top view of the calculation, forget the details like finding the intersection between them and then changing to polar coordinates etc

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[sage vine]

how does taking the double integral of z = x^2+y^2 give you the red marked volume? shouldn't it give the volume above z = x^2+y^2?

[sage vine]

thought it gave this

ahh wait I think I get it

double integral gives the volume under a surface

am I thinking right about the volumes given from these double integrals?
this would make sense that blue volume minus red volume gives the volume between the two paraboloids

<@&286206848099549185>

[lone heart]

@sage vine Has your question been resolved?

[golden sequoia]

yes. The upper parabola - the lower parabola gives the volume

It'll probably be easier to compute this using polar coordinates

[sage vine]

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[alpine sable]

Is $\frac{a}{b}c=\frac{ac}{b}$ true? If so, how could I prove it?

[ocean seal]

3317

[elfin snow]

yes it is true

and idk if you can really prove that

iirc it's a definition or convention

[alpine sable]

Or demonstrate it, anything.

[marsh rapids]

you can prove it when you've defined what's a fraction and the rule for multiplying fractions together, and what fraction an integer is

[alpine sable]

Not numerically oc

[alpine sable]

Take the "standard" ring of reals then.

[marsh rapids]

field you mean

[alpine sable]

Oh yeah, sorry

[marsh rapids]

reals are much more painful to handle than rationals

[alpine sable]

All I want is something to demonstrate this thingie, doesn't need to be paper grade

[marsh rapids]

a proof only makes sense as a deduction from previously known properties and definitions. What definition of real numbers do you choose and how do you define multiplication and division on them ?

[alpine sable]

Multiplication on reals is a pain in the ass to define, never really thought about it.

Say rationals then.

Indeed

Wait, that was dumb

Lol the whole question was dumb

Just multiply both sides by $b$

[ocean seal]

3317

[alpine sable]

It works, doesn't it? Even if not rigorously.

[marsh rapids]

there's no b

of course it works

[alpine sable]

Whatcha mean

[marsh rapids]

nvm, can't read

[alpine sable]

Ooh

[marsh rapids]

But what level of study are you at ? I'm very confused by you asking whether this is true and knowing what a field is at the same time

[alpine sable]

Wellll

i guess you could say high school, but I live in Brazil

I love mathematics, that's my life's passion

I'm obsessed with "proofs", demonstrations and such

[marsh rapids]

have you ever seen a construction of the real numbers ?

[alpine sable]

Not really, just the naturals

By peano oc

[marsh rapids]

not ready for the reals then I'd say

[alpine sable]

I can't disagree

[marsh rapids]

a quotient ring of Cauchy sequences is a much more complex object to comprehend

[alpine sable]

Oof, I can imagine

[marsh rapids]

constructing Z and Q is alright, constructing R is on a whole other level

proving the property of the upper bound and completeness is another step up

[alpine sable]

What about you, what level of study are you at?

[marsh rapids]

first year past high school

[alpine sable]

Oofie

One day I'll be like you

[marsh rapids]

studying at the highest level of math study you can have in France

i.e. arguably best school + math major

[alpine sable]

Oh wow

I'm genuinely happy for you, man

Also thanks for your help

God bless : )

.close

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[wild jolt]

This is a new topic for me, I don't know how to solve this one. I would be happy if someone could teach me each step

[marsh rapids]

you should know that 1/(1+x²) integrates to arctan

see if you can use substitution to get back to that

[scenic wren]

Just put x²=√3tank

[lone heart]

@wild jolt Has your question been resolved?

[wild jolt]

I can't get any from it

[wild jolt]

like this?

[scenic wren]

No

tank*√3

Like this

[wild jolt]

this one?

[scenic wren]

Yep

[wild jolt]

okay what I do next?

oh wait, is that answer?

[scenic wren]

Just do the substitution everything will cancel out nicely, don't forget to change the lower and upper bounds tho.

[wild jolt]

ohhh okay okay, I will try more of those questions

how about this one?

[scenic wren]

5x+3=t?

4/5 ig

[wild jolt]

what does t represent?

[scenic wren]

Dummy variable

Take anything else doesn't matter

[wild jolt]

ohhh okay okay, so what do I do next?

[scenic wren]

I'm doing a substitution

5x+3=t

[wild jolt]

u-sub?

[scenic wren]

Yup

Change the lower and upper bounds

[wild jolt]

what to do with this?

[scenic wren]

Answer

[wild jolt]

ohh

i thought it was part of equation hahahaha

[scenic wren]

No no

[wild jolt]

thanks a lot sir!! I wanna be as smart as you someday

[wild jolt]

haha thanks for the help!!

[scenic wren]

I'm not smart XD.

[scenic wren]

Yw :)

[wild jolt]

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[austere fable]

According to a survey, about 45% of people ages 18 and older in a city have a dog. You ask 6 randomly chosen people (ages 18 and older) whether they have a dog. What is the probability that at most 4 people have a dog? Round your answer to the nearest hundredth.

[austere fable]

We're on the unit of probabilities, and I'm not too sure how to do this

[fervent timber]

omg jamew7 I am sub

[austere fable]

I'm fake jamew7, but any idea how to do?

[fervent timber]

oof

[fervent timber]

hmm

[alpine sable]

at most 4 people= either 0, 1, 2, 3 or 4 people. therefore it's the union of the events : H_k:"k people have a dog" for k=0,1,2,3,4. the probability of each H_k should be easy to calculate. then you just sum the probabilities

or you can instead calculate the complementary probability of 5 people or more having a dog, this should be a geometric sum or something, but not necessary

[austere fable]

I'm confused on like the 45% having a dog thing. My guess was just 0.45 x 4/6

[alpine sable]

how do you get the 4/6

[austere fable]

At most 4 people have dogs out of the 6

[alpine sable]

but setting 4/6 looks more like exactly 4 people, not at most 4 people. I think here thinking about the binomial distribution is even better. where n: is the number of successes (e.g. owning a dog here), the probability p of success is 45%

sorry n is the total number (population)

P(X=k) : the number of k successes among n trials where X follows the binomial

[austere fable]

I don't think we got up to binomial distribution yet

it's a later section in the unit

[alpine sable]

Ah I see. but anyway the probability of 4 people among 6 succeding involves the binomial coefficient $\binom{6}{4}$

[ocean seal]

AimaneSN

[alpine sable]

$\binom{6}{4} \cdot (\frac{45}{100})^4 \cdot (1 - \frac{45}{100})^2$

[ocean seal]

AimaneSN

[alpine sable]

2 is 6-4

[austere fable]

What what does the 6 4 represent

the first section

[alpine sable]

the number of combinations of 4 elements chosen from a total of 6 elements

[alpine sable]

this is the probability for exactly 4 people having a dog

we need to add to it the probability of exactly 3 people having a dog, then 2 people having a dog, then 1, then 0

anyway that's my solution, maybe there are simpler ways

[austere fable]

Okay, thank you

[alpine sable]

[austere fable]

.close

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[spiral citrus]

Hello! How to find this integral with Weierstrass substitution?

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[dusky niche]

For this why are the equilibrium points at x=4,-4 and not y=4,-4

[dusky niche]

So the function ends up being dx/dt = 16-x^2

And autonomous equation is of the form dy/dt = f(y)

[lone heart]

@dusky niche Has your question been resolved?

[solemn grove]

dx/dt = (4-x)(4+x) and when dx/dt = 0, x = +_4

[dusky niche]

but why? i really dont get it. on the graph it looks that the equillibirum is at y=4,-4

[solemn grove]

the graph is for dy/dx I reckon

not dy/dt or dx/dt

different graph

[solemn grove]

Your question is pretty vague,

[dusky niche]

is there a way that you can tell that i am missing lol, this is all very new to me. we just had one lecture on it

[solemn grove]

ok,

[dusky niche]

yeah sadly, i am a little confused. but i believe that you are right with the dx/dt part

[solemn grove]

for starters, you know how differentiation works rightv

?

[dusky niche]

yes

[solemn grove]

so instead of having value y depending on value of x directly

we have two functions that depend on t

[dusky niche]

yes

and for autonomous equations the right side does not depend explicitly on the ind var

[solemn grove]

that’s what parametrising is all about

[solemn grove]

Yeah well, basically there are infinite eqns that solve a differential

eqn

That’s why we have slope fields to show how they behave

[dusky niche]

so would the equillibirum solutions not be y=4, and y =-4?

since there is the straight line there

[solemn grove]

for certain points of x and y

[dusky niche]

and we can see the f-increasing and f-decreasing

[solemn grove]

yeah no, I need the whole question to tell you what is actually going on

[dusky niche]

that is the whole question

[solemn grove]

You might be right because dy/dx does look constant

[dusky niche]

thats why this problem is confusing

[solemn grove]

Just a slope field without any eqns?

[dusky niche]

yes, in question a we are supposed to find the differential equation that explains it

and b is too find the equillibrium solutions from the direction field

"connect the graph to the differential equation"

ive seen online that the answer is dx/dt = 16-x^2 but i just didnt understand it

[solemn grove]

Nah, that makes no sense cuz we don’t have x

You’re prolly looking at the wrong place

@dusky niche

[dusky niche]

it is the same problem and all of the multiple answers says that

but i think we are right

for 16-y^2 to be it

[solemn grove]

It would make more sense

[dusky niche]

because then the equillibirum soilutions would also be y=4,-4

[solemn grove]

and here y= +_4

[dusky niche]

which makes logical sense

[solemn grove]

Yeah

[dusky niche]

yes

[solemn grove]

there is no parameter t in this question whatsoever

[dusky niche]

also autonomous functions are of the form dy/dt=f(y)

thats where the t is from

that is given in our notes

[solemn grove]

Ah

[dusky niche]

so dy/dt=(16-y^2) makes sense

typo

since y=4,-4 and dy/dt would be zero there

[solemn grove]

but that still doesn’t explain how t is related to x tho?

but according to what you said, what is x tho? like t = x?

[dusky niche]

yeah i guess that is the hole in my logic i really dont know, i thought i understood from the notes and lecture but this doesnt really make sense

[solemn grove]

Yeah no, trust me

there is something not right about it

[dusky niche]

yeah i trust you on that

[solemn grove]

there is nothing to parametrise here

it’s just dy/dx = that

you can check on desmos

It’s pretty handy for polar coords, slope fields and graphs in general tbf

geogebra if you wanna go try plotting on 3d

Other than that

Wolfram Alpha ftw

[dusky niche]

okay that sounds good i will check it

[solemn grove]

Ye, worth it

[dusky niche]

i believe that it is right to say dy/dx=(16-y^2)

[solemn grove]

Yeah, ^ you can just tell it looks identical

[dusky niche]

it does

[solemn grove]

yeah

[dusky niche]

thank you for your help!

[solemn grove]

No worries man, happy to help

if you have any worksheets

please do send help

[dusky niche]

i will if i get any

sounds good

[solemn grove]

If you dm me, I’ll try my best to respond

It’s exam season here,

so please be a tad bit considerate

[dusky niche]

im done with exams, i just started summer classes

[solemn grove]

but I’ll try my best

[dusky niche]

good luck on exams!

[solemn grove]

Ahh, lucky you

[solemn grove]

Thanks man

[dusky niche]

i mean yeah, but its very fast paced lol

[solemn grove]

I tend to like fast paced learning

[dusky niche]

its nice though because my grade doesnt matter because this is a transferred class, but i still want to do well

[solemn grove]

Fair plays

[dusky niche]

i do too but i also have to work so its hard lol and i have a chemistry course right now too

it gets busy

[solemn grove]

Ahh, I feel ya

[dusky niche]

okay im gonna finish this up, have a good one1

!*

[solemn grove]

you too mate

[dusky niche]

and ill close the chat

[solemn grove]

cya

[dusky niche]

cya

.close

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[mild drum]

can anyone help please

<@&286206848099549185>

ty bro and do u know the @alpine cape

yea im really stuck

im so confused

so it is possible to form sector

@alpine cape

hmm ok

.close

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[rugged dust]

Hello! I'm taking a course on Logic

My question is about notation. Does

∀x∀y∀z = ∀x,y,z

?

[rugged dust]

I've never seen it written like that before and it's messing with me a little

[marsh rapids]

Yes, for all x, y, z is a very common abuse of notation equivalent to forall x, forall y, forall z. How could they be different?

[rugged dust]

Alright, thank you. I was just uh.. a little worried because I've never seen it written like that, maybe there was something I was missing

.close 🙂

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[dusky elk]

ik that the answer is 3sqrt(35)

[dusky elk]

but how can i prove that this is a right angle

[alpine sable]

becayse US is diagonal

[dusky elk]

wdym?

any mathematical explanation to this?

[tacit arch]

All chords that go through the center are diameters

[dusky elk]

if a chord goes through the center does it autmatically make it a right angle triangle?

[tacit arch]

https://en.m.wikipedia.org/wiki/Thales's_theorem

[dusky elk]

ohh thx

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[serene moon]

can anyone help me caculus vc

[serene moon]

Calculus curve sketching and optimization

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[alpine sable]

hii can anyone double check that these are right. its ib math applications 2

[alpine sable]

i think i may have switched the steps

pic i meant to send ^

in 4 i used a compound interest formula, in 5 i used the financial app on ti-84 but i feel like it may be the other way around

or one for both

[last ether]

Is that what's it's called?

Oh shit that's an old post

Oops

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@alpine sable Has your question been resolved?

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@alpine sable Has your question been resolved?

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@alpine sable Has your question been resolved?

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[crimson ether]

does anyone know a fast way to divide with irrational number as the answer, say 64 is divided by 39?

[vale wigeon]

i think you misunderstand what "irrational" means

the number 64/39 is not irrational

[lone heart]

@crimson ether Has your question been resolved?

[crimson ether]

oh sorry i meant something with very long answer

do u know a faster way other than by checking one by one what number 39 can be multiplied with to be just below the number in question

or is doing the brute way the only way?

[vale wigeon]

do you want to get a decimal approximation? if so then yes you will have to do the long division

may help to write out the first nine multiples of 39 for that purpose

otherwise the fraction 64/39 is irreducible so there is little else that can be done to it (except maybe writing it as a mixed number or something)

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[alpine sable]

[
\int_1^\infty \frac{1}{x(1+\tan^2 x)} \text{ dx}
]

[ocean seal]

mahmooz

[alpine sable]

been stuck on this for 2 days

i need to prove that this integral diverges

i tried direct comparison test, limit comparison test, Dirichlet's test

none worked, any ideas please?

[vale wigeon]

what did you compare with?

[alpine sable]

1/x

1/x diverges

but that doesnt tell us anything about this integral

[lethal stump]

1 + tan^2(x) -> trig identity?

[alpine sable]

1/x is bigger

[alpine sable]

huh

[lethal stump]

1 + tan^2(x) = sec^2(x)

[vale wigeon]

this is what the integral looks like

[lethal stump]

simplifies integral to cos^2(x)/x

[alpine sable]

yes but that helps me how

[vale wigeon]

yeah definitely do that it'll make it easier to think about

[alpine sable]

so it converges?

[vale wigeon]

no

i didn't say that

[alpine sable]

h wait

it diverges

lol

my bad

[vale wigeon]

and had no intention of doing so

[alpine sable]

on the left

[lethal stump]

cos^2(x) = (1/2)(1 + cos(2x)) maybe?

[alpine sable]

might grow to infinity on the left

[vale wigeon]

you're integrating from 1 to infinity

[alpine sable]

yes

then it doesnt necessarily diverge

[vale wigeon]

so it's more like this

you can bound it from below by a step function

[lethal stump]

i got it down to

[vale wigeon]

or by a triangular spiky function

with humps that have area dropping off as 1/n

skipping over many details here

[lethal stump]

wait they changed desmos to blue?

[vale wigeon]

guess so

[alpine sable]

how does this help

[lethal stump]

idk i got it down to integral from 1 to inf of

1/2x + cos(2x)/2x

[vale wigeon]

let me try to be more explicit

[weary wyvern]

maybe write cos^2(x) in terms of cos(2x)

[weary wyvern]

oh yea this

[lethal stump]

ig you could use some integration tricks to evaluate this integral

[alpine sable]

hmmm 1/2x diverges

[vale wigeon]

you can show that $\int_{(k-\tfrac12)\pi}^{(k+\tfrac12)\pi} \frac{\cos^2(x)}{x} \dd{x}\geq \frac{c}{k}$

[lethal stump]

but my brain is too fried rn

[vale wigeon]

HOLY SHIT DESMOS CAN DO IMPROPER INTEGRALS NOW???

[weary wyvern]

what's c?

[lethal stump]

desmos always did improper integrals

[alpine sable]

whats k lol

[vale wigeon]

oh nope

[lethal stump]

yea its kinda fire

[vale wigeon]

@weary wyvern @alpine sable k is a natural number, c is a constant to be figured out

oh but

[lethal stump]

what if you summed from k = 1->inf or smth like that

[vale wigeon]

rip

[vale wigeon]

that's my idea yes

[ocean seal]

Ann

[vale wigeon]

there we go typo fixed

[weary wyvern]

ngl maybe you could just evaluate the integral with dutis

[vale wigeon]

perhaps

[alpine sable]

no idea what tudis is

so probs not supposed to be using it

[weary wyvern]

differentiate under the integral sign

its how you do sinx/x

[alpine sable]

this question is a part of my Dirichlet's test homework

but that test doesnt work here

https://blog.mahmoodsheikh.com/math/calculus2_homework9/

[vale wigeon]

you can try to bound from below the area of one "hump"

[alpine sable]

dirichlet worked on all questions except the last

[vale wigeon]

and show that these areas drop off no more quickly than 1/k

exploiting the divergence of the harmonic series

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[uneven oyster]

im trying to find the "correct" formula to count this but ive hit a wall

there are these formulas about like

First / Second Central Difference Approx.
First Forward Difference Approx.
First Backward Difference Approx.
Second Forward Difference Approx.
Second Backward Difference Approx.

[lone heart]

@uneven oyster Has your question been resolved?

[uneven oyster]

Can someone explain to me what's the difference between using First and Second difference approx. and when to use them? thanks!

<@&286206848099549185>

[last ether]

You can estainate f'(x) by taking the slope between each data point; you can estimate f"(x) by taking the slope of the slopes

[lone heart]

@uneven oyster Has your question been resolved?

[uneven oyster]

just did that, thanks a lot!!1 :))

[lone heart]

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[alpine sable]

Hello guys I need help on this

[alpine sable]

Can someone help me I am not understanding how to even start

[tender dew]

250*750/10000

[marsh rapids]

find what 1 cm² is equivalent to, then find how many hectares 3 cm² are

[alpine sable]

Ok so 1cm sq is equivalent to 62500 m sq

So 3 cm sq is equivalent to 187500 m sq

And if we divide by 10000 we get 1.875

[marsh rapids]

looks good to me

[alpine sable]

Ok thank you

This one

[marsh rapids]

A ratio of 3:2 for a quantity means you get 3 / (3+2) * m where m is the base amount of money. What would be next ratio have to be to get the same amount of money?

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

Well I DK sorry

Oh wait I got it I think
So
So let money be x
So man gets 3x/5
So woman gets 2x/5
If money is doubled
2x
2x - 3x/5
So now we get the amount the woman gets right

[marsh rapids]

say you share $5, then you get 3 and she gets 2. Now you share $10, you still get $3. How much does she get ? What ratio does that make ? See if the answer is the same in the general case

[alpine sable]

7

[marsh rapids]

that also works

[alpine sable]

Thanks!

[marsh rapids]

though 7 is not a ratio

[alpine sable]

So 3:7

[marsh rapids]

yes, 3/10 * 2x = 6x/10 = 3x/5 = 3/5 x

[alpine sable]

Cya

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[timber veldt]

1 in a 1 million but it's 6x more common 1 in what would it be? i need this to find out how much i gonna grind in game.

[worn fox]

Divide a million by 6

[timber veldt]

can you provide the answer thanks

[worn fox]

I'm sure you have access to a calculator

[timber veldt]

ok

oh im that dumb nvm

thanks

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[alpine sable]

Hello, not sure if this is the right server for it but I'll try to ask anyways
While I understand that
B->C, C->D ==> B->D
C->D, CD->B ==> C->B

I dont get how to "decompose" A->BCD
I think that A -> BCD can be decomposed in A->B, A->C, A->D , correct?

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

<@&286206848099549185>

[lone heart]

@alpine sable Has your question been resolved?

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[unkempt umbra]

can someone help me solve this?

[violet orchid]

what does generator mean here

[unkempt umbra]

@violet orchid

[lone heart]

@unkempt umbra Has your question been resolved?

[unkempt umbra]

<@&286206848099549185>

[lone heart]

@unkempt umbra Has your question been resolved?

[lone heart]

@unkempt umbra Has your question been resolved?

[lone heart]

@unkempt umbra Has your question been resolved?

[lone heart]

@unkempt umbra Has your question been resolved?

[native granite]

@unkempt umbra Have you tried some calculations yet ?

I made some calculations on my own with like brute force (never did that before), and it seems that there is multiple solutions
I feel like this is just a game of setting down some notations for the generator l and interpret the conditions you have (l and pi parallel, l is included in P)

One thing that seems to simplify calculations is to work with a normal vector for the plane pi (it is easier to interpret the condition l and pi are parallel, imo)

[oak perch]

I also don’t know what generator means in your context. But if you mean find point on the surface and a line containing that point that is contained in the surface and parallel to x-z+1=0. Then any point we have at most 2 lines satisfy this condition:
Any (u,v,w) on this surface, if 25u-5v-3 doesn’t equal 0, then let x=3/(25u-5v-3) y=5x. Then this line {(u,v,w)+t(x,y,x+1):t} satisfies the condition. If 25u+5v-3 doesn’t equal 0, then let x’=3/(25u+5v-3), y’=-5x. Then this line {(u,v,w)+t(x’,y’,x’+1):t} satisfies the condition.
only those points (u,v,w) such that 25u+/-5v-3=0 we can’t find a line satisfying the condition, there is only one such point (3/25,0,3/50)

[lone heart]

@unkempt umbra Has your question been resolved?

[unkempt umbra]

where did yo get 25u-5v-3 from?

[unkempt umbra]

.close

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[velvet sparrow]

so i solved this and got -4.050 but that should result in undefined right?

log(-4.050 + 4) = undefined

is this question just messed up?

[bitter vault]

it works as a solution if you assume a complex valued log

[dense sleet]

I mean writing it this way makes no sense but writing it like
$$log_3\left(\frac{3x}{x + 4}\right)$$

[ocean seal]

Pluton

[dense sleet]

Makes sense

[lone heart]

@velvet sparrow Has your question been resolved?

[velvet sparrow]

hm okay

[lone heart]

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[silent forge]

i know its a dumb question but why x = 7 and not x = 6.4?

[dawn birch]

how did you get x = 6.4 @silent forge

[silent forge]

'6x + y = 47' plus 'x - y = 2' is 7x=45,,, so i did 45 divided by 7 and got 6.4

is the website wrong or am i tripping

[dawn birch]

um

how is 47 + 2 = 45

lmao

[silent forge]

oh

bruh ☠️

let me just

retry

oh i see it

49 divided by 7 is 7

i feel so dumb thank you 😭

[dawn birch]

no problem

[silent forge]

its 1am anyway

im not that stupid 🤕

[tacit arch]

You're just tired. Go to sleep and come back when you're rested

[silent forge]

my huuuuge math exam is on monday

i like studying so

i'll sleep soon tho

thank you

[lone heart]

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[weary tartan]

hi

[weary tartan]

so I had my maths exam today

there was a question with a point E with coordinates (e,e)

There was some stuff given and we could form a quadratic equation and solve for e

And so I did

My answers came 2 and 8 so I was wondering if the coordinates of E(e,e) would be (2,8) and (8,2) or (2,2) and (8,8)

eh nvm dumb question

.close

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[harsh scaffold]

So I am completely new to the concept of Sigma other than knowing it means the sum of all the terms and that k I think means constant

[harsh scaffold]

I don't really know how I would approach this problem, anyone have some tips?

[alpine kiln]

so this question is really deceiving

the sequence is a_n not the sigma

[harsh scaffold]

so...

[alpine kiln]

so the third term would be 2*3+4

whats the nth term

[harsh scaffold]

how does the sigma affect the n above it?

and what does the k=1 mean? a difference of 1?

[serene junco]

The k=1 at the bottom means that you start with k=1 in the first term

[harsh scaffold]

ok

[serene junco]

Sometimes it might be k=0 instead

[harsh scaffold]

so they substitute k for n?

[serene junco]

Or something else, but 1 and 0 are most common

Yes

[harsh scaffold]

so just a different format?

[serene junco]

Yeah I suppose

Wow, Induction was right, the wording of this question is kind of misleading

[harsh scaffold]

So when it asks for the nth term how would I translate the above into a formula I could use?

[serene junco]

I'm going to try to clarify the terms they're using here first if that's okay

[harsh scaffold]

sure

im new to thi

this

[serene junco]

They say a_n is a sequence, just a list of numbers. Maybe something like 6, 8, 10, 12, ...

If we wanted to come up with a formula for this sequence, we might come up with 2n+4.

Because 2(1)+4 gives us the first term 6

2(2)+4 gives us the second term 8

etc

Does that make sense so far?

[harsh scaffold]

yes, so the 3rd term would be 2(3)+4=10

[serene junco]

Exactly

[harsh scaffold]

making d=2

for a new equatin

equation

[serene junco]

Ahh yes, the common difference is 2

[harsh scaffold]

So I could write it like, An=6+(n-1)2 so far?

[serene junco]

Sure, that's equivalent. If you distribute and combine like terms, you'd get 2n+4

[harsh scaffold]

right

[serene junco]

So, the nth partial sum of the sequence is the sum of the first n terms.

For example the 3rd partial sum is 6+8+10 = 24

[harsh scaffold]

yea i got that much

[serene junco]

This is where the sigma comes from

[harsh scaffold]

yes

So sigma is just S_n?

[serene junco]

The $n$th partial sum can be given by $\sum_{k=1}^n 2k+1$

[ocean seal]

tatpoj

[serene junco]

Yes

[harsh scaffold]

the only thing that confuses me still is the n above and k below

[serene junco]

The k below just means k is the "index" variable for the summation.

We plug in 1 for k to get the first term, then 2 for the second, all the way up until we plug in n for k for the last term

That's what the n at the top means

A simpler example might be $\sum_{k=1}^4 \frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}$

[ocean seal]

tatpoj

[serene junco]

For example

[harsh scaffold]

so if it were k=2 then the equation would differ to 2k(2)+1?

or 4k(1)+1?

[serene junco]

if it had k=2 at the bottom, that means you'd plug in 2 for k for the first term, then 3, then 4, etc.

up to whatever is written at the top

[harsh scaffold]

ok

[serene junco]

Does this equation make sense to you?

[harsh scaffold]

yes

because n=4 meaning it goes to the 4th term with k=1

[serene junco]

Right, the 4 at the top and the 1 at the bottom just means you're going to have a term where k=1, one where k=2, then k=3, then k=4

So

In your problem, they wrote out the partial sum in sigma notation

[harsh scaffold]

right

[serene junco]

But the question actually asks you for the nth term of the sequence a_n. Not the nth partial sum.

[harsh scaffold]

so in the question, it would go 6, 8, 10, 12, ...

I guess the last part I need to understand is the ending

I assume the answer is C but what does the (2n+4) mean?

just the equation so forth on?

[serene junco]

C is the nth partial sum

Its the sum of the first n terms

[harsh scaffold]

like 2(5)+4)

[serene junco]

This is the 5th term

n could be anything

[harsh scaffold]

yes

just as example

[serene junco]

But what is, in general, the nth term

in terms of n

[harsh scaffold]

uhm

B?

[serene junco]

yes

[harsh scaffold]

because it shows the way to get the term

ok

[serene junco]

The reason this question is so confusing is because they show you the formula for the partial sum

[harsh scaffold]

yea

[serene junco]

And then ask you about the sequence itself instead

If it asked you for the nth partial sum, then C would be correct

[harsh scaffold]

Alright

thank you

[serene junco]

sure thing

[harsh scaffold]

i may have some more questions as time goes on

ill just post it here if need be

[serene junco]

I know I was kind of long winded but I wanted to try to clear this up in general lol

Alright cool, good luck

[harsh scaffold]

yea thanks it helped me understand what I was doing

this would be a partial sum right?

I would start off with needing to find the 31st term

I assume

An=0.25+(31-1)0.25

[serene junco]

Yes, the 31st partial sum

Times four

[harsh scaffold]

Why times 4?

[serene junco]

Four friends will each put in 25 cents the first day

etc

[harsh scaffold]

oh right

forgor

[serene junco]

lol

[harsh scaffold]

so the first term would be 1?

[serene junco]

Yeah sure

I meant you could multiply the whole thing by 4 when you're done but that works too

[harsh scaffold]

true

lemme just try it with 1 rq

An=1+(31-1)1
An=31$?

for the final term

[serene junco]

yeah

[harsh scaffold]

And then I could plug the first and last into partial sum formula

So it would be S_31=(31/2)(1+31)

[serene junco]

right

[harsh scaffold]

494 total moneys accumilated

right?

[serene junco]

I get 496

[harsh scaffold]

or not accumilated

typo

[serene junco]

Yep, that's right

[harsh scaffold]

it wouldnt be accumilated it would be partial sum

so how would you say that in money terms

because the end result of money wouldnt be 496

it would be 31

[serene junco]

No, the total amount of money accumulated at the end is 496.

They add 31 dollars on the final day

[harsh scaffold]

ohhh

yea

ok

I see

thank you

[serene junco]

You're welcome

[harsh scaffold]

false?

because a_1 would just be 8?

or

well maybe

because n_0 would be 3

+5

[serene junco]

Don't worry about any specific term.

$a_n = a_{n-1} + 5$

[ocean seal]

tatpoj

[serene junco]

This means, the nth term is equal to the previous term, plus 5

[harsh scaffold]

so true

[serene junco]

Yes

[harsh scaffold]

so what would it mean by 8, a_n

[serene junco]

Ahh I didn't see that part

This is another way to describe a sequence, called the recursive rule.

They gave you the first term $a_1 = 8$, and then told you how to get from one term to the next, $a_n = a_{n-1} + 5$.

[ocean seal]

tatpoj

[serene junco]

In other words, start with 8, then add 5 repeatedly

[harsh scaffold]

right

isnt this the really inefficient one

that can only find the one right infront of it

[serene junco]

Yeah, a rule like a_n = 5n + 3 is better

But not all sequences can be written so nicely

[harsh scaffold]

mhm

[kindred jackal]

you can add an auxiliary sequence to help you write it nicer

But it's usually provided with the exercise

[harsh scaffold]

so end result would be true?

since they both represent the same terms

[serene junco]

yes the statement is true

[harsh scaffold]

ok

arithmetic means adds a constant natural number each term right

or subtracts

[serene junco]

Doesn't have to be a natural number

but yes

[harsh scaffold]

A: is geometric
B: +130
C: +1
D: +0

[serene junco]

I agree

[harsh scaffold]

epic

100% 😎

[serene junco]

🥳

[harsh scaffold]

im doing a pre cal course

first sem is alg 2 pretty much and 2nd is trig

I was trying to get ahead and do trig 3 years early but they didnt have a just trig course so I gotta suffer through algebra

[serene junco]

Yeah, precalc is basically a hodgepodge of seemingly unrelated topics

[harsh scaffold]

yea

[serene junco]

But they're related by the fact that you'll need them for calculus

[harsh scaffold]

yup

which I guess is useful because after this thats where I go

I really like geometry because it made sense when you look at it but algebra is just confusing because its like number form of geometry

[serene junco]

I get that

Trust me, these sequences and series are very important for calc though

[harsh scaffold]

ive gone through like khan academy, brainly, and some other courses to try and understand alg 2 but I always got lost at factoring and functions

lik the U and V stuff made no sense

[serene junco]

It's kind of cliche to say, but it's true: algebra is the reason most students struggle with calculus

Brushing up on those skills before calc rolls around is probably a good idea

[harsh scaffold]

yea I figured though that if I never get algebra down ill struggle with everything else

and since I'm trying to go into a career that I figure would probably use algebra and calculus a lot I am really trying to nail it down fast

anyways on to my next lesson ill be back in like 15 minutes after studying

[serene junco]

Good luck 😄

[harsh scaffold]

thanks

[fluid eagle]

good luck from a lurker as well

[harsh scaffold]

actually its like 8 so I think ill pick this up in the morning

adios

.close

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[stable portal]

I am bad at proofs and learning Linear Algebra, found this proofs-like question in my textbook which I suck at, and was hoping for some assistance.

[stable portal]

This is the example the textbook refers to but it didnt really help clear up anything on my end

[lone heart]

@stable portal Has your question been resolved?

[stable portal]

<@&286206848099549185>

[weary wyvern]

substitute B in the first equation

[stable portal]

for A^t

?

[weary wyvern]

in the equation A = ...

[stable portal]

something along the lines of A = pqA?

[weary wyvern]

yes

[stable portal]

pq = 1

im more confused about

the A = 0 = B

portion

[weary wyvern]

write this as kA = 0

[stable portal]

kA = 0 ?

[weary wyvern]

for some number k

[stable portal]

ah

is there a theorem that allows that?

[weary wyvern]

distributivity I guess, but that's more of an axiom

[stable portal]

ah ok

[weary wyvern]

bring everything to one side of the equation

[stable portal]

so A = pqA changes into kA = 0 from theorem/axiom

[weary wyvern]

sure... but it's quite simple

[weary wyvern]

.

[stable portal]

bring A = 0/k?

and then A = 0?

[weary wyvern]

ok you don't know what k is

k is not defined

[stable portal]

oh

its not a constant then?

[weary wyvern]

not yet, we haven't found it

[stable portal]

arbitrary real numbers?

[stable portal]

ok

[weary wyvern]

A - pqA = 0

do you agree?

[stable portal]

A = pqA

A - pqA = 0

yes

I agree

[weary wyvern]

ok

now factor out the A

but be careful A is a matrix, not a number

[stable portal]

oh

hm

[weary wyvern]

what's 2A + 3A

[stable portal]

5A

[weary wyvern]

yes

use that same idea here

[stable portal]

-pqA = 0

?

[weary wyvern]

it should be an equation

[stable portal]

-pqA = 0

?

[weary wyvern]

no

[stable portal]

hm

A(1 -pq) = 0

well its a matrix so i'm assuming that doesn't work?

work

[weary wyvern]

no

(1)A + (-pq)A = 0

this is what we have

[stable portal]

hm

ok because pq is a constant

ok

[weary wyvern]

wait did you just edit this

this is right

[stable portal]

yeah i forgot the 1

[weary wyvern]

ok now use this

[finite flax]

@weary wyvern are we able to assume cases where A is invertible and not invertible?

[weary wyvern]

wdym "are we able to assume"

sure you can assume if you want, but why

[finite flax]

because then we can use matrix algebra and inverse matrices to show pq=1

if A is invertible

[weary wyvern]

I don't think we require A to be invertible

[finite flax]

if it's not, this would only be true if A=B=0

[finite flax]

oh.

ok

[weary wyvern]

the zero matrix isn't even invertible, which is what we're trying to show it could equal

[finite flax]

I know

[stable portal]

A(1-pq) = 0

aij = 0

if pq = 0, A(1) = 0

[weary wyvern]

use the result

what's k in this case?

[stable portal]

negative?

or rather

1-pq

[weary wyvern]

no what is it precisely

[weary wyvern]

yes

[stable portal]

if pq != 0, then aij = 0 => all i and j are zero

?

[finite flax]

good job

[weary wyvern]

not quite...

[stable portal]

hm

[finite flax]

write out the statements, line by line

[stable portal]

if (1-pq) does not equal 0

[weary wyvern]

you just need the result, not the solution

[stable portal]

oh...

[weary wyvern]

if kA = 0 then k=0 or A=0

[stable portal]

oh

so because its the format

kA = 0

?

[weary wyvern]

yes we got our equation to that form

[stable portal]

ok

and since we know

A = 0

the same applies

to B = 0?

[weary wyvern]

no you don't

[finite flax]

how would you show that?

[stable portal]

wait we didnt show that A = 0?

[weary wyvern]

no

[finite flax]

you already have this

a_ij=0, where p, q, and the i,j th elements are scalars

[weary wyvern]

A could very well be non zero

[finite flax]

what conditions make that possible?

[stable portal]

i and j are zero or (1-pq) are zero?

[finite flax]

(*always possible)

no, i and j are just indices

you mean a_ij or (1-pq) are all zero

should be "either"

one or the other (or both, I guess) would have to be 0

[stable portal]

hm

so i have to prove thats the case then?

[weary wyvern]

you don't really need to worry about indices

[finite flax]

you already know that is the case (?)

from Algebra class in high school.

zero-product property of equality

[weary wyvern]

(1 - pq) A = 0 so just apply the result to this

[stable portal]

you mean 1 - pq = 0

?

[weary wyvern]

no

[finite flax]

you are over-thinking

[stable portal]

definitely.

[finite flax]

in high school, you learned that if you have a product of factors equal to zero, then AT LEAST ONE of the factors must be 0.

[stable portal]

yes

so therefore either A = 0 or k = 0 then?

[finite flax]

you can cite this as a fact.

[weary wyvern]

yes

[finite flax]

k?

we are talking about pq 1-pq

[stable portal]

1 - pq*

[finite flax]

k was the example

[finite flax]

yes, touche

[stable portal]

hm

[weary wyvern]

so what do we know

[stable portal]

either A = 0 or 1-pq = 0?

[weary wyvern]

yes

if 1 - pq = 0 then what

[stable portal]

either p or q is 0?

[finite flax]

yes, but need extra explanation for A, because the writeup is in terms of the i,j'th element of A, a_ij

[weary wyvern]

no you don't

[weary wyvern]

no

[finite flax]

he doesn't need to say every element a_ij need be 0 for this to hold?

[stable portal]

pq = 1?

[weary wyvern]

not really, that was already proved in the example which I believe he cas use without justification

[weary wyvern]

yes

[stable portal]

hm

so either pq = 1 or A = 0?

[weary wyvern]

precisely

if A = 0 then what

[finite flax]

ok, he can just cite the theorem when he replaces A with an arbitrary element a_ij?

[stable portal]

B = 0?

since the same can be applied?

[weary wyvern]

ig

[weary wyvern]

how

[stable portal]

following the same process?

[finite flax]

I think showing B is 0 is easier, just write it down and do it

$B=qA^T$

[weary wyvern]

if we follow the same process then by symmetry we will get either B = 0 or pq = 1 which gives what we want so yes

[ocean seal]

Disorganized

[stable portal]

therefore A = 0 = B or pq = 1?

[finite flax]

A is 0...
0 transpose is 0...
a constant times 0 is 0...
B is zero.

[weary wyvern]

yes

[weary wyvern]

but this is easier

[stable portal]

hm ok

alright i think i get the gist of it

i believe i was overthinking too much

[finite flax]

np, that's part of doing math

.close

[stable portal]

.close

[lone heart]

Channel closed

Closed by @stable portal

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[soft needle]

Can I get help with this?

[tacit arch]

there's hella stuff there

pick something to focus on

[soft needle]

I missed 2 days of class

But I'm a quick learner so I just need someone to explain to me how it works

[tacit arch]

ok but that's helpful to helping you

[soft needle]

Alright, how should I do the first 4?

[tacit arch]

pick one problem and try it

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

[lone heart]

@soft needle Has your question been resolved?

[soft needle]

@tacit arch

<@&286206848099549185>

[lone heart]

@soft needle Has your question been resolved?

[soft needle]

Unfortunately, no, Bot.

[finite flax]

@soft needle it doesn't have to be like this

look up the definition of a linear sequence

you are only struggling because you don't understand what the symbols mean