help-17

[twin meteor]

Frosst

[lunar lagoon]

ask in a different channel

[lament arch]

create a channel go to #❓how-to-get-help

[plain field]

I did.

[lament arch]

wait

[plain field]

alright

[rugged orchid]

You should notice that doing $(g\circ f)(x)$ and $(f\circ g)(x)$ is different

[twin meteor]

Frosst

[rugged orchid]

One of them isn’t quite the inverse of the other

[lament arch]

what about the square root

[rugged orchid]

If you aren’t comfortable with function composition notation

You can just write f(g(x)) and g(f(x)) as well

[lunar lagoon]

so then you square root

i didnt want to just write the answer

[rugged orchid]

If you get x in both cases then they are indeed inverses

But they aren’t in this case

Well, one of them

[lament arch]

so f(g(x)) = x is indeed right

[lament arch]

just not the oppsite

[rugged orchid]

Yes

But g(f(x)) = |x|

Which is a bit of a problem

[lament arch]

Gotcha ok

[lament arch]

so all my math here checks out

[rugged orchid]

It looks terrible I think that’s a 9 on row 2

I have no idea what you are writing

What’s inside the square root

Why isn’t the square root covering everything

Why does it just say x on 1 side

Why is there a 9

Why does it randomly say x = y

Where did f^-1 come from

How is it equal to x

[lament arch]

huh?

[rugged orchid]

There are so many problems with that working out that no, your math does not check out

[lament arch]

at the end y is the inverse of the function

that's hwy its f^-1 = x

[rugged orchid]

If $f^{-1}(x)=x$ then this implies $f(x)=x$

[twin meteor]

Frosst

[rugged orchid]

So no, that is not correct

[rugged orchid]

Btw if this was your working out you’d get 0 marks

So maybe consider try practice better working out habits

[lament arch]

I'm confused with g(f(x)

the q is (f(x)

I plugged the left into the right

[rugged orchid]

What the frick is happening

[lament arch]

bro

[rugged orchid]

Where did 9 come from

[lament arch]

do you see a q?

if you don't see a q

[rugged orchid]

That’s a q?

[lament arch]

then you can assume thats what it is

[rugged orchid]

Ok then where did x come from

[lament arch]

the x is q

[rugged orchid]

Then why is q still in the equation

And why does it just say x on the left side

[lament arch]

the q is f(x)

...

q = f(q)

[rugged orchid]

Ok so q is f(x)

[lament arch]

q = q

[rugged orchid]

what?????

[lament arch]

.....

im just swapping bro

[rugged orchid]

What exactly are you swapping

[lament arch]

you start out as f(q) = q

[rugged orchid]

No

[lament arch]

dude

[rugged orchid]

That is wrong

[lament arch]

oh my god

im getting cancer

[rugged orchid]

We don’t have f(q) = q

[rugged orchid]

You’re writing like cancer

[lament arch]

ignore everything else

that is f(q) = q

ok now swap them

[rugged orchid]

That says g

[lament arch]

I like f

[rugged orchid]

You already have f

[lament arch]

q = f(q)

[rugged orchid]

You can’t use f

[lament arch]

change f(q) to q

and q to x

[rugged orchid]

What the fuck

[lament arch]

you can ur still solvilng for the inverse

[random peak]

I don't think you know how functions work

[lament arch]

ur just finding the answer

it doesnt matter if theres another function thats f

[rugged orchid]

Ok now we have q(x) = x² - 3 and f(x) = sqrt(x+3)

Now we’ve swapped f for q

And g for f

And q for x

[lament arch]

I'm just trying to find the answer

im using a mouth to draw

notation can fuk off tbh

f(x) = sqrt(x+3)

[rugged orchid]

Notation will bite you in this ass if not here then soon

Although it’s already biting you in the ass here

Now if you put f into q

[lament arch]

here

9 is q

and +-q

since we square rooted

[rugged orchid]

Ok why is the square root over the - 3

[lament arch]

aren't you plugging q^2 - 3 into q

q is under square root

[rugged orchid]

Yes

You get

[torpid linden]

Sorry to interrupt. I did my own calculation and it is inverse, but this definition and x>=-3

[rugged orchid]

$f(g(q))=(\sqrt{q+3})^2-3$

[twin meteor]

Frosst

[lament arch]

im doing the opposite

g(f(q)

I already did that and got that its an inverse

[rugged orchid]

$g(f(q))=\sqrt{q^2-3+3}$

[twin meteor]

Frosst

[rugged orchid]

You have a random square on the top right

And your left side says g(q)

Rather than g(f(q))

[lament arch]

right

[lament arch]

I did that to remove the square root

whats wrong w it

[rugged orchid]

Because you can’t just magically make a square out of nothing

You can’t just add bits

If you squared the right side

[lament arch]

do u not see a square in the function

[rugged orchid]

The left side gets squared as well

[lament arch]

it says

g(q) = sqrt{q + 3}

[rugged orchid]

$(g(f(q)))^2=\left(\sqrt{q^2-3+3}\right)^2$

[twin meteor]

Frosst

[lament arch]

what inthe

[rugged orchid]

If you put a square on the right side

You need to do the same to the left side

But we don’t even want or need that

[rugged orchid]

You’ve just added a ^2 for no reason

[lament arch]

the left side is squared

too

[rugged orchid]

No it’s not

[lament arch]

ok so what do we do

[rugged orchid]

The left side isn’t even correct

As I said earlier

[lament arch]

dude

oh my god

why are you still on notation

fuck

[rugged orchid]

It should say g(f(q))

I’m saying 2 + 2 = 4

[lament arch]

ok u already established that

[rugged orchid]

You’re saying 2 - 2 = 4

And you’re claiming the difference is notation

[lament arch]

whats wrong with me doing g(f(q))^2

on the left

[rugged orchid]

Nothing is wrong with that

It just doesn’t help to show they are inverses

[lament arch]

so what am I supposed to do

[rugged orchid]

$g(f(q))=\sqrt{q^2-3+3}$

[twin meteor]

Frosst

[rugged orchid]

$g(f(q))=\sqrt{q^2}$

[twin meteor]

Frosst

[rugged orchid]

$g(f(q))=|q|$

[twin meteor]

Frosst

[rugged orchid]

Therefore, f is not the inverse of g

Done.

[lament arch]

how can you turn sqrt q^2 to absolute value q

[rugged orchid]

That’s the definition of absolute value

[lament arch]

what mathematical function

[rugged orchid]

$\sqrt{x^2}=|x|$

[twin meteor]

Frosst

[lament arch]

you do exponent to remove square root but what about this

ok so just substitution

[rugged orchid]

It is literally just how we define absolute value

Like, it literally is how we know what absolute value means

[lament arch]

Whats stops someone from

suaring the right side and left side

[rugged orchid]

Nothing

[lament arch]

then square rooting again

you get that they are inverses

[rugged orchid]

Nope

[rugged orchid]

Cos now the absolute value sign goes on the left side

[lament arch]

can you show me the math for doing this method

[rugged orchid]

$g(f(q))=\sqrt{q^2}$

[twin meteor]

Frosst

[rugged orchid]

$(g(f(q)))^2=\left(\sqrt{q^2}\right)^2$

[twin meteor]

Frosst

[rugged orchid]

$(g(f(q)))^2=q^2$

[twin meteor]

Frosst

[rugged orchid]

Right?

$\sqrt{(g(f(q)))^2}=\sqrt{q^2}$

[twin meteor]

Frosst

[rugged orchid]

$|g(f(q))|=|q|$

[twin meteor]

Frosst

[rugged orchid]

Still doesn’t work

[rugged orchid]

Even if from here you do the inside bit for q first you end up like this

[lament arch]

is +- q

the same as absolute vlaue q

[rugged orchid]

$(g(f(q)))^2=|q|^2$

[twin meteor]

Frosst

[rugged orchid]

$(g(f(q)))^2=q^2$

[twin meteor]

Frosst

[rugged orchid]

You take square root still doesn’t work

[rugged orchid]

Not quite

[lament arch]

why isn't sqrt{q^2} = +- q

and is absolute value

you remove the exponent

then you addd plus minus

[rugged orchid]

Because $\sqrt{(-3)^2}=|-3|=3$

[twin meteor]

Frosst

[rugged orchid]

Negative numbers become positive first

It’s quite interesting actually that the order in which the square and the square root does matter when the thing inside is negative

The square root and square functions don’t commute

$\sqrt{(a)^2}\neq (\sqrt{a})^2$ if $a<0$

[twin meteor]

Frosst

[lament arch]

But q isn't negative

[rugged orchid]

How do you know

[lament arch]

I hate math

[rugged orchid]

Exactly you don’t know that q isn’t negative

So you can’t assume so

[lament arch]

this is

crazy

[rugged orchid]

Thus you arrive here

And you say

Nope, that doesn’t work

f doesn’t invert what g did to x

f is not the inverse of g

[torpid linden]

I am sorry.

[lament arch]

I hope I never need to know this for calculus

[rugged orchid]

You will need much more for calculus

[lament arch]

[rugged orchid]

When you do calculus, you’re expected to be able to freely manipulate algebra stuff without any issues

When you’re doing algebra, you’re expected to be able to freely do arithmetic without any issues

Calculus will also introduce a whole new bunch of notation

If you ever do multivariate calculus, having bad notation will actually instantly end you

There are so many letters

[torpid linden]

f(q) , q is all real numbers and g(q), q is greater than equal to -3

[torpid linden]

Well, this is what I got. Inverse for q greater and equal to -3

[vocal sleet]

@lament arch Has your question been resolved?

[vocal sleet]

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[lost bear]

Top image, find radius c

I’m unsure why b is given at all also because isn’t there enough information from just a

[lost bear]

How

[vocal sleet]

@lost bear Has your question been resolved?

[vocal sleet]

@lost bear Has your question been resolved?

[vocal sleet]

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[vast shale]

Need help with integral

[vast shale]

Basic one

[merry osprey]

Send it

[vast shale]

Integral of x^-4

[merry osprey]

You know how to do it for x^n?

[vast shale]

I get -1/3x^3

My answer sheet says it’s not negative

[merry osprey]

It is

[obtuse sierra]

this is the correct answer

[vast shale]

Really?

[merry osprey]

Yep

[obtuse sierra]

,w integrate x^(-4)

[twin meteor]

Results provided by WolframAlpha

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

Requested by oldieidlog

[vast shale]

What the heck

Answer sheet wrong

Anger

Hold on

It’s part of a larger problem

Can u guys take a look at my work

[merry osprey]

Sure

[vast shale]

The pink part is where I don’t know what went wrong

It’s like I lost a negative somewhere

[merry osprey]

It’s multiplied by -1

The integral

[vast shale]

Oh my got

[vast shale]

Tyty, been forever since I took calc

.close

[vocal sleet]

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[merry osprey]

No worries

[vocal sleet]

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[vast shale]

Hello, I want to check if my answer is correct or not

[vast shale]

It says that: If ...., calculate the value of M

Something that worries me is if by taking the square root of cos^2sen^2, I need to put that is the absolute value
Idk if that would be important in this case

Because in the case that one of them is negative, then I think the answer should be -1/8 ??

[untold surge]

No missed one answer

[vast shale]

Oh okay thanks

[untold surge]

2 is correct, you need another one

remember x^2=a, x=+/- sqrt(a)

[vast shale]

mm yes, I remember that

[untold surge]

So fix your 5th line

[dull bear]

Do they phrase it as the value (singular) in how they ask the question?

[untold surge]

Wait a second I misread something

[vast shale]

Mm yep

I think

[untold surge]

Nvm I didn’t

[untold surge]

So continue

[vast shale]

that because M = + cube root

oh okay, so it shoukd be +-

[untold surge]

Yeah

[dull bear]

If they phrase it like that and give no other restrictions, then poorly phrased question

[untold surge]

+/- 1/8

[vast shale]

mm do you mean that they should have put M= +-cube root

mm but its cube

[untold surge]

Just replace your 1/8 with +/- 1/8 in your 5th line then continue

[dull bear]

Like they should e.g. state something like "M is a positive number such that...", or otherwise e.g. "find the possible values of M"

[vast shale]

ohh okay, I get it

(sorry for continuing)
But (-2)(-2)(-2) is -8, not 8

or is there something I'm not understanding well?

because I want to understand

or are you reffering that I should put + - 1/8

and take the cube root of a negative number also ?

ooh okay, I get it
so I need to take into account that by taking the square root of cos^2sen^2, the result could also be +-1/8

[dull bear]

Yea you could have gotten ±1/8 out of that intermediate step, unless they force you to pick one (which they haven't, hence me asking about how the question was phrased)

[vast shale]

Mm yes, it could have been (-)(+) and viceversa, so yes, you are right, thanks a lot ((:

[dull bear]

I'm very much assuming whoever wrote the question is assuming you'd take the positive answer (the +2 that you got in the end) but -2 is also valid under how they wrote it

[vast shale]

Yes, thanks ((:

.close

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[glossy mango]

hi, i'm having a lot of trouble with this rollercoaster project for algebra II

[glossy mango]

heres the directions

it's kinda confusing me out on how to do 5 curves

im using vertical as feet

[vocal sleet]

@glossy mango Has your question been resolved?

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@glossy mango Has your question been resolved?

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[ocean anvil]

Hello, ive been trying for so long for this exercise ; im down bad for some help

Let $f$:$\mathbb{R}$ $\rightarrow$ $\mathbb{R}$ such that $\forall x \in \mathbb{R}, f(f(x)) + x = 2 \times f(x)$. Suppose $\forall (a, b) \in \mathbb{R}, f(a) = f(b) \Rightarrow a = b$ (injectivity).
Show that $\forall (a, b) \in \mathbb{R}, a < b \Rightarrow f(a) < f(b)$

[twin meteor]

demiryolu mühendisi

[ocean anvil]

i try absurd

suppose $a < b$ and $f(b) < f(b)$. Then i get to $ b - a < f(f(a)) - f(f(b))$

[twin meteor]

demiryolu mühendisi

[vocal sleet]

@ocean anvil Has your question been resolved?

[silk kindle]

Mmm try replacing x by f^(-1) (x) 🤔

Do you think that's allowed?

Then you follow the rule given for the function

I found it in mathstack exchange xd

Also given that it's injective, it's either an increasing or a decreasing function

[vocal sleet]

@ocean anvil Has your question been resolved?

[vocal sleet]

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[modern geyser]

Does someone know if you are allowed to take out a factor from an absolute value by dividing/multiplying both sides?

[rugged orchid]

Yes

$|ab|=|a||b|$

[modern geyser]

that doesnt mess with anything?

[twin meteor]

Frosst

[rugged orchid]

Then you just divide by |a|

That’s always positive so the inequality doesn’t change

Well

Unless a = 0

In which case you can’t

But clearly, 2 ≠ 0 so it’s fine

[modern geyser]

why doesnt 0 work?

[rugged orchid]

Can’t divide by 0

[modern geyser]

oh

ty

.close

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[frigid pivot]

Well, to close off limits for the night, I have no idea how to deal with limits in which infinity is raised to the power of infinity

[glossy maple]

Either use x = e^(lnx) or try to transform it into a limit with resembles (1 + 1/x)^x

[silent quartz]

Try (1+1/n)^n

[frigid pivot]

I am unfamiliar with this formula

[glossy maple]

$\lim_{x \to \infty} \parens{1 + \frac{1}{x}}^x = e$

[twin meteor]

RedstonePlayz09

[frigid pivot]

But why is only one of the numbers divided by x?

[glossy maple]

Well, that's how it is. Try to change how your current limit is written

[frigid pivot]

Alright

[frigid pivot]

I am having some trouble doing this since I don't really get it

Huh?

[glossy maple]

That's not quite right

Denominator should be 3x + 5 and it needs to be -4

[frigid pivot]

Where does the 4 come from, though?

And what about the other 3x?

[silent quartz]

Yes but me no skillo in latex
So
I chsnged ((3x+5)/(3x+1))^x-3/x^3-2x

[glossy maple]

$\frac{3x + 1}{3x + 5} = \frac{3x + 5 - 4}{3x + 5} = 1 - \frac{4}{3x + 5}$

[twin meteor]

RedstonePlayz09

[glossy maple]

By the way, what limit rules can you use that are related to exponentiation?

Can you use this:

$\lim_{x \to c} {\pmap{f}{x}}^{\pmap{g}{x}} =\parens{\lim_{x \to c} \pmap{f}{x}}^{\lim_{x \to c} \pmap{g}{x}}$

[twin meteor]

RedstonePlayz09

[frigid pivot]

Now, I understand this formula

[glossy maple]

(Assuming some conditions of course)

[frigid pivot]

At least somewhat

[glossy maple]

I'm asking if you're allowed to use it? Have you seen it in your book etc?

If not, then I'd recommend using the e^(ln(x)) method.

[frigid pivot]

I think I have seen it but never used it

[glossy maple]

Can you verify whether you're allowed to use it as a known theorem?

[frigid pivot]

Not like the teacher could prohibit the use of this formula

[glossy maple]

Well okay

[frigid pivot]

So I am guessing that it is another way to solve it?

[glossy maple]

No, you just need it for this method

Try to get your limit to the form:

$\lim_{x \to \infty} \parens{\parens{1 + \frac{1}{t}}^t}^k$

[twin meteor]

RedstonePlayz09

[glossy maple]

Where t and k are some expressions involving x.

And if you manage to do that, you will find that:

$\lim_{x \to \infty} t = \infty$

[twin meteor]

RedstonePlayz09

[glossy maple]

So the whole (1 + 1/t)^t thing becomes just e according to this.

[frigid pivot]

And I would but I am really struggling to wrap my head around this formula

[glossy maple]

Which one?

[frigid pivot]

Even if I could pull it off this time I certainly will not be able to apply it to any other limit

[glossy maple]

Alright, I'll try to help

So first, let's look at the part inside the parenthesis:

[frigid pivot]

Right

[glossy maple]

As we said, we have this:

Now, let's change it up a bit more. We want to have 1 + 1/something

So I'll write it like this:

$\frac{3x + 1}{3x + 5} = 1 + \frac{1}{-\frac{3x+5}{4}}$

[twin meteor]

RedstonePlayz09

[glossy maple]

So far do you understand?

[frigid pivot]

Where does the 4 come from?

[glossy maple]

.

Continuing from the last step there.

I just flipped the fraction

a/b = 1/(b/a)

[frigid pivot]

Even in the attachment it appears out of nowhere, though

[glossy maple]

Look carefully

From the second to third step, we split the fraction.

We have (3x + 5)/(3x + 5) + (-4)/(3x + 5)

[frigid pivot]

Yeah

[glossy maple]

Get it now?

[frigid pivot]

Not really...

[glossy maple]

Which part?

$\frac{3x + 1}{3x + 5} = \frac{3x + 5 - 4}{3x + 5} = 1 - \frac{4}{3x + 5}$

I'll write it again

[twin meteor]

RedstonePlayz09

[glossy maple]

So you have 3x + 1

That is the same as 3x + 5 - 4 because 5 - 4 = 1

Ok?

[frigid pivot]

Ah, I see

But why put it that way?

[silent quartz]

@glossy maple u need kick minus

[glossy maple]

I need what?

[silent quartz]

Change it, to delete minus

[glossy maple]

Because now we have 3x + 5 + something

in the numerator

[glossy maple]

No, the minus is correct

[glossy maple]

@frigid pivot Now look at the last step here

We split the fraction, into 2 fractions

[frigid pivot]

Wait

[glossy maple]

One with 3x + 5 in the numerator, and one with -4 in the numerator

$\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$

[twin meteor]

RedstonePlayz09

[frigid pivot]

But why did the 4 become positive?

[glossy maple]

Well I just put the negative in front

You can also write it as:

$1 + \frac{-4}{3x + 5}$

[twin meteor]

RedstonePlayz09

[frigid pivot]

Ah, alright I get all of it now

[glossy maple]

Ok, now you also get this?

[frigid pivot]

Dunno if enough to pull it off on my own but I get the principle

[glossy maple]

?

[frigid pivot]

Yeah

[glossy maple]

Okay, now let's move into the exponent.

[frigid pivot]

Except for why the 4 is the denominator

[glossy maple]

We flipped the fraction

[frigid pivot]

What for?

[glossy maple]

$\frac{-4}{3x+5} = \frac{1}{\frac{3x+5}{-4}}$

[twin meteor]

RedstonePlayz09

[glossy maple]

We are trying to get our limit to look like:

(1 + 1/n)^n

Except we'll have some expression with x instead of n

For example, if we get:

$\lim_{x \to \infty} \parens{\parens{1 + \frac{1}{2x + 1}}^{2x + 1}}^{\frac{x+1}{2x}}$

[twin meteor]

RedstonePlayz09

[glossy maple]

Look carefully, our limit will become something similar.

We'll have first something like

(1 + 1/?)^?

We'll show this approaches e.

Then, we'll have some extra exponent, which will have a finite limit (I hope)

The limit of (x+1)/(2x) is 1/2

So the answer here is e^(1/2)

[frigid pivot]

I see it awfully clearly

Almost seems easy

[glossy maple]

Yes, the hard part is getting it to look like this.

[frigid pivot]

Haha yeah

[glossy maple]

You have to manipulate the expression you already have very carefully.

So, now we have the 1 + 1/something part

Now we need to take care of the exponent.

So our exponent right now is:

$\frac{x^3 - 2x}{x - 3}$

[twin meteor]

RedstonePlayz09

[frigid pivot]

Right

[glossy maple]

So we'll split it into 2 parts.

The first is the exponent we NEED, which is 3x + 5.

[frigid pivot]

Why?

Why do we need it as exponent?

[glossy maple]

To use that the limit is e, we need to raise the 1 + 1/? to the same power as what we're dividing by.

So we write it as:

[glossy maple]

Sorry, meant (3x + 5)/(-4) here

[frigid pivot]

Ok

[glossy maple]

$\frac{x^3 - 2x}{x - 3} = \frac{3x + 5}{-4} \cdot \parens{\frac{x^3 - 2x}{x - 3} \cdot \frac{-4}{3x + 5}}$

Just tell me if you understand this

I multiplied by (3x + 5)/(-4)

and then multiplied by (-4)/(3x + 5)

I haven't changed the value at all, since that's the same as multiplying by 1

Everything cancels out

[frigid pivot]

What confuses me is why it all = x^3 - 2x/ x - 3

[glossy maple]

Oh

I have a mistake there

[frigid pivot]

And why you needed both versions of 3x + 5

[twin meteor]

RedstonePlayz09

[glossy maple]

Now it's correct, look again

It's kind of hard to explain in text, I'm hoping that once you see the final step it will become clear.

[frigid pivot]

Hopefully

[glossy maple]

In the end, everything comes down to bringing our limit to something SIMILAR to this.

[glossy maple]

Something of this form

[frigid pivot]

I see

And to get there we gotta get through 1 + 1/n

[glossy maple]

Yes

Now our limit becomes:

[glossy maple]

Also, let me simplify what we have here

$\frac{3x + 5}{-4} \cdot \frac{-4(x^3 - 2x)}{(x-3)(3x+5)}$

[twin meteor]

RedstonePlayz09

[glossy maple]

Now, I can write our limit as:

$\lim_{x \to \infty} \parens{1 + \frac{1}{\frac{3x+5}{-4}}}^{\frac{3x + 5}{-4} \cdot \frac{-4(x^2 - 2x)}{(x-3)(3x+5)}}$

So far good?

Also it seems that the limit diverges

[frigid pivot]

Kinda, because I get it

[glossy maple]

Oh, wait

[frigid pivot]

But the next step is unclear

?

[glossy maple]

I must have a mistake somewhere since wolfram says it doesn't diverge

[frigid pivot]

What do you exactly mean by "diverge"?

You mean more than 1 result?

[glossy maple]

I mean it goes to infinity

Like, to me it seems like it does

Maybe this exponent rule isn't true for this case

One sec

I hope we didn't do this for nothing

If you want we can change the exponent a little for it to work,

or we can try a different way

That's my bad

[silent quartz]

limit= 0 ?

[frigid pivot]

Sure, it's fine

[glossy maple]

So if we had x^2 instead of x^3

[frigid pivot]

Go ahead

[glossy maple]

We would have this

[twin meteor]

RedstonePlayz09

[silent quartz]

Mean e^0

[glossy maple]

It's of the form e^infinity, in which case our limit law for exponentiation doesn't apply

Anyways, now we have:

$\lim_{x \to \infty} \parens{\parens{1 + \frac{1}{\frac{3x+5}{-4}}}^{\frac{3x + 5}{-4}}}^{\frac{-4(x^2 - 2x)}{(x-3)(3x+5)}}$

[twin meteor]

RedstonePlayz09

[frigid pivot]

Wait a minute

Actually nevermind

My bad

[glossy maple]

Now, since (3x + 5)/(-4) goes to -infinity as x goes to infinity, and:

$\lim_{x \to -\infty} \parens{1 + \frac{1}{x}}^x = e$

[twin meteor]

RedstonePlayz09

[glossy maple]

We can conclude the inside part tends to e as x goes to infinity.

And the exponent of the top, which is

$\frac{-4(x^2 - 2x)}{(x-3)(3x+5)}$

[twin meteor]

RedstonePlayz09

[frigid pivot]

Quick question

[glossy maple]

goes to -4/3 as x goes to infinity

Yes?

[frigid pivot]

What is e?

[glossy maple]

Well it's a very important constant, and there are several ways to define it.

Sometimes you even define it USING this limit of (1 + 1/x)^x

[frigid pivot]

Ah, I see, a constant, alright

[glossy maple]

It's about 2.7

[frigid pivot]

Anyways please continue

[glossy maple]

(Irrational of course)

[glossy maple]

So this goes to -4/3

You can see it by comparing the coefficients of x^2 in both the numerator and denominator

On the top, the coefficent of x^2 is -4

and on the bottom it's 3

Ok?

[frigid pivot]

Yeah I see that

[glossy maple]

And we finally conclude that this equals e^(-4/3)

[frigid pivot]

Where is the divergence, though?

[glossy maple]

Oh one sec

Actually I'll change our original problem a LITTLE more

Just add a negative to the exponent

[frigid pivot]

Alright

[glossy maple]

What I did isn't good since we need a positive exponent aswell (another one of the restrictions to the f(x)^g(x) limit thing)

So we'll have that the exponent tends to 4/3 as x goes to infinity

and our answer will be e^(4/3)

Ok and let me see how we can do your original problem

Probably LHopitals if I'm being honest

[frigid pivot]

Well I heard of that earlier

Seemed terribly complex

[glossy maple]

$\lim_{x \to \infty} \parens{\frac{3x+1}{3x+5}}^{\frac{x^3 - 2x}{x-3}} = \lim_{x \to \infty} e^{\frac{x^3 - 2x}{x-3} \cdot \ln \parens{\frac{3x+1}{3x+5}}}$

[twin meteor]

RedstonePlayz09

[glossy maple]

We write our limit like this

[frigid pivot]

Right

[glossy maple]

Now, we can just take the limit of the exponent (since our base is a positive constant)

$\lim_{x \to \infty} \frac{x^3 - 2x}{x-3} \cdot \ln \parens{\frac{3x+1}{3x+5}}$

[twin meteor]

RedstonePlayz09

[glossy maple]

One last trick now:

This is of the form infinity * 0

Which is indeterminate

So we take the part that's infinity, which is the (x^3 - 2x)/(x - 3)

And move it to the denominator:

$\lim_{x \to \infty} \frac{\ln \parens{\frac{3x+1}{3x+5}}}{\frac{x-3}{x^3 - 2x}}$

[twin meteor]

RedstonePlayz09

[glossy maple]

So far do you understand?

[frigid pivot]

I see

[glossy maple]

Now this is of the form infinity/infinity

And you can use LHopitals

Do you think you can continue from here? This is just simple (but long) calculation now

[frigid pivot]

So you divide everything by x^3

I'd assume so

[glossy maple]

I just took the (x^3 - 2x)/(x - 3) part

and Flipped the numerator and denominator, but moved the whole thing to the denominator

Same thing as writing a/b as 1/(b/a)

Like before

[frigid pivot]

Ah, then I can probably finish it myself

You've helped plenty already

[glossy maple]

No problem

Also my fault for not trying it out first and seeing the first approach wouldn't work.

[frigid pivot]

And for someone who doesn't get paid to deal with people objectively dumber than you, you are awfully patient

[glossy maple]

It does work in other cases, but you should stick to LHopitals if I'm being honest

[frigid pivot]

I probably will have to

[glossy maple]

Remember that LHopitals only applies in cases like 0/0 and infinity/infinity

Sometimes to have to use tricks/manipulations to get our limit to look like that

[frigid pivot]

But definitely won't be able to do it on time for the retake of the retake of the retake of the retake of my maths

Well, thanks for the help

I owe you one

See ya

[glossy maple]

No problem lol

Good luck!

[frigid pivot]

I will need it

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[digital skiff]

Hi, looking to kind of get my thought process checked with this combinatorics problem. ```
a) (5 choose 1) * (14 choose 3),
b) (10 choose 2) * (5 choose 2),
c) (10 choose 3) * (5 choose 1 )+ 10 choose 4

[digital skiff]

basically any sanity check is heavily appreciated

[vocal sleet]

@digital skiff Has your question been resolved?

[digital skiff]

<@&286206848099549185>

[livid tapir]

For a) got right idea but says at least one non fiction rip

[digital skiff]

Yeah at least, so there is a minimum of one which means choosing one from 5

then the remaining 3 can be chosen arbitrarily from the remaining

[livid tapir]

ah u right

Looks all good then

wait hm

actually maybe it can't be right, I think this is gonna count some cases multiple times on a)

e. G. b1 (b2 b3 b4)

And b2 (b3 b4 b1)

[digital skiff]

i dont think a can count multiple cases because we are removing the book taken from the pile

[livid tapir]

Yeah but it's counting the case where you first remove say book 1 and then remove say books 2-4 as different as first removing book 2 and then the other 3 books. (assume for this example b1 and b2 are non-fiction)

[digital skiff]

given order does not matter here i dont see how that is working the way you say it is

[livid tapir]

Yeah it shouldn't matter if they were independent choices but they're not independent choices

[digital skiff]

how are they not independent?

youre picking arbitrary objects out of a pile

[livid tapir]

They're still distinct objects. If they weren't distinct then there would be one way to take out 4 books - by taking out 4 books

But even restricted to type they're distinct

Otherwise there'd still only be one way to like, take out 2 objects out of 5 non fiction

[digital skiff]

maybe. i think this is overthinking territory almost

there are 10 fiction books which are different objects, but not distinct in the sense that order matters for a book report

a student writing 4 papers does not care about the order

[livid tapir]

ok so I found a similar problem were the top answer is getting at the same thing I tried to get it with regards to overcounting https://math.stackexchange.com/questions/1645347/choosing-a-combination-of-books-under-given-restrictions

maybe they can explain it better than me cus I sux lol

[digital skiff]

oh so we could literally just subtract the possibility where no non-fiction is chosen?

[livid tapir]

yeah I think so

[digital skiff]

trynna think of what that would be rn

i really appreciate your help by the way, not trying to be argumentative - this stuff be confusing sometimes

[livid tapir]

oh no problem lol it's making me realize how vague the multiplication rule is about doing things in one way and then another

[digital skiff]

i guess this is 15 choose 4 - 5 choose 0

well, actually there would be the case where we did 10 choose 4

so this should probably be 15 choose 4 - 10 choose 4

[livid tapir]

yeah that sounds right

[digital skiff]

much appreciated, i will definitely be thinking about this in future problems

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[open granite]

9x^2-12xy+4y^2

[open granite]

when i factor this i keep getting (3x-2)(-3x+2)

[compact shadow]

you mind if i try to latex this?

[open granite]

nope

[twin meteor]

Editmond

[compact shadow]

right?

[open granite]

yep!

[vocal sleet]

Show your work, and if possible, explain where you are stuck.

[compact shadow]

this is a perfect sqaure

[twin meteor]

Editmond

[compact shadow]

wait i left out a y

$(ax-by)^2 = a^2x^2-abxy+b^2y^2$

[twin meteor]

Editmond

[compact shadow]

try using the above

[open granite]

thanks!

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[vast shale]

How do I differentiate $$\frac{\sqrt{2x}}{5}$$?

[vast shale]

hmmm wait

[twin meteor]

Sir Edgar

[vast shale]

ah yes

[twin meteor]

Dyssrupt

[obtuse sierra]

separate it

[vast shale]

wdym constant value?

[twin meteor]

Dyssrupt

[obtuse sierra]

what is the derivative of 5x?

[vast shale]

5

wait let me type out what I tried

[obtuse sierra]

yeah, you dont differentiate the a in ax

a is constant

[vast shale]

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

[twin meteor]

Sir Edgar

[vast shale]

this is what I've tried

[midnight igloo]

try $\sqrt{2x}=\sqrt{2}\sqrt{x}$

[twin meteor]

Luke

[vast shale]

oh wait, I thought 2x was a single term

like binded

single object

I can separate it?

[obtuse sierra]

yes you can.

[vast shale]

I see

thanks

I suppose I can close this

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[vast shale]

what do i do for this problem?

[vast shale]

knowns are Vi = 0, displacement = 9.8 t = 1

equation to use

im not sure

can anyone give me a hint

[dreamy viper]

find velocity at t=1

use that as v initial

or.I guess you could find distance from 0 to 1

and subtract it from 19.6

a=-9.81, gravity

[vast shale]

oh right i forgot

[dreamy viper]

you don't know displacement

thats what youre trying to find

velocity isn't constant its increasing

[vast shale]

whats up with the total distance then

[dreamy viper]

so it isn't 19.6/2

[dreamy viper]

youre finding distance from 0 to 1 seconds

not 0 to 2 seconds

[vast shale]

i dont get it still but ill figure it out eventually

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[stable whale]

Can a Diophantine equation have more than one particular solution?

[stable whale]

If so, does that mean that every particular solution it has is correct and thus every generalised solution generated from all of them are correct?

[proven garden]

wdym by every generalised solution generated from all of them

[stable whale]

I mean as in

If an LDE has two unknowns (x, y) then there are infinite solutions that could be used as the (x0, y0)?

And thus infinite ways to portray the generalised solution?

[proven garden]

there are infinite solutions yes

I'm not sure what you mean by portray the generealised solution

[stable whale]

The generalised solution is constructed using (x0, y0), right?

But if there’s infinite pairs of (x,y), how can I tell which one exactly out of them I should use as the (x0, y0)?

[proven garden]

hmm

I think all x0,y0 works

but lemme check

ok

so

if you have a generalized solution

something like x=3m+2 y=2m+1

you can substitute m=n+1

and get x=3n+5 y=2n+3

[stable whale]

Ohhh

What I mean is that like

For example 5x + 3y =4. One set of solutions is (2,-2), another is (5, -7). A generalised solution formed by the first is x=2+3n, y=-2-5n, and the second: x=5+3m, y=-7-5m

Does that mean that either of these generalised solutions is correct?

[proven garden]

if you let m=n-1 then you get the same x,y value

so every generalized solution is just a change of variable

thus they are all equivalent

[stable whale]

So yes?

[proven garden]

yes

[stable whale]

I see thanks

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[steel shoal]

Is it wrong to just insert the numbers into x here?

[steel shoal]

x=0 in (a), x=1 in (b), x=3 in (c)

or can you do that

[copper rapids]

Don't you see any problem in doing that, like getting undefined values?

[vast shale]

well you will have a singularity there

so that's kind of the issue

[steel shoal]

ohhhh right

my bad sorry

[worthy citrus]

if you can do that with no issues like dividing by zero, then you definitely should, so its worth checking

[steel shoal]

omg how do you solve that again

I guess just trying around until it works

[worthy citrus]

yeah you need to manipulate the expressions until direct substitution is fine

[steel shoal]

ok thanks

play around with binomial formulas I guess

I think (b) you can cancel the denominator like that

(c) maybe too

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.reopen

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✅

[worthy citrus]

yeah in b you wanna factor the bottom

and c factor the top

[steel shoal]

nice ty

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[weak halo]

Im stuck here, am I doing the right thing?

[weak halo]

idk how to isolate the x

nvm

nvm the nvm

wait

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[covert granite]

Is a forest juts multiple trees? Do they need to be connected in any way or is the graph below a single forest?

[paper depot]

a forest is a graph consisting of one or more trees yes

so yes that picture depicts a forest

[covert granite]

a single forest right? or am i missing something

[paper depot]

yes sure

can i see the original problem

[covert granite]

there was no original problem, just clarifying an explanation

thank you ann

wait i will be back in 2 mins

with a question

[paper depot]

k

ping me

[covert granite]

what is a bipartite graph?

[paper depot]

as in you want me to rattle off the definition?

[covert granite]

yes, but in a way i can understand it

[paper depot]

there are two equivalent definitions:

1. a bipartite graph is a graph whose vertices may be partitioned into two sets A and B such that no edge connects vertices from the same set.
2. a bipartite graph is a graph that is 2-colorable.

(i.e. colorable using only 2 colors)

another way to rephrase it is:

a graph is bipartite if there's a way to color its vertices red and blue in such a way that there are no red-red or blue-blue edges.

[covert granite]

Thank you!

is " 2-colorable." a widely used term

[paper depot]

sure is

[covert granite]

and do the sets have to have equal length?

[paper depot]

no

a k-coloring is a proper coloring with k colors

a proper coloring is a coloring where no edge connects vertices of the same color

a k-colorable graph is a graph that has at least one k-coloring

[covert granite]

i dont understand what you mean by this

[paper depot]

which word(s) do you not understand

[covert granite]

"k-coloring"

[paper depot]

look two messages above

a k-coloring is a proper coloring with k colors

[covert granite]

what is a "proper coloring"?

[paper depot]

a proper coloring is a coloring where no edge connects vertices of the same color

[covert granite]

im sorry

so would these all be appropriate definitions

Forest = a collection of trees

Eulerian circuit = a path that uses every edge of a graph exactly once

Bipartite graph = a graph whose vertices can be divided into two sets in which no edge connects vertices from the same set```

@paper depot

[paper depot]

Leaf = a vertex with a single edge connected to it
don't like this wording

[covert granite]

should i say "only"

to be more specific?

[paper depot]

"a vertex with only a single edge connected to it" maybe,
but i think the best wording is "a vertex whose degree is 1"

that way there is zero ambiguity

[covert granite]

kk ty

ur right

are the rest all correct?

[paper depot]

the last two are ok but i'm a bit iffy with the one for "forest". but on the other hand i don't know of one that would be better and not heavy.

[covert granite]

Can you give me a heavy one for "forest" so i can have it for reference?

or link me to one

[paper depot]

a forest is a graph in which every connected component is a tree
equivalently, a forest is a graph without cycles (whether it be connected or not).

[covert granite]

thank you so much

is there a !rep thing here

or nah

[paper depot]

don't know of one + don't really care about those things either

[covert granite]

oh ok

ty again

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[vast shale]

DF is parallel to AC and DE is parallel to BC. in deed to prove the buttom line

[vast shale]

yo anyone

[proven garden]

Maybe the similar triangles can help

[vast shale]

yeah but i should use Talles

[proven garden]

yes

that's the property of similar triangles

[glossy maple]

Try doing talles in triangles AED and BDF.

Then, try to combine these two results with some manipulations.

Hint: ||Whenever you have CF, you can write it as BC - BF for example, which could help||

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[static barn]

hio

[static barn]

hi

can anyone help with this

are the graphs correct/

[proven garden]

yes

[static barn]

how can i explain the second part

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@static barn Has your question been resolved?

[static barn]

anyonre <@&286206848099549185>

[static barn]

ok

.clos

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[severe thistle]

How did we go from rhs to lhs

[hidden kelp]

they multiplied by sqrt(65)/sqrt(65)

[severe thistle]

I know, and calculating it seems valid, but how did they express that

I don't know why it's true or why it works to write it like that

[hidden kelp]

what is sqrt(65)/sqrt(65) equal to?

[severe thistle]

OHHHH okay okay okay

I'm udmb lol

My bad

thank you <3

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[hidden kelp]

no you're not. Brainfarts happen to everyone. Come back if you have more questions :)

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[half yew]

.

[half yew]

For question one it just seems a little too easy to be right but my main question lies with two

I used determinants & cross products to find one orthogonal vector but I’m unsure how to find another

Would it be using projections? If so how would I start that.

[full hatch]

You can just multiply your vector by lampta

Note that the direct sum of U and U_orthogonal = R^3. Dimensionwise this means that dim(U_orthogonal) = 1. Therefore, it can be expressed as the span of the vector you just found

[half yew]

Ohhh you’re right. I didn’t think about just using scalars. So all I’d need to do is multiply u x v by 2 for example?

[full hatch]

It'd be great if you could use 42 instead, but yes

[half yew]

Ha, sounds good. Thank you!

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[full hatch]

You're welcome

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[half yew]

For #2 did I make a mistake? The numbers seem very messy

[half yew]

That I believe has a lengthy of one if I normalized it correctly so if that work is right I should just need to double it

[proven garden]

length of 2

not 1

yes double the unit vector

[half yew]

But the unit vector is right?

[proven garden]

yep

[half yew]

Sounds good, thanks

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[velvet cove]

can someone check my work for 2c

[velvet cove]

wait the question is cut off it just says “is tangent to the curve at point P.”

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[elfin moon]

How to solve this one?

[proven garden]

you can check for small n

[elfin moon]

I didn't understand the notation

Suppose i choose n=3

,w solutions x^2=1mod n n>2

[twin meteor]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by arjunn5589

[proven garden]

U(n) is a group

of coprime natural numbers less than n

[elfin moon]

What will i put for x?

0,1,2,3...?

[proven garden]

x is element of U(n)

[elfin moon]

I took n=10
So U(n) = {1,3,7,9} or {3,7,9}

Am I right?

[proven garden]

you include the 1

as it's under multiplication mod n

[elfin moon]

What do I do next?

3^2 /10 =9
1^2/10=1

7^2/10=9

9^2/10=1

So 2 solutions?

[proven garden]

1^2=1
3^2=9
7^2=9
9^2=1

so 2

do you know the lagrange theorem?

[elfin moon]

What langrange theorem?

[proven garden]

the order of a group

and the order of its subgroup

their relationship

[elfin moon]

It will divide?

[proven garden]

yes

say a is an element of G

consider the subgroup generated by a, let's denote it as <a>

|<a>|=order of a

[elfin moon]

Order is 4

I guess

[proven garden]

so order of a divides |G|

a^2 != 1 means that |<a>| must not divide 2

[elfin moon]

Ahhh hard to understand