help-27

[woven radish]

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Requested by .bonk__

[iron kindle]

you prob fucked something up with your signs

[summer summit]

can you look at my work? ill send a picture of it here

[devout snow]

@summer summit Has your question been resolved?

[summer summit]

i sent gpt a photo of this and it said -16 as well (yes i know gpt isnt trustworthy but i got the same)

[iron kindle]

!nogpt

[summer summit]

did i switch the bounds?

[summer summit]

yes i know bro

[iron kindle]

bot ded

smh

[summer summit]

lol

[iron kindle]

from -2 to 0 should be -2x+x^2-2

[summer summit]

yeah i switched the bounds of the integral it should be the opposite

so this is wrong?

[iron kindle]

yes

[summer summit]

ok i gotta email them then lmao

because this is practice for an integration bee

[iron kindle]

lmao

[summer summit]

i would assume for a competition solely on integrals you would want the answers to be right.

[wind mason]

Which one?

[iron kindle]

the answer is correct, the derivation is just wrong

[summer summit]

yeah thats what i meant

[iron kindle]

can you give the full context?

[summer summit]

like which college?

[wind mason]

I guess? I just want the name of the bee.

[summer summit]

literally just the answers no work

[summer summit]

depaul university; its their own thing, only for undergrads

[wind mason]

Ok

[iron kindle]

yeah they should switch the bounds between those two

answer is still correct though

[devout snow]

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[reef tendon]

How do I find for a symmetric A an orthogonal matrix Q such that $Q^T A Q = diag(A_1, A_2, \dots, A_k)$, where $A_k$ is a 1x1 or 2x2 matrix?

This has something to do with spectral theorem

[woven radish]

Lambda

[reef tendon]

Let's say a matrix like this one

What I really need is the name of the process I need to use, I can look it up on youtube, but don't know what phrase to use

[stone stump]

(orthogonal) diagonalization

[reef tendon]

Thanks

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[grim breach]

Is there any reason why a cube forms when I display a set and it's subsets like this??

[fierce ember]

The number of vertices in an n dimensional "cube" is 2^n

same thing as the the number of subsets

[grim breach]

oh interesting

ive never thought of that or came across it

i appreciate it

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[granite anvil]

g(x) = 10sin(6x) +20

but when im solving with calculator im gettting sin6x = -4/5

6x = -0.927

x = -0.155

since sin is periodic i added pi, but thats only getting me 2.987

which is not in the domain of x

so what am i doing wrong

this is what the ms says but im still lost

[supple knot]

,w arcsin(-4/5) / 6

[woven radish]

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[supple knot]

this looks right

[supple knot]

you added after dividing by 6

,tex .reflect trig/sin

[woven radish]

riemann

[supple knot]

the one on the right

[granite anvil]

ok okk

ty

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[hexed plinth]

My professor presented orthogonal trajectories in a very weird way.
He said that for an equation F(x, y, c) = 0, with c as a constant, you can find orthogonal trajectories to it. He made it so that he could explain how to find orthogonal trajectory for x^2 - y^2 = C

The point I didn't get was this weird notation of F(x, y, c) = 0.
If you say there are three inputs for one output, then I imagine this function could be modeled in the fourth dimension (i.e., F(x, y, c) = w). Ignoring the fact that c was included, then I could imagine that our function is such as F(x, y) = z. So then what he meant by F = 0, is that the z values are zero? That is confusing.

[gaunt quiver]

the notation is a bit ... confusing albeit standard, ill give you that

[midnight echo]

You said it yourself. C is a constant. You have 2 variables in total (including output and input vars) so your function will be modeled in the 2nd dimension. Your prof's notation is a bit odd but not inherently wrong.

[hexed plinth]

So, F = 0, essentially stands out for z = 0 in this case?

I agree with both of you this notation is weird smh.
How are you usually presented with this concept then, if you could please tell me ?

[midnight echo]

No not really... its a bit confusing to try and "reverse engineer" your profs notation

[hexed plinth]

Want me to give you the way he presented this?

[gaunt quiver]

if we take function f(x,y) = x^2 - y^2, the function defines an infinite set of curves in the third dimensions, in which x^2 - y^2 = C
is a specific curved for the specific chosen constant C.
when trying to look for an orthogonal trajectory one usually rewrites it as x^2 - y^2 - c = 0.

but since the process of finding orthogonal trajectory is by differentiation, the constant doesn't really matter that much

[gaunt quiver]

personally, I've never seen it noted as if C is an extra input.
its not really a "third" input as C specifies a specific 2d curve in a 3d space, in this example at least

its been a while since i tackled this though i might be rusty

[hexed plinth]

Right! So what you are saying makes a lot more sense tbh.
In such a case, the function could just be defined as F(x, y) = c, and thus F is some specific function for that c value

[hexed plinth]

It's fine you've explained so well!

I don't know if this helps at all, but he presented it like this

If we have a family of curves C1, C2, … , Cn and another curve C, we say that C is orthogonal to this family if it is orthogonal (perpendicular) to every curve in the family.

Let F(x, y, C) = 0 be the equation of a family of curves. By differentiating and eliminating the parameter C, we obtain the differential equation of the family.
If F(x ,y ,y′) = 0 is the differential equation of the family, then:
F(x ,y ,−1/y′) = 0 will be the differential equation of the orthogonal trajectories. Solving this last equation gives the orthogonal trajectories.

Two families of curves are said to be mutually orthogonal if each curve in one family is orthogonal to every curve in the other. Such families form an orthogonal net.

well, thank you! @gaunt quiver @midnight echo

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[supple light]

Need help being able to quickly decipher these type of problems idk why but looking at them always confuses me and i have to really meditate on understanding what side of the z score im calculating for. i know too that i can't really "read" it and think thats part of why im struggling.

[devout snow]

@supple light Has your question been resolved?

[supple light]

I need help solving the socnd part as well i thought i understood .

What i did was solved for the z score of 209.8

209.8-218.6 (mue) / standard error which is 81.6/squareroot of 169

i ended up with -8.8/6.276923077 = -1.401960784

i inserted that into the normalcdf function as my lower limit

but got the wrong answer

<@&286206848099549185>

[midnight echo]

CLT

[supple light]

right thats the subject

[midnight echo]

Did you get the first part right?

[supple light]

yessir

[midnight echo]

Cause that doesn't apply clt, it's just straight normal

Alright

Well your initial std is 81.6, to apply the CLT compresson you divide it by sqrt(169)

[supple light]

i showed my work above just unsure what i missed

for the standard error yes i did that

i got 6.276923077 when i did

then to calculate the z score i took that qoutient to dived -8.8 which was the difference between x and the mean

then i plugged that as my lower limit in normalcdf

[midnight echo]

yup, so you get P(z>-1.402)

What's the numeric output you got?

[supple light]

i wasn't sure if i was upposed to use a standad distribution or not but i tried both. . 9966492 for when i kept the normal distribution and .1595055 for when i used standard 0 and 1 for the mean and SD which seemed more accurate given the mean being 209.8

[midnight echo]

Both are wrong

[supple light]

i saw lol

[midnight echo]

You are supposed to use the standard normal distribution though

[supple light]

is normalccdf the wrong function?

[midnight echo]

The whole point of your transformation and z score (-1.402) is to transform your problem to one with a mean of zero and std of 1

what langauge & inputs?

[supple light]

im not sure what you mean but i'm using a ti 84 if that helps

[midnight echo]

What's your inputs?

Like lower, upper, mu, sigma, etc

[supple light]

normalcdf(-1.4019, 1e99, 0,1)

i think isee where i went wrong .

[midnight echo]

Looks correct, and it outputs?

[supple light]

when i re entered i got .9195273

[midnight echo]

yeah thats right lmao

[supple light]

okay great yea i used ans as my input intially and it populated something else glad my math was decent thank you !

the question doesnt specify rounding so i was trying to avoid typing all that up but thanks again

[midnight echo]

Np. You could have also used (209.8, 1e99,218.6 ,6.2769...)

Basically instead of having to standardize, you wouldve just modified the std according to CLT and chucked the values in directly

Which in natural language would be "The prob of being between 209 and pinf with a mean of 218.6 and std of 6.2769"

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@supple light Has your question been resolved?

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[past fossil]

I got a question 🙂

[past fossil]

The area of a shape is 6x*2-2x

one side length is 2x

and is the other one

the shape is a square

[lime plaza]

What is the question?

[lime plaza]

.

[past fossil]

I said it

[mystic scarab]

I don't see any question mark, which usually denotes a question 😅

[past fossil]

I was stating what the question said lol

Ill say it again

[mystic scarab]

Mmh alright

[past fossil]

The area of a square is 6x*2-2x
the one side length is givin as 2x
find the missing side length

I cant draw it out bc im on computer

IT make sense?

<@&286206848099549185>

[mystic scarab]

Yeah all good

[past fossil]

So ik 2x is gonna have to be times by 2

so thats 4x for 2/4 sides

of the square

[mystic scarab]

Wait, are you sure it's a square?

!original

[devout snow]

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.

[past fossil]

lemme try

im on computer tho

it shows a square right

and one side is labelled

2x

and the total area of the square is 6x squared -2x

?

[mystic scarab]

...

[past fossil]

yes

im sure

[lime plaza]

if it's a square all sides should be equal

[past fossil]

well maybe its just a shape

yea prob

It said shape on the question

the picture showed smth close to a circle tho

I guess it threw me off

Lets call it a rectangle

imma go to bed this taking so long and im tired

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[grand warren]

$M$ is a subvariety of dimension $p$ of $\mathb{R}^n$.
\ \
For any $a$ of $M$, there exists an open $U$ of $\mathbb{R}^n$ containing $a$ and a submersion $g \colon U \to \mathbb{R}^{n-p}$ such that $U \cap M = g^{-1}(0)$.
\ \
For any $a$ of $M$, there exists an open $U$ of $\mathbb{R}^n$ containing $a$, an open $\Omega$ of $\mathbb{R}^p$ containing $0$, and an application $h \colon \Omega \to \mathbb{R}^n$ which is both an immersion in $\mathbb{R}^n$ and a homeomorphism of $\Omega$ onto $U \cap M$.
\ \
For any $a$ of $M$, there exists an open $U$ of $\mathbb{R}^n$ containing $a$, an open $V$ of $\mathbb{R}^p$ containing $(a_1, \dots, a_p)$ and a smooth application $G$ from $V$ into $\mathbb{R}^{n-p}$ such that, after any permutation of coordinates, $U \cap M$ is equal to the graph of $G$.

[woven radish]

tm
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[grand warren]

between these 4 equivalences, which is the one most often used to show that a set is a sub manifold?

bc i want to show that the torus defined by \ $T^n = \left{ z \in \mathbb{C}^n , \ |z_1|^2 =…=|z_n|^2 =1 \right}$ is a sub manifold

[woven radish]

tm

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[viral lynx]

Part (b)

1 moment I will draw a diagram

If im not mistaken they look like this

Line m can be rewritten into y = mx+c
-4 = (3/4)(-1) + c
c = -13/4
y = 3x/4 - 13/4

So the slopes of the other lines are also 3/4

Im not really sure what to do next though

[hollow ice]

Equation of the tangent line is gonna be y = (3/4)x+c

so, for some constants c1 and c2, the line is gonna be r distance away from the center, and for those values of c, the line would be a tangent

[hollow ice]

So what you need to do is find the values of c such that the distance between the line and the center of the circle is same as the radius of the circle

[viral lynx]

Im guessing we using |ax + by - c|/sqrt(a^2 + b^2)?

[hollow ice]

yea

that would give you the 2 values of c

[viral lynx]

With (3, -1) as x and y

[hollow ice]

yea

[viral lynx]

with
a = 3/4
b = -1
c = c

All equal to 5

[hollow ice]

yea

|ax + by - c|/sqrt(a^2 + b^2) = 5

[cinder nova]

your slope is flipped, it should be 4/3

[viral lynx]

Oh oops! Thanks 😅

[viral lynx]

I was always a bit confused about this formula, like I kinda understand it but not 100%. If we just forget about the original question for 1 second...

Just so that I understand, which corresponds to which

Where is the point (x, y)

And which line is ax + by + c

[hollow ice]

x,y is the center of that blue circle

and the red line is ax+by+c

[viral lynx]

And is the circle even necessary? or can we use this formula without it

[hollow ice]

nah circle is not necessary

[viral lynx]

So this equation is for finding the distance between a point and a line, right?

Aka the green line in this case

[hollow ice]

yea

but tbh, if I were asked a figure for the formula, I'd draw something like this. Like that circle thing is unrelated to the stuff in the formula

[viral lynx]

Ahhh OK, I think I fully understand it now 🤩

Thank you so much for your help!!

❤️

[hollow ice]

npnp

[viral lynx]

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[vital dagger]

Help please I can't find X

[tight canyon]

38?

[vital dagger]

How do i calculate that thooo

Yes its 38 in the answers

[tight canyon]

angle x is sharing same arc with the angle CAO=76

[vital dagger]

Yes

[tight canyon]

i just know that x is half of the arc angle

[vital dagger]

Back sorry wifi issues

[vital dagger]

I got 76 and I thought it was equal to x but its jot

Not

[vital dagger]

BUT ITS NOT LIKE

It's not opposite it

It's not like this

Nvm I googled it

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[ebon rose]

I tried doing this question using both separable and linear DE for practice but I’m messing up my signs somewhere

[ebon rose]

$$40+Ce^{-\frac{t}{20}}$$ vs $$40-Ce^{-\frac{t}{20}}$$

[woven radish]

Thyroxine

[ebon rose]

also for the first part ignore x, its t

[rugged sparrow]

what is your problem

[lime plaza]

$\frac{dy}{dx}=2-\frac{y}{20}$

[woven radish]

devthemasked

[lime plaza]

Is this your question?

[ebon rose]

yeah

[rugged sparrow]

the step where you raised both sides with base e

e^(-x+c)≠e^(-x)+e^c

[ebon rose]

the e^-x/20 in the first way?

[ebon rose]

oh okay thank you!

does it matter if the C is + or -?

[rugged sparrow]

no

[ebon rose]

okay tyy

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[sand isle]

i just have a question abt this image

so

a being a vector has a1 and a2 coordinates and b being a vector has b1 and b2 coordinates

easy simple whatever

now geometrically it tells me to visualise it using head to tail rule

but in the graph it is kind of evident that a1 is the same as b1

however it isn't in the question?

slightly confusing

[ripe grove]

b1 and b2 aren't coordinates

It's the magnitude and direction of the vector

So b1=3 means that vectors b has an x component that is 3 units long

And b2=-2 means that the vector b2 has a y component of -2 (2 units down)

So b1 and b2 can easily be read as 3 units right 2 units down

[sand isle]

interesting

[ripe grove]

Very interesting

I have a good note that explains it if you'd be fine with me sending it

[sand isle]

Sure

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[past fossil]

Got a question

[smoky gyro]

whats your question

[smoky gyro]

,rccw

[past fossil]

I just don’t know what to do here

[pseudo basin]

ok let's see

do you know how to find the volume of a box in general

[past fossil]

Yea it’s 18x^2+54x

[pseudo basin]

no, i said in general.

[past fossil]

Like simplified

[pseudo basin]

no.

forget about this problem

let's say you have a box with a known length, width and height

how do you find its volume

[past fossil]

Times

Them all together

[pseudo basin]

yes, V = l * w * h

[past fossil]

Yea

[pseudo basin]

ok, good.

in this question you know V, l and h

and you need to find w

[past fossil]

So you times L and H to get the total of them, then see the difference between the volume and L*H

[pseudo basin]

no you are overthinking it

[past fossil]

And that would be W?

[pseudo basin]

and not the difference

[pseudo basin]

take this formula, and:

• replace V with 18x^2+54x
• replace l by 6x
• replace h by 3

DO NOT simplify

show what you get

[past fossil]

18x^2+54x= 6x* w * 3

I symplified

And got the right side to be 18x*width

And ik to get width im going to have to divide the left side by the right side

Let me do that not

Now

18x^2+ 54x

---

18x

Divide them

And let me see what I whet

Get

I got x+3

[mystic scarab]

Yep correct

[past fossil]

Ty!

.solved

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[upbeat stone]

How do i do this with Cauchy

I did it with AM-GM but i know there is a Cauchy solution but i cannot seem to find a way to make the lhs into a product of sums

x, y, z > 0

[supple knot]

Did you try (y, z, x) and (1/x, 1/y, 1/z)

[devout snow]

@upbeat stone Has your question been resolved?

[upbeat stone]

I thought about it but there are many other terms other than the 3 on lhs so idk how it helps

[devout snow]

@upbeat stone Has your question been resolved?

[devout snow]

@upbeat stone Has your question been resolved?

[supple knot]

did you try it

[upbeat stone]

Yeah but i just noticed i forgot to square the sum of the products of the elements of the sequences and that the remaining terms might cancel in both sides 😳

[devout snow]

@upbeat stone Has your question been resolved?

[devout snow]

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[heavy ibex]

Hi all, I have a math assignment in Set Theory, it goes as follows:

If (G, *) is a group and a,b,c in G, solve the following equations while listing the characteristics which are used:

a) x * a * b * x * c = b * x * a  (under the assumption that (G, *) is an Abelian Group)
b) a * x * b * c * x = a * b * x  (under the assumption that (G, *) is not an Abelian Group)

For a), I got that both sides are equal when x = c^(-1); for b), I got that both sides are equal if and only if b and c are commutative (and x = c^(-1) here too)

Meaning, that for a) both sides are indeed equal, and for b), they are not unless b and c are commutative.

Did I do this correctly?

And if there's any more information I need to provide, please do let me know. I'm new to discrete mathematics.

[graceful cosmos]

Stating that "both sides are equal" is a strange thing to do. Of course both sides are equal. The question starts by telling you they are.

Your solution to a) is good.

You don't really have a solution to b) here.

Assume nothing is commutative. What does the equation simplify to?

[heavy ibex]

For b) I got it down to c^(-1) * b = b * c^(-1), and since it's not an Abelian group, I assume that this is not a solution since the operation is not commutative?

I could also rewrite my entire solution in English so it's easier to read and just send it here, if that'll help anything

[graceful cosmos]

I can get it down to xbc = b

Just by multiplying a few inverses in on the left or right

Other than rearranging things, this is as simple as we can do

[heavy ibex]

This is how I did it

Ah nevermind I see my mistake

Should be this then

[graceful cosmos]

There's really not a lot of flexibility to general groups. All we can really do is take a off the left and take x off the right

[heavy ibex]

I defined y even though the operation is not commutative in this context, so y = b * x can't be applied to both x * b and b * x

[heavy ibex]

Yeah, I understand; although the thing that kind of stumps me is what they want me to list whilst I'm doing this; what kind of characteristics

[graceful cosmos]

I like that you're saying what you're multiplying by. Common terms are "left-multiply" and "right-multiply"

When the term ends up on the left or right of both sides

You may want to include that x•x^-1 = e, the identity. That e•x = x. You use these a few times without mentioning, maybe they want that

[heavy ibex]

Like this, you mean?

or is it better to write forall k in G, k * e = k since I'm doing c * e and x * e here (in order to not write ... * e = ... every single time?)

[graceful cosmos]

I wouldn't go that far haha

The "forall" I mean

[heavy ibex]

So it is just sufficient to write one example of it

Anyway, thank you, that's all 👍

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[robust bobcat]

for this proof can I just assume x and y are irrational where is the additive inverse of x. the sum of these two is 0 which is not irrational so it is false

[iron kindle]

its saying always, so simply giving a counterexample is enough

[robust bobcat]

ok thx for clearing that up

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[smoky sentinel]

.

[smoky sentinel]

i understant how to graph the region and how to get the limits of integration, but can someone pls explain how they got the fuction f(y,z) = 2

[lyric hornet]

because x is from 0 to 2

imagine it as the "height"

volume is base times height

[smoky sentinel]

hmm ok

@lyric hornet can i ask u another question?

how did they get the f(x,y) for this graph the 'height' are u different fuctions?

[lyric hornet]

hmm my explanation wasn't so ideal... that's my bad, but the reason there's a y now is because the height is different at different y-values

at any given (y,z) the "x-height" was always 2

so we intergrated over that

but now any (y,z) changes as y does because of the way we constructed our shape

[devout snow]

@smoky sentinel Has your question been resolved?

[smoky sentinel]

hmm ok

[devout snow]

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[nimble sable]

so i got x - x^2/2 + x^3/3 -x^4/4

but now what

deepseek did this but i dont quite understand how it got that

[acoustic leaf]

can you write the general term for the sum?

[nimble sable]

this thing right?

[acoustic leaf]

that doesn't seem to be the correct formula for this sum

[nimble sable]

i would right the general term by looking at the pattern here right @acoustic leaf

(ignoring that deepseek found it)

write*

[acoustic leaf]

yes, or derive it from a known series

[nimble sable]

so (-1)^n+1*x^n+1/(n+1)!

nvm there is no factorial so i can just do /n

[nimble sable]

so i take the absolute value which gives me x^(n+1)/n

[acoustic leaf]

that formula doesn't match the sum

the denominator and exponent should match

[nimble sable]

is n indexed from 0 or 1

if its maclaurin its indexed from n = 0?

[acoustic leaf]

it doesn't matter, use whichever one gives a simpler formula

[nimble sable]

okay so (-1)^(n+1)*(x^n)/n

[acoustic leaf]

sure

[nimble sable]

lol

well what now

[acoustic leaf]

is there any aspect of the series we can use to get an easy error bound?

what error bound theorems do you know?

[nimble sable]

alternating series theorem

[acoustic leaf]

that works

[nimble sable]

i dont fully understand this theorem tbh but i plug 0.001 into my Rn?

[acoustic leaf]

well we want R_N < 0.001 so we should have R_N <= a_(N+1) < 0.001

so if we can get our upper bound on R_N less than 0.001 then R_N will definitely be less than 0.001

[nimble sable]

ok...how do we apply this here

[acoustic leaf]

so you want the error on your sum to be less than 0.001

you know that the error on your sum is at most the next term in the series

so if you can find an N such that the next term in the series (after N) is less than 0.001, then the remainder (which is smaller) is also less than 0.001

[acoustic leaf]

it should be in absolute values

[nimble sable]

oh right

so when you say "error on my sum" you mean the difference between the approximated value and the exact value

we want to know how far that is from ln(1.35)?

[nimble sable]

well can i omit the absolute value if i "get rid of" the (-1)^n+1

[acoustic leaf]

well that's what you get when you apply the abs val, yes

[acoustic leaf]

yes, we want it to be less than 0.001 away

[nimble sable]

hmm. okay so the point of doing the maclaurin polynomial was to recognize a pattern for the general series. now using the absolute value of that series, we plug in n values until we get a number less than 0.001

so if n = 4
i get 0.001050 which still > 0.001
if n = 5
i get 3.06377E^-4(which is 0.0003 i think) which is < 0.001

so the smallest n value that can satisfy the error bound is n = 5

is my assessment correct? is that how i approach these problems?

[acoustic leaf]

that would imply that if you were to sum 5 terms of the maclaurin polynomial it would be good to 3 decimal places

[nimble sable]

im missing your point

[acoustic leaf]

just interpreting your result

[nimble sable]

so...am i on the nose here

[acoustic leaf]

seems fine to me

[nimble sable]

okay sick

[nimble sable]

can i bother you with one more problem similar to this

[acoustic leaf]

what is it?

[nimble sable]

lmk if my work doesnt make sense im the only one who can understand it tbh

but im pretty sure the pattern is x - x^2/6

wait ^3 nvm

[acoustic leaf]

sin(x) has a pretty famous maclaurin series

[nimble sable]

0 + x + -0x^2 -x^3/6

ugh

there

are 0s for every sin so idk whether to make it - or +

[acoustic leaf]

well it doesn't really matter for the 0's

[nimble sable]

im kinda struggling with the pattern: x - x^3/3!

[acoustic leaf]

tbh this is one that's worth memorizing

[nimble sable]

would it be cheating if i just googled the series expansion for it lmao

[acoustic leaf]

well the pattern for the derivatives is fairly straightforward, nut making a formula is a bit tricky

[nimble sable]

it jumps from x^1 to x^3

[acoustic leaf]

so i don't think there's anything wrong with just looking up the formula here

[nimble sable]

okay x^2n+1

/ (2n+1)

[acoustic leaf]

that's just a way of getting only odd numbers

[nimble sable]

well isnt that what the expansion is?

its only odd

[acoustic leaf]

yes, that's how they put the expansion into a formula

[nimble sable]

ok so lets go with that

[acoustic leaf]

so the procedure from here is very similar to the previous

[nimble sable]

??

the only value that works here is 2

[acoustic leaf]

the n you solve for with the alternating series test is the number of terms, not necessarily the degree

[nimble sable]

so that means i need a different series

[acoustic leaf]

no

[nimble sable]

err general formula

[acoustic leaf]

just need to interpret the n value correctly

so if you plug in n = 2, then that is the maximum error if the n = 2 term is the first term not summed

so the highest term is the n = 1 term

but what is the power of x for n = 1?

[nimble sable]

3

i need it to equal 1?

since the first term is x^1/1!

we start indexing from 1 right

[acoustic leaf]

for this formula we should start indexing from 0, since plugging in n = 0 gives you x

if you wanted to index from 1 you would use 2n - 1 instead

[nimble sable]

okay so why isnt 2n+1 working

[acoustic leaf]

the question is asking for the degree of the maclaurin polynomial

you need to be able to convert between a value of n and the degree of the corresponding polynomial

[nimble sable]

hm and degree is different from nth term

[nimble sable]

well how do i find that

[acoustic leaf]

well the degree is the highest power of x right

and the exponent on x is 2n + 1

[nimble sable]

right

[nimble sable]

so what do i do

[acoustic leaf]

so if you have a value of n corresponding to an exponent, and a formula for the exponent corresponding to that value of n...

[nimble sable]

youre talking about a(n+1)?

so my a(n) is 2n+1

so a(n+1) would be 2(n+1)+1

is that wym

[acoustic leaf]

so the first value you plugged in to the sequence that works is 2

so that's you're n + 1 right

[nimble sable]

i suppose so

[acoustic leaf]

so then the highest power of x in the sequence is corresponding to n = 1, so x^(2n + 1) = x^(2*1 + 1)...

[nimble sable]

ohhhhh oh my god i finally get it

it only fulfills the condition when its on x^3

which means third degree

this whole time i kept associating it with the n value

ty

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[solid osprey]

how do i get that f is linear by subbing x=a+b on part e?

[solid osprey]

im p sure its related to theorem 3.8 at the top

so what we know:
f is odd
f(0)=0
f(x+y)=f(x)+f(y)

oh wait i forgot these have a solution at the bottom

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[inner ibex]

can i have some help for part B

[devout snow]

@inner ibex Has your question been resolved?

[pseudo basin]

the direction vector of your line should be:

• perpendicular to the direction vector of the given line
• parallel to the yz plane -- and thus perpendicular to i

do you understand why this is so?

[inner ibex]

so i only care about

j and k

[pseudo basin]

mmmmmmmm

i guess ISH??

but not how i would have thought about it.

[inner ibex]

oh

well when i think ab it

i just think it turns 2d now and we throw away i

[pseudo basin]

ig in this case that is true, but

here is a more general question

how do you find a vector that is perpendicular to two given vectors?

[inner ibex]

dot product = 0

oh....................

[pseudo basin]

reread what im asking carefully

[inner ibex]

uh

u use vector equation

like

idk how to explain it

so first

i would find the vector where the dot producti s 0

[pseudo basin]

i think you're overthinking it then.

[inner ibex]

and then

i put starting point and add it

[pseudo basin]

the cross product is what i was looking for from you.

[inner ibex]

oh

idk cross product

[pseudo basin]

you what

💀

[inner ibex]

my teacher said we learn that after the holidays

[pseudo basin]

okay

well then i guess yeah you have to solve for it the hard way

[inner ibex]

its in the chapter after

[inner ibex]

oh

whats the hard way?

[pseudo basin]

the "let your vector be yj + zk and declare it perpendicular to 2j-k and solve for y and z manually" way

which means your shortcut of "oh the i component is automatically 0" is justified

[inner ibex]

oh i think thats what i was triyng to explain

[pseudo basin]

ok yeah

i did not know you didn't yet have access to cross product

[inner ibex]

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[restive river]

whenever

[restive river]

i rearrange

part bi

i get it wrong

nvm

[devout snow]

@restive river Has your question been resolved?

[devout snow]

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[rugged sparrow]

fauji

[devout snow]

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[sly oyster]

x = ± √(8) * i , ± √(2) the sum of |x|^2 is?

[sly oyster]

I wrote it as -8+(-8)+2+2= -12 which is wrong

[pure flower]

this is not readable..

[sly oyster]

Uhh ok

I will write it on paper then

-12 is wrong and idk why

[knotty sage]

$|a+bi|=\sqrt{a^2+b^2}$

[woven radish]

yoboiqimmah

[sly oyster]

Oooohhh I got it!

I just presumed that the mod just mean for changing ng -ve to +ve

I am an idiot

Thanks

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[knotty sage]

nah it isn't obvious taking absolute value of a complex number

[sly oyster]

But it should have been, I was literally doing a complex number question

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[weary olive]

are these two answers the same answer? they just got switched up in the exclude... is it allowed to exclude with - ?

[meager rock]

are you talking about the last term in the first image?

this one?

[weary olive]

ye

thats was my solution

[meager rock]

its equal to the answer

take - common

$$(-4x+y) / (-2x^2 + 3y^2)$$

[weary olive]

so it would be the same? just so my prof. doesnt blame me haha

[meager rock]

you can take -1 outside

$$-1(4x-y) / -1(2x^2 -3y^2)$$

[weary olive]

give me a second, ill send a pic

[woven radish]

ℝ𝕚𝕥𝕧𝕚𝕜

ℝ𝕚𝕥𝕧𝕚𝕜

[meager rock]

-1 cancels out

$$(4x-y) / (2x^2 -3y^2)$$

[woven radish]

ℝ𝕚𝕥𝕧𝕚𝕜

[meager rock]

so ya, it is the same answer

[knotty sage]

you can use \frac{a}{b} btw would be better imo

[meager rock]

ohh, i didn't know that

[weary olive]

i dont know how to convert...

[meager rock]

ohh

[weary olive]

like

[meager rock]

one sec

first we will do numerator

and then denominator

[weary olive]

is it okay to leave the answer with the opposite operators?

or do i have to convert it

[meager rock]

depends

[meager rock]

generally, you have to convert it

[weary olive]

hmm okay

[meager rock]

so in the numerator

[weary olive]

is this correct?

[meager rock]

we have -4x + y

[weary olive]

ye

[meager rock]

that is equal to -1 x 4x + (-1) (-1) y

because -1 * -1 = 1

[knotty sage]

maybe multiplying by -1/-1 would be a bit better/easier imo

[meager rock]

from there, we can take -1(4x-y)

@weary olive did u understand?

[weary olive]

not really

[meager rock]

do you know how to take things common in an equation?

[weary olive]

not sure haha

[knotty sage]

maybe

[weary olive]

isnt it the thing i did?

or does it lead to a wrong answer

[meager rock]

ya, but not exatly

[knotty sage]

you tried factoring here and did it wrongly

[knotty sage]

you can continue from here multiply by -1/-1

[meager rock]

you can do this also

[weary olive]

what do you mean by -1/-1

could you paint me that or smth haha

i have to visually see this 🥲

[meager rock]

like multiply -1 to numerator

and -1 to denominator

[meager rock]

one sec

$$ \frac{-1}{-1} * \frac{-4x+y}{-2x^2 + 3y^2} $$

[woven radish]

ℝ𝕚𝕥𝕧𝕚𝕜

[meager rock]

like that

now just multiply it

[weary olive]

ahh okay

i got it

but why do i have to do it this way?

not the other way i did

[meager rock]

you did not change the signs

that was the problem

when you multiply -1 with (-4x + y)

you get 4x-y

when you multiply you will get the answer

[weary olive]

but this allowed, right?

to exclude like this

[meager rock]

i was talking about the way i told u

[weary olive]

yeye haha 🙂

[meager rock]

you can multiply and divide anything by the same thing

so if you have any fraction

$$\frac{a}{b}$$

[woven radish]

ℝ𝕚𝕥𝕧𝕚𝕜

[meager rock]

you can multiple it and divide it by any number and it will still be equal to a/b

[weary olive]

ohh okay, so if the bottom part of the / cant be done cuz of the - then i have to straight on go with the +?

[meager rock]

i didnt understand

[weary olive]

if it would be -8x³

[meager rock]

did u write -4x?

or -1

[weary olive]

wdym

[meager rock]

the arrow is pointing to the - sign

after the sign what is there

[weary olive]

you mean -4x?

[weary olive]

ye -4x

[meager rock]

ok, but you dont have to multiple it by 4x

you will complicate the answer

[weary olive]

wdym

[meager rock]

One sec

I will show everything visually

wait

[weary olive]

instead of - go with +?

[meager rock]

no

go with - only

but dont write 4x

one sec wait

$$ \frac{-4x+y}{-2x^2+3y^2} = \frac{-1}{-1} * \frac{-4x+y}{-2x^2+3y^2} = \frac{4x-y}{2x^2-3y^2}$$

[woven radish]

ℝ𝕚𝕥𝕧𝕚𝕜

[meager rock]

thats how u do it

[knotty sage]

would prefer using brackets when multiplying these

[meager rock]

you can

its too much work to keep using brackets

so i didn't use

but you can

[weary olive]

aight

but ty guys ^^

thanks for helping :D

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[cinder moss]

We consider x and y two real numbers. Prove (using scalar product) that x+y < (1+x²)(1+y²)

[cinder moss]

Lesser or equal

[drifting mauve]

maybe

[cinder moss]

But how to remove the absolute value

Ok

So with absolute value it’s correct ?

Thank you

[oak bane]

wait a second

i read it wrong sorry

x+y <= |x+y|

@cinder moss if its true for abs it should be true without abs

[cinder moss]

Im stuck here

[drifting mauve]

$x+y \leq (x^2+1)\cdot (y^2+1)$ are you sure its less or equal?

[woven radish]

parabolicinsanity

[drifting mauve]

and not just less?

[cinder moss]

Let me check

Less or equal

[stone stump]

what can you say about |x+y| vs |x+y|^2

[cinder moss]

Well it’s not always true that |x+y|² is greater than |x+y|

[stone stump]

when is it not

[cinder moss]

If x+y=0.5 then (x+y)²=0.25

For example

[stone stump]

for example, yes

and in general?

[cinder moss]

When it’s a whole number ?

[stone stump]

0.5 is not a whole number

[cinder moss]

When it’s less than 1

[drifting mauve]

i guess equality can hold

[stone stump]

less than 1, good

what can you tell me about the right hand side vs 1

[drifting mauve]

$|x+y|^2=(x+y)^2\geq x+y$ maybe this is the way

[woven radish]

parabolicinsanity

[cinder moss]

Well 1+x² and 1+y² are both greater than 1 so their product is also greater than 1

[stone stump]

yes

put those things together

[stone stump]

the last inequality isnt always true which is the problem here

[cinder moss]

(x+y)² is less than x+y when the latter is less than 1

The right side is greater than 1

So if x+y is greater than 1 then the inequality is correct

[stone stump]

yes

[cinder moss]

But how to prove it when x+y<1 ?

[stone stump]

x+y <= 1 <= (1+x^2)(1+y^2)

x+y <= (x+y)^2 <= (1+x^2)(1+y^2)

those are the two inequalities you need

[cinder moss]

Thank you so much

Not only did you help me but you didn’t give me the answer straight away and im thankful for that

[stone stump]

thats the point of the server

sadly not everyone knows that

[cinder moss]

Yeah

.close

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[sick valley]

can someone recognize

if the numerator is e^(it) or e^(iz) ?

or something else

[drifting mauve]

$f(z)=\frac{e^{iz}}{(z+i)^2(z-i)^2}$

correct?

[sick valley]

isnt the denominator (z-i)^2 (z+i)^2 ?

[drifting mauve]

oop yes

[woven radish]

parabolicinsanity

[sick valley]

i also think its e^(iz)

thx

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[hardy tendon]

I cannot, for the life of me, understand this method of getting the proof. please help.

[hardy tendon]

ping pls

[devout snow]

@hardy tendon Has your question been resolved?

[fickle slate]

Which part is the problem

[hardy tendon]

the red is what i cant get, but it is to prove that the top line is equal to the bottom line

[fickle slate]

Try to convert tan sec cosec in sin and cos

[hardy tendon]

i dont follow..

[fickle slate]

???

[hardy tendon]

tan sec cosec? where is that?

[fickle slate]

I meant like in the second expression

Convert tanA into sinA and cosA

Similarly with sec and cosecA

[hardy tendon]

yeah thats done its the ones in red that i cant get

[fickle slate]

TanA is sin/cos

You got csc/cos

Same with cot it is cos/sin

[hardy tendon]

the sin actually worked idk how i didnt get that @fickle slate

so did the cos

/sin

thank you so much i understand how to do the proof but my mind was mashed by the way its asked in this webwork

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[eternal aspen]

Tangents confuse me

[livid geyser]

Take derivative to find tangent slope

[pure flower]

!status

[devout snow]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[eternal aspen]

I have I found that -π/2 + sin(xo) = -cos(xo)•(π/2) + xo•cos(xo)

How do I continue

[devout snow]

@eternal aspen Has your question been resolved?

[livid geyser]

I think move everything to one side and treat it as f(x_0) = 0. plug in for a few convenient values of x_0, I think you can get both solutions by just guess and checking?

Then take derivative to show behavior of the graph. itll only have one critical between [0, pi], and if there were a another solution, that would imply the existence of a second critical point in that domain

[livid geyser]

I think this seems right

There's a critical point between every two roots due to MVT

[devout snow]

@eternal aspen Has your question been resolved?

[devout snow]

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[silk herald]

how can I finish the equation?

[silk herald]

sry I already asked before but the request expired

[pure flower]

whyd you multiply this

[silk herald]

i multiplied both sides by this

to cancel with the numerator on the right of the equation

so then I used (a-b) x (a+b) = a^2 + b^2

[pure flower]

im not convinced you're writing stuff very nicely

[silk herald]

if that makes sense, correct me if im wrong

[pure flower]

im massively confused

[silk herald]

sorry, on the second line I found the common denominator for the left side of the equation

then I factored (x2 - roo2) in front

I can rewrite it if it's difficult to read?

[pure flower]

whered the thing disappear

[silk herald]

(a-b) x (a+b)

I used this formula with the term next to it

idk if it was called term, I meant with this

turns into x^2 - 2 under the cube root, no?

[pure flower]

holy shit okay

you need to work on showing your work properly

but yeah this is fine

[silk herald]

good to know ty

so with that I was stuck at this

i don't even know if I wrote it correctly, like does the cube root fall or stay there above x^2 - 2?

[pure flower]

what?

anyway here letting the $x^2-2$ be $z$, you have $z z^{\frac13}$

[woven radish]

Percy

[silk herald]

oh nice

so from here I find common denominator again?

[pure flower]

hmmmmmmmmm

yeah no this sucks

im not suree how youd solve this properly loll

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@silk herald Has your question been resolved?

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[lofty monolith]

When I differentiate the lifetime utility expression with respect to C1

Does the 1/1+delta disappear

[upper schooner]

Assuming that delta is a constant (or at least wrt C1), you multiply the derivative of sqrt{C1} by 1/(1 + delta)

(Also hiii )

[lofty monolith]

Hellooo

[lofty monolith]

O ok had to check I ain’t crazy

How ru chartbit

[upper schooner]

I’m good, been a bit busy but for right now, I’m chilling at least

[upper schooner]

How about you?

[lofty monolith]

Ye I’m good

Just busy with schl

[upper schooner]

Awwww, happy you’re good hope everything stays good (though I’m sure I’ll still see you )

[lofty monolith]

Yeah 🫡

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@lofty monolith Has your question been resolved?

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[verbal prairie]

This should be a quick one, I dont trust myself so I ask: Is this correct?

[mortal crypt]

not quite

$a^2 = b^2 + c^2 -2bc \cdot\cos(A)$

cos (alpha) instead of alpha

[verbal prairie]

Oh so its the cos of alpha

yeah

Okay thanks! ^^

[mortal crypt]

yw

[verbal prairie]

Have a nice day

[mortal crypt]

thnx u 2

[woven radish]

Yeatte

[mortal crypt]

.close

,close

.close

.open

wutevr

[muted mural]

It's .close but it needs to be op who does it

[mortal crypt]

mm i c

[verbal prairie]

Oh wait before you leave, how do I get alpha?

[mortal crypt]

in this case, it would be arccos( the formula you had before)

$\arccos(\frac{yy-xx-length^2}{-2xy}) = \alpha$

[verbal prairie]

I hope its a build in function (I am coding something)

[woven radish]

Yeatte

[verbal prairie]

Thanks again :D

[mortal crypt]

arccos is just the same as inverse cos

it may sometimes be in the program as just acos

[verbal prairie]

It is ^^

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[brittle inlet]

My question is regarding the standard topology on R, I was to prove that any given interval is connected(and then that no other subsets of R are connected, but that implication is somewhat straightforward)

[brittle inlet]

thing is, I wanted to outline my idea of a proof so someone can check whether it is right:

given an arbitrary interval J, we assume it is not connected, then, we can find a continuous surjection f:J->{-1,1}, since f is surjective, there exist some a,b such that f(a)=-1 and f(b)=1

then, by the intermediate value theorem, we know there must exist some some c ∈ (a,b) such that f(c)=0, however, since we know J is an interval, and a,b∈J, necessarily, c∈J, thus, our given function maps to {-1,0,1}, not {-1,1}, which implies the stated surjection f does not exist, and hence J is connected

[placid rover]

sounds reasonable

i can't remember what is used to prove IVT.

just in case there's a circular issue

[brittle inlet]

isn´t IVT a result previous to the definition of the notion of connectedness though?

[placid rover]

if it is, then good

[brittle inlet]

alright, thank you!

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[granite anvil]

for this question can i js set sin2x = sin(3x-pi)

2x = 3x-pi

x = pi?

[lunar harbor]

You're missing 5 other solutions

[granite anvil]

oh

[lunar harbor]

You're better off finding the general solution

and using that to determine all the solutions in the interval

[granite anvil]

hm ok

may i get a hint

[upper ermine]

$\sin(a) = \sin(b) \implies a = b + 2k\pi \quad k \in \mathbb{Z}$

[woven radish]

.RODATA

[upper ermine]

$2x = 3x - \pi + 2k\pi \quad k \in \mathbb{Z}$

[woven radish]

.RODATA

[upper ermine]

find multiple solutions in range [0, 2π)

[granite anvil]

ummmmm

im still confused lol

[lunar harbor]

Or $a=\pi -b+2k \pi$

[woven radish]

Civil Service Pigeon

[granite anvil]

x = -2kpi?

so when k = 1 x = -2pi which is extraneous, when k = -1, x = 2pi also extraneous, when k = 0 x = 0

and k can only be integers

so

im confused

[lunar harbor]

.

[granite anvil]

2x = 3x + 2kpi

x = -2kpi

what am i doing wrong

[devout snow]

@granite anvil Has your question been resolved?

[granite anvil]

i've done

sin2x = -sin3x

sin2x +sin3x = 0

[devout snow]

@granite anvil Has your question been resolved?

[devout snow]

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[sand isle]

chat

[sand isle]

i have a question

ok so

is it possible to write u = [2,2] as a linear combination of v=[-1,1] and x=[3,-3]

assuming u v and x are vectors

what my teach did was like

2 = c1(-1) + c2(3) and 2=c1(1) + c2(-3)

and basically just added it

right

but why

why did she add the equations

it makes sense visually but i didnt understand that part

[acoustic leaf]

once you have a system of linear equations, a common method of solving those equations is adding them together

[sand isle]

makes sense yeah

[graceful cosmos]

When you add these two, c1 disappears