help-49

[ripe gulch]

Write the whole thing out with y instead of x

[hearty nexus]

okk

y(y^2-9y+14)

im sorry if this is wrong

[ripe gulch]

It’s right except the first y you could just leave that as x

[hearty nexus]

ah alright

[ripe gulch]

So: x(y^2 - 9y + 14)

Now just factor the expression inside the brackets

[hearty nexus]

(y-7)(y-2)

[ripe gulch]

Good now substitute x^2 back in for y

[hearty nexus]

OH

(x^2 - 7)(x^2 - 2)

[onyx copper]

One more step 🙂

[ripe gulch]

And don’t forget the x at the beginning

[hearty nexus]

x((x^2 - 7)(x^2 - 2))

like that??

[ripe gulch]

Yes but the brackets around aren’t necessary

[hearty nexus]

oh true

wait that really helped THANK YOU 🙏

[ripe gulch]

You’re welcome

[onyx copper]

Does the problem expect you to factor out x^2 - 7 further?

[hearty nexus]

lemme js screenshot all this

[hearty nexus]

lemme check

[ripe gulch]

How would you factor that further?

[hearty nexus]

nope it doesnt

[onyx copper]

A^2 - B^2

(a-b)(a+b)

[ripe gulch]

7 isn’t a square number

[onyx copper]

Yeah but couldn't you use square roots

[hearty nexus]

have u been in this server for 6 years 😨

[ripe gulch]

I don’t think anything would require you to do that if B isn’t a perfect square

[onyx copper]

Oh ok

Idk what the context of the question is

[ripe gulch]

Unless it explicitly tells you

[hearty nexus]

anyways tysmmm

can i close?

.close

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[obtuse totem]

how to calculate when theres two singularities enclosed in a closed loop

[obtuse totem]

I understand for the case where theres one singularity that is outside the loop, you would include it in the f(z)

[midnight plank]

@obtuse totem Has your question been resolved?

[fallow scarab]

You just sum the residue separately

https://math.libretexts.org/Bookshelves/Analysis/Complex_Variables_with_Applications_(Orloff)/09%3A_Residue_Theorem/9.04%3A_Residues

[midnight plank]

@obtuse totem Has your question been resolved?

[obtuse totem]

oh i see

ok thank you

[obtuse totem]

@fallow scarab what should you do if the curve is clockwise and traversed two times?

for clockwise, I’m assuming you just take the negative integral of counterclockwise

i’ll send the question

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.clsoe

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[olive herald]

how to find the function for this graph

[olive herald]

a(x+4)(x-1.5)(x-2.9)

is the trick I learned

0/3.1

3.1= a *4 *-1.5 *-2.9

a60/29 = 3.1

divide by 60/29

1.498333333333

It just isn't quite it

[chilly cobalt]

i have no idea how to explain this but this isnt a third degree polynomial function

[olive herald]

So my trick only works on third degree?

[chilly cobalt]

yes

for example your trick wouldnt work with x^2 + 1

[olive herald]

how would I do it with those?

I think the picture above is 5th degree

[chilly cobalt]

it is

[olive herald]

basically how do I do it with 2nd degree and anything above 3?

[chilly cobalt]

im not really sure, someone else probably knows, sorry

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[dense arch]

can anyone help me with a ? im trying to solve using first order approximations

[subtle zinc]

have you got the total resistance?

[dense arch]

yeah

[subtle zinc]

which is?

[dense arch]

in series 790

parallel 59,97

thats simple

the confusing part is finding the tolerance

[subtle zinc]

find the min and max resistance for each resistor

[dense arch]

overall tolerance

[subtle zinc]

basically each resistor has a range of possible resistances, ie an interval

add the min and max for each interval to get the range for R

[dense arch]

100 ohm -> 95 - 105, 220 ohm -> 209 - 231, 470 ohm -> 423 - 517

i add 95 209 and 423?

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[flint anchor]

how do i find pmf

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[spice perch]

can anyone help me with transportation problems

[spice perch]

this is how they look like

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[hardy shuttle]

After splitting apart an integral like: $5\cdot\int(2x+3x)dx$, the number infront has to be multiplied across everything like this: $5\bigg(\int(2x)dx+\int(3x)dx\bigg)$, right?

[grand pond]

dingypine

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[nimble copper]

right

[hardy shuttle]

good 2 know

thank u

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[austere quest]

how do you do this?

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.reopen

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✅

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[neat canyon]

How do I do f+g for piecewise functions

and what would the domain be

[proud scaffold]

i think that the domain would be the intersection of the domain of both

because we need both to be defined at the values x's

then we can add as usual

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[crisp obsidian]

for part a, i think its true, but im unsure how to prove

as I reach the point where -e < bn - Lb < e

and -e < an -La < e

but i cant show that an < ab from La < Lb

i am here

sorry i was looking at problem

oh ok

how so?

no.

nevermind

[rose trout]

You can choose epsilon strategically

In the definition

[crisp obsidian]

it means that there lmit of the sqeucne appraoches a value

[crisp obsidian]

hmm

but epsilon must be greater than 0

are you saying choose two differnet epsilon?

[rose trout]

It's similar to how you prove that a limit can converge only to at most one value.

[rose trout]

No

[crisp obsidian]

i get an - e < La

[crisp obsidian]

and Lb < bn + e

i dont think it has to do with geomtric progresion

[hard umbra]

this question quite literally has nothing to do with GPs

[crisp obsidian]

im just lost on how to do this inequlity. Liike since epsilon is always positive i cant really show an < bn

[hard umbra]

perhaps one should draw a picture

[rose trout]

The idea is that both sequences approach a different value, so they eventually aren't close to each other. You should try a sketch

[crisp obsidian]

yes... but this is what I am getting |an - La| < e and |bn - Lb| < e

like i see it intuitevly

but stuck on how to show using epsilon defintion

[hard umbra]

how big should e be so that they're separated from each other?

how big is the gap between La and Lb

[crisp obsidian]

shouldn

shouldn't it just be Lb - La?

since Lb > La

[hard umbra]

yes

[crisp obsidian]

hmm that doesn't seem to work tho

[hard umbra]

so to separate Lb from La

you need to keep the an terms on the left half of that gap

and the bn terms on the right half of that gap

notice how i said half

[crisp obsidian]

half of gap... wait hmm

but the problem is the bn terms could be on both sides of the gap...

i think

[hard umbra]

that's not how limits work

[crisp obsidian]

doesnt it only matter |bn - Lb| < e

where Lb is limit of bn

[hard umbra]

if bn were consistently on the left half of the gap, they would be no limit

yes but e can be any positive number

as small as you want

[crisp obsidian]

yeah

[hard umbra]

and all bn past a certain point must satisfy that inequality

[crisp obsidian]

all -e < bn - Lb < e

[hard umbra]

so all bn past a certain point must be as close to Lb as you desire

for example, closer than half the gap

[crisp obsidian]

hmm, im not quite seeing this in terms of inequlities tho

[hard umbra]

set e = (Lb - La)/2

-e < bn - Lb rewrites into (La + Lb)/2 < bn

an - La < e rewrites into an < (La + Lb)/2

[crisp obsidian]

hmmm wait a sec

[crisp obsidian]

hmm but how come this doesnt work with generalized values of epsilon?

[hard umbra]

what

what is "generalised values of epsilon"

[crisp obsidian]

like any epislon > 0

[hard umbra]

it works for any epsilon < (Lb - La)/2

and that's all that matters

you only need one good choice of epsilon to do the proof

[crisp obsidian]

wait but how come when i plug in epsilon for an - e < bn + e i dont quite get the same answer...?

[rose trout]

You do the reverse.
a_n < a_n + e < bn - e < b_n

[crisp obsidian]

ig, what was missing whwen i just directly did an - e < La < Lb < bn + e

[hard umbra]

you're not comparing an and bn in that chain

you won't get anywhere with that

[crisp obsidian]

but i end up with an < bn + 2e

[hard umbra]

not very helpful is it

[crisp obsidian]

yeah

[hard umbra]

which is why you should be using the other side of the inequalities

[crisp obsidian]

hmmm, like which side again sry

[hard umbra]

as was just mentioned by azyrstufficantspell

[crisp obsidian]

oh hmm

[hard umbra]

right here

[crisp obsidian]

yeah i dont see how a_n + e < b_n - e

[hard umbra]

well that's the point

you need to choose some good values of e so that it is true

i did that for you earlier

[hard umbra]

right about here

[crisp obsidian]

dang okay. I hmm. Like did you figure that out from looking at a number line?

[hard umbra]

well what could you possibly choose

[hard umbra]

i explained what you have to do here

[crisp obsidian]

im not sure. Im just confused on how to even figure that out

[hard umbra]

and all i did was put that into symbols

[crisp obsidian]

yeah. Ig i didnt really see hmm

[hard umbra]

all you had to do was draw a picture and look at it

it doesn't even have to be 1/2

you could take 69/420 on one side

and 351/420 on the other

or take 69/420 on both sides it doesn't matter

as long as you divide the gap into two bits that don't overlap

[crisp obsidian]

like the gap is between La, Lb

[hard umbra]

yes

Lb - La is the gap

make the an terms be stuck on one half and the bn terms be stuck on the other

now they'll never see each other ever again

[crisp obsidian]

oh like the epsilon needs ot be so that La + e < Lb - e?

[hard umbra]

yes

[crisp obsidian]

ohhh wait. Because i only had the condoiiton that e > 0. But i needed anotehr condition that e < Lb + La /2 ?

therefore i didnt have enough information to make a conclusion

[hard umbra]

that's not a necessary condition

you just need to choose an epsilon that works

[hard umbra]

i gave examples here

[crisp obsidian]

hmm. Is that the ratio of La to Lb?

or wait, do we need to chose differnet epsilon i think for a_n and b_n?

[hard umbra]

you can

but you don't have to

if you choose the same epsilon, it had better be at most half the gap

[crisp obsidian]

like how would it worko with 351/420? Because we are using the same epsilon im not seeing on the diagram how that would owrk

[hard umbra]

but if you use different epsilon, you can choose them to sum to at most the gap

an < La + 69/420 (La - Lb)

Lb - 351/420 (La - Lb) < bn

[crisp obsidian]

ah, so the 351/420 (like any number not 1/2) would require two different epsilon

[hard umbra]

well

that should be clear if you draw a picture

[crisp obsidian]

yeah, it seems the epsilon need to be difernet

since one extends 351/420 of La - Lb

while another extends less than that

oh wait. If i choose two differnet epsilons ohh wait... number line really helps i see. Since as long as the epsiilon less than Lb-La/2

[crisp obsidian]

as long as any epsilon chosen so it doesn't intersect like you said then it works. However i was wondeirng, is there a way to do so without number line? Or like figuring out with inequality? Im just curious in case there coems a times when i can no logner use number line

[hard umbra]

well

that's like asking "can i do geometry without drawing diagrams"

like yes but good luck

[crisp obsidian]

oh dang alr. sry i ddint know much. Its cuz i plan on taking analyssi in the future

but manifoldy stuf

but like i see the diagrams for that

and i am jsut confused

also for part b this is my proof

i think its false

because if the limits can be equal

then i can find a number n for all N in naturals such an > bn

[hard umbra]

sure

[crisp obsidian]

so proof seems alr then

[hard umbra]

ye

[midnight plank]

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[midnight plank]

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[near hazel]

implicit differentiation

what I did was cancel the dy/dx on the 8 since the y is gone when you derive it

but I assume that is dumb so idk what to do

[neon swift]

may i see your solution? because the 8 should be in the denominator

@near hazel

[near hazel]

what how

I skipped a bunch of steps but its just deriving everything and then moving everything to one side

[neon swift]

oh the y is not really gone

dy/dx(8y) = 8(dy/dx)

[near hazel]

well, time to re do it rq

[neon swift]

yea y is gone but there is still dy/dx there

then u factor out the dy/dx

[near hazel]

so um

im stuck a bit

[neon swift]

uhmm wheree

cos u get this wait

[grand pond]

Drei-kun

[near hazel]

yeah I got that

[neon swift]

now tranpose the 30x^2 to the right side

so only the dy/dx remains on the left side

[grand pond]

Drei-kun

[near hazel]

30x^2/18y * 8

?

[neon swift]

after this, you factor out the dy/dx in 18y and 8

[neon swift]

the final ans? not quitee

[near hazel]

hhmmm

[neon swift]

soo try thiss

[near hazel]

oh

[neon swift]

then after that, you divide both sides by 18y + 8 to get your dy/dx

lmk if u got it or u got confused on smth

[near hazel]

I GET IT WAIT WAIT

[neon swift]

NICEEE

[near hazel]

(-30x^2)/(18y + 8)

?

[neon swift]

YUP

THATS THE ANSWER

[near hazel]

HELL YEAH

THANKS BROOOO

[neon swift]

u js forgot that 8y will still have dy/dx

alr gotchu bru

[near hazel]

🔥

[near hazel]

i forgor abt factoring

thats what was confusing me

thats how you turn it all into one

[neon swift]

yup all the dy/dx remains on the left side

then js factor it

[near hazel]

man thats great, time to get stuck on more questions 💪

.close

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[opaque kestrel]

How do i set up the equations pls

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[weary gate]

It’s not given that KRU is 90 degrees

You can even see in the diagram that there’s a little tilt (though you shouldn’t be going off the diagram(

It can be a scalene triangle

(and is)

Which again you can see visually, but you shouldn’t be going off of the diagram. It’s just a clue in

If it was isosceles you’d be correct

[tawdry laurel]

dude he is trying his best

be grateful someone is trying to help you

because i have other stuff to do...?

we arent obliged to help you

[sinful trout]

wtf?

[tawdry laurel]

hi ren

[sinful trout]

camarozzz please do not ping moderators because you are not receiving help

[sinful trout]

yo

[sinful trout]

like death here said, we are not obliged to help you

and just because you didn't instantaneously get the assistance you required from a helper doesn't mean you call them out to mods of all people

please be patient

[midnight plank]

@spiral ravine Has your question been resolved?

[fallow scarab]

.close

[midnight plank]

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[fallow scarab]

Got banned

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[low tiger]

Please help with questions 11 and 11

[low tiger]

,rotate

[brazen moth]

Is the answer B

?

@low tiger

[low tiger]

Idk

No answer key

[brazen moth]

That slightly black region is the intersection

Of all the regions

@low tiger

[low tiger]

Ooh

[brazen moth]

It's in first and second

Goes to infinity

[low tiger]

But how second quadrant? X>=0

[brazen moth]

Where does it say x>0

@low tiger

[low tiger]

,rotate

[brazen moth]

y>0

Oh

X too

[low tiger]

Yaaa

[brazen moth]

Sorry didn't see that

[low tiger]

Np np np

But what does unbounded and bounded mean?

[brazen moth]

Does the region have finite area and it doesn't extend forever

It's bounded

[low tiger]

Ooh

Okk

[brazen moth]

This one is bounded

[low tiger]

Okkk

Thnx!

[brazen moth]

Welcome

[low tiger]

.close

[midnight plank]

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[low tiger]

. reopen

.reopen

[midnight plank]

✅

[low tiger]

Dumb me, I need help with the next question too pls

[brazen moth]

Sure

12th?

[low tiger]

Yea

!rotate

!rotate

[brazen moth]

Determinants

[low tiger]

Yea

[brazen moth]

Are numbers associated with square matrix

[low tiger]

Accosted?

[brazen moth]

So c

[low tiger]

Okkk

[brazen moth]

Because determinant can only be calculated for n×n matrix

[low tiger]

Okk

Thank you again!

[brazen moth]

Ur welcome

Again!

[low tiger]

:).

.close

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[abstract cape]

chatgpt told me is

[
\text{Total Arrangements} = 5! \times \binom{6}{4} \times 4! = 120 \times 15 \times 24 = 43,200
]

[grand pond]

Mushfiqur

[abstract cape]

but I have doubt with this answer

[midnight plank]

@abstract cape Has your question been resolved?

[onyx copper]

i would also not trust chatgpt for math, but what is your specific doubt with this answer? (I have not yet confirmed if it is right or wrong)

[misty carbon]

chatgpt is correct

[midnight plank]

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[twilit field]

I started by assuming v is an eigenvector and a is the corresponding eigenvalue

so $(T-\lambda I)(v)=0$

[grand pond]

A dense set(Ping when reply)

[twilit field]

so $av -a\lambda v =0$

[grand pond]

A dense set(Ping when reply)

[twilit field]

now what

I can conclude that $a> a\lambda$

[grand pond]

A dense set(Ping when reply)

[exotic pelican]

you know how p(x) = det(T - x I) is a polynomial?

[twilit field]

no idea

[hard umbra]

this is from axler so you can solve it without any mention of determinants

[twilit field]

$T(v) - \lambda I(v)=0$

yes

[grand pond]

A dense set(Ping when reply)

[hard umbra]

you need to answer my question

how many λs are there where T - λI isn't invertible

[twilit field]

equal to the dimension of $n$

[grand pond]

A dense set(Ping when reply)

[hard umbra]

false

[twilit field]

of V

sorry

[hard umbra]

T = 0 only has a single eigenvalue

so definitely not equal to dimV

[twilit field]

Why does it only have a single eigenvalue

[hard umbra]

Tv = 0v = 0

[twilit field]

yes

okay, so 1 eigevalue

[twilit field]

$\lambda_2 v - \lambda v=0$

[grand pond]

A dense set(Ping when reply)

[hard umbra]

that doesnt answer the question

T could have many more eigenvalues than just 1

the crux of this question lies in determining how many there can be

[twilit field]

there are finitely many

I've already proven these results

[hard umbra]

okay so you know there are only finitely many

now you are given α

where will the eigenvalues be positioned relative to α

[twilit field]

they can be less than or more than or equal to

[hard umbra]

do you know that F has an ordering?

[twilit field]

yes

I'm assumig it's R

[hard umbra]

even when F = C?

it is not necessary

[twilit field]

when it's C , no ordering, no

[hard umbra]

where can the eigenvalues be placed relative to α?

[twilit field]

No idea

[hard umbra]

i mean they can be just about anywhere

but there is certainly finitely many of them

[twilit field]

yes

[hard umbra]

so can they be arbitrarily close to α?

[twilit field]

no

[hard umbra]

because?

[twilit field]

there's no measure of closness without orderiing

[hard umbra]

false

you have the absolute value

you can certainly measure distances in C

lets say the eigenvalues of T are λ₁, λ₂, ..., λₙ, and for now assume that α does not coincide with any of these

how close can the eigenvalues get to α?

[twilit field]

arbitrarily close

[hard umbra]

you just said no

give me a bound on the distance

[twilit field]

I'm confused

I think this is too much for now

[hard umbra]

the λ₁, λ₂, ..., λₙ are fixed, and α is also fixed

your job is to show that you can find a small disc around α that doesn't contain any eigenvalues

[twilit field]

hmm

Yeah, I'm lost

[hard umbra]

in this picture, can you find a disc around α not containing a single λ?

[twilit field]

Yes

[twilit field]

Something like that

[hard umbra]

so thats what you're supposed to do in this problem

λ₁, λ₂, ..., λₙ, and α are given

find a small disc around α that doesnt contain any λs

[twilit field]

hmm

is it fine if I do this later

[hard umbra]

[twilit field]

Thanks

.close

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[last slate]

this is the solution for the first D.E

[sinful trout]

very readable mhm

[last slate]

i feared it

[sinful trout]

what's the second bit

[last slate]

1 min i will write down myslef and send it

[grand pond]

ren

[last slate]

i hope this is readable

ll

lol

solution for the second D.E

solution for the first D.E

Then he said Homogenous solution

there is no problem in adding a 0

[visual tiger]

yep

compute this for both the left and right term

one of them is solution of the driven equation

so it's gonna be equal to

the other is solution of the undriven equation

so it's 0

[sweet wing]

Newton notation wow

[visual tiger]

when you add a solution of the undriven equation, the total drive doesn't change

[last slate]

because its 0 anyways

what does he mean by homogenous solution over here?

adding solutions of 2 different D.E's ?

[last slate]

but this can't be a solution for the undriven D.E

why so ?

[last slate]

isn't it supposed to be like this ?

i saw a similar process while using complex numbers for proving smth else in SHM

hehe too many questions sorry :P

i just had to look up what homogenous solutions are 😅

.solved

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[last slate]

the physical interpretation makes much more sense

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[low tiger]

Question 18

[low tiger]

Please help

,rotate

[low tiger]

Please ping

[thick jolt]

@low tiger

What have you tried

[midnight plank]

@low tiger Has your question been resolved?

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[low tiger]

Help with question 18 pls

[next rover]

,rccw

[thick jolt]

Whatt did u try?

[low tiger]

Is it's a line joining a and b its b- a right?

[thick jolt]

Yep

[low tiger]

I get 2i+2j-2k?

Ab I mean

[thick jolt]

Yes

[low tiger]

So b?

[thick jolt]

Yes

[low tiger]

But y not c?

[thick jolt]

Yes that's also correct

Every option with the ratios 1:1:-1 is right

[low tiger]

I took the coefficients of I, j and k as Dr's and taking 2 and lambda

Okie thx

[thick jolt]

Anytime

[low tiger]

That is what I was confused about

[thick jolt]

Okk

[low tiger]

Bye

[thick jolt]

Bye

[low tiger]

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[haughty rampart]

Sketch the graph of a function that is increasing between
points (0, 0) and (3, 8) and decreasing between points (3, 8)
and (9, –3) how to draw it

@everyone

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[twilit field]

let $log_2(3)=x$. Then $2^x=3$ , so $2^{\frac{p}{q}} =3$ Thus $2^p=3^q$ which is absurd, as $3^q$ is always odd, and $2^p$ is always even, thus $x$ can't be rational

[runic hamlet]

why is that absurd

[grand pond]

A dense set(Ping when reply)

[runic hamlet]

p=0?

[twilit field]

That would require q=0, which isn't possible

[nova pike]

you can just directly assume that p and q are natural numbers considering that x>0, and therefore the LHS is also even whereas the RHS is always odd in the equation 2^p = 3^q, therefore a contradiction and showing x cannot be rational

[twilit field]

mhm

[twilit field]

Thanks!

[nova pike]

no problem :))

[twilit field]

I think contradiction works here

Let $x= \frac{p}{q}$ be a solution. where p,w are co-prime
$\frac{p^3}{q^3}+ \frac{p}{q}=-3$ so $p^3+q^2p=-3q^3$
\
We start by dividing across by $q^2$
\
That would mean that $\frac{p^3}{q^2}+p=-3q$

[grand pond]

A dense set(Ping when reply)

[twilit field]

that would mean $q^2 \mid p^3$ which isn't posible as p,q are co-prime

we have thus arrived at a contradiction

[grand pond]

A dense set(Ping when reply)

[nova pike]

yes, this looks correct

[twilit field]

thanks

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[twilit field]

We thus wish to show $3 \mid a^3-a \implies 3 \mid [ a(a-1)(a+1)]$let $a= 3k$, in this case we're done as $3\mid 3k$, let $a = 3k+1$, we're done here too as then $3 \mid a-1$. Lastly, let $a=3k+2$ then $3\mid a+1$, and we're done.

[grand pond]

A dense set(Ping when reply)

[twilit field]

Is this fine

[midnight plank]

@twilit field Has your question been resolved?

[nova pike]

Yes, this looks fine

[twilit field]

thanks

[nova pike]

I’m pretty sure you can just say that by pigeonhole principle, given this factorisation is the product of three consecutive integers, there must be one of them that is divisible by 3.

[twilit field]

That's smart

ngl

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[dreamy lichen]

Q = 1?

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[rose steppe]

how to calculate this

[lethal path]

you can also observe that 6^(n + 4) dominates in the numerator, and that 11^(n = 1) dominates in the denominator

hence the limit equals $\lim_{n \to \infty} \frac{6^4 \cdot 6^n}{11 \cdot 11^n} = \frac{6^4}{11} \cdot 0$

[grand pond]

south, just south

[rose steppe]

i figured out it's 0

[lethal path]

dominates as in when you divide the numerator and denominator by the leading terms, the other terms in the numerator and denominator will go to 0

[rose steppe]

but I need formal proof

[lethal path]

no need

these exercises just want you to calculate it

I know

[midnight plank]

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[twilit field]

so here T,S^-1TS have different eigenvalues,?

[twilit field]

*eigenvectors

So we have T(v)= \lambda v

[carmine void]

What have you tried so far?

[twilit field]

I thought their eigenvectors were the same

[carmine void]

No

[twilit field]

ah

[zealous schooner]

No, but they can be expressed nicely

[twilit field]

okay

hmm

[carmine void]

I would actually try to do b) before a). Try to find an eigenvalue for S^(-1)TS if v is an eigenvalue for T

This shouldn't be so hard. The second thing I tried worked

[twilit field]

hmm

[zealous schooner]

Would the eigenvectors of S-1TS just be ||Sv||?

[twilit field]

let me try

Let $v$ be an eigenvector of $T$, Let's find $S^{-1}TS(S^{-1}(T(v)))$

[grand pond]

A dense set(Ping when reply)

[zealous schooner]

S^-1(T(v))?

[twilit field]

yes

[visual tiger]

I mean it works, but since T(v) = some scalar multiple of v, why not just use v?

[twilit field]

oh

[visual tiger]

also if v is an eigenvector for eigenvalue 0...

[twilit field]

right

so $S^{-1}TS(S^{-1}(v))$

[grand pond]

A dense set(Ping when reply)

[twilit field]

so that gives us $S^{-1}T(v)$

[grand pond]

A dense set(Ping when reply)

[twilit field]

so $S^{-1}(\lambda(v)) = \lambda S^{-1}(v)$

[grand pond]

A dense set(Ping when reply)

[carmine void]

Do you mean S^(-1)v?

[zealous schooner]

I don't think so

[twilit field]

so the eigenvector of S^{-1}TS is S^{-1}(v)?

[zealous schooner]

i think @carmine void agrees with you

but I got something different for some reason

Let $v'$ be an eigenvector of $S^{-1}TS$ with eigenvalue $\lambda$, then $S^{-1}TSv'=\lambda v'$ or $TSv'=\lambda Sv'$

[grand pond]

kheerii

[zealous schooner]

so Sv' = (constant) * v, where v is a corresponding eigenvector of T with eigenvalue lambda

[zealous schooner]

oh wait yeah

my bad

my method matches

[twilit field]

Now this shows they have the same eigenvalues

[zealous schooner]

does it

[twilit field]

yes

[zealous schooner]

hmmmm

interesting

because we showed a one-one correspondence between the eigenvectors?

[twilit field]

I think so

[zealous schooner]

that's pretty sick

[twilit field]

I'm not sure though

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[manic bison]

No clue how to start

Yes

Did nothing

f(k).f(3) <0

Kya kar lunga iska

Ha

Itna hi hota hai

Exactly one root ka condition

D > 0 lagake intersection lega fir pata chalega D>0 useless hai

[manic bison]

Already isme aa jata hai

[manic bison]

Wo rehne de answer kaise aayega bata

Nah

Nvm

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[chilly adder]

k appears to be 5

im not sure how rigorous my solution is

[manic bison]

Aa gaya

[manic bison]

Sahi hai ye

.close

[chilly adder]

how

[manic bison]

Condition diya hai usme wo f(4) + a hai

[chilly adder]

hm

[manic bison]

Waise karke usko f(3) + kuch convert kar

[chilly adder]

f(3) + 8a + b?

[manic bison]

Nah

[manic bison]

Ha

But

[chilly adder]

then?

[manic bison]

Hame f(x).f(3) pe comment karna hai

[manic bison]

f(3) ko -(8a+b) likh sakte hai

[chilly adder]

aight

so

(8a+b)f(x) > 0?

[chilly adder]

yess

[manic bison]

Ha

[chilly adder]

now?

[manic bison]

17a +4b +c is zero

Same thing

Kya farak

[chilly adder]

alright

[manic bison]

f(5) ke terms mai try kar ab

Kuch pattern dikha?

[chilly adder]

f(5) = 25a + 5b + c = 8a + b, ah

[manic bison]

Ha

Ab mai so raha hu ping mat karna

[chilly adder]

so (8a+b)^2>0 which is true

nice

[manic bison]

.close

[chilly adder]

i used IVT and reached $\operatorname{sgn}\left[(k^2-17)a+(k-4)b\right]=\operatorname{sgn}(8a+b)$ and got stuck lol thanks

[grand pond]

@chilly adder

[chilly adder]

seems intuitive to compare the coefficients but thats not rigorous

[manic bison]

.

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[last slate]

Quick question to anyone familiar with mean sea level. Is mean sea level a point on a vertical datum, or a surface

[carmine sigil]

There are a lot of asterisks about the term "mean sea level." But the surface itself is called the geoid. The mean sea level is just a standardized value applicable to a specific coast

I hope this answers your question, but let me know if I misinterpreted it.

@last slate

[last slate]

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[umbral scroll]

https://www.amazon.fr/Art-Craft-Problem-Solving/dp/0471789011

[umbral scroll]

Does someone know where i could find a french translation of this book

[fallow scarab]

don't use help channels for books

[umbral scroll]

or know where i could find a book that explains math problem solving and math in general that could recomment me ?

[umbral scroll]

any book about math you recommend ?

[fallow scarab]

#book-recommendations

[umbral scroll]

i love you : )

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[exotic beacon]

Hey everyone i am on section d of this question.
I have plotted A and B on my circle however i cannot figure out where the point O comes from?

[exotic beacon]

is the perpendicular bisector of AB and where that meets the circle point A?

[livid python]

Origin I guess

Like point of origin

[midnight plank]

@exotic beacon Has your question been resolved?

[exotic beacon]

that didnt maker a triangle though

oh wait maybe

ill try both

[midnight plank]

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[radiant remnant]

I'm studying the continuity of

[radiant remnant]

$f(x,y)=x+y+\frac{x^3y}{x^4+y^2}$

[grand pond]

Miguel

[radiant remnant]

And 0 if not defined

I checked continuity, it's continuous. I calculated partial derivatives (1 and 1)

So for derivability I need

$\lim_{x,y\to (0,0)}\frac{f(x,y)-x-y}{\sqrt{x^2+y^2}}$

[grand pond]

Miguel

[radiant remnant]

$= \lim_{x,y\to (0,0)}\frac{x^3y}{(x^4+y^2)\sqrt{x^2+y^2}}$

[grand pond]

Miguel

[radiant remnant]

And I don't know how to calculate this limit

I feel it doesn't exist, but I'm not sure

Oh, they're also asking me if the directional derivatives are continuous, but let's check this first

[last slate]

the limit is 0

[radiant remnant]

How do I prove it?

[last slate]

l

l'hopital

[radiant remnant]

I've not studied L'hôpital with multivariate

I gotta do it bounding it by something that tends to 0

[midnight plank]

@radiant remnant Has your question been resolved?

[midnight plank]

@radiant remnant Has your question been resolved?

[prime hornet]

it might help to notice that since $(x^2 - y)^2 \geq 0$, $x^4 - 2x^2y + y^2 \geq 0$, so $\frac{x^4 + y^2}{2} \geq x^2y$. then if we multiply both sides by $x$ and take absolute values (to preserve this inequality), we can write $\frac{|(x^4 + y^2)x|}{2} \geq |x^3y|$

[grand pond]

higher!

[prime hornet]

then $f(x, y) \leq x + y + \frac{|(x^4 + y^2)x|}{2(x^4 + y^2)} = x + y + \frac{|x|}{2}$

[grand pond]

higher!

[prime hornet]

and the RHS goes to 0 as (x, y) \to (0, 0)

[midnight plank]

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[floral apex]

I need help confirming that one of these is invariant

I am having a really hard tracking the substitution

say we take xy'' = y y'

and lets call it x -> az where z is the new iv

[grand pond]

jan Niku

jan Niku

jan Niku

[floral apex]

but i know this is wrong

[carmine sigil]

d/dx(f(ax)) = df/d(ax) d(ax)/dx = a f'(ax). No?

[floral apex]

im not sure what you mean. why are you looking at f(ax)?

[carmine sigil]

True, sorry

[floral apex]

i know the answer is supposed to be something like at least $\frac 1a \dv{y}{f} = \dv{y}{x}$

[grand pond]

jan Niku

[floral apex]

but i cannot get that to come out

[midnight plank]

@floral apex Has your question been resolved?

[fallow scarab]

it'd help to think of it in terms of $x \mapsto ax$. $y(x) \mapsto y(ax)$. $y'(x) \mapsto ay'(ax)$ from chain rule. $y''(x) \mapsto a^2 y''(ax)$ can you write the entire equation in terms of before $x$ and after $\mapsto ax$ ?

[grand pond]

riemann

[floral apex]

before: $$xy'' = yy'$$ after: $$ax \qty( a^2 y''(ax) ) = y(ax) a y'(ax)$$

[grand pond]

jan Niku

[floral apex]

@fallow scarab

this is my problem: this is incorrect

[fallow scarab]

why

[floral apex]

because its not invariant

[fallow scarab]

divide both sides by a

[floral apex]

sure, you get $$a (ax) y''(ax) = y(ax) y'(ax)$$

[grand pond]

jan Niku

[floral apex]

this is wrong

[fallow scarab]

oh your after you multiplied by a onto x

[floral apex]

i dont know what that means

[fallow scarab]

that a shouldn't be there

[floral apex]

arent we taking x -> ax

it wont matter

[fallow scarab]

yea but you don't have it to start

[floral apex]

what do you mean

im not sure i follow

maybe you have a suggestion of another way to write this

[fallow scarab]

Try $t = ax$ so that $x = t/a$ and abuse $\dot{y} = \frac{dy}{dt}$ and express $\dot{y}$ in terms of $y'$ etc.

[grand pond]

riemann

[fallow scarab]

your "equidimensional" ODE should be the same with just dots instead of primes

[floral apex]

oh, I'm just dumb

man I'm having the thickest brained few days here

sorry riemann, appreciate it

.close

[midnight plank]

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[fallow scarab]

all good! i didn't really express it well until the end

[midnight plank]

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[paper owl]

does this look right? the answer im supposed to get is -5/8

[paper owl]

im given that dx/dt = 10

[mighty rock]

Well you know the value of y at x=8 right?

[paper owl]

oh

wait

its just 8y=4

ok and y = 1/2 plug in

[mighty rock]

Yep!

[paper owl]

.close

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[midnight plank]

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[obtuse willow]

This is correct SO far right?

How I dissected the problem?

[patent fractal]

yes

now keep rolling

[obtuse willow]

Okay.

[patent fractal]

do you know how to proceed

[obtuse willow]

If I recall this correctly…

It’s solving for a number in the upper and lower bounds, subtracting them, and adding them all together right?

[last slate]

Hello, I am wondering if any of you could solve this problem, it's a function: "3x+1" and x times 2. if the number is odd, you multiply it by 3, add 1, if the number is even, you divide it by 2. Have fun 😄 (you can start with any number you wish)

[obtuse willow]

Um…

[patent fractal]

bruh

[obtuse willow]

This is for MY problem.

[patent fractal]

im not really sure what you mean

you need to split up the integral now

[obtuse willow]

Yeah that’s what I mean.

One other thing.

[last slate]

wdym "bruh"?

[obtuse willow]

You’re in the wrong chat room.

[patent fractal]

just ignore them

[last slate]

woops

[patent fractal]

whats ur q

[last slate]

sorry, im new here

[obtuse willow]

How do I find the anti derivative of 12?

[patent fractal]

turn $\cos^2$ into something easier

[grand pond]

rain

[obtuse willow]

Lol what?

[obtuse willow]

How would I do that?

[patent fractal]

well do you know the antiderivative of sec^2

[obtuse willow]

That would be 1/cos^2x right?

[patent fractal]

that is just

equal to sec^2(x)

im asking if you know the integral of sec^2

[obtuse willow]

No.

No I do not.

[midnight plank]

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[void rose]

can someone help me with this two column proof

[brazen moth]

Hey @void rose

[void rose]

hi

[brazen moth]

Since 5 and 6 are equal

3 and 4 will be equal too

Correct?

[void rose]

yeah

[brazen moth]

And LR is common side for both triangle

And is hence equal for both of them

Is this enough of a hint?

I think you can manage it from this point on

[void rose]

uhh yeah i got it thank you

[brazen moth]

Anything else

?

[void rose]

nope thank you for helping me

.close

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[midnight plank]

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[hard karma]

<@&286206848099549185>

[strange river]

What part of the problem are you stuck on?

[hard karma]

tension part

im not really good at physics so i cant really tell the difference between the magnitude and the tension

[odd parcel]

Are you trying to find it via vectors or component method

[hard karma]

i dont know 😦

[weak sparrow]

funny i just had an exam on this and forgot but i can help. so youd want to split the triangle into 2 right angle triangles

so then youre able to find tension in AC and BC seperately

you can use trigonometry to find the tension in each cable

so the hypotenuse gives the T in AC or BC

oh wait we dont have an angle

oh but you have two sides

[midnight plank]

@hard karma Has your question been resolved?

[midnight plank]

@hard karma Has your question been resolved?

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[twilit field]

I would like to prove that $gcd(a,b) = gcd(a+nb,b)$

[grand pond]

A dense set(Ping when reply)

[twilit field]

I was thinking that induction would be a good idea

[nova yoke]

not really necessary, you can do a direct proof

[twilit field]

hmm

[nova yoke]

show that if d divides a and b, then it divides a + nb
then show that if d divdes a+nb and b, then it divides a

that's all you need to do

[twilit field]

ah right

let gcd(a,b)=.d

[nova yoke]

can let d be any divisor

[twilit field]

sure

[nova yoke]

the argument i sketched shows that {a,b} and {a+nb, b} have exactly the same set of common divisors

therefore they have the same gcd

[twilit field]

we then have $d \mid b$ trivially . We also have $d \mid a \land d \mid nb \implies d \mid a+nb$

[grand pond]

A dense set(Ping when reply)

[twilit field]

we now prove if $d \mid b \land d \mid a+nb \implies d \mid a$

[grand pond]

A dense set(Ping when reply)

[twilit field]

Not sure what to do from here

hmm

oh

[nova yoke]

try expressing a in terms of b and a+nb

[twilit field]

$b=dk$

[grand pond]

A dense set(Ping when reply)

[twilit field]

$a+nb = dk_1$

[grand pond]

A dense set(Ping when reply)

[twilit field]

so $nb=ndk$

[grand pond]

A dense set(Ping when reply)

[twilit field]

this tells us $d\mid a$

[grand pond]

A dense set(Ping when reply)

[twilit field]

Anyway, have a class now

gtg

thanks

[midnight plank]

@twilit field Has your question been resolved?

[midnight plank]

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