help-49

[lyric charm]

could be some stupid arithmetic error

[molten bay]

A little bit

Give me some time

@lyric charm

[lyric charm]

where is -1/2 * -2 * x * x^3/6 from the expansion of -1/2(x - x^3/6)^2?

oh wait no i do see the x^4/6

it disappeared afterwards tho it looks like

wait a minute...

[molten bay]

Ohh hang on

1/6

I got it

[lyric charm]

i mean the x^6 term should have been trashed also

[molten bay]

@lyric charm

[lyric charm]

you trashed too much but this time you were lucky

[molten bay]

What do you mean by trashed?

If i take only one term of it then

(1-x^2/2-1+x^2/2)/x^4 which is 0

.close

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[hybrid sage]

can someone help me do just the number and algebra row in the table provided given the instructions included about statistics Measures of Position and Central Tendency

[lyric charm]

have you done anything thus far

[hybrid sage]

i've did the median

[lyric charm]

have you sorted these things in ascending order

[hybrid sage]

but i'm not sure if its right since i've only tried

[hybrid sage]

yea

[lyric charm]

ok im gonna take your word for it that the median is correct

so now what else do they want you to find

the quartiles?

[hybrid sage]

how about the table?

can you help me correct the table

[lyric charm]

id like to see the original data sorted in ascending order and ungrouped

[hybrid sage]

sure

[lyric charm]

dont rly wanna do it myself cause it's a bit energy-consuming

[hybrid sage]

do i still have to tally it?

Ascending Order: 20,28,28,32,32,32,36,36,36,36,44,44,44,48,48,48,48,48,48,52,52,56,56,56,60,60,60,64,64,64,68,68,72,72,76,80,84,84

is that good enough??

what's next??

[lyric charm]

ok so that's 28 data points

20,28,28
that's 3 points that go in the 20-29 class interval

and you put 2 as that interval's frequency

[hybrid sage]

oh yea mb

its confusing 😭

what else did i miss

wait

is it correct that i can do up until 89 if the highest interval is 84

[lyric charm]

the highest score you mean?

[hybrid sage]

yea

i corrected it

[lyric charm]

ok sorry im finding myself too unfocused to continue looking at this

[hybrid sage]

its okay

20, 28, 28, 32, 32, 32, 36, 36, 36, 44, 44. 44, 48, 48, 48, 48, 52, 52, 52, 56, 56, 60, 60, 64, 64, 64, 68, 72, 72, 76, 80, 84, 84

cf = 33

i fixed it

[midnight plank]

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[stable halo]

this is gonna sound completely stupid, and i know this is more data sci but idk whhere elsr to ask, but if i have a bunch of data points as shown above, how would i get the x coord at which the straight line becomes now a line with a gradient.?

[winged maple]

uh, do you mean roughly

as in if you drew a curve of best fit, the point when it stops being "basically straight" and starts being "basically curved down"?

[stable halo]

Yea would have to be roughly

i dont want to just eye it as it would be cheating

[winged maple]

and I assume you want more than just eyeballing it

[stable halo]

even though trhat would be pretty easy

[winged maple]

yeah okay
fit a quadratic (or other) curve to the data, get the equation
differentiate it, solve for when the gradient becomes sufficiently negative (i.e. it's curving down enough to be "actually decreasing")

[stable halo]

Yea that could work

thanks

but like

there is so much noise

how would i fit a curve to that data

[winged maple]

honestly I would just not bother with that last step - fit a curve and eyeball it

choosing a threshold as to when you deem it sufficiently "curved" is eyeballing anyway

[stable halo]

Yea exactly

my guess is i might need to use some machine learning stuff

[winged maple]

yes

[stable halo]

which is way out of my grade

[winged maple]

are we talking, excel/google sheets level of paygrade here haha

because you just want a (quadratic) regression

[stable halo]

wym, like where this data came from?

[winged maple]

as in yes, how do you have this data

[stable halo]

or can excel do stuff like this

[winged maple]

yes

[stable halo]

for my physics project, atomic microscope needle and voltage stuff

its for my undergad bsc

so pretty mathy

[winged maple]

curve of best fit in excel or quadratic regression in scipy

[stable halo]

probably

use scipy

[winged maple]

yeah, it'd be better (for academic project purposes) to do this thoroughly

[stable halo]

Yea

thats also why ML is out of my range like i probably could learn not worth the effort though

[winged maple]

yeah I'm suggesting reaching for "ML" in teh very simple sense - regression curves

no need to train anything or w/e

[stable halo]

Yea i can do regression

i did non linear regrssion using euler newton method

[winged maple]

fit the curve with scipy, calculate the x values that are deemed sufficientyl curved for a range of threshold values, and then report your findings as that interval in some way. makes it more mathy than just "I picked this gradient because idk"

[stable halo]

for my group project

[winged maple]

even better.

you know more than me, go for it

[stable halo]

ahah that was a mess to leanr as a physics student

but thanks for the help

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[molten bay]

Any ideas?

[hasty haven]

do u know ur double angle for cos? @molten bay

[molten bay]

Yeah

[hasty haven]

we’re gonna wanna use u-sub to make it simpler

what form of double angle would allow for a simpler du? @molten bay

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[dusty portal]

Let $a_1, a_2\ldots a_n$ be a strictly increasing sequence such that $a_n>0$ and $a_{2n}=a_n+n$. It is known that if $a_n$ is prime, then $n$ is prime. Find, with proof, the value of $a_{1993}$.

[grand pond]

;(

[dusty portal]

!status

[midnight plank]

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

[dusty portal]

1

Literally all I know is that a_0 is 0

a_2=a_1+1 is prime so that means a_1 is composite

Literally nothing else

[runic hamlet]

the condition a_2n = a_n + n combined with the fact that the sequence is strongly increasing is very limiting

[runic hamlet]

how do you know that a_2 is prime

[dusty portal]

That’s the problem

I don’t know if I can use the converse or not

[runic hamlet]

why should you be able to

thats not how that works

[dusty portal]

Exactly 💀

So I’m essentially stuck

[runic hamlet]

.

[dusty portal]

What do you mean by that

“Very limiting”

[runic hamlet]

it heavily restricts the possibilites of what a_n could be

[dusty portal]

Oh, ok

But I can’t find it

[runic hamlet]

what does it say about a_2, a_3, a_4

ah wait I assumed that the a_ns are integers

I suppose thats not given

[dusty portal]

Nope

[runic hamlet]

well they probably fucked up when writing the problem

[dusty portal]

Figured out that a_n=n+c is a general solution

Wait nevermind lol

It says positive integers

[runic hamlet]

surprise

[dusty portal]

Valid?

[runic hamlet]

yes

[dusty portal]

I’m in class so I can probably explore this

I’ll come back in god knows how long

I have a lab to do

[runic hamlet]

you are basically done

you just need to find c

[dusty portal]

Ok

0?

a_0=0

Nevermind

Yeah I’ll figure it out

[runic hamlet]

(a_0 doesnt even exist)

[sharp wave]

is a0 even defined?

[dusty portal]

Yeah

Aight I think my brain is fried but I got it

I think a_n=n is a solution anyways

If a_n is prime then n is prime

So a_{1993}=1993

Bruh gg

[runic hamlet]

why is a_n=n the only solution

why not a_n=n+12037017569123123

[hasty haven]

bc 2(n+c) ≠ 2n + c + n

well

it is

but only if c=0

[runic hamlet]

a_2n = 2n+c, an+n = n+c+n

[hasty haven]

that’s so true actually

[sharp wave]

you can easily disprove this, the statement says if n is prime then an is prime, put n = 3 you get a3 = some even number.

[runic hamlet]

it does not say that

it says if a_n is prime then n is prime

[sharp wave]

oops, but i think you can find a non prime n for which an becomes prime

[midnight plank]

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[warm island]

Let f: AB -> CD be the projection of points in AB to CD
Let x be the point where pojections are made in AB, for each point x in AB we can draw a line to CD which intersects a unique point Y.

Thus we have f(x) = y. this is one to one since if we have x1 and x2, and they mapped to the same y on CD then the projection would make no sesnse. this is also onto since all points in AB will project to CD. Therefore this is a bijective mapping

did i answer this? lol

[midnight plank]

@warm island Has your question been resolved?

[placid spoke]

projected how? if you just draw a line from AB straight down to CD, f is not onto

since the length of CD is larger than the length of AB

[midnight plank]

@warm island Has your question been resolved?

[warm island]

Yeah you make a good point

I'm guessing there's some sort of relationship I can make idk

[placid spoke]

you can actually pretty easily make a linear relationship between the two.

if you assume x goes from 0 to L and y goes from 0 to M (with L < M) construct the line that goes from (0,0) to (L,M)

[warm island]

alright ill think about it im doing this just for fun '

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[true egret]

Totally stuped on this problem from Real Analysis. I've got two vectors x and y, both in R^n. I've got an affine function Ax+b, with some matrix A and bias vector b that I don't know. I'm supposed to optimze A and b so that | |y-(Ax+b)| |^2 is minimized. I've got no clue how to do this. I don't know how to differentiate with respect to a vector or a matrix. Need some direction on what I'm supposed to be doing here. Book is useless and I don't want to spend 2 days barking up the wrong tree researching this.

[true egret]

no

[wary thorn]

Well

If you get to choose A and b

Choose A so that Ax=y has a solution and let b be the zero vector

Then it's minimized

I assume that that's not the intention of the question though, so what do you mean "optimize A and b"?

the extrema of f(x)=||y-(Ax+b)||^2
will have grad f (x) =0

aswell

[midnight plank]

@true egret Has your question been resolved?

[true egret]

What they literally ask is "find the affine function g(x)=ax+b such that the sum (j ranging from 1 to the number of dimensions) of (y_i-g(x_j))^2 is minimzed." I think that's equivalent to minimizing | | y-(Ax+b) | | ^2

But then it's such a stupid question... that can't be what they're fishing for

I don't know how to take the gradient of a function that has a matrix in it

[midnight plank]

@true egret Has your question been resolved?

[midnight plank]

@true egret Has your question been resolved?

[humble wolf]

you can use linear algebra to solve this

[humble wolf]

https://math.libretexts.org/Bookshelves/Linear_Algebra/Interactive_Linear_Algebra_(Margalit_and_Rabinoff)/06%3A_Orthogonality/6.5%3A_The_Method_of_Least_Squares

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[upbeat plinth]

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[lethal path]

you want to add up $\sqrt{dx^2 + dy^2} = \sqrt{1 + (dy/dx)^2} \sqrt{dx^2}$ and $\sqrt{dx^2} = dx$

[grand pond]

south

[upbeat plinth]

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[polar star]

,w derivative of 2x^(5/4) wrt x

[grand pond]

Results provided by WolframAlpha

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Requested by _mia_34

[lethal path]

that's not what the arc length formula says

yes you should be doing sqrt(1 + (y')^2) instead

[upbeat plinth]

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[lethal path]

yes and then try taking sqrt(1/4) = 1/2 out of the square root

[upbeat plinth]

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[lethal path]

so $\sqrt{ \frac{4 + 25 x^{1/2}}{4} } \cdots$

[grand pond]

south

[upbeat plinth]

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[lethal path]

write 1 = 4/4

[upbeat plinth]

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[lethal path]

yeah if it's more natural to you, you can multiply the square root by sqrt(4) = 2, then divide by 2 on the outside

^ this is how I would do it

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[steep mirage]

Ik one of the factors is (x-1)

But idk if I gotta find a new other one to know what to divide this by or what I do from here

[nimble copper]

If you haven't been taught any methods to solve a cubic then you have to find another factor

[steep mirage]

Alr ty

[lethal path]

yes the technique is always to divide by linear factors until you get a quadratic

except if you can spot if the cubic is a perfect cube or some other special case, mainly that and a^3 + b^3 or a^3 - b^3

[steep mirage]

We were not taught that

[lethal path]

don't worry if you can't spot these special cases though

[steep mirage]

So prob shouldn't do on exam

[lethal path]

yep exactly

exam just wants you to guess till you get p(x) = 0 again

[steep mirage]

Ok

What does the p stand for

Btw

This random but yk

[lethal path]

p stands for polynomial!

[steep mirage]

Fancy

[lethal path]

f stands for function

g is also a function and h usually

[steep mirage]

Ig that makes sense 💀

[lethal path]

cause of the alphabet haha

[steep mirage]

Yeah 💀

Idk

I did one of these questions to study

Got it first try

Idk if I wanna do more

[steep mirage]

For this 7a

Would I just set x has 2

And make it = 0

The remainder theorem I think it's called

[lethal path]

correct!

so that's actually called the factor theorem, but that's just the remainder theorem with remainder = 0

[steep mirage]

Ok

Another question actually

For 11

Would I just start by diving by -6

Theater that factor it

To get my other 2 factors

Which is the L and W

[lethal path]

,calc (-6)^3 + 14(-6)^2 + 63(-6) + 90

[grand pond]

Result:

0

[lethal path]

yes so you would need to divide the polynomial by x - (-6)

or x + 6

[steep mirage]

Okay

Yep

Ty

All for now

[lethal path]

np!!

[dim solar]

but will that result is zero right ?

[lethal path]

yeah like just making sure the question wasn't broken

I know that it has x + 6 already

[dim solar]

oh okay

[steep mirage]

Math fun

When you actually understand it

But I also hate studying late

So idk

But Ty guys

[dim solar]

it's about the journey not the goal bro

just enjoy when things getting hard that what i mean

and the best times is when getting the answer

if you don't need the channel to forget to close it @steep mirage

don't*

[steep mirage]

Actually

Got 1 more thing 💀

How do you find the multiplicity

Wtf is multiplicity

Can someone give an example

[dim solar]

bro try to revisit the lesson first 💀

[steep mirage]

I did

But no where does it say how you find the multiplicity

My teachers notes suck ass

I was gone at an appointment

[dim solar]

i see

i'am in a middle of a session i will finish it and try to figure it out with you

[steep mirage]

Actually

[dim solar]

can i ask smth personal tho

[steep mirage]

Do you just simply look at it to find the multiplicity

[dim solar]

u are at g12 right

[steep mirage]

I just don't understand how you find the multiplicity when you aren't given a picture

[steep mirage]

11, but this is a grade 12 course

[dim solar]

yeah because of that i see these questions in my Papers

is that SAT or just regular school work /

[steep mirage]

Regular

Test tmr

This is the hardest math class my school offers

And it ain't even bad

[dim solar]

oh bad situation you in bro

[steep mirage]

Nah it'll be aight

[dim solar]

lol that's a regular shit in my SAT Course

without so much yapping im gonna finish quickly and try to solve it with you

[steep mirage]

Okay

Just wist 1 moment

[midnight plank]

@steep mirage Has your question been resolved?

[dim solar]

sorry bro i didn't figure it out myself

@steep mirage

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[median ridge]

im sorry- what does that mean?

like using x measured in radians??

nvm im freaking dumb

.close

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[lethal path]

yes it's not true in degrees or any other unit

[median ridge]

i mean degrees isnt really like a proper thing i'd say

like that sounds weird

but if u think of it with the unit circle, it makes no sense to use degrees

[solid iris]

degree and radian are unitless

[lethal path]

exactly radians are the natural units for all trig functions

[lethal path]

correct so more like natural conversion factor

[median ridge]

.reopen

[midnight plank]

✅

[median ridge]

ok but now im a little bit

with this limit

sinx/x

[normal prawn]

not unitless but dimensionless

[median ridge]

am i supposed to be imagining a graph here or is it just an expression?

[lethal path]

both

well the graph of (sin x)/x is pretty tricky

[median ridge]

yeah

thats why im kinda

confused

[solid iris]

hmm yes dimensionless is better

for the physics minded

[median ridge]

if its the graph of it, it makes me confused but expression wise is different

[lethal path]

what do you mean by expression wise

[median ridge]

look idk

but with the graph, there is a point where x = 0

which is 0/0 but its still there in an actual graph so it confuses me to use radians

because x = pi is an actual point on the graph

[normal prawn]

Actually, there is a hole at (0,1)

I mean, at x=0 function is not defined

[median ridge]

obviously i can scroll forever but shouldnt there be a more obvious circle to show that theres a hole there?

[lethal path]

Desmos isn't very good at that unfortunately

[lethal path]

if you hold and click on the graph at exactly x = 0

[normal prawn]

Loll

[lethal path]

it does show the white circle

but not otherwise

[normal prawn]

truee

[median ridge]

oh okay

i get it now thank you

do you think it was kinda a bad example to use orrr?

[lethal path]

no!

it's really good you're thinking about limits and how that differs from if a function is defined at that point or not

that was the whole point of your original message

what your textbook or notes wrote there

[median ridge]

okay

they put a photo of the graph below but i still wouldve liked to see a hole lmao

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[prime garden]

Expand a^x.

[prime garden]

As a sum of powers of a.

[wary thorn]

What

What is your doubt

[normal prawn]

Give us an example

[lyric charm]

!xy

[midnight plank]

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.

[prime garden]

Let a = 5.
We need to expand 5^x.
Let 5^x = k
Then (k - 1)/(a - 1) = 1 + a + a^2 + a^3 + ........... + a^(x-1)
where k = 5
a - 1 = 4
(k-1) = 4 + 4a + 4a^2 + 4a^3 + 4a^4.................... + 4a^(x-1)
k = 5 + 4 * 5^1 + 4 * 5^2 + 4 * 5^3 + 4 * 5^4.............. + 4 * 5^(x-1)
So 5^x = 5 + 4 * 5^1 + 4 * 5^2 + 4 * 5^3 + 4 * 5^4.............. + 4 * 5^(x-1)

[lyric charm]

???

what are you REALLY trying to do?

[wary thorn]

he asked the writign sir to do this

i can tell

[lyric charm]

https://tenor.com/view/shut-up-shut-up-be-quiet-funny-gif-19370091

[wary thorn]

he truly did if you ask it it it also gives this weird line of reasoning of a-1 expansions

[prime garden]

I guess one can find 5^x from 5^(x-1)

[normal prawn]

Ok so do you want to write the expansion of a^x using the geometric sum formula?

[prime garden]

No

[normal prawn]

That's exactly what u did

[prime garden]

It's useless

[normal prawn]

bruh

then?

[prime garden]

!status quo

[midnight plank]

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

[lyric charm]

!done

[midnight plank]

If you are done with this channel, please mark your problem as solved by typing .close

[prime garden]

Ikr

.close

[midnight plank]

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[midnight plank]

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[empty ivy]

Hi I dont get part i like how do you start

[sage helm]

hmm

[lethal path]

probably use the fact that $\sin^2 u = \frac{1 - \cos 2u}{2}$

[sage helm]

[ \f{\sin a + \sin b}{2} \geq \sqrt{\sin a \sin b} ]

[grand pond]

neon

south

[sage helm]

on a restricted domain

[lethal path]

oml it rearranges to that? cool

[sage helm]

no it doesn't lol

it's just the AM-GM

[sage helm]

.

[lethal path]

ah I know ok let's stop here and let them think then'

[sage helm]

this probably works

[lethal path]

nah your idea's better

[tawdry laurel]

am gm yay

Hi south

[lethal path]

ah but you need to be careful cause sin a, sin b need to be nonnegative here

[sage helm]

.

[lethal path]

the way to fix that is to use the rearrangement, $(\sin a + \sin b)^2 \ge 4 \sin a \sin b$

[grand pond]

south

[sage helm]

yeah

that's always true

actually this might be it

sum of sines inside

[lethal path]

yeah then Jensen's or something

[grand pond]

neon

[empty ivy]

Wait whats that

[lethal path]

Jensen's inequality, go search it up

it's useful for situations like this where you want to compare f(x + y) and f(x) + f(y)

[empty ivy]

Idk whats going on in that 😭

[slim eagle]

watch a yt vid

u gotta see the graphs

[empty ivy]

Kk

Okay thanks everyone

.close

[midnight plank]

Channel closed

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[idle glen]

you can go ahead and try to apply Jensen's but I think that is an overkill, you can start from here and prove 1 from algebraic manipulation and then prove 2 from 1 and some more inequalities manipulation
(I feel this is the expected solution)

[midnight plank]

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[novel dune]

how do i graph this?

[winged maple]

q3?

[novel dune]

q2

[winged maple]

break it into 2 pieces, and differentiate each piece

[novel dune]

um

how do i do that

[winged maple]

well it has a straight part and a parabola part, right

[novel dune]

yes

[winged maple]

so the gradient for the straight part would be...

[flat spire]

dr du

[novel dune]

4

[novel dune]

💔 yeah

[flat spire]

yr 11?

[novel dune]

yeah

[flat spire]

i c

u gonna do 4u?

[novel dune]

prolly not if im struggling on this

[flat spire]

u got this gng

[novel dune]

😭

[north robin]

Find the equation for the straight line graph and the parabola separately

[novel dune]

do i need the equation

or can i do with just like points

[pale vapor]

how will you do it with just points

[novel dune]

no idea

[pale vapor]

how will you so it with equation

[novel dune]

also no idea

[pale vapor]

oh

do you know what derivatives are

[novel dune]

yeah

the gradient of the tangent of a line

[pale vapor]

yeah

do you know how to find derivative of a given equation

[novel dune]

yes

i just dont really get how to turn a graph into the derrivative of the graph

[pale vapor]

so you can find the eqn of the parabola and differentiate it

[novel dune]

okay

lemme find it rq

y= x^2-2x?

[pale vapor]

yeah

[novel dune]

and then the derivative is 2x - 2

[pale vapor]

yes

now the break the graph into two parts

-infinity to -2 and -2 to infinity

[novel dune]

huh

[pale vapor]

basically the line and parabola

[novel dune]

um

[topaz urchin]

did you find slope of the straight line

[novel dune]

yes

its 4

[topaz urchin]

yeah so

from x = -infinity to x = -2

it's a straight line right

[novel dune]

uh

yeah

oh yeqh

i get it

yeah

[topaz urchin]

and the slope is 4

so you will plot y = 4 from x = -infinity to x = -2

[novel dune]

do i put open circle at -2

[novel dune]

or closed

[topaz urchin]

a sec

back

[topaz urchin]

dont think about that for now, first plot the graph of derivative of parabola

[novel dune]

is it just a straight line 2x-2

[pale vapor]

yes

[novel dune]

would this be right

[pale vapor]

plot y = 2x - 2 from x=-2 to x=infinity

[pale vapor]

yeah seems right

idk about the circle part though

[novel dune]

o

[topaz urchin]

should be open

[novel dune]

whats the curve called when its like all wiggly

[topaz urchin]

wdym by that

[novel dune]

like this

[topaz urchin]

that's the graph for a cubic function

[novel dune]

oh

thanks

how would i do this

😭

[north robin]

What is a stationary point

[novel dune]

1,0

[north robin]

You’re question is for part a right?

[novel dune]

both a and b

[north robin]

Let’s go over a first

What do you think the question is asking you?

[novel dune]

prove that the stationary point of infelction is at x=1

[north robin]

In reference to the graph

[novel dune]

yes

can i just say that the vertex of the parabola is 1,0

[north robin]

You need to justify the fact that the the stationary point is actually a stationary by referencing the graph

So you need to consider the gradient function’s behaviour at the point (1,0)

And link that to the graph

[novel dune]

wel

the gradient is 0

at 1,0

since its just a straight line on the x axis

[north robin]

Therefore

[novel dune]

since the parabola is concave down its maximum turning point is at 1,0 and since the gradient at 1,0 is 0 its also a stationary point

?

[north robin]

Pretty much yeah

[novel dune]

how would i graph it

?

[north robin]

You talking about b?

[novel dune]

yes

[north robin]

Consider the derivative right

We know it’s going to be an increasing function up until the stationary point then it’s gonna start decreasing

[novel dune]

uhh

yeah?

[north robin]

Wait nvm the derivative is constantly negative right?

So how do you think the original graph is going to look like?

[novel dune]

isnt the original grpah the one they gave?

[north robin]

No that’s the derivative of the original

[novel dune]

oh

i dont know

[north robin]

It’s negative so the values of the original function are always going down right?

[novel dune]

yes

[north robin]

Ok and we know that the original function has the value -3 at 1

[novel dune]

yes

[north robin]

We know that the derivative is quadratic, so what type of function must the original be?

[novel dune]

um

decreasing

idk

[north robin]

Right but what type of polynomial

It’s a cubic function

The graph that you said was all wiggly

Since we know that the original is cubic and it has the point of inflection at that point.

Also it’s constantly decreasing you can put these together to make a sketch of the graph

[novel dune]

😭

[north robin]

What do you not understand?

[novel dune]

if its constanly

decreasing

do i just start from otp left

[north robin]

Yes

Then it has a point of inflection at the point that was mentioned

[novel dune]

and then wat

is it just 1 point of inflection

[north robin]

Yes since it’s the only stationary point

You know what a point of inflection is right?

[novel dune]

its like

where the concavity changes

so would i draw it

from the top left

and then it passes through

1,0

and hten to the bottom right

[north robin]

Not 1,0

That’s for the derivative

[novel dune]

oh

for 1,-3

it it like

idk

can u just give me the asnwer atp

[north robin]

This is all right though

You have the answer

[novel dune]

okat

okay

is this right

[north robin]

Yeah, the slopes aren’t that steep but since it’s a sketch it’s fine

[novel dune]

okay

[novel dune]

how do i determine the nature of a stationary point?

[north robin]

You need to look at the second derivative or take values of the derivative of either side of the stationary point

[novel dune]

well

i did the 2nd derivative

and i have the x values of the stationary points

what do i do now

[north robin]

Plug the x value into the second derivative then use this

[novel dune]

okay

[last arch]

I don't think that last one is correct

The one for stationary inflection points

[novel dune]

holy crap man why this hw so hard

[north robin]

Oh right, if it’s zero it doesn’t necessarily mean it’s a inflection point

[last arch]

Yea it's inconclusive

Then you'd have to use the first derivative test

[midnight plank]

@novel dune Has your question been resolved?

[novel dune]

no

[north robin]

Sorry I am a bit busy as I have a test in like 45 mins

[novel dune]

oh

goodluck

[topaz urchin]

which one

[novel dune]

q1

[topaz urchin]

do you know the condition

for which a function is increasing

[novel dune]

if its greater than 0

?

[topaz urchin]

yeah if the derivative

is greater than 0

so differentiate that equation

[novel dune]

i did

i got the x values

of

-2,-1 and 2

but idk what to do nwo

[topaz urchin]

do you know how to solve

(x+2)(x+1)(x-2) > 0

[novel dune]

no

idk

[topaz urchin]

hmm you'll have to look up that first

because that's how you'll get the values of x

[novel dune]

well

dont u just equate

it

to

0?

[topaz urchin]

no, the condition is that derivative should be greater than 0

not equal

[novel dune]

oh

well

is it just

x > -2

x > -1

and x > 2

[topaz urchin]

no no that's not how you solve these

[novel dune]

oh

how do u solve them

[topaz urchin]

so draw a number line

[novel dune]

okay

done

[topaz urchin]

our factors are

i mean

our equation

is

12x^3 + 12x^2 - 48x - 48 > 0

right

[novel dune]

yes

[topaz urchin]

yeah so if we divide it by 12 we get

x^3 + x^2 - 4x - 4 > 0

[novel dune]

yeah

[topaz urchin]

so you get

(x+2)(x+1)(x-2) > 0

right

[novel dune]

yes

[topaz urchin]

yeah so plot -2, -1 and 2 on the number line

[novel dune]

okay

i did it

[topaz urchin]

now we have 2 on the right side

if you put x greater than 2

[topaz urchin]

this whole thing is positive

[novel dune]

yes

[topaz urchin]

so we want x = 2 to infinity

now we'll look at

x between -1 and 2

[novel dune]

yes

[midnight plank]

@novel dune Has your question been resolved?

[topaz urchin]

Sorry I had to go lol

You can try it yourself first

Then we can cross check

[midnight plank]

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[stoic garnet]

Can't we have well defined functions which are not expressible as a expression but instead is better found in other ways

[prime garden]

an expression*

[lyric charm]

$\int_0^x e^{-t^2}\dd{t}$

[grand pond]

Ann

[lyric charm]

famously this integral is not expressible via elementary functions of x

[prime garden]

Has it been proved? Proved that it's not possible?

[stoic garnet]

But it's still expressed as an integral right? But like functions which are descriptive in english

But can't be expressed using any symbols whatsoever without actually making like a lot of cases

[lyric charm]

https://en.wikipedia.org/wiki/Risch_algorithm?wprov=sfla1

seems so

tho i have not looked into it

[prime garden]

Ok

[lyric charm]

"can't be expressed using any symbols whatsoever" is a tall order!

[prime garden]

give an example

[stoic garnet]

Hmm makes sense! But what about like what if we want to know if we can go from input to output always

Given input and a description of the function

How to find output

[midnight plank]

@stoic garnet Has your question been resolved?

[stoic garnet]

Is my question off base

<@&286206848099549185>

.close

[midnight plank]

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[spark gazelle]

how could i do this on a calculator?

[polar star]

R_3 is just for n = 3

you can easily compute it using 3 subintervals

but regardless, can u shwo the whole question?

u may not have to use a calculator at all

[spark gazelle]

this is the whole question

[polar star]

"which of the following statements is true"

show the statements

[spark gazelle]

oh sorry

idk my test is with calc so i thought i might as well as try to use it to the best of my ability

ik its an overestimate

bc its a right riemann sum

[polar star]

and because it is increasing

[spark gazelle]

ig itd be 13.133?

[polar star]

nah use calculator for that

[spark gazelle]

how tho

[polar star]

its hard to tell just by inspection

u got casio?

[spark gazelle]

uhh

no

ti89 titanimu

[polar star]

okay fine

u will have to do it manually then i guess

[spark gazelle]

yeah we're not allowed to use any other calculator

[polar star]

do u know an expression for the right riemann sum

its gonna be ∑f(a+kdeltax)deltax

[spark gazelle]

uhh

[polar star]

u compute that from k = 1 to 3

so u just add (f(a+deltax)+f(a+2deltax)+f(a+3deltax))deltax

a being lower bound, delta x being (b-a)/3

[spark gazelle]

tf

uh

[polar star]

do u know what riemann sums are?

[spark gazelle]

i mean vaguely

i just kind of know its the area under a curve

[polar star]

do u know how to convert that integral to its riemann sum?

[spark gazelle]

nope

oh wait

i can just plug the equation into

a plot and do y function

[limpid vigil]

[\int_{a}^{b} f(x)dx =\lim_{n \to \infty } \sum_{i=1}^{n} f(x_i)\Delta x ] where (\Delta x =\frac{b-a}{n} ,x_i=a+i\Delta x)

[grand pond]

HeavensJester

[spark gazelle]

oh

ok so the definition i get

now how to plug into a calculator to actualy solve though?

[limpid vigil]

use as approximation like instead of n \to \infty let like n=1000 or something

[spark gazelle]

nvm i just put it into plot graph

and did it from there

i think doing summation is just overcomplicating it

[limpid vigil]

so something like [\sum_{i=1}^{1000} f\left(a+ i\cdot\left(\frac{b-a}{1000}\right)\right) \frac{b-a}{1000} ]

[grand pond]

HeavensJester

[spark gazelle]

i think

that might be overcomplicating the problem a bit

.close

[midnight plank]

Channel closed

Closed by @spark gazelle

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[limpid vigil]

not really in your case let the 1000(I chose 1000 as it would be a better approximation for the actual integral) be 3 as the

[midnight plank]

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[lament fox]

I don't know how to approach this

[cedar mason]

i mean

one would assume the numerator factors out

,wolf give factors of (3x^5 - x^4 + 2x^3 -12x^2 -2x +1)

[grand pond]

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[cedar mason]

thanks wolffram

yeah that's weird

[cedar coral]

this seems like you need to decompose it to simple fractions

start by factoring the denominator for that

you can start with (x^3-1) = (x-1)(x^2+x+1), so you'll have 4 fractions

[midnight plank]

@lament fox Has your question been resolved?

[lament fox]

that would be a gnarly partial fraction decomposition no?

[cedar coral]

so?

[lament fox]

I don't want to do all of that

I doubt they would make me do that

[cedar mason]

yeah

,wolf expand (1 + x + x^2) (1 + x + x^3)

[grand pond]

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[cedar mason]

okay thats not it

cmon wolf send help

[tough shale]

,w (3x^5-x^4+2x^3-12x^2-2x+1)/(x^3-1)

[grand pond]

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[tough shale]

C'mon, I need the quotient and remainder

,w quotient and remainder of (3x^5-x^4+2x^3-12x^2-2x+1)/(x^3-1)

[grand pond]

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[midnight plank]

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[chilly cobalt]

Let $a$ be a real number satisfying $\$
img

If the set of possible values of $a$ is $(x,y]$ for real numbers $x,y$. What is the value of y - x? (correct to 4 decimal places.)

[grand pond]

Copter
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[chilly cobalt]

can someone help me? i have no idea where to begin

please dear math gods, bless me with thee knowledge😭

[cedar mason]

i dont understand what the question means

the latex is weird too

[small jasper]

Hints:

• 2009-1=2008
• 2552-1=2551
• 2008+2551=4559

[chilly cobalt]

find the set of real number a that satisfies the equation, probably

[small jasper]

Yeah that’s the interpretation that I went with

[chilly cobalt]

yeah i did notice that...

im not good with this type of stuff xd

[small jasper]

._. Why didn’t you say that then

But what can you conclude about the floors (as in what are their values)

[chilly cobalt]

a>1?

[small jasper]

I’m asking about the values of these

[chilly cobalt]

oh

floor(2009/a) = 2008 and floor(2552/a) = 2551

[small jasper]

mhm

[chilly cobalt]

so like

next thing might be inequalities, i think?

[small jasper]

Yup

[chilly cobalt]

aaa ive never done this😭

2009 > 2009/a >= 2008, maybe?

YIPPE

give me a moment

1 < a <= 2552/2551

,w 2552/2551

[grand pond]

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[chilly cobalt]

.

[small jasper]

,calc 2552/2551

[grand pond]

Result:

1.000392003136

[chilly cobalt]

so the answer is 0.0004

[small jasper]

Mhm

[chilly cobalt]

tyy!!

,close

.close

[midnight plank]

Channel closed

Closed by @chilly cobalt

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[midnight plank]

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[ruby willow]

yo can i get help

[ruby willow]

<@&286206848099549185>

[mossy moat]

arcsine(0.59)

[ruby willow]

?

sorry im slow in the head

[lyric charm]

your calculator should have a button that says sin^-1 or something similar.

[ruby willow]

oh yeah

[lyric charm]

use it to calculate $\sin^{-1}(0.59)$.

[grand pond]

Ann

[mossy moat]

this will give one answer

[lyric charm]

and also make sure the calculator is in degree mode.

[ruby willow]

its asking for 2 answers tho

[lyric charm]

do what i said first. that's going to be your part a.

[ruby willow]

ok i gt 36.1570082

[limber grove]

!15m also

[midnight plank]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[lyric charm]

round to nearest whole.

[ruby willow]

36

[ruby willow]

i used it once

[limber grove]

uh

did you read the second part of the sentence

[lyric charm]

right

[ruby willow]

oh sorry

new to server

[lyric charm]

look at the graph now and see that it crosses the line y=0.59 in two points

one of them is the x=36° you just found

[ruby willow]

thats gotta be the one closer to 0

[lyric charm]

the other is symmetric to it across the line x=90°

[ruby willow]

yes

how do i calculate the other one

[lyric charm]

in other words the other solution will be 180-36

[ruby willow]

how

ok i got it ty

[lyric charm]

sin(180°-x) = sin(x) always

[ruby willow]

alr

[midnight plank]

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[vocal briar]

can someone explain to me why my logic is wrong when I see, integral of cosx/sinx

[vocal briar]

im not sure if i aksed this before bc i forget but i js did integral of (tanx)^-1

but this doesn't get the right ans of ln|sinx| + C

cuz integral of tanx^-1 is js ln|tanx| * (secx)^2 + C i think

the correct way is to use u sub which i realize now

but why is this way wrong?

im aware it gives me the deriv as 1/ sinxcosx which is obv not cosx/sinx

but i still dont see what is wrong with converting cosx/sinx to (tanx)^-1

[vocal briar]

even this is wrong acc in case i didn't make that clear

[bitter tide]

why not just do a substitution

[vocal briar]

no like ik you can

but the first time i did this question i didn't

so like im js not sure what is wrong with the method

[bitter tide]

well u do get cotx, but thats just overcomplicating sum trivial

[vocal briar]

ok but integral of cotx and tanx^-1 is diff tho

like the method i mean

not the ans

[bitter tide]

well tan^-1 is arctan. if u mean 1/tanx, then thats cotx

[vocal briar]

yeah

but like i wanted to use ln

cuz like 1/x would've been lnx

[bitter tide]

well yea ud needa sub and then u get ugly stuff

[vocal briar]

yeah but like can u show me how to do it that way anyway 😭😭

im kind of confused on what to do after i get integral of tanx^-1

[pale vapor]

can you send your work

[bitter tide]

ok lemme get sum paper, im rusty on integrals

[vocal briar]

i js wrote

$ln|tanx|\cdot\frac{1}{cos^2{x}} +C$

[grand pond]

gamer75431

[vocal briar]

which im not sure if thats right

[pale vapor]

no send your wor

k

[vocal briar]

i js know that when you differentiate this you dont get cosx/sinx

[pale vapor]

i can tell you where you went wrong

[vocal briar]

heres what i did

im not sure its much better than what i sent tho lol

[bitter tide]

ok well i ended up after sub u=tanx with integral du/u(1-u^2)

ah u cant apply ln to that

it needs linear function on bottom

[vocal briar]

ln doesn't apply to trig func?

[bitter tide]

nope

[vocal briar]

oh

[pale vapor]

integral of 1/tanx dosent work like that

onlt integrak of 1/x i lnx

only*

[vocal briar]

ohh

[bitter tide]

can be generalized to 1/ax+b

ln(ax+b)/a

[vocal briar]

oh okk

[pale vapor]

yeah but just u sub there

[bitter tide]

well yea but u can skip that

[pale vapor]

yeah

[vocal briar]

right that makes sense since deriv would've been a/ln(ax+b)

of F(x) = (ax+b)^-1

[bitter tide]

deriv of ax+b is a

[vocal briar]

no sorry i forgot to type it right lmao

[bitter tide]

u dont sub the whole thing. the linear function to get 1/u

then u can ln that

[vocal briar]

wdym

[vocal briar]

like is this wrong?

[bitter tide]

yeah

ok lets say u have 1/(ax+b)

integrate that