help-39

[outer drift]

?

[thin lynx]

x = x^1

square root x is x^.5

Exponent rules

[outer drift]

oh

[gloomy hare]

$x\sqrt{x}=x^{\frac{1}{2}}x^{1}=x^{\left(1+\frac{1}{2}\right)}$

[jolly parrot]

Combustion

[outer drift]

and what happens to the 78

poofed?

[gloomy hare]

it's a constant

so yeah 0

[outer drift]

so the answer is just that? x^(1+.5)?

[gloomy hare]

no

power rule

nx^(n-1)

[outer drift]

okay

um

lemmes ee

[feral leaf]

This was combining xsqrt(x) to have one exponent

[outer drift]

i got

1.5x^0.5

but that seems wrong?

[gloomy hare]

why is it divided by x^0.5

[outer drift]

or is it right

not divided

sry

typo

[gloomy hare]

then it's right

[outer drift]

ah okay

and a cleaner way to write it would be 3x^1/2
/2

[outer drift]

how about this one

i understand that I'm supposed to use chain rule

and power rule

[gloomy hare]

yeah

[outer drift]

but not sure how to start

[feral leaf]

Rewrite the sqrt as an exponent

[outer drift]

do i turn the bottom into ^2 and remove SQRT

ye

but then agter that it's still 1/rest of equation

[gloomy hare]

$\frac{1}{u^{a}}=u^{-a}$

[jolly parrot]

Combustion

[outer drift]

so ^-2

[gloomy hare]

-1/2

square root is 1/2

[outer drift]

so i got this

[gloomy hare]

.

[outer drift]

did I do something wrong here

oh

oaky

lemme redo

[gloomy hare]

also

chain rule was wrong

should've been (3x^2 + 2)

[outer drift]

How come

Aren’t I supposed to take eh derivative of the inside

[gloomy hare]

yeah

[outer drift]

Which is 2

[gloomy hare]

what's the derivative of x^3

[outer drift]

Oh

Oml

I’m sos Horus

Stupid

[gloomy hare]

happens

[outer drift]

Okay lemme fix it up

[mild socket]

heLLO how can i help

ooh calculus

like derivates and integrals and thingies

sounds like fun

[outer drift]

How about now

[gloomy hare]

yep that's good

[outer drift]

Should I go furthe r

Or just leave it

[gloomy hare]

leave it

[outer drift]

Okay

[gloomy hare]

or rewrite it as $-\frac{1}{2\left(x^{3}+2x+7\right)^{\frac{3}{2}}}\left(3x^{2}+2\right)$

[jolly parrot]

Combustion

[outer drift]

Sure

seems like this is how mathway rights it

but oh well

[gloomy hare]

that's not equal to what we achieved

[outer drift]

I made another error

I didn’t include 7

In sqrt

We did it right

[outer drift]

how about this one:

mathway literalyl says it can't solv eit

do I just solve it separetely

both parts

[gloomy hare]

product rule

• chain

• whatever

[outer drift]

aight ill use prduce rule first

[outer drift]

How tf do I simply this mess

or should I just leave it

[gloomy hare]

can't read what you wrote but i don't see the e^(....) in the left side

[outer drift]

It’s there

F’(x) g(x)

• g’(x) f(x)

[gloomy hare]

oh i meant the right side

[outer drift]

The derivative of both are there

[gloomy hare]

there should be an e^(....) in the right side

[outer drift]

Why

[gloomy hare]

derivative of e^f(x) is just f'(x)e^f(x)

[outer drift]

There should be o nly 1

[gloomy hare]

nope

should be 2

unless you factor it out

[outer drift]

oh I wrote the wrong seivative

Omg

It’s so long

Jesus

[pearl pond]

@outer drift Has your question been resolved?

[pearl pond]

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[trail vine]

Hi, I want to ask for this question:

Determine the truth value of the following propositions.

1. ∃x ∀y (x ≤ y^2), where the domain is the set of positive real numbers R+.

I dont understand the solution where we have to set y = sqrt(x/2)

[trail vine]

<@&286206848099549185>

[pearl pond]

@trail vine Has your question been resolved?

[feral sedge]

for all x, there exists a y (sqrt (x/2)) that is a counterexample

[trail vine]

yea but how can we assume to set y that

oh wait i think i figured it out

thanks

.close

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[near echo]

this feels really intuitive and i would like a hint if possible. let (\varepsilon > 0), and let (I_\varepsilon = (1-\varepsilon, 1 + \varepsilon)).\
i want to show that for any constant (C > 0), there exists an (n\in\mathbb{N}) such that (f(x) = x^n) satisfies
[(f')^{-1}(I_\varepsilon) = (x_0, x_1)\text{ where }x_1 - x_0 < C]
geometrically it's just asking for (f) to get unboundedly-steep, which seems quite obvious. it's also clear that for (n\ge 2), (f') is in (I_\varepsilon) for an open and connected interval, the algebra is just kicking me though

[jolly parrot]

maximo

[near echo]

(x_0, x_1) is an interval btw

[brisk pike]

I_epsilon is also an interval then?

[near echo]

yup

[brisk pike]

So we have to show the inverse of the derivative of f(x) evaluated in the interval I_epsilon will always be in the interval (x0,x1)?

Where 0 < x1-x0 < C

[near echo]

hm i don't think that's it

let me reread what you sent

ok yes i think i do agree

like

we can for sure say there exists x_0 and x_1 such that

(nx_0^{n-1} = 1 - \varepsilon,\ nx_1^{n-1} = 1 + \varepsilon)

[jolly parrot]

maximo

[brisk pike]

Ah

[near echo]

we somehow need to argue that we can get these x_0 and x_1 arbitrarily close

oh

[brisk pike]

We have to prove forall C > 0 there exists x0 and x1 such that they satisfy the above restrictions as well as there difference is less than C

[near echo]

is this as easy as solving for x_0 and x_1 and taking a limit?

yeah

i think it is just a limit which should be easy

[brisk pike]

Should be that

[near echo]

[x_0 = \sqrt[n-1]{\frac{1-\varepsilon}{n}}]

[jolly parrot]

maximo

[brisk pike]

I am assuming u will need to use epsilon delta definition?

[near echo]

no this is for a paper

[brisk pike]

'paper' as in research paper or exam

[near echo]

this was a loose end

research

i just said "for large enough n this works" and never fixed it lol

[brisk pike]

Now take the limit as x_0 approaches x_1, right?

[brisk pike]

I see. What's the research paper on?

[near echo]

it should be enough to show this goes to 1, which i hope it does

[near echo]

this is a special case of functionals on sets of paths connecting 2 points. the paper is about what we can say about the minimizer to a function that measures the energies of these paths
this particular example is about cases where there are no minimizers at all

[brisk pike]

What's a minimizer?

[near echo]

in the calc 1 sense

well

let (\mathcal{P}) be the set of paths connecting poitns (p,q\in\mathbb{R}^n). then you consider the energy functional
[E:\mathcal{P}\to\mathbb{R},\quad E(\gamma) = \int_I\mathcal{L}(\dot\gamma(t)), dt]
where (\mathcal{L}) is the Lagrangian of this energy functional. a minimizer is a path (\gamma_0) such that
[\gamma_0 = \argmin{E(\gamma)\mid \gamma\in\mathcal{P}}]

[jolly parrot]

maximo

[feral sedge]

nth root of n goes to 1 is well known, this should follow

[near echo]

like literally takes the smallest value

[near echo]

figures, thank you!

[brisk pike]

Okay you have lost me therefore I will not ask anything. Too many weird symbols

[near echo]

the paper is about characterizing extremizers to these functionals E, or instead characterizing the functionals given information about what kind of functions minimize E itself

regardless, thanks for the help kraken getting talking made it clear

.close

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[rich timber]

how do you compare graphs to the parent function?

[rich timber]

?

[plush bramble]

!original

[pearl pond]

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.

@rich timber Has your question been resolved?

[rich timber]

I don’t have an equation

.close

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[rigid mist]

Point M moves along the circle $(x-4)^2+(y-8)^2=20$. Then it broke away from it and moving along a tangent to the circle, cuts the x-axis at the point $(-2,0)$. The coordinates of the point on the circle at which the moving point broke away can be

[jolly parrot]

🐱!Yajat! 【Catfan1398】🐱

[sudden pumice]

whats the issue u are facing

[rigid mist]

just dont know how to begin

ik its center

slope m tangent formula

but cant do much

[sudden pumice]

do u know how to get equation of tangent if u know its point of contact with circle?

[rigid mist]

no ig

[sudden pumice]

umm then idk how to explain

[rigid mist]

telling the formula might just help

[sudden pumice]

do u know T=0

[rigid mist]

yes

[sudden pumice]

yes so to get the tangent at a point P on the circle , we do T=0 with respect to the pointP

[rigid mist]

what would be the equation then

[sudden pumice]

the equation of circle is x^2 + y^2 - 8x - 16y +60=

if we do T=0 here then what would u get?

[rigid mist]

like using the transformations and putting x1 and y1 as -2 and 0?

this?

[sudden pumice]

no

[rigid mist]

that would give us the equation of the tangent

[sudden pumice]

yes

[rigid mist]

thats what i meant

[sudden pumice]

well u could do it like that but i had something else in mind

[rigid mist]

okay in this method

after getting the equation of the tangent

what should i do

?

find the value of x and y?

that is thier intersection ppint

[sudden pumice]

yes

[rigid mist]

yeaa

ok

thanks

.close

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[finite lily]

im trying to find out how to get the speed of the block before it hits the spring

[regal herald]

energy conservation, the added elastic potential energy when the spring stops compressing should be the same as the blocks kinetic energy prior to collision, i think anyway, i could be wrong

[midnight haven]

use energy conservation and find the height from which the block was dropped.
mgh = (1/2) kx²

as the block is dropped, initial velocity = 0
use v² - u² = 2as equation
v² = 2gh
v = sqrt(2gh)

[finite lily]

the height would be 0.19m?

[midnight haven]

lemme check

,calc 0.5 * 238 * (0.19)^2

[jolly parrot]

Result:

4.2959

[midnight haven]

,calc 0.335 * 9.8

[jolly parrot]

Result:

3.283

[midnight haven]

,calc 4.2959/3.283

[jolly parrot]

Result:

1.3085287846482

[midnight haven]

it's 1.3 m

@finite lily

did you convert grams into kilograms before substituting the value of m in the equation?

[finite lily]

yeah

i just thought the height was 0.19 m because thats what the spring compression was

so the speed would be ,calc sqrt(2 * 9.8 * 1.3085)

[midnight haven]

ah, nope the amount spring compresses need not to be equal to height of the block

[midnight haven]

yep

bot commands won't work in middle of a message

[finite lily]

,calc sqrt(2 * 9.8 * 1.3085)

[jolly parrot]

Result:

5.0642472293521

[midnight haven]

,sqrt(2 * 9.8 * 1.3085)

L

[midnight haven]

yep

[finite lily]

hmm, doesnt seem to work

[midnight haven]

that isn't the correct answer?

[finite lily]

no

[midnight haven]

use g = 10 m/s²

[finite lily]

still doesnt work

[midnight haven]

wait

we can't use energy conservation here ig so, cause the question says "becomes attached to the spring" and in such cases energy isn't conserved

can use work-energy theorem tho

work done by spring force + work done by gravity = ∆KE = 0 - (1/2)mv²

@finite lily

[midnight haven]

perfectly inelastic collision takes place between the block and the spring's head sorta thing

and in perfectly inelastic collisions, after the collision both the particles move with common velocity

and energy isn't conserved

but momentum is conserved (it's conserved for every type of collision)

[midnight haven]

-4.2959 + 0.62377 = -(1/2)mv²

,calc (4.2959 - 0.62377) * 2

[jolly parrot]

Result:

7.34426

[midnight haven]

7.34426 = 0.335 * v²

,calc sqrt(7.34426/0.335)

[jolly parrot]

Result:

4.6822178696751

[midnight haven]

can you check whether that's correct?

[finite lily]

sorry about that

yes thats correct

[midnight haven]

awesome

[midnight haven]

here's the reason why energy conservation didn't work

[finite lily]

ok that makes sense

Thank you very much

[midnight haven]

you're welcome! the wording of the question was quite tricky haha

if it was a MCQ question and the answer which we got using energy conservation was one of the option

im pretty sure most of the students are gonna tick that 💀

including me 💀

[finite lily]

lol

.close

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[pulsar lark]

,rccw

[pulsar lark]

!status

[pearl pond]

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

@hidden swift Has your question been resolved?

[pearl pond]

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[undone bison]

So I have expressed $\frac{\partial ^2 z}{\partial x^2}$ as $(z_s s_x)_x = (2z_s)x=2z{sx}$

[jolly parrot]

george clooney real account

[undone bison]

and got the overall expression $2z_{sx}+3z_{tx}-4z_{sy}-9z_{ty}=0$

[jolly parrot]

george clooney real account

[undone bison]

now i am stuck

[pearl pond]

@undone bison Has your question been resolved?

[pearl pond]

@undone bison Has your question been resolved?

[pearl pond]

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[ornate drum]

how do i simplify (1 + sqrt3/3) / (1 - sqrt 3/3)

[ornate drum]

so that I can solve for this tangent

[robust sentinel]

hmmm conjugate?

[ornate drum]

ah okay, so multiply by 1 + sqrt3/3

let me try

did not help nw I have (1 + (2sqrt3/3)+1/9) / (1 - (2sqrt3/3) - 1/3)

which is basically just as confusing

the /3's in the denominator can be multiplied to the numerator?

what about the denominator's in the numerator?

[robust sentinel]

i'd multiply 1+sqrt3/3 by (1-sqrt3/3)/(1-sqrt3/3)

and do the same for the bottom part

[ornate drum]

.close

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[pearl pond]

@scenic cedar Has your question been resolved?

[pearl pond]

@scenic cedar Has your question been resolved?

[pearl pond]

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[rustic owl]

i have to prove this combinatorially

[rustic owl]

is my reasoning correct?

left hand side:
we choose k vertexes from the set of n vertexes and color them blue or red

right hand side:
we choose 0 vertexes from the set of n vertexes that are going to be blue and k that are going to be red

[opal moss]

yup

[rustic owl]

ect.

[opal moss]

you don’t need to type the rest

[rustic owl]

thanks 🫡

[worldly glacier]

ty IV

[rustic owl]

.close

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[earnest pendant]

Need help with this

[earnest pendant]

So far what I did was
(a) part :-

dV/dt = 2

V = 1/3 πr²h
dV/dt = 2/3 πr dh/dt
2 = 2/3 π(5) dh/dt
dh/dt = 3/5π

Is this correct?
Or did I make a mistake?

[limber glacier]

That's not how u differentiate the product of r²h

They are both functions of time

[earnest pendant]

Oh

[limber glacier]

You could try to express the radius in terms of the height

Or vice versa

So that you get an expression in just one variable for the volume

[earnest pendant]

How ?

[limber glacier]

Draw a picture

The radius of the conical volume of water remaining in the tank is automatically determined by its height

[earnest pendant]

Oh k

[earnest pendant]

How would you find
dr/dt in an equation with both
dh/dt and dV/dt ?

Radius can be found but what about dr/dt ?

[limber glacier]

If u have an expression for the radius in terms of h

[earnest pendant]

r² = 3V/πh

[limber glacier]

Well yes but not involving the volume

Use a simple geometric relationship

[earnest pendant]

Hm don't know those

[limber glacier]

Okay so u know the original cone has radius 5 and height 14

Let's say u cut it into two at a point where the height is 7

Can u find the radius of the resulting smaller cone

[earnest pendant]

I get what you are saying
But making the expression is confusing me a bit

[limber glacier]

Have u drawn a picture

[earnest pendant]

.close

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[polar steppe]

I do not understand this question

I barely understand 2. I recall drawing slopes for questions like these but it might not be the same.

[pearl pond]

@polar steppe Has your question been resolved?

[polar steppe]

<@&286206848099549185>

[near haven]

slope fields?

[polar steppe]

yes!

[polar steppe]

Yes slope fields

[near haven]

yeah just

use them

start at initial condition of p = 1.1

[polar steppe]

Wouldn't I need the general solution firsT?

[near haven]

don't think so?

[polar steppe]

hmm

so give me the starting point here

[near haven]

you could start at any t value

dp/dt depends on p only, not t

and the starting p value should be 1.1

[polar steppe]

okay I follow that I think

what would P(t) be

[near haven]

at the initial point it would be 1.1

[polar steppe]

hmmm

[near haven]

${y}^{4}-10,{y}^{3}+36,{y}^{2}-56,y=C,\left({e}^{6,x},{y}^{4}-10,{e}^{6,x},{y}^{3}+36,{e}^{6,x},{y}^{2}-54,{e}^{6,x},y+27,{e}^{6,x}\right)-32,;;{y=3},;;{y=1}$

meh

that's the solution

[polar steppe]

I'm not computing the lim as t approaches infin of a constant 1.1 am I?

[near haven]

no

okay let's start

at (0,1.1)

the slope at p = 1.1 is -0.5

[polar steppe]

how did you get that

[near haven]

dp/dt at p = 1.1

[polar steppe]

So, are you doing:
(1.1 - 1)(1.1-2).. ect?

[near haven]

yeah

[polar steppe]

huh

okay

[near haven]

so P(t) is decreasing

[polar steppe]

oh wait

nvm

continue

[near haven]

okay

[polar steppe]

yes I agree that it's decreasing

[near haven]

so let's say it's now at p = 1.05

the slope would still be negative, aka decreasing

so we go down to p = 1

[polar steppe]

the slope is 0

[near haven]

yep

so p(t) isn't changing

[polar steppe]

cause of multiplication of 0

[near haven]

roughly like this

so

as we said, p(t) isn't changing because the rate of change (slope) is 0

so p stays at 1

and as t goes to infinity, it'll be 1

[polar steppe]

I see that

[near haven]

the answer should be 1

[polar steppe]

I follow

[near haven]

yep so just do the same for initial condition of 1.9

[polar steppe]

some of the steps I'm not too sure about though

[near haven]

like?

[polar steppe]

Let me try.

I'm pretty sure I know where I'll get stuck

[near haven]

alright

[polar steppe]

So, P(1.9) would give us a negative slope -0.2079

would we just chose another number arbitarly close to the one given and see what happens?

I wasn't sure why you picked p = 1.05

[near haven]

yeah

[polar steppe]

I assumed it was because it was close to what was given and you were looking for something

are we looking for where the slope is 0

[near haven]

also

this helps you ALOT

[near haven]

pretty much looking at where the curve stagnates

[polar steppe]

I see I see

Well

[near haven]

you can see that at 1<p<2 f(p) (the slope of p(t)) is negative

[polar steppe]

why wouldn't it be the case that if P_0 = 1.9, lim n approaches infin P(t) = 2?

[polar steppe]

true

[polar steppe]

wait are you saying it's negative because it's unders the x axis

[near haven]

yeah

honestly don't know how to explain this well because my knowledge is based upon some videos on khan, i didn't take this yet lmao

[polar steppe]

oooo okie

well you've fooled me haha. You seem to know what you're talking about so

[near haven]

slope of p(t) is negative at 1.9, so it's decreasing, so it'll go to a value lower than 1.9 and the slope is negative when 1<p<2 so it'll keep going down until it reaches p = 1 and stays there, so unless 3>p => 2 it can't logically reach 2

[near haven]

barely

[polar steppe]

I'll have to think about that answer.

[near haven]

actually it'll never go to 2 unless the starting condition is 2

[polar steppe]

the limit will not be 2 unless the starting condition is 2

[near haven]

yeah

[polar steppe]

why not 1.5 or something

why 1

[near haven]

because at P = 1 the slope is 0, the rate of change is 0, P won't change as t changes, it'll stay at P = 1, so it'll stay frozen

[near haven]

at p = 1.5, the slope is negative, it won't stay there it'll go down

[polar steppe]

but doesn't the slope change from 1 - 2?

huh

okay

[near haven]

it does, but it's always negative in 1<p<2

means at any starting condition in 1<P<2, it'll always go to 1 as t goes to infinity

here's a good way to visualize it https://www.geogebra.org/m/W7dAdgqc

[polar steppe]

Alrighty

Ty for all your help here

.close

[pearl pond]

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[fluid gulch]

how would I find the force tension of two masses on an inlcine place with a pully

[pearl pond]

Please don't occupy multiple help channels.

[sudden pumice]

umm u can make their acceleration equal if they are connected by a string

[midnight haven]

draw the FBD and write the force equations and solve them.

if the inclined plane is frictionless plane then f = 0

also don't forget the tension force, it isn't there in above pic

[pearl pond]

@fluid gulch Has your question been resolved?

[pearl pond]

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[random cipher]

e

[random cipher]

bro

this is my help

[fossil drum]

💀

[random cipher]

U just came out of no where bro

[signal atlas]

He left already

Post ur question

[random cipher]

confuse

<@&286206848099549185>

[signal atlas]

What r u confused about it just looks like an equation atm

Also dont ping helpers for first 15 min

[random cipher]

man

oconfuse with this part

https://tenor.com/view/cats-happy-face-big-smile-battle-cats-the-battle-cats-gif-12631153

i give up

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[random cipher]

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[pearl pond]

✅

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[random cipher]

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[pearl pond]

✅

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✅

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[random cipher]

w u

w u

wu

uwu

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[pearl pond]

✅

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[sweet junco]

3^m * 3^n = 3^(m+n)

3^m / 3^n = 3^(m-n)

Use that

[hidden swift]

Idk how to do it

[pearl pond]

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[leaden tangle]

Can someone explain

[leaden tangle]

3d

[rich granite]

Can you visualize the graph for h(x)?

[near haven]

$h\left(x\right)=\int_{x}^{3}f\left(t\right)dt=-\int_{3}^{x}f\left(t\right)dt$

[jolly parrot]

CH3

[midnight haven]

(3^6 * 3^5) / 3^7 = 3^(6+5) / 3^7 = 3^11 / 3^7 = 3^(11 - 7) = 3^(4) / 1 = 81 / 1 = 81 : 1

[near haven]

$h'\left(x\right)=-f\left(x\right)$

[jolly parrot]

CH3

[near haven]

you now know when h(x) is increasing/decreasing

[pearl pond]

@leaden tangle Has your question been resolved?

[leaden tangle]

isn't x smaller tho

[leaden tangle]

not really

like if 3 is bigger

shouldn't it be at top

[near haven]

yeah

well

you can still flip it

[leaden tangle]

so why would you want to flip it

don't you want smaller

at the bottom

[near haven]

not always

okay let's say

we don't flip it

[pearl pond]

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[near haven]

type .reopen

[pearl pond]

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[rich granite]

This is how your h(x) looks like

[rich granite]

Sorry
I didn't see it got closed

.close

[pearl pond]

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[trim crane]

hello

[trim crane]

I am stuck on how to find the time it takes to fill a inverted circular cone

this is related rates in calculus

the work I did to find the time to fill up the circuler cone is on the right

would that be correct?

<@&286206848099549185>

[pearl pond]

@trim crane Has your question been resolved?

[pearl pond]

@trim crane Has your question been resolved?

[pearl pond]

@trim crane Has your question been resolved?

[pearl pond]

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[vast holly]

Why do they use a set for eigenspace in the first but Span({...}) for eigenspace the second?

[timid spindle]

just different notation for the same thing

the span of one vector is the same as all multiples of it

[vast holly]

So they are both correct? Or is there a preferred way for me to write it?

[wary jacinth]

<@&286206848099549185> i am in year 7 and need help with ratio

[vast holly]

!help

[pearl pond]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[timid spindle]

not really a preferred way, either is fine

[shadow jasper]

Open your own channel dont use someone elses

[wary jacinth]

how to open chanel

[shadow jasper]

Also year 7? How old are you?

[wary jacinth]

12

[fluid spindle]

b&

[rigid mist]

lmao

ded

[oblique schooner]

for 1y until they are 13 lmao

unfortunate

[rigid mist]

how r all the mods here

[vast holly]

[vale jay]

L+ratio

[rigid mist]

lets just not flood her channel with all these

[vast holly]

.close

[pearl pond]

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[frank goblet]

Hey

[frank goblet]

,rotate

[frank goblet]

In the third line we have an equation that i have to solve for t

E and h are constants so we can ignore them if we just want to find a way to solve it

.close

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[vast holly]

What is this exactly supposed to mean

they are equal to each other and then take inverse

what ?!

[shrewd basin]

It’s looking for the inverse matrix of the identity map

[vast holly]

I found the first change-of-basis matrix

but I don't know what I should input into sympy to get the second one @shrewd basin

[vast holly]

so its not a spelling mistake?

[shrewd basin]

Have you computed what they told you to compute

[vast holly]

yes

I computed it and then took the inverse of it and got a matrix:

i/2, 1/2 - i/2
-i/2, 1/2+i/2

Not sure if its correct

theres no answer here

<@&286206848099549185>

[humble lintel]

Yes it's a typo, the inverse applies to just the change matrix on the right hand side

[pearl pond]

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[languid plinth]

what am i supposed to do here😭😭

[pearl pond]

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[pearl pond]

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[timid path]

Need help on this question:
a and b are natural numbers st.
a/b > sqrt(2).
Prove that a/b - 1/(2ab) > sqrt(2)

[pearl pond]

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[pearl pond]

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[jovial mango]

Let non-negative numbers $x,y$ satisfiy $\sqrt{2+\sqrt 3}=x\sqrt 2+y\sqrt 6$. Calculate the value of $A=x^2+y^2$

[jolly parrot]

ChemicalMendeleev

[shrewd basin]

what have you trie

[jovial mango]

Square both sides

[earnest stratus]

There's a better way...

[jovial mango]

How?

[earnest stratus]

Try rewriting everything inside sqrt as a perfect square.

(a+b)^2 = a^2 + 2ab + b^2

[jovial mango]

$\frac{\sqrt 6+\sqrt 2} 2=x\sqrt 2+y\sqrt6$

[jolly parrot]

ChemicalMendeleev

[earnest stratus]

You can now compare...

[jovial mango]

Compare what?

[earnest stratus]

The two sides...

[shrewd basin]

Should x and y be nonnegative rational numbers?

[jovial mango]

Rational

[shrewd basin]

Then just compare the “coefficients” or sqrt

[jovial mango]

They are the same

[pearl pond]

@jovial mango Has your question been resolved?

[jovial mango]

.close

[pearl pond]

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[pearl pond]

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[warm osprey]

I was just playing around with the dot product formula to understand it. Why does what I have written here not work? Like if you chose values v and u, the LHS is not = RHS. Why?

[sharp vigil]

v • u is a scalar, you can't take the dot product of a scalar and a vector

[warm osprey]

So is the dot product not associative?

[bold ledge]

no it's not even defined for three vectors

[warm osprey]

Fair enough

Thanks

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[terse cedar]

do u know polynomial division ??

[bleak plover]

yes you mean dx/dy?

bruh, nah but i know polynomial divison

[bleak plover]

yes yes

ik it

[terse cedar]

https://tenor.com/view/just-do-it-words-of-wisdom-encouragement-advice-life-lessons-gif-4413390

[bleak plover]

okey do that, then integrate the asnwer?

[terse cedar]

yeah it helps

[bleak plover]

ty, will start

@terse cedar

[terse cedar]

correct

can you solve the integral now ?

[bleak plover]

Would like some help

[terse cedar]

did u solve indef integral of 2x ?

[bleak plover]

dont know where to start, so no

[terse cedar]

$\int \frac{2x^3-4x^2-x-3}{x^2-2x-3}dx = \int 2x+\frac{5x-3}{x^2-2x-3}dx \ =\int 2x \ dx +\int \frac{5x-3}{x^2-2x-3}dx$

[jolly parrot]

Adam Chebil

[bleak plover]

okey

so do 2x by its self, and - both sides at the end?

[terse cedar]

wdym?

- both sides ?

[bleak plover]

2x = 2 intg x

and then solve (5x-3)/(x^2 -2x-3)

[terse cedar]

correct

[bleak plover]

like that? okey

[pearl pond]

@bleak plover Has your question been resolved?

[pearl pond]

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[hollow warren]

Hi can anyone help me with proving y^5 = x^2 + 2017 has no integer solutions

[hollow warren]

my first thought was a difference between 2 squares and I think it makes sense but then I also thought about using modular arithmetic but I’m not sure about this claim ‘y^5 is congruent to 0 mod 11’ for integer solutions I used it but don’t understand it

[pearl pond]

@hollow warren Has your question been resolved?

[hollow warren]

<@&286206848099549185>

[pearl pond]

@hollow warren Has your question been resolved?

[pearl pond]

@hollow warren Has your question been resolved?

[pearl pond]

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[sullen leaf]

I am being asked this problem: Let the points P = (-1,2,1,), Q = (2,1,3), and R = (1,6,-3). Find an equation for a plane containing P, Q, and R.

[pearl pond]

@sullen leaf Has your question been resolved?

[sullen leaf]

<@&286206848099549185>

[pearl pond]

@sullen leaf Has your question been resolved?

[pearl pond]

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[graceful wasp]

i know how to find the sum but i dont get how im finding a "partial sum"

[pearl pond]

@graceful wasp Has your question been resolved?

[graceful wasp]

<@&286206848099549185>

[graceful wasp]

still dont get how im finding a "partial sum"

whats a "partial sum"

[regal herald]

a sum going up to a finite term

could say its part of an infinite sum with the same expression, hence the 'partial'

[graceful wasp]

what are you talkin abt

[regal herald]

just dont overthink it

the sum could have an infinite number of terms

the one shown in the question could just be a part of that sum but only going up to a certain term

so its a partial sum

like if i have the sum from n=1 to infinity of n
then 1+2+3+4 is a partial sum
the whole sum is 1+2+3+4+5+6+7+... forever

[graceful wasp]

so what am i doing with this question

if i got 2,775 as my sum

[regal herald]

if thats what you got, then thats the answer

[graceful wasp]

well apparently thats wrong lol

[regal herald]

what number did you use for n

[graceful wasp]

37

[regal herald]

should have been 38

[graceful wasp]

why 38

[regal herald]

how did you get the 37?

[graceful wasp]

uhhh

lemme see

resending this for myself

149-1 / 4

= 37

[regal herald]

when you did that you neglected '1' as a term

you just counted how many steps after it to get to 149

[graceful wasp]

ohh okay

so 150 * 19

2,850?

[regal herald]

seems that way

[graceful wasp]

can u tell me what im supposed to do with the 5-4n

[regal herald]

5-4n is the general term of the sum basically
you can do this again, with just n=n

pretty sure anyway

(sorry didnt mean to ping)

youll have the sum in terms of n

[graceful wasp]

like

the 5-4n would be plugged in where?

for this

[regal herald]

a_n

[graceful wasp]

alright ill try it

[graceful wasp]

i cant find the number of terms the same way i did before with this can i?

[regal herald]

it would just be n

[graceful wasp]

okay

[regal herald]

when n=1, you get 5-4=1 when n=2 you get 5-8=-3
when n=n you just get 5-4n, its the nth term

youll basically end up with this

just unfactorised

[graceful wasp]

i understand that but like

is there even a sum?

ive been staring at this shit hoping it would click and it hasnt lol

[regal herald]

it is a sum
ig you could write it like this:

$\sum_{k=1}^{n}{5-4k}$

[jolly parrot]

AℤØ

[regal herald]

same thing

[graceful wasp]

you fine with me pinging you if i end up needing to open a new channel?

ill try putting that in in a bit

[regal herald]

sure, i cant guarantee ill be awake though, its 6am

[graceful wasp]

dayum

where are you from

[regal herald]

uk

[graceful wasp]

gotcha

[pearl pond]

@graceful wasp Has your question been resolved?

[pearl pond]

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[pearl pond]

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[midnight haven]

hey, i need help with some homework can anyone help me? I have been trying to solve it but i can't. This is the homework question: Determine all integers a, b, and c with 0 < a ≤ b ≤ c such that the measure of the
volume of a cuboid whose edge lengths have the dimensions a, b and c
is like the measure of the sum of all edge lengths of the cuboid. The edge lengths are
where measured in centimeters, the volume in cubic centimeters

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

<@&286206848099549185>

can anyone please help me?

i don't even know where to begin solving this question

.close

[pearl pond]

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• 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!

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[midnight haven]

Help

[blissful cloak]

[sick scaffold]

or post the question

[midnight haven]

Give me a sec

I don’t understand the steps for it

[sick scaffold]

y= -2

[midnight haven]

The question and equation. Is like

[sick scaffold]

if they just run at the same y

just say y = c

[midnight haven]

Y=MX+B

is what he’s looking for

It’s graphing

Questions

[sick scaffold]

its y= 0x -2

[midnight haven]

how do you get the 0x

[sick scaffold]

thats why i write y = -2

[midnight haven]

oh

[sick scaffold]

u can put any x but y still be -2

[midnight haven]

k

Thanks

Makes more sense figuring out the y

B

[sick scaffold]

graph gonna be parrarel to x-axis

[midnight haven]

Thanks!

How do I close

[sick scaffold]

. close

[midnight haven]

!close

[sick scaffold]

use doty

[midnight haven]

Oh

.close

[pearl pond]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[forest vine]

Can someone explain this question to me? I don't quite understand the notation and how to approach it.

[forest vine]

I'm supposed to check the functions f1 to f5 for injectivitiy and surjectivity

But I don't understand what it mean when it says R^2 -> R

or vice versa

[ember cloak]

Do you have a specific questions for any of these?

[forest vine]

oh hey what's up man

not really.

I understand injectivity and surjectivity

but Idk how I can check these funciton for them.

and what does it mean when it goes R^2 to R or vice versa?

[ember cloak]

The first one

[forest vine]

does that mean that the elements from R^2 are two elements in one and the elements in R are just single elements?

[ember cloak]

It is expressed very badly, but sort of. R^2 = R×R. So it basically means that both x and y are in R

Do you want to do the first one together?

[forest vine]

so f_1(1,2)=3?

[forest vine]

yes that'd be great

[ember cloak]

Exactly

[forest vine]

but that doesn't make sense to me

how come the inputs are 2 elements and the output only one? how do you even check something like this for injectivity and surjectivity

[ember cloak]

Let's start with surjectivity

Let z be an arbitary real number

Can you express z as f1(x,y) for x,y in R?

[forest vine]

it woulkd be f(z,z) = z+z?

[ember cloak]

z+z = 2z, which is not usually the same as z

But that's close to what I'd do

Can you change the second argument a bit so that it works?

[forest vine]

what do you mean by argument here?

[ember cloak]

Like the y in (x,y)

[forest vine]

y = 0

right?

[ember cloak]

Exactly!

So every z in R can be expressed as f(z,0). So we are surjective

[forest vine]

ohhh

since the obermenge is just R

ok I understand this

thanks

what about injectivity tho?

wait

its not injectiv

[ember cloak]

Yes !

[forest vine]

cuz f(0,1) = f(1,0)

right?

[ember cloak]

Well done

[forest vine]

Dopeee!!

thanks for the tip

so to prove this

but how would I mathmatically show the proof?

[ember cloak]

Are you in the "Vorkurs"?

[forest vine]

I guess for the injectivity I can just use the counterexample

[ember cloak]

Yes

[forest vine]

no I'm not

I missed it :(

[ember cloak]

Ah, we had this one back then

[forest vine]

these are just from our weekly homework sheets

[ember cloak]

It will get harder soon 🙂

[forest vine]

Damn, Ig i shoul've done vorkurs

[shrewd basin]

Ayo bloodborne enjoyer

[forest vine]

broo

[forest vine]

AYEEE

[shrewd basin]

[forest vine]

have you also done CoMa?

[ember cloak]

Computerorientierte Mathematik?

[forest vine]

ja

[ember cloak]

Das ist bei uns fürs dritte Semester angesetzt. Also sitze ich gerade drin

[forest vine]

achso

[ember cloak]

Aber mit Python kenne ich mich schon aus

[forest vine]

ja das ist zum großen vorteil

[ember cloak]

Also falls du Fragen hast, die nicht über ein gewisses Niveau hinausgehen...

[forest vine]

müsst ihr auch eure aufgaben in gruppen abgeben?

[ember cloak]

Dürfen wir nicht

[forest vine]

ohh

in meiner uni sind wir verplichtet die hausaufgaben in gruppen abzugeben

[ember cloak]

Aber er hat eh noch kein Übungsblatt hochgeladen, obwohl er es seit 2 Wochen angekündigt hat

[forest vine]

dann ist es bei euch viel schwieriger

[forest vine]

xDDD

[ember cloak]

Bei uns wird es angeraten. Nur der in CoMa und in Numerik erlauben es nicht

[forest vine]

achso

es ist erst 3 Wochen her und alle meine gruppenmitglieder im CoMa haben das modul schon schon verlassen

die machen nur Ana und Lina im ersten Semester

[ember cloak]

Das ist am Anfang normal. 50% der Leute sind bis zum Ende weg

[forest vine]

ja

höffentlich bin ich nicht teil dieses 50%

💀

[ember cloak]

Fairerweise ist CoMa bis jetzt auch unnötig

[forest vine]

Ja

die machen jetzt nur ana und lina

was hast du im erseten semester für module gemacht?

[ember cloak]

Nein, bist du nicht. Mathematik ist nicht unbedingt schwer. Man muss nur Bock drauf haben. Und ich habe schon den Eindruck, dass du es hast

[ember cloak]

Lin-A, Ana, Propädeutikum

[forest vine]

danke bro, das gibt mir höffnung

was ist Propädeutikum?

[ember cloak]

*Hoffnung

[forest vine]

👍

[ember cloak]

Wollte es nur ergänzen, weil du ja gesagt hast du lernst noch

[ember cloak]

Offiziell: Einführung in das Mathematische Schreiben. Inoffiziell: Lin-A Widerholung. War bei uns ziemlich unnötig. Da haben wir meistens Schafkopf oder so gespielt

[forest vine]

hahaha

also ähnlich wie vorkurs?

[ember cloak]

Der Vorkurs hat sogar etwas gebracht

[forest vine]

lol

ok bro, ich versuche den rest der frage mit deine hinweisen zu machen und wenn ich wieder fragen hätte, schreibe ich wieder

∀z ∈ R : f1( x , 0 ) = f1( 0 , y ) = z @ember cloak geht es, die surjektivität von f_1 so zu beschreiben?

[pearl pond]

@forest vine Has your question been resolved?

[ember cloak]

Für alle z in R gilt f(z,0) = z

So wie du es beschrieben hast ist das nicht so schön, weil nicht klar gesagt ist wie du x,y wählst

[pearl pond]

@forest vine Has your question been resolved?

[pearl pond]

Channel closed

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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.

[boreal crescent]

f an odd function and 1-periodic, how to show that f(0) = 0 and f(1/2) = 0

[boreal crescent]

like if it's 1 periodic how can f(1/2) = f(0) ?

[ember cloak]

f(1/2) = f(-1/2) by the fact that it is 1 periodic

[boreal crescent]

yes sure

[ember cloak]

What does odd mean for a function?

[boreal crescent]

that it rotate of an angle pi over O, like f(-x) = -f(x)

[ember cloak]

Ok, can you plug in 0 for x?

[boreal crescent]

yes sure f(-0) = -f(0) so

[ember cloak]

f(0) = -f(0)

[boreal crescent]

yes

okay so 2f(0) = 0 so f(0) = 0

ohh okay

and we do the same for f(1/2) and f(-1/2) so it's equal to 0

[glass meadow]

Only true if f is continuous at 0 and 1/2 though

[boreal crescent]

ok ty 🙂

[ember cloak]

I'm sorry that I have to ask but, where is continuity needed (assuming the function exists for x=0)?

[glass meadow]

Defined, same thing in this case

[ember cloak]

So defined is enough? Continuity is not required?

[glass meadow]

If it's defined at 0 and odd then it's continuous at 0

[ember cloak]

Well done ! Can you make like the same argument for f(1/2)?

[boreal crescent]

ye already done that like we have 2f(1/2) = 0 so f(1/2) = 0

[glass meadow]

Never mind actually, could be discontinuous

Just needs to be defined

[ember cloak]

How about f(x) = 1/x for every x, but 0 and f(0)=0? It's odd but not continous

[glass meadow]

Yes

[ember cloak]

Nice, gg 🙂

[glass meadow]

I had in mind that if it's continuous on one side and odd then it's continuous on both sides

[boreal crescent]

it can be discontinuous so and f(1/2) doesnt exists, sorry if it's a dumb question but like can it be 1-periodic and defined on 1,2,3,4...

and also -1,-2,-3...

[glass meadow]

Yeah it can

[boreal crescent]

so in this case f(1/2) doesn't exists ?

[ember cloak]

So there is a tiny error in the question. Well spottet Nel 🙂

[glass meadow]

Yes

f being odd and 1-periodic is not a sufficient condition for the value at any x to exist

That's the gist of it

[boreal crescent]

if it's odd f(0) = 0 exists but not all the others point