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[ember oar]

Gotchat

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[obtuse pebble]

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[white hinge]

quick question:
in a probability distribution table, can x be something other than a number? Like yes or No?

[white hinge]

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[twin oracle]

hi anyone able to help a bit with latex

[twin oracle]

trying to make a tittle page for lab report

[languid tinsel]

Hello, I can't get a hang of the function domain

[timid silo]

#latex-help for your latex questions

[twin oracle]

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[merry girder]

hi

[merry girder]

@sonic rapids

[obtuse pebble]

@merry girder Has your question been resolved?

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[steel goblet]

a little confused on how to do this

[steel goblet]

do i calculate the partial derivative with respect to z, then with respect to x, divide them (z/x) and then plug in 1 for x, y, z?

[warm canopy]

uh i think you can just implicitly differentiate

like apply d/dx to both sides of the equation

[steel goblet]

so 2y+z^3 = 0?

and then would i plug in 1 for y and z?

[warm canopy]

hmm

no

you have no dz/dx present which is the thing you want to solve for

d/dx(z^3x) is not z^3

and d/dx(2yz) is not 0

[steel goblet]

im confused

how is not z^3? Aren't you treating z^3 as a constant? like d/dx(3x) is just 3

[warm canopy]

no, z is a function of x and y

[steel goblet]

okay i understand that now. So I implicitly differentiate, right?

[warm canopy]

yeah

[steel goblet]

i got the problem wrong lol so it gave me a different one. Same concept but just different numbers

so im not entirely sure i remember how to implicity differentiate. So im differentiating with respect to x, right?

so d/dx(9xy) + d/dx(z^4x)...etc

so to start, d/dx(9xy) is 9, right?

[steel goblet]

how exactly do i implicity differentiate?

cuz there are three variables

[warm canopy]

y is constant with respect to x so d/dx(9y*x) is just 9y

z^4x is now a product of two functions of x

[steel goblet]

oh okay. so is d/dx(z^4x) = z^4 * dx/dz?

[warm canopy]

you're forgetting both the product rule and the chain rule

[steel goblet]

so then is it 4xz^3+z^4?

[warm canopy]

forgetting chain rule when you differentite z^4

z is a function of x

[steel goblet]

im just confused because z does not look like a function.

like i know chain rule is d/dx [f(g(x))] is f'(g(x)) * g'(x)

but i don't really see how chain rule can be applied here

[warm canopy]

z = z(x)

f is ^4

[steel goblet]

so its really z(x)^4 x, right?

[warm canopy]

yeah, we just dont write it all the time

well sorry really z = z(x,y)

like the question says

but importantly its still a function of z so chain rule must be applied if youre partial diffing wrt x

[steel goblet]

im still kinda confused. Like is this how the derivative would be:

z'(x,y)

[warm canopy]

here f(x) = x^4 and g(x) = z(x) [really z(x,y) but for this purpose its fine]

[steel goblet]

so then is g'(x) = z'(x)?

and is that it?

[warm canopy]

yeah but now these derivatives are really partial derivatives

i wouldNT use ' to denote them

[steel goblet]

wait so how would you denote d/dx (z^4)?

[warm canopy]

$\frac{\dd}{\dd z}(z^4) \cdot \pdv{z}{x}$

[warm shale]

ΣΑCu

[steel goblet]

so would this be correct ?

[warm canopy]

uh no

[warm canopy]

you didnt finish evaluating this

[steel goblet]

im so lost

so d/dz(z^4) is that just 4z^3?

[warm canopy]

yes

[warm canopy]

and the derivative you have here should really be a partial derivative

[steel goblet]

so is this right?

[warm canopy]

yeah

[steel goblet]

and d/dx (-2yz) is that just -2y dx/dz?

[warm canopy]

dz/dx

[steel goblet]

so now do i just solve for dz/dx but moving everything to the other side of the equation?

[warm canopy]

yeah rearrange for it

[steel goblet]

i got dz/dx = (-9y-z^4)/(4xz^3-2y)

oh awesome. Yes and then i just plug in 1 for all the variables. Thank you so much @warm canopy

really helped me out

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[unreal storm]

I have a tridiagonal matrix with dimensions n x n, which is a symmetrical matrix. I know the eigenvalues of this matrix are real.

Now I multiply the first and last row of this matrix with a scalar. Apparently the eigenvalues of this matrix are also real. Is there a proof of this, or a theorem that says this?

[unreal storm]

I have tried to verify this using examples, and it seems to hold true for my case. However, I can’t seem to prove this myself, or find a proof of this online

[alpine bison]

yes

symmetrical matrix has always real eigen values

[unreal storm]

Well if I multiply the first and last rows of a symmetric matrix with a scalar it is not symmetric anymore

[alpine bison]

ah sorry

[unreal storm]

My professor said that doing this is some sort of transformation, and since this transformation can be reversed, and the resulting matrix has real eigenvalues (symmetric again), the non-symmetric matrix has real eigenvalues.

[alpine bison]

mh,.. let me think

no its false

for example if you multiply only the first raw for a scalar ..then you could get a matrix that has not real eigenvalue

$\begin{matrix}\lambda &&2\lambda \ 2 && -1 \end{matrix}$

[warm shale]

everg

[alpine bison]

for lambda=-1

$\begin{matrix}-1 &&-2 \ 2 && -1 \end{matrix}$

[warm shale]

everg

[unreal storm]

I was thinking of proving it using something like this:

Let A be a symmetric matrix with values a b 0 … 0 in the first column. Then the characteristic polynomial is equal to (a-lambda)d1 -bd2, where d1 and d2 are determinants. Let lambda be an eigenvalue. Then lambda = (ad1-bd2)/d1.

Since the eigenvalues of a symmetric matrix are real, the RHS is real too.

Now let B be matrix A after multiplying the first row with a real scalar c. Then its characteristic polynomial is (ca-lambda)d1-cbd2, where a, b, d1 and d2 are the same as before, but lambda is an arbitrary eigenvalue of this matrix B. Then we can express lambda as lambda=(cad1-cbd2)/2=c (ad1-bd2)/d1. Which is a multiplication of a real value and a real value, which must be real.

Doesn’t this hold up?

[alpine bison]

has det>0 and trace <0 ...impossible for real eigenvalue

you did a proof for a symmetric matrix and then you change only the first row by multiplication ?

[unreal storm]

Well we can also assume A to be an arbitrary tridiagonal matrix with real eigenvalues. Point is that (ad1 + ad2)/d1 is real, and a real value multiplied by a real value is always real

Or did I do something wrong?

[alpine bison]

sorry i can t follow ..now i am very tired ...maybe there are more energic guys here

[unreal storm]

No worries, thanks for the effort

[obtuse pebble]

@unreal storm Has your question been resolved?

[unreal storm]

<@&286206848099549185>

[obtuse pebble]

@unreal storm Has your question been resolved?

[obtuse pebble]

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[fair girder]

can i get some help with this

[fair girder]

idk where to start

[bronze mica]

I assume you have learned how to actually find an explicit antiderivative?

[obtuse pebble]

@fair girder Has your question been resolved?

[fair girder]

is that the one with numbers on top and bottom of the S

[bronze mica]

S?

$\int$?

[warm shale]

Edward II

[fair girder]

that thing

yea

or is it an f

whats it called

[bronze mica]

originally an s, but not really anymore

I usually call it an integral sign

[fair girder]

alr

so explicit antiderivative is this right

[bronze mica]

anyway

[fair girder]

with an interval

integral with interval

[bronze mica]

uh yeah except I'm questioning the x^2

[fair girder]

?

[bronze mica]

definite integral

[fair girder]

thats just some example i found

[bronze mica]

well

technically indefinite here

[fair girder]

explicit antiderivative is definite integral?

[bronze mica]

I'm going to assume you've not learnt the fundamental theorem of calculus then

hmm

[fair girder]

yea i have

[bronze mica]

oh

then it gets much easier

so $\int_0^9f(x)dx=F(9)-F(0)$ and rearrange, and the integral is the area under f

[warm shale]

Edward II

[fair girder]

yes

[bronze mica]

and now f is really nice to calculate the area under because it's all triangles, rectangles, and a missing semicircle

(unimportant: by find an explicit antiderivative I meant FTC and forgot the theorem's name 💀)

[fair girder]

its fine

so now i just have to find the area under the curve?

[bronze mica]

yes

[fair girder]

i know how to find the first two

the triangle and rectangle

how would i find the area outside the semicircle?

from 4 to 6

those curved triangles

riemann

?

[bronze mica]

well, if the semi circle wasn't there it would be a rectangle

[fair girder]

true

[bronze mica]

and we've 'cut' a piece out

[fair girder]

alr one sec

ah i see

so i get 1 for triangle then 3 for rectangle and then circle is pi*r^2 so pi(1)^2

and then from 6 to 9

they cancel out cuz they are symmetrical

so the second rectangle is 3 but we cut out the semicircle which is pi/2 so 3-pi/2

plus the 4 from the triangle and first rectangle

7-pi/2

so thats F(9)

and then -F(0) which is 2

so 5-pi/2

[fair girder]

right?

oh wait

we dont need to subtract the F(0)

its just asking for F(9)

[bronze mica]

uhh you've made two mistakes that cancel out

the rectangles do not have area 3

[fair girder]

oh wait its area 2

yea just realized

[bronze mica]

and on rearranging this you need to add the F(0) to the LHS that is the area

you overestimated the area by two and didn't add

[fair girder]

why do i have to add the F(0)?

[bronze mica]

because we're rearranging for F(9)

so it's from bringing F(0) to the LHS

[fair girder]

whats lhs

left hand side

//

??

i dont understand why we rearrange

[bronze mica]

we want just F(9) right

[fair girder]

yea

[bronze mica]

so we have 'area' = F(9) - F(0), so F(9) = area + F(0)

[fair girder]

oh yea makes sense

cuz this is the area of whole thing

ok i got it

alr ty again

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[unreal whale]

are these correct?

im pretty sure they are

<@&286206848099549185>

<@&286206848099549185>

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[white mulch]

I need help with this question

[white mulch]

Its integral of (-3cscx times cotx) dx

[obtuse pebble]

@white mulch Has your question been resolved?

[white mulch]

.closer

.close

[obtuse pebble]

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[timid silo]

hello

[timid silo]

The amount of a certain medicine, in milligrams, in a patient’s body t hours after an initial dose can be modeled by the exponential decay function f given by f(t) = ab^t. An initial dose of the medicine is 10 milligrams, and after 2 hours the amount in the body is 5 milligrams. At what time will the amount in the body be 0.01 milligram? (Assume no additional doses of the medicine are given after the initial dose.)

i don't understand this problem at all

[tepid sedge]

have you learned half life

[tepid sedge]

.

[timid silo]

huh

what's wrong

[tepid sedge]

Have you learned half life?

[timid silo]

what is that

[tepid sedge]

it means every set number of time, something will be halfed, in this care, its the medicine

every 2 hrs

so how many times of 2 hrs do it need to pass to have only 0.01 milligrams left?

[timid silo]

idk

[tepid sedge]

https://www.youtube.com/watch?v=IhJ4ofr2glM&ab_channel=Mario'sMathTutoring

that will help you understand how to do ur problem

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[timid silo]

Trying to figure out the correct domain for f(g(x))

[honest sail]

first, what is f(g(x))?

[timid silo]

F(g(x)) is also (f•g)(x)

[honest sail]

can you write it out in a formula form with $x$ and numbers in it?

[warm shale]

Dee3Cay

[honest sail]

not just $f\circ g(x)$

[warm shale]

Dee3Cay

[timid silo]

I believe I wrote down 1/(1/(x-9)+(3/1) for that

[honest sail]

ok now you'll need $x-9$ to not be 0, also, $\frac{1}{x-9} +3$ cannot be 0 too, why?

[warm shale]

Dee3Cay

[timid silo]

I do not know

[honest sail]

here you wrote $f\circ g(x) = \frac{1}{\frac{1}{x-9}+3}$

[warm shale]

Dee3Cay

[honest sail]

so now do you know why?

[timid silo]

Yes because I subsitituted g(x) for x in the equation of f(x)

[honest sail]

yes you did that

but can we have $\frac{1}{x-9}+3 = 0$?

[warm shale]

Dee3Cay

[timid silo]

Yes we can

We can subtract 3 on both sides

Then have 1/(x-9)=-3

[honest sail]

look at this equation

what will happen to the denominator?

[timid silo]

Oh shit. Is it actually 1/(x-9)+3/1=0

[honest sail]

yes

[timid silo]

So combined I got 1+3/(x-9+1)=0

I hope that’s correct for now

4/(x-8)=0

[obtuse pebble]

@timid silo Has your question been resolved?

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[obtuse pebble]

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[junior jay]

help

[junior jay]

now I know this would lead to an improper fraction

and then turned into a mixed fraction

wait bruh Im stupis

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[obtuse pebble]

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[junior jay]

.reopen

[obtuse pebble]

✅

[junior jay]

so the mixed fraction would b

but how do I make this into a %

<@&286206848099549185>

[dire plinth]

just write 2 as a percentage, 3/5 as a percentage then just add them together

[royal basin]

@junior jay what is 3/5 as a percentage?

[obtuse pebble]

@junior jay Has your question been resolved?

[obtuse pebble]

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[timid silo]

have you learnt about free variables

[unique tendon]

Yes

[timid silo]

okay can you identify which

[unique tendon]

They are the one's not with leading 1's

[timid silo]

the non leading ones

[timid silo]

that will be when its in reduced echelon form

[unique tendon]

So x2 and x4?

[timid silo]

Could anyone help me with this sheet of work, I am year 7 and I cannot wrap my head around it.

[timid silo]

look at x4 again

[timid silo]

!help

[obtuse pebble]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[unique tendon]

Sorry x2 and x5

[timid silo]

very good

[unique tendon]

What do you mean by this

I mean i know what reduced echelon form is

Isn't this in reduced?

[timid silo]

apparently we're expecting 3 free variables? hmm

[timid silo]

have you seen the form where its only

1 0 0
0 1 0
0 0 1

[unique tendon]

Ok i get it this is in echelon form but not reduced

I tried reading what to do next but i don't quite get it

It says we make the leading variables subject of formula?

[timid silo]

anyways lets proceed

[timid silo]

yeah

so who will be s and t

[unique tendon]

x2 and x5

[timid silo]

okay, now write the equations for this augmented matrix

[timid silo]

with those substitutions

[unique tendon]

Like this?

[timid silo]

yes

so what is x1

x2, x3, and so on

[unique tendon]

Wait so if we substitute x2 with s how will we get x2

[timid silo]

good question

what did we x2 was again

[unique tendon]

S

[timid silo]

exactly

[unique tendon]

Or?

[timid silo]

we'll be expressing x1,...x5 in terms of s and t

[unique tendon]

Hmm okay well let me work it out gimme like 3 then you'll confirm?

[timid silo]

np

$$\begin{bmatrix}
x_1 \
x_2 \
x_3 \
x_4 \
x_5 \
\end{bmatrix} = s \begin{bmatrix}
? \
? \
? \
? \
? \
\end{bmatrix} + t \begin{bmatrix}
? \
? \
? \
? \
? \
\end{bmatrix} + \begin{bmatrix}
? \
? \
? \
? \
? \
\end{bmatrix}$$

[timid silo]

ping when you're done take your time

[unique tendon]

Here's what i got

[timid silo]

uh x2 and x5 are wrong

we already know what they are

[timid silo]

.

[unique tendon]

Hmm

I don't get it

[timid silo]

we started by identifying the free variables right

which were x2 and x5

[unique tendon]

Y es

[timid silo]

then we let them s and t s,t in R

[unique tendon]

So meaning X2 and X5 is s and t

[timid silo]

now now want to find the other variables x1, x3,... in terms of these

[timid silo]

yeah

[unique tendon]

Hmm

[timid silo]

first equation

3x1 -6s + 5t = -1

agree?

[timid silo]

which you did fine here

all thats left is to sub the s and t

[timid silo]

still not clear?

[unique tendon]

A friend of mine

[timid silo]

did what

[unique tendon]

1

[unique tendon]

2

Where is she wrong

[timid silo]

are these her free variables

[unique tendon]

Yes

[timid silo]

in this matrix do we have x3 and x4 are free variables

[unique tendon]

https://www.chegg.com/homework-help/questions-and-answers/question-2-consider-matrix-left-begin-array-cccccc-3-6-0-0-5-1-0-0-2-0-4-9-0-0-0-3-7-9-0-0-q107719675

There's a chegg entry to this as well

[timid silo]

is that the matrix your friend was working with

[unique tendon]

I have no idea tbh

[unique tendon]

Same as mine

[timid silo]

you gave me a link to this matrix?

[timid silo]

if its the same as yours then its plain wrong the beginning

[unique tendon]

Yes

[timid silo]

eh

what are the free variables again

[unique tendon]

Im new to this topic 🥲

Ok so the free variables are the one's in not leading 1's

So i was wondering here there are no 1's

[timid silo]

non pivot ones

[unique tendon]

I don't know what that means🥲 iv seen it somewhere tho

[timid silo]

or if you want to check with that method of leading ones then you have to further reduce the matrix

[timid silo]

eh okay lets do this then

[timid silo]

do row operations on this augmented matrix

until you have leading ones

[unique tendon]

I reduce it right

Okay lemme do that

[timid silo]

to reduced row echelon form

I end up with
$$\begin{pmatrix}
1 & -2 & 0 & 0 & \frac{5}{3} & -\frac{1}{3}\
0 & 0 & 1 & 0 & 2 & \frac{9}{2}\
0 & 0 & 0 & 1 & \frac{7}{3} & 3\
0 & 0 & 0 & 0 & 0 & 0
\end{pmatrix} $$

[timid silo]

look at these two and tell me what changed

[timid silo]

this

[unique tendon]

Omg same

[timid silo]

do you now see how you can get the free variables

with or without the reduced row echelon form

no need for the leading 1s definition

[unique tendon]

Wait

[timid silo]

okay then now identify for here

[timid silo]

yeah

[unique tendon]

So

[unique tendon]

Here X2 is and x5 is s and t and the other x1 34 are in terms of s and t

[timid silo]

x1, x3, x4 yes

[timid silo]

im about to show you why and a cleaner way to show that

[unique tendon]

Ok perfect thank you so much , are there any resources i can refer to which can further explain to me

[unique tendon]

Ok please do so

[timid silo]

hmm

[timid silo]

give me the x1, x2, x3, x3, x5 first

[unique tendon]

Also is there a shorter way than this for getting the reduced row echelon form

[timid silo]

:Xd: i learnt from many places checkout #book-recommendations for a linalg book

[unique tendon]

x1 = 2s-5/3t-1/3

x2 😒

[timid silo]

row operations are the way to go

[timid silo]

[unique tendon]

x3=9/2-2t

x4=3-7/3t

x5=t

[timid silo]

now fill in
$$\begin{bmatrix}
x_1 \
x_2 \
x_3 \
x_4 \
x_5 \
\end{bmatrix} = s \begin{bmatrix}
? \
? \
? \
? \
? \
\end{bmatrix} + t \begin{bmatrix}
? \
? \
? \
? \
? \
\end{bmatrix} + \begin{bmatrix}
? \
? \
? \
? \
? \
\end{bmatrix}, s, r \in \mathbb{R}$$

copy paste code

[warm shale]

now fill in ![catKing](https://cdn.discordapp.com/emojis/812386361297862747.webp?size=128 "catKing") 
$$\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5 \\
\end{bmatrix} = s \begin{bmatrix}
? \\
? \\
? \\
? \\
? \\
\end{bmatrix} + t  \begin{bmatrix}
? \\
? \\
? \\
? \\
? \\
\end{bmatrix} + \begin{bmatrix}
? \\
? \\
? \\
? \\
? \\
\end{bmatrix}, s, r \in \mathbb{R}$$

[unique tendon]

now fill in ![catKing](https://cdn.discordapp.com/emojis/812386361297862747.webp?size=128 "catKing") 
$$\begin{bmatrix}
x_1 \\2s-(5/3)t-(1/3)
x_2 \\s
x_3 \\(9/2)-2t
x_4 \\3-(7/3)t
x_5 \\t
\end{bmatrix} = s \begin{bmatrix}
? \\
? \\
? \\
? \\
? \\
\end{bmatrix} + t  \begin{bmatrix}
? \\
? \\
? \\
? \\
? \\
\end{bmatrix} + \begin{bmatrix}
? \\
? \\
? \\
? \\
? \\
\end{bmatrix}$$

[warm shale]

iamlegendd

[timid silo]

💀 just fill in the "?"

[unique tendon]

Oh no😂

[timid silo]

lmao

[unique tendon]

Where

Sorry im kinda new ro discord

[timid silo]

s \begin{bmatrix}
? \\
? \\
? \\
? \\
? \\
\end{bmatrix} + t  \begin{bmatrix}
? \\
? \\
? \\
? \\
? \\
\end{bmatrix} + \begin{bmatrix}
? \\
? \\
? \\
? \\
? \\
\end{bmatrix```

these ones replace the ?

[unique tendon]

With?

[timid silo]

uh

[timid silo]

.

if its faster do on paper

this is a nice way to combine all the equations into one

[unique tendon]

Tbh i have no idea what you're asking of me

Im sorry

[timid silo]

x1 = 2s-5/3t-1/3
x2 = s
x3=9/2-2t
x4=3-7/3t
x5=t

right

[unique tendon]

Yes

[timid silo]

x2 can be written as
x2= s + 0t + 0 right

[unique tendon]

Yes

[timid silo]

what about x3

[unique tendon]

0s -2t+9/2

[timid silo]

good

what about x5

[unique tendon]

0s+t+0

[timid silo]

okay good
so now we have
x1 = 2s-5/3t-1/3
x2 = 0s -2t+9/2
x3 = 2s-5/3t-1/3
x4 = 0s - 7/3t + 0
x5 = 0s+t+0

right

[unique tendon]

Yessir

[timid silo]

now look at this again

[timid silo]

see we are repeating the same thing over and over

whats the neat way to group all of that

[unique tendon]

What same thing ..like the s and the t?

[timid silo]

yes

how can i write those as a vector

[unique tendon]

Ooh mhm

[timid silo]

this is the biggest clue ever

[unique tendon]

You mean like this?

[timid silo]

yeah can you now fill in

[unique tendon]

So in the s we like write

2
0
2
0
0

Right?

Ok so how can we use this code

😬

[timid silo]

x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5 \\
\end{bmatrix} = s \begin{bmatrix}
? \\
? \\
? \\
? \\
? \\
\end{bmatrix} + t  \begin{bmatrix}
? \\
? \\
? \\
? \\
? \\
\end{bmatrix} + \begin{bmatrix}
? \\
? \\
? \\
? \\
? \\
\end{bmatrix}, s, r \in \mathbb{R}$$```

just copy paste

[unique tendon]

s \begin{bmatrix}
? \\2
? \\0
? \\2
? \\0
? \\0
\end{bmatrix} + t  \begin{bmatrix}
? \\
? \\
? \\
? \\
? \\
\end{bmatrix} + \begin{bmatrix}
? \\
? \\
? \\
? \\
? \\
\end{bmatrix```

[timid silo]

replace the ? with your answer :Xd:

[unique tendon]

Like this?

x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5 \\
\end{bmatrix} = s \begin{bmatrix}
2 \\
0 \\
2 \\
0 \\
0 \\
\end{bmatrix} + t  \begin{bmatrix}
? \\
? \\
? \\
? \\
? \\
\end{bmatrix} + \begin{bmatrix}
? \\
? \\
? \\
? \\
? \\
\end{bmatrix}, s, r \in \mathbb{R}$$```

[timid silo]

continue

[warm shale]

$$\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5 \\
\end{bmatrix} = s \begin{bmatrix}
2 \\
0 \\
2 \\
0 \\
0 \\
\end{bmatrix} + t  \begin{bmatrix}
? \\
? \\
? \\
? \\
? \\
\end{bmatrix} + \begin{bmatrix}
? \\
? \\
? \\
? \\
? \\
\end{bmatrix}, s, r \in \mathbb{R}$$

[unique tendon]

$$\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5 \\
\end{bmatrix} = s \begin{bmatrix}
2 \\
0 \\
2 \\
0 \\
0 \\
\end{bmatrix} + t  \begin{bmatrix}
5 \\
-2 \\
-5 \\
-7 \\
1 \\
\end{bmatrix} + \begin{bmatrix}
-1/3 \\
9/2 \\
-1/3 \\
0 \\
0 \\
\end{bmatrix}, s, r \in \mathbb{R}$$

[warm shale]

iamlegendd

[unique tendon]

Yay

It this it?

[timid silo]

yay

[timid silo]

good job

[unique tendon]

OmG thank you so much!

[timid silo]

yw

[unique tendon]

So is there like a place where we get this codes from?

[timid silo]

you mean latex?

https://www.overleaf.com/learn

if the linalg you'll see it in your course soon

[unique tendon]

Hmm okay

One last question

Why were they saying there are 3 variables in the question

[timid silo]

did we see 3?

[unique tendon]

Nope

[timid silo]

probably a trick question

[unique tendon]

Did you check this?

[timid silo]

i dont have access to that site

the only thing i saw is the matrix

[unique tendon]

Ohk

Well thank you so much

Is there any other information i could get to know

[timid silo]

my only advice is to look up linalg parametric solutions questions and answers

or just get a decent textbook recommendation from #book-recommendations

[unique tendon]

Great 😊

How do i close this thread

[timid silo]

.close

[unique tendon]

.close

[obtuse pebble]

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[brazen raptor]

Hey I need help solving this function: f(x)=ln(1+3(x−1)^2). Im having trouble since the argument seems to never be verified because delta is always negative. Any help?

[royal basin]

wym by "solving this function"

[brazen raptor]

i need to find the largest intervals in which it is invertible and finding the equations for those inverted functions

sorry english isnt my first language

the solution starts by saying that this function is defined for all Real numbers

but i dont get how

since for a natural logarithm to even exist, its argument needs to be greater than 0

[timid silo]

well thing is, that (x-1)^2 changes stuff

your domain is all of R

[brazen raptor]

so i shouldnt try to solve for x

[timid silo]

i'd say start by getting a good idea of how the function looks like

the thing is, that 1 + inside also changes things quite a bit

3(x-1)^2 is never negative

the lowest value 3(x-1)^2 can be is 0

which means the lowest value f(x) can be is also 0, because ln(1) = 0

so this eliminates you having to deal with any negative values for f(x)

[brazen raptor]

right

so this

[timid silo]

yeah so

[brazen raptor]

is totally futile

[timid silo]

I mean, i wouldnt really be needing to solve anything if i was doing it myself

because intuition is all you need here, probably

[brazen raptor]

im just having trouble visualizing the function

i guess im not really sure what the single things actually do on the plain

[timid silo]

yeah we will get to that

[brazen raptor]

i know that x^2 basically copies and flips the function and makes it symmetrical right?

[timid silo]

yeah

do you know about the graph of $\m \ln{\abs x}$ @brazen raptor

[brazen raptor]

so now i have something like this

[timid silo]

because in reality your function is a slightly altered version to what theat looks like

[timid silo]

yea that pretty much is it

[brazen raptor]

ok

next step i think would be to multiply by 3

[timid silo]

the rest doesnt matter

multiplying by 3 and adding 1 doesnt really change the shape

[brazen raptor]

oh

but the position

[timid silo]

you need to know three things

[timid silo]

1- your graph is in this shape

2- your function intersects the x axis at x = 1

[brazen raptor]

because of the +1?

[timid silo]

yes

[brazen raptor]

right

[timid silo]

so you already know that the "branching" begins at x= 1

[royal basin]

does it have a cusp tho

idt it does

[timid silo]

idts

[timid silo]

im guessing this was just an approximate drawing

[brazen raptor]

yes

shouldnt the function sort of compress if i multiply it by a number greater than 1?

[timid silo]

yeah but like this doesnt matter for your question

[brazen raptor]

right right

[timid silo]

you want to find the largest interval for which it is invertible

[brazen raptor]

yes

[timid silo]

scaling (x-1)^2 doesnt change that

[brazen raptor]

so one is smaller than x=1 and one is greater than x=1?

talking about the intervals in which it is invertible

[timid silo]

ye

both are invertible intervals

[brazen raptor]

yes

and noi i just need to express what happens in those intervals

and since i have basically two normal logarithms but flipped and slid to the right in both intervals i should represent that in the form of 2 equations

but theyre flipped so i have to do ^-1 too

how did e get there

[obtuse pebble]

@brazen raptor Has your question been resolved?

[brazen raptor]

.close

[obtuse pebble]

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[long jacinth]

Find all the clicks on the graphs in the image

someone pls

i cant find anywhere what clicks are or cliques

[obtuse pebble]

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[heavy bramble]

Will it be going north leaning east at a 80 degrees angle at 608km/h ?

[heavy bramble]

bc the hypotenuse is 608

and tan(600/100) incerse is 80 degrees

[obtuse pebble]

@heavy bramble Has your question been resolved?

[obtuse pebble]

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[drifting roost]

hey guys, how to solve x< x^2 < |x|

[drifting roost]

x<x^2 means x-x^2<0 or x(1-x) < 0

this only holds between (-inf,0) and (1,inf)

the other part i forgot how

[rich herald]

And x<|x| only holds between (-inf,0)

[drifting roost]

what about x^2 < |x|

that holds for (0,1)?

actually it holds for (-1,0) and (0,1)

okay but how do i formally solve it though?

[rich herald]

You could divide everything by x which would give 1<x<1 when x is positive and 1>x>-1 when x is negative

The first one has no solutions

[drifting roost]

u can't do that?

u don't know if x is positive or negative

dividing without knowing its sign is kinda undefined

also |x|/x is sgn(x)

[rich herald]

Yeah but I consider when x is positive and when x is negative

Seperatly

[drifting roost]

Hmm okay sure

[obtuse pebble]

@drifting roost Has your question been resolved?

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[frigid gale]

e

[royal basin]

!status

[obtuse pebble]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[frigid gale]

1

[royal basin]

are you familiar with the coordinate plane and graphing linear equations (and maybe inequalities) on it?

[frigid gale]

maybe not

[royal basin]

then this problem is impossible for you right now, unfortunately.

[frigid gale]

k

.close

[obtuse pebble]

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[wise relic]

who please know the definition to do it

[fluid snow]

You can just cancel u_2 + 7

Then u_3 + 7 etc.

Imagine cancelling a lot of diagonals

Which two terms will be left?

[wise relic]

it will left

(u2007 +7)/u1+7

@fluid snow i understanded

Broo Thanks Man

❤️

[covert zealot]

type .close to close this help channel. : )

[obtuse pebble]

@wise relic Has your question been resolved?

[obtuse pebble]

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[timid silo]

Hello, I have a question about Taylor series, a Taylor series has this form:
$f(x)=\sum_{k=0}^{\infty} a_k (x-a)^k$ However, this is only true when x is between a-R and a+R, where $R = \lim_{n \to \infty} \frac{a_{n+1}}{a_n}$

[warm shale]

Şêro

[timid silo]

Now the part I don’t understand is to find R

Can someone give me maybe an example?

Also I know how to find a_k

It’s just the kth derivative evaluated at x=a divided by k!

[viscid gull]

Do u have a specific problem in mind

[timid silo]

Well let’s say f(x)=sin(x) and I choose a to be 0

Then what’s R

I just need to know what that limit means

How do I calculate that

What’s $a_{n+1}$

[warm shale]

Şêro

[viscid gull]

in that case you use the next nonzero term relative to a_n for a_(n + 1)

So for instance $a_n$ would be $\frac{x^{2n + 1}}{(2n + 1)!}$

[fossil crag]

1/R is the limit you wrote, with absolute values

[timid silo]

Oh yeah I forgot the absolute value

[fossil crag]

(and rather use limsup but that's out of the point for now)

[timid silo]

But it’s when n approaches infinity

So what would be the next nonzero term

[viscid gull]

Ye im getting to that

[timid silo]

Can you just give an example

[viscid gull]

$a_{n + 1} = \frac{x^{2n + 3}}{(2n + 3)!}$

[warm shale]

992qqoloy

[timid silo]

From where?

[warm shale]

992qqoloy

[viscid gull]

From the mclauren series of sin(x)

[fossil crag]

$a_{2n} = 0, a_{2n+1} = \frac{(-1)^n}{(2n+1)!}$

[warm shale]

rafilou2003

[viscid gull]

Right woops

Wait actially

[timid silo]

What’s the McLauren series

I just know the Taylor formula

[viscid gull]

Taylor series centered at 0

[fossil crag]

yes it's the taylor formula, $a_n = \frac{f^{(n)}(a)}{n!}$

[warm shale]

rafilou2003

[timid silo]

So what’s n+1?

Just plug-in n+1

?

[fossil crag]

yes

[timid silo]

But how can we find such a complicated limit

I mean we divide that all by the n+1 term too

And then take the limit as n approaches infinity

[fossil crag]

So R is the limit of |an/an+1|

[timid silo]

Yea as n approaches infinity

But how can we find that

[fossil crag]

but here, since all even terms are 0

we only take it for odd terms

[timid silo]

I mean what’s the Infinitiv derivative evaluated at 0

[fossil crag]

there is no infinite derivativ

[timid silo]

But we wrote $f^{(n)}$

[warm shale]

Şêro

[timid silo]

And n approaches infinity

[fossil crag]

we take the lim of f^(n)/f^(n+1)

f^(inf) doesn't have to exist

[timid silo]

Huh

But I thought we take the limit of a_n+1 / a_n

[fossil crag]

as I said, that would be in order to find 1/R

if you want R, you take an/an+1

[timid silo]

Okay yeah

But $a_n = \lim_{n \to \infty} \frac{f^(n)(x=a)}{n!}$

[warm shale]

Şêro

[fossil crag]

lim an = that

but who said lim an had to exist?

[timid silo]

Huh

But that would mean R doesn’t exist

[fossil crag]

r can exist

(-1)^n / (-1)^n = 1, so the limit is 1, but does lim of (-1)^n exist?

[timid silo]

Yeah but what’s R in this case

I mean we take the infinith derivative

[fossil crag]

we don't

we take a limit of ratios

lemme show you

[timid silo]

Yea but in that ratio is a_n and a_n is defined by the nth derivative of f evaluated at x=a divided by n!

[fossil crag]

as I said, lim (an/an+1) can exist without lim(an) existing

[fossil crag]

back to this

[timid silo]

Nope

It doesn’t

[fossil crag]

?

[timid silo]

We n goes to infinity (-1)^n doesn’t exist

It changes between 1 and -1

[fossil crag]

yes, so lim (-1)^n doesn't exist

[timid silo]

Okay I see

[fossil crag]

ok so we don't care about lim(an) because it probably won't exist

we want lim(|an/an+1|)

[timid silo]

What about n+1

[timid silo]

Yes how do we get that

[fossil crag]

we don't care about lim(an+1) either

[timid silo]

Okay

[fossil crag]

well, we compute for a given n, an/an+1

[timid silo]

So does something cancel out in the ratio or what happens

[fossil crag]

Here as I said, since even terms are 0, we discard them

[timid silo]

But in our case n approaches infinity

[fossil crag]

yes, so before we approach infinity, we compute it for a fixed n

[timid silo]

Huh

Why that

[fossil crag]

and THEN

we make the quantity we found approach for n -> infinity

[timid silo]

Is that even allowed

[fossil crag]

yes

if an = bn

then lim(an) = lim(bn)

[timid silo]

Hmm

[fossil crag]

so before taking the limit

[timid silo]

I see your point

[fossil crag]

we compute an/an+1 for any given n

[timid silo]

So we choose like an easier n

Compute it

And then find the limit as n approaches infinity of that

[fossil crag]

yes, while keeping the n general

so, for any n, an/an+1 = ...

we will use the formula $a_{n} = \frac{(-1)^n}{(2n+1)!}$

[warm shale]

rafilou2003

[fossil crag]

(we discarded terms = 0)

did you compute an/an+1?

[timid silo]

.close

[obtuse pebble]

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[harsh goblet]

Hi! How do I calculate this?

[harsh goblet]

(without a calculator)

[timid silo]

all of the roots are like some powers of 4 or 5

you can represent all of them in exponent form and try to see good simplifications

[harsh goblet]

I got here but its probably wrong

So I get them inside the roots and I get this:

[harsh goblet]

this is good

[harsh goblet]

but this isn't

wait why does this give false?

[jagged falcon]

Can you convert everything in terms of some powers of 2 and 5?

[harsh goblet]

like what?

[jagged falcon]

For example, (0.625)^1/3=(625/1000)^1/3=(5^4/5^3*2^3)^1/3=(5/2^3)^1/3=cbrt(5)/2

[harsh goblet]

Could you write this in wa and send the picture so I can understand what's happening here?

[obtuse pebble]

@harsh goblet Has your question been resolved?

[obtuse pebble]

@harsh goblet Has your question been resolved?

[obtuse pebble]

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[weary saffron]

there are 30 students in a class

[weary saffron]

the teacher asked every student in which day they were born

as in sunday, monday etc

and the year and month dont matter

what is true to say?

A/ every day of the week at least one student was born

B/ at least in one day of the eek there were born less than 3 students

C/ at least in one day of the week there were born more than 4 students

D/ in two day sof the week there were born at least 5 students

[jagged onyx]

A is obviously False

[weary saffron]

i agree

[jagged onyx]

WTF

[weary saffron]

💀

[jagged onyx]

anyways

uhh

[weary saffron]

couldnt they be born in like the same day

all of them

[jagged onyx]

yea

[weary saffron]

how am i supposed to know

[past sand]

Distribute 30 things into 7 buckets

[weary saffron]

maybe by elimintation

[past sand]

You'll have at least floor(30/7) in one bucket

[jagged onyx]

Dude

what

[past sand]

Yes, and that eliminates two of the options

And for B/ you can just find a distribution where it's false

[jagged onyx]

how does that eliminate 2 options?

[weary saffron]

yea

[weary saffron]

how

[jagged onyx]

it only eliminates A

[past sand]

If all students are born on Monday, then D is false too since all other days have no birthday

D would need for example 25 on Monday and 5 on Tuesday

[jagged onyx]

yea ig

[weary saffron]

how do i choose between B and C

i think its C

[jagged onyx]

what is the avg distribution of students?

[past sand]

I practically gave you the answer

[jagged onyx]

yea

[weary saffron]

oh i see

thanks

.close

[obtuse pebble]

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[forest ridge]

if it is 3, how so?

[forest ridge]

bc i thought it was 5

since it says consecutive

bc like (5,6,7)

but i realised it says that AC and BC are consecutive

so 6 has nothing to do with this whole thing

nvm got it thank you

.close

[obtuse pebble]

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[wind topaz]

for any t(placeholder for an angle) number. sin(-t)+cos(-t)=

[wind topaz]

and the answers are the following

sint-cost; -sint+cost; cost+sint; -sint-cost

[ashen dove]

Cos (-t)=cos t
Do u know this

[wind topaz]

so so

sleep deprivatin set me back like a week in math

[wind topaz]

so logically the answer

is

sint+cost

[ashen dove]

sin (-t) isn't sin t

[wind topaz]

uhoh

so -sint?

which makes it -sint+cost

[timid silo]

yes

[wind topaz]

aightt

[pine sail]

But once you're told it isn't the other three alternatives, of course you'd know which one it is...

[wind topaz]

my genius is coming back

[wind topaz]

party pooper ahh

[pine sail]

Whatever you say, you can ignore me and close this if you think so.

Though of course, the ideal thing to do would be look at the graphs of these functions and figure out for yourself which one's even and which one's odd.

[wind topaz]

i remember cos being odd

and sin being even

[pine sail]

It's the other way.

[wind topaz]

rightg

mb

[pine sail]

Thought your genius was back.

[wind topaz]

you're funny

[wind topaz]

it's coming back

let it cook

[viscid gull]

Jesse it's time to cook

[wind topaz]

help me man

[viscid gull]

oh I thought u had it already

[viscid gull]

.

[wind topaz]

i do

but i need help on like

8 more of these tiny lil things

[pine sail]

You can share them, no help without looking at the question.

[wind topaz]

soz i was

don smth

for any t number, for which there's cos=/0, we have tan(-t)=

i can't cook

nah i can't even light the gas on this

[pine sail]

Again, if you'd just read what odd and even functions mean and their graphs, it'd be much easier.

Because it seems to me all these problems are solely focused on that.

[wind topaz]

ya

they arwa

i should get more sleep

i'll cook for the next week

trust

[viscid gull]

Well done Walter white

Sab and Sab the 2 best cooks north of the border

[wind topaz]

tyy

.close

[obtuse pebble]

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[royal basin]

is that actually uniquely determined

[pine sail]

You need to know if your answer is right?

It is.

[gaunt birch]

no 61 is the right answer

i just don't know how to get it

[pine sail]

Yeah it is.

Well, 31 is prime, how about start with that.

[gaunt birch]

yea

[pine sail]

What do you know about prime numbers?

[royal basin]

ah, yeah, that tracks.

[gaunt birch]

no factors other than 1

bro i forgot what prime numbers were..

[pine sail]

Are you sure?

[pine sail]

This isn't right.

[gaunt birch]

oh

idk

[polar fossil]

(5 is a factor of 5)

[gaunt birch]

it cannot be divided

by anything other than itself

[polar fossil]

yeah so 1 and itself are the factors

[gaunt birch]

okay

so it's a prime number

what about it

i remember my teacher going like a is 31

but i can't rememebr

[pine sail]

You have that prime number as a product of two numbers. You also know that prime numbers only have 2 factors.

[gaunt birch]

1 and 31

[royal basin]

if two natural numbers multiply to give a prime, that could ONLY have happened if one of the numbers was 1 and the other was that prime.

[gaunt birch]

so

b is

1

[pine sail]

No?

[gaunt birch]

or is it the opposite way

bro idk i'm trynna remember when my teacher said..

i think she got the answer wrong, she ended up with 31 i think

[pine sail]

I mean, you can totally figure it out freshly rn.

The answer's right, pretty sure.

[gaunt birch]

but i don't get it

[pine sail]

.

[royal basin]

so b is 1
no

the two numbers being multiplied aren't a and b but a and a-b

and a-b is what's equal to 1 (bc it's known to be smaller than a)

[gaunt birch]

huh

[royal basin]

it says a(a-b)=31

[gaunt birch]

so a-b is supposed to be 1

[royal basin]

yes a-b is 1 that's what i said

[gaunt birch]

okay

OMG I GOT IT

SO A IS 31 AND B IS 30?

OMG YOU'RE A GENIUS

BRO I'VE BEEN STUCK ON THIS QUESTION FOR THE PAST HOUR

GOD BLESS YOUR SOUL

YOU SAVED ME

THANK YOU

.close

[obtuse pebble]

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[timid silo]

(i) of (P10)

Why do we have this condition here?

Oh I under

Understand

a is not an element of P

.close

[obtuse pebble]

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[obtuse pebble]

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[eager pivot]

how would i even start to solve this problem 😭

[rich herald]

To get x_2, you need to draw the line tangent to f at x=x_1

x_2 is the intersection of this tangent and the x-axis

And you can do this again to find x_3

[obtuse pebble]

@eager pivot Has your question been resolved?

[obtuse pebble]

@eager pivot Has your question been resolved?

[obtuse pebble]

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[dense tinsel]

Dont know how to solve

[lone echo]

are the 86º angles EDA and EBC?

[dense tinsel]

sorry that's an E not a G

yeah you're right

i tried everything, I dont know how to properly solve it

[lone echo]

okay so.
Angles:
DAE + AED + 86º = 180º
CDE + EDA = 180º
CDE + DEB + 86º + BCD = 360º

[dense tinsel]

YES CHEF

[lone echo]

you also have that on the big ass triangle:
sin86º/(x+8cm) = sinBCD/(x+5cm)

[dense tinsel]

stop

stop

stop

what is sen