help-13

[cedar kiln]

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[livid hound]

show work

[civic geyser]

wait i think i know the problem

nvm thats not it

[dire geode]

,rotate

[livid hound]

sign error near the end

you turned the -36/2 into +36/2

[civic geyser]

Ohh ok

it says my answer is wrong

I got 33/4

<@&286206848099549185>

[livid hound]

show full work

show exactly what you're doing that leading up to that value

[livid hound]

,rotate

[livid hound]

leave out no details

[civic geyser]

to get the area you do b-a

right

s[b-a]

[livid hound]

that's all fine

how are you going from
$$\br{\frac 34 - \frac92 + 6} - \br{\frac{48}{4} - \frac{36}{2} -12}$$
to your value

[wraith dagger]

ℝam()n()v

[livid hound]

how are you simplifying that to get 33/4

[civic geyser]

i used a graphing calculator

😄

[cedar kiln]

@civic geyser Has your question been resolved?

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[crimson sedge]

Hello, A = 23!
In how many different combinations will A be a product of consecutive numbers?

[crimson sedge]

The answer seems to be 3, but how?

[sudden pivot]

23 fact. ?

[crimson sedge]

Yep

[sudden pivot]

what does combination mean here

[crimson sedge]

like 1.2.3....23 and 2.3.4......23

do you think the question lacks something?

the other one seems to be 5.6.7.....24 but how do we determine if these are all

[sudden pivot]

honestly dont know

can you give some context please

[crimson sedge]

sure

do you know what consecutive is

[sudden pivot]

yes

n, n+1,n+2?

[crimson sedge]

yep

so we'll need to write 23! as the multiplication of consecutive numbers like 2.3....23 and 1.2.3....23

[sudden pivot]

these are same

i think

[crimson sedge]

if we take the 1.2.3.4 as 24 in the beginning and add it to the end it will be 5.6.7...23.24

[sudden pivot]

Oh ok

[crimson sedge]

yes so these make 3 combinations, and I'm asking how do we determine there's just 3

[sudden pivot]

intuitively these are the only possible

[crimson sedge]

haha yeah

[sudden pivot]

23!/1, 23!, 24!/4!

new combination must have int. bigger than 23

[crimson sedge]

yes

[sudden pivot]

Otherwise it is less than 23!

[crimson sedge]

I'll go eat real quick brb

[sudden pivot]

ok

so we nneed to prove that 23!=m!/n! has only one solution if m and n are int

oh 24!/4! is 1.2.3. 4.5.../1.2.3.4=5.6.7.8.9...
But it is also
1.2.3.4.5.../1.2.3.4=1.2.3.4.5.../24 = 23!

we remove some largest nums using ONE factorial

Btw i noticed

24!/4!=(4!)!/4!

it is equal to (4!-1)!

Wow

It doesnt help

Nvm

Product of removed numbers from start MUST equal product of removed nums from end

So since we use only one factorial we cant remove things like 24.25(4!.5.5)

That means that with one factorial we can only remove 24 but not 25

YES

My "proof" is a mess but hopefully you can see an idea

@crimson sedge

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

back

[crimson sedge]

exactly

@sudden pivot thank you so much for your help

if I recall correctly someone here tried to solve it with polynomials and derivatives, they were probably chai trex

I will work on what you said aol

[cedar kiln]

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[wary hamlet]

how do i do this

[spice kraken]

Δ = 18abcd – 4b³d + b²c² – 4ac³ – 27a²d².

[wary hamlet]

wat

[twilit knot]

What class/year is this?

Depending on that, you could try guessing with remainder/factor theorem.

[wary hamlet]

yr12

idk what class

its a hsc question

[twilit knot]

Do you use factor/remainder theorem

Basically trying in numbers until it works and then factorising the factor out

[wary hamlet]

ye

long division

[twilit knot]

Yes, but before dividing you try numbers

Well, if you haven't been taught another method, I'd suggest you try that

[wary hamlet]

ye

but theres a value a in it

[twilit knot]

OH

mb

Idk, what's the discriminant for cubics?

[twilit knot]

Can you explain this

[spice kraken]

I googled it

[wary hamlet]

nice

<@&286206848099549185>

[vagrant elbow]

Do you know how to work with derivatives?

[wary hamlet]

yes

[vagrant elbow]

Impose a condition on a such that the cubic must have two extrema, the values of which are of opposite signs

[wary hamlet]

extrema?

max min

ok

[vagrant elbow]

the general term for a maxima/minima

indeed

[wary hamlet]

so to find max min

first diff=0

[spice kraken]

you should find two local max/min points

[wary hamlet]

0,-2a/3

idk what to do after

cus they aint even x intercepts

[spice kraken]

,w graph x^3+x^2

[wraith dagger]

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[spice kraken]

,w graph x^3+x^2-0.05

[wraith dagger]

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Upgrade to WolframAlpha Pro!

Requested by _wherewolf_

[spice kraken]

do you see how to ensure 3 x-intercepts?

[wary hamlet]

yes

but the equation is x^3+ax^2-1

and we only get min max points

which aint x intercepts

[spice kraken]

notice the min max points of the graph

and their relationship with x intercepts

[wary hamlet]

are they in the middle?

[spice kraken]

yep

the second x intercept must be betweeen the min/max points

how do we use that to solve the problem?

[wary hamlet]

0+-2a/3

divided by 2

so one intercept is -a/3

[spice kraken]

wait no

the intercept is between

but not necessary middle

[wary hamlet]

what is it

[spice kraken]

,w graph 10(x-1)(x-2.5)(x-3)

[wary hamlet]

k

[wraith dagger]

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[spice kraken]

it just have to be between

how do we ensure it pass through x=0 between the min/max points?

this will be easy to understand using the intermediate theorem

[wary hamlet]

whats that

[spice kraken]

it basically said if you have two points of a function a,f(a) and b,f(b)

there exist point c between a and b such that f(c) is between f(a) and f(b) for any f(c) between f(a) and f(b)

[wary hamlet]

so what would the middle intercept be?

[spice kraken]

it depend on a

and we just need to make sure middle intercept exist

the middle intercept must be y=0

we want y=0 between two min/max points

let's consider three cases

1. What happens if both min/max points are positive?

2. what happens if both min/max points are negative?

3. What happens it one is positive and other is negative?

[wary hamlet]

we dont know a

thats the thing

[spice kraken]

don't care about a for now

try drawing a cubic graph for the three cases

you'd see how the min/max value affect the middle intercept

[wary hamlet]

ok

[spice kraken]

btw by positive and negative I mean of the y value

not x value

[wary hamlet]

got the answeer

i understand now

👍

.close

[cedar kiln]

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[cedar kiln]

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[wanton summit]

please help with iii

[wanton summit]

<@&286206848099549185>

[cedar kiln]

@wanton summit Has your question been resolved?

[cedar kiln]

@wanton summit Has your question been resolved?

[cedar kiln]

@wanton summit Has your question been resolved?

[lunar lynx]

Use the fact that in terms of its argument $\theta$, z can be written $e^{i\theta}$.

[wraith dagger]

Enemagneto

[wanton summit]

already did this

sorry

[lunar lynx]

Actually don't even need that.

[wanton summit]

i just used the fact that its a rhombus

so theta becomes theta on 2

[lunar lynx]

Show then.

[wanton summit]

perpendicular bisector means theres a right angle

at

at OMA

where M is the midpoint of OC

[lunar lynx]

Why do you wanna do it with geometry ?

[wanton summit]

its a lot more elegant

[lunar lynx]

It's a cinch with algebra.

[wanton summit]

ok how do you do it algebriaclally

its asking about the modulus

not the argument

[lunar lynx]

You mean... in general, geometry is lot more elegant than algebra ?

[wanton summit]

of course

geometric solutions are more fun to find too haha

[lunar lynx]

Yeah. I can read. Lol

[wanton summit]

no no

not insulting you im

just confused

HAHA

[lunar lynx]

Many would disagree. Nevertheless, we digress.

[wanton summit]

really

[lunar lynx]

What's your approach geometrically ?

[wanton summit]

well maybe this level of algebra

well complete the rhombus draw the lines

[lunar lynx]

With algebra, I can do it in my head. 😐

[wanton summit]

we have proven BOA is theta from b

and a rhombus splits theta in 2 from a diagonal

let M be the midpoint of OC

diagonals bisect perpendicular on a rhombus

so OMA is a right angle

we work out costheta with right angled trig

and multiply by two

for the magnitude of OC

im not sure how to do it algebraically

algebraically thats twice now

[lunar lynx]

Wait... I'm confused. So you have done it geometrically? What's your doubt then?

[lunar lynx]

Just put z = cos(x) + i sin(x), i have replaced theta with x for ease of writing.

In |z+z^2|.

[wanton summit]

Wdym whats my doubt i already said i’ve done it i just forgot to close the channel

sorry about that

[wanton summit]

what about the radius

or modulus

[lunar lynx]

Gosh. Lmao.

[lunar lynx]

That just means that you take mod at end.
anyway, your |z| = 1

[wanton summit]

oh

right yes

how do you arrive at the half angle expansion for cos though

in your head in particular

[lunar lynx]

|z+z^2| = |z| |1+z| = 1 | 1 + cosx + i sinx| = Sqrt(1 + cos^2(x) + 2cos(x) + sin^2(x)|
= sqrt(2(1+cosx))

Now, use that 1+cosx = 2cos^2(x/2)

There's your answer.

Now, i gotta go. Close this after you are done.

[wanton summit]

.close

[cedar kiln]

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[cedar kiln]

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[fair crown]

a1! * a2! * ... * an! <= (a1+a2+...+an)! It should be proven by the method of induction, so we have the base for n=1 and it's trivial and it's equal a1! = a1!, and we have a hypothesis that it is correct for n=k, so we should prove it for n=k+1

[fair crown]

I think we should probably use the a_k+1 ! member and say, a_k+1! = a_k+1 * a_k * a_k-1 * ... * a2 * a1, and the hypothesis covers the a_k-1 * ... * a2 * a1 part

But I am not sure, it's just an idea

so it's a_k+1 * a_k!

.close

[cedar kiln]

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[bitter wadi]

hi i have a quick question on parametrization

[bitter wadi]

so lets say I have a surface S: x+y^2-z=0

and lets say that $0 \leq x \leq 1$ and $0 \leq y \leq 2$

[wraith dagger]

jriver

[bitter wadi]

and i want to parameterize this in terms of $u$ and $v$, where $x = u$ and $y=v$

[wraith dagger]

jriver

[bitter wadi]

what would my position function $\vec{r}(u,v)$ look like?

[wraith dagger]

jriver

[bitter wadi]

thanks in advance for the help

btw in case i did not phrase this clearly, this problem is from: https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/surface-integrals-introduction/v/surface-integral-example-part-1-parameterizing-the-unit-sphere

im just confused on how he created his $\vec{r}$ function

[wraith dagger]

jriver

[cedar kiln]

@bitter wadi Has your question been resolved?

[cedar kiln]

@bitter wadi Has your question been resolved?

[bitter wadi]

rip

[dire geode]

Can you post screenshots

Or timestamp

[bitter wadi]

its pretty much what i wrote above but sure

10:02 is timestamp

nvm i get it now

.close

[cedar kiln]

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[cedar kiln]

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[white latch]

Hello there, so I am currently learning calculus and am hell bent on the idea of learning the proofs, whilst learning the material as I find it fascinating. Anyways, so I am in the midst of learnign the power rule for the derivative where we have x^n, differentiate it and we have nx^(n-1). Now I know that if I implement the definition of the derivative I will ultimately get the same result as nx^(n-1), after going through some nasty algebra. After this, I noticed another way of proving it which was using the binomial expansion theorem, which I have no knowledge of as my Pre Calculus 12 Course did not cover combinatronics. With this said for learning future calc proofs and calculus will the fact that I don't knwo combinatronics come back to haunt me? Should I learn it now or perhaps later in my own time?

[dire geode]

It can't hurt

This isn't a math problem. Go to #discussion for study tips

[white latch]

May I speak with you in DMS for a second?

[dire geode]

No

[white latch]

May I speak with you in discussion?

[dire geode]

Anyone in #discussion can help you

[white latch]

Right, and not discrediting anyone, but I've seen you on this server and you have a great deal of knowledge according to what I've seen, so I would appreciate if you had the time it would be nice, but if not no worries and have a good day or night. Heading to the discussions channel.

[dire geode]

No, anyone else can be just as helpful

[white latch]

Alright thanks. Cya.

[crystal raptor]

Just at the very least learn binomial expansion

[cedar kiln]

@white latch Has your question been resolved?

[cedar kiln]

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[fair crown]

the original question is to look at arrays of length 6, and to answer if we have more arrays that the sum of all digits is 27 or that the sum of first three digits is equal to the sum of last three.
the first answer is binomial coefficient 6 + 27 - 1 choose 27. i'm trying to solve the other part now, i tried writing the numbers to try and see some kind of logic but just from 0,0,0 to 1,1,1 we have like 38 combinations or more, maybe i missed some, from (0,0,0),(0,0,1),(0,1,0),(1,0,0) combinations to (9,9,9)
that was my initial idea, and we have to restrict a,b,c,d,e,f to digits from 0 to 9

[fair crown]

abcdef where a+b+c=d+e+f

.close

[cedar kiln]

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[hot python]

7:30 i think

[fathom night]

Do you know what the minute hand is?

short line = hour
long line = minute

[cedar kiln]

@hot python Has your question been resolved?

[hot python]

oh no i cant read

let me

so like

3?

1:30

7:30

10:15

10:45
but these are my options

[obsidian coral]

Do you know which way is counter clockwise?

First do you know how to read an analog clock?

[steel seal]

Lol

[cedar kiln]

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[crimson sedge]

i am stuck on these ! could someone pls help ?

[hearty arch]

uh, which question

[crimson sedge]

all of them, idk how to do it

[hearty arch]

is there something in particular that is confusing you?

why don't you give it a shot and then if you get stuck somewhere come back with your work so we can help you through it

look up a formula for the average rate of change, that first question should be fairly straight forward from there

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

i have, but it didnt rlly gelp idk 😭

[hearty arch]

i mean you use 2 coordinates to get the rate of change right?

$rate of change: \frac{y_2-y_1}{x_2-x_1}$

[wraith dagger]

Wumbo

[hearty arch]

then plug the coordinates into this equation to get your rate of change

[crimson sedge]

ohh

wait

i got

0.5

[crimson sedge]

%

^

or 1/2

would that be it then ?

[hearty arch]

it's not a percentage, it's just a value

which one were you doing?

[crimson sedge]

i didnt mean to put thst sign

the first

a)

[hearty arch]

what are the 2 given points on that graph?

[crimson sedge]

(1,6) and (-1,2)

[hearty arch]

i think that 6 should be a 5

[crimson sedge]

OH

so 1?

[hearty arch]

so you have (5-2)/(1-(-1))

right?

[crimson sedge]

yes

[hearty arch]

okay so then what's the average rate of change over the interval [-1, 1]

[crimson sedge]

oh wait

3/2

[hearty arch]

yesss

[crimson sedge]

oh yayy

[hearty arch]

okay try the next one

[crimson sedge]

oki

[hearty arch]

same thing different coordinates

[crimson sedge]

7/3

[hearty arch]

correct!

the third one is only slightly trickier

but it's the same thing, different coordinates

just mind your signs

[crimson sedge]

2/-1

but

that dosent seem right ?

wait

[hearty arch]

should be -2/1 but yes

[crimson sedge]

yes

[hearty arch]

yes

[crimson sedge]

i meant

thats the answer then?

[hearty arch]

mhm!

the graph is going down, so the rate of change is decreasing

[crimson sedge]

ohh okay

[hearty arch]

between that interval

[crimson sedge]

what abt the number 4 question?

[hearty arch]

so for this one, it's the same question as before, but instead of a physical graph, it gives you the function of the graph

and the interval

use the given choices to figure out which interval gives you a negative rate of change

[crimson sedge]

uhh

[hearty arch]

you can use the function itself to find the coordinates by plugging in the intervals

[crimson sedge]

how could i do this

[hearty arch]

so for the first one you have -3 and 4

[crimson sedge]

ohhhh

okay

[hearty arch]

plug those into your equation to find their y counterparts

yesss

then find the rate of change

and figure out which interval has a negative rate of change

[crimson sedge]

would the other one bc 3,5 then?

be*

[hearty arch]

the first coordinate would be (-3,5), yes

[crimson sedge]

oki oki

wait negative 3?

[hearty arch]

you're talking about 4a?

[crimson sedge]

ni just

like

hold on

for the given

like

(3,5) and (-3,4)?

or uh

idk

[crimson sedge]

^

[hearty arch]

for a your x-coordinates should just be the intervals

[crimson sedge]

i dont understand

[hearty arch]

-3 should be the x1 coordinate and 4 should be the x2 coordinate

[crimson sedge]

ohh

[hearty arch]

you have an interval for the function right? -3<t<4

[crimson sedge]

yes

[hearty arch]

those are your x values that you're using

for your 2 coordinates

[crimson sedge]

yes that makes sense

[hearty arch]

plug -3 into h(x) and it spits out a value, that's your y1 coordinate, and you do the same with x=4 to get your y2 coordinate

[crimson sedge]

ohhh

oki

uh

[crimson sedge]

i got 5 and 54

is this how yo do it ?

@hearty arch

<@&286206848099549185> 😭

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[nova pendant]

Is the answer 3/2 or 3/root10

[nova pendant]

Idk if I’m right or the teacher is right

Seems off

[kindred storm]

Show your work.

What have you done so far?

[nova pendant]

This is her work but if I follow it then idk where 2 came from in the bottom

I get root 10

[kindred storm]

OK, let me try.

[nova pendant]

3 squared is 9 plus 1 is ten

[kindred storm]

(x - 1)^2 + (3x + 1 - 1)^2 = x^2 - 2x + 2 + 9x^2 = 10 x^2 - 2x + 2
20 x - 2 = 0
20 x = 2
x = 1/10

(1/10 - 1)^2 + (3(1/10))^2 = 81/100 + 9/100 = 90/100 = 9/10

sqrt(9/10) = 3/sqrt(10).

So, I think you're right.

[nova pendant]

😭

So annoying this TA keeps getting stuff wrong and she’s the one grading my exams

Also one more thing I don’t remember how to solve

[gentle flower]

happened to me once, a quick email to the professor fixed everything

[kindred storm]

Plus, you can prove it's not 3/2 by using x = 0, so you get (x, 3x + 1) = (0, 1), which is 1 from (1, 1).

So, the minimum distance is less than or equal to 1.

[nova pendant]

Where do I even start here

[gentle flower]

derivative of the inverse

[nova pendant]

Ok

So flip the variables of f(x)=16x+cos(2pix)

And solve for y?

[tropic oxide]

❌

gonna be hard

[nova pendant]

Idk any other way

[cedar kiln]

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[nova pendant]

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[crimson sedge]

I have a question about this

For the 15 degree angle in the semi circle

to find the area of it

do we plugin 30 degrees into the sector formula

30/360 x pi3^2

or 15 degrees

because its an inscribed angle

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

<@&286206848099549185>

[royal plume]

You find the area with the 15° and then subtract it from the area of the big sector (the 30°)

I’m not sure about the use of the semi circle

[crimson sedge]

thats not my question

i just wanted to ask if we input 30 degrees or 15 degrees into the sector formula

as it is an inscribed angle

[royal plume]

Yeah I mentioned those two because I wasn’t sure which sector you were talking about

For the smaller one, 15
Bigger one, 30

[crimson sedge]

lol still not my question

okay lemme explain

for that small 15 degree angle in the circle

since its an inscribed angle

do we say 30 degrees?

as it means the actual sector is 30 degrees of the circle?

[royal plume]

Ahh you mean should we use the central angle or not

Uh I don’t think so (?) Every sector you solve you always take the inscribed angle

[crimson sedge]

ok its fine

thanks anyways

ima head to sleep gn

.close

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[stuck walrus]

jstn please be patient because i was literally trying to help you figure it out two channels ago

[civic meteor]

sorry my bad

[stuck walrus]

have you tried starting from the right... that looks horrendous

[civic meteor]

ah yes

[stuck walrus]

Hold up slow down

I think you should try this instead

[civic meteor]

where abouts

[stuck walrus]

sin(2A) + sin(2B) maybe

[civic meteor]

should i phase shift cos into sin using pi/2

[stuck walrus]

definitely not?? dont do things that you dont have to

i.e. dont introduce a pi into the mix

its worse enough

[civic meteor]

ok

[cedar kiln]

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[raw wren]

why can't P(A)+ P(B) exceed 1?'

as in independant even if it exceeds p(A*B) will won't allow to go beyond 1 as a whole

[clear berry]

I am not so sure what if A and B both are guaranteed to happen then won't the sum be 2?

[deep oriole]

then the probability of A and B is 1

still doesnt ever exceed 1

[clear berry]

like take A to be the event of getting 1-6 on a dice roll and B be the event of getting either a head or a tail

then p(A)=1 and p(B)=1, then the sum is 2?

[deep oriole]

the probability of getting both 1-6 on a dice AND either a head or tails on a coin is still 100%

not any greater

oh wait i understand you, p(A) + p(B), not p(A and B)

[cedar kiln]

@raw wren Has your question been resolved?

[clear berry]

yep, it's just given that they are independent NOT that they're mutually exclusive

[deep oriole]

yeah sorry i was misunderstanding the question, im pretty sure p(A) + p(B) can be independant and still greater than 1

[raw wren]

so can it be greater than one

[deep oriole]

yeah

[raw wren]

answer is D though

[deep oriole]

yeah because you cant figure out anything just from that

[raw wren]

so does it mean it can be greater than 1 also less than 1 is possible

[deep oriole]

just like if you know x + y = 1

you know nothing of x or y

[raw wren]

but in the explaination they said it can't exceed 1

it can be less than that but can;t exeed they said

that confused me

[deep oriole]

im pretty sure it cant exceed 1 if they are mutually exclusive, but if they are independant they can be

[raw wren]

is the explaination wrong

[deep oriole]

probably

how do they explain it

[raw wren]

exactly

[deep oriole]

it can exceed 1

idk what the solution is saying

[raw wren]

yeah man

.close

[cedar kiln]

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[cedar kiln]

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[still smelt]

Quick question: does anyone know why reversing the process of differentiation over a specific interval yields the area below the curve at that interval?

[crimson delta]

you might want to check out 3b1b on youtube

he made a video on that

[slate lintel]

it's called the fundamental theorem of calculus

[cedar kiln]

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[glossy inlet]

f : R --> R is a non-constant twice differentiable function such that f(x) = f(4 - x), and f(x) = 0 has atleast two distinct repeated roots in (2, 4). What is the minimum number of roots for f''(x) = 0 in [0, 4]?

[glossy inlet]

heavy machinery, this server

[dire geode]

!show

[cedar kiln]

Show your work, and if possible, explain where you are stuck.

[glossy inlet]

i would if i had any

i cant seem to wrap my head around how id go about doing this

[dire geode]

What is the minimum number of roots for f''(x) = 0 in [0, 4]?
is the question about roots of f(x) or f''(x) ?

[glossy inlet]

f''(x)

second derivative yes

[dire geode]

do you recognize symmetry of f?

[glossy inlet]

about x = 4?

[dire geode]

nah

[glossy inlet]

[dire geode]

f(x) = f(4-x) implies f is symmetric about x=2

[glossy inlet]

oh

yeah mb

[dire geode]

so whatever repeated roots in f(x) on (2, 4), it also has those on (0,2)

[glossy inlet]

I see

which means theres a total of 4 distinct roots in (0, 4)?

or more specifically (0, 2) u (2, 4)

if they're repeated it implies the curve is tangent to the axis at that point, right?

[dire geode]

yes

[glossy inlet]

right uh so f'(x) has 4 roots as well?

[dire geode]

you're on the right track

[glossy inlet]

is it necessary for those roots to be repeated as well?

cuz if I take a function that has repeated roots and has the axis as a tangent like x^3

0 has multiplicity 3

and the derivative has multiplicity 2

[junior dome]

if f(x) has a root repeated n times then its derivative has same root repeated n-1 times

[glossy inlet]

oh is that a property?

[junior dome]

well more of a result

[glossy inlet]

interesting, which means none of the roots are repeated

i.e. the graph cuts the axis

[glossy inlet]

i can see it for polynomials

i suppose it only happens for polynomials though 😅

@dire geode is the answer 3? (sorry for the ping)

[junior dome]

its 2 i think

[glossy inlet]

4 roots, none repeated, implies theres 2 minima and a maxima right?

[junior dome]

4 repeated roots of f(x)

3 repeated roots of f'(x)

2 repeated roots of f''(x)

[glossy inlet]

wait a minute let me process

there are 4 DISTINCT roots yes

each with multiplicity 2

the derivative is 0 at those points since the x axis is tangent to the curve

which implies f'(x) has 4 roots?

wheres the fallacy here?

[junior dome]

no no

the graph cuts x axis

that doesnt mean x axis is tangent to graph

if f(a) = 0 that doesnt mean f'(a) = 0

[glossy inlet]

it can be both right, tangent and cutting

[junior dome]

no

[glossy inlet]

yeah but repeated roots implies tangency?

[glossy inlet]

what about the tangent to y = x^3 at x = 0?

it is the x axis

tangent, and crossing

[junior dome]

not with every graph

[glossy inlet]

was riemann incorrect though

[junior dome]

idk

[glossy inlet]

or am i simply being narrowminded and viewing this whole concoction through the eyes of polynomial functions

[junior dome]

a tangent at a point on the curve may cross the curve at that point

its not a must

[glossy inlet]

but f''(x) = 0 implies an inflection point for f(x)

and isnt it standard nature for the tangent to cross the graph at an inflection point?

[junior dome]

yes

[glossy inlet]

when talking about "repeated roots", what other types of functions come to mind?

surely only polynomials, i cannot think of any others

[junior dome]

yeah

[glossy inlet]

so why again am I incorrect to assume that the x axis is tangent to the curve

regardless of wether or not it cuts the curve

[junior dome]

,w plot x^2 + 2x + 1

[glossy inlet]

yes my point exactly imagine 4 of those bumps

,w plot y = ((x - 1)(x - 2)(x - 3)(x - 4))^2

[wraith dagger]

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Requested by jewels9006

[glossy inlet]

this sortathing

[junior dome]

oh wait, you talking about just repeated roots?

[glossy inlet]

um

[glossy inlet]

i wanted to know if this was incorrect

,w plot y = d/dx(((x - 1)(x - 2)(x - 3)(x - 4))^2)

[wraith dagger]

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Requested by jewels9006

[glossy inlet]

oh boy

yeah so notice how this particular one CUTS the axis at all those points?

but since we're talking about minimum...

(in my question)

ill figure something out

thank you @junior dome and @dire geode ❤️

.closed

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[austere oxide]

How do i continue?

[dire geode]

is this supposed to be generating functions?

[austere oxide]

yes

[austere oxide]

apologies for the writing 🥲

[cedar kiln]

@austere oxide Has your question been resolved?

[cedar kiln]

@austere oxide Has your question been resolved?

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@austere oxide Has your question been resolved?

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[crimson sedge]

Please explain the red highlighted part:

[crystal raptor]

what is the definition of a (cumulative) distribution function

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

That I know.

[crimson sedge]

I can't understand why it's 1 if x > 2pi

[crystal raptor]

Explain what the definition is and then perhaps it will become clear

[crimson sedge]

I think, this sums it well 😀

[crimson sedge]

@crystal raptor Sorry, but I still don’t understand why it’s 1 if x > 2 * pi.

[dire geode]

You have only shown the density P(x). you need the "cumulative distribution function"

find that in your notes/book

[crimson sedge]

Thank you for this hint.

[dire geode]

take that, @crystal raptor

[crimson sedge]

I am sorry, I still don’t understand it.

I understand for continuous random variable but not for the discrete random variable.

Please help.

[dire geode]

show the cumulative distribution function

[dire geode]

this is a continuous random variable

[dire geode]

you gotta point out where you don't understand the definition of distribution function

[crimson sedge]

Red underlined part.

[crimson sedge]

I read this article: https://www.ncl.ac.uk/webtemplate/ask-assets/external/maths-resources/statistics/distribution-functions/cumulative-distribution-function.html

[dire geode]

that's a specific distribution function

correct

calculate this

for three cases of x

[crimson sedge]

Yes, I understood the first two cases.

I can’t understand why it’s 1 for the third case?

[dire geode]

Did you calculate P(X <= 2pi)?

Do you know how to calculate P(X <= x) using integrals?

[crimson sedge]

Okay, yes. Now it makes sense.

Thank you for being patient and explaining me this.

.close

[cedar kiln]

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[cedar kiln]

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[vapid sand]

is cos3AcosA the same as cos(3A+A)?

[vapid sand]

full context btw

is trig functions not on today for most people?

[zenith sail]

no, cos(x)cos(y) is not generally equal to cos(x+y)

[vapid sand]

I see

[zenith sail]

I'm guessing you've been studying double angle identities

and sum/difference identities?

[vapid sand]

yes

I assume I should split cos3A into cos(2A+A)

[zenith sail]

I haven't tried this yet but that seems like the most straightforward approach, yeah

[vapid sand]

alright, btw quick question, why is cos 3A not the same with CosA

[zenith sail]

not sure what you mean by 'not the same'

[vapid sand]

maybe I should clarify, why can't I merge them onto cos4A

[zenith sail]

oh, well, basically same answer as this

cos(3A)cos(A) is not the same as cos(3A+A)

[vapid sand]

oh is it beccause if we were to assume A is a number, 3A and A are not the same number

hence we can't just put them together?

[zenith sail]

even if they were the same number, cos(A)cos(A) is not the same as cos(2A)

[vapid sand]

Ah

Wouldn't it actually become (CosA)^2?

[zenith sail]

I'd use brackets, but yes

(CosA)^2 or Cos^2(A)

[vapid sand]

ah I see

alright

thanks for the help then

.close

[cedar kiln]

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[zenith sail]

no prob 👍

[cedar kiln]

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[shy crescent]

can someone help me

[lament elbow]

Use linearity probably

[shy crescent]

Im just confused how you go from

E[x1+x2+...+xk]/E[x1+x2+...+Xn] to anything

[violet flume]

i'm not certain you can even apply this

the expectation of a ratio isnt the ratio of expectations is it?

i cant solve it but there is some discussion https://math.stackexchange.com/questions/2136004/expectation-of-ratio-of-sums-of-i-i-d-random-variables-whats-wrong-with-the-s

[cedar kiln]

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[buoyant latch]

So I’m trying to learn about sufficiency and completeness of statistics.

A statistic is just a function of my RV’s.

A sufficient statistic is one that contains enough information about the sample to create a MVUE? Or just an unbiased estimator of my parameters

[buoyant latch]

I’m given both that sufficient statistic corresponds to particular parameters and also a sufficient statistic could be a vector

[royal loom]

.coose

[cedar kiln]

@buoyant latch Has your question been resolved?

[cedar kiln]

@buoyant latch Has your question been resolved?

[cedar kiln]

@buoyant latch Has your question been resolved?

[cedar kiln]

@buoyant latch Has your question been resolved?

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[crimson sedge]

Suppose $T$ and $U$ are linear transformations from $\R^n$ to $\R^n$ such that $\map T{U\vb x} = \vb x$ for all $\vb x$ in $\R^n$. Is it true that $\map U{T\vb x}= \vb x$ for all $\vb x$ in $\R^n$?

[crimson sedge]

I'm trying to prove this, but I am quite lost on how to word it exactly

Can I just say something like "Let A be the standard matrix of U, then the transformation U is invertible iff A is invertible"

I already proved that earlier, so taking it as fact, if U is invertible and T is inverse, then it should be symmetric because AA^-1 = I and A^-1A = I?

Am I bullshitting this too much or

[crimson delta]

well you have to use that the matrices are square

the claim wouldn't be true otherwise

this mostly is a question of what theorems you already know

[crimson sedge]

So

Prove that AA^-1 = I is only true for n x n matrices?

[crimson delta]

no

[crimson sedge]

And like, the opposite

[crimson delta]

that it implies A^-1 A=I

well the notation is bad because we are already assuming its the inverse

[crimson sedge]

Is it common to just branch out this much

[crimson delta]

you have to show AB=I implies BA=I

wdym with branch out

[crimson sedge]

Like to prove x you have to prove y and to prove y you to have to prove z

[crimson delta]

well proofs get longer

although so far you haven't really done anything except use that linear transformations and matrices are essentially the same thing

you just translated it, but all the next steps you could also write in terms of linear transformations directly

[crimson sedge]

This feels like something I should know

Can't you just say like

A = B^-1

By right hand multiplication

Then like

[mighty drift]

I already found myself answering a question in 2 pages in exams. Sometimes it's the simplest way, sometimes it's the expected way or even the only way

[crimson sedge]

B(B^-1) = I

[crimson delta]

you dont know that B^-1 exists

in terms of injective and surjective, you have to show that if A is surjective, then it is injective

(prove that AB=I means A is surjective)

[crimson sedge]

Well

This amounts to proving that A is a n x n matrix of linearly independent columns

[crimson delta]

yeah at some point it gets natural. "oh we just have to show this and that. oh whoops that already took 2 pages"

[crimson sedge]

You can build a counterexample?

[crimson delta]

do you know the big matrix invertibility theorem?

[crimson sedge]

Suppose there exists a m x n matrix A such that AD = I_m, but this just implies that A is square either way

[crimson delta]

with like 10 different statements which are all equivalent to a matrix being invertible?

[crimson sedge]

Yeah

,rccw

[crimson delta]

that's the theorem you want

specifically points j and k

[crimson sedge]

So prove j -> k

?

[crimson delta]

yes

well and show C=D

[crimson sedge]

This feels easy to do but somehow I am always getting it wrong

Idk hmm

I mean if a is true, then A^-1 works for C, so (a) -> (j) and (k)

But we are not supposing A is invertible

Yeah sorry I will close this channel I'm just massively confused at the moment

Maybe I'll get it soon

.close

[cedar kiln]

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[hoary spoke]

any idea how to simplify it?

what should I do?

I am not sure how will partial fractions work here

I have to find it's laplace inverse transform

Anyone?

[hoary spoke]

don't worry about this part, Simplifying it is all what matters now

[hoary spoke]

just help me factorise this fraction somehow

[brittle lynx]

try divisors of 2 to factorise the denominator

I think s=-1 is a root

[hoary spoke]

oh

Why didn't I try finding its roots

okay good idea

wow it works

thanks

.close

[cedar kiln]

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[cedar kiln]

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[halcyon adder]

I have to look for asymptotes

Do periodic functions not have a non vertical asymptote?

[empty kestrel]

tan(x) is periodic and has vertical asymptotes for example

[halcyon adder]

I mean non vertical

[empty kestrel]

Oh sorry, then the only case would be a constant function

[halcyon adder]

Ok thanks

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[crimson sedge]

Hi

[crimson sedge]

I have a dumb question

🙂

I want to transform the next general conic equation to the equation of a hyperbole

$x^2-y^2+6x+8=0$

[wraith dagger]

Kangaroo

[crimson sedge]

So I factorize it

(x+3)^2-9+8+6=0

$(x+3)^2-9+8+6=0$

$(x+3)^2-9+8+6x=0$

[wraith dagger]

Kangaroo

[crimson sedge]

$(x+3)^2+6x-1=0$

[wraith dagger]

Kangaroo

[crimson sedge]

oh

[sick gazelle]

where is the y²?

[crimson sedge]

sorry I noticed it now

haha

$(x+3)^2-y^2+6x-1=0$

[sick gazelle]

why is that 6x there?

u factorized it to (x+3)² right?

[crimson sedge]

$(x+3)^2-y^2-1=0$

[wraith dagger]

Kangaroo

[crimson sedge]

Sorry

I copy it from and note book and that was not my first try

So I mixed it

[sick gazelle]

then first try it by urself dude

[crimson sedge]

why you say that ? I already tried, I am stuck here

[crimson sedge]

Talk me with respect, no body is asking for your help

.close

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[long arrow]

it's geometric series where $a = 1, r = 4$, sum is:
$$a\frac{1-r^n}{1-r}=1 \cdot \frac{1-4^n}{1-4}=\frac13 (4^n-1)$$

[wraith dagger]

Modus

[cedar kiln]

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[unborn pike]

I need to show that this limit exists and then find the limit

[unborn pike]

My intuition says 1 but

I gotta use concepts ive learnt in class, which include epsilon n or sandwich theorem

[cedar kiln]

@unborn pike Has your question been resolved?

[junior dome]

you can use l'hopital

[solid juniper]

write n^(1/n) = e^(ln(n^(1/n))) = e^(ln(n)/n)

[cedar kiln]

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[winged lagoon]

hey i need help with math

[umbral dew]

?

[winged lagoon]

please can you help me with math

[umbral dew]

maybe

just ask

[junior dome]

[umbral dew]

there are lots of people

here to help u

[frosty ocean]

Nobody can help, until you ask.

[umbral dew]

look, here is the old guy

[winged lagoon]

i cant do linear equations

[umbral dew]

old guy

[winged lagoon]

HELP PLEASE

[crystal raptor]

ask a specific question

[south tundra]

We can't help if you don't tell us what to help with

[cedar kiln]

@winged lagoon Has your question been resolved?

[crystal raptor]

this was one of the help channels of all time

[dire geode]

read all the advice people gave and open a new channel

.close

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[halcyon adder]

How do i find this

[dire geode]

,w solve 1/2 - xe^x = 0

[halcyon adder]

Wihtout calculator

[wraith dagger]

Results provided by WolframAlpha

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

Requested by rie.mann

[dire geode]

learn Lambert W function

[halcyon adder]

Ok

[cedar kiln]

@halcyon adder Has your question been resolved?

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[crimson sedge]

I need help with this question. I initially god 3/5 as an answer.

I know that since they are similar triangles and that the sides correspond to one another, they would share the same triginometric values.

Answer is B though

[slate lintel]

triangle is suspicious

[deep oriole]

that diagram doesnt make sense

[crimson sedge]

How so?

[slate lintel]

how do you know where the angles are

in particular where the right angle is

[deep oriole]

how can hypoteneuse not be the biggest side?

[crimson sedge]

i dont even know tbh'

[crimson sedge]

ohh

so hypotenuse is 5?

[deep oriole]

yeah

[slate lintel]

in a right triangle the hypotenuse is always the longest side

[crimson sedge]

oh wait yeah how even am i supposed to do this question without where the angles are located?

[crimson sedge]

ok

[slate lintel]

there's a very famous theorem about right triangles

probably the most famous theorem

[crimson sedge]

ok what is it?

[slate lintel]

the Pythagorean theorem, have you heard of it?

[crimson sedge]

lol

yes

[slate lintel]

yeah so it applies here

[crimson sedge]

I don’t understand how it would apply in this scenario

[slate lintel]

ah, see it works both ways

if you have a triangle that satisfies the Pyth theorem then that triangle is a right triangle

[crimson sedge]

Okay

So because it’s a 3 4 5 triangle it’s a right triangle

Got it

I just don’t get how that would take us a step closer to finding the answer

[slate lintel]

well, now we know it's a right triangle

and we know which side is which

in particular we know that BC is the hypotenuse and is 5

[crimson sedge]

Yeah

Okay

@slate lintel

got it!

thx

.close

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[raven fox]

I need help it says find each value if g(x)=x^2-x and they give me g(5n) to plug in how do I solve

[cedar kiln]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[wanton marsh]

oh apologies

[jaunty mural]

And your solution was wrong at that.

[wanton marsh]

it wa a substitution, not a solution

[jaunty mural]

I'm saying your solution, as in the answer to the question, was wrong.

[jaunty mural]

x, the parameter of the equation is a number.
5n is also a number.

[zealous compass]

How would you find g(2)

[wanton marsh]

yes, and i'm saying it wasn't an answer

it was a mere substitution

[raven fox]

Plug in 2 for all the x’s

[raven fox]

Oh ok

[jaunty mural]

and 2 is also a number

do show the answer once you've got it.

[zealous compass]

So what would you do for g(5n)

[raven fox]

Would the equation be g(5n)=5n^2-5n

?

[jaunty mural]

No, consider this:

Rather than 2

substitute in 1+1

[zealous compass]

(5n)^2 but yes

[jaunty mural]

and tell us what you get

[zealous compass]

??? How is that incorrect guys

[jaunty mural]

!nosols

[cedar kiln]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[jaunty mural]

Can't you see what I'm doing

[zealous compass]

He got it?