precalculus

[viscid thistle]

@uncut mulch

[rugged linden]

hello i don't understand the last sentance, why are we applying this for $A(n + n_1 -1)$?

[obsidian monolith]

mirzathecutiepie:

[echo wagon]

Let B(n) = A(n+n1 - 1)

Then B(1) corresponds to A(n1)

And B(2) is A(n1 + 1)

So you do induction on B(n) starting at n = 1, and that proves A(n) for n >= n1

[rugged linden]

ahhhhhh

i seee

one more thing

apostol defines an inductive set to be an inductive set if and only if the element 1 is in the set, and if k is in the set then k + 1 is too

And then concludes the smallest possible inductive set is the positive numbers

but isn't this wrong?

because if the element 1 is in the set, by setting this equal to k + 1 we get k + 1 = 1 and k = 0, and so on

therefore the smallest possible inductive set has to the be integers as a whole, not the positive integers

[arctic grotto]

I need help with my test

[willow bear]

are you in the test right now?

[sick steppe]

@arctic grotto if it's a test then no

[willow bear]

or are you looking for review BEFORE the test?

[arctic grotto]

Yes

[willow bear]

so which is it

[quick mirage]

g3n0 acting sus

[cobalt storm]

When solving the trig equation, y = csc^2x−cscx, you will get 3 solutions if 0 ≤ x ≤ 2π . True or false?

I converted both ratios to sine

[viscid thistle]

I dont see the equation

[cobalt storm]

y = csc^2x−cscx

is this not an equation?

oh maybe its a trick question?

[viscid thistle]

y = 0?

are you talking about roots?

[cobalt storm]

solving trig equations

yeah for the x values

[viscid thistle]

y = csc^2x−cscx is a function

[cobalt storm]

oh

its not an equation

bc theres no value to = to ?

like if it were 5 = csc^2x - cscx

that would be an equation right

[viscid thistle]

that would be an equation

[cobalt storm]

but if its just y = csc etc

its not an equation

its function

oh yeah cuz then i would need to substitute y

so its not an equation

yeah this is a trick question then

thanks

Is it possible to actually solve quadratic trig equations without seeing a graph

Or actually going and solving it

It would right because what if I had sin^x = 2

Well I know sinx can never = 2

Like the base function

[sick steppe]

@cobalt storm sin^x isnt a thing

[quaint mason]

idk how to approach this

[viscid thistle]

First, it isn't -sqrt(10) it's -sqrt(x^10)

[junior sable]

that's what i meant, mb

oh you know what, nvm i get it

ty tho

[quaint mason]

wym?

[viscid thistle]

@ripe adder here

,tex \laws

[ripe adder]

Ln(45/26) = a(ln(o.6))

[viscid thistle]

Yes

Wait

Yes

[ripe adder]

ok so I now need to isolate

[viscid thistle]

Preferribly ln and not Ln

[ripe adder]

I can remove ln by multiplying by e

?

[viscid thistle]

When you have an equation of the type 6=5x

What do you do

[ripe adder]

divide whole quation by 5

[viscid thistle]

Then why do you think dividing by ln(0.6) on both sides won't suffice our problem

[ripe adder]

no idea tbh

should I simplify the left hand 45/26 before applying the x 0.6?

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

There we go

That's my list

[quick mirage]

Pretty nice

[ripe adder]

ill save that rq

thank you

[viscid thistle]

All g

should I simplify the left hand 45/26 before applying the x 0.6?
I mean you can

But i'd do it after

Both ways work either

[ripe adder]

ok

I got 2.88461

= a

as (45/26) / .6 = 2.88461

[viscid thistle]

If you want an approximation okay then

Where's the ln at

[ripe adder]

it was

[viscid thistle]

,w (ln(45/26))/(ln(0.6))

[quaint mason]

Can I get some help on this problem

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Al𝟛dium#4084

[ripe adder]

ln(45/26) / ln(.6)

I cancelled out hte ln's

[viscid thistle]

The ln's don't cancel each other

[ripe adder]

fuck

so do I do the whole

[quaint mason]

yes

[ripe adder]

ln(45/26) / ln(.6)

for final?

[viscid thistle]

Preferribly, no

I'd simplify

[quaint mason]

45/26. then find ln of that

divide by ln of .6

[quick mirage]

How did you simplify ln(0.6^x)

[ripe adder]

by using al3diums law

[viscid thistle]

It's not mine

[ripe adder]

to separate it

or well

whoevers law

[quaint mason]

lmao

[ripe adder]

ok so uhhh

[viscid thistle]

You can approximate here if you want

[ripe adder]

I dont know how to simplify ln(45/26)

I can do it through a algebra calc

but Idk what process its using or what the ln does

[viscid thistle]

Is it asking you for the exact answer or an approximation

[quick mirage]

wait huh, how did you go from ln(45/26) = ln(0.6^x) to ln(45/26)/ln(0.6)

[ripe adder]

exact solution

[quick mirage]

I think my brain is dead

[ripe adder]

its technically ln(45/26)/ln(0.06) = a

because after separating a we moved it all to the left

[viscid thistle]

Yeah

[quick mirage]

ln = log e (x) so you cant take out a that way

[viscid thistle]

Huh?

[ripe adder]

im also confused

[quick mirage]

I think im missing something critical here, what did you guys do so far

[ripe adder]

so far we are at ln(45/26)/ln(0.06) = a rn

and we need exact solution

[viscid thistle]

$$\ln(\frac{45}{26})=\ln(0.6^a)=a\ln(0.6)$$ $$\frac{\ln(\frac{45}{26})}{\ln(0.6)}=a$$

[quick mirage]

and the original equation was ln(45/26) = ln(0.6^a) right?

[ripe adder]

yes

[obsidian monolith]

Al𝟛dium:

[quick mirage]

Ohhh nevermind i am braindead

[viscid thistle]

so far we are at ln(45/26)/ln(0.06) = a rn
@ripe adder anyways

I'd leave it like that

[quick mirage]

i was working with something different earlier on, really messing me up

sorry

[viscid thistle]

Because otherwise we'd have a bunch of terms

[ripe adder]

@viscid thistle @quick mirage when I just input it unsimplified

its correct1

thanks guys

[viscid thistle]

It's not unsimplified

[ripe adder]

or

[viscid thistle]

It's simplified

[ripe adder]

yes

thanks again @viscid thistle

you have helped me on math since junior year

true homie

[quick mirage]

#goals

[viscid thistle]

Yw!

[quaint mason]

can anyone please help?

[velvet blade]

Isn't this supposed to be fg(x^2)?

what am I missing?

$(f+g)(x)= f(x)+g(x)$ how is this true? Or is it defined like this?

[obsidian monolith]

lazypawtato:

[somber yew]

I mean, this is how you define it.

What else do you expect it to be?

[velvet blade]

Oh okay, is it defined using like maths outside the scope of this book?

[somber yew]

No, this notation is fairly standard.

[velvet blade]

Or is it just defined like this to just work?

Oh okay

[somber yew]

You'll likely come across this again in Spivak maybe.

[velvet blade]

He didn't say functions are distributive

He just said commutative and associative properities

[somber yew]

It isn't meaningful to talk about "distributivity" of functions, (f+g)(x) is simply notation for two functions f and g being added acting on x.

[velvet blade]

Okayy

[somber yew]

Though the way it works it looks like function distributes over x haha.

[velvet blade]

Also I thought f(x)g(x)=fg(x^2)

[somber yew]

Mmm f(x)=x, g(x)=x^2. f(x)g(x)=x^3 or f(x)g(x)=fg(x^2)=x^6?

@velvet blade

[velvet blade]

Yeah I'm here I'm just... contemplating this

[somber yew]

Lmao it is the first one obviously

Just multiply the images

[velvet blade]

Okayy

Got it

[somber yew]

f(x) means the image of x under f, same goes for g(x), so f(x)g(x) is interpreted as the product of images under f and g

[velvet blade]

Okayy

if f(x)=x^2 , g(x)= x^3 , then fg(x)= x^5?

[somber yew]

Correct.

[velvet blade]

Okay got it. So the terminology means multiplying the output of the functions instead of the functions themselves?

[willow bear]

addition and multiplication on functions are both done pointwise

[velvet blade]

pointwise?

[willow bear]

pointwise, meaning at every point (input) independently

f+g is by defn the function which sends an input x to f(x)+g(x)

[velvet blade]

Okayy

[willow bear]

f*g is by defn the function which sends x to f(x)*g(x)

[velvet blade]

Got it

[willow bear]

i would advise against writing a product of functions by just concatenating their names

it might get misread as composition in the wrong context

[velvet blade]

Got it. Thanks.

[velvet blade]

How do I show this?

[willow bear]

they give you a hint for what to use as the even component

but really you could just write down, f(x) = E(x) + O(x) where E is the "even part" and O is the "odd part"

then you have f(-x) = E(-x) + O(-x) = E(x) - O(x)

so $\begin{cases} E(x) + O(x) = f(x) \ E(x) - O(x) = f(-x) \end{cases}$ and from here you can solve for $E(x)$ and $O(x)$

[obsidian monolith]

Ann:

[velvet blade]

then you have f(-x) = E(-x) + O(-x) = E(x) - O(-x)
How can O(-x) be -O(-x)

[willow bear]

typo

[velvet blade]

Okay got it

Is this the correct explanation for sum of the odd functions?

[echo wagon]

Don't use O for both functions

It looks like you're just adding an odd function to itself

[velvet blade]

Oh okayy

I didn't understand the highlighted sentence

[proud hound]

Consider that a light signal propagates in a vacuum with constant speed. Knowing that this signal was emitted from here on Earth towards a Planet called Bongo, where during the journey the signal spent a time t expressed in hours that is given by the solution of the equation multiplied by one million

Considering the information, determine the distance between the Earth and the planet Bongo in light years

Heeelp ;-;

a matter of physics, but it has more math than physics

[smoky pagoda]

Note first that RHS is $\log_y(y^{2t}) = 2t$. Then note that $\log_{10}(0.001) = -3$. Third, notice that $-\log_t(\frac{1}{t}) = 1$. And finally note that $6\log_2(2^{\frac{1}{3}}) = 2$. So in the end the equation boils down to $t^2 - 3 = 2t$ which is a quadratic equation.

They just made it look a lot more daunting than it is

[obsidian monolith]

Little Narwhal:

[smoky pagoda]

@proud hound

[proud hound]

plus it has two solutions one is 3 and the other -1

Only the real solution can only be -1

there is no negative time

[smoky pagoda]

$t^2 -2t - 3 = 0 <=> (t-3)(t+1) = 0 <=> t = 3 , t= -1$ So only solution is t = 3 since there can be no negative time

[obsidian monolith]

Little Narwhal:

[smoky pagoda]

so if light travelled for 3,000,000 hours

how many light years did it travel

[proud hound]

so I have to know first how much is 1 light year?

have any formula to find out? @smoky pagoda

[smoky pagoda]

what's the definition of a light year

[proud hound]

The Light Year is a measure of length that corresponds to the distance covered by light in a year.

my head: 🤯

9,46*10¹²km (one light year)

@smoky pagoda

[smoky pagoda]

right that's the distance covered by light in a year

Now

you have a measure of how much time light has travelled

Say instead of 3,000,000 hours the answer for t had been 1 year

then the answer would be 1 light year

if it had been 2 years

the answer wouldve been 2 light years

so now we need to figure out the ratio between 3,000,000 hours and 1 year is

for another example if t was 24h

then we wouldve travelled 1/365 light years

so how many hours are in a year?

[proud hound]

8760 hours?

[smoky pagoda]

right

so now

what is 3,000,000 / 8760

[proud hound]

about 342

😳

[smoky pagoda]

so how far is the planet Bongo?

[proud hound]

342 light-year

wow

so that was it ?

[smoky pagoda]

yeah normally

[proud hound]

I love you

Thank you very much

[smoky pagoda]

👍

[willow bear]

@velvet blade if the same polynomial function could admit two different expressions with different sequences of coefficients then the degree of a polynomial function would not be well-defined

because if you could have the same function be described by two different expressions - say, one where the highest power of x is 9 and another where it's 7, both with nonzero coefficients - then what would the degree even be? 7? 9? something else based on an expression you didn't even think of?

[velvet blade]

Got it, thanks again

[quaint mason]

kinda still stuck on this problem if anyone can help pls and ty

[sick steppe]

@quaint mason what have you tried

[quaint mason]

ive tried the first one and

it didnt make any sense to me xD

ppl told me to just graph it but idek how to

[sick steppe]

You can isolate for m

[quaint mason]

i can isolate for m

wym by that

like get m by itself using algebra? @sick steppe

[sick steppe]

yes

[quaint mason]

i tried that already

start by subtracting 52.55 both sides

[sick steppe]

oh wait it says via graphing

[quaint mason]

that gives right side -8.55

ye idk how to graph it tho, ppl say just change m to x

which is what i did on desmos

and it didnt work

also from the graphs, how would u get the values for m?

[sick steppe]

I graphed it fine so...

[quaint mason]

How did u graph it? tf

[sick steppe]

just plugged it into desmos

[quaint mason]

i did

[sick steppe]

ok, so when does the graph equal 44?

[quaint mason]

lemme plug it back into desmos again

when does the graph equal 44? do u mean like what is the value of y when x is 44 or?

[sick steppe]

m cant be 44, so y value

[quaint mason]

72.252?

[sick steppe]

I get 3.624 and 10.736

[quaint mason]

wait how did u get those values tho

??

[sick steppe]

i graphed the function and 44 together and clicked the intersection points on desmos

[quaint mason]

ohhh and then u took the values only between 0<m<12?

[sick steppe]

yes

[quaint mason]

ah i see

tyvm mate

[quaint mason]

is horizontal shift based on phase shift?

[junior sable]

What happened in the last step? why did the 4/x^6 turn to zero?

[steel tulip]

They evaluated the limit from x to -inf

[junior sable]

I'm supposed to take a limit of 4/x^6 and come up with zero?

[steel tulip]

lim x -> -inf (k/(x^n)) = 0 where k is a constant

It's also stated in the image

[junior sable]

when you take a limit of a function, are you also taking the limit of every part of that function too?

[steel tulip]

Im not sure what you mean

[junior sable]

lim x -> -inf (k/(x^n)) = 0. It looks like I'm supposed to take a limit of 4/x^6 and come up with zero, but you can only change 4/x^6 into 0 if you take a limit of it.

limits seem like delicate tools and I don't feel comfortable applying them everywhere

[steel tulip]

Yes, that's true. You cannot solve 4/(x^6) = 0 "normally"

They're an intuitive but nuanced idea, yeah

[junior sable]

is lim x -> inf (k/(x^n)) = 0 true too?

[steel tulip]

Yes

Keep in mind this is only for n greater than 0

[junior sable]

okay

when you take a limit of a function, are you also taking the limit of everything in the function? for instance, can I rewrite my original problem as lim of 9x^6/ lim of sqrt(9x^12 +4x^6)

[steel tulip]

Yes you could, since the limit is working on every x

[junior sable]

and the limits of ordinary numbers like 5 is 5

right?

[steel tulip]

Yep, because they have nothing to with a variable like x. Makes sense?

[junior sable]

yes, it makes a lot more sense now. thank you!

[steel tulip]

Np!

[quaint mason]

How do u solve for this

[earnest jungle]

Take the standard form for the cos function: $f(\theta) = a \text{cos}(b\theta + c) + d$. The phase shift is defined as: $-\frac{c}{b}$.

[obsidian monolith]

LifeSource:

[earnest jungle]

@quaint mason

[quaint mason]

so first one is -0/1 so ur sayin its undefined?
@earnest jungle

[earnest jungle]

No

0/1 is not undefined

it is 0

[quaint mason]

first one is 0 degrees?

[lean thistle]

i need help with this

i think it’s infinity just by looking at it

[fallen salmon]

May I ask you guys a quick combinations/permutations question?

My question is: How many combinations is there when using 16 unique letters, none are repeating

I know there are calculcators for that, but I can't remember what's ''r'' for that

[alpine scaffold]

Must all 16 be used?

[fallen salmon]

yeah

[alpine scaffold]

Isn't it just 16! then?

,calc 16!

[obsidian monolith]

Result:

2.0922789888e+13

[alpine scaffold]

Haven't done counting in ages

but

It's sort of difficult to explain why over text

I think of it as choosing a letter out of the 16 and then for the next spot you only have 15 letters, and then 14 letters, so on and so you get
16 x 15 x 14 x ... x 3 x 2 x 1 = 16!

[fallen salmon]

Yeah that'll do, I tried with smaller numbers like 3, and it has 6 combinations, 123 132 213 231 321 312

which means that with 16 is 16!

Thanks a lot! Kinda in hurry, so didn't have time to think hah

[alpine scaffold]

No problem

[viscid thistle]

could anyone show me the working out on this problem, I am a little lost

[uncut mulch]

expand and equate real and imaginary parts

[viscid thistle]

and then i would solve it simultaneously or?

[echo wagon]

Yes

[viscid thistle]

i cant isolate the (a+bi), because i will always end up with it having to be multiplied by a fraction on that side

[echo wagon]

(1+3i)(a+bi) expands into what? @viscid thistle

[viscid thistle]

yeah but doesnt that complicate things

ohhhh

(a + 3ai + bi -3b)

[echo wagon]

So what's the real part of that?

And what's the imaginary part?

[viscid thistle]

Re: a -3b

and Im: 3a +b

?

[echo wagon]

Yes

Now equate that to the real and imaginary parts on the other side

[viscid thistle]

would i use the original equation for the simultaneous part

or

oh nvm

i got it

thank you!

[somber yew]

What is the rationale behind comparing terms with same coefficients on both sides? Does it need deeper justification?(I want to prove that the additive inverse of a complex number is unique, and I am free to assume properties of reals, but the last bit seems to boil down to "equating like terms")

[echo wagon]

Do you mean equating the real and imaginary parts?

[somber yew]

Yes

[echo wagon]

Then, it is by definition. Two complex numbers are equal iff their real parts are equal and their imaginary parts are equal.

[somber yew]

I just want to be sure "equating.." isn't hand wavy for a proof

Ah oki

Makes sense.

Thanks!

[echo wagon]

Np

[viscid thistle]

in the same way can we say for an equation of the form a + bsqrt(c) = x+ysqrt(c), a = x and b = y?

[harsh smelt]

no

a = x and b = y implies that a + bsqrt(c) = x+ysqrt(c)

but not vice versa

e.g x=y=0, a = -bsqrt(c) !=0

[viscid thistle]

ah I see thank you

[harsh smelt]

but in algebra you can for example adjust to Q sqrt(2) and then sqrt(2) will play the same role as i in complex numbers

[viscid thistle]

sorry I'm not completely following. What is Q?

[harsh smelt]

rationals

i mean this is far not precalc so you do not have to care

[viscid thistle]

So if you had x + a(sqrt2) = 1 + 2(sqrt2) can we conclude that x = 1 and a = 2?

[harsh smelt]

if you get this from rationals with sqrt(2) adjusted yes

i mean

this will be true only in particular field

not in general

[viscid thistle]

I guess I'm a bit confused about when we can equate coefficients

[harsh smelt]

@viscid thistle ok let me first speak about points on plane

[viscid thistle]

ok

[harsh smelt]

(x,y) is typical point

it is true that two points are equal iff corresponding coords are equal

[viscid thistle]

I agree with that

[harsh smelt]

so in complex numbers

we can treat a+ib as point (a,b)

that's why we speak about equality of two complex iff their real and imaginary parts are equal

[viscid thistle]

ohh I see

i learned it a while ago but the graph interprtation of a point is to do with polar trig right?

It's escaped me by now so I can only speak of it vaguly

[harsh smelt]

ye (x,y) can be transformed to polar form

[viscid thistle]

that's a very cool way of explaning why we can equate real and imaginary

[harsh smelt]

analogously, when i adjust sqrt(2) to rationals. I mean now i speak about points (a,b) where a denotes the "rational part of my number" and b denotes "sqrt(2) part"

the point is here that i is not part of real numbers so 1+i != 2 or any real number and the same for sqrt(2) and rationals

[viscid thistle]

ohhh

so basically its (rational, irrational) = (rational, irrational)

[harsh smelt]

ye like that

i mean later idea would become more clear

i am quite far from explaining it rigorously

the point is here that i is not part of real numbers so 1+i != 2 or any real number and the same for sqrt(2) and rationals
@harsh smelt tho, we could agree on certain rules, but it will change arithmetics in general

in fact i am speaking now about abstract algebraic structure

[viscid thistle]

heh I have no clue what that is

I do have an example problem though i need a few mins to find it

about the equating stuff

[harsh smelt]

generalisation of school algebra to more abstract setup

[viscid thistle]

I believe this is one where you'd need to do equating

i need another minute to show my working

sorry uploading the pic

that took longer than expected

im out of practice

[harsh smelt]

uh wait

if i am not wrong there are no such integer values

otherwise it imples 3^m = even

[viscid thistle]

uh on my test I got m = 5 and n = -5

this is from a year ago tho

[harsh smelt]

,w 5+5log_3(2)=10log_9(6)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Commander Vimes#7532

[harsh smelt]

hmhm

then i am wrong ye

[viscid thistle]

but why can we equate the two sides?

I did it intuitively without really thinking about it

[harsh smelt]

oh wait i am idiot, it would be false in naturals

since 3^m=2^n6^5 is contradiction i arrived at

but in integers it is alright

well i guess it is in your particular problem that equating on both sides helped

[blissful ridge]

After you simplify a bit more
You'll arrive at

3^{m-5}=2^{n+5}

[harsh smelt]

ye, godfather

but i firstly thought i was in N

[viscid thistle]

so we can equate the two sides because of the fact that the equation is (Integer + non integer = Integer + non integer)?

[harsh smelt]

well i guess it is in your particular problem that equating on both sides helped
@harsh smelt i mean if you have a+b=c+d, a = c ad b = d will be always a solution but not always all solutions

well in your case you had similar structure yes

so your equating helped

oh, ye, lol log_3(2) is irrational

[viscid thistle]

oh lol I didn't know that

how did you approach it with exponents?

[harsh smelt]

i just rewritten m as log_3(3^m)

[viscid thistle]

oh okay

so in summary you equate when the number types on both sides of the equation are strictly complements of each other?

[harsh smelt]

why complements

irrationals are not complements of integers

but in summary i can equate if i know that a+b=c+d cannot be satisfied other way

[viscid thistle]

ah okay

makes sense thank you so much!

[harsh smelt]

yw

[quaint mason]

am i doing this right

[blissful kayak]

Hmm...

There seems to be a small problem

We don't know what g(-1) is...

Oh wait...

I'm stupid...

It's periodic

g(-1)=g(6)

Since it's periodic

[quaint mason]

ah i got it but ty

[strong ermine]

im lost and dont know where to start

[coarse rapids]

Hello, im a bit triped up on this last problem was wondering if someone could help.

[viscid thistle]

can someone help me with this

how did he go from this to this

[velvet blade]

I didn't understand how we're able to write the whole function as the highlighted part without adding the product of +c in ((x-c)+c)^k

[willow bear]

$( (x-c) + c)^k = \sum_{j=0}^k \binom{k}{j} c^{k-j} (x-c)^j$

[obsidian monolith]

Ann:

[willow bear]

binomial theorem.

the exact values of the constants are given by $b_j = \binom{k}{j} c^{k-j}$ but you're not really interested in them for this purpose

[velvet blade]

This hasn't been taught to me yet

Oh

[obsidian monolith]

Ann:

[willow bear]

oh god

i mean ok like

do you agree that you could in principle expand (z+c)^k into a sum of powers of z

even if you can't yet say exactly what the coefficients will be

[velvet blade]

yes I believe so, for some particular coefficients

[willow bear]

yeah so

replace z with (x-c)

( (x-c) + c)^k can likewise be expanded into a sum of powers of (x-c)

[velvet blade]

Ohh got it

[willow bear]

which is what they did here

[velvet blade]

Thanks

[velvet blade]

Do I learn that highest degree terms have coefficient of 1 in a binomial in the theorem?

[willow bear]

that sure is a bunch of words you just said

[velvet blade]

I don't understand what you mean

[viscid thistle]

I think you should re-formulate the question

[velvet blade]

I was asking if the term of highest degree in a binomial has the quotient as 1, if not, how is $a_n=b_n$

[obsidian monolith]

lazypawtato:

[velvet blade]

<@&286206848099549185>

[willow bear]

I was asking if the term of highest degree in a binomial has the quotient as 1,
you have misused at least one word here

i think you meant coefficient and not quotient

but if you expand (x+c)^n then the highest term does have a coefficient of 1, i.e. it is just x^n

[velvet blade]

Sorry, yes. I meant coefficient

So a_n = 1?

Again, I'm sorry for wrong wording

[willow bear]

no, nobody said a_n was 1

when we expand f(x) = sum[j=0,n] a_j x^j in powers of (x-c), the only term which gives us (x-c)^n upon expansion is x^n

[velvet blade]

Oh

[viscid thistle]

How to do this?

Am I supposed to take just 2 points and solve to find equation

?

[viscid thistle]

or something else.==>

hello

if anyone discovers this question plz ping with ans. Good Night

[pine wing]

It just wants you to use a calculator to solve all of these

@viscid thistle

[viscid thistle]

Am I supposed to take just 2 points and solve to find equation
?

[pine wing]

That is one way to do it

But if you use a TI -84 or something similar you can just input the table and it will do all of the work for you.

I'm assuming a Ti-84 is what they mean by "graphing utility"

[viscid thistle]

I use desmos's online graphing calculator at this moment

ty

[viscid thistle]

no way i got this worng

wrong*

[bold meadow]

it is correct

[sour hemlock]

no it is not

first term is 1

last term is 100

use arithmetic series formula

n(n+1)/2

oh wait

yes it's correct

sorry

I got confused by your handwriting

tell your teacher to mark it as correct

but I guess he crossed out your answer because you wrote n(sum of last term + first term) / 2

[sick steppe]

Gauss be like

[onyx agate]

Alright I have no idea what channel this question should go in cause it's weird

A farmer owns X acres of land. She profits P1 dollars per acre of corn and P2 dollars per acre of oats. Her team has Y hours of labor available. The corn takes H1 hours of labor per acre and oats require H2 hours of labor per acre. How many acres of each can be planted to maximize profits?

I think I've already solved it, it's just tripping me out because I was expecting it to be an optimization problem but it didn't end up involving any calculus

But you're maximizing P = P1X1 + P2X2 with constraints X = X1 + X2 and
Y >= H1X1 + H2X2

All linear

You just put X2 in terms of X1 and solve the inequality for X1, then the highest possible value of X1 is what's gonna give you maximum profit

Right?

[tropic arch]

Whats the difference between

limit and end behavior?

[astral mantle]

we use limit notation to describe end behavior

[tropic arch]

so they're the same thing basically?

[astral mantle]

not really

[viscid thistle]

Hello, can someone help me solve this problem?

[potent imp]

@viscid thistle

let the angles be a, b, and c

with a>b>c

we have a=3c+10, a-20=b+c, and a+b+c=180

use this system of 3 equations and 3 variables to solve for a b and c

[outer halo]

Guys how would I do this

[echo wagon]

Part a?

Can you tell me what the equation for the volume and surface area of a rectangular prism is?

[outer halo]

Length x breadth x height

Surface area is 2L( B+H+w)

[outer halo]

Then ?

[willow bear]

there are 4 parts to this problem, nzxzy

which one(s) do you need help on

@outer halo

[outer halo]

A and B

[willow bear]

uh huh

[echo wagon]

Sorry for ghosting btw, I didn't see you reply

[willow bear]

okay so first things first please don't use x for multiplication

there is a variable here called x

i do not want to risk confusion on that front no matter how minimal the chance

anyway, your box is x by x by y

when calculating its surface area you should keep in mind it is an open top box, so the top face should not be counted in the surface area.

does this much make sense to you? Y/N

or rather, "yes it makes sense"/"no it does not make sense"/"give me some time to process this"

@outer halo

[outer halo]

Give me some time to process this

[willow bear]

ok, take your time

[outer halo]

So it’s basically X by x by y which is equal to the A

[willow bear]

i

that's way too vague for me to agree with

[viscid thistle]

Ann could you help me with this, I’m new to calc

[willow bear]

@viscid thistle go to #calculus or a questions channel please. i'm a bit busy with nzxzy right now

[viscid thistle]

Ok

[willow bear]

@outer halo x by x by y is simply me repeating what the dimensions of the box are

[outer halo]

Okay

[willow bear]

with me having reiterated them once more, it should be clear that:
\begin{itemize}
\item the \textbf{volume} of your box is $x^2y$
\item the \textbf{surface area} of your box is $x^2 + 4xy$
\end{itemize}

[obsidian monolith]

Ann:

[outer halo]

Okay

[willow bear]

is that clear thus far? Y/N/need-time

[outer halo]

So in A part I gotta substitute it with the surface area with A to an answer ?

[willow bear]

please answer my question

[outer halo]

Yes thank you , I’m understanding better

[willow bear]

ok

so now, the volume of the box is known and fixed; it is equal to 32 (cubic meters)

thus, $x^2 y = 32$ and therefore $y = \frac{32}{x^2}$

[obsidian monolith]

Ann:

[willow bear]

and therefore the surface area can be expressed entirely in terms of $x$ using the above, which gives: $$A(x) = x^2 + 4x \cdot \frac{32}{x^2}$$

[obsidian monolith]

Ann:

[willow bear]

does this make sense to you? Y/N/time

[outer halo]

Yes thank you

I understood very well

Thank you for helping me

[willow bear]

ok, can you continue on your own from here?

[outer halo]

Yes

Thank you

[willow bear]

ok good

[outer halo]

Hi Ann ,
Im sorry I have another question, the second part says A dash

Does that mean it’s the inverse

[willow bear]

no

$A'$ is the derivative of $A$

[obsidian monolith]

Ann:

[willow bear]

it could've been written as $\dv{A}{x}$ instead.

[obsidian monolith]

Ann:

[outer halo]

Okay

Got it

[lean thistle]

i got this on my test a few hours ago and i still dont know how to do it

[willow bear]

express $x_{n+2}$ in terms of $x_n$ i think thatd be my first step

[obsidian monolith]

Ann:

[willow bear]

or wait no

that will be horrible nevermind that idea

[lean thistle]

yeah its too messy

[blissful ridge]

Can we say that lim (x->∞) x_n=0??

[willow bear]

mb find the limit of x_n+1/x_n

@blissful ridge can we? doesnt seem plausible to me

[harsh smelt]

mb find the limit of x_n+1/x_n
is this +1 inside index or this is fraction

[lean thistle]

index i think

[blissful ridge]

Looking at iterations of x_1,
It's decreasing, at very large n, it should approach zero

express x_{n+2}$ in terms of x_n i think thatd be my first step

If it is indeed zero, then after doing this and some simplification, we can have the desired result

[lean thistle]

how do i prove that it goes to 0

[blissful ridge]

Start from x_1, you'll see the next terms are decreasing

So eventually it'll reach zero

[obsidian monolith]

Commander Vimes:

Commander Vimes:

$\frac{x_{n+2}}{x_{n+1}}=\frac{x_{n+1}^4+3}{4}$ and for second quotient just change index

[lean thistle]

so lim x_{n+1}/x_{n} = 3/4?

[harsh smelt]

no

https://media.discordapp.net/attachments/363224154469826562/775682740073791548/694042020100046990.png?width=960&height=116

finally you have

[obsidian monolith]

Commander Vimes:

[harsh smelt]

and once you have shown that limit of x_n is zero you have desired result

[lean thistle]

but how do i show that it goes to 0?

[harsh smelt]

well is it intuitively clear for you?

[lean thistle]

hmm

kinda

[harsh smelt]

i guess you can try and play with algebra so you can use monotone sequence theorem

[lean thistle]

is it decreasing?

[harsh smelt]

kinda
@lean thistle you have initial term in (0,1) and it gots raised to power of 5 and then divided by 4

how is it not obvious that limit is zero?

[lean thistle]

well im not smart lol

[harsh smelt]

neither am i

[lean thistle]

shouldnt i plug in some values?

actually just 1

[harsh smelt]

wym

[lean thistle]

to see if it decreases

[harsh smelt]

well you can

[blissful ridge]

If you plug in 1 it'll just stay 1

[lean thistle]

well it decreases

indeed

[blissful ridge]

Increases?

[lean thistle]

sorry

decreases*

i plugged in 0,5

with a calculator

[blissful ridge]

5??

[lean thistle]

sorry omg

0.5

but i think every x_{n} is between 0 and 1

[harsh smelt]

well 0 will be constant 0

[lean thistle]

if i prove that it is bounded by 0 and 1

[blissful ridge]

but i think every x_{n} is between 0 and 1

No, only x_1 is between (0,1)

[harsh smelt]

no need, just prove that it is bounded below and that it is monotonic

then you can prove that limit is 0

(by solving L = (L^5+3L)/4)

[lean thistle]

No, only x_1 is between (0,1)
@blissful ridge if its not bounded by 0 it doesnt make sense

(by solving L = (L^5+3L)/4)
@harsh smelt L is 0 in R

[harsh smelt]

well, this equation has two solutions 0 and 1

but until you show boundedness below and that sequence is monotonically decreasing you are not allowed to do this

[velvet blade]

how did he get $b_n=a_n$?

[obsidian monolith]

lazypawtato:

[velvet blade]

<@&286206848099549185>

[jagged glade]

By comparing coefficients of (x-c)^n term of f(x) and (x-c)g(x)?

[velvet blade]

Ok, so I had this doubt a little cleared by ann, but I wanted to ask again just to be sure.$b_n(x-c)^n = a_n(x-c+c)^n$ right?

[obsidian monolith]

lazypawtato:

[velvet blade]

So both sequences a and b are supposed to be different right?

Did I do this right?

[jagged glade]

From my understanding of the 2 pages, it should be saying $b_n(x-c)^n = a_n(x-c)^n$
But I think your reasoning is also correct.

[obsidian monolith]

Biscuit:

[velvet blade]

But what about every other term with lower degrees? it's when when we're considering terms with -c we get difference in a_n and b_n right?

[jagged glade]

The other terms would be mixed with the -c and binomial coefficients. So it's just the b_n and b_0 that can be determined that straightforwardly

[velvet blade]

Okay got it. Thanks for the help.

[velvet blade]

I don't understand how the remaining constant g_n(x)= a_n

wait I got it

the last constant =$a_n (x^(n-n))$

[obsidian monolith]

lazypawtato:

lazypawtato:

[willow bear]

x^{n-n}

[obsidian monolith]

Galactic:

[viscid thistle]

What have you tried so far

[small forge]

I'm pretty sure e, b, a, are equivalent

[viscid thistle]

I can't see the letters, but i assume you are saying 1, 2 and 5 are equivalent

If you say so, then yes, so far so good

If you think 5 is equivalent, why are doubting on 3

They simply used $\log(a^b)=b\log(a)$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

@small forge

[small forge]

thanks!

[viscid thistle]

???

We are not even done

[small forge]

i got it lol

thanks tho

[viscid thistle]

Alright

[small forge]

This should the last one. I know I have to use the exponential growth formula to find the value after 20 years. I did that and it came out to 37287.66. I'm just confused on the second part when using the exponential decay formula

The formula I came up with is 10000 = 37287.66(1 - 0.120)^n but I don't know where to go from there

[small forge]

<@&286206848099549185>

[sick steppe]

solve for n @small forge

[small forge]

Yeah that's the problem I'm having I always mess up when isolating for an exponent

[sick steppe]

logs

you'll get $N = (1-.12)^n \implies ln(N) = ln(.88^n)$

[obsidian monolith]

moshill1:

[hexed granite]

can anyone help with simple precalculus problems

I have about 15 of them

for homework that he assigned and is checked for correctness

[small forge]

@sick steppe i got .72 which doesn't make much sense

[sick steppe]

it should be over 10

[small forge]

I'm not sure if i did it properly. I did 37287.66(1 - 0.12)^n and got 32812.56^n. then i used log32812.56 base 10000 and got that

[sick steppe]

what

how did you get 32812.56^n??

[small forge]

i evaluated the earlier part in order to get the log i needed

[polar mica]

How could I estimate this graph's equation?

[cerulean cedar]

It's strange that they didn't give you any numbers

[polar mica]

My teacher is a bit of a you-know-what

So yes

[cerulean cedar]

F

I'm not too sure how to do it without numbers, sorry

[polar mica]

Well the numbers can be estimated

The roots look could be at -7, -3, -1, 4

[cerulean cedar]

Alright

[polar mica]

The assignment is just to "estimate"

[cerulean cedar]

The first thing is to find out how the graph behaves around each root

For the root -7, it goes right up to it and bounces back down kind of like a parabola

[polar mica]

First root has multiplicity 2, second root has multiplicity 1, third root has multiplicity 1, fourth root has multiplicity 3

Based on these behaviors, right

[cerulean cedar]

Yah

[hexed granite]

The only ones i dont have an answer to are 4, and 6

[polar mica]

@cerulean cedar So I graphed it out, and it looks super weird on Desmos

[cerulean cedar]

Yah that's another thing

[trail burrow]

you can add a scaling factor if it's too large

[cerulean cedar]

So something like this would work, you write down the roots and raise it to the multiplicity

[viscid thistle]

@hexed granite for 4 you just have to set up the roots into an equation. -> (x-0)(x-A)(x-B)

[polar mica]

That's close enough

What I need for is the assignment is just to show I understand how to identify the properties of the graph

[cerulean cedar]

In this case multiplying it by 1/1000 makes it look good so yah

[polar mica]

So it should work, thanks

[hexed granite]

@viscid thistle it has to be completely multiplied out

is it that simple?

[viscid thistle]

yeah just set it up like that then multiply it out

I can do the multiplication rq if you want

[hexed granite]

if you can

[polar mica]

@cerulean cedar I can't get it that exact...

[hexed granite]

A=10 B=29

[viscid thistle]

okay

[cerulean cedar]

I mean your teacher didn't give you numbers or anything so idk what they expected

[polar mica]

Why would she make the answer that compex

[cerulean cedar]

Not sure, usually on tests and stuff they'll make it as simple as possible

1/1000 is an estimation, 1/3000 looks a bit nicer (these are just random values I tested btw)

[polar mica]

I feel like I'd lose marks for going that exact

[cerulean cedar]

Yeah maybe, I'm not sure

[polar mica]

But there's no whole numbers

[viscid thistle]

are you allowed to ask help for a question you have on hw or no? i am lost and feel dumb asf rn

[cerulean cedar]

You can ask it if you want

ima go to class now adios

[viscid thistle]

@hexed granite

[hexed granite]

Thank you

[viscid thistle]

np

[pastel pecan]

Does anyone know how to graph a tan function?

[viscid thistle]

4(x-2)^2=252 if anyone can help me with this i would appreciat it

[polar mica]

I'll get you

hold up

[viscid thistle]

@viscid thistle Are you solving for x?

it just says solve the equation idk man

okay

sorry

np

When you're to "solve" an equation with only one vairable, "solve" just means solve for the variable

[pastel pecan]

4x^2-16x+16=252

4x^2-16x-236

then use quadratic formula to get your anwse

If anyone knows how to do this lmk

[hexed granite]

B=29

[viscid thistle]

@hexed granite

[hexed granite]

thanks

[pastel pecan]

ok I got it

[hexed granite]

A=10 B=29

[viscid thistle]

assuming you already know those values

what have you tried so far

[hexed granite]

I plugged in the numbers and said the height was 1000 bc 100(10)

i hate word problems

[viscid thistle]

what, why 100(10)

[hexed granite]

the last term is the height no?

[viscid thistle]

okay, let's build some intuition

[patent beacon]

The entire equation is the height, haha

[hexed granite]

fuck

[viscid thistle]

so

do you know how a parabola with a<0 looks like?

[hexed granite]

-16t^2 +29t+1000

[viscid thistle]

in general

graphically, i mean.

[hexed granite]

upward?

U

[viscid thistle]

all the way around

downwards

,w plot -x^2

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Al𝟛dium#4084

[viscid thistle]

this is the looking

general looking

now, do you notice where does this type of parabola, have a maximum, ie where does it reach it's highest point?

(by looking at the general graph above)

[hexed granite]

0

[viscid thistle]

okay, yeah in this case it's 0, but it has it's name

does "vertex" ring a bell?

[hexed granite]

yea

[viscid thistle]

do you then understand that the vertex of a parabola with a<0, will be the maximum and hence, what we are looking for?

[hexed granite]

yea

[viscid thistle]

oky

so do you know the formula for the vertex

[hexed granite]

y=a(x-h)^2+k

i think

or its ax^2 +bx +c

[viscid thistle]

i mean yes, you can complete the square to get the vertex, but when you have the form y=ax^2+bx+c, you can also do this to find the vertex

$V_x=\frac{-b}{2a}$

[tender beacon]

Hello, who can help me in my homework

[viscid thistle]

not here

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

where b and a are constants found by y=ax^2+bx+c

but both methods work, you can complete the square to find the vertex, or use the formula above to find the vertex too

[hexed granite]

So when you look at the eqn, it would be V=-29/2(-16)?

[viscid thistle]

yes but use parens

-29/( 2(-16))

and this will give us the x-coordinate of the vertex, which is the time at where it reaches the highest height

[hexed granite]

so -29/-32?

[viscid thistle]

but we want the y-coordinate, so to find it, you can do h(29/32) and you are done.

so -29/-32?
yes

cancel the minus out to get 29/32 basically

[hexed granite]

thats the y?

[viscid thistle]

29/32 is the x-coordinate of the vertex, ie the time t where it reaches the highest height

but we want the height not the time (so as h is the function that describes height, t the time, h(t) the height in terms of t), hence we do h(29/32)

[hexed granite]

oh so 29/32 is the t

[viscid thistle]

yes

[hexed granite]

so i plug it into og eqn

[viscid thistle]

yes, basically find h(29/32)

[hexed granite]

ok cool thx

[tender beacon]

Hey, who can help me in any part of my homework

I dont understand anything

[hexed granite]

put a problem in they are more than happy to help

A=10 B=29 Last two i have

[trail burrow]

you can substitute x²=z and write the equation as y = z² + (4 - A)z - 4A and just solve it like a quadratic equation

if you input that into the the quadratic equation (p=4-A and q=-4A for x_1,2 = -p/2 +- sqrt((p/2)² - q)), you get
z_1 = A/2 - 2 + sqrt((A/2-2)² + 4A)
and
z_2 = A/2 - 2 - sqrt((A/2-2)² + 4A)

actually, this seems easier if you use the different quadratic equation, x_1,2 = -p +- sqrt(p² -4q), because there the terms become
z_1 = A - 4 + sqrt((A - 4)² + 4A)
and
z_2 = A - 4 - sqrt((A - 4)² + 4A)
where you don't need any fractions anymore

and then you can solve (A - 4)² as A² - 8A + 16

I'm actually not sure if the A² - 4A + 16 you get as the result inside the square root can be simplified any further, but at this point you can undo the substitution and get the positive and negative roots of both z values to get all 4 x values

[viscid thistle]

Could anyone help me A and possibly B?

[trail burrow]

does a still continue to the right? it looks like it's been cut off

[viscid thistle]

no he messed up on that

its 252

[trail burrow]

when you want to solve a quadratic equation like this where every variable is inside the square, you should try to transform the equation in a way that everything on one side is in the square and everything else is on the other side of the equal sign

so in this case, you divide by 4 on both sides so that you have (x - 2)² = 252/4

[quick mirage]

In this context, I would assume that doing sqrt(x) here would imply that you want to do +- sqrt(x)

Especially considering how the second question is literally a quadratic equation

[viscid thistle]

oh would i then do +2 and x equals whatever the answer is from adding it

[trail burrow]

yes

[viscid thistle]

ty

how about B could you help with that aswell please

[trail burrow]

that's slightly more complicated

do you know the quadratic formula?

[quick mirage]

Subtracting 5x from both sides gives you
0 = 3x^2 - 5x + 1

^^

[viscid thistle]

oh find the factor then right?

and do that formual thingy i forgot the name

oh you said it

[quick mirage]

If you can, otherwise you would need to use the quadratic formula

[viscid thistle]

got it

ty both appreciate it

[hexed granite]

A=10

[viscid thistle]

i got 65 for A is that correct?

[cloud hemlock]

What’s the easiest way to remember the Unit circle (radians,degrees, and points)

[quick mirage]

@cloud hemlock

cosine is relative to x

sine is relative to y

tan is relative to sin / cos (y/x)

cosine is x/r, so if x < 0 (when the angle is between 90 and 270 degrees), it is negative

and the opposite is true when the angle is 0- 90 or 270 - 360

sine is y/r, so y < 0 when the angle is between 180 and 360 degrees

and the opposite is true when the angle is between 0-180 degrees

tan, well, is just y/x, so in the first and third quadrant its positive, and the second and fourth quadrant its negative

[cloud hemlock]

Ok and for the radians, like I’m looking at the circle. And how are the radians solved for like pi/6 goes to 5pi/6 like how does that happen

[quick mirage]

think of 2pi as 360 degrees

pi is 180 degrees with that in mind

so pi/6 is like 180/6 = 30

but you shouldn't depend on that relationship

[cloud hemlock]

I’m trying to memorize the whole circle because I have a quiz on Thursday and I still don’t understand the whole circle, like how it works if that makes sense

[quick mirage]

try to picture that in mind you go counterclockwise the circle, pi gets bigger

You don't have to do that

Just have to think of 2pi as a full angle

[cloud hemlock]

?

[quick mirage]

2pi is the entire circle in essence

and if you can just understand what pi/2, pi/3, pi/4, pi/5, pi/6 are like angles you know of, like 90 degrees, 60 degrees, then you don't need to memorize the whole thing

Just think of 5pi/6 as like p/6 just multiplied by 5

and it gets much simpler

[cloud hemlock]

So then do I multiply by what to get 7pi or 11pi

I’m really bad at trig I’m sorry lol

[quick mirage]

well, I can tell you that they are coterminal with 180 degrees

because they are multiples of pi (180 degrees)

but not with 2pi (360 degrees)

Even then

7 pi = 3*2pi + pi

so its like 3 complete rotations of the circle

and one half rotation

[cloud hemlock]

Ok

Thank you so much

[quick mirage]

No problem

[cobalt storm]

I’m having trouble with #10

I get solutions of 2 and -5

But the textbook gets 2 and 2.5?

[patent beacon]

You're solving (x + 2)(x - 1) = 8 - 2x?

[sour hemlock]

hmm should be 2 and -5

[cobalt storm]

Yes

Idk why my text book gets 2.5 as a solution

I got 2 and -5 which should be right

[valid estuary]

Could anyone help me out with this problem? - The infinite geometric series totals 81. If the sum of the first four terms is 65, what is the first term of the series?

[hexed otter]

helpp meee

proofing mathematical induction

i tried doing the first one and still failed

[patent lance]

I am having some trouble

wth this

if anyone could help

[viscid thistle]

Im lost can someone help please?

[thorn moat]

Letg(y)=3y/y+4 find (g(y))^2 does this mean to pass y through as an input compute and then apply the second power?

[velvet granite]

@viscid thistle did you factor?

[viscid thistle]

i managed to get it ty tho

[loud marsh]

do exponential equations have inadmissible solutions the same how log equations do

[astral mantle]

you mean like an extraneous solution?

it depends on the exponential equations

[loud marsh]

like inadmissible roots

i guess its the same

[astral mantle]

you can have them because to solve them you might need to use logs

[round quest]

what is the sum of the integers from -10 to 50

[dim jungle]

20(61) = 1220

[pastel pecan]

does anyone understand how this works?

like if cos^2(a)-sin^2(a)=cos(2a)

why cant the opposite be true?

[harsh smelt]

wym

[pastel pecan]

Well Im just stumped

I know that this is a double angle formula

But I was wondering how the math checks out

I cant seem to find a proper explanation reasoning as to why it does

Dunno if I should as this question in proofs and logic?

shoot might of asked it in the worng place but i wsa unsure

I've heard of it mentioned yes

ok

[round quest]

how do i find the sum of the first 100 multiples of 4 from 4 to 400, inclusive.

[uncut mulch]

look up arithmetic series

[pastel pecan]

I got it

thanks

[astral mantle]

use the arithmetic sum $\frac{n}{2}(a_1 + a_n)$

[obsidian monolith]

Star_:

[sleek ruin]

So if we use ever use the inverse to find an angle, we have to check for the second angle and if it applies. Do we have to do that for the Law of Cosines. I'm just checking how people be doing their exercises and they never do calculations to find the possible second angle.

[willow bear]

what context is this in?

triangles?

any angle in a triangle is between 0 and 180°, and cos is one-to-one on that interval.

so there simply is no second angle

[sleek ruin]

triangles yeah

How about sin?

[willow bear]

sin isn't

[peak badge]

Triangles in this context make no sense here.

[sleek ruin]

There are options such as Two solutions, one solutions, or no solution

[willow bear]

that's why if you're using the law of sines to find a missing angle

[sleek ruin]

Might be saying something wrong

[willow bear]

you might end up with ambiguity