precalculus

[viscid thistle]

how'd you get that?

here's a question: what's the asymptote of 2^(x-2)? and how can that give you the asymptote of 2^(x-2) + 1, your function?

[alpine basin]

Wouldnt I have to solver for x 2^x-2 = 0?

Honestly I just put it in my calculator and graphed, but it doesnt show the asymptote

[viscid thistle]

ok first, can you tell me what an asymptote is

in your own words

(here's the graph of 2^(x-2))

[alpine basin]

the place that the line rides along but never touches

[viscid thistle]

that's one way to look at it, sure

so looking at the graph here, what's the asymptote of 2^(x-2) ?

[alpine basin]

maybe 4.5

[viscid thistle]

let me rephrase

what's the horizontal asymptote of 2^(x-2)? (there are no vertical or slant asymptotes so there is no ambiguity in leaving out the adjective, but ill put it in for brevity)

[alpine basin]

o

0 right?

[viscid thistle]

yeah, y = 0

[alpine basin]

wait

How can you tell theres no vertical one

[viscid thistle]

because an exponential function doesn't "blow up" at a point; it's defined for all x and keeps increasing

by "blow up" i mean "exhibit behavior like the function 1/x near 0"

(that's a really rough way of looking at it but i'd rather stick to intuition rn)

anyways, with the (horizontal) asymptote of 2^(x-2), can you tell me the horizontal asymptote of 2^(x-2) + 1?

i encourage you to think in terms of what the graph of the new function will look like

[alpine basin]

1

[viscid thistle]

yes

[alpine basin]

so when it comes to exponention functions

you're always looking for the horizontal asymptote?

[viscid thistle]

yep

an exponential function a^x, for a > 0, can only have horizontal asymptotes (if a = 1 this is a bit silly)

it might look like there's a vertical asymptote because exponential functions grow really fast, but i can guarantee you there will not be

[alpine basin]

ok

so the horizontal asymptote of this function would be -2 right?

[viscid thistle]

a more analytical way to see that the asymptote of your function 2^(x-2) + 1 is y = 1 is to notice that 2^(x-2) is increasing and goes to 0 as x goes to negative infinity

but i think you can get away with thinking in terms of graphs for now

for your next question

what does the graph of the logarithm function look like?

you're right in that there will be an asymptote involving -2, but im not sure that it's going to be a horizontal one

[viscid thistle]

i mean you can just graph it using desmos, yeah

[alpine basin]

oh my b

[viscid thistle]

but you can also find the rough shape of the graph just by playing with the graph of log_2(x)

what is the asymptote of log_2(x)?

[alpine basin]

So log functions can have vertical asymptotes

0

but it has a horizontal one as well right?

[viscid thistle]

no

here's a nice way to think about it in terms of the graphs

you know that exponential functions have horizontal asymptotes and no vertical asymptotes, right?

[alpine basin]

ye

[viscid thistle]

since logarithmic functions are inverses of exponentials, their graphs look like those of exponential functions but reflected over the line y = x

what should that reflection do to the asymptote?

[alpine basin]

make it so it only has a vertical asymptote?

[viscid thistle]

exactly!

(there are better ways to see this, and also, something like log(x^2 + 1) won't have any asymptotes)

[alpine basin]

hmm

so log2(x) asymptote is x = 0?

[viscid thistle]

yeah

now can you find the vertical asymptote of -log_2(x + 2)?

just by thinking about reflecting/translating the graph of log_2(x)

[alpine basin]

Uhm

It's -2 right?

[viscid thistle]

yeah

[alpine basin]

okok

[viscid thistle]

x = -2

[alpine basin]

yeah

[viscid thistle]

you wanna be precise here

[alpine basin]

Okay

Can I ask you another question thats not really related to asymptotes? xD

[viscid thistle]

i guess so

[alpine basin]

So im trying to simplify this into a single log

and the middle part, 1/3ln(x^3y^6) simplified would be ln(xy^2) right?

[viscid thistle]

$\ln(a)+\ln(b)=\ln(a\cdot b)$

[obsidian monolith]

Al𝟛dium:

[alpine basin]

Yeah

[viscid thistle]

aledium if you want to take it from here ill let you do so

[alpine basin]

Thanks for helpin tho Ttera

[viscid thistle]

aledium if you want to take it from here ill let you do so
@viscid thistle i mean idc i was just pointing it out, if you want to continue go with it

If you are busy, i can take it if you want

ill be back in 10-15 minutes so go ahead, ty

[alpine basin]

I was just gonna ask, would it be incorrect for me to simplify the middle log first

[viscid thistle]

You can't simplify further in the ln(x³y⁶)

[alpine basin]

well it would be ln(x^3y^6)^1/3 right?

[viscid thistle]

Yep. But its not useful at all

[alpine basin]

wym

[viscid thistle]

1/3 is better out there as a factor than on the exponent

Its considered simplified

So the 1/3 leave it there and write it as a single ln

[alpine basin]

1/3ln((y^3(x^3y^6)/(y^5))

[viscid thistle]

Tho the 5lny may be better to write it as ln(y⁵)

ln((1/3y^3(x^3y^6)/(-5y))
Where does the 1/3 come from?

[alpine basin]

you said leave it out there

[viscid thistle]

It should be out of the ln

[alpine basin]

oh

so 1/3ln((y^3(x^3y^6)/(y^5))

[viscid thistle]

Yes.

[finite wraith]

sir

1/3 goes as power now

which is third root

[viscid thistle]

Now go ahead and simplify the inner ln

[finite wraith]

thank you for having my attention

👋

[viscid thistle]

???

[alpine basin]

would it be ln(x^3y^4)?

[viscid thistle]

Yes.

1/3

Don't forget the 1/3 tho

$\frac{1}{3}\ln(x³y⁴)$

[alpine basin]

yeah

[obsidian monolith]

Al𝟛dium:

[alpine basin]

Hmm

the simplified answer is lnx and Im not sure how Id get that

[viscid thistle]

Wait a sec

[alpine basin]

when cause wouldnt y^4 change to y^4/3

ok

[viscid thistle]

Hm?

Can you post the supposed answer?

Is it really ln(x) alone?

Bc wolfram doesn't seem to agree neither

[alpine basin]

yeah

[muted granite]

Not sure where to start. Any help would be great.

[viscid thistle]

bro the channel's occupied

Are you not decent enough to see that this channel is taken?

@viscid thistle what do you think tho?

for what?

The solution says its ln(x)

i got that

lemme double check

I have fucked somewhere

wolfram says it's equal to ln(x) as well, maybe you fudged the input

Yeah prolly did

(at least for positive x,y, but who cares lol)

also @muted granite that should go in #linear-algebra or in one of the question channels, not #precalculus (even if this question is from a precalculus class, it is not really precalc)

,w simplify ln(y³)+(1/3) ln(x³y⁶)-5ln(y)

Hm.

click the link

[alpine basin]

oh ye it says alternate

assuming theyre positive

[viscid thistle]

spam!

Uh

Well ig you can show the way to ln(x)

Can you?

Im gonna try find where i have fucked

i took a look at your conversation with buckets

simplifying the middle term actually makes this pretty straightforward

if you throw the (1/3) to the exponent inside the ln in the middle term and them do some simplifying on the last term, the final answer pops out cleanly

Ok just realised

[alpine basin]

simplifying the middle term is ln(xy^2) right?

[viscid thistle]

yes

[alpine basin]

then

hold on lemme write it out

[viscid thistle]

I was working out a different thing

On my paper

Thats what made me go oof

[alpine basin]

ln((y^3(x^3y^6))/y)?

[viscid thistle]

Nah to get to ln(x) you actually don't have to write it into a single log wtf

As far as i have done it

[alpine basin]

idk the directions say write to single log and simplify, and yeah simplifying that single log doesnt look like it'd be lnx

[viscid thistle]

Uhhh okay so

Lets ignore the "write into a single ln"

Back to the start

$\ln(y³)+\frac{1}{3}\ln(x³y⁶)-5\ln(y)$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

So on the first ln simplify it

@alpine basin

Taking the 3

[alpine basin]

uh

It would be ln(y^3*the 2nd ln) right?

[viscid thistle]

Nono we won't write it onto a single log this time

[alpine basin]

but I think I have to tho

[viscid thistle]

Look at the first ln

Hmm

[alpine basin]

when she looks at my work or whatever

[viscid thistle]

I don't think there is a way of simplifying it to ln(x) without not writing it onto a single ln

there is if you skip a step lol

As far as i have done it

[alpine basin]

hmm

wym skip a step

[viscid thistle]

there is if you skip a step lol
@viscid thistle can you dm me your work lol

Or here idc

yeah one second lemme write it up again

sent

Yeah

Its not the way id have done it but it gets what you want

@viscid thistle you want me or you to explain it to him

i can explain how i did it

Ok ill dip

you have $$ \ln(y^3) + \frac{1}{3} \ln(x^3y^6) - 5\ln(y) $$

[obsidian monolith]

TTerra:

[viscid thistle]

Y'all spent 40 minutes on this?

i didnt

I actually went the wrong way

So yeah lol

[alpine basin]

lol

[viscid thistle]

Not the wanted way

one way to simplify this is to get (+1) in front of each log, and then use the fact that $\ln(ab) = \ln(a) + \ln(b)$

[obsidian monolith]

TTerra:

[viscid thistle]

Which was correct but not wanted

you already figured out how to simplify the middle term to $\ln(xy^2)$

[obsidian monolith]

TTerra:

[alpine basin]

yes sir

[viscid thistle]

what about the last term, $-5\ln(y)$?

[obsidian monolith]

TTerra:

[viscid thistle]

how can you simplify that?

i really hope you're not typing up the answer

^

Lol.

[alpine basin]

well

wouldnt it be ln(y^5)

[viscid thistle]

close

[alpine basin]

but its like divided or some shit

[viscid thistle]

can you be just a little more precise?

i know what you mean, i just want you to write it down

[alpine basin]

ok

Altogether I got

ln((y^3(xy^2)/y^5)

[viscid thistle]

hey im only talking about simplifying -5 ln(y) right now

but

yes

that is correct

lets format it a little better: $\ln(\frac{y^3 x y^2}{y^5})$

[obsidian monolith]

TTerra:

[viscid thistle]

can you simplify the thing inside the ln?

[alpine basin]

yes

[viscid thistle]

to?

[alpine basin]

uh one sec

omg

ln(x)

[viscid thistle]

ya

[alpine basin]

bruh how did I not see that

[viscid thistle]

I had a brainfart tbh

it happens to all of us, dw bout it

[alpine basin]

welp

thanks xD

[viscid thistle]

basically all logarithm simplifications boil down to ln(ab) = ln(a) + ln(b)

so given a problem like this, you'll probably get far just by hitting it with a hammer until it's in a form to which you can apply that

[alpine basin]

ok

[finite wraith]

are u guys still doing the same problem?

holy cow

[viscid thistle]

no

[alpine basin]

Could someone help me solve this problem please?

[viscid thistle]

@alpine basin What's up? How is the problem going?

[alpine basin]

@viscid thistle Oh my bad, I skipped it since I didnt really know how to solve it

[viscid thistle]

Have you drawn it? @alpine basin

[alpine basin]

nah

howd I draw it?

[viscid thistle]

Start by a circle

Trace the the radius of 5 inches

[alpine basin]

oh cause radius is half the diameter

[viscid thistle]

An draw the sector after

Yeah

Do you know how a sector looks like

??

[alpine basin]

Nah

Not sure

[viscid thistle]

https://images.app.goo.gl/wNPC8rDGZnp7K1TTA

Click and that's the formula and the look of it

Basically just plug what you have, nothing more pretty much

$A_{\text{sector of a circle}}=\frac{\theta}{360}πr²$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

@alpine basin

Where $\theta$ is in degrees

[obsidian monolith]

Al𝟛dium:

[alpine basin]

Ok lemme see

so theta is 60 right?

[viscid thistle]

Yeah

[alpine basin]

k one sec plugging it in

Ok

I got 13.08in^2

[viscid thistle]

,w (60)/(360)π*5²

Yup.

[alpine basin]

Thank you

How would I get the length of the arc though

[viscid thistle]

$s=r\cdot \theta$ where s is the length of the arc and $\theta$ is ON RADIANS

[obsidian monolith]

Al𝟛dium:

[alpine basin]

hmm

so Ihave to convert 300 degrees to radians?

yep I think that worked

Thank you

[viscid thistle]

Yeah

Np

[rare zephyr]

Can you distinguish the two objects?
@barren sapphire sorry for being very late to answer but yes

[viscid thistle]

@rare zephyr are you here?

[rare zephyr]

Ye

[viscid thistle]

So

Think about this: do we care about order? @rare zephyr

[rare zephyr]

Yeah

[viscid thistle]

Logs question

[finite wraith]

Logs answer:

[viscid thistle]

Can we re-state ln(x^2/cos(x)) as ln(x^2*sec(x))

[finite wraith]

i mean yeah

1/cosx is secx

ye u can do that

[viscid thistle]

thank you

[viscid thistle]

is anyone available?

@viscid thistle what have you tried

i tried to use symmetry

and I got stuck

Then I tried dividing by z to make z+1/z = z^

z^6

Well $(z+1)⁷=z⁷$ look at the exponents

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Consider the sqrt

its 7

so could i just do this?

Mb idk how you say it in english

Consider the root of base 7

$\sqrt{(z+1)^7} = \sqrt{z^7}$

[obsidian monolith]

Disabled_Skooter:

[viscid thistle]

Almost

but 7 as a radical

$\sqrt[7]{(z+1)^7} = \sqrt[7]{z^7}$

Yeha

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

i just dont know how to do it

oh okay

After that is it simple algebra

so then we get $(z+1) = z$

[obsidian monolith]

Disabled_Skooter:

[viscid thistle]

Yeah

Remove the parens after and move the z terms to one side and the 1 to the other and pretty much you shouldn't have more problems with it

i dont follow

@viscid thistle you here?

i dont follow
@viscid thistle oh yeah

Okay so

$$(z+1)=z$$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

$$z+1=z$$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

@viscid thistle now move the terms with z to one side and the "1" to the other

right

Wait a seccc

[willow bear]

is z meant to be complex

[obsidian monolith]

Disabled_Skooter:

[willow bear]

no, this proves that $z \notin \bR$.

[obsidian monolith]

Ann:

[viscid thistle]

ohhhh

how would i prove $z\ne 0$

[obsidian monolith]

Disabled_Skooter:

[viscid thistle]

also @willow bear yea they are complex numbers

[willow bear]

are you asking how you would prove z=0 is not a solution of (z+1)^7 = z^7?

[viscid thistle]

yes

Oh you should have told lol

thats only part of it

Are you really asking why z=0 is not a solution?

Or you just invented it

no im really asking why

how can i prove that $z\ne 0$

[obsidian monolith]

Disabled_Skooter:

[viscid thistle]

I mean you can see that the solutions for that $z\notin \bR$ bc the z vanishes at the last step

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

right

[willow bear]

i mean

idk like

[viscid thistle]

Or simply by plugging z=0, you'll get 1=0 which is blatantly false

yea

[willow bear]

0 isn't a solution since 0 ≠ 1

unless you refuse to believe 0 ≠ 1

[viscid thistle]

Lol

haha

you never know

okay

now continuing my question how do i use the roots of unity to find the solutions?

[steel tulip]

Is z supposed to be a real number?

[viscid thistle]

i dont think so

Is z supposed to be a real number?
@steel tulip $z\notin\bZ$

[obsidian monolith]

Al𝟛dium:

[steel tulip]

I know

[viscid thistle]

right

how do i find the numbers that are z?

[willow bear]

well you may first consider taking the absolute value of both sides

and then taking the 7th root

this will get you |z+1| = |z|

so you know that z must lie on the perpendicular bisector of 0 and -1

so z = -1/2 + it where t is real

[jade heron]

there aren't any

oh chat scrolled down

lol

[viscid thistle]

yea i got that

hmm

[formal edge]

https://math.stackexchange.com/questions/2905854/how-to-solve-the-equation-z17-z7-for-all-z

[compact crystal]

$z=cos(x)+i\cdot sin(x)$

[obsidian monolith]

BeatriceBernardo:

[compact crystal]

$RHS = cos(7x)+i sin(7x)$

[obsidian monolith]

BeatriceBernardo:

[compact crystal]

$z+1=cos(x)+i\cdot sin(x)$

[obsidian monolith]

BeatriceBernardo:

[compact crystal]

or maybe that would be easier, can't remember

$(1+cis(x))^7$

[obsidian monolith]

BeatriceBernardo:

[compact crystal]

$(1+cos(x) + i sin(x))^7$

[obsidian monolith]

BeatriceBernardo:

[viscid thistle]

$\cos(0)$

[obsidian monolith]

Disabled_Skooter:

[viscid thistle]

$(cos(0)+cos(x) + i sin(x))^7 =cos(x)+i\cdot sin(x)$

[obsidian monolith]

Disabled_Skooter:

[compact crystal]

z^7 + 7 z^6 + 21 z^5 + 35 z^4 + 35 z^3 + 21 z^2 + 7 z + 1

binomial theorem

[viscid thistle]

$z^7=(cos(7x) + isin(7x)$

[obsidian monolith]

Disabled_Skooter:

[compact crystal]

$(cos(0)+cos(x) + i sin(x))^7 =cos(7x)+i\cdot sin(7x)$

[obsidian monolith]

BeatriceBernardo:

[compact crystal]

$cos(0)+cos(x)$

[obsidian monolith]

BeatriceBernardo:

[viscid thistle]

cos(x)+1

[compact crystal]

https://www.cliffsnotes.com/study-guides/trigonometry/trigonometric-identities/product-sum-and-sum-product-identities

[viscid thistle]

$(2cos^2(x/2)+isin(x))^7 = (cos(7x) + isin(7x)$

[obsidian monolith]

Disabled_Skooter:

[compact crystal]

https://www.dummies.com/education/math/calculus/trig-identities-for-pre-calculus/

z

$\mathbb{Z}$

[obsidian monolith]

BeatriceBernardo:

[compact crystal]

Represents the set of integers. (The Z is for Zahlen, German for "numbers", and zählen, German for "to count".)

https://www.wolframalpha.com/input/?i=(1%2Bz)^7%3Dz^7

Well, I could be wrong, but my instinct says: use the trig idents., gtg now, good luck!

Another approach is to use log on both side

[compact crystal]

$$(1+cos(x)+isin(x))^7 = cos(7x) + isin(7x)$$
$$(cos(0)+isin(0)+cos(x)+isin(x))^7 = cos(7x) + isin(7x)$$
$$(cos(0)+cos(x)+isin(0)+isin(x))^7 = cos(7x) + isin(7x)$$
$$(cos(0)+cos(x)+i(sin(0)+sin(x)))^7 = cos(7x) + isin(7x)$$
$$(2cos^2(x/2)+2isin(x/2)cos(x/2))^7 = cos(7x) + isin(7x)$$
$$(2cos(x/2) cos(x/2)+isin(x/2))^7 = cos(7x) + isin(7x)$$
$$2^7cos^7(x/2) (cos(x/2)+isin(x/2))^7 = (cos(7x) + isin(7x)$$
$$2^7cos^7(x/2) (cos(7x/2)+isin(7x/2)) = (cos(7x) + isin(7x)$$

[obsidian monolith]

BeatriceBernardo:

[compact crystal]

@viscid thistle the rest is just trig now

[viscid thistle]

what is that????

@compact crystal

i fixed it

[viscid thistle]

I am sensing (e^ix) to have applications in that bunch of equalities

Given e^ix = cos(x) + isin(x)

[viscid thistle]

So I am a little stuck at the basics, I found a question in a grade 10 Canadian pre-calc textbook, which led me down a rabbit hole.

I am applying exponent laws, and I get 2 different answers.

I understand why there are two different answers, but I do not understand why I am able to apply the exponent laws without having to declare "for x<0 for LHS, and for x>= 0 for RHS".

By "without having to declare .." I mean, that every time I had to prove something in university, the conditions were very clear and obvious. But here, I don't see the "obviousness"

square roots can be hairy

some textbooks would define the square root of a number to only be the positive solution

it makes sense from complex number point of view, but w.o it, its quite mind boggling. I might be just brain derping

otherwise, the operation is not a function

if you define sqrt(x) to be any numbers a satisfying a^2=x then both the positive and negative answers satisfy

in that case, sqrt((x)^2)=|x|

[uncut mulch]

although not explicitly stated, the exponent laws are for cases when the base is positive.

[viscid thistle]

although not explicitly stated, the exponent laws are for cases when the base is positive.
@uncut mulch

Ah, that makes sense.

[remote veldt]

it's (part) of the reason that sqrt(-1 * -1) is not the same as sqrt(-1) * sqrt(-1)

[wide ocean]

how would one be able to figure this out?

k= -5 for sure

[viscid thistle]

Try plugging in some values for x

[wide ocean]

Ah like the values on the graph

[viscid thistle]

And think about the x-shift

[wide ocean]

oh right I forgot about the x shift

3 to the right so (x-3)

B, C, or D

then just plug in and see if X matches Y in the equations and graphs?

[viscid thistle]

Yes

Try x=0

This should be enough

[willow bear]

it's one of the possible definitions of the determinant

not rly

i mean

i guess the determinant itself has a geometric visual behind it

like if you have a matrix $A$ and you start with the unit cube $Q = {(x_1, x_2, \dots, x_n) \mid 0 \leq x_i \leq 1}$ then $A(Q)$ will be a parallelepiped with volume $\absv{\det(A)}$

[obsidian monolith]

Ann:

[wide ocean]

can someone explain why it's A?

i thought: reflection on x axis

verticle compression by 1/5 on y axis

[obsidian monolith]

Xoc:
Compile Error! Click the reaction for details. (You may edit your message)

[harsh smelt]

can someone explain why it's A?
because - reflects around x-axis

and 1/5 stretch it along x-axis

[wide ocean]

@harsh smelt oh I see. also why isn't the stretch along y axis?

because 1/5 isn't squared or?

so x becomes 5x

[harsh smelt]

it would stretch along y axis if 1/5 would be in argument

but it is in function

[willow bear]

@rotund kindle no, they're real numbers. (x1, x2, ..., xn) is a vector in R^n.

and no, A is a matrix. it does not have a volume.

[wide ocean]

cheers, also how do you differentiate function and argument? in this case?

-1/5x^2 like which part do you mean is argument vs function

[harsh smelt]

-1/5x^2 like which part do you mean is argument vs function
apply power rule

[uncut mulch]

the transformation can be viewed in both ways.

vertical shrink or horizontal stretch

$g(x) = -\frac15 f(x) = -f \br{\frac{x}{\sqrt{5}}} $

[obsidian monolith]

ramonov:

[rare zephyr]

Help with no6

[harsh smelt]

ok so

[willow bear]

is it just me or is the answer 0 as stated

[harsh smelt]

bad wording prolly

[willow bear]

the only line that can be drawn passing through 3 named points is the line on which A, B, C, D, E and F sit

[harsh smelt]

if it is meant that 1 line should pass through 3 points then 0

but if it is meant that just any triangles with vertices at these points then answer is not zero

bruh

[willow bear]

of course

but wording is important

[alpine basin]

How do I solve number 1 for this? I think I got the amplitude correct, which is 2, but Im having trouble understanding how to get the rest of what it asks

[viscid thistle]

The 2 influences only the amplitude

[alpine basin]

so it would still be period 2pi, phase and vertical shift 0?

[viscid thistle]

Yes

[alpine basin]

How is the phase shift of this pi/4? I thought it was pi/2

cause of this

[viscid thistle]

You want isolate the x

2(x+pi/4)

b=2

[alpine basin]

ah okay thanks

[alpine basin]

for these problems I can graph these, but what does it mean continue after one complete cycle, like how would I know when once cycle is complete on the graph

[viscid thistle]

Draw more than one period

[alpine basin]

ye how would I know how much is one tho like, for 9 the period is pi right?

but what is that in terms of the graph visually, idk

[viscid thistle]

Period of sin and cos is 2pi

You should draw like 3

At least one with negative x

[alpine basin]

how is 2pi measured on a normal graph tho

[uncut mulch]

pi is just a number

you can scale your x-axis in fractional multiples of pi

[alpine basin]

Okay

thanks

[viscid thistle]

PRECAL uwu

[tawny nacelle]

no

[bold ether]

how would you go about finding the sum of a series with a non-constant common ratio or differance?

Like writing it as a formula.

[serene heath]

whats your series

[bold ether]

something like this

[hexed ermine]

Ok so since the sub sequences turns into a constant that means you can write your general form as a polynomial

[serene heath]

its a quadratic sequence

[hexed ermine]

Here it would be a quadratic like

So you can decompose each part, and use what you know on summing up to n on for k^2 and k and what not

[bold ether]

how would that be written

am gonna try it wait

[hexed ermine]

You know it's the form of ax^2+bx+c

[bold ether]

am not sure how to write it as a formula though

what i wanted to find the 20th term

[hexed ermine]

Well does the sequence start with a_0 or a_1

I'm assuming a_1

[bold ether]

with a 1

[hexed ermine]

If that's the case plug in 1 for x and set equal to the first term

And do that 3 times

Because theres 3 unknowns

You can make a system out ofnit

[alpine basin]

How does this equal pi/4? I got 3 pi/4 when referring to the unit circle but idk

[acoustic harbor]

3pi/4 is not in the domain

[uncut mulch]

the range of arcsin is [-pi/2,pi/2]

[acoustic harbor]

yea range my bad

[alpine basin]

but 3pi/4 has a sin for sqrt(2)/2 as well as pi/4

so how would I know which one to refer to

[acoustic harbor]

but its not in the range

3pi/4 is not in [-pi/2,pi/2]

so u have to use pi/4

[alpine basin]

where did uget that range tho

[uncut mulch]

definition of arcsin

[alpine basin]

wat

isnt there a sin for every degree on the unit circle

[uncut mulch]

the range of arcsin is restricted so that it is a function

and well defined

[alpine basin]

is that the only way to tell

idk im making zero connections to this and the question

like this one is -pi/6 but looking at the unit circle to me i thought it'd be 7pi/6

[acoustic harbor]

7pi/6 not in the range

[uncut mulch]

,w graph arcsin(x)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by ramonov#0921

[maiden pebble]

https://i.imgur.com/9RO1J9T.png

Dont they equal - and + Infinity?

[uncut mulch]

be more specific

[maiden pebble]

lim h(x) as x--> -2^- exists at infinity

lim h(x) as x--> -2^+ exists at -infinity

[uncut mulch]

= doesn't quite indicate equality here

[maiden pebble]

True, but I dont know what otyher symbol to use

[uncut mulch]

its generally accepted to write it like that though

the limits can be represented by -inf and inf
but technically the limits don't actually exist since it doesn't approach a fixed value

[maiden pebble]

😠

[uncut mulch]

$\lim_{x\to 2^-} h(x) = \infty$ indicates that:
as $x \to 2^-, h(x) \to \infty$

[obsidian monolith]

ramonov:

[maiden pebble]

but since u can never get to infinity

uggggg

[uncut mulch]

how you answer depends on context

what are the other options?

[maiden pebble]

I went on so I dont have a picture but it was
lim x--> -2 exists
h(x) is continuous at -2
h(x) is defined at -2

none of the above

and obv "none of the above" was the answer

[uncut mulch]

yeh

tldr the limit can be represented by -inf or inf, but doesn't exist

[maiden pebble]

well i read it all, but thanks

[viscid thistle]

any help would be nice

[visual finch]

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

guys how do i set this up

I'm forgetting the way i do it

[upbeat bone]

Construct the problem into geometric series @visual finch

[harsh cipher]

Hello

can I ask a question ? not related to math

I think you guys might be able to help me

[willow bear]

then why are you asking it in the precalc channel

[harsh cipher]

Hi Ann!!!

because people in this channel are from different parts of the world

and their time zone is different compared to members in the other channel

[pale bison]

lmao what?

[willow bear]

so what's your question

[viscid thistle]

?

Lmao

[hard hornet]

because people in this channel are from different parts of the world
and their time zone is different compared to members in the other channel
holy fuck this is big brain as hell

[harsh cipher]

other channels aren't filled ppl

[uncut mulch]

i mean it doesn't explain why the question was posted here in any way whatsoever

[harsh cipher]

but nm found my answer on youtube 🙂

[pale bison]

nice delete, i saw

[viscid thistle]

thx

[tawny nacelle]

[viscid thistle]

is this correct

[willow bear]

no

[viscid thistle]

i didnt learn what the ( [ stuff means

[willow bear]

( means the endpoint is not included

[ means that it is

[viscid thistle]

i only know the <x< one

oh alright

thank you

but would it be

-1 < x < 5 ?

xER

[willow bear]

no

-1 < x ≤ 5.

[viscid thistle]

Ok

Would this be correct

[grizzled hull]

Have you either consulted your notes or attempted to graph this?

[viscid thistle]

no but now i will

[reef jasper]

how do i find all the x values

[viscid thistle]

,rotate

[reef jasper]

oh tnx

[upbeat bone]

Yeah, since you want those two vectors to be perpendicular, that means the dot product of those two vectors are equal to 0

Just apply dot product to your equation and you should be fine @reef jasper

[reef jasper]

yeah but there can be so many x values

oh i should solve it out?

[upbeat bone]

Well you're asked to give all the possible values of x right?

[reef jasper]

yeah

[upbeat bone]

In other word, they're asking you to solve your equation

Try applying dot product first, what'll you get from that?

[reef jasper]

-x^2 +3x -6

so just factor

[upbeat bone]

Yeah

[reef jasper]

k tnx

[upbeat bone]

Wait, you're multiplying that wrong no? @reef jasper

[reef jasper]

oh yeah i just found out lol

[upbeat bone]

Me too lol

[reef jasper]

wait how else then

[upbeat bone]

It should've been $$-x^2+3x+6=0$$

[obsidian monolith]

Wilston Lynx:

[upbeat bone]

You can apply quadratic formula I think

[reef jasper]

yeah ok i’ll try that

[reef jasper]

,rotate

[reef jasper]

wtf does this even mean

[barren sapphire]

@reef jasper For a), $(0,-2)$ is the rectangular coordinates of a point (which you drew). Give a polar coordinate representation for the point that uses a positive $r$ like $(r,\theta)=(2,?)$. And then also give polar coordinate representation with a negative $r$. For b), $(-5,-24)$ is the rectangular coordinates of a point. Represent in some way all of the polar coordinate representations that use a positive $r$.

[obsidian monolith]

dirib:

[rare sentinel]

Wow TeXit 🤩

[reef jasper]

but how do u find theta

[rare sentinel]

Think about it

How many degrees are you going around to reach (0, -2)

[reef jasper]

is it jsut pi/2 and 3pi/2

[rare sentinel]

Do u know what it means by polar representation

[reef jasper]

yes i think jsut (r,theta)

?

[rare sentinel]

Ok so

For a single point on the polar coordinate system, you can have infinitely many ways to write it

[reef jasper]

(2,3pi/2)

[rare sentinel]

In your case, you can do (2, 3pi/2) OR (-2, pi/2)

Whoops yeah

[reef jasper]

,rotate

[reef jasper]

i feel like i messes up somewhere

[viscid thistle]

What did you do as last step?

[reef jasper]

is multiply 2 by 2sin

so i get cos@-1=sin@

so i just thought i could set them both =to 0

[viscid thistle]

Why?

lol

[reef jasper]

lol idk

what else can i do

[viscid thistle]

Graph them

so i just thought i could set them both =to 0
It's not the only solution

Wait

Those aren't even solutions

Sin(x) on the denominator

Careful

[reef jasper]

oh ok

[viscid thistle]

You could try $\sin (x) =\sqrt{1-\cos^2(x)}$, then $\cos(x) =t$

[obsidian monolith]

HoboSas:

[reef jasper]

how do i get rid of the 2 so it can = 0

[viscid thistle]

Did you even read what I wrote?

Is this right

[uncut mulch]

no

[viscid thistle]

can i get some help with it

[uncut mulch]

the notation is also bad

but it should be interpreted as

$(f\circ g)(-2)$ or$f(g(-2))$

[obsidian monolith]

ramonov:

[uncut mulch]

you are dealing with function compositions

[viscid thistle]

oh

so it would be $((-2)^2+2)-1)$

[obsidian monolith]

alfred:

[viscid thistle]

idk

$((-2)^2+1)$

[obsidian monolith]

alfred:

[uncut mulch]

that will get you what you need, but you should be using abs values

due to how f is defined

[viscid thistle]

so $((-2)^2+3)$ ?

[obsidian monolith]

alfred:

[uncut mulch]

no I mean

$|(-2)^2+2)-1|$

[obsidian monolith]

ramonov:

[viscid thistle]

oh

k i get it

does it need brackets after the abs value symbols?

[uncut mulch]

if you mean:$(|(-2)^2+2)-1|)$, no

[obsidian monolith]

ramonov:

[viscid thistle]

kk thank you

[uncut mulch]

and then simplify

[viscid thistle]

5

K I tried the second part of that question

Can I get a tutorial for this question

so for a would it be $$|({0,1,2}+{0,1,2})(2)|$$

[obsidian monolith]

alfred:

[alpine basin]

How does 1 + sin^2x/cos^2x combine to (cos^2x+sin^2)/cos^2x) ?

Wouldnt it be cos^2x+sin^2x + sin^2x/cos^2x

[viscid thistle]

Can you show?

[alpine basin]

they just took 1+sin^2x/cos^2x and changed it to just (sin^2x+cos^2x)/cos^2x

[viscid thistle]

...

Common denominator

[alpine basin]

common denominator in cos^2x+sin^2x + (sin^2x/cos^2x)?

wat

[viscid thistle]

Common denominator inside the parenthesis

[alpine basin]

idk what u mean

they changed 1 to sin^2x + cos^2x

[viscid thistle]

No really

[alpine basin]

but where did the other sin^2x go

[viscid thistle]

sin^2 was already there

[alpine basin]

there should be two of them tho right?

[viscid thistle]

$1+\frac{\sin^2(x)}{\cos^2(x)}=\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}$

[obsidian monolith]

HoboSas:

[alpine basin]

how

[viscid thistle]

???

Common denominator

What's wrong with that

[alpine basin]

oh fuck im retarded

i see it now

[viscid thistle]

Nice

How do I restrict the domain

Just cut the function

oh

but how do u do that

i just state it in the domain thing

Just erase

Yes

Like that

oh k

ty

What's the domain

Of f^-1

idek honestly

the green is the inverse

is 1<x<2 the correct restriction

Nope

That's so smol

Can you make it bigger

which

Domain

like that

...

The green one

Look at it

What's its range

what about it

The green is the inverse

yeah

y>1

Domain of a function is the range of its inverse

so if x>=1, you can construct f^-1

oh

Either x>=1 or x<= 1 is fine because it's symmetrical

i see

alright

You want f(x) to be injective

injective?

Yes

oh nvm

k

f(a)!=f(b) for every a!=b

[stuck lark]

you might’ve seen a synonym “1 to 1”

[viscid thistle]

^

[alpine basin]

How would I rewrite this?

[fleet yew]

@alpine basin sum formoola

[alpine basin]

I got sqrt(3)/2sinx + 1/2cosx

but the answer is, howd they get -?

[viscid thistle]

this is pretty much just algebra but

y\ =\ \frac{-1}{.05x+1}+1

fuck

this is general form of the equation i want

i want the horizontal asymptote at 1

i have a dataset for it, but i don't think polynomial regression is the best idea for this cuz otherwise it wouldn't be super accurate unless i input everything from the table

basically trying to find the equation used in this calculator

and i can't find it after parsing through all the javascript from inspect element

[barren sapphire]

@viscid thistle I don't understand. You wrote "this is the general form" but there were no variables in your form

[viscid thistle]

forget that

i just meant a rational equation

[barren sapphire]

Any rational function? Or specifically ones with a horizontal asymptote at 1? With numerator and denominator of degree less than something (3?)?

[viscid thistle]

d[0,inf) r[0,1)

[barren sapphire]

I don't understand your notation

[uncut mulch]

@alpine basin 11pi/6 is in Q4, and the sin of that is negative

[viscid thistle]

domain and range

there i fixed the brackets

[barren sapphire]

rational functions don't do that

[viscid thistle]

welp

[barren sapphire]

Like the maximal domain of a rational function is everything except the handful of zeros in the denominator

you might only care about positive inputs

but that's different

[viscid thistle]

alright

so what's the next steps

[barren sapphire]

You need to refine your problem a bit. Like if you want a horizontal asymptote of 1, that would help. But maybe you don't want that?

[viscid thistle]

yes

is this better

d[0,inf) r[0,1)

[barren sapphire]

Again, no rational function has a domain like that, but that doesn't matter

[viscid thistle]

hmm ok

i'm being stupid

yeah don't mind the domain range shit, i just wanted all positive values

and horiztonal asymp at 1

[barren sapphire]

Well, this form doesn't necessarily guarantee positive values on positive inputs, but it does (basically) guarantee a horizontal asymptote at 1. How about we look for something of the form $1+\dfrac{ax^2+bx+c}{dx^3+ex^2+fx+g}$?

[obsidian monolith]

dirib:

[viscid thistle]

alright

[barren sapphire]

Doing that by hand is an absolute nightmare, and I assume you don't want to do that

And it doesn't sound like you need the theory behind it either

[viscid thistle]

yeah lol

is it some taylor polynomial stuff

[barren sapphire]

worse

[viscid thistle]

lol

[barren sapphire]

Under those assumptions, let's ask Mathematica to find a fit to that form for the data in your Desmos image: https://tio.run/##lY8xj8IwDIX3/oo3Ame1SdyE5A8wM7ChnhTaAh3onaBDpaq/vbgR0s3nwfqen2U9P@Jwbx9x6Oq4LBpf2MTd@G0ELrtRer0tNo1MWLj9ONfk3LYocmQ4dH1z6Ibz6Rn71@/Pqz1PkyYYAhNKgiU4wp7gCYGg1UyYVK5K70VmWEtksCxS5ZqDMwm8dn4FY7RLlrFhXyYINvAKzIrt3xF2rNImh9LwPFfJ@udXki4SLoSa0BBawpVwk9BjVRTHZ9cPy/IG

It says:

,w plot 1 + (-2468.22 + 0.139744 x + 0.460248 x^2)/(2468.22 + 123.588 x + 2.63208 x^2 + 0.0289341 x^3) for x from 0 to 10

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by dirib#6336

[barren sapphire]

That passes through those points.

[viscid thistle]

alright

[barren sapphire]

Now before you get excited, the huge numbers like 2468.22 suggest that this was probably not the right kind of function to use, and this is not a good thing to try to do

I would recommend getting someone to figure out the code instead

[viscid thistle]

yeah i suspect the extrapolation closer to 1

will be fucked

[barren sapphire]

Well, glad I could be of no help I guess

[viscid thistle]

lmao it's alright

so to figure it out analytically

that's probably beyond me

[barren sapphire]

I mean, if the context involves some algorithm and/or some randomness, there's probably a known approximation to whatever it is you're calculating

[viscid thistle]

right

i can't find the approximation tho lol

welp guess ill just have to do it the dirty way

thanks anyway

[barren sapphire]

What I was trying to say was: if you can't figure out the code or what a good approximation to the code would be, maybe someone else can

probably not me though

[viscid thistle]

oh, guess i'll try a programming server then

my python

is very iffy

[barren sapphire]

I thought you said it was in javascript

[viscid thistle]

yeah the website is in javascript

i know java well enough

[barren sapphire]

I'm imagining a mathematician reads the javascript, and then says "oh, this is calculating the [blah blah] numbers, which have known approximation [something]"

[viscid thistle]

yeah the thing is

the javascript only had the design elements

i beleive it was locked behind admin account

to prevent editing, so i couldn't take a peak at the equations

[barren sapphire]

oh

then ignore me

[viscid thistle]

no worries

[livid willow]

can you help with these problems ?

[viscid thistle]

@barren sapphire y = 1.000002041(1-exp(-0.050125173x))

i found a neat website that did non linear regression

http://www.colby.edu/chemistry/PChem/scripts/lsfitpl.html

this is epic

[viscid thistle]

@livid willow if a plane is parallel to the plane we want, their normal vectors will be parallel as well. So we can use the normal vector of x-3y+7z-11=0 and a point to find the wanted plane

[zealous latch]

Can someone check if my set notation on correct?

[viscid thistle]

What is k?

[zealous latch]

k would be any integer >= 0 where would i define this?

[pale bison]

just write it out

$k\in\N$ or "where $k$ is a positive integer"

[obsidian monolith]

Publius:

[harsh smelt]

or $k \in \bZ^{+}$

[obsidian monolith]

Commander Vimes:

[pale bison]

@native sequoia

[obsidian monolith]

Publius:

Commander Vimes:

[pale bison]

no that's illegal

[frozen needle]

the "0 is in N" vs "0 isn't in N" thing is kinda like the "arrays start at 0" vs "arrays start at 1" thing

[pale bison]

clearly array starts at 0

the other choice is illegal

ok i'm just joking don't call me being pedantic

[harsh smelt]

clearly array starts at 0
@pale bison hm, but what is zero?

haha

[viscid thistle]

is anyone on that thinks they can help?

[viscid thistle]

12 / .75

12/1

3/4

12*4

3*1

48/3

16

[junior sable]

ty @viscid thistle !

[viscid thistle]

cake = 1

and we have 12 people

they all want their cake in half peices

im going to need a sec

12/12 = 1 = cake

24 people

1 = 2 people

0.5

$\frac{12}{12}$

[obsidian monolith]

Disabled_Skooter:

[viscid thistle]

$\frac{24}{2}$

[obsidian monolith]

Disabled_Skooter:

[barren sapphire]

@viscid thistle That site is more convenient to use, but Mathematica like I linked can handle basically any form you can write down, not just the 25 on that site

[viscid thistle]

will note for the future, thanks

[alpine basin]

how do I solve number 11 for this? Help appreciated

[viscid thistle]

@alpine basin u there?

[alpine basin]

ye

[viscid thistle]

So well first lets isolate the $\sin(2\theta)$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

$\sin(2\theta)=\frac{1}{2}$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

@alpine basin

[alpine basin]

ye

Thats how far I got

[viscid thistle]

Take the arcsin on both sides

[alpine basin]

the wot

[viscid thistle]

The arcsin, never heard of it?

[alpine basin]

nah dont think we've learned that yet pretty sure

[viscid thistle]

Do you know it as $\sin^{-1}(x)$ maybe?

[obsidian monolith]

Al𝟛dium:

[alpine basin]

yea

the inverse?

[viscid thistle]

Yeah

[alpine basin]

hmm

[viscid thistle]

What do you get

[alpine basin]

honestly not sure

[viscid thistle]

Write it

Ill feel otherwise like if im making it to you

[alpine basin]

What do you want me to write

[viscid thistle]

What do you get
After you arcsin both sides

[alpine basin]

sin^-1(2theta) = sin^-1(1/2)

?

[viscid thistle]

$2\theta=\arcsin(\frac{1}{2})$.

Hm

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

@alpine basin ok?

Just took the inverse of sine on both sides

[alpine basin]

Okay

[viscid thistle]

Now look at the RHS, note that we are not given an angle, 1/2 isnt an angle here bc its the inverse

Do you know how to look that value on the unit circle? @alpine basin

[alpine basin]

uh

You said its not an angle tho

what part of the unit circle should I look at

[viscid thistle]

At the value of the opposite side

What you would usually get by looking the value with an angle

[alpine basin]

ye so sin of 1/2?

[viscid thistle]

??

Like if you were looking for the angle on the case $\sin(x)=\frac{1}{2}$

[obsidian monolith]

Al𝟛dium:

[alpine basin]

so 30 degrees?

and 150?

[viscid thistle]

Please use radians form

[alpine basin]

pi/6 5pi/6

[viscid thistle]

Yup

So will divide into 2 cases

$2\theta=\frac{π}{6}\ 2\theta=\frac{5π}{6}$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Note that we won't add periodicity here bc we are on a given interval

Now just simply divide by 2 on both sides and you are done

@alpine basin

[alpine basin]

ok one sec

pi/3 and 4pi/3?\

[viscid thistle]

No.

[alpine basin]

o

pi/12

and 5pi/12

[viscid thistle]

Yeah

Yes

[alpine basin]

Hmm the answer says this tho

[viscid thistle]

Wth

Hold up

Wolfram gives my answer

Symbolab too

Photomath as well

...