help-28

[slate shuttle]

bro canu draw (if u want)

[raw quartz]

I think it’s orthogonal to the plane

Not an orthogonal plane

[analog shale]

suppose that OB=r, find O

[slate shuttle]

Hmmmm🤔 but it seems like its imposible we to find exact o we have to find a range

but ima try this

Hold on

[analog shale]

oh we want to find O in terms of r

we are just assuming the radius is r for now

we will try and find what r is later

[slate shuttle]

Like this?

[analog shale]

justify why OA=OB

[analog shale]

set OB=x, find the position of O

[slate shuttle]

Hmm i think i wroted the reasons can you tell me where you didnt get it

[slate shuttle]

🤔🤔🤔🤔🤔🤔hmmmmmm

Fine imatry

[analog shale]

um can you explain I don't see it

[hidden fractal]

ok i think i got the answer to this so i can maybe help try to explain now??

[analog shale]

yeah sure, but try and not give the solution away when you are explaining

[hidden fractal]

okk so in order to minimise the radius we assume (11,9) is a point on the circumference

[analog shale]

why can we make that assumption?

[hidden fractal]

um if its not on the circumference the circle can be smaller i would have thought

i could be tripping though, it is late at night right now

[analog shale]

while your claim is correct, it might be a bit hard to prove directly

[hidden fractal]

ok fine i dont need that step for right now

anways i would call the center C

and try to work out the line that both b and c are on

[slate shuttle]

The distance of point a from the line
y-x=0
Lets say this d
Let's replace y and x with the coordinates of a.
d=|9-11|/root 2
d=root 2
Thats why ıts root 2

İf you get this i ll explain others

[torn jolt]

√(4²+4=? Both should be square no

[slate shuttle]

Wdym

[analog shale]

yes but i mean why is A on the circle?

[slate shuttle]

İ forgot write its the (11,9)

[analog shale]

you didn't prove that

[analog shale]

yeah but (11,9) is inside the circle, it might not be on the circle?

[torn jolt]

look at your solution

[slate shuttle]

Lies inner region of a circle

*lines

Am inwrong?

[analog shale]

yeah but it is possible for this to happen

where A is not on the circumference of the circle

[torn jolt]

yh OA≠OB cause OA isn't the radius of the circle

[analog shale]

i think the question says it is possible for OA=OB (minimum value of the radius is achieved there, although simply claiming OA=OB is unfounded).

[slate shuttle]

Yes, you are right, but wouldn't it make our job easier to accept it like this because it asks us for its smallest value?

[analog shale]

well, you would have to show that smallest value => A on the circle

you can see that there is a leap in your argument, by claiming that A lies on the circumference of the circle

[torn jolt]

possibility is there but the question says A is inside the circle

not on the circle

[analog shale]

yeah i think by "inside" they definitely meant including the boundaries.

but yeah you got the same point as mine

OP is assuming it lies on the circle

[hidden fractal]

inside would mean approaching the boundary i presume

so we can count touching the boundary

[analog shale]

the question would be invalid if "inside" don't include the boundary

[slate shuttle]

What would you do if I asked you what is the smallest value of x+3, where x is an integer greater than 5?

Think this

[analog shale]

because a minimum value does not exists

[torn jolt]

don't you think we are making it more complicated?there must be an easy way to do

[analog shale]

x+3>=5, x>=2, so the minimum is 2

[hidden fractal]

i am sure that A lies on the circumference, but i doubt it even matters

have u worked out the line that O must be on

@slate shuttle

[torn jolt]

yh

[analog shale]

to some extent, you are not solving for an equality but rather an inequality

[slate shuttle]

You write x+3>8, that is, you take it as if it is equal to 8, then if it is greater than 8, you say the smallest 9, right?

[analog shale]

Also I get the point you are trying to tell me. Indeed, you can just solve for x+3=5 in this case, and you can prove this. But in the circle case, it is not obvious why

[slate shuttle]

1min

[torn jolt]

I think we should retranslate the que to verify if it's right or not

[hidden fractal]

if ur looking for integer values for x+3

[slate shuttle]

Hmmmm

[hidden fractal]

yea its painful

you can see something so clearly

but u just can't prove it

[slate shuttle]

Yes i cant probe it

İm thinking how to explain it hmm

[hidden fractal]

just start by doing this i think

it seems v useful

[analog shale]

I think it might be better to work with the assumption that OA<=OB rather than OA=OB. (becuase for OA<=OB you are not assuming anything)

[slate shuttle]

Okay guys letme look at it again.

[slate shuttle]

somewords may not exatly translate

[analog shale]

@hidden fractal is your answer on the options they gave?

[slate shuttle]

Lets accept a is near by the point you located

[hidden fractal]

yes

[analog shale]

fr? hm

[slate shuttle]

I got something

[analog shale]

are they asking for radius or diameter?

[hidden fractal]

please tell me we know the equation of the purple line now

it is extremely useful

[analog shale]

this can't be right, the radius on the diagram is 5.9

but no radius less then 6 is on the option list

they must be asking for diameter

[hidden fractal]

what diagram

is the radius 5.9

[analog shale]

this

[slate shuttle]

Radiua

[hidden fractal]

the squares represent length 2 tho

[analog shale]

oh wait

i messed up one bit

[analog shale]

nvm

[slate shuttle]

İ found this..

[hidden fractal]

niceeee

i get r>sqrt(50) as well

[slate shuttle]

Haha wp bro

[hidden fractal]

i think my method was less clean than this tho

[analog shale]

right that should be correct

[slate shuttle]

Bro ofc you used computer😂

[hidden fractal]

no

i used it to draw diagrams

[analog shale]

but you are still making the assumption that A lies on the circumference of the circle?

[hidden fractal]

i literally cant draw a circle for my life

[slate shuttle]

hmm wdym

[slate shuttle]

İ aggre your way way more clenaer bro

[hidden fractal]

no your method was a lot faster

i did a really stupid method

[analog shale]

aren't you assuming OA=OB?

[slate shuttle]

No no yours better❤️

[analog shale]

when you got r=5\sqrt 2?

[slate shuttle]

Bro its the only possible choice you can try if u want

i didnt find any other

[analog shale]

I mean, again, your intuition is right, but the justification is not there

[hidden fractal]

considering its not an olympiad level problem where ur getting rigoursly marked i think making this assumpition is safe enough

[slate shuttle]

hmm

Bro you realy right

[analog shale]

typically one need to be rigorous in math, and in their justification. Because many of the time where mistakes are made is due to some incorrect intuition.

[analog shale]

like to make sure you fully understand the subject you have to justify these intuitions.

[slate shuttle]

Yea ...

but like i said only 5root2 fit there

[torn jolt]

use the info of B

[analog shale]

still, this is more or less just intuition

[torn jolt]

y=x

[raw quartz]

$B\left(7,7\right)$ and Tangent line $T = 14-k$ such that center $C\left(k,14-k\right)$ then radius $r = \mathcal{D}\left(C, B\right)$ such that $\mathcal{D}\left(A,C\right)<r$ then, $r = \mathcal{D}\left(C, B\right)= \sqrt{2}\left|k-7\right|$ and $\mathcal{D}\left(A,C\right) = \sqrt{\left|11-k\right|^2 + \left|k-5\right|^2}$ such that $2\left(k^2 -14k + 49\right)>\left(121-22k+k^2\right)+ \left(k^2-10k+25\right)$

[slate shuttle]

Bro i tried your way but failed maybe you can show us😁

[torn jolt]

okay

[slate shuttle]

Man you right

[analog shale]

this is the solution i had in mind too

[hidden fractal]

hm does anyone want to do a fun number theory problem then

[slate shuttle]

but it's intuition using a little bit of logic, after all, the 3-4-5 triangle should come to mind there.

[slate shuttle]

İm done bro💀

[analog shale]

prob best to not occupy a help channel for this, maybe #math-discussion

[hidden fractal]

yes - i think we should close this then

[slate shuttle]

Didnt realy get it ...

[analog shale]

@slate shuttle try and see if you follow Shoshir's solution

[analog shale]

i can explain it, which part did you not get?

[torn jolt]

How you wrote√(11-k)²+(k-5)²??

[slate shuttle]

Where

[torn jolt]

Nah I got that you simplify it

[analog shale]

distance between A and C , where A=(11,9), and C=(k,14-k)

[slate shuttle]

İ dont think im able do this

[analog shale]

I can help you understand their solution if there is any point you dont get

just lmk

[slate shuttle]

Why r=D(C,B)

[glossy valve]

Shoshir

[raw quartz]

made a mistake with greater than srry

[slate shuttle]

Np

[torn jolt]

r=|k-(14-k|/√1²+1² = |2h-14|/√2 = √2|h-7|

[analog shale]

we defined C to be the center of the circle

[torn jolt]

don't take OA on circle

[slate shuttle]

yeah makes sense🤔

[analog shale]

so since the circle is tangent to y=x, CB is the radius

so r=D(C,B) (here we set radius to r)

[torn jolt]

now compare the prev eq and value of r as √2(h-7)

[slate shuttle]

Go on

Pls contunie

[torn jolt]

you'll get h=12

put the value of h in r=√2(h-7)

youll get the integer

[slate shuttle]

makes sense🤔

[analog shale]

why did the k change into h

[torn jolt]

dude I took as (h,k)

he wrote differently

[slate shuttle]

this problem was hard xd

[raw quartz]

Simplified, $2k^2 - 28k + 98 > 2k^2-32k+146$ then $-28k>-32k+48$ and $4k>48$ so therefore $k>12$ and $r=\sqrt{2}\left|k-7\right|$, so $\min\left\lbrace \mathcal{D}\left(C,B\right)\right\rbrace = \lim_{k\to 12} \sqrt{2}\left|k-7\right| + \epsilon $ therefore $\min\left\lbrace \mathcal{D}\left(C,B\right)\right\rbrace = 5 \sqrt{2} + \epsilon$

[glossy valve]

Shoshir

[analog shale]

wait what is your h?

you didnt define it

[slate shuttle]

@analog shale see he even used limit😁

[torn jolt]

the center as (x,y)= (h,k)

the coordinates of center

[slate shuttle]

Nice

[analog shale]

yeah, limits are fine, you dont quite need it though

[slate shuttle]

Bro how can i know 😂

need limit in this circle

İm not that smart

I think I'll go with intuitive😵‍💫

[torn jolt]

it is assuming A is on the circle that's why the answer is 5√2

[analog shale]

then how did you get this formula with r?

[raw quartz]

I just did it because $k\in \left(12,\infty\right)$ isn't a compact interval

[glossy valve]

Shoshir

[torn jolt]

distance formula

[analog shale]

you mean closed?

[raw quartz]

https://en.wikipedia.org/wiki/Compact_space#Open_cover_definition

[analog shale]

closed and bounded below gives inf in the set

[analog shale]

yeah im aware of that

im just saying u compact is too strong for this

[raw quartz]

I've been studying too much real analysis

[slate shuttle]

Agree

[raw quartz]

Because... Closed and compact are the same thing for a subset of $\mathbb{R}^n$

[analog shale]

oh you are assuming the right triangle is isosceles to get that

[glossy valve]

Shoshir

[analog shale]

[n,infty) is closed but not compact

heine borel

[raw quartz]

fuck

I forgot about the bounded part

[slate shuttle]

guys 1 more bullet?

[torn jolt]

i am not assuming anything

[raw quartz]

Sure

[torn jolt]

it is the point to line distance formula

[slate shuttle]

nvm its hard

[raw quartz]

send it

[slate shuttle]

sure?

[raw quartz]

yep

[analog shale]

thats $r=|k-7-(h-7)|/\sqrt 2$ no?

[glossy valve]

qwertytrewq

[analog shale]

where did you get $(2h-7)/\sqrt 2$?

[glossy valve]

qwertytrewq

[torn jolt]

I eliminated k in the form of h as 14-h

[slate shuttle]

A polynomial P(x) of degree 4 with real coefficients satisfies the inequality P(x)>=x for every real number x.
P(4)=?

[slate shuttle]

@raw quartz

[analog shale]

oh ok yeah, it looks the same just with k h flipped

u essentially just did (14-h,h)

[torn jolt]

yup (h,14-h)

[analog shale]

isnt that the same question?

[slate shuttle]

yeah maybe he solve it by different way i wanna see

[raw quartz]

$P_4\left(x\right) \geqslant x, \quad \forall x\in \mathbb{R}$

[glossy valve]

Shoshir

[slate shuttle]

Guys i can make anothee solution if you want

[analog shale]

the way you figured out is the fastest i can think of

[torn jolt]

nope

[slate shuttle]

ok😒😔

[raw quartz]

$P\left(1\right)=1 \\ P\left(2\right) = 4 \\ P\left(3\right)=3$

[glossy valve]

Shoshir

[slate shuttle]

😮

[slate shuttle]

Ye

[analog shale]

much much better than solving a system of linear equations with 5 unknowns

[slate shuttle]

😮

i aggre

But sadly there something i cant explained to u

maybe its because i trNSLATED it wrong idk

[analog shale]

the translation of the problem is fine, is there some part of the solution you didnt fully prove?

[slate shuttle]

Yeah..

[raw quartz]

$a+b+c+d+e = 1 \\ 16a+8b + 4c +2d + e = 4 \\ 81a+27b+9c + 3d + e = 3$

[glossy valve]

Shoshir

[slate shuttle]

İ still looking a simple way to explain it to you

but there is still something i didnt realy understand about the question too because the turkish is problematic too

[analog shale]

the proof is along the lines of rolles theorem

[slate shuttle]

İdk what is rolles therom. Sory

[slate shuttle]

Bro just draw a graph it would be easiee

[analog shale]

reasonable

[slate shuttle]

😄😃

[raw quartz]

$a+b+c+d = 1-e \\ 8a+4b+2c+d = 2 - \frac{e}{2} \\ 27a+9b+3c+d = 1 - \frac{e}{3}$

[glossy valve]

Shoshir

[torn jolt]

can I give him a hint?

[slate shuttle]

Ofc why even asking me

[raw quartz]

$\exists x \in \left[2,3\right], \quad \partial P\left(x\right)<0$

[slate shuttle]

ignore it

[glossy valve]

Shoshir

[slate shuttle]

One of them should be 28

[torn jolt]

Q(x)=P(x)-x

Q(1)=P(1)-1=1-1=0

check for 1,2 and 3

youll find the next step then

[raw quartz]

$\int_{\left[2,3\right]} P\left(x\right),dx ,= -1$

[glossy valve]

Shoshir

[torn jolt]

?,what are you doing

[raw quartz]

mean value theorem

[analog shale]

u can use the fact that P(3)=3 and P(x)\geq x to deduce something about the derivative (inverse function theorem)

[slate shuttle]

@raw quartz
Hint:

[analog shale]

or some version of rolles by looking at P(x)-x

[raw quartz]

$P^{-1}\left(3\right) = P\left(3\right) \ P^{-1}\left(4\right) = \frac{P\left(2\right)}{2} \ P^{-1}\left(1\right) = P\left(1\right)$

[glossy valve]

Shoshir

[slate shuttle]

@raw quartz

[analog shale]

not that but if P'(3)\neq dx/dx at 3, then local injectivity say that there exists a small region for which P(x) will dip below x

[raw quartz]

$A\left(x\right) = P\left(x\right)-x \ A\left(x\right) = \left(x-3\right)\left(x-1\right)\left(x-r\right)\left(x-s\right)$

[glossy valve]

Shoshir

[analog shale]

u forgot constant multiple

[slate shuttle]

Why?

[crude kayak]

1 is a root for P(x) - x

[slate shuttle]

Ahh

Mb

[crude kayak]

how can you be sure that A has four roots?

[slate shuttle]

Hoursmof slepness..

[analog shale]

given

[slate shuttle]

1minutw

[analog shale]

degree 4

[analog shale]

factoring in complex im assuming

[crude kayak]

oic thx idk the language (turkish?)

[analog shale]

me neither

[crude kayak]

google translate is a great invention

[raw quartz]

$A\left(2\right) = 2r + 2s - rs - 4 \ P\left(2\right) = 2r+2s - rs - 2 \ 2r+2s-rs-2 = 16a+8b+4c+2d+e$

[glossy valve]

Shoshir

[slate shuttle]

Thats ıt..

[analog shale]

this is still not enough, u can inverse func for A at 3

or rolle

[slate shuttle]

@crude kayak

[analog shale]

by inv function i mean the "local injectivity" part

[crude kayak]

the choices aren't fractions

[raw quartz]

$A\left(4\right) = 3rs - 12r - 12s + 48 \ P\left(4\right) = 3rs - 12r - 12s + 52$

[glossy valve]

Shoshir

[analog shale]

thats an underlined 22

[slate shuttle]

whats wrong?

İt should be true

[analog shale]

your underline looks like 22/7

[slate shuttle]

💀

Fr?

[analog shale]

yeah you drew a 7 under the 22

[slate shuttle]

🤨🤨🤨🤨

Bro

its eniugh today ig

[raw quartz]

$A\left(4\right) = 3\left(s-4\right)\left(r-4\right)$

[glossy valve]

Shoshir

[slate shuttle]

Bro what are u doin

[raw quartz]

$P\left(4\right) = 3\left(s-4\right)\left(r-4\right) + 4 \geq 0$

[glossy valve]

Shoshir

[slate shuttle]

😑

Did you even look at what i send?

Guys we are here lik 3 hours

[raw quartz]

$P\left(x\right) = ax^4 + bx^3 + cx^2 + x\left(d+1\right) $

[analog shale]

,texsp ||Set $A(x):= P(x)-x$, then $A(x)\geq 0$. Since $A(1)=0$, we clain that $A'(1)=0$, suppose $A'(1)\neq 0$, then local injectivity tells us that there exists a neighborhoud $(1-\epsilon,1+\epsilon)$ such that $A$ is injective. But this implies that either $A$ will be negative somewhere, which is a contradiction. Therefore $A'(1)=A(1)=0$ (one can also directly compute the limit to see that $A'(1)=0$). This means that $1$ is a double root of $A$. SImilarly $A'(3)=A(3)=0$ so $3$ is a double root of $A$. Since $A$ is degree $4$ it follows that $A(x)=c(x-1)^2(x-3)^2$. Plugging in $A(2)=2$ yields $c=2$ and thus $P(4)=22$.||

[slate shuttle]

İma go sleep bye

[analog shale]

bye

[glossy valve]

qwertytrewq

[raw quartz]

$\frac{P\left(4\right)}{3s-12} + \frac{8}{3}= r$

[glossy valve]

Shoshir

[raw quartz]

$P\left(4\right) = P\left(4\right) + 8s - 28$

[glossy valve]

Shoshir

[raw quartz]

$s=\frac{7}{2}$

[glossy valve]

Shoshir

[analog shale]

how did you get that?

[analog shale]

from here, you for got that A(x) is a constant multiple of (x-3)(x-1)(x-r)(x-s)

[raw quartz]

$P\left(4\right) = 3\left(s-4\right)\left(\frac{P\left(4\right)}{3s-12}+\frac{8}{3}\right)-4$

[glossy valve]

Shoshir

[analog shale]

so it should be c(x-3)(x-1)(x-r)(x-s)

[raw quartz]

why

[analog shale]

roots at 3, 1 implies $A(x)= c(x-3)(x-1)(x-r)(x-s)$

[glossy valve]

qwertytrewq

[analog shale]

why is $c=1$?

[glossy valve]

qwertytrewq

[analog shale]

it is not given that $P,A$ is monic

[glossy valve]

qwertytrewq

[raw quartz]

$\frac{-3P\left(4\right)}{2}+3 = 0$

[glossy valve]

Shoshir

[raw quartz]

$P\left(4\right) = 2 C$

[glossy valve]

Shoshir

[analog shale]

Why is this? this simplifies to $0=8(s-4)-4$

[glossy valve]

qwertytrewq

[analog shale]

want to find $3 c r s - 12 c r - 12 c s + 48 c$, given $-c r s + 2 c r + 2 c s - 4 c=2$

[glossy valve]

qwertytrewq

[analog shale]

I am pretty sure lagrange interpolation say that there are infinitely many set of solutions

[full forum]

@slate shuttle Has your question been resolved?

[full forum]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
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[thick hedge]

Trying to find the extrema using Lagrange multipliers

[onyx glen]

so you have $\mathcal L(x,y,\lambda) = x^2 + y^2 - \lambda(xy - 1)$, yes?

[glossy valve]

Ann

[thick hedge]

wait

how did you get that

[onyx glen]

or plus λ(xy-1)... i don't actually remember what the convention it but it also might not matter

yeah sorry plus

other than that tho it is a lagrangian like any other

[thick hedge]

I was thinking $\grad{x^2+y^2} = \lambda\grad{xy}$

[glossy valve]

What a wonderful world !

[austere cove]

You solve xy = 1 for 0, because LMs only work with constraints set to 0, iirc

[onyx glen]

is that how you were taught to do it

[thick hedge]

This is what I was taught

[thick hedge]

I don't follow

[onyx glen]

that remark doesnt apply in your form

but i mean ok this is basically just the same thing restated in a different format

also

suggest writing gradients and shit as columns sometimes

$\div(x^2 + y^2) = \bmqty{2x \ 2y}$

[glossy valve]

Ann

[onyx glen]

\\ for newline inside a matrix

[umbral dome]

it doesn't really matter if you require it = const. or = 0 since the constant will go away after differentiating

[thick hedge]

so $\bmqty{2x \ 2y}= \lambda \bmqty{y \ x}$

[austere cove]

Fair enough!

[glossy valve]

What a wonderful world !

[thick hedge]

Which is sort of sus

Wait no

$2x = \lambda y ; 2y = \lambda x; xy=1$

[glossy valve]

What a wonderful world !

[thick hedge]

I have to solve this system of equations

[onyx glen]

yes

[thick hedge]

so $4(xy) = \lambda(xy)$.
\
So $\lambda =4$
\
We thus have
$2x= 4y$
\
$2y=4x$
\
Whose only solution is $(0,0)$.
\
However, $0 \cdot 0 \neq 1$

[glossy valve]

What a wonderful world !

[onyx glen]

hold up

intuition says we did miss something

oh yes you did

it's $4xy = \lambda^2 xy$

[glossy valve]

Ann

[thick hedge]

ah

[onyx glen]

whence $\lambda = \pm 2$ not 4.

[glossy valve]

Ann

[thick hedge]

yea

but I think this doesn't change anything

[onyx glen]

it changes a lot actually

you actually get solutions this time

[thick hedge]

wait, I get 2x=2y, so x=y

so the entire line x=y as a solution

That gives me $2x^2$

no?

[glossy valve]

What a wonderful world !

[thick hedge]

wait, no

[onyx glen]

what is the intersection if any between the line x = y and the curve xy = 1

[thick hedge]

the constraint gives me x^2=1

[thick hedge]

(1,1)

[onyx glen]

that's it?

it consists of just one point?

[thick hedge]

that and (-1,-1)

[onyx glen]

it consists of just those two points?

[thick hedge]

yes

[onyx glen]

are you sure?

[thick hedge]

yea

[onyx glen]

ok yes you're correct

[thick hedge]

Thanks!

[onyx glen]

now look at the other case

i.e. λ = -2

[thick hedge]

hmm

well, that's x=-y

which has no solutions

[full forum]

@thick hedge Has your question been resolved?

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[austere cove]

@thick hedge

[thick hedge]

.reopen

[full forum]

✅

[austere cove]

This problem should have a solution

[thick hedge]

Yeah, It does, found one

[austere cove]

Specifically at (1, 1) just from a geometric analysis

Oh.

I'm late to the party

[onyx glen]

we found (±1, ±1) yes

[austere cove]

Oooh, I just read the bottom

My bad

.close

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[eternal dock]

hey, I have two functions F and f. f=1/(x^2+1), and F'=f, I'm technically not supposed to be able to find F(x) and I'm supposed to prove F(x)+F(-x)=0 for each x in R

[gilded matrix]

@eternal dock

is R for real number?

[eternal dock]

yes

[onyx glen]

is F(0) fixed as 0 though

[eternal dock]

yes

[onyx glen]

right

[eternal dock]

sorry forgot to mention

[gilded matrix]

?

[onyx glen]

anyway $F(x) \overset?= -F(-x)$ seems reasonably easy to prove by integral manipulation

[glossy valve]

Ann

[gilded matrix]

I can do it but I m not getting the question

[eternal dock]

oh

my mind didn't go there

[gilded matrix]

it's given that F'=f but we have to find for F

[eternal dock]

is there no way to prove that F is an odd function

[gilded matrix]

?

[eternal dock]

F is arctan however I'm not supposed to know that

[onyx glen]

no, we have to do the question WITHOUT finding F.

[onyx glen]

there is...?

[rugged void]

derivative of an even func is always an odd func

🫠

[onyx glen]

but how do you know that the antiderivative of an even function is always odd

[rugged void]

is it not, smh?

[eternal dock]

well there's a proof somewhere probably

[rugged void]

oh frick i misread it

that is absolutely mb

[eternal dock]

well if the derivative of an even function is odd

that means that for f which is the derivative of F, and even, F has to be odd?

[gilded matrix]

I think we have to use integration at some point

[onyx glen]

are we allowed to refer to integrals in any way, shape or form at all or no @eternal dock

[eternal dock]

yes

just not solve integral of 1/x^2+1dx

we haven't learnt tan sub

[eternal dock]

I just solved it with integral manipulation, I'm curious about the thing the other guy mentioned

[onyx glen]

brackets

[onyx glen]

what thing

[eternal dock]

.

[rugged void]

that's... fairly elementary to see?

also like you can literally just... show that if F' is even, F is odd

just... write the statement down for F' being even and integrate

[eternal dock]

we cannot use things we aren't taught without proof and I'm not really sure how I'd prove that

[rugged void]

try it

what'd the statement for "F' is even" be

[eternal dock]

f(-x)=-f(x)

[rugged void]

mmm no

that's an odd function

and we know that f is even

[eternal dock]

yea not the -

my brain isn't working

prove that d/dx F(x)=d/dx F(-x)

from that we get f(x)=f(-x)

therefore F is odd

?

[eternal dock]

since this is true for f(x)=1/(x^2+1)

[rugged void]

this is the statement, yes

now try to integrate both sides--left side is easy, but the RHS will require a very basic application of the chain rule

[eternal dock]

we're not supposed to be able to integrate that yet since we haven't learnt tan substitution

[rugged void]

...

not in explicit form dude

but we know that F'(x) = f(x)

and therefore, F'(x) = F'(-x)

you can integrate that

[eternal dock]

F(x)=-F(x)

which is what we want to prove

[rugged void]

that's literally it, yes

two/three lines

smh

😭

[eternal dock]

fair enough

thank you

.close

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[thorny shuttle]

how do i proceed i tried replacing x-> -x

[onyx glen]

why not put x = -1

[rugged void]

i was so confused when i saw this because i was like is this not just that

[thorny shuttle]

not really

i am pretty sure i gotta take limits

[onyx glen]

what

what do you mean "not really"

are you saying that putting x=-1 is illegal somehow or what

[thorny shuttle]

try putting x =1

hol up wait i can do -1

[rugged void]

not

x = 1

yeah-

[thorny shuttle]

for 1 i gotta put limit

right

got it

[rugged void]

😭

[thorny shuttle]

cuz 0/0 indeterminate

nvm

[rugged void]

...

[thorny shuttle]

complex numver moment

[onyx glen]

-1 is not the same thing as 1

not even in R

[thorny shuttle]

i mean yeah

i didnt consider that

lmao

[sudden cargo]

are you supposed to make a generating function?

[full forum]

@thorny shuttle Has your question been resolved?

[full forum]

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[cursive trout]

help me pls

[cursive trout]

im new in matrices

A is 2*2 matrix = [2 1 -3 4]

B is 2*1 matrix [7 6]

A*X = B

find X

help ples

[pure basin]

Uh

[cursive trout]

yeaah

?

[pure basin]

ok so

ax=b

[cursive trout]

yes

yes

[pure basin]

x=a^-1b

[cursive trout]

hmm

[pure basin]

Find a inverse

[cursive trout]

idkk

man i am in standard 10th

indi

[pure basin]

Then why are you doing this

[cursive trout]

india

icse

[pure basin]

Learn the basics?

[cursive trout]

this is in our syllabus

[cursive trout]

yes

addition

subtraction

multiplication

[pure basin]

You don't know inverse of a?

[cursive trout]

but dont know division

nope

[pure basin]

There is no division

[cursive trout]

waht is inverse

aahh whatever

[pure basin]

For a 2x2 matrix

[ 1 2
3 4]

Do you know determinant?

[cursive trout]

no

[pure basin]

(1 * 4) -(2 * 3) gives us the determinant

[cursive trout]

okay

[pure basin]

I assume you have no 3x3 matrix

[cursive trout]

and then

[cursive trout]

no

only 2x2 and 2x1 in syllabus

[pure basin]

a^-1 = adjoint(a)/determinant(a)

[cursive trout]

adjoint?

[pure basin]

Yes

For a 2x2 matrix there is a trick

To easily find it

[cursive trout]

okay waht is that

[pure basin]

But you need to know what is adjoint

[cursive trout]

yes

what is it?

[pure basin]

Or I think you will learn next year

[cursive trout]

hmm

so what to do now

[pure basin]

So the trick is basically

[4 -2
-3 1]

You swap the diagonal and add negative to the other 2

So adjoint is the transpose of a cofactor matrix

[cursive trout]

okaay, i am pretty much unable to understand whatever u are saying

[pure basin]

Ok

[rose zephyr]

$$
{\begin{bmatrix}
a & b \
c & d \
\end{bmatrix}}^{-1}

\frac{1}{a d - b c}
\begin{bmatrix}
d & -b \
-c & a \
\end{bmatrix}
$$

[pure basin]

Woah

so its bmatrix for a matrix ok

[rose zephyr]

I missed the determiant

One se

[pure basin]

Oh yea

What u said is adjoint

@cursive trout you don't know cofactors?

[glossy valve]

StrangeQuarkAL

[cursive trout]

nah nothing

[pure basin]

Then how can you even solve that question

Wait I think we need to

Yeah I think it's that way we need to guess it

So do you know what is order of a matrix

No need to find inverse

[cursive trout]

no easy method with simple arithmetics

[pure basin]

There is

[cursive trout]

yes

[pure basin]

Ok

so you gave a 2x2 matrix

[cursive trout]

hmm

[pure basin]

and we need to find x which is a x b matric

now the answer is a 2x1 matrix right?

[cursive trout]

yes

[pure basin]

Which means we need t multiply with a 2x1 matrix?

[cursive trout]

so X much be 2x 1

yes

[pure basin]

So just put [a b]

And solve ur linear equation

in 2 variables

[cursive trout]

i cant understand what u saying

[pure basin]

Write down

[cursive trout]

ax = b

[pure basin]

yes

[cursive trout]

x = b/a

[pure basin]

multiply a matrix no

Division is NOT POSSIBLE

MATRIX division NO

[cursive trout]

what to mutiply witj

[pure basin]

So the best way to find AX=B

Would be X=A^-1B

But you don't know what is A^-1

So we multiply AX

[cursive trout]

yes

ax

[pure basin]

Multiply AX

[cursive trout]

how we multiplay ax

[pure basin]

What is ur A

[cursive trout]

[2 1
-3 4]

[pure basin]

Yes

Now your X=[a b]

[rose zephyr]

Do you not know matrix multiplication @cursive trout ?

[pure basin]

wait did i see it right

[cursive trout]

ewin?

[cursive trout]

i know

[pure basin]

[7 6] is a 2x1 ok

[cursive trout]

okay

[pure basin]

ok then multiply [ 2 1 -3 4] with [a b] = [7 6]

[cursive trout]

what do u mean by multiplaying [ 2 1 -3 4] with [ a b]

[rose zephyr]

$$
\begin{bmatrix}
2 & 1 \
-3 & 4 \
\end{bmatrix}
\begin{bmatrix}
a \
b \
\end{bmatrix}
$$

[pure basin]

MULTIPLICATIon???

[glossy valve]

StrangeQuarkAL

[pure basin]

U just said u know matrix multiplication

[cursive trout]

yes

[pure basin]

Then do it

[cursive trout]

but how did

[pure basin]

?

[cursive trout]

X = [a
b]

[rose zephyr]

Simplify this with matrix multiplication

[pure basin]

because we don't know x

but we know it has to be a 2x1 matrix

[cursive trout]

ohh

okayy

okay okay

understood

thankssssssssssss

[pure basin]

For 3x3 matric the a^-1b would be best but 2x2 you can just solve a linear equation

[cursive trout]

thnkkk u so muchh

thnksssss

[pure basin]

if done .close

[cursive trout]

.close

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[elder saddle]

The probability of at least 3 . Would be the inverse of what . In the problem there’s 6 experiments I was thinking using binomcdf 3 but I’m not sure

[gritty flax]

?

[elder saddle]

What would be the way to solve for the probability of at least 3

I can show the problem big that helps

[gritty flax]

P(X >= 3) ?

Its the same as 1 - P(X<3)

[elder saddle]

So what would I enter binomcdf 2?

That doesn’t sound right

I know to use the compliment rule

[gritty flax]

.

Thats complement rule

And yeah P(X < 3) = P(X <= 2)

[elder saddle]

Can you elaborate I mean cause P(X<3) could also be 1 or P (X<=1 ) too right

[gritty flax]

?

I don't get what you said but no P(X < 3) = P(X<= 2) cuz in terms of integers

n < 3 => n <= 2

[elder saddle]

Why do you use 2 wouldn’t 1 suffice as well

[gritty flax]

?

[elder saddle]

Bro

[gritty flax]

Why would 1 be sufficient?

[elder saddle]

I’m asking

[gritty flax]

Yeah but why do you think so ?

[elder saddle]

Because it’s less than 3 if you’re saying the P(X<3)

[gritty flax]

Yeah cuz at least include the 3

So complement of P(X>= 3)

Is P(X < 3)

And since the number are integers

X < 3 => X <= 2

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[formal quiver]

There are four circles and four straight lines how many number of highest intersect point possible?

[celest tapir]

what are the types of intersections you expect?

[full forum]

@formal quiver Has your question been resolved?

[formal quiver]

A line intersect a circle at one point

8×4=32

@celest tapir

[celest tapir]

there are more

[formal quiver]

Could you tell me how?

[full forum]

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

@formal quiver Has your question been resolved?

[full forum]

@formal quiver Has your question been resolved?

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[main scroll]

How do we ! Decimal numbers

[main scroll]

Ex:
$3 3.14!$

[glossy valve]

Kakapo | Kakapo

[soft quiver]

why the factorial?

[main scroll]

I meant how do we factorial decimals

[soft quiver]

i don't get it

[umbral dome]

there is an extension to the factorial called the gamma function which can accept decimal inputs

[soft quiver]

true but does the op mean the same?

[main scroll]

How does it work then?

[umbral dome]

the gamma function can be computed with an integral

,, \Gamma(x) = (x-1)! = \int_0^\infty t^{x-1} e^{-t} \odif t

[glossy valve]

cloud

[main scroll]

Do we do like
$1.11.21.31.41.5$ for factorials

[glossy valve]

Kakapo | Kakapo

[soft quiver]

no

[soft quiver]

this

[main scroll]

Ty

.close

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[scenic brook]

quick question

[scenic brook]

can we not differentiate this as we would to b^f(x) ?

[onyx glen]

no because the base x is not constant

[scenic brook]

oh ok

so how would we go about differentiating this

[amber robin]

Do you know about logarithmic differentiation?

[scenic brook]

oh

yes

log both sides?

[amber robin]

Yes

[celest tapir]

you can write it as $e^{\sqrt{x}\ln x}$ and use the chain rule

[glossy valve]

dyxn

[scenic brook]

oh that too

right right

okok thanks

[celest tapir]

logarithmic differentiation works too

you'll get the same thing after subbing in y

[amber robin]

Or you can also do this:

$(\sqrt{x})x^{\sqrt{x}-1} + x^{\sqrt{x}}(\ln{\sqrt{x}})$

[glossy valve]

Suika

[celest tapir]

lol

[amber robin]

x^n formula + a^x formula

[celest tapir]

multivariable chain rule

[amber robin]

It works yk

[celest tapir]

yes

[twin wolf]

so extra

[full forum]

@scenic brook Has your question been resolved?

[full forum]

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[graceful pollen]

im still so confused, i understand it but still get the wrong answer
need help with c and D

[amber robin]

They asked for the longest path or the shortest path?

[graceful pollen]

honeslty i do not knoq

know

[amber robin]

What did u get for A and B?

[graceful pollen]

a and b i got right

this is the answer

c and d wrong

[amber robin]

Wtf is this?

[graceful pollen]

idk ask my school why they teaching us this

this is the chapter subject

i think i know how to do it

by looking at the answers

let me see

yeah nvm.

[full forum]

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[slate shuttle]

Now that we have equality.
What is the value of the definite integral?

[celest tapir]

Apply it once

What do you get?

[slate violet]

tan(0 + pi/2 - x) is...

[slate shuttle]

İ stuck here

[celest tapir]

what has bro written

Call the integral I

[slate shuttle]

Bro im trying

[glossy valve]

dyxn

[celest tapir]

yep I get it

[celest tapir]

apply the rule you just learnt

replace x with 0 + π/2 - x

[slate shuttle]

🤔🤔🤔🤔🤔🤔🤔🤔🤔hmmmm

[celest tapir]

so what does tan (0 + π/2 - x) become?

[slate shuttle]

İdk im stuck

Help

[celest tapir]

do you not know what tan (π/2 - x) is

[slate shuttle]

İk

But

İts not helping

[celest tapir]

it will help

what is it?

[slate shuttle]

🤯

😵‍💫

[celest tapir]

come on man, trust the process

if you're being uncooperative it's gonna affect you

[slate shuttle]

Like this?

[celest tapir]

Yes!

okay

okay good

now add those two representations of A

[slate shuttle]

Hmmmmmmm🤔🤔🤔🤔🤔🤔

Pi/4 ?

[celest tapir]

yes

[slate shuttle]

Bro what would you rate this?

[celest tapir]

rate what?

[slate shuttle]

Problem

[celest tapir]

it's fairly standard, you'll encounter this method of adding the integral to itself after manipulation many times

idk man like 7/10 for concept

[slate shuttle]

Bro can u help again

[celest tapir]

with?

[slate shuttle]

Another

[celest tapir]

depends on the problem

[slate shuttle]

f'(2)=?

İdk where to start

[celest tapir]

hmm

okay yeah

so first partially differentiate wrt x

[glossy valve]

dyxn

[celest tapir]

now if you substitute x and y to be 1 you'll have f'(2) = f'(1) + 2

oh wait nvm

[slate shuttle]

Hmmmmmmm🤔🤔🤔🤔🤔🤔🤔🤔

[celest tapir]

tbf if you sub x = 0 you have

f'(y) = f'(0) + 2y

and you can solve for f now

f(y) = f'(0)y + y^2 + C

[slate shuttle]

İ didnt gwt it

[celest tapir]

do you get how I got here

[slate shuttle]

Whaat😶‍🌫️

[slate shuttle]

Yes

[celest tapir]

now replace all x with 0

[glossy valve]

dyxn

[celest tapir]

okay?

[slate shuttle]

You mean f'(x+y)=f'(x)?

[celest tapir]

huh?

[slate shuttle]

Hmmmmm🤔🤔🤔

[celest tapir]

take this equation and wherever there's an x put a 0 in there

[celest tapir]

you will end up with this

which you can integrate both sides of wrt y

[glossy valve]

dyxn

[celest tapir]

all good?

[slate shuttle]

😵‍💫

[slate shuttle]

Why we did this

[celest tapir]

we're finding f(x)

[slate shuttle]

Hmmmmmm🤔🤔🤔 okey

Aaaaaa i find f(2)=3t then

[celest tapir]

Oh wait bro we're given f'(0) = 3

you didn't tell me that

😭

[slate shuttle]

Oh sry

Yes

[celest tapir]

well anyways

[slate shuttle]

İ find f'(2)=3

[celest tapir]

now we know that $f(x) = C + 3x + x^2$

[glossy valve]

dyxn

[slate shuttle]

Hmmmmmmmm🤔🤔🤔

[celest tapir]

I think that might be incorrect

[slate shuttle]

Oh okey😓

[celest tapir]

from the original plug in 0 everywhere and you get f(0) =-1

[slate shuttle]

Hmmmmmm🤔🤔🤔🤔

[celest tapir]

so we know that $f(x) = x^2 + 3x - 1$

[glossy valve]

dyxn

[celest tapir]

now all you gotta do is differentiate it once and plug in 2

[slate shuttle]

2x+3=7

[celest tapir]

yes

[slate shuttle]

Thx

But

İ would skip this if i saw something like that

[celest tapir]

lol that's up to you

[slate shuttle]

What would you ratebthis

[celest tapir]

8/10

[slate shuttle]

Can you help me again

[celest tapir]

open a new channel

frankly i'm quite tired i don't know if I can

[slate shuttle]

.close

[full forum]

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[stuck loom]

Find the roots of x^{9}-x^{8}-8x^{7}+8x^{6}+22x^{5}-22x^{4}-23x^{3}+23x^{2}+6x-6.

[stuck loom]

$x^{9}-x^{8}-8x^{7}+8x^{6}+22x^{5}-22x^{4}-23x^{3}+23x^{2}+6x-6$

[glossy valve]

Mathematician

[celest tapir]

,w Find the roots of x^{9}-x^{8}-8x^{7}+8x^{6}+22x^{5}-22x^{4}-23x^{3}+23x^{2}+6x-6.

[glossy valve]

Results provided by WolframAlpha

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Requested by dyxn06

[cerulean cargo]

lol wtaf

[celest tapir]

oof

[onyx glen]

well, i suppose this is quite the speedy solution to that.

but OP won't like this.

[celest tapir]

i wanted to check if it was doable

[onyx glen]

it is.

[cerulean cargo]

how do u know just by looking?

[celest tapir]

symmetric nonomial

[onyx glen]

x=1 is guessable (or you can spot the obvious term grouping) and once you factor it out you're left with a quartic in x^2

[cerulean cargo]

oh ok i see

[winter rune]

but how would you solve the quartic?

[drowsy ivy]

,w 1+1

[analog shale]

it might be a diffference of squares

i suspect

[onyx glen]

#bots

[onyx glen]

i will deliberately hold off on telling you that until @stuck loom shows signs of life again.

[winter rune]

all quartics are, but how would you find which two sqquares have that difference?

[analog shale]

||because im guessing its probably in the form of (x^4+ax^2+c)||

[winter rune]

the quartic looks randomish to me

not like that

and ik there is a general solution for quartics, but i dont think that is expected

[analog shale]

||it will be (x^2+e)^2-dx^2 for some d,e||

[fervent owl]

as in when you factor out x-1, you are left with a quartic in x^2

[winter rune]

ik that

[winter rune]

its not that sort of quartic

[analog shale]

oh you meant the quartic in x^2

[winter rune]

its
||x^4 - 8x^3 + 22x^2 - 23x + 6||

[fervent owl]

judging by the Wolframalpha output, ||x^2=2 and x^2=3|| are guessable roots

[analog shale]

that you just use ||rational root theorem||

[winter rune]

||yeh, but i couldnt guess them||

[fervent owl]

so then you can factor out ||x^2-2 and x^2-3||

[analog shale]

it provides all rational roots, if there is one

and there is.

[fervent owl]

||what you have left you can solve with repeated application of quadratic formula||

[winter rune]

||ah, so case bash is sensible in this case||

[analog shale]

not many cases just ||+- 1,2,3,6||, but yeah

[winter rune]

||makes sense||

[full forum]

@stuck loom Has your question been resolved?

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[slender rose]

How do i exactly calculate sin and cos of numbers?

[hot herald]

depends

[sand escarp]

of numbers?

[grave elm]

What do you even mean by "exactly"?

[onyx glen]

and which numbers

[onyx glen]

!xy

[full forum]

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.

[slender rose]

I want to know how sine waves even work in the first place

[slender rose]

Any

[slate violet]

from the unit circle definition, it's the height of the point which makes an angle theta

[sand escarp]

https://tenor.com/view/math-curve-circle-sine-cosine-gif-4839795

[onyx glen]

https://www.youtube.com/watch?v=1m9p9iubMLU

[wild sleet]

you know how they work

find x, draw a vertical line, you get y

nobody calculates them on their own, it's a calculator thing

you can do this though

[sand escarp]

Yes, this is pretty much how cos or sine of any number is calculated, together with trig identities I assume

[slate violet]

this is called a Taylor series by the way and you need calculus for it

[wild sleet]

only to come up with it, there's no calculus in using it

[slate violet]

ah true