help-23

[earnest rain]

After

Yes but I need integral pratice

Need volume for it anyway

[final loom]

I don't understand your variable substitutions. Did you use x = rcos t, y = rsin t, z = z?

The limits then, for r would be [-R/h(h-z), R/h(h-z)], t would be [0, 2π], z would be [0, h/2]

[earnest rain]

No, dxdy jacobi only p

There is no x or y inside the integrals

And t i put -pi/2 to pi/2 because it says x>0

Why do you say r goes from negative? R is the radius so starts from 0

<@&286206848099549185>

[final loom]

Yeah you're right. r > 0 ✓

[earnest rain]

So why result wrong?

[final loom]

Let me check

The limits (-R/2) to (R/2) is for what?

Shouldn't it be (-π/2) to π/2 if you substituted x = cos t, y = sin t

You should be having $\ \int_0^{h/2} \int_{-\pi /2}^{\pi /2} \int_{0}^{R(h-z)/h} \rho , \cdot \dd \rho \dd \theta \dd z = \frac{\pi R^2}{2h^2} \int_0^{h/2} (h - z)^2 \dd z$

@earnest rain

[flat frigate]

Arya

[final loom]

This'll give you 7πR²h/48

[earnest rain]

So what i do wrong?

[final loom]

??

Idek how you managed to send a "z" term to the denominator in line 2

[earnest rain]

Its a 2

[earnest rain]

Its h/2

[final loom]

Oh, and what's the limits -R/2 to R/2 for?

[earnest rain]

Its h/2

[final loom]

Oh

[earnest rain]

The z limits

Wait no

[final loom]

But z limit is from 0 to h/2

[earnest rain]

Its -pi/2

Pi/2

[final loom]

Hmm

[earnest rain]

Its as u wrote

[final loom]

So you're basically getting this in line 2 ig

[earnest rain]

Ye

[hardy monolith]

,rccw

[final loom]

Denom should be 4

@earnest rain

[earnest rain]

Oh shit

Thx

.close

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[mighty sandal]

hi guys!! me again. I really have confusion on how to make a venn diagram for these questions so can someone help me along with an explanation:

[mighty sandal]

which might help me in future examples as well...

[frail pumice]

You have two sets:
C - Set of sodas that have high amounts of caffeine
S - Set of sodas that have high amounts of sugar

So, in the Venn diagram, you have to draw 2 circles

[mighty sandal]

yes and their intersection is 6

[frail pumice]

Exactly

[mighty sandal]

so, do I do:

only Caffeine= 8-6 = 2
only Sugar = 12-6= 6
Intersection = 6

6 + 6 + 2 = 14 -> Here is my problem

[frail pumice]

I think the problem here is that you think that all the 24 sodas have one of the properties, that is, the union of the sets is the total of the sodas

But instead, there are 24 - 14 = 10 sodas that doesn't have either high amount of caffeine or sugar

[mighty sandal]

oh alright, so- can be 10 sodas could be placed in the universal set? like within the box, but outside the venn diagram

[frail pumice]

Yes :P

[mighty sandal]

ohhh alright thank you so much

[frail pumice]

you're welcome

[mighty sandal]

now, the next follow up is calculate the probablity that the soda picked at random is high in sugar and but not in caffiene

so in terms of the venn diagram, would I need to calculate only Sugar

or will the intersection count?

[frail pumice]

That's right!

[frail pumice]

Nope, because the intersection has sugar and caffeine

And the question wants no caffeine

[deep rapids]

jee type question

[mighty sandal]

ohhh so it's a part of caffiene which the question doesn't want

[deep rapids]

@mighty sandal u in 11th?

[mighty sandal]

I'm in 10th lol

[deep rapids]

oh cool

[mighty sandal]

dream is IIT but...

I'm really messing up in probability, set and venn diagrams

also tysm for your help @frail pumice !!

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[inland temple]

Hello, I understand all the working here, but I'm struggling to see why they decided to do the whole thing with u = sin(theta) instead of just using the double angle formula for costheta sintheta?

[inland temple]

Is it just a matter of preference? How do I know when I should do the d(sin(theta)) thing?

[vague slate]

It's a choice
You can go either way

If you go with double angle formula, you still gotta do a sub

[inland temple]

Ah I see, thanks
So presumably once you get good at spotting it it's a useful tool

[vague slate]

Yeah it's the best tool

[inland temple]

Oh? Do you not just replace it with 0.5sin2theta and then you can integrate that normall?

*normally

[vague slate]

You get 0.5 sin 2theta

So then you gotta sub the 2theta with some thing
I mean yeah the sub will not be a long one but still

[inland temple]

Alright cool thanks

.close

:)

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[fickle trail]

I don't understand the logic of f(x+h) how it suddenly becomes f(x+h)^2

[fickle monolith]

it's (x+h)^2

[hardy monolith]

they filled in f(x)=x^2

f(x+h)=(x+h)^2

[fickle monolith]

f(x)=x²
so f(x+h)= (x+h)²

just replace x with x +h

[lean otter]

If f (x) = x^2
Then f(x+h) is (x+h)^2

[fickle trail]

f(x+h) - f(x) / h
we start with this

and i get that f(x) = x^2

so f(x+h) - x^2 / h is also correct

but if f(x) = x^2 why does f(x+h) = (x+h)^2

we are just replacing x with x+h?

thats all we are doing?

putting a different input into the function

[fickle monolith]

basically yes

you can for example think of x+h as a separate variable

to get more clear

[fickle trail]

what if the function was f(x) = x^2 - h

f(x+h) would be x^2?

[fickle monolith]

no

[fickle trail]

(x+h)^2 - h?

[fickle monolith]

yeah

[fickle trail]

hmmm

[fickle monolith]

to get more clear, you can like assume u=x+h and substitute it into f(x+h)

to see what's going on

[fickle trail]

what about this?
f(h) = h^2 - h
(x+h)^2 - h

[fickle monolith]

what?

[fickle trail]

just different variable names

not in this same context

imagine it as a new question

[fickle monolith]

i mean what do you want to mean in the last line

[fickle monolith]

in this

[fickle trail]

i'm not sure lol

just trying to make sense of replacing inputs

the bracket for a function is like a jail

different from regular bracket

[fickle monolith]

don't worry too much it's just change of input variables

[fickle trail]

f(x) = x^2
f(x+h) = (x+h)^2

hmmmm

interesting

i suppose that makes sense

wow, that's a crapton of work that derivative power rule does so easily for us

to find instantaneous rate of change for f(x) = x^2

[plucky elk]

hopefully that means you'll appreciate using derivative rules and how much simpler they are

[fickle trail]

yes

they are amazing

do we still use limits in higher calc? or no need

[plucky elk]

yes

limits are like the algebra of calculus

[fickle trail]

ah OK, i was thinking limits would be used more for understanding correlation between average of secant to slope of tangeant

i just learned difference of squares method by drawing out a 4 piece square today

I would normally use FOIL on that

FOIL prob faster, this looks like way more writing / drawing
but kinda cool to see it done this way

[plucky elk]

that's equivalent to foil

[fickle trail]

funny I never saw it explained like this until today

been using it for years now

almost feels like matrix algebra, maybe there is some correlation here

btw, when someone says "Matrix Algebra" and "Linear Algebra", do they both mean exactly the same thing? if so, why the two names?

or would there be subtle differences between both definitions

kinda like perpendicular and orthogonal for example, 99% similar but tiny difference

[plucky elk]

this has nothing to do with matrices

[fickle trail]

oh

[plucky elk]

tables look the same as matrices but have nothing to do with each other and you should just infer from context what it is

[fickle trail]

Linear algebra is a broad field that studies vector spaces, linear transformations, and abstract mathematical structures, while matrix algebra is a subset of linear algebra that specifically focuses on matrices and their operations, such as addition, multiplication, inversion, and decomposition. Although matrices are a fundamental tool in linear algebra, the latter extends beyond matrices to more general concepts like vector spaces and eigenvalues. The distinction is analogous to the difference between a general theory and its practical computational tools.

[plucky elk]

or ask

[fickle trail]

now I gotta look up with "analogous" means

`Analogous means that two things are similar in certain ways, allowing for a comparison between them. They share certain features, functions, or relationships, even though they might be different in other aspects or in their specific nature.

Analogous in Everyday Use: You might say, “A library is analogous to the internet in that both provide access to information.”`

ty @plucky elk - always here to help 🙂

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[last falcon]

I can`t understand how to get rid of the root here

[last falcon]

This is what I started with :

,, \int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\sqrt{\dfrac{64,{e}^{\frac{8,x}{3}}}{9}+2,{e}^{\frac{4,x}{3}}}}{;\mathrm{d}x}

[flat frigate]

Ulerand

[last falcon]

,, \dfrac{3}{4}\int{\sqrt{\dfrac{64{e}^{2u}}{9}+2{e}^{u}}}{\mathrm{d}u}

[flat frigate]

Ulerand

[last falcon]

Now I'm here, also u = 4x/3

[hardy monolith]

notice that 64/9=(8/3)^2

[last falcon]

Yes, but I can't really take it out of the root

because of +2e^u

[loud wren]

question, is this even alowed?

oh never mind

i saw that

[last falcon]

That?

[loud wren]

factor out an e^u

or wait

idt this is a perfect square case

so factor an e^u

[last falcon]

,, \dfrac{3}{4}\int{\sqrt{{e}^{u}(\dfrac{64{e}^{u}}{9}+2})}{\mathrm{d}u}

[flat frigate]

Ulerand

[last falcon]

Like that?

[loud wren]

yes

now it looks uglier

im stuck

[cedar owl]

where did the bounds go

[loud wren]

💀

[last falcon]

I just didn't write them, anyway first I need to integrate it

[last falcon]

Maybe I can split on the two roots it now?

,, \dfrac{3}{4}\int{\sqrt{{e}^{u}}*\sqrt{(\dfrac{64{e}^{u}}{9}+2})}{\mathrm{d}u}

[flat frigate]

Ulerand

[last falcon]

If I could only get rid of the root at all..

[empty gyro]

Maybe try $v=\sqrt{e^u}$

[flat frigate]

SWR

[azure delta]

If you sub v=e^u, you can follow it up with this

https://www.wolframalpha.com/input?i=integral+of+root(5%2B2%2Fx)

Swr's suggestion might be easier though

[last falcon]

Hm, I'll try this first then

[last falcon]

Also in this case, $du = 2v/v^2 dv$ ?

[flat frigate]

Ulerand

[last falcon]

,, \dfrac{3}{4}\int{v\sqrt{(\dfrac{64{v}^{2}}{9}+2})}\dfrac{2v}{{v}^{2}} {\mathrm{d}v}

[flat frigate]

Ulerand

[safe radish]

@last falcon Has your question been resolved?

[autumn kite]

have u tried taking e^4x/3 = v?

@last falcon

[last falcon]

So what about dx? Is it gonna be dx = 4/3 * 2v/v^2 ?

[autumn kite]

wont dv= 4/3 ( e^4x/3 ) dx

like factor out e^4x/3 in the original

[last falcon]

Hm, maybe you are right in this one, yeah

[autumn kite]

yea its coming like that

the bounds look gay asf ngl

[last falcon]

Okay I'll try this substitution in this one

[autumn kite]

u getting it?

[last falcon]

Not really, there still a root that stands in the way

,, \dfrac{3}{4}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\sqrt{\dfrac{64,{v}}{9}+2,{v}^}}{;\mathrm{d}x}

[flat frigate]

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

[last falcon]

I can factor out v the same as I did earlier

But it doesn't really gets me anywhere

[autumn kite]

u get [64/9 v + 2] dv

right?

inside the root

[last falcon]

dv right

[autumn kite]

ye

then again consider that term as t

u feel me?

[last falcon]

so like [64/9 v + 2] dv = t?

[autumn kite]

ye

not the dv part

[64/9 v + 2]= t

[last falcon]

ah

,, \dfrac{3}{4}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\sqrt{t}{;\mathrm{d}t}

[flat frigate]

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

[last falcon]

,, \dfrac{3}{4}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\sqrt{t}}{;\mathrm{d}v}

[flat frigate]

Ulerand

[last falcon]

and $t = 64/9 v + 2$

[flat frigate]

Ulerand

[autumn kite]

the limits change

its not from -pi/2 to pi/2

[last falcon]

Why so?

[autumn kite]

do uk how to convert the limits when u use substitution

[last falcon]

Aren't when u change the symbol from + to - ?

[autumn kite]

no it changes

limits change

but in this case i wudnt wanna change the limits

i wud solve this question indefinetly and then replace the value of t and then replace the value of v

cuz like when we change the limits, it gets uglier

[last falcon]

,, \dfrac{3}{4}\int\limits_{\frac{\pi}{2}}^{-\frac{\pi}{2}}{\sqrt{t}}{\mathrm{d 64/9v +2}

,, \dfrac{3}{4}\int\limits_{\frac{\pi}{2}}^{-\frac{\pi}{2}}{\sqrt{t}}{;\mathrm{d} 64/9 v + 2}

[flat frigate]

Ulerand

[autumn kite]

https://www.khanacademy.org/math/calculus-1/cs1-integrals/cs1-integrating-with-u-substitution/a/u-substitution-definite-integrals#:~:text=There's actually another thing you,you calculate the definite integral.&text=Remember%3A When using ‍ -substitution,for the limits of integration.
check this out and tell me if u understand

[autumn kite]

what u doin bro?

[last falcon]

Well this is explanation of the substitution ?

[autumn kite]

ye it explains why limits change

[last falcon]

Okay, I guess I understand why it changes, tho I didn't really know about this

But as u said, we can just do the indefinetly and then replace it

[autumn kite]

im having a lil doubt in that cuz when u integrate indefinetly u get the constant C

[last falcon]

But the question is how to solve it even indefinetly

[autumn kite]

u mean definitely?

[last falcon]

Well we will replace the value t to v then to x and with old limits we can solve this

[autumn kite]

no the constant C wud surely make a difference

we will get constants 2 times

since we used substitution twice

[last falcon]

Can't I ignore constant ? or what I shall do then

[autumn kite]

see either we can change the limit twice

or we wud need help from someone else regarding that indefininte integration constant stuff

cuz im not sure waht to o

[last falcon]

I never done const in defined integrals, even with the substitution, i just straight igored it and returned value (for example t) to x

[autumn kite]

even i nvr faced the problem, since whenever i do substition i change the limits and it gets more simplified

[last falcon]

I found this method of getting rid of root, but I don't really understand how it coud be. Like were the "2" before the e^u goes with the denominator and what +(0) is

[safe radish]

@last falcon Has your question been resolved?

[plucky elk]

without more context that's wrong. they just disappeared a e^u term

[last falcon]

Hmm.. I thought so, then I really have no clue how to deal with this

[last falcon]

aight, I give up on this one. Maybe another time

.close

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[night bay]

how to prove by mathematical induction that a polygon of n sides has n(n-3)/2 diagonals?

[safe radish]

@night bay Has your question been resolved?

[flint fiber]

Its for n => 3 right?

First you have to show that n = 3 is right

[night bay]

yep

[night bay]

what's the next step

[flint fiber]

Ok, we need to find first a pattern how those diagonals really work

This would serve as our basis

For example,
3 sided = 0
4 sided = 2
5 sided = 5
6 sided = 9
7 sided = 14

Can you see a pattern?

[night bay]

yeah kinda

what would the next step be?

[flint fiber]

What do you think is the pattern

[night bay]

the only pattern I see is the diagonals have a difference of 2,3,4,5

[flint fiber]

Thats partially correct but specifically its diagonal of n+1 gon = diagonal of n gon + (n-2)

So we will use this equation to continue our mathematical induction

[flint fiber]

So suppose k exist such that k(k-3)/2 is the diagonal of a k sided polygon

So we need to show that its true for k+1

So the equation would become (k+1)(k-2)/2

[night bay]

Can you explain this more? I don't get this part

[safe radish]

@night bay Has your question been resolved?

[safe radish]

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[rare vale]

Can someone pls tell me what i got wrong here?

[rare vale]

there shouldnt be a +3la at the end

it showed the answer earlier

[rare vale]

how is that wrong?

Wait i see it

i mixed up the terms lmao

thanks for pointing it out!

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[lean otter]

I’m stuck

I’ve never had an equation with a division sign like this

I’m trash

[rugged arch]

you can write it as 2L/6

,, L + 2 = \frac{2L}{6}

[flat frigate]

mmmm7

[rugged arch]

more familiar now?

[lean otter]

I kinda did that but I still suck

Cause the answer is 3 but I get 4

[rugged arch]

show work maybe?

2/6 simplifies to 1/3 so i’m guessing you got till $L + 2 = \frac L3$

[flat frigate]

mmmm7

[lean otter]

Is that incorrect

[rugged arch]

no

6 multiplies to (L + 2)

u only multiplied it to 2

[lean otter]

L becomes 6L?

[rugged arch]

yes

[lean otter]

How do you avoid getting a negative answer? It keeps coming out negative

[rugged arch]

u don’t

because the correct answer is a negative one

[lean otter]

That gives me an answer of 1. My thing told me L equals 3

The answer sheet

The question don’t even make sense though

Am I high or does that not make sense

[lean otter]

How can L be 6x older than Benjamin and L is 3 years old.

[rugged arch]

,w L = 6B and L + 2 = 2(B + 2)

[flat frigate]

Results provided by WolframAlpha

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Requested by mmmm70177

[rugged arch]

this is not what u wrote though

[rugged arch]

when B was 6 months old L was 3 years old

[rugged arch]

,w L + 2 = 2L/6 lol

[flat frigate]

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[rugged arch]

just write it out as follows:

Currently, denote lauren’s age as L and benjamin’s age as B

in 2 years surely their age would’ve increased by 2

so in two years, their ages are L + 2 and B + 2 respectively

now use what your question says to form equations:

currently Lauren is six times as old as benjamin:

L = 6B

in two years, Lauren is twice as old as Benjamin:

(L+2) = 2(B + 2)

then solve those two equations simultaneously

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

Okay I see. The answer being negative didn’t make sense to me

It still kinda doesn’t not gonna lie

.close

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[trail otter]

is coposition of two monotonic sequences monotonic?

[digital leaf]

Yes,
One can look at the different cases easily
If u is nondecreasing and v is nonincreasing
x ≤ y
u(x) ≤ u(y)
v(u(x)) ≥ v(u(y))

note that it works for any mapping, not only sequences

https://en.m.wikipedia.org/wiki/Monotonic_function#:~:text=The composite of two monotone mappings is also monotone.

The composite of two monotone mappings is also monotone.

[desert pasture]

What about $a_x=e^{x^3}$

that isn't monotonic

[flat frigate]

What a wonderful world!

[trail otter]

wait could you please elaborate
why when i plug nondecreasing mapping into a nonincreasing it preserves its momotonicity?

[desert pasture]

,w graph e^{x^3}

[flat frigate]

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[desert pasture]

,w is e^{x^3} monotonic

[flat frigate]

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[desert pasture]

huh

sorry

[trail otter]

this seems to work

[desert pasture]

yeah

[digital leaf]

Consider u and v be monotonic

Either u is constant or nondecreasing or nonincreasing

Same for v

So you have exhaustively 9 cases to study

Study when u is constant, v constant
Then u constant, v nonincreasing etc

I only wrote one of the nine cases, for u nondecreasing and v nonincreasing, but you easily see that the reasoning is similar up to changing the symbol ≥≤=

you first consider any value (or integer for sequences) x ≤ y
You then apply u, then v over this inequality
v(u( • )) is your composite of monotonic
if then v(u( x )) = v(u( y )), the composite is constant hence monotonic
If v(u( x )) ≤ v(u( y )), your composite is nondecreasing hence monotonic
and so on

[trail otter]

ah alright i think i see it now

thanks

[safe radish]

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[lean otter]

It seems wrong…

[fathom adder]

hmm

its 5 balls out of how many ?

[lean otter]

nvm actually, I figured it out 😅

.close

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[merry herald]

PCM problem solving, how do i start solving such problems

[merry herald]

the graph:

[safe radish]

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@merry herald Has your question been resolved?

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[bleak steppe]

can you help me ?

[safe radish]

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[worldly umbra]

J ai du mal avec la 4.b

[worldly umbra]

Car comment je peut en deduire ?

[safe radish]

@worldly umbra Has your question been resolved?

[worldly umbra]

🙏🏻

[rustic goblet]

it's hard for people to help you if you don't tell them which part you're struggling with

it's also a little hard because not everyone here speaks French

[worldly umbra]

I said

It's 4.b

[rustic goblet]

ah, my bad

[worldly umbra]

Yes you're right

But how can i translate all

[rustic goblet]

okay well, it sounds like you need to verify that $V_n = M^nV_0$?

[flat frigate]

fallenstars

[worldly umbra]

Yess

[rustic goblet]

okay, part 4a) states that $V_{n + 1} = MV_n$

[flat frigate]

fallenstars

[rustic goblet]

I'll write a pattern out, and hopefully you can see the reason why 4b is true

[worldly umbra]

Yes it's done

[worldly umbra]

I'll try

[rustic goblet]

\begin{align*}
V_1 & = MV_0 \ V_2 & = MV_1 = M(MV_0) = M^2V_0 \ V_3 & = MV_2 = M(M^2V_0) = M^3V_0
\end{align*}

[flat frigate]

fallenstars

[rustic goblet]

see a pattern?

[worldly umbra]

Yesss

But how do you know MV1=M(MV0)

[worldly umbra]

Ahahaha yesss

A yesss

[rustic goblet]

V1 = MV0 comes from part 4a

[worldly umbra]

I didn't see this 😭

Thxxxx

[rustic goblet]

everything else goes by induction

[rustic goblet]

no problem

[worldly umbra]

Can i ask you a last thing

🙏🏻

[rustic goblet]

sure, but I can't promise to be able to answer

[worldly umbra]

You can you are smarttt

I don't understand the 5.b

The start

[rustic goblet]

what confuses you?

the problem is claiming that V_n = this matrix for all n in N

[worldly umbra]

express Un as a function of n

It's not already done????

[rustic goblet]

I'm not exactly sure what it means to express U_n as a function of n

[worldly umbra]

😭

I use google translate

[rustic goblet]

nono, I understand what you're saying

I just don't understand the question

[worldly umbra]

Whyy

[rustic goblet]

maybe they want you to write U_n as an explicit vector

like they did with V_n

I'm not entirely sure

[worldly umbra]

But it's already done?

[rustic goblet]

how is it done?

[worldly umbra]

😭

I'm confuseddddd

[rustic goblet]

why do you think it's already done?

[worldly umbra]

@fathom adder can 🥲

Understand

[fathom adder]

haha

[worldly umbra]

Becauseeeee

[fathom adder]

let me read

[worldly umbra]

😭

C est moi encoree

Désolée du dérangement

[rustic goblet]

Yakubros is here to say the day

[fathom adder]

haha

[worldly umbra]

😭

[fathom adder]

matrices sequences

[worldly umbra]

Ouii

Mais enft j ai pas compris

[fathom adder]

thank you fallenstars for what you've done till here

[fathom adder]

pour quelle question

[worldly umbra]

La 5.b on nous dit d exprimer Un en fonction de n mais c est pas deja fait??

C est ça que je comprends pas

[rustic goblet]

this was an exercise in my very limited ability to read French

[worldly umbra]

sorry🙏🏻

[rustic goblet]

I haven't read French properly in quite a few years now

[rustic goblet]

issok

[worldly umbra]

So It's a good exercise

[fathom adder]

où est-ce que tu l'aurais déjà fait ?

[worldly umbra]

Bah Un(an, bn)

On la un peu exprimer en fonction de n?

[fathom adder]

Vn = Un - U

Un = Vn + U

tu as Vn

et tu as U

[worldly umbra]

Oui?

Enft c est je crois je comprends pas la question

[fathom adder]

pourquoi donc ?

exprimez Un en fonction de n

[worldly umbra]

Bah jsp ce que je doit faire

[fathom adder]

avant tu n'avais pas d'expression de Vn en fonction de n

sauf qu'il te la donne gentillement

[worldly umbra]

Un = Vn + U donc je doit trouver ca en fonction de n?

[fathom adder]

exactemement

surement que U est de la même forme que Vn

pour pouvoir faire la somme

[worldly umbra]

Petite question bete mais comment sait on que un c est ça

Genre juste en changeant me signe de u

Mais justement la question après ne suis pas

On me dit ma limite de an il y a pas de rapport

[worldly umbra]

A si

Enft

[fathom adder]

on te dit quoi dans la question 4 ?

[fathom adder]

hereusement

[worldly umbra]

Oui mais on nous donne la suite Vn=Un-U

[fathom adder]

eh bien ajoute U des deux côtés

[worldly umbra]

A oui

[fathom adder]

Un = (an, bn)

[worldly umbra]

Mais ça doit etre penilble d exprimer Un en fonction de n

[fathom adder]

c'est quoi la valeur de U

?

[worldly umbra]

Att je retrouve la feuille

U(380,270)

C est ce que j ai trouvé

[fathom adder]

ok

et bien ajoute les deux matrices comme convenu

[worldly umbra]

Comment ça ?

[fathom adder]

Vn + U

[worldly umbra]

Att pour la 5.b?

[fathom adder]

oui

[worldly umbra]

Avec Un=Vn+U?

[fathom adder]

oui pour trouver Un en fonction de n

faut simplifier à droite

[worldly umbra]

Donc on remplace U par la matrice

Je remplace Vn par M^nVo?

[fathom adder]

non

par la valeur qu'il te donne

[worldly umbra]

Ahaha juste ça?

[fathom adder]

oui

[worldly umbra]

Aahahah

[fathom adder]

et après pour la suite c'est assez tout droit

[worldly umbra]

Donc j ai -100/30,8^n-140/30,5^n+380

Pour la première ligne dcp

[fathom adder]

en effet

et donc limite de an ?

[worldly umbra]

380

Dcp

[fathom adder]

parfait

je te laisse marquer la phrase de conclusion

[worldly umbra]

Et la question c ???

[fathom adder]

interprétation de limite

[worldly umbra]

On a pas deja conclu?

[worldly umbra]

Oui

Fin la c et la b c est la meme chose

[fathom adder]

la réponse est une phrase en français où le seul chiffre est la limite de an

[worldly umbra]

Oui

Juste ça pour la c?

Ça a l air trop simple

[fathom adder]

oui tu explique ce que veux dire le resultat de la 4b dans le contexte de l'exercice

[worldly umbra]

Oh ok

[fathom adder]

pourquoi cela devrait être compliqué ?

[worldly umbra]

C est simple alors

[worldly umbra]

J avoue dis comme ça

Merci vrmt 🙏🏻

Tu me sauves toujours la vie

[fathom adder]

haha on est pas nombreux à parler français en même temps

pas de soucis

[worldly umbra]

Oui c est dommage

D ailleur ça te prend pas trop de temps pour aider les autres?

[fathom adder]

quand je sais ce qu'il faut faire non

puis c'est comme je veux si je veux pas le faire je le fais pas

là avec gg trad pour un exo comme ça tu aurais galéré

[worldly umbra]

Oe j avoue

Mais c est chaud de traduire en anglais

[fathom adder]

surtout qu'on a parfois pas les mêmes visions sur les maths

y'a des choses qu'ils font qu'on ne fait pas du tout

et vice versa

[worldly umbra]

Oe c'est ça qui est nul

C'est dommage

[fathom adder]

des fois c'est pas plus mal

comme i = sqrt(-1)

[worldly umbra]

Oui?

Aaa oui

Nous c est ✓-1

[fathom adder]

nan nous c'est i² = -1

le probleme de racine de -1 c'est le symbole racine

[worldly umbra]

Oui?

Pas eux?

[fathom adder]

eux c'est ✓-1 justement

ce qui n'est pas à notre goût

[worldly umbra]

A oui je vois

Bon bah merci pour ton aide vrmt

[fathom adder]

ça marche, si tu as tout compris c'est bon pour moi

[worldly umbra]

Oui merci

Passe une bonne soirée

[fathom adder]

de même

[worldly umbra]

.close

[safe radish]

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• Show your work, and if possible, explain where you are stuck.
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[tender horizon]

Can someone please help me with this question

[tender horizon]

I thought of writing it out as steps

So for the 1st they could have gotten SSS (Scissor, Scissor, Scissor)

2nd they could have gotten RRR (rock, rock, rock)

3rd they could have gotten PPP (paper, paper, paper)

and 4th they could have gotten SRP (scissor, rock, paper)

I also know that there is a 1/3 chance of either choosing scissor, rock, or paper

but idk how to calculate the chances of losing/winning with this info

[marsh walrus]

So, you know the like

you know the total number of possibilities right

lets just think about 1 round

[marsh walrus]

how many possible outcomes, TOTAL, in a single round?

[tender horizon]

6 possible outcomes?

cause 3!=6

[marsh walrus]

this seems waayyy too low

it seems like, take some person is first

they can choose 1 of the 3

then for each of those 3 possibilities, the second person can choose 1 of 3

and the same for the third

so for each of the 3, another 3, and then another 3

[tender horizon]

oh i thought you meant something else mb

[tender horizon]

so 27 cause 3^3

[marsh walrus]

yea

[tender horizon]

and 27*4=108 total possible outcomes

cause there are 4 rounds

[marsh walrus]

before you go all the way out

do you know the probability that no one wins in the 1-round case

its not so bad to think out

[tender horizon]

Well there is a 1/9 chance two players pick the same thing

and there is a 1/3 chance the third player wins

so 1/27?

[marsh walrus]

hope the color comes through

not quite

does the layout of the diagram make sense?

this is a single round

i have a feeling im overloading the diagram

here's circled in blue all the outcomes wwhere no one wins

[tender horizon]

oh so 4/9?

[marsh walrus]

6/27

[tender horizon]

You also have to include where all choices are different

[marsh walrus]

i have

oh, shit

i missed some

[tender horizon]

lol no worries

[marsh walrus]

so

9/27

how do you feel about that conclusion?

[tender horizon]

cool

So there is a 9/27 chance that they lose a single match

[marsh walrus]

no

there is a 9/27 chance that no one wins

[tender horizon]

so 1/3 chance

[marsh walrus]

yea

[tender horizon]

it said it was wrong

[marsh walrus]

its not correct as the final answer

its correct for a single round

we need to use more thinking to get to the final answer

right, and were looking at a single match

you dont have to delete anything lol its all good

i should have maybe made it more clear the process i was gonna go through

than jumping right into numbers

It's important to agree that each round is independent, that's the last big thing

I mean, if someone wins some round, that doesnt change the outcome of the next round

it has no influence

[tender horizon]

yeah so we multiply all of the events up to get the ans

[marsh walrus]

if you win a round of rock paper scissors that doesnt make you more or less likely to win the next round

agreed?

you got it

[tender horizon]

yeah i agree

[marsh walrus]

,w (1/3)^4

[flat frigate]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by janniku

[tender horizon]

thanks for breaking it down

[marsh walrus]

np

[tender horizon]

.close

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[lapis breach]

can someone please help me understand rieman sums and sigma notation

[lapis breach]

ill type out a specific question give me 1 min

im having trouble expressing it as sigma notation

right now im working on the midpoint

this is my understanding so far

I feel like i can do left ones

And probably right ones

but i keep messing up translating the difference between them

[rigid inlet]

The only major difference between them is exactly what the x_i are

[lapis breach]

yea ik that but how do i write that difference

in the sigma notation

[rigid inlet]

You divide your interval into n(=4) subintervals. The midpoint sum involves evaluating the function at the midpoint of each subinterval

Do you need to?

[lapis breach]

i need to write the sigma notation for the right left and midpoint

then plug it into a calculator and have the answers

professor didnt teach anything so ive been here for for 5 hours so far tryiung to teach myself

im at the point where i understand the concepts just dont know how to write it

i think

[rigid inlet]

I don't love what you wrote for M but the fact that decimals are included means that there isn't really a clean way to write it

[lapis breach]

well im not sure if thats right

i dont think it is

cause idk how else to represent that its a different x_n value

its x_0 = 1.5

x_1 =2.5

etc

right

[rigid inlet]

Indeed

[lapis breach]

so how do i write that as a sigma

[rigid inlet]

$\sum_{i=0}^3 f(x_i)$ with your list of $x_i$s is what I would do

[flat frigate]

Steakanator

[lapis breach]

what does that mean

[rigid inlet]

What does what mean

[lapis breach]

why three and 0

i thought riemans always had to start with 1]

[rigid inlet]

You're the one who wrote x_0 = 1.5

Your index can start with whatever you want

[lapis breach]

yea but when i plug it into calculator it doesnt work

says its undefined

unless i use 1 and 4

or 2 and 5 etc

[rigid inlet]

Please show me exactly what you're putting into this calculator

[lapis breach]

i think i need to just start over again ive made no progress in my understanding

[rigid inlet]

I can't stop you but I'm not sure if such drastic action is needed

[rigid inlet]

Once more

[lapis breach]

so wait if we go back to this one

would you be able to help me understand

how to go backwards from the sigma notation i have

back to the other parts

so i knownwhat part comes from what

[rigid inlet]

Keep this list at the front of your memory

[lapis breach]

okay

[rigid inlet]

$\sum_{x=1}^4 f(x)$ as you wrote is equivalent to $\sum_{i=0}^3 f(x_i)$

[flat frigate]

Steakanator

[rigid inlet]

In the first we're moving through each number (which we can do since they're all integers and sigma notation is really meant for exclusively integers). In the second we're moving through each index

[lapis breach]

okay

so one is using the index names and one is using the actual values

[rigid inlet]

The second is f(x_0) + f(x_1) + etc. = f(1) + f(2) + etc. Because x_0 = 1, x_1 = 2, etc.

[lapis breach]

okay

so can i do it either way

writiing it via the index or writing it via the values

[rigid inlet]

Yes and no

[lapis breach]

how do I know when to use which

[rigid inlet]

You can do either way in this case because your values are integers separated by 1

The index form is universal

[lapis breach]

so ill always be able to use the index form?

in every case?

[rigid inlet]

In every case

[lapis breach]

ok so would you be able to teach me how to go from into the index form

so then i have a method i can apply to any problem

[rigid inlet]

Just read off this

[lapis breach]

okay 1 sec

okay so for this i see where the zero is coming from

why is it three should it not be 4

cause its looping through 5 different values

from index 0 to 4

[rigid inlet]

You're only covering left endpoints in that example

x_4 is the right endpoint of the 4th subinterval

[lapis breach]

so there is only 4 parts

that hit

on the left side

which indexes are those?

indexes 1 through 4?

indexes 0-3

[rigid inlet]

Those are the right

[lapis breach]

are the left

[rigid inlet]

0-3 are the left

[lapis breach]

Okay.

so x_4 = 5 isnt needed for this scenario

[rigid inlet]

Not for the left sum

[lapis breach]

so this is the left sum

because deltax is 1

cause 5-1/4 = 1

the right sum

[rigid inlet]

Having i as your sum index but x in the summand is bad form

[lapis breach]

what should i be

k?

[rigid inlet]

You need your variables to agree

[lapis breach]

so it should be 4/i+3?

[rigid inlet]

If you're summing through values of i then you need something involving i within your summand

Sure

[lapis breach]

so like this?

[rigid inlet]

There should normally be brackets around 4/i + 3, but since deltax is 1 it technically doesn't matter this time

[lapis breach]

ok

so now giuve me 1 second to write soemthign

Okay so now for this problem

to find the right left and middle

it would be this

correct?

cause same interval and n still = 4

so numbers should be the sam,e

for the left and right

[rigid inlet]

You really need to write the variable you're indexing through at the bottom of the sum

[lapis breach]

okay

[rigid inlet]

Because for L you've written the correct numbers for i, while for R you've written the equivalent values of x

[lapis breach]

wdym]

[rigid inlet]

5 is not an appropriate value of i

i stops at 4

[lapis breach]

how can you tell

[rigid inlet]

From the starting list

i is the index. Your index goes from 0 to 4

[lapis breach]

is the index not 0 --> 5?

[rigid inlet]

No

[lapis breach]

even though it says 1,5

[rigid inlet]

That is x

[lapis breach]

so i is the little number

x is the right side

[rigid inlet]

Sure

[lapis breach]

x is going from 1-5

[rigid inlet]

No

Yes

[lapis breach]

i is going from 0-4

so it should be like this

WAIT

it should start at 1

left excludes the last one

right excludes the first

so then how do i do the middle one?

[rigid inlet]

You need a different list, most simply

[lapis breach]

so it would be th9is right

left hand is the red

right hand is the blue

left side is the values right side is the index

but idk what to do for midpoint

[rigid inlet]

Again you need a new list

Your first subinterval is [1,2]. What is the midpoint?

[lapis breach]

would be this right

?

wait no it would be shifted down by one

starting at 0.5

is this correct?

[rigid inlet]

0.5 is not the midpoint of any subinterval

[lapis breach]

okay so the first one i had was correct

im just subtracting the left from the right yeah

@rigid inlet im looking for the index values of those right?

sorry im just really struggling to understnad

been working for hours

[rigid inlet]

Sorry I have to go, good luck

[lapis breach]

<@&286206848099549185>

!close

.close

[safe radish]

Channel closed

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[lapis breach]

.reopen

[safe radish]

✅

[stoic dune]

Those are the midpoints of the rectangles

Remember, you should have 4

[lapis breach]

why

like how do i figure out what those are

[stoic dune]

Break [1,5] into 4 parts.
Report back the center of each part

[lapis breach]

because n=4?

[stoic dune]

The 4 parts are
[1,2] [2,3] [3,4] [4,5]

Their centers are
1.5, 2.5, 3.5, 4.5

[lapis breach]

okay thanks so much

now how do i put that into a sigma

notation

[stoic dune]

Note that n + 0.5 gets those points, for n = 1,2,3,4

[lapis breach]

which is the right one

[stoic dune]

f(n + 0.5) is the height of each rectangle

1×f(n + 0.5) is the area

Where f is x² + 3 in this case

[lapis breach]

so are those generally true for all cases?

[stoic dune]

But, imo, if you're struggling, do it without sigma notation at first

[lapis breach]

i need to do it with sigma

because my professor told me to

she "taught" us integrals, sigma notation and this shit in 30 minutes and expected us to understand it all

[stoic dune]

Right. You definitely should work your way up to sigma notation, but it's a lot to learn all at once

[lapis breach]

I just straight up dont have time

class is so fast paced for no reason

but what do i do for the sigma cause it cant be 4 and then i=1

because thats already the R

right one

and it cant be 3 and i=0

and it cant be i=0.5 and 4.5

cause appearently you cant use decimals

]

would this work

cause its the same as the LEFT one

but every i value is 1/2 higher

@stoic dune

ok so that is correct except it is 4 and 1 for some reason

great now i gotta figure out how to do it with a table

cause that didnt take me long enough

[stoic dune]

First, identify the points. Which values of x are we sampling from?

There's a table of 11 values there. We're going to use 10 of them

[stoic dune]

The process is similar to this

[lapis breach]

we will be using the first 10 for the left

and all but the first value for the right

correct>

[lapis breach]

so is this correct? formatting wise?

im trying to make sure i understand sigma notation

[stoic dune]

That is a way to write 1 + 2 + ... + 10 in sigma notation

[lapis breach]

ok awesome

[stoic dune]

This question is all calculation, and sigma notation won't be very helpful for this one

Wait sorry, you want to write i, not xi

[lapis breach]

why

cause isnt i referring the the index

and x is the value associated with the index

[stoic dune]

i is the value, 1, 2, 3...

xi is some value that depends on i

[lapis breach]

so for every value of i there is an x

i is the index value no?

x is the value that we actually care about

right>

the area under the curve

[stoic dune]

xi is going to be the "sampling point". Left, midpoint, or right, depending on the question

[lapis breach]

okay so i just found these formulas from my professor what is the difference between this and what ive been doing? Would it be easier to judt plug and chug using these for each type?

[stoic dune]

No difference. Use which you feel more confident with

[lapis breach]

this might be whats not making sense to me

i was under the impression that we have
x_i
x_0=1
x_1=2
x_2=3
x_3=4
x_4 = 5

[lapis breach]

okay ill just contine what im doing cause i dont get thi

[stoic dune]

For the example above,
x1 = 0, x2 = 0.5, x3 = 1...

It's where we "pin" each rectangle

[lapis breach]

where each rectangle hits the graph

at the left mid or right point

right>

would you be able to walk me through setting this one up

so i can just know how to do it and can relicate that for future cases

[safe radish]

@lapis breach Has your question been resolved?

[lapis breach]

no

[safe radish]

@lapis breach Has your question been resolved?

[lapis breach]

https://cdn.discordapp.com/attachments/903481106819612714/1327819868296314940/IMG_5568.jpg?ex=6784741e&is=6783229e&hm=ff71586df523dd02bafd1e0e4b88eaf3122f592af9a23a9399570659b5463c25&

Hello <@&286206848099549185> im working on trying to put this table of values into sigma notation so i can become more familiar with it. Ive gotten to the answer without it for this problem but using the formula above how do i get those values to put it in notation

my k value would be 0

my n value would be 10

delta x = 0.5

i believe those are correct but where do get the orther values

Would this be as far as i can get without plugging in all the tables values in for a because im not explicitly told what f(x) is?

well actually now ive just got to express sums in sigma notation regardless

i guess id want to start by finding deltax

which is 2

[hoary saddle]

me before scrolling up knows whats going on

a is just n+2
b is just n+4
c is just n+7
d is just 1/nx(n+1)

or am i suppose to explain it idk

[safe radish]

@lapis breach Has your question been resolved?

[lapis breach]

aaaaa

can u exolain how to do these

[stiff kindle]

So, I assume you know the definition of left Riemann sum?

or you can refer to it?

[lapis breach]

wdym

these?

[stiff kindle]

yeah so R_L I presume is the left Riemann sum. So can you write the problem in this form?

[lapis breach]

yep

i know how to get the k value and the n value obviously deltax is b-a/n but what is a

oh i see

[stiff kindle]

starting point

[lapis breach]

is a the initial vlalye

yea

ok so i just plug all those values in and is that it

[stiff kindle]

yep

[lapis breach]

oh ok

thanks

im just so tired been doing calc fr 11 hoirs

i shjould have been able to figure that one out

[stiff kindle]

no worries, takes a bit to get the hang of it...

[lapis breach]

yea its hard cause my professor desnt actually teach us anything

so i gotta teach myself all of it

[stiff kindle]

if you practice enuf times it'll come easy as it's not hard just need to get familiar

[lapis breach]

yea

i just keep losing track of what everything is

[stiff kindle]

understandable

[lapis breach]

cause im learning integrals sigma notation and rieman sums at same time

and they all build off eachother

[stiff kindle]

yep

[lapis breach]

for this its just do area of square + triangle

and do that for all the sections

that it asks for

displacement = area under velocity curve right

[stiff kindle]

yes

[lapis breach]

dude calc 2 is so much harder then calc 1

i mean i think ts also cause my professor sucks but like calc 2 just seems like more stuff to remember and formulas and stuff

well ig thats calc 1 also with derivatives

but

idk

just seems less intuitive

[stiff kindle]

yeah integrals are probably more hairy although it depends I guess. It's worth grinding it out, it will all be second nature. Every one has to go through it 😅

Intuition often times comes with practice as well.