help-27

[cedar crane]

mmmm

[meager yarrow]

Remember that you just need to find secx

[cedar crane]

yeah

do you just bring the secant over?

oh but that would make it a negative

idk

[meager yarrow]

You should try to turn tan^2(x) into sec^2(x)

[cedar crane]

uhhhh

do you just bring the tangent over to the secant?

or is there some factoring stuff cause I'll be honest I don't remember anything about factoring except that you can eliminate things in common

[meager yarrow]

Actually you can try adding 1 to both sides of the equation

[cedar crane]

I thought that was a chemistry thing

[meager yarrow]

x=y
x+1=y+1

are they the same thing

[cedar crane]

I'd say no

idk about algebraically but at least graphed those are different right?

[meager yarrow]

Uhh maybe that’s no good example

x+1-1=x correct?

[cedar crane]

yeah

[meager yarrow]

Add 1 to both sides
And it’s correct right?

[cedar crane]

like as in would it still = to x?

[meager yarrow]

Are both sides still equal

[cedar crane]

I suppose

should I be writing this on paper

oh

yeah

it is equal

[meager yarrow]

Adding and subtracting numbers to both sides of the equation
It is still true

Back to the question:

$\tan^2{\theta} = 3$

[woven radish]

Afi

[cedar crane]

okay

uhh

[meager yarrow]

What happens when we add 1 to both sides

[cedar crane]

equation would still be unbalanced right

1+tan^2 = 4?

[meager yarrow]

yes

[cedar crane]

yeah

so it'll still be unbalanced as it is right now

[meager yarrow]

$sec^2\theta = 1 + tan^2\theta$

[woven radish]

Afi

[cedar crane]

if you add to something you have to add to both sides?

1+sec^2=2+tan^2

oh wait

now are you allowed to subtract?

it'll be negative though

uh

[meager yarrow]

$sec^2\theta = 1 + tan^2\theta$

$1+tan^2\theta = 4$

[woven radish]

Afi

[meager yarrow]

What conclusion can you make from these?

[cedar crane]

uhhh

idk but they look incomplete to me

[meager yarrow]

Remember you just want to find the value of secx

Try combining the two equations

[cedar crane]

wdym combining

oh like

taking a variable from one side to the other?

[meager yarrow]

$sec^2\theta = 1 + tan^2\theta = 4$

[woven radish]

Afi

[meager yarrow]

We have

$\sec^2\theta = 4$

Correct?

[woven radish]

Afi

[cedar crane]

uhhh

I think so

where did the tangent go though

[meager yarrow]

We can just ignore the tangent

[cedar crane]

mmm okay

[meager yarrow]

Since sec^2 x = 1 + tan^2 x

[cedar crane]

okay

[meager yarrow]

1+tan^2 = 4?

We essentially replace the whole left hand side with sec^2

[cedar crane]

uh

[meager yarrow]

Ok I wanna ask which part do you not understand

[cedar crane]

like a lot of it honestly

Feel like the numbers are jumping from one thing to another

I don't understand how it went from sec^2 x = 1+tan^2 to 1+tan^2=4

since it seems like the right went to the left and sec^2 replaced by a 4

[meager yarrow]

Lemme ask you

If x=y and y=z
Is x=z?

[cedar crane]

er

I'm not a mathematician so I don't really know but assuming that by = you mean they're the same functions then yeah

they're just variables

[meager yarrow]

This applies to your case
sec^2 = 1 + tan^2 and 1 + tan^2 = 4
Then we have sec^2 = 4

x is sec^2
y is 1+tan^2
z is 4

[cedar crane]

okay let me think

oh I see what you mean

and we have 2 y values?

[meager yarrow]

Wdym by y values

[cedar crane]

1+tan^2

oh wait did you mean that like as an equation or something

[meager yarrow]

Yes

[cedar crane]

sec^2=1 + tan^2

ah

ok

[meager yarrow]

Sorry for the ambiguity

[cedar crane]

no problem

[meager yarrow]

Any questions?

[cedar crane]

nope I think I understand it a lot better now but I think i'd still like to polish it a little bit by watching some videos on it

what would you call what we just did

?

[meager yarrow]

Transitivity of equality

[cedar crane]

Ah

[cedar crane]

oh yeah it's been a while since i've done anything like that

but thanks

[meager yarrow]

Ok so we have $\sec^2\theta = 4$

[woven radish]

Afi

[meager yarrow]

Solve for secant

[cedar crane]

alright

sec^2-tan^2=1 is what I need right?

wait hold up

[cedar crane]

do you want to find this

or this

sec^2-tan^2=1 and tan^2=3 so isn't that really just 3-1?

[meager yarrow]

What we have done:

tan^2x=3

We add 1 to both sides of the equation and get

1+tan^2x = 4

Rewrite 1+tan^2x as sec^2x (since 1+tan^2x = sec^2x)

sec^2x = 4

[cedar crane]

ohhh I see

I fully understand the first 3 lines

trying to understand the 4th rn

[cedar crane]

is the 4th line plugging in an equation

[meager yarrow]

Correct

[cedar crane]

and what equation is that

sec^2-tan^2?

[meager yarrow]

This identity

$sec^2\theta = 1 + tan^2\theta $

[fossil locust]

Divide the Pythagorean identity by cos^2 to see it

[cedar crane]

sec^2(theta) = 1+tan^2(theta) ?

[meager yarrow]

Yes

[cedar crane]

alright

[cedar crane]

kinda dumb question but where did the secant thing originally come in from cause i think that's the thing causing a lot of confusion

[meager yarrow]

Try to do this

[cedar crane]

divide the formula or what we have right now

[meager yarrow]

This identity:
Sin^2x + Cos^2x =1

[cedar crane]

oh I see

[meager yarrow]

Remind you that secx is 1/cosx

[cedar crane]

would the new cos^2 come in from the side with the = 1?

or do we use the cos^2 we have

and divide the 1 by it

oh but then it's squared

so don't we have to convert it back to cosx

[cedar crane]

ohhh

I see

okay

1/cos^2 turns into secant and sin^2/cos^ turns into tangent

cos^2/cos^2 turns into 1 because they're like terms right?

idk how this ends out but does secant subtract tangent to get 1

I'm new to the the trig function things so I don't really know

tan^2-secant^2+1

wait no

sec^2-tan^2

?

since that is the formula from earlier

[meager yarrow]

sec^2-tan^2=1 is true

[cedar crane]

idk how dependable it is

[cedar crane]

okay neat

and uhh

2=1 so

wait

[meager yarrow]

?

[cedar crane]

wdym

[cedar crane]

the secant goes to the other side right

with the tangent

does it turn into a 1?

[meager yarrow]

Wdym by that

[cedar crane]

the formula i knew prior to this

[meager yarrow]

State the formula

[cedar crane]

sec^2-tan^2=1

or did i get it wrong

[meager yarrow]

No no this is true

[cedar crane]

oh okay

can it be used in this situation though?

[meager yarrow]

You don’t need to

[cedar crane]

Mmm

[meager yarrow]

tan^2x=3

We add 1 to both sides of the equation and get

1+tan^2x = 4

Rewrite 1+tan^2x as sec^2x (since 1+tan^2x = sec^2x)

sec^2x = 4

We just replace 1+tan^2 with sec^2

[cedar crane]

oh I see

[meager yarrow]

How would you solve for secant after these steps?

[cedar crane]

uhhh

square root both?

[meager yarrow]

Write it down

Don’t forget the negative value

[cedar crane]

what negative value?

[meager yarrow]

Given x^2=4

What are the values of x?

[cedar crane]

oh

uhhh

does it have to do something with the secant?

[meager yarrow]

Nope it’s just all the possible values that satisfy the equation

Since if we call secx = -2
Then sec^2x is still 4

[cedar crane]

oh I see

[meager yarrow]

So -2 is a possible value for secx

Now what is the other value for secx

[cedar crane]

not sure but I'm pretty sure it's only asking for one value which is 2

let me see

oh

yeah

[meager yarrow]

Hmm

anymore questions?

[cedar crane]

nope but do you mind writing a script for all of this so I can use this in reference I understood it but I think I just need practice on it

I don't want to miss something then screw it up because of that lol

[meager yarrow]

Perhaps we can do that in DM
Do you mind?

[cedar crane]

oh yeah

sure

tell me when to close this channel in dms too

[devout snow]

@cedar crane Has your question been resolved?

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[balmy root]

ok so i understand 15. and that it's false

because for example, sin (390 degrees) is positive and half of it is 195 degreees

and sin(195 degrees) is negative

right

but for 16., the answer is also false

but im not able to find an example like that

hmm

[vague zealot]

Take anything between 90 to 180 degrees

[balmy root]

one sec lemme try

lets do cos(120)

which is equal to -1/2

and then lets now do

half of it

cos(60)

1/2

oh i see lol

that was an easy proof

[balmy root]

thank you!

.close

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[pulsar relic]

i'm stuck trying to find the points of inflection

[pulsar relic]

because i run into a divide by zero when plugging in the interval values into f(x). (they are asymptotes) how do i show my work in finding points of inflection? if there even are any? can there just be none?

[devout snow]

@pulsar relic Has your question been resolved?

[pulsar sand]

You need to look at what your definition of an inflection point is. Usually you only call a point (c,f(c)) an inflection point if f(x) is continuous at c. I don't know if any books do it any other way.

If you have a discontinuity like at -2, then even though it may change concavity, it won't be an inflection point.

[pulsar relic]

inflection point is when the concavity changes direction (?). what you said sounds more right though

[pulsar sand]

That's not a mathematical definition

[pulsar relic]

i havent taken any classes on proof-ing things i just regurgitate things i read in the course material 😭 and this wasnt really explained in words or examples

[pulsar sand]

It probably was and you ignored it

[pulsar relic]

thanks for the help

still need someone to explain it to me though how to answer the problem

[vivid estuary]

I believe an inflection point is when the second derivative is equal to 0 or undefined and it changes sign or the function changes concavity

[pulsar relic]

i've calculated the undefined intervals, and the concavity changes between those intervals but there isnt exactly a point of inflection for this function. I dont know how to show that though

[vivid estuary]

Hmm I’m not sure

Oh

[vivid estuary]

I did some research and I think I got the answer now

For an inflection point to exist the original function must be continuous

At x = 2 and -2 the original function is not continuous

Therefore it has no inflection points I believe

[pulsar relic]

alright, i'll use this explanation, confirmed by 2 people makes me feel more confident about it 😅

[vivid estuary]

Here is where I looked at

https://socratic.org/questions/can-a-point-of-inflection-be-undefined#:~:text=Explanation%3A,which the function is undefined.

[pulsar relic]

i kinda get it more now with that extra graph example, ty 🙂

.close

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[unreal stone]

🔥🔥

😅😅

3

Np 😅😅

[dusky drift]

<@&268886789983436800>

[polar flower]

it would be better if you post relevant questions.

[dusky drift]

He's trolling

[devout snow]

@odd mantle Has your question been resolved?

[dusky drift]

<@&268886789983436800> ??

[winter torrent]

2+2 = 4 day mute, please do not troll in help channels

they already left I hate it here

[devout snow]

Channel closed

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[celest dock]

(banned)

[devout snow]

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[onyx helm]

Hi, i want to reciprocate this function:

Sum( 1 to m-1 ) + n + 1 with m>n

if necessary i can prove that this function is bijective.
i tried to expand it, and work on it like a 2nd degree polynom but i have no idea how to use the inegality in it.

[stone stump]

what do you mean with reciprocate? also can you post a picture of the function instead?

[onyx helm]

be reciprocate i mean f^-1(x)

with m>n

[onyx helm]

i want to found a function that can found m and n from an x

[devout snow]

@onyx helm Has your question been resolved?

[onyx helm]

<@&286206848099549185>

[devout snow]

@onyx helm Has your question been resolved?

[finite briar]

so if i get u correctly

u want a function f(x)

that gives u m and n

[finite briar]

such that its of this form?

[onyx helm]

yes

exactly

[finite briar]

well first of all

u need to find a transformation (function) that gives two unique m and n (integers ig) for some x

[finite briar]

we can then put that in this to get the function u want

do u understand that?

[onyx helm]

nop

[finite briar]

...

well u want to get 2 numbers from 1 number

and u need to do that in a unique way

because else it wont work for all m and n

[finite briar]

thats called the inverse

[onyx helm]

yes but my function is bijective

[onyx helm]

okay i think i get it !

[finite briar]

good

[finite briar]

that is the sum

[finite briar]

u need a bijective function that takes 2 numbers and spits out 1 number

does that make sense?

[onyx helm]

no i'm confuse

my function is a bijection that spits 1 number from 2

if i understant you tell me to fount a g(x) so i can have g(f(x))
that take 1 and spits 1 ?

[finite briar]

oh no no

rn ur writing

$f(x) = \left(\sum_{k=1}^{m-1} k\right) + n + 1$

[woven radish]

カナヴ

[finite briar]

correct?

[onyx helm]

yes

[finite briar]

but u arent providing m or n

so u need a function

that takes x

and gives u m and n uniquely

[onyx helm]

it's for every m and n in N, m > n

[onyx helm]

so a function that do that ?

[finite briar]

thats just the domain

[onyx helm]

but how i do a function that have 2 out ?

[finite briar]

inverse of a 2-in 1-out function using vectors

what do u even need this for tho

[onyx helm]

for a program i'm doing

[finite briar]

if u can provide me more context, i might be able to help find a better solution

[devout snow]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[onyx helm]

yes, my "proof" of bijectivity tell the context:

Consider two list the first one is construct by associatting Letter to the natural number (the alphabet is for the exemple, but consider it infinite) so you associate A to 0, B to 1, C to 2 and so on...
The second is construct associatting every couple of letter to a natural.
we procede be expend the first list progresively
so when there is only A in the first list we can't form a list so the neutral élément is associated to 0
(that first part is not nessesary but it's most convenient)
BxA -> 1
CxA -> 2
CxB -> 3
DxA -> 4
DxB -> 5
and so on...

we took two letter from the first list and we assosiate m to the max value and n to the min value.
and we found that the function Sum( 1 to m-1) + n + 1, tell the position of the couple in the second list.

[finite briar]

lemme read thru tbat

[onyx helm]

np

[finite briar]

couple as in pair?

[onyx helm]

yep

i'm not fluent english i might don't use the right word

[finite briar]

wait why does it need to be infinite?

wahts ur native language then?

[onyx helm]

French

[onyx helm]

it's for pair a list of element

[finite briar]

kk

[onyx helm]

(i'm looking at Vector-valued function, my solution might be here i guess)

[onyx helm]

ah no mb
f(m,n)

[finite briar]

BRUH 💀

[onyx helm]

the inverse i search is a g(x) so

[devout snow]

@onyx helm Has your question been resolved?

[devout snow]

@onyx helm Has your question been resolved?

[finite briar]

yeah

well not me this person

but i suppose it isnt needede now

[spark sleet]

@onyx helm Salut, tu as encore besoin d'aide ?

[onyx helm]

salut, oui je veux bien

[spark sleet]

Ok, peux-tu expliquer en français ce que tu recherches plus exactement? Ça me paraissait flou

Tu cherches la réciproque d'une fonction?

[onyx helm]

oui

[spark sleet]

Tu peux écrire clairement sa définition (en incluant les arguments du domaine d'abord)?

Par exemple, f(m,n) si j'ai bien compris ?

[spark sleet]

Ok, et donc étant donné un nombre naturel x, tu cherches un couple (m,n), avec m>n, qui fera en sorte que f(m,n) = x?

[onyx helm]

oui exactement

[spark sleet]

Et tu as déjà démontré que c'était bijectif

[onyx helm]

oui

[spark sleet]

Ok ok

[onyx helm]

(je suis plus dispo dans 30m a moitier occuper atm)

[spark sleet]

Si tu prends un nombre x naturel (plus grand strictement que 0), il se situera forcément entre deux termes de la suite des sommes des entiers naturels: 1, 3, 6, 10, 15, 21, etc......

Tu dois prendre le plus grand entier de cette suite qui est plus petit ou égal à x, et ça te donnera la valeur m correspondante

ensuite, il faut aller ''chercher'' x avec le n

Ça peut s'écrire formellement mais c'est crucial d'abord de comprendre l'idée

[onyx helm]

oui tout a fait, c'est d'ailleurs l'idée que j'avais pour le résoudre avec un recursion de manière algorithmique, mais je sais pas comment faire de manière mathématiques

[spark sleet]

Tu veux dire, tu sais pas comment bien l'écrire?

[onyx helm]

nan, si je le fait avec un programme avec des boucles avec un condition qui check que l'entier est plus petit que x, ça je sais faire.
mais je veux trouvé une solution qui m'évite de faire des récursion, car c'est une fonction que je vais beaucoup appeler

[spark sleet]

En gros, étant donné x, tu veux une formule explicite pour m et n

sans faire de ''boucles''

[onyx helm]

exactement

[spark sleet]

Pas évident hein !

[onyx helm]

nop ^^

[spark sleet]

Pourquoi tu veux pas faire de boucles ?

[onyx helm]

par ce que c'est une fonction asser primaire de mon programe, et que ça va être appeler en masse et ça va alourdire le programe énormement

[spark sleet]

@onyx helm : si tu définis la fonction f(y) = y(y+1)/2
Alors étant donné x, la valeur m correspondante est donnée par la partie entière de f^(-1)(x)

et ça se calcule, f^(-1)(x)

f défini sur les réels positifs

et la valeur de n est ainsi donnée par x - f(m)

et tu as tes 2 formules

[onyx helm]

ouai je vais essayer, mais me semble qu'il faut quand même une condition pour vérifier une valeur

[spark sleet]

C'est-à-dire ?

[onyx helm]

un if qui check que f(m) est plus petit que l'entier x

[spark sleet]

Par définition de m, f(m) sera plus petit que x

toujours

pas besoin de if

ou égal à x

[onyx helm]

je vais essayer comme ça alors

[spark sleet]

Si tu calcules f^(-1)(x), tu devrais obtenir:

$f^{-1}(x) = \sqrt{2x + 1/4} - 1/2, \qquad x \geq 0$

Donc $m = \lfloor \sqrt{2x + 1/4} - 1/2 \rfloor$

[woven radish]

all matrices are invertible

all matrices are invertible

[onyx helm]

.cancel

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[stiff mist]

Hello

[stiff mist]

So my current thing is

So far I think I have one option correct

I tested f(-5) for it and it came out to 1 divided by 0 which is undefined

Im a little unsure about the rest

[winter patrol]

but f(-5) isn't undefined

pay close attention to the conditions on the right
for which piece to use

[stiff mist]

hmm

I believe the first box is correct, because as x approached -5 the denominator appraoched 0 making it lean towards infinity?

[winter patrol]

yes

[stiff mist]

hmm alright...

gimme a second to do some calculations

Wait I have a question

How is f(-5) defined?

I plugged in -5 for the top expression 1 / x + 5

[winter patrol]

why are you plugging -5 into the top expression

[stiff mist]

oh wait

[winter patrol]

pay close attention to the conditions on the right
for which piece to use

[stiff mist]

it says not equal to -5

oh wait no its the opposite

it is

thats what makes it defined

[winter patrol]

wdym by opposite

[stiff mist]

hold on just a moment

no nevermind, i just meant that when the function is equal to 1 when x = -5

[winter patrol]

yeh

[stiff mist]

im not sure what makes the limit finite, is it when it has a finite number it stops at?

hmmm

I feel the answer would be none of the above no?

wait nevermind

.close

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[mossy nymph]

How does R2 dot R1 + R1 dot R2 = 0 get to 2R2 dot R1 = 0?

[supple knot]

Dot product is commutative

a dot b = b dot a

number / 2 = 0 implies number = 0

[mossy nymph]

ah that's right

so i wonder why it's not 2R2 dot 2R1

[acoustic leaf]

in real number multiplication, $2(ab) = (2a)b = a(2b) \ne(2a)(2b)$, similar reasoning applies here

[woven radish]

cloud

[mossy nymph]

thats true, i guess i still don't get what makes it 2R2 dot R1 and not 2R1 dot R2

unless those are the same thing?

[analog trellis]

They are the same thing.

[mossy nymph]

oh wait, its the whole dot product times 2?

[analog trellis]

Yeah.

[mossy nymph]

that makes so much more sense lol

thanks yall

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[fiery yoke]

If I have a solution set like that, is my dimension still 4?

[torn vessel]

dimension of what?

[fiery yoke]

The solution set?

And i think it should be 1 because we only have one free variable right?

[devout snow]

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[fiery yoke]

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[glacial dragon]

Do you guys think this is solvable

Or is it too like ambiguous

[dense lynx]

this definitely seems solvable yes

unless i grossly misinterpret the problem

that is, by "iteration" i assume the problem refers to function composition

[glacial dragon]

Yeah

[glacial dragon]

What approach would you take?

Not the solution, i just need an approach bc im not sure where to start.

[dense lynx]

i probably poke around and examine the domain and range of the first few iterations

[glacial dragon]

👍 thats smart. Thank you so much!

[dense lynx]

welcome 🙂

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[hasty lance]

How would I do 1200 is deposited into an account that earns 7% compounded monthly. How long would it take to reach 1398.23

[hasty lance]

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[restive river]

trying to solve this by setting the simplest term equal to a variable and changing the rest in terms of that same variable

[restive river]

only problem is I lack that math skill

so for example im trying to let sqrt(x+1) = a

what would the rest of the terms be equals to

in respect to a?

if it has a name or something so I can learn it, but I assume its basic algebra

[near trout]

i don't think that works necessarily

can you just conjugate?

[restive river]

I'm assuming that wouldn't work

[near trout]

probably hit it with a taylor series then

or just l'hopital if you can use that

[lost laurel]

Even the binomial approximation may work here

[near trout]

do you know the binomial expansion?

yeah

[restive river]

not allowed to (apparently not in the course)

[near trout]

i mean the binomial series is just a taylor series

[restive river]

first time hearing of a binomial series lol

wt is that

[lost laurel]

then use binomial approximation.

[restive river]

taylor i meant

but how does it look like

[restive river]

woah ok intimidating to look at

[lost laurel]

$(1+x)^a= 1-ax$ is the binomial approximation

[woven radish]

Why am. I here

[lost laurel]

try that

[restive river]

I remember my teacher going over this but instead having the numerator and denominator with a bunch of square roots

[restive river]

hm ok

so

sqrt(1+x) would be

1-0.5(x)?

[lost laurel]

wait, my bad

it should be $1+ax$

[woven radish]

Why am. I here

[restive river]

1+0.5x

that would certainly make it easier

[lost laurel]

yes, pretty much, unless I'm tripping

,w binomial approximation of $(1+x)^(0.5)$

[woven radish]

Wolfram Alpha doesn't understand your query!
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Requested by physicsrocks

[restive river]

hm

$\sqrt{x+1}=(x+1)^{\frac{1}{2}}=1+\frac{1}{2}x?$

[woven radish]

mj

[lost laurel]

I think so , going by wikipedia's defn

[restive river]

$\frac{1+0.5x-1+x}{4+1.5x-4+x}$

[woven radish]

mj

[restive river]

$\frac{1+0.5x-(1+x)}{4+1.5x-(4+x)}$

[woven radish]

mj

[restive river]

$\frac{-0.5x}{0.5x}$

this is equals to 1

wait i definetly did something wrong

ans is -2

[lost laurel]

,w limit of $\frac{\left(\sqrt{x+1}-\sqrt{\left(2x+1\right)}\right)}{\sqrt{3x+4}-\sqrt{2x+4}}$

[woven radish]

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[lost laurel]

yup, it's right!

[restive river]

no i meant

im wrong

by getting 1

it's supposed to be -2

[lost laurel]

oh

[restive river]

who's taylor and what is this series

[woven radish]

mj

[restive river]

ok noticed a math error

but still -1

closer

[lost laurel]

it's 4+ 3x/2

oh

nvm, that's 1.5

ok, sorry I have to go now, sorry if I misguided you

[restive river]

no worries lol

i figured out how to do this question

the intended way

just thought this was

interesting

[lost laurel]

Using rationalisation?

[restive river]

no using this taylor series

[lost laurel]

ah,ok

[restive river]

neat trick, will look more into it

thanks though

[lost laurel]

Just check if you've made a calculation error anywhere

[arctic field]

this is not the right taylor series

[restive river]

how come

[arctic field]

,, (1 + x)^\alpha \approx 1 + \alpha x

[arctic field]

but this says nothing about (4 + x)^α

[restive river]

so your saying this only applies to

(1+x) equations?

[arctic field]

what equations

[restive river]

or am I intended to factor out

what makes it not that

and then multiply

after

[arctic field]

,, (4 + x)^{1/2} \approx 2 + \f14 x

[restive river]

$4(1+x)^{0.5}=4(1+0.5x)=4+2x?$

[woven radish]

mj

[arctic field]

i don't like your use of the equals sign here

[restive river]

apologies

[arctic field]

the first = is most certainly not =

[somber plume]

what would the answer be to this?

[restive river]

.occupied

[arctic field]

!help

[devout snow]

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[restive river]

yea but how come

we're getting

different answers

[arctic field]

because your method is wrong

[restive river]

how come?

[arctic field]

,, (4 + 3x)^{1/2} \approx 2 + \f34x

[arctic field]

using the taylor series approximation

[restive river]

3*(1/2) is what your doing here?

ok tell me this

what do you do to get from point a to point b

starting with the 4 to 2

[arctic field]

no you factor out the 4 to apply the (1 + x) approximation

[restive river]

ok great

so you h ave no

now

but wait

you have to apply the exponent

first

$(2+3x^{1/2})$

[woven radish]

mj

[restive river]

ah im confused now lol

[arctic field]

what did you do

[restive river]

all I did w as

apply the sqrt

to the inside

of the bracket

[arctic field]

that doesn't make sense

[restive river]

mb

[arctic field]

you can't "apply" a sqrt to an expression like (4 + 3x)

[restive river]

ok but how would you factor out a 4

from something that has a power

being applied to it

[arctic field]

,, \sqrt {4 + 3x} = \sqrt {4(1 + \tf34 x)}

[restive river]

shouldn't you apply the power first before factoring?

[restive river]

oh okay okay

I see

lol yea that makes more sense now that i think about it

so in our case $\sqrt{4+2x}=\sqrt{4(1+ \tfrac12 x)}$

wait a minute no

[arctic field]

use \tfrac

[woven radish]

mj

[restive river]

hm

okay

$\sqrt{4(1+ \tfrac12 x)}=4(1+ \tfrac12 x)^{1/2}$

[woven radish]

mj

[restive river]

$4(1+0.5(0.5)(x))$

[woven radish]

mj

[restive river]

$(4(1+\tfrac14 x))$

[woven radish]

mj

[restive river]

$4+x$

[woven radish]

mj

[restive river]

,w taylor series sqrt(2x+4)

@arctic field does that look right or

[woven radish]

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[restive river]

lol def

not

ok nvm cool method hopefully I'll get to learn it

later

[arctic field]

you have to square root the 4 here

it'll be a 2

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[fierce osprey]

When you graph a polynomial function, does the turning points must equal 1 less then the degree? or is 1 less then the degree the max amount of turning points the graph can have?

[dense lynx]

consider polynomial x^4

what is its degree?

how many turning points does it have?

[shrewd thistle]

x^4 works, too

[fierce osprey]

1

[shrewd thistle]

or even just x... a flat line

which has... no turning points

[dense lynx]

and what is the degree of x^4 ?

[fierce osprey]

4

[dense lynx]

does this answer your question, then?

[fierce osprey]

yeah

thanks

[dense lynx]

welcome 🎉

[fierce osprey]

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[gloomy valve]

Can someone explain part B

We have this theorem

We would move f(x) and p(x) to the LHS

and take absolute values

I was more confused on how we would bound the term on the LHS tho

that taylors expression term

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[compact vigil]

how to do this problem

[compact vigil]

i tried to start but got to nowhere

i tried to get the torque or weight of the beam or something and then i didnt know what to do i dont know if i was meant to do

[turbid raptor]

is T=1700N in the conditions?

[compact vigil]

the question is right here

yes

[turbid raptor]

and the length of that beam is 4.2m

[compact vigil]

yeah

[turbid raptor]

so this point

do force analysis u can get

[compact vigil]

is that the rotation point

[turbid raptor]

no, that tit up point

[compact vigil]

?

[turbid raptor]

here

[compact vigil]

what do i do with that point what r u saying im confused

[turbid raptor]

the component force of T that paralled with this pink line u drew, let's say F1

satisfied F1*4.2m = 1300Nx3.8m

[compact vigil]

r u using the torques

[turbid raptor]

yes maybe

and u can get F1

θ=arccos(F1/T)

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[craggy whale]

hi

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@compact vigil Has your question been resolved?

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[jagged swan]

i think i messed up somewhere but im not sure what since there is no answer key

can someone check my work

[winter patrol]

a) is fine,
but then why are you using -1/12 for the slope in b)

[jagged swan]

idk i just remember seeing my teacher do the negative recipricol when doing the equation of the tangent

[winter patrol]

you misremembered

that's for equation of the normal

[jagged swan]

uh oh

so for the tangent line would it stay at 12

[winter patrol]

as those are perpendicular to the tangent

[jagged swan]

ohhh

[winter patrol]

yes. the slope of the tangent line is the same as the derivative at that point

[jagged swan]

we arent at derivatives yet i dont even know what that means

so the equation would be

12(x-2) + 15

o i think it worked

yipee

[winter patrol]

we arent at derivatives yet
you're pretty much being asked to apply the limit definition of the derivative to get the slope
not sure why they're not mentioning derivatives here

[jagged swan]

im in hs right now and we just started limits im pretty sure we do derivatives next

[winter patrol]

its worth looking up "limit definition of the deriative"

[jagged swan]

alright

thank you

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[heady tulip]

quick question

[heady tulip]

can you define abs(x) as sqrt(x^2) instead of a piecewise function

... i mean it still being somehow a piecewise function

but still, is it valid?

[fossil locust]

Yes

That's absolutely valid

[heady tulip]

okokokokok thanx

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[deep vortex]

This is 17.6

@stone stump

surely you have done similar questions of the type "given are some vector which span a space, find a basis of that space" before?
Nope, but is the first step getting it to RREF?

[stone stump]

yes

keep track of which rows you switch

[deep vortex]

Why is 17.6 needed at all

[stone stump]

well existence proof

always good to know that the thing you are looking for actually exists

[deep vortex]

Ah

\begin{align*} \mr{1 & 1 & 1 & 0 & 3 \ -1 & 0 & 1 & 1 & -1 \ 0 &1 & 2& 1 & 2 \ 1 & 2 & 3 & 2 & 4} &\sim
\mr{1 & 1 & 1 & 0 & 3 \
0 & 1 & 2 & 1 & 2 \
0 & 1 & 2 & 1 & 2 \
1 & 2 & 3 & 2 & 4} &&(I_2 \to I_2 + I_1)
\ &\sim
\mr{1 & 1 & 1 & 0 & 3 \
0 & 1 & 2 & 1 & 2 \
0 & 1 & 2 & 1 & 2 \
0 & 1 & 2 & 2 & 1} &&(I_4 \to I_4 - I_1)
\ &\sim
\mr{1 & 1 & 1 & 0 & 3 \
0 & 1 & 2 & 1 & 2 \
0 & 0 & 0 & 0 & 0 \
0 & 1 & 2 & 2 & 1} &&(I_3 \to I_3 - I_2)
\ &\sim
\mr{1 & 1 & 1 & 0 & 3 \
0 & 1 & 2 & 1 & 2 \
0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 1 & -1} &&(I_4 \to I_4 - I_2)
\ &\sim
\mr{1 & 1 & 1 & 0 & 3 \
0 & 1 & 2 & 0 & 3 \
0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 1 & -1} &&(I_2 \to I_2 - I_4)
\ &\sim
\mr{1 & 1 & 1 & 0 & 3 \
0 & 1 & 2 & 0 & 3 \
0 & 0 & 0 & 1 & -1 \
0 & 0 & 0 & 0 & 0} &&(I_3 \leftrightarrow I_4)
\ &\sim
\mr{1 & 0 & -1 & 0 & 0 \
0 & 1 & 2 & 0 & 3 \
0 & 0 & 0 & 1 & -1 \
0 & 0 & 0 & 0 & 0} &&(I_1 \to I_1 - I_2)
\end{align*}

[deep vortex]

So now, we pretty much found the base of the row space, right?

It's the individual rows

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.reopen

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✅

[deep vortex]

#1195800467880558672 message

[devout snow]

@deep vortex Has your question been resolved?

[deep vortex]

Why?

[devout snow]

@deep vortex Has your question been resolved?

[compact vigil]

I don't get that

[turbid raptor]

u dont get what?torques?

[devout snow]

@deep vortex Has your question been resolved?

[devout snow]

@deep vortex Has your question been resolved?

[devout snow]

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[quaint ether]

this should be equal to exp(x) no??

[granite arch]

yes

[quaint ether]

so thers a mistake

[granite arch]

by using the property lim h->0 (e^h-1)/h = 1, u get e^x

[granite arch]

yes

[quaint ether]

okay thanks !!

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[prime narwhal]

is this right>?

[prime narwhal]

and what about this one

[winter patrol]

can you show your work

[prime narwhal]

i erased it

😭

[winter patrol]

do it again

[prime narwhal]

nvm i re calced

i got it now

dont need help

.close

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[devout snow]

Please don't occupy multiple help channels.

[strong fiber]

I'm still not sure what's wrong here

[cinder bobcat]

take -pi/3 and add pi to it to get the correct angle, since that's a half rotation

that's 2pi/3, not 5pi/3

[strong fiber]

How did you know that you have to add pi? Isn't that gonna land in the 2nd quadrant?

[cinder bobcat]

yes, but if you're in Q2, and r is negative, then you land in Q4

it flips the line 180 degrees

[strong fiber]

Oh nice

So, when picking the angle, i have to keep an eye on where the line is gonna land

And pick the angle that will make the line land on the quadrant that matches the cartesian coordinate

Right?

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[compact vigil]

Why is those equal why doing that

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[restive river]

this question giving me a serious headache

[restive river]

what approach do you take?

[misty crest]

what have u tried

do u understand how to evaluate limits

[restive river]

well, i can tell there is an infinite discontinuity at 0

[misty crest]

yes that’s true

because vertical asymptote at x=0

[restive river]

or is that a jump one?

[misty crest]

infinite

[restive river]

how do you tell that

[misty crest]

division by zero with the numerator not ≠0

if the numerator =0 as well then it’d be a removable hole

jump occurs with piecewise functions

[restive river]

ah, the nmerator ok

[misty crest]

think of it like a limit

if u divide any number by let’s say 0.000001

it’s going to be very large

and the smaller the denominator gets the larger the quotient gets

[restive river]

I get that

so to solve this, i just plug in until i see pattern?

[misty crest]

so are u aware of any algebraic techniques

[misty crest]

wdym

[restive river]

im confused how to do this analytically

[misty crest]

like guess and check?

no

i mean

im sure that works

but

it’s not a good habit

[restive river]

so what do I need to do

[misty crest]

well u know it’s infinite correct

so it’s either going to go to positive infinity or negative infinity

[restive river]

I think I need to split it up and get limits of smaller portions, then get limit of the whole thing?

[misty crest]

so what u can do

is determine the sign of the function before the discontinuity

which will determine if it approaches +infinity or negative infinity at the left or at the right

of the asymptote

[restive river]

AHHH

because numerator is not 0

[misty crest]

now the numerator will always be positive correct

because e^x >0

[restive river]

I was so confused about this, thank you

[misty crest]

so adding them will always be>0

so we only need to look at the sign of the denominator

also note

how would u determine the sign of the denominator

so e^2x-e^3x

[restive river]

wait, plug in 0 to numerator its 0?

[misty crest]

wdym

[restive river]

if you plug in 0, numerator is 0 denominator is 0

[misty crest]

the numerator is always >0

no

e^0 is 1

so 5+9=14

[restive river]

ahh

so its 14/0

[misty crest]

yes

[restive river]

therefore infinite

[misty crest]

because the denominator is 1-1=0

so

e^2x-e^3x

will this be positive or negative for x>0

[restive river]

so its positive infinity from right

[misty crest]

no

[restive river]

its positive from right

[misty crest]

think of it

[restive river]

negative left

[misty crest]

like

(e^x)^2-(e^x)^3

the cubed will be larger than the square for x>0

if x<0 then it’s a negative exponent which makes it 1/e^2x + 1/e^3x

[restive river]

so its negative?

[misty crest]

and here since the terms r in the denominator and e^3x>e^2x when u take the reciprocal or do 1/ it will be positive

so from the left it’s positive

from the right it’s negative

u could even plug in values in ur head if it makes more sense

let’s say x=1 which is >0

e^2 -e^3 is clearly negative

but if we did x=-1 which is <0

1/e^2 -1/e^3

since e^3>e^2 the reciprocal will be smaller

so 1/e^2>1/e^3 therefore it’s positive

[restive river]

alright so answers are

-\inf, \inf

what about approaching infinity

[misty crest]

hmm

from the left it’s positive lol

and right it’s negative

i’m not trying to just give u the answer

i’m trying to help u

understand it

[restive river]

aye

[misty crest]

ur just tryna do ur homework

i get it

[restive river]

I want to understand it

so approaching -\inf

[misty crest]

from the right

[restive river]

yeah I don't even know where to begin for this

[misty crest]

which is 0+

or r u talking about the next question

the limits at infinity

[restive river]

yeah

[misty crest]

which correspond to horizontal asymptotes

well plug in infinity for x

u get infinity in the numerator

and it won’t work

because u get -infinity in the denominator

so u have to use algebra

to simplify

to do this

divide by the highest degree

so divide each term in the fraction by e^3x

to simplify this to just constants

tell me what u get

for each term

when dividing each term by e^3x

[restive river]

highest in denominator?

[misty crest]

wdym

think of e^x as some other variable we can call u

then e^2x=u^2

[restive river]

highest in all terms? just the denominator?

[misty crest]

and e^3x=u^3

the highest degree of all the terms

which is e^3x

[misty crest]

^

you’ll see why we do this in a bit

just simplify first

like what’s 5e^x/e^3x

and 9e^3x/e^3x

and the same thing for the terms in the denominator

[restive river]

how do you divide exponents?

[misty crest]

use the u substitution i told u it’ll make more sense

what’s u/u^3

1/u^2 correct

[restive river]

yeah i see

[misty crest]

u have 3 u’s in the denominator and only one up top

ok so

tell me

[misty crest]

^

[misty crest]

^

and the same for the denominator terms

[restive river]

i have no idea, the terms are different

5 on top, 1 on bottom

[misty crest]

5 is a constant multiple

[restive river]

also bottom has a constant in the exponent

[misty crest]

…

it’s the same thing as saying e^x cubed

recall

(a^n)^m=a^n*m

[restive river]

so the answer i 5/e^2x?

[misty crest]

yes

5/e^2x

[restive river]

ahh

but thats a fraction

[misty crest]

we want it to be

just wait until u do all of them

i’ll explain after

once u have a new fraction

with everything simplified

[restive river]

so the 9 term is then just 9 after the division

[misty crest]

mhm

and what about the denominator terms

[restive river]

e^2x becomes 1/e^x

then e^3x becomes 1

[misty crest]

so we have (5/e^2x +9)/(1/e^x -1) correct

now

[restive river]

yeah

[misty crest]

what happens if we take the limit as x goes to infinity for those fraction terms

[restive river]

5/e^2x gets larger?

[misty crest]

r u sure

e^2x goes to infinity

but 5/infinity is not infinity

what’s 5/100000000

it’s near zero yes

the larger the denominator gets the smaller the quotient gets

[restive river]

so its 9

?

[misty crest]

divided by -1

so -9

[restive river]

-9

[misty crest]

because the two fraction terms go to zero

u get 9/-1 or -9

now what happens for negative infinity

with our same expression

with the fractions

[restive river]

I see, this is weird but i see

last one is 9?

no wait you have to do this again

[misty crest]

no?