help-49

[spice scroll]

but then why did we change H_n to be this expression:
H_n (i assume) = y'' +ay' + by

[twilit jetty]

@spice scroll that questoin is too far ahead

also, thats wrong

stick with me because this gets convoluted

[spice scroll]

im struggling to follow

[twilit jetty]

hey

stop typing

Im still going

[spice scroll]

can we go over the goal of this function?

[twilit jetty]

ok?

[spice scroll]

okay

[twilit jetty]

you dont follow by asking harder questions

you follow by understanding easier ones

[spice scroll]

can we start over?

[twilit jetty]

we already have

[spice scroll]

i think i lost you

[twilit jetty]

we

already have

[spice scroll]

alright

[twilit jetty]

this is the starting over

[spice scroll]

great

ty

[twilit jetty]

since we already started over, you now know what solving an equation by y'' + 3y' + 5y = 0 means

[spice scroll]

can you say it?

as in, literally the meaning

[twilit jetty]

.

[spice scroll]

yes

[spice scroll]

alright

go on

[twilit jetty]

the method to solve equations of the form y'' + (number) y' + (number) y = 0 starts with an educated guess

[spice scroll]

wait wait

for y=e^x this is not the case

it's 9e^x=0

[twilit jetty]

so unfortunately we cant use y = e^x

big woop

[spice scroll]

never true

[twilit jetty]

we go with a different y than just y = e^x

you know we couldve also tried e^(2x), e^(-0.5x), etc.
so youve discounted y = e^x so far and we know that wont work

among the general "e^(rx)", lets see what happens when we plug that in

[spice scroll]

so we are looking for a form of "e^rx" for this equation, alright

to make this equality to 0 hold true

[twilit jetty]

not that form necessarily, but thats pretty much it

[spice scroll]

or trying to use them specifically due to e^rx 's nature

as a deriviative

i think i understand

[twilit jetty]

thats more like it, e^(rx) has conveniences attached to it that lets it be the only kind of "shape" involved here

[spice scroll]

yes

[twilit jetty]

now as before, you can plug in y = e^(rx) into the equation, and see if you can solve for r

[spice scroll]

r^2 * e^rx + 3r * e^rx + 5 * e^rx = 0

[twilit jetty]

brb

[spice scroll]

we can let T = e^rx

and then we have a polynomial fucntion of the form 2T^2 + 3T + 5 = 0

which is 2 complex numbers

but then T=complex number doesnt make sense, since e^rx is over the real numbers i think in our case usually

[twilit jetty]

@spice scroll back

[spice scroll]

ysz

[twilit jetty]

you likely dont expect that $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ and $\cos(x)=\frac{e^{ix}+e^{-ix}}2$

[grand pond]

mtt

[spice scroll]

not at all

[twilit jetty]

but the fact that theyre true means we cant discount complex roots, because those roots can be used to combine otherwise complex solutions into sines and cosines

we'll get to that in a bit

[twilit jetty]

you dont exactly have this be the polynomial

[spice scroll]

i understand

[twilit jetty]

instead the polynomial is r^2 + 3r + 5 = 0

[twilit jetty]

you can think of it as us dividing by e^(rx)

[spice scroll]

yes i understand

[twilit jetty]

now lets say you solve r^2 + 3r + 5 = 0

[spice scroll]

so we will just solve for r

[twilit jetty]

yea

[spice scroll]

2 complex numbers

theyre not too pretty

[twilit jetty]

yea theyre not, I didnt have the foresight to choose better numbers

try r^2 + r - 12 = 0 instead

(this would from trying y = e^(rx) on y'' + y' - 12y = 0)

[spice scroll]

r = 3, -4

great numbers

and reals

[twilit jetty]

now with r = 3 and r = -4 being plausible, that means what two functions can fit y'' + y' - 12y = 0?

[spice scroll]

are y=e^3x and y=e^-4x

[twilit jetty]

yep those work

however those arent the only ones that can fit

if you look closer, 2e^(3x) and e^(3x) - e^(-4x) also work

in general, you can multiply a solution by a constant and also add/subtract two solutions together to still have solutions

this allows for any function of the form y = Ae^(3x) + Be^(-4x) to be a valid solution
y = Ae^(3x) + Be^(-4x) is "the" solution to y'' + y' - 12y = 0

@spice scroll stuck on anything so far

[spice scroll]

nope

i just had to go for a while sorry

[twilit jetty]

yea np

since youre back, we can move on to the next point which connects us to solving this

[spice scroll]

great

[twilit jetty]

so far we've gone over the solutions to y'' + ay' + by = 0

(if you get two real roots as a solution)

(one real root / two complex roots will result in these other cases)

[spice scroll]

this can be generalized as a linear equation of the row matirx (1 a b )x = 0 right?

[twilit jetty]

not necessarily

we're going to use a different idea to solving u(n+2) = a u(n+1) + b u(n) instead

[spice scroll]

how can we solve for something of increasing indexes? i mean we can use recursion for decreasing indexes like G_n = 2G_(n-1), but i dont understand for increasing ones

[twilit jetty]

increasing indexes just means you can replace n with n+1 or with n+2, then rearrange the equation to be decreasing instead

G(n+1) = 2G(n) is the same as G(n) = 2G(n-1)

[spice mason]

What’s up

[spice scroll]

oh i see

[twilit jetty]

yea

[spice scroll]

small world

[spice mason]

Uhh are you guys doing like complex analysis or something?

I don’t understand what you guys are talking about

[twilit jetty]

he found out a method he hasnt heard before to solve equations of the form un = a u(n-1) + b u(n-2)

we've gone through solving a very similar y'' + ay' + by = 0 so far

[spice scroll]

yes

[spice mason]

Double deravative with respect to y….

I’ve intruded to much, sorry lol

[twilit jetty]

nw

youre solving the equation for a possible function y where adding y'', ay', and by add up to 0

its a bit different than solving for values

[spice mason]

I’m only calc bc rn

[spice scroll]

yes but then in our case it isnt 0 but H_n right?

[spice scroll]

me too

[spice mason]

Whaaa

[twilit jetty]

not really

[spice mason]

Then why don’t I understand this

Hm

[twilit jetty]

this is differential equation

[spice scroll]

because i dont either

[spice mason]

Is this like infinite series

Ahhh

Okay

Diff

Diff kinda annoying tv

Tbh

[spice scroll]

so how the relation be created?

[spice mason]

It’s fun though (but I haven’t learned any of it)

[spice scroll]

as in, we want to connect
y''+ay`+by=0
to
H_n = lambda^n

[twilit jetty]

you can get from solve y'' = a y' + by to solve H(n+2) = a H(n+1) + Hn with a trick involving an infinite series

@spice scroll avoid looking at that lambda^n for now

now do you know about taylor series

[spice scroll]

alright

[spice scroll]

ive seen some equation regarding it but i havent learned it yet

we did do some series stuff although

[twilit jetty]

whats important is that some functions can be written as a "power series"

so $f(x)$ can be a $\sum_{n=0}^\infty a_nx^n$

[grand pond]

mtt

[spice scroll]

i dont think i follow

i do understand this can be a function of course

[twilit jetty]

yea thats all

[spice scroll]

alright

lets move on

[twilit jetty]

now lets pretend that $h(x)=\sum_{n=0}^\infty\frac{H_n}{n!}x^n$

[grand pond]

mtt

[spice scroll]

also here, can H_n+2 and H_n be swapped maybe?

[twilit jetty]

not like that, we'll get to that

[spice scroll]

alright

[twilit jetty]

now if you simplify h'(x)

[spice scroll]

same thing but without the sum symbol?

[twilit jetty]

dont know what you mean but try it

[spice scroll]

now lets pretend that $h(x)=n=\frac{H_n}{n!}x^n$

[twilit jetty]

you dont just erase the sum symbol

[grand pond]

Ayanokoji (ALWAYS PING ME)

[twilit jetty]

no not like that either

[spice scroll]

then i dont know how to integrate it

[twilit jetty]

Im asking to differentiate it

[spice scroll]

i mean derivitate

[twilit jetty]

what you hopefully meant is to do:

,,\dv{}{x}h(x)=\dv{}{x}\sum_{n=0}^\infty\frac{H_n}{n!}x^n=\sum_{n=0}^\infty\dv{}{x}\frac{H_n}{n!}x^n

[grand pond]

mtt

[spice scroll]

i dont understand this

i mean you want to integrate it that way?

[twilit jetty]

differentiate it that wya

[spice scroll]

d/dx is usually for that reason there

[spice scroll]

ik but why the notation

[twilit jetty]

so that you know which part Im differentiating

,,\dv{}{x}\qty(h(x))=\dv{}{x}\qty(\sum_{n=0}^\infty\frac{H_n}{n!}x^n)=\sum_{n=0}^\infty\dv{}{x}\qty(\frac{H_n}{n!}x^n)

[grand pond]

mtt

[spice scroll]

oh wow

well usually in high school we just did (y)'=y'=...

[twilit jetty]

yea seeing the ' in the corner I figured wouldnt be clear but I thought wrong

,,h'(x)=\qty(\sum_{n=0}^\infty\frac{H_n}{n!}x^n)'=\sum_{n=0}^\infty\qty(\frac{H_n}{n!}x^n)'

[grand pond]

mtt

[spice scroll]

yes alright

so we deriviate for x

[twilit jetty]

yea

(differentiate for x)

[spice scroll]

but i honestly dont know if i can keep up with this

[twilit jetty]

yea you dont have to

[spice scroll]

alright

[twilit jetty]

I can just continue from here, and youll eventually see how it all fits together

[spice scroll]

alright

[twilit jetty]

the derivative of x^n is usually nx^(n-1)

[spice scroll]

i dont know how to do the differentiation, you probably know

[twilit jetty]

that works for n=0, n=1, n=2, etc.

[spice scroll]

true

[twilit jetty]

so (x^n)' should just be nx^(n-1)

[spice scroll]

yes

[twilit jetty]

now Hn and n! are both constants

[spice scroll]

oh i get it

[twilit jetty]

yep

[spice scroll]

so since its just a poly sum we can just change that

[twilit jetty]

,,h'(x)=\sum_{n=0}^\infty\frac{nH_n}{n!}x^{n-1}

[grand pond]

mtt

[spice scroll]

i get it

lets continue

[twilit jetty]

now for some simplifying

n/n! is 1/(n-1)!

[spice scroll]

true

[twilit jetty]

,,h'(x)=\sum_{n=0}^\infty\frac{H_n}{(n-1)!}x^{n-1}

[grand pond]

mtt

[twilit jetty]

now keep in mind that when n = 0, this is just 1

d/dx of that is just 0

,,h'(x)=\sum_{n=1}^\infty\frac{H_n}{(n-1)!}x^{n-1}

[grand pond]

mtt

[spice scroll]

yes

i dont see how this function leads anywhere yet but i get this part

[twilit jetty]

the trick is going to come

,,h'(x)=\sum_{n=0}^\infty\frac{H_{n+1}}{n!}x^n

[grand pond]

mtt

[spice scroll]

you mean if we defrenciate again h''(x)=0?

[twilit jetty]

not really

wait I think I should do this an easier way

[spice scroll]

why did you assume H_n+1 =n*H
_n

[twilit jetty]

,,h(x)=H_0+H_1x+H_2\frac{x^2}{2!}+H_3\frac{x^3}{3!}+H_4\frac{x^4}{4!}+\cdots
\h'(x)=H_1+H_2x+H_3\frac{x^2}{2!}+H_4\frac{x^3}{3!}+\cdots

[grand pond]

mtt

[twilit jetty]

so I think this should be clearer

you d/dx each term and you end up with that for h'(x)

[spice scroll]

i think i see it

but then again this is an infinite series...

[twilit jetty]

so far we're assuming h(x) works

so then h'(x) should work as well

[spice scroll]

alright

go on

[twilit jetty]

,,h(x)=H_0+H_1x+H_2\frac{x^2}{2!}+H_3\frac{x^3}{3!}+H_4\frac{x^4}{4!}+H_5\frac{x^5}{5!}+\cdots
\h'(x)=H_1+H_2x+H_3\frac{x^2}{2!}+H_4\frac{x^3}{3!}+H_5\frac{x^4}{4!}+H_6\frac{x^5}{5!}+\cdots
\h''(x)=H_2+H_3x+H_4\frac{x^2}{2!}+H_5\frac{x^3}{3!}+H_6\frac{x^4}{4!}+H_7\frac{x^5}{5!}+\cdots

[grand pond]

mtt

[spice scroll]

well i see what youre doing, and it is true

lets continue

[twilit jetty]

now we try something interesting

we know that H(n+2) = H(n-1) - 6Hn

[spice scroll]

yes

[twilit jetty]

now consider h'', h', and h here

term by term, you know that H2 = H1 - 6 H0

then H3 = H2 - 6 H1

(so H3 x = H2 x - 6 H1 x)

[spice scroll]

but can we rally take n=0 to defined?

[twilit jetty]

oh I forgot completely about that

the sequence begins at 1 instead of 0

alr thats a minor change

,,h(x)=H_1x+H_2\frac{x^2}{2!}+H_3\frac{x^3}{3!}+H_4\frac{x^4}{4!}+H_5\frac{x^5}{5!}+\cdots
\h'(x)=H_2x+H_3\frac{x^2}{2!}+H_4\frac{x^3}{3!}+H_5\frac{x^4}{4!}+H_6\frac{x^5}{5!}+\cdots
\h''(x)=H_3x+H_4\frac{x^2}{2!}+H_5\frac{x^3}{3!}+H_6\frac{x^4}{4!}+H_7\frac{x^5}{5!}+\cdots

[grand pond]

mtt

[spice scroll]

yes

[twilit jetty]

there we go

[twilit jetty]

so h''(x) = h'(x) - 6 h(x) and without an H0 anywhere

does that make sense

[spice scroll]

alright awesome

[spice scroll]

i dont really understand how we can say that

[twilit jetty]

use Hn = H(n-1) - 6H(n-2)

[spice scroll]

im trying, but taking H_n or H_n-1 or H_n-2 into one side is kind of hard to try and make this equality with all sides

no i just dont see it

[twilit jetty]

,,h''(x)=H_3x+H_4\frac{x^2}{2!}+H_5\frac{x^3}{3!}+H_6\frac{x^4}{4!}+\cdots
\=(H_2-6H_1)x+(H_3-6H_2)\frac{x^2}{2!}+(H_4-6H_3)\frac{x^3}{3!}+(H_5-6H_4)\frac{x^4}{4!}+\cdots

[grand pond]

mtt

[twilit jetty]

now you factor this a different way:

[spice scroll]

this i agree on

[twilit jetty]

,,h''(x)=H_3x+H_4\frac{x^2}{2!}+H_5\frac{x^3}{3!}+H_6\frac{x^4}{4!}+\cdots
\=(H_2-6H_1)x+(H_3-6H_2)\frac{x^2}{2!}+(H_4-6H_3)\frac{x^3}{3!}+(H_5-6H_4)\frac{x^4}{4!}+\cdots
\=H_2x+H_3\frac{x^2}{2!}+H_4\frac{x^3}{3!}+H_5\frac{x^4}{4!}+H_6\frac{x^5}{5!}+\cdots
\-6H_1x-6H_2\frac{x^2}{2!}-6H_3\frac{x^3}{3!}-6H_4\frac{x^4}{4!}-6H_5\frac{x^5}{5!}+\cdots

[grand pond]

mtt

[spice scroll]

i dont understand the bottom line

[twilit jetty]

I expanded it

[spice scroll]

when you say "expand" do you mean open parenthases?

[twilit jetty]

yep, in english that is what it means

[spice scroll]

oh i understand now

[twilit jetty]

yea

[spice scroll]

i was confused because i wouldve expected these to be side by side

as in i thought this was another "equals to" step with "=" omitted

i get it

[twilit jetty]

yea that was an issue

Im not really able to show this at its clearest

[spice scroll]

isnt this what taylor expansion is?

[twilit jetty]

no this is a power series

[spice scroll]

i think i once saw a video about something similiar

[twilit jetty]

it resembles a taylor series and we'll treat it like that, but for now its just a power series

[spice scroll]

alright great

[twilit jetty]

,,h''(x)=H_3x+H_4\frac{x^2}{2!}+H_5\frac{x^3}{3!}+H_6\frac{x^4}{4!}+\cdots
\=(H_2-6H_1)x+(H_3-6H_2)\frac{x^2}{2!}+(H_4-6H_3)\frac{x^3}{3!}+(H_5-6H_4)\frac{x^4}{4!}+\cdots
\=H_2x+H_3\frac{x^2}{2!}+H_4\frac{x^3}{3!}+H_5\frac{x^4}{4!}+H_6\frac{x^5}{5!}+\cdots
\-6H_1x-6H_2\frac{x^2}{2!}-6H_3\frac{x^3}{3!}-6H_4\frac{x^4}{4!}-6H_5\frac{x^5}{5!}+\cdots
\=H_2x+H_3\frac{x^2}{2!}+H_4\frac{x^3}{3!}+H_5\frac{x^4}{4!}+H_6\frac{x^5}{5!}+\cdots
\-6(H_1x+H_2\frac{x^2}{2!}+H_3\frac{x^3}{3!}+H_4\frac{x^4}{4!}+H_5\frac{x^5}{5!}+\cdots)

[grand pond]

mtt

[twilit jetty]

the next line is to "factor" the -6 "out of" the second row

[spice scroll]

i understand

[spice scroll]

yes

[twilit jetty]

then, finally:

,,h''(x)=H_3x+H_4\frac{x^2}{2!}+H_5\frac{x^3}{3!}+H_6\frac{x^4}{4!}+\cdots
\=(H_2-6H_1)x+(H_3-6H_2)\frac{x^2}{2!}+(H_4-6H_3)\frac{x^3}{3!}+(H_5-6H_4)\frac{x^4}{4!}+\cdots
\=H_2x+H_3\frac{x^2}{2!}+H_4\frac{x^3}{3!}+H_5\frac{x^4}{4!}+H_6\frac{x^5}{5!}+\cdots
\-6H_1x-6H_2\frac{x^2}{2!}-6H_3\frac{x^3}{3!}-6H_4\frac{x^4}{4!}-6H_5\frac{x^5}{5!}+\cdots
\=H_2x+H_3\frac{x^2}{2!}+H_4\frac{x^3}{3!}+H_5\frac{x^4}{4!}+H_6\frac{x^5}{5!}+\cdots
\-6(H_1x+H_2\frac{x^2}{2!}+H_3\frac{x^3}{3!}+H_4\frac{x^4}{4!}+H_5\frac{x^5}{5!}+\cdots)
\=h'(x)-6h(x)

[grand pond]

mtt

[spice scroll]

i agree

[twilit jetty]

nice

[spice scroll]

so we get
h''(x)=h'(x)-6*h(x)

[twilit jetty]

yep

[spice scroll]

lets continue

[twilit jetty]

so then h''(x) - h'(x) + 6 h(x) = 0

[spice scroll]

yes

[twilit jetty]

now we know how to solve stuff like this

we assume h(x) = e^(rx)

to essentially get r^2 - r + 6 = 0

then we solve for r

[spice scroll]

but can we really? we saw it isnt neccsarily like that

or do i not understand e^x properly?

[twilit jetty]

remember from earlier we made an "educated guess"

this is that educated guess

it is not the final h(x)

it is a guess

[spice scroll]

alright

[twilit jetty]

you can also think of it as

[spice scroll]

yes

[twilit jetty]

"we know e^(rx) works"

"we try out e^(rx) inside"

[spice scroll]

we assume rather? or check?

[twilit jetty]

we check

[spice scroll]

alright

but can e^rx = h(x)
as some sort of a property of a power series?

[twilit jetty]

in this case, not really

[spice scroll]

alright

[twilit jetty]

in the most general case, we can have:

[spice scroll]

so it might not be equal to what we are subsitituing as?

[twilit jetty]

yea

as you saw before, what we substitute in at first is not the final solution

[spice scroll]

yes

lets continue

[twilit jetty]

,,y''-y'-12y=0\implies y=Ae^{4x}+Be^{-3x}

[grand pond]

mtt

[spice scroll]

i dont know how to do this in between step

[twilit jetty]

no Im just showing the final solution

[spice scroll]

alright

so ill not worry about the algebra of those mid steps? i think i could make do it with more time

but it could be hard

[twilit jetty]

put it this way

if we have $y''+ay'+by=0$,

[grand pond]

mtt

[twilit jetty]

you plug in $e^{rx}$ to check when it can work

[grand pond]

mtt

[twilit jetty]

this results in $r^2+ar+b=0$

[grand pond]

mtt

[twilit jetty]

assume that you can get two different values for $r$: $m$ and $n$

[grand pond]

mtt

[twilit jetty]

so we now know $e^{mx}$ and $e^{nx}$ work

[grand pond]

mtt

[twilit jetty]

are you with me so far

[spice scroll]

yes

completely

[twilit jetty]

now $Ae^{mx}+Be^{nx}$ also works

[grand pond]

mtt

[spice scroll]

as in, equal to y?

[twilit jetty]

yea

[spice scroll]

i mean i dont think thats the case neccsarily

it's like saying if both x=2 and x=3 are a solution for something like:
x^2-3x+6=0
then
2^2-3*3+6=0

[twilit jetty]

youre not thinking about what kind of problem we're doing here

we know that $(e^{mx})''+a(e^{mx})'+b(e^{mx})=0$

[grand pond]

mtt

[twilit jetty]

also that $(e^{nx})''+a(e^{nx})'+b(e^{nx})=0$

[grand pond]

mtt

[twilit jetty]

now if you multiply one of the solutions by a constant, you can factor that constant out and still get 0

[spice scroll]

yes true

[spice scroll]

true

[twilit jetty]

,,(Ae^{mx})''+a(Ae^{mx})'+b(Ae^{mx})=0
\A(e^{mx})''+Aa(e^{mx})'+Ab(e^{mx})=0
\A((e^{mx})''+a(e^{mx})'+b(e^{mx}))=0

[grand pond]

mtt

[spice scroll]

yes

lets continue

[twilit jetty]

this also works for Ae^(mx) + Be^(nx) as follows

[spice scroll]

yes true

[twilit jetty]

,,(Ae^{mx}+Be^{nx})''+a(Ae^{mx}+Be^{nx})'+b(Ae^{mx}+Be^{nx})=0
\\text{expand}
\(Ae^{mx})''+(Be^{nx})''+a(Ae^{mx})'+a(Be^{nx})'+b(Ae^{mx})+b(Be^{nx})=0
\\text{rearrange}
\(Ae^{mx})''+a(Ae^{mx})'+b(Ae^{mx}))+(Be^{nx})''+a(Be^{nx})'+b(Be^{nx})=0
\\text{factor}
\A((e^{mx})''+a(e^{mx})'+b(e^{mx}))+B((e^{nx})''+a(e^{nx})'+b(e^{nx}))=0

[grand pond]

mtt

[spice scroll]

i understand

[twilit jetty]

nice

so you can see Ae^(mx) + Be^(nx) works since e^(mx) and e^(nx) do

in general, A f(x) + B g(x) would work if f(x) and g(x) do

you see how that works, right?

[spice scroll]

yes

[spice scroll]

yes

[twilit jetty]

this property of the differential equation is called "linear"

this is a "linear differential equation"

[spice scroll]

like in linear algebra right?

[twilit jetty]

yep thats the same linear

just like with vector spaces

[spice scroll]

great

[twilit jetty]

very good

this also means e^(mx) and e^(nx) are your "basis vectors" for the solutions

[spice scroll]

we just started vector spaces so i dont really know "basis bectors" as a term

i mean i do know stuff like 2 reals
(1
1)
can represent a 2 dimensional space if i remeber correct, like y=x

[twilit jetty]

its a similar idea to "any of A (1, 0) + B (0, 1) is within the 2D space R^2"

[spice scroll]

this i think i get

lets cotntinue

[twilit jetty]

nice

now we have the $y''-y'+12y=0$

[spice scroll]

i need to go to sleep soon since i have lectures in the morning

so lets continue

[twilit jetty]

alr

didnt know thats what you meant by continuing

using what we learned, we know that the solution $Ae^{mx}+Be^{nx}$ works

[grand pond]

mtt

mtt

[spice scroll]

yes true

[twilit jetty]

oh wait

the equation's different

oops

[spice scroll]

yes

i forgot about the 7 too

[twilit jetty]

well $H_n=7H_{n-1}-6H_{n-2}$ will mean that $h''(x)=7h'(x)-6h(x)$

[grand pond]

mtt

[spice scroll]

yes

therefore
$h''(x)-7h'(x)+
6h(x)=0$

[grand pond]

Ayanokoji (ALWAYS PING ME)

[twilit jetty]

yep

then?

[spice scroll]

and we can do the same thing we did earlier

oh i need to find the values of r for this case too right?

[twilit jetty]

yea

[spice scroll]

r= 7+-sqrt(43)

[twilit jetty]

try doing that again

[spice scroll]

i dont see how it's wrong it's
r^2 -7 * 2r + 6 = 0

[twilit jetty]

2r?

[spice scroll]

oh right

[spice scroll]

r= 6,1

[twilit jetty]

yep

[spice scroll]

oh thats also the result they had

[twilit jetty]

yea $h(x)=Ae^{1x}+Be^{6x}$

[grand pond]

mtt

[twilit jetty]

we're not done yet though

[spice scroll]

oh they did a similiar polynomial of power 2 with lambda instead of r

[spice scroll]

yes

[twilit jetty]

now A and B need to be found

[spice scroll]

and A and B are actually these constants

but question would be how to get h(x)=H_n

[twilit jetty]

well how that method works is for a different reason

its a shortcut

[spice scroll]

how do we do it?

[twilit jetty]

you just follow along what they did

[spice scroll]

for A and B they just took the base cases H_1 and H_2

[twilit jetty]

yea its essentially a different method

[spice scroll]

but we dont know why h(x)=H_n

[twilit jetty]

thats not true

thats not true at all

you are mistaking their method with mine

what we're going through is a justification that will eventually lead to their shortcuts

[spice scroll]

alright

lets continue

[twilit jetty]

now we know about H1 and H2

what are they, again

[spice scroll]

H_1 = 7, H_2 = 43

[twilit jetty]

wait somethings off here

[spice scroll]

@twilit jetty im feeling pretty dizzy, i think i need to get some sleep

[twilit jetty]

alr you can go for that

[spice scroll]

im very sorry for wasting your time

[twilit jetty]

Ill try to figure out how the rest of this plays out, I forgot the necessary steps

[spice scroll]

i think i understand better the method they used

but still not agree on it

[twilit jetty]

you cant really agree on their method

you cant trust half of the stuff its written down, you have to think of it as scratch work

you can think of it as "we know how the solutions usually look, so we only write down the bare minimum to remember what we're doing"

[spice scroll]

as in this is just a a pattern people use for this problems?

[twilit jetty]

yea

[spice scroll]

seems kind of weird to use a pattern we never understand

[twilit jetty]

it is

[spice scroll]

especially in a class for mathemticians

well i dont know what to think

but i understand part of something

again, thank you very much for your help mate!

[twilit jetty]

np

[spice scroll]

goodnight

[twilit jetty]

cya

[spice scroll]

.close

[midnight plank]

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[midnight plank]

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[tough jungle]

Is this correct?

[astral shuttle]

i dont think so

[spare tiger]

du is incorrect, since your sin and cosine have different inputs

[tough jungle]

Oh right mb

.close

[midnight plank]

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[midnight plank]

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[golden mortar]

Need help for this

Apparently it's not B

[small jasper]

$\lim_{x \to 1000^{-}}$ refers to the limit from the left only

[grand pond]

Civil Service Pigeon

[small jasper]

you're right that $\lim_{x \to 1000} D(x)$ doesn't exist, but that's not what's being asked

[grand pond]

Civil Service Pigeon

[winter lily]

oh

my bad david

[golden mortar]

I'm confused

Lol

I'm so close to finishing this hw too

[winter lily]

what's being asked?

[small jasper]

#help-49 message

[golden mortar]

) Use​ one-sided limits to determine the limit of​ D(x) as x approaches​ $1,000 and as x approaches​ $3,000. First, find the​ one-sided limits and the limit as x approaches​ $1,000.
Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

[winter lily]

thanks

i see

[golden mortar]

there's a chart on the top right

and a section above

2

[winter lily]

i thought the limit from the left didn't exist because it could be either x or 0.97x depending on where you are, right?

[small jasper]

????

1000 is neither an endpoint nor included in the domain of the x part

[golden mortar]

I'm so lost

[winter lily]

i might be reading this wrong, but, if we're coming from the left side of 1000, wouldn't either of the first two solutions apply?

[small jasper]

k uh

pretend this is the best picture you've ever seen

[golden mortar]

ok

[small jasper]

we're taking the one-sided limit from the left

meaning we're going from this direction

[golden mortar]

ok

[small jasper]

now which part of the function does this correspond to?

"this" being the left hand limit

[winter lily]

this part?

why couldn't it also be this?

[small jasper]

see picture above

[winter lily]

isn't this also coming from the left?

[golden mortar]

I'm a bit lost ngl

[winter lily]

oh, so whatever is closest?

OH

OHHHH

[small jasper]

[winter lily]

💀 💀 💀 💀 💀 💀

i get it now

thank you

[golden mortar]

Bro

[winter lily]

because i was STRUGGLINGGGG

[golden mortar]

Pleaseeeee explain

[small jasper]

Well what part of the function are we going along to approach x=1000 from the left

[golden mortar]

0.97 right?

[small jasper]

0.97x*

but yeah

[golden mortar]

ok

[small jasper]

so that's the part of the function we take the limit of

$\lim_{x \to 1000^{-}} D(x)=\lim_{x \to 1000^{-}} (0.97x)$

[grand pond]

Civil Service Pigeon

[golden mortar]

WAIT

IT WAS 0.97X

the whole time

[winter lily]

i think i was thinking about the limit going up as x goes up

[small jasper]

k you guys lmk if you need me

[golden mortar]

lol

ok

[winter lily]

alright

thanks!

[golden mortar]

damn

[winter lily]

david is that the last problem?

[golden mortar]

it was 970

[winter lily]

oh

it wanted you to solve

oh.

[golden mortar]

this is the next one

Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.
A.
ModifyingBelow lim With x right arrow 1000 Superscript plus Upper D left parenthesis x right parenthesisequals
  
enter your response here ​(Simplify your​ answer.)
B.
The limit does not exist.

[winter lily]

can you send a picture

it's easier to read that way'

never mind i see what it's saying

[winter lily]

alright

[golden mortar]

so 910?

[winter lily]

@small jasper would it be 950 or 910?

[small jasper]

take this as you will

[winter lily]

i would try 950 then

yay!

thank you, civil service pigeon

[golden mortar]

yayyyy

ok

same problem

[winter lily]

okay

[golden mortar]

imagine a repeat of the same question

That is what I got

[winter lily]

can you send a picture?

it might not be the same problem

[golden mortar]

last one was 1000+

this is regular 100

1000*

[winter lily]

does it have *exactly the same data in the table?

and in the piecewise function?

[winter lily]

oh, okay

thanks

the limit doesn't exist

well

[golden mortar]

ok

lets go

its right

[winter lily]

okay good

any other problems?

[winter lily]

try 950

[golden mortar]

That's not it

[winter lily]

.

[golden mortar]

@small jasper We need your help bro

[winter lily]

one sec

[golden mortar]

lol

[winter lily]

im confused

[small jasper]

isn't this legit

the same exact thing as last time

[winter lily]

yes

and i tried to follow the rules from last time

but it didn't work

unless i did something wrong

[golden mortar]

ye

idk what is wrong

[winter lily]

FRICK

I KNOW

IT'S SUPPOSED TO ME 0.95(3000)

NOT 0.95(1000)

TRY 2850

😭 😭 😭 😭 😭

[golden mortar]

2850

got you

Aye

you got it

[winter lily]

phew

i was STRESSINGGGG

[golden mortar]

now its 3000+

lol

[winter lily]

okay

dne

i think

it's the same as last time

NO

NONOONONON

i misread it

type in 2730

[golden mortar]

nice job bro :O

[winter lily]

thanks

[golden mortar]

this reg 3000

[winter lily]

and thanks pigeon

DNE

[golden mortar]

now

[winter lily]

does not exist

[golden mortar]

ok

thx kai

[winter lily]

np

[golden mortar]

If you want to stick around I am going back to understand what I got wrong on my other submissions

[winter lily]

alright

[golden mortar]

nice I'm up to a 90% now

[winter lily]

awesome!

[winter lily]

1 sec

graph c

[golden mortar]

thx btw

[winter lily]

np

[golden mortar]

Tbh I figured it out

Although you lot helped me out

so much

[winter lily]

great!

[neon swift]

i was kinda curious, whats the difference again between a shaded circle and an unshaded one in the graph?

[winter lily]

the reason why i said it was graph c is because we know f(0) is -1

so the circle is shaded anyway

[neon swift]

ohh yeaaa cos if it is graph a, -1 is not shaded, alr tyy

[winter lily]

but since the limit at 0- is unshaded, it remains unshaded

[winter lily]

because that's the actual value at f(0)

[neon swift]

yupp, got itt

[winter lily]

i probably just gave you false information

but that's what i got from the graph

[golden mortar]

Yo

[winter lily]

i guess

well no

yes

i'm sure

not io guess

[neon swift]

y = 1?

[golden mortar]

I tried that

Didn't work

[winter lily]

and it didn't work?

...

isn't y=1 the only horizontal line that passes through (0,1)

oh

type y=1

wait

yes

y=1

[golden mortar]

I tried 1

didnt work

[winter lily]

i mean

put the y= in too

so type in

y=1

[golden mortar]

...

okay so it was that

[winter lily]

okay

don't feel bad, it confused me too

[winter lily]

this one idk

sorry 🙁

[midnight plank]

@golden mortar Has your question been resolved?

[strange jacinth]

-3

[strange jacinth]

y=1

[strange jacinth]

C

[midnight plank]

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[golden mortar]

.reopen

[midnight plank]

✅

[golden mortar]

We back

[slender walrus]

we can't see the graph

[golden mortar]

Sorry bro

[slender walrus]

what's your issue here?
do you know which part of the graph you should be looking at?

[golden mortar]

Idk what to look for no

[slender walrus]

do you understand the limit notation

it looks like you did a few similar questions earlier

[golden mortar]

I'm trying to understand but losing the main idea

[slender walrus]

the - or + when present in superscript
indicates the left or right side limit respectively

so here since there's a + in superscript ,
follow the curve towards x=-2 from the right side

[golden mortar]

I'm losing this

[slender walrus]

which part

the notation is the same as what you applied here
#help-49 message

English pls, also

[pearl idol]

This help channel is occupied, please don't talk in here unless you're helping Davisdavid

[golden mortar]

Idk bro this is my first time looking at this. I got online class

[slender walrus]

refer to the one you did earlier

that you seemingly understood

[midnight plank]

@golden mortar Has your question been resolved?

[midnight plank]

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[neon swift]

u js substitute it right?

[midnight plank]

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[wide ore]

Any hints?

[sage olive]

write the general term as (-1)^n(n+1)^2 / 5^n

[stray rose]

Yeah it’ll be an alternating geometric series

[sage olive]

maybe you can find a taylor series or manipulate into a geometric series or something

[wide ore]

They haven't taught us Taylor series yet

[sage olive]

there isn't a fixed common ratio

[wide ore]

@stray rose hisoka 💀

@sage olive alright

[stray rose]

There is

Should be n+1/5

Nvm I’m dumb

Forgot common ratio has to be just a number

[wide ore]

Ig I will be able to do it

Will ask if I don't get it

Thank you

.close

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[strong pebble]

I have a question regarding orienation in vector spaces

[strong pebble]

What does "arbitrary choice" of bases being positively or negatively oriented?

Our formal definition in our course in analytical geometry states

say e and b are two bases of some vector space V

e is positively oriented with b if det(A)>0 where A is the transition matrix from base e to b

does that mean that I just choose if det(A) is >0 or <0 ?

[midnight plank]

@strong pebble Has your question been resolved?

[strong pebble]

<@&286206848099549185>

[pulsar basin]

@stiff heart

[midnight plank]

@strong pebble Has your question been resolved?

[strong pebble]

<@&286206848099549185>

[mighty hamlet]

"Arbitrary choice" means you can decide how you define "positive". In your screenshot, it said that right-hand bases are typically assigned positive. However, you don't need to follow that standard. You can use the definition in your class to decide what is considered 'positively' oriented.

[runic hamlet]

instead of the usual determinant, you could define a new function which you could call for example dit such that dit(A)=-det(A). then dit satisfies all the usual properties of the determinant. and then you could say stuff is positive if dit(A)>0.

[strong pebble]

like, for example, -det(A) is positively oriented?

[mighty hamlet]

see this

in other words: yes. Positively orientated doesn't need to equate to positive determinent. Just as in the screeen shot, orientation is just a cycle/direction it's turning.

[midnight plank]

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[midnight plank]

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[last slate]

i started off with p E 4k, 4k+1, 4k+2, 4k+3

and with some manual labour, it seems 4k+3 seems to satisfy the asked condition

the issue is with the options

according to my solution, B and C are correct

can anyone help (please ping while replying)

[next rover]

@last slateit all sounds wrong

[last slate]

i see

[next rover]

what's 4k+3

[last slate]

can you please explain

[last slate]

ah

i should have added more details

[next rover]

like why do you think 4k+3 is special

[last slate]

because it seems to satisfy the condition

[next rover]

(−i) ^ 15 is not

working

ok i'm wrong

[last slate]

yeah..

[next rover]

nevermind no idea

[last slate]

if it works for 3 it works for 4k+3 numbers

[last slate]

😭

thanks, i would appreciate some help

the options are very tricky

[next rover]

well the problem is 4k+3 is not (C)

[last slate]

huh?

[next rover]

C means 4k+3 always works, and 4k+1 always works and 4k+2 always works

[last slate]

hm

i got it

they assumed all the 4 values of p

i was solving for just 4k+3

[next rover]

B also doesn;t work

[last slate]

yeah i got it

[next rover]

i can;t do it in my head i just use calculator

[last slate]

its a specific case in those 3

question has beauty i must say

thanks for your help frog

.close

[midnight plank]

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[next rover]

you;re ignoring "sufficient"

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[vast bone]

i used b^2-4ac=0

but it said i was wrong

[cyan pendant]

the system of equations has exactly one solution, not the quadratic

[cyan pendant]

you're assuming that there is 1 root (x-int), whereas a system of equations is an intersection between multiple equations

[vast bone]

ohhh so i only use the quadratic if its like intersect or something?

then what should i dooo

[cyan pendant]

try to do some rearranging and see if you can solve a simultaneous equation

hint: substitution

[vast bone]

LMAO

okay i saw that

and i get it now

[cyan pendant]

that's exactly what i said... but he gave it away

[vast bone]

so basically plug it in first before doing method

[vast bone]

:(

thanks tho

also matching pfp kind of

[midnight plank]

@vast bone Has your question been resolved?

[vast bone]

why is this not C

i put C and it said i got it wrong

[cyan pendant]

the graph needs to be continous, consider the functions when a is less than or equal to 50 and when a is greater than 50

[fallen aurora]

function in option C does not cover all the possible number of points earned by students

[cyan pendant]

when a is less than or equal to 50, what is the function?

[vast bone]

uhh then thats 5a ?

[vast bone]

? wait wdym

[cyan pendant]

y = 5a, from 0 to 50 inclusive

[vast bone]

i still dont get why its not C tho since we dont know how many students but you alr know the first 50 got 5 points each so the rest is just 50*5 + 3a

[cyan pendant]

but when a is greater than 50, its y = 3a + c

[vast bone]

mhm

[cyan pendant]

the graph needs to be continuous or it would mean that the total points drop or increase after a is 50

ill show you;

[vast bone]

okay

[cyan pendant]

at a = 50, the total points drop which does not make sense in this context

[vast bone]

wait okay i kind of get it but like

how do you find c 😭 (sorry)

[cyan pendant]

well, you know that the first graph is y = 5a

[midnight plank]

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[fallen aurora]

Here's a simple explanation for option c not being the correct option:-
According to the data for 50 assignments a student should earn 250 points,
now if you analyse g(50) for option C will definitely not going to give you 250
The only two options that give 250 for g(50) are B and D

[midnight plank]

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[vast bone]

ohh

[midnight plank]

Please don't occupy multiple help channels.

[fallen aurora]

For an additional assignment a student earns 253 points, as given in question
so the required function will be that which have g(51)=253

got it?

[vast bone]

OHH

wait yes i think i get it now

okay thx

so its B then

[fallen aurora]

yeah

[midnight plank]

@vast bone Has your question been resolved?

[midnight plank]

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[whole wasp]

Please help me

I have no clue how to find the equation of the ellipse

<@&286206848099549185>

please

i really

need help

asap

im not trying to be rude or anything

but im actually gonna be in a lot of trouble if i don't get help

please

😭

[midnight plank]

@whole wasp Has your question been resolved?

[whole wasp]

@robust cloak

PLEASE HELPERS

<@&286206848099549185>

PLEASEEE

PLEASEEEEEEEEEEEE

of the lines yes

but with the ellipse

idk how to

nope

I didn' t get the foci

L2 and L3= (17,-14)

no

[whole wasp]

L1 and L3= (1,2)

L1 and L2 are parallel

whats delta

the area of the triangle?

i mean how would the circumcircle help tho we need to find the ellipse don't we

right right

oh shiiit

i read it wrong

my bad my bad

would you be down

to help me

in other questions

they're all related to ellipses

its alright

thanks

oh wait no

i still need to know

the focus right

of the ellipse

which I do not know

we need to find the circumcentre of rqs

we dk rs

s is the focus

yes but we need to know the location of the focus don't we

to use the distance formula

alright

take your time

yes

yep

c=ae

right

but here

the ellipse

is oblique

there'll be an xy term in it's equation

so it won't fit the general form

how old are you if i may ask

its alright

[whole wasp]

are you in high school or college?

you seem to be pretty smart

thanks for

trying to help

[midnight plank]

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[midnight plank]

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[whole wasp]

.reopen

[midnight plank]

✅

[whole wasp]

<@&286206848099549185>

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[whole wasp]

no

[midnight plank]

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[midnight plank]

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[steady rose]

. The sum of the digits of the positive integer n is 123. The sum of the digits of 2n is 66. The
digits of n include two 3s, six 7s, p 5s, q 6s and no other digits. What is p^2 + q^2?
(A) 106 (B) 160 (C) 72

[half ingot]

Bro I swear this was in help 3

[steady rose]

it was

but it wasnt pinned

[half ingot]

😭😭

Now we wait ig

[steady rose]

yeah

[gleaming spear]

honestly your best bet might be to just find integer solutions for p and q

based on the answer choices

[steady rose]

maybe

[gleaming spear]

yea that works

each answer choice only has one square, so its pretty straightforward

[steady rose]

i dont get the "The sum of the digits of 2n is 66" part

[gleaming spear]

ngl you don't need that

you can just cheese it with the answer choices

what is this for? competition or school hw

[flat veldt]

The digit x 2 then add all the numbers in the new digit obtained from multiplication by 2

[steady rose]

k

[steady rose]

random summer class thing

i still have no idea how u do it D:

[gleaming spear]

just using the answer choices

you know that p & q are integers

[steady rose]

r they

[gleaming spear]

can you have a decimal number of digits?

[steady rose]

o im dumb

ur right

[gleaming spear]

its a very cheeky solution, and wouldnt work if there were multiple choices

theres prolly a legit way to solve it but they're all much harder

[steady rose]

so its 106?

k thx

.close

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[steady rose]

.reopen

[midnight plank]

✅

[steady rose]

?

[small jasper]

5p+6q=75, so the only possibilities are (p,q)=(15,0), (9,5), and (3,12)

It’s decently trivial to finish from here given the digit sum of 2n

Anyway I gtg 👋

[steady rose]

k bye

.close

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[gleaming spear]

yea just same diophantine method with less cheese ig

also easier lol

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[livid crystal]

Can someone help me get started here? Idk what the first step is :(

[livid crystal]

We are learning about area rn

Do I use area of triangle here?

But for which triangle

[last slate]

use pythagoras theorem

4^2 - (3.2)^2 = x^2

[frail burrow]

I think he's supposed to use triangle similarity

Try giving the vertices names (A, B, C, ...)

And finding two triangles which are similar

Then, use that to build an equation and solve for x.

[midnight plank]

@livid crystal Has your question been resolved?

[chilly helm]

is it marked as right angle by the teacher? if so you can find x with a^2+b^2=c^2

[livid crystal]

Yes

[chilly helm]

once you find x, take the area of a rectangle (base*height, not using the longest side) and divide by 2

[livid crystal]

What method are they using here? 🥲

[chilly helm]

the same one but for both triangles at the same time!

here they are treating the long side of the combined shape as the base and using x as the height

[livid crystal]

Ohhh

[chilly helm]

it looks like they solved for x algebraically by using the given measurements of the big triangle (3 and 4)

[livid crystal]

I see

[midnight plank]

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[stone sapphire]

can somebody fact check this

[next rover]

that would be my guess