help-27

[devout snow]

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[static pollen]

Can you cancel out imaginary numbers like this?

[bleak tide]

are you using j for sqrt(-1) here?

[static pollen]

Yeah

the answer is correct

[bleak tide]

Yeah that looks correct to me, what makes you think it could be wrong?

[static pollen]

because

e.g we have 5/3i

then you would multiply with i/i to get rid of the i in the denominator right?

so (5/3i) *(i/i)

[bleak tide]

Yeah or just note i^4 = 1. so i^{-1} = i^{4}i^{-1} = i^{4-1} = i^3 = -i

so move the i up as -i

[static pollen]

?

[bleak tide]

x/i would be -ix

[static pollen]

yes

but if i would do it that way i end up with jvo =jv2

which isnt correct

[static pollen]

look at the skribbled answer

[bleak tide]

the 2nd scribbled answer has an i magically appear on the leftmost v_0 term

[static pollen]

yeah

but

[bleak tide]

sorry I misread

[static pollen]

j(vo+vo/j) =jvo-jvo right?

[bleak tide]

is v0 not vo?

[static pollen]

yeah typo

[bleak tide]

$$j\left(v_o -\frac{ v_o}{j}\right)=jv_o-v_o$$

[woven radish]

nHail

[bleak tide]

remember to distribute fully

[static pollen]

its supossed to be a plus

[bleak tide]

Nope

[static pollen]

it then becomes jvo/j^2 which is jvo/-1

what do i not understand?

[bleak tide]

your correct line of reasoning up here was correct

I was explaining what you did wrong in the scribbled out part

[static pollen]

.close

[devout snow]

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[devout snow]

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[agile dew]

this would be a half circle, correct?

[agile dew]

well

the top half of an ellipse

with the cartesian equation being

[lusty sapphire]

Why just the top half?

[idle oasis]

because of the bounds of t

[lusty sapphire]

Oh I missed the bounds

[idle oasis]

yup but im pretty sure y≥-4 is a condition

since u only want the top half

[devout snow]

@agile dew Has your question been resolved?

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[ionic swift]

Hi everyone!
I managed to confuse myself to the point were I would appreciate some help with understanding the ramsey number R(k,l) , a lot.
I am supposed to show that R(3,4) = 9.
The definition in my book tells me "R(k,l) is the smallest n in N such that every two coloring of the K_n contains a red k cliqure or a blue l clique"
So we know that R(3,3) = 6 already. If we tried to colorize the K_6 such that we have neither a red 3-clique nor a blue 4 clique then R(3,3) already tells us that this will be impossible for the K_6, doesnt it ? Since every two coloring of the K_6 has to contain a red or blue 3 clique. So by that logic R(3,4) would have to be 6 as well. But apparently thats not the case.
Can someone spot where I went wrong in my thoughtprocess ?

[devout snow]

@ionic swift Has your question been resolved?

[devout snow]

@ionic swift Has your question been resolved?

[devout snow]

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[restive river]

Hey, I checked my answer with Wolfram but don't know where I made my mistake

[restive river]

Is multiplying sum notations together to find the Taylor series of the product wrong?

[acoustic leaf]

it doesn't work like that unfortunately

[restive river]

Should I expand

Then multiply the polynomials

That's probably a safer bet right

[acoustic leaf]

yes

[restive river]

Okay, let me try that

[crisp niche]

you should read about multiplication of serieses

[woven radish]

XO

[crisp niche]

no,

[restive river]

Yup

I'm fixing my mistake right now

[acoustic leaf]

you can see that (1 + 2) (3 + 4) ≠ (1*3) + (2*4), so it follows that more complex sums would also not multiply like that

[crisp niche]

$\left( \sum_{n=0}^{\infty }a_{n} \right)\cdot \left( \sum_{n=0}^{\infty }b_{n} \right)=\sum_{n=0}^{\infty }c_{n}\text{ }\text{ }\text{, where:}\\c_{n}=\sum_{k=0}^{n}\left( a_{k}\cdot b_{n-k} \right)$

[woven radish]

Joanna Angel

[devout snow]

@restive river Has your question been resolved?

[restive river]

,w (1/3! - 1/4!)

[woven radish]

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[restive river]

,w -1/(2!3!) + 1/6!

[woven radish]

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[restive river]

,w Taylor cos(x)sin(-x^2) about 0

[woven radish]

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[restive river]

Okay cool

I can confirm that polynomial expansion and term by term multiplication worked for me

Thank you guys

.close

[devout snow]

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[lyric hornet]

How would I show $\binom{n}{k}\approx\frac{n^k}{k!}$ for significantly large $n$?

[woven radish]

PajamaMamaLlama

[tiny panther]

context?

in brief, $\binom nk = {n(n-1)\dots(n-k+1)\over k!}$

[woven radish]

Edward II

[crisp niche]

so in other words, you want to prove Binomia distribution convergges to Possion distribution

[lyric hornet]

it's part of a bigger problem 😅

[tiny panther]

and then those ks in the numerator don't matter because the ns dominate

ok so brief will do

[lyric hornet]

oh wait yea fair enough

[lyric hornet]

tysm

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[devout snow]

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[noble shoal]

Hi is this where I go for help

[devout snow]

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[floral tulip]

I require assistance

[devout snow]

@floral tulip Has your question been resolved?

[devout snow]

@floral tulip Has your question been resolved?

[devout snow]

@floral tulip Has your question been resolved?

[devout snow]

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[hidden kiln]

the fact that the limit of the difference quotient is f'(c)

[weak cove]

When they write f’(c)

That is when

If f was not differentiable you couldn’t make that step

[hidden kiln]

yes

f'(c) would not exist otherwise

[torn vessel]

if h=0 then f(c+h) = f(c), but you don't want that, you need the limit to prove continuity.
so they assume h =/= 0, and since h is not 0 you can multiply by h/h

no, that doesn't show continuity. that shows f(c) = f(c)

to show continuity you need f(c+h) = f(c) as h approaches 0

i.e. you need $\lim_{h\to 0} f(c+h) = f(c)$ for continuity

[woven radish]

Zybikron

[devout snow]

@restive river Has your question been resolved?

[torn vessel]

lim f(x) as x approaches c is saying "what's happening to f(x) around x = c", if f(x) is continuous then around c, f(x) should basically be f(c)
lim f(c+h) as h approaches 0 is saying "what's happening to f(c+h) around h=0", if f(x) is continuous then around h=0, f(c+h) should basically just be f(c)

when you calculated limits at some point you plugged in values around that value of x, right?
So for a limit as x approaches 1 you'd plug in .9, .99, .999, 1.1, 1.01, 1.001, etc. yeah?
that's the same idea as plugging in 1 - .1, 1-.01, 1-.001, 1+.1, 1+.01, 1+.001
so instead of thinking of it as f(x) as x -> 1, you can think of it as f(1+h) as h-> 0 because you're just plugging in values of h 'close to 0'

[weak cove]

Let x=c+h

[devout snow]

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[high turret]

Can someone guide me through these, thank you!

[restive river]

what are f and g

[high turret]

My bad

Same as the other one

I did not draw g correctly :0

[restive river]

idk if I'm missing something but you can't just ignore an element in the domain

[flint egret]

https://www.youtube.com/watch?v=KJGp0pyPoVo&list=PLDesaqWTN6EQ2J4vgsN1HyBeRADEh4Cw-&index=11&ab_channel=ProfessorLeonard Can you look at 2:08:11 - Why is the 1/2 moved over if theta = tan inverse x/2 wouldn't it be 2taninverse(x/2)

[high turret]

Should be like this

[restive river]

oh wait

oh wait I see

the mapping is done from right to left

I see yeah it is a function

[restive river]

just not a surjective one

yeah okay

[high turret]

It’s injective though right

[restive river]

yeah

[high turret]

Yeah sorry!!

Drew it wrong

Ok so

[high turret]

I will watch rn

Wait

Lmao

I thought that was a reply to me

This channel is occupied rn so pls find an open one if you don’t mind

How do I compute the composition? Or like what does it mean

Ima restart

.close

[devout snow]

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[devout snow]

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[sacred plank]

whats the difference for e) and g)

[sacred plank]

what do i do differently

as theres a number with x^2 this time

[drifting sierra]

Just factorize it out from the whole thing first

[sacred plank]

divide?

[drifting sierra]

I mean yeah but you still need it somewhere

[sacred plank]

so everything would get divided by 3?

[drifting sierra]

Yes

[sacred plank]

x^2 -4x -12

is that how it would look after dividing?

[drifting sierra]

Yes

Inside parentheses

With a 3 in front

[sacred plank]

would i always do that

[drifting sierra]

3x^2 - 12x - 36 is not equal to x^2 - 4x - 12, just like 9 is not equal to 3...

9 is equal to 3*3 however

[sacred plank]

ahh okay ty!

.close

[devout snow]

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[smoky gyro]

how do i do this using u-sub? ive tried doing weird subs but ive been getting no where $\int\frac{x+3}{x^{2}+6x+2}dx$

[prime egret]

is that meant to be 6x

[smoky gyro]

oh ye

[woven radish]

water beam

[prime egret]

Let (\mr u = \mb{x^2 + 6x+2}) should work

[woven radish]

Pure

[prime egret]

But you can do this immediately by inspecting

( \df12\il{}{}{\df{2x+6}{x^2+6x+2}x)

( \df12 \il{}{}{\df{2x+6}{x^2+6x+2}}x )

[woven radish]

Pure

[smoky gyro]

hm

oh

i see

yeah

[prime egret]

[smoky gyro]

got the sub

[prime egret]

\fact{In general
[\il{}{}{\df{a\mr{f’(x)}}{\mb{f(x)}}}x = a\log\abs{\mb{f(x)}} + C]
}

[smoky gyro]

F

[prime egret]

\fact{In general
[\il{}{}{\df{a\mr{f’(x)}}{\mb{f(x)}}}x = a\log\abs{\mb{f(x)}} + C]
}

No response

Not even time out

[smoky gyro]

https://tenor.com/view/walter-white-walter-crying-walter-dying-gif-20964719

the bot rn

[prime egret]

lol it’s down again

[smoky gyro]

@woven radish

grrrrrrrrr

[prime egret]

\fact{In general
[\il{}{}{\df{a\mr{f’(x)}}{\mb{f(x)}}}x = a\log\abs{\mb{f(x)}} + C]
}

[woven radish]

Pure

[prime egret]

lol okay

@smoky gyro

Works now

[smoky gyro]

woohooo

whats the a here

in my case

[prime egret]

Constant

1/2

[smoky gyro]

noice

[prime egret]

( \mo{\df12} \il{}{}{\df{\mr{2x+6}}{\mb{x^2+6x+2}}}x = \mo{\df12}\log\abs{\mb{x^2+6x+2}} +C)

[woven radish]

Pure

[smoky gyro]

oo i see

interesting

[prime egret]

[smoky gyro]

thanks for the help

[prime egret]

[smoky gyro]

.close

[devout snow]

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[bold venture]

i don’t understand how to set up the equation

[mystic oak]

To begin, it's the distance between the man and the lightpost that is changing by 7' per second, not the length of the shadow

[vagrant steeple]

write a function for his shadow then differentiate and put values

[warm lantern]

if the comments by joachim and ayushch are not clear enough, I'm going to modify your diagram a bit

Now your goal is to try and define the function S(x)

where x is the distance between the person and the lamppole

and S is the length of the shadow

[devout snow]

@bold venture Has your question been resolved?

[devout snow]

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[amber oriole]

Hi. I don't understand how to find the partial derivative of a nultivariable function, because I don't know which variable has precedence as being treated as a constant. For example, consider: f(x,y) = x^3 + x^2 * y^3 - 2y^3. If we want to differentiate with respect to x, which y-variable is considered a constant and which isn't?

[wintry jasper]

all of them are treated as constants

so the derivative would become 3x^2 + y^3 * 2x - 0

[amber oriole]

What about x^2 * y + 2y: With respect to x; why is the stand-alone y equal to 0 and the equation = 2xy?

[wintry jasper]

imagine instead of y

it's a constant a

x^2 * y + 2y -> ax^2 + 2a

the derivative would be a * 2x + 0 right?

[amber oriole]

Yes

[wintry jasper]

now do the same thing with y and you'll get 2xy

just treat y as a constant always

(if you are taking the derivative wrt x)

if the function is f(x, y) = xy
then the partial derivative wrt x is 1.y
if the function is f(x, y) = x + y
then the partial derivative wrt x is 1 + 0

[devout snow]

@amber oriole Has your question been resolved?

[bold venture]

i got ds/dt = 21/8 ft/s

@vagrant steeple @mystic oak

[warm lantern]

very well done mate

[bold venture]

thank you

[mystic oak]

!noping

[devout snow]

Please do not ping individual helpers unprompted.

[mystic oak]

but well done

[bold venture]

oop

[warm lantern]

yeah you can dm us instead

[mystic oak]

also you lost your channel due to inactivity

[warm lantern]

actually wait I shouldn't say that

[bold venture]

also is 40 just not used

x = 40ft

[warm lantern]

no it seems it wasn't used

because S is a linear function

[mystic oak]

Make a new help channel if you have more questions, this one is occupied by someone else now

[warm lantern]

also ^

[amber oriole]

.close

[devout snow]

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[bright juniper]

Need help

[bright juniper]

With 179

[devout snow]

@bright juniper Has your question been resolved?

[bright juniper]

<@&286206848099549185>

[forest rapids]

What did you try?

[bright juniper]

I don’t even know where to start

<@&286206848099549185>

[restive river]

bonjour j'ai des soucis en maths

[bright juniper]

need help with 179

[restive river]

I SPEAK ENGLISH

[smoky eagle]

ask in a different channel

#❓how-to-get-help

[forest rapids]

What did you try?

[bright juniper]

i cant start

[restive river]

https://tenor.com/view/clap-clapping-applause-applaud-cheerleader-gif-12591824

[forest rapids]

There are lots of roots. Any way to try to make them go away?
HOw did you do the other ones?

[bright juniper]

square them?

[forest rapids]

Worth trying

[restive river]

ok ok faster

[forest rapids]

? You datoo3 to think faster? Is that fair?

@bright juniperDid that look as if it is going to be useful?

[restive river]

yes

[forest rapids]

$\sqrt((4+\sqrt(16-x))/2) + \sqrt((4-\sqrt(16-x))/2) = \sqrt(4+\sqrt(x)) + \sqrt(16-x)$

[woven radish]

G. Spark

[forest rapids]

What does it look like now?

$\sqrt((4+\fract{\sqrt(16-x))}{2}) + \sqrt((4-\sqrt(16-x))/2) = \sqrt(4+\sqrt(x)) + \sqrt(16-x)$

My TEX is bad

[bright juniper]

(a+b)^2??

[forest rapids]

Yep, you will get some terms like that

[bright juniper]

ok so first what do i do?

[forest rapids]

Well, you could try squaring the left side and squaring the right side.
If z = u, then z^2 = u^2, so still correct

[bright juniper]

can you try solving it?

[forest rapids]

YOu mean, can I do it to see whether it will work?

[bright juniper]

yea

[forest rapids]

The left hand side looks to have two terms that are really similar.

[bright juniper]

so worth doing?

first i multypyd the equation by 2

now i will square it

[forest rapids]

$\sqrt((4+\frac{\sqrt(16-x))}{2}) + \sqrt((4-\frac{\sqrt(16-x))}{2}) = \sqrt(4+\sqrt(x)) + \sqrt(16-x)$

[woven radish]

G. Spark

[forest rapids]

$2\sqrt((4+\frac{\sqrt(16-x))}{2}) + 2\sqrt((4-\frac{\sqrt(16-x))}{2}) = 2\sqrt(4+\sqrt(x)) + 2\sqrt(16-x)$

[woven radish]

G. Spark

[forest rapids]

Did the 2 help?

[bright juniper]

it got rid of the /

in the firs equation

[forest rapids]

Inside the root?

[bright juniper]

oooooooooo

forgot that

[forest rapids]

I'd like to see what you got one just one term - we should check that one

[bright juniper]

so multypying by 2 is not good

[forest rapids]

Not seeing that as helping at this time.

[bright juniper]

ill just square the first side

[forest rapids]

$\sqrt((\frac{4+\sqrt(16-x))}{2}) + \sqrt((\frac{4-\sqrt(16-x))}{2}) = \sqrt(4+\sqrt(x)) + \sqrt(16-x)$

$\sqrt((\frac{4+\sqrt(16-x))}{2}) + \sqrt((\frac{4-\sqrt(16-x))}{2}) = \sqrt(4+\sqrt(x)) + \sqrt(16-x)$

[woven radish]

G. Spark

[bright juniper]

Got this

By squaring the firs side

[forest rapids]

$(\frac{4+\sqrt(16-x)}{2}) + (\frac{4-\sqrt(16-x)}{2}) + 2\sqrt(\frac{4+\sqrt(16-x)}{2})\sqrt(\frac{4-\sqrt(16-x)}{2}=$

[woven radish]

G. Spark

[bright juniper]

yes i got that

[forest rapids]

wrong before - seems getting rid of the 2 might have been a good idea; but needed to compensate for the root.
So, squaring the RHS?

[bright juniper]

i get get rid of the 2 now

no?

[forest rapids]

How?

Just checking.

[bright juniper]

its not in the root anymore

[forest rapids]

YOu have to do the same thing to the LHS and RHS. What did you do to get rid of the 2?

[bright juniper]

idk man i give up

im not solving this

[forest rapids]

$\sqrt(\frac{4+\sqrt(16-x)}{2})) + \sqrt(\frac{4-\sqrt(16-x)}{2}) + = \sqrt(4+\sqrt(x)) + \sqrt(16-x)$

[woven radish]

G. Spark

[forest rapids]

So the two is under the root.

[bright juniper]

square it

then get rid of the 2

[forest rapids]

Either way
square first, or 2 first.

[bright juniper]

can yoou solve this you think?

[forest rapids]

$\sqrt(\frac{4+\sqrt(16-x)}{2}))=\frac{\sqrt(4+\sqrt(16-x))}{\sqrt(2}$

[woven radish]

G. Spark

[forest rapids]

So what do we do to get rid of the root(2)?

[bright juniper]

idk

[forest rapids]

Beacuse if we do that, we must do the same thing to all terms, including the RHS

a/b ; how do we get rid of the b?

a/root(2) ; How do we get rid of the root(2)?

[bright juniper]

.ცლოსე

.close

[devout snow]

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[strong willow]

number 48

my work

which doesn't works

[untold lodge]

wut u mean it dosnt works?

[strong willow]

terms does not cancel

@untold lodge

.close

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[restive river]

Pls help

[restive river]

<@&286206848099549185>

[final vine]

yo

[red blade]

I can help.

[wintry jasper]

whats your question?

[red blade]

That’s what I was about to ask.

[restive river]

I need to work out the equation in the form of y=mx+c, where M and C are integers or fractions in their simplest forms.

[restive river]

.

[red blade]

Do you know what the question is asking?

[wintry jasper]

is this for a test

[red blade]

I’m trying to figure out what’s getting you stuck.

[wintry jasper]

if it is we're not allowed to help

[restive river]

Yeah

[restive river]

It's homework

[red blade]

Looks like homework.

[wintry jasper]

nvm its not

yea

[red blade]

Are you aware for what m and c mean here?

[restive river]

Yes

[red blade]

Do you know what c is?

[restive river]

M is gradient

C is y intercept

[red blade]

Yes, but do you know the value of c?

[restive river]

Yeah

[red blade]

So your question was how to find m, correct?

[restive river]

No

[red blade]

What is stopping you from answering the problem, then?

[red blade]

Ohh you’re trying to simply a fraction.

What’s the form you currently have.

[red blade]

Yes, but what is the answer you got? You do know how to find the answer, right?

[restive river]

For image 1, I have 6x - 4

[red blade]

How did you get 6x-4?

[restive river]

image 2, I have 3x + 5

[red blade]

How did you get that?

[restive river]

😶

[red blade]

I do not know what you mean by this.

[restive river]

The gradient for Question 3D is 6

The gradient for Question 3E is 3

I give up

[red blade]

Ah, I see. Recount the squares in 3D

I think you just miscounted.

[restive river]

Oh ok

Gradient = 5?

[red blade]

Those gradients are correct. Now, how did you get the c values?

[restive river]

Question 3D, C = 4

Question 3E, C = 5

[red blade]

Where did you get 5?

[red blade]

I see where you got 4, but what about 5?

Oh the numbers weren’t visible to me. Yeah, that’s right.

[restive river]

Now please simplify it somehow 😭

[red blade]

Oh, you got the gradient wrong.

[restive river]

Please just tell me the amswer

I need to revise for exams in a bit

[red blade]

You’re not going to learn it if I just tell you the answer.

Take the intersection with x=1

[restive river]

Omg

I'll learn it at school

[sharp garnet]

Chixen is right

[red blade]

You clearly didn’t.

[restive river]

I did but I forgot

I came here for a reminder

[red blade]

I’m reminding you now

[restive river]

Not a 20 minute walkthrough

[red blade]

It’s the most effective way to teach. People remember things so much better when they figure it out themselves.

[restive river]

I don't care

I could have been doing something else but you've been taking ages to type and 20 minutes gone by.

I just want the answers so I can leave the group

[red blade]

I’m not that slow of a typer. I’m just on my phone.

Anyway, I’m not just going to give you answers. I don’t think anyone you’ll find in here will.

[restive river]

Waste of time

🖕

Bye bitch

[red blade]

Alright.

[sharp garnet]

Damn

[thin inlet]

people are ungrateful as hell lmao

[sharp garnet]

@red blade I'm new here, do you guys have to deal with stuff like these on a regular basis?

[red blade]

I mostly talk in #discussion

But from my experience, most people are grateful.

[sharp garnet]

Atleast that's nice to hear

[red blade]

A lot of people just want the answer initially, but when you walk them through it, they thank you. This guy was just a dick.

[sharp garnet]

Yeah I agree

I really liked your approach of helping

[devout snow]

@mystic heath Has your question been resolved?

[devout snow]

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[devout snow]

Please don't occupy multiple help channels.

[timber pebble]

oh, yea

.close

[devout snow]

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[dapper wyvern]

hey

[dapper wyvern]

how must i solve these two

[red blade]

We need more context. How are i,j,k defined?

Is this linear algebra?

[dapper wyvern]

that is all there was in the test

in the questions i mean

[dapper wyvern]

correct it is linear

[red blade]

Oh ok. “Value” means the magnitude of the vector, yes?

Or quaternions?

[dapper wyvern]

yea magnitude

[dapper wyvern]

i dont know what that is i probably didnt study it yet

[red blade]

Ok then no. Not quaternions.

[dapper wyvern]

yea

[red blade]

Wait but we’re multiplying vectors? It doesn’t specify cross or dot product. What are i j and k??

[dapper wyvern]

well the first one is getting multiplied in brackets so i think that means its dot product

[dapper wyvern]

they're unit vectors

if I recall correctly

[red blade]

but how are you multiplying them?

What vector multiplication are you doing?

[dapper wyvern]

do you mean the type of multiplication if yes then

its the scalar or the

. product

[red blade]

Ok dot product.

[dapper wyvern]

ya

[red blade]

When finding the magnitude of the product of vectors, the magnitude function actually distributes among multiplication.

[dapper wyvern]

oh

[red blade]

|ab|=|a| |b|

[dapper wyvern]

alright

[red blade]

So for |3j(k+i)| you get 3 |j| |k+i|

[dapper wyvern]

yea

[red blade]

Since i, j, and k are unit vectors, |j|=1, and addition can be done with pythagorean theorem. |i+k|=sqrt(|i|^2+|k|^2)=sqrt(1^2+1^2)=sqrt(2)

[dapper wyvern]

alright yea

[red blade]

You should get it from here.

[dapper wyvern]

Thanks

what must i do in the second one

.close

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[ivory quail]

In sin2x = 0.8, why doesn’t 2.68 rads work when 2.68 rads is in the second quadrant, thus qualifying CAST rule?

[trail eagle]

2.68 is in the second quadrant but 2*2.68 is in the fourth quadrant ..

[devout snow]

@ivory quail Has your question been resolved?

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[frigid creek]

so we have a right triangle

[frigid creek]

the smaller sides are 6 and 2

we need to find the angles between the medians of the sharp angles

[rugged sparrow]

are you allowed to use co-ordinate geometry

[devout snow]

@frigid creek Has your question been resolved?

[frigid creek]

no

[frigid creek]

forget it i alr closed the

book

too sleepy for allat

.close

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[muted lintel]

tan(v) = 1/2

Find the exact value for sin (v) for 0°<v<90°

[misty crest]

draw a right triangle

[muted lintel]

alr

[misty crest]

and label one of the angles as v

then u know tan v equals opposite side/adjacent side

which is 1/2

then from there u should be able to find some

sine*

using pythagoras

[muted lintel]

ooo i get it

thanks

ill respond if i get it

nope, im still kinda stuck

ill work on it more and ill ping you if im still stuck

Thank you so much @misty crest

ily

.close

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[inner thunder]

I don't understand why this problem doesn't seem to take into account N(0) when calculating the population at t >= 0

[inner thunder]

I thought the formula was N(t) = N(0) + integral N(x)

In this case, since the initial population was 200, wouldn't N(0) be 200?

[hollow kelp]

hi

im new to this server

where can i ask for help?

[inner thunder]

Usually you ask in an available help channel and it'll assign the channel to you

[hollow kelp]

OH

TY

[inner thunder]

Still trying to figure out why they seem to disregard the initial 200 at N(0)

Even in the book, a similar problem doesn't use +C

So I'm wondering if this is just a mistake on the question

[acoustic leaf]

they don't disregard it, they set N(0) = 200 = (antiderivative + C) and solve for C

[acoustic leaf]

the book seems to do something similar but gloss over the step where they actually do it. notice how the 100 somehow turns into 1000; the 1000 is C and can be found the same way as shown above

[inner thunder]

ohhh I see

thanks

.close

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[analog furnace]

where does the 1/((n+1)(n+2)) come from?

[analog furnace]

an = 1 / (n^2+n)

[devout snow]

@analog furnace Has your question been resolved?

[analog furnace]

I guess the question is more how to find an+1

oh i think i got it

I think it is an = 1/n(n+1) therefore, an+1 = 1/((n+1)(n+1+1) which = 1/((n+1)(n+2)

but im trying to think why that works

.close

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[devout snow]

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[high forum]

how to solve this to make
r.h.s=l.h.s
without multiplying with cos²x

[high forum]

by two sides if u can

[devout snow]

@high forum Has your question been resolved?

[wraith horizon]

from the left side, I'd start by writing tan^2 as sin2/cos2 and sin2 +1 in terms of cos2 and cancel

[devout snow]

@high forum Has your question been resolved?

[devout snow]

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[short iron]

[\int_{0}^{\frac{\pi}{4}}\sqrt{(2\sin x)^2 - (2\cos x)^2}dx]
[\int_{0}^{\frac{\pi}{4}}\sqrt{(-2( \cos(2x))}dx]

[woven radish]

dopediscorduser

[short iron]

Stuck on integrating this

Ignore redundant parenth under radical

[trail eagle]

I don't think there is a primitive to this with elementary functions.

[short iron]

Maybe I made a larger mistake then

[trail eagle]

Is it the arclength?

[short iron]

Asked to find arclength x=2cosx
y=2sinx over 0 to pi/4

Yes

$x' = 2sinx$
$y' = -2cosx$

[woven radish]

dopediscorduser

[trail eagle]

On

Hh

The minus sign

Shouldn't be there.

This is just a circle so you can sort of guess the answer, but it's (y')^2 so no minus

[short iron]

Oh whoops

[trail eagle]

This makes the cos^2 and sin^2 cancel to 1

[short iron]

$\int 1 dx$

?

[woven radish]

dopediscorduser

[short iron]

Like that

Except not indefinite

[trail eagle]

Well you'll get a factor of 2 I think

[short iron]

Factor out two?

[trail eagle]

No but once you have (2cos(x))^2 + (2sin(x))^2

That gives

4

Then sqrt

So 2

[short iron]

Oh I see

So integrated it'd just be 2x?

[trail eagle]

Yes

[short iron]

Alright, thank you for your help

.close

[devout snow]

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[proud hemlock]

trying to solve this ODE but i don tunderstnad how to find the particular integral

[proud hemlock]

roots are -j and +j

so CF is yH = Acos(t) + Bsin(t)

but where do i go from here to get particular integral

[worthy talon]

Do you know how yP will look like?

[proud hemlock]

no

im guessing it will be in the form of the CF?

[worthy talon]

Hold up this isnt my native language i gotta polish up on the terms

[proud hemlock]

ok nw

[worthy talon]

You know that your solution will be in the form of yH+yP right @proud hemlock

[proud hemlock]

oh the general solution yh

[worthy talon]

Its easy to find the yH from the characteristic equation

p²+1=0

[proud hemlock]

yep

[worthy talon]

You did that already

Have you ever studied the method of undetermined coefficients

For solving ODE

[proud hemlock]

sounds familiar

probably used it but no idea what it is

[worthy talon]

From the non-homogeneous part of your DE, you can take a good guess on what your yP will look like

[proud hemlock]

i dont know

not a cue

clue

[lusty sapphire]

If the particular solution would be a homegenous solution, try multiplying the particular solution by t

[proud hemlock]

what is the particular solution

[lusty sapphire]

That is, the particular solution should have been of the form $x_p(t)=C_1\cos{t}+C_2\sin{t}$, but that's your homegenous solution. So try $x_p(t)=C_1t\cos{t}+C_2t\sin{t}$, and see which values of $C_1$ and $C_2$ will be your particular solution.

[lusty sapphire]

It's what I think you meant by particular integral. Unless particular integral means something else.

[proud hemlock]

yes i fked the terminology

hhmm

[woven radish]

SWR

[proud hemlock]

no

this is og question

theyr on about particular integral aswell

so how do u find the particular integral boss

[lusty sapphire]

Looks like they mean the same thing

[proud hemlock]

okay sorry boss my mistake

but

so u determined the particular integral form from the CF?

looks to me like u just slapped the CF and gave it a new name and mutlipled by t am i correct

[lusty sapphire]

yes but its an intuition strategy

[proud hemlock]

ok new question

this one boss

[lusty sapphire]

same idea

[proud hemlock]

the CF is Ae^-t + tBe^-t

[lusty sapphire]

Then multiply by t again

[proud hemlock]

u sure bout that boss

[lusty sapphire]

so particular integral is of the form $y_p(t)=C_1t^2e^{-t}$, and solve for $C_1$

[woven radish]

SWR

[proud hemlock]

hold onn

how did u get that

from that CF?

herro??

[lusty sapphire]

get what

[proud hemlock]

its correct but idk how u got it

that PI

because using ur previous method it would be t(Ae^-t + tBe^-t)

[lusty sapphire]

It's an intuition strategy I learned from my class

Say you have some generic ODE of the form $Ay''+By'+Cy=f(x)$

[woven radish]

SWR

[lusty sapphire]

The particular integral will depend on $f(x)$, and you can take guesses depending on what $f(x)$ is.

[woven radish]

SWR

[lusty sapphire]

If $f(x)=\alpha e^{\beta t}$, then PI is of the form $\gamma e^{\beta t}$, and you need to solve for $\gamma$.

[woven radish]

SWR

[proud hemlock]

hmmm

and the other form is sincos

pls no ban

but yh if it was 8cost

then u use the Acos() + Bsin()

[lusty sapphire]

If $f(x)=\alpha \cos{\omega t}+\beta\sin{\omega t}$, then the PI is of the form $\gamma \cos{\omega t}+\delta\sin{\omega t}$, and you'll need to solve for $\gamma$ and $\delta$.

[woven radish]

SWR

[proud hemlock]

so roots have nothing to do with the assumpting that u make

[lusty sapphire]

But here's the problem. Suppose the particular solution is one of the homogenous solutions. Then you need to multiply PI by t to make it independent.

[lusty sapphire]

yes and no

[proud hemlock]

yh im rlly good at that part

no problem in that regard

[proud hemlock]

huh

[lusty sapphire]

I need the roots to get the homegenous solution. CF as you call it

Depending on the order of your DE, you may end up with more than one CF

If your PI is a constant multiple of a CF, then you need to multiply your PI by t to make it independent from the CF

[proud hemlock]

strictly speaking max 2 orders

[lusty sapphire]

You did a similar thing here. Your CF was e^-t twice, but you must have two distinct CFs since the DE's order was 2. So you multiplied one of the CFs by t. And because your PI is equal to your CF, and because one of the CFs is aleady multiplied by t, you need to multiply the PI by t a second time to make it distinct from both CFs

[proud hemlock]

yep yep ypepy epypepypyp

u r a big yep

thank you very much sir

i don tknow how i can repay

you

thank you sir

.close

[devout snow]

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[copper bramble]

.reopen

part b?

[gleaming socket]

Parallel lines and angles

[devout snow]

@copper bramble Has your question been resolved?

[devout snow]

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[static ether]

guys welp i have an exam tomorrow and i cant figure out the last 2 lines

[neon frost]

The answer sheet says there is only one solution, pi/2. What am I doing wrong?

[crisp niche]

no al si wrong

plz use substitution t = sinx

then you get quadratic equation for t

solve it

and then you get value fo rsinx

unless you can see trinomial at once, but if not use t = sinx

[neon frost]

t^2-2t+1?

[crisp niche]

yes

= 0

ofc

[neon frost]

OH IT WORKS

[crisp niche]

🙂

$t^{2}-2t+1=\left( t-1 \right)^{2}=0\Leftrightarrow t=1$

[neon frost]

THANK YOU SO MUCH. The factors were (t-1)^2

[woven radish]

Joanna Angel

[crisp niche]

yvw 🙂

then sinx = 1

and you find one solution, which is super easy

on yoru nterval

[neon frost]

This discord is literally a lifesaver. 🤣 I'm studying for my trig test tonight and there's no one to help me

[crisp niche]

maybe ppl thought it was too easy to deal with lol

🙂

[devout snow]

@static ether Has your question been resolved?

[devout snow]

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[devout snow]

@restive river Has your question been resolved?

[devout snow]

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[dreamy bobcat]

Hello guys - i have been tasked with finding the derivative of (6-y)/(6y^2+x)

[dreamy bobcat]

and substituting for (1,1)

this is what i got for the derivative

This is what i got when i substituted x = 1 and y = 1

is my method/working correct?

[devout snow]

@dreamy bobcat Has your question been resolved?

[dreamy bobcat]

<@&286206848099549185>

[devout snow]

@dreamy bobcat Has your question been resolved?

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[quartz shale]

the 1000 is actually 10000

but either way

what i have rn is P=I(2)^(t/n)

so 10000=2^(t/40)

i think i went wrong somewhere but idk

[dense jay]

a bit yeah

start off with the doubling every 40 minutes, so t=2/3
if P=A* 2^{kt}
2=2^{2/3 k}

find k

[devout snow]

@quartz shale Has your question been resolved?

[quartz shale]

So then 2/3/(log 2/log 2) =k

Right

[dense jay]

seems like a complicated process you used there, but no

3/2/(log 2/log 2) =k

i would just directly compare the exponents though, since theyre base both on 2 anyway

1=2/3 k

[devout snow]

@quartz shale Has your question been resolved?

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[restive river]

Can anyone help me out in this trigonometry question

[surreal mulch]

what’s the question

[restive river]

I dont know how to approach this 😭😭

I dont even know if I should have filled out the angles or are they useless in solving the question

[surreal mulch]

do you know the answer

because i got 11

here’s how

[restive river]

It said 13.0??

But please explain!

Any approach is better then none

[surreal mulch]

oh then idk

ok here’s what i did

i’ll write it out

@restive river

[restive river]

OHHH

You used sine law?

[surreal mulch]

yea

[restive river]

AYYY

Thank you for your help!! 😄

[surreal mulch]

love sin law❣️❣️

[surreal mulch]

ofc!

[restive river]

Its so baby ❤️

.close

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[ornate jacinth]

help

[ornate jacinth]

<@&286206848099549185>

[jade oak]

Both intercepts are same distance from vertex

[ornate jacinth]

-2

[jade oak]

Yeah

[ornate jacinth]

ohh

thanks

[devout snow]

@ornate jacinth Has your question been resolved?

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[rough night]

How do I do this

[restive river]

Please show your working

[rough night]

I dont even know how to get started

[restive river]

Alright. Do you know reimann sum?

[rough night]

Yeah I do

[restive river]

Alright

First divide both sides by 3

[rough night]

Okay

[restive river]

Simplify

[rough night]

Okay

Im not sure how to simplify this

[restive river]

12/3 = 4

[rough night]

The f is kinda throwing me ff

off

[nova sun]

how do u define the function ?

[restive river]

I'm assuming it to be reimann integrable

[rough night]

X is 4+4/j

[restive river]

x_j = 4 + 4j/n

No the 3 is gone in division with 12

[rough night]

Ohh

Okay

[restive river]

Yes

[rough night]

oops

[restive river]

It's alright I understand

[restive river]

Now you need to convert this to an integral from 4 to 8

Can you convert this expression into a reimann sum?

[rough night]

No

I dont know how

[restive river]

Well what's 8-4

[rough night]

Can I just remove the f

4

[nova sun]

so u have to find the primitive of the function

[restive river]

You can't denotes the function

[nova sun]

no

[rough night]

Okay

[restive river]

And you're integrating from 4 to 8

[rough night]

Okay so 4 is the mid point?

[restive river]

Recall the definition of reimann sum (can be left or right depending on behaviour of f(x))

[rough night]

Yeah

[nova sun]

so youy have to calculate F(8) - F(4)

[rough night]

Or I need to find F

Which I dont know

[nova sun]

you can calculate the derivative of f

[crisp niche]

no need to know neither f, nor f'

[rough night]

f is 4+j/n?

[restive river]

$\frac{b-a}{n} \sum_{j=1}^{n} f(a + \frac{(b-a)j}{n})$

[woven radish]

𝓘 , 𝓣𝓵𝓸𝓻𝓽𝓱

[restive river]

This is the definition of reimann sum

[rough night]

Okay so b is 8

and a = 4

[restive river]

Yep

[nova sun]

well…

Thats easier

[restive river]

And you write 4 as 8-4 in your expression

[rough night]

Wait I get the same as this

[restive river]

And when you take the limit as n goes to infinity, it converts to a reimann integral which you want to evaluate

[restive river]

Yep

Even the 4 outside

[nova sun]

but, then, how to evaluate the riemann integral ?

[restive river]

Well this is the definition

You take the limit

[restive river]

Uh what

[rough night]

You said n goes to infinity

Oh wait no

[nova sun]

The limit equals to 3

[rough night]

So its gonna be

f(4)

[crisp niche]

no 3

3 has been reduced

[nova sun]

the answer is in the question

[crisp niche]

no

[nova sun]

?

[crisp niche]

$\lim_{n \to \infty } \sum_{k=1}^{n}\frac{12}{n}f\left( 4+k\cdot \frac{4}{n} \right)=3\Leftrightarrow \\3\cdot \lim_{n \to \infty }\frac{8-4}{n}\sum_{k=1}^{n}f\left( 4+k\cdot \frac{8-4}{n} \right)=3\\\int_{4}^{8}f\left( x \right)\text{ }dx=1$

[woven radish]

Joanna Angel

[restive river]

Yes but you are also adding an infinite amount of infinitesimal strips (don't forget we are summing these strips)

So it's equal to an integral

[nova sun]

hum you just created a 3 out of nowhere

[restive river]

It's factored out

[crisp niche]

12 = 3*4

[nova sun]

oh yeah i didnt see

[rough night]

.close

[devout snow]

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[devout snow]

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[restive river]

$$\text{prove that if } {\epsilon_i}_{i=1}^n \text{is basis for } V^n, \text{then } \det(\epsilon_1, \ldots, \epsilon_n) \neq 0.$$ How can i do that? 🥺

[woven radish]

Maria

[devout snow]

@restive river Has your question been resolved?

[devout snow]

@restive river Has your question been resolved?

[noble niche]

Tbh i would like to say something smart

But im so tired that i give up

The lame out of the blue solution is to

Prove by contradiction

That if it were 0 then the derived formula of determinant for 2x2 matrices for example wouldnt be the same as it is

Wait nvm

I figured it out im jusr too tired to brain function

[restive river]

why not?

[noble niche]

so precisely

Ooh

Hold up

[noble niche]

Nvm that

It wouldnt be0 because

e1,...en are column vectors of identity matrix

[restive river]

i had an idea to choose such set (v1, ... , vn) that det(v1, ... , vn) = 0 and prove that it is linearly dependent, but it doesn't seem to be good 😦

[noble niche]

Wait

Look

[restive river]

yes

[noble niche]

The action of a linear transformation on j'th basis vector is given by: T(ej) = Aijei (summation over repeated indices) , so if Aij = Iij , then for example T(e1) = Ii1ei , which holds true because Iij = 1 iff i = j , so I11 = 1 , otherwise Ii1 = 0

Doesnt this show that the column vectors of Iij are the basis

det(I) = 1 is a property

[restive river]

is Aij scalar?

[noble niche]

well

yes

Iij = 1 iff i = j

Otherwise its 0

[restive river]

what is l in your formula?

[noble niche]

Identity matrix

[noble niche]

This is a property of determinants

It has to satisfy that

[restive river]

yes

[noble niche]

Well i have to go so i cant finish my cerebration

But try showing that det(e1,...en) = det(I)

That seems to be the main point here

Ig

Or maybe not*

<@&286206848099549185>

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[devout snow]

@restive river Has your question been resolved?

[noble niche]

so @restive river still got no idea?

[devout snow]

@restive river Has your question been resolved?

[wild helm]

If you have a basis for a vectorial space V, then $\forall x \in V$ we know that x can be expressed as a linear combination of the vectors of the basis and THIS COMBINATION IS UNIQUE

[woven radish]

milo_schwartz

[wild helm]

In conclusion x is the solution of a nxn linear system where the matrix of coefficient is the same matrix you have to show has determinant not equal to 0

But is well know that if a nxn linear system has a unique solution then it must be determinant not 0

So there is really nothing to prove there

[devout snow]

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[orchid leaf]

I want to complete this table of types of proofs, how they are stated in symbolic logic, the first step of the proof and the goal you have to reach with the proof. For the types and statements I think that I have included most types of proofs and their respective statement in symbolic logic. I am not sure about contradiction. Is the statement of a contradiction the one I have written?

[orchid leaf]

I want to fill in the rest of the table so whoever can help it would be greatly appreciated

P(x) and Q(x) are part of the implication P(x) => Q(x)

Well it depends on the type of proof but P(x) and Q(x) are arbitrary statements

<@&286206848099549185>

Is my question so nonstandard as to not being able to respond to it, even after I have pinged helpers?

[devout snow]

@orchid leaf Has your question been resolved?

[orchid leaf]

<@&286206848099549185>

[forest rapids]

Yes. And time of day. People are sleeping.

Been a while since I did any formal propositional stuff. but, yes I was struggling to understand what you wanted.
It seems a good idea, but I can't quite get what the idea is, or the structure, so I thought to leave it to others younger and more current in the field. Sorry

Hold the faith. Someone who knows all about this will appear and viola - help.

[frosty cradle]

ask in #proofs-and-logic

[devout snow]

@orchid leaf Has your question been resolved?

[devout snow]

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[hexed apex]

So I got this differential equation $\frac{d^2}{dt^2}e^t^a^n^x+\frac{1}{(1+sin(x))^4}=0$ and have been told to linearize it, so I now got $\frac{d^2}{dt^2}(1+x)+(1-4x)=0$

[devout snow]

Please don't occupy multiple help channels.

[woven radish]

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

[hexed apex]

I can turn the second equation into $\frac{d^2}{dt^2}(1+x)-4x=-1$