help-0

[grave quartz]

I mean U can use the same technique you would use to find a quartic normally

right for understanding purposes u have 2 sets of complex conjugates

[patent topaz]

yes

[grave quartz]

to work out the polynomial you can simply

$$(x-2i)(x+2i)(x-(2-i))(x-(2+i))$$

[ocean seal]

SWF ;)

[grave quartz]

because you have 4 roots

[patent topaz]

yeah, that's my problem- expanding the brackets leaves me with $(0-4i^2)$, right?

[ocean seal]

WayneTundra

[grave quartz]

eh

[patent topaz]

rather, the first difference of squares

[grave quartz]

it shouldn't

[patent topaz]

sorry, i'm misunderstanding then. x-2i: the x component is zero?

[grave quartz]

what

ye

[patent topaz]

(x-2i)(x+2i)=(0x-2i)(0x+2i)?

[grave quartz]

The polynomial should be $$x^4 -4x^3 +9x^2 -16x +20$$

[ocean seal]

SWF ;)

[grave quartz]

here

just expand the brackets as I put them up there

follow that working

[patent topaz]

huh. thanks for the help, it's very much appreciated.

is the minus in $x-(2\pm{i})$personal preference?

[ocean seal]

WayneTundra

[patent topaz]

.close

[lone heart]

Channel closed

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Use .reopen if this was a mistake.

[lone heart]

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[quartz trout]

hi, i have a problem in understanding tangents and normals using derivatives i dont understand how to proceed, i understand its basics to some extent but i get stuck really easily,i dont know if these kind of questions are entertained but any kind of suggestion would help

[vapid shuttle]

I was hoping someone could verify my work for this

$f(x,a)=\frac{ax}{a^2+x^2}$

[ocean seal]

Austiin

[vapid shuttle]

ax is a product of continous functions

a^2+x^2 is a sum of continous functions

their quotient is continous unless the denominator equals 0

so that is

$a^2+x^2=0 \implies a=x=0$

[ocean seal]

Austiin

[vapid shuttle]

So really we only need to show continuity at (0,0)

elsewhere it is garuanteed by our theorems for sum/product/quotient of continous functions

so we consider

$\lim_{(x,a)\to (0,0)} \frac{ax}{a^2+x^2}$

[ocean seal]

Austiin

[vapid shuttle]

this is equal to

$\lim_{x\to 0} \frac{0}{x^2}$

[ocean seal]

Austiin

[vapid shuttle]

since a is a constant, not a variable

if it approaches 0 it must be 0

so this limit is just 0

now we have to compare this to our function evaluated at (x,0)

$f(x,0)=\frac{0}{x^2}=0$

[ocean seal]

Austiin

[vapid shuttle]

so our f(x,a) is continous at (0,0) so it is continous everywhere

and the exact same work for f(a,y)

is that good enough?

[worn fox]

lemme just rejig your argument a little

if a = 0, f(x,a) is the zero function to everything is fine nothing to do there

[vapid shuttle]

same for f(a,y)

[worn fox]

for a non zero, its just $\lim_{x \to 0} \frac{ax}{a^2+x^2}$

[ocean seal]

ΣΑCu

[worn fox]

you dont have to worry about two variable limits here

[vapid shuttle]

which is 0/a^2=0

[worn fox]

yep

[worn fox]

it was just this part that was a bit iffy and not needed

[young finch]

@vapid shuttle

[vapid shuttle]

what are the points of these limits

[young finch]

oh wait youre actually austin

[vapid shuttle]

if we do not compare them to the function at that point

[quartz trout]

Only solvable questions are to be posted here, or these kind of questions are okay too?

[rose sigil]

@vapid shuttle

[young finch]

#❓how-to-get-help

[worn fox]

well you can and should

[vapid shuttle]

these are fine but this channel is occupied. Open an available one and see #❓how-to-get-help for guides

[quartz trout]

I read it

[worn fox]

but atleast for f the limit is easy bc theres no indeterminant form to finding the limit does just amount to plugging the limit point in

[young finch]

no you didnt

[worn fox]

lol help-0 moment

[vapid shuttle]

you're interrupting my channel @quartz trout I have posted my question in here, that is why you need to go to an available help channel instead of this one

[vapid shuttle]

yes so what is the point of taking the limits?

[quartz trout]

Oh sorry I just realise u put ur question right before me I didn't know that would instantly make a new channel sorry

[vapid shuttle]

no worries

try #help-1

[worn fox]

if you just found the value at x=0 then you haven't checked the limit, it just happens in this case that the limit is easy and amounts to the same procedure

[vapid shuttle]

Okay can we restart a little here

I'm getting a bit confused

we have this

we have immediately that it is cont. everywhere except possibly (0,0)

that is all we need to show

[worn fox]

not (0,0)

[vapid shuttle]

what do you say is the procedure

[worn fox]

these are functions of one variable now

[vapid shuttle]

oh

f(x,a)

[worn fox]

a is arbitary but fixed

[vapid shuttle]

yes

okay

so

f(x)=(ax)/(a^2+x^2)

we wish to show is continous everywhere

the only case where it might not be, is a=0 and x=0

so now our procedure?

compare limit x->0 to f(0) ?

but f(0) is undefined if a=0

so I am stuck

[worn fox]

ah no so

if a=0, then f(x,a) = 0 for all x

[vapid shuttle]

what about x=0

[worn fox]

the function is the zero function the second you decided a was 0

[vapid shuttle]

hmm

I don't see how that works

if x is 0 that is still division by 0

[worn fox]

think about the family of functions a/x

if i pick the function where a=0, thats just the zero function

[vapid shuttle]

unless x=0

is what I would think

because isn't that saying

0/x = 0

for all x

but what about x=0

it is undefined

@fallen verge

[fallen verge]

What have you done now

[vapid shuttle]

i divided by 0

[fallen verge]

Oh god we're doing stuff with poles

[vapid shuttle]

poles?

[fallen verge]

No im being stupid

Ignore me

[worn fox]

yeah i think you're right im being dumb

[fallen verge]

What

[vapid shuttle]

no time for games

[fallen verge]

You know i dont understand mv, why did you call me here

[vapid shuttle]

as sigma pointed out

it isn't multivariable

a is arbitrary but fixed

[fallen verge]

So what are you trying to prove

[vapid shuttle]

f(x,a) continous for all x

let $a\in \mathbb{R}$ and $$f(x)=\frac{ax}{x^2+a^2}$$ we need to show $f(x)$ is continous

[ocean seal]

Austiin

[worn fox]

my issue (or perhaps confusion) was that proving this limit is equal to f(0,0) would imply continuity of f(x,y) at (0,0) as you've just changed the name of y

[vapid shuttle]

no it wouldn't

because I just set a=0

you can't just set y=0

y can range

the paths

remember the curves

[fallen verge]

Have you considered the fact that it doesnt not necessarily approach 0?

[worn fox]

you were letting a go to 0 in that latex

[vapid shuttle]

the fact that it does not not necessarily approach 0

[fallen verge]

I meant one not

[vapid shuttle]

I was considering f(x,a) but a is a constant, so this limit is just the same as x->0 and letting a=0

not letting a->0

[fallen verge]

Should approach 1/2 if you approach it when a=x

if you approach it on any line a=cx, you get the limit=c/(1+c^2)

[vapid shuttle]

uh

[fallen verge]

So the limit does not exist

[vapid shuttle]

a is constant

it can't equal x

[worn fox]

ah, take the limit as x goes to zero first

and then set a = 0

[vapid shuttle]

this is 0/a^2=0 (unless a=0)

[vapid shuttle]

then this makes div by 0

[worn fox]

idk maybe just cry then

[vapid shuttle]

😭

[fallen verge]

lim y->a f(x,y) is not always equal to f(x,a)

[vapid shuttle]

eric pls save me

I know you're out there

[worn fox]

okay well f(x,0) is zero everywhere except maybe x=0

have a good night

[vapid shuttle]

you too

[worn fox]

the issue here isn't the limit right, its what the value at x=0 is

[fallen verge]

I think that its not always the case that $$\lim_{(x,y) \rightarrow (0,0)} f(x,y)= \lim_{x \rightarrow 0} \lim_{y \rightarrow 0} f(x,y)$$

[ocean seal]

GarlicBredFr

[vapid shuttle]

holy crap holy crap holy crap im an idiot im an idiot im an idio

[fluid kettle]

austiin

[vapid shuttle]

this question is so stupid

[fallen verge]

Because balls

[worn fox]

what have you done

[vapid shuttle]

it doesn't involve all relevant information

in the question

it says

[worn fox]

fuck SAKE

[vapid shuttle]

f(x,y) as in example blah

but then gives f(x,y) anyways

but

in example blah

h(0,0)=0

[worn fox]

AAAAAAAAAAAAAAA

[vapid shuttle]

AAAAAAAAAAAAAAAAAAAAAAAAAAAA

[fallen verge]

🔋

[worn fox]

would be a tiny battery

[vapid shuttle]

ok

so literally all we care about are those limits then

f(x,0)= 0 unless x=0, but in this case it is f(0,0) which is defined to equal 0

so now we just need to care about the lim x->0 of f(x,0)

which is 0

ty lhopitalsrule

[worn fox]

so in the case where a=0, the function is zero literally everywhere but x=0, you could just do epsilon shite

[vapid shuttle]

same for the f(a,y)

and we are done

[vapid shuttle]

no at x=0 it is the case of evaluating f(0,0) which is defined to be 0 in the example

[fallen verge]

Im still confused

[worn fox]

im talkin about the limit

[vapid shuttle]

it is single variable limit

just use lhopitals

[fallen verge]

Austin i want my i back

[vapid shuttle]

ok

[worn fox]

you are literally the epsilon delta guy

omg he got automuted

[fallen verge]

Lmao what

[worn fox]

he missed the l on like

[fallen verge]

Oh well, i guess we'll get back to it at some point

Should i coose?

@vapid shuttle welcome back

[vapid shuttle]

I'm back

[worn fox]

watch those fingers

[vapid shuttle]

oopsie

okay

@worn fox

may I present to you

hopefully

something good

$f(x,a)=\frac{ax}{a^2+x^2}$

[worn fox]

the alt j song?

[ocean seal]

Austin

[vapid shuttle]

tontinous everywher except a=x=0 immediately from composition/product/sum of continous functions theorems

[fallen verge]

Thanks for the i

[vapid shuttle]

so now we consider

let a=0

and take lim x->0

this is

$\lim_{x\to 0} \frac{0}{x^2}=0$

[ocean seal]

Austin

[vapid shuttle]

and if we are considering the case of a=x=0

this is defined in our function the same as f(0,0)

which we defined to be 0

[fallen verge]

You can use epsdel here right

[vapid shuttle]

so we get the limit equal to the value

and it is continous there

please tell me that is good ???

[worn fox]

yeah but you have considered the limit in the case a =0

[vapid shuttle]

I just did!!!>!??!?!?!?!?

[worn fox]

i think your woes about dividing by 0 were valid

[vapid shuttle]

dear god

where is the division by 0 happening now

the f(0,0) is defined by the function

so no worries there

the limit comes from lhopital

I should be fine?

[worn fox]

oh youre lhopitaling fine

honestly im just scared of you after the automute

whatever you say

[vapid shuttle]

wait a minute...

what if I was onto something big

and that is why the mods muted me

to stop me from my realization

[fallen verge]

We want your fields medal

[vapid shuttle]

the mods wanted my fields medal

[vapid shuttle]

can we do eps delta instead

I feel bad now

you made me feel bad

[worn fox]

its half 1 AM i do not have the brain capacity for eps del lol

i think what you did was fine and im just slow

[vapid shuttle]

how can we do eps delta for x=a if f(a) is undefined

wait wtf am I doing

no don't say anything

don't

mods mute him

[worn fox]

you should have 0 < |x-0| < del implies |f(x,0) - f(0,0)| < eps

[worn fox]

or that

[vapid shuttle]

so

I need

[worn fox]

honestly this probs sucks

[vapid shuttle]

$|x|<\delta \implies |f(x)|<\varepsilon$

[ocean seal]

Austin

[worn fox]

0 < |x|

[vapid shuttle]

let epsilon>0

thanks sherlock

[worn fox]

its important

[vapid shuttle]

ik I'm jk

uhh

here

if

$0<|x|<\delta$

[ocean seal]

Austin

[vapid shuttle]

then x is nonzero

then f(x) is 0

then since eps>0

|f(x)|<eps

[worn fox]

yeye

[vapid shuttle]

so literally just choose any delta

WOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO

https://tenor.com/view/brilliant-smart-clever-impressive-patrick-j-adams-gif-11836232

https://tenor.com/view/suits-dance-dancing-jam-music-gif-5174541

https://tenor.com/view/catshakira-cat-dance-catdance-catdancing-gif-22503034

[worn fox]

im more confortable with that than the other way

feels better

[vapid shuttle]

Final draft

Are we good?

.close

[lone heart]

Channel closed

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vapid steppe]

is it true that if you get (nonzero)/0 when doing direct substitution into a limit, that the limit is one of the infinities?

[upbeat hornet]

what do you mean?

[vapid steppe]

like lets say you have $\lim_{x\to c}\frac{f(x)}{g(x)}$

[ocean seal]

Jash

[vapid steppe]

and you know f(c) does not equal 0

and g(c)=0

will the limit have to be either plus or minus infinity?

[mystic vortex]

Are you also implying that the limit at c of g is 0?

[vapid steppe]

yea but like just when you do direct substitution

[upbeat hornet]

you can only do that if the function is continuous

[vapid steppe]

im trying to see if this statement is always true: If $\lim_{x\to 0}{\frac{f(x)}{x}}=1$, then $f(0)=0$

[ocean seal]

Jash

[mystic vortex]

No, f could be piecewise

Do you have the condition that f is continuous?

[vapid steppe]

i know that f is defined for all real numbers

not necessarily continuous tho

[mystic vortex]

Wait lemme think

Right, so no

[vapid steppe]

is it true that If $\lim_{x\to 0}{\frac{f(x)}{x}}=1$, then $\lim_{x\to 0}{f(x)}=0$

[ocean seal]

Jash

[mystic vortex]

You could have f(x) = 0 for all x not equal to 0, f(x) = 1 for x = 0

Hmm

Pretty sure yeah

[vapid steppe]

but when you do direct substitution and get like 2/0 or something doesnt that mean its an asymptote

[mystic vortex]

Assuming the direct substitution was valid I believe so yeah

Oh

I think instead of an asymptote it could also be DNE

not sure actually

I'm probably not the person who should be helping you rn

[vapid steppe]

is lim as x->c of f(x)/g(x) always the same of lim of f / lim of g

even if the limit dne

[mystic vortex]

I recall there being some condition for that, lemme check

[upbeat hornet]

which limit dne

[mystic vortex]

The condition is that the limit of g(x) is not equal to 0

[vapid steppe]

orignal f/g

[near apex]

I don't think so. For example, that a question you just solved before this.

[vapid steppe]

side question does $\lim_{x\to 0}{\sin{\frac{\pi}{x}}}$ exist

[ocean seal]

Jash

[near apex]

No

[vapid steppe]

how do you know

is it cuz its not 0/0

since sin is continuous can you like bring the sin out of the limit

[near apex]

Because argument is either positive infinity or negative infinity and sine is an oscillatory function.

[vapid steppe]

and just evaluate limit of pi/x

[vapid steppe]

then take sin of this

[mystic vortex]

So that's sin of undef

[vapid steppe]

oh yea lim of pi/x doesnt exist on both sides of 0

[upbeat hornet]

no, it wiggles more and more as it approaches the origin

infinitely many times

[vapid steppe]

oh so dne

[upbeat hornet]

not when both sides don’t exist

[near apex]

I don't think that even that suffices. Maybe, i'm missing something.

[mystic vortex]

Oh it also gave the condition that both limits exist

Of f and g

[near apex]

Ah. Makes sense now.

I was getting stuck with infinity/infinity case.

[rocky loom]

could someone help with geometry?

[near apex]

Where individual limits don't exist but f/g does.

[rocky loom]

or is this not for geometry

[vapid steppe]

i just dont get why this is not always true

[near apex]

#❓how-to-get-help

[mystic vortex]

I think it also specified it being the limit approaching a real number

[vapid steppe]

cuz lets assume that f(0)=0, then u would have 0/0 implying to use lhopitals rule which then the limit exists

[mystic vortex]

No

[vapid steppe]

oh is it cuz 0/0 doesnt necessarily imply the limit exists

[mystic vortex]

F(0) is not the same as the limit approaching 0

[upbeat hornet]

not this channel, take an empty help channel like #help-13

[near apex]

No, but being 0/0, they can apply l'hopital. Right?

[vapid steppe]

yea

[rocky loom]

ye ty :)

[upbeat hornet]

#❓how-to-get-help describes some things you should know

[vapid steppe]

but i dont think that necessarily means the limit will exist at the end

[near apex]

Yes. So, what does that give you?

[mystic vortex]

But the limit might not be 0/0 just cause f(0) = 0

[vapid steppe]

so its not never true

[mystic vortex]

It is stated nowhere that f is continuous

[near apex]

Of course not. That'll depend on f'(x).

[mystic vortex]

I gtg

[vapid steppe]

but also like limit as x->0 of x/x^2 dne

even if its 0/0

[near apex]

Oh. Yes. Good point.

[vapid steppe]

chatgpt said consider f(x)=x+1 and evaluate lim as x to 0 of f(x)/x

the limit is 1

but f(0)=1

[vapid steppe]

wait this limit dne tho lol

[near apex]

Lol. Same thought.

[vapid steppe]

so like back to basics for a limit to exist that means that the left hand limit equals the right hand limit

[near apex]

Yes

[vapid steppe]

im trying to think of counterexamples

[near apex]

Well, tag me if you come up with something.

[upbeat hornet]

you always have to check for yourself if it is right

[vapid steppe]

it isnt lol cuz that limit dne

[vapid steppe]

how about like f(x)=(x^2+x)/x

wait nvm

[lone heart]

@vapid steppe Has your question been resolved?

[lone heart]

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• Show your work, and if possible, explain where you are stuck.
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[sand pendant]

If the second and fourth terms of a G.P. are 8 and 32 respectively, what is the sum of the first four terms?

[ivory fern]

!status

[lone heart]

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

[young finch]

what's a GP

[violet bear]

geometric progression

[young finch]

ah

i figured it was geometric something

[young finch]

@sand pendant

[lone heart]

@sand pendant Has your question been resolved?

[lone heart]

Channel closed

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[lone heart]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wary wyvern]

hy

help mee

[mental flame]

This channel is also occupied scooby

Read #❓how-to-get-help for directions on opening your own

[wary wyvern]

ohh what i need to know answers?

plss im new to b dicord

should u help mee

[left isle]

read this

click on the link and read it

[mental flame]

Nobody is going to help you until you create your own Channel

[wary wyvern]

ahh

so i need to quit

[young finch]

assuming F(x) is the anti derivative

[wary wyvern]

?

[young finch]

so that means F'(x)=f(x)

so whats your issue

just a property of derivatives

f'(x)=0 for certain x values where f(x) has extrema

not really backwards

[placid zinc]

When the function hits a minimum, the slope is 0 there. You can see that as a root of the derivative

[young finch]

if you just shift it over to make F(x) f(x), then f(x) is f'(x)

.close

[lone heart]

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[devout hare]

It is known that a conic section is the geometric locus of points in a plane α for which the ratio between the distances to a fixed point F (focus) and a fixed line d (directrix) is equal to a non-negative constant e (eccentricity of the conic). This concept extends to the definitions of ellipse, hyperbola, and parabola, with the first two being constructed from a pair of perpendicular directrices to the support of the major axis (or real axis in the case of the hyperbola, 2a units of length), located at a distance of a/e units from the center of the curve, where "e" is the eccentricity. In this sense, how is it possible to deduce that the eccentricity of both an ellipse and a hyperbola is necessarily c/a, where c is the focal distance and a half of its major/real axis length, from its definition?

[devout hare]

I don't know where to begin

[lone heart]

@devout hare Has your question been resolved?

[devout hare]

<@&286206848099549185>

[lone heart]

@devout hare Has your question been resolved?

[devout hare]

.close

[lone heart]

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[oak perch]

No

$f(x)^{2}+3f(x)=g(f(x))=x^{2}-3x$ therefore $|f(x)+\frac{3}{2}|=|x-\frac{3}{2}|$

[ocean seal]

Cogwheels of the mind

[oak perch]

So it’s definitely not unique. Infinitely many. If you require f to be continuous then 4 many, if you require f to be differentiable then 2 many.

[deep quartz]

ah I see, that makes a lot more sense, thank you for your help

[oak perch]

Np

[deep quartz]

.close

[lone heart]

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[flat roost]

Is it possible to find the area bounded by the lines y = x+3 and y = 6-2x without looking at the graphs with integration?

[upbeat hornet]

from where to where?

[flat roost]

"and the x-axis"

[west girder]

Yes

[young finch]

set the equations equal to each other and solve

that'll give left bound and I think right bound will be obvious

[upbeat hornet]

looking at the graphs helps intuition

[flat roost]

do i add both equations up when integrating

[flat roost]

yep the graph gives a triangle

[young finch]

in this case you have to split it up into two integrals

or put them in terms of y

[tardy stag]

tbh if i were doing this integral I'd probably do a very rough sketch first

[flat roost]

wait what if

i find the intersection point

of the two lines

and then integrate from left bound to that point, then that point to right bound

[young finch]

thats what I said by splitting it up into two integrals

[flat roost]

i would too but i dont have a graphing calculator yet

[flat roost]

ohh

[tardy stag]

yeah i mean by hand

[young finch]

you don't need a graphing calculator to sketch two linear lines

[upbeat hornet]

the lines are affine

[cedar juniper]

The equations are for straight lines, you dont need to integrate

[flat roost]

yeah that works

[young finch]

what is affine

[upbeat hornet]

linear means f(x + y) = f(x) + f(y)

[upbeat hornet]

But it can be done by integration

[flat roost]

if the equation is y = x+ constant its linear right

straight line

[upbeat hornet]

Ignore this, it isn’t relevant to the question

[upbeat hornet]

for your purposes this is right

[flat roost]

i see

thanks guys\

.close

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[west girder]

If we have an infinite double cone and a plane, how do we know the cross section is a parabola or hyperbola?

[west girder]

We can easily see if the cross section is an ellipse, but a parabola and hyperbola share the same features in the scenario of an infinite double cone

[oak perch]

By comparing two angles

[west girder]

Which?

[wind cloak]

if you tilt it hard enough it's a hyperbola

hyperbolas have two branches so you can figure it out that way

[oak perch]

α<β ellipsis, α=β parabola, α>β hyperbola

[west girder]

Is the big X the double cone?

[oak perch]

Yes, a slice of it

[west girder]

Oh I see

That’s very cool

[oak perch]

It is

Ancient Greek people discovered it I think

[west girder]

Is there a name for this fact so I can look into it?

[oak perch]

Can’t help with that. I only started reading math in English since undergraduate level.

[west girder]

It’s alright I’ll probably be able to find if I look hard enough

Thanks!!

[oak perch]

So I don’t know what it’s called in English. And I don’t even remember this has a name.

Np

[wind cloak]

Just look for "geometric visualization of conic sections" or something like that

[west girder]

Ok thanks!

.close

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[west girder]

Oh I just realized, the parabola won’t look the same

It just clicked that the plane is parallel to the edge of the double cone in a parabola

The fact is intuitive

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[spare fern]

How would I differentiate this?

[spare fern]

$$-6xe^{-2x}$$

[ocean seal]

Lex1729

[flat vale]

product rule

[hushed locust]

product rule + chain rule

[spare fern]

so is u and v both negative?

for example is it u = -6x and v = e^-2x

or u = -6x and v = -e^-2x

[hushed locust]

pick one to be negative

[spare fern]

alright so u = -6x and v = e^-2x

so u' = -6 and v'= -2e^-2x

is this correct?

[hushed locust]

so far yes

[spare fern]

hold on I'll send the full question

Here is my working

Just not sure where I messed up haha

[spark sky]

Shouldn't this be -12e^-2x?

[spare fern]

but it is -6x * -2e^-2x

which is positive

[spark sky]

no i meant the first part, it should be -12e^-2x + 12xe^-2x
New to differential equations, so I might be pointing at the wrong thing.

[spare fern]

I don't understand

are you talking about the part of the screenshot you sent me?

[spark sky]

yeah

[spare fern]

but the original gradient function is -6xe^-2x -e^-2x

we would need to use the product rule to find the second derivative of that

[spark sky]

Sorry I don't think I qualify helping you with this since I am literally new to these concepts

[flat roost]

@spare fern

the second line here is an extension of the first, right?

[spare fern]

yes

[flat roost]

then it is wrong

[spare fern]

Oh I see now

[flat roost]

it should be -6xe^2x - e^-2x

if you take out the e^-2x, you should get e^-2x (-2(3x+2)+3)

simplified to e^-2x(-6x-1)

[spare fern]

Thanks for pointing that out I'll redo the problem again from the start

[flat roost]

you are welcome, good luck

[bold magnet]

solve for values of x in range 0<=x<=360 Q.3sec.sqr.x-2tanx=8

[flat roost]

can you express it better

thanks

[bold magnet]

how can i type square in keyboard

[flat roost]

square root?

[bold magnet]

neh a square like 2sqr

[flat roost]

huh

[bold magnet]

power

[flat roost]

^2?

OH

do you mean

sec^2x-2tanx =8

[bold magnet]

yeh

tnks

yes

[flat roost]

ok here is what i thought of

sec^2x - 2tanx = 8

use the 1+tan^2x=sec^2x identity for the sec^2x

1+tan^2x-2tanx = 8

tan^2x-2tanx-7=0

now we have a quadratic

[bold magnet]

yeah I got it thnks a lott

[flat roost]

👍

[lone heart]

@spare fern Has your question been resolved?

[lone heart]

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[steel olive]

What's wrong in my soln

I'm getting 5

Correct answer is 17

[oak perch]

You can order numbers of this form like this:

(a,b,c): a>=b>=c
(a,b,c)<(a’,b’,c’) iff a<a’, or a=a’, b<b’, or a=a’,b=b’, c<c’

So there are (a+1)(a+2)/2 many of them of the form (a, - , -)

[steel olive]

oh ya got it i didnt count all cases

[oak perch]

Don’t close it yet

[steel olive]

yea,so basically like the numbers wherw a = 0,b = 0 and c=1 and their permutations,they will be counted as a single integer

right?

and they are asking for 101st smallest integer

[oak perch]

Yeah. Next calculate maximum $m$ such that $\sum_{a=0}^{m}\frac{(a+1)(a+2)}{2} \leq 99$, so $\frac{m^{3}}{6}+m^{2}+\frac{11m}{6}+1 \leq 99$

[ocean seal]

Cogwheels of the mind

[oak perch]

m=6

So the 86 th minimal number is (7,0,0)

There are b+1 many numbers of the form (7, b, -)

Similar method or direct calculation

The 101 th minimal number is

(7,5,0)

=2431

Then

Done, 17 indeed

[oak perch]

Here I mean find maximum k such that Σ(b+1): 0<=b<=k =(k^2)/2+(3k/2)+1<=15, k=4 so the 101th number is (7,5,0)

[steel olive]

could u explain this part once again

[oak perch]

b+1 many of the form (a,b,-) right?

From (a,b,0) to (a,b,b)

So how many numbers are of the form (a, -, -)? Sum of numbers of numbers of the form (a, b, -): b from 0 to a, so Σ(b+1): b from 0 to a

(a+1)(a+2)/2 many

BTW, at the beginning, I define order in that form because a<a’, we can have (a,b,c)<(a’,b’,c’) because divide by 3^a on both sides, we have (a,b,c)/3^a= 1+3^(b-a)+3^(c-a)<=3<3+3^(b’-a)+3^(c’-a)<=(a’,b’,c’)/3^a

The rest is similar

Got it?

[lone heart]

@steel olive Has your question been resolved?

[steel olive]

ya got it thanks a ot

loy

lot

.close

[lone heart]

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[oak perch]

Okay , np

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[sick kayak]

Can anyone provide and explain the solution for this

[tardy tapir]

,rotate

[oak perch]

For n large enough

2^(1/2)-2^(1/(2n+1)) converges to sqrt(2)-1

Which is smaller than 1

So it will converge to 0

[sick kayak]

Ooh

Yes yes thank you

[oak perch]

Np

[wind cloak]

,w solve product from k = 1 to infinity 2^(1/2) - 2^(1/(2k + 1))

[ocean seal]

Results provided by WolframAlpha

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Requested by neonperseus

[wind cloak]

seems legit

[oak perch]

Of course it is, it is actually
0<=a_n<=a_N (sqrt(2)-1)^(n-N)

For some N

[sick kayak]

.close

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[formal rune]

Help factorise g

[vale wigeon]

what's troubling you?

[formal rune]

How to break brachet

Exponent is throwing me off, has been a while bc we don’t do easy math at higher

[azure tapir]

well

[vale wigeon]

you don't want to expand here just yet.

[azure tapir]

is your question $5(x+1)^2-4(x+1)^3

$5(x+1)^2-4(x+1)^3$

[ocean seal]

Sunset

[formal rune]

Yes

[azure tapir]

do you know the

bodmas rule

?

[vale wigeon]

@azure tapir are you about to suggest that op expand this entire thing?

cause thats gonna be both painful and obstructive

[azure tapir]

no

just this

$5(x^2+1)-4(x^3+1)$

[ocean seal]

Sunset

[vale wigeon]

this is incorrect lmfao

(x+1)^2 ≠ x^2 + 1

and (x+1)^3 ≠ x^3 + 1

[formal rune]

Yes

[azure tapir]

peter dont lisen to me

[formal rune]

Ok

[vale wigeon]

anyway...

[formal rune]

Help

[vale wigeon]

for factorization you will want to pull out (x+1)^2 from both terms here.

[formal rune]

How

[vale wigeon]

you know the distributive law, yes?

[formal rune]

Yes

[vale wigeon]

and you also know (x+1)^3 can be written as (x+1) * (x+1)^2, yes?

[formal rune]

Yes

[vale wigeon]

$5 \cdot (x+1)^2 - 4(x+1) \cdot (x+1)^2 = [5 - 4(x+1)] \cdot (x+1)^2$

[ocean seal]

Ann

[formal rune]

Can you dumb it down pls

[ocean seal]

Sunset

[formal rune]

More step

[vale wigeon]

there was never any 5(x^2+1) !!!!

[azure tapir]

no

it was a miss splet

[vale wigeon]

ok but why did your latex end up here

[azure tapir]

spelt

[vale wigeon]

you said you know the distributive law.

can you state the distributive law for me?

[formal rune]

Is like

A(b + c) = ab + ac

Right

[vale wigeon]

no capitalization on the a

but yes

now

this law goes BOTH ways.

so from ab + ac you can get a(b+c)

this is called "factoring out a" or "pulling out a"

do you understand this yes or no

[formal rune]

Yeah

[vale wigeon]

yeah so that is what i am doing here

[formal rune]

Like tht

[formal rune]

Where

[vale wigeon]

-4(x+1) isn't equal to (-4x+1)

and i am not even touching that anyway

[formal rune]

Oh

[vale wigeon]

$5 \cdot (x+1)^2 - 4(x+1) \cdot (x+1)^2 = [5 - 4(x+1)] \cdot (x+1)^2$

[ocean seal]

Ann

[vale wigeon]

i am factoring out (x+1)^2

despite ie being written with more than one letter it is still a thing that in principle can get factored out

[formal rune]

Where did the five go

[vale wigeon]

it didn't go anywhere...

[formal rune]

G is my end goal btw

Why is this not clicking for me 😭

[vale wigeon]

i really don't know how to 'dumb it down' even further

[formal rune]

Can you do it step by step

[vale wigeon]

i am trying to show you one step

this is one step

[formal rune]

Oh

[vale wigeon]

it cannot be broken down any further

[formal rune]

So the textbook is wrong ?

[vale wigeon]

no, the textbook is not wrong.

we just won't be finished after this one step.

[formal rune]

But if it can’t be broken down any further

Oh

[vale wigeon]

THIS ONE STEP can't be broken into smaller sub-steps.

[formal rune]

I’m confused, so this step can’t be used to get the answe in the textbook?

[vale wigeon]

qe4rjgjfsdrnhqearjhsrth

this is one step. towards the solution.

[formal rune]

Oh 😭

[vale wigeon]

i didn't show you the steps that FOLLOW it.

because you're saying that this one step by itself confuses you.

so we can't move on yet.

[formal rune]

I’m confused on how this becomes the second thing

[vale wigeon]

it's the distributive law!!!

ac - bc = (a-b)*c

[formal rune]

What do I need to learn to make me not get confused

Oh

Let me process this

So in this scenario

C = (x+1)^2

?

[vale wigeon]

yes

lowercase c but yes

[formal rune]

But what about the other (x+1)

[vale wigeon]

wym

for the purposes of our step it can just stay as is

we'll handle it in the next step

[formal rune]

Got it thanks

This was correct, right ? @vale wigeon

[vale wigeon]

no

you didn't factor it correctly

i showed you exactly how to do it and you messed it up

[formal rune]

😭

Which line

Wait

I see

So far so good ? @vale wigeon

[vale wigeon]

no

b is 4(x+1) not just 4.

[formal rune]

Oh

Ohh

@vale wigeon like that

[vale wigeon]

yes now you've reproduced my thing

[formal rune]

Yessss

This is right ?

[vale wigeon]

yes now it is

[formal rune]

Yay thx

[olive ice]

no hate man but your paper looks like someone shot it with a sawed-off shotgun with all the therefore symbols 🤣

[formal rune]

😭

[lone heart]

@formal rune Has your question been resolved?

[lone heart]

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[severe stump]

Can someone explain how the distributive property works and how can I apply in math problems

[desert elm]

couuld somebody possibly explain and help me for a maths problem?

[vale wigeon]

@desert elm the other person got here first -- please get your own channel. #❓how-to-get-help for instructions.

@severe stump
a * (b + c) = a*b + a*c

the applications of the distributive property are very broad and numerous so it's impossible to just summarize it all in one discord chat

but applying the distributive property is often called "expanding" or "factoring" depending on which direction you go

[lone heart]

@severe stump Has your question been resolved?

[severe stump]

how can I apply it to my bookwork?
It says here
X=x+1

How am I supposed to get the value of x

[vale wigeon]

eh?

wait what

[severe stump]

Wait

[vale wigeon]

are capital X and lowercase x meant to be the same

maybe send a picture of you're in doubt

[severe stump]

2y^2-3y+4

1. y=y+1

Sadly I didn't get a picture of it

[vale wigeon]

ok now i'm even more confused at what you're supposed to do.

are you able to send a picture?

[severe stump]

I need to simplify it

[severe stump]

I just remembered what my teacher says

😔

[vale wigeon]

do you have perfect memory

[severe stump]

But is it possible to apply the distributive property here?

[vale wigeon]

i think you don't

right now it's neither possible nor impossible bc we don't have a clearly written problem statement

[severe stump]

Given equation:
2y^2-3y+4

question:

1. y=y+1

[severe stump]

Thank you for your time though, I appreciate it

.close

[lone heart]

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[lone heart]

@quasi trail Has your question been resolved?

[lone heart]

@quasi trail Has your question been resolved?

[azure tapir]

tip

make your paper clear and readable

[lone heart]

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[vital wyvern]

ever considered using an equals sign

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[alpine sable]

I have this graph with the above and I'm tasked to find the range of a in terms of p

[alpine sable]

where do I start?

what does f(x) and g(x) even have in common?

and what does a, b, c correspond to in a graph?

<@&286206848099549185>

[echo socket]

The sign of a dictates which way the parabola will be directed and the absolute value of a determines how wide/thin the parabola will look like

[alpine sable]

i see

[echo socket]

If it's positive, the parabola goes up, and if it's negative, it goes down

[alpine sable]

so p here will definitely be smaller than a

and p will be positive

or not

[echo socket]

That would be correct if the greater values or a corresponded with wider parabolas

But that is not the case

The greater the absolute value of a, the thinner the parabola is

[alpine sable]

ohhh I see

[echo socket]

So here we deduce that a is positive and less than p

[alpine sable]

is there an explanation to why is that the case?

[echo socket]

Greater leading coefficients make the polynomials increase/decrease faster

So the slopes tangent to the graph become more vertical

Hence it looks thinner

[alpine sable]

ahh

makes sense

let me write this down

[echo socket]

The final bit would be to write this as a double inequality

[alpine sable]

right

now what does b and c here represent?

[echo socket]

c is basically the elevation, you will see the graph be lifted as you increase the value of c

And b is kind of hard to explain in this case, I think it would be a bit more clear if you played around with its value in desmos

[alpine sable]

i see

alright

thanks a lot

.close

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[placid spire]

laplace for this one?

this is the solution

I understand the top part

but not the bottom

They used this formula sheet:

number 3 and 15

15 is used on the bottom part

But i dont see how they can split it up in two different ones?

number 15 I dont understand tho

it says e^-at * f(t)

what is f(t)?

I assume f(t) is t^2 / 2

[uncut torrent]

yeah

f(t)=t^2/2

so in the form e^(-at)*f(t), a=2 right

they're using formula 15 because they found the laplace transform for f(t)

[placid spire]

I see

what about this one

they use 22 and 5

I dont understand when they used 22

They got that L{te^-t}

how did they get that?

that should be the F(s) from 22, but where does it come from

we have that e^-(t-3) but that isn't te^-t?

[tacit arch]

Do you know what u(t-d) is

[placid spire]

not sure

[tacit arch]

It's defined in your book

Look for it

[lone heart]

@placid spire Has your question been resolved?

[placid spire]

it doesnt say what u(t-d) is

I see second shift theorem tho

but that doesnt explain how e^-(t-e) turns into te^-t

[lone heart]

@placid spire Has your question been resolved?

[lone heart]

@placid spire Has your question been resolved?

[tacit arch]

this should explain what u(t) is

[tacit arch]

do you at least know what u(t) is?

u(t - d) is a composition of functions

i recommend reviewing those

,tex .transformation rules

[ocean seal]

riemann

[hazy torrent]

hi

how to solve 10-5

[tacit arch]

!help

[lone heart]

Please read #❓how-to-get-help

[lone heart]

@placid spire Has your question been resolved?

[lone heart]

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[young finch]

im a bit confused on how unfixed eigenvectors can be, when computing them im seeing that it can go one of two ways, setting one variable as the free variable and writing the eigenvector based off that, or setting the other variable as the free variable

[young finch]

i feel like my question does not make sense

like here, couldnt it [1,-1,0] or [-1,1,0]?

<@&286206848099549185>

[pallid scarab]

The first vector is -1* the second one so they're linearly dependent

[young finch]

right

[vapid shuttle]

it doesn't, what is your question?

[young finch]

but i still dont understand any logical reasoning to make it [-1,1,0] instead of [1,-1,0]

[young finch]

why [-1,1,0] instead of [1,-1,0] for eigenvectors

[pallid scarab]

Because we just chose it

[vapid shuttle]

show original matrix

[pallid scarab]

Both are valid eigenvectors

[pallid scarab]

Here is original matrix on the left

[young finch]

so there are multiple eigenbasis that are all the same thing?

[vapid shuttle]

those are not eigenvectors of that matrix

those are vectors in the nullspace of that matrix

[pallid scarab]

Which are eigenvectors for 0

[vapid shuttle]

no

[pallid scarab]

AX = 0 = 0X

[vapid shuttle]

yes

hah

but anyways he wrote lambda=-2 so that isn't what he meant

[vapid shuttle]

input <1,-1,0> into this

[pallid scarab]

Right :)

[vapid shuttle]

Ax will be <2,-4,0>

which is not -2 * x

however it is 2 * x

so the difference is a change in sign

input <-1,1,0> instead

[pallid scarab]

Uh no

[vapid shuttle]

Ax will get <-2,4,0>

oh shit

yeah

I was thinking about only the first row

my bad idk what I'm doing

[young finch]

[pallid scarab]

[1,-1,0] and [-1,1,0] are both valid eigenvectors for lambda = -2

But since they're linearly dependent

You can only choose one of them for your eigenbasis

They chose [-1,1,0] in the solution

You're equally valid to choose [1,-1,0] instead

[young finch]

okay

is this the same for eigenvalues with a multiplicity of 2?

when solving for htem

[vapid shuttle]

sometimes you can choose 2 independent ones in that case

sometimes you can't

depends on if the eigenvalues are degenerate

[young finch]

im a degenerate

[pallid scarab]

Honestly same

[young finch]

okay so if i can write it as x1=cx2 then it can also be x_2=cx_1

but theres certain obvious cases where it has a set eigenvector for the eigenvalue

okay i think i get it

thank you!

.close

[lone heart]

Channel closed

Closed by @young finch

Use .reopen if this was a mistake.

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[proven haven]

dont mind that i put quadratic

alg 2

[vapid shuttle]

I mind

[proven haven]

it was accident

lol

[vapid shuttle]

are you sorry

[proven haven]

yes

so how do you solve this

[grave quartz]

plot the points lol

I forgot to say the last time

[proven haven]

how to plot points

[grave quartz]

it willl be obvious

well y = f(x) here

[proven haven]

oh

okay

[grave quartz]

so think of the right column as y values and the left is x

[proven haven]

okok