help-23

[steel gulch]

Pretty much what i said here

[keen ginkgo]

Ohh okay

[steel gulch]

Otherwise you’d have to fuck around with integrals

[keen ginkgo]

Nope I refuse integrals in a stats paper

I think I get it

Thank you

[steel gulch]

No

*np

[keen ginkgo]

Lmao

[steel gulch]

Stupid auto correct

[keen ginkgo]

I was so taken aback 💀

[safe radish]

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[viscid ore]

Does anyone know how to solve this and explain to me how to get it right every time?

<@&286206848099549185>

[wintry locust]

Your approach?

Lets say for the first one

[viscid ore]

Let me write it out

@wintry locust

[wintry locust]

So whats the confusion?

You got it right

[viscid ore]

are all the answers right or do I need to present them on a number line?

[wintry locust]

For the right side, it would be 1/sqrt(2)

Ideally, for clarity you should represent them on a number line

Basically the solution space would be $(-\infty, -1] \cup (-\frac{1}{\sqrt2}, \frac{1}{\sqrt2}) \cup [1, \infty)$

[flat frigate]

fukwerint

[viscid ore]

why are we excluding -+ 1/root of 2

they should be greater than or equal?

[wintry locust]

Your solution is either |x| >= 1 or |x| <= 1/sqrt(2), right?

When did I exclude those points?

[viscid ore]

[-1/root of 2, 1/root of 2] ?

I will try doing the logarithmic one now

How do I solve of the apsolute values, should i write them as x2-2x and -x2 + 2x

as more branches and solve for each one?

or can I use a table?

@wintry locust

[wintry locust]

Oh yes, I missed the square brackets, my bad

No

How can the input to log function be negative?

log(x) >= 0 implies x >= 1

Not x<=-1

[viscid ore]

oh so the apsolute values need to be positive only?

[wintry locust]

No, for breaking the absolute values, just take cases around the zeroes of the expressions inside the abs

For example

Your function contains abs around (x+2) and (x^2 - 2x)

So you can break your solution into 4 cases, around x = -2, 0, 2

And open the absolute values accordingly in 4 cases

[viscid ore]

-2 and 2 are for |x+2| right?
where did I get the 0

the apsolute value of a quadratic function is always positive right?, and since I don’t have C or the free number, I get 0 ?

@wintry locust

[wintry locust]

No x+2 = 0 only for x = -2

x^2 - 2x = x(x-2), so x = 0 and x = 2 are for this

[viscid ore]

So i plug these values into first case since the second one cant be negative for the log

[safe radish]

@viscid ore Has your question been resolved?

[wintry locust]

Yup

And break the first equation (on the left) into the cases I mentioned and solve

[safe radish]

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[hybrid pier]

can someone go over with me on how to find the derivative?

[hybrid pier]

a is 3

[fickle pendant]

quotient rule

[hybrid pier]

u'v-uv'/v^2

[fickle pendant]

or, if you prefer, product rule where the fraction becomes (...)^-1

[hybrid pier]

this 0ne

[fickle pendant]

yep, so what is u and what is v (and therefore what is u' and what is v')

[hybrid pier]

is this right?

[hybrid pier]

my u is x^2-9 and my v is sqrt(x+a)

[fickle pendant]

hard to say without working it out myself without seeing each of the 4 values 😛

okay yep

u' is obvious, and v'?

[hybrid pier]

\frac{1}{2\sqrt{x+3}}\left(x-3\right)

[fickle pendant]

uh, not sure where the x-3 is coming from?

[hybrid pier]

1/(2*Sqrt(x+3))(1)

[hybrid pier]

.

[fickle pendant]

d/dx of (x+3)^(1/2) is (1/2)*(1/sqrt(x+3))

[proud galleon]

wait so you're doing the derivative of $\frac{x^2 - 9}{\sqrt{x+3}}$?

[hybrid pier]

yes

no

its -9

[flat frigate]

poudel

[proud galleon]

okay

alr

okay w quotient rule

[fickle pendant]

$\frac{1}{2(sqrt(x+3))}$

[flat frigate]

PrettyPrincessKitty FS

[hybrid pier]

how do i write like that too?

[hybrid pier]

thats v'

[fickle pendant]

in which case I think you have got it correct

[proud galleon]

when you have $\frac{u}{v}$, the derivative is $\frac{u'v - uv'}{v^2}$

[flat frigate]

poudel

[fickle pendant]

I haven't done the simplification to double check but it looks about right

[hybrid pier]

mine is right?

[proud galleon]

is this your answer??

[hybrid pier]

yeah

after trying to simlify

[proud galleon]

wait let me quickly do it lol

[hybrid pier]

i want it to be always positive on the down tho

thats what a calc gave me

i want it like that

idk what i did wrong

[fickle pendant]

you can do some factoring of the top

3 is common to the top, then you can probably factorise into (x-b)(x+3) and then the x+3 cancels

[hybrid pier]

wait how

here i took the (x^2-9)to the top

[proud galleon]

ohh wait

here's a much better way

of approaching this q

I think

start by simplfying the given expression

instead of instantly applying quotient rule

$\frac{x^2 - 9}{\sqrt{x+3}}
\ \ = \frac{(x+3)(x-3)}{\sqrt{x+3}}
\ \ = \sqrt{x+3}(x-3)$

[flat frigate]

poudel

[proud galleon]

then just apply product rule from here

,ask differentiate \frac{x^2 - 9}{\sqrt{x+3}}

[flat frigate]

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Requested by poudel

[hybrid pier]

what is that?

[proud galleon]

differentiate just means 'find the derviative'

[hybrid pier]

what is that

[proud galleon]

this is the answer to ur question

like the solution

yk ur working out

[hybrid pier]

i need to 'go pro'

[proud galleon]

can you send each of your steps

what dy wanna learn how to type math 💀💀

[proud galleon]

the handwriting is wild lmao

[hybrid pier]

):

[proud galleon]

icl im not sure what u did here

[hybrid pier]

can we vc and i will show ya?

[proud galleon]

how??

id rather not vc tbh

[hybrid pier]

you dont need to talk

just look at the ss

ok?

[proud galleon]

yeah alr

[safe radish]

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[spiral saddle]

find $f : \left(1, +\infty\right) \to \mathbb{R}\$ such that $f(2) = 7$ and $\(x^2 -x)f(x) = 14 + \int_{2}^{x} (9t-4)f(t)dt$

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[faint seal]

You can differentiate both sides of this equation wrt x to form a differential equation in x

[spiral saddle]

,, (2x -1)f'(x) = \int_{2}^{x} (9t-4)f(t)dtdx

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

something like this?

[faint seal]

no not at all

for the left side you need to use the product rule

[spiral saddle]

oh god

[faint seal]

$\frac{d}{dx}(f(x)\cdot g(x))\ne f'(x)\cdot g'(x)$

[flat frigate]

kheerii

[faint seal]

and for the left side you need to use the fundamental theorem of calculus part 1

[spiral saddle]

yes sorry

[spiral saddle]

a moment

what?

[spiral saddle]

right side

you mean?

[faint seal]

yes this

[faint seal]

yeah sorry the right side

[spiral saddle]

,, (2x-1) \cdot f(x) + (x^2 -x) \cdot f'(x)

[faint seal]

yes indeed

[spiral saddle]

,, (2x-1) \cdot f(x) + (x^2 -x) \cdot f'(x) = \frac{d}{dx} \int_{2}^{x} (9t-4)f(t)dt

bruh

I missclicked

[spice wing]

Can i know the question ?

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

this is the question

but its translated to english

[spice wing]

For right side apply newton leibnitz

You only need to differentiate ?

[spiral saddle]

but wait I cannot use FTC1 if the integral is with respect towards t

no?

[spice wing]

Why dont you differentiate rhs

[spiral saddle]

how to

[spice wing]

newton leibnitz

Im writing the formula

For one variable

d/dx int u(x) to v(x) of f(t) dt

Will be

v'(x)×f(v(x)) - u'(x)×f(u(x))

In this question it would be

(9x-4)(f(x)) on differentiation

[spice wing]

Lower limit u(x) upper v(x)

[spiral saddle]

,, (2x-1) \cdot f(x) + (x^2 -x) \cdot f'(x) = (9x-4)f(x)

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spice wing]

Yeah was this your question?

[spiral saddle]

how do I

find f

[spice wing]

ahh okay !

[spiral saddle]

why are they giving away f(2) = 7

[spice wing]

f'x(x²-x) = f(x)(7x+3)

[spiral saddle]

I agree

[spice wing]

Or f'(x) = f(x)(7x+3)/(x²-x)

[spiral saddle]

okay

now what

[spice wing]

one min

Hey

f'(x)/f(x) = 7x+3/x²-x

now integrate both sides

ln f(x) = ....

@spiral saddle ...

I hope you can solve the rest

Using f(2) = 7 will help you find the value of integration constant

[spiral saddle]

lowkey this is the hard part

how do I differentiate this diff eq

,, \int \frac{f'(x)}{f(x)} dx = \int \frac{7x+3}{x^2 -x} dx

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[faint seal]

partial fractions will work for the right side integral

[spiral saddle]

by using partial fractions you mean this $\frac{7x+3}{x(x - 1)} = \frac{a}{x} + \frac{b}{x-1}$

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[faint seal]

yes indeed

[spiral saddle]

,w solve 7x + 3 = a(x-1) + bx

[faint seal]

just compare the coefficients man

[spiral saddle]

7x + 3 = x(a + b) -a
a+b = 7
a = -3

[spiral saddle]

im new to it man, relax

b = 10

so $\frac{-3}{x} + \frac{10}{x-1}$

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

-3lnx + 10ln(x-1) + c

,w integrate -3/x + 10/(x-1)

[flat frigate]

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Upgrade to WolframAlpha Pro!

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

but question is how do I integrate lhs

[safe radish]

@spiral saddle Has your question been resolved?

[craggy sedge]

Shouldn't it be 7x-3 tho

[spiral saddle]

you are right

,, (x^2 -x) \cdot y' = (9x-4-(2x-1))y

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

,, (x^2 - x) \cdot y' = (7x -3)y

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[craggy sedge]

Did you get the answer

[spiral saddle]

still no

lowkey I am getting confused since there are multiple constants C

,, \frac{dy}{dx} \cdot \frac{1}{y} = \frac{7x-3}{x^2 -x} \cdot \frac{dx}{dy}

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

,, \ln(y) = 3\ln(x) + 4\ln(x-1) + C

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

this is what we get after partial fraction decomposition

,, \frac{7x-3}{x^2 -x} = \frac{a}{x} + \frac{b}{x-1}

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

then we solve for a(x-1) + bx = 7x- 3

ax - a +bx = 7x-3

x(a + b) - a = 7x-3

7 = a+ b

a = 3

b = 4

then integral of 3/x is 3lnx and integral 4/(x-1) is 4ln(x-1)

[spiral saddle]

now that we are here

is where things get tough

when I try to calculate C there are multiple C's

[craggy sedge]

There's only one Value

To evaluate it plug x=2

[spiral saddle]

,, y = e^{3\ln(x) + 4\ln(x-1) + C}

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

,, 7 = e^{3\ln(2) + 4\ln(1) + C}

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

,, 7 = e^{3\ln(2)+C} = e^{3\ln(2)} \cdot e^{C} = 7

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

,, e^c = \frac{7}{e^{3\ln(2)}}

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

$e^{3 \cdot \ln(2)} = e^{\ln\left(2^3\right)}$

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

$e^c = \frac{7}{8}$

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

$y = x^3 + (x-1)^4 + e^c$

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

this is from this $y = e^{3\ln(x) + 4\ln(x-1) + C}$

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

and finally $f(x) = \frac{7}{8} \cdot \left(x^3 \cdot (x-1)^4\right)$

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

.close

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[lament crescent]

Im not the best at math so please be patient. This question is on slope intercept form, but the line isnt linear/straight and its asking me to find the slope?? How do I find the slope if it’s not linear?

[spice furnace]

,rotate

[topaz tree]

you need to find approx slope

So treat the line as approximately linear and find the slope...

[old plover]

here the slope isn't linearly increasing you'd notice it immediately right

[spice furnace]

a more precise answer is there is a strong linear relationship. saying no is not that meaningful. While it is true the graph is not linear that does not mean we can;t approximate it using a straight line

[old plover]

So, graph ain't linear

approx slope can be found out with differentiation easily

or you can just find it at the lowest collision

just draw a tangent

bruh stop reacting with skull emotions

it's true, differentiation takes like 2 secs

[lament crescent]

so would i be using (1, 280) and (7, 300) ?

[spice furnace]

yes

[safe radish]

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[lone arch]

#linear-algebra message

[lone arch]

.close

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[gleaming matrix]

i needto know howto solve second part

[safe radish]

@gleaming matrix Has your question been resolved?

[gleaming matrix]

<@&286206848099549185>

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@gleaming matrix Has your question been resolved?

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@gleaming matrix Has your question been resolved?

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[ornate osprey]

I'm a bit confused on why this problem is divergent

[ornate osprey]

-ln|7-6.999999999| doesn't cause issues?

[restive niche]

What happens to the argument of the logarithm as k-->7

[restive niche]

No, it doesn't

[ornate osprey]

I guess -ln(0) is undefined

ah that's probably why

[restive niche]

As k--->7-, the argument 7-k approaches 0+

So the ln term grows arbitrarily large

Hence, it is said to be divergent

[ornate osprey]

i see

so ln(0) is technically a massive number?

[restive niche]

Ln(0) isnt a number

[ornate osprey]

right

[restive niche]

There is no number such that e^(that number)=0

[ornate osprey]

true

[restive niche]

As x approaches 0, ln(x) goes to - infinity

[ornate osprey]

oh i see

thank you!

.solved

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[vocal glen]

can somebody explain how this happened

[old chasm]

Factoring?

[vocal glen]

wdym

[old chasm]

Like, $\frac{1}{4\pi\varepsilon_0}$

[flat frigate]

print("NAME")

[old chasm]

Got factored out, yeeted to zero

[wintry locust]

Multiply both sides by |r-r1|^3

[vocal glen]

Wdym factored out could you please explain

[vocal glen]

let me try

[wintry locust]

Multiply both sides by $4\pi\epsilon_0$ too then

[flat frigate]

fukwerint

[vocal glen]

that works

[old chasm]

Yeah, and that should be it

[vocal glen]

but i want to learn about the factoring thing print name is talking about

[old chasm]

Like ab + ac = a(b+c)

That kind of factoring

[vocal glen]

how are you using that in here?

[old chasm]

Well, both terms are multiplied by $\frac{1}{4\pi\varepsilon}$

[flat frigate]

print("NAME")

[vocal glen]

shouldnt we multiply it by

instead of

[old chasm]

you should. But like, my point is it is currently multiplied by $\frac{1}{4\pi\varepsilon}$

[flat frigate]

print("NAME")

[vocal glen]

alright thanks man!

[old chasm]

so you can say $\frac{q_1}{4\pi\varepsilon} + \frac{q_2}{4\pi\varepsilon} = \frac{1}{4\pi\varepsilon} (q_1+q_2)$ for example

[flat frigate]

print("NAME")

[vocal glen]

we took it common!

[old chasm]

yeah, that is the more straightforward way of saying it

I usually call that factoring because you got two things multiply to each other lol

[vocal glen]

alright thank you so much man!

.close

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[upper pecan]

is (e^x)/x quasi polynomial?

[safe radish]

@upper pecan Has your question been resolved?

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@upper pecan Has your question been resolved?

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[tribal inlet]

!help

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

[tribal inlet]

why does there need to be a space

between the 11 and i

else its wrong

ohh

i figured it out

[fathom adder]

you can write it like 2i * sqrt(11) too

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[fathom plume]

trying to represent AC interms of BC or AB

rn all i have is AC-BC=AB

TC = tan(beta)BC

cant find expression for AC

[wintry locust]

AC = AB + BC

[fathom plume]

but then to represent AB i still need AC - BC

[wintry locust]

And TC/(AB + BC) = tan alpha

So tan(beta)BC/(AB + BC) = tan(alpha)

[severe slate]

try to represent all the tans as lengths

and do some algebra

[fathom plume]

oh ok ty

.close

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[crude shell]

Can someone tell me what am i doing wrong here ? So CBB’ is a triangle that has both sides equal

[crude shell]

But if we apply law of sines CBB’ and CB’B is 114

And that exceeds the angle of 180

[wintry locust]

sin(alpha) = k

Then sin(pi - alpha) is also k

[thin bridge]

you've mixed up your angles

[wintry locust]

So since alpha you found is not possible

Then pi-alpha must be the value you are looking for

[thin bridge]

<AB'B will be the obtuse case
and <ABC the acute case

[crude shell]

yeah but idk how i found it that way

[crude shell]

oh

[crude shell]

hows this even possible though

[thin bridge]

unit circle, properties of sine

[crude shell]

oh yeah

forgot that

so sin (180-angle) = sin(angle) ?

[thin bridge]

yes

[crude shell]

okay but if its the same then what is the matter whether i called B' alpha or π-alpha

like i dont understand its the same thing anyways

[thin bridge]

sin(alpha) and sin(pi - alpha) are the same
alpha and pi-alpha themselves are different

[crude shell]

i think i got it

bc of the acute angle thing we gotta call it π-alpha right

to make sure it AB'B is obtuse we call it π-alpha right?

[thin bridge]

lets separate these triangles to make it a bit more clear

[wintry locust]

Yes, assuming your alpha is less than pi/2

[crude shell]

i did this as the final btw

before u continue

sorry to interrupt though

am i going right

[wintry locust]

Yup

[hardy kernel]

How do I denote minute in calculator

[crude shell]

tbh idk wither

either

[wintry locust]

1 degree has 60 minutes

So convert to decimal

[thin bridge]

$\boxed{\deg ' "}$ or DMS

[flat frigate]

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

[crude shell]

i use desmos, couldnt figure out yet so i am converting it

[crude shell]

i did this using desmos, didnt work for me

[wintry locust]

If an angle is 30 degrees and 40 minutes, then it can be written as 30 + 40/60 ~ 30.67 degrees

[crude shell]

alr

[hardy kernel]

Why is the value of sinO coming greater then1

[wintry locust]

Show

Your calculator needs to be in degree mode btw

If its in radian mode then convert the degrees to radians

[crude shell]

btw my question is done

i got it right

im closing the chat

thanks for the help yall

.close

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[lethal musk]

the question says: "sum these 6-bit two's complement binary numbers, indicate whether or not the sum overflow a 6-bit result"
if I have 27 + 31
the result will be 111010
so it's negative
but there's no overflow bit, is this considered an overflow?

[buoyant shadow]

i'm not sure

cuz yeah you could argue either way

it didn't overflow "physically" but something bad happened still

no way to guess

i would guess yes

[safe radish]

@lethal musk Has your question been resolved?

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[long otter]

I am struggling to continue from here

[split ether]

Are you sure f^2 means f(x)f(x) rather than f(f(x))?

[long otter]

Oh

Hold on be right back

Found it thank you @split ether

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[pulsar marsh]

Find the sum of all integer values of the a, for each of which the equation [image] has positive x.

Here's what I did:
n = x.
4ⁿ - 15 × 2ⁿ – 30a + 4a² = 0
2²ⁿ - 15 × 2ⁿ – 30a + 4a² = 0
t = 2ⁿ
t² - 15t – 30a + 4a² = 0
t1 + t2 = 15
t1t2 = 4a² – 30a
D = 225 - 16a² + 120a
t = (–225 ± sqrt(225 – 16a² + 120a))/2

[small ravine]

well why set n=x

anyway sure

we need D to be positive or 0

[pulsar marsh]

Mathematically, this does not make any sense.

[small ravine]

wdym

[pulsar marsh]

I can't explain because of a bad translator.

[small ravine]

anyway forget the n and x thing. You just need D>=0 in order for the equation to have a solution

[pulsar marsh]

I do not know. I want to find all values of a for which x > 0. I don't know how to apply D here.

[small ravine]

oh wait i misunderstood the question because of yeah the translation

[pulsar marsh]

Ooоо, I got it!

[small ravine]

you first need the equation to have a solution

for that

you need D>=0 ok?

[pulsar marsh]

225 ‐ 16a² + 120a >= 0
-16a² + 120a + 225 >= 0

And I don't know what to do next, because solving for a does not work out because of very strange numbers.

[small ravine]

alr so is the question that all solutions x should be positive or just one of the solutions must be positive

[pulsar marsh]

Ok, I'll think about it later today. I'm busy, so I want to close the chat. Thank you for helping.

[small ravine]

well, let me know if you need help, you can ping me

[pulsar marsh]

Okе, thanks again

.close

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[glacial sonnet]

we go agane

[glacial sonnet]

i came across a random comment that gave a procedure to compute floor(sqrt(x)), where x is an integer. "find the highest bit set, left shift 1 half this many bits as the initial value, then few iterations of newtons method, cast to int"
say MSB of x is at pos i, are they choosing n = x << (i / 2) as their initial value? if so is what is the particular reason for this choice?

second, i was wondering what the computational complexities of the nth order householder methods and potra ptaks two step method. i haven't given much thought to this yet but for sufficiently smooth f, first order, newtons, we'll have $x_{n + 1} = x_{n} - \frac{f(x_n)}{f'(x_n)}$ and second order, halleys, $x_{n+1} = x_n - \frac{2f(x_n)f'(x_n)}{2(f'(x_n))^2 - f(x_n)f''(x_n)}$. computationally what exactly is the disctinction? does having fewer functioanl evaluations improve performance?
in potra ptaks two step method, the third order is given by $z_n = x_n - \frac{f(x_n) + f(y_n)}{f'(x_n)}$ where $y_n = x_n - \frac{f(x_n)}{f'(x_n)}$, then perform netwons method to get a method of order four, next rewrite $f(z_n)$ as its taylor series and perform a sequence of substitutions back to get a method of order 8, why exactly does this sequence of substitutions reduce the computational complexity

[flat frigate]

catgirl pee

[fathom jewel]

woah

[glacial sonnet]

[peak estuary]

if x has for example biggest bit set like 24, then x is roughly size 2^24, so the root would roughly be 2^12

[shadow sparrow]

thats a nice name

[glacial sonnet]

thank you very much, would you like to become my fan

[shadow sparrow]

i am a big fan

[glacial sonnet]

right but what is significant about choosing n = x << (i / 2) as the initial guess?

[glacial sonnet]

bless, may I become a fan of yours as well

[shadow sparrow]

u may

[glacial sonnet]

[peak estuary]

well it will be roughly that size

[glacial sonnet]

OH

wouldn't I want to right shift by i / 2 then

not left

[safe radish]

@glacial sonnet Has your question been resolved?

[plucky elk]

Such a weird and creative name...

[glacial sonnet]

does it bother you?

if so why don't you cryaboutit

[plucky elk]

[glacial sonnet]

nvm I see

they're choosing n = 1 << (i/2)

[safe radish]

@glacial sonnet Has your question been resolved?

[glacial sonnet]

[lean otter]

I was just intrigued by this name

[glacial sonnet]

are you a fan of catgirl pee

[lean otter]

yes

[glacial sonnet]

excellent

another fan to my collection

[lean otter]

true

[glacial sonnet]

.close

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[tiny frost]

Can someone explain what they want me to do when they say to "decide the X matrice"?

[tiny frost]

$$X \begin{bmatrix}
-5 & 1 & 4 \
6 & 9 & -1 \
2 & -7 & 3
\end{bmatrix}=\begin{bmatrix}
-35 & -23 & 38
\end{bmatrix}$$

[flat frigate]

Totalani

[safe radish]

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.

[tiny frost]

that is the original

[granite idol]

oh, it's asking you to find the matrix that, when multiplied with the given one, gives you the RHS

[tiny frost]

so X is essentially $$\begin{bmatrix}
x \
y \
z
\end{bmatrix}$$ and I multiply this in and solve?

[flat frigate]

Totalani

[granite idol]

not quite, but that's the idea

[tiny frost]

well how would you do it?

[granite idol]

you need to make sure the dimensions work

[tiny frost]

aha

$$\begin{bmatrix}
a & b & c \
d & e & f \
g & h & i
\end{bmatrix}$$

[flat frigate]

Totalani

[tiny frost]

no

oh i got it

$$\begin{bmatrix}
x & y & z
\end{bmatrix}$$ has to be this since the answer is this wya

[flat frigate]

Totalani

[granite idol]

yup

[tiny frost]

one thing I dont get is how I distribute it, I looked at an example in the book but I still dont get it

let me show you

$$XA=\begin{bmatrix}
a & b \
c & d
\end{bmatrix}\begin{bmatrix}
-4 & 2 & 1 \
0 & -3 & -5
\end{bmatrix}=\begin{bmatrix}
-4a & 2a & -3b & a & -5b \
-4c & 2c & -3b & c & -5d
\end{bmatrix}$$

[flat frigate]

Totalani

[tiny frost]

??

[granite idol]

you can just do regular matrix multiplication

[tiny frost]

oh

wait

$$\begin{bmatrix}
-5x & x & 4x \
6y & 9y & -y \
2z & -7z & 3z
\end{bmatrix}$$ im clearly doing something wrong

[flat frigate]

Totalani

[granite idol]

the product will be 1 x 3

[tiny frost]

oh I just understood the example the book shows

I appreciate it, ima rest now beore my head exwplodes

.close

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[ebon elbow]

Can someone explain the reason for the thing they did from 2nd to 3rd line

[rigid inlet]

Well factoring out the 2 simplifies the integrand, and adding and subtracting 2 allowed the split they did in the next equality

[ebon elbow]

So you can make up numbers in the exam or is there a rule to it

[rigid inlet]

Well as long as you add and subtract the same thing it's a valid operation

They chose 2 particularly because of the denominator

[safe radish]

@ebon elbow Has your question been resolved?

[solid fjord]

how did you complete the square

say for example

you have x² - 4x

and you know x² - 4x + 4 is (x - 2)²

add 4 and subtract 4 and you have

(x - 2)² - 4

same case for this

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[safe radish]

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[hybrid pier]

i did a question and i got 2
(
sin(a)sin(90−2a)
R
2
sin(180−2a)
2
sin(90+a)
​
)
but the answer is R
2
cos
2
(a)tan(2a) are they the same?

[hybrid pier]

(R^2sin(180-2a)^2sin(90+a))/sin(a)*sin(90-2a))/2

but the answer is R^2*cos^2(a)*tan(2a)

are they the same?

@glacial sonnet did you finish watching that episode?

[glacial sonnet]

I will watch it later

now is not the time for me

[hybrid pier]

i am just warning your brain will never be the same after this

[glacial sonnet]

good thing it's smooth already

[hybrid pier]

do you know how can i write this in the cool way where it shows it normally?

[glacial sonnet]

It looks too messy and my eyes hurti

[hybrid pier]

oh ik

[hybrid pier]

is this =

[glacial sonnet]

I suppose you should play with some sin angle sum identities

[hybrid pier]

i mean if its the same do i need to that or will my tester accept this

sin(90+a)=cos a right?

[glacial sonnet]

[lean otter]

Yes

[hybrid pier]

yes to what?

[hybrid pier]

to this?

[hybrid pier]

this?

[hybrid pier]

or this

[lean otter]

This

[hybrid pier]

oh ty (:

and sin 90-a is also cos a?

[lean otter]

Also sin(180-b)=sin(b)

[hybrid pier]

yeah that 1 ik

[hybrid pier]

is this right?

[lean otter]

Yes

[hybrid pier]

how

sin 90-a = sin 90+a?

no way it does

[lean otter]

Yes

[hybrid pier]

how tf?

u sure?

[lean otter]

Yes

[hybrid pier]

i need to write it down

so sin(90+-a) = cos(a) thats what i need to write right?

[lean otter]

Look at this graph

[hybrid pier]

i have my finals tommorow are you able to learn with me for a bit?

pls?

[lean otter]

Pardon my "straight line"

But clearly the points on the either side of the 90° mark are equal, correct

Therefore sin(90+x)=sin(90-x)

Does that make sense?

[hybrid pier]

oh

ok ty

[hybrid pier]

so?@lean otter

[lean otter]

Whereas for 180° it is also equal in magnitude but different in sign

Therefore sin(180+x)=-sin(180-x)

Does that make sense?

[lean otter]

Yes

[hybrid pier]

yay

[safe radish]

@hybrid pier Has your question been resolved?

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[scarlet comet]

What are the elements of Q(pi)? Let E = Q(pi) and F = Q(pi^3). Why is E/F a finite extension? Why is [E : F] = 3?

[peak estuary]

because E=F(pi). find a minimal polynomial of pi over F

[safe radish]

@scarlet comet Has your question been resolved?

[peak estuary]

the elements of Q(pi) look like rational functions with rational coefficients evaluated in pi. similar for Q(pi^3), but evaluated in pi^3

so stuff like pi+17pi^5 or (1-3pi^2)/(pi-32pi^5)

[scarlet comet]

why is e/f a finite extension?

@peak estuary

[scarlet comet]

<@&286206848099549185>

[scarlet comet]

.close

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[void onyx]

Can someone help with Question 1 Part D, Question 2 Part A, and B, and Question 3

[fathom jewel]

Is there a way to send a better and clearer picture?

[void onyx]

yeah sorry

1 sec

@fathom jewel Can you see this one clearly?

[fathom jewel]

yeah thanks

so 1D)

[void onyx]

yeah

[fathom jewel]

,,\lim_{x \to 0} f(2-x^2)

[flat frigate]

𝔸dωn𝓲²s

[void onyx]

yes

[fathom jewel]

this is basically f(2)

and what's f(2)?

[void onyx]

f(2) as in i go on the graph and see like the black point at 2?

so like -2?

right

[fathom jewel]

yea I agree

[void onyx]

ok

[fathom jewel]

I think black dot is defined

[void onyx]

yes

[fathom jewel]

2A and 2B

[void onyx]

wait so -2 is the answer?

[fathom jewel]

ok first step you already factorized

[void onyx]

for Q1D

like theres no x^2?

[fathom jewel]

I believe so

[void onyx]

wait so

if it said lim x -> 0 f(3-x^2)

would i do f(3) there? @fathom jewel

[fathom jewel]

yea

[void onyx]

so what does the -x^2 mean

[fathom jewel]

it's basically like a function decomposition I'd say

If g(x) = 3-x² then you would have f(g(x)) here

[void onyx]

hm alr

ok so -2s the answer

ok so uhh

Q2 part A

[fathom jewel]

What's on the left written?

First step is always...

direct sub?

[void onyx]

It said "First step is always direct substitution" yes

i think thats wrong

the question is just what is the first step

idk though

[fathom jewel]

hmm

[void onyx]

i felt like it was just factor

[fathom jewel]

my first step is to analyze it

and if you were to sub

you would divide by 0

[void onyx]

wait

if you were to sub it

[fathom jewel]

so direct sub doesnt work (yet)

yea

[void onyx]

wouldnt it be (2)^3 + (2)^2 - 4(2) - 4

/ (2 - 2) (2 + 3)

oh yeah ur right

it would

ok

[fathom jewel]

yeah but what about the denominator

ok

[void onyx]

alr nvm

scratch that

[fathom jewel]

yea

[lime dust]

[fathom jewel]

that's how I interpret A

[void onyx]

ok so

its factor right?

[fathom jewel]

yea

that's the consequence for B

to simplify things maybe

so that maybe (x-2) may cancel out

[void onyx]

what

wym

[fathom jewel]

that factor

it could cancel out

if we write the numerator also as factors

[void onyx]

in part B?

[fathom jewel]

yea

[void onyx]

what cancels in (x+2)(x+1) / (x+3)

[fathom jewel]

oh you did that already

you're fast haha

ok

now Part C is selfexplanatory

direct sub works now

[void onyx]

yea i got C right

but im tryna figure out why B was wrong

caps

like uh

what was i supposed to right instead

"what would you show when solving this limit?"

[fathom jewel]

haha ehm

that you can cancel out (x-2) the critical factor?

How were you taught limits

[void onyx]

oh yeah prolly

but like

[void onyx]

idk how to explain this

[fathom jewel]

I mean like

there is something that has been taught

step-wise

when you were introduced to limits

so maybe there is the answer

[void onyx]

yeah u right

but tbh i forgot

[fathom jewel]

as otherwise it seems vague

to me

[void onyx]

just tryna refresh before calc bc next yr

[fathom jewel]

ok good

Q3

tbh i dont really get the question but I have an idea what they might want

I think they are referring to "jumps"

like these

because f(x) = x is continuous so that wouldn't really make sense with instant change but I might be wrong tho

[void onyx]

hm

alr

.close

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[scarlet comet]

Any help with finding the base case and recurrence for IN[i] and OUT[i]?

[safe radish]

@scarlet comet Has your question been resolved?

[glacial sonnet]

you should consider what exactly it is you're tabulating

if you can determine what that is, the base case and recurrance is quite simple

[scarlet comet]

I did some work so far I believe the base cases are the leaves

[glacial sonnet]

i also dislike this approach to the tree max path sum problem but iiwii

[scarlet comet]

Not sure about the recurrence

[scarlet comet]

Do you think so too?

[glacial sonnet]

yes that is what the base case will be

[scarlet comet]

I think for the in case we just add up the value of the current node along with the in's of the left and right child?

I am not sure about the out case

[glacial sonnet]

well

for a given vertex v, out[v] is storing the maximum path sum in v's subtree, where such a path does not contain v itself

what would such a path look like?

[scarlet comet]

^

are you saying we take the maximum of both in cases?

of the children?

actually i have a feeling that we take the max of both in AND out cases

since the in case might not be the max

[scarlet comet]

just tested this on an example input and it's not working either

[scarlet comet]

this doesnt seem simple to me lol

[glacial sonnet]

here's a picture

lets say im considering blue vertex's subtree

so blue_left and blue_right

lets say that the orange path is the maximum path sum that in tracks

and red is max path sum tracked by out

uh i should have done something for the right subtree also

[scarlet comet]

we add another ^ right, and yeah that's the issue I was facing too, we can't include the red and the orange at the same time, so it's not simply the parent value + the invalue of both children

[glacial sonnet]

sorry i confused myself

that's not what out should look like

one sec

ok here

ok so lets say we cached blue.left and blue.right already

now when we go up a level and look at blue

out[blue.left] is the red path sum

and out[blue.right] is the purple path sum

how would you update out[blue] now

[scarlet comet]

max of in[blue.left], in[blue.right]

[glacial sonnet]

and something else

[scarlet comet]

not sure

[glacial sonnet]

you'll want to add blue's value as well

[scarlet comet]

???

[glacial sonnet]

these should be outs

[scarlet comet]

why do we add blue's value if we are doing out[blue] which excludes the blue vertex

[scarlet comet]

not sure what you mean by this either

if we do max of out[blue.left], out[blue.right]

we would just take the max of those 4 nodes which is unoptimal

the only way i can see a solution to this is if we also consider height

[scarlet comet]

can u explain this?

[safe radish]

@scarlet comet Has your question been resolved?

[glacial sonnet]

Sorry meant out[blues parent] here

For updating the left subtree

[safe radish]

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@scarlet comet Has your question been resolved?

[safe radish]

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[rain vault]

Can someone explain to me how I did this problem wrong? I’m so tired of doing these wrong..

[sharp tundra]

when you multiplied the left side by 2(x-2), you forgot to also multiply the right side by that

[rain vault]

i multiplied the right side by 2

since the x-2 cancels out right?

[sharp tundra]

not the right fraction, the right side of the equation. you should get $(x-2)(2x+5)-2(3x)=x\color{red}(2)(x-2)\color{black}$

[flat frigate]

evelyn

[sharp tundra]

you forgot the part in red

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[wind pier]

Use the graph to find the solution of the simultaneous equation

[thin narwhal]

whats the problem

[wind pier]

The graph one above

[thin narwhal]

its the point where the two things intersect

[wind pier]

So it would be 1,2

[thin narwhal]

wrong way around

[wind pier]

Ok thanks

[thin narwhal]

when youre done close the channel

[wind pier]

How do I do that?

[thin narwhal]

.close

[wind pier]

.close

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[lean otter]

3^5^11^2024! divided by 19 it slike power over power can anyone help me FINDREMAINDER

[lean otter]

ITS a ioqn and olympiad q

m

[lusty spear]

if I am not mistaken you could use modular exponentiation for this

[brave wolf]

olympiad question?

Is this from on-going olympiad?

[lean otter]

yes

but pls give me the explanationa nd working

[lean otter]

yes

[lean otter]

plz explanation

[lusty spear]

should you not solve this yourself if this is an ongoing competition ?

I am sorry, rn I have no time to go into detail

there is plenty material on this topic on the internet

[brave wolf]

<@&268886789983436800>

[eternal carbon]

mODs

[brave wolf]

.

[lean otter]

help me

plz

[brave wolf]

Do the olympiad yourself, if you cant do that and you wanna know the solution, come ask once it ends

[lean otter]

noo

its over

the test is over

[thin narwhal]

but you just said it was ongoing

[brave wolf]

You said it's on-going olympiad

[lean otter]

i came to tell that it was recently completed

few days back

i promise'

[thin narwhal]

but why say is ongoing

[lean otter]

in our country there is no online olympiad all are made to wrrite in classroom

[lean otter]

sorry

it recent;y got concluded

i swear

[brave wolf]

Sorry for the mod ping then

[edgy cloud]

"Ongoing" was from a leading question, and I think the possibility of language barriers leads to enough doubt there.

[lean otter]

ok guyz it was my mistake i mistook it

[brave wolf]

alr, do you know about fermats little theorem then?

[lean otter]

yess

ik

a power pie of p equals 1 mod p

[brave wolf]

Alright, so from that we know that 3^18 = 1 (mod 19)

[lean otter]

ya

[brave wolf]

so it might be useful to find 5^11^2024! mod 18

[lean otter]

yep

now how do u like simplify the powers'

that is the challenge then only we can get an indea

its power over power

[edgy cloud]

You've already simplified it somewhat -- now you have a tower of height 3 instead of one of height 4.

[lean otter]

ya so 2024 fac ends in 0

can we fins the rm for 3 power 5'

then power 11

and then power 2024 fac

[edgy cloud]

No.

The power tower is supposed to be understood as 3^(5^(11^(2024!)))).

That's why Meth said you now need to find 5^11^2024! modulo 18 (rather than modulo 19).

[lean otter]

ya ok

now what to do next?

[edgy cloud]

This is a power tower problem again, but it's shorter than the one we started with, so we're making progress.

The modulus is now 18, which is not a prime, so we turn to Euler's theorem instead of Fermat's little, but otherwise the next step goes just as before.

And since you've been responding just "ya" and "yep" to Meth's and my explanations, I know that you understood that first step perfectly, so you should be able to go the rest of the way for yourself.

[brave wolf]

I'm really sorry, my parents called me so i had to go for a moment (thanks for taking over tropo)

[brave wolf]

this is the euler's theorem, fermats little theorem is a special case of it.
Apply it and you'll get power tower of height 2, then just continue like this until you get to the top

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

iam not abl eto simplify the tower

tahts the issue

u dont get it top

whtever u told at first it was basic that swhy i told ye ya

i also did those steps

when u solve it fully there s many issues

try an say if u feel it sthat easy

[brave wolf]

I did try it :]

So what's your current progress?

You here @lean otter

The idea is to continuously get smaller and smaller towers

then build up the tower back by back-substituting

ping me once you come

[lean otter]

am back

lemme know the ans iam asking bcoz am not getting sorry for the trouble

[brave wolf]

Alright

So

3^18 = 1 (mod 19)

we know that 5^11^2024! could be written as 18k + l

and then it would simplify to 3^(18k + l) mod 19

but that's just 3^l mod 19

so good idea would be to find the l, such that 5^11^2024! = 18k + l

but that's just 5^11^2024! mod 18

so we reduced the problem of finding 3^5^11^2024! mod 19 to finding 5^11^2024! mod 18

now we shall reduce it again

so note that 5^6 = 1 (mod 18)

because of euler's theorem

now we can apply same argument

5^(6k + l) = 5^l (mod 18)

so it suffices to find 11^2024! mod 6

...

ill let you do this one yourself

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

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[lean otter]

i sew thanks u very much

[lean otter]

git it

[safe radish]

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[safe radish]

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[thorn roost]

If i want to see what percent bigger 2.43 billion acres is than 60,000,000 acres how would i do that

[thorn roost]

Is this simple
division with big numbers

[dense knoll]

yea basically

2.43b/60m

[thorn roost]

I think I have discalcula So my brains always fuzzy with math even though i really like it.

Thanks

.close

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[jaunty trench]

Why when working with % it’s always given 2 places after decimal place, e.g. 31.98%?

[rapid blade]

...

GOOD QUESTION