help-36

[warped bluff]

This is easy I just didn’t think about that

Thanks

I will solve it and send the answer

[raw pike]

Np!

[warped bluff]

I got the blue base is 1.054092559

I got the ratio is 0.72 with the 2 reacurring

[raw pike]

I haven’t done it myself. Can you show your reasoning?

[raw pike]

Wait, where did you calculate the blue triangle area?

[warped bluff]

Oh my bad

[raw pike]

Is the area of the blue triangle just 5/6?

[raw pike]

Alright that looks good, good job! If you want/need an exact answer then you can try the similar triangle method which doesn’t involve any trig

Or does the 13/18 happen to be exact?

[warped bluff]

What do you mean by exact

[raw pike]

Like is it actually just 13/18

[warped bluff]

So the pentagon area is 13/6 and the area of the rectangle is 3 so you multiply each by 6 and get 13:18

So yeah it is exact

[raw pike]

Oh sweet alright

[warped bluff]

Thanks for your help

[raw pike]

You’re welcome!

[warped bluff]

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[valid mantle]

number 16

help

i tought it has but the answers say it doesnt

why does the limit not exist

cant we do h*(2+h)/h = 2+h=2+0=2

[celest crane]

Try taking the limit from both the positive and negative side and see what results you get.

[valid mantle]

but why cant i do this

[celest crane]

Did you try doing what I said to do?

[valid mantle]

okay okay

so

if we take something negative a little less than zero

the limit will be negative and if we take something a bit positive it will be positive?

[celest crane]

Yes, but how does that affect the limit from each direction in this particular problem?

[valid mantle]

does this show it doesnt exist?

[celest crane]

It does, but it's important for you to understand why it is DNE.

[valid mantle]

yeah thats what im tryong to do

so the limit is not equal from both sides so the function is not differentiable?

do i check from both sides all the limits that go to a number?

[celest crane]

You don't always have to. Note that h in the denominator cannot equal 0, so in that case you should check because that is what the limit is approaching.

Although you can technically eliminate the h in the denominator mathematically, you must always consider the original equation/expression.

[valid mantle]

i see

thanks

[celest crane]

👍

[valid mantle]

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[sullen totem]

radius and interval of convergence?

[sullen totem]

i think lim is

radius = 0 and interval = {0}??

help

<@&286206848099549185>

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[tranquil pine]

hi

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@tranquil pine Has your question been resolved?

[vital crag]

Don't open a help channel without a question

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[tame tusk]

How do i continue from here

Topic: Solving Trigo. Functions

[vivid walrus]

,rccw

[tame tusk]

Oh sorry

[vivid walrus]

what's the question?

oh wait

it's 4?

this seems like a proof

[tame tusk]

Which among the numbers in the set is the solution of the following

Like we need to do the solution

For the following Trigo equation

[vivid walrus]

isn't this trig identities?

[tame tusk]

Yea

[vivid walrus]

you use pythagorean trigonometric identities

not sure what you did above

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[clever locust]

Hi.
I need help with this:
"Let A = {1, 2, 4, a, b, c}. Identify the following as true or false.
c) c∉A
e) { } ∉ A"
I got these wrong. I said that "c" and "e" are false. The book says the opposite.
Isn't c an element of A?
The book also says that for any set A, since there are no elements of Ø that are not in set A, then Ø ⊆ A. Is "e" right because you can't say that Ø is an element?

[blissful meadow]

Neither are right. Maybe they mixed up true/false in the negation?

[humble marlin]

I think e) is true since you don't really refer to an empty set as an element of a set unless you are talking about a power set. In this case, you would say the empty set is an element of the power set. Otherwise, you have to say it is a subset of A because an empty set is a subset of all sets, at least that's how it is defined. For c, however, I have no clue.

[blissful meadow]

Oh wait yes e is true. It's a bit annoying with the negation haha

But yeah the empty set is not an element of A, so { } ∉ A is a true statement.

As for c, it is pretty clear that c is in A, so c∉A should be false.

[clever locust]

oh. so would Ø also be considered a set?

[humble marlin]

well it's called an empty set

so yes it is a set

[clever locust]

gotcha. still didn't register that well that it is actually a set

[clever locust]

thanks! still hadn't gotten that it was actually a set. sometimes i understand things but usually forget what they're called

[clever locust]

i feel like it should be completely false. but then i get worried that it was something I didn't understand

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[thick estuary]

so I need help with this one

[thick estuary]

I cant seperate it like so but is there a way I could seperate it?

ik the answer is 2/3 x^3 + 9/2x^2 + 4x + c im just wondering if there is some way I can seperate them (even though they arent constants) rather than multiplying them

[fossil geyser]

Not really. You could do integration by parts, if you've learned that

but the simplest method, by far, is just multiplying

[thick estuary]

ok, thanks. I think we learn integration by parts next chapter. Thanks

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[tranquil pine]

So I factored it but im not sure what to do from here $\frac{5x(x-3)(x-2)}{3(x-2)(x+1)}$

[soft zealot]

Matt

[lyric summit]

you can start with:

1. domain

2. limits on ends of intervals of th doman

3. asymptoes, their equations

4 zeros of fucntions

also notice that (x-2) is reduced

ti iwl make yoru funciton easdier

but still x cant be equal to 2

so it is hole there

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[tranquil pine]

ok i get the rest but im not quite sure

how to find the asymptotes for this equation yk

since i graphed it and its slanted

theres no vertical asymptotes so thats whats confusing me yk

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[rich tide]

do you know the formula for cos2x

in terms of cos and in terms of tan

[radiant hazel]

Complete the Square.

[jagged inlet]

? @trim sigil

[radiant hazel]

tanx= sin/cos

[rich tide]

well go by my method

its way too short

[radiant hazel]

And below: Cos²(x) +sin²(x)=1

[rich tide]

$2\cos^{2}x-1=\cos2x=\frac{1-\tan^{2}x}{1+\tan^{2}x}$

[soft zealot]

B-eard

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[trim sigil]

Guys idk where to start
What method?

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[lethal pebble]

what's the sum of the roots of this equation

[lyric summit]

it's very easy, but time-consuming, I don't know if anyone will want to do it here, you have to put each denominator into a product form, and then multiply both sides of the equation by the common denominator, etc.

[lethal pebble]

there is an easier way

faster

[lyric summit]

you can use partial fractions sure

so if you know, for what you publish it here >?

[lethal pebble]

why do you care

[lyric summit]

🙂

[lethal pebble]

I'm not forcing anyone to solve it

[lyric summit]

got it you jsut wanted to share it, )

[lethal pebble]

i see

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[fickle apex]

he just wants the answer and is lazy to solve it bruh

[fickle apex]

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[elfin venture]

Do i use binomial

[elfin venture]

since it looks like for a binomial distribution

I thought i should use n trials 100, k be 3 and probability be 0.02

so then find P(X=3)

is this right?

<@&286206848099549185> Hi, just need some small help

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[tranquil pine]

<tikz;siunitx;enumitem>
\usetikzlibrary{decorations.pathmorphing}
Three identical \qty{8.50}{kg} masses are hung by three identical springs. Each spring has a force constant of \qty{7.80}{kN/m} and was \qty{12.0}{cm} long before any masses were attached to it.
\env{enumerate}{[a)]
\ii Draw a free-body diagram of each mass. 
\ii How long is each spring when hanging as described? 
}
My sketch of the problem is as follows:
\env{center}{
  \tikz{
    \tikzset{
      spring/.style={thick, decorate, decoration={zigzag, pre       length=0.1cm, post length=0.1cm, segment length=6}},
      rect/.style={fill=orange!70, thick, draw=orange},
    }
    \draw[thick, black] (0,0) -- (2,0); 
    \foreach \y/\label in {0/A, -2/B, -4/C} {
      \draw[spring] (1,\y) -- (1,{\y-1});
      \draw [rect] (0.5, \y-2) rectangle (1.5,{\y-1}) node[midway,text=black]{\label};
        }
    }
}
And my free-body diagrams are as follows: 
\env{center}{
  \tikz{
    \tikzset{
      rect/.style={fill=orange!70, thick, draw = orange},
      weight/.style={->, thick, red},
      sf/.style={->, thick, blue}
       }
    \foreach \z/\label/\nextlabel in {0/A/B, 2/B/C, 4/C}{ 
      \draw [rect] (\z, 0) rectangle (\z+1, 1) node[midway, text = black]{\label};
    \ifnum\z<4:
        \draw[weight] (\z+0.3, 0) -- ++(0, -1) node[left, anchor=east] {$w_{\label}$};
    \else:
        \draw[weight] (\z+0.5, 0) -- ++(0, -1) node[left, anchor=east] {$w_{\label}$};
    \fi
      \draw[sf] (\z+0.5, 1) -- ++(0, 1) node[right, anchor=west] {$F_{\label}$};
    \ifnum\z<4:
      \draw[sf] (\z+0.6, 0) -- ++(0, -1) node[right, anchor=west] {$F_{\nextlabel}$};
    \fi
       }
    }
}

[tranquil pine]

My questiion is, are my free-body diagrams correct? And how do i go about calculating the extended length?

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@tranquil pine Has your question been resolved?

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@tranquil pine Has your question been resolved?

[native spindle]

I think they are okay

I suppose you just have to find the raw values for each force now

[tranquil pine]

hm

so lets take our origin to be the ceiling

we have [
F_A = kx_1\
F_B = kx_2 \
F_C = kx_3
]

[tranquil pine]

i guess?

but for the x's what are we counting as the elongation?

no wait

[
w_C-F_C = 0 \
F_C + w_B - F_B = 0\
w_A + F_B - F_A = 0
]

[native spindle]

I'm sorry for saying this now of all times, alex, but although I can interpret force diagrams, the last time I dealt with spring forces was 5 years ago. So I'll have to spend some time rejogging my memory

[tranquil pine]

[
mg = kx_3 \implies x_3 = \f{mg}k\
kx_2 = mg + mg = 2mg \implies x_2 = \f{2mg}k \
kx_1 = 2mg + mg = 3mg \implies x_1 = \f{3mg}k
]

[tranquil pine]

would this be correct then i guess

[tranquil pine]

oh no worries

[native spindle]

well, you've assumed that the entire system is at rest right?

nothing is accelerating anywhere?

therefore the sum of all forces is zero for objects A, B and C?

[tranquil pine]

indeed

[native spindle]

then I will trust you on this one

For a more concrete verification I'm going to do a gigachad move

which requires swallowing my pride

<@&286206848099549185> I need someone to take my place to help with calculating simple forces involving springs

[tranquil pine]

its okay i think that should be it

i hope

makes sense at least

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[zealous parcel]

hi, is the answer for this

[zealous parcel]

is "-ln | sec z | + C" ?

if not can anyone show me the step by step on how to solve this

[shell oriole]

How did u get -ln""

Show ur work first

@zealous parcel basically 1+Cot^2 is csc^2 right

[zealous parcel]

yes

[shell oriole]

And then cancel one csc

So that's csc cot

[zealous parcel]

wait

so

[shell oriole]

Yea?

[shell oriole]

Cos there csc^2 on top

And csc in denominator

[zealous parcel]

ooh it will become csc (z) cot (z)

[shell oriole]

Ya

[zealous parcel]

whats the next step?

[shell oriole]

Express everything in sin cos

[shell oriole]

Express it as cos/sin^2

@zealous parcel any more doubts?

[zealous parcel]

still confused huhu

[shell oriole]

Csc is 1/sin

Cot is cos/sin

[shell oriole]

Substitute man

[zealous parcel]

isnt cot is 1/tan?

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[thick grotto]

(lg(a))^(1/2) + (lg(b))^(1/2) + (lg(a)^(1/2)) + (lg(b)^(1/2)) = 100.

ab = ?

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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.

[thick grotto]

Ok

Also, all terms are integers*

<@&286206848099549185>

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@thick grotto Has your question been resolved?

[warm forge]

What's "lg a"? Is this a new way to write log(a)?

$\sqrt{log(a)} +\sqrt{log(b)} + log(\sqrt{a}) +log(\sqrt{b}) =100$?

[soft zealot]

G. Spark

[warm forge]

This does not match your description at the top.

But that is incorrect, so struggling for the problem statement.

Maybe possible to get rid of the logs?

[thick grotto]

lg(a) = log₁₀(a)

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@thick grotto Has your question been resolved?

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@thick grotto Has your question been resolved?

[pliant shore]

lg(a) can also mean log base 2 (a)

depends on field

[thick grotto]

But here lg(a) means log₁₀(a). This is a simple decimal logarithm.

I don't rule out that it could be a binary logarithm (which I doubt), but even so.. I'm already interested in finding an answer to the question, is it possible to find a solution to this problem with a decimal logarithm?

[pliant shore]

yeah so $\sqrt{\log a} + \sqrt{\log b} + \frac{1}{2} \left(\log a + \log b \right) = 100$

[soft zealot]

south

[pliant shore]

or $p + q + \frac{1}{2} \left(p^2 + q^2 \right) = 100$

[soft zealot]

south

[pliant shore]

$p^2 + q^2 + 2p + 2q = 200$

[soft zealot]

south

[pliant shore]

ah but $p^2 = \log a \implies 10^{p^2}$ so we need to find $10^{p^2+q^2}$ for $ab$

[soft zealot]

south

[pliant shore]

the only integer solutions should be p = 8, q = 10

and like p = 10, q= 8

I'm not sure how to prove this however

but then if all terms are integers, the solution to p^2 + q^2 + 2p + 2q = 200 will give all the integer solutions of the original equation

ah right, add +1+1 to both sides to get $(p + 1)^2 + (q + 1)^2 = 202$

[soft zealot]

south

[pliant shore]

if it helps we only need to check the squares from 1 to 100

like q + 1 = 11 is the same as p + 1 = 9

nice problem

[rustic wedge]

Its easy to observe 0 <= p,q <= 11

[rustic wedge]

So u don't have many values to test @thick grotto

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@thick grotto Has your question been resolved?

[thick grotto]

Thank you for your responses. I think this is enough for me to try to find a more accurate explanation of the solution. Anyway, if I don't succeed, I will ask this question again. Goodbye.

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[verbal steppe]

I’m sort of stuck

On

$\frac{dy}{dx}e^{-x^2} -2xye^{-x^2} = sin(2x)$

[soft zealot]

Merineth

[verbal steppe]

Now I’m supposed to take the primitive of each term, simplify and then put in x = 0 and y = 1

However how tf do I take the integral of

$\int \frac{dy}{dx}e^{-x^2}$

[soft zealot]

Merineth

[verbal steppe]

This is the solution provided but

I don’t understand where -2x goes..

[quasi lotus]

apply the integrating factor

[opal plinth]

Product rule or integration by parts

[quasi lotus]

I(x) = e^integral(P(x))dx

then y = 1/I(x) integral(I(x)Q(x))dx
Q(x) is -e^x^2(sin2x)

[verbal steppe]

omg I didn’t realize

Integration by parts

Let me give it a try

[quasi lotus]

lol

a u-sub might be possible too, but do it however u feel like it

[verbal steppe]

$\int y’ * \int e^{-x^2}$

Is that what you mean

[soft zealot]

Merineth

[quasi lotus]

no

[verbal steppe]

I tried utilizing this

On

[verbal steppe]

That

But I just got an infinite loop of it

[quasi lotus]

just use the tabular method (DI method)

[verbal steppe]

Idk what that is

So no can do

[quasi lotus]

an easier method of IBP

[verbal steppe]

IBP?

[quasi lotus]

i never learnt the the original equation for IBP (integration by parts)

[verbal steppe]

Well

I’m doing original

[quasi lotus]

use the DI method, way easier

[verbal steppe]

No

[quasi lotus]

Okay, you do you

[verbal steppe]

I won’t unlearn it and start using another method

[quasi lotus]

lol u will take less time to learn the DI method than u will take to learn the original IBP equation

[verbal steppe]

Ok, what is Di method?

[quasi lotus]

A shortcut method for IBP

watch a video of blackpenredpen using it

he explains it in full detail

[quasi lotus]

sorry i made a mistake here (incorrectly read the question)

@verbal steppe this should be the final answer, hope this helped lol (sorry if i made a mistake in my solution i tend to make stupid mistakes)

[verbal steppe]

Okay the DI method seems pretty alright

Just looked it up

[quasi lotus]

yeah its easy

[quasi lotus]

also this solution is (probably) wrong, its supposed to be e^-x^2 not e^x^2 (for the final part)

[verbal steppe]

That is from a math exam

[quasi lotus]

nvm i forgot to change my incorrect answer to the correct one LMAO

[verbal steppe]

I’d be surprised if it was wrong

[quasi lotus]

yeah it IS correct

[quasi lotus]

no its not

i just forgot the negative sign

@verbal steppe heres the general formula to solve these linear differential equations

[verbal steppe]

What is the integral of

$e^{-x^2}$

[soft zealot]

Merineth

[verbal steppe]

It’s divided, right?

[quasi lotus]

its undefined

[verbal steppe]

So it can’t be solved?

[quasi lotus]

without special functions? no

if u add the limits from -inf to +inf, then u will get sqrt(pi)

[verbal steppe]

How does he remove -2x?

[quasi lotus]

wdym

[quasi lotus]

yee i dont understand that method, i used my own method i learnt

[verbal steppe]

@opal plinth does the smartest man alive get it?

[crude spoke]

i concur...nel is the smartest man alive

[crude spoke]

they expressed it in a product rule form?

after rewriting they can just reduce the entire expression to:

,, {\color{green}\qty(e^{-x^2}y)'} = e^{-x^2} y' + \qty(e^{-x^2})' y

[soft zealot]

kanna

[final saddle]

@verbal steppe Has your question been resolved?

[opal plinth]

What kanna said

[verbal steppe]

Yeah !! I get it now

Took a while

Bless kanna

Second smartest man alive ❤️

[opal plinth]

Also it's very wrong to say that the integral of e^{-x^2} is undefined

It's complicated but certainly defined

However you don't need that if you want to do integration by parts

Take u'(x) = -2xe^{-x^2}, so u(x) = e^{-x^2}
You have u(x)y'(x) + u'(x)y(x)
The integral of that is int [ u(x)y'(x) dx] + int [ u'(x)y(x) dx]
= int [ u(x)y'(x) dx] + u(x)y(x) - int [ u(x)y'(x) dx]
= u(x)y(x)
= e^{-x^2} y

And on the other side you have the integral of -sin(2x)

@verbal steppe yeah?

[verbal steppe]

I’m here :3

We are just wondering

$\int (e^{-x^2}y)’ dx = -\int sin(2x)dx$

Each and every time we solve this

And put in f(0) = 1

We get that C = -1/2

But the solutions says C = 1

What is going wrong?

[soft zealot]

Merineth

[verbal steppe]

C=1

[opal plinth]

Do you have the C inside or outside the parentheses

Never mind you shouldn't get -1/2 anyway

[opal plinth]

What's the next step from this?

[verbal steppe]

Can I come back with an answer in like 10 min? I have to quickly go to the store

[opal plinth]

No worries

[verbal steppe]

[final saddle]

@verbal steppe Has your question been resolved?

[verbal steppe]

Okay i'm back :3

sorry it took a while

$e^{-x^2}y + C_1 = \frac{cos2x}{2} + C_2$

@opal plinth This is my first step

[soft zealot]

Merineth

[opal plinth]

Ok, now remove the C1 and just use C

and isolate y

[verbal steppe]

oh

so it's necessary to isolate y?

can't i just plug in y and x now?

[opal plinth]

Well ok let's do that

$e^{-0^2} \cdot 1 = \frac{\cos(2 \cdot 0)}{2} + C$

[soft zealot]

Nel

[verbal steppe]

LHS = 1
RHS = 1/2 + C

isolate C and i'm at -1/2

[opal plinth]

Wrong

[verbal steppe]

oh

$1 \cdot 1 = \frac12 + C$

[soft zealot]

Merineth

[verbal steppe]

OH

uh

C = 1/2

[opal plinth]

Right

[verbal steppe]

But according to him

C = 1

[verbal steppe]

here

[opal plinth]

Because it's factored by 1/2

[verbal steppe]

OHHHHHH

[opal plinth]

Use 1/2 in this and you'll get the same final answer

[verbal steppe]

Nel, the math god

[opal plinth]

(well, 1/2 as C2 and 0 as C1)

[verbal steppe]

$xy'-3y=\frac{x^4}{1-x^2}$

[soft zealot]

Merineth

[verbal steppe]

if i have something like this

i'm having trouble getting rid of the x from the first term

do i divide by x?

$y' -\frac{3y}{x} = \frac{\frac{x^4}{1-x^2}}{x}$

[soft zealot]

Merineth

[verbal steppe]

Will that allow me to use

[opal plinth]

Yes

[verbal steppe]

Do you have any advice for finding primitive to

hmm

$\int \frac{-3}{x}dx$

[soft zealot]

Merineth

[verbal steppe]

Can that be written as

$-3\int\frac1x dx$

[soft zealot]

Merineth

[half salmon]

Yes it can

[verbal steppe]

so when i "move" out -3 i am actually in fact factorising it out from the function?

[opal plinth]

Yes

[verbal steppe]

Perfect

I'll be sure to remember that

[opal plinth]

It works because of the linearity of differentiation, which I mentioned before, I think yesterday

If you think of the integration as a function f, you have f(-3/x), and because that function is linear, it's equal to -3f(1/x)

Which you can think of as "factorizing" out

[verbal steppe]

\begin{align}
y' -\frac{3y}{x} &= \frac{\frac{x^4}{1-x^2}}{x} \
\text{Multiply with IF:} e^{G(x)} &= e^{-3ln|x|} \
y'e^{-3ln|x|} - \frac{3ye^{-3ln|x|}}{x} &= \frac{x^4e^{-3ln|x|}}{x+x^3} \
(ye^{-3ln|x|})' &= \frac{x^4e^{-3ln|x|}}{x+x^3}
\end{align}

I'm not sure how i use align

[opal plinth]

\begin{align}
y' -\frac{3y}{x} &= \frac{\frac{x^4}{1-x^2}}{x} \\
\text{Multiply with IF:} e^{G(x)} &= e^{-3ln|x|} \\
y'e^{-3ln|x|} - \frac{3ye^{-3ln|x|}}{x} &= frac{x^4e^{-3ln|x|}}{x+x^3} \\
(ye^{-3ln|x|})' &= \frac{x^4e^{-3ln|x|}}{x+x^3}
\end{align}

[soft zealot]

Merineth

[opal plinth]

You need to simplify e^{-3 ln x}

[verbal steppe]

Oh

i didn't realize that could be done

Becuase i'm having trouble on step 4 on RHS trying to do an integral of that

[opal plinth]

Well, also the x + x^3 is wrong, and even if it were correct, it should have been simplified too

[verbal steppe]

Is it?

Isn't it just inverse multiplication

[opal plinth]

Yeah

[verbal steppe]

x multiplied with entire denominator

1-x^3

woops

i made a + sign

[opal plinth]

You have x^4 / (x (1-x^2))

[verbal steppe]

aaah

i can simplify

hold on let me try!

my best guess would be to

$e^{-3lnx} = x^{-3}$

[soft zealot]

Merineth

[verbal steppe]

giving the simplified version of

$\frac{1}{1-x^2}$

[soft zealot]

Merineth

[opal plinth]

Yes

[verbal steppe]

<3

$\int (y'x^3)'dx = \int \frac{1}{1-x^2}$

[soft zealot]

Merineth

[verbal steppe]

$y'x^3 + C = \int \frac{1}{1-x^2}$

[soft zealot]

Merineth

[verbal steppe]

$y'x^{-3} + C_1 = -\frac{tan^{-1}(x)}{x}+C_2$

[soft zealot]

Merineth

[verbal steppe]

And since we had f(1) = 0

$0\dot 1^{-3} + C_1 = -\frac{tan^{-1}(1)}{1}+C_2$

[soft zealot]

Merineth

[opal plinth]

Wait how did you get y'x^3

[verbal steppe]

Hm

i had $ye^{-3lnx}$

[soft zealot]

Merineth

[verbal steppe]

yx^-3

i think i keep making small mistakes like that

but it should be y

and not y'

[opal plinth]

Still incorrect

[verbal steppe]

really?

$e^{xlna} = a^x$

[soft zealot]

Merineth

[verbal steppe]

ohgod

no wait

I still think it's right?

x is -3

a is x

x^-3

[opal plinth]

(yx^-3)' = y'x^-3 - 3yx^-4

Yeah ok

So yx^-3 + c_1 = ...

Integral of 1/(1-x^2) is not -tan^-1(x)/x

[verbal steppe]

Let me double check

where x is -x and a is 1

[opal plinth]

No

[verbal steppe]

1 is sqrt 1?

[opal plinth]

(-x)^2 = x^2

[verbal steppe]

hmm

[opal plinth]

You can't use that if you have -x^2

[verbal steppe]

$\int \frac{1}{(-x)^2 + 1^2}$

[soft zealot]

Merineth

[verbal steppe]

So not like this?

[opal plinth]

That's not what you have

[verbal steppe]

oh wait hold on, can't i move the - infront of the

integral

[opal plinth]

Which one

[verbal steppe]

$-\int \frac{1}{x^2 + 1^2}$

[soft zealot]

Merineth

[opal plinth]

Absolutely not

[verbal steppe]

is that because it is

-x * -x

[opal plinth]

No

[verbal steppe]

hmm then i'm not really sure how i will utilize the formula

$\int \frac{1}{1-x^2}$

[soft zealot]

Merineth

[verbal steppe]

this is what i had from the start

[opal plinth]

$\frac{1}{1 - x^2} = -\frac{1}{- 1 + x^2} \neq -\frac{1}{1 + x^2}$

[verbal steppe]

but i need a + between them for the formula to work

oh yeah that seems silly lol

[soft zealot]

Nel

[opal plinth]

Rearrange the terms and find another formula

[verbal steppe]

i was thinking

that i swap x^2 and 1

$\int \frac{1}{-x^2+1}$

[soft zealot]

Merineth

[verbal steppe]

like so

[opal plinth]

Sure, then what?

[verbal steppe]

I now use the formula

[opal plinth]

No

Again, -x^2 != (-x)^2

[verbal steppe]

Hmm

[opal plinth]

$\int \frac{1}{1-x^2} = -\int \frac{1}{x^2-1}$

[soft zealot]

Nel

[opal plinth]

You must have a formula that looks like this RHS

[opal plinth]

I guess you don't

[verbal steppe]

No, sadly

That's why i'm trying to make arctan to work

[opal plinth]

Ok but you can factorize x^2 - 1

[verbal steppe]

(x-1)(x+1)

[opal plinth]

Right, now let me think

Ok, can you do partial fraction decomposition?

[verbal steppe]

I retried it

what about -tan^-1(x)

$\int \frac{1}{x^2+a^2}dx = \frac1a tan^{-1}(\frac{x}{a} ) + C$

[soft zealot]

Merineth

[verbal steppe]

where a = 1, x= -x

$\frac11 tan^{-1}(\frac{-x}{1} ) + C$

[soft zealot]

Merineth

[opal plinth]

.

[verbal steppe]

uh

i'm so confused by what he is doing

i think

i fucked up

I have swapped to -

instead of +

somewhere in my solution

[opal plinth]

Are you sure it's xy'-3y=\frac{x^4}{1-x^2} ?

[verbal steppe]

Yes i found it..

i accidentally swapped from + to minus

in the denominator somewhere while

solving

it should be 1/1+x^2

[opal plinth]

[verbal steppe]

i'll redo the whole thing properly ..

brb

Okay

so

C= -pi/4

Is what i got

[opal plinth]

Can you state the full question?

[verbal steppe]

$y = x^3(tan^{-1}(x) - \frac{pi}{4})$

Yes i will

one sec

$xy'-3y=\frac{x^4}{1+x^2}, y(1)=0$

[soft zealot]

Merineth

Merineth

[verbal steppe]

this was the original

[verbal steppe]

that would be my answer

[opal plinth]

,w {xy'-3y=\frac{x^4}{1+x^2}, y(1)=0}

[soft zealot]

Results provided by WolframAlpha

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

Requested by nelfeal

[opal plinth]

Looks ok

[verbal steppe]

Other than a lot of carelessness with signs

i think i'm getting the hang of diff equations

[opal plinth]

[verbal steppe]

one tiny mistake however.. will cost me dearly..

[opal plinth]

You'll be better at it than me at this rate

[verbal steppe]

press X for doubt

❌

It would be nice if i could ace one tho

without any help

If i have

$Ae^{4x}-10Be^{4x}+25Ce^{4x} = e^{4x}$

If i understnad it correctly

A has to be 1 in order to get e^4x

but then remains -10Be^4x and +25e^4x

they both have to cancel eachother out

[opal plinth]

No, you can just divide everything by e^(4x)...

[verbal steppe]

How am i missing this?

[opal plinth]

Always try to factorize stuff that has x in it by stuff that doesn't have x in it

[verbal steppe]

i missed the C

[soft zealot]

Merineth

[verbal steppe]

sorry..

[opal plinth]

Same thing

[verbal steppe]

i tried doing it alone but i seem to always get stuck when trying to determine A B and C

$A-10B+25C = 1$

[opal plinth]

Yeah

No

= 1

[soft zealot]

Merineth

[verbal steppe]

ofc

[opal plinth]

What is this supposed to solve?

[verbal steppe]

I tried to solve a diff equation by myself

let me type it up

$y''-10y'+25y=e^{4x}, y(0) = 0, y'(0)=-2$

[soft zealot]

Merineth

[verbal steppe]

my first initial thought was that this is solved with second order ODE

so i do the y(x) = yh(x) + yp(x)

my answer to yh(x) is

$yh(x)=(C_0+C_1 x)e^{5x}$

[soft zealot]

Merineth

[verbal steppe]

I'm fairly confident that is right

[outer grove]

$5-10

[opal plinth]

Ok

[verbal steppe]

So then now i just have to figure out yp(x)

so i rewrite it in a way that gives me an idea of how the solution should be

i.e

ohgod

I think i solved it by explain it

[soft zealot]

Merineth

[opal plinth]

Happens

[verbal steppe]

There is no C

y is just e^4x

[opal plinth]

Well in that case there is no B or A either

[verbal steppe]

Well

$y''-10y'+25y=e^{4x}$

[soft zealot]

Merineth

[verbal steppe]

I have this

now i just have to replace y'', y' and y with

what i imagine there to be

y(x) = e^4x
y'(x) = Be^4x
y''(x) = Ae^4x

[opal plinth]

No you misunderstand

You need to take y = Ae^(4x) because the RHS is of that form

Then you calculate y' and y''

There is no B or C

y' and y'' are fully defined by y (with A)

[verbal steppe]

y(x) = Ae^{4x}
y'(x) = Be^4x
y''(x) = Ce^4x

is that what you mean

[opal plinth]

Yeah but you're making your life harder

Just differentiate Ae^{4x}

[verbal steppe]

I'm not sure i understand what you mean by that

[opal plinth]

f(x) = Ae^{4x}, what is f'(x) ?

[verbal steppe]

4Ae^4x

[opal plinth]

Right, so that's your y'

[verbal steppe]

Ah

I see what you mean

funny thing is

i wrote that down on my paper

and never actually used it

$16e^{4x} -10(4e^{4x}) +25(e^{4x}) = e^{4x}$

[soft zealot]

Merineth

[opal plinth]

Yeah

[verbal steppe]

And now simplify?

[opal plinth]

I mean you're missing A but that's still true because A=1

[verbal steppe]

where would A be?

[opal plinth]

.

[opal plinth]

.

[verbal steppe]

oh

$16Ae^{4x} -10(4Ae^{4x}) +25(e^{4x}) = e^{4x}$

[soft zealot]

Merineth

[opal plinth]

Still missing one

[verbal steppe]

I think i have to repeat this part because i'm not following

i thought i always had A B and C

[opal plinth]

Not sure where you got that from

[verbal steppe]

the yp(x) part

Is there a method which i can remember by?

step 1: guess what the solution should be?

[opal plinth]

If the RHS looks like pe^{kx} + qe^{mx} + re^{nx} then you'll have to take y = Ae^{kx} + Be^{mx} + Ce^{nx}

Maybe that's why you thought about A,B, and C

[verbal steppe]

i was thinking

x^2+x+C

i'm not sure

[opal plinth]

?

[verbal steppe]

idk

[opal plinth]

I don't know what you're referring to

[verbal steppe]

isn't there a method

or a step by step guide on how to find the particular solution?

in my case

shouldn't it be

i figure out what f(x), f'(x) and f''(x) is

[opal plinth]

Not that I know of

[verbal steppe]

and then substitute my function with those answers

[opal plinth]

Isn't that what we are doing?

[verbal steppe]

yeah but there are no variables

the solution is just

f(x) = e^4x
f'(x) = 4e^4x
f''(x) = 16e^4x

after i put those in

don't i just simplify and we are done?

[opal plinth]

If you put these in and this equation is satisfied, then you are done, yes

But in general that kind of equation won't be satisfied if you just take y = e^{4x} instead of y = Ae^{4x}

(this one is, but it's just by chance - or rather because the one who made the exercise made it so)

[verbal steppe]

So how many A:s should there be?

because when i simplify i get "1"

well i solve the equation but there isn't any variable

i assume that should be A

[opal plinth]

$16Ae^{4x} - 10(4Ae^{4x}) + 25(Ae^{4x}) = e^{4x}$

[soft zealot]

Nel

[opal plinth]

This is what the equation should look like

From there, you solve for A and find A = 1

So you got your y_p = e^{4x}

[verbal steppe]

aaah i see

$y(x) = (C_0 + C_1 x)e^{5x} + e^{4x}$

So that would be my final answer on f(x)

now i just need to put in y and x?

[opal plinth]

y(x) not f(x)

[verbal steppe]

sorry yeah y(x)

[soft zealot]

Merineth

[opal plinth]

Yeah, solve that with y(0) = 0 and y'(0) = -2

[verbal steppe]

what do i substitute y(x) for?

0?

or i mean

if x=0 and y=0

that would give me C + 1

[opal plinth]

C_0

[verbal steppe]

Yeah C_0

[opal plinth]

Yes

[verbal steppe]

So C_0 + 1?

is the answer?

[opal plinth]

= 0

[verbal steppe]

C_0 = -1?

[opal plinth]

Yes

[verbal steppe]

What made you determine it shoåuld be = 0? y(x)

Sorry that was braindead question lol

i get why

hm i tried to solve the f'(x)

I tried my best at least..

[opal plinth]

Product rule where u(x) = (C_0 + C_1 x) and v(x) = e^{5x} ?

[verbal steppe]

woops i tried chain rule on it

[opal plinth]

With what functions?

[verbal steppe]

$y'(x) = C_1 e^5x+(C_0+C_1x)5e^{5x}+4e^{4x}$

[soft zealot]

Merineth

[verbal steppe]

that is with product rule

$-2 = C_1 e^5x+(C_0+C_1x)5e^{5x}+4e^{4x}$

[soft zealot]

Merineth

[opal plinth]

Yeah ok

(assuming you meant C_1 e^{5x})

[verbal steppe]

yea

can't for the life of me solve it..

$C_1e^{0}+(C_0+C_1x)5e^0 + 4e^0 = -2$

[soft zealot]

Merineth

[verbal steppe]

$C_1+5C_0 + 4 = -2$

[soft zealot]

Merineth

[opal plinth]

Yes

[verbal steppe]

so now i have to determine C_0 and C_1

$C_1+5C_0 + 6 = 0$

[soft zealot]

Merineth

[verbal steppe]

C_1 = -1?

[opal plinth]

Yes

[opal plinth]

You already had C_0

[opal plinth]

You could've put it in here even before differentiating

[verbal steppe]

ohgod i did not realize that

that i could use C_0 from f(x)

;-;

I feel like it's break time but got anxiety for upcoming test lmao

[opal plinth]

[verbal steppe]

I've been on math since like 14:00

[opal plinth]

Call chartbit for some hugs

[verbal steppe]

I'm not sure if i dare

I've bothered you and her enough as is :(

[opal plinth]

Do you have the final answer for this?

[verbal steppe]

I think so

isn't it the y(x) = yh(x) + yp(x)

[opal plinth]

Yea

[verbal steppe]

Yes i do!

$y(x) = (C_0 + C_1x)e^{5x}+e^{4x}$

[soft zealot]

Merineth

[opal plinth]

Well that's not very final is it

[verbal steppe]

Oh i have to give C_0 and C1?

[opal plinth]

Of course, why else did you calculate them?

[verbal steppe]

$y(x) = (-1 -x)e^{5x}+e^{4x}$

[soft zealot]

Merineth

[opal plinth]

,w {y''-10y'+25y=e^{4x}, y(0) = 0, y'(0)=-2}

[soft zealot]

Results provided by WolframAlpha

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

Requested by nelfeal

[opal plinth]

[verbal steppe]

[opal plinth]

We wouldn't come back if it truly bothered us

[verbal steppe]

chartbit

hasn't typed since yesterday ;-;

so i clearly shooed her away hahah

[opal plinth]

Lol

Likely just busy

[verbal steppe]

I still have to learn / practice complex equations, integrals and something i don't know the name of

in fact

[opal plinth]

[verbal steppe]

it has been so long that i can't remember anything from complex equations

There will be a total of 10 questions with 5p each, and i need at least 15 to pass

The past 3 days i've literally dreamt of differential equations

this is awful haha

[opal plinth]

So full points on just 3 questions?

[verbal steppe]

Yes

Something similar to this will appear on the test

[opal plinth]

Seems easy enough if you're careful about silly mistakes

[verbal steppe]

With 2 additional questions

Yeah..

i have never done a silly mistake

[opal plinth]

(3) and (4) are easy

[verbal steppe]

maclaurin

Yeah i thought maclaurin was pretty easy

[opal plinth]

(5) you can do by now

[verbal steppe]

just plug into formula

(5) and (7) is something we have done the past days

A question tho, what the fuck is a hyperbeln

hyperbola i think eng term is

[opal plinth]

x^2 - y^2 = 1 is the equation of a hyperbola

It's a conic section, like ellipses and parabolas

[verbal steppe]

Oh i see

so the question isn't around hyperbola but rather

it's extreme point?

[opal plinth]

It's asking about the point (0,1), right?

[verbal steppe]

8. Which is the shortest distance from the hyperbola x2 − y2 = 1 to the point (0, 1)?

[opal plinth]

Oh shortest distance

That kinda sucks

[verbal steppe]

Extremely

[opal plinth]

Well if you can graph the thing it's easier

But then, I'm not sure, maybe just vector maths

[verbal steppe]

x2-y2=1

isn't that a linear equation?

How can that produce a hyperbola

[opal plinth]

No it's squares

[verbal steppe]

oooh

lol

[opal plinth]

You have a point P on the hyperbola, and you want the distance from P to (0,1) to be minimal

That means the tangent to the hyperbola at P is perpendicular to the line from P to (0,1)

[verbal steppe]

So anywhere on the green line?

[opal plinth]

Yeah

[verbal steppe]

Hmm

i think i have an idea

of how i'd solve it

[opal plinth]

It means P = (P_x, P_y) such that P_x^2 - P_y^2 = 1

[verbal steppe]

I'm fairly sure this question involves derivatives

[opal plinth]

Yeah for the tangent

[verbal steppe]

the derivative of the function provides the incline/decline on the function

but

[opal plinth]

Or maybe you can calculate the distance as a function and minimize it by just finding the min through the derivative

[verbal steppe]

that sounds

like the solution

[opal plinth]

Both ways pretty much as complex

[verbal steppe]

however i'm not certain how it would look

or even to begin

[opal plinth]

Well you'd need a parametric equation of the hyperbola

If you've never worked with that before, it's not going to be very fun

[verbal steppe]

Yeah i'm looking at the solution right now..

Pretty sure this question is

advanced

I do understand tho that

$d= \sqrt{(x-0)^2 + (y-1)^2}$

[soft zealot]

Merineth

[verbal steppe]

Is the utilization of pythagoras

Actually this isn't too bad

Once i get started with that equation

Anyhow it's def time to take a break

Thanks for tn nel

[opal plinth]

[verbal steppe]

Let chart know she is mieeed :P

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[uneven halo]

Hi

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[scarlet elm]

Display an infinite sequence $a_1,a_2,...$ of integers with both of the following properties:\

No term in the sequence is equal to 0\
For all n positive integers, $a_n + a_{2n} + ... + a_{2023n} = 0$

[soft zealot]

Monkagoras

[scarlet elm]

Could anyone give me a hint on what do I need to know to independently solve this?

[ocean stratus]

Maybe just have it so that each pair cancels like

(1,-1,1,-1,...)

Would that work or is there some limitation

[scarlet elm]

That wouldn't work for n = 2 per example

wait, I think I have misinterpreted the question

nevermind, it wouldn't really work if n were to be equal two

I was thinking of some function in which $a_n = -a_{(2023-n)}$

however as I developed that idea I got to an invalid result

[soft zealot]

Monkagoras

[scarlet sequoia]

Think about another way which restricts you further but in a positive way

What if I asked you to make the a_n as simple as possible, for example make as many of the a_n = 1

If I asked you to do this for the first 2023 terms for example, how many of those a_n can you set equal to 1 without having further problems with the conditions you have?

[scarlet elm]

If I have not misunderstood the question I would need to have 1011 or 1012 terms set equal to 1

[scarlet sequoia]

There is really no way you can set more equal to 1?

[scarlet elm]

I could set 2022 terms equal to 1 and only a single term equal to -2022

[scarlet sequoia]

Yes exactly

Conveniently, do you know which term is gonna be equal to -2022 ?

[scarlet sequoia]

In any case, this is basically the strategy I would adopt

[scarlet elm]

The first?

Oh, nevermind, I thought it started at a_0

It would be the last in the case that n = 1

however how could I describe the last term of sequences for other n?

[scarlet sequoia]

I'll help you a bit further :
Suppose every n non-multiple of 2023 has a_n = 1

[scarlet elm]

I thought about that but if every multiple of 2023 was equal to -2022 it would be wrong for n = 7 as 7 is a factor of 2023

[scarlet sequoia]

Ofc, you can't have all a_(2023n) equal to -2022

However you can define that sequence by induction

[scarlet elm]

Sorry, I can't really figure it out, I just got to if all a_n = 1 then there would be no such constant sequence because anywhere that I place the -2022 there are infinite numbers which are multiples of that place and got stuck here. Could you give me your resolution?

[scarlet sequoia]

So my place of reasoning is you just let a_(2023n) = -a_n - a_(2n) - ... - a_(2022n)

[scarlet elm]

OH

I see

that's clever

ty

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[snow flame]

hi i need help with this question. am i wrong in thinking when the matrix is raised to an infinitely high power, it converges to a direction of (1,1) and then i should take unit vector of that? can someone show me a step by step of how to do this cause my answer is incorrect and i have no clue what else to do. thanks in advance

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@snow flame Has your question been resolved?

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@snow flame Has your question been resolved?

[raw pike]

!help

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To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[fallen remnant]

Help to solve this question

[raw pike]

!help

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To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[raw pike]

Open your own channel

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[slow wing]

How can I resolve this ?

[slow wing]

The 19

[raw pike]

,ccw

Forgor the command

[raw pike]

Raise both sides in exponents to get rid of the logs

First do 2^(everything) on both sides

[slow wing]

Ok

Thks

[lyric summit]

It is enough to even use the definition of logarithm three times, and the result is already there

[slow wing]

Ok thks

[lyric summit]

check this : x = 125

if you get the same result

[slow wing]

I've got the fix and it's the answer, thank you.

[lyric summit]

yw )

[slow wing]

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[tranquil pine]

Let V be the set of all ordered pairs of real numbers. Consider the following addition and scalar multiplications operations on u = (u1, u2) and v =(v1, v2):
u+v = (u1 + v1, u2 + v2) and ku = (0,ku2). Explain why V is closed under addition and scalar multiplication

[tranquil pine]

I understand why it is closed under addition because the result gives a pair of real number

but why is it closed under scalar multiplcation

let' s say u = (-1, 2) and k = 2, then ku = (-2,4)

but it says that ku = (0,ku2)