help-36

[amber holly]

1/x * x = 0 * x

1 = 0 ?

[ashen shadow]

Nope

[amber holly]

Hence the contradiction

Meaning there are no solutions

[ashen shadow]

Ok but if it was 1/x=1 or any number less than or greater than zero then there would be only one solution right?

[amber holly]

1/x = k has a solutions for all k except 0

So yeah

This comes from the fact that 1/x = k is equivalent to 1 = kx

And kx = 1 has a solution iff k is nonzero

Wtf is wrong with my typing

[ashen shadow]

So in this graph we could see that if
x=0 then the solutions would be -infty and +infty but if this fraction has only one solution then x=0 is not possible

Can I explain 1/0 with this contradiction

[amber holly]

Wdym by explaining 1/0?

[ashen shadow]

Means explaining, 1/0 is undetermined

[amber holly]

Assume 1/0 = a
Then, by definition, 1 = a * 0

But that leads to 1 = 0

By the definition of 0

[ashen shadow]

Hmmm

And what about

$\frac{1}{0^{-}}$

[soft zealot]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[amber holly]

How is you 0^- defined?

[ashen shadow]

And
$\frac{1}{0^{+}}$

[soft zealot]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[ashen shadow]

It is used in limits

[amber holly]

Ik, but you need to define it

[ashen shadow]

A number which is less then 0 but really really really close to 0

[amber holly]

If you mean $\lim_{x\to{0^-}}\frac{1}{x}$, then it's $= -\infty$, but technically the limit doesn't exist

[soft zealot]

A Lonely Bean

[ashen shadow]

Means it is undefined too

[amber holly]

Yeah

[ashen shadow]

Ok thanks

Cause my teacher

Asked me when I was at quiz

She asked that

What's 1/infinity

I said undefined

But she said that it is zero

And I got negative marking too

🥲

Thank you btw

[amber holly]

We don't define infinity so 1/infinity isn't defined as well

But the limit of 1/x as x approaches infinity or minus infinity is 0

[ashen shadow]

Oh ok

Means 1/infty =0

And my another teacher said 1/0=infinity

[ashen shadow]

I knew it was totally wrong

[amber holly]

And infinity isn't defined

As a number

You can say that the limit is infinity

But the limit technically still does not exist

[ashen shadow]

Um ok

I could understand

Tnx

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[quaint mauve]

Hi, how can use a taylor polynomial to find an aproximate solution of an equation

[quaint mauve]

I had x*e^y-ln(y)=0 and I had to find the second order taylor polynomial for x_0=0

I got P_2(x)=1+e*x+(e^2)/(2) x^2

and I have to find a solution for the equation e^y - 50*ln(y)=0 with it

[shell mountain]

What have you tried

[quaint mauve]

nothing yet, I dont really know how they are related to each other

I mean, both equations, the one I found the taylor for, and the second I have to solve are pretty similar

[shell mountain]

Well, what happens if you multiply xe^y - ln(y) = 0 by 50?

[quaint mauve]

ye I get sth similar

[shell mountain]

What do you get?

[quaint mauve]

but Ill have 50*x*e^y-50ln(y)

[shell mountain]

And we want e^y - 50ln(y)

[quaint mauve]

oh so for x=50?

[shell mountain]

So what value of x makes 50xe^y = e^y

[quaint mauve]

x=1 sorry

[shell mountain]

x = 1 gives 50e^y - 50ln(y)

[quaint mauve]

oh wait Im dumb

dont have to answer that fast

1/50

[shell mountain]

Yes

So if you multiply the taylor polynomial by 50 and plug in x = 1/50, you'll have your approximate value of y

[quaint mauve]

so solve the polynomial for 1/50=x

oh ok, Ill try that

makes sense, ty

ye solved, thx again

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[tranquil pine]

for a) I would just set up the characteristic polynomial

right

from which I got x^2 - x - a

but for what values of that polynomial does it have a double root

look at its discriminant

or complete the square

ah ok

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[rugged yew]

Does anyone know how to do this?

[rugged yew]

We have to use the properties of logarithms to solve the problem

But I don't know how to solve this specific question

the "e" is throwing me off

[tulip coyote]

Don't forget that $\ln=\log_{e}$

[soft zealot]

chartbit

[rugged yew]

okay

[stoic mural]

ln(a)=b means
e equals a if taken to the power of b

so in our case
e equals e^4.5 if taken to which Power?

from this we can also see that e^x and ln(x) are inverse of each other

[rugged yew]

i still don't really get it

[rugged yew]

i understand that those r inverses

[stoic mural]

if f and g are inverse of each other then f(g(x))=x

[rugged yew]

ok

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[loud mountain]

This is just a practice test

But I’m confused on how I should approach this question

<@&286206848099549185>

[frank sparrow]

consider factoring

can you factor anything inside the square roots?

[loud mountain]

no?

[frank sparrow]

@soft zealot $x^3 +x^2$

do u see how?

[loud mountain]

yes, you can factor x^2 into both?

[soft zealot]

joyousbanana

[frank sparrow]

yup, and you get $x^2 (x+1)$

[soft zealot]

joyousbanana

[frank sparrow]

So now, you have a perfect square, you can factor that out of the square root

and remember the condition given

that $x>0$

[soft zealot]

joyousbanana

[frank sparrow]

See if you can do it from here

[loud mountain]

yep, I'm still lost

[frank sparrow]

ok

so

$\sqrt{x^3 + x^2} = \sqrt{x^2 (x+1)}$, since we just found that out

[soft zealot]

joyousbanana

[frank sparrow]

we simply just factored that

[loud mountain]

so the x^2 would cancel? since its inside the square root?

[frank sparrow]

exactly

it's a perfect square

[loud mountain]

oh.

[frank sparrow]

so this would simplify to

$x \sqrt{x+1}$

[soft zealot]

joyousbanana

[frank sparrow]

Try to do the same thing for the other square root

[loud mountain]

$4 \sqrt{x+1}$

[soft zealot]

KermitWilks

[loud mountain]

i got this

[frank sparrow]

nice, but u have to take the square root of 4

[loud mountain]

2

[frank sparrow]

yup, so u get $2 \sqrt{x+1}$

[soft zealot]

joyousbanana

[loud mountain]

yep, I just missed typed

[frank sparrow]

ok

so now can u simplify $x \sqrt{x+1} - 2\sqrt{x+1}$?

[soft zealot]

joyousbanana

[loud mountain]

would the 2 sqaure roots cancel?

[frank sparrow]

not exactly

they would combine

but not cancel out

how bout this

let $y = \sqrt{x+1}$

[soft zealot]

joyousbanana

[frank sparrow]

Now you have xy - 2y

try to finish from here

[loud mountain]

im lost

[frank sparrow]

factor

[loud mountain]

y(x-2) ?

so the answer would be

$(x-2) \sqrt{x+1}$

[soft zealot]

KermitWilks

[frank sparrow]

you got it!

[loud mountain]

wow

what did i do with the x>0??

[frank sparrow]

so basically

if x<0

the square root would have to become negative

but since it's positive

[loud mountain]

oh I get it

[frank sparrow]

we can keep it positive

[loud mountain]

This really jogged my memory😅

Thanks!!

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[frank sparrow]

no problem

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[digital surge]

Hey, I know that you can have complement of a binary number but does decimal numbers also have this?

[tranquil pine]

sure, you can do something similar. Not sure if it has many uses though

instead of doing 2^n-num, you do 10^n-num

[digital surge]

Wdym something similar, how would it work, since I don't find anything about it on google

[tranquil pine]

https://www.geeksforgeeks.org/10s-compliment-of-a-decimal-number/

[digital surge]

Oh, I thought it would work something like 25 = -25

[tranquil pine]

thats not true in binary either

its just a convention for compsci

(because computers represent negative numbers using the first bit or something)

[digital surge]

Hmm I see, thanks gonna look up on that :). Is there a mathematical operation for changing the + to a -, if so what is it called?

[tranquil pine]

multiplying by -1?

[digital surge]

No native ones, like for example absolute of a number?

[tranquil pine]

wdym native?

[digital surge]

😅 sorry been doing some programming lately, I mean like a special operator specially for that operation

[tranquil pine]

-1*

or -sign*abs

[digital surge]

😛 Thanks haha

Thanks for the help, all I needed to know 🙂

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[tranquil pine]

Can two complex vectors be parallel? If yes, how can I find whether they are?
(With complex vectors I mean vectors with complex components)

[final saddle]

@tranquil pine Has your question been resolved?

[tranquil pine]

My assumption is that the condition of parallelism between two complex vectors exists, since the orthogonality condition also exists. But the orthogonality condition relies on the dot product, which has different definitions for complex spaces, while the parallelism condition doesn’t so I’m somewhat unsure. For 3D vectors it should be possible to check the parallelism condition by using the complex cross product formula; but what about other directions?

[tranquil pine]

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[worldly vale]

You can just see if the inner product is 1

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[tranquil pine]

oh

[tranquil pine]

So that works

So all I need is to decide which dot product definition to use?

[worldly vale]

Well yes, bc that determines what you mean by parallel

The usual notion of parallel is just the one that comes from the usual euclidean inner product

But the word carries over to any inner product

[tranquil pine]

But doesn’t this hold only if both vectors are normalized? (we can continue in #linear-algebra btw, mb)

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[digital surge]

I don't get this, why is there a sum of one product term and a product of four sum terms?

[digital surge]

And how come this is neither sop/pos?

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[digital surge]

<@&286206848099549185>

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[tranquil pine]

how could you apply non-Euclidean geometry to space? im looking at possibly creating a mathematical model to represent an elliptical orbit of a planet in space.

[kind mirage]

i dont get your qstn sir/mam

u wanna apply this in space where there is no curvature ?

[tranquil pine]

i dont know much about the topic, but what do you think of it?

[kind mirage]

well as far as i know u need a sphere to continue the non-euclidean geometry

by sphere i mean there should be a curve

[worldly vale]

what do you exactly mean by "apply non-Euclidean geometry to space?"

what space?

[tranquil pine]

like the universe in general

[worldly vale]

well 4D space time is non-euclidean

[kind mirage]

well that seems like a topic to not discuss haha

[worldly vale]

even 3D space is non-euclidean with enough mass present

the application here is just the application of differential geometry in physics

[tranquil pine]

could a mathematical model be made to display this?

[worldly vale]

well yes and thats literally what they do

[kind mirage]

uummmm...yeah possibly

[tranquil pine]

where do yall recommend i start from then

like researching wise

[worldly vale]

differential geometry and how its applied in general relativity

[kind mirage]

the BASIC basic u know

[tranquil pine]

alright thank you to the both of you! Have a great day!

[kind mirage]

u too

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[plain roost]

Need help with this

[plain roost]

<@&286206848099549185>

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@plain roost Has your question been resolved?

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[tranquil pine]

Okay @thorn zodiac

This is going to take a bit but

First things first

Apply addition formula of $\sin(A+B) = \sin(A)\cos(B)+\sin(B)\cos(A)$

[soft zealot]

♡LexQa♡

[tranquil pine]

What do you have now

[thorn zodiac]

from arccos I got a=1 b= sqrt3 and c = 2 so what would I do with this? im really unfamiliar with arc stuff

[tranquil pine]

What you wrote doesn't make much sense to me. Can you try to write the thing into this form?

If you need help with that I can guide you

[thorn zodiac]

yes please

[tranquil pine]

What is our A and B in this

[tranquil pine]

A and B in context to this

[thorn zodiac]

would A=arccos1/2 and B= arctan 1/5

[tranquil pine]

Correct!

So first thing sin(A)

what is that

Just write it out

[thorn zodiac]

sin(arccos1/2)

[tranquil pine]

Correct!

Now do it for the rest

[tranquil pine]

Until you have this form

Can you write it out fully to me

You there?

[thorn zodiac]

Yes one sec

[tranquil pine]

Alright

[thorn zodiac]

(sin(arccos1/2) x cos(arctan1/5)) + (sin(arctan 1/5) x cos(arccos 1/2))

[tranquil pine]

Yes

Alright

So let's evaluate them one by one

First, sin(arccos(1/2))

Let's find what arccos(1/2) is first

Do you have an idea? @thorn zodiac

It's basically asking for what angle of cos that would give you 1/2

[thorn zodiac]

pi/3

[tranquil pine]

Correct

[thorn zodiac]

oh 60

[tranquil pine]

No keep it radians

[thorn zodiac]

oh okay

[tranquil pine]

Okay next

This means sin(pi/3)

What's that

[thorn zodiac]

0?

[tranquil pine]

No

[thorn zodiac]

sqrt3/2

[tranquil pine]

Correct

So we have

Sqrt(3)/2 x cos(arctan1/5)) + (sin(arctan 1/5) x cos(arccos 1/2))

Next

cos(arctan(1/5))

This is going to be a bit trickier to explain. But I think the geometric explanation is best

I want you to draw a right triangle @thorn zodiac

[thorn zodiac]

I have drawn a right triangle

[tranquil pine]

Alright

Now

I want you to assume $\theta = \arctan(\frac{1}{5})$

[soft zealot]

♡LexQa♡

[tranquil pine]

And draw a theta at one of the two angles

[thorn zodiac]

yes

[tranquil pine]

Also assume $x = \frac{1}{5}$

[soft zealot]

♡LexQa♡

[tranquil pine]

So you have $\theta = \arctan(x)$

Oop

Bot slow

[tranquil pine]

Anyways, I want you to take tan of both sides of this, what do you have

[thorn zodiac]

im not understanding

[tranquil pine]

$tt$

Ffs it had to die now

Okay whatever

[tranquil pine]

We have theta = arctan(x) correct ?

[thorn zodiac]

this is saying that arctan(1/5) = arctan (1/5) so im confused

yes

[tranquil pine]

Yeah that's why it is an equality is it not ahaha

You will see why in a bit

[tranquil pine]

What do you have after taking tan of both sides?

[thorn zodiac]

1/5?

[tranquil pine]

Keep it as x, I am deriving a general formula for you

It's the longer way, but I'm trying to make you understand how to do any future problem like this

So anyways, tan(theta) = x is what we have now right? @thorn zodiac

[thorn zodiac]

yes

[tranquil pine]

Okay so actually

What does tan mean?

What does it give?

[thorn zodiac]

opp/adj

[tranquil pine]

Exactly

We have tan(theta) = x/1

Our opposite is x and our adjacent is 1

Can you write those down

On the triangle u drew

[thorn zodiac]

Yes

[tranquil pine]

Now, we want to find the hypotenuse

Apply Pythagorean theorem

[thorn zodiac]

so a=1 b=x which is 1/5 and i can just use path theorem for c

Yes

[tranquil pine]

Yes, again keep it as x for now

So what is our hypotenuse

[thorn zodiac]

2x

[tranquil pine]

No

That's improper usage of Pythagorean theorem

You have $c^2 = a^2 + b^2$ as the Pythagorean theorem, from that we say $c^2= 1 + x^2$

Pain this bot

But do you understand?

[thorn zodiac]

sqrt 1^2 + x^2 = c

[tranquil pine]

Yes

Exactly

[thorn zodiac]

thats what i did

[tranquil pine]

It's sqrt(1+x^2)

[thorn zodiac]

oh i dont add them together

[tranquil pine]

Now we want to find cos(theta)

[thorn zodiac]

okay i understand

[tranquil pine]

What is that from the triangle?

[thorn zodiac]

1/2?

[tranquil pine]

Nono

Your adjacent is 1

Your hypotenuse is sqrt(1+x^2)

Cos(theta) = adj/hyp

This means..

[thorn zodiac]

oh yes cos(theta) = 1/sqrt1+x^2

[tranquil pine]

Exactly

Now what is theta?

We defined it at the very beginning

[thorn zodiac]

arctan(1/5)

[tranquil pine]

What is x?

[thorn zodiac]

1/5

[tranquil pine]

Right. Now can you substitute those values back in the equation

And we can do it now

[thorn zodiac]

( sin(pi/3) x cos(theta) ) + ( sin(theta) x cos (pi/3) )

[tranquil pine]

Yes

But we are doing cos(theta)

So

Cos(arctan(1/5)) = 1/sqrt(1+(1/5)^2) right?

[thorn zodiac]

Yes

[tranquil pine]

Okay.

Evaluate it please

Make it a simple fraction with the square root in the denominatorif possible

[thorn zodiac]

= 1/ sqrt1.04

[tranquil pine]

Keep it in fraction form...

The thing inside the square root

$==$

[thorn zodiac]

1/sqrt 26/25

[tranquil pine]

Yep

No wait

It's the opposite

[thorn zodiac]

yes typo my bad

[tranquil pine]

Okay so now we have

sqrt(3)/2(sqrt(26/25) + sin(arctan(1/5)) * cos(pi/3)

For the sin one, refer back to the triangle u made

What is it going to be

[thorn zodiac]

sin(sqrt26/25) ?

[tranquil pine]

No..

What's our hypotenuse and opposite

In the triangle

[thorn zodiac]

sin = (1/5) / sqrt 1+x^2

[tranquil pine]

Yes!!

Well just make it 1/5 for x in the square root as well

Now compute that

Keep it in fraction form

Tell me what u get

[thorn zodiac]

1/5 / sqrt26/25

do I go further

[tranquil pine]

Yes that's gonna be 1/5(sqrt(26/25)

Okay

Finally last thing

Cos(pi/3)

Whats that

[thorn zodiac]

1/2

[tranquil pine]

Right

So we have

sqrt(3)/2(sqrt(26/25) + 1/(10(sqrt(26/25))

Now it's simplifying time

[thorn zodiac]

how exciting

[tranquil pine]

We can evaluate the square root of 25

What is that

[thorn zodiac]

5

[tranquil pine]

Okay

So

sqrt(3)/2(sqrt(26))/5+ 1/2(sqrt(26)

Is that fine by u

$hh$

[thorn zodiac]

at the end shouldnt it be sqrt(26) / 5

[tranquil pine]

Yeah

But there is 10 right

So 10/5

2

U with me so far?

[thorn zodiac]

uh yes

[tranquil pine]

Anyways we take the 5 to the numerator because complex fractions to get

5sqrt(3)/2(sqrt(26) + 1/2(sqrt(26)

For both, you can rationalize the denominator

But multiplying by conjugate

But this answer is also good enough

Up to u

[thorn zodiac]

Yes, thank you for your help

i have more problems i need help with if youre up for it

[tranquil pine]

I'm so dead tired I need to sleep

[thorn zodiac]

go to sleep

[tranquil pine]

Just open a new help channel I'm sure someone will help u

And yeah I will

Cya

[thorn zodiac]

cya

.close

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[polar urchin]

Trying to solve a question with the method below(picture), except my math equation is:

35x ≡ 1 mod 3

When I want to start, I get:

x ≡ 1/35 (mod 3)
x ≡ 35^-1 (mod 3)
but then,

3 ≡ ... ? How do I continue if my mod is a smaller number than 35

[desert mantle]

well first it would be helpful to reduce 35 mod 3

but ok I guess you don't want to?

you can still do the euclidean algorithm as usual

3 = 0*35 + 3

35 = 11*3 + 2

etc

[polar urchin]

oh so multiplying with 0 is fine

ok thanks!

[soft zealot]

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

[polar wing]

did u blocke me

what did i even do

[final saddle]

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[elfin bloom]

E infinite dimensions normed vector space, F a subspace of
finite dimensions
if x is in E and not F, can i find y in F such that d(x,F)=d(x,y)?
and why?

[desert mantle]

F is closed

[elfin bloom]

yes

[desert mantle]

use that

d(x,F) is defined as the infimum of something

[elfin bloom]

does it suffice to take the orthogonal projection of x onto F?

[desert mantle]

get a sequence of elements in F which approach that infimum

[elfin bloom]

i dont think that would work

[desert mantle]

well if you have an inner product, sure

[elfin bloom]

if i have an inner product it works?

[desert mantle]

otherwise talking about anything "orthogonal" doesn't make sense

[elfin bloom]

AH

[desert mantle]

why not

[elfin bloom]

ok so i take a sequence of elements yn such that d(x,yn)-> d(x,F)

right???

[elfin bloom]

now if yn converges i am done

the problem is that i have no clue if it does

except.... is it because my space is complete that it converges?

[elfin bloom]

Ok orhtogonal doesnt make sense but cant i just take a basis for F and stick it with the corresponding coeficients of my x vector, and call it y

without "projection"

and then y surely exists

[final saddle]

@elfin bloom Has your question been resolved?

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[grand anchor]

The difference between the area of ​​a rectangle and the area of ​​a square is 72
cm2. If we know that the length of the rectangle is twice the side of the
square, and the width exceeds it by 3 cm, what are the dimensions of the
two parallelograms?

[oblique portal]

what exactly are you struggling on

?

[grand anchor]

how to write the difference of cm² between the area of ​​a rectangle and the area of ​​a square

[oblique portal]

youd just put it in an equation

[grand anchor]

how?

[final saddle]

@grand anchor Has your question been resolved?

[grand anchor]

.close

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[tranquil bone]

y^2 = xarctan(1/x) how to sketch this?

[wraith crater]

[tranquil bone]

[wraith crater]

whens your interview?

[tranquil bone]

8;40 tmr

[grim nebula]

faq finally doing some graph sketching

[tranquil bone]

[wraith crater]

I know said you won't get in but try your best you got this. Youll buy me a meal if you get in.

anyway

[tranquil bone]

Wow so wholesome

[wraith crater]

xarctan(1/x) is even

so we have a graph which is symmetrical about the x and y axis

also

arctan(x) = tan(x) = x for small x

ie. arctan(1/x) = 1/x for large x

[grim nebula]

[wraith crater]

so y^2 approaches 1 as x appraoches infinity

[grim nebula]

arctan(1/x) + arctan(x) = sign(x) pi/2

[wraith crater]

[tranquil bone]

Right

It’s

Like a reflected paragola

That approaches 1 and -1

Near 0 it acts like y=+-sqrt(pi/2x)

[wraith crater]

yes a paragola

[tranquil bone]

Okay okay

Calm

give me last minute tips for interview

[wraith crater]

gives him tips

[grim nebula]

i don have any tips

[tranquil bone]

Panic

Got it

[grim nebula]

be confident

believe in yourself

[tranquil bone]

Okay calm

[wraith crater]

when you have a graph sketching problem its often helpful to look at the first few terms of expansion

the interviewers are also likely to ask for the behaviour near 0

[tranquil bone]

So that’s where Taylor

Comes in

[wraith crater]

Yeah or just

lets say you have

sin^2(x)/x

whats the gradient like near 0

[tranquil bone]

It’s like

oh

[grim nebula]

[tranquil bone]

Sinx = x

Near 0

[wraith crater]

[tranquil bone]

So it’s like y=x near 0

Calm

[wraith crater]

Also if you see a problem that youve seen before don't do it like instantly

act like you struggle a little bit

it's deceitful but a good trick

[tranquil bone]

Why

[wraith crater]

they want to see how you approach a problem youve never seen before

so if you know the questions and do them instantly

theyll ask harder questions

so

[tranquil bone]

[tranquil pine]

and if they do give you harder questions (that you arent sure about), talk through your thought process (eg do some examples etc)

[tranquil bone]

What if I have no idea at all

Like blank

[oblique portal]

make something up

[wraith crater]

waffle

[tranquil pine]

solve a simpler problem that looks similar

[wraith crater]

if you have no idea just ask for hints

[tranquil bone]

Hmm I see

[wraith crater]

is it an online intervieqw

[tranquil bone]

Yes

[wraith crater]

idk how much practice youve done but youve been doing a lot of number theory (MO) problems

[tranquil pine]

quietly type your question here and spam ping pure

[wraith crater]

i dont think thosell help (at all)

[tranquil bone]

Yes I’ve just been doing those

[wraith crater]

calm

[tranquil bone]

Can’t tho

I need to have 2 cameras

In

On

[wraith crater]

bros actually considering it

[tranquil pine]

damn

[tranquil bone]

Let me ask you last question

Maths(question

A natural number is selected from 1 to 1000000 at random what’s the probability that it’s cube ends in 11

[wraith crater]

any idea?

[tranquil bone]

Noope

[tranquil pine]

try figuring out a number which satisfies that property first

[wraith crater]

we can express these numbers as 100q + r for r<100

[tranquil bone]

Binomial

Then

Ern

The last 2 digits of n^3 will come from the last 2 digits of r^3

We need to check r = 11,21,…,91

?

[wraith crater]

[grim nebula]

how to check mod 100 without checking mod 100

[tranquil bone]

But how would I check this

Just by hand?

[tranquil pine]

100=2^2*5^2, so you could check mod 4

[tranquil bone]

Got that 71^3 ends in 11

Only 71

[tranquil bone]

Hm

[tranquil pine]

https://cdn.discordapp.com/emojis/690630274907897887.gif?size=32&quality=lossless

[tranquil bone]

Is the answer

0.01?

[wraith crater]

your cat clap

is small

and blurry

[tranquil pine]

fake nitro

[grim nebula]

[wraith crater]

lol

[tranquil bone]

Okay calm

[wraith crater]

Good luck for the interview

[final saddle]

@tranquil bone Has your question been resolved?

[final saddle]

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[swift stag]

I'm not sure which method to use

should I just try a u-sub or something

or is this one of those trig sub problems

[vital crag]

$4+x^2 = 4(1+ ?)$

[soft zealot]

riemann

[swift stag]

x^2

[dusty harbor]

no

[swift stag]

(1/4)x^2

[vital crag]

let me be even more explicit

$4+x^2 = 4(1+ (?)^2)$

[soft zealot]

riemann

[swift stag]

(1/4)x

[vital crag]

((1/4) x)^2 = x^2 / 16

[swift stag]

uhm

[vital crag]

$\left(\frac{a}{b}\right)^2 = \frac{a^2}{b^2}$

[soft zealot]

riemann

[swift stag]

I'm not seein' it

looks like you factored a 4 out

basically

somehow

4 x ? = x^2

I have no idea

[vital crag]

yes i did factor a 4 out

$4\left(1 + \frac{x^2} {4}\right)$

[soft zealot]

riemann

[swift stag]

huh, ok

[vital crag]

multiply through to show that it's equal to 4+x^2

[swift stag]

seems to be

[vital crag]

can you fill in the question mark now?

[vital crag]

using this rule

[swift stag]

x/2

[vital crag]

right

you can factor out 4 from the integral using the constant rule

then use a simple u-sub to get to an integral you're familiar with

[swift stag]

interesting

this isn't a trig problem is it

[tranquil pine]

screams arctan

[vital crag]

define "trig problem"

[swift stag]

involving trig functions

trig subs

[vital crag]

it's only a trig sub if you haven't been taught it before

but you should have been taught

,w integral 1/(1+x^2)

[soft zealot]

Results provided by WolframAlpha

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

Requested by riemann#2529

[swift stag]

ah

[vital crag]

if you haven't seen this, then yes you need to derive it

[swift stag]

imteresting

this is in my textbook's table of integrals

so if I factor out the a

it's just arctan(u)

or arctan(x)

[tranquil pine]

a=1

[swift stag]

right

[tranquil pine]

$\int \frac{du}{1^2 + u^2} = \frac{1}{1}arctan(\frac{u}{1} ) + C$

[soft zealot]

hibyehibye

[tranquil pine]

$\int \frac{du}{1 + u^2} = arctan(u) + C$

[soft zealot]

hibyehibye

[swift stag]

ic

thx

is that right?

I think I did something wrong

[vital crag]

,w diff 1/4 * arctan(x/2)

[soft zealot]

Results provided by WolframAlpha

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

Requested by riemann#2529

[vital crag]

you have to modify your dx to du when doing your u-sub

[swift stag]

oh, I forgot to u-sub

[vital crag]

you can also just straight use this from the beginning

$4 + x^2 = (?)^2 + x^2$

[soft zealot]

riemann

[swift stag]

aight

[final saddle]

@swift stag Has your question been resolved?

[final saddle]

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[void needle]

Hello, does the solution set for csc^2x + cot^2x = 0 include two answers, or just one?

[void needle]

I looked the problem up on mathway and symbo lab and the answer is only giving a solution set for when cotx = -1

I solved the problem by using trig identities to sub csc^2x for 1 + cot^2x and solved from there

so I had one side = -1 and the other = 0

[dry light]

Think about the range of each term

[void needle]

What do you mean

The interval notation im working with is 0, 360 FWI

so my solution set should be {0, 180, 135, 315}

[void needle]

oh yeah u right

cot cannot be 0

undefined at zero

wait

no

thats not right

cotangent is all real numbers

[final saddle]

@void needle Has your question been resolved?

[void needle]

@dry light range is irrelevant when given an interval

because your range is the interval

there are places on the unit circle between 0-360 degrees where cot is 0

so im probably doing the problem wrong

[dry light]

The range will tell you whether or not it can even equal 0

The problem is this

csc^2(x)'s range is y >= 1

cot^2(x)'s range is y >= 0

@void needle you understand this right

Because that means the range of cot^2(x) + csc^2(x)'s range is y >= 1

So it'll never equal 0

[final saddle]

@void needle Has your question been resolved?

[final saddle]

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[last stone]

I'm doing a final review sheet for calc 2. This problem is a bit challenging:

[shell mountain]

What have you tried

[last stone]

I get to the part of setting up the integral, but I am unsure how to evaluate it.
Right now I have this:

[last stone]

I tried to factor in the square and I end up with:

then combining terms I get

but this seems really complicated

[shell mountain]

Notice √(x⁶) can be simplified and x¹² + 2x⁶ + 1 can be factored

And √(4) can be simplified

Also, are you sure that's correct? WA says that's a different function from the original

Nvm

It's correct

I typed it wrong

[shell mountain]

Refer back to this

[last stone]

so the 4x^6 (what is the markup syntax for superscripts?) can be turned to 2x^3 right

[shell mountain]

You mean latex syntax? It's what you wrote, 4x^6

And yes

[last stone]

did you use latex here?

[shell mountain]

On my phone, if I press and hold the numbers, I can write them

[grim nebula]

there's a quick way to do this

[last stone]

ah

[grim nebula]

notice how when you expand the square

the middle term you get is -1/2

but once you add on the 1

it becomes 1/2

[last stone]

right

[grim nebula]

what that amounts to is switching the - to a + in the original square

[grim nebula]

so the bracket here

so you can refactor the whole thing

[last stone]

so it now becomes 1 + (x^3/2 + 1/2x^3)?

[grim nebula]

and get (x^3/2 + 1/2x^3)^2

no the 1 gets eaten

[last stone]

oh right

[grim nebula]

to make the - into a +

[last stone]

interesting

[grim nebula]

and now you can just root the whole thing

[last stone]

is that always the case for these problems?

[grim nebula]

uh

no

i assume this is by design

to make your life easy

[last stone]

that is a really nice shortcut

[grim nebula]

when you root don't forget to add absolute values

and carefully consider your domain of integration to see if you can remove them

in this case you're good but sometimes the thing you're squaring might be negative

[last stone]

yeah I'm not sure how to even factor the x^12 + 2x^6 + 1

[grim nebula]

if you make a substitution of u = x^6 it just becomes u^2 + 2u + 1

that you should know how to factor

[last stone]

ah yeah that is a lot easier

@grim nebula I got -15/12 with the method you showed me

I think the answer is supposed to be 33/16

I figured it out, thanks you two

.close

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[copper urchin]

I have a question regarding the binomial distribution.

While graphing things out in Desmos, I am realizing that the more trials there are, regardless of what p is, the distribution starts tending to be more and more symmetric:

Can someone explain to me why that is? Trying to find any intuition, pure English, algebraically, etc. are fine.

[copper urchin]

And of course, ignoring the line going jank onto the left (or the right if p > 0.5)

Intuitively, it doesn't make that much sense. A binomial distribution for a weighted coin intuitively should be skewed even under infinite trials, but the graphs don't seem to represent that, and I'm wondering where in my intuition am I wrong.

It seems like I'm not the only one coming to this conclusion...

https://faculty.elgin.edu/dkernler/statistics/ch06/6-2.html

[hearty zephyr]

it depends on what you mean by 'skewed'
for larger n the binomial becomes symmetric about the mean, which is np
so it moves with more trials

[copper urchin]

np meaning?

[hearty zephyr]

n times p

[copper urchin]

Ah

[hearty zephyr]

with a few trials it starts heavily skewed, with more trials it starts to even out and become more symmetric with extremes becoming equally unlikely

[trail mango]

central limit theorem might be worth looking at but i’m not sure how much intuition it would give

[hearty zephyr]

^^^^

[copper urchin]

I will. Thanks for all the help!

!close

.close

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[tranquil pine]

Where did the e disappear

In the derivative

And just eliminate it with the Numerator and take out constants

[final saddle]

@grand axle Has your question been resolved?

[tranquil pine]

You also erased the -0.1 from the derivative

You have 100 ÷ 0.1

That's not

10

It is 100 ÷ -0.1

Just I was speaking about the actual division

that's why I omitted the negative sign

Just

Take another substitution

v = 1 + u

And take out the -1000 out

[tranquil pine]

You can just say $\int c\cdot f(x)dx = c\int f(x)dx$

[soft zealot]

♡LexQa♡

[tranquil pine]

Where c is some constant

[final saddle]

@grand axle Has your question been resolved?

[vocal spire]

I think you should've substitute $v=1+u$ instead of $v=\left( 1+u \right) ^{2}$

[soft zealot]

Welf

[vocal spire]

so next you'll have $\int \frac{1}{v^{2}}\text{d}v$

[soft zealot]

Welf

[tranquil pine]

That's what I meant originally indeed

It's just $-1000\int\frac{1}{v^2}dv$

[soft zealot]

♡LexQa♡

[tranquil pine]

Which you can rewrite as $-1000\int v^{-2}dv$

Proceed from here

[soft zealot]

♡LexQa♡

[tranquil pine]

You are like so close to it my guy

[tranquil pine]

Integrate this

It's just the power rule for polynomials

Remove the integral sign because u already got it

Comes with experience I guess? You just realise what to do

Becuz for example

you were able to eliminate the numerator with that derivative

Which helps

[final saddle]

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[rustic thunder]

I’m not sure how I find the within groups sum of squares?

[misty nymph]

Greetings

[rustic thunder]

Hello

[final saddle]

@rustic thunder Has your question been resolved?

[rustic thunder]

.close

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[polar wing]

why is this 0, it feels like its indeterminante

because its 0/0

but i think if its like x/x^2

it diverges somehow

[tranquil pine]

x^2/x = x (x/x)

and lim of x/x = 1

[polar wing]

oh i see

thx

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[near gull]

I have 44 socks in my drawer, each either red or black. In the dark I randomly pick two socks, and the probability that they do not match is 192/473. How many of the 44 socks are red?

[near gull]

i dont know whre to start

[dusty harbor]

the probability is larger than 1

[near gull]

lol true

oh wait

the quesation

its

192/473

[dusty harbor]

aight so

[vital crag]

Fun problem

[dusty harbor]

we have x red socks

you can work it out algebraically

44-x black socks

then its probably easier to calculate the probability that you grab matching socks

and do 1 - that

[near gull]

wait what i lost u afterthis

[cloud zephyr]

probably not so much if you end up running to work in an unmatched pair of socks

[dusty harbor]

,w x^2 -44x + 946 = 384

[soft zealot]

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[dusty harbor]

oh

[drowsy scroll]

complex socks

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[drowsy scroll]

,w x^2 - 88x + 768 = 0

[soft zealot]

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[drowsy scroll]

okay so that’s definitely not it haha

this is such a fun problem

[near gull]

im so confused lol

[drowsy scroll]

okay so I made some progress

44C2 is 946, so there are 946 possible combinations of socks, right?

now, if you multiply the numerator and denominator of 192/473 by 2, you get 384/946

so there are 384 possible non-matching pairs of socks

therefore, 562 possible matching pairs of socks

the thing is that they could be or both red or both blue

hmm

what if we tried to find how many possible matching pairs of socks there are IN TERMS of how many red socks there are, and then solve for how many red socks there are

yup this is just what I was trying earlier

no idea

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[near gull]

i kinda get what ur tryna do but im still confused

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[lime steppe]

i will really like some help pls i need understand for test tomorrow

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[lime steppe]

anyone help pls

[modest birch]

do you know what reflect across the y-axis means

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[fickle geode]

i need help with this question

[dry light]

Unit circle

[grand bronze]

$\sin\theta=y$-value

[soft zealot]

Duh Hello

[fickle geode]

is sin an x or y value

[dry light]

Idk man read what they said

[fickle geode]

OH

i didnt even read it lol

my bad

but theres multiple y values of the -√3/2

[dry light]

Yup

That's why there are multiple answers

[fickle geode]

got it

thank you

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[steep lodge]

Having trouble with this one. I think I'm calculating the flux through z=0 and z=2 wrong, because when I use the flux of the two z and the closed surface to find the flux through x^2 + y^2 = 9, I get 54pi, but when I specifically solve for the flux through x^2 + y^2 = 9 to verify, I get 0

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[steep lodge]

<@&286206848099549185>

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[steep lodge]

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[tardy goblet]

question 10, i don't know what i'm supposed to do, is there even an inverse to that function?

[real needle]

First factorize it I think

x⁴(1-x³)

[tardy goblet]

Right

Could you plz walk me through it, thanks

[real needle]

Then factorisation?

[tardy goblet]

With 1^3-x^3?

We're trying to get the inverse first right

[real needle]

x^4 is a common factor in both x^4 and x^7 so you can take it out of the equation and factorize
x^4 * 1 = x^4 and
x^4 * x^3 = x^7

[real needle]

Yea but I think first you have to get it where only one variable is x

I think

[tardy goblet]

Yeah i get that