help-0

[vestal yarrow]

Hi there, im stuck on how they differentiated z with respect to v

[winter light]

Recall that f^g = e^(g·ln(f))

And then think of u as a constant, such as 2, 3 or any other number

[vestal yarrow]

oh thanks yes

.close

[lone heart]

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[winter light]

You're welcome

[lone heart]

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[burnt wren]

Find domain of f(x)
I have tried solving the problem but the method is too long and confusing
Can anyone solve this in a simpler way

[sharp thorn]

What is domain of square root function

[burnt wren]

All non negative integers ig

[sharp thorn]

0 to infinitely

[burnt wren]

Yes

[sharp thorn]

Infinity

Numbers below e, will give negative results when we take log

I think there is no such domain I'm not sure

[rigid smelt]

log(x) is negative when x<1

not when x<e

[burnt wren]

Yes

[sharp thorn]

Yes okay wait so log will be negative is it's argument is lesser than 1

And sinx should be positive

Or else it will log 0 and thatisnt defines

[burnt wren]

The answer is pi/6 to pi/2
But i understood nothing from the solution
I couldn't find in internet too

[sharp thorn]

Listen

Sinx can't be negative

Do you understand why

[burnt wren]

Yes

[sharp thorn]

And it will just be 2sinx

[burnt wren]

Because we can simplify the function log to log of 2sinx

Yea

[sharp thorn]

And as I said earlier 2sinx must be greater than 1

So that log doesn't give negative results

[burnt wren]

Yes

[rigid smelt]

2sin(x) can be 1 and less than 1

[burnt wren]

Oh yea

The base can also be less than 1 then

Right

¿

[rigid smelt]

also, you should not simplify |sinx|+sinx to 2sinx at all, this is not true for all x

[sharp thorn]

Yes but we consider where sinx is positive so it can't be less 0

Base of log is more than 1

Like always

[burnt wren]

Wait we can since it's inside square root which is defined for positive
For now

[sharp thorn]

For now yes

It ain't legal but it works

[rigid smelt]

you didn't say this, you said 2sin(x)>1 which is not true

[burnt wren]

Since is greater than 0

[sharp thorn]

I meant 2sinx should be greater than 1 so we get positive numbers adter log

[rigid smelt]

again, you tried to simplify something without actually giving any restrictions, this is going to make everything harder for you to keep track

[burnt wren]

*sinx

[sharp thorn]

If we consider 2sinx must be greater than 1, then x must be pi\6

Greater than pi\6

[burnt wren]

But how do you consider it must be greater than 1

[sharp thorn]

Or else , after taking log it will give you negative number right

Which cannot be because we have sqrt

[burnt wren]

Sinx can be equal to one

[sharp thorn]

Yes

[rigid smelt]

don't forget about the denominator, either both the numerator and denominator are negative or both or positive

[burnt wren]

Yess

So we cannot say since greater than1

[rigid smelt]

also, are you sure the answer is [pi/6,pi/2]? What was the original question

[sharp thorn]

This question has more logic than solving

[burnt wren]

Yes the answer is correct

[rigid smelt]

no, it isn't

[sharp thorn]

Yeaa, couldn't it also have some values in the second quadrant?

[burnt wren]

Do you want me to say the procedure i understood in half

[rigid smelt]

is the question to state the domain of the function you took a picture of?

[burnt wren]

I mean for sqft function defined for positive and 0

Yes

[sharp thorn]

It can't have any values in 3 or 4 wuadrant

[burnt wren]

Find the domain

[rigid smelt]

I can show you some x values outside of that interval which still makes your function defined

again, like I said, your domain is not quite correct

[burnt wren]

Example ?

[sharp thorn]

5pi\6 to pi?

[rigid smelt]

take some value that is larger than pi/2 and less than 3pi/4 for example

something like x=2pi/3

[burnt wren]

Okay..

Should I check 2pi/3

Wait

[sharp thorn]

Is the denominator inside log or outside

[rigid smelt]

sure, either way, your function is defined there

[burnt wren]

Its not denominator ¿¿

[rigid smelt]

then what was it supposed to be?

[burnt wren]

Maybe that's why you were confused

[rigid smelt]

perhaps so

[sharp thorn]

What is your question

[burnt wren]

That's the base

[sharp thorn]

Ohhhhh

[burnt wren]

Yea ma bad

[sharp thorn]

Use log_a b = log b\ log a

[burnt wren]

I d k if that will be any useful
Won't it make more complex lol

[sharp thorn]

It will give 2 logs to work with

Will be easier ig

You can split the roots too

[burnt wren]

I am not getting it
I am confused again with all cases
Could you please show it on paper if possible

[rigid smelt]

there's not much cases. Try looking at a simpler expression log(u)/log(v)
You want to find u and v such that this expression is >=0. Can you state all of the possible cases?

[burnt wren]

Ok wait
Let me see

Here we can take it as log of 2sinx right

[rigid smelt]

also, once again, the domain is not [pi/6, pi/2). This simply because your function is periodic with a period of 2pi. Meaning there will be values of x outside of this interval that will also define your function. What you should rather have is
[pi/6+2kpi, pi/2 + 2kpi) for integer k.

[burnt wren]

And since there is no base
Or more like there could be any base by splitting it
We will take sinx + cosx between 0 and 1 or greater than 1

[burnt wren]

Yea i probabaly missed that part x is between 0 and pi
Included.

[rigid smelt]

no, like I have said. Do not try to simplify it. It will only makes stuff more complicated for you.

[rigid smelt]

anyway, I just want you to state the cases, using u and v

[burnt wren]

So like mod sin x plus since is what you want me to take

[rigid smelt]

yes, still waiting for the cases

[burnt wren]

Mod of sinx plus sinx
Case 1 is its between 0 and 1 included
Case 2 is it's greater than 1

Next is
For sinx plus cosx

Sinx plus cosx gbetween 0 and 1

Sinx plus cosx greater than 1

Also I belive sinx plus 90 is not equal to sin 45

If we take sinx plus cosx greater than 1 case

And also x cannot be 0 also
If we use first case

[rigid smelt]

great. So all in all it's just
C1:
0<|sinx|+sinx<=1 and 0<sinx+cosx<1
C2:
|sinx|+sinx>=1 and sinx+cosx>1
Now basically, if you do a little inspection, you can see that the first case is never possible. Which leaves you with the second case only

[burnt wren]

Ahh ues

[rigid smelt]

can you see why the first case is not possible though?

[burnt wren]

Kinda

Is it because 9f minimum or maximum stuff idk

No that's completely unrelated i feel

Wait lemme try to work it out once

[rigid smelt]

yeah kind of. Well here is how the inspection would play out
So |sinx|+sinx is either 0 when sinx<0 or is equal to 2sin(x) otherwise
Which means 0<|sinx|+sinx<=1 is satisfied when 0<sin(x)<=1/2. but when this happens, sinx+cosx>=1 or sin(x)+cos(x)<0

This can be shown in the unit circle. And this happens in the first and second quadrant

[burnt wren]

Oh wait

Just realised in one it's lesser than 1 or greater than 1

I think I got it

I lost internet tho : ((

[rigid smelt]

also, by doing the same inspection, you can also show that sinx+cosx>1 for x's that lies in the first quadrant. Basically, most of the inequalities can be solved using unit circles. But if you want to do it algebraically then it's fine too.

[burnt wren]

Yeah got it

I can solve from here

I think 😂

[rigid smelt]

sure

[burnt wren]

Ok thank you very much waller and bettim : )

.close

[rigid smelt]

np

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[warped topaz]

If I do double integration do I need to use the same term for u and dv both times?

[cursive badger]

Yes

[warped topaz]

Like can I use the du from the first integration as the dv for the 2nd integration?

[cursive badger]

Integration by parts you mean

[warped topaz]

yeye

[cursive badger]

If you switch you’ll go in circles

[warped topaz]

okok

[cursive badger]

Or something along those lines

[warped topaz]

Yeah I think I get you

Alright thanks!

❤️

.close

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[edgy cave]

All they added to the task: "Solve the system of equations". At the same time, I found that the value of x=2, and d=4. What should I write in the window below?

[gray isle]

is that the only solution you found?

I suppose just enter that as
(2 ; 4)
if there are no other solutions, leave blank

[tardy stag]

i see a quadratic

i am suspicious that there is only one solution

[gray isle]

^

[edgy cave]

I also managed to get that d=1, x=1, but if I substitute them into the first equation, it will be incorrect

[gray isle]

incorrect how?

show how you're subbing/simplifying to reach that conclusion

[edgy cave]

log2(1+3-2) - log2(1) = 1
1 - log2(1) = 1
log2(1)+1 = 1
log2(1) = 0

OOOOOOOUCH. For some reason, I thought that the logarithm cannot = 0, because I confused it with another definition. Yes, this decision must also be correct.

Well, thank you. I will try to enter these data in the window, although I do not know in what order.

[gray isle]

assume x comes first

[edgy cave]

.close

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[gray isle]

or it probably told you at the start

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[warped topaz]

https://i.imgur.com/A8VZmhQ.png

[warped topaz]

How is it possible to solve this, even with repeated integration the sin will just keep changing to cos, then to sin, then to cos forever

Same with the cos, they will never cancel or anything

[worn fox]

The general idea is to have the original integral show up again after doing by parts 2 or more times

In which case you have an equation involving your integral I which you can just rearrange for I

[warped topaz]

I dont get what you mean sry

[worn fox]

Do by parts twice and see what you get

[vale wigeon]

wait uh

theres a better way lmfao

$\sin(a)\cos(b) = \frac12 (\sin(a+b) + \sin(a-b))$

[ocean seal]

Ann

[vale wigeon]

yours, then, is $\frac12 \int [\sin(6x) - \sin(2x)] \dd{x}$

[ocean seal]

Ann

[worn fox]

Well I just figure the title of the question is "Repeated integration by parts"

[warped topaz]

I kinda want to learn both ways, but right now I have questions about both haha

[warped topaz]

In the 2nd integral for the last question I tried I got (-9sin(3x)(-cos(2x)/4)

[vale wigeon]

uhh

yeah no i'd say that like

[warped topaz]

Like that was my '- integral v*du' bit

[vale wigeon]

learning how to use a tool also includes learning when its appropriate to use and when not

and here i would say it's not

or like

you sort of CAN use it but it'll be very very painful and unenlightening

[warped topaz]

Ah oke I see

[worn fox]

I agree with you, I just think it would be good for them to do both, especially if they've never seen these types of IBP questions where you loop back around (if this is one of those; I haven't checked)

[warped topaz]

Does this also work with cos and cos? Or sin and sin?

[lone heart]

@warped topaz Has your question been resolved?

[warped topaz]

.close

[lone heart]

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[primal galleon]

A geometric sum is equal to 215. The first term is 5, and the last term is 320. Find the number of terms in the sum.

[near apex]

!status

[lone heart]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[primal galleon]

i'm stuck

215 = a_1 * (r^n - 1) / (r - 1)

215 = 5 (r^n - 1) / r-1)

[west girder]

You can find r in terms of n and vice versa

[primal galleon]

how can you show me

[west girder]

Use the fact that the first term is 5 and the last term is 320

[primal galleon]

215 = 5 * (r^n - 1) / (r - 1)
320 = a * r^(n-1)
320 = 5 * r^(n-1)

320 / 215 = (5 * r^(n-1)) / (5 * (r^n - 1) / (r - 1))

Like that?

5 = 320 * k^(1-m)

[near apex]

Use the last term as well.

[near apex]

What is k and m ?

[primal galleon]

I'm just using k to stand for ratio (in my native tongue)

[near apex]

Oh. You are close here.

try writing r^n in terms of r^(n-1).

[near apex]

Basically so that there is a term of r^(n-1).

Okay. Try thinking this way.

You have this: 215 = 5 * (r^n - 1) / (r - 1)

and this: 320 = 5 * r^(n-1)
What if you multiply r on both sides in this one?

You should get something which you can substitute.

[primal galleon]

I get the ratio to be -2, do you get the same?

[lone heart]

@primal galleon Has your question been resolved?

[lone heart]

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[lone heart]

@random sable Has your question been resolved?

[random sable]

<@&286206848099549185>

[random sable]

Any idea anyone?

<@&286206848099549185>

.close

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[alpine sable]

Exercise 4.17. Suppose x ∈ R with x > −1. Prove that, for every n ∈ N,
1 + nx ≤ (1 + x) n .

[alpine sable]

Here is how far, i have gotten solving this problem:

Our base case is (1 + 0 \leq (1 + 0)^1 = 1 \leq 1) and via the base case we can assume that if (x \in \mathbb{R}) with (x > - 1) for some (n \in \mathbb{N}) we can: (1 + nx \leq (1 + x)^n) (IH) we can now begin by adding (+ 1) so (1 + (n + 1) \times x) and we get (nx + x + 1) and since (n \in \mathbb{N}) we can (nx + x + 1 \leq nx + x + n) or (n(x + 1) + x)

[ocean seal]

scholablade

[alpine sable]

I also tried tackling the problem this way

Our base case is (1 + 0 \leq (1 + 0)^1 = 1 \leq 1) and via the base case we can assume that if (x \in \mathbb{R}) with (x > - 1) for some (n \in \mathbb{N}) we can: (1 + nx \leq (1 + x)^n) (IH) we can now begin by adding (+ 1) so (1 + (n + 1) \times x) and since (1 + n \not= 0) since (n \in \mathbb{N}) we have (1 + (n + 1) \times \frac{1}{- n - 1}) or (1 + \frac{n + 1}{- n - 1})

[ocean seal]

scholablade

[lone heart]

@alpine sable Has your question been resolved?

[heavy moss]

Anybody know how to solve e^-2 02 without a calculator?

[lone heart]

Please read #❓how-to-get-help

[alpine sable]

I recommend getting your own room in the MATH HELP (AVAILABLE)

[junior jetty]

did you get help yet

[alpine sable]

nope

[junior jetty]

okay i recommend you join gohar’s guide on discord

[heavy moss]

how do I get my own room in math help?

[alpine sable]

Do you see the channel with the math help availaible?

Pick one of the rooms like help-5

[tacit arch]

Why are you spamming other people's channels

[tacit arch]

They already linked you how

#❓how-to-get-help

[junior jetty]

i'm not spamming

[tacit arch]

.

[junior jetty]

i'm just trying to help someone

that's not spamming

[tacit arch]

Advertising is a form of spamming.

[junior jetty]

oh i didn't know sorry

[alpine sable]

#rules

[alpine sable]

why did you use x=0 in your base case?

x should be arbitrary

n should be the base case instead

its probably easier to go from right to left (ie start with (1+x)^(n+1))

[alpine sable]

Did you make a typo in the problem? Because the RHS simpliefies to be n+nx, so the problem becomes prove that $1\leq n$ which is trivial because n is natural number

[ocean seal]

ryanstaal2006

[alpine sable]

based on context, I would guess op meant rhs= (1+x)^n

Oh, yeah my bad the problem is Suppose $x \in \mathbb{R}$ and $x > - 1$ Prove that for every n \in \mathbb{N} \
$1 + nx \leq (1 + x)^n$

[ocean seal]

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

[alpine sable]

Yes

[alpine sable]

so if we have $(1+x)^{n+1}$, try using the IH and some appropriate bounding to get $1+(n+1)x$

[ocean seal]

toby____

[alpine sable]

Okay

Didn't know that was allowed to go from right to left

$a \geq b \iff b\leq a$, so its fine to go either way

[ocean seal]

toby____

[alpine sable]

That makes sense

Not finished but wanted to share progress

Our base case is (1 + x \leq (1 + x)^1 ) and via the base case we can assume that if (x \in \mathbb{R}) with (x > - 1) for some (n \in \mathbb{N}) (IH), we can now begin by adding (+ 1) so ((1 + x)^{n + 1}), since (n \in \mathbb{N}), therefore (n \geq 0), so ((1 + x)^{n + 1} \geq 1 + x)

[ocean seal]

scholablade

[alpine sable]

This is true but youre looking for something stronger

$(1+x)^{n+1}=(1+x)(1+x)^{n}$

[ocean seal]

ryanstaal2006

[alpine sable]

Now how can you use youre IH?

I don't see it but am thinking about it

And you should also use that x>-1 later at some point

Our base case is (1 + 0 \leq (1 + 0)^1 = 1 \leq 1) and via the base case we can assume that if (x \in \mathbb{R}) with (x > - 1) for some (n \in \mathbb{N}) we can: (1 + nx \leq (1 + x)^n) (IH) we can now begin by adding (+ 1) so ((1 + x)^{n + 1}) and via the induction hypothesis we can (1 + nx \leq (1 + x)(1 + x)^n).

That's the connection, i think

[ocean seal]

scholablade

[alpine sable]

This is not always true. If for example x=-1/2 that means 1+x=1/2. And you can take for example 4\leq 6 but 4\leq 6*1/2 isnt true

Do you want a hint how to use the IH?

Maybe a bit more time, i would like to solve this

If you have no issue with that

Alright

Ping me when you react

For sure

@alpine sable

Our base case is (1 + x \leq (1 + x)^1 = 1 \leq 1) and via the base case we can assume that if (x \in \mathbb{R}) with (x > - 1) for some (n \in \mathbb{N}) we can: (1 + nx \leq (1 + x)^n) (IH) we can now begin by adding (+ 1) so ((1 + x)^{n + 1}) or ((1 + x)(1 + x)^n) and via the induction hypothesis we can ((1 + x) \times 1 + nx), therefore giving us (nx + x + 1) or (1 +(n + 1) \times x)

[ocean seal]

scholablade

[alpine sable]

Though i didn't use x < - 1, it looks good imo

Yes looks good, you only didnt write nx^2 after expanding the brackets. Where did that go?

[alpine sable]

Thanks, about the problem am not sure where it is

Because there should be brackets around the (1+nx)

So you get nx+x+1+nx^2

Oh, yes you are right

But you have already done most of the proof. You only have to prove that the $nx+x+1+nx^2\geq nx+x+1$

[ocean seal]

ryanstaal2006

[alpine sable]

Will do that

Done

@alpine sable

Our base case is (1 + x \leq (1 + x)^1 = 1 \leq 1) and via the base case we can assume that if (x \in \mathbb{R}) with (x > - 1) for some (n \in \mathbb{N}) we can: (1 + nx \leq (1 + x)^n) (IH) we can now begin by adding (+ 1) so ((1 + x)^{n + 1}) or ((1 + x)(1 + x)^n) and via the induction hypothesis we can ((1 + x) \times (1 + nx)), therefore giving us (nx^2 + nx + x + 1), since a square of a (\mathbb{R}) number always gives us a positive number and (n \in \mathbb{N}) therefore (nx + x + 1 + nx^2 \geq nx + x + 1) or ((1 + x)^{n + 1}), therefore by induction for every (n \in \mathbb{N}), (1 + nx \leq (1 + x)^n).

[ocean seal]

scholablade

[alpine sable]

Yes seems good to me. Now do you see where you used the fact that x>-1? Because it is important for the proof

Nope

It seems useless information

Since a squared R number will always be positive regardless if it's more than - 1

You go from $1+nx\leq (1+x)^n$ to $(1+nx)(1+x)\leq (1+x)^n(1+x)$. So you multiply both sides with $1+x$. If you multiply both sides of an inequality with a negative number the inequality sign flips. So $1+x$ cant be negative

[ocean seal]

ryanstaal2006

[alpine sable]

Oh, yeah that property

Tricky

Thanks for the help!

I'l close this room now

.close

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[alpine sable]

Can someone help me find the image of this function?

[alpine sable]

I asked this on another channel but my phone turned off

I want without calculus

I was thinking AM-GM

[tacit arch]

Why can't you use calculus

[alpine sable]

We can but I think it is more interesting without

[finite flax]

rube question: why isn't the function definition enough to define the image?

like, do they want a sequence?

[tacit arch]

How do you know it's possible without calculus

I mean sure am-gm is possible too but a proof of am-gm uses calculus

[alpine sable]

With AM GM: f(x)>=2sin(x)^n cos(x)^n=2^(n-1) (sin(2x))^n

It is a problem for 10th graders

Sorry not 2^(n-1)

2^(1-n)

I think the image is [1/2^(n-1); 1]

But we need to find the minimum of 2^(1-n)(sin(2x))^n that is touched by f(x)

[arctic salmon]

10th grade in America?

[alpine sable]

No

Romania

[finite flax]

cosine and sine both have ranges in [-1,1]

[arctic salmon]

Oh ok

[finite flax]

so hypothetically if all of R can go in

we can get a minimum out of (0)^(2n) + 0^(2n) = 0 this is wrong

and a maximum of 1^(2n) + 1^(2n) = 1 also wrong

[0,1]

[alpine sable]

for n=1: the img is just 1

[finite flax]

the image is a set, if n=1 you still have infinitely many outcomes for infinitely many x

do you have access to the answer key?

[alpine sable]

Nope

[finite flax]

ok, well let me try to explain again

you and I both know that sine and cosine have ranges of [-1,1]

[alpine sable]

Yes

[finite flax]

if we take EVEN POWERS of these outcomes, the highest we can get out is +1 and the lowest we can get out is 0

right?

because an even number of factors is always a positive product

[alpine sable]

Ok right

[finite flax]

ok, I'm wrong about the maxes and the min

[alpine sable]

:))

I was confused

[finite flax]

yeah, because the same input of x for sine and cosine will be offset in the output

there's a couple of ways to analyse this then

you can graph or do a substitution

[alpine sable]

The image is [2^(1-n);1]

If we look at 2^(1-n)(sin(2x))^n in x=pi/4 we get the minimum that is touched by f

But how we can show this is the minimum of f

In pi/4

*for the restructiom [-pi,pi]

restriction*

[finite flax]

you sound pretty sure

did you find the solution?

looks correct

I'm checking a graph in desmos

and that looks right

wish I could animate this

[tacit arch]

The max of sin(2x) is at pi/4 which is the min of f if you can plug in pi/4 to f

[alpine sable]

[2^(1-n);1]. sum of periodic functions we get periodic function the local maximum are the same and the local minimum are the same. In pi/4 the derivative =0 and done. This is with calc

[finite flax]

https://www.desmos.com/calculator/5bssujaaer

the local minima are what are changing

the maximum is constant, yes

[finite flax]

good job

oh, I see what you are saying though

[alpine sable]

Yes but I dont know how to finish it with only AM-GM without calc

[finite flax]

wait, so do you already know calculus at 10th grade?

[alpine sable]

Maybe if we suppose that there is a value smaller than 2^(1-n) that is in the im

I am 11th but I friend asked me this and he does not know calc

And I am stuck:)))

[finite flax]

...but you DO know calculus?

at least through differentiation?

[alpine sable]

Yes. I prepare for the next year olympiad and now I do putnam integral inequalities problems and group theory now Jordan Hölder form

But I hate trig

[finite flax]

neat.

[alpine sable]

Yeah @tacit arch is right the AM-GM idea does not work without calc

But I am sure there is a method using trig identities

This function f reminds me about Newton Binomial and the trig form of A complex number. Some ideas?

[finite flax]

I'm still wrapping my head around AM-GM (you mean "AM-GM Inequality", yes?)

[alpine sable]

Yes

[finite flax]

proof https://www.youtube.com/watch?v=FMqZY03scZ4
math olympiad example https://www.youtube.com/watch?v=NupnO6HXKWw

[alpine sable]

Yes I know the proof

[finite flax]

So I think in your case, what you are saying, is that it is most useful to examine the minimum at x=pi/4, so you would like a sequence of numbers where n is ascending

[alpine sable]

The inequality works too well for this function f

[finite flax]

and do something with AM-GM

[alpine sable]

I dont want any sequence

The image of f as n goes inf is [0,1]

[finite flax]

yes, I see that

um, is 0 included though?

or is it asymptotically approaching 0

[alpine sable]

Hmm

No it is not

[finite flax]

you need AM-GM to spit out [2^(1-n),1]

[alpine sable]

I used AM GM to have the form 2^(1-n) (sin(2x))^n

It works too well

[finite flax]

hm

well I can't catch up to you fast enough to be any use here

I'm not familiar with the use of AM-GM

<@&286206848099549185>

[alpine sable]

Ah

It is so simple

Omg

[finite flax]

?

Far as I got

[alpine sable]

No i was thinking i did it without calculus

I think it works by induction

Without AM GM

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[sullen cradle]

Can someone tell me if this is actually correct and if not, where i went wrong?

[sonic ridge]

theres a couple things wrong here

[sullen cradle]

Ok im all eyes

[sonic ridge]

step 6 is wrong, dividing both sides by -1 would not make the right side 0

[sullen cradle]

When did i do that?

Ill label my steps hold up

[alpine sable]

Please don't use × to denote multiplication, especially with an x in your equations

[sonic ridge]

also, -(x*-y) does not equal (-x*y)

it just equals (x*y)

[sullen cradle]

What shall i use?

[sonic ridge]

you can just write xy or (x)(y)

[sullen cradle]

Can i use “.”?

[sonic ridge]

sure

from step 5 to step 6

[sullen cradle]

Oh. What does it become?

[sonic ridge]

how did you make the right hand side = 0

[sullen cradle]

Hmm ur right. Thats my mistake

[sonic ridge]

just (xy)

[sullen cradle]

But the - went to the numerator

And then both were affected by it

[sonic ridge]

no, youre thinking of addition, where you distribute the negative

[sullen cradle]

Hence they change sides right?

[sullen cradle]

Ohhh

So that was the fatal flaw in my idea

Thank you so much

[sonic ridge]

np

see #❓how-to-get-help , this channel is taken

[lone heart]

@sullen cradle Has your question been resolved?

[lone heart]

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[unique forge]

how do i solve this?

[lone heart]

@unique forge Has your question been resolved?

[unique forge]

<@&286206848099549185>

[upbeat hornet]

Do you understand the question?

[lone heart]

@unique forge Has your question been resolved?

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[tender anchor]

how do i find h

[lone heart]

@tender anchor Has your question been resolved?

[surreal meadow]

if you have w then you have h

write A in terms of w, h, and T

[lone heart]

@tender anchor Has your question been resolved?

[tender anchor]

.close

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[cunning trout]

so i was expanding (9/10)^6

[cunning trout]

where did i go wrong?

[sullen meteor]

Pretty sure you didn’t use the right thing: it’s 1 6 and so on not 1 5

[cunning trout]

wait what?

[sullen meteor]

Yeah

Cause think of it

(A+b)^2 is a^2+2ab+b^2

So 3 terms

It’s always one more term

[cunning trout]

but look at the coefficients here:

[sullen meteor]

Yeah

But you did to the power of 6

[sullen meteor]

You write 6

[cunning trout]

oh crap

…

[sullen meteor]

So uh that might be the only problem

[cunning trout]

yeah lol

thanks

[sullen meteor]

Np

[cunning trout]

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[bold panther]

please help

i tried solving it but i dont think my answer is correct

[tacit arch]

Show your work

[bold panther]

are you still free to help me(?)

[tardy stag]

did you actually get an answer? i don't see one

[bold panther]

no because when i tried to factor the equation i got at the end, it wouldnt come out even, so i assumed that i got a part wrong somewhere

[tardy stag]

oh and are you considering the 4 (shouldn't it be 2?) in the middle at all?

[bold panther]

OH

Yes i wrote that wrong

I will resolve it now with the correct number

[tardy stag]

i got a surprising answer

[bold panther]

:o

[bold panther]

i got an unfactorable equation again :(

[tardy stag]

wait yeah i had an arithmetic error

[bold panther]

Oh

[tardy stag]

what did you get for sintheta or costheta/

or equation etc

[bold panther]

I did this but the equation is still unfactorable

[tardy stag]

is this the eqn of a trapezoid?

,tex .plane geom

[bold panther]

Yes

[ocean seal]

Hayley

[bold panther]

That is the (a+b)/2 part 2costheta is the height

Wait a minute

I changed one of the 4s but forgot the other

[tardy stag]

2costheta is the height yeah

[tardy stag]

ahh ok

[bold panther]

ok i finally got a factorable equation :D

[bold panther]

Would the answer be 5pi/6?

(Sorry I sent wrong pic)

[tardy stag]

well it's either pi/6 or 5pi/6 right

[bold panther]

Yes

[bold panther]

Would 5pi/6 not work since the cosine would be negative?

[tardy stag]

yeah think about what angles are actually physically possible here

[bold panther]

ahh i see

so only pi/6 would be the answer right?

[tardy stag]

well i'm a bit confused about that

because you're showing that the derivative is negative at 0 meaning increasing the angle makes it worse

[bold panther]

oh

[tardy stag]

but intuitively that doesn't really make sense to me i think?

[bold panther]

maybe angle should be 0?

[tardy stag]

what's your expression for dA/dt?

as simplified as you can get it

i think it's that line starting with -8, right?

[bold panther]

2sin^2(x) +sinx-1 =0

[tardy stag]

that's an equation that describes when the derivative is equal to 0

[bold panther]

yes

[tardy stag]

but the left hand side of that is not the derivative

[bold panther]

it is the derivative i got for the area

[tardy stag]

no it isn't

[bold panther]

oh

[tardy stag]

you had negative signs in there and 8s and stuff

what happened to those

[bold panther]

i divided by -4

[tardy stag]

so... the left hand side is the derivative divided by -4?

[bold panther]

yes

[tardy stag]

ok

so if you were to multiply that by -4 you'd get the actual dA/dt

earlier we only cared when it equaled 0 because we wanted critical points

but we do actually care when it's above/below zero

seems to me that this is dA/dt

[bold panther]

oh

but since the right hand side is equal to 0, cant i just divide by -4?

[tardy stag]

if you're trying to find when it equals 0, sure

but like... take the function f(x) = -x^2
its derivative is -2x
when is f increasing / decreasing?

if you divide by -2 you'll flip the result

[bold panther]

ohhh

[tardy stag]

it should be pretty clear to see that dA/dt is above 0 at t=0; in particular it's 4

(i'm using t for theta...)

[bold panther]

ohh i see

so i messed up when i divided by -4, and flipped the results. so it really is pi/6

[tardy stag]

yeah

[bold panther]

ohh i see

[tardy stag]

i made the same mistake btw

and many others

so don't feel bad about that i think

[bold panther]

thank you for helping me :D

.close

[lone heart]

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[meager bone]

Hello

i need help with this and idk where to start

I think i have to substitute X^2 for U

and then solve using the quadratic formula

[tardy stag]

that is not what you have to do

you should solve it by grouping and factoring

[meager bone]

okie

now how i do that lmaooo

im sorry

[tardy stag]

group the first two terms together and factor their GCD out of them

do the same with the last two terms

[meager bone]

okieeee

how do i get the GCD?

i grouped them

[tardy stag]

okay, look for common factors

like

from that first one you can pull out an x, right?

you can probably pull out more than that

[meager bone]

maybe 2x?

ahhh idkkkk

[tardy stag]

you're looking at $3x^3 - 6x^2$, right?

[ocean seal]

Hayley

[meager bone]

yes

and then -27x+54

[tardy stag]

we'll get to that one later

[meager bone]

okie

[tardy stag]

if we write 6 as 2*3

and expand the x's out

[meager bone]

okay let me write it down and see how it turns out

i have 3 times x times x times x - 2 times 3 times x times x

[tardy stag]

yes

now go through and underline something from the first term that's also in the second one

(and underline it in the second one as well)

[meager bone]

okay

[tardy stag]

write that off to the side, and do it again

and do it again, until you can't anymore

(don't underline the same thing twice)

[meager bone]

How’s this so far?

[tardy stag]

good

i think that's all of them

[meager bone]

yup

[tardy stag]

now write it down like this:
(all the underlined parts, multiplied together) * (non-underlined stuff from the left + non-underlined stuff from the right)

that's kind of hard to explain here

[meager bone]

i think i got it

ill send a pic so youll see how i do

[tardy stag]

this is what i did, you can check yourself against it

[meager bone]

oh

i got 3x^4 (x-2)

[tardy stag]

where'd you get four x's from

[meager bone]

because thers 4 x's

[tardy stag]

yeah but we only write it down once for each set

[meager bone]

oh

[tardy stag]

what we're doing here is the opposite of the distributive property

[meager bone]

so one from left and one from right

[tardy stag]

$a(b+c) = ab + ac$

[ocean seal]

Hayley

[meager bone]

yup

okay

[tardy stag]

no, one pair from both and another pair from both

look at my colors

[meager bone]

ohh okay

sorry

i got it

[tardy stag]

do you see that the 3 is red and there's a green x and a blue x

kk

[meager bone]

yes

green goes with green and blue goes with blue and you take one of each

thats just my thinking

[tardy stag]

yeeeee

[meager bone]

okieeeee

[tardy stag]

and then the stuff that doesn't get paired lives in teh parentheses

ok now do the same for the second pair

[meager bone]

okay

ill show you what i get

-27(x+54)?

[tardy stag]

hmm not quite, 54 was unchanged there

if you think about what 27*54 is, that's a big number and that doesn't show up in your original equation

[meager bone]

but i do multiply them?

so -27x-1458?

[tardy stag]

what? no

[tardy stag]

can you show me how you got this?

[meager bone]

okay

[tardy stag]

uh, okay, that was less enlightening than i had hoped

you decided on the factor of 27

or -27 rather

and you successfully factored it out of the first term
but you didn't factor it out of the second term, the 54

[meager bone]

not sure what im supposed to do to it?

im so sorry

[tardy stag]

ok

we want to find something that is both a factor of 27 and of 54

and then we do the transformation $ab + ac = a(b + c)$

[ocean seal]

Hayley

[meager bone]

9 and 6?

or just 9 i think

because 9 times 3 and 6 times 9

[tardy stag]

9 would be one, what do you get when you factor a 9 out of both of those terms?

[meager bone]

idrk how to factor

like im not good at it

3 and 6?

[tardy stag]

ok you desperately, urgently need to practice factoring especially simpler things

but

[meager bone]

i kinda know how

[tardy stag]

yes, 3 and 6, so our resulting equation is 9(-3x + 6)

but do you see that there might be another thing we can factor out of that?

[meager bone]

yea 1 and 2?

so 9(-x+2)?

[tardy stag]

you can't just throw away that 3

you're pulling it out as a common factor

[meager bone]

is it that thing where its like i have to find something that adds up to -3 and one that multiplies to 6?

[tardy stag]

no, that's solving a quadratic

which we're not doing

[meager bone]

oh

ok

[tardy stag]

anyway

[meager bone]

yea ill find some sample stuff to factor online

[tardy stag]

so 9 * (-3x + 6)

ignore the 9 for a sec, we want to factor that 3 out of the -3x + 6

so we end up with 9 * 3 * (-x + 2)

do you understand this?

[meager bone]

yes actually

for once i get it lmaoooo

lemme write it donw myself rq

like you said

were doing reverse distributiomn

[tardy stag]

you should do all of these when you're done with this problem https://www.cusd80.com/cms/lib6/AZ01001175/Centricity/Domain/3342/GCF key.pdf

[meager bone]

okay

[tardy stag]

second page is the key

[meager bone]

thanks

so do i keep the 9 and 3 like that or do i multply them?

[tardy stag]

you can multiply them together to get 27

[meager bone]

okay

then i distribute?

[tardy stag]

if you distribute won't you end up right back where you started

[meager bone]

oh ur right

sorry

so i keep it like that then

[tardy stag]

yes we're almost done here

[meager bone]

okay

thank you

[tardy stag]

now remember at the start we had (that first bit) + (that second bit), and we've found equivalent expressions for those

[meager bone]

yes

[tardy stag]

here let me write it out

[meager bone]

ok

[tardy stag]

$3x^2(x-2) - 27(x - 2)$

[ocean seal]

Hayley

[tardy stag]

now we get to do the reverse-distribution thing one more time

[meager bone]

oh okay

i was gonna ask if i should substiture U for x-2

[tardy stag]

you could i guess

[meager bone]

but sure we can do that

idk thats just what we did in class

[tardy stag]

that might make it easier to see but you can also just see that there's an (x-2) in both of them

[meager bone]

then we do quad equations

so you combine the x-2?

[tardy stag]

yes, we factor out the (x-2)

[meager bone]

okay

[tardy stag]

it's like we underlined the whole (x-2) in red

[meager bone]

ok

let me write it donw

ok i wrote it and underlined

ready to proceed

[tardy stag]

yes okay so factor it out like we did before with the 3s and the x's

[meager bone]

ok

i split it first right?

into 2 different parts

so the ones in common are x-2

let me show you

[tardy stag]

no

keep the (x-2) as a unit

if you have to call it U then do that

[meager bone]

ok

Like this?

This literally feels like SpongeBob getting his boating license 😭😭😭😭

[tardy stag]

let me show you...

[meager bone]

0k

[meager bone]

ok i get it like that

the colours help alot

[tardy stag]

figured they would

[meager bone]

thanks

now do i distribute the 3?

[tardy stag]

no don't distribute you would just be undoing the thing that we spent so much effort doing

we want to undistribute

which is called factoring

[meager bone]

ok

[tardy stag]

so....

the original problem asked for you to solve $3x^3 - 6x^2 - 27x + 54 = 0$

[ocean seal]

Hayley

[meager bone]

yes

[tardy stag]

do you know how to do that at this point?

[meager bone]

i think yes

split, factor, bring them together, factor

thats what i have written so far

[tardy stag]

err yes that

[meager bone]

yeeeeee

[tardy stag]

what i would do with that worksheet is do like two of them and then carefully scroll down to make sure you're doing it right before doing the rest

[meager bone]

the one you sent?

[tardy stag]

yeah

[meager bone]

ok

wait so we done?

like wtih the one i sent

[tardy stag]

um well did you find solutions?

[meager bone]

oh

nah

[tardy stag]

i asked you if you knew how to do that

[meager bone]

nah im sorry

i thought you were asking for a recap

like with steps

[tardy stag]

okay, please, like, answer the questions that I ask

[meager bone]

sorry

[tardy stag]

so...

the original problem asked for you to solve $3x^3 - 6x^2 - 27x + 54 = 0$

[ocean seal]

Hayley

[meager bone]

yes

[tardy stag]

and we showed that the left hand side was equal to that factored expression

do you know why we factored the left side?

apart from "the problem told us to"

[meager bone]

imma be honet no

i cant tell you why

because it was equal to the factored expression?

[tardy stag]

okay...
0 is a wonderfully special number because if you have an equation like $A\cdot B = 0$ then you know that either $A = 0$ or $B = 0$ or both

[ocean seal]

Hayley

[meager bone]

yea i remember that

[tardy stag]

do you understand that when we say we "factored the left side" the new expression we got is equal to the original?

so $3(x-2)(x^2-9) = 0$

[ocean seal]

Hayley

[meager bone]

so i solve it like 3(x-2)=0 and x^2-9=0?

[meager bone]

partially

yes

[tardy stag]

meaning that at least one of these is true:

• 3 = 0
• x - 2 = 0
• x^2 - 9 = 0

[meager bone]

and i need to solve to find that out right?

[tardy stag]

yes

[meager bone]

ok

gimme a sec

[tardy stag]

you would try each one of those in turn

and see if they produce any solutions

[meager bone]

so the last 2 gave me solutions which were x=2 and x= plus minus 3

[tardy stag]

yep

[meager bone]

and thats my final answer

[tardy stag]

yesh

[meager bone]

https://tenor.com/view/spongebob-drivers-license-gif-7626540

[tardy stag]

took a hot minute

[meager bone]

thank you so much

[tardy stag]

should only take about 1-2 minutes for you to do that i think, so practicing is essential

so yes do that worksheet i sent

[meager bone]

sure

thanks

.close

[lone heart]

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[dull lily]

can someone help me with 2A, im just kind of lost overall

[surreal meadow]

do you know what they are asking

[dull lily]

yeah so the problem starts off by telling us to assume that the set W is a subset of R5

and x1,x2,x3,x4,x5 are the values within the set

[surreal meadow]

no

[dull lily]

oh

rip

then no, im lost besides that

[surreal meadow]

the elements in the set are tuples

of the form

(x1,x2,x3,x4,x5)

[dull lily]

mhm i see

the issue is i dont evem know how to start the proble,, because it asks to show that the 0 vector sub 5 would fall in as a subset

for part a

right?

[surreal meadow]

yes

what is 0_5

[dull lily]

wait

im sorry what do you mean by

0_5

[surreal meadow]

(\vec{0}_5)

[ocean seal]

maximo

[surreal meadow]

do you know what this means

it's coming straight from the question

[dull lily]

yeah isnt it just a 0 vector

[surreal meadow]

yes

what does it look like

[dull lily]

X sub 5 vector

?

[surreal meadow]

0_5 = <0,0,0,0,0>

[dull lily]

ohhhhh

wait i see it now

so just for clarification

if it was 0_3

[surreal meadow]

show these values for x1,x2,...,x5 satisfy the conditions imposed

[dull lily]

it would be <0,0,0>

[surreal meadow]

yes

[dull lily]

awesome

so then would the next step just be to plug in 0 for all

the conditions

[surreal meadow]

this vector has
x1 = 0
x2 = 0
...
x5 = 0

check whether that satisfies the conditions mentioned for W

[dull lily]

and that would be this basically

right

[surreal meadow]

yes

[dull lily]

well since its a zero vector

then every value plugged in would be

0=0

therefore all conditions are satisfied

right?

[surreal meadow]

yes

[dull lily]

awesome, that makes sense!

my next question would be what is u vector and v vector for part B

u and v vectors are never introduced

[surreal meadow]

they are arbitrary

have you taken a course on proofs before?

[dull lily]

i have once a while ago

this is more of a reintroduction for me

but i see what your saying by arbitrary

basically the question asks if we have 2 random vectors v and u, that satisfy all the given conditions, then u and v added together would also satisfy the given conditions

right?

[surreal meadow]

yes

that's what the question says

[dull lily]

relating to the first question

we know that the 0 vector

satisfies all given conditions

does that neccesarily mean that u and v are 0 vectors?

[surreal meadow]

no

u and v are arbitrary

all we know is that they satisfy the conditions given

[dull lily]

right

how would i prove that arbitrary + arbitrary = arbitrary

?

[surreal meadow]

that's not what you're proving

[surreal meadow]

read what you sent here again

[dull lily]

right

were proving that if 2 values satisfy w, if those two values are added, then the results would be arbitrary

?

[surreal meadow]

we are proving that if u and v satisfy the conditions to be in W

then the vector given by

u + v

also satisfies the conditions to be in W

[dull lily]

right, but im struggling to find out how i would even start this proof

[surreal meadow]

let's do an easier one

W = { (x1,x2) | x2 = 0 }

let's take two vectors u and v satisfying x2 = 0

and we let them be arbitrary

so they look as follows:
u = <a,0>
v = <b,0>

[dull lily]

right

[surreal meadow]

what is u + v?

[dull lily]

would u + v just be 0

[surreal meadow]

no

do you know how to add vectors?

[dull lily]

<a,0> + <b,0>

right

[surreal meadow]

yes

do you know how to add vectors?

[dull lily]

yeah

oh

<a+b, 0>

?

[surreal meadow]

yes

[dull lily]

my bad

[surreal meadow]

does <a+b,0> satisfy x2 = 0?

[dull lily]

yes because of the 0 in <a+b, 0>

?

[surreal meadow]

yes

so we've just proved u + v is in W

[dull lily]

right so same process but for a bigger set

right'?

[surreal meadow]

yes

let's look at our conditions

[dull lily]

mhm

[surreal meadow]

this is saying

oof ok i was going to do something that probably isn't helpful

give me one second

[dull lily]

lol youre good

[surreal meadow]

ok let's take two vectors u, v

we can write them as:
u = <u1,u2,u3,u4,u5>
v = <v1,v2,v3,v4,v5>

and u + v would be what?

[dull lily]

right

<u1+v1> <u2+v2> etc <u5,v5>

[surreal meadow]

u + v = <u1 + v1, ..., u5 + v5>

[dull lily]

bad notation on my part but yeah

[surreal meadow]

let's call this vector w

[dull lily]

makes sense

[surreal meadow]

w = u + v

[dull lily]

mhm

[surreal meadow]

and then
w1 = u1 + v1, ...

[dull lily]

w2 = u2+v2

[surreal meadow]

now let's see if the conditions are satisfied

w3 = u3 + v3

[dull lily]

right

[surreal meadow]

and since u and v satisfy the conditions of W

[dull lily]

are we still using the conditions that were given in the OG problem

[surreal meadow]

u3 = 2u1 - 3u4
v3 = 2v1 - 3v4

[surreal meadow]

yes

[dull lily]

ok cool

[surreal meadow]

we can assume this since u and v satisfy the conditions of W

[dull lily]

uhhh sorry

[dull lily]

how does this satisfy the conditions w

[surreal meadow]

these are the conditions to be in W

look at the first equation

if u is in W

[dull lily]

right

[surreal meadow]

then u3 = 2u1 - 3u4

[dull lily]

right makes sense

[surreal meadow]

same goes for v

[dull lily]

right we can just sub them out right