help-28

[torn jolt]

Sheeeesh

That's roughhhh values

[sturdy sparrow]

fr

[torn jolt]

Uh my work will start in like 7 mins

I should better leave

[sturdy sparrow]

oh ok

ty

i got the first one thx to u

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[spare latch]

can someone solve this for me
1/8 = 60*(1/2)^(t/360)

[spare latch]

for t

[gritty rose]

use logarithms

[spare latch]

i need someone to double check for me

[brittle steeple]

what's your work?

[devout kraken]

i always rewrite stuff like (1/2)^(t/360) as e^(ln(1/2) t/360) = e^(-ln(2) t/360) and take the natural log of both sides, after i move the 60 to the other side.

[spare latch]

ok im good

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[warm pulsar]

can someone explain me this recursive definition

[warm pulsar]

and also give an example

whats been said here looks very different

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@warm pulsar Has your question been resolved?

[warm pulsar]

oh wait lol im dumb

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[solid saffron]

hey guys

[solid saffron]

help me with this one please:

0.5(3x-4) = (x+1)/4

[ember shadow]

what have you tried

[solid saffron]

1.5x - 2 = (x + 1)/4

1.5x - (x+1)/4 = 2

[ember shadow]

there's a simpler way than your last step

think of what else you could do

[solid saffron]

Have no idea

Tell me

[ember shadow]

Make a guess. Better to struggle here than on an exam. What else could you do besides subtract "(x+1)/4"

[solid saffron]

1.5x -2 = (x + 1)/4 | *4
6x - 8 = x + 1

[solid saffron]

I really have no idea

Wait no

[ember shadow]

you have the right idea, but how do you cancel out the 4 in the denominator on the right?

[solid saffron]

5x = 9

x = 9/5 ?

[ember shadow]

not 9.5, but 5x=9 is right

[solid saffron]

Wrong ch, 9/5

[ember shadow]

yes!

[solid saffron]

Thank you

)

Help me one more please

[ember shadow]

ok

[solid saffron]

(x^2 - x + a)/2x + 3 = 0

I should find a(min) when this equation has only 1 solution

[ember shadow]

a(min)?

[solid saffron]

minimal value of a

[ember shadow]

well how could you simplify that equation? i.e., get rid of the fraction if possible?

$\frac{x^2 - x + a}{2x} + 3 = 0$

[glossy valve]

cwatson

[solid saffron]

How did you do that?

Ah

Nvm

[ember shadow]

That's just the original problem still

[solid saffron]

2x + 3 is not equal to 0

x is not equal to -(3/2)

[ember shadow]

Oh should the whole denominator be "2x + 3"?

[solid saffron]

Yes

[ember shadow]

$\frac{x^2 - x + a}{2x + 3} = 0$

[glossy valve]

cwatson

[ember shadow]

But yes, you're correct that $x \neq -\frac{3}{2}$

[glossy valve]

cwatson

[ember shadow]

So use the quadratic formula for the numerator

[solid saffron]

x^2 - x + a = 0

D = x^2 - 4a

But what's next?

[ember shadow]

What is "D"?

[solid saffron]

Discriminant

[ember shadow]

$b = -1$ here, so $\sqrt{b^2 - 4ac} = \sqrt{(-1)^2 - 4a}$, meaning...

[glossy valve]

cwatson

[solid saffron]

Or do you mean x(x - 1) + a = 0?

[solid saffron]

1 - 4a

[ember shadow]

Yes, so what condition do you put on the discriminant?

[solid saffron]

That's what I don't know

[ember shadow]

If we had $a = 1$, what would the answer be?

[solid saffron]

0?

[ember shadow]

Yes!

[solid saffron]

No, then D = -3, empty value

[ember shadow]

Correct, so solve the equation for the condition you correctly got

$1 - 4a \geq 0$

[glossy valve]

cwatson

[solid saffron]

a =< 1/4

Why >=?

[ember shadow]

Because the square root of a negative number is not real (if you've learned about real vs. complex numbers)

[solid saffron]

Might 1-4a > 0 not be possible?

[ember shadow]

It includes the ">". It means "greater than OR equal to 0"

[solid saffron]

Sure, I know

[solid saffron]

So, here we are

a <= 1/4

[ember shadow]

No you already got it. $a \leq \frac{1}{4}$

[glossy valve]

cwatson

[solid saffron]

So, should I draw a line with x axe?

[ember shadow]

Actually let's back up. When will the equation have only 1 solution?

[solid saffron]

If a = 1/4

[ember shadow]

Yes

So that's actually the answer, if I understand the problem correctly

[solid saffron]

Yes, thank you so much! )

Have a nice day!

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[opal nexus]

P(x)=B(x)*(x-2)^3 + (x+1)^2
P(x)=C(x)*(x-2)^2 + K(x)

K(x) = ?

[opal nexus]

note: answer is polynomial, not number

[brittle steeple]

answer is a polynomial, you mean?

[opal nexus]

yeah

[brittle steeple]

equation is something with an equal sign

[opal nexus]

okay

step 1.

[brittle steeple]

looks like a question about polynomial remainders

[opal nexus]

yeah

[brittle steeple]

so let's rephrase it in terms of those

[opal nexus]

how?

[brittle steeple]

what is the remainder of P(x) mod (x-2)^3

[opal nexus]

(x+1)^2

[brittle steeple]

Right

Then the question is

What's the remainder of P(x) mod (x-2)^2

[opal nexus]

yes

[brittle steeple]

(also, does K(x) have to be degree 1?)

[opal nexus]

yes

[brittle steeple]

let me give a different problem along the same lines

the remainder of n when divided by 1000 is 287

what's the remainder of n when divided by 100

[opal nexus]

87?

[brittle steeple]

Right

[opal nexus]

hmm

n = B*10^3 + 287
n = C*10^2 + 87

B*10^3 + 200 = B*10^2
B*10 + 2 = C
B*10 = C - 2
B = (C - 2)/10

oof

[brittle steeple]

That process could work

[opal nexus]

I couldn't figure out in my mind

[brittle steeple]

essentially the point is though

a+b mod n = a mod n + b mod n

[opal nexus]

Is this a theorem?

[brittle steeple]

so P(x) mod (x-2)^2 = [B(x)(x-2)^3 mod (x-2)^2] + [(x+1)^2 mod (x-2)^2)]

yeah probably

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@opal nexus Has your question been resolved?

[opal nexus]

wait-

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[opal nexus]

well

.reopen

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✅

[opal nexus]

ah wait

I have another problems too

but I think I should try to solve again.

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[tame rune]

what is an easy without calculator way to find this ?

[onyx glen]

try to establish comparisons between the sums

like for example you can see that sum I is greater than sum II because each term in I is greater than the corresponding term in II

[tame rune]

okay thanks. Thats what i was thinking

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[last locust]

Can someone pls lmk if this is correct?

[gritty rose]

you can just check with a calculator?

,calc 3^(log(9)/ (log(3)+log(9)))

,calc 9^(1 - log(9)/ (log(3)+log(9)))

nope

oh that's a 9

[glossy valve]

Result:

2.0800838230519

Result:

2.0800838230519

[gritty rose]

yup!

[last locust]

Ok thanks

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[modern wagon]

Does anyone know how to solve this?: s(9)=4x+40

[onyx glen]

is that exactly how the question was given to you?

symbol for symbol?

or did you abbreviate something?

[modern wagon]

This is what was given to me and my teacher told me to replace x with 9 and solve

[onyx glen]

okay see

you didn't replace all instances of x with 9

you should have written s(9) = 4*9 + 40

[modern wagon]

Oh ok

So how do I solve?

[onyx glen]

s(9) = 4*9 + 40

do you see the part i bolded

[modern wagon]

Yes

[steady hornet]

A round table meeting is held between eight persons A B C D E F G and H. In how many ways can they be seated so that A and H always sit diagonally opposite to each other?

Please tell me that the answer is 720.

[modern wagon]

Huh

[onyx glen]

@steady hornet please open your own channel. read #❓how-to-get-help for instructions.

[onyx glen]

okay, good. are you able to calculate 4*9 + 40?

[modern wagon]

Yes

[steady hornet]

Already tried. It's been 2 hrs and yet not accepted

[onyx glen]

no reason to intrude on other people's channels.

@modern wagon do any questions remain from you?

[modern wagon]

No I think I'm ok thank you

[onyx glen]

ok then you can close this

[modern wagon]

.close

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[crude ridge]

can you explain the last line, how did the possible abelian grousp become Z8xZ2xZ2 or Z8xZ4

[spice orchid]

What's the question

[crude ridge]

q5

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[hidden siren]

Confused how I am supposed to input the last one. Or if any errors were made

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[hidden siren]

<@&286206848099549185>

?

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[narrow rose]

How do i find the other values of sin? I forgot

[viral jasper]

$\sin{\theta}=\sin(2\pi+\theta)$

[glossy valve]

SWR

[narrow rose]

?

[devout valley]

and also $\sin(\theta) = \sin(\pi - \theta)$

[glossy valve]

@devout valley

[narrow rose]

so do I keep adding 2pie for the other values whenever its sin?

[viral jasper]

You could do $2k\pi$ where $k$ is any integer

[glossy valve]

SWR

[narrow rose]

so the next value would be 7/3 pie?

alr ty

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Channel closed

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[wanton mesa]

.reopen

[spice orchid]

ya fucked it

open a new one

[wanton mesa]

lol

k

[dapper minnow]

.close

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[torn jolt]

can someone explain me how to solve this task

[gritty rose]

too grainy a photo

just type it out

and type your work

[torn jolt]

type my work? what does that mean

anyways, ill just write it out in word

[gritty rose]

your work in attempting the problem

[torn jolt]

i cannot attempt the problem

because i don't know how to even solve it

[safe trench]

Which part is your problem?

[torn jolt]

the entire task

i don't know how to solve it

[safe trench]

The first step that is done is to substitute u=x^2

That way you get to solve a quadratic equation instead of quartic

[torn jolt]

what?

why did we write x^4 = u^2 and x^2 = u?..

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[hidden geode]

I don't understand hwo this could be -a

$\sqrt{a^2}$

[glossy valve]

Jshy <3

[hidden geode]

Is only a right?

it cant be negative if it is squared?

[glossy valve]

rkky

[hidden geode]

I don't really get that logic though

-(-2) is still just 2

[slate rose]

yes, but here your a = -2

[hidden geode]

If I miss out on that as an answer, would it be losing marks?

[hidden geode]

Right... okay I kinda get it

yeah

[slate rose]

and that’s why for negative a you get -a for sqrt(a^2)

[hidden geode]

Got it, I understand

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[hollow moth]

so in equivalence classes like this one, can the modulo ever be negative?

[hollow moth]

like can it ever be -17?

[spice orchid]

sure but it'll just be the same as 17

[hollow moth]

wait so 7 % -17 is the same as 7 % 17?

[spice orchid]

well i think the modulo operator works a bit differently

,calc 7 % -17

[glossy valve]

Result:

-16.93

[spice orchid]

okay thats not what i wanted

the point is that the equivalence classes mod -17 will be the same as the equivalence classes mod 17

$[a]{17} = [b]{17} \iff (a-b) | 17 \ \iff (b-a) | -17 \iff [a]{-17} = [b]{-17}$

[glossy valve]

ΣAC

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[cunning plover]

how does du/e^x turn into e^-xdu

[west sequoia]

remember that $\frac{1}{x^a}=x^{-a}$

[glossy valve]

baro | awake

[cunning plover]

ok thanks

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[barren iris]

OKAY

I just want to walk through this with someone

So, from (0,2)U(2,3) it's slowing down

thoughts?

[meager dew]

why isn't it slowing down at t=2?

[barren iris]

oh

should I use brackets?

is that what you mean

the slope of the tangent line is 0 there

it's for sure slowing down at 2

[meager dew]

?

[barren iris]

is it not?

[meager dew]

i mean why split the interval up?

[barren iris]

hmm

should I go from (0,3)?

The particle is going towards a tangent slope of 0

[meager dew]

oh you were talking about at t=3

[barren iris]

I guess I don't have to split it up.

[pulsar gale]

It’s speeding up on (1,2)

I think it would help to draw the tangent line at every integer value of x

[barren iris]

Dang

I'm getting closer

I'll try that

Is it because the slope gets steeper starting at 1 to 2?

[full forum]

@barren iris Has your question been resolved?

[pulsar gale]

Yes

[barren iris]

Ok

I feel ot

I'm so much closer to getting this

I'm gonna draw the tangent lines and post it here

[pulsar gale]

Did you do it

[barren iris]

Not yet. Had to do something

[pulsar gale]

Shouldn’t take that long

[pulsar gale]

Yeah

[barren iris]

on (2,3) it's very steep, but it's going toward a slope of 0

right?

[pulsar gale]

Yes

@barren iris so

[barren iris]

so, (0,1) is increasing, (1,2) is increasing, (2,3) is slowing down, (3,4) is increasing

[pulsar gale]

What do you mean by increasing

[barren iris]

speeding up*

[pulsar gale]

So you said that (0,2) is speeding up?

[barren iris]

Yes

[pulsar gale]

Draw the other tangent lines

0,1

[barren iris]

that one looks flatter

[pulsar gale]

What does that mean

[barren iris]

Like the slope is flatter than the one from (1,2)

[pulsar gale]

Draw at 0 too

Make the inner lines

Thinner

[barren iris]

wait

at zero?

um

one sec

[pulsar gale]

Uh

First one is wrong

[barren iris]

oh

what did I do wrong?

[pulsar gale]

How is it 0 at t=0

[barren iris]

idk what else it would be tbh

I don't think I can justify why it's 0

idk what the slope would be other than 0 there

[pulsar gale]

Just draw the tangent line

Like at 0.000001

[barren iris]

?

[pulsar gale]

Yeah

Well no

It’s steeper

[barren iris]

I don't think I know it

[pulsar gale]

youre overthinking it

its just that

@barren iris

[barren iris]

huh...

[pulsar gale]

so

[pulsar gale]

from 0 to 4: large neg, small neg, large neg, 0, large neg.

sure

youve been labeling at every 0.5 though

i said every integer val

@barren irisi need to go soon

if you want to finish this you gotta respond a bit faster

[barren iris]

gotcha

If you've gotta go, that's totally fine. I'll get it eventually. I'm not super confident on this

[pulsar gale]

okay

so (0,1)

speeding up or slowing down

[barren iris]

Speeding up.

very steep.

[pulsar gale]

😔

it goes from steep to not steep

so it slows down

[barren iris]

but it's not like flat at 1 though

[pulsar gale]

but its less steep than at 0

[barren iris]

like it is at 3

okay

[pulsar gale]

it doesnt need to decrease to 0

[barren iris]

so

it's because it's starting from some point, to a smaller slope?

[pulsar gale]

yes

[barren iris]

so like a similar idea to the thing you pointed out before?

[pulsar gale]

yes

okay now (1,2)

i or d

[barren iris]

from flat to steep

flater*

so, increase?

[pulsar gale]

yes

[barren iris]

https://tenor.com/view/spongebob-squarepants-begging-pretty-please-beg-on-your-knees-pray-for-mercy-gif-26344462

oh bet

[pulsar gale]

(2,3)

[barren iris]

decreasing

[pulsar gale]

try not to use terminology like "decreasing" and "increasing"

[barren iris]

okie

[pulsar gale]

that is ambiguious

[barren iris]

it's slowing down

[pulsar gale]

yes

(3,4)

[barren iris]

speeding up

[pulsar gale]

if youre going to use the word increasing say "increasing in magnitude"

because increasing can be from -5 to 0

but that is slowing down

alright so altogether

[barren iris]

(0,1) slowing down
(1,2) speeding up
(2,3) slowing down
(3,4) speeding up

[pulsar gale]

yes!

[barren iris]

FINALLY

[pulsar gale]

@barren iris is it right

[barren iris]

Putting in everything now

yes

everything was correct

[pulsar gale]

yay

[barren iris]

yayayayay

thank you so much for your time

[pulsar gale]

youre welcome

good luck with the next ones

you got this

[barren iris]

❤️

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[spark yew]

any idea how to make this limit?

[full forum]

@spark yew Has your question been resolved?

[full forum]

@spark yew Has your question been resolved?

[full forum]

@spark yew Has your question been resolved?

[gritty rose]

did you do the integration? maybe a substitution will help

[spark yew]

i couldnt think of a good substitution

or any at all

[gritty rose]

then try it for m=1 first. then for m=2 try doing integration by parts. for m=3, try to find some inductive hypothesis

[spark yew]

okay sure ill come back after im done

[spark yew]

The first one gave me 0

The second one is

and the third one i think it's 0 cus it's an odd function with opposite values at the limit (-1 and 1)

How can this help me solve this limit tho?

[full forum]

@spark yew Has your question been resolved?

[brave basin]

@spark yew For odd values of m, it's gonna be 0, you're right.

It's an odd function

When m is even, without integrating at all, can you guess what the integral approaches? Take m very large, what do you think happens?

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[flint thicket]

Let y = 5cos[5(x+pi/5)]

[flint thicket]

Okay so two main questions

The square brackets are very unusual here because no other question has them and this isn't in the textbook or the lecture

Secondly

[vast fossil]

Treat them like regular parenthesis

[flint thicket]

It's asking me what the phase shift is

is that just another way of saying horizontal shift?

I looked it up online and that's what it said but pi/5 wasn't the answer

so I'm unsure of if I'm misunderstanding the question or the answer

[vast fossil]

No, phase shift is different

[vast fossil]

Weird, cuz it must be it

What does the answersheet say?

[flint thicket]

one sec

lumen just says that its wrong

lemme get it wrong again

okkkk

I see

so pi/5 is 0.62831...

So lumen sometimes gives the answer differently

so maybe the answer here was -pi/5

maybe phase shift means like

the whole shift with the direction

okay yeah that works

ty

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[full forum]

@craggy sandal Has your question been resolved?

[main vortex]

yeah they have not responded to my question as well and I have been waiting for almost 3 hours

[full forum]

@craggy sandal Has your question been resolved?

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@craggy sandal Has your question been resolved?

[polar valve]

https://math.stackexchange.com/questions/1292213/for-n-1-odd-prove-that-phi-2nx-phi-n-x

[polar valve]

It would have been helpful to have had an indication of which theorems were not available.

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[void widget]

,rccw

[void widget]

So yeah the question is in my copy
Also
K.COS THETA/K.SIN THETA
WHERE SIN=K(√2-1) AND COS = K

The actual question

[full forum]

@void widget Has your question been resolved?

[void widget]

<@&286206848099549185> I would give you blessings if you help me 😊

[elfin stream]

nope

try squaring both sides

then subtracting 4ab

a = sin(theta), b = cos(theta)

[void widget]

I'm saying this prove is right?

[elfin stream]

nope

[void widget]

Where is the problem?

[drifting hatch]

I’m stuck on the line sec^2(theta) = 1 + tan^2(theta)

[elfin stream]

how did you get k*tan(theta)/k

[drifting hatch]

How does that follow from what’s above?

[void widget]

K.COS THETA/K.SIN THETA
WHERE SIN(theta)=K(√2-1) AND COS(theta)= K

I think I had to mention that

Before giving the question

[elfin stream]

yes 😂

[void widget]

Yupp 😂

[elfin stream]

Okay yes same

[void widget]

I was trying to see that is
Sec²x=1+tan²x is valid here

Let us say theta = x

Cause I don't have theta in my keyboard anyways

So I got that
sec²x≠1+tan²x

In this question

And that can't be possible

Also if
cosx=1
Then sec²x=1 too

And I put an assumption in the answer that
"K=1"

[drifting hatch]

Wait a sec 😂😂 sec^2 = 1+tan^2 is just an identity. My mistake

[void widget]

Yes

[drifting hatch]

Oh I got it, it’s just not true when cosx = 1 that sinx + cosx = sqrt(2)cosx

[void widget]

sec²x-tan²x≠1

And that's a catch

That I got in the question

And It is not possible right

[drifting hatch]

So the k=1 assumption doesn’t make sense since you’re taking k to be cosine

[void widget]

K could be any constant

Cause it's a ratio

Oh yeah

[drifting hatch]

Then tell me how you changed cos^2(x) into 1

[void widget]

K=1
cosx=1
cos²x=1

Right?

[drifting hatch]

I think I misled you here, the identity is always true I just forgot it

[void widget]

Hm

[drifting hatch]

Why is cosx = 1?

[void widget]

It's an assumption

[drifting hatch]

That means x is a multiple of pi, but then you can check that the original identity you’re assuming doesn’t hold

A multiple of 2pi*

The sin+cos = sqrt2(cos) only holds for certain values of theta

[void widget]

How would you explain this
when k = 1
cos x - sin x
= 1-(√2-1)
= 1-√2+1
= 2-√2
Taking √2 common
=√2(√2-1)
=√2sinx ( this is the thing which we have to prove)

[drifting hatch]

How did you get cosx - sinx = 1-(sqrt2 - 1)?

When cosx = 1, sinx has to be 0

[void widget]

I already showed you that
cos x= 1
sin x = √2-1

[void widget]

That's also a point

😢

[drifting hatch]

I don’t think you did show me that. Can you tell me where you think you justified it?

[void widget]

Wait

This step

Sin x=√2-1
cos x =1
If k = 1

I'm fking confused

[drifting hatch]

That step is coming from the sin + cos = sqrt(2)cos

[void widget]

Yes

[drifting hatch]

The cosx = 1 is something you made up after

[void widget]

Yupp

[drifting hatch]

It’s incompatible with that formula

[drifting hatch]

This doesn’t apply to all x

[void widget]

So what's the conclusion from there?

[drifting hatch]

The step where you turn cos^2x into 1 isn’t justified

[void widget]

Ohk

[void widget]

But then why that gives the right answer

[drifting hatch]

Because sinx = sqrt(2) - 1 isn’t true for all x

[rotund sail]

hi

[void widget]

Hmm

[drifting hatch]

Sqrt2 - 1 is a constant and sinx isn’t

[rotund sail]

hi

hi arnab

[void widget]

@rotund sail hi

[drifting hatch]

Hello

[void widget]

Hmm

[rotund sail]

tell ur problems

am here

[void widget]

@rotund sail

But it's resolved

@drifting hatch thanks

[drifting hatch]

So what I said makes sense now?

[void widget]

Yupp

[drifting hatch]

Great, have a good one

[rotund sail]

give me not resolved

[rotund sail]

hey

[void widget]

Find the exact sum of
Zeta function

From n=0

[drifting hatch]

Hello plz don’t ping specific helpers without permission

[void widget]

Ok thanks

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[rotund sail]

what

[void widget]

Nothing I don't have any question in my mind

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[torn jolt]

My question is, how to solve this integral? My work here, I'm not gettting a way further from this.

[atomic blade]

You have to use IBP twice

[torn jolt]

sorry, i suck at math. did not I use twice

[hot herald]

is that 2 the question number

[torn jolt]

Yeah

I got my other questions but stuck here

[hot herald]

note that the original integral reappears after two applications of IBP

[drifting hatch]

Looks pretty good. See how you have the same integral you stated with? Solve for it like a variable

[hot herald]

solve the resultant equation for your original integral

helps to let I = original integral

[torn jolt]

am I not trying to solve that equation lol I did not understand what you all meant, so sorry

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[full forum]

Please don't occupy multiple help channels.

[fervent escarp]

what exactly is confusing here

[lucid glade]

idk how to solve it

[fervent escarp]

do you understand the very first notation?

[lucid glade]

not really

[fervent escarp]

triangle XYZ is similar to triangle BAC which means that their sides are "proportional"

for example side BC could be equal to 6, side XZ is equal to12(given)

that would imply that every other side of the triangle BAC is also shrinked down by 2 from XYZ

angles however stay the same

[lucid glade]

oh okok

[fervent escarp]

which is what the question wants you to do

since angles stay the same, all the trig functions of angle C are equal to trig functions of angle Z

[lucid glade]

how to get the tanC

and cosC

[fervent escarp]

just how you would get cosZ and tanZ

[lucid glade]

ok

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[sharp quartz]

Hi

[sharp quartz]

can anyone explain how you can get x from here

-1<log2(X^2/2)<1

where 2 is the base actually

ik very basic doubt

[glacial pasture]

you can 'undo' the log. by putting everything as an exponent of 2
so 2^{-1}<x^2/2<2^{1}

[sharp quartz]

yes yes

got it

thanks

[frigid pecan]

you can also do log2((x^2)/2) = log2(x^2) - log2(2) = log2(x^2) - 1

[sharp quartz]

Oh right yes

[grave elm]

=2log2(x)-1

[sharp quartz]

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[tropic helm]

hi

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[slate pike]

Anyone knows how to prove: ?

[slate pike]

<@&286206848099549185>

[tropic helm]

hi

what do you need

[slate pike]

Hi, I cannot figure out how to prove the above

[tropic helm]

here

this is one

this will help

i wrote this down like 2 ears ago

years

oh wait

wrong one

is this the full image?

[slate pike]

yep

[tropic helm]

what grade are you in?

[slate pike]

First year university

[tropic helm]

oh

sorrry

i forgot how to do this

idk how

[slate pike]

No problem, thanks

I'll just ask prof on monday then

[carmine minnow]

what does the bracket mean, stirling numbers?

[slate pike]

The second kind

[carmine minnow]

how are they defined

"number of ways to partition a set of n objects into k non-empty subsets"?

[slate pike]

Number of possible partitions of n-set of k-blocks

[carmine minnow]

so it seems that we are selecting m objects to partition into k non-empty subsets and doing something with the other n-m things

that's what the sum is suggesting

[slate pike]

I suppose so,

[carmine minnow]

can you spot what is being done with the other n-m things

[slate pike]

I mean the intepretation of the right side for me is we choose m elements and then arrange them into k blocks, and we do it for all ms between 0 and n

but why would it give us all combinations for n+1 set of k+1 partitions

also there must exist an algebraic way to proof this

[carmine minnow]

if you have a combinatorial interpretation, then you can't really look for algebraic ways

unless you have an algebraic expression for the stirling numbers of the second kind

then you can consider an algebraic way to prove this

[slate pike]

well I know there is no algebraic expression for the stirlings numbers but they have the 'pascal' like recursive property, so I thought maybe something with that

[carmine minnow]

you'll need to pick some algebraic expression

(it probably would contain some summation as part of it)

[carmine minnow]

we are choosing m from n items, which means we are doing something with the n+1 th item

[slate pike]

For each item in the n+1 set we take away one and then choose m items?

sorry I don't get it

[carmine minnow]

the n+1 th item wasn't chosen, since the expression has n choose m, not n+1 choose m

so we are doing something with the n+1 th item, and then something with the n-m items we didn't choose

[slate pike]

well we put them in one partition I suppose

[carmine minnow]

so now we need to show that this process matches up one thing on the left hand side with one thing on the right hand side

Given one way to partition n+1 things into k+1 subsets, show that it corresponds to one way to choose m items from n items and then putting the m items in k subsets (for some choice of m)

(and in the backwards direction)

[slate pike]

I see makes sense now, thanks a lot!

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[vast fossil]

!status

[full forum]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[bronze magnet]

1

im pretty new to breaking down a function into composite functions

[vast fossil]

4 - x^7 is inside the ( )^3

So it means we can pick 4 - x^7 to be the inner function (g(x))

Leaving the outer function to be u^3 (f(u))

So (4 - x^7)^3 = f(g(x)) where g(x) = 4 - x^7 and f(u) = u^3

Are you good with applying the chain rule after the setup or do you want me to continue?

[bronze magnet]

so my first part should be u^3 , 4-x^7 ?

[vast fossil]

Yeah

[bronze magnet]

ok then yes continue

[vast fossil]

So, according to chain rule the derivative of f(g(x)) is f'(g(x)) * g'(x)

Let's first get f'(x)

f(x) = x^3 so, according to the power rule, f'(x) = 3x^2

So f'(g(x)) is simplified to 3(g(x))^2

We can also plug in the expression for g(x) here

Yielding 3(4 - x^7)^2

Alright, that's the first term of the product, now we need to get g'(x) to finish things up

g(x) = 4 - x^7 so, applying the power rule yet again, we get g(x) = -7x^6

Eventually leaving us with d/dx [f(g(x))] = 3(4 - x^7)^2 * (-7x^6)

Or simply -21x^6(4 - x^7)^2

[bronze magnet]

did you do this without calc?

[vast fossil]

Calculus or calculator?

[bronze magnet]

calculator

[vast fossil]

Yes, I did that without a calculator

[bronze magnet]

mind walking me through the steps?

i dont start with distributing 3 right?

(12-3x^7)^2

[vast fossil]

No, you can't distribute like that

3t^2 =/= (3t)^2

[bronze magnet]

so i should square everything first>?

[vast fossil]

You don't have to square the polynomial, just leave it like that

[bronze magnet]

3(16-x^9)

oh

[vast fossil]

Nor is that correct

Squaring 4 - x^7 yields 16 - 8x^7 + x^14

[bronze magnet]

how did you get

when you multiply

[vast fossil]

Multiplied 3 by -7

[bronze magnet]

doesnt it go to everything

[vast fossil]

And just moved the x^6 to the front

Thanks to the commutativity of multiplication

[bronze magnet]

communitivity?

spelling?

[vast fossil]

a * b = b * a

[bronze magnet]

i mean the spelling of that word

[vast fossil]

No, I spelled it correctly

[bronze magnet]

oh

interesting

how did you know we would use chain rule

[vast fossil]

Chain rule comes in whenever you have to differentiate a composition of two functions

That was the case here

[bronze magnet]

ok

ill see if i can do this one

f(u) = sqrtx g(x) = x^9 + 5

then i get derivative of both?

[vast fossil]

Correct (although it's f(u) = sqrt(u))

[bronze magnet]

how is that different than x

sqrt u = 1/2sqrtu

x^9 + 5 = 9x^8

[vast fossil]

Right, now you plug in

[bronze magnet]

the 9x^8 into the sqrt of other one?

[vast fossil]

No, x^9 + 5 goes under the sqrt here

Cuz it's f'(g(x)) * g'(x)

[bronze magnet]

but we got derivative of g(x)

which is 9x^8

[vast fossil]

But the thing inside f' is g(x), not g'(x)

So we have f'(g(x)) * g'(x) = 1/2sqrt(x^9 + 5) * 9x^8

[bronze magnet]

im not sure how to multiply those

[vast fossil]

You don't, you just leave that as 9x^8/2sqrt(x^9 + 5)

[bronze magnet]

the 9x^8 goes on top?

[vast fossil]

Yes

The rest on the bottom

[bronze magnet]

wow epic

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[inland fjord]

I'm not sure how to begin questions d and e

[inland fjord]

For question b I used monotonic convergence theorem and for question c I used the ratio test. For these two sequences I don't believe I can use either of those tests.

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[inland fjord]

<@&286206848099549185>

[polar valve]

sin x < x?

[hidden tree]

?????

[inland fjord]

I don't understand. Are you saying that sin(1/n) is upper bounded by n?

[inland fjord]

Is my question unclear?

[polar valve]

sin x is upper bounded by x. -> sin (1/n) < (1/n).

[inland fjord]

Ok. What does that tell me about it's convergence?

[polar valve]

what do you know about the converence from 1/n?

[inland fjord]

As we take the limit of 1/n to infinity it will approach zero.

[polar valve]

sin (1/n) is lower then 1/n but >0.

so what does this mean?

[inland fjord]

Ah ok. So sin(1/n) must converge to zero.

Is this the squeeze theorem? Using the upper bound of 1/n and the lower bound of zero?

[polar valve]

yes

[inland fjord]

Gotcha. Thank you 🙂

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[modest sail]

Ok so for 5

I would

Ok thats easy

Gg

Ok

Actually

I know how to get 36

But idk how to get other variables with 6

6^?= x^2

Huh

[wide sundial]

What’s log ab

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[dusk turret]

Difference between (a,b) € R² and (a,b) € R ? All I know is that R is a set of real numbers, but today I saw R² in class and I don't know what (a,b) € R² means

[ivory cairn]

R^2 is RxR, or ordered pairs of real numbers, or coordinates on the xy-plane.
So (a,b) in R^2 is an ordered pair.

(a,b) in R would be an interval from a to b

[dusk turret]

Aah got it, well explained thank you

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[tame rune]

Ada says Bruna lies, Bruna says Charlotte lies, Charlotte says Ada and Bruna Lies. Who is telling the truth ?

[tame rune]

How do i set up truth table for this kind of problem ?

[neon crystal]

like this?

 A | B | C
---+---+---
 F | T | F
   |   |
... ... ...

or maybe

 A | B | C | valid
---+---+---+-------
 F | F | F |   F
 F | T | F |   T
... ... ...

[tame rune]

wait i'll show on paper

@neon crystal

[neon crystal]

,rcw

[neon crystal]

aren't the first three and last three columns identical?

[tame rune]

for first statement, what should i consider Bruna lies or the fact that Ada says ?

[neon crystal]

I'd have a column for "Ada is telling the truth"

[neon crystal]

and so on

[tame rune]

i saw this problem,

[neon crystal]

here p is "Ada is telling the truth", q is "Bruna is telling the truth", and r is "Charlotte is telling the truth"

so the valid situations are those where the left three columns are the same as the right three columns

because "Ada is telling the truth" should be true if and only if what Ada says is true

[tame rune]

would be please wait a couple minutes so i can try the problem myself ?

[neon crystal]

I actually have to go now, but I'm sure someone else can help

[tame rune]

all right thanks

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[hot herald]

illegally

[worthy topaz]

This is a quadratic equation and leave it at the simplest surd?

Burh did I do something illegal?

[vast fossil]

8/(-8) is not 1

[worthy topaz]

Whole question for context?

[hot herald]

8/(-8) is not 1
that's your issue

[worthy topaz]

Okay thanks 👍

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[dim geyser]

Is there function of real that is continuous on rational and discontinuous on irrational?

[silver thorn]

Apparently not, it follows from a couple theorems in analysis

https://www.quora.com/Is-there-any-function-which-is-continuous-at-all-rational-points-and-is-discontinuous-at-all-irrational-points?ch=17&oid=17745000&share=59fd67b2&srid=gImEU&target_type=question

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[tiny atlas]

Hello, is this a valid operation? cancelling the 'h'

[fast peak]

yes

[tiny atlas]

ok thank you

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[slender elk]

can someone help me answer this

[simple totem]

lmao what

does similarity in volume relate to height just like that

with like any shape

[atomic blade]

Yeah

[simple totem]

damn that’s pretty cool

[slender elk]

lol

any1 know how to do it?

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[polar valve]

think about this: if you double a length, how would
a) the area of a square
b) the volume of a cube
grow?

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[gilded peak]

Help, I'm still confused with it but I answer a few parts.

[gilded peak]

Operation: Number System Fraction

<@&286206848099549185>

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[void warren]

hexadecimal is it?

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[compact vault]

How to do this?

[craggy basin]

what do u wanna do

like

factorise ?

[compact vault]

Yes

[vapid barn]

try 1

or -1

[compact vault]

Ok

[vapid barn]

-1 works for sure, so you know (x+1) is a factor

now just do the synthetic division

[compact vault]

👍

What is the solution of Non-deterministic Polynomial Complete?

[onyx glen]

this question does not make sense

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[fast peak]

well not that part but the -S(P)+S(P) and the -sigma+sigma

[torn jolt]

Ahhhhh shit

I see it

Lol tyty

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[idle blade]

Problem number 12

I’m sorry if the question isn’t clear because of my random calculations

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[idle blade]

<@&286206848099549185>

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[limber flicker]

!status

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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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[verbal river]

i am able to get it in the form A/x +B/X^2 +CX+D/X^2+1 but i do not know how to get the coefficents for A,B,C,D

i am only able to get B=-3 when i se x=0

but i cannot get a,c,d

however

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[torn jolt]

How can I calculate r value of exponential dataset?

[torn jolt]

r-value is pearson correlation for linear graphs

excel... I think I figured it out

BIG BRAINN

HAHAHAHA

https://tenor.com/view/clapping-awesome-gif-26321471

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[inland latch]

how can i find the local low and highpoints of the graph f(x) based of the graph in the picture f'(x)