help-23

[glacial cairn]

It's just that the choice of the substitution is convenient

You should read up on this https://en.wikipedia.org/wiki/Integration_by_substitution
(or find some tutorial online for integration by substitution)

[safe parrot]

I just don’t understand how that applies here

For example

Where does x^2 go?

It turns into du lol

[glacial cairn]

Yes, because of this

Or, if it's easier to understand, dx = du / (6x^2)

Note the 1/6 that goes in front of the integral

[safe parrot]

du divided by 6x^2?

[glacial cairn]

Yeah

If u = f(x) then du/dx = f'(x)

That's differentiation with respect to x

[safe parrot]

I’m honestly trying my best but this makes no sense

I have

$\int \frac{sinx}{u^2 +1}$ where u = cosx

[flat frigate]

Merineth

[safe parrot]

Right?

[glacial cairn]

No, you're missing dx

[safe parrot]

What is dx

My original function?

[glacial cairn]

No

[safe parrot]

Cosx dx?

[glacial cairn]

You need to go back to the definition of an integral, you're missing basic knowledge here

[safe parrot]

Clearly, this is not something we have been taught

[glacial cairn]

Take f(x) = sqrt(x)

[safe parrot]

<- Nell

[glacial cairn]

The integral of f is, roughly speaking, the "area under the curve"

Right?

[safe parrot]

Yes sir

[glacial cairn]

(and that area is signed of course)

The definite integral, or antiderivative, is that area between 0 and x

[safe parrot]

Yes

[glacial cairn]

So here, if you look at the green rectangles, they form an approximation of the area under the curve between 0 and 1

[safe parrot]

Yea

[glacial cairn]

What's the area of each individual rectangle?

[safe parrot]

Well they are also integrals, if you aren’t referring to base * height

[glacial cairn]

I am referring to base * height

That's why we use rectangles

[safe parrot]

Ok!

Well it would be base * height for each rectangle

Between 0 and x

[glacial cairn]

Right, and what's height?

[safe parrot]

y

[glacial cairn]

Take just one rectangle, note the x coordinate of its left side t

What's the length of that left side?

[safe parrot]

x2 - x1

[glacial cairn]

[safe parrot]

From origo to x2 - from origo to x1

= distance between x1 and x2

[glacial cairn]

Ok I guess you chose x1 and x2 instead

But that would be its bottom side, wouldn't it?

[safe parrot]

Yes

[glacial cairn]

So the base

[safe parrot]

Yea

[glacial cairn]

Now what's the height?

[safe parrot]

That would be the f(x1) = sqrt(x1)

[glacial cairn]

sqrt(x1)

You need to keep the same variable...

[safe parrot]

My bad. Yes x1

[glacial cairn]

Ok, do you agree that the sum of the area of all those rectangles forms an approximation of the area under sqrt(x)?

[safe parrot]

Yes, sir

Totally agree

[glacial cairn]

Do you also agree that it becomes a better approximation when the base of each rectangle becomes smaller

(which also means more rectangles)

[safe parrot]

Yes I do

[glacial cairn]

Well, the integral is the limit of this sum

For the height, we have sqrt(x_n)

For the base, we have some small number d_n

As we pick d_n to be smaller and smaller, the approximation becomes better

But we also need the number of rectangles

x is the right edge of the area we want to calculate

So the number of rectangles is n = x/d_n

Does that make sense?

[safe parrot]

Let us process, one sec

[glacial cairn]

Here's a graph to play around with
https://www.desmos.com/calculator/5gkwx2oyuj

It only shows the top side of the rectangles

[safe parrot]

So when x -> 0

lim x -> 0

The approximate becomes better

[glacial cairn]

d_n, not x

[safe parrot]

Okay we think we get it

The smaller the d_n the more rectangles

The better the approximation

[glacial cairn]

Yes

So let's write down the sum now

We have the number of rectangles x/d_n

The base of each rectangle d_n

And the height, f(i*d_n) with i an integer between 0 and x/d_n

[safe parrot]

We are lost

[glacial cairn]

x_1 for the i'th rectangle is i*d_n

[safe parrot]

d_n is the length from x1 to x2?

[glacial cairn]

Yes

[safe parrot]

and x is y?

[glacial cairn]

No?

[safe parrot]

x is y?

Actually I think it’s best if we just watch some Yt videos on it

We are hopeless

[glacial cairn]

i is just 0, 1, 2, 3, ...

@safe parrot we're almost there...

[safe parrot]

Okay ;-;

That picture makes sense

Why aren’t we just naming them x1 x2?

[glacial cairn]

Because we have too many rectangles

It's not the same x1 and x2 for every rectangle...

x2-x1 however is the same, that's d_n

[safe parrot]

Ok

[glacial cairn]

So do you understand all of these points?

• left side of the full area at 0
• right side of the full area at x
• base of rectangle i: d_n
• height of rectangle i: f(i*d_n)
• number of rectangles: n = x/d_n

[safe parrot]

Yes that makes sense

[glacial cairn]

So the sum is: $\sum_{i=0}^{\frac{x}{d_n}} f(i \cdot d_n) \cdot d_n$

[flat frigate]

Nel

[glacial cairn]

Right?

[safe parrot]

What the fuck is that hahaha

I know it’s the sum sign

[glacial cairn]

I just put each of my points in the sum

[safe parrot]

We have never dealt with the sum sign ever in our life lmao

[glacial cairn]

Damn what is your school teaching

[safe parrot]

I don’t know to be honest but it’s clear that it’s dog water bad

No wonder we are struggling like hell

[glacial cairn]

Ok let's not deal with the sum sign I guess

[safe parrot]

My so says you must live in Japan since you are so Knowledgeable haha

[glacial cairn]

$f(0 \cdot d_n) d_n + f(1 \cdot d_n) d_n + f(2 \cdot d_n) d_n + ... + f(\frac{x}{d_n} \cdot d_n) d_n$

[flat frigate]

Nel

[safe parrot]

That makes sense

I get that

[glacial cairn]

Just summing the area of each rectangle one by one

[glacial cairn]

Well that's the exact same thing

[glacial cairn]

The issue with this notation is that the number of terms is unclear

And as we take the limit as d_n goes to 0, the number of terms increases to infinity

That's more or less the definition of the integral

It's the limit of that sum as d_n goes to 0

Here's the graph from earlier, with the antiderivative of sqrt(x) in black and this sum in green:
https://www.desmos.com/calculator/xmq7s0mccm

You can play around with d_n

Actually it's not quite right, it shouldn't be that precise, but I'm not sure what I did wrong

Anyway it's just for intuition

[safe parrot]

I’m just wondering how knowing about that applies to substituting

[glacial cairn]

I'm coming to that

[safe parrot]

Okay!

[glacial cairn]

As d_n becomes closer and closer to 0, it becomes infinitesimal and we write it dx

The number of rectangles becomes infinite, but because dx is infinitesimal, the sum still makes sense

It's like summing a very large number of very small things

[safe parrot]

Makes sense

[glacial cairn]

The rest is f(i*d_n) which is just the function at each point

So the integral is written $\int f(x) ,dx$, where $\int$ roughly means "infinite sum" and $,dx$ roughly means "infinitely small steps"

[flat frigate]

Nel

[safe parrot]

Okay

[glacial cairn]

So the dx is important, you can't just throw it away, otherwise the infinite sum will just yield infinity

[safe parrot]

Makes sense

[glacial cairn]

Now where you make a substitution, you change the variable from x to u, so you also need to change the step from dx to du

For example if u = 2x then u grows twice as fast as x so the steps du are twice as large as the steps dx

So du = 2dx

Take the example from Wikipedia:

$\int (2x^3+1)^7 (x^2) ,dx$

[flat frigate]

Nel

[glacial cairn]

If you set u to what's on the left in the parentheses: u = 2x^3 + 1, then you want to know the step size du so you can just integrate a function of u instead of a function of x

Does that make sense?

[safe parrot]

So now we have dx and du?

[glacial cairn]

Well, if you set u = 2x^3+1, then you get

$\int (u^7) (x^2) ,dx$

The goal is to get replace all instances of x with u

[vagrant ice]

what's du/dx (2x^3 + 1)?

[flat frigate]

Nel

[safe parrot]

So the goal is to replace all x with u?

[vagrant ice]

and all dx with du

yes

[glacial cairn]

Yes but it needs to match what you set, u = 2x^3+1

[safe parrot]

Okay

[glacial cairn]

Now, u is like a function

u(x) = 2x^3+1

You can find du/dx by differentiating this, because that's pretty much the definition

du is the vertical step size of u(x), and dx is the horizontal step size

du/dx is just the slope of u(x)

Like in this thing:

"change in x", at the limit, is dx

"change in y", at the limit, is dy (or du in our case)

[safe parrot]

[safe parrot]

So what happens after this step

[glacial cairn]

First you find du

The derivative of u(x) = 2x^3+1 is u'(x) = 6x^2

du/dx = u'(x) = 6x^2

So du = 6x^2 dx or dx = du/(6x^2)

[safe parrot]

So it’s just the derivatives

After we found du, then?

[glacial cairn]

da/db is the derivative of a with respect to b

You replace dx with du according to what you found

So x^2 dx = du/6

[safe parrot]

So we get (u^7)(x^2) * u’

[glacial cairn]

No

1/u'

[safe parrot]

Okay so that is how you get -1/sinx

[glacial cairn]

$\int (2x^3+1)^7 (x^2) ,dx = \int u^7 \frac{du}{6}$ where $u=2x^3+1$ and $du = 6x^2 dx$

[flat frigate]

Nel

[glacial cairn]

Yes

[safe parrot]

So dx is actually

Just

Derivative of x

[glacial cairn]

No it's an infinitesimal change in x

A derivative is of a function, not a variable

[safe parrot]

Ok

So what is the next step?

[glacial cairn]

Where?

[safe parrot]

Never mind

We are having trouble with reading

Or rather

Understanding by reading

[glacial cairn]

So is there anything still unclear to you?

[safe parrot]

Just to be clear

Whenever I substitute with u on something

I have to multiply it with the derivative

Of u

[glacial cairn]

No

In the context of an integral, you have to deal with the dx

[safe parrot]

I think we need to do some examples I guess later

Because we are so lost

[glacial cairn]

$\int f(x) ,dx = \int u(x) \frac{du}{u'(x)}$

[flat frigate]

Nel

[glacial cairn]

The point is to choose u correctly

[safe parrot]

I’m sorry but we really don’t get it

[safe parrot]

Not even this

It’s just so weird and inconsistent

[glacial cairn]

.

[safe parrot]

I don’t learn by reading 😭

I read it over and over and over

It doesn’t stick

The way everything is formulated and explained in text for math is not how normal people speak and understand

[glacial cairn]

Maybe try this
https://www.youtube.com/playlist?list=PLZHQObOWTQDMsr9K-rj53DwVRMYO3t5Yr

[safe parrot]

So I don’t understand how people are expected to understand it

[glacial cairn]

I don't think it talks about integration by substitution but it might help with the fundamentals

[safe parrot]

I gusss I’ll hvave to watch a video of it

And see if it makes more sense

Because reading just doesn’t work

[safe radish]

@safe parrot Has your question been resolved?

Channel closed

Closed by @safe parrot

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[bitter matrix]

So I've been taught at a level how to solve second order differential equations in the form $a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + c = 0$. but now my professor wants us to solve them with higher orders and mainly with x(t) multiplied. I'm currently doing a question where I have to solve $\frac{d^2}{dt^2}x(t) + 2\frac{d}{dt}x(t) + x(t) = 0$ and it would be fine if it wasn't for everything being multiplied by x(t). How do I go about solving this? My professor does a shit job at explaining it

[flat frigate]

Solaris (firecatto)

[raven heart]

what

wdym it's multiplied by x(t)

d/dt x(t) means the derivative of x(t) with respect to t

it's just an alternate notation

@bitter matrix

[bitter matrix]

so they are the same thing?

damn

[raven heart]

what would d/dt mean on it's own anyway

[bitter matrix]

idk lol

it's just calculus notation that confuses me

[desert juniper]

on the first, you have that y is a function of x.
on the second, you got that x is a function of t.
In both cases, you got two derivatives of a function with respect to its variable

[bitter matrix]

when solving the general solution and it's complex I was taught that it'd be in the form $e^p^x(Acos(qx) + B(sin(qx))$ where the roots where $p+-qi$, but now he is asking for this form...?

[flat frigate]

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

[bitter matrix]

I know it's a third order differential equation but what

[raven heart]

complex exponentials

the cos and sin you have when you're interested in real solutions come from complex exponentials

[bitter matrix]

so how would I do it in that form?

how does it translate?

I was thinking about euler's formula but sin doesn't have an i infront of it

[raven heart]

so you have exp(0t) (a constant), exp(2it), exp(-2it)

[raven heart]

well the idea is, if you have a complex solution U(t) + iV(t) to a linear diffeq (U and V are real functions), then U(t) and V(t) themselves are also solutions

it's not hard to show

so that's how you translate from complex solutions to real solutions

[light stone]

there are 3 solutions

e^1, e^2 and e^3

or e^2i

what you wrote

[bitter matrix]

ok, i see, thanks

so in the first example, with the equation $\frac{d^2}{dt^2}x(t) + 2\frac{d}{dt}x(t) + x(t) = 0$, i've gotten the general soltuion $x(t)=(A+Bt)e^-^t$. Now i'm given the boundary conditions $x(0)=A$ and $\frac{dx}{dt}(0)=0$. The first seems to be completely useless cause when I enter t=0 into my general solution it just gives $x(0)=A$ which is what the boundary condition said. However when I differentiate my general solution and then compute t=0 I end up with $A=B$, which still doesn't let me state what A and B are

What am I doing wrong?

[flat frigate]

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

[safe radish]

@bitter matrix Has your question been resolved?

[safe radish]

@bitter matrix Has your question been resolved?

[safe radish]

@bitter matrix Has your question been resolved?

[safe radish]

@bitter matrix Has your question been resolved?

[safe radish]

@bitter matrix Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[frozen current]

how many decimal zeroes does this have?

[grim patrol]

Isnt the answer zero

as 99999 over 99999 to the same power is bassically 1

its like 1/1=1

It would be like

1/(1-1)

1/0

which is undefined

Dont believe me im like not a maths expert

[frozen current]

wolfram alpha says differently

and this isnt even the entirety of it

i think our math professor trolled us

by telling us to find the number of decimal zeroes

or im just stupid

[safe radish]

@frozen current Has your question been resolved?

Channel closed

Closed by @frozen current

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[rich tree]

I forgot what X\Y means

[thin bridge]

context?

[rich tree]

Sets

[thin bridge]

elements in X but not in Y

[rich tree]

Ah wait it’s ummm like without X without Y I can’t remember the name now

[thin bridge]

set minus

[rich tree]

Ahhh thank you!

.close

[safe radish]

Channel closed

Closed by @rich tree

Use .reopen if this was a mistake.

[grizzled topaz]

its like x/(x-1) you can't just assume x is 1 so i don't think you can assume that its undefined

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[grizzled topaz]

Tbh not really sure where to start

[safe radish]

Please don't occupy multiple help channels.

[grizzled topaz]

I've just shuffled things around

[grizzled topaz]

click on the 1st image to see it in full

[safe radish]

@grizzled topaz Has your question been resolved?

[grizzled topaz]

<@&286206848099549185>

[vital steppe]

Remember that $\sin{(\pi - \theta)} = \sin{\theta} $ so the left side becomes (I'm talking about the first exercise) $2\sin{\frac{\pi}{5}}\cos{\frac{\pi}{5}}$

And then you're done, because the last term I wrote is equal to the right side, just remember $\sin{(2\theta)} = 2\sin{\theta}\cos{\theta}$

For the second one you have to use similar reasoning

[flat frigate]

milo_schwartz

milo_schwartz

[safe radish]

@grizzled topaz Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tacit pendant]

how could i apply chain rule here

[tacit pendant]

$e^{5tanx} + 5^{2x}$

[flat frigate]

odokawa

[tacit pendant]

chain rule for the left hand and chain rule for the right hand or both in the same operation

[weary osprey]

Step 1: Find the derivative of $e^{5 \tan x}$

[flat frigate]

Dubs

[tacit pendant]

it should be

$5sec^2{x} (e^{5tanx})$

correct me if I am wrong

[weary osprey]

Fix the typos

[tacit pendant]

what's a typo

[weary osprey]

Fix the typographical error

[tacit pendant]

mmmmm

probably you mean this

[weary osprey]

$5\sec^2x e^{5 \tan x}$

[flat frigate]

Dubs

[weary osprey]

You shouldn’t leave the x of secx

[tacit pendant]

oaky oaky

[flat frigate]

odokawa

[weary osprey]

It’s correct, now the right side

Let’s take $y= 5^{2x}$

[tacit pendant]

okay okay

i think that´s easier

[flat frigate]

Dubs

[tacit pendant]

mmm it would be

$10^2$

[flat frigate]

odokawa

[weary osprey]

Nope

How do you get that?

[tacit pendant]

because i drop the 2 and 2 * 5 is 10
and the derivative of 2x is 2

[weary osprey]

Nope it doesn’t work like that

If you have x^2, then you can apply power rule

Notice the base is a costant like 5

it’s not a variable

[tacit pendant]

let me find the power rule rq

[weary osprey]

$x^2$ and $2^x$ are different

[tacit pendant]

potencia means power

[flat frigate]

Dubs

[tacit pendant]

but here there are 2 of them

[weary osprey]

For x^2, you can apply the power rule given above

[rocky rune]

you get this right?

sorry for interrupting

[weary osprey]

but for 2^x, you cannot bring that x down and take away 1 from the exponent

[weary osprey]

Jeez

[tacit pendant]

i think it´s something like that but i dont know how to do that actually

[weary osprey]

My purpose is defeated

[tacit pendant]

well

[rocky rune]

sorry big man even i had doubt in this one thats why i asked

[tacit pendant]

i think this is too hard for me

[weary osprey]

Let’s take $y= 5^{2x}$

[tacit pendant]

how can i search in youtube

[flat frigate]

Dubs

[tacit pendant]

something that explain me this

[weary osprey]

Can you apply log both sides?

Natural log

base e

[tacit pendant]

why? there is a rule or something¿

[weary osprey]

We are going to figure out a rule to differentiate the above

Clearly you cannot apply the power rule

Power rule only holds if you have something of the form x^n

[weary osprey]

As you can see here

[tacit pendant]

okay

in this case is the oppositive

we have

n^x

[weary osprey]

But y=$5^{2x}$ is not in that form right?

[flat frigate]

Dubs

[tacit pendant]

we have the oppositive

$n^x$

[weary osprey]

Apply natural log both sides?

[flat frigate]

odokawa

[weary osprey]

Yes

[tacit pendant]

yes but why, i would like to know if there is a rule or something i can based on

[weary osprey]

Let’s see how to differentiate such a case

[weary osprey]

Our goal is to use the idea that derivative of lnx is 1/x

to tackle this problem

That’s the monologue i can give

Can you apply natural log both sides and tell what you get?

[tacit pendant]

okay oaky

[weary osprey]

$\ln y= \ln(5^{2x})$

[flat frigate]

Dubs

[weary osprey]

There you go the first step

[tacit pendant]

$ln(e^{5tanx}) + ln (5^{2x})$

[weary osprey]

Let’s take $y= 5^{2x}$

[flat frigate]

Dubs

odokawa

[tacit pendant]

okay oka y

[weary osprey]

You have already found the derivative of the first term

[tacit pendant]

lets work with right hand

[weary osprey]

Now you have to only find the derivative of 5^2x

[tacit pendant]

sorry i didnt understand whay did you mean by both sides

oaky okay

$ln(5^{2x})$

[flat frigate]

odokawa

[tacit pendant]

here we go

[weary osprey]

Let me wrap everything real quick

So we are on track

We have to find the derivative of $e^{5 \tan x} + 5^{2x}$

[flat frigate]

Dubs

[tacit pendant]

yes

[weary osprey]

Step 1: Find the derivative of $e^{5 \tan x}$ Done. \
Step 2: Find derivative of $5^{2x}$ \
Step 3: set $y= 5^{2x}$, differentiate \
Step 4: Add both of them to get final answer

[flat frigate]

Dubs

[weary osprey]

Is there a confusion?

[tacit pendant]

one question
in the step 2 can I apply this?

because i think ln is too advanced for me

[weary osprey]

Notice here the base is e, but yours it’s 5

It’s not going to the too advance? let’s see how it goes

$\ln y= \ln(5^{2x})$

[flat frigate]

Dubs

[weary osprey]

Can you solve the right side ?

Apply log rules

[tacit pendant]

yes

i can

[weary osprey]

Try?

[tacit pendant]

give me 1 minute

to do it

[weary osprey]

sure

[tacit pendant]

i practiced like 30 minutes ago

[weary osprey]

great

[tacit pendant]

$\frac{1}{5^{2x}} dx(5^{2x})$

[flat frigate]

odokawa

[tacit pendant]

@weary osprey

all right, sorry

for wasting your time

im shit

[weary osprey]

Not an issue

[tacit pendant]

oh

[weary osprey]

Take your time

[tacit pendant]

but

[tacit pendant]

i thought
log(x) was 1/x * x'

that's why i write that

.close

[safe radish]

Channel closed

Closed by @tacit pendant

Use .reopen if this was a mistake.

[tacit pendant]

.close

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[solar cosmos]

hi i need HELP

[solar cosmos]

im crying

how do i do this

[lean otter]

do you know about the limit [
\lim_{x \to 0} \f{\m\sin x}x = 1
]

[potent seal]

lim sinx /x as x tends to 0 is 1

[potent seal]

Also note that the term within sin and the denominator need to be the same

[solar cosmos]

so like

sin(x)/x = 1?

[potent seal]

When the limit is 0

Yeah

[solar cosmos]

so what do i put on the calculator

to get 1

so how do i do it

i have my exam in 2 days and im stressing out

[potent seal]

The method is above

Try it once

If you're not able to get it then come back

[solar cosmos]

yes i didnt really get it

[tacit pendant]

you can do it, i also have examen tomorrow about derivatives, and ive been in this discord server for the last 15 hours without rest, sometimes is about stress, you can man ❤️

[potent seal]

since the term within sin and denominator need to be the same, how about you multiply ax to both numerator and denominator?

[solar cosmos]

appreciate it bro, gl on ur exam 🔥

[solar cosmos]

so we can just cancel them out?

we can cancel sin x and x

[potent seal]

Well it becomes 1

[solar cosmos]

i still dont get how it becomes 1 ;-;

can u write it on paint or something

[potent seal]

There's a proof for that

It's seperate

Unless you're supposed to show the working of that proof for the question above

[solar cosmos]

nope im nto supposed to show the working

[potent seal]

Oh ok

[solar cosmos]

the answer is a/b, is that also equal to 1

[potent seal]

https://youtu.be/Ve99biD1KtA?feature=shared

You can refer to this if you want to know why

[solar cosmos]

i just need to know how to solve it....

so how can i solve it.....

[potent seal]

You could start my multiplying ax to numerator and denominator

As I said

[solar cosmos]

that'll confuse me more

[potent seal]

...

[solar cosmos]

can u show me how

[potent seal]

That is supposed to be the way but alr

But x is tending to 0

And the term within sin should be the same as denominator

sin(x)/x

So sin(ax)/ax

[safe radish]

@solar cosmos Has your question been resolved?

[solar cosmos]

I wish I can say yes 😭

<@&286206848099549185>

.close

[safe radish]

Channel closed

Closed by @solar cosmos

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[rose steeple]

Hello i need help with a calculus problem: i want to calculate the area inside the loop of the curve from the picture but don't know how to.

[upper mulch]

Find the intersections, find the relation between the two (which one has a bigger value), then use integral..

[rose steeple]

My intersections don't match the model answer i have 0 and 0.908... and this is the model answer:

[safe radish]

@rose steeple Has your question been resolved?

[upper mulch]

That's weird.

You should probably ask your teacher/proffesor about this.

[rose steeple]

It should be an easy excercise in the book so prob a apllication of formule would suffice, thanks anyway I'll look into it more

[upper mulch]

Alright.

[safe radish]

@rose steeple Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[azure wigeon]

is this correct

[azure wigeon]

$Z:=:2\sqrt{3}+2i::=>:\left|z\right|=\sqrt{\left(2\sqrt{3}\right)^2+\left(2i\right)^2}:=>:\left|z\right|=\sqrt{12-4}:=>:\left|z\right|=\sqrt{8}$

[flat frigate]

Crawling Ham

[icy lance]

almost

dont include the i

just sqrt( (2root3)^2+(2)^2 )

[azure wigeon]

oh

[halcyon carbon]

,, z = \mr{a} + i\mb{b} \implies \abs{z} = \sqrt{\mr{a}^2 + \mb{b}^2}

[flat frigate]

Pure

[halcyon carbon]

i just want to latex

[azure wigeon]

thanks. that's all.

.close

[safe radish]

Channel closed

Closed by @azure wigeon

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wide cloud]

im confused for this circled area

[icy lance]

the f o g is wrong, but whats your confusion?

or is that the problem

[wide cloud]

yea im pretty sure the 16 and 9 are right but idk what the 5 is supposed ot be

[icy lance]

the 9 is wrong

show me f o g in its factored form before you expanded anything

[civic cave]

I don't understand on how to do this

[icy lance]

this channel is in use

#❓how-to-get-help

[civic cave]

Alr thx

[wide cloud]

ok

mb restroom

[icy lance]

(4x+3)^2+5 right, all sqrt'd

the thing is (a+b)^2 is not a^2+b^2 like you wrote there

well, kind of

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

[wide cloud]

ok so like...

1 sec

[icy lance]

not quite no, your x disappeared to some far away land

[wide cloud]

oops it always does

[icy lance]

and that middle term wouldnt be a +

[icy lance]

2ab not 2(a+b)

[wide cloud]

k

[icy lance]

not quite

the first term would be (4x)^2 not 4x^2 and the last would be 3^2 not 4^2

middle term is okay though

[wide cloud]

Ok

[icy lance]

not quite

2(4x)(3) isnt 2(4x)+2(3)

also whats that LHS?

[wide cloud]

the software?

sorry im takin so many tries

[icy lance]

no, the left hand side of that equation, why is there a 16x+9 over there

[wide cloud]

oh

[icy lance]

(4x+3)^2=16x^2+24x+9 though yes

[wide cloud]

thought we needed it its just a trinomial ok

makes it easier

oh

😭

thought i was done

[icy lance]

forgot the +5

[wide cloud]

oh 16x^2+24x+14

[icy lance]

indeed

[wide cloud]

Thank you

.close

[safe radish]

Channel closed

Closed by @wide cloud

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[safe radish]

Please don't occupy multiple help channels.

[weary osprey]

.close

[safe radish]

Channel closed

Closed by @weary osprey

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[edgy thistle]

How do you know which one it is as they both give you the same answer?

[icy lance]

'first positive angle'

[radiant kettle]

it says first positive value so u have to take the one before 5pi/3

[dark cargo]

sin(a) = sin(pi-a) and sin(a) = -sin(-a)

[edgy thistle]

How do you know which is closer to the positive?

[potent seal]

sin is negative in 3rd and 4th quadrants

So the angle in 3rd quadrant would be smaller than the one in the next quad

[edgy thistle]

Oh so you just look at where the function is negative in the quadrants and choose the smallest one?

[potent seal]

That's one way to do it, yes

[edgy thistle]

Kk thanks a lot

.close

[safe radish]

Channel closed

Closed by @edgy thistle

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[short sparrow]

Use the "Schrankensatz" (idk what its called in english) to show that
$$
|A^2-B^2| \leq 2 |A - B |,
$$
for all $A,B \in \mathbb{R}^{n,n}$ with $|A|,|B| < 1$, where $|\cdot |$ is the operator norm on $\mathbb{R}^{n,n}$.\
\
NOTE: First show that $(Dq(A))(B)=AB + BA$ for $A,B \in \mathbb{R}^{n,n}$ and
$$
q : \mathbb{R}^{n,n} \rightarrow \mathbb{R}^{n,n}, \ A \mapsto A \cdot A.
$$

[flat frigate]

Levens

[short sparrow]

any help?

[raven heart]

what's the Schrankensatz then ?

@short sparrow

[safe radish]

@short sparrow Has your question been resolved?

[short sparrow]

Let $\mathcal{V},\mathcal{W}$ be finite-dimensional Banach spaces, $U \subseteq \mathcal{V}$ open, and $f:U \rightarrow \mathcal{W}$ differentiable. Also, let $a,b \in U$, so that the "connecting distance" $\overline{ab}$ between $a$ and $b$ in $U$ is within $U$. Then holds
$$
|f(b) - f(a)| \leq \sup_{x \in \overline{ab}} |Df(x)| \cdot |b-a|,
$$
where $\mathcal{L}(\mathcal{V},\mathcal{W})$ is provided with the operator norm regarding the norms in $\mathcal{V}$ and $\mathcal{W}$.\
\
\
idk what the "connecting distance" is in english again, so here's the definition of that:\
\
Let $\mathcal{V}$ be a vector space and let $a,b \in \mathcal{V}$. Then $\overline{ab} := {a +t(b-a) \ | \ t \in [0,1]}$ is called the "connecting distance".

[flat frigate]

Levens

[raven heart]

ok yeah

you tried anything already ? @short sparrow

[short sparrow]

I did this
[q(A + tB) = (A + tB) \cdot (A + tB) = A \cdot A + 2tA \cdot B + t^2B \cdot B.]
[\lim_{{t \to 0}} \frac{1}{t}(q(A + tB) - q(A)) = \lim_{{t \to 0}} \frac{1}{t}(A \cdot A + 2tA \cdot B + t^2B \cdot B - A \cdot A).]
[\lim_{{t \to 0}} \frac{1}{t}(2tA \cdot B + t^2B \cdot B) = 2A \cdot B + B \cdot B.]

and then
[||A^2 - B^2|| = ||q(A) - q(B)||.]
[||q(A) - q(B)|| \leq ||Dq(A)|| \cdot ||A - B||.]
Since (||A||, ||B|| < 1), (||AB + BA|| \leq 2||A|| \cdot ||B|| < 2.)

[flat frigate]

Levens

[short sparrow]

this is without all the words lol

but maybe u can imagine

[raven heart]

1st line, you don't know that AB = BA

so tAB + tBA doesn't have to equal 2tAB

also the t^2/t B^2 vanishes when taking the limit

so that indeed the derivative of q at A in the direction of B is AB+BA

[short sparrow]

oh okay so what could i do

[raven heart]

nah it's just a correction of the first part

you used AB+BA afterwards anyway

however

how does that follow from your Schrankensatz ?

how do you know ||Dq(A)|| is the sup of the ||Dq(x)|| for x in the segment/'connecting distance' AB ?

[short sparrow]

idk tbh i jsut wanted to write something down

[raven heart]

well let's look at the sup then

,tex \begin{align*} \sup_{M\in \overline{AB}} \|Dq(M)\| &= \sup_{t\in [0,1]} \|Dq((1-t)A + tB)\| \\ &= \sup_{t\in [0,1]} \|(1-t)Dq(A) + tDq(B)\|\end{align*}

[flat frigate]

aPlatypus

[raven heart]

we can notice that Dq itself is linear here, so we can split that thing up

from triangular inequality we get that ||(1-t)Dq(A) + tDq(B)|| <= (1-t) ||Dq(A)|| + t ||Dq(B)||

and you can show that the operator norm of Dq(M) is 2 ||M|| for all M (that's what you tried to show in your second part)

,tex \begin{align*} \sup_{M\in \overline{AB}} \|Dq(M)\| &= \sup_{t\in [0,1]} \|Dq((1-t)A + tB)\| \\ &= \sup_{t\in [0,1]} \|(1-t)Dq(A) + tDq(B)\| \\ &\leq \sup_{t\in [0,1]} (1-t)\|Dq(A)\| + t\|Dq(B)\| \\ &=\sup_{t\in [0,1]} 2(1-t)\|A\| + 2t\|B\|\end{align*}

[flat frigate]

aPlatypus

[raven heart]

using $|A| \leq 1$ and $|B| \leq 1$, we indeed get that $\sup_{M\in \overline{AB}} |Dq(M)| \leq 2$

[flat frigate]

aPlatypus

[raven heart]

and now we can use your Schrankensatz

@short sparrow

[short sparrow]

mmmm i see i see

thanks, ill have a go

.close

[safe radish]

Channel closed

Closed by @short sparrow

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[raw pivot]

hi I need help proofing that the gamma function is smooth I have already proven that the gamma function is differentiable 1 time if that helps

[solar hazel]

yes that helps a lot because smooth and differentiable are equivalent for complex functions

[raw pivot]

I have proven that I can switch the integral and the derivative for the first derivative

[raw pivot]

for me its only defined for all Real values greater than 0

[solar hazel]

@junior smelt

[raw pivot]

$\Gamma'(x) = \frac{d}{dx} \int_0^\infty t^{x-1} e^{-t} , dt$

[flat frigate]

tobi

[raw pivot]

$\Gamma'(x) = \int_0^\infty t^{x-1} e^{-t} \log(t) , dt$

[flat frigate]

tobi

[raw pivot]

I have proven that I can switch differentiation and integral for the first derivative

[safe radish]

@raw pivot Has your question been resolved?

Channel closed

Closed by @raw pivot

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lean otter]

help me

[lean otter]

How did my math teacher go from

x^-15/6 y^5/6 to

[slender tusk]

think about the exponent rules

(x^a)^b - x^ab
x^a/x^b = x^(a-b)

[flat frigate]

Pure

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lean otter]

for this one tell me if im on the right track
so just writing out the top part i got
((x+h)^2 - 9(x+h)+12) - (x^2+9x+12)

[lean otter]

am i right so far?

[hollow pagoda]

yes

dont forget to divide by h

[lean otter]

nice alr

ye ye

just making sure i got the long part right so far

[hollow pagoda]

yh

[lean otter]

so for the 2nd part - F(x)

i just swap the symbols?

or do i only swap the minus to positives

because rn i got (X^2 + h^2 - 9x - 9h + 12) - (x^2 + 9x + 12)

like would these 2 12's cancel out?

or is the +12 in the 2nd part supposed to be -12

[hollow pagoda]

get rid of x^2 and 12

[lean otter]

should the last 12 be -12?

[hollow pagoda]

u will get something like this:
h^2 -18x -9h

divided by h

[lean otter]

ik but yk because how its - (x^2 -9x +12) and i swapped the -9x to +9x

would i do the same with the +12 and that would be -12

and then they cancel out

[hollow pagoda]

u can ye

[lean otter]

alr

wait what

i got sum way dif

(X^2 + h^2 - 9x - 9h + 12) + (x^2 + 9x - 12)

did i do sum wrong?

.close

[safe radish]

Channel closed

Closed by @old yacht

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lean otter]

if the gradient is -y-x/3y+x

[lean otter]

and i need to find out the points P,Q at a vertical tangent, so im guessing when the gradient is underfined how do i do that?

i thought about making the gradient equal to zero but i literally dont know what to do

.close

[safe radish]

Channel closed

Closed by @real sparrow

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

@wind siren Has your question been resolved?

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[coral sail]

This statement is confusing me, what is U? what does it mean when it says U does not equal R^n? I need to show if this statement is true or false, but I cant figure out what it means

[fickle pendant]

U is some subspace and it's..not N-dimensional euclidean space over the real numbers

[stoic dune]

U is a subspace of Rⁿ

Note we're excluding the case where U is the entire space, which is normally a subspace.

[coral sail]

ah I see, so U is not specifically defined, it is only a general subspace of R^n as long as it is not all of R^n

[stoic dune]

Yeah exactly

[coral sail]

ok, thanks!

.close

[safe radish]

Channel closed

Closed by @coral sail

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wise schooner]

having trouble understand this, why would k/2 +2 = 0 lead to 2 basic eigenvectors?
there are no leading 1's in any of the rows, doesn't that mean that there are 4 parameters and therefore 4 basic eigenvectors?

[safe radish]

@wise schooner Has your question been resolved?

[safe radish]

@wise schooner Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[devout harbor]

I need help solving a system of 4 equations

[devout harbor]

a1 + a3 = x10

b1+b3 = x20

sqrt(3)a2 +a4=x30

sqrt(3)b2+b4=x40

they answers should be in terms of x

for example b1 = 5x10 - 5/2x20

that is what I got but it does not work with the back of the book

[fleet condor]

none of this notation makes sense

what is "5x10"

and you have at least 8 variables in your 4 equations

[devout harbor]

sorry for the confusion

this is a differential equations

and so I had intial conditions where x1(0) = x10

it's just the notation the book choose

[icy lance]

could you take a picture of said question

[devout harbor]

so I subbed in zero and got that present equation

yes for sure

[royal kiln]

!original

[safe radish]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[royal kiln]

thanks

[devout harbor]

so the difference between mine and this example problem

is that my x1 has a sqrt(3) everywhere where there is a 3

because of how the eigenvalues worked out

my problem follows the example up until the eigenvalues

this is the problem statement

number 3

[fleet condor]

looks like you switch it all to b's then use your preferred method of simultaneous linear equation solving

[devout harbor]

gotcha

so rn I know that a1 = 7/10b1, a2=7/10b2, a3=1/2b3, a4=1/2b4

and so I just sub in for ai's?

rn I have....
7/10b1 +1/2b3=x10
b1 + b3 = x20
sqrt(3) 7/10b2 + 1/2b4 = x30
sqrt(3) b2 + b4 = x40

but when I plug this into wolfram alpha I get a different answer from the book :\

This is the answer in the book

this is what I get 😦

x2 is represented in terms b1 cost + b2sint and so on

[safe radish]

@devout harbor Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[subtle robin]

Mr. Earl E. Bird leaves his house for work at exactly 8:00 A.M. every morning. When he averages 40 miles per hour, he arrives at his workplace three minutes late. When he averages 60 miles per hour, he arrives three minutes early. At what average speed, in miles per hour, should Mr. Bird drive to arrive at his workplace precisely on time?

[subtle robin]

i did average speed = total distance / total time

s = 2d/d/40 + d/60

is this correct

o then i got d=48

so i substitute it back into my original equation

to solve for s

and i got s=48

.close

[safe radish]

Channel closed

Closed by @subtle robin

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fallen ether]

can someone help me with this step by step? I watched a video on the shell method but still don't know how to start.

[plucky elk]

Do a relevant example
https://tutorial.math.lamar.edu/classes/calci/volumewithcylinder.aspx

[fallen ether]

is the 4th example the closest to my problem/

[plucky elk]

Try it

[fallen ether]

wait no they have x values

i dont have any x values

[junior smelt]

...but you can find them! "...region bounded by $y =\sqrt{x} + 1$ and $y = x^2 + 1$"

[flat frigate]

@junior smelt

[fallen ether]

yes i will make them equal to each other and solve

[junior smelt]

Yep

[fallen ether]

would i just have sqrt(x) - x^2 = 0?

[junior smelt]

That would be what you're trying to solve yea

[safe radish]

@fallen ether Has your question been resolved?

[fallen ether]

i got my x values and i plugged in 0.5 into both original equations but now i dont know how to move forward

[junior smelt]

probably would have to leave someone else to explain the shell method, I'm not too great at explaining it

[fallen ether]

<@&286206848099549185>

[safe radish]

@fallen ether Has your question been resolved?

[fallen ether]

.close

[safe radish]

Channel closed

Closed by @fallen ether

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[opaque lintel]

Symbolab shows the top 2 as thet answer, and I did it on pather and got pi/4 myself

Idk how they got 5pi/12

but even then

The website isn't accepting the solutions

[quiet plume]

Notice that cos(5pi/4) = -sqrt(2)/2 as well.

[opaque lintel]

Is it because I have to ignore the 3 ?

until later

[quiet plume]

No.

It`s because cos(x) = y has 2 solutions almost everywhere.

[opaque lintel]

Okay but

We're
cos(3x) = sqrt(2)/2

I have that 3 in there still

so I have to acos

to isolate x

[quiet plume]

Yes

But

When you do arccos, it only gives the solution in [0,pi]

[opaque lintel]

acos can't go past pi/2

Yes

So then I have to ignore the 3 while finding initial solutions then?

[drowsy karma]

$2cos3x=-\sqrt{2}\Leftrightarrow cos3x=-\frac{\sqrt{2}}{2}\Leftrightarrow \\3x=\frac{3\pi}{4}+2k\pi\text{ }\text{ }\vee \text{ }\text{ }3x=\frac{5\pi}{4}+2k\pi\Leftrightarrow \\x=\frac{\pi}4{}+\frac{2}{3}k\pi\text{ }\text{} \vee \text{ }\text{ }x=\frac{5}{12}\pi+\frac{2}{3}k\pi$

[quiet plume]

The other solution is 2pi - arccos(...)

[flat frigate]

Joanna Angel

[drowsy karma]

and now you need to sleect k

[drowsy karma]

yes i give you more

all general solutions

and now

you must sleect k

to mke solutions go to yoru interval

[opaque lintel]

What

[drowsy karma]

do you understand me ?

[opaque lintel]

No

[drowsy karma]

i gave you all solutions

of yoru euqation

in a geenral way

and now you have to establish values of k

[opaque lintel]

I've already included those

[drowsy karma]

to less

cant you see?

[opaque lintel]

No

[drowsy karma]

k = 0, 1, 2, 3, 4, etc

[opaque lintel]

Anything past 0 will break out of [0,2pi)

[drowsy karma]

you need to choose the value of k so that the root falls within your range

[opaque lintel]

What

[drowsy karma]

$2cos3x=-\sqrt{2}\Leftrightarrow cos3x=-\frac{\sqrt{2}}{2}\Leftrightarrow \\3x=\frac{3\pi}{4}+2k\pi\text{ }\text{ }\vee \text{ }\text{ }3x=\frac{5\pi}{4}+2k\pi\Leftrightarrow \\x=\frac{\pi}4{}+\frac{2}{3}k\pi\text{ }\text{} \vee \text{ }\text{ }x=\frac{5}{12}\pi+\frac{2}{3}k\pi$

[quiet plume]

The point is, there are more solutions that you would usually get in [0, 2pi) because you have 3x, not x. You can think of it as your cos wave being squished on the y axis by a factor of 3, so if you had 2 solutions, you should end up with 6 in a way because it does the whole thing 3 times in the same range.

[flat frigate]

Joanna Angel

[drowsy karma]

yes

[opaque lintel]

OHHHH

[drowsy karma]

for k = 0, 1, 2

i gave you

[opaque lintel]

So then I should have 4 four solutions?

pi/4
11pi/12
5pi/4
23pi/2

[drowsy karma]

$x\in \left{\frac{\pi}{4},\frac{11\pi}{12},\frac{19\pi}{12},\frac{5\pi}{12},\frac{13\pi}{12},\frac{7\pi}{4} \right}$

[flat frigate]

Joanna Angel

[quiet plume]

So in general, if you must solve cos(3x) = y, you would do :

3x = arccos(y) + 2pin
3x = 2pi - (arccos(y)) + 2pin

Then solve for x by dividing by 3 and check where your solutions land by adding progressively larger n

[drowsy karma]

6 solutions as Azyra also confirmed that

[opaque lintel]

Wow, that is a lot more work than our class usually does. I guess that's the downside of the professor using a website with everything done outside of the class

[drowsy karma]

🙂 well that happens

[opaque lintel]

\

[drowsy karma]

🙂

[opaque lintel]

Normally with this professor it doesn't, just since he started using this website

I do wish he'd give us some harder problems so that we can carry it over to other classes

He makes all the exams/in-class really simple problems

[drowsy karma]

yes an di suspect, he was not verifying this website difficutly level

andnow

you ahve all surprises

[opaque lintel]

Can I use the same room for a new question or do I need to close and then re-open in a new channel?

[drowsy karma]

i thikn no need

to create other room

is yours

i never created a room yet 🙂

[drowsy karma]

your trig equation ebhaves liek quadratic equation

[opaque lintel]

Not sure how to get rid of the 1

Ohhhhh

[drowsy karma]

so plz substitue u = cost

pay attention that you only accept :

$\left| u \right|\le 1$

[flat frigate]

Joanna Angel

[opaque lintel]

cos(t) - 2
and
cos(t) + 1

so just -1

[drowsy karma]

no

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

[flat frigate]

Joanna Angel

[opaque lintel]

oh

That's right, I forgot the 2

[drowsy karma]

🙂

you must find 4 solutions in your given interval

[opaque lintel]

Yeah

cos(t) = 1
cos(t) = -1/2

[drowsy karma]

yes and each fo the gives you two solutions

[opaque lintel]

Wait, wouldn't it be 3 solutions?

[drowsy karma]

no

cost = 1 has two solutions

[opaque lintel]

cos(t) is only 1 at one point

[drowsy karma]

two

[opaque lintel]

Really?

[drowsy karma]

0

[opaque lintel]

Oh, 2pi

[drowsy karma]

and

[opaque lintel]

but

[drowsy karma]

2Pi

[opaque lintel]

2pi not included

so only 3 solutions

[drowsy karma]

ah ok

i did not see it )

small photo

so 3 solutins yes

[opaque lintel]

0
2pi/3
4pi/3

This one was much simplier

[drowsy karma]

i agree 🙂

[drowsy karma]

congrats 🙂

[opaque lintel]

This one looks weird

i guess move the 2 over to 2sin?

[safe radish]

@opaque lintel Has your question been resolved?

[opaque lintel]

I have no idea how to factor this

<@&286206848099549185>

[eager plover]

yes

also please do not forget to put the angle theta (the 0 with a dash in the middle) or x

[opaque lintel]

I intentionally leave it out, sorry

and no, I'm not changing that unless it's for exams

Only because they dock points

[eager plover]

it hurts ny brain

you have to specify the angke

always

[opaque lintel]

Don't know what to say. Either way, I gave up on the problem and just used a website to do it for me

Which sucks because Idk what it did to get those answers

[eager plover]

i can help you factor thats no problem

[opaque lintel]

This is what it suggests

Idk how the hell it got that

[opaque lintel]

This is as far as I got and I don't even know what's left to go on

[eager plover]

gimme a second to copy down the problem ok?

[opaque lintel]

Okay

[young nexus]

its simple ab+a+b+1=(a+1)(b+1)

[opaque lintel]

I'm going to start working on the next problem in the meantime

[opaque lintel]

Ah hell, it's one of those rules I probably forgot

but how come one of them is 2 and the other one is 1

[opaque lintel]

Where can I learn more about this trick/rule

[eager plover]

the co -efficient 2 is on the cosine not the 1

[eager plover]

(sin x + 1)*(2cos x + 1)

[young nexus]

practice

[opaque lintel]

??

[opaque lintel]

I get that, but idk how to get it there

[opaque lintel]

Yes, the website told me that here: @eager plover

[eager plover]

FOIL

[opaque lintel]

I am confused

I'm just going to move on. Thank you though

[eager plover]

im sending you a photo now

1* 2cos x = 2 cos x but 1*1 = 1

[opaque lintel]

The FOIL part does not matter to me.

I need to know how to get to that step

[eager plover]

It’s just the formula ab+a+b+1 = (a+1)(b+1)

It’s a memorization thing

[opaque lintel]

I'll have to assume so, it's hard for me to see it

Either way I'm stuck on a new problem now

[eager plover]

In math there’s a lot of don’t question it just memorize it formulas

[opaque lintel]

I am stuck on this mess now:

I did everything by hand and then used a calculator to find those unfamiliar angles sin=1/6

but according to this website one of the answers is -arcsin(1/6)+2pi